Russell C. Hibbeler - Mechanics of Materials, Eighth Edition - Instructor\'s Solutions Manual
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1–1. Determine the resultant internal normal force acting on the cross section through point A in each column. In (a), segment BC weighs 180 lb>ft and segment CD weighs 250 lb>ft. In (b), the column has a mass of 200 kg>m. (a) + c ©Fy = 0;
5 kip
8 kN
B
200 mm
200 mm
6 kN
6 kN
FA - 1.0 - 3 - 3 - 1.8 - 5 = 0 10 ft
FA = 13.8 kip
8 in.
Ans.
3m
8 in. 200 mm
(b) + c ©Fy = 0;
3 kip
FA - 4.5 - 4.5 - 5.89 - 6 - 6 - 8 = 0 FA = 34.9 kN
3 kip
200 mm
4.5 kN
4.5 kN
C
Ans. 4 ft
A
A
1m
4 ft D (a)
1–2. Determine the resultant internal torque acting on the cross sections through points C and D. The support bearings at A and B allow free turning of the shaft. ©Mx = 0;
A 250 Nm 300 mm
TC - 250 = 0 TC = 250 N # m
©Mx = 0;
(b)
400 Nm
200 mm
Ans.
TD = 0
150 Nm
C
150 mm
Ans.
200 mm
B
D
250 mm 150 mm
1–3. Determine the resultant internal torque acting on the cross sections through points B and C. A
©Mx = 0;
TB = 150 lb # ft ©Mx = 0;
600 lbft B 350 lbft
TB + 350 - 500 = 0 Ans.
TC - 500 = 0 TC =
3 ft
C 500 lbft
1 ft
500 lb # ft
Ans.
2 ft 2 ft
1
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*1–4. A force of 80 N is supported by the bracket as shown. Determine the resultant internal loadings acting on the section through point A.
0.3 m A 30 0.1 m
80 N
Equations of Equilibrium: +
Q©Fx¿ = 0;
NA - 80 cos 15° = 0 NA = 77.3 N
a+ ©Fy¿ = 0;
Ans.
VA - 80 sin 15° = 0 VA = 20.7 N
a+
©MA = 0;
Ans.
MA + 80 cos 45°(0.3 cos 30°) - 80 sin 45°(0.1 + 0.3 sin 30°) = 0 MA = - 0.555 N # m
Ans.
or a+
©MA = 0;
MA + 80 sin 15°(0.3 + 0.1 sin 30°) -80 cos 15°(0.1 cos 30°) = 0 MA = - 0.555 N # m
Ans.
Negative sign indicates that MA acts in the opposite direction to that shown on FBD.
2
45
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•1–5.
Determine the resultant internal loadings in the beam at cross sections through points D and E. Point E is just to the right of the 3-kip load.
3 kip 1.5 kip/ ft
A D 6 ft
Support Reactions: For member AB a + ©MB = 0; + : ©Fx = 0; + c ©Fy = 0;
9.00(4) - A y(12) = 0
A y = 3.00 kip
Bx = 0 By + 3.00 - 9.00 = 0
By = 6.00 kip
Equations of Equilibrium: For point D + : ©Fx = 0; + c ©Fy = 0;
ND = 0
Ans.
3.00 - 2.25 - VD = 0 VD = 0.750 kip
a + ©MD = 0;
Ans.
MD + 2.25(2) - 3.00(6) = 0 MD = 13.5 kip # ft
Ans.
Equations of Equilibrium: For point E + : ©Fx = 0; + c ©Fy = 0;
NE = 0
Ans.
- 6.00 - 3 - VE = 0 VE = - 9.00 kip
a + ©ME = 0;
Ans.
ME + 6.00(4) = 0 ME = - 24.0 kip # ft
Ans.
Negative signs indicate that ME and VE act in the opposite direction to that shown on FBD.
3
E
B 6 ft
4 ft
4 ft
C
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1–6. Determine the normal force, shear force, and moment at a section through point C. Take P = 8 kN.
B
0.1 m
Support Reactions: a + ©MA = 0;
8(2.25) - T(0.6) = 0
0.5 m C
T = 30.0 kN
+ : ©Fx = 0;
30.0 - A x = 0
A x = 30.0 kN
+ c ©Fy = 0;
Ay - 8 = 0
A y = 8.00 kN
0.75 m
0.75 m
A 0.75 m
P
Equations of Equilibrium: For point C + : ©Fx = 0;
- NC - 30.0 = 0 NC = - 30.0 kN
+ c ©Fy = 0;
Ans.
VC + 8.00 = 0 VC = - 8.00 kN
a + ©MC = 0;
Ans.
8.00(0.75) - MC = 0 MC = 6.00 kN # m
Ans.
Negative signs indicate that NC and VC act in the opposite direction to that shown on FBD.
1–7. The cable will fail when subjected to a tension of 2 kN. Determine the largest vertical load P the frame will support and calculate the internal normal force, shear force, and moment at the cross section through point C for this loading.
B
0.1 m
0.5 m C
Support Reactions: a + ©MA = 0;
0.75 m
P(2.25) - 2(0.6) = 0 P
P = 0.5333 kN = 0.533 kN + : ©Fx = 0;
2 - Ax = 0
+ c ©Fy = 0;
A y - 0.5333 = 0
Ans.
A x = 2.00 kN A y = 0.5333 kN
Equations of Equilibrium: For point C + : ©Fx = 0;
- NC - 2.00 = 0 NC = - 2.00 kN
+ c ©Fy = 0;
Ans.
VC + 0.5333 = 0 VC = - 0.533 kN
a + ©MC = 0;
Ans.
0.5333(0.75) - MC = 0 MC = 0.400 kN # m
Ans.
Negative signs indicate that NC and VC act in the opposite direction to that shown on FBD.
4
0.75 m
A 0.75 m
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*1–8. Determine the resultant internal loadings on the cross section through point C. Assume the reactions at the supports A and B are vertical.
6 kN 3 kN/m
Referring to the FBD of the entire beam, Fig. a, a + ©MB = 0;
1 - A y(4) + 6(3.5) + (3)(3)(2) = 0 2
+ c ©Fy = 0; a + ©MC = 0;
C
A y = 7.50 kN
NC = 0
D 1.5 m
0.5 m 0.5 m
Referring to the FBD of this segment, Fig. b, + : ©Fx = 0;
B
A
1.5 m
Ans.
7.50 - 6 - VC = 0
VC = 1.50 kN
MC + 6(0.5) - 7.5(1) = 0
Ans.
MC = 4.50 kN # m
Ans.
•1–9.
Determine the resultant internal loadings on the cross section through point D. Assume the reactions at the supports A and B are vertical.
6 kN 3 kN/m
Referring to the FBD of the entire beam, Fig. a,
B
A
a + ©MA = 0;
By(4) - 6(0.5) -
1 (3)(3)(2) = 0 2
By = 3.00 kN 0.5 m 0.5 m
Referring to the FBD of this segment, Fig. b, + : ©Fx = 0; + c ©Fy = 0;
ND = 0 VD -
1 (1.5)(1.5) + 3.00 = 0 2
a + ©MD = 0; 3.00(1.5) -
C
Ans. VD = - 1.875 kN
Ans.
1 (1.5)(1.5)(0.5) - MD = 0 MD = 3.9375 kN # m 2 = 3.94 kN # m
5
Ans.
D 1.5 m
1.5 m
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1–10. The boom DF of the jib crane and the column DE have a uniform weight of 50 lb/ft. If the hoist and load weigh 300 lb, determine the resultant internal loadings in the crane on cross sections through points A, B, and C.
D 2 ft
F
A
B 8 ft
3 ft
5 ft C 300 lb 7 ft
E
Equations of Equilibrium: For point A + ; © Fx = 0; + c © Fy = 0;
NA = 0
Ans.
VA - 150 - 300 = 0 VA = 450 lb
a + ©MA = 0;
Ans.
- MA - 150(1.5) - 300(3) = 0 MA = - 1125 lb # ft = - 1.125 kip # ft
Ans.
Negative sign indicates that MA acts in the opposite direction to that shown on FBD. Equations of Equilibrium: For point B + ; © Fx = 0;
NB = 0
+ c © Fy = 0;
VB - 550 - 300 = 0
Ans.
VB = 850 lb a + © MB = 0;
Ans.
- MB - 550(5.5) - 300(11) = 0 MB = - 6325 lb # ft = - 6.325 kip # ft
Ans.
Negative sign indicates that MB acts in the opposite direction to that shown on FBD. Equations of Equilibrium: For point C + ; © Fx = 0; + c © Fy = 0;
VC = 0
Ans.
- NC - 250 - 650 - 300 = 0 NC = - 1200 lb = - 1.20 kip
a + ©MC = 0;
Ans.
- MC - 650(6.5) - 300(13) = 0 MC = - 8125 lb # ft = - 8.125 kip # ft
Ans.
Negative signs indicate that NC and MC act in the opposite direction to that shown on FBD.
6
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1–11. The force F = 80 lb acts on the gear tooth. Determine the resultant internal loadings on the root of the tooth, i.e., at the centroid point A of section a–a.
a
F 80 lb 30
Equations of Equilibrium: For section a–a +
Q©Fx¿ = 0;
VA - 80 cos 15° = 0
0.23 in.
VA = 77.3 lb a+ ©Fy¿ = 0;
Ans. A
NA - 80 sin 15° = 0
0.16 in.
NA = 20.7 lb a + ©MA = 0;
Ans.
- MA - 80 sin 15°(0.16) + 80 cos 15°(0.23) = 0 MA = 14.5 lb # in.
45
Ans.
*1–12. The sky hook is used to support the cable of a scaffold over the side of a building. If it consists of a smooth rod that contacts the parapet of a wall at points A, B, and C, determine the normal force, shear force, and moment on the cross section at points D and E.
a
0.2 m
0.2 m B
0.2 m
0.2 m D
E
Support Reactions: + c ©Fy = 0;
NB - 18 = 0
d+ ©MC = 0;
18(0.7) - 18.0(0.2) - NA(0.1) = 0
0.2 m
0.3 m
NB = 18.0 kN
A
NA = 90.0 kN + : ©Fx = 0;
NC - 90.0 = 0
0.3 m
NC = 90.0 kN
Equations of Equilibrium: For point D + : © Fx = 0;
18 kN
VD - 90.0 = 0 VD = 90.0 kN
+ c © Fy = 0;
Ans.
ND - 18 = 0 ND = 18.0 kN
d+ © MD = 0;
Ans.
MD + 18(0.3) - 90.0(0.3) = 0 MD = 21.6 kN # m
Ans.
Equations of Equilibrium: For point E + : © Fx = 0; 90.0 - VE = 0 VE = 90.0 kN + c © Fy = 0; d + © ME = 0;
Ans.
NE = 0
Ans.
90.0(0.2) - ME = 0 ME = 18.0 kN # m
Ans.
7
C
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•1–13.
The 800-lb load is being hoisted at a constant speed using the motor M, which has a weight of 90 lb. Determine the resultant internal loadings acting on the cross section through point B in the beam. The beam has a weight of 40 lb>ft and is fixed to the wall at A.
M 1.5 ft A D 4 ft
+ : ©Fx = 0;
3 ft
3 ft
4 ft
0.25 ft
Ans.
VB - 0.8 - 0.16 = 0 VB = 0.960 kip
a + ©MB = 0;
B
- NB - 0.4 = 0 NB = - 0.4 kip
+ c ©Fy = 0;
4 ft
C
Ans.
- MB - 0.16(2) - 0.8(4.25) + 0.4(1.5) = 0 MB = - 3.12 kip # ft
Ans.
1–14. Determine the resultant internal loadings acting on the cross section through points C and D of the beam in Prob. 1–13.
M 1.5 ft A D 4 ft
4 ft
C
B
3 ft
3 ft
4 ft
For point C: + ; ©Fx = 0; + c ©Fy = 0; a + ©MC = 0;
NC + 0.4 = 0;
NC = - 0.4kip
VC - 0.8 - 0.04 (7) = 0;
Ans.
VC = 1.08 kip
Ans.
- MC - 0.8(7.25) - 0.04(7)(3.5) + 0.4(1.5) = 0 MC = - 6.18 kip # ft
Ans.
ND = 0
Ans.
For point D: + ; ©Fx = 0; + c ©Fy = 0; a + ©MD = 0;
VD - 0.09 - 0.04(14) - 0.8 = 0;
VD = 1.45 kip
Ans.
- MD - 0.09(4) - 0.04(14)(7) - 0.8(14.25) = 0 MD = - 15.7 kip # ft
Ans.
8
0.25 ft
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1–15. Determine the resultant internal loading on the cross section through point C of the pliers. There is a pin at A, and the jaws at B are smooth.
20 N 40 mm
120 mm
15 mm C
+ c ©Fy = 0; + : ©Fx = 0; +d ©MC = 0;
- VC + 60 = 0;
VC = 60 N
Ans.
NC = 0 - MC + 60(0.015) = 0;
MC = 0.9 N.m
D
Ans. 80 mm 20 N
*1–16. Determine the resultant internal loading on the cross section through point D of the pliers.
B
A
Ans.
30
20 N 40 mm
120 mm
15 mm
R+ ©Fy = 0;
VD - 20 cos 30° = 0;
VD = 17.3 N
Ans.
+b©Fx = 0;
ND - 20 sin 30° = 0;
ND = 10 N
Ans.
+d ©MD = 0;
MD - 20(0.08) = 0;
MD = 1.60 N.m
Ans.
C A D 80 mm 20 N
9
30
B
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•1–17.
Determine resultant internal loadings acting on section a–a and section b–b. Each section passes through the centerline at point C.
5 kN
B
b a
Referring to the FBD of the entire beam, Fig. a, a + ©MA = 0;
NB sin 45°(6) - 5(4.5) = 0
1.5 m
NB = 5.303 kN
C
Referring to the FBD of this segment (section a–a), Fig. b, b
+b©Fx¿ = 0;
Na - a + 5.303 cos 45° = 0
Na - a = - 3.75 kN Va - a = 1.25 kN
Ans.
+a ©Fy¿ = 0;
Va - a + 5.303 sin 45° - 5 = 0
a + ©MC = 0;
5.303 sin 45°(3) - 5(1.5) - Ma - a = 0 Ma - a = 3.75 kN # m Ans.
Ans.
Referring to the FBD (section b–b) in Fig. c, + ; ©Fx = 0;
Nb - b - 5 cos 45° + 5.303 = 0 Nb - b = - 1.768 kN = - 1.77 kN
+ c ©Fy = 0; a + ©MC = 0;
Vb - b - 5 sin 45° = 0
Vb - b = 3.536 kN = 3.54 kN
Ans. Ans.
5.303 sin 45° (3) - 5(1.5) - Mb - b = 0 Mb - b = 3.75 kN # m
10
Ans.
A
45
3m
45
1.5 m a
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1–18. The bolt shank is subjected to a tension of 80 lb. Determine the resultant internal loadings acting on the cross section at point C.
C 6 in. 90
A
B
Segment AC: + : ©Fx = 0;
NC + 80 = 0;
+ c ©Fy = 0;
VC = 0
NC = - 80 lb
Ans. Ans.
MC = - 480 lb # in.
Ans.
1–19. Determine the resultant internal loadings acting on the cross section through point C. Assume the reactions at the supports A and B are vertical.
6 kip/ft
a + ©MC = 0;
MC + 80(6) = 0;
6 kip/ft
A
C 3 ft
Referring to the FBD of the entire beam, Fig. a, a + ©MB = 0;
1 1 (6)(6)(2) + (6)(6)(10) - A y(12) = 0 A y = 18.0 kip 2 2
Referring to the FBD of this segment, Fig. b, + : ©Fx = 0;
NC = 0
+ c ©Fy = 0;
18.0 -
a + ©MC = 0;
Ans. 1 (3)(3) - (3)(3) - VC = 0 2
MC + (3)(3)(1.5) +
VC = 4.50 kip
Ans.
1 (3)(3)(2) - 18.0(3) = 0 2
MC = 31.5 kip # ft
Ans.
11
B
D 3 ft
6 ft
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*1–20. Determine the resultant internal loadings acting on the cross section through point D. Assume the reactions at the supports A and B are vertical.
6 kip/ft
Referring to the FBD of the entire beam, Fig. a,
A
a + ©MB = 0;
+ c ©Fy = 0; a + ©MA = 0;
C
1 1 (6)(6)(2) + (6)(6)(10) - A y(12) = 0 A y = 18.0 kip 2 2
Referring to the FBD of this segment, Fig. b, + : ©Fx = 0; ND = 0 18.0 -
1 (6)(6) - VD = 0 2
MD - 18.0 (2) = 0
6 kip/ft
3 ft
B
D 3 ft
6 ft
Ans. VD = 0
Ans.
MD = 36.0 kip # ft
Ans.
The forged steel clamp exerts a force of F = 900 N on the wooden block. Determine the resultant internal loadings acting on section a–a passing through point A.
•1–21.
200 mm F 900 N
Internal Loadings: Referring to the free-body diagram of the section of the clamp shown in Fig. a, ©Fy¿ = 0;
900 cos 30° - Na - a = 0
Na - a = 779 N
Ans.
©Fx¿ = 0;
Va - a - 900 sin 30° = 0
Va - a = 450 N
Ans.
a + ©MA = 0;
900(0.2) - Ma - a = 0
Ma - a = 180 N # m
12
Ans.
a
30 a
A
F 900 N
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1–22. The floor crane is used to lift a 600-kg concrete pipe. Determine the resultant internal loadings acting on the cross section at G.
0.2 m 0.2 m
0.4 m
0.6 m
G E B
F 0.3 m
H
C 0.5 m 75
A
Support Reactions: We will only need to compute FEF by writing the moment equation of equilibrium about D with reference to the free-body diagram of the hook, Fig. a. a + ©MD = 0;
FEF(0.3) - 600(9.81)(0.5) = 0
FEF = 9810 N
Internal Loadings: Using the result for FEF, section FG of member EF will be considered. Referring to the free-body diagram, Fig. b, + : ©Fx = 0; + c ©Fy = 0; a + ©MG = 0;
9810 - NG = 0
NG = 9810 N = 9.81 kN
VG = 0
Ans. Ans.
MG = 0
Ans.
13
D
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1–23. The floor crane is used to lift a 600-kg concrete pipe. Determine the resultant internal loadings acting on the cross section at H.
0.2 m 0.2 m
0.4 m
0.6 m
G E B
F 0.3 m
H
C 0.5 m 75
A
Support Reactions: Referring to the free-body diagram of the hook, Fig. a. a + ©MF = 0;
Dx(0.3) - 600(9.81)(0.5) = 0
Dx = 9810 N
+ c ©Fy = 0;
Dy - 600(9.81) = 0
Dy = 5886 N
Subsequently, referring to the free-body diagram of member BCD, Fig. b, a + ©MB = 0;
FAC sin 75°(0.4) - 5886(1.8) = 0
+ : ©Fx = 0;
Bx + 27 421.36 cos 75° - 9810 = 0 Bx = 2712.83 N
+ c ©Fy = 0;
27 421.36 sin 75° - 5886 - By = 0
FAC = 27 421.36 N
By = 20 601 N
Internal Loadings: Using the results of Bx and By, section BH of member BCD will be considered. Referring to the free-body diagram of this part shown in Fig. c, + : ©Fx = 0;
NH + 2712.83 = 0
NH = - 2712.83 N = - 2.71 kN
Ans.
+ c ©Fy = 0;
- VH - 2060 = 0
VH = - 20601 N = - 20.6 kN
Ans.
a + ©MD = 0;
MH + 20601(0.2) = 0
MH = - 4120.2 N # m = - 4.12 kN # m Ans.
The negative signs indicates that NH, VH, and MH act in the opposite sense to that shown on the free-body diagram.
14
D
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*1–24. The machine is moving with a constant velocity. It has a total mass of 20 Mg, and its center of mass is located at G, excluding the front roller. If the front roller has a mass of 5 Mg, determine the resultant internal loadings acting on point C of each of the two side members that support the roller. Neglect the mass of the side members. The front roller is free to roll.
2m
G
C
B
A
1.5 m
Support Reactions: We will only need to compute NA by writing the moment equation of equilibrium about B with reference to the free-body diagram of the steamroller, Fig. a. a + ©MB = 0; NA (5.5) - 20(103)(9.81)(1.5) = 0
NA = 53.51(103) N
Internal Loadings: Using the result for NA, the free-body diagram of the front roller shown in Fig. b will be considered. + ; ©Fx = 0; 2NC = 0
NC = 0
+ c ©Fy = 0; 2VC + 53.51(103) - 5(103)(9.81) = 0
VC = - 2229.55 N = - 2.23 kN
Ans.
Ans.
a + ©MC = 0; 53.51(103)(2) - 5(103)(9.81)(2) - 2MC = 0 MC = 4459.10 N # m = 4.46 kN # m Ans.
15
4m
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z
•1–25.
Determine the resultant internal loadings acting on the cross section through point B of the signpost. The post is fixed to the ground and a uniform pressure of 7 lb>ft2 acts perpendicular to the face of the sign.
3 ft 2 ft
©Fx = 0;
(VB)x - 105 = 0;
(VB)x = 105 lb
©Fy = 0;
(VB)y = 0
Ans.
©Fz = 0;
(NB)z = 0
Ans.
Ans. 3 ft 7 lb/ft2
Ans.
©Mx = 0;
(MB)x = 0
©My = 0;
(MB)y - 105(7.5) = 0;
©Mz = 0;
(TB)z - 105(0.5) = 0;
(MB)y = 788 lb # ft
6 ft
Ans.
(TB)z = 52.5 lb # ft
B
Ans. A
4 ft
y
x
1–26. The shaft is supported at its ends by two bearings A and B and is subjected to the forces applied to the pulleys fixed to the shaft. Determine the resultant internal loadings acting on the cross section located at point C. The 300-N forces act in the z direction and the 500-N forces act in the x direction. The journal bearings at A and B exert only x and z components of force on the shaft.
z
A
400 mm 150 mm 200 mm C 250 mm
x
300 N
300 N B
500 N 500 N y
©Fx = 0;
(VC)x + 1000 - 750 = 0;
©Fy = 0;
(NC)y = 0
©Fz = 0;
(VC)z + 240 = 0;
(VC)x = - 250 N
Ans. Ans. Ans.
(VC)z = - 240 N (MC)x = - 108 N # m
©Mx = 0;
(MC)x + 240(0.45) = 0;
©My = 0;
(TC)y = 0
©Mz = 0;
(MC)z - 1000(0.2) + 750(0.45) = 0;
Ans. Ans.
(MC)z = - 138 N # m Ans.
16
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1–27. The pipe has a mass of 12 kg>m. If it is fixed to the wall at A, determine the resultant internal loadings acting on the cross section at B. Neglect the weight of the wrench CD.
z
A
300 mm
200 mm B 60 N
x
D 400 mm
60 N 150 mm C
©Fx = 0;
(NB)x = 0
Ans.
©Fy = 0;
(VB)y = 0
Ans.
©Fz = 0;
(VB)z - 60 + 60 - (0.2)(12)(9.81) - (0.4)(12)(9.81) = 0 Ans.
(VB)z = 70.6 N ©Mx = 0;
(TB)x + 60(0.4) - 60(0.4) - (0.4)(12)(9.81)(0.2) = 0 (TB)x = 9.42 N # m
©My = 0;
Ans.
(MB)y + (0.2)(12)(9.81)(0.1) + (0.4)(12)(9.81)(0.2) - 60(0.3) = 0 (MB)y = 6.23 N # m
©Mz = 0;
Ans. Ans.
(MB)z = 0
17
150 mm
y
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z
*1–28. The brace and drill bit is used to drill a hole at O. If the drill bit jams when the brace is subjected to the forces shown, determine the resultant internal loadings acting on the cross section of the drill bit at A.
O
x
Internal Loading: Referring to the free-body diagram of the section of the drill and brace shown in Fig. a, ©Fx = 0; ©Fy = 0; ©Fz = 0; ©Mx = 0; ©My = 0; ©Mz = 0;
A VA B x - 30 = 0
A NA B y - 50 = 0 A VA B z - 10 = 0
A MA B x - 10(2.25) = 0
A TA B y - 30(0.75) = 0
A MA B z + 30(1.25) = 0
A VA B x = 30 lb
Ans.
A NA B y = 50 lb
Ans.
A VA B z = 10 lb
Ans.
A MA B x = 22.5 lb # ft A TA B y = 22.5 lb # ft
A MA B z = - 37.5 lb # ft
Ans. Ans. Ans.
The negative sign indicates that (MA)Z acts in the opposite sense to that shown on the free-body diagram.
18
3 in. 9 in.
Fx 30 lb
A
Fz 10 lb 9 in.
6 in.
6 in.
6 in.
Fy 50 lb y
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•1–29. The curved rod has a radius r and is fixed to the wall at B. Determine the resultant internal loadings acting on the cross section through A which is located at an angle u from the horizontal.
B
A r
U
Equations of Equilibrium: For point A R+ ©Fx = 0;
P
P cos u - NA = 0 NA = P cos u
Q+ ©Fy = 0;
Ans.
VA - P sin u = 0 VA = P sin u
d+ ©MA = 0;
Ans.
MA - P[r(1 - cos u)] = 0 MA = Pr(1 - cos u)
Ans.
19
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1–30. A differential element taken from a curved bar is shown in the figure. Show that dN>du = V, dV>du = - N, dM>du = - T, and dT>du = M.
M dM V dV
©Fx = 0; (1)
N
©Fy = 0; N sin
du du du du - V cos + (N + dN) sin + (V + dV) cos = 0 2 2 2 2
(2)
©Mx = 0; T cos
du du du du + M sin - (T + dT) cos + (M + dM) sin = 0 2 2 2 2
(3)
©My = 0; du du du du - M cos + (T + dT) sin + (M + dM) cos = 0 2 2 2 2 du du du du Since is can add, then sin , cos = = 1 2 2 2 2 T sin
Eq. (1) becomes Vdu - dN +
dVdu = 0 2
Neglecting the second order term, Vdu - dN = 0 dN = V du Eq. (2) becomes Ndu + dV +
QED dNdu = 0 2
Neglecting the second order term, Ndu + dV = 0 dV = -N du Eq. (3) becomes Mdu - dT +
QED dMdu = 0 2
Neglecting the second order term, Mdu - dT = 0 dT = M du Eq. (4) becomes Tdu + dM +
QED dTdu = 0 2
Neglecting the second order term, Tdu + dM = 0 dM = -T du
N dN
M V
du du du du + V sin - (N + dN) cos + (V + dV) sin = 0 N cos 2 2 2 2
QED
20
(4)
T
T dT
du
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1–31. The column is subjected to an axial force of 8 kN, which is applied through the centroid of the cross-sectional area. Determine the average normal stress acting at section a–a. Show this distribution of stress acting over the area’s cross section.
8 kN 75 mm 75 mm
10 mm
10 mm
70 mm
10 mm
70 mm a a
A = (2)(150)(10) + (140)(10) = 4400 mm2 = 4.4 (10-3) m2 s =
8 (103) P = 1.82 MPa = A 4.4 (10 - 3)
Ans.
*1–32. The lever is held to the fixed shaft using a tapered pin AB, which has a mean diameter of 6 mm. If a couple is applied to the lever, determine the average shear stress in the pin between the pin and lever.
B 12 mm A 250 mm 20 N
a + ©MO = 0; tavg =
- F(12) + 20(500) = 0;
F = 833.33 N
V 833.33 = p 6 = 29.5 MPa 2 A 4 (1000 )
Ans.
21
250 mm 20 N
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•1–33.
The bar has a cross-sectional area A and is subjected to the axial load P. Determine the average normal and average shear stresses acting over the shaded section, which is oriented at u from the horizontal. Plot the variation of these stresses as a function of u 10 … u … 90°2.
P
P u A
Equations of Equilibrium: R+ ©Fx = 0;
V - P cos u = 0
V = P cos u
Q+ ©Fy = 0;
N - P sin u = 0
N = P sin u
Average Normal Stress and Shear Stress: Area at u plane, A¿ =
s =
A . sin u
P sin u N P = = sin2 u A A¿ A sin u
tavg =
Ans.
V P cos u = A A¿ sin u =
P P sin u cos u = sin 2u A 2A
Ans.
1–34. The built-up shaft consists of a pipe AB and solid rod BC. The pipe has an inner diameter of 20 mm and outer diameter of 28 mm. The rod has a diameter of 12 mm. Determine the average normal stress at points D and E and represent the stress on a volume element located at each of these points.
4 kN
D
At D: sD =
4(103)
P = A
p 2 4 (0.028
P = A
p 4
- 0.02 2)
= 13.3 MPa (C)
Ans.
At E: sE =
8(103) (0.012 2)
B
A
= 70.7 MPa (T)
Ans.
22
6 kN 6 kN E
C
8 kN
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1–35. The bars of the truss each have a cross-sectional area of 1.25 in2. Determine the average normal stress in each member due to the loading P = 8 kip. State whether the stress is tensile or compressive.
B
3 ft
A 4 ft
P
Joint A: sAB =
FAB 13.33 = = 10.7 ksi A AB 1.25
(T)
Ans.
sAE =
FAE 10.67 = = 8.53 ksi A AE 1.25
(C)
Ans.
(C)
Ans.
Joint E: sED =
FED 10.67 = = 8.53 ksi A ED 1.25
sEB =
FEB 6.0 = = 4.80 ksi A EB 1.25
C
Ans.
(T)
Joint B: sBC =
FBC 29.33 = 23.5 ksi = A BC 1.25
(T)
Ans.
sBD =
FBD 23.33 = = 18.7 ksi A BD 1.25
(C)
Ans.
23
E 0.75 P
4 ft
D
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*1–36. The bars of the truss each have a cross-sectional area of 1.25 in2. If the maximum average normal stress in any bar is not to exceed 20 ksi, determine the maximum magnitude P of the loads that can be applied to the truss.
B
3 ft
A 4 ft
P
Joint A: + c ©Fy = 0;
3 - P + a b FAB = 0 5
FAB = (1.667)P + : ©Fx = 0;
4 - FAE + (1.667)Pa b = 0 5 FAE = (1.333)P
Joint E: + c ©Fy = 0;
FEB - (0.75)P = 0 FEB = (0.75)P
+ : ©Fx = 0;
(1.333)P - FED = 0 FED = (1.333)P
Joint B: + c ©Fy = 0;
3 3 a b FBD - (0.75)P - (1.667)Pa b = 0 5 5 FBD = (2.9167)P
+ : ©Fx = 0;
4 4 FBC - (2.9167)Pa b - (1.667)P a b = 0 5 5
FBC = (3.67)P The highest stressed member is BC:
sBC =
C
(3.67)P = 20 1.25
P = 6.82 kip
Ans.
24
E 0.75 P
4 ft
D
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•1–37.
The plate has a width of 0.5 m. If the stress distribution at the support varies as shown, determine the force P applied to the plate and the distance d to where it is applied.
4m P d x
s (15x1/2) MPa
The resultant force dF of the bearing pressure acting on the plate of area dA = b dx = 0.5 dx, Fig. a, 1
1
dF = sb dA = (15x2)(106)(0.5dx) = 7.5(106)x2 dx + c ©Fy = 0;
L
dF - P = 0 4m 1
L0
7.5(106)x2 dx - P = 0
P = 40(106) N = 40 MN
Ans.
Equilibrium requires a + ©MO = 0;
L
xdF - Pd = 0
4m 1
L0
x[7.5(106)x2 dx] - 40(106) d = 0 d = 2.40 m
Ans.
25
30 MPa
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1–38. The two members used in the construction of an aircraft fuselage are joined together using a 30° fish-mouth weld. Determine the average normal and average shear stress on the plane of each weld. Assume each inclined plane supports a horizontal force of 400 lb. N - 400 sin 30° = 0;
N = 200 lb
400 cos 30° - V = 0;
V = 346.41 lb
A¿ =
1.5 in.
30
1 in. 1 in.
800 lb
800 lb 30
1.5(1) = 3 in2 sin 30°
s =
N 200 = = 66.7 psi A¿ 3
Ans.
t =
V 346.41 = = 115 psi A¿ 3
Ans.
1–39. If the block is subjected to the centrally applied force of 600 kN, determine the average normal stress in the material. Show the stress acting on a differential volume element of the material.
150 mm 600 kN
150 mm 150 mm 150 mm
The cross-sectional area of the block is A = 0.6(0.3) - 0.3(0.2) = 0.12 m2. savg =
600(103) P = = 5(106) Pa = 5 MPa A 0.12
Ans.
The average normal stress distribution over the cross-section of the block and the state of stress of a point in the block represented by a differential volume element are shown in Fig. a
26
50 mm 100 mm 100 mm 50 mm
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*1–40. The pins on the frame at B and C each have a diameter of 0.25 in. If these pins are subjected to double shear, determine the average shear stress in each pin.
3 ft 500 lb
3 ft A 3 ft
Support Reactions: FBD(a) a + ©Mg = 0;
500(6) + 300(3) - Dy (6) = 0 Dy = 650 lb
+ ; ©Fx = 0;
500 - Ex = 0
Ex = 500 lb
+ c ©Fy = 0;
650 - 300 - Ey = 0
Ey = 350 lb
a + ©MB = 0;
Cy (3) - 300(1.5) = 0
Cy = 150 lb
+ c ©Fy = 0;
By + 150 - 300 = 0
By = 150 lb
1.5 ft
From FBD (b) 150(1.5) + Bx(3) - 650(3) = 0 Bx = 575 lb From FBD (c), + : ©Fx = 0;
Cx - 575 = 0
Cx = 575 lb
Hence, FB = FC = 2 5752 + 1502 = 594.24 lb Average shear stress: Pins B and C are subjected to double shear as shown on FBD (d) (tB)avg = (tC)avg =
1.5 ft
300 lb D
From FBD (c),
a + ©MA = 0;
C
B
V 297.12 = p 2 A 4 (0.25 ) = 6053 psi = 6.05 ksi
Ans.
27
3 ft E
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•1–41.
Solve Prob. 1–40 assuming that pins B and C are subjected to single shear.
3 ft 500 lb
3 ft A 3 ft C
B
Support Reactions: FBD(a) a + ©Mg = 0;
1.5 ft
500(6) + 300(3) - Dy (6) = 0
300 lb D
Dy = 650 lb + ; ©Fx = 0;
500 - Ex = 0
Ex = 500 lb
+ c ©Fy = 0;
650 - 300 - Ey = 0
Ey = 350 lb
From FBD (c), a + ©MB = 0; + c ©Fy = 0;
Cy (3) - 300(1.5) = 0 By + 150 - 300 = 0
Cy = 150 lb By = 150 lb
From FBD (b) d+ ©MA = 0;
150(1.5) + Bx(3) - 650(3) = 0 Bx = 575 lb
From FBD (c), + : ©Fx = 0;
Cx - 575 = 0
Cx = 575 lb
Hence, FB = FC = 2 5752 + 1502 = 594.24 lb Average shear stress: Pins B and C are subjected to single shear as shown on FBD (d) (tB)avg = (tC)avg =
1.5 ft
594.24 V = p 2 A 4 (0.25 ) = 12106 psi = 12.1 ksi
Ans.
28
3 ft E
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1–42. The pins on the frame at D and E each have a diameter of 0.25 in. If these pins are subjected to double shear, determine the average shear stress in each pin.
3 ft 500 lb
3 ft A 3 ft C
B 1.5 ft
1.5 ft
Support Reactions: FBD(a) a + ©ME = 0;
300 lb
500(6) + 300(3) - Dy(6) = 0
D
Dy = 650 lb + ; ©Fx = 0;
500 - Ex = 0
Ex = 500 lb
+ c ©Fy = 0;
650 - 300 - Ey = 0
Ey = 350 lb
Average shear stress: Pins D and E are subjected to double shear as shown on FBD (b) and (c). For Pin D, FD = Dy = 650 lb then VD = (pD)avg =
VD = AD
FD z
= 325 lb
325 p 2 4 (0.25)
Ans.
= 6621 psi = 6.62 ksi For Pin E, FE = 2 5002 + 3502 = 610.32 lb then VE = (tE)avg =
Fg z
= 305.16 lb
VE 305.16 = p 2 AE 4 (0.25 ) = 6217 psi = 6.22 ksi
Ans.
29
3 ft E
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1–43. Solve Prob. 1–42 assuming that pins D and E are subjected to single shear.
3 ft 500 lb
3 ft A
Support Reactions: FBD(a) a + ©ME = 0;
3 ft
500(6) + 300(3) - Dy(6) = 0
+ ; ©Fx = 0;
500 - Ex = 0 650 - 300 - Ey = 0
+ c ©Fy = 0;
C
B
Dy = 650 lb
1.5 ft
Ex = 500 lb Ey = 350 lb
1.5 ft
300 lb D
Average shear stress: Pins D and E are subjected to single shear as shown on FBD (b) and (c). For Pin D, VD = FD = Dy = 650 lb (tD)avg =
VD = AD
650 p 2 4 (0.25 )
Ans.
= 13242 psi = 13.2 ksi For Pin E, VE = FE = 2 5002 + 3502 = 610.32 lb (tE)avg =
VE 610.32 = p 2 AE 4 (0.25 ) = 12433 psi = 12.4 ksi
Ans.
*1–44. A 175-lb woman stands on a vinyl floor wearing stiletto high-heel shoes. If the heel has the dimensions shown, determine the average normal stress she exerts on the floor and compare it with the average normal stress developed when a man having the same weight is wearing flat-heeled shoes. Assume the load is applied slowly, so that dynamic effects can be ignored. Also, assume the entire weight is supported only by the heel of one shoe. Stiletto shoes: A =
1.2 in.
1 (p)(0.3)2 + (0.6)(0.1) = 0.2014 in2 2
0.3 in. 0.1 in. 0.5 in.
P 175 lb = s = = 869 psi A 0.2014 in2
Ans.
Flat-heeled shoes: A =
1 (p)(1.2)2 + 2.4(0.5) = 3.462 in2 2
s =
P 175 lb = = 50.5 psi A 3.462 in2
Ans.
30
3 ft E
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•1–45.
The truss is made from three pin-connected members having the cross-sectional areas shown in the figure. Determine the average normal stress developed in each member when the truss is subjected to the load shown. State whether the stress is tensile or compressive.
500 lb
C
sBC =
FBC 375 = = 469 psi A BC 0.8
(T)
Ans.
(T)
Ans.
4 ft
Joint A: FAC 500 = = 833 psi A AC 0.6
.2 in
A
1–46. Determine the average normal stress developed in links AB and CD of the smooth two-tine grapple that supports the log having a mass of 3 Mg. The cross-sectional area of each link is 400 mm2. + c ©Fy = 0;
1.5
Ans.
(C)
AB
FAB 625 = = 417 psi A AB 1.5
A
sAB =
AAC 0.6 in.2
ABC 0.8 in.2
Joint B:
œ sAC =
3 ft
A
C
20 B
E
D
2(F sin 30°) - 29.43 = 0 0.2 m
F = 29.43 kN a + ©ME = 0; P cos 20°(0.2) - (29.43 cos 30°)(1.2) + (29.43 sin 30°)(0.4 cos 30°)
1.2 m
= 0 P = 135.61 kN
30
30 0.4 m
s =
135.61(103) P = 339 MPa = A 400(10 - 6)
Ans.
31
B
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1–47. Determine the average shear stress developed in pins A and B of the smooth two-tine grapple that supports the log having a mass of 3 Mg. Each pin has a diameter of 25 mm and is subjected to double shear. + c ©Fy = 0;
A
C
20 E
B
D
2(F sin 30°) - 29.43 = 0 0.2 m
F = 29.43 kN 1.2 m
a + ©ME = 0; P cos 20°(0.2) - (29.43 cos 30°)(1.2) + (29.43 sin 30°)(0.4 cos 30°) = 0 P = 135.61 kN tA = tB =
V = A
135.61(103) 2 p 2 (0.025) 4
30
30 0.4 m
= 138 MPa
Ans.
*1–48. The beam is supported by a pin at A and a short link BC. If P = 15 kN, determine the average shear stress developed in the pins at A, B, and C. All pins are in double shear as shown, and each has a diameter of 18 mm.
P
4P
4P
2P
0.5m
0.5 m 1m
1.5 m
1.5 m
C 30 B
For pins B and C: 82.5 (103) V tB = tC = = p 18 2 = 324 MPa A 4 (1000 )
A
Ans.
For pin A: FA = 2 (82.5)2 + (142.9)2 = 165 kN tA =
82.5 (103) V = p 18 2 = 324 MPa A 4 (1000 )
Ans.
32
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•1–49. The beam is supported by a pin at A and a short link BC. Determine the maximum magnitude P of the loads the beam will support if the average shear stress in each pin is not to exceed 80 MPa. All pins are in double shear as shown, and each has a diameter of 18 mm.
P
4P
4P
2P
0.5m
0.5 m 1m
1.5 m
1.5 m
C 30 B
a + ©MA = 0;
A
2P(0.5) + 4P(2) + 4P(3.5) + P(4.5) - (TCB sin 30°)(5) = 0
TCB = 11P + : ©Fx = 0;
A x - 11P cos 30° = 0
A x = 9.5263P + c ©Fy = 0;
A y - 11P + 11P sin 30° = 0
A y = 5.5P FA = 2 (9.5263P)2 + (5.5P)2 = 11P Require; t =
V ; A
80(106) =
11P>2 p 2 4 (0.018)
P = 3.70 kN
Ans.
33
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1–50. The block is subjected to a compressive force of 2 kN. Determine the average normal and average shear stress developed in the wood fibers that are oriented along section a–a at 30° with the axis of the block.
50 mm a
150 mm
2 kN
2 kN 30
a
Force equilibrium equations written perpendicular and parallel to section a–a gives +Q©Fx¿ = 0;
Va - a - 2 cos 30° = 0
Va - a = 1.732 kN
+a©Fy¿ = 0;
2 sin 30° - Na - a = 0
Na - a = 1.00 kN
The cross sectional area of section a–a is A = a
0.15 b (0.05) = 0.015 m2. Thus sin 30°
(sa - a)avg =
1.00(103) Na - a = = 66.67(103)Pa = 66.7 kPa A 0.015
Ans.
(ta - a)avg =
1.732(103) Va - a = = 115.47(103)Pa = 115 kPa A 0.015
Ans.
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1–51. During the tension test, the wooden specimen is subjected to an average normal stress of 2 ksi. Determine the axial force P applied to the specimen. Also, find the average shear stress developed along section a–a of the specimen.
P
a
4 in.
a 2 in.
1 in.
Internal Loading: The normal force developed on the cross section of the middle portion of the specimen can be obtained by considering the free-body diagram shown in Fig. a. + c ©Fy = 0;
P P + - N = 0 2 2
4 in.
N = P
Referring to the free-body diagram shown in fig. b, the shear force developed in the shear plane a–a is + c ©Fy = 0;
P - Va - a = 0 2
Va - a =
P 2
Average Normal Stress and Shear Stress: The cross-sectional area of the specimen is A = 1(2) = 2 in2. We have savg =
N ; A
2(103) =
P 2
P = 4(103)lb = 4 kip
Ans.
4(103) P = = 2(103) lb. The area of the shear plane is 2 2 = 2(4) = 8 in2. We obtain
Using the result of P, Va - a = Aa - a
A ta - a B avg =
2(103) Va - a = = 250 psi Aa - a 8
Ans.
35
P
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*1–52. If the joint is subjected to an axial force of P = 9 kN, determine the average shear stress developed in each of the 6-mm diameter bolts between the plates and the members and along each of the four shaded shear planes.
P
P 100 mm
Internal Loadings: The shear force developed on each shear plane of the bolt and the member can be determined by writing the force equation of equilibrium along the member’s axis with reference to the free-body diagrams shown in Figs. a. and b, respectively. ©Fy = 0; 4Vb - 9 = 0
Vb = 2.25 kN
©Fy = 0; 4Vp - 9 = 0
Vp = 2.25 kN
Average Shear Stress: The areas of each shear plane of the bolt and the member p are A b = (0.0062) = 28.274(10 - 6)m2 and A p = 0.1(0.1) = 0.01 m2, respectively. 4 We obtain
A tavg B b =
2.25(103) Vb = 79.6 MPa = Ab 28.274(10 - 6)
A tavg B p =
Vp Ap
=
Ans.
2.25(103) = 225 kPa 0.01
Ans.
36
100 mm
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•1–53.
The average shear stress in each of the 6-mm diameter bolts and along each of the four shaded shear planes is not allowed to exceed 80 MPa and 500 kPa, respectively. Determine the maximum axial force P that can be applied to the joint.
P
P 100 mm 100 mm
Internal Loadings: The shear force developed on each shear plane of the bolt and the member can be determined by writing the force equation of equilibrium along the member’s axis with reference to the free-body diagrams shown in Figs. a. and b, respectively. ©Fy = 0;
4Vb - P = 0
Vb = P>4
©Fy = 0;
4Vp - P = 0
Vp = P>4
Average Shear Stress: The areas of each shear plane of the bolts and the members p are A b = (0.0062) = 28.274(10 - 6)m2 and A p = 0.1(0.1) = 0.01m2, respectively. 4 We obtain
A tallow B b =
Vb ; Ab
80(106) =
P>4 28.274(10 - 6)
P = 9047 N = 9.05 kN (controls)
A tallow B p =
Vp Ap
;
500(103) =
Ans.
P>4 0.01
P = 20 000 N = 20 kN
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1–54. The shaft is subjected to the axial force of 40 kN. Determine the average bearing stress acting on the collar C and the normal stress in the shaft.
40 kN
30 mm
C
40 mm
Referring to the FBDs in Fig. a, + c ©Fy = 0;
Ns - 40 = 0
Ns = 40 kN
+ c ©Fy = 0;
Nb - 40 = 0
Nb = 40 kN
Here, the cross-sectional area of the shaft and the bearing area of the collar are p p A s = (0.032) = 0.225(10 - 3)p m2 and A b = (0.04 2) = 0.4(10 - 3)p m2. Thus, 4 4
A savg B s =
40(103) Ns = 56.59(106) Pa = 56.6 MPa = As 0.225(10 - 3)p
Ans.
A savg B b =
40(103) Nb = 31.83(106)Pa = 31.8 MPa = Ab 0.4(10 - 3)p
Ans.
38
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1–55. Rods AB and BC each have a diameter of 5 mm. If the load of P = 2 kN is applied to the ring, determine the average normal stress in each rod if u = 60°.
A u P B
Consider the equilibrium of joint B, Fig. a, + : ©Fx = 0;
2 - FAB sin 60° = 0
+ c ©Fy = 0;
2.309 cos 60° - FBC = 0
FAB = 2.309 kN
C
FBC = 1.155 kN
The cross-sectional area of wires AB and BC are A AB = A BC =
p (0.0052) 4
= 6.25(10 - 6)p m2. Thus,
A savg B AB =
2.309(103) FAB = 117.62(106) Pa = 118 MPa = A AB 6.25(10 - 6)p
Ans.
A savg B BC =
1.155(103) FBC = 58.81(106) Pa = 58.8 MPa = A BC 6.25(10 - 6)p
Ans.
39
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*1–56. Rods AB and BC each have a diameter of 5 mm. Determine the angle u of rod BC so that the average normal stress in rod AB is 1.5 times that in rod BC. What is the load P that will cause this to happen if the average normal stress in each rod is not allowed to exceed 100 MPa?
A u P B
Consider the equilibrium of joint B, Fig. a, FAB cos u - FBC = 0
+ c ©Fy = 0; + : ©Fx = 0;
(1)
P - FAB sin u = 0
(2)
p (0.0052) 4 = 6.25(10 - 6)p m2. Since the average normal stress in rod AB is required to be
The cross-sectional area of rods AB and BC are A AB = A BC =
1.5 times to that of rod BC, then
A savg B AB = 1.5 A savg B BC FAB FBC = 1.5 a b A AB A BC FAB 6.25(10 - 6)p
= 1.5 c
FBC 6.25(10 - 6)p
d
FAB = 1.5 FBC
(3)
Solving Eqs (1) and (3), u = 48.19° = 48.2°
Ans.
Since wire AB will achieve the average normal stress of 100 MPa first when P increases, then FAB = sallow A AB = C 100(106) D C 6.25(10 - 6)p D = 1963.50 N Substitute the result of FAB and u into Eq (2), P = 1.46 kN
Ans.
40
C
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•1–57.
The specimen failed in a tension test at an angle of 52° when the axial load was 19.80 kip. If the diameter of the specimen is 0.5 in., determine the average normal and average shear stress acting on the area of the inclined failure plane. Also, what is the average normal stress acting on the cross section when failure occurs? +
b © Fx = 0;
52 0.5 in.
V - 19.80 cos 52° = 0 V = 12.19 kip
+a © Fy = 0;
N - 19.80 sin 52° = 0 N = 15.603 kip
Inclined plane: s¿ =
P ; A
œ tavg =
s¿ =
V ; A
15.603 p(0.25)2 sin 52°
œ tavg =
Ans.
= 62.6 ksi
12.19 p(0.25)2 sin 52°
Ans.
= 48.9 ksi
Cross section: s =
P ; A
tavg =
s =
V ; A
19.80 = 101 ksi p(0.25)2
Ans.
tavg = 0
Ans.
1–58. The anchor bolt was pulled out of the concrete wall and the failure surface formed part of a frustum and cylinder. This indicates a shear failure occurred along the cylinder BC and tension failure along the frustum AB. If the shear and normal stresses along these surfaces have the magnitudes shown, determine the force P that must have been applied to the bolt.
P
A 45
45 50 mm
Average Normal Stress: For the frustum, A = 2pxL = 2p(0.025 + 0.025) A 2 0.05 + 0.05 2
2
B
3 MPa
3 MPa B
= 0.02221 m2 s =
P ; A
3 A 106 B =
4.5 MPa
F1 0.02221
C 25 mm 25 mm
F1 = 66.64 kN Average Shear Stress: For the cylinder, A = p(0.05)(0.03) = 0.004712 m2 F2 V ; 4.5 A 106 B = tavg = A 0.004712 F2 = 21.21 kN Equation of Equilibrium: + c ©Fy = 0;
P - 21.21 - 66.64 sin 45° = 0 P = 68.3 kN
Ans.
41
30 mm
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1–59. The open square butt joint is used to transmit a force of 50 kip from one plate to the other. Determine the average normal and average shear stress components that this loading creates on the face of the weld, section AB.
50 kip 30 30 2 in.
Equations of Equilibrium: a+ © Fy = 0; +
Q© Fx = 0;
N - 50 cos 30° = 0 - V + 50 sin 30° = 0
B
N = 43.30 kip
A 6 in.
V = 25.0 kip
50 kip
Average Normal and Shear Stress: A¿ = a s = tavg
2 b (6) = 13.86 in2 sin 60°
N 43.30 = = 3.125 ksi A¿ 13.86 V 25.0 = = = 1.80 ksi A¿ 13.86
Ans. Ans.
*1–60. If P = 20 kN, determine the average shear stress developed in the pins at A and C. The pins are subjected to double shear as shown, and each has a diameter of 18 mm.
C
Referring to the FBD of member AB, Fig. a a + ©MA = 0;
30
FBC sin 30° (6) - 20(2) - 20(4) = 0
FBC = 40 kN A
+ : ©Fx = 0;
A x - 40 cos 30° = 0
+ c ©Fy = 0;
A y - 20 - 20 + 40 sin 30°
A x = 34.64 kN A y = 20 kN
2m
P
Thus, the force acting on pin A is FA = 2 A x 2 + A y 2 = 2 34.64 2 + 202 = 40 kN Pins A and C are subjected to double shear. Referring to their FBDs in Figs. b and c, FA FBC 40 40 VA = VC = = = 20 kN = = 20 kN 2 2 2 2 p The cross-sectional area of Pins A and C are A A = A C = (0.0182) 4 = 81(10 - 6)p m2. Thus tA =
20(103) VA = 78.59(106) Pa = 78.6 MPa = AA 81(10 - 6)p
Ans.
tC =
20(103) VC = 78.59(106) Pa = 78.6 MPa = AC 81(10 - 6)p
Ans.
42
B 2m
2m P
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•1–61. Determine the maximum magnitude P of the load the beam will support if the average shear stress in each pin is not to allowed to exceed 60 MPa. All pins are subjected to double shear as shown, and each has a diameter of 18 mm.
C 30
Referring to the FBD of member AB, Fig. a, a + ©MA = 0; + : ©Fx = 0;
FBC sin 30°(6) - P(2) - P(4) = 0 A x - 2P cos 30° = 0
A x = 1.732P
A y - P - P + 2P sin 30° = 0
+ c ©Fy = 0;
A
FBC = 2P
FA = 2 A x 2 + A y 2 = 2 (1.732P)2 + P2 = 2P All pins are subjected to same force and double shear. Referring to the FBD of the pin, Fig. b, 2P F = = P 2 2
The cross-sectional area of the pin is A = tallow =
V ; A
60(106) =
2m
P
Ay = P
Thus, the force acting on pin A is
V =
B 2m
p (0.0182) = 81.0(10 - 6)p m2. Thus, 4
P 81.0(10 - 6)p
P = 15 268 N = 15.3 kN
Ans.
43
2m P
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1–62. The crimping tool is used to crimp the end of the wire E. If a force of 20 lb is applied to the handles, determine the average shear stress in the pin at A. The pin is subjected to double shear and has a diameter of 0.2 in. Only a vertical force is exerted on the wire.
20 lb
C
E A
Support Reactions:
5 in.
From FBD(a) a + ©MD = 0;
B D
1.5 in. 2 in. 1 in.
20(5) - By (1) = 0
+ : ©Fx = 0;
20 lb
By = 100 lb
Bx = 0
From FBD(b) + : ©Fx = 0; a + ©ME = 0;
Ax = 0 A y (1.5) - 100(3.5) = 0 A y = 233.33 lb
Average Shear Stress: Pin A is subjected to double shear. Hence, Ay FA VA = = = 116.67 lb 2 2 (tA)avg =
VA 116.67 = p 2 AA 4 (0.2 )
= 3714 psi = 3.71 ksi
Ans.
1–63. Solve Prob. 1–62 for pin B. The pin is subjected to double shear and has a diameter of 0.2 in.
20 lb
Support Reactions:
a + ©MD = 0; + : ©Fx = 0;
C
E
From FBD(a)
A
20(5) - By (1) = 0
By = 100 lb
5 in.
Bx = 0
1.5 in. 2 in. 1 in.
Average Shear Stress: Pin B is subjected to double shear. Hence, By FB VB = = = 50.0 lb 2 2 (tB)avg =
B D
VB = AB
p 4
50.0 (0.2 2)
= 1592 psi = 1.59 ksi
Ans.
44
20 lb
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*1–64. The triangular blocks are glued along each side of the joint. A C-clamp placed between two of the blocks is used to draw the joint tight. If the glue can withstand a maximum average shear stress of 800 kPa, determine the maximum allowable clamping force F.
50 mm
F glue
45
25 mm
Internal Loadings: The shear force developed on the glued shear plane can be obtained by writing the force equation of equilibrium along the x axis with reference to the free-body diagram of the triangular block, Fig. a. + : ©Fx = 0;
F cos 45° - V = 0
V =
F
2 2 F 2
Average Normal and Shear Stress: The area of the glued shear plane is A = 0.05(0.025) = 1.25(10 - 3)m2. We obtain
tavg =
V ; A
2 2 F 2 800(103) = 1.25(10 - 3) F = 1414 N = 1.41 kN
Ans.
•1–65.
The triangular blocks are glued along each side of the joint. A C-clamp placed between two of the blocks is used to draw the joint tight. If the clamping force is F = 900 N, determine the average shear stress developed in the glued shear plane.
50 mm 45
F glue 25 mm
Internal Loadings: The shear force developed on the glued shear plane can be obtained by writing the force equation of equilibrium along the x axis with reference to the free-body diagram of the triangular block, Fig. a. + : ©Fx = 0;
900 cos 45° - V = 0
V = 636.40 N
Average Normal and Shear Stress: The area of the glued shear plane is A = 0.05(0.025) = 1.25(10 - 3)m2. We obtain tavg =
V 636.40 = 509 kPa = A 1.25(10 - 3)
Ans.
45
F
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1–66. Determine the largest load P that can be a applied to the frame without causing either the average normal stress or the average shear stress at section a–a to exceed s = 150 MPa and t = 60 MPa, respectively. Member CB has a square cross section of 25 mm on each side.
B
Analyse the equilibrium of joint C using the FBD Shown in Fig. a, + c ©Fy = 0;
4 FBC a b - P = 0 5
FBC = 1.25P
2m
a
Referring to the FBD of the cut segment of member BC Fig. b. + : ©Fx = 0;
+ c ©Fy = 0; The
3 Na - a - 1.25Pa b = 0 5 4 1.25Pa b - Va - a = 0 5
cross-sectional
area
of
section
Na - a = 0.75P a A
Va - a = P a–a
is
A a - a = (0.025) a
0.025 b 3>5
= 1.0417(10 - 3)m2. For Normal stress, sallow =
Na - a ; Aa - a
150(106) =
0.75P 1.0417(10 - 3)
P = 208.33(103) N = 208.33 kN For Shear Stress Va - a ; tallow = Aa - a
60(106) =
P 1.0417(10 - 3)
P = 62.5(103) N = 62.5 kN (Controls!)
Ans.
46
C 1.5 m P
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1–67. The prismatic bar has a cross-sectional area A. If it is subjected to a distributed axial loading that increases linearly from w = 0 at x = 0 to w = w0 at x = a, and then decreases linearly to w = 0 at x = 2a, determine the average normal stress in the bar as a function of x for 0 … x 6 a.
w0
x a
a
Equation of Equilibrium: + : ©Fx = 0;
1 w0 1 a x + w0 b(a - x) + w0a = 0 2 a 2
-N +
N =
w0 A 2a2 - x2 B 2a
Average Normal Stress: N s = = A
w0 2a
(2a2 - x2) =
A
w0 A 2a2 - x2 B 2aA
Ans.
*1–68. The prismatic bar has a cross-sectional area A. If it is subjected to a distributed axial loading that increases linearly from w = 0 at x = 0 to w = w0 at x = a, and then decreases linearly to w = 0 at x = 2a, determine the average normal stress in the bar as a function of x for a 6 x … 2a.
w0
x a
a
Equation of Equilibrium: + : ©Fx = 0;
-N +
1 w0 c (2a - x) d(2a - x) = 0 2 a N =
w0 (2a - x)2 2a
Average Normal Stress: s =
N = A
w0 2a
(2a - x)2 =
A
w0 (2a - x)2 2aA
Ans.
The tapered rod has a radius of r = (2 - x>6) in. and is subjected to the distributed loading of w = (60 + 40x) lb>in. Determine the average normal stress at the center of the rod, B.
•1–69.
A = pa 2 -
r w (60 40x) lb/ in. x r = (2 — ) in. 6
3 2 b = 7.069 in2 6
x B
6
© Fx = 0;
N -
L3
(60 + 40x) dx = 0;
3 in.
N = 720 lb
720 s = = 102 psi 7.069
Ans.
47
3 in.
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1–70. The pedestal supports a load P at its center. If the material has a mass density r, determine the radial dimension r as a function of z so that the average normal stress in the pedestal remains constant. The cross section is circular.
P r1
z r
Require: P + W1 P + W1 + dW s = = A A + dA P dA + W1 dA = A dW P + W1 dW = = s dA A
(1)
dA = p(r + dr)2 - pr2 = 2p r dr dW = pr2(rg) dt From Eq. (1) pr2(rg) dz = s 2p r dr r rg dz = s 2 dr z r rg dr dz = 2s L0 Lr1 r
rg z r = ln ; r1 2s
p
r = r1 e(2a)z
However, s =
P p r21
r = r1 e
(p r12rg )z
Ans.
2P
48
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1–71. Determine the average normal stress at section a–a and the average shear stress at section b–b in member AB. The cross section is square, 0.5 in. on each side.
150 lb/ft
Consider the FBD of member BC, Fig. a, a + ©MC = 0;
B
4 ft
C
60 a
FAB sin 60°(4) - 150(4)(2) = 0
FAB = 346.41 lb a
Referring to the FBD in Fig. b, +
b©Fx¿ = 0;
Na - a + 346.41 = 0
b
Na - a = - 346.41 lb
Referring to the FBD in Fig. c. + c ©Fy = 0;
Vb - b - 346.41 sin 60° = 0
b
Vb - b = 300 lb
The cross-sectional areas of section a–a and b–b are A a - a = 0.5(0.5) = 0.25 in2 and 0.5 b = 0.5 in2. Thus A b - b = 0.5 a cos 60° Na - a 346.41 Ans. = = 1385.64 psi = 1.39 ksi sa - a = Aa - a 0.25 tb - b =
Vb - b 300 = = 600 psi Ab - b 0.5
Ans.
49
A
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•1–73. Member B is subjected to a compressive force of 800 lb. If A and B are both made of wood and are 38 in. thick, determine to the nearest 14 in. the smallest dimension h of the horizontal segment so that it does not fail in shear. The average shear stress for the segment is tallow = 300 psi.
tallow = 300 =
B
307.7
13
(32)
12
h h
3 in. 4
Ans.
1–74. The lever is attached to the shaft A using a key that has a width d and length of 25 mm. If the shaft is fixed and a vertical force of 200 N is applied perpendicular to the handle, determine the dimension d if the allowable shear stress for the key is tallow = 35 MPa. a + ©MA = 0;
5
A
h = 2.74 in. Use h = 2
800 lb
a A
d a 20 mm
500 mm 200 N
Fa - a (20) - 200(500) = 0 Fa - a = 5000 N
tallow =
Fa - a ; Aa - a
35(106) =
5000 d(0.025)
d = 0.00571 m = 5.71 mm
Ans.
1–75. The joint is fastened together using two bolts. Determine the required diameter of the bolts if the failure shear stress for the bolts is tfail = 350 MPa. Use a factor of safety for shear of F.S. = 2.5.
30 mm
30 mm
350(106) = 140(105) 2.5
40 kN 3
tallow = 140(106) =
20(10 ) p 4
80 kN
40 kN
d2
d = 0.0135 m = 13.5 mm
Ans.
50
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*1–76. The lapbelt assembly is to be subjected to a force of 800 N. Determine (a) the required thickness t of the belt if the allowable tensile stress for the material is (st)allow = 10 MPa, (b) the required lap length dl if the glue can sustain an allowable shear stress of (tallow)g = 0.75 MPa, and (c) the required diameter dr of the pin if the allowable shear stress for the pin is (tallow)p = 30 MPa.
800 N 45 mm
t dl
dr 800 N
Allowable Normal Stress: Design of belt thickness. 10 A 106 B =
P ; A
(st)allow =
800 (0.045)t
t = 0.001778 m = 1.78 mm
Ans.
Allowable Shear Stress: Design of lap length. VA 400 ; 0.750 A 106 B = (tallow)g = A (0.045) dt dt = 0.01185 m = 11.9 mm
Ans.
Allowable Shear Stress: Design of pin size. VB 400 ; 30 A 106 B = p 2 (tallow)P = A 4 dr dr = 0.004120 m = 4.12 mm
Ans.
•1–77.
The wood specimen is subjected to the pull of 10 kN in a tension testing machine. If the allowable normal stress for the wood is (st)allow = 12 MPa and the allowable shear stress is tallow = 1.2 MPa, determine the required dimensions b and t so that the specimen reaches these stresses simultaneously. The specimen has a width of 25 mm.
10 kN
Allowable Shear Stress: Shear limitation tallow =
V ; A
1.2 A 106 B =
t
A
b
3
5.00(10 ) (0.025) t
t = 0.1667 m = 167 mm
Ans.
Allowable Normal Stress: Tension limitation sallow =
P ; A
12.0 A 106 B =
10 kN
10(103) (0.025) b
b = 0.03333 m = 33.3 mm
Ans.
51
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1–78. Member B is subjected to a compressive force of 600 lb. If A and B are both made of wood and are 1.5 in. thick, determine to the nearest 1>8 in. the smallest dimension a of the support so that the average shear stress along the blue line does not exceed tallow = 50 psi. Neglect friction.
600 lb
3
5 4
B A
Consider the equilibrium of the FBD of member B, Fig. a, + : ©Fx = 0;
4 600 a b - Fh = 0 5
Fh = 480 lb
Referring to the FBD of the wood segment sectioned through glue line, Fig. b + : ©Fx = 0;
480 - V = 0
V = 480 lb
The area of shear plane is A = 1.5(a). Thus, tallow =
V ; A
50 =
480 1.5a
a = 6.40 in Use a = 612 in.
Ans.
52
a
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1–79. The joint is used to transmit a torque of T = 3 kN # m. Determine the required minimum diameter of the shear pin A if it is made from a material having a shear failure stress of tfail = 150 MPa. Apply a factor of safety of 3 against failure.
100 mm T
A
T
Internal Loadings: The shear force developed on the shear plane of pin A can be determined by writing the moment equation of equilibrium along the y axis with reference to the free-body diagram of the shaft, Fig. a. ©My = 0; V(0.1) - 3(103) = 0
V = 30(103)N
Allowable Shear Stress: tfail 150 tallow = = = 50 MPa F.S. 3 Using this result, tallow =
V ; A
50(106) =
30(103) p d 2 4 A
dA = 0.02764 m = 27.6 mm
Ans.
*1–80. Determine the maximum allowable torque T that can be transmitted by the joint. The shear pin A has a diameter of 25 mm, and it is made from a material having a failure shear stress of tfail = 150 MPa. Apply a factor of safety of 3 against failure.
100 mm T
Internal Loadings: The shear force developed on the shear plane of pin A can be determined by writing the moment equation of equilibrium along the y axis with reference to the free-body diagram of the shaft, Fig. a. ©My = 0; V(0.1) - T = 0
T
V = 10T
Allowable Shear Stress: tfail 150 tallow = = = 50 MPa F.S. 3 The area of the shear plane for pin A is A A =
p (0.0252) = 0.4909(10 - 3)m2. Using 4
these results, tallow =
V ; AA
50(106) =
10T 0.4909(10 - 3)
T = 2454.37 N # m = 2.45 kN # m
Ans.
53
A
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•1–81.
The tension member is fastened together using two bolts, one on each side of the member as shown. Each bolt has a diameter of 0.3 in. Determine the maximum load P that can be applied to the member if the allowable shear stress for the bolts is tallow = 12 ksi and the allowable average normal stress is sallow = 20 ksi.
60 P
P
N - P sin 60° = 0
a+ ©Fy = 0;
P = 1.1547 N
(1)
V - P cos 60° = 0
b+ ©Fx = 0;
P = 2V
(2)
Assume failure due to shear: tallow = 12 =
V (2) (0.3)2 p 4
V = 1.696 kip From Eq. (2), P = 3.39 kip Assume failure due to normal force: sallow = 20 =
N (2) p4 (0.3)2
N = 2.827 kip From Eq. (1), P = 3.26 kip
Ans.
(controls)
54
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1–82. The three steel wires are used to support the load. If the wires have an allowable tensile stress of sallow = 165 MPa, determine the required diameter of each wire if the applied load is P = 6 kN.
A C
The force in wire BD is equal to the applied load; ie, FBD = P = 6 kN. Analysing the equilibrium of joint B by referring to its FBD, Fig. a, + : ©Fx = 0;
FBC cos 30° - FAB cos 45° = 0
(1)
+ c ©Fy = 0;
FBC sin 30° + FAB sin 45° - 6 = 0
(2)
Solving Eqs. (1) and (2), FAB = 5.379 kN For wire BD, FBD ; sallow = A BD
FBC = 4.392 kN
165(106) =
6(103) p 2 4 dBD
dBD = 0.006804 m = 6.804 mm Use dBD = 7.00 mm For wire AB, FAB ; sallow = A AB
165(106) =
Ans.
5.379(103) p 4
dAB 2
dAB = 0.006443 m = 6.443 mm Use dAB = 6.50 mm For wire BC, FBC ; sallow = A BC
165(106) =
Ans.
4.392(103) p 4
dBC 2
dBC = 0.005822 m = 5.822 mm dBC = 6.00 mm
Ans.
55
45
B
30
D P
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1–83. The three steel wires are used to support the load. If the wires have an allowable tensile stress of sallow = 165 MPa, and wire AB has a diameter of 6 mm, BC has a diameter of 5 mm, and BD has a diameter of 7 mm, determine the greatest force P that can be applied before one of the wires fails.
A C
45
The force in wire BD is equal to the applied load; ie, FBD = P. Analysing the equilibrium of joint B by referring to its FBD, Fig. a, + : ©Fx = 0;
FBC cos 30° - FAB cos 45° = 0
(1)
+ c ©Fy = 0;
FBC sin 30° + FAB sin 45° - P = 0
(2)
Solving Eqs. (1) and (2), FAB = 0.8966 P For wire BD, FBD sallow = ; A BD
FBC = 0.7321 P
165(106) =
p 4
P (0.0072)
P = 6349.94 N = 6.350 kN For wire AB, FAB ; sallow = A AB
165(106) =
0.8966 P (0.0062)
p 4
P = 5203.42 N = 5.203 kN For wire BC, FBC sallow = ; A BC
165(106) =
0.7321 P (0.0052)
p 4
P = 4425.60 N = 4.43 kN (Controls!)
Ans.
56
B
30
D P
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*1–84. The assembly consists of three disks A, B, and C that are used to support the load of 140 kN. Determine the smallest diameter d1 of the top disk, the diameter d2 within the support space, and the diameter d3 of the hole in the bottom disk. The allowable bearing stress for the material is 1sallow2b = 350 MPa and allowable shear stress is tallow = 125 MPa.
140 kN d1
20 mm A
10 mm
B C d3 d2
Solution Allowable Bearing Stress: Assume bearing failure for disk B. (sb)allow =
P ; A
350 A 106 B =
140(103) p 4
d21
d1 = 0.02257 m = 22.6 mm Allowable Shear Stress: Assume shear failure for disk C. tallow =
125 A 106 B =
V ; A
140(103) pd2 (0.01)
d2 = 0.03565 m = 35.7 mm
Ans.
Allowable Bearing Stress: Assume bearing failure for disk C. 140(103) P ; 350 A 106 B = (sb)allow = p A A 0.035652 - d23 B 4
d3 = 0.02760 m = 27.6 mm
Ans.
Since d3 = 27.6 mm 7 d1 = 22.6 mm, disk B might fail due to shear. t =
140(103) V = = 98.7 MPa 6 tallow = 125 MPa (O. K !) A p(0.02257)(0.02) d1 = 22.6 mm
Therefore,
Ans.
•1–85.
The boom is supported by the winch cable that has a diameter of 0.25 in. and an allowable normal stress of sallow = 24 ksi. Determine the greatest load that can be supported without causing the cable to fail when u = 30° and f = 45°. Neglect the size of the winch.
B
u 20 ft
s =
P ; A
24(103) =
p 4
T ; (0.25)2
A
T = 1178.10 lb d
+ : ©Fx = 0;
- 1178.10 cos 30° + FAB sin 45° = 0
+ c ©Fy = 0;
- W + FAB cos 45° - 1178.10 sin 30° = 0 W = 431 lb
Ans.
FAB = 1442.9 lb
57
f
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1–86. The boom is supported by the winch cable that has an allowable normal stress of sallow = 24 ksi. If it is required that it be able to slowly lift 5000 lb, from u = 20° to u = 50°, determine the smallest diameter of the cable to 1 the nearest 16 in. The boom AB has a length of 20 ft. Neglect the size of the winch. Set d = 12 ft.
B
u 20 ft f
A
Maximum tension in cable occurs when u = 20°. sin c sin 20° = 20 12
d
c = 11.842° + : © Fx = 0;
- T cos 20° + FAB cos 31.842° = 0
+ c ©Fy = 0;
FAB sin 31.842° - T sin 20° - 5000 = 0 T = 20 698.3 lb FAB = 22 896 lb
P ; s = A
Use
20 698.3 p 2 4 (d) d = 1.048 in.
24(103) =
d = 1
1 in. 16
Ans.
1–87. The 60 mm * 60 mm oak post is supported on the pine block. If the allowable bearing stresses for these materials are soak = 43 MPa and spine = 25 MPa, determine the greatest load P that can be supported. If a rigid bearing plate is used between these materials, determine its required area so that the maximum load P can be supported. What is this load?
P
For failure of pine block: s =
P ; A
25(106) =
P (0.06)(0.06)
P = 90 kN
Ans.
For failure of oak post: s =
P ; A
43(106) =
P (0.06)(0.06)
P = 154.8 kN Area of plate based on strength of pine block: 154.8(10)3 P s = ; 25(106) = A A A = 6.19(10 - 3)m2
Ans.
Pmax = 155 kN
Ans.
58
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*1–88. The frame is subjected to the load of 4 kN which acts on member ABD at D. Determine the required diameter of the pins at D and C if the allowable shear stress for the material is tallow = 40 MPa. Pin C is subjected to double shear, whereas pin D is subjected to single shear.
4 kN 1m E
1.5 m C
45 D 1.5 m
Referring to the FBD of member DCE, Fig. a, a + ©ME = 0;
Dy(2.5) - FBC sin 45° (1) = 0
(1)
+ : ©Fx = 0
FBC cos 45° - Dx = 0
(2)
B 1.5 m
Referring to the FBD of member ABD, Fig. b, a + ©MA = 0;
4 cos 45° (3) + FBC sin 45° (1.5) - Dx (3) = 0
(3)
Solving Eqs (2) and (3), FBC = 8.00 kN
Dx = 5.657 kN
Substitute the result of FBC into (1) Dy = 2.263 kN Thus, the force acting on pin D is FD = 2 Dx 2 + Dy 2 = 2 5.6572 + 2.2632 = 6.093 kN Pin C is subjected to double shear white pin D is subjected to single shear. Referring to the FBDs of pins C, and D in Fig c and d, respectively, FBC 8.00 = = 4.00 kN VC = VD = FD = 6.093 kN 2 2 For pin C, tallow =
VC ; AC
40(106) =
4.00(103) p 4
dC 2
dC = 0.01128 m = 11.28 mm Use dC = 12 mm For pin D, VD ; tallow = AD
40(106) =
Ans.
6.093(103) p 4
dD 2
dD = 0.01393 m = 13.93 mm Use
dD = 14 mm
Ans.
59
A
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•1–89. The eye bolt is used to support the load of 5 kip. Determine its diameter d to the nearest 18 in. and the required thickness h to the nearest 18 in. of the support so that the washer will not penetrate or shear through it. The allowable normal stress for the bolt is sallow = 21 ksi and the allowable shear stress for the supporting material is tallow = 5 ksi.
1 in. h
d
5 kip
Allowable Normal Stress: Design of bolt size 5(103) P sallow = ; 21.0(103) = p 2 Ab 4 d d = 0.5506 in. Use d =
5 in. 8
Ans.
Allowable Shear Stress: Design of support thickness 5(103) V ; 5(103) = tallow = A p(1)(h) Use h =
3 in. 8
Ans.
60
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1–90. The soft-ride suspension system of the mountain bike is pinned at C and supported by the shock absorber BD. If it is designed to support a load P = 1500 N, determine the required minimum diameter of pins B and C. Use a factor of safety of 2 against failure. The pins are made of material having a failure shear stress of tfail = 150 MPa, and each pin is subjected to double shear.
P A
Internal Loadings: The forces acting on pins B and C can be determined by considering the equilibrium of the free-body diagram of the soft-ride suspension system shown in Fig. a. a + ©MC = 0;
FBD = 5905.36 N + : ©Fx = 0;
Cx - 5905.36 cos 60° = 0
+ c ©Fy = 0;
5905.36 sin 60° - 1500 - Cy = 0 Cy = 3614.20 N
Cx = 2952.68 N
Thus, FC = 2 Cx 2 + Cy 2 = 2 2952.682 + 3614.202 = 4666.98 N Since both pins are in double shear, FB 5905.36 VB = = = 2952.68 N 2 2
VC =
FC 4666.98 = = 2333.49 N 2 2
Allowable Shear Stress: tfail 150 tallow = = = 75 MPa F.S. 2 Using this result, VB tallow = ; AB
75(106) =
2952.68 p 2 d 4 B
dB = 0.007080 m = 7.08 mm tallow =
VC ; AC
75(106) =
Ans.
2333.49 p 2 d 4 C
dC = 0.006294 m = 6.29 mm
Ans.
61
30 mm
B 60
1500(0.4) - FBD sin 60°(0.1) - FBD cos 60°(0.03) = 0
FB = FBD = 5905.36 N
100 mm 300 mm
C
D
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1–91. The soft-ride suspension system of the mountain bike is pinned at C and supported by the shock absorber BD. If it is designed to support a load of P = 1500 N, determine the factor of safety of pins B and C against failure if they are made of a material having a shear failure stress of tfail = 150 MPa. Pin B has a diameter of 7.5 mm, and pin C has a diameter of 6.5 mm. Both pins are subjected to double shear.
P A
Internal Loadings: The forces acting on pins B and C can be determined by considerning the equilibrium of the free-body diagram of the soft-ride suspension system shown in Fig. a. + ©MC = 0;
1500(0.4) - FBD sin 60°(0.1) - FBD cos 60°(0.03) = 0 FBD = 5905.36 N
+ : ©Fx = 0;
Cx - 5905.36 cos 60° = 0
+ c ©Fy = 0;
5905.36 sin 60° - 1500 - Cy = 0 Cy = 3614.20 N
Cx = 2952.68 N
Thus, FB = FBD = 5905.36 N
FC = 2 Cx 2 + Cy 2 = 2 2952.682 + 3614.202 = 4666.98 N
Since both pins are in double shear, VB =
FB 5905.36 = = 2952.68N 2 2
VC =
FC 4666.98 = = 2333.49 N 2 2
Allowable Shear Stress: The areas of the shear plane for pins B and C are p p A B = (0.00752) = 44.179(10 - 6)m2 and A C = (0.00652) = 33.183(10 - 6)m2. 4 4 We obtain
A tavg B B =
VB 2952.68 = 66.84 MPa = AB 44.179(10 - 6)
A tavg B C =
VC 2333.49 = 70.32 MPa = AC 33.183(10 - 6)
Using these results, tfail = (F.S.)B = A tavg B B tfail (F.S.)C = = A tavg B C
100 mm 300 mm
150 = 2.24 66.84
Ans.
150 = 2.13 70.32
Ans.
62
30 mm
B 60
C
D
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*1–92. The compound wooden beam is connected together by a bolt at B. Assuming that the connections at A, B, C, and D exert only vertical forces on the beam, determine the required diameter of the bolt at B and the required outer diameter of its washers if the allowable tensile stress for the bolt is 1st2allow = 150 MPa and the allowable bearing stress for the wood is 1sb2allow = 28 MPa. Assume that the hole in the washers has the same diameter as the bolt.
2 kN 1.5 kN 1.5 m 1.5 m 1.5 m
3 kN 2m
2m
1.5 m C
D
A B
From FBD (a): a + ©MD = 0;
FB(4.5) + 1.5(3) + 2(1.5) - FC(6) = 0 4.5 FB - 6 FC = - 7.5
(1)
From FBD (b): a + ©MD = 0;
FB(5.5) - FC(4) - 3(2) = 0 5.5 FB - 4 FC = 6
(2)
Solving Eqs. (1) and (2) yields FB = 4.40 kN;
FC = 4.55 kN
For bolt: sallow = 150(106) =
4.40(103) p 2 4 (dB)
dB = 0.00611 m = 6.11 mm
Ans.
For washer: sallow = 28 (104) =
4.40(103) p 2 4 (d w
- 0.006112)
dw = 0.0154 m = 15.4 mm
Ans.
•1–93.
The assembly is used to support the distributed loading of w = 500 lb>ft. Determine the factor of safety with respect to yielding for the steel rod BC and the pins at B and C if the yield stress for the steel in tension is sy = 36 ksi and in shear ty = 18 ksi. The rod has a diameter of 0.40 in., and the pins each have a diameter of 0.30 in.
C
4 ft
For rod BC: s =
A
1.667 P = 13.26 ksi = p 2 A 4 (0.4 )
F. S. =
sy s
=
36 = 2.71 13.26
Ans. 3 ft
For pins B and C:
w
0.8333 V = 11.79 ksi = p t = 2 A 4 (0.3 ) F. S. =
ty t
=
B
1 ft
18 = 1.53 11.79
Ans.
63
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1–94. If the allowable shear stress for each of the 0.30in.-diameter steel pins at A, B, and C is tallow = 12.5 ksi, and the allowable normal stress for the 0.40-in.-diameter rod is sallow = 22 ksi, determine the largest intensity w of the uniform distributed load that can be suspended from the beam.
C
4 ft
Assume failure of pins B and C: tallow = 12.5 =
1.667w
A
p 2 4 (0.3 )
w = 0.530 kip>ft
Ans.
(controls)
B
Assume failure of pins A: 3 ft
FA = 2 (2w)2 + (1.333w)2 = 2.404 w tallow
w
1.202w = 12.5 = p 2 4 (0.3 )
1 ft
w = 0.735 kip>ft Assume failure of rod BC: sallow = 22 =
3.333w p 2 4 (0.4 )
w = 0.829 kip>ft
1–95. If the allowable bearing stress for the material under the supports at A and B is 1sb2allow = 1.5 MPa, determine the size of square bearing plates A¿ and B¿ required to support the load. Dimension the plates to the nearest mm. The reactions at the supports are vertical. Take P = 100 kN.
40 kN/m
Referring to the FBD of the bean, Fig. a
1.5 m
A
a + ©MA = 0;
NB(3) + 40(1.5)(0.75) - 100(4.5) = 0
NB = 135 kN
a + ©MB = 0;
40(1.5)(3.75) - 100(1.5) - NA(3) = 0
NA = 25.0 kN
For plate A¿ , NA (sb)allow = ; A A¿
1.5(106) =
25.0(103) a2A¿
aA¿ = 0.1291 m = 130 mm
Ans.
For plate B¿ , sallow =
NB ; A B¿
1.5(106) =
135(103) a2B¿
aB¿ = 0.300 m = 300 mm
Ans.
64
P
A¿
B¿ 3m
B
1.5 m
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40 kN/m
*1–96. If the allowable bearing stress for the material under the supports at A and B is 1sb2allow = 1.5 MPa, determine the maximum load P that can be applied to the beam. The bearing plates A¿ and B¿ have square cross sections of 150 mm * 150 mm and 250 mm * 250 mm, respectively.
A
P
A¿
B¿ 3m
1.5 m
B
1.5 m
Referring to the FBD of the beam, Fig. a, a + ©MA = 0;
NB(3) + 40(1.5)(0.75) - P(4.5) = 0
NB = 1.5P - 15
a + ©MB = 0;
40(1.5)(3.75) - P(1.5) - NA(3) = 0
NA = 75 - 0.5P
For plate A¿ , NA (sb)allow = ; A A¿
1.5(106) =
(75 - 0.5P)(103) 0.15(0.15)
P = 82.5 kN For plate B¿ , NB ; (sb)allow = A B¿
1.5(106) =
(1.5P - 15)(103) 0.25(0.25)
P = 72.5 kN (Controls!)
Ans.
•1–97.
The rods AB and CD are made of steel having a failure tensile stress of sfail = 510 MPa. Using a factor of safety of F.S. = 1.75 for tension, determine their smallest diameter so that they can support the load shown. The beam is assumed to be pin connected at A and C.
B
D 6 kN 5 kN
4 kN
Support Reactions: a + ©MA = 0;
FCD(10) - 5(7) - 6(4) - 4(2) = 0 A
FCD = 6.70 kN a + ©MC = 0;
4(8) + 6(6) + 5(3) - FAB(10) = 0 FAB = 8.30 kN
Allowable Normal Stress: Design of rod sizes For rod AB sallow =
sfail FAB = ; F.S A AB
510(106) 8.30(103) = p 2 1.75 4 d AB dAB = 0.006022 m = 6.02 mm
Ans.
For rod CD sallow =
FCD sfail = ; F.S A CD
C 2m
510(106) 6.70(103) = p 2 1.75 4 d CD
dCD = 0.005410 m = 5.41 mm
Ans.
65
2m
3m
3m
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1–98. The aluminum bracket A is used to support the centrally applied load of 8 kip. If it has a constant thickness of 0.5 in., determine the smallest height h in order to prevent a shear failure. The failure shear stress is tfail = 23 ksi. Use a factor of safety for shear of F.S. = 2.5.
A
Equation of Equilibrium: + c ©Fy = 0;
V - 8 = 0
V = 8.00 kip
Allowable Shear Stress: Design of the support size tallow =
tfail V = ; F.S A
23(103) 8.00(103) = 2.5 h(0.5)
8 kip
h = 1.74 in.
Ans.
1–99. The hanger is supported using the rectangular pin. Determine the magnitude of the allowable suspended load P if the allowable bearing stress is (sb)allow = 220 MPa, the allowable tensile stress is (st)allow = 150 MPa, and the allowable shear stress is tallow = 130 MPa. Take t = 6 mm, a = 5 mm, and b = 25 mm.
20 mm 75 mm 10 mm a
a
b
Allowable Normal Stress: For the hanger (st)allow =
P ; A
150 A 106 B =
P (0.075)(0.006)
P
P = 67.5 kN
37.5 mm
Allowable Shear Stress: The pin is subjected to double shear. Therefore, V = tallow =
130 A 106 B =
V ; A
P 2
P>2 (0.01)(0.025)
P = 65.0 kN Allowable Bearing Stress: For the bearing area (sb)allow =
P ; A
220 A 106 B =
37.5 mm
t
P>2 (0.005)(0.025)
P = 55.0 kN (Controls!)
Ans.
66
h
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*1–100. The hanger is supported using the rectangular pin. Determine the required thickness t of the hanger, and dimensions a and b if the suspended load is P = 60 kN. The allowable tensile stress is (st)allow = 150 MPa, the allowable bearing stress is (sb)allow = 290 MPa, and the allowable shear stress is tallow = 125 MPa.
20 mm 75 mm 10 mm a
a
37.5 mm
t P 37.5 mm
Allowable Normal Stress: For the hanger (st)allow =
P ; A
150 A 106 B =
60(103) (0.075)t
t = 0.005333 m = 5.33 mm
Ans.
Allowable Shear Stress: For the pin tallow =
125 A 106 B =
V ; A
30(103) (0.01)b
b = 0.0240 m = 24.0 mm
Ans.
Allowable Bearing Stress: For the bearing area (sb)allow =
P ; A
290 A 106 B =
30(103) (0.0240) a
a = 0.00431 m = 4.31 mm
Ans.
67
b
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•1–101. The 200-mm-diameter aluminum cylinder supports a compressive load of 300 kN. Determine the average normal and shear stress acting on section a–a. Show the results on a differential element located on the section.
300 kN
a
Referring to the FBD of the upper segment of the cylinder sectional through a–a shown in Fig. a, +Q©Fx¿ = 0;
Na - a - 300 cos 30° = 0
+a©Fy¿ = 0;
Va - a - 300 sin 30° = 0
30
Na - a = 259.81 kN
a
Va - a = 150 kN
0.1 Section a–a of the cylinder is an ellipse with a = 0.1 m and b = m. Thus, cos 30° 0.1 b = 0.03628 m2. A a - a = pab = p(0.1)a cos 30°
A sa - a B avg =
259.81(103) Na - a = = 7.162(106) Pa = 7.16 MPa Aa - a 0.03628
Ans.
A ta - a B avg =
150(103) Va - a = = 4.135(106) Pa = 4.13 MPa Aa - a 0.03628
Ans.
d
The differential element representing the state of stress of a point on section a–a is shown in Fig. b
1–102. The long bolt passes through the 30-mm-thick plate. If the force in the bolt shank is 8 kN, determine the average normal stress in the shank, the average shear stress along the cylindrical area of the plate defined by the section lines a–a, and the average shear stress in the bolt head along the cylindrical area defined by the section lines b–b.
8 mm
a 7 mm
b
8 kN
18 mm b a
P = ss = A
8 (103)
= 208 MPa
Ans.
(tavg)a =
8 (103) V = = 4.72 MPa A p (0.018)(0.030)
Ans.
(tavg)b =
8 (103) V = = 45.5 MPa A p (0.007)(0.008)
Ans.
p 4
(0.007)2
68
30 mm
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1–103. Determine the required thickness of member BC and the diameter of the pins at A and B if the allowable normal stress for member BC is sallow = 29 ksi and the allowable shear stress for the pins is tallow = 10 ksi.
C
1.5 in.
Referring to the FBD of member AB, Fig. a, a + ©MA = 0;
2(8)(4) - FBC sin 60° (8) = 0
+ : ©Fx = 0;
9.238 cos 60° - A x = 0
+ c ©Fy = 0;
9.238 sin 60° - 2(8) + A y = 0
FBC = 9.238 kip
60 B
A x = 4.619 kip
8 ft
A y = 8.00 kip 2 kip/ft
Thus, the force acting on pin A is FA = 2 A 2x + A 2y = 2 4.6192 + 8.002 = 9.238 kip Pin A is subjected to single shear, Fig. c, while pin B is subjected to double shear, Fig. b. FBC 9.238 VA = FA = 9.238 kip VB = = = 4.619 kip 2 2 For member BC FBC ; sallow = A BC
29 =
9.238 1.5(t)
t = 0.2124 in.
Use t = For pin A, VA ; tallow = AA
10 =
9.238 p 2 4 dA
1 in. 4
Ans.
dA = 1.085 in. 1 Use dA = 1 in 8
For pin B, VB ; tallow = AB
10 =
4.619 p 2 4 dB
Ans.
dB = 0.7669 in Use dB =
13 in 16
Ans.
69
A
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*1–104. Determine the resultant internal loadings acting on the cross sections located through points D and E of the frame.
150 lb/ft
Segment AD:
1.5 ft
+
: ©Fx = 0;
ND - 1.2 = 0;
+ T ©Fy = 0;
VD + 0.225 + 0.4 = 0;
a + ©MD = 0;
ND = 1.20 kip
Ans.
VD = - 0.625 kip
B
E
Ans. 2.5 ft
MD + 0.225(0.75) + 0.4(1.5) = 0 MD =
A 4 ft
D
C
- 0.769 kip # ft
Ans.
3 ft
5 ft
Segment CE: Q+ ©Fx = 0;
NE + 2.0 = 0;
R+ ©Fy = 0;
VE = 0
a + ©ME = 0;
NE = - 2.00 kip
Ans. Ans.
ME = 0
Ans.
•1–105.
The pulley is held fixed to the 20-mm-diameter shaft using a key that fits within a groove cut into the pulley and shaft. If the suspended load has a mass of 50 kg, determine the average shear stress in the key along section a–a. The key is 5 mm by 5 mm square and 12 mm long.
a + ©MO = 0;
a
75 mm
F (10) - 490.5 (75) = 0
F = 3678.75 N tavg =
a
3678.75 V = = 61.3 MPa A (0.005)(0.012)
Ans.
70
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1–106. The bearing pad consists of a 150 mm by 150 mm block of aluminum that supports a compressive load of 6 kN. Determine the average normal and shear stress acting on the plane through section a–a. Show the results on a differential volume element located on the plane.
6 kN
a
Equation of Equilibrium: +Q©Fx = 0;
Va - a - 6 cos 60° = 0
Va - a = 3.00 kN
a+ ©Fy = 0;
Na - a - 6 sin 60° = 0
Na - a = 5.196 kN
30 a
150 mm
Averge Normal Stress And Shear Stress: The cross sectional Area at section a–a is 0.15 b (0.15) = 0.02598 m2. A = a sin 60° sa - a =
5.196(103) Na - a = = 200 kPa A 0.02598
Ans.
ta - a =
3.00(103) Va - a = = 115 kPa A 0.02598
Ans.
1–107. The yoke-and-rod connection is subjected to a tensile force of 5 kN. Determine the average normal stress in each rod and the average shear stress in the pin A between the members.
5 kN
40 mm
For the 40 – mm – dia rod: s40
30 mm
5 (103) P = p = = 3.98 MPa 2 A 4 (0.04)
Ans.
For the 30 – mm – dia rod:
5 kN
3
s30 =
5 (10 ) V = p = 7.07 MPa 2 A 4 (0.03)
Ans.
Average shear stress for pin A: tavg =
A 25 mm
2.5 (103) P = p = 5.09 MPa 2 A 4 (0.025)
Ans.
71
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*1–108. The cable has a specific weight g (weight>volume) and cross-sectional area A. If the sag s is small, so that its length is approximately L and its weight can be distributed uniformly along the horizontal axis, determine the average normal stress in the cable at its lowest point C.
A
s C L/2
Equation of Equilibrium: a + ©MA = 0;
Ts -
gAL L a b = 0 2 4 T =
B
gAL2 8s
Average Normal Stress: gAL2
gL2 T 8s s = = = A A 8s
Ans.
72
L/2
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2–1. An air-filled rubber ball has a diameter of 6 in. If the air pressure within it is increased until the ball’s diameter becomes 7 in., determine the average normal strain in the rubber. d0 = 6 in. d = 7 in. e =
pd - pd0 7 - 6 = = 0.167 in./in. pd0 6
Ans.
2–2. A thin strip of rubber has an unstretched length of 15 in. If it is stretched around a pipe having an outer diameter of 5 in., determine the average normal strain in the strip. L0 = 15 in. L = p(5 in.) e =
L - L0 5p - 15 = = 0.0472 in.>in. L0 15
Ans.
2–3. The rigid beam is supported by a pin at A and wires BD and CE. If the load P on the beam causes the end C to be displaced 10 mm downward, determine the normal strain developed in wires CE and BD.
D
E
4m
P
¢LBD ¢LCE = 3 7
A
3 (10) = 4.286 mm 7 ¢LCE 10 = = = 0.00250 mm>mm L 4000
3m
¢LBD = eCE
eBD =
B
Ans.
¢LBD 4.286 = = 0.00107 mm>mm L 4000
Ans.
1
C
2m
2m
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*2–4. The two wires are connected together at A. If the force P causes point A to be displaced horizontally 2 mm, determine the normal strain developed in each wire.
C 300
œ = 23002 + 22 - 2(300)(2) cos 150° = 301.734 mm LAC
eAC = eAB
œ - LAC LAC 301.734 - 300 = = = 0.00578 mm>mm LAC 300
mm
30⬚
Ans.
30⬚
300
A
P
mm
B
•2–5. The rigid beam is supported by a pin at A and wires BD and CE. If the distributed load causes the end C to be displaced 10 mm downward, determine the normal strain developed in wires CE and BD.
E D 2m 1.5 m 3m
2m A
B
C
w
Since the vertical displacement of end C is small compared to the length of member AC, the vertical displacement dB of point B, can be approximated by referring to the similar triangle shown in Fig. a dB 10 = ; dB = 4 mm 2 5 The unstretched lengths of wires BD and CE are LBD = 1500 mm and LCE = 2000 mm. dB 4 Ans. = = 0.00267 mm>mm A eavg B BD = LBD 1500
A eavg B CE =
dC 10 = = 0.005 mm>mm LCE 2000
Ans.
2
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2–6. Nylon strips are fused to glass plates. When moderately heated the nylon will become soft while the glass stays approximately rigid. Determine the average shear strain in the nylon due to the load P when the assembly deforms as indicated.
y 2 mm P
3 mm 5 mm 3 mm 5 mm 3 mm
g = tan - 1 a
2 b = 11.31° = 0.197 rad 10
x
Ans.
2–7. If the unstretched length of the bowstring is 35.5 in., determine the average normal strain in the string when it is stretched to the position shown. 18 in.
6 in. 18 in.
Geometry: Referring to Fig. a, the stretched length of the string is L = 2L¿ = 2 2182 + 62 = 37.947 in. Average Normal Strain: eavg =
L - L0 37.947 - 35.5 = = 0.0689 in.>in. L0 35.5
Ans.
3
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u
*2–8. Part of a control linkage for an airplane consists of a rigid member CBD and a flexible cable AB. If a force is applied to the end D of the member and causes it to rotate by u = 0.3°, determine the normal strain in the cable. Originally the cable is unstretched.
D
P 300 mm
B
AB = 24002 + 3002 = 500 mm
300 mm
AB¿ = 2400 + 300 - 2(400)(300) cos 90.3° 2
2
A
C
= 501.255 mm eAB =
AB¿ - AB 501.255 - 500 = AB 500
400 mm
= 0.00251 mm>mm
Ans.
•2–9.
Part of a control linkage for an airplane consists of a rigid member CBD and a flexible cable AB. If a force is applied to the end D of the member and causes a normal strain in the cable of 0.0035 mm>mm, determine the displacement of point D. Originally the cable is unstretched.
u D
300 mm
B
AB = 23002 + 4002 = 500 mm
300 mm
AB¿ = AB + eABAB A
= 500 + 0.0035(500) = 501.75 mm
C
501.752 = 3002 + 4002 - 2(300)(400) cos a a = 90.4185°
400 mm
p (0.4185) rad u = 90.4185° - 90° = 0.4185° = 180° ¢ D = 600(u) = 600(
P
p )(0.4185) = 4.38 mm 180°
Ans.
4
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2–10. The corners B and D of the square plate are given the displacements indicated. Determine the shear strains at A and B.
y A
16 mm D
B
3 mm 3 mm 16 mm
16 mm
Applying trigonometry to Fig. a f = tan - 1 a
13 p rad b = 39.09° a b = 0.6823 rad 16 180°
a = tan - 1 a
16 p rad b = 50.91° a b = 0.8885 rad 13 180°
By the definition of shear strain,
A gxy B A =
p p - 2f = - 2(0.6823) = 0.206 rad 2 2
Ans.
A gxy B B =
p p - 2a = - 2(0.8885) = -0.206 rad 2 2
Ans.
5
C
16 mm
x
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2–11. The corners B and D of the square plate are given the displacements indicated. Determine the average normal strains along side AB and diagonal DB.
y A
16 mm D
B
3 mm 3 mm 16 mm
16 mm
Referring to Fig. a, LAB = 2162 + 162 = 2512 mm LAB¿ = 2162 + 132 = 2425 mm LBD = 16 + 16 = 32 mm LB¿D¿ = 13 + 13 = 26 mm Thus,
A eavg B AB =
LAB¿ - LAB 2425 - 2512 = = -0.0889 mm>mm LAB 2512
Ans.
A eavg B BD =
LB¿D¿ - LBD 26 - 32 = = -0.1875 mm>mm LBD 32
Ans.
6
C
16 mm
x
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*2–12. The piece of rubber is originally rectangular. Determine the average shear strain gxy at A if the corners B and D are subjected to the displacements that cause the rubber to distort as shown by the dashed lines.
y 3 mm
C D
2 = 0.006667 rad 300 3 u2 = tan u2 = = 0.0075 rad 400 u1 = tan u1 =
400 mm
gxy = u1 + u2
A
= 0.006667 + 0.0075 = 0.0142 rad
Ans.
•2–13.
The piece of rubber is originally rectangular and subjected to the deformation shown by the dashed lines. Determine the average normal strain along the diagonal DB and side AD.
f = tan
B 2 mm
3 mm
C D
400 mm
3 b = 0.42971° a 400
AB¿ = 2(300)2 + (2)2 = 300.00667 w = tan - 1 a
x
y
AD¿ = 2(400)2 + (3)2 = 400.01125 mm -1
300 mm
A
2 b = 0.381966° 300
a = 90° - 0.42971° - 0.381966° = 89.18832° D¿B¿ = 2(400.01125)2 + (300.00667)2 - 2(400.01125)(300.00667) cos (89.18832°) D¿B¿ = 496.6014 mm DB = 2(300)2 + (400)2 = 500 mm 496.6014 - 500 = -0.00680 mm>mm 500 400.01125 - 400 = = 0.0281(10 - 3) mm>mm 400
eDB =
Ans.
eAD
Ans.
7
300 mm
B 2 mm
x
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2–14. Two bars are used to support a load. When unloaded, AB is 5 in. long, AC is 8 in. long, and the ring at A has coordinates (0, 0). If a load P acts on the ring at A, the normal strain in AB becomes PAB = 0.02 in.>in., and the normal strain in AC becomes PAC = 0.035 in.>in. Determine the coordinate position of the ring due to the load.
y
B
C
60⬚
5 in.
8 in.
A
x
P
Average Normal Strain: œ = LAB + eAB LAB = 5 + (0.02)(5) = 5.10 in. LAB œ = LAC + eACLAC = 8 + (0.035)(8) = 8.28 in. LAC
Geometry: a = 282 - 4.33012 = 6.7268 in. 5.102 = 9.22682 + 8.282 - 2(9.2268)(8.28) cos u u = 33.317° x¿ = 8.28 cos 33.317° = 6.9191 in. y¿ = 8.28 sin 33.317° = 4.5480 in. x = -(x¿ - a) = -(6.9191 - 6.7268) = -0.192 in.
Ans.
y = -(y¿ - 4.3301) = -(4.5480 - 4.3301) = -0.218 in.
Ans.
8
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2–15. Two bars are used to support a load P. When unloaded, AB is 5 in. long, AC is 8 in. long, and the ring at A has coordinates (0, 0). If a load is applied to the ring at A, so that it moves it to the coordinate position (0.25 in., -0.73 in.), determine the normal strain in each bar.
y
B
C
60⬚
5 in.
8 in.
A
x
P
Geometry: a = 282 - 4.33012 = 6.7268 in. LA¿B = 2(2.5 + 0.25)2 + (4.3301 + 0.73)2 = 5.7591 in. LA¿C = 2(6.7268 - 0.25)2 + (4.3301 + 0.73)2 = 8.2191 in. Average Normal Strain: eAB =
=
eAC =
=
LA¿B - LAB LAB 5.7591 - 5 = 0.152 in.>in. 5
Ans.
LA¿C - LAC LAC 8.2191 - 8 = 0.0274 in.>in. 8
Ans.
9
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*2–16. The square deforms into the position shown by the dashed lines. Determine the average normal strain along each diagonal, AB and CD. Side D¿B¿ remains horizontal.
y 3 mm D¿
B¿
B
D
Geometry: AB = CD = 2502 + 502 = 70.7107 mm 53 mm
C¿D¿ = 2532 + 582 - 2(53)(58) cos 91.5°
50 mm 91.5⬚
= 79.5860 mm C
B¿D¿ = 50 + 53 sin 1.5° - 3 = 48.3874 mm
A
x
C¿
AB¿ = 2532 + 48.38742 - 2(53)(48.3874) cos 88.5°
50 mm 8 mm
= 70.8243 mm Average Normal Strain:
eAB =
=
eCD =
=
AB¿ - AB AB 70.8243 - 70.7107 = 1.61 A 10 - 3 B mm>mm 70.7107
Ans.
C¿D¿ - CD CD 79.5860 - 70.7107 = 126 A 10 - 3 B mm>mm 70.7107
Ans.
•2–17.
The three cords are attached to the ring at B. When a force is applied to the ring it moves it to point B¿ , such that the normal strain in AB is PAB and the normal strain in CB is PCB. Provided these strains are small, determine the normal strain in DB. Note that AB and CB remain horizontal and vertical, respectively, due to the roller guides at A and C.
A¿
B¿
A
B
L
Coordinates of B (L cos u, L sin u)
u
Coordinates of B¿ (L cos u + eAB L cos u, L sin u + eCB L sin u)
C¿ D
LDB¿ = 2(L cos u + eAB L cos u) + (L sin u + eCB L sin u) 2
2
LDB¿ = L 2cos2 u(1 + 2eAB + e2AB) + sin2 u(1 + 2eCB + e2CB) Since eAB and eCB are small, LDB¿ = L 21 + (2 eAB cos2 u + 2eCB sin2 u) Use the binomial theorem, LDB¿ = L ( 1 +
1 (2 eAB cos2 u + 2eCB sin2 u)) 2
= L ( 1 + eAB cos2 u + eCB sin2 u) Thus, eDB =
L( 1 + eAB cos2 u + eCB sin2 u) - L L
eDB = eAB cos2 u + eCB sin2 u
Ans.
10
C
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2–18. The piece of plastic is originally rectangular. Determine the shear strain gxy at corners A and B if the plastic distorts as shown by the dashed lines.
y 5 mm 2 mm 2 mm
B
4 mm
C 300 mm
Geometry: For small angles, 2 mm D
2 a = c = = 0.00662252 rad 302 b = u =
A
x
400 mm 3 mm
2 = 0.00496278 rad 403
Shear Strain: (gB)xy = a + b = 0.0116 rad = 11.6 A 10 - 3 B rad
Ans.
(gA)xy = -(u + c) = -0.0116 rad = -11.6 A 10 - 3 B rad
Ans.
2–19. The piece of plastic is originally rectangular. Determine the shear strain gxy at corners D and C if the plastic distorts as shown by the dashed lines.
y 5 mm 2 mm 2 mm
B
4 mm
C 300 mm 2 mm D
A 400 mm 3 mm
Geometry: For small angles, 2 = 0.00496278 rad 403 2 = 0.00662252 rad b = u = 302 Shear Strain: a = c =
(gC)xy = -(a + b) = -0.0116 rad = -11.6 A 10 - 3 B rad
Ans.
(gD)xy = u + c = 0.0116 rad = 11.6 A 10 - 3 B rad
Ans.
11
x
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*2–20. The piece of plastic is originally rectangular. Determine the average normal strain that occurs along the diagonals AC and DB.
y 5 mm 2 mm 2 mm
Geometry:
B
4 mm
C
AC = DB = 24002 + 3002 = 500 mm
300 mm
DB¿ = 24052 + 3042 = 506.4 mm
2 mm D
A¿C¿ = 2401 + 300 = 500.8 mm 2
2
x
A 400 mm 3 mm
Average Normal Strain: eAC =
A¿C¿ - AC 500.8 - 500 = AC 500
= 0.00160 mm>mm = 1.60 A 10 - 3 B mm>mm eDB =
Ans.
DB¿ - DB 506.4 - 500 = DB 500
= 0.0128 mm>mm = 12.8 A 10 - 3 B mm>mm
Ans.
•2–21.
The force applied to the handle of the rigid lever arm causes the arm to rotate clockwise through an angle of 3° about pin A. Determine the average normal strain developed in the wire. Originally, the wire is unstretched.
D
600 mm
Geometry: Referring to Fig. a, the stretched length of LB¿D can be determined using the consine law, A
LB¿D = 2(0.6 cos 45°)2 + (0.6 sin 45°)2 - 2(0.6 cos 45°)(0.6 sin 45°) cos 93°
B
= 0.6155 m Average Normal Strain: The unstretched length of wire BD is LBD = 0.6 m. We obtain eavg =
C
45⬚
LB¿D - LBD 0.6155 - 0.6 = 0.0258 m>m = LBD 0.6
Ans.
12
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2–22. A square piece of material is deformed into the dashed position. Determine the shear strain gxy at A.
y 15.18 mm B
Shear Strain: (gA)xy =
89.7° p - ¢ ≤p 2 180°
C
15.24 mm
15 mm
= 5.24 A 10 - 3 B rad
Ans.
89.7⬚ A
2–23. A square piece of material is deformed into the dashed parallelogram. Determine the average normal strain that occurs along the diagonals AC and BD.
15 mm 15.18 mm
x
D
y 15.18 mm B
C
15.24 mm
15 mm 89.7⬚ A
Geometry: AC = BD = 2152 + 152 = 21.2132 mm AC¿ = 215.182 + 15.242 - 2(15.18)(15.24) cos 90.3° = 21.5665 mm B¿D¿ = 215.182 + 15.242 - 2(15.18)(15.24) cos 89.7° = 21.4538 mm Average Normal Strain: eAC =
eBD
AC¿ - AC 21.5665 - 21.2132 = AC 21.2132
= 0.01665 mm>mm = 16.7 A 10 - 3 B mm>mm
Ans.
= 0.01134 mm>mm = 11.3 A 10 - 3 B mm>mm
Ans.
B¿D¿ - BD 21.4538 - 21.2132 = = BD 21.2132
13
15 mm 15.18 mm
D
x
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*2–24. A square piece of material is deformed into the dashed position. Determine the shear strain gxy at C.
y 15.18 mm B
C
15.24 mm
15 mm 89.7⬚ A
(gC)xy =
15 mm 15.18 mm
x
D
p 89.7° - ¢ ≤p 2 180° = 5.24 A 10 - 3 B rad
Ans.
u ⫽ 2⬚
•2–25.
The guy wire AB of a building frame is originally unstretched. Due to an earthquake, the two columns of the frame tilt u = 2°. Determine the approximate normal strain in the wire when the frame is in this position. Assume the columns are rigid and rotate about their lower supports.
u ⫽ 2⬚
B
Geometry: The vertical displacement is negligible 3m
xA
2° = (1) ¢ ≤ p = 0.03491 m 180° A
2° xB = (4) ¢ ≤ p = 0.13963 m 180°
1m
x = 4 + xB - xA = 4.10472 m A¿B¿ = 232 + 4.104722 = 5.08416 m AB = 232 + 42 = 5.00 m Average Normal Strain: eAB =
=
A¿B¿ - AB AB 5.08416 - 5 = 16.8 A 10 - 3 B m>m 5
Ans.
14
4m
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2–26. The material distorts into the dashed position shown. Determine (a) the average normal strains along sides AC and CD and the shear strain gxy at F, and (b) the average normal strain along line BE.
y 15 mm C
25 mm D
10 mm B E
75 mm
90 mm
A
Referring to Fig. a, LBE = 2(90 - 75)2 + 802 = 26625 mm LAC¿ = 21002 + 152 = 210225 mm LC¿D¿ = 80 - 15 + 25 = 90 mm f = tan-1 ¢
25 p rad ≤ = 14.04° ¢ ≤ = 0.2450 rad. 100 180°
When the plate deforms, the vertical position of point B and E do not change. LBB¿ 15 = ; LBB¿ = 13.5 mm 90 100 LEE¿ 25 = ; 75 100
LEE¿ = 18.75 mm
LB¿E¿ = 2(90 - 75)2 + (80 - 13.5 + 18.75)2 = 27492.5625 mm Thus,
A eavg B AC =
LAC¿ - LAC 210225 - 100 = = 0.0112 mm>mm LAC 100
Ans.
A eavg B CD =
LC¿D¿ - LCD 90 -80 = = 0.125 mm>mm LCD 80
Ans.
A eavg B BE =
LB¿E¿ - LBE 27492.5625 - 26625 = = 0.0635 mm>mm LBE 26625
Ans.
Referring to Fig. a, the angle at corner F becomes larger than 90° after the plate deforms. Thus, the shear strain is negative. 0.245 rad
Ans.
15
80 mm
F
x
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2–27. The material distorts into the dashed position shown. Determine the average normal strain that occurs along the diagonals AD and CF.
y 15 mm
25 mm D
C 10 mm
The undeformed length of diagonals AD and CF are
B E
LAD = LCF = 280 + 100 = 216400 mm 2
2
The deformed length of diagonals AD and CF are
75 mm
90 mm
LAD¿ = 2(80 + 25) + 100 = 221025 mm 2
2
LC¿F = 2(80 - 15)2 + 1002 = 214225 mm A
Thus,
A eavg B AD =
LAD¿ - LAD 221025 - 216400 = = 0.132 mm>mm LAD 216400
Ans.
A eavg B CF =
LC¿F - LCF 214225 - 216400 = = -0.0687 mm>mm LCF 216400
Ans.
*2–28. The wire is subjected to a normal strain that is 2 defined by P = xe - x , where x is in millimeters. If the wire has an initial length L, determine the increase in its length.
80 mm
P ⫽ xe⫺x
L
2
dL = e dx = x e-x dx L 2
L0
x e-x dx
L 1 1 1 2 2 = - c e-x d 冷 = - c e-L - d 2 2 2 0
=
x
2
x x
¢L =
F
1 2 [1 - e-L ] 2
Ans.
16
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•2–29. The curved pipe has an original radius of 2 ft. If it is heated nonuniformly, so that the normal strain along its length is P = 0.05 cos u, determine the increase in length of the pipe.
e = 0.05 cos u ¢L =
L
2 ft
e dL
=
A
u
90°
(0.05 cos u)(2 du)
L0
90°
= 0.1
90°
cos u du = [0.1[sin u] 0冷 ] = 0.100 ft
L0
Ans.
Solve Prob. 2–29 if P = 0.08 sin u.
2–30.
dL = 2 due = 0.08 sin u ¢L =
e dL
L
90°
=
2 ft
(0.08 sin u)(2 du)
L0
= 0.16
L0
90°
sin u du = 0.16[-cos u] 0冷 = 0.16 ft
Ans.
2–31. The rubber band AB has an unstretched length of 1 ft. If it is fixed at B and attached to the surface at point A¿, determine the average normal strain in the band. The surface is defined by the function y = (x2) ft, where x is in feet.
y y ⫽ x2
A¿
Geometry: 1 ft
L =
L0
A
1 + a
dy 2 b dx dx
However y = x2 then
1 ft
dy = 2x dx
B
1 ft
L =
=
L0
A
u 90°
21 + 4 x2 dx
1 1 ft C 2x21 + 4 x2 + ln A 2x + 21 + 4x2 B D 冷0 4
= 1.47894 ft Average Normal Strain: L - L0 1.47894 - 1 eavg = = = 0.479 ft>ft L0 1
Ans.
17
A 1 ft
x
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*2–32. The bar is originally 300 mm long when it is flat. If it is subjected to a shear strain defined by gxy = 0.02x, where x is in meters, determine the displacement ¢y at the end of its bottom edge. It is distorted into the shape shown, where no elongation of the bar occurs in the x direction.
y
⌬y x 300 mm
Shear Strain: dy = tan gxy ; dx
dy = tan (0.02 x) dx 300 mm
¢y
dy =
L0
L0
tan (0.02 x)dx
¢y = -50[ln cos (0.02x)]|0300 mm = 2.03 mm
Ans.
The fiber AB has a length L and orientation u. If its ends A and B undergo very small displacements uA and vB , respectively, determine the normal strain in the fiber when it is in position A¿B¿.
•2–33.
y B¿ vB B L
Geometry: LA¿B¿ = 2(L cos u - uA) + (L sin u + yB) 2
2
A
= 2L3 + u2A + y2B + 2L(yB sin u - uA cos u) Average Normal Strain: LA¿B¿ - L eAB = L =
A
1 +
2(yB sin u - uA cos u) u2A + y2B + - 1 L L2
Neglecting higher terms u2A and y2B 1
eAB
2(yB sin u - uA cos u) 2 = B1 + R - 1 L
Using the binomial theorem: eAB = 1 +
=
2uA cos u 1 2yB sin u ¢ ≤ + ... - 1 2 L L
yB sin u uA cos u L L
Ans.
18
uA A¿
u x
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2–34. If the normal strain is defined in reference to the final length, that is, Pnœ = lim a p : p¿
¢s¿ - ¢s b ¢s¿
instead of in reference to the original length, Eq. 2–2, show that the difference in these strains is represented as a second-order term, namely, Pn - Pnœ = PnPnœ .
eB =
¢S¿ - ¢S ¢S
œ = eB - eA
¢S¿ - ¢S ¢S¿ - ¢S ¢S ¢S¿
¢S¿ 2 - ¢S¢S¿ - ¢S¿¢S + ¢S2 ¢S¢S¿ 2 2 ¢S¿ + ¢S - 2¢S¿¢S = ¢S¢S¿ =
=
(¢S¿ - ¢S)2 ¢S¿ - ¢S ¢S¿ - ¢S = ¢ ≤¢ ≤ ¢S¢S¿ ¢S ¢S¿
= eA eBœ (Q.E.D)
19
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•3–1.
A concrete cylinder having a diameter of 6.00 in. and gauge length of 12 in. is tested in compression. The results of the test are reported in the table as load versus contraction. Draw the stress–strain diagram using scales of 1 in. = 0.5 ksi and 1 in. = 0.2110-32 in.>in. From the diagram, determine approximately the modulus of elasticity.
Stress and Strain: s =
P (ksi) A
e =
dL (in./in.) L
0
0
0.177
0.00005
0.336
0.00010
0.584
0.000167
0.725
0.000217
0.902
0.000283
1.061
0.000333
1.220
0.000375
1.362
0.000417
1.645
0.000517
1.768
0.000583
1.874
0.000625
Modulus of Elasticity: From the stress–strain diagram Eapprox =
1.31 - 0 = 3.275 A 103 B ksi 0.0004 - 0
Ans.
1
Load (kip)
Contraction (in.)
0 5.0 9.5 16.5 20.5 25.5 30.0 34.5 38.5 46.5 50.0 53.0
0 0.0006 0.0012 0.0020 0.0026 0.0034 0.0040 0.0045 0.0050 0.0062 0.0070 0.0075
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3–2. Data taken from a stress–strain test for a ceramic are given in the table. The curve is linear between the origin and the first point. Plot the diagram, and determine the modulus of elasticity and the modulus of resilience. Modulus of Elasticity: From the stress–strain diagram E =
33.2 - 0 = 55.3 A 103 B ksi 0.0006 - 0
S (ksi)
P (in./in.)
0 33.2 45.5 49.4 51.5 53.4
0 0.0006 0.0010 0.0014 0.0018 0.0022
S (ksi)
P (in./in.)
0 33.2 45.5 49.4 51.5 53.4
0 0.0006 0.0010 0.0014 0.0018 0.0022
Ans.
Modulus of Resilience: The modulus of resilience is equal to the area under the linear portion of the stress–strain diagram (shown shaded). ut =
1 lb in. in # lb (33.2) A 103 B ¢ 2 ≤ ¢ 0.0006 ≤ = 9.96 2 in. in in3
Ans.
3–3. Data taken from a stress–strain test for a ceramic are given in the table. The curve is linear between the origin and the first point. Plot the diagram, and determine approximately the modulus of toughness. The rupture stress is sr = 53.4 ksi. Modulus of Toughness: The modulus of toughness is equal to the area under the stress–strain diagram (shown shaded). (ut)approx =
lb in. 1 (33.2) A 103 B ¢ 2 ≤ (0.0004 + 0.0010) ¢ ≤ 2 in. in + 45.5 A 103 B ¢ +
1 lb in. (7.90) A 103 B ¢ 2 ≤ (0.0012) ¢ ≤ 2 in. in +
= 85.0
lb in. ≤ (0.0012) ¢ ≤ in. in2
1 lb in. (12.3) A 103 B ¢ 2 ≤ (0.0004) ¢ ≤ 2 in. in
in # lb in3
Ans.
2
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*3–4. A tension test was performed on a specimen having an original diameter of 12.5 mm and a gauge length of 50 mm. The data are listed in the table. Plot the stress–strain diagram, and determine approximately the modulus of elasticity, the ultimate stress, and the fracture stress. Use a scale of 20 mm = 50 MPa and 20 mm = 0.05 mm>mm. Redraw the linear-elastic region, using the same stress scale but a strain scale of 20 mm = 0.001 mm>mm. Stress and Strain: s =
dL P (MPa) e = (mm/mm) A L 0
0
90.45
0.00035
259.9
0.00120
308.0
0.00204
333.3
0.00330
355.3
0.00498
435.1
0.02032
507.7
0.06096
525.6
0.12700
507.7
0.17780
479.1
0.23876
Modulus of Elasticity: From the stress–strain diagram (E)approx =
228.75(106) - 0 = 229 GPa 0.001 - 0
Ans.
Ultimate and Fracture Stress: From the stress–strain diagram (sm)approx = 528 MPa
Ans.
(sf)approx = 479 MPa
Ans.
3
Load (kN)
Elongation (mm)
0 11.1 31.9 37.8 40.9 43.6 53.4 62.3 64.5 62.3 58.8
0 0.0175 0.0600 0.1020 0.1650 0.2490 1.0160 3.0480 6.3500 8.8900 11.9380
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3–5. A tension test was performed on a steel specimen having an original diameter of 12.5 mm and gauge length of 50 mm. Using the data listed in the table, plot the stress–strain diagram, and determine approximately the modulus of toughness. Use a scale of 20 mm = 50 MPa and 20 mm = 0.05 mm>mm. Stress and Strain: s =
P dL (MPa) e = (mm/mm) A L 0
0
90.45
0.00035
259.9
0.00120
308.0
0.00204
333.3
0.00330
355.3
0.00498
435.1
0.02032
507.7
0.06096
525.6
0.12700
507.7
0.17780
479.1
0.23876
Modulus of Toughness: The modulus of toughness is equal to the total area under the stress–strain diagram and can be approximated by counting the number of squares. The total number of squares is 187. (ut)approx = 187(25) A 106 B ¢
N m ≤ a 0.025 b = 117 MJ>m3 m m2
Ans.
4
Load (kN)
Elongation (mm)
0 11.1 31.9 37.8 40.9 43.6 53.4 62.3 64.5 62.3 58.8
0 0.0175 0.0600 0.1020 0.1650 0.2490 1.0160 3.0480 6.3500 8.8900 11.9380
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3–6. A specimen is originally 1 ft long, has a diameter of 0.5 in., and is subjected to a force of 500 lb. When the force is increased from 500 lb to 1800 lb, the specimen elongates 0.009 in. Determine the modulus of elasticity for the material if it remains linear elastic. Normal Stress and Strain: Applying s =
s1 =
s2 =
¢e =
0.500 p 2 4 (0.5 )
1.80 p 2 4 (0.5 )
dL P and e = . A L
= 2.546 ksi
= 9.167 ksi
0.009 = 0.000750 in.>in. 12
Modulus of Elasticity: E =
¢s 9.167 - 2.546 = = 8.83 A 103 B ksi ¢e 0.000750
Ans.
3–7. A structural member in a nuclear reactor is made of a zirconium alloy. If an axial load of 4 kip is to be supported by the member, determine its required cross-sectional area. Use a factor of safety of 3 relative to yielding. What is the load on the member if it is 3 ft long and its elongation is 0.02 in.? Ezr = 14(103) ksi, sY = 57.5 ksi. The material has elastic behavior. Allowable Normal Stress: F.S. =
3 =
sy sallow 57.5 sallow
sallow = 19.17 ksi sallow =
P A
19.17 =
4 A
A = 0.2087 in2 = 0.209 in2
Ans.
Stress–Strain Relationship: Applying Hooke’s law with e =
0.02 d = = 0.000555 in.>in. L 3 (12) s = Ee = 14 A 103 B (0.000555) = 7.778 ksi
Normal Force: Applying equation s =
P . A
P = sA = 7.778 (0.2087) = 1.62 kip
Ans.
5
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*3–8. The strut is supported by a pin at C and an A-36 steel guy wire AB. If the wire has a diameter of 0.2 in., determine how much it stretches when the distributed load acts on the strut.
A
60⬚ 200 lb/ft
a + ©MC = 0;
1 FAB cos 60°(9) - (200)(9)(3) = 0 2
9 ft
FAB = 600 lb
The normal stress the wire is sAB =
FAB = AAB
p 4
600 = 19.10(103) psi = 19.10 ksi (0.22)
Since sAB 6 sy = 36 ksi, Hooke’s Law can be applied to determine the strain in wire. sAB = EPAB;
19.10 = 29.0(103)PAB PAB = 0.6586(10 - 3) in>in
9(12) The unstretched length of the wire is LAB = = 124.71 in. Thus, the wire sin 60° stretches dAB = PAB LAB = 0.6586(10 - 3)(124.71) = 0.0821 in.
Ans.
6
B
C
Here, we are only interested in determining the force in wire AB.
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The s –P diagram for a collagen fiber bundle from which a human tendon is composed is shown. If a segment of the Achilles tendon at A has a length of 6.5 in. and an approximate cross-sectional area of 0.229 in2, determine its elongation if the foot supports a load of 125 lb, which causes a tension in the tendon of 343.75 lb.
•3–9.
s =
s (ksi) 4.50 A
3.75 3.00 2.25 1.50
P 343.75 = = 1.50 ksi A 0.229
125 lb
0.75 0.05
From the graph e = 0.035 in.>in. d = eL = 0.035(6.5) = 0.228 in.
0.10
P (in./in.)
Ans.
s (ksi)
3–10. The stress–strain diagram for a metal alloy having an original diameter of 0.5 in. and a gauge length of 2 in. is given in the figure. Determine approximately the modulus of elasticity for the material, the load on the specimen that causes yielding, and the ultimate load the specimen will support.
105 90 75 60
From the stress–strain diagram, Fig. a,
45
60 ksi - 0 E = ; 1 0.002 - 0 sy = 60 ksi
E = 30.0(103) ksi
Ans.
30 15
su>t = 100 ksi
0
Thus, PY = sYA = 60 C p4 (0.52) D = 11.78 kip = 11.8 kip
Ans.
Pu>t = su>t A = 100 C p4 (0.52) D = 19.63 kip = 19.6 kip
Ans.
7
0 0
0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.001 0.002 0.003 0.004 0.005 0.006 0.007
P (in./in.)
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s (ksi)
3–11. The stress–strain diagram for a steel alloy having an original diameter of 0.5 in. and a gauge length of 2 in. is given in the figure. If the specimen is loaded until it is stressed to 90 ksi, determine the approximate amount of elastic recovery and the increase in the gauge length after it is unloaded.
105 90 75 60 45 30 15 0
From the stress–strain diagram Fig. a, the modulus of elasticity for the steel alloy is E 60 ksi - 0 = ; 1 0.002 - 0
E = 30.0(103) ksi
when the specimen is unloaded, its normal strain recovered along line AB, Fig. a, which has a gradient of E. Thus Elastic Recovery =
90 90 ksi = 0.003 in>in = E 30.0(103) ksi
Ans.
Thus, the permanent set is PP = 0.05 - 0.003 = 0.047 in>in Then, the increase in gauge length is ¢L = PPL = 0.047(2) = 0.094 in
Ans.
8
0 0
0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.001 0.002 0.003 0.004 0.005 0.006 0.007
P (in./in.)
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*3–12. The stress–strain diagram for a steel alloy having an original diameter of 0.5 in. and a gauge length of 2 in. is given in the figure. Determine approximately the modulus of resilience and the modulus of toughness for the material. The Modulus of resilience is equal to the area under the stress–strain diagram up to the proportional limit. sPL = 60 ksi
PPL = 0.002 in>in.
Thus, (ui)r =
1 1 in # lb sPLPPL = C 60(103) D (0.002) = 60.0 2 2 in3
Ans.
The modulus of toughness is equal to the area under the entire stress–strain diagram. This area can be approximated by counting the number of squares. The total number is 38. Thus,
C (ui)t D approx = 38 c 15(103)
lb in in # lb d a0.05 b = 28.5(103) 2 in in in3
s (ksi) 105 90 75 60 45 30 15 0
0 0
0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.001 0.002 0.003 0.004 0.005 0.006 0.007
P (in./in.)
9
Ans.
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•3–13.
A bar having a length of 5 in. and cross-sectional area of 0.7 in2 is subjected to an axial force of 8000 lb. If the bar stretches 0.002 in., determine the modulus of elasticity of the material. The material has linear-elastic behavior.
8000 lb
8000 lb 5 in.
Normal Stress and Strain: 8.00 P = = 11.43 ksi A 0.7
s =
e =
dL 0.002 = = 0.000400 in.>in. L 5
Modulus of Elasticity: E =
s 11.43 = = 28.6(103) ksi e 0.000400
Ans.
3–14. The rigid pipe is supported by a pin at A and an A-36 steel guy wire BD. If the wire has a diameter of 0.25 in., determine how much it stretches when a load of P = 600 lb acts on the pipe.
B
Here, we are only interested in determining the force in wire BD. Referring 4 ft to the FBD in Fig. a a + ©MA = 0;
FBD A 45 B (3) - 600(6) = 0
FBD = 1500 lb
A
sBD
3 ft
1500 = 30.56(103) psi = 30.56 ksi p 2 (0.25 ) 4
Since sBD 6 sy = 36 ksi, Hooke’s Law can be applied to determine the strain in the wire. sBD = EPBD;
D C
The normal stress developed in the wire is FBD = = ABD
P
30.56 = 29.0(103)PBD PBD = 1.054(10 - 3) in.>in.
The unstretched length of the wire is LBD = 232 + 42 = 5ft = 60 in. Thus, the wire stretches dBD = PBD LBD = 1.054(10 - 3)(60) = 0.0632 in
Ans.
10
3 ft
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3–15. The rigid pipe is supported by a pin at A and an A-36 guy wire BD. If the wire has a diameter of 0.25 in., determine the load P if the end C is displaced 0.075 in. downward.
B
4 ft
P
A
D C 3 ft
Here, we are only interested in determining the force in wire BD. Referring to the FBD in Fig. a FBD A 45 B (3) - P(6) = 0
a + ©MA = 0;
FBD = 2.50 P
The unstretched length for wire BD is LBD = 232 + 42 = 5 ft = 60 in. From the geometry shown in Fig. b, the stretched length of wire BD is LBD¿ = 2602 + 0.0752 - 2(60)(0.075) cos 143.13° = 60.060017 Thus, the normal strain is PBD =
LBD¿ - LBD 60.060017 - 60 = = 1.0003(10 - 3) in.>in. LBD 60
Then, the normal stress can be obtain by applying Hooke’s Law. sBD = EPBD = 29(103) C 1.0003(10 - 3) D = 29.01 ksi Since sBD 6 sy = 36 ksi, the result is valid. sBD =
FBD ; ABD
29.01(103) =
2.50 P (0.252)
p 4
P = 569.57 lb = 570 lb
Ans.
11
3 ft
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s (MPa)
*3–16. Determine the elongation of the square hollow bar when it is subjected to the axial force P = 100 kN. If this axial force is increased to P = 360 kN and released, find the permanent elongation of the bar. The bar is made of a metal alloy having a stress–strain diagram which can be approximated as shown.
500 600 mm P
250
50 mm 5 mm
0.00125
Normal Stress and Strain: The cross-sectional area of the hollow bar is A = 0.052 - 0.042 = 0.9(10 - 3)m2. When P = 100 kN, s1 =
100(103) P = 111.11 MPa = A 0.9(10 - 3)
From the stress–strain diagram shown in Fig. a, the slope of the straight line OA which represents the modulus of elasticity of the metal alloy is E =
250(106) - 0 = 200 GPa 0.00125 - 0
Since s1 6 250 MPa, Hooke’s Law can be applied. Thus s1 = Ee1; 111.11(106) = 200(109)e1 e1 = 0.5556(10 - 3) mm>mm Thus, the elongation of the bar is d1 = e1L = 0.5556(10 - 3)(600) = 0.333 mm
Ans.
When P = 360 kN, s2 =
360(103) P = 400 MPa = A 0.9(10 - 3)
From the geometry of the stress–strain diagram, Fig. a, e2 - 0.00125 0.05 - 0.00125 = 400 - 250 500 - 250
e2 = 0.0305 mm>mm
When P = 360 kN is removed, the strain recovers linearly along line BC, Fig. a, parallel to OA. Thus, the elastic recovery of strain is given by s2 = Eer;
400(106) = 200(109)er er = 0.002 mm>mm
The permanent set is eP = e2 - er = 0.0305 - 0.002 = 0.0285 mm>mm Thus, the permanent elongation of the bar is dP = ePL = 0.0285(600) = 17.1 mm
Ans.
12
0.05
P (mm/mm) 50 mm
P 5 mm
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3–16. Continued
13
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s (ksi)
3–17. A tension test was performed on an aluminum 2014-T6 alloy specimen. The resulting stress–strain diagram is shown in the figure. Estimate (a) the proportional limit, (b) the modulus of elasticity, and (c) the yield strength based on a 0.2% strain offset method.
70 60 50 40 30 20 10 0
0.02 0.002
Proportional Limit and Yield Strength: From the stress–strain diagram, Fig. a, spl = 44 ksi
Ans.
sY = 60 ksi
Ans.
Modulus of Elasticity: From the stress–strain diagram, the corresponding strain for sPL = 44 ksi is epl = 0.004 in.>in. Thus, E =
44 - 0 = 11.0(103) ksi 0.004 - 0
Ans.
Modulus of Resilience: The modulus of resilience is equal to the area under the
14
0.04 0.004
0.06 0.006
0.08 0.008
0.10 0.010
P (in./in.)
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s (ksi)
3–18. A tension test was performed on an aluminum 2014-T6 alloy specimen. The resulting stress–strain diagram is shown in the figure. Estimate (a) the modulus of resilience; and (b) modulus of toughness.
70 60 50 40 30 20 10 0
0.02 0.002
0.04 0.004
0.06 0.006
0.08 0.008
0.10 0.010
P (in./in.)
stress–strain diagram up to the proportional limit. From the stress–strain diagram, spl = 44 ksi
epl = 0.004 in.>in.
Thus,
A Ui B r = splepl = (44)(103)(0.004) = 88 1 2
1 2
in # lb in3
Ans.
Modulus of Toughness: The modulus of toughness is equal to the area under the entire stress–strain diagram. This area can be approximated by counting the number of squares. The total number of squares is 65. Thus,
C A Ui B t D approx = 65 B 10(103)
lb in. in # lb c0.01 d = 6.50(103) 2R in. in in3
Ans.
The stress–strain diagram for a bone is shown, and can be described by the equation
3–19. The stress–strain diagram for a bone is shown, and can be described by the equation P = 0.45110-62 s ⫹ 0.36110-122 s3, where s is in kPa. Determine the yield strength assuming a 0.3% offset.
P
s
P ⫽ 0.45(10⫺6)s + 0.36(10⫺12)s3 P
e = 0.45(10-6)s + 0.36(10-12)s3, dP = A 0.45(10-6) + 1.08(10-12) s2 B ds E =
ds 1 2 = = 2.22 MPa dP 0.45(10 - 6)
Ans.
s=0
15
P
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*3–20. The stress–strain diagram for a bone is shown and can be described by the equation P = 0.45110-62 s ⫹ 0.36110-122 s3, where s is in kPa. Determine the modulus of toughness and the amount of elongation of a 200-mmlong region just before it fractures if failure occurs at P = 0.12 mm>mm.
P
s
P ⫽ 0.45(10⫺6)s + 0.36(10⫺12)s3 P
When e = 0.12
120(103) = 0.45 s + 0.36(10-6)s3 Solving for the real root: s = 6873.52 kPa 6873.52
ut =
LA
dA =
L0
(0.12 - e)ds
6873.52
ut =
L0
(0.12 - 0.45(10-6)s - 0.36(10-12)s3)ds 6873.52
= 0.12 s - 0.225(10-6)s2 - 0.09(10-12)s4|0 = 613 kJ>m3
Ans.
d = eL = 0.12(200) = 24 mm
Ans.
16
P
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•3–21.
The stress–strain diagram for a polyester resin is given in the figure. If the rigid beam is supported by a strut AB and post CD, both made from this material, and subjected to a load of P = 80 kN, determine the angle of tilt of the beam when the load is applied. The diameter of the strut is 40 mm and the diameter of the post is 80 mm.
B
2m P
A
C 0.75 m 0.75 m
D
0.5 m
From the stress–strain diagram, E =
32.2(10)6 = 3.22(109) Pa 0.01
s (MPa) 100 95
Thus,
70 60
40(10 ) FAB = p = 31.83 MPa 2 AAB 4 (0.04)
sAB =
eAB
50
31.83(106) sAB = 0.009885 mm>mm = = E 3.22(109)
20 0
7.958(106) sCD = 0.002471 mm>mm = E 3.22(109)
dAB = eABLAB = 0.009885(2000) = 19.771 mm dCD = eCDLCD = 0.002471(500) = 1.236 mm Angle of tilt a: tan a =
18.535 ; 1500
tension
40 32.2
40(103) FCD = p = 7.958 MPa 2 ACD 4 (0.08)
sCD =
eCD =
compression
80 3
a = 0.708°
Ans.
17
0
0.01 0.02 0.03 0.04
P (mm/mm)
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3–22. The stress–strain diagram for a polyester resin is given in the figure. If the rigid beam is supported by a strut AB and post CD made from this material, determine the largest load P that can be applied to the beam before it ruptures. The diameter of the strut is 12 mm and the diameter of the post is 40 mm.
B
2m P
Rupture of strut AB: sR =
FAB ; AAB
50(106) =
P>2
A
; p 2 4 (0.012)
0.75 m 0.75 m
P = 11.3 kN (controls)
D
0.5 m
Ans. s (MPa)
Rupture of post CD: FCD ; sR = ACD
C
95(10 ) =
100 95
P>2
6
p 2 4 (0.04)
compression
80 70 60
P = 239 kN
50 tension
40 32.2 20 0
0
0.01 0.02 0.03 0.04
P (mm/mm)
s (ksi)
3–23. By adding plasticizers to polyvinyl chloride, it is possible to reduce its stiffness. The stress–strain diagrams for three types of this material showing this effect are given below. Specify the type that should be used in the manufacture of a rod having a length of 5 in. and a diameter of 2 in., that is required to support at least an axial load of 20 kip and also be able to stretch at most 14 in.
15 P unplasticized 10
copolymer
flexible
5
(plasticized)
Normal Stress:
P
P s = = A
20 p 2 = 6.366 ksi (2 ) 4
0
0
Normal Strain: e =
0.25 = 0.0500 in.>in. 5
From the stress–strain diagram, the copolymer will satisfy both stress and strain requirements. Ans.
18
0.10
0.20
0.30
P (in./in.)
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*3–24. The stress–strain diagram for many metal alloys can be described analytically using the Ramberg-Osgood three parameter equation P = s>E + ksn, where E, k, and n are determined from measurements taken from the diagram. Using the stress–strain diagram shown in the figure, take E = 3011032 ksi and determine the other two parameters k and n and thereby obtain an analytical expression for the curve.
s (ksi) 80 60 40 20 0.1
0.2
0.3
0.4
0.5
P (10–6)
Choose, s = 40 ksi,
e = 0.1
s = 60 ksi,
e = 0.3
0.1 =
40 + k(40)n 30(103)
0.3 =
60 + k(60)n 30(103)
0.098667 = k(40)n 0.29800 = k(60)n 0.3310962 = (0.6667)n ln (0.3310962) = n ln (0.6667) n = 2.73
Ans.
k = 4.23(10-6)
Ans.
•3–25.
The acrylic plastic rod is 200 mm long and 15 mm in diameter. If an axial load of 300 N is applied to it, determine the change in its length and the change in its diameter. Ep = 2.70 GPa, np = 0.4. s =
P = A
elong =
300 p 2 4 (0.015)
300 N
300 N 200 mm
= 1.697 MPa
1.697(106) s = 0.0006288 = E 2.70(109)
d = elong L = 0.0006288 (200) = 0.126 mm
Ans.
elat = -Velong = -0.4(0.0006288) = -0.0002515 ¢d = elatd = -0.0002515 (15) = -0.00377 mm
Ans.
19
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3–26. The short cylindrical block of 2014-T6 aluminum, having an original diameter of 0.5 in. and a length of 1.5 in., is placed in the smooth jaws of a vise and squeezed until the axial load applied is 800 lb. Determine (a) the decrease in its length and (b) its new diameter.
800 lb
800 lb
a) s =
P = A
elong =
p 4
800 = 4074.37 psi (0.5)2
s -4074.37 = -0.0003844 = E 10.6(106)
d = elong L = -0.0003844 (1.5) = -0.577 (10 - 3) in.
Ans.
b) V =
-elat = 0.35 elong
elat = -0.35 (-0.0003844) = 0.00013453 ¢d = elat d = 0.00013453 (0.5) = 0.00006727 d¿ = d + ¢d = 0.5000673 in.
Ans.
s(MPa)
3–27. The elastic portion of the stress–strain diagram for a steel alloy is shown in the figure. The specimen from which it was obtained had an original diameter of 13 mm and a gauge length of 50 mm. When the applied load on the specimen is 50 kN, the diameter is 12.99265 mm. Determine Poisson’s ratio for the material.
400
Normal Stress: s =
P = A
50(103) p 4
(0.0132)
= 376.70 Mpa 0.002
Normal Strain: From the stress–strain diagram, the modulus of elasticity 400(106) = 200 GPa. Applying Hooke’s law E = 0.002 elong =
elat =
376.70(106) s = 1.8835 A 10 - 3 B mm>mm = E 200(104)
d - d0 12.99265 - 13 = = -0.56538 A 10 - 3 B mm>mm d0 13
Poisson’s Ratio: The lateral and longitudinal strain can be related using Poisson’s ratio. V = -
-0.56538(10 - 3) elat = 0.300 = elong 1.8835(10 - 3)
Ans.
20
P(mm/mm)
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s(MPa)
*3–28. The elastic portion of the stress–strain diagram for a steel alloy is shown in the figure. The specimen from which it was obtained had an original diameter of 13 mm and a gauge length of 50 mm. If a load of P = 20 kN is applied to the specimen, determine its diameter and gauge length. Take n = 0.4.
400
Normal Stress: s =
P = A
20(103) p 4
(0.0132)
= 150.68Mpa 0.002
P(mm/mm)
Normal Strain: From the Stress–Strain diagram, the modulus of elasticity 400(106) E = = 200 GPa. Applying Hooke’s Law 0.002 elong =
150.68(106) s = 0.7534 A 10 - 3 B mm>mm = E 200(109)
Thus, dL = elong L0 = 0.7534 A 10 - 3 B (50) = 0.03767 mm L = L0 + dL = 50 + 0.03767 = 50.0377 mm
Ans.
Poisson’s Ratio: The lateral and longitudinal can be related using poisson’s ratio. elat = -velong = -0.4(0.7534) A 10 - 3 B
= -0.3014 A 10 - 3 B mm>mm
dd = elat d = -0.3014 A 10 - 3 B (13) = -0.003918 mm d = d0 + dd = 13 + ( -0.003918) = 12.99608 mm
Ans.
•3–29.
The aluminum block has a rectangular cross section and is subjected to an axial compressive force of 8 kip. If the 1.5-in. side changed its length to 1.500132 in., determine Poisson’s ratio and the new length of the 2-in. side. Eal ⫽ 10(103) ksi. s =
elat =
2 in.
8 kip
8 kip 3 in.
P 8 = = 2.667 ksi A (2)(1.5)
elong =
v =
1.5 in.
s -2.667 = -0.0002667 = E 10(103)
1.500132 - 1.5 = 0.0000880 1.5
-0.0000880 = 0.330 -0.0002667
Ans.
h¿ = 2 + 0.0000880(2) = 2.000176 in.
Ans.
21
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3–30. The block is made of titanium Ti-6A1-4V and is subjected to a compression of 0.06 in. along the y axis, and its shape is given a tilt of u = 89.7°. Determine Px, Py, and gxy.
y
Normal Strain: ey =
4 in. u
dLy Ly
=
-0.06 = -0.0150 in.>in. 4
Ans.
Poisson’s Ratio: The lateral and longitudinal strain can be related using Poisson’s ratio.
x
5 in.
ex = -vey = -0.36(-0.0150) = 0.00540 in. >in.
Ans.
Shear Strain: b = 180° - 89.7° = 90.3° = 1.576032 rad gxy =
p p - b = - 1.576032 = -0.00524 rad 2 2
Ans.
3–31. The shear stress–strain diagram for a steel alloy is shown in the figure. If a bolt having a diameter of 0.75 in. is made of this material and used in the double lap joint, determine the modulus of elasticity E and the force P required to cause the material to yield. Take n = 0.3.
P/2 P/2
P
t(ksi) 60
The shear force developed on the shear planes of the bolt can be determined by considering the equilibrium of the FBD shown in Fig. a + ©F = 0; : x
V + V - P = 0
V = =
g(rad)
P
0.00545
2
From the shear stress–strain diagram, the yield stress is ty = 60 ksi. Thus, ty =
Vy A
;
60 =
P>2
p 4
A 0.752 B
P = 53.01 kip = 53.0 kip
Ans.
From the shear stress–strain diagram, the shear modulus is G =
60 ksi = 11.01(103) ksi 0.00545
Thus, the modulus of elasticity is G =
E ; 2(1 + y)
11.01(103) =
E 2(1 + 0.3)
E = 28.6(103) ksi
Ans.
22
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*3–32. A shear spring is made by bonding the rubber annulus to a rigid fixed ring and a plug. When an axial load P is placed on the plug, show that the slope at point y in the rubber is dy>dr = -tan g = -tan1P>12phGr22. For small angles we can write dy>dr = -P>12phGr2. Integrate this expression and evaluate the constant of integration using the condition that y = 0 at r = ro. From the result compute the deflection y = d of the plug.
P
h
ro
y
d
ri r y
Shear Stress–Strain Relationship: Applying Hooke’s law with tA =
g =
P . 2p r h
tA P = G 2p h G r
dy P = -tan g = -tan a b dr 2p h G r
(Q.E.D)
If g is small, then tan g = g. Therefore, dy P = dr 2p h G r
At r = ro,
y = -
dr P 2p h G L r
y = -
P ln r + C 2p h G
0 = -
P ln ro + C 2p h G
y = 0
C =
Then, y =
ro P ln r 2p h G
At r = ri,
y = d d =
P ln ro 2p h G
ro P ln ri 2p h G
Ans.
23
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•3–33.
The support consists of three rigid plates, which are connected together using two symmetrically placed rubber pads. If a vertical force of 5 N is applied to plate A, determine the approximate vertical displacement of this plate due to shear strains in the rubber. Each pad has cross-sectional dimensions of 30 mm and 20 mm. Gr = 0.20 MPa.
C
B
40 mm
40 mm
A
tavg =
g =
V 2.5 = = 4166.7 Pa A (0.03)(0.02)
5N
t 4166.7 = 0.02083 rad = G 0.2(106)
d = 40(0.02083) = 0.833 mm
Ans.
3–34. A shear spring is made from two blocks of rubber, each having a height h, width b, and thickness a. The blocks are bonded to three plates as shown. If the plates are rigid and the shear modulus of the rubber is G, determine the displacement of plate A if a vertical load P is applied to this plate. Assume that the displacement is small so that d = a tan g L ag.
P
d A
h
Average Shear Stress: The rubber block is subjected to a shear force of V =
P . 2
P
t =
V P 2 = = A bh 2bh
Shear Strain: Applying Hooke’s law for shear P
g =
t P 2bh = = G G 2bhG
Thus, d = ag = =
Pa 2bhG
Ans.
24
a
a
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s (ksi)
3–35. The elastic portion of the tension stress–strain diagram for an aluminum alloy is shown in the figure. The specimen used for the test has a gauge length of 2 in. and a diameter of 0.5 in. When the applied load is 9 kip, the new diameter of the specimen is 0.49935 in. Compute the shear modulus Gal for the aluminum.
70
0.00614
From the stress–strain diagram,
P (in./in.)
70 s = = 11400.65 ksi e 0.00614
Eal =
When specimen is loaded with a 9 - kip load, s =
P = A
p 4
9 = 45.84 ksi (0.5)2
s 45.84 = = 0.0040208 in.>in. E 11400.65
elong =
0.49935 - 0.5 d¿ - d = = - 0.0013 in.>in. d 0.5
elat =
V = -
Gal =
elat -0.0013 = 0.32332 = elong 0.0040208
11.4(103) Eat = = 4.31(103) ksi 2(1 + v) 2(1 + 0.32332)
Ans.
s (ksi)
*3–36. The elastic portion of the tension stress–strain diagram for an aluminum alloy is shown in the figure. The specimen used for the test has a gauge length of 2 in. and a diameter of 0.5 in. If the applied load is 10 kip, determine the new diameter of the specimen. The shear modulus is Gal = 3.811032 ksi. P s = = A
70
0.00614
10 = 50.9296 ksi p 2 (0.5) 4
From the stress–strain diagram E =
70 = 11400.65 ksi 0.00614
elong =
G =
s 50.9296 = = 0.0044673 in.>in. E 11400.65
E ; 2(1 + v)
3.8(103) =
11400.65 ; 2(1 + v)
v = 0.500
elat = - velong = - 0.500(0.0044673) = - 0.002234 in.>in. ¢d = elat d = - 0.002234(0.5) = - 0.001117 in. d¿ = d + ¢d = 0.5 - 0.001117 = 0.4989 in.
Ans.
25
P (in./in.)
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s(psi)
3–37. The s – P diagram for elastic fibers that make up human skin and muscle is shown. Determine the modulus of elasticity of the fibers and estimate their modulus of toughness and modulus of resilience.
55
11 1
E =
11 = 5.5 psi 2
Ans.
ut =
1 1 (2)(11) + (55 + 11)(2.25 - 2) = 19.25 psi 2 2
Ans.
ut =
1 (2)(11) = 11 psi 2
Ans.
3–38. A short cylindrical block of 6061-T6 aluminum, having an original diameter of 20 mm and a length of 75 mm, is placed in a compression machine and squeezed until the axial load applied is 5 kN. Determine (a) the decrease in its length and (b) its new diameter.
a)
s =
-5(103) P = p = - 15.915 MPa 2 A 4 (0.02)
s = E elong ;
- 15.915(106) = 68.9(109) elong elong = - 0.0002310 mm>mm
d = elong L = - 0.0002310(75) = - 0.0173 mm b)
v = -
elat ; elong
0.35 = -
Ans.
elat -0.0002310
elat = 0.00008085 mm>mm ¢d = elat d = 0.00008085(20) = 0.0016 mm d¿ = d + ¢d = 20 + 0.0016 = 20.0016 mm
Ans.
26
2 2.25
P(in./in.)
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3–39. The rigid beam rests in the horizontal position on two 2014-T6 aluminum cylinders having the unloaded lengths shown. If each cylinder has a diameter of 30 mm, determine the placement x of the applied 80-kN load so that the beam remains horizontal. What is the new diameter of cylinder A after the load is applied? nal = 0.35. a + ©MA = 0;
FB(3) - 80(x) = 0;
a + ©MB = 0;
-FA(3) + 80(3 - x) = 0;
FB =
80 kN x
A
80x 3 FA =
B
210 mm
220 mm
(1) 80(3 - x) 3
3m
(2)
Since the beam is held horizontally, dA = dB s =
P ; A
d = eL = a
P
e = P A
E
dA = dB ;
s A = E E
bL =
PL AE
80(3 - x) (220) 3
80x 3 (210)
=
AE
AE
80(3 - x)(220) = 80x(210) x = 1.53 m
Ans.
From Eq. (2), FA = 39.07 kN sA =
39.07(103) FA = 55.27 MPa = p 2 A 4 (0.03 )
elong =
55.27(106) sA = -0.000756 = E 73.1(109)
elat = -velong = -0.35(-0.000756) = 0.0002646 œ dA = dA + d elat = 30 + 30(0.0002646) = 30.008 mm
Ans.
*3–40. The head H is connected to the cylinder of a compressor using six steel bolts. If the clamping force in each bolt is 800 lb, determine the normal strain in the 3 bolts. Each bolt has a diameter of 16 in. If sY = 40 ksi and 3 Est = 29110 2 ksi, what is the strain in each bolt when the nut is unscrewed so that the clamping force is released?
C L
H
Normal Stress: s =
P = A
800
A B
p 3 2 4 16
= 28.97 ksi 6 sg = 40 ksi
Normal Strain: Since s 6 sg, Hooke’s law is still valid. e =
28.97 s = 0.000999 in.>in. = E 29(103)
Ans.
If the nut is unscrewed, the load is zero. Therefore, the strain e = 0
27
Ans.
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•3–41. The stone has a mass of 800 kg and center of gravity at G. It rests on a pad at A and a roller at B. The pad is fixed to the ground and has a compressed height of 30 mm, a width of 140 mm, and a length of 150 mm. If the coefficient of static friction between the pad and the stone is ms = 0.8, determine the approximate horizontal displacement of the stone, caused by the shear strains in the pad, before the stone begins to slip. Assume the normal force at A acts 1.5 m from G as shown. The pad is made from a material having E = 4 MPa and n = 0.35.
0.4 m
B
Equations of Equilibrium: a + ©MB = 0; + ©F = 0; : x
FA(2.75) - 7848(1.25) - P(0.3) = 0
[1]
P - F = 0
[2]
Note: The normal force at A does not act exactly at A. It has to shift due to friction. Friction Equation: F = ms FA = 0.8 FA
[3]
Solving Eqs. [1], [2] and [3] yields: FA = 3908.37 N
F = P = 3126.69 N
Average Shear Stress: The pad is subjected to a shear force of V = F = 3126.69 N. t =
V 3126.69 = = 148.89 kPa A (0.14)(0.15)
Modulus of Rigidity: G =
E 4 = = 1.481 MPa 2(1 + v) 2(1 + 0.35)
Shear Strain: Applying Hooke’s law for shear g =
148.89(103) t = 0.1005 rad = G 1.481(106)
Thus, dh = hg = 30(0.1005) = 3.02 mm
Ans.
28
P
G
1.25 m
0.3 m 1.5 m
A
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3–42. The bar DA is rigid and is originally held in the horizontal position when the weight W is supported from C. If the weight causes B to be displaced downward 0.025 in., determine the strain in wires DE and BC. Also, if the wires are made of A-36 steel and have a cross-sectional area of 0.002 in2, determine the weight W.
E 3 ft 2 ft D
3 ft B
5 3 = 0.025 d
A
4 ft
d = 0.0417 in eDE =
C
0.0417 d = = 0.00116 in.>in. L 3(12)
Ans. W
3
sDE = EeDE = 29(10 )(0.00116) = 33.56 ksi FDE = sDEADE = 33.56 (0.002) = 0.0672 kip a + ©MA = 0;
-(0.0672) (5) + 3(W) = 0
W = 0.112 kip = 112 lb
Ans.
sBC =
W 0.112 = = 55.94 ksi ABC 0.002
eBC =
sBC 55.94 = 0.00193 in.>in. = E 29 (103)
Ans.
3–43. The 8-mm-diameter bolt is made of an aluminum alloy. It fits through a magnesium sleeve that has an inner diameter of 12 mm and an outer diameter of 20 mm. If the original lengths of the bolt and sleeve are 80 mm and 50 mm, respectively, determine the strains in the sleeve and the bolt if the nut on the bolt is tightened so that the tension in the bolt is 8 kN. Assume the material at A is rigid. Eal = 70 GPa, Emg = 45 GPa.
50 mm A
30 mm
Normal Stress: 8(103)
sb =
P = Ab
p 2 4 (0.008 )
ss =
P = As
p 2 4 (0.02
= 159.15 MPa
8(103) - 0.0122)
= 39.79 MPa
Normal Strain: Applying Hooke’s Law eb =
159.15(106) sb = 0.00227 mm>mm = Eal 70(109)
Ans.
es =
39.79(106) ss = 0.000884 mm>mm = Emg 45(109)
Ans.
29
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*3–44. The A-36 steel wire AB has a cross-sectional area of 10 mm2 and is unstretched when u = 45.0°. Determine the applied load P needed to cause u = 44.9°.
A
400 mm
u 400 m
m
B P
¿ LAB 400 = sin 90.2° sin 44.9° ¿ = 566.67 mm LAB
LAB =
e =
400 = 565.69 sin 45°
¿ - LAB LAB 566.67 - 565.69 = = 0.001744 LAB 565.69
s = Ee = 200(109) (0.001744) = 348.76 MPa a + ©MA = 0 P(400 cos 0.2°) - FAB sin 44.9° (400) = 0
(1)
However, FAB = sA = 348.76(106)(10)(10 - 6) = 3.488 kN From Eq. (1), P = 2.46 kN
Ans.
30
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•4–1. The ship is pushed through the water using an A-36 steel propeller shaft that is 8 m long, measured from the propeller to the thrust bearing D at the engine. If it has an outer diameter of 400 mm and a wall thickness of 50 mm, determine the amount of axial contraction of the shaft when the propeller exerts a force on the shaft of 5 kN. The bearings at B and C are journal bearings. A
Internal Force: As shown on FBD.
B
C
D
5 kN
Displacement:
8m
dA =
PL = AE
-5.00 (103)(8) p 4
(0.42 - 0.32) 200(109)
= -3.638(10 - 6) m = -3.64 A 10 - 3 B mm
Ans.
Negative sign indicates that end A moves towards end D.
4–2. The copper shaft is subjected to the axial loads shown. Determine the displacement of end A with respect to end D. The diameters of each segment are dAB = 3 in., dBC = 2 in., and dCD = 1 in. Take Ecu = 1811032 ksi.
50 in.
A
p The cross-sectional area of segment AB, BC and CD are AAB = (32) = 2.25p in2, 4 p p ABC = (22) = p in2 and ACD = (12) = 0.25p in2. 4 4 Thus, PCD LCD PiLi PAB LAB PBC LBC = + + AiEi AAB ECu ABC ECu ACD ECu 2.00 (75)
6.00 (50)
=
(2.25p) C 18(10 ) D 3
+
p C 18(10 ) D 3
-1.00 (60)
+
(0.25p) C 18(103) D
= 0.766(10 - 3) in.
Ans.
The positive sign indicates that end A moves away from D.
122
60 in.
2 kip
6 kip
The normal forces developed in segment AB, BC and CD are shown in the FBDS of each segment in Fig. a, b and c respectively.
dA>D = ©
75 in.
B 2 kip
1 kip C
3 kip
D
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4–3. The A-36 steel rod is subjected to the loading shown. If the cross-sectional area of the rod is 50 mm2, determine the displacement of its end D. Neglect the size of the couplings at B, C, and D.
1m
A 9 kN B
The normal forces developed in segments AB, BC and CD are shown in the FBDS of each segment in Fig. a, b and c, respectively. The
cross-sectional areas of all 2 1 m b = 50.0(10 - 6) m2. A = A 50 mm2 B a 1000 mm dD = ©
the
segments
are
PiLi 1 = a PAB LAB + PBC LBC + PCD LCD b AiEi A ESC 1
=
50.0(10 ) C 200(109) D -6
c -3.00(103)(1) + 6.00(103)(1.5) + 2.00(103)(1.25) d
= 0.850(10 - 3) m = 0.850 mm
Ans.
The positive sign indicates that end D moves away from the fixed support.
123
1.5 m
1.25 m
C
4 kN
D
2 kN
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*4–4. The A-36 steel rod is subjected to the loading shown. If the cross-sectional area of the rod is 50 mm2, determine the displacement of C. Neglect the size of the couplings at B, C, and D.
1m
1.5 m
1.25 m
C
A 9 kN B
4 kN
D
2 kN
The normal forces developed in segments AB and BC are shown the FBDS of each segment in Fig. a and b, respectively. The cross-sectional area of these two segments 2 1m are A = A 50 mm2 B a b = 50.0 (10 - 6) m2. Thus, 10.00 mm dC = ©
PiLi 1 = A P L + PBC LBC B AiEi A ESC AB AB 1
=
50.0(10 - 6) C 200(109) D
c -3.00(103)(1) + 6.00(103)(1.5) d
= 0.600 (10 - 3) m = 0.600 mm
Ans.
The positive sign indicates that coupling C moves away from the fixed support.
4–5. The assembly consists of a steel rod CB and an aluminum rod BA, each having a diameter of 12 mm. If the rod is subjected to the axial loadings at A and at the coupling B, determine the displacement of the coupling B and the end A. The unstretched length of each segment is shown in the figure. Neglect the size of the connections at B and C, and assume that they are rigid. Est = 200 GPa, Eal = 70 GPa.
dB =
PL = AE
dA = ©
12(103)(3) p 4
12(103)(3) p 4
2
A
B
18 kN 6 kN 3m
= 0.00159 m = 1.59 mm
(0.012)2(200)(109)
PL = AE
C
2m
Ans.
18(103)(2) 9
(0.012) (200)(10 )
+
p 2 9 4 (0.012) (70)(10 )
= 0.00614 m = 6.14 mm
Ans.
4–6. The bar has a cross-sectional area of 3 in2, and E = 3511032 ksi. Determine the displacement of its end A when it is subjected to the distributed loading.
x
w ⫽ 500x1/3 lb/in.
A
4 ft x
P(x) =
L0
w dx = 500
x
L0
1
x3 dx =
1500 43 x 4
L
dA =
4(12) P(x) dx 1 3 1 1500 4 1500 b a b(48)3 = x3 dx = a AE 4 (3)(35)(108)(4) 7 (3)(35)(106) L0 L0
dA = 0.0128 in.
Ans.
124
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4–7. The load of 800 lb is supported by the four 304 stainless steel wires that are connected to the rigid members AB and DC. Determine the vertical displacement of the load if the members were horizontal before the load was applied. Each wire has a cross-sectional area of 0.05 in2.
E
F
4 ft H
D
C 2 ft
Referring to the FBD of member AB, Fig. a
5 ft
4.5 ft
a + ©MA = 0;
FBC (5) - 800(1) = 0
FBC = 160 lb
a + ©MB = 0;
800(4) - FAH (5) = 0
FAH = 640 lb
800 lb A
B 1 ft
Using the results of FBC and FAH, and referring to the FBD of member DC, Fig. b a + ©MD = 0;
FCF (7) - 160(7) - 640(2) = 0
a + ©MC = 0;
640(5) - FDE(7) = 0
FCF = 342.86 lb FDE = 457.14 lb
Since E and F are fixed, dD =
457.14(4)(2) FDE LDE = = 0.01567 in T A Est 0.05 C 28.0 (106) D
dC =
342.86 (4)(12) FCF LCF = = 0.01176 in T A Est 0.05 C 28.0 (106) D
From the geometry shown in Fig. c, dH = 0.01176 +
5 (0.01567 - 0.01176) = 0.01455 in T 7
Subsequently, dA>H =
640(4.5)(12) FAH LAH = = 0.02469 in T A Est 0.05 C 28.0(106) D
dB>C =
160(4.5)(12) FBC LBC = = 0.006171 in T A Est 0.05 C 28.0(106) D
Thus,
A + T B dA = dH + dA>H = 0.01455 + 0.02469 = 0.03924 in T
A + T B dB = dC + dB>C = 0.01176 + 0.006171 = 0.01793 in T From the geometry shown in Fig. d, dP = 0.01793 +
4 (0.03924 - 0.01793) = 0.0350 in T 5
125
Ans.
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*4–8. The load of 800 lb is supported by the four 304 stainless steel wires that are connected to the rigid members AB and DC. Determine the angle of tilt of each member after the load is applied.The members were originally horizontal, and each wire has a cross-sectional area of 0.05 in2.
E
F
4 ft H
D
C 2 ft
Referring to the FBD of member AB, Fig. a,
5 ft
4.5 ft
a + ©MA = 0;
FBC (5) - 800(1) = 0
FBC = 160 lb
a + ©MB = 0;
800(4) - FAH (5) = 0
FAH = 640 lb
800 lb A
B 1 ft
Using the results of FBC and FAH and referring to the FBD of member DC, Fig. b, a + ©MD = 0;
FCF (7) - 160(7) - 640(2) = 0
FCF = 342.86 lb
a + ©MC = 0;
640(5) - FDE (7) = 0
FDE = 457.14 lb
Since E and F are fixed, dD =
457.14 (4)(12) FDE LDE = = 0.01567 in T A Est 0.05 C 28.0(106) D
dC =
342.86 (4)(12) FCF LCF = = 0.01176 in T A Est 0.05 C 28.0(106) D
From the geometry shown in Fig. c dH = 0.01176 +
u =
5 (0.01567 - 0.01176) = 0.01455 in T 7
0.01567 - 0.01176 = 46.6(10 - 6) rad 7(12)
Ans.
Subsequently, dA>H =
640 (4.5)(12) FAH LAH = = 0.02469 in T A Est 0.05 C 28.0(106) D
dB>C =
160 (4.5)(12) FBC LBC = = 0.006171 in T A Est 0.05 C 28.0(106) D
Thus,
A + T B dA = dH + dA>H = 0.01455 + 0.02469 = 0.03924 in T
A + T B dB = dC + dB>C = 0.01176 + 0.006171 = 0.01793 in T From the geometry shown in Fig. d f =
0.03924 - 0.01793 = 0.355(10 - 3) rad 5(12)
Ans.
126
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4–8. Continued
127
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•4–9.
The assembly consists of three titanium (Ti-6A1-4V) rods and a rigid bar AC. The cross-sectional area of each rod is given in the figure. If a force of 6 kip is applied to the ring F, determine the horizontal displacement of point F.
D
4 ft
C
ACD ⫽ 1 in2 2 ft E
Internal Force in the Rods: a + ©MA = 0; + ©F = 0; : x
FCD(3) - 6(1) = 0
FCD = 2.00 kip
6 - 2.00 - FAB = 0
FAB = 4.00 kip
AAB ⫽ 1.5 in2 6 ft
B
1 ft F
6 kip 2 1 ft AEF ⫽ 2 in
A
Displacement: dC =
2.00(4)(12) FCD LCD = 0.0055172 in. = ACD E (1)(17.4)(103)
dA =
4.00(6)(12) FAB LAB = 0.0110344 in. = AAB E (1.5)(17.4)(103)
dF>E =
6.00(1)(12) FEF LEF = 0.0020690 in. = AEF E (2)(17.4)(103)
œ dE 0.0055172 = ; 2 3
œ dE = 0.0036782 in.
œ dE = dC + dE = 0.0055172 + 0.0036782 = 0.0091954 in.
dF = dE + dF>E = 0.0091954 + 0.0020690 = 0.0113 in.
Ans.
4–10. The assembly consists of three titanium (Ti-6A1-4V) rods and a rigid bar AC. The cross-sectional area of each rod is given in the figure. If a force of 6 kip is applied to the ring F, determine the angle of tilt of bar AC.
D
C
4 ft ACD ⫽ 1 in
2
2 ft E
Internal Force in the Rods: a + ©MA = 0;
FCD(3) - 6(1) = 0
FCD = 2.00 kip
+ ©F = 0; : x
6 - 2.00 - FAB = 0
FAB = 4.00 kip
AAB ⫽ 1.5 in2 B
Displacement: dC =
2.00(4)(12) FCD LCD = 0.0055172 in. = ACD E (1)(17.4)(103)
dA =
4.00(6)(12) FAB LAB = 0.0110344 in. = AAB E (1.5)(17.4)(103)
u = tan - 1
dA - dC 0.0110344 - 0.0055172 = tan - 1 3(12) 3(12) = 0.00878°
Ans.
128
6 ft
A
1 ft F
6 kip 2 1 ft AEF ⫽ 2 in
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4–11. The load is supported by the four 304 stainless steel wires that are connected to the rigid members AB and DC. Determine the vertical displacement of the 500-lb load if the members were originally horizontal when the load was applied. Each wire has a cross-sectional area of 0.025 in2.
E
F
G
3 ft 5 ft
H
D
C 1 ft
2 ft
1.8 ft I
Internal Forces in the wires: A
FBD (b) FBC(4) - 500(3) = 0
+ c ©Fy = 0;
FAH + 375.0 - 500 = 0
FAH = 125.0 lb
a + ©MD = 0;
FCF(3) - 125.0(1) = 0
FCF = 41.67 lb
+ c ©Fy = 0;
FDE + 41.67 - 125.0 = 0
FBC = 375.0 lb
FBD (a)
FDE = 83.33 lb
Displacement: dD =
83.33(3)(12) FDELDE = 0.0042857 in. = ADEE 0.025(28.0)(106)
dC =
41.67(3)(12) FCFLCF = 0.0021429 in. = ACFE 0.025(28.0)(106) œ dH = 0.0014286 in.
dH = 0.0014286 + 0.0021429 = 0.0035714 in. dA>H =
125.0(1.8)(12) FAHLAH = 0.0038571 in. = AAHE 0.025(28.0)(106)
dA = dH + dA>H = 0.0035714 + 0.0038571 = 0.0074286 in. dB =
375.0(5)(12) FBGLBG = 0.0321428 in. = ABGE 0.025(28.0)(106)
dlœ 0.0247143 = ; 3 4
1 ft 500 lb
a + ©MA = 0;
œ dH 0.0021429 = ; 2 3
B 3 ft
dlœ = 0.0185357 in.
dl = 0.0074286 + 0.0185357 = 0.0260 in.
Ans.
129
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*4–12. The load is supported by the four 304 stainless steel wires that are connected to the rigid members AB and DC. Determine the angle of tilt of each member after the 500-lb load is applied. The members were originally horizontal, and each wire has a cross-sectional area of 0.025 in2.
E
F
G
3 ft 5 ft
H
D
C 1 ft
2 ft
1.8 ft I
Internal Forces in the wires:
A
FBD (b) FBG(4) - 500(3) = 0
+ c ©Fy = 0;
FAH + 375.0 - 500 = 0
FAH = 125.0 lb
a + ©MD = 0;
FCF(3) - 125.0(1) = 0
FCF = 41.67 lb
+ c ©Fy = 0;
FDE + 41.67 - 125.0 = 0
FBG = 375.0 lb
FBD (a)
FDE = 83.33 lb
Displacement: dD =
83.33(3)(12) FDELDE = 0.0042857 in. = ADEE 0.025(28.0)(106)
dC =
41.67(3)(12) FCFLCF = 0.0021429 in. = ACFE 0.025(28.0)(106)
œ dH 0.0021429 = ; 2 3
œ dH = 0.0014286 in.
œ + dC = 0.0014286 + 0.0021429 = 0.0035714 in. dH = dH
tan a =
0.0021429 ; 36
dA>H =
125.0(1.8)(12) FAHLAH = 0.0038571 in. = AAHE 0.025(28.0)(106)
a = 0.00341°
Ans.
dA = dH + dA>H = 0.0035714 + 0.0038571 = 0.0074286 in. 375.0(5)(12) FBGLBG = 0.0321428 in. = ABGE 0.025(28.0)(106)
tan b =
1 ft 500 lb
a + ©MA = 0;
dB =
B 3 ft
0.0247143 ; 48
b = 0.0295°
Ans.
130
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•4–13. The bar has a length L and cross-sectional area A. Determine its elongation due to the force P and its own weight.The material has a specific weight g (weight>volume) and a modulus of elasticity E.
d =
=
L P(x) dx 1 = (gAx + P) dx AE L0 L A(x) E
L
gAL2 gL2 1 PL a + PLb = + AE 2 2E AE
Ans.
P
4–14. The post is made of Douglas fir and has a diameter of 60 mm. If it is subjected to the load of 20 kN and the soil provides a frictional resistance that is uniformly distributed along its sides of w = 4 kN>m, determine the force F at its bottom needed for equilibrium.Also, what is the displacement of the top of the post A with respect to its bottom B? Neglect the weight of the post.
20 kN A y w
2m
Equation of Equilibrium: For entire post [FBD (a)] + c ©Fy = 0;
F + 8.00 - 20 = 0
B
F = 12.0 kN
Ans.
Internal Force: FBD (b) + c ©Fy = 0;
-F(y) + 4y - 20 = 0 F(y) = {4y - 20} kN
Displacement: L
dA>B =
2m F(y)dy 1 = (4y - 20)dy AE L0 L0 A(y)E
=
1 A 2y2 - 20y B AE
= -
冷
2m 0
32.0 kN # m AE 32.0(103)
= -
p 2 4 (0.06 )
13.1 (109)
= - 0.8639 A 10 - 3 B m Ans.
= - 0.864 mm Negative sign indicates that end A moves toward end B.
131
F
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4–15. The post is made of Douglas fir and has a diameter of 60 mm. If it is subjected to the load of 20 kN and the soil provides a frictional resistance that is distributed along its length and varies linearly from w = 0 at y = 0 to w = 3 kN>m at y = 2 m, determine the force F at its bottom needed for equilibrium. Also, what is the displacement of the top of the post A with respect to its bottom B? Neglect the weight of the post.
20 kN A y w
2m
B F
Equation of Equilibrium: For entire post [FBD (a)] + c ©Fy = 0;
F + 3.00 - 20 = 0
F = 17.0 kN
Ans.
Internal Force: FBD (b) + c ©Fy = 0;
-F(y) +
1 3y a b y - 20 = 0 2 2
3 F(y) = e y2 - 20 f kN 4 Displacement: L
dA>B =
2m F(y) dy 1 3 = a y2 - 20b dy AE L0 4 L0 A(y)E
=
2m y3 1 a - 20y b 2 AE 4 0
= -
38.0 kN # m AE 38.0(103)
= -
p 2 4 (0.06 )
13.1 (109)
= -1.026 A 10 - 3 B m Ans.
= -1.03 mm Negative sign indicates that end A moves toward end B.
132
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*4–16. The linkage is made of two pin-connected A-36 steel members, each having a cross-sectional area of 1.5 in2. If a vertical force of P = 50 kip is applied to point A, determine its vertical displacement at A.
P A
2 ft B
C 1.5 ft
Analysing the equilibrium of Joint A by referring to its FBD, Fig. a, + ©F = 0 ; : x
+ c ©Fy = 0
The
3 3 FAC a b - FAB a b = 0 5 5 4 -2Fa b - 50 = 0 5
initial
FAC = FAB = F
F = -31.25 kip
length
of members AB and AC is 12 in b = 30 in. The axial deformation of members L = 21.52 + 22 = (2.50 ft)a 1 ft AB and AC is
d =
(-31.25)(30) FL = = -0.02155 in. AE (1.5) C 29.0(103) D
The negative sign indicates that end A moves toward B and C. From the geometry 1.5 shown in Fig. b, u = tan - 1 a b = 36.87°. Thus, 2
A dA B g =
d 0.02155 = = 0.0269 in. T cos u cos 36.87°
Ans.
133
1.5 ft
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•4–17.
The linkage is made of two pin-connected A-36 steel members, each having a cross-sectional area of 1.5 in2. Determine the magnitude of the force P needed to displace point A 0.025 in. downward.
P A
2 ft
Analysing the equilibrium of joint A by referring to its FBD, Fig. a + ©F = 0; : x
3 3 FAC a b - FAB a b = 0 5 5
+ c ©Fy = 0;
4 -2Fa b - P = 0 5
The
initial
B
1.5 ft
F = -0.625 P
of members AB and AC are 12 in b = 30 in. The axial deformation of members L = 21.5 + 2 = (2.50 ft)a 1 ft AB and AC is 2
length
2
d =
C
FAC = FAB = F
-0.625P(30) FL = = -0.4310(10 - 3) P AE (1.5) C 29.0(103) D
The negative sign indicates that end A moves toward B and C. From the geometry 1.5 shown in Fig. b, we obtain u = tan - 1 a b = 36.87°. Thus 2 (dA)g =
d cos u
0.025 =
0.4310(10 - 3) P cos 36.87°
P = 46.4 kips
Ans.
134
1.5 ft
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4–18. The assembly consists of two A-36 steel rods and a rigid bar BD. Each rod has a diameter of 0.75 in. If a force of 10 kip is applied to the bar as shown, determine the vertical displacement of the load.
C
A 3 ft 2 ft
B
Here, FEF = 10 kip. Referring to the FBD shown in Fig. a,
1.25 ft
a + ©MB = 0;
FCD (2) - 10(1.25) = 0
FCD = 6.25 kip
a + ©MD = 0;
10(0.75) - FAB(2) = 0
FAB = 3.75 kip
The cross-sectional area of the rods is A = A and C are fixed,
3.75 (2)(12) FAB LAB = = 0.007025 in. T A Est 0.140625p C 29.0(103) D
dD =
6.25(3)(12) FCD LCD = = 0.01756 in T A Est 0.140625p C 29.0(103) D
From the geometry shown in Fig. b dE = 0.007025 +
1.25 (0.01756 - 0.00725) = 0.01361 in. T 2
Here, dF>E =
10 (1) (12) FEF LEF = = 0.009366 in T A Est 0.140625p C 29.0(103) D
Thus,
A + T B dF = dE + dF>E = 0.01361 + 0.009366 = 0.0230 in T
Ans.
135
D
0.75 ft F
p (0.752) = 0.140625p in2. Since points 4
dB =
E
10 kip
1 ft
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4–19. The assembly consists of two A-36 steel rods and a rigid bar BD. Each rod has a diameter of 0.75 in. If a force of 10 kip is applied to the bar, determine the angle of tilt of the bar.
C
A
Here, FEF = 10 kip. Referring to the FBD shown in Fig. a, 3 ft
a + ©MB = 0;
FCD (2) - 10(1.25) = 0
FCD = 6.25 kip
a + ©MD = 0;
10(0.75) - FAB (2) = 0
FAB = 3.75 kip
2 ft
B
p The cross-sectional area of the rods is A = (0.752) = 0.140625p in2. Since points 4 A and C are fixed then,
dB =
3.75 (2)(12) FAB LAB = = 0.007025 in T A Est 0.140625p C 29.0(103) D
dD =
6.25 (3)(12) FCD LCD = = 0.01756 in T A Est 0.140625p C 29.0(103) D
0.01756 - 0.007025 = 0.439(10 - 3) rad 2(12)
Ans.
136
D
0.75 ft F
10 kip
From the geometry shown in Fig. b, u =
1.25 ft
E
1 ft
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*4–20. The rigid bar is supported by the pin-connected rod CB that has a cross-sectional area of 500 mm2 and is made of A-36 steel. Determine the vertical displacement of the bar at B when the load is applied.
C
3m
45 kN/m
Force In The Rod. Referring to the FBD of member AB, Fig. a a + ©MA = 0;
3 1 1 FBC a b (4) - (45)(4) c (4) d = 0 5 2 3
4m
Displacement. The initial length of rod BC is LBC = 232 + 42 = 5 m. The axial deformation of this rod is dBC =
50.0(103)(5) FBC LBC = = 2.50 (10 - 3) m ABC Est 0.5(10 - 3) C 200(109) D
3 From the geometry shown in Fig. b, u = tan - 1 a b = 36.87°. Thus, 4 (dB)g =
2.50(10 - 3) dBC = = 4.167 (10 - 3) m = 4.17 mm sin u sin 36.87°
137
B
A
FBC = 50.0 kN
Ans.
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•4–21.
A spring-supported pipe hanger consists of two springs which are originally unstretched and have a stiffness of k = 60 kN>m, three 304 stainless steel rods, AB and CD, which have a diameter of 5 mm, and EF, which has a diameter of 12 mm, and a rigid beam GH. If the pipe and the fluid it carries have a total weight of 4 kN, determine the displacement of the pipe when it is attached to the support.
F B
D
k G 0.75 m
0.75 m k H
E
A
C
Internal Force in the Rods: 0.25 m 0.25 m
FBD (a) a + ©MA = 0;
FCD (0.5) - 4(0.25) = 0 FAB + 2.00 - 4 = 0
+ c ©Fy = 0;
FCD = 2.00 kN
FAB = 2.00 kN
FBD (b) FEF - 2.00 - 2.00 = 0
+ c ©Fy = 0;
FEF = 4.00 kN
Displacement: dD = dE =
FEFLEF = AEFE
dA>B = dC>D =
4.00(103)(750) p 4
(0.012)2(193)(109)
PCDLCD = ACDE
= 0.1374 mm
2(103)(750) p 4
(0.005)2(193)(109)
= 0.3958 mm
dC = dD + dC>D = 0.1374 + 0.3958 = 0.5332 mm Displacement of the spring dsp =
Fsp k
=
2.00 = 0.0333333 m = 33.3333 mm 60
dlat = dC + dsp = 0.5332 + 33.3333 = 33.9 mm
Ans.
138
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4–22. A spring-supported pipe hanger consists of two springs, which are originally unstretched and have a stiffness of k = 60 kN>m, three 304 stainless steel rods, AB and CD, which have a diameter of 5 mm, and EF, which has a diameter of 12 mm, and a rigid beam GH. If the pipe is displaced 82 mm when it is filled with fluid, determine the weight of the fluid.
F B
D
k G 0.75 m
0.75 m k H
E
Internal Force in the Rods: A
C
FBD (a) a + ©MA = 0;
FCD(0.5) - W(0.25) = 0
FCD =
W - W = 0 2
W 2
FAB +
+ c ©Fy = 0;
FAB =
0.25 m 0.25 m
W 2
FBD (b) FEF -
+ c ©Fy = 0;
W W = 0 2 2
FEF = W
Displacement: dD = dE =
FEFLEF = AEFE
W(750) p 2 9 4 (0.012) (193)(10 )
= 34.35988(10 - 6) W dA>B = dC>D =
FCDLCD = ACDE
W 2
(750)
p 2 9 4 (0.005) (193)(10 )
= 98.95644(10 - 6) W dC = dD + dC>D = 34.35988(10 - 6) W + 98.95644(10 - 6) W = 0.133316(10 - 3) W Displacement of the spring dsp =
W 2
Fsp k
=
60(103)
(1000) = 0.008333 W
dlat = dC + dsp 82 = 0.133316(10 - 3) W + 0.008333W W = 9685 N = 9.69 kN
Ans.
139
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4–23. The rod has a slight taper and length L. It is suspended from the ceiling and supports a load P at its end. Show that the displacement of its end due to this load is d = PL>1pEr2r12. Neglect the weight of the material. The modulus of elasticity is E.
r(x) = r1 +
A(x) =
r2
r1L + (r2 - r1)x r2 - r1 x = L L
L
p (r1L + (r2 - r1)x)2 L2 r1
L
PL2 Pdx dx d = = 2 A(x)E pE [r L + (r L0 L 1 2 - r1)x] = -
L 1 PL2 c dƒ p E (r2 - r2)(r1L + (r2 - r1)x) 0
= -
=
= -
P
1 1 PL2 c d p E(r2 - r1) r1L + (r2 - r1)L r1L
r1 - r2 PL2 1 1 PL2 c d = c d p E(r2 - r1) r2L r1L p E(r2 - r1) r2r1L
r2 - r1 PL2 PL c d = p E(r2 - r1) r2r1L p E r2r1
QED
*4–24. Determine the relative displacement of one end of the tapered plate with respect to the other end when it is subjected to an axial load P.
P d2 t
w = d1 +
d1 h + (d2 - d1)x d2 - d1 x = h h
h P(x) dx P = d = E L0 [d1h L A(x)E
h
dx + ( d 2 - d1 )x ] t h
h
=
Ph dx E t L0 d1 h + (d2 - d1)x
d1 P
h
dx Ph = E t d1 h L0 1 + d2 -
h d1 h d2 - d1 Ph d1 = a b cln a1 + xb d ƒ E t d1 h d2 - d1 d1 h 0 d1 h x
=
d2 - d1 d1 + d2 - d1 Ph Ph cln a1 + bd = cln a bd E t(d2 - d1) d1 E t(d2 - d1) d1
=
d2 Ph cln d E t(d2 - d1) d1
Ans.
140
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4–25. Determine the elongation of the A-36 steel member when it is subjected to an axial force of 30 kN. The member is 10 mm thick. Use the result of Prob. 4–24.
20 mm 30 kN
30 kN 75 mm 0.5 m
Using the result of prob. 4-24 by substituting d1 = 0.02 m, d2 = 0.075 m t = 0.01 m and L = 0.5 m. d = 2c = 2c
d2 PL ln d Est t(d2 - d1) d1 30(103) (0.5) 9
200(10 )(0.01)(0.075 - 0.02)
ln a
0.075 bd 0.02
= 0.360(10 - 3) m = 0.360 mm
Ans.
4–26. The casting is made of a material that has a specific weight g and modulus of elasticity E. If it is formed into a pyramid having the dimensions shown, determine how far its end is displaced due to gravity when it is suspended in the vertical position.
b0
b0
L
Internal Forces: + c ©Fz = 0;
P(z) -
1 gAz = 0 3
P(z) =
1 gAz 3
Displacement: L
d =
P(z) dz L0 A(z) E L1 3
=
gAz
L0 A E
dz
=
L g z dz 3E L0
=
gL2 6E
Ans.
141
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4–27. The circular bar has a variable radius of r = r0eax and is made of a material having a modulus of elasticity of E. Determine the displacement of end A when it is subjected to the axial force P.
L
Displacements: The cross-sectional area of the bar as a function of x is A(x) = pr2 = pr0 2e2ax. We have
x
B
L
d =
L P(x)dx P dx = 2 A(x)E pr0 E L0 e2ax L0
r0 r ⫽ r0 eax
L 1 P 2 c d = pr0 2E 2ae2ax 0
= -
A
P a1 - e - 2aL b 2apr0 2E
P
Ans.
*4–28. The pedestal is made in a shape that has a radius defined by the function r = 2>12 + y1>22 ft, where y is in feet. If the modulus of elasticity for the material is E = 1411032 psi, determine the displacement of its top when it supports the 500-lb load.
y
500 lb 0.5 ft
r⫽
2 (2 ⫹ y 1/ 2)
4 ft
d =
=
P(y) dy L A(y) E y
4 dy 500 3 14(10 )(144) L0 p(2 + y2-1
2
)
2
1 ft
4 -3
= 0.01974(10 )
L0
r
1 2
(4 + 4y + y) dy
4 2 3 1 = 0.01974(10 - 3)c4y + 4 a y2 b + y2 d 3 2 0
= 0.01974(10 - 3)(45.33) = 0.8947(10 - 3) ft = 0.0107 in.
Ans.
142
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•4–29.
The support is made by cutting off the two opposite sides of a sphere that has a radius r0 . If the original height of the support is r0>2, determine how far it shortens when it supports a load P. The modulus of elasticity is E.
P
r0
Geometry: A = p r2 = p(r0 cos u)2 = p r20 cos2 u y = r0 sin u;
dy = r0 cos u du
Displacement: L
P(y) dy L0 A(y) E
d =
= 2B
When y =
u u r0 cos u du P P du = 2 R B R E L0 p r20 cos2 u p r0 E L0 cos u
=
u 2P [ln (sec u + tan u)] 2 p r0 E 0
=
2P [ln (sec u + tan u)] p r0 E
r0 ; 4
u = 14.48°
d =
=
2P [ln (sec 14.48° + tan 14.48°)] p r0 E 0.511P p r0 E
Ans.
Also, Geometry: A (y) = p x2 = p (r20 - y2) Displacement: L
d =
P(y) dy L0 A(y) E 0
0
r0 + y p 2P 1 2P p dy = ln = B R 2 2 2 pE L0 r0 - y p E 2r0 r0 - y 0 =
P [ln 1.667 - ln 1] p r0 E
=
0.511 P p r0 E
Ans.
143
r0 2
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4–30. The weight of the kentledge exerts an axial force of P ⫽ 1500 kN on the 300-mm diameter high strength concrete bore pile. If the distribution of the resisting skin friction developed from the interaction between the soil and the surface of the pile is approximated as shown, and the resisting bearing force F is required to be zero, determine the maximum intensity p0 kN>m for equilibrium. Also, find the corresponding elastic shortening of the pile. Neglect the weight of the pile.
P p0
12 m
Internal Loading: By considering the equilibrium of the pile with reference to its entire free-body diagram shown in Fig. a. We have 1 p (12) - 1500 = 0 2 0
+ c ©Fy = 0;
p0 = 250 kN>m
Ans.
Thus, p(y) =
250 y = 20.83y kN>m 12
The normal force developed in the pile as a function of y can be determined by considering the equilibrium of a section of the pile shown in Fig. b. 1 (20.83y)y - P(y) = 0 2
+ c ©Fy = 0;
P(y) = 10.42y2 kN
Displacement: The cross-sectional area of the pile is A =
p (0.32) = 0.0225p m2. 4
We have L
d =
12 m P(y)dy 10.42(103)y2dy = 0.0225p(29.0)(109) L0 L0 A(y)E 12 m
=
L0
5.0816(10 - 6)y2dy
= 1.6939(10 - 6)y3 冷 0
12 m
= 2.9270(10 - 3)m = 2.93 mm
Ans.
144
F
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4–31. The column is constructed from high-strength concrete and six A-36 steel reinforcing rods. If it is subjected to an axial force of 30 kip, determine the average normal stress in the concrete and in each rod. Each rod has a diameter of 0.75 in.
4 in. 30 kip
Equations of Equilibrium: 6Pst + Pcon - 30 = 0
+ c ©Fy = 0;
3 ft
[1]
Compatibility: dst = dcon Pcon(3)(12)
Pst(3)(12) p 4
(0.752)(29.0)(103)
=
[p4 (82) - 6(p4 )(0.75)2](4.20)(103)
Pst = 0.064065 Pcon
[2]
Solving Eqs. [1] and [2] yields: Pst = 1.388 kip
Pcon = 21.670 kip
Average Normal Stress: sst =
scon =
Pst = Ast
Pcon = Acon
1.388 p 2 4 (0.75 )
= 3.14 ksi
21.670
p 2 4 (8 )
- 6 A p4 B (0.752)
Ans.
Ans.
= 0.455 ksi
*4–32. The column is constructed from high-strength concrete and six A-36 steel reinforcing rods. If it is subjected to an axial force of 30 kip, determine the required diameter of each rod so that one-fourth of the load is carried by the concrete and three-fourths by the steel.
4 in. 30 kip
Equilibrium: The force of 30 kip is required to distribute in such a manner that 3/4 of the force is carried by steel and 1/4 of the force is carried by concrete. Hence Pst =
3 (30) = 22.5 kip 4
Pcon =
1 (30) = 7.50 kip 4 3 ft
Compatibility: dst = dcon PstL Pcon L = AstEst Acon Econ Ast =
22.5Acon Econ 7.50 Est
3 C p4 (82) - 6 A p4 B d2 D (4.20)(103) p 6 a bd2 = 4 29.0(103) d = 1.80 in.
Ans. 145
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•4–33.
The steel pipe is filled with concrete and subjected to a compressive force of 80 kN. Determine the average normal stress in the concrete and the steel due to this loading. The pipe has an outer diameter of 80 mm and an inner diameter of 70 mm. Est = 200 GPa, Ec = 24 GPa.
80 kN
Pst + Pcon - 80 = 0
+ c ©Fy = 0;
(1)
500 mm
dst = dcon Pcon L p 2 (0.07 ) (24) 4
Pst L p 2 4 (0.08
- 0.072) (200) (109)
=
(109)
Pst = 2.5510 Pcon
(2)
Solving Eqs. (1) and (2) yields Pst = 57.47 kN sst =
Pst = Ast
scon =
Pcon = 22.53 kN 57.47 (103)
p 4
(0.082 - 0.072)
Ans.
= 48.8 MPa
22.53 (103) Pcon = 5.85 MPa = p 2 Acon 4 (0.07 )
Ans.
4–34. The 304 stainless steel post A has a diameter of d = 2 in. and is surrounded by a red brass C83400 tube B. Both rest on the rigid surface. If a force of 5 kip is applied to the rigid cap, determine the average normal stress developed in the post and the tube.
5 kip
B B A 8 in.
Equations of Equilibrium: + c ©Fy = 0;
3 in.
Pst + Pbr - 5 = 0[1]
Compatibility:
d
dst = dbr Pst(8) p 2 3 4 (2 )(28.0)(10 )
Pbr(8) =
p 2 4 (6
- 52)(14.6)(103)
Pst = 0.69738 Pbr
[2]
Solving Eqs. [1] and [2] yields: Pbr = 2.9457 kip
Pst = 2.0543 kip
Average Normal Stress: sbr =
sst =
Pbr = Abr
2.9457 = 0.341 ksi - 52)
Ans.
p 2 4 (6
Pst 2.0543 = p 2 = 0.654 ksi Ast 4 (2 )
Ans.
146
0.5 in.
A
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4–35. The 304 stainless steel post A is surrounded by a red brass C83400 tube B. Both rest on the rigid surface. If a force of 5 kip is applied to the rigid cap, determine the required diameter d of the steel post so that the load is shared equally between the post and tube.
5 kip
B B
A
A 8 in.
Equilibrium: The force of 5 kip is shared equally by the brass and steel. Hence
3 in.
Pst = Pbr = P = 2.50 kip Compatibility:
d
0.5 in.
dst = dbr PL PL = AstEst AbrEbr Ast = p a b d2 = 4
AbrEbr Est p 4
(62 - 52)(14.6)(103) 28.0(103)
d = 2.39 in.
Ans.
*4–36. The composite bar consists of a 20-mm-diameter A-36 steel segment AB and 50-mm-diameter red brass C83400 end segments DA and CB. Determine the average normal stress in each segment due to the applied load. + ©F = 0; ; x
250 mm
D
FC - FD + 75 + 75 - 100 - 100 = 0 FC - FD - 50 = 0
+ ;
(1)
0 = ¢ D - dD 0 =
50(0.25)
150(0.5) p 2 9 4 (0.02) (200)(10 )
-
-
FD(0.5) p 2 9 4 (0.05 )(101)(10 )
p 2 9 4 (0.05 )(101)(10 )
-
500 mm
50 mm
FD(0.5) p 2 9 4 (0.02 )(200)(10 )
FD = 107.89 kN From Eq. (1), FC = 157.89 kN sAD =
107.89(103) PAD = 55.0 MPa = p 2 AAD 4 (0.05 )
Ans.
sAB =
42.11(103) PAB = 134 MPa = p 2 AAB 4 (0.02 )
Ans.
sBC =
157.89(103) PBC = 80.4 MPa = p 2 ABC 4 (0.05 )
Ans.
147
250 mm
20 mm 75 kN 100 kN A
75 kN
100 kN B
C
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•4–37.
The composite bar consists of a 20-mm-diameter A-36 steel segment AB and 50-mm-diameter red brass C83400 end segments DA and CB. Determine the displacement of A with respect to B due to the applied load.
250 mm
D
+ ; 0 =
-
500 mm
50 mm
250 mm
20 mm 75 kN 100 kN A
75 kN
100 kN B
C
0 = ¢ D - dD 150(103)(500)
50(103)(250)
p 2 9 4 (0.02 )(200)(10 )
FD(500) p 2 9 4 (0.05 )(101)(10 )
-
-
p 2 9 4 (0.05 )(101)(10 )
FD(500) p 2 9 4 (0.02) (200)(10 )
FD = 107.89 kN Displacement: dA>B =
42.11(103)(500) PABLAB = p 2 9 AABEst 4 (0.02 )200(10 )
= 0.335 mm
Ans.
4–38. The A-36 steel column, having a cross-sectional area of 18 in2, is encased in high-strength concrete as shown. If an axial force of 60 kip is applied to the column, determine the average compressive stress in the concrete and in the steel. How far does the column shorten? It has an original length of 8 ft.
16 in.
Pst + Pcon - 60 = 0
+ c ©Fy = 0; dst = dcon ;
60 kip
Pst(8)(12) 18(29)(103)
(1)
Pcon(8)(12) =
[(9)(16) - 18](4.20)(103)
Pst = 0.98639 Pcon
(2)
Solving Eqs. (1) and (2) yields Pst = 29.795 kip; sst =
Pcon = 30.205 kip
Pst 29.795 = = 1.66 ksi Ast 18
scon =
Ans.
Pcon 30.205 = = 0.240 ksi Acon 9(16) - 18
Ans.
Either the concrete or steel can be used for the deflection calculation. d =
29.795(8)(12) PstL = 0.0055 in. = AstE 18(29)(103)
Ans.
148
9 in.
8 ft
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4–39. The A-36 steel column is encased in high-strength concrete as shown. If an axial force of 60 kip is applied to the column, determine the required area of the steel so that the force is shared equally between the steel and concrete. How far does the column shorten? It has an original length of 8 ft.
60 kip 16 in.
9 in.
8 ft
The force of 60 kip is shared equally by the concrete and steel. Hence Pst = Pcon = P = 30 kip dcon = dst;
Ast =
PL PL = Acon Econ Ast Est
[9(16) - Ast] 4.20(103) AconEcon = Est 29(103) = 18.2 in2
d =
Ans.
30(8)(12) PstL = 0.00545 in. = AstEst 18.2(29)(103)
Ans.
*4–40. The rigid member is held in the position shown by three A-36 steel tie rods. Each rod has an unstretched length of 0.75 m and a cross-sectional area of 125 mm2. Determine the forces in the rods if a turnbuckle on rod EF undergoes one full turn. The lead of the screw is 1.5 mm. Neglect the size of the turnbuckle and assume that it is rigid. Note: The lead would cause the rod, when unloaded, to shorten 1.5 mm when the turnbuckle is rotated one revolution.
B
D
0.75 m E A
0.5 m
0.5 m
C 0.75 m
a + ©ME = 0;
-TAB(0.5) + TCD(0.5) = 0
F
TAB = TCD = T + T ©Fy = 0;
(1)
TEF - 2T = 0 TEF = 2T
(2)
Rod EF shortens 1.5mm causing AB (and DC) to elongate. Thus: 0.0015 = dA>B + dE>F 0.0015 =
T(0.75) -6
2T(0.75) 9
(125)(10 )(200)(10 )
+
(125)(10 - 6)(200)(109)
2.25T = 37500 T = 16666.67 N TAB = TCD = 16.7 kN
Ans.
TEF = 33.3 kN
Ans.
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•4–41.
The concrete post is reinforced using six steel reinforcing rods, each having a diameter of 20 mm. Determine the stress in the concrete and the steel if the post is subjected to an axial load of 900 kN. Est = 200 GPa, Ec = 25 GPa.
900 kN 250 mm
375 mm
Referring to the FBD of the upper portion of the cut concrete post shown in Fig. a Pcon + 6Pst - 900 = 0
+ c ©Fy = 0;
(1)
Since the steel rods and the concrete are firmly bonded, their deformation must be the same. Thus 0 con = dst Pcon L Pst L = Acon Econ Ast Est
C 0.25(0.375) -
Pcon L 6(p4 )(0.022)
D C 25(10 ) D
Pst L =
9
(p4 )(0.022)
C 200(109) D
Pcon = 36.552 Pst
(2)
Solving Eqs (1) and (2) yields Pst = 21.15 kN
Pcon = 773.10 kN
Thus, scon =
sst =
773.10(103) Pcon = 8.42 MPa = Acon 0.15(0.375) - 6(p4 )(0.022) 21.15(103) Pst = 67.3 MPa = p 2 Ast 4 (0.02 )
Ans.
Ans.
150
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4–42. The post is constructed from concrete and six A-36 steel reinforcing rods. If it is subjected to an axial force of 900 kN, determine the required diameter of each rod so that one-fifth of the load is carried by the steel and four-fifths by the concrete. Est = 200 GPa, Ec = 25 GPa.
900 kN 250 mm
375 mm
The normal force in each steel rod is Pst =
1 5
(900) 6
= 30 kN
The normal force in concrete is Pcon =
4 (900) = 720 kN 5
Since the steel rods and the concrete are firmly bonded, their deformation must be the same. Thus dcon = dst Pcon L Pst L = Acon Econ Ast Est 720(103) L
30(103)L
C 0.25(0.375) - 6(p4 d2) D C 25(109) D
=
49.5p d2 = 0.09375
p 4
d2 C 200(109) D
d = 0.02455 m = 24.6 mm
Ans.
4–43. The assembly consists of two red brass C83400 copper alloy rods AB and CD of diameter 30 mm, a stainless 304 steel alloy rod EF of diameter 40 mm, and a rigid cap G. If the supports at A, C and F are rigid, determine the average normal stress developed in rods AB, CD and EF.
300 mm
450 mm 40 kN
A
B
E
30 mm
F 40 mm
C
Equation of Equilibrium: Due to symmetry, FAB = FCD = F. Referring to the freebody diagram of the assembly shown in Fig. a, + ©F = 0; : x
2F + FEF - 2 C 40(103) D = 0
(1)
Compatibility Equation: Using the method of superposition, Fig. b, + B 0 = -d + d A: P EF 0 = -
40(103)(300) p 2 9 4 (0.03 )(101)(10 )
+ cp
FEF (450) 2
9
4 (0.04 )(193)(10 )
+
A
B
FEF>2 (300) d p 2 9 4 (0.03 )(101)(10 )
FEF = 42 483.23 N Substituting this result into Eq. (1), F = 18 758.38 N
151
30 mm
40 kN
D G
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4–43.
Continued
Normal Stress: We have, sAB = sCD =
sEF =
F 18 758.38 = 26.5 MPa = p 2 ACD 4 (0.03 )
Ans.
FEF 42 483.23 = 33.8 MPa = p 2 AEF 4 (0.04 )
Ans.
152
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*4–44. The two pipes are made of the same material and are connected as shown. If the cross-sectional area of BC is A and that of CD is 2A, determine the reactions at B and D when a force P is applied at the junction C.
B L – 2
Equations of Equilibrium: + ©F = 0; ; x
FB + FD - P = 0
[1]
Compatibility: + B A:
0 = dP - dB 0 =
0 =
P A L2 B
2AE
- C
FB
A L2 B
AE
FB +
A L2 B
2AE
S
3FBL PL 4AE 4AE
FB =
P 3
Ans.
From Eq. [1] FD =
C
2 P 3
Ans.
153
D P L – 2
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•4–45.
The bolt has a diameter of 20 mm and passes through a tube that has an inner diameter of 50 mm and an outer diameter of 60 mm. If the bolt and tube are made of A-36 steel, determine the normal stress in the tube and bolt when a force of 40 kN is applied to the bolt. Assume the end caps are rigid.
160 mm
40 kN
Referring to the FBD of left portion of the cut assembly, Fig. a + ©F = 0; : x
40(103) - Fb - Ft = 0
(1)
Here, it is required that the bolt and the tube have the same deformation. Thus dt = db Ft(150)
p 2 4 (0.06
- 0.05 ) C 200(10 ) D 2
Fb(160) =
9
p 2 4 (0.02 )
C 200(109) D
Ft = 2.9333 Fb
(2)
Solving Eqs (1) and (2) yields Fb = 10.17 (103) N
Ft = 29.83 (103) N
Thus, sb =
10.17(103) Fb = 32.4 MPa = p 2 Ab 4 (0.02 )
st =
Ft = At
29.83 (103) p 2 4 (0.06
- 0.052)
40 kN 150 mm
Ans.
= 34.5 MPa
Ans.
154
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4–46. If the gap between C and the rigid wall at D is initially 0.15 mm, determine the support reactions at A and D when the force P = 200 kN is applied. The assembly is made of A36 steel.
Equation of Equilibrium: Referring to the free-body diagram of the assembly shown in Fig. a, 200(103) - FD - FA = 0
(1)
Compatibility Equation: Using the method of superposition, Fig. b, + B A:
d = dP - dFD 0.15 =
200(103)(600) p 2 9 4 (0.05 )(200)(10 )
- Cp
FD (600)
2 9 4 (0.05 )(200)(10 )
0.15 mm
P A
+ ©F = 0; : x
600 mm
600 mm
+
FD (600) S p 2 (0.025 )(200)(109) 4
FD = 20 365.05 N = 20.4 kN
Ans.
Substituting this result into Eq. (1), FA = 179 634.95 N = 180 kN
Ans.
155
50 mm
D B
25 mm
C
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4–47. Two A-36 steel wires are used to support the 650-lb engine. Originally, AB is 32 in. long and A¿B¿ is 32.008 in. long. Determine the force supported by each wire when the engine is suspended from them. Each wire has a crosssectional area of 0.01 in2.
B¿ B
A¿ A
TA¿B¿ + TAB - 650 = 0
+ c ©Fy = 0;
(1)
dAB = dA¿B¿ + 0.008 TA¿B¿ (32.008)
TAB (32) (0.01)(29)(106)
=
(0.01)(29)(106)
+ 0.008
32TAB - 32.008TA¿B¿ = 2320 TAB = 361 lb
Ans.
TA¿B¿ = 289 lb
Ans.
*4–48. Rod AB has a diameter d and fits snugly between the rigid supports at A and B when it is unloaded. The modulus of elasticity is E. Determine the support reactions at A and B if the rod is subjected to the linearly distributed axial load.
p⫽
A
1 p L - FA - FB = 0 2 0
+ ©F = 0; : x
(1)
Compatibility Equation: Using the method of superposition, Fig. b, + B A:
0 = dP - dFA L
0 =
L0
FA (L) P(x)dx AE AE
L
0 =
L0
B x
Equation of Equilibrium: Referring to the free-body diagram of rod AB shown in Fig. a,
P(x)dx - FAL
156
p0
p0 x L
L
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4–48.
Continued
Here, P(x) =
p0 2 1 p0 a xb x = x . Thus, 2 L 2L 0 =
L p0 x2 dx - FAL 2L L0
0 =
p0 x3 L ¢ ≤ ` - FAL 2L 3 0
FA =
p0L 6
Ans.
Substituting this result into Eq. (1), FB =
p0L 3
Ans.
157
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•4–49.
The tapered member is fixed connected at its ends A and B and is subjected to a load P = 7 kip at x = 30 in. Determine the reactions at the supports. The material is 2 in. thick and is made from 2014-T6 aluminum.
A
B 3 in.
P
6 in.
x 60 in.
y 1.5 = 120 - x 60 y = 3 - 0.025 x + ©F = 0; : x
FA + FB - 7 = 0
(1)
dA>B = 0 30
-
L0
60 FA dx FBdx + = 0 2(3 - 0.025 x)(2)(E) L30 2(3 - 0.025 x)(2)(E) 30
-FA
L0
60
dx dx + FB = 0 (3 - 0.025 x) L30 (3 - 0.025x)
60 40 FA ln(3 - 0.025 x)|30 0 - 40 FB ln(3 - 0.025x)|30 = 0
-FA(0.2876) + 0.40547 FB = 0 FA = 1.40942 FB Thus, from Eq. (1). FA = 4.09 kip
Ans.
FB = 2.91 kip
Ans.
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4–50. The tapered member is fixed connected at its ends A and B and is subjected to a load P. Determine the location x of the load and its greatest magnitude so that the average normal stress in the bar does not exceed sallow = 4 ksi. The member is 2 in. thick.
A
B 3 in.
P
6 in.
x 60 in.
y 1.5 = 120 - x 60 y = 3 - 0.025 x + ©F = 0; : x
FA + FB - P = 0
dA>B = 0 x
-
60 FA dx FBdx + = 0 Lx 2(3 - 0.025 x)(2)(E) L0 2(3 - 0.025 x)(2)(E) x
-FA
60
dx dx + FB = 0 L0 (3 - 0.025 x) Lx (3 - 0.025 x)
FA(40) ln (3 - 0.025 x)|x0 - FB(40) ln (3 - 0.025x)|60 x = 0 FA ln (1 -
0.025 x 0.025x ) = -FB ln (2 ) 3 1.5
For greatest magnitude of P require, 4 =
FA ; 2(3 - 0.025 x)(2)
4 =
FB ; 2(3)
FA = 48 - 0.4 x
FB = 24 kip
Thus, (48 - 0.4 x) ln a 1 -
0.025 x 0.025 x b = -24 ln a2 b 3 1.5
Solving by trial and error, x = 28.9 in.
Ans.
Therefore, FA = 36.4 kip P = 60.4 kip
Ans.
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4–51. The rigid bar supports the uniform distributed load of 6 kip>ft. Determine the force in each cable if each cable has a cross-sectional area of 0.05 in2, and E = 3111032 ksi.
C
6 ft
6 kip/ft A D
B 3 ft
a + ©MA = 0; u = tan - 1
TCB a
2 25
b(3) - 54(4.5) + TCD a
2 25
b9 = 0
(1)
6 = 45° 6
L2B¿C¿ = (3)2 + (8.4853)2 - 2(3)(8.4853) cos u¿ Also, L2D¿C¿ = (9)2 + (8.4853)2 - 2(9)(8.4853) cos u¿
(2)
Thus, eliminating cos u¿ . -L2B¿C¿(0.019642) + 1.5910 = -L2D¿C¿(0.0065473) + 1.001735 L2B¿C¿(0.019642) = 0.0065473 L2D¿C¿ + 0.589256 L2B¿C¿ = 0.333 L2D¿C¿ + 30 But, LB¿C = 245 + dBC¿ ,
LD¿C = 245 + dDC¿
Neglect squares or d¿ B since small strain occurs. L2D¿C = (245 + dBC)2 = 45 + 2 245 dBC L2D¿C = (245 + dDC)2 = 45 + 2 245 dDC 45 + 2245 dBC = 0.333(45 + 2245 dDC) + 30 2 245 dBC = 0.333(2245 dDC) dDC = 3dBC Thus, TCD 245 TCB 245 = 3 AE AE TCD = 3 TCB From Eq. (1). TCD = 27.1682 kip = 27.2 kip
Ans.
TCB = 9.06 kip
Ans.
160
3 ft
3 ft
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*4–52. The rigid bar is originally horizontal and is supported by two cables each having a cross-sectional area of 0.05 in2, and E = 3111032 ksi. Determine the slight rotation of the bar when the uniform load is applied.
C
See solution of Prob. 4-51.
6 ft
TCD = 27.1682 kip dDC =
TCD 245 0.05(31)(103)
27.1682245 = 0.1175806 ft 0.05(31)(103)
=
6 kip/ft A D
B
Using Eq. (2) of Prob. 4-51, 3 ft
3 ft
3 ft
(245 + 0.1175806)2 = (9)2 + (8.4852)2 - 2(9)(8.4852) cos u¿ u¿ = 45.838° Thus, ¢u = 45.838° - 45° = 0.838°
Ans.
•4–53.
The press consists of two rigid heads that are held together by the two A-36 steel 12-in.-diameter rods. A 6061T6-solid-aluminum cylinder is placed in the press and the screw is adjusted so that it just presses up against the cylinder. If it is then tightened one-half turn, determine the average normal stress in the rods and in the cylinder. The single-threaded screw on the bolt has a lead of 0.01 in. Note: The lead represents the distance the screw advances along its axis for one complete turn of the screw.
12 in.
2 in.
10 in.
+ ©F = 0; : x
2Fst - Fal = 0 dst = 0.005 - dal
Fst(12) p ( 4 )(0.5)2(29)(103)
= 0.005 -
Fal(10) p(1)2(10)(103)
Solving, Fst = 1.822 kip Fal = 3.644 kip srod =
Fst 1.822 = p = 9.28 ksi Ast ( 4 )(0.5)2
Ans.
scyl =
Fal 3.644 = = 1.16 ksi Aal p(1)2
Ans.
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4–54. The press consists of two rigid heads that are held together by the two A-36 steel 12-in.-diameter rods. A 6061T6-solid-aluminum cylinder is placed in the press and the screw is adjusted so that it just presses up against the cylinder. Determine the angle through which the screw can be turned before the rods or the specimen begin to yield. The single-threaded screw on the bolt has a lead of 0.01 in. Note: The lead represents the distance the screw advances along its axis for one complete turn of the screw.
12 in.
2 in.
10 in.
+ ©F = 0; : x
2Fst - Fal = 0 dst = d - dal Fst(12)
(p4 )(0.5)2(29)(103)
= d-
Fal(10)
(1)
p(1)2(10)(103)
Assume steel yields first, sY = 36 =
Fst (p4 )(0.5)2
;
Fst = 7.068 kip
Then Fal = 14.137 kip; sal =
14.137 = 4.50 ksi p(1)2
4.50 ksi 6 37 ksi steel yields first as assumed. From Eq. (1), d = 0.01940 in. Thus, 0.01940 u = 360° 0.01 u = 698°
Ans.
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4–55. The three suspender bars are made of A-36 steel and have equal cross-sectional areas of 450 mm2. Determine the average normal stress in each bar if the rigid beam is subjected to the loading shown.
A 2m
1m
+ c ©Fy = 0; a + ©MD = 0;
FAD + FBE + FCF - 50(103) - 80(103) = 0 FBE(2) + FCF(4) - 50(103)(1) - 80(103)(3) = 0
(1) (2)
Referring to the geometry shown in Fig. b, dBE = dAD + a dBE =
dCF - dAD b(2) 4
1 A d + dCF B 2 AD
FBE L FCF L 1 FADL = a + b AE 2 AE AE FAD + FCF = 2 FBE
(3)
Solving Eqs. (1), (2) and (3) yields FBE = 43.33(103) N
FAD = 35.83(103) N
FCF = 50.83(103) N
Thus, sBE =
43.33(103) FBE = 96.3 MPa = A 0.45(10 - 3)
Ans.
sAD =
35.83(103) FAD = 79.6 MPa = A 0.45(10 - 3)
Ans.
sCF = 113 MPa
Ans.
163
80 kN
50 kN E
D
Referring to the FBD of the rigid beam, Fig. a,
C
B
1m
1m
F 1m
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*4–56. The rigid bar supports the 800-lb load. Determine the normal stress in each A-36 steel cable if each cable has a cross-sectional area of 0.04 in2.
C
12 ft
800 lb B A
5 ft
Referring to the FBD of the rigid bar, Fig. a, FBC a
a + ©MA = 0;
12 3 b(5) + FCD a b (16) - 800(10) = 0 13 5
(1)
The unstretched length of wires BC and CD are LBC = 2122 + 52 = 13 ft and LCD = 2122 + 162 = 20 ft. The stretches of wires BC and CD are dBC =
FBC (13) FBC LBC = AE AE
dCD =
FCD(20) FCD LCD = AE AE
Referring to the geometry shown in Fig. b, the vertical displacement of the points on d 12 3 the rigid bar is dg = . For points B and D, cos uB = and cos uD = . Thus, cos u 13 5 the vertical displacement of points B and D are
A dB B g =
FBC (13)>AE dBC 169 FBC = = cos uB 12>13 12AE
A dD B g =
FCD (20)>AE dCD 100 FCD = = cos uD 3>5 3 AE
The similar triangles shown in Fig. c give
A dB B g 5
=
A dD B g 16
1 169 FBC 1 100 FCD b = b a a 5 12 AE 16 3AE FBC =
125 F 169 CD
(2)
Solving Eqs. (1) and (2), yields FCD = 614.73 lb
FBC = 454.69 lb
Thus, sCD =
FCD 614.73 = 15.37(103) psi = 15.4 ksi = ACD 0.04
Ans.
sBC =
FBC 454.69 = = 11.37(103) psi = 11.4 ksi ABC 0.04
Ans.
164
D 5 ft
6 ft
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4–56.
Continued
165
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•4–57.
The rigid bar is originally horizontal and is supported by two A-36 steel cables each having a crosssectional area of 0.04 in2. Determine the rotation of the bar when the 800-lb load is applied.
C
12 ft
800 lb B A
Referring to the FBD of the rigid bar Fig. a, a + ©MA = 0;
FBC a
12 3 b(5) + FCD a b (16) - 800(10) = 0 13 5
5 ft
(1)
The unstretched length of wires BC and CD are LBC = 2122 + 52 = 13 ft and LCD = 2122 + 162 = 20 ft. The stretch of wires BC and CD are dBC =
FBC (13) FBC LBC = AE AE
dCD =
FCD(20) FCD LCD = AE AE
Referring to the geometry shown in Fig. b, the vertical displacement of the points on d 12 3 the rigid bar is dg = . For points B and D, cos uB = and cos uD = . Thus, cos u 13 5 the vertical displacement of points B and D are
A dB B g =
FBC (13)>AE dBC 169 FBC = = cos uB 12>13 12AE
A dD B g =
FCD (20)>AE dCD 100 FCD = = cos uD 3>5 3 AE
The similar triangles shown in Fig. c gives
A dB B g 5
=
A dD B g 16
1 169 FBC 1 100 FCD a b = a b 5 12 AE 16 3 AE FBC =
125 F 169 CD
(2)
Solving Eqs (1) and (2), yields FCD = 614.73 lb
FBC = 454.69 lb
Thus,
A dD B g =
100(614.73)
3(0.04) C 29.0 (106) D
= 0.01766 ft
Then u = a
0.01766 ft 180° ba b = 0.0633° p 16 ft
Ans.
166
D 5 ft
6 ft
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4–58. The horizontal beam is assumed to be rigid and supports the distributed load shown. Determine the vertical reactions at the supports. Each support consists of a wooden post having a diameter of 120 mm and an unloaded (original) length of 1.40 m. Take Ew = 12 GPa.
18 kN/m
A
B
C 1.40 m
2m
a + ©MB = 0; + c ©Fy = 0;
FC(1) - FA(2) = 0
(1)
FA + FB + FC - 27 = 0
dB - dA dC - dA = ; 2 3
1m
(2)
3dB - dA = 2dC
3FBL FAL 2FCL = ; AE AE AE
3FB - FA = 2FC
(3)
Solving Eqs. (1)–(3) yields : FA = 5.79 kN
Ans.
FB = 9.64 kN
Ans.
FC = 11.6 kN
Ans.
4–59. The horizontal beam is assumed to be rigid and supports the distributed load shown. Determine the angle of tilt of the beam after the load is applied. Each support consists of a wooden post having a diameter of 120 mm and an unloaded (original) length of 1.40 m. Take Ew = 12 GPa. a + ©MB = 0; c + ©Fy = 0;
18 kN/m
A
FC(1) - FA(2) = 0
2m
3FB - FA = 2FC
(3)
Solving Eqs. (1)–(3) yields : FA = 5.7857 kN;
FB = 9.6428 kN;
FC = 11.5714 kN
3
dA =
5.7857(10 )(1.40) FAL = 0.0597(10 - 3) m = p 2 9 AE (0.12 )12(10 ) 4
dC =
11.5714(103)(1.40) FCL = 0.1194(10 - 3) m = p 2 9 AE (0.12 )12(10 ) 4
tan u =
1.40 m
(2)
3dB - dA = 2dC
3FBL FAL 2FCL = ; AE AE AE
C
(1)
FA + FB + FC - 27 = 0
dB - dA dC - dA = ; 2 3
B
0.1194 - 0.0597 (10 - 3) 3
u = 0.0199(10 - 3) rad = 1.14(10 - 3)°
Ans.
167
1m
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*4–60. The assembly consists of two posts AD and CF made of A-36 steel and having a cross-sectional area of 1000 mm2, and a 2014-T6 aluminum post BE having a crosssectional area of 1500 mm2. If a central load of 400 kN is applied to the rigid cap, determine the normal stress in each post. There is a small gap of 0.1 mm between the post BE and the rigid member ABC.
400 kN 0.5 m
A
Equation of Equilibrium. Due to symmetry, FAD = FCF = F. Referring to the FBD of the rigid cap, Fig. a, FBE + 2F - 400(103) = 0
(1)
Compatibility Equation. Referring to the initial and final position of rods AD (CF) and BE, Fig. b, d = 0.1 + dBE F(400)
1(10 ) C 200(10 ) D -3
9
= 0.1 +
FBE (399.9)
1.5(10 - 3) C 73.1(109) D
F = 1.8235 FBE + 50(103)
(2)
Solving Eqs (1) and (2) yield FBE = 64.56(103) N
F = 167.72(103) N
Normal Stress. sAD = sCF =
sBE =
B
C 0.4 m
D
+ c ©Fy = 0;
0.5 m
167.72(103) F = 168 MPa = Ast 1(10 - 3)
Ans.
64.56(103) FBE = 43.0 MPa = Aal 1.5(10 - 3)
Ans.
168
E
F
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•4–61.
The distributed loading is supported by the three suspender bars. AB and EF are made of aluminum and CD is made of steel. If each bar has a cross-sectional area of 450 mm2, determine the maximum intensity w of the distributed loading so that an allowable stress of 1sallow2st = 180 MPa in the steel and 1sallow2al = 94 MPa in the aluminum is not exceeded. Est = 200 GPa, Eal = 70 GPa. Assume ACE is rigid.
1.5 m
1.5 m
B al
D st
A
F al
C
2m
E
w
a + ©MC = 0;
FEF(1.5) - FAB(1.5) = 0 FEF = FAB = F
+ c ©Fy = 0;
2F + FCD - 3w = 0
(1)
Compatibility condition : dA = dC FCDL FL = ; 9 A(70)(10 ) A(200)(109)
F = 0.35 FCD
(2)
Assume failure of AB and EF: F = (sallow)al A = 94(106)(450)(10 - 6) = 42300 N From Eq. (2) FCD = 120857.14 N From Eq. (1) w = 68.5 kN>m Assume failure of CD: FCD = (sallow)st A = 180(106)(450)(10 - 6) = 81000 N From Eq. (2) F = 28350 N From Eq. (1) w = 45.9 kN>m
(controls)
Ans.
169
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4–62. The rigid link is supported by a pin at A, a steel wire BC having an unstretched length of 200 mm and crosssectional area of 22.5 mm2, and a short aluminum block having an unloaded length of 50 mm and cross-sectional area of 40 mm2. If the link is subjected to the vertical load shown, determine the average normal stress in the wire and the block. Est = 200 GPa, Eal = 70 GPa.
C 200 mm
B 100 mm
D
450(250) - FBC(150) - FD(150) = 0
50 mm
750 - FBC - FD = 0
[1]
Compatibility: dBC = dD FD(50)
FBC(200) 22.5(10 - 6)200(109)
=
40(10 - 6)70(109)
FBC = 0.40179 FD
[2]
Solving Eqs. [1] and [2] yields: FD = 535.03 N
FBC = 214.97 N
Average Normal Stress: sD =
sBC =
150 mm
450 N
Equations of Equilibrium: a + ©MA = 0;
A 150 mm
FD 535.03 = 13.4 MPa = AD 40(10 - 6)
Ans.
FBC 214.97 = 9.55 MPa = ABC 22.5(10 - 6)
Ans.
170
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4–63. The rigid link is supported by a pin at A, a steel wire BC having an unstretched length of 200 mm and crosssectional area of 22.5 mm2, and a short aluminum block having an unloaded length of 50 mm and cross-sectional area of 40 mm2. If the link is subjected to the vertical load shown, determine the rotation of the link about the pin A. Report the answer in radians. Est = 200 GPa, Eal = 70 GPa.
C 200 mm
B 100 mm
D
450(250) - FBC(150) - FD(150) = 0
50 mm
750 - FBC - FD = 0
[1]
Compatibility: dBC = dD FD(50)
FBC(200) -6
9
22.5(10 )200(10 )
=
40(10 - 6)70(109)
FBC = 0.40179 FD
[2]
Solving Eqs. [1] and [2] yields : FD = 535.03 N
FBC = 214.97 N
Displacement: dD =
535.03(50) FDLD = 0.009554 mm = ADEal 40(10 - 6)(70)(109)
tan u =
150 mm
450 N
Equations of Equilibrium: a + ©MA = 0;
A 150 mm
dD 0.009554 = 150 150
u = 63.7(10 - 6) rad = 0.00365°
Ans.
171
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*4–64. The center post B of the assembly has an original length of 124.7 mm, whereas posts A and C have a length of 125 mm. If the caps on the top and bottom can be considered rigid, determine the average normal stress in each post. The posts are made of aluminum and have a cross-sectional area of 400 mm2. Eal = 70 GPa.
800 kN/m
A
a + ©MB = 0;
100 mm
B
C
-FA(100) + FC(100) = 0 (1) 2F + FB - 160 = 0
(2)
dA = dB + 0.0003 F (0.125)
FB (0.1247)
400 (10 - 6)(70)(106)
=
400 (10 - 6)(70)(106)
+ 0.0003
0.125 F - 0.1247FB = 8.4
(3)
Solving Eqs. (2) and (3) F = 75.762 kN FB = 8.547 kN sA = sC =
sB =
75.726 (103) 400(10 - 6)
8.547 (103) 400 (10 - 6)
= 189 MPa
Ans.
= 21.4 MPa
Ans.
•4–65.
The assembly consists of an A-36 steel bolt and a C83400 red brass tube. If the nut is drawn up snug against the tube so that L = 75 mm, then turned an additional amount so that it advances 0.02 mm on the bolt, determine the force in the bolt and the tube. The bolt has a diameter of 7 mm and the tube has a cross-sectional area of 100 mm2.
L
Equilibrium: Since no external load is applied, the force acting on the tube and the bolt is the same. Compatibility: 0.02 = dt + db 0.02 =
P(75)
P(75) -6
9
100(10 )(101)(10 )
+
125 mm
800 kN/m
FA = FC = F + c ©Fy = 0;
100 mm
p 2 9 4 (0.007 )(200)(10 )
P = 1164.83 N = 1.16 kN
Ans.
172
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4–66. The assembly consists of an A-36 steel bolt and a C83400 red brass tube. The nut is drawn up snug against the tube so that L = 75 mm. Determine the maximum additional amount of advance of the nut on the bolt so that none of the material will yield. The bolt has a diameter of 7 mm and the tube has a cross-sectional area of 100 mm2.
L
Allowable Normal Stress: (sg)st = 250 A 106 B =
Pst p 2 4 (0.007)
Pst = 9.621 kN (sg)br = 70.0 A 106 B =
Pbr 100(10 - 6)
Pbr = 7.00 kN Since Pst 7 Pbr, by comparison he brass will yield first. Compatibility: a = dt + db 7.00(103)(75) =
100(10 - 6)(101)(109)
7.00(103)(75) +
p 2 9 4 (0.007) (200)(10 )
= 0.120 mm
Ans.
173
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4–67. The three suspender bars are made of the same material and have equal cross-sectional areas A. Determine the average normal stress in each bar if the rigid beam ACE is subjected to the force P. a + ©MA = 0;
D
d FCD(d) + FEF(2d) - Pa b = 0 2 FCD + 2FEF =
+ c ©Fy = 0;
B
F
L
P
P 2
(1)
FAB + FCD + FEF - P = 0
A
C d 2
(2)
d 2
E d
dC - dE dA - dE = d 2d 2dC = dA + dE 2FCDL FABL FEFL = + AE AE AE 2FCD - FAB - FEF = 0
(3)
Solving Eqs. (1), (2) and (3) yields P 3
P 12
FAB =
7P 12
sAB =
7P 12A
Ans.
sCD =
P 3A
Ans.
sEF =
P 12A
Ans.
FCD =
FEF =
*4–68. A steel surveyor’s tape is to be used to measure the length of a line. The tape has a rectangular cross section of 0.05 in. by 0.2 in. and a length of 100 ft when T1 = 60°F and the tension or pull on the tape is 20 lb. Determine the true length of the line if the tape shows the reading to be 463.25 ft when used with a pull of 35 lb at T2 = 90°F. The ground on which it is placed is flat. ast = 9.60110-62>°F, Est = 2911032 ksi.
P
P 0.2 in. 0.05 in.
dT = a¢TL = 9.6(10 - 6)(90 - 60)(463.25) = 0.133416 ft d =
(35 - 20)(463.25) PL = 0.023961 ft = AE (0.2)(0.05)(29)(106)
L = 463.25 + 0.133416 + 0.023961 = 463.41 ft
Ans.
174
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•4–69.
Three bars each made of different materials are connected together and placed between two walls when the temperature is T1 = 12°C. Determine the force exerted on the (rigid) supports when the temperature becomes T2 = 18°C. The material properties and cross-sectional area of each bar are given in the figure.
Copper Steel Brass Est ⫽ 200 GPa Ebr ⫽ 100 GPa Ecu ⫽ 120 GPa ast ⫽ 12(10⫺6)/⬚C abr ⫽ 21(10⫺6)/°C acu ⫽ 17(10⫺6)/⬚C Ast ⫽ 200 mm2
300 mm
+ ) (;
Acu ⫽ 515 mm2
Abr ⫽ 450 mm2
200 mm
100 mm
0 = ¢T - d
0 = 12(10 - 6)(6)(0.3) + 21 (10 - 6)(6)(0.2) + 17 (10 - 6)(6)(0.1) F(0.3) -
-6
F(0.2) 9
200(10 )(200)(10 )
-
-6
F(0.1) 9
-
450(10 )(100)(10 )
515(10 - 6)(120)(109)
F = 4203 N = 4.20 kN
Ans.
k ⫽ 1000 lb/in.
4–70. The rod is made of A-36 steel and has a diameter of 0.25 in. If the rod is 4 ft long when the springs are compressed 0.5 in. and the temperature of the rod is T = 40°F, determine the force in the rod when its temperature is T = 160°F.
k ⫽ 1000 lb/ in.
4 ft
Compatibility: + B A:
x = dT - dF x = 6.60(10 - 6)(160 - 40)(2)(12) -
1.00(0.5)(2)(12) p 2 3 4 (0.25 )(29.0)(10 )
x = 0.01869 in. F = 1.00(0.01869 + 0.5) = 0.519 kip
Ans.
4–71. A 6-ft-long steam pipe is made of A-36 steel with sY = 40 ksi. It is connected directly to two turbines A and B as shown. The pipe has an outer diameter of 4 in. and a wall thickness of 0.25 in. The connection was made at T1 = 70°F. If the turbines’ points of attachment are assumed rigid, determine the force the pipe exerts on the turbines when the steam and thus the pipe reach a temperature of T2 = 275°F.
6 ft A
Compatibility: + B A:
0 = dT - dF 0 = 6.60(10 - 6)(275 - 70)(6)(12) -
F(6)(12) p 2 4 (4
- 3.52)(29.0)(103)
F = 116 kip
Ans. 175
B
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*4–72. A 6-ft-long steam pipe is made of A-36 steel with sY = 40 ksi. It is connected directly to two turbines A and B as shown. The pipe has an outer diameter of 4 in. and a wall thickness of 0.25 in. The connection was made at T1 = 70°F. If the turbines’ points of attachment are assumed to have a stiffness of k = 8011032 kip>in., determine the force the pipe exerts on the turbines when the steam and thus the pipe reach a temperature of T2 = 275°F.
6 ft A
B
Compatibility: x = dT - dF x = 6.60(10 - 6)(275 - 70)(3)(12) -
80(103)(x)(3)(12) p 2 4 (4
- 3.52)(29.0)(103)
x = 0.001403 in. F = k x = 80(103)(0.001403) = 112 kip
Ans.
•4–73.
The pipe is made of A-36 steel and is connected to the collars at A and B. When the temperature is 60° F, there is no axial load in the pipe. If hot gas traveling through the pipe causes its temperature to rise by ¢T = 140 + 15x2°F, where x is in feet, determine the average normal stress in the pipe. The inner diameter is 2 in., the wall thickness is 0.15 in.
A
Compatibility: L
0 = dT - dF 0 = 6.60 A 10 - 6 B
Where
dT =
L0
8ft
L0
(40 + 15 x) dx -
0 = 6.60 A 10 - 6 B B 40(8) +
a ¢T dx F(8) A(29.0)(103)
15(8)2 F(8) R 2 A(29.0)(103)
F = 19.14 A Average Normal Stress: s =
B 8 ft
19.14 A = 19.1 ksi A
Ans.
176
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4–74. The bronze C86100 pipe has an inner radius of 0.5 in. and a wall thickness of 0.2 in. If the gas flowing through it changes the temperature of the pipe uniformly from TA = 200°F at A to TB = 60°F at B, determine the axial force it exerts on the walls. The pipe was fitted between the walls when T = 60°F.
A
B 8 ft
Temperature Gradient: T(x) = 60 + a
8 - x b140 = 200 - 17.5x 8
Compatibility: 0 = dT - dF 0 = 9.60 A 10 - 6 B
Where
dT = 1 a¢Tdx
2ft
0 = 9.60 A 10 - 6 B
L0
[(200 - 17.5x) - 60] dx 2ft
L0
(140 - 17.5x) dx -
F(8) p 2 4 (1.4
- 12)15.0(103)
F(8) p 2 4 (1.4
- 12) 15.0(103)
F = 7.60 kip
Ans.
4–75. The 40-ft-long A-36 steel rails on a train track are laid with a small gap between them to allow for thermal expansion. Determine the required gap d so that the rails just touch one another when the temperature is increased from T1 = -20°F to T2 = 90°F. Using this gap, what would be the axial force in the rails if the temperature were to rise to T3 = 110°F? The cross-sectional area of each rail is 5.10 in2.
d
40 ft
Thermal Expansion: Note that since adjacent rails expand, each rail will be d required to expand on each end, or d for the entine rail. 2 d = a¢TL = 6.60(10 - 6)[90 - (-20)](40)(12) Ans.
= 0.34848 in. = 0.348 in. Compatibility: + B A:
0.34848 = dT - dF 0.34848 = 6.60(10 - 6)[110 - (-20)](40)(12) -
d
F(40)(12) 5.10(29.0)(103)
F = 19.5 kip
Ans.
177
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*4–76. The device is used to measure a change in temperature. Bars AB and CD are made of A-36 steel and 2014-T6 aluminum alloy respectively. When the temperature is at 75°F, ACE is in the horizontal position. Determine the vertical displacement of the pointer at E when the temperature rises to 150°F.
0.25 in.
A
3 in.
C
E
1.5 in.
Thermal Expansion:
A dT B CD = aal ¢TLCD = 12.8(10 - 6)(150 - 75)(1.5) = 1.44(10 - 3) in.
B
D
A dT B AB = ast ¢TLAB = 6.60(10 - 6)(150 - 75)(1.5) = 0.7425(10 - 3) in. From the geometry of the deflected bar AE shown Fig. b, dE = A dT B AB + C = 0.7425(10 - 3) + B
A dT B CD - A dT B AB 0.25
S(3.25)
1.44(10 - 3) - 0.7425(10 - 3) R (3.25) 0.25
= 0.00981 in.
Ans.
•4–77. The bar has a cross-sectional area A, length L, modulus of elasticity E, and coefficient of thermal expansion a. The temperature of the bar changes uniformly along its length from TA at A to TB at B so that at any point x along the bar T = TA + x1TB - TA2>L. Determine the force the bar exerts on the rigid walls. Initially no axial force is in the bar and the bar has a temperature of TA.
+ :
x A TA
0 = ¢ T - dF
(1)
However, d¢ T = a¢ T dx = a(TA +
TB - TA x - TA)dx L
L
¢T = a
= ac
L
TB - TA TB - TA 2 x dx = ac x d冷 L 2L L0 0 TB - TA aL Ld = (TB - TA) 2 2
From Eq.(1). 0 =
FL aL (TB - TA) 2 AE
F =
a AE (TB - TA) 2
B
Ans.
178
TB
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4–78. The A-36 steel rod has a diameter of 50 mm and is lightly attached to the rigid supports at A and B when T1 = 80°C. If the temperature becomes T2 = 20°C and an axial force of P = 200 kN is applied to its center, determine the reactions at A and B.
0.5 m
FB - FA + 200(103) = 0
(1)
When the rod is unconstrained at B, it has a free contraction of dT = ast ¢ TL = 12(10 - 6)(80 - 20)(1000) = 0.72 mm. Also, under force P and FB with unconstrained at B, the deformation of the rod are dP =
dFB =
PLAC = AE
FB LAB = AE
200(103)(500) p 2 4 (0.05 )
C 200(109) D
FB (1000) p 2 4 (0.05 )
C 200(109) D
= 0.2546 mm = 2.5465(10 - 6) FB
Using the method of super position, Fig. b, + B A:
B
P
Referring to the FBD of the rod, Fig. a + ©F = 0; : x
C
A
0 = -dT + dP + dFB 0 = -0.72 + 0.2546 + 2.5465(10 - 6) FB FB = 182.74(103) N = 183 kN
Ans.
Substitute the result of FB into Eq (1), FA = 382.74(103) N = 383 kN
Ans.
179
0.5 m
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4–79. The A-36 steel rod has a diameter of 50 mm and is lightly attached to the rigid supports at A and B when T1 = 50°C. Determine the force P that must be applied to the collar at its midpoint so that, when T2 = 30°C, the reaction at B is zero.
C
A
B
P 0.5 m
0.5 m
When the rod is unconstrained at B, it has a free contraction of dT = ast ¢TL = 12(10 - 6)(50 - 30)(1000) = 0.24 mm. Also, under force P with unconstrained at B, the deformation of the rod is dP =
PLAC = AE
P(500) p 2 4 (0.05 )
C 200(109) D
= 1.2732(10 - 6) P
Since FB is required to be zero, the method of superposition, Fig. b, gives + B A:
0 = -dT + dP 0 = -0.24 + 1.2732(10 - 6)P P = 188.50(103) N = 188 kN
Ans.
*4–80. The rigid block has a weight of 80 kip and is to be supported by posts A and B, which are made of A-36 steel, and the post C, which is made of C83400 red brass. If all the posts have the same original length before they are loaded, determine the average normal stress developed in each post when post C is heated so that its temperature is increased by 20°F. Each post has a cross-sectional area of 8 in2.
A
Equations of Equilibrium: a + ©MC = 0; + c ©Fy = 0;
FB(3) - FA(3) = 0
FA = FB = F
2F + FC - 80 = 0
[1]
Compatibility: (dC)F - (dC)T = dF
(+ T) FCL 8(14.6)(103)
- 9.80 A 10 - 5 B (20)L =
FL 8(29.0)(103)
8.5616 FC - 4.3103 F = 196
[2]
Solving Eqs. [1] and [2] yields: F = 22.81 kip
FC = 34.38 kip
average Normal Sress: sA = sB =
sC =
F 22.81 = = 2.85 ksi A 8
Ans.
FC 34.38 = = 4.30 ksi A 8
Ans.
180
C
B
3 ft
3 ft
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•4–81.
The three bars are made of A-36 steel and form a pin-connected truss. If the truss is constructed when T1 = 50°F, determine the force in each bar when T2 = 110°F. Each bar has a cross-sectional area of 2 in2.
t
5f
5f
t
A
4 ft
B
D 3 ft
(dT ¿)AB - (dF ¿)AB = (dT)AD + (dF)AD
(1)
œ cos u; However, dAB = dAB
œ dAB =
dAB 5 = dAB cos u 4
Substitute into Eq. (1) 5 5 (dT)AB - (dF)AB = (dT)AD + (dF)AD 4 4 FAB(5)(12) 5 d c6.60(10 - 6)(110° - 50°)(5)(12) 4 2(29)(103) = 6.60(10 - 6)(110° - 50°)(4)(12) +
FAD(4)(12) 2(29)(103)
620.136 = 75FAB + 48FAD + ©F = 0; : x
3 3 F - FAB = 0; 5 AC 5
+ c ©Fy = 0;
4 FAD - 2a FAB b = 0 5
(2) FAC = FAB
(3)
Solving Eqs. (2) and (3) yields : FAD = 6.54 kip
Ans.
FAC = FAB = 4.09 kip
Ans.
181
C 3 ft
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4–82. The three bars are made of A-36 steel and form a pinconnected truss. If the truss is constructed when T1 = 50°F, determine the vertical displacement of joint A when T2 = 150°F. Each bar has a cross-sectional area of 2 in2.
t
5f
5f
t
A
4 ft
(dT ¿)AB - (dF ¿)AB = (dT)AD + (dF)AD
(1)
œ However, dAB = dAB cos u;
œ dAB =
B
dAB 5 = dAB cos u 4
3 ft
Substitute into Eq. (1) 5 5 (d ) - (dT)AB = (dT)AD + (dF)AD 4 T AB 4 FAB(5)(12) 5 d c6.60(10 - 6)(150° - 50°)(5)(12) 4 2(29)(103) = 6.60(10 - 6)(150° - 50°)(4)(12) +
FAD(4)(12) 2(29)(103)
239.25 - 6.25FAB = 153.12 + 4 FAD 4 FAD + 6.25FAB = 86.13 + © F = 0; : x
3 3 F - FAB = 0; 5 AC 5
+ c © Fy = 0;
4 FAD - 2 a FAB b = 0; 5
(2) FAC = FAB
FAD = 1.6FAB
(3)
Solving Eqs. (2) and (3) yields: FAB = 6.8086 kip:
FAD = 10.8939 kip
(dA)r = (dT)AD + (dT)AD = 6.60(10 - 6)(150° - 50°)(4)(12) +
D
10.8939(4)(12) 2(29)(103)
= 0.0407 in. c
Ans.
182
C 3 ft
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4–83. The wires AB and AC are made of steel, and wire AD is made of copper. Before the 150-lb force is applied, AB and AC are each 60 in. long and AD is 40 in. long. If the temperature is increased by 80°F, determine the force in each wire needed to support the load. Take Est = 29(103) ksi, Ecu = 17(103) ksi, ast = 8(10-6)>°F, acu = 9.60(10-6)>°F. Each wire has a cross-sectional area of 0.0123 in2.
40 in. 60 in.
45⬚
45⬚
A 150 lb
Equations of Equilibrium: + ©F = 0; : x
FAC cos 45° - FAB cos 45° = 0 FAC = FAB = F 2F sin 45° + FAD - 150 = 0
+ c ©Fy = 0;
[1]
Compatibility: (dAC)T = 8.0 A 10 - 6 B (80)(60) = 0.03840 in. (dAC)Tr =
(dAC)T 0.03840 = = 0.05431 in. cos 45° cos 45°
(dAD)T = 9.60 A 10 - 6 B (80)(40) = 0.03072 in. d0 = (dAC)Tr - (dAD)T = 0.05431 - 0.03072 = 0.02359 in. (dAD)F = (dAC)Fr + d0 F(60)
FAD (40) 6
0.0123(17.0)(10 )
=
0.0123(29.0)(106) cos 45°
C
D
B
+ 0.02359
0.1913FAD - 0.2379F = 23.5858
[2]
Solving Eq. [1] and [2] yields: FAC = FAB = F = 10.0 lb
Ans.
FAD = 136 lb
Ans.
183
60 in.
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*4–84. The AM1004-T61 magnesium alloy tube AB is capped with a rigid plate E. The gap between E and end C of the 6061-T6 aluminum alloy solid circular rod CD is 0.2 mm when the temperature is at 30° C. Determine the normal stress developed in the tube and the rod if the temperature rises to 80° C. Neglect the thickness of the rigid cap.
25 mm a
Section a-a
E
B
A
20 mm
C
25 mm
a 0.2 mm 300 mm
Compatibility Equation: If tube AB and rod CD are unconstrained, they will have a free expansion of A dT B AB = amg ¢TLAB = 26(10 - 6)(80 - 30)(300) = 0.39 mm and
A dT)CD = aal ¢TLCD = 24(10 - 6)(80 - 30)(450) = 0.54 mm. Referring deformation diagram of the tube and the rod shown in Fig. a, d =
to
the
C A dT B AB - A dF B AB D + C A dT B CD - A dF B CD D
0.2 = C 0.39 -
F(300)
p A 0.025 - 0.02 B (44.7)(10 ) 2
2
9
S + C 0.54 -
F(450)
p 4
A 0.0252 B (68.9)(109)
S
F = 32 017.60 N Normal Stress: sAB =
F 32 017.60 = = 45.3 MPa AAB p A 0.0252 - 0.022 B
sCD =
F 32 017.60 = = 65.2 MPa p 2 ACD 4 A 0.025 B
Ans.
Ans.
F = 107 442.47 N
184
450 mm
D
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•4–85. The AM1004-T61 magnesium alloy tube AB is capped with a rigid plate. The gap between E and end C of the 6061-T6 aluminum alloy solid circular rod CD is 0.2 mm when the temperature is at 30° C. Determine the highest temperature to which it can be raised without causing yielding either in the tube or the rod. Neglect the thickness of the rigid cap.
25 mm a
Section a-a
E
B
A
20 mm
C
25 mm
a 0.2 mm 300 mm
Then sCD =
F 107 442.47 = = 218.88MPa 6 (sY)al p 2 ACD 4 A 0.025 B
(O.K.!)
Compatibility Equation: If tube AB and rod CD are unconstrained, they will have a free expansion of A dT B AB = amg ¢TLAB = 26(10 - 6)(T - 30)(300) = 7.8(10 - 6) (T - 30) and
A dT B CD = aal ¢TLCD = 24(10 - 6)(T - 30)(450) = 0.0108(T - 30).
Referring to the deformation diagram of the tube and the rod shown in Fig. a, d =
C A dT B AB - A dF B AB D + C A dT B CD - A dF B CD D
0.2 = C 7.8(10 - 3)(T - 30) -
+ C 0.0108(T - 30) -
107 442.47(300)
p A 0.0252 - 0.022 B (44.7)(109)
107 442.47(450)
p 4
A 0.0252 B (68.9)(109)
S
S
T = 172° C
Ans.
185
450 mm
D
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4–86. The steel bolt has a diameter of 7 mm and fits through an aluminum sleeve as shown. The sleeve has an inner diameter of 8 mm and an outer diameter of 10 mm. The nut at A is adjusted so that it just presses up against the sleeve. If the assembly is originally at a temperature of T1 = 20°C and then is heated to a temperature of T2 = 100°C, determine the average normal stress in the bolt and the sleeve. Est = 200 GPa, Eal = 70 GPa, ast = 14(10-6)>°C, aal = 23(10-6)>°C.
A
Compatibility: (ds)T - (db)T = (ds)F + (db)F 23(10 - 6)(100 - 20)L - 14(10 - 6)(100 - 20)L =
p 2 4 (0.01
FL + - 0.0082)70(109)
FL p 2 9 4 (0.007 )200(10 )
F = 1133.54 N Average Normal Stress: ss =
F = As
sb =
F 1133.54 = 29.5 MPa = p 2 Ab 4 (0.007 )
1133.54 = 40.1 MPa - 0.0082)
Ans.
p 2 4 (0.01
Ans.
4–87. Determine the maximum normal stress developed in the bar when it is subjected to a tension of P = 8 kN.
5 mm 40 mm
20 mm P
P
For the fillet:
r ⫽ 10 mm 20 mm
r 10 = = 0.5 h 20
w 40 = = 2 h 20 From Fig. 10-24. K = 1.4 smax = Ksavg = 1.4 a
8 (103) b 0.02 (0.005)
= 112 MPa For the hole: r 10 = = 0.25 w 40 From Fig. 4-25. K = 2.375 smax = Ksavg = 2.375 a
8 (103) b (0.04 - 0.02)(0.005)
= 190 MPa
Ans.
186
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*4–88. If the allowable normal stress for the bar is sallow = 120 MPa, determine the maximum axial force P that can be applied to the bar.
5 mm 40 mm
20 mm P
P
Assume failure of the fillet.
r ⫽ 10 mm 20 mm
r 10 = = 0.5 h 20
w 40 = = 2; h 20 From Fig. 4-24. K = 1.4
sallow = smax = Ksavg 120 (106) = 1.4 a
P b 0.02 (0.005)
P = 8.57 kN Assume failure of the hole. r 10 = = 0.25 w 20 From Fig. 4-25. K = 2.375 sallow = smax = Ksavg 120 (104) = 2.375 a
P b (0.04 - 0.02) (0.005)
P = 5.05 kN (controls)
Ans.
•4–89.
The member is to be made from a steel plate that is 0.25 in. thick. If a 1-in. hole is drilled through its center, determine the approximate width w of the plate so that it can support an axial force of 3350 lb. The allowable stress is sallow = 22 ksi.
0.25 in. w
3350 lb
sallow = smax = Ksavg
1 in.
3.35 d 22 = K c (w - 1)(0.25) w =
3.35K + 5.5 5.5
By trial and error, from Fig. 4-25, choose
w =
r = 0.2; w
K = 2.45
3.35(2.45) + 5.5 = 2.49 in. 5.5
Since
r 0.5 = = 0.2 w 2.49
3350 lb
Ans.
OK
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4–90. The A-36 steel plate has a thickness of 12 mm. If there are shoulder fillets at B and C, and sallow = 150 MPa, determine the maximum axial load P that it can support. Calculate its elongation, neglecting the effect of the fillets.
r = 30 mm 120 mm r = 30 mm 60 mm P A
w 120 = = 2 h 60
and
60 mm P D
B
Maximum Normal Stress at fillet: r 30 = = 0.5 h 60
C
800 mm
200 mm
200 mm
From the text, K = 1.4 smax = sallow = Ksavg 150(106) = 1.4 B
P R 0.06(0.012)
P = 77142.86 N = 77.1 kN
Ans.
Displacement: d = ©
PL AE 77142.86(800)
77142.86(400) =
9
+
(0.06)(0.012)(200)(10 )
(0.12)(0.012)(200)(109)
= 0.429 mm
Ans.
4–91. Determine the maximum axial force P that can be applied to the bar. The bar is made from steel and has an allowable stress of sallow = 21 ksi.
0.125 in. 1.25 in.
1.875 in.
P
Assume failure of the fillet. r 0.25 = = 0.2 h 1.25
P
w 1.875 = = 1.5 h 1.25 0.75 in.
From Fig. 4-24, K = 1.73 sallow = smax = Ksavg 21 = 1.73 a
P b 1.25 (0.125)
P = 1.897 kip Assume failure of the hole. r 0.375 = = 0.20 w 1.875 From Fig. 4-25, K = 2.45 sallow = smax = Ksavg 21 = 2.45 a
P b (1.875 - 0.75)(0.125)
P = 1.21 kip (controls)
Ans.
188
r ⫽ 0.25 in.
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*4–92. Determine the maximum normal stress developed in the bar when it is subjected to a tension of P = 2 kip.
0.125 in. 1.25 in.
1.875 in.
At fillet: P
r 0.25 = = 0.2 h 1.25
P
w 1.875 = = 1.5 h 1.25 0.75 in.
From Fig. 4-24, K = 1.73 smax = Ka
r ⫽ 0.25 in.
P 2 d = 22.1 ksi b = 1.73 c A 1.25(0.125)
At hole: r 0.375 = = 0.20 w 1.875 From Fig. 4-25, K = 2.45 smax = 2.45 c
2 d = 34.8 ksi (1.875 - 0.75)(0.125)
(Controls)
Ans.
•4–93.
Determine the maximum normal stress developed in the bar when it is subjected to a tension of P = 8 kN.
5 mm 60 mm P
Maximum Normal Stress at fillet: r 15 = = 0.5 h 30
P ht
= 1.4 B
8(103) R = 74.7 MPa (0.03)(0.005)
Maximum Normal Stress at the hole: r 6 = = 0.1 w 60 From the text, K = 2.65 smax = K savg = K
P (w - 2r) t
= 2.65 B
8(103) R (0.06 - 0.012)(0.005)
= 88.3 MPa
P r = 15 mm 12 mm
w 60 = = 2 h 30
and
From the text, K = 1.4 smax = Ksavg = K
30 mm
(Controls)
Ans.
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4–94. The resulting stress distribution along section AB for the bar is shown. From this distribution, determine the approximate resultant axial force P applied to the bar. Also, what is the stress-concentration factor for this geometry?
0.5 in. A P
4 in. 1 in.
B 12 ksi
P =
L
3 ksi
sdA = Volume under curve
Number of squares = 10 P = 10(3)(1)(0.5) = 15 kip savg =
K =
Ans.
15 kip P = = 7.5 ksi A (4 in.)(0.5 in.)
smax 12 ksi = = 1.60 savg 7.5 ksi
Ans.
4–95. The resulting stress distribution along section AB for the bar is shown. From this distribution, determine the approximate resultant axial force P applied to the bar. Also, what is the stress-concentration factor for this geometry?
0.5 in. A 0.6 in. 0.8 in.
Number of squares = 28 P = 28(6)(0.2)(0.5) = 16.8 kip savg
P 16.8 = = = 28 ksi A 2(0.6)(0.5)
K =
smax 36 = = 1.29 savg 28
0.6 in.
Ans. B
6 ksi
36 ksi
Ans.
*4–96. The resulting stress distribution along section AB for the bar is shown. From this distribution, determine the approximate resultant axial force P applied to the bar. Also, what is the stress-concentration factor for this geometry?
10 mm A 20 mm 80 mm B 5 MPa
Number of squares = 19
30 MPa
6
P = 19(5)(10 )(0.02)(0.01) = 19 kN savg =
K =
P
0.2 in.
Ans.
19(103) P = = 23.75 MPa A 0.08(0.01)
smax 30 MPa = = 1.26 savg 23.75 MPa
Ans.
190
P
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•4–97.
The 300-kip weight is slowly set on the top of a post made of 2014-T6 aluminum with an A-36 steel core. If both materials can be considered elastic perfectly plastic, determine the stress in each material. Aluminum 1 in. 2 in. Steel
Equations of Equilibrium: + c ©Fy = 0;
Pst + Pal - 300 = 0
[1]
Elastic Analysis: Assume both materials still behave elastically under the load. dst = dal Pst L p 2 (2) (29)(103) 4
Pal L =
p 2 4 (4
- 22)(10.6)(103)
Pst = 0.9119 Pal Solving Eqs. [1] and [2] yields: Pal = 156.91 kip
Pst = 143.09 kip
Average Normal Stress: sal =
Pal = Aal
156.91 - 22)
p 2 4 (4
(OK!)
= 16.65 ksi 6 (sg)al = 60.0 ksi sst =
Pst 143.09 = p 2 Ast 4 (2 ) = 45.55 ksi 7 (sg)st = 36.0 ksi
Therefore, the steel core yields and so the elastic analysis is invalid. The stress in the steel is sst = (sy)st = 36.0 ksi
Ans.
p Pst = (sg)stAst = 36.0a b A 22 B = 113.10 kip 4 From Eq. [1] Pal = 186.90 kip sal =
Pal = Aal
186.90 = 19.83 ksi 6 (sg)al = 60.0 ksi - 22)
p 2 4 (4
Then sal = 19.8 ksi
Ans.
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4–98. The bar has a cross-sectional area of 0.5 in2 and is made of a material that has a stress–strain diagram that can be approximated by the two line segments shown. Determine the elongation of the bar due to the applied loading.
A
B 5 ft
8 kip
C 5 kip
2 ft
s(ksi) 40
20
Average Normal Stress and Strain: For segment BC sBC =
0.001
PBC 5 = = 10.0 ksi ABC 0.5
10.0 20 = ; eBC 0.001
eBC =
0.001 (10.0) = 0.00050 in.>in. 20
Average Normal Stress and Strain: For segment AB sAB =
PAB 13 = = 26.0 ksi AAB 0.5
40 - 20 26.0 - 20 = eAB - 0.001 0.021 - 0.001 eAB = 0.0070 in.>in. Elongation: dBC = eBCLBC = 0.00050(2)(12) = 0.0120 in. dAB = eAB LAB = 0.0070(5)(12) = 0.420 in. dTot = dBC + dAB = 0.0120 + 0.420 = 0.432 in.
Ans.
192
0.021
P (in./in.)
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4–99. The rigid bar is supported by a pin at A and two steel wires, each having a diameter of 4 mm. If the yield stress for the wires is sY = 530 MPa, and Est = 200 GPa, determine the intensity of the distributed load w that can be placed on the beam and will just cause wire EB to yield. What is the displacement of point G for this case? For the calculation, assume that the steel is elastic perfectly plastic.
E
D
800 mm A
B
C G w
400 mm
Equations of Equilibrium: a + ©MA = 0;
FBE(0.4) + FCD(0.65) - 0.8w (0.4) = 0 0.4 FBE + 0.65FCD = 0.32w
[1]
Plastic Analysis: Wire CD will yield first followed by wire BE. When both wires yield FBE = FCD = (sg)A p = 530 A 106 B a b A 0.0042 B = 6.660 kN 4 Substituting the results into Eq. [1] yields: w = 21.9 kN>m
Ans.
Displacement: When wire BE achieves yield stress, the corresponding yield strain is eg =
sg E
530(106) =
200(109)
= 0.002650 mm>mm
dBE = eg LBE = 0.002650(800) = 2.120 mm From the geometry dBE dG = 0.8 0.4 dG = 2dBE = 2(2.120) = 4.24 mm
Ans.
193
250 mm
150 mm
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*4–100. The rigid bar is supported by a pin at A and two steel wires, each having a diameter of 4 mm. If the yield stress for the wires is sY = 530 MPa, and Est = 200 GPa, determine (a) the intensity of the distributed load w that can be placed on the beam that will cause only one of the wires to start to yield and (b) the smallest intensity of the distributed load that will cause both wires to yield. For the calculation, assume that the steel is elastic perfectly plastic.
E
D
800 mm A
B
C G w
400 mm
Equations of Equilibrium: a + ©MA = 0;
FBE(0.4) + FCD(0.65) - 0.8w (0.4) = 0 0.4 FBE + 0.65 FCD = 0.32w
[1]
(a) By observation, wire CD will yield first. p Then FCD = sg A = 530 A 106 B a b A 0.0042 B = 6.660 kN. 4 From the geometry dCD dBE = ; 0.4 0.65
dCD = 1.625dBE FBEL FCDL = 1.625 AE AE FCD = 1.625 FBE
[2]
Using FCD = 6.660 kN and solving Eqs. [1] and [2] yields: FBE = 4.099 kN w = 18.7 kN>m
Ans.
(b) When both wires yield FBE = FCD = (sg)A p = 530 A 106 B a b A 0.0042 B = 6.660 kN 4 Substituting the results into Eq. [1] yields: w = 21.9 kN>m
Ans.
194
250 mm
150 mm
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•4–101.
The rigid lever arm is supported by two A-36 steel wires having the same diameter of 4 mm. If a force of P = 3 kN is applied to the handle, determine the force developed in both wires and their corresponding elongations. Consider A-36 steel as an elastic-perfectly plastic material.
P
450 mm 150 mm 150 mm 30⬚ A
C
300 mm B
Equation of Equilibrium. Refering to the free-body diagram of the lever shown in Fig. a, FAB (300) + FCD (150) - 3 A 103 B (450) = 0
a + ©ME = 0;
2FAB + FCD = 9 A 103 B
(1)
Elastic Analysis. Assuming that both wires AB and CD behave as linearly elastic, the compatibility equation can be written by referring to the geometry of Fig. b. dAB = a
300 bd 150 CD
dAB = 2dCD
(2)
FAB L FCD L = 2a b AE AE FAB = 2FCD
(3)
Solving Eqs. (1) and (3), FCD = 1800 N
FAB = 3600 N
Normal Stress. sCD =
FCD = ACD
sAB =
FAB = AAB
1800
p 4
A 0.0042 B
p 4
A 0.0042 B
3600
= 143.24 MPa 6 (sY)st
(O.K.)
= 286.48 MPa 7 (sY)st
(N.G.)
Since wire AB yields, the elastic analysis is not valid. The solution must be reworked using FAB = (sY)st AAB = 250 A 106 B c
p A 0.0042 B d 4 Ans.
= 3141.59 N = 3.14 kN Substituting this result into Eq. (1), FCD = 2716.81 N = 2.72 kN sCD =
Ans.
FCD 2716.81 = = 216.20 MPa 6 (sY)st p 2 ACD 4 A 0.004 B
(O.K.)
195
D
E
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4–101.
Continued
Since wire CD is linearly elastic, its elongation can be determined by dCD =
FCD LCD = ACD Est
2716.81(300)
p 4
A 0.0042 B (200) A 109 B Ans.
= 0.3243 mm = 0.324 mm From Eq. (2), dAB = 2dCD = 2(0.3243) = 0.649 mm
Ans.
196
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4–102. The rigid lever arm is supported by two A-36 steel wires having the same diameter of 4 mm. Determine the smallest force P that will cause (a) only one of the wires to yield; (b) both wires to yield. Consider A-36 steel as an elastic-perfectly plastic material.
P
450 mm 150 mm 150 mm 30⬚ A
C
300 mm B
Equation of Equilibrium. Refering to the free-body diagram of the lever arm shown in Fig. a, a + ©ME = 0;
FAB (300) + FCD (150) - P(450) = 0 2FAB + FCD = 3P
(1)
Elastic Analysis. The compatibility equation can be written by referring to the geometry of Fig. b. dAB = a
300 bd 150 CD
dAB = 2dCD FAB L FCD L = 2a b AE AE FCD =
1 F 2 AB
(2)
Assuming that wire AB is about to yield first, FAB = (sY)st AAB = 250 A 106 B c
p A 0.0042 B d = 3141.59 N 4
From Eq. (2), FCD =
1 (3141.59) = 1570.80 N 2
Substituting the result of FAB and FCD into Eq. (1), P = 2618.00 N = 2.62 kN
Ans.
Plastic Analysis. Since both wires AB and CD are required to yield, FAB = FCD = (sY)st A = 250 A 106 B c
p A 0.0042 B d = 3141.59 N 4
Substituting this result into Eq. (1),
197
D
E
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4–103. The three bars are pinned together and subjected to the load P. If each bar has a cross-sectional area A, length L, and is made from an elastic perfectly plastic material, for which the yield stress is sY, determine the largest load (ultimate load) that can be supported by the bars, i.e., the load P that causes all the bars to yield. Also, what is the horizontal displacement of point A when the load reaches its ultimate value? The modulus of elasticity is E.
B L u
C L
P
u L
D
P = 3141.59 N = 3.14 kN
A
Ans.
When all bars yield, the force in each bar is, FY = sYA + ©F = 0; : x
P - 2sYA cos u - sYA = 0
P = sYA(2 cos u + 1)
Ans.
Bar AC will yield first followed by bars AB and AD. dAB = dAD =
dA =
FY(L) sYAL sYL = = AE AE E
dAB sYL = cos u E cos u
Ans.
*4–104. The rigid beam is supported by three 25-mm diameter A-36 steel rods. If the beam supports the force of P = 230 kN, determine the force developed in each rod. Consider the steel to be an elastic perfectly-plastic material.
D
F
E
600 mm P
Equation of Equilibrium. Referring to the free-body diagram of the beam shown in Fig. a, + c ©Fy = 0;
FAD + FBE + FCF - 230 A 103 B = 0
(1)
FBE + 3FCF = 460 A 103 B
(2)
FBE(400) + FCF(1200) - 230 A 103 B (800) = 0
a + ©MA = 0;
400 mm
Elastic Analysis. Referring to the deflection diagram of the beam shown in Fig. b, the compatibility equation can be written as dBE = dAD + a dBE =
dCF - dAD b(400) 1200
2 1 d + dCF 3 AD 3
FBEL 2 FCDL 1 FCF L = a b + a b AE 3 AE 3 AE FBE =
2 1 F + FCF 3 AD 3
(3)
Solving Eqs. (1), (2), and (3) FCF = 131 428.57 N
FBE = 65 714.29 N FAD = 32 857.14 N
198
A
B
400 mm
C
400 mm
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4–104.
Continued
Normal Stress. sCF =
FCF 131428.57 = = 267.74 MPa 7 (sY)st p 2 ACF 4 A 0.025 B
(N.G.)
sBE =
FBE 65714.29 = = 133.87 MPa 6 (sY)st p 2 ABE 4 A 0.025 B
(O.K.)
sAD =
FAD 32857.14 = = 66.94 MPa 6 (sY)st p 2 AAD 4 A 0.025 B
(O.K.)
Since rod CF yields, the elastic analysis is not valid. The solution must be reworked using FCF = (sY)st ACF = 250 A 106 B c
p A 0.0252 B d = 122 718.46 N = 123 kN 4
Ans.
Substituting this result into Eq. (2), FBE = 91844.61 N = 91.8 kN
Ans.
Substituting the result for FCF and FBE into Eq. (1), FAD = 15436.93 N = 15.4 kN
Ans.
sBE =
FBE 91844.61 = = 187.10 MPa 6 (sY)st p 2 ABE 4 A 0.025 B
(O.K.)
sAD =
FAD 15436.93 = = 31.45 MPa 6 (sY)st p 2 AAD 4 A 0.025 B
(O.K.)
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•4–105.
The rigid beam is supported by three 25-mm diameter A-36 steel rods. If the force of P = 230 kN is applied on the beam and removed, determine the residual stresses in each rod. Consider the steel to be an elastic perfectly-plastic material.
D 600 mm
P A
400 mm
Equation of Equilibrium. Referring to the free-body diagram of the beam shown in Fig. a, + c ©Fy = 0;
FAD + FBE + FCF - 230 A 103 B = 0
(1)
FBE + 3FCF = 460 A 103 B
(2)
FBE(400) + FCF(1200) - 230 A 103 B (800) = 0
a + ©MA = 0;
Elastic Analysis. Referring to the deflection diagram of the beam shown in Fig. b, the compatibility equation can be written as dBE = dAD + a dBE =
dCF - dAD b(400) 1200
2 1 d + dCF 3 AD 3
(3)
FBE L 2 FCD L 1 FCF L = a b + a b AE 3 AE 3 AE FBE =
2 1 F + FCF 3 AD 3
(4)
Solving Eqs. (1), (2), and (4) FCF = 131428.57 N
FBE = 65714.29 N
FAD = 32857.14 N
Normal Stress. sCF =
FCF 131428.57 = = 267.74 MPa (T) 7 (sY)st p 2 ACF 4 A 0.025 B
(N.G.)
sBE =
FBE 65714.29 = = 133.87 MPa (T) 6 (sY)st p 2 ABE 4 A 0.025 B
(O.K.)
sAD =
FAD 32857.14 = = 66.94 MPa (T) 6 (sY)st p 2 AAD 4 A 0.025 B
(O.K.)
Since rod CF yields, the elastic analysis is not valid. The solution must be reworked using sCF = (sY)st = 250 MPa (T) FCF = sCF ACF = 250 A 106 B c
F
E
p A 0.0252 B d = 122718.46 N 4
200
B
400 mm
C
400 mm
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4–105.
Continued
Substituting this result into Eq. (2), FBE = 91844.61 N Substituting the result for FCF and FBE into Eq. (1), FAD = 15436.93N sBE =
FBE 91844.61 = = 187.10 MPa (T) 6 (sY)st p 2 ABE 4 A 0.025 B
(O.K.)
sAD =
FAD 15436.93 = = 31.45 MPa (T) 6 (sY)st p 2 AAD 4 A 0.025 B
(O.K.)
Residual Stresses. The process of removing P can be represented by applying the force P¿ , which has a magnitude equal to that of P but is opposite in sense, Fig. c. Since the process occurs in a linear manner, the corresponding normal stress must have the same magnitude but opposite sense to that obtained from the elastic analysis. Thus, œ sCF = 267.74 MPa (C)
œ sBE = 133.87 MPa (C)
œ sAD = 66.94 MPa (C)
Considering the tensile stress as positive and the compressive stress as negative, œ = 250 + (-267.74) = -17.7 MPa = 17.7 MPa (C) (sCF)r = sCF + sCF
Ans.
œ = 187.10 + (-133.87) = 53.2 MPa (T) (sBE)r = sBE + sBE
Ans.
œ (sAD)r = sAD + sAD = 31.45 + (-66.94) = -35.5 MPa = 35.5 MPa (C)
Ans.
201
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4–106. The distributed loading is applied to the rigid beam, which is supported by the three bars. Each bar has a cross-sectional area of 1.25 in2 and is made from a s(ksi) material having a stress–strain diagram that can be approximated by the two line segments shown. If a load of 60 w = 25 kip>ft is applied to the beam, determine the stress in each bar and the vertical displacement of the beam.
4 ft A
B
5 ft
36
A
0.0012
a + ©MB = 0;
0.2
FC(4) - FA(4) = 0; FA = FC = F
+ c ©Fy = 0;
2F + FB - 200 = 0
(1)
Since the loading and geometry are symmetrical, the bar will remain horizontal. Therefore, the displacement of the bars is the same and hence, the force in each bar is the same. From Eq. (1). F = FB = 66.67 kip Thus, sA = sB = sC =
66.67 = 53.33 ksi 1.25
Ans.
From the stress-strain diagram: 60 - 36 53.33 - 36 = : e - 0.0012 0.2 - 0.0012
e = 0.14477 in.>in.
d = eL = 0.14477(5)(12) = 8.69 in.
Ans.
202
4 ft
∋ (in./in.)
B
C
w
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4–107. The distributed loading is applied to the rigid beam, which is supported by the three bars. Each bar has a cross-sectional area of 0.75 in2 and is s(ksi) made from a material having a stress–strain diagram that can be approximated by the two line segments 60 shown. Determine the intensity of the distributed loading w needed to cause the beam to be displaced 36 downward 1.5 in.
0.0012
a + ©MB = 0; + c ©Fy = 0;
FC(4) - FA(4) = 0;
4 ft A
A
0.2
(1)
Since the system and the loading are symmetrical, the bar will remain horizontal. Hence the displacement of the bars is the same and the force supported by each bar is the same. From Eq. (1), FB = F = 2.6667 w
(2)
From the stress-strain diagram: e =
1.5 = 0.025 in.>in. 5 (12)
60 - 36 s - 36 = ; 0.025 - 0.0012 0.2 - 0.0012
s = 38.87 ksi
Hence F = sA = 38.87 (0.75) = 29.15 kip From Eq. (2), w = 10.9 kip>ft
Ans.
203
B
5 ft
FA = FC = F
2F + FB - 8 w = 0
4 ft
∋ (in./in.)
B
C
w
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*4–108. The rigid beam is supported by the three posts A, B, and C of equal length. Posts A and C have a diameter of 75 mm and are made of aluminum, for which Eal = 70 GPa and 1sY2al = 20 MPa. Post B has a diameter of 20 mm and is made of brass, for which Ebr = 100 GPa and 1sY2br = 590 MPa. Determine the smallest magnitude of P so that (a) only rods A and C yield and (b) all the posts yield.
P
A
B
FA = FC = Fal Fat + 2Fat - 2P = 0
+ c ©Fy = 0;
(1)
(a) Post A and C will yield, Fal = (st)alA = 20(104)(pa )(0.075)2 = 88.36 kN (Eal)r =
(sr)al 20(104) = 0.0002857 = Eal 70(104)
Compatibility condition: dbr = dal = 0.0002857(L) Fbr (L) p 2 (0.02) (100)(104) 4
= 0.0002857 L
Fbr = 8.976 kN sbr =
8.976(103) p 3 4 (0.02 )
= 28.6 MPa 6 sr
OK.
From Eq. (1), 8.976 + 2(88.36) - 2P = 0 P = 92.8 kN
Ans.
(b) All the posts yield: Fbr = (sr)brA = (590)(104)(p4 )(0.022) = 185.35 kN Fal = 88.36 kN From Eq. (1); 185.35 + 2(88.36) - 2P = 0 P = 181 kN
Ans.
204
C br
al 2m
©MB = 0;
P
2m
2m
al 2m
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•4–109.
The rigid beam is supported by the three posts A, B, and C. Posts A and C have a diameter of 60 mm and are made of aluminum, for which Eal = 70 GPa and 1sY2al = 20 MPa. Post B is made of brass, for which Ebr = 100 GPa and 1sY2br = 590 MPa. If P = 130 kN, determine the largest diameter of post B so that all the posts yield at the same time.
P
A
B
2(Fg)al + Fbr - 260 = 0
(1)
(Fal)g = (sg)al A = 20(106)(p4 )(0.06)2 = 56.55 kN From Eq. (1), 2(56.55) + Fbr - 260 = 0 Fbr = 146.9 kN (sg)br = 590(106) =
146.9(103) p 3 4 (dB)
dB = 0.01779 m = 17.8 mm
Ans.
205
C br
al 2m
+ c ©Fy = 0;
P
2m
2m
al 2m
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4–110. The wire BC has a diameter of 0.125 in. and the material has the stress–strain characteristics shown in the figure. Determine the vertical displacement of the handle at D if the pull at the grip is slowly increased and reaches a magnitude of (a) P = 450 lb, (b) P = 600 lb.
C 40 in.
A
D
B 50 in.
30 in. P
s (ksi)
Equations of Equilibrium: a + ©MA = 0;
FBC(50) - P(80) = 0
(a) From Eq. [1] when P = 450 lb,
[1] 80 70
FBC = 720 lb
Average Normal Stress and Strain: sBC =
FBC = ABC
720 p 2 4 (0.125 )
P (in./in.)
= 58.67 ksi
0.007
From the Stress–Strain diagram 58.67 70 = ; eBC 0.007
eBC = 0.005867 in.>in.
Displacement: dBC = eBCLBC = 0.005867(40) = 0.2347 in. dBC dD = ; 80 50
dD =
8 (0.2347) = 0.375 in. 5
(b) From Eq. [1] when P = 600 lb,
Ans.
FBC = 960 lb
Average Normal Stress and Strain: sBC =
FBC = ABC
960 p 2 4 (0.125)
= 78.23 ksi
From Stress–Strain diagram 78.23 - 70 80 - 70 = eBC - 0.007 0.12 - 0.007
eBC = 0.09997 in.>in.
Displacement: dBC = eBCLBC = 0.09997(40) = 3.9990 in. dD dBC = ; 80 50
dD =
8 (3.9990) = 6.40 in. 5
Ans.
206
0.12
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4–111. The bar having a diameter of 2 in. is fixed connected at its ends and supports the axial load P. If the material is elastic perfectly plastic as shown by the stress–strain diagram, determine the smallest load P needed to cause segment CB to yield. If this load is released, determine the permanent displacement of point C.
P A
C
2 ft
B
3 ft
s (ksi)
20
0.001
When P is increased, region AC will become plastic first, then CB will become plastic. Thus, FA = FB = sA = 20(p)(1)2 = 62.832 kip + ©F = 0; : x
FA + FB - P = 0
(1)
P = 2(62.832) = 125.66 kip P = 126 kip
Ans.
The deflection of point C is, dC = eL = (0.001)(3)(12) = 0.036 in. ; Consider the reverse of P on the bar. FB ¿(3) FA ¿(2) = AE AE FA ¿ = 1.5 FB ¿ So that from Eq. (1) FB ¿ = 0.4P FA ¿ = 0.6P dC ¿ =
0.4(P)(3)(12) 0.4(125.66)(3)(12) FB ¿L = 0.02880 in. : = = AE AE p(1)2(20>0.001)
¢d = 0.036 - 0.0288 = 0.00720 in. ;
Ans.
207
P (in./in.)
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*4–112. Determine the elongation of the bar in Prob. 4–111 when both the load P and the supports are removed.
P A
C
2 ft
B
3 ft
s (ksi)
20
0.001
When P is increased, region AC will become plastic first, then CB will become plastic. Thus, FA = FB = sA = 20(p)(1)2 = 62.832 kip + ©F = 0; : x
FA + FB - P = 0
(1)
P = 2(62.832) = 125.66 kip P = 126 kip
Ans.
The deflection of point C is, dC = eL = (0.001)(3)(12) = 0.036 in. ; Consider the reverse of P on the bar. FB ¿(3) FA ¿(2) = AE AE FA ¿ = 1.5 FB ¿ So that from Eq. (1) FB ¿ = 0.4P FA ¿ = 0.6P The resultant reactions are FA ¿¿ = FB ¿¿ = -62.832 + 0.6(125.66) = 62.832 - 0.4(125.66) = 12.568 kip When the supports are removed the elongation will be, d =
12.568(5)(12) PL = 0.0120 in. = AE p(1)2(20>0.001)
Ans.
208
P (in./in.)
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s
•4–113.
A material has a stress–strain diagram that can be described by the curve s = cP1>2. Determine the deflection d of the end of a rod made from this material if it has a length L, cross-sectional area A, and a specific weight g. L 1 2
A
s2 = c2 e
s = ce ;
s2(x) = c2e(x)
P
(1) d P(x) ; A
However s(x) =
e(x) =
dd dx
From Eq. (1), P2(x)
= c2
A2
P2(x) dd = dx A2c2
dd ; dx
L
d =
1 1 P2(x) dx = 2 2 (gAx)2 dx 2 2 Ac L A c L0 g2
=
d =
L
c2 L0
x2 dx =
g2 x3 L 冷 c2 3 0
g3L3
Ans.
3c2
4–114. The 2014-T6 aluminum rod has a diameter of 0.5 in. and is lightly attached to the rigid supports at A and B when T1 = 70°F. If the temperature becomes T2 = -10°F, and an axial force of P = 16 lb is applied to the rigid collar as shown, determine the reactions at A and B.
A
B P/2 P/2 5 in.
8 in.
+ 0 = ¢ - ¢ + d : B T B 0 =
0.016(5) p 2 3 4 (0.5 )(10.6)(10 )
- 12.8(10 - 6)[70° - (-10°)](13) +
FB(13) p 2 (0.5 )(10.6)(103) 4
FB = 2.1251 kip = 2.13 kip + ©F = 0; : x
Ans.
2(0.008) + 2.1251 - FA = 0 FA = 2.14 kip
Ans.
4–115. The 2014-T6 aluminum rod has a diameter of 0.5 in. and is lightly attached to the rigid supports at A and B when T1 = 70°F. Determine the force P that must be applied to the collar so that, when T = 0°F, the reaction at B is zero. + :
A P/2 5 in.
0 = ¢ B - ¢ T + dB 0 =
P(5) p 2 3 4 (0.5 )(10.6)(10 )
B P/2
- 12.8(10 - 6)[(70)(13)] + 0
P = 4.85 kip
Ans.
209
8 in.
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*4–116. The rods each have the same 25-mm diameter and 600-mm length. If they are made of A-36 steel, determine the forces developed in each rod when the temperature increases to 50° C.
C
600 mm 60⬚ B
A
60⬚
600 mm
Equation of Equilibrium: Referring to the free-body diagram of joint A shown in Fig. a, FAD sin 60° - FAC sin 60° = 0
+ c ©Fx = 0; + ©F = 0; : x
FAC = FAD = F
FAB - 2F cos 60° = 0 FAB = F
(1)
Compatibility Equation: If AB and AC are unconstrained, they will have a free expansion of A dT B AB = A dT B AC = ast ¢TL = 12(10 - 6)(50)(600) = 0.36 mm. Referring to the initial and final position of joint A, dFAB - A dT B AB = a dT ¿ b
AC
- dFAC ¿
Due to symmetry, joint A will displace horizontally, and dAC ¿ = a dT ¿ b
AC
dAC = 2dAC. Thus, cos 60°
= 2(dT)AC and dFAC ¿ = 2dFAC. Thus, this equation becomes
dFAB - A dT B AB = 2 A dT B AC - 2dAC FAB (600)
p 4
A 0.025 B (200)(10 ) 2
9
- 0.36 = 2(0.36) - 2 C
F(600)
p 4
A 0.0252 B (200)(109)
FAB + 2F = 176 714.59
S (2)
Solving Eqs. (1) and (2), FAB = FAC = FAD = 58 904.86 N = 58.9 kN
Ans.
210
D
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•4–117.
Two A-36 steel pipes, each having a crosssectional area of 0.32 in2, are screwed together using a union at B as shown. Originally the assembly is adjusted so that no load is on the pipe. If the union is then tightened so that its screw, having a lead of 0.15 in., undergoes two full turns, determine the average normal stress developed in the pipe. Assume that the union at B and couplings at A and C are rigid. Neglect the size of the union. Note: The lead would cause the pipe, when unloaded, to shorten 0.15 in. when the union is rotated one revolution.
B
A 3 ft
C 2 ft
The loads acting on both segments AB and BC are the same since no external load acts on the system. 0.3 = dB>A + dB>C 0.3 =
P(2)(12)
P(3)(12) 3
0.32(29)(10 )
+
0.32(29)(103)
P = 46.4 kip P 46.4 = = 145 ksi A 0.32
sAB = sBC =
Ans.
4–118. The brass plug is force-fitted into the rigid casting. The uniform normal bearing pressure on the plug is estimated to be 15 MPa. If the coefficient of static friction between the plug and casting is ms = 0.3, determine the axial force P needed to pull the plug out. Also, calculate the displacement of end B relative to end A just before the plug starts to slip out. Ebr = 98 GPa.
100 mm
B
15 MPa
P - 4.50(106)(2)(p)(0.02)(0.1) = 0 P = 56.549 kN = 56.5 kN
Ans.
Displacement: PL dB>A = a AE 0.1 m
56.549(103)(0.15) =
2
9
p(0.02 )(98)(10 )
+
L0
P
A
Equations of Equilibrium: + ©F = 0; : x
150 mm
0.56549(106) x dx p(0.022)(98)(109)
= 0.00009184 m = 0.0918 mm
Ans.
211
20 mm
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4–119. The assembly consists of two bars AB and CD of the same material having a modulus of elasticity E1 and coefficient of thermal expansion a1, and a bar EF having a modulus of elasticity E2 and coefficient of thermal expansion a2. All the bars have the same length L and cross-sectional area A. If the rigid beam is originally horizontal at temperature T1, determine the angle it makes with the horizontal when the temperature is increased to T2.
D
B
L
A
C
d
Equations of Equilibrium: a + ©MC = 0; + c ©Fy = 0;
FAB = FEF = F FCD - 2F = 0
[1]
Compatibility: dAB = (dAB)T - (dAB)F
dCD = (dCD)T + (dCD)F
dEF = (dEF)T - (dEF)F From the geometry dCD - dAB dEF - dAB = d 2d 2dCD = dEF + dAB 2 C (dCD)T + (dCD)F D = (dEF)T - (dEF)F + (dAB)T - (dAB)F 2 B a1 (T2 - T1) L + = a2 (T2 - T1) L -
FCD (L) R AE1 F(L) F(L) + a1 (T2 - T1) L AE2 AE1
[2]
Substitute Eq. [1] into [2]. 2a1 (T2 - T1) L +
4FL FL FL = a2 (T2 - T1)L + a1 (T2 - T1)L AE1 AE2 AE1
F 5F + = a2 (T2 - T1) - a1 (T2 - T1) AE1 AE2 F¢
5E2 + E1 b = (T2 - T1)(a2 - a1) ; AE1E2
F =
AE1E2 (T2 - T1)(a2 - a1) 5E2 + E1
(dEF)T = a2 (T2 - T1) L (dEF)F =
AE1E2 (T2 - T1)(a2 - a1)(L) E1 (T2 - T1)(a2 - a1)(L) = AE2 (5E2 + E1) 5E2 + E1
dEF = (dEF)T - (dEF)F =
a2 L(T2 - T1)(5E2 - E1) - E1L(T2 - T1)(a2 - a1) 5E2 + E1
(dAB)T = a1 (T2 - T1) L (dAB)F =
AE1E2 (T2 - T1)(a2 - a1)(L) E2 (T2 - T1)(a2 - a1)(L) = AE1 (5E2 + E1) 5E2 + E1
dAB = (dAB)T - (dAB)F =
F
a1 L(5E2 + E1)(T2 - T1) - E2 L(T2 - T1)(a2 - a1) 5E2 + E1
212
E
d
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4–119.
Continued
dEF - dAB =
L(T2 - T1) [a2 (5E2 + E1) - E1 (a2 - a1) - a1 (5E2 + E1) 5E2 + E1 + E2 (a2 - a1)]
=
L(T2 - T1) C (5E2 + E1)(a2 - a1) + (a2 - a1)(E2 - E1) D 5E2 + E1
=
L(T2 - T1)(a2 - a1) (5E2 + E1 + E2 - E1) 5E2 + E1
=
L(T2 - T1)(a2 - a1)(6E2) 5E2 + E1
u =
3E2L(T2 - T1)(a2 - a1) dEF - dAB = 2d d(5E2 + E1)
Ans.
*4–120. The rigid link is supported by a pin at A and two A-36 steel wires, each having an unstretched length of 12 in. and cross-sectional area of 0.0125 in2. Determine the force developed in the wires when the link supports the vertical load of 350 lb.
12 in. C 5 in. B
Equations of Equilibrium: a + ©MA = 0;
4 in. A
-FC(9) - FB (4) + 350(6) = 0
[1]
Compatibility:
6 in.
dC dB = 4 9
350 lb
FC(L) FB (L) = 4AE 9AE 9FB - 4FC = 0‚
[2]
Solving Eqs. [1] and [2] yields: FB = 86.6 lb
Ans.
FC = 195 lb
Ans.
213
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•5–1.
A shaft is made of a steel alloy having an allowable shear stress of tallow = 12 ksi. If the diameter of the shaft is 1.5 in., determine the maximum torque T that can be transmitted. What would be the maximum torque T¿ if a 1-in.-diameter hole is bored through the shaft? Sketch the shear-stress distribution along a radial line in each case.
T T¿
Allowable Shear Stress: Applying the torsion formula tmax = tallow =
12 =
Tc J T (0.75) p 2
(0.754)
T = 7.95 kip # in.
Ans.
Allowable Shear Stress: Applying the torsion formula tmax = tallow =
12 =
T¿c J T¿ (0.75) p 2
(0.754 - 0.54)
T¿ = 6.381 kip # in. = 6.38 kip # in. tr = 0.5 in =
T¿r = J
6.381(0.5) p 2
(0.754 - 0.54)
Ans.
= 8.00 ksi
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5–2. The solid shaft of radius r is subjected to a torque T. Determine the radius r¿ of the inner core of the shaft that resists one-half of the applied torque 1T>22. Solve the problem two ways: (a) by using the torsion formula, (b) by finding the resultant of the shear-stress distribution.
r¿ r
T
a)
tmax =
t =
Since t =
r¿ =
Tc Tr 2T = p 4 = J p r3 2 r
(T2 )r¿ p 2
(r¿)4
=
T p(r¿)3 T r¿ 2T = a b r pr3 p(r¿)3
r¿ t ; r max r 1
= 0.841 r
Ans.
24 r 2
b)
r¿
dT = 2p
L0 r 2
r¿
dT = 2p
L0 r 2
L0
tr2 dr
L0
r tmax r2 dr L0 r r¿
dT = 2p
r 2T 2 a 3 br dr L0 r pr
r¿
4T T = 4 r3 dr 2 r L0 r¿ =
r 1
Ans.
= 0.841r
24
215
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5–3. The solid shaft is fixed to the support at C and subjected to the torsional loadings shown. Determine the shear stress at points A and B and sketch the shear stress on volume elements located at these points.
10 kN⭈m C A 50 mm
The internal torques developed at Cross-sections pass through point B and A are shown in Fig. a and b, respectively. The polar moment of inertia of the shaft is J =
p (0.0754) = 49.70(10 - 6) m4. For 2
point B, rB = C = 0.075 Thus,
tB =
4(103)(0.075) TB c = 6.036(106) Pa = 6.04 MPa = J 49.70(10 - 6)
Ans.
From point A, rA = 0.05 m. tA =
TArA 6(103)(0.05) = 6.036(106) Pa = 6.04 MPa. = J 49.70 (10 - 6)
216
Ans.
B
75 mm 4 kN⭈m 75 mm
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*5–4. The tube is subjected to a torque of 750 N # m. Determine the amount of this torque that is resisted by the gray shaded section. Solve the problem two ways: (a) by using the torsion formula, (b) by finding the resultant of the shear-stress distribution.
75 mm
100 mm 750 Nm 25 mm
a) Applying Torsion Formula: tmax =
Tc = J
750(0.1) p 2
(0.14 - 0.0254)
tmax = 0.4793 A 106 B =
= 0.4793 MPa
T¿(0.1) p 2
(0.14 - 0.0754)
T¿ = 515 N # m
Ans.
b) Integration Method: r t = a b tmax c
dA = 2pr dr
and
dT¿ = rt dA = rt(2pr dr) = 2ptr2 dr 0.1m
T¿ =
L
2ptr2 dr = 2p
r tmax a br2 dr c L0.075m
=
0.1m 2ptmax r3 dr c L0.075m
=
2p(0.4793)(106) r4 0.1 m c d2 0.1 4 0.075 m
= 515 N # m
Ans.
5–5. The copper pipe has an outer diameter of 40 mm and an inner diameter of 37 mm. If it is tightly secured to the wall at A and three torques are applied to it as shown, determine the absolute maximum shear stress developed in the pipe.
tmax =
Tmax c = J
A
30 N⭈m
90(0.02) p 2
4
4
(0.02 - 0.0185 )
20 N⭈m
= 26.7 MPa
Ans.. 80 N⭈m
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5–6. The solid shaft has a diameter of 0.75 in. If it is subjected to the torques shown, determine the maximum shear stress developed in regions BC and DE of the shaft. The bearings at A and F allow free rotation of the shaft.
F E D C B
(tBC)max =
35(12)(0.375) TBC c = 5070 psi = 5.07 ksi = p 4 J 2 (0.375)
Ans.
(tDE)max =
25(12)(0.375) TDE c = 3621 psi = 3.62 ksi = p 4 J 2 (0.375)
Ans.
A
35 lb⭈ft
5–7. The solid shaft has a diameter of 0.75 in. If it is subjected to the torques shown, determine the maximum shear stress developed in regions CD and EF of the shaft. The bearings at A and F allow free rotation of the shaft.
F E D C B
(tEF)max =
TEF c = 0 J (tCD)max =
25 lb⭈ft 40 lb⭈ft 20 lb⭈ft
Ans.
A
25 lb⭈ft 40 lb⭈ft 20 lb⭈ft
35 lb⭈ft
15(12)(0.375) TCD c = p 4 J 2 (0.375)
= 2173 psi = 2.17 ksi
Ans.
218
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300 Nm
*5–8. The solid 30-mm-diameter shaft is used to transmit the torques applied to the gears. Determine the absolute maximum shear stress on the shaft.
500 Nm
A 200 Nm
Internal Torque: As shown on torque diagram. C
Maximum Shear Stress: From the torque diagram Tmax = 400 N # m. Then, applying torsion Formula.
400 Nm 300 mm
abs = tmax
Tmax c J 400(0.015)
=
p 2
(0.0154)
Ans.
= 75.5 MPa
The shaft consists of three concentric tubes, each made from the same material and having the inner and outer radii shown. If a torque of T = 800 N # m is applied to the rigid disk fixed to its end, determine the maximum shear stress in the shaft.
500 mm
T 800 Nm
ri 20 mm ro 25 mm
2m
p p p ((0.038)4 - (0.032)4) + ((0.030)4 - (0.026)4) + ((0.025)4 - (0.020)4) 2 2 2 -6
ri 26 mm ro 30 mm
4
J = 2.545(10 ) m tmax =
B
400 mm
•5–9.
J =
D
800(0.038) Tc = 11.9 MPa = J 2.545(10 - 6)
Ans.
219
ri 32 mm ro 38 mm
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5–10. The coupling is used to connect the two shafts together. Assuming that the shear stress in the bolts is uniform, determine the number of bolts necessary to make the maximum shear stress in the shaft equal to the shear stress in the bolts. Each bolt has a diameter d.
T R
r
n is the number of bolts and F is the shear force in each bolt. T
T F = nR
T - nFR = 0; T
tavg =
F 4T nR = p 2 = A ( 4 )d nRpd2
Maximum shear stress for the shaft: tmax =
Tc Tr 2T = p 4 = J pr3 2r 4T 2T = nRpd2 p r3
tavg = tmax ;
n =
2 r3 Rd2
Ans.
5–11. The assembly consists of two sections of galvanized steel pipe connected together using a reducing coupling at B. The smaller pipe has an outer diameter of 0.75 in. and an inner diameter of 0.68 in., whereas the larger pipe has an outer diameter of 1 in. and an inner diameter of 0.86 in. If the pipe is tightly secured to the wall at C, determine the maximum shear stress developed in each section of the pipe when the couple shown is applied to the handles of the wrench.
C
B
A
15 lb 6 in.
tAB =
tBC
Tc = J
Tc = = J
8 in.
210(0.375) p 2
(0.3754 - 0.344)
Ans. 15 lb
210(0.5) p 2
= 7.82 ksi
(0.54 - 0.434)
= 2.36 ksi
Ans.
220
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*5–12. The motor delivers a torque of 50 N # m to the shaft AB. This torque is transmitted to shaft CD using the gears at E and F. Determine the equilibrium torque Tⴕ on shaft CD and the maximum shear stress in each shaft. The bearings B, C, and D allow free rotation of the shafts.
A 50 mm 30 mm
Equilibrium:
B
a + ©ME = 0; a + ©MF = 0;
50 - F(0.05) = 0
F = 1000 N
35 mm T¿
T¿ - 1000(0.125) = 0 T¿ = 125 N # m
C
E
125 mm D
F
Ans.
Internal Torque: As shown on FBD. Maximum Shear Stress: Applying torsion Formula. (tAB)max =
50.0(0.015) TAB c = 9.43 MPa = p 4 J 2 (0.015 )
Ans.
(tCD)max =
125(0.0175) TCDc = 14.8 MPa = p 4 J 2 (0.0175 )
Ans.
•5–13. If the applied torque on shaft CD is T¿ = 75 N # m, determine the absolute maximum shear stress in each shaft. The bearings B, C, and D allow free rotation of the shafts, and the motor holds the shafts fixed from rotating.
A 50 mm
Equilibrium:
30 mm
a + ©MF = 0;
75 - F(0.125) = 0;
a + ©ME = 0;
600(0.05) - TA = 0
B
F = 600 N
35 mm T¿
TA = 30.0 N # m Internal Torque: As shown on FBD. Maximum Shear Stress: Applying the torsion formula (tEA)max =
30.0(0.015) TEA c = 5.66 MPa = p 4 J 2 (0.015 )
Ans.
(tCD)max =
75.0(0.0175) TCDc = 8.91 MPa = p 4 J 2 (0.0175 )
Ans.
221
C
E
125 mm D
F
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250 N⭈m
5–14. The solid 50-mm-diameter shaft is used to transmit the torques applied to the gears. Determine the absolute maximum shear stress in the shaft.
75 N⭈m
A
325 N⭈m 150 N⭈ m
B 500 mm
The internal torque developed in segments AB , BC and CD of the shaft are shown in Figs. a, b and c.
C D
400 mm 500 mm
The maximum torque occurs in segment AB. Thus, the absolute maximum shear stress occurs in this segment. The polar moment of inertia of the shaft is p J = (0.0254) = 0.1953p(10 - 6)m4. Thus, 2
A tmax B abs =
250(0.025) TAB c = 10.19(106)Pa = 10.2 MPa = J 0.1953p(10 - 6)
222
Ans.
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5–15. The solid shaft is made of material that has an allowable shear stress of tallow = 10 MPa. Determine the required diameter of the shaft to the nearest mm.
15 N⭈m 25 N⭈m A
30 N⭈m B
60 N⭈m C
70 N⭈m D E
The internal torques developed in each segment of the shaft are shown in the torque diagram, Fig. a. Segment DE is critical since it is subjected to the greatest internal torque. The polar p d 4 p 4 moment of inertia of the shaft is J = a b = d . Thus, 2 2 32
tallow
TDE c = ; J
d 70a b 2 10(106) = p 4 d 32 d = 0.03291 m = 32.91 mm = 33 mm
223
Ans.
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*5–16. The solid shaft has a diameter of 40 mm. Determine the absolute maximum shear stress in the shaft and sketch the shear-stress distribution along a radial line of the shaft where the shear stress is maximum.
15 N⭈m 25 N⭈m A
30 N⭈m B
The internal torque developed in each segment of the shaft are shown in the torque diagram, Fig. a.
60 N⭈m C
70 N⭈m D E
Since segment DE subjected to the greatest torque, the absolute maximum shear p stress occurs here. The polar moment of inertia of the shaft is J = (0.024) 2 = 80(10 - 9)p m4. Thus,
tmax =
70(0.02) TDE c = 5.57(106) Pa = 5.57 MPa = J 80(10 - 9)p
Ans.
The shear stress distribution along the radial line is shown in Fig. b.
•5–17.
The rod has a diameter of 1 in. and a weight of 10 lb/ft. Determine the maximum torsional stress in the rod at a section located at A due to the rod’s weight.
4.5 ft B
Here, we are only interested in the internal torque. Thus, other components of the internal loading are not indicated in the FBD of the cut segment of the rod, Fig. a. ©Mx = 0;
TA - 10(4)(2) = 0
TA = 80 lb # ft a
The polar moment of inertia of the cross section at A is J =
12in b = 960 lb # in. 1ft
p (0.54) = 0.03125p in4. 2
Thus
tmax =
960 (0.5) TA c = = 4889.24 psi = 4.89 ksi J 0.03125p
Ans.
224
4 ft
A 1.5 ft
1.5 ft
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5–18. The rod has a diameter of 1 in. and a weight of 15 lb/ft. Determine the maximum torsional stress in the rod at a section located at B due to the rod’s weight.
4.5 ft B
4 ft
Here, we are only interested in the internal torque. Thus, other components of the internal loading are not indicated in the FBD of the cut segment of the rod, Fig. a. ©Mx = 0;
TB - 15(4)(2) = 0
TB = 120 lb # ft a
12 in b = 1440 lb # in. 1ft
p The polar moment of inertia of the cross-section at B is J = (0.54) 2 = 0.03125p in4. Thus,
tmax =
1440(0.5) TB c = = 7333.86 psi = 7.33 ksi J 0.03125p
Ans.
225
A 1.5 ft
1.5 ft
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5–19. Two wrenches are used to tighten the pipe. If P = 300 N is applied to each wrench, determine the maximum torsional shear stress developed within regions AB and BC. The pipe has an outer diameter of 25 mm and inner diameter of 20 mm. Sketch the shear stress distribution for both cases.
P
B
Internal Loadings: The internal torque developed in segments AB and BC of the pipe can be determined by writing the moment equation of equilibrium about the x axis by referring to their respective free - body diagrams shown in Figs. a and b. ©Mx = 0; TAB - 300(0.25) = 0
TAB = 75 N # m
TBC = 150 N # m
Allowable Shear Stress: The polar moment of inertia of the pipe is p J = A 0.01254 - 0.014 B = 22.642(10 - 9)m4. 2
A tmax B AB =
75(0.0125) TAB c = 41.4 MPa = J 22.642(10 - 9)
A tAB B r = 0.01 m = A tmax B BC =
Ans.
TAB r 75(0.01) = 33.1 MPa = J 22.642(10 - 9)
150(0.0125) TBC c = 82.8 MPa = J 22.642(10 - 9)
A tBC B r = 0.01 m =
Ans.
TBC r 150(0.01) = 66.2 MPa = J 22.642(10 - 9)
The shear stress distribution along the radial line of segments AB and BC of the pipe is shown in Figs. c and d, respectively.
226
A 250 mm P
And ©Mx = 0; TBC - 300(0.25) - 300(0.25) = 0
250 mm
C
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*5–20. Two wrenches are used to tighten the pipe. If the pipe is made from a material having an allowable shear stress of tallow = 85 MPa, determine the allowable maximum force P that can be applied to each wrench. The pipe has an outer diameter of 25 mm and inner diameter of 20 mm.
P 250 mm
C B
A 250 mm
Internal Loading: By observation, segment BC of the pipe is critical since it is subjected to a greater internal torque than segment AB. Writing the moment equation of equilibrium about the x axis by referring to the free-body diagram shown in Fig. a, we have ©Mx = 0; TBC - P(0.25) - P(0.25) = 0
TBC = 0.5P
Allowable Shear Stress: The polar moment of inertia of the pipe is p J = A 0.01254 - 0.014 B = 22.642(10 - 9)m4 2 tallow =
TBC c ; J
85(106) =
0.5P(0.0125) 22.642(10 - 9)
P = 307.93N = 308 N
Ans.
227
P
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•5–21.
The 60-mm-diameter solid shaft is subjected to the distributed and concentrated torsional loadings shown. Determine the absolute maximum and minimum shear stresses on the outer surface of the shaft and specify their locations, measured from the fixed end A.
A
2 kN⭈m/m
1.5 m 1200 N⭈m C
The internal torque for segment BC is Constant TBC = 1200 N # m, Fig. a. However, the internal for segment AB varies with x, Fig. b. TAB - 2000x + 1200 = 0
TAB = (2000x - 1200) N # m
The minimum shear stress occurs when the internal torque is zero in segment AB. By setting TAB = 0, 0 = 2000x - 1200
x = 0.6 m
Ans.
And d = 1.5 m - 0.6 m = 0.9 m
Ans.
tmin = 0
Ans.
The maximum shear stress occurs when the internal torque is the greatest. This occurs at fixed support A where d = 0
Ans.
At this location, (TAB)max = 2000(1.5) - 1200 = 1800 N # m The polar moment of inertia of the rod is J =
tmax =
p (0.034) = 0.405(10 - 6)p. Thus, 2
(TAB)max c 1800(0.03) = 42.44(106)Pa = 42.4 MPa = J 0.405(10 - 6)p
228
Ans.
B 0.8 m
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5–22. The solid shaft is subjected to the distributed and concentrated torsional loadings shown. Determine the required diameter d of the shaft to the nearest mm if the allowable shear stress for the material is tallow = 50 MPa.
A
2 kN⭈m/m
1.5 m 1200 N⭈m C
The internal torque for segment BC is constant TBC = 1200 N # m, Fig. a. However, the internal torque for segment AB varies with x, Fig. b. TAB - 2000x + 1200 = 0 TAB = (2000x - 1200) N # m For segment AB, the maximum internal torque occurs at fixed support A where x = 1.5 m. Thus,
A TAB B max = 2000(1.5) - 1200 = 1800 N # m Since A TAB B max 7 TBC, the critical cross-section is at A. The polar moment of inertia p d 4 pd4 of the rod is J = . Thus, a b = 2 2 32 tallow =
Tc ; J
50(106) =
1800(d>2) pd4>32
d = 0.05681 m = 56.81 mm = 57 mm
229
Ans.
B 0.8 m
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*5–24. The copper pipe has an outer diameter of 2.50 in. and an inner diameter of 2.30 in. If it is tightly secured to the wall at C and a uniformly distributed torque is applied to it as shown, determine the shear stress developed at points A and B. These points lie on the pipe’s outer surface. Sketch the shear stress on volume elements located at A and B.
B A
C
125 lbft/ft
4 in. 9 in. 12 in.
Internal Torque: As shown on FBD. Maximum Shear Stress: Applying the torsion formula tA =
TA c J 125.0(12)(1.25)
=
tB =
p 2
(1.254 - 1.154)
Ans.
= 3.02 ksi
Ans.
TB c J 218.75(12)(1.25)
=
= 1.72 ksi
p 2
(1.254 - 1.154)
•5–25.
The copper pipe has an outer diameter of 2.50 in. and an inner diameter of 2.30 in. If it is tightly secured to the wall at C and it is subjected to the uniformly distributed torque along its entire length, determine the absolute maximum shear stress in the pipe. Discuss the validity of this result.
B A
C
125 lbft/ft
4 in. 9 in.
Internal Torque: The maximum torque occurs at the support C. Tmax = (125 lb # ft>ft)a
12 in.
25 in. b = 260.42 lb # ft 12 in.>ft
Maximum Shear Stress: Applying the torsion formula abs = tmax
Tmax c J 260.42(12)(1.25)
=
p 2
(1.254 - 1.154)
Ans.
= 3.59 ksi
According to Saint-Venant’s principle, application of the torsion formula should be as points sufficiently removed from the supports or points of concentrated loading.
230
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5–26. A cylindrical spring consists of a rubber annulus bonded to a rigid ring and shaft. If the ring is held fixed and a torque T is applied to the shaft, determine the maximum shear stress in the rubber.
ro
ri
T h
T r
T F t = = = A 2prh 2p r2 h Shear stress is maximum when r is the smallest, i.e. r = ri. Hence, tmax =
T 2p ri 2 h
Ans.
300 N⭈m
5–27. The A-36 steel shaft is supported on smooth bearings that allow it to rotate freely. If the gears are subjected to the torques shown, determine the maximum shear stress developed in the segments AB and BC. The shaft has a diameter of 40 mm.
100 N⭈m A
The internal torque developed in segments AB and BC are shown in their respective FBDs, Figs. a and b. The polar moment of inertia of the shaft is J =
A tAB B max
200 N⭈m B
p (0.024) = 80(10-9)p m4. Thus, 2 C
300(0.02) TAB c = 23.87(106)Pa = 23.9 MPa = = J 80(10-9)p
A tBC B max =
200(0.02) TBC c = 15.92(106) Pa = 15.9 MPa = J 80(10-9)p
231
Ans.
Ans.
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300 N⭈m
*5–28. The A-36 steel shaft is supported on smooth bearings that allow it to rotate freely. If the gears are subjected to the torques shown, determine the required diameter of the shaft to the nearest mm if tallover = 60 MPa.
100 N⭈m
The internal torque developed in segments AB and BC are shown in their respective FBDs, Fig. a and b
A 200 N⭈m B
Here, segment AB is critical since its internal torque is the greatest. The polar p d 4 pd4 moment of inertia of the shaft is J = . Thus, a b = 2 2 32 C
tallow
TC = ; J
60(106) =
300(d>2) pd4>32
d = 0.02942 m = 30 mm
Ans.
•5–29.
When drilling a well at constant angular velocity, the bottom end of the drill pipe encounters a torsional resistance TA . Also, soil along the sides of the pipe creates a distributed frictional torque along its length, varying uniformly from zero at the surface B to tA at A. Determine the minimum torque TB that must be supplied by the drive unit to overcome the resisting torques, and compute the maximum shear stress in the pipe. The pipe has an outer radius ro and an inner radius ri . TA +
TB B
L
1 t L - TB = 0 2 A tA
2TA + tAL TB = 2
Ans.
Maximum shear stress: The maximum torque is within the region above the distributed torque. tmax =
tmax =
Tc J (2TA + tAL) ] (r0) 2 p 4 4 (r r i) 2 0
[
(2TA + tAL)r0 =
Ans.
p(r40 - r4i )
232
A
TA
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5–30. The shaft is subjected to a distributed torque along its length of t = 110x22 N # m>m, where x is in meters. If the maximum stress in the shaft is to remain constant at 80 MPa, determine the required variation of the radius c of the shaft for 0 … x … 3 m. x
x
T =
L
t dx =
Tc t = ; J
L0
10 x2dx = 6
80(10 ) =
3m c
10 3 x 3 t ⫽ (10x2) N⭈m/m
3 (10 3 )x c p 2
c4
c3 = 26.526(10-9) x3 c = (2.98 x) mm
Ans.
5–31. The solid steel shaft AC has a diameter of 25 mm and is supported by smooth bearings at D and E. It is coupled to a motor at C, which delivers 3 kW of power to the shaft while it is turning at 50 rev>s. If gears A and B remove 1 kW and 2 kW, respectively, determine the maximum shear stress developed in the shaft within regions AB and BC. The shaft is free to turn in its support bearings D and E.
TC =
3(103) P = = 9.549 N # m v 50(2p)
TA =
1 T = 3.183 N # m 3 C
3 kW
2 kW 25 mm
1 kW
A D
(tAB)max =
3.183 (0.0125) TC = 1.04 MPa = p 4 J 2 (0.0125 )
Ans.
(tBC)max =
9.549 (0.0125) TC = 3.11 MPa = p 4 J 2 (0.0125 )
Ans.
233
B
E
C
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*5–32. The pump operates using the motor that has a power of 85 W. If the impeller at B is turning at 150 rev>min, determine the maximum shear stress developed in the 20-mm-diameter transmission shaft at A.
150 rev/min A
Internal Torque: v = 150
rev 2p rad 1 min = 5.00p rad>s ¢ ≤ min rev 60 s
P = 85 W = 85 N # m>s T =
P 85 = = 5.411 N # m v 5.00p
Maximum Shear Stress: Applying torsion formula tmax =
Tc J 5.411 (0.01)
=
p 4 2 (0.01 )
= 3.44 MPa
Ans.
•5–33.
The gear motor can develop 2 hp when it turns at 450 rev>min. If the shaft has a diameter of 1 in., determine the maximum shear stress developed in the shaft. The angular velocity of the shaft is v = ¢ 450
rev 2p rad 1 min ≤ ¢ ≤ ¢ ≤ = 15p rad>s min 1 rev 60 s
and the power is P = 2 hp ¢
550 ft # lb>s ≤ = 1100 ft # lb>s 1 hp
Then T =
P 1100 12 in = = 23.34 lb # ft a b = 280.11 lb # in v 15p 1ft
The polar moment of inertia of the shaft is J =
tmax =
p (0.54) = 0.03125p in4. Thus, 2
280.11 (0.5) Tc = = 1426.60 psi = 1.43 ksi J 0.03125p
Ans.
234
B
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5–34. The gear motor can develop 3 hp when it turns at 150 rev>min. If the allowable shear stress for the shaft is tallow = 12 ksi, determine the smallest diameter of the shaft to the nearest 18 in. that can be used. The angular velocity of the shaft is v = a 150
rev 2p rad 1 min ba ba b = 5p rad>s min 1 rev 60 s
and the power is P = (3 hp) a
550 ft # lb>s b = 1650 ft # lb>s 1 hp
Then T =
P 1650 12 in = = (105.04 lb # ft)a b = 1260.51 lb # in v 5p 1 ft
The polar moment of inertia of the shaft is J =
tallow =
Tc ; J
12(103) =
p d 4 pd4 a b = . Thus, 2 2 32
1260.51 (d>2) pd4>32
d = 0.8118 in. =
7 in. 8
Ans.
5–35. The 25-mm-diameter shaft on the motor is made of a material having an allowable shear stress of tallow = 75 MPa . If the motor is operating at its maximum power of 5 kW, determine the minimum allowable rotation of the shaft. Allowable Shear Stress: The polar moment of inertia of the shaft is p J = A 0.01254 B = 38.3495(10-9) m4. 2 tallow =
Tc ; J
75(106) =
T(0.0125) 38.3495(10-9)
T = 230.10 N # m Internal Loading: T =
P ; v
230.10 =
5(103) v
v = 21.7 rad>s
Ans.
235
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*5–36. The drive shaft of the motor is made of a material having an allowable shear stress of tallow = 75 MPa. If the outer diameter of the tubular shaft is 20 mm and the wall thickness is 2.5 mm, determine the maximum allowable power that can be supplied to the motor when the shaft is operating at an angular velocity of 1500 rev>min. Internal Loading: The angular velocity of the shaft is v = a 1500
rev 2p rad 1 min ba ba b = 50p rad>s min 1 rev 60 s
We have T =
P P = v 50p
Allowable Shear Stress: The polar moment of inertia of the shaft is p J = A 0.014 - 0.00754 B = 10.7379(10-9) m4. 2
tallow =
Tc ; J
75(106) =
a
P b(0.01) 50p
10.7379(10-9)
P = 12 650.25 W = 12.7 kW
Ans.
•5–37.
A ship has a propeller drive shaft that is turning at 1500 rev>min while developing 1800 hp. If it is 8 ft long and has a diameter of 4 in., determine the maximum shear stress in the shaft caused by torsion.
Internal Torque: v = 1500
rev 2p rad 1 min a b = 50.0 p rad>s min 1 rev 60 s
P = 1800 hpa
T =
550 ft # lb>s b = 990 000 ft # lb>s 1 hp
990 000 P = = 6302.54 lb # ft v 50.0p
Maximum Shear Stress: Applying torsion formula tmax =
6302.54(12)(2) Tc = p 4 J 2 (2 ) = 6018 psi = 6.02 ksi
Ans.
236
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5–38. The motor A develops a power of 300 W and turns its connected pulley at 90 rev>min. Determine the required diameters of the steel shafts on the pulleys at A and B if the allowable shear stress is tallow = 85 MPa.
60 mm 90 rev/min
A B
150 mm
Internal Torque: For shafts A and B vA = 90
rev 2p rad 1 min a b = 3.00p rad>s min rev 60 s
P = 300 W = 300 N # m>s P 300 = = 31.83 N # m vA 3.00p
TA =
vB = vA a
rA 0.06 b = 3.00pa b = 1.20p rad>s rB 0.15
P = 300 W = 300 N # m>s
TB =
P 300 = = 79.58 N # m vB 1.20p
Allowable Shear Stress: For shaft A tmax = tallow = 85 A 106 B =
TA c J 31.83 A d2A B
A B
p dA 4 2 2
dA = 0.01240 m = 12.4 mm
Ans.
For shaft B tmax = tallow = 85 A 106 B =
TB c J 79.58 A d2B B
A B
p dB 4 2 2
dB = 0.01683 m = 16.8 mm
Ans.
237
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5–39. The solid steel shaft DF has a diameter of 25 mm and is supported by smooth bearings at D and E. It is coupled to a motor at F, which delivers 12 kW of power to the shaft while it is turning at 50 rev>s. If gears A, B, and C remove 3 kW, 4 kW, and 5 kW respectively, determine the maximum shear stress developed in the shaft within regions CF and BC. The shaft is free to turn in its support bearings D and E.
v = 50
3 kW 4 kW A
B
D
C
E
F
rev 2p rad c d = 100 p rad>s s rev
TF =
12(103) P = = 38.20 N # m v 100 p
TA =
3(103) P = = 9.549 N # m v 100 p
TB =
4(103) P = = 12.73 N # m v 100 p
(tmax)CF =
38.20(0.0125) TCF c = 12.5 MPa = p 4 J 2 (0.0125 )
Ans.
(tmax)BC =
22.282(0.0125) TBC c = 7.26 MPa = p 4 J 2 (0.0125 )
Ans.
*5–40. Determine the absolute maximum shear stress developed in the shaft in Prob. 5–39.
v = 50
12 kW
5 kW 25 mm
3 kW 4 kW
rev 2p rad c d = 100 p rad>s s rev
TF =
12(103) P = = 38.20 N # m v 100p
TA =
3(103) P = = 9.549 N # m v 100p
TB =
4(103) P = = 12.73 N # m v 100p
A D
Tmax = 38.2 N # m 38.2(0.0125) = 12.5 MPa tabs = Tc = max p 4 J 2 (0.0125 )
Ans.
238
B C
From the torque diagram,
12 kW
5 kW 25 mm
E
F
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•5–41.
The A-36 steel tubular shaft is 2 m long and has an outer diameter of 50 mm. When it is rotating at 40 rad> s, it transmits 25 kW of power from the motor M to the pump P. Determine the smallest thickness of the tube if the allowable shear stress is tallow = 80 MPa.
P
M
P
M
The internal torque in the shaft is T =
25(103) P = = 625 N # m v 40
The polar moment of inertia of the shaft is J =
tallow =
Tc ; J
80(106) =
p (0.0254 - Ci 4). Thus, 2
625(0.025) p 4 2 (0.025
- Ci 4)
Ci = 0.02272 m So that t = 0.025 - 0.02272 = 0.002284 m = 2.284 mm = 2.5 mm
Ans.
5–42. The A-36 solid tubular steel shaft is 2 m long and has an outer diameter of 60 mm. It is required to transmit 60 kW of power from the motor M to the pump P. Determine the smallest angular velocity the shaft can have if the allowable shear stress is tallow = 80 MPa. The polar moment of inertia of the shaft is J =
tallow =
Tc ; J
80(106) =
p (0.034) = 0.405(10-6)p m4. Thus, 2
T(0.03) 0.405(10-6)p
T = 3392.92 N # m P = Tv ;
60(103) = 3392.92 v v = 17.68 rad>s = 17.7 rad>s
Ans.
239
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5–43. A steel tube having an outer diameter of 2.5 in. is used to transmit 35 hp when turning at 2700 rev>min. Determine the inner diameter d of the tube to the nearest 1 8 in. if the allowable shear stress is tallow = 10 ksi.
v =
2700(2p) = 282.74 rad>s 60
d 2.5 in.
P = Tv 35(550) = T(282.74) T = 68.083 lb # ft tmax = tallow = 10(103) =
Tc J 68.083(12)(1.25) p 4 2 (12.5
- ci 4)
ci = 1.2416 in. d = 2.48 in. Use d = 212 in.
Ans.
*5–44. The drive shaft AB of an automobile is made of a steel having an allowable shear stress of tallow = 8 ksi. If the outer diameter of the shaft is 2.5 in. and the engine delivers 200 hp to the shaft when it is turning at 1140 rev>min, determine the minimum required thickness of the shaft’s wall.
v =
B
1140(2p) = 119.38 rad>s 60
P = Tv 200(550) = T(119.38) T = 921.42 lb # ft tallow = 8(103) =
Tc J 921.42(12)(1.25) p 4 2 (1.25
- r4i )
,
ri = 1.0762 in.
t = ro - ri = 1.25 - 1.0762 t = 0.174 in.
Ans.
240
A
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•5–45.
The drive shaft AB of an automobile is to be designed as a thin-walled tube. The engine delivers 150 hp when the shaft is turning at 1500 rev>min. Determine the minimum thickness of the shaft’s wall if the shaft’s outer diameter is 2.5 in. The material has an allowable shear stress of tallow = 7 ksi.
v =
B
A
1500(2p) = 157.08 rad>s 60 P = Tv 150(550) = T(157.08) T = 525.21 lb # ft
tallow = 7(103) =
Tc J 525.21(12)(1.25) p 4 2 (1.25
- r4i )
ri = 1.1460 in.
,
t = ro - ri = 1.25 - 1.1460 t = 0.104 in.
Ans.
5–46. The motor delivers 15 hp to the pulley at A while turning at a constant rate of 1800 rpm. Determine to the nearest 18 in. the smallest diameter of shaft BC if the allowable shear stress for steel is tallow = 12 ksi. The belt does not slip on the pulley.
B
C 3 in.
The angular velocity of shaft BC can be determined using the pulley ratio that is vBC
1.5 in.
rA 1.5 rev 2p rad 1 min = a b vA = a b a1800 ba ba b = 30p rad>s rC 3 min 1 rev 60 s
A
The power is P = (15 hp) a
550 ft # n>s b = 8250 ft # lb>s 1 hp
Thus, T =
P 8250 12 in. = = (87.54 lb # ft)a b = 1050.42 lb # in v 30p 1 ft
The polar moment of inertia of the shaft is J =
tallow =
Tc ; J
12(103) =
p d 4 pd4 a b = . Thus, 2 2 32
1050.42(d>2) pd4>32
d = 0.7639 in =
7 in. 8
Ans.
241
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5–47. The propellers of a ship are connected to a A-36 steel shaft that is 60 m long and has an outer diameter of 340 mm and inner diameter of 260 mm. If the power output is 4.5 MW when the shaft rotates at 20 rad>s, determine the maximum torsional stress in the shaft and its angle of twist.
T =
4.5(106) P = = 225(103) N # m v 20
tmax =
f =
225(103)(0.170) Tc = 44.3 MPa = p J [(0.170)4 - (0.130)4] 2
Ans.
225 A 103 B (60) TL = 0.2085 rad = 11.9° = p JG [(0.170)4 - (0.130)4)75(109) 2
Ans.
*5–48. A shaft is subjected to a torque T. Compare the effectiveness of using the tube shown in the figure with that of a solid section of radius c. To do this, compute the percent increase in torsional stress and angle of twist per unit length for the tube versus the solid section.
T c 2
T
c
Shear stress: For the tube, (tt)max =
c
Tc Jt
For the solid shaft, (ts)max =
Tc Js
% increase in shear stress =
=
(ts)max - (tt)max (100) = (tt)max Js - Jt (100) = Jt
p 2
Tc Jt
-
Tc Js
Tc Js
(100)
c4 - [p2 [c4 - (p2 )4]] p 2
[c4 - (p2 )4]
(100) Ans.
= 6.67 % Angle of twist: For the tube, ft =
TL Jt(G)
For the shaft, fs =
TL Js(G)
% increase in f =
ft - fs (100%) = fs
=
Js - Jt (100%) = Jt
TL Jt(G)
-
TL Js(G)
TL Js(G) p 2
(100%)
c4 - [p2 [c4 - (p2 )4]] p 2
[c4 - (p2 )4]
(100%)
= 6.67 %
Ans. 242
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•5–49.
The A-36 steel axle is made from tubes AB and CD and a solid section BC. It is supported on smooth bearings that allow it to rotate freely. If the gears, fixed to its ends, are subjected to 85-N # m torques, determine the angle of twist of gear A relative to gear D. The tubes have an outer diameter of 30 mm and an inner diameter of 20 mm. The solid section has a diameter of 40 mm.
400 mm 250 mm 400 mm B
fND = ©
TL JG
A
(85)(0.25)
2(85)(0.4) =
p 2
4
4
9
(0.015 - 0.01 )(75)(10 )
+
p 2
85 Nm
(0.024)(75)(109)
= 0.01534 rad = 0.879°
Ans.
5–50. The hydrofoil boat has an A-36 steel propeller shaft that is 100 ft long. It is connected to an in-line diesel engine that delivers a maximum power of 2500 hp and causes the shaft to rotate at 1700 rpm. If the outer diameter of the shaft is 8 in. and the wall thickness is 38 in., determine the maximum shear stress developed in the shaft. Also, what is the “wind up,” or angle of twist in the shaft at full power?
100 ft
Internal Torque: v = 1700
rev 2p rad 1 min a b = 56.67p rad>s min rev 60 s
P = 2500 hp a T =
550 ft # lb>s b = 1 375 000 ft # lb>s 1 hp
P 1 375 000 = = 7723.7 lb # ft v 56.67p
Maximum Shear Stress: Applying torsion Formula. tmax =
Tc J 7723.7(12)(4)
=
p 2
(44 - 3.6254)
Ans.
= 2.83 ksi
Angle of Twist: f =
TL = JG
7723.7(12)(100)(12) p 2
(44 - 3.6254)11.0(106)
= 0.07725 rad = 4.43°
Ans.
243
C
D 85 Nm
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5–51. The engine of the helicopter is delivering 600 hp to the rotor shaft AB when the blade is rotating at 1200 rev>min. Determine to the nearest 18 in. the diameter of the shaft AB if the allowable shear stress is tallow = 8 ksi and the vibrations limit the angle of twist of the shaft to 0.05 rad. The shaft is 2 ft long and made from L2 steel. A
B
1200(2)(p) = 125.66 rad>s v = 60 P = Tv 600(550) = T(125.66) T = 2626.06 lb # ft Shear - stress failure tallow = 8(103) =
Tc J
2626.06(12)c p 2
c4
c = 1.3586 in. Angle of twist limitation f =
0.05 =
TL JG 2626.06(12)(2)(12) p 2
c4(11.0)(106)
c = 0.967 in. Shear - stress failure controls the design. d = 2c = 2 (1.3586) = 2.72 in. Use d = 2.75 in.
Ans.
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*5–52. The engine of the helicopter is delivering 600 hp to the rotor shaft AB when the blade is rotating at 1200 rev>min. Determine to the nearest 18 in. the diameter of the shaft AB if the allowable shear stress is tallow = 10.5 ksi and the vibrations limit the angle of twist of the shaft to 0.05 rad. The shaft is 2 ft long and made from L2 steel. A
v =
B
1200(2)(p) = 125.66 rad>s 60 P = Tv
600(550) = T(125.66) T = 2626.06 lb # ft Shear - stress failure tallow = 10.5(10)3 =
2626.06(12)c p 2
c4
c = 1.2408 in. Angle of twist limitation f =
0.05 =
TL JG 2626.06(12)(2)(12) p 2
c4 (11.0)(106)
c = 0.967 in. Shear stress failure controls the design d = 2c = 2 (1.2408) = 2.48 in. Use d = 2.50 in.
Ans.
245
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•5–53. The 20-mm-diameter A-36 steel shaft is subjected to the torques shown. Determine the angle of twist of the end B.
A
Internal Torque: As shown on FBD.
D
Angle of Twist:
C B
TL fB = a JG
30 Nm 600 mm
200 mm
20 Nm 800 mm
1 [-80.0(0.8) + (-60.0)(0.6) + (-90.0)(0.2)] = p 4 9 (0.01 )(75.0)(10 ) 2 = -0.1002 rad = | 5.74° |
80 Nm
Ans.
5–54. The assembly is made of A-36 steel and consists of a solid rod 20 mm in diameter fixed to the inside of a tube using a rigid disk at B. Determine the angle of twist at D. The tube has an outer diameter of 40 mm and wall thickness of 5 mm.
A
B
The internal torques developed in segments AB and BD of the assembly are shown in Fig. a and b
0.4 m
C 0.1 m
p The polar moment of inertia of solid rod and tube are JAB = (0.024 - 0.0154) 2 p = 54.6875(10 - 9)p m4 and JBD = (0.014) = 5(10 - 9)p m4. Thus, 2 fD = ©
Ti Li TAB LAB TBD LBD = + Ji Gi JAB Gst JBD Gst -60 (0.4)
90(0.4) =
54.6875(10 - 9)p [75(109)]
+
5(10 - 9)p [75(109)]
= -0.01758 rad = 1.01°
Ans.
246
150 N⭈m
D 0.3 m 60 N⭈m
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5–55. The assembly is made of A-36 steel and consists of a solid rod 20 mm in diameter fixed to the inside of a tube using a rigid disk at B. Determine the angle of twist at C. The tube has an outer diameter of 40 mm and wall thickness of 5 mm.
A
B
The polar moment of inertia = 54.6875 (10 - 9)p m4. Thus,
fC = ©
of
the
tube
is
J =
150 N⭈m
0.4 m
The internal torques developed in segments AB and BC of the assembly are shown in Figs. a and b.
C 0.1 m
p (0.024 - 0.0154) 2
D 0.3 m 60 N⭈m
Ti Li TAB LAB TBC LBC = + Ji Gi JGst J Gst =
1 C 90(0.4) + 150(0.1) D 54.6875(10 )p [75(109)] -9
= 0.003958 rad = 0.227°
Ans.
*5–56. The splined ends and gears attached to the A-36 steel shaft are subjected to the torques shown. Determine the angle of twist of end B with respect to end A. The shaft has a diameter of 40 mm.
fB>A = ©
300 N⭈m
500 N⭈m
A
200 N⭈m
-300(0.3) 200(0.4) 400(0.5) TL = + + JG JG JG JG
C
400 N⭈m
190 = = JG
300 mm
190 p 4 (0.02 )(75)(109) 2
D B
400 mm
= 0.01008 rad = 0.578°
Ans. 500 mm
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•5–57.
The motor delivers 40 hp to the 304 stainless steel shaft while it rotates at 20 Hz. The shaft is supported on smooth bearings at A and B, which allow free rotation of the shaft. The gears C and D fixed to the shaft remove 25 hp and 15 hp, respectively. Determine the diameter of the shaft to the nearest 18 in. if the allowable shear stress is tallow = 8 ksi and the allowable angle of twist of C with respect to D is 0.20°.
External Applied Torque: Applying T =
TM =
40(550) = 175.07 lb # ft 2p(20)
TD =
15(550) = 65.65 lb # ft 2p(20)
A
D 10 in.
6 in.
25(550) = 109.42 lb # ft 2p(20)
Internal Torque: As shown on FBD. Allowable Shear Stress: Assume failure due to shear stress. By observation, section AC is the critical region. Tc J
tmax = tallow =
175.07(12) A d2 B
8(103) =
p 2
A d2 B
4
d = 1.102 in.
Angle of Twist: Assume failure due to angle of twist limitation. fC>D = 0.2(p) = 180
TCDLCD JG 65.65(12)(8) p 2
B 8 in.
P , we have 2pf TC =
C
A d2 B (11.0)(106) 4
d = 1.137 in. (controls !) 1 Use d = 1 in. 4
Ans.
248
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5–58. The motor delivers 40 hp to the 304 stainless steel solid shaft while it rotates at 20 Hz. The shaft has a diameter of 1.5 in. and is supported on smooth bearings at A and B, which allow free rotation of the shaft. The gears C and D fixed to the shaft remove 25 hp and 15 hp, respectively. Determine the absolute maximum stress in the shaft and the angle of twist of gear C with respect to gear D.
A
C D
10 in.
B 8 in. 6 in.
External Applied Torque: Applying T =
TM =
40(550) = 175.07 lb # ft 2p(20)
TD =
15(550) = 65.65 lb # ft 2p(20)
P , we have 2pf TC =
25(550) = 109.42 lb # ft 2p(20)
Internal Torque: As shown on FBD. Allowable Shear Stress: The maximum torque occurs within region AC of the shaft where Tmax = TAC = 175.07 lb # ft. abs = tmax
175.07(12)(0.75) Tmax c = 3.17 ksi = p 4 J 2 (0.75 )
Ans.
Angle of Twist: fC>D =
TCD LCD JG 65.65(12)(8)
=
p 2
(0.754)(11.0)(106)
= 0.001153 rad = 0.0661°
Ans.
249
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5–59. The shaft is made of A-36 steel. It has a diameter of 1 in. and is supported by bearings at A and D, which allow free rotation. Determine the angle of twist of B with respect to D.
A
B
60 lb⭈ft C
2 ft 60 lb⭈ft
2.5 ft
The internal torques developed in segments BC and CD are shown in Figs. a and b.
D 3 ft
p The polar moment of inertia of the shaft is J = (0.54) = 0.03125p in4. Thus, 2 TiLi TBC LBC TCD LCD = + FB/D = a JiGi J Gst J Gst -60(12)(2.5)(12) =
(0.03125p)[11.0(106)]
+ 0
= -0.02000 rad = 1.15°
Ans.
*5–60. The shaft is made of A-36 steel. It has a diameter of 1 in. and is supported by bearings at A and D, which allow free rotation. Determine the angle of twist of gear C with respect to B.
A
B
60 lb⭈ft C
2 ft 60 lb⭈ft
2.5 ft
The internal torque developed in segment BC is shown in Fig. a
D 3 ft
p The polar moment of inertia of the shaft is J = (0.54) = 0.03125p in4. Thus, 2 fC>B =
-60(12)(2.5)(12) TBC LBC = J Gst (0.03125p)[11.0(106)] = -0.02000 rad = 1.15°
Ans.
250
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•5–61.
The two shafts are made of A-36 steel. Each has a diameter of 1 in., and they are supported by bearings at A, B, and C, which allow free rotation. If the support at D is fixed, determine the angle of twist of end B when the torques are applied to the assembly as shown.
D
10 in.
C
80 lbft A
30 in.
40 lbft
8 in. 10 in. 12 in.
Internal Torque: As shown on FBD. Angle of Twist: TL fE = a JG 1 =
p 2
(0.54)(11.0)(105)
[-60.0(12)(30) + 20.0(12)(10)]
= -0.01778 rad = 0.01778 rad fF =
6 6 f = (0.01778) = 0.02667 rad 4 E 4
Since there is no torque applied between F and B then fB = fF = 0.02667 rad = 1.53°
Ans.
251
4 in.
6 in. B
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5–62. The two shafts are made of A-36 steel. Each has a diameter of 1 in., and they are supported by bearings at A, B, and C, which allow free rotation. If the support at D is fixed, determine the angle of twist of end A when the torques are applied to the assembly as shown.
D
10 in.
C
80 lbft A
30 in.
40 lbft
8 in. 10 in. 12 in.
Internal Torque: As shown on FBD. Angle of Twist: TL fE = a JG 1 =
p 2
(0.54)(11.0)(106)
[-60.0(12)(30) + 20.0(12)(10)]
= -0.01778 rad = 0.01778 rad
fF =
6 6 f = (0.01778) = 0.02667 rad 4 E 4
fA>F =
TGF LGF JG -40(12)(10)
=
p 2
(0.54)(11.0)(106)
= -0.004445 rad = 0.004445 rad fA = fF + fA>F = 0.02667 + 0.004445 = 0.03111 rad = 1.78°
Ans.
252
4 in.
6 in. B
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5–63. The device serves as a compact torsional spring. It is made of A-36 steel and consists of a solid inner shaft CB which is surrounded by and attached to a tube AB using a rigid ring at B. The ring at A can also be assumed rigid and is fixed from rotating. If a torque of T = 2 kip # in. is applied to the shaft, determine the angle of twist at the end C and the maximum shear stress in the tube and shaft.
12 in. 12 in.
B
T
1 in. A
0.5 in. C
Internal Torque: As shown on FBD. Maximum Shear Stress: (tBC)max =
2.00(0.5) TBC c = 10.2 ksi = p 4 J 2 (0.5 )
TBA c = J
(tBA)max =
2.00(1) p 2
(14 - 0.754)
Ans.
= 1.86 ksi
Ans.
Angle of Twist: fB =
TBA LBA JG (2.00)(12)
=
p 2
4
(1 - 0.754)11.0(103)
fC>B =
TBC LBC JG 2.00(24)
=
= 0.002032 rad
p 2
(0.54)11.0(103)
= 0.044448 rad
fC = fB + fC>B = 0.002032 + 0.044448 = 0.04648 rad = 2.66°
Ans.
253
0.75 in.
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*5–64. The device serves as a compact torsion spring. It is made of A-36 steel and consists of a solid inner shaft CB which is surrounded by and attached to a tube AB using a rigid ring at B. The ring at A can also be assumed rigid and is fixed from rotating. If the allowable shear stress for the material is tallow = 12 ksi and the angle of twist at C is limited to fallow = 3°, determine the maximum torque T that can be applied at the end C.
12 in. 12 in.
B
T
1 in. A
0.5 in. C
Internal Torque: As shown on FBD. Allowable Shear Stress: Assume failure due to shear stress. tmax = tallow =
12.0 =
TBC c J T (0.5) p 2
(0.54)
T = 2.356 kip # in tmax = tallow =
12.0 =
TBA c J T (1) p 2
(14 - 0.754)
T = 12.89 kip # in Angle of Twist: Assume failure due to angle of twist limitation. fB =
TBA LBA = JG
T(12) p 2
(14 - 0.754) 11.0(103)
= 0.001016T fC>B =
TBC LBC = JG
T(24) p 2
(0.54)11.0(103)
= 0.022224T (fC)allow = fB + fC>B 3(p) = 0.001016T + 0.022224T 180 T = 2.25 kip # in (controls !)
Ans.
254
0.75 in.
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•5–65.
The A-36 steel assembly consists of a tube having an outer radius of 1 in. and a wall thickness of 0.125 in. Using a rigid plate at B, it is connected to the solid 1-in-diameter shaft AB. Determine the rotation of the tube’s end C if a torque of 200 lb # in. is applied to the tube at this end. The end A of the shaft is fixed supported.
B C 200 lb⭈in.
4 in.
A 6 in.
fB =
TABL = JG
fC>B =
200(10) p 2
TCBL = JG
(0.5)4(11.0)(106)
= 0.001852 rad
-200(4) p 2
4
(1 - 0.8754)(11.0)(106)
= -0.0001119 rad
fC = fB + fC>B = 0.001852 + 0.0001119 = 0.001964 rad = 0.113°
Ans.
255
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5–66. The 60-mm diameter shaft ABC is supported by two journal bearings, while the 80-mm diameter shaft EH is fixed at E and supported by a journal bearing at H. If T1 = 2 kN # m and T2 = 4 kN # m, determine the angle of twist of gears A and C. The shafts are made of A-36 steel.
E A
600 mm D 100 mm H
T2 600 mm B 75 mm 900 mm
Equilibrium: Referring to the free - body diagram of shaft ABC shown in Fig. a ©Mx = 0; F(0.075) - 4(103) - 2(103) = 0
F = 80(103) N
Internal Loading: Referring to the free - body diagram of gear D in Fig. b, ©Mx = 0; 80(103)(0.1) - TDH = 0
TDH = 8(103)N # m
Also, from the free - body diagram of gear A, Fig. c, ©Mx = 0; TAB - 4(103) = 0
TAB = 4 A 103 B N # m
And from the free - body diagram of gear C, Fig. d, ©Mx = 0; -TBC - 2 A 103 B = 0
TBC = -2(103) N # m
Angle of Twist: The polar moment of inertia of segments AB, BC and DH p of the shaft are JAB = JBC = and A 0.034 B = 0.405(10 - 6)p m4 2 p 4 -6 4 JDH = A 0.04 B = 1.28(10 )p m . We have 2 fD =
8(103)(0.6) TDH LDH = 0.01592 rad = JDHGst 1.28(10 - 6)p(75)(109)
Then, using the gear ratio, fB = fD a
rD 100 b = 0.02122 rad b = 0.01592a rB 75
Also, fC>B =
-2(103)(0.9) TBC LBC = -0.01886 rad = 0.01886 rad = JBCGst 0.405(10 - 6)p(75)(109)
fA>B =
4(103)(0.6) TABLAB = 0.02515 rad = JAB Gst 0.405(10 - 6)p(75)(109)
Thus, fA = fB + fA>B fA = 0.02122 + 0.02515 = 0.04637 rad = 2.66°
Ans.
fC = fB + fC>B fC = 0.02122 + 0.01886 = 0.04008 rad = 2.30°
Ans.
256
T1
C
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5–66.
Continued
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5–67. The 60-mm diameter shaft ABC is supported by two journal bearings, while the 80-mm diameter shaft EH is fixed at E and supported by a journal bearing at H. If the angle of twist at gears A and C is required to be 0.04 rad, determine the magnitudes of the torques T1 and T2. The shafts are made of A-36 steel.
E A
600 mm D 100 mm H
T2 600 mm B 75 mm 900 mm
Equilibrium: Referring to the free - body diagram of shaft ABC shown in Fig. a ©Mx = 0; F(0.075) - T1 - T2 = 0
T1
F = 13.333 A T1 + T2 B
Internal Loading: Referring to the free - body diagram of gear D in Fig. b, ©Mx = 0; 13.333 A T1 + T2 B (0.1) - TDE = 0
TDE = 1.333 A T1 + T2 B
Also, from the free - body diagram of gear A, Fig. c, ©Mx = 0; TAB - T2 = 0
TAB = T2
and from the free - body diagram of gear C, Fig. d ©Mx = 0; TBC - T1 = 0
TBC = T1
Angle of Twist: The polar moments of inertia of segments AB, BC and DH p of the shaft are and JAB = JBC = A 0.034 B = 0.405(10 - 6)pm4 2 p JDH = A 0.044 B = 1.28(10 - 6)pm4. We have 2 fD =
1.333 A T1 + T2 B (0.6) TDE LDH = 2.6258(10 - 6) A T1 + T2 B = JDE Gst 1.28(10 - 6)p (75)(109)
Then, using the gear ratio, fB = fD ¢
rD 100 b = 3.5368(10 - 6) A T1 + T2 B ≤ = 2.6258(10 - 6) A T1 + T2 B a rB 75
Also, fC>B =
T1(0.9) TBC LBC = 9.4314(10 - 6)T1 = JBC Gst 0.405(10 - 6)p(75)(109)
fA>B =
T2(0.6) TAB LAB = 6.2876(10 - 6)T2 = JAB Gst 0.405(10 - 6)p(75)(109)
Here, it is required that fA = fC = 0.04 rad. Thus, fA = fB + fA>B 0.04 = 3.5368(10 - 6) A T1 + T2 B + 6.2876(10 - 6)T2 T1 + 2.7778T2 = 11309.73
(1)
fC = fB + fC>B 0.04 = 3.5368(10 - 6) A T1 + T2 B + 9.4314(10 - 6)T1 3.6667T1 + T2 = 11309.73
(2)
258
C
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5–67.
Continued
Solving Eqs. (1) and (2), T1 = 2188.98 N # m = 2.19 kN # m
Ans.
T2 = 3283.47 N # m = 3.28 kN # m
Ans.
*5–68. The 30-mm-diameter shafts are made of L2 tool steel and are supported on journal bearings that allow the shaft to rotate freely. If the motor at A develops a torque of T = 45 N # m on the shaft AB, while the turbine at E is fixed from turning, determine the amount of rotation of gears B and C.
A
45 Nm
B 1.5 m
50 mm
D C
Internal Torque: As shown on FBD. 0.5 m
Angle of Twist: fC =
TCE LCE JG 67.5(0.75)
=
p 2
(0.0154)75.0(103)
= 0.008488 rad = 0.486° fB =
75 50
Ans.
fC = 0.729°
Ans.
259
E 75 mm 0.75 m
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•5–69.
The shafts are made of A-36 steel and each has a diameter of 80 mm. Determine the angle of twist at end E.
0.6 m A
B 150 mm 10 kN⭈m
C
D
200 mm 0.6 m
E
Equilibrium: Referring to the free - body diagram of shaft CDE shown in Fig. a, ©Mx = 0; 10(103) - 2(103) - F(0.2) = 0
150 mm
F = 40(103) N
0.6 m 2 kN⭈m
Internal Loading: Referring to the free - body diagram of gear B, Fig. b, ©Mx = 0;
-TAB - 40(103)(0.15) = 0
TAB = -6(103) N # m
Referring to the free - body diagram of gear D, Fig. c, ©Mx = 0; 10(103) - 2(103) - TCD = 0
TCD = 8(103) N # m
Referring to the free - body diagram of shaft DE, Fig. d, ©Mx = 0;
-TDE - 2(103) = 0
Angle of Twist: The polar p J = A 0.044 B = 1.28(10 - 6)p m4. 2
TDE = -2(103) N # m moment
of
inertia
of
the
shafts
are
We have
fB =
-6(103)(0.6) TAB LAB = -0.01194 rad = 0.01194 rad = JGst 1.28(10 - 6)p(75)(109)
Using the gear ratio, fC = fB ¢
rB 150 b = 0.008952 rad ≤ = 0.01194 a rC 200
fE>C = ©
TiLi TCD LCD TDE LDE = + JiGi JGst JGst
Also,
0.6 =
-6
1.28(10 )p(75)(109)
b 8(103) + c -2(103) d r
= 0.01194 rad Thus, fE = fC + fE>C fE = 0.008952 + 0.01194 = 0.02089 rad = 1.20°
Ans.
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5–69.
Continued
261
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5–70. The shafts are made of A-36 steel and each has a diameter of 80 mm. Determine the angle of twist of gear D.
0.6 m
Equilibrium: Referring to the free-body diagram of shaft CDE shown in Fig. a, 3
3
A
©Mx = 0; 10(10 ) - 2(10 ) - F(0.2) = 0
F = 40(10 ) N
-TAB - 40(103)(0.15) = 0
150 mm 10 kN⭈m
C
Internal Loading: Referring to the free - body diagram of gear B, Fig. b, ©Mx = 0;
B
3
TAB = -6(103) N # m
D
200 mm 0.6 m
E
Referring to the free - body diagram of gear D, Fig. c, ©Mx = 0; 10(103) - 2(103) - TCD = 0
150 mm
TCD = 8(103) N # m
0.6 m 2 kN⭈m
Angle of Twist: The polar moment p J = A 0.044 B = 1.28(10 - 6)p m4. We have 2 fB =
of
inertia
of
the
shafts
are
-6(103)(0.6) TAB LAB = -0.01194 rad = 0.01194 rad = JGst 1.28(10 - 6)p(75)(109)
Using the gear ratio, fC = fB ¢
rB 150 b = 0.008952 rad ≤ = 0.01194a rC 200
Also, fD>C =
8(103)(0.6) TCD LCD = 0.01592 rad = JGst 1.28(10 - 6)p(75)(109)
Thus, fD = fC + fD>C fD = 0.008952 + 0.01592 = 0.02487 rad = 1.42°
Ans.
262
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*5–72. The 80-mm diameter shaft is made of 6061-T6 aluminum alloy and subjected to the torsional loading shown. Determine the angle of twist at end A.
0.6 m 0.6 m C
10 kN⭈m/m B A 2 kN⭈m
Equilibrium: Referring to the free - body diagram of segment AB shown in Fig. a, ©Mx = 0;
-TAB - 2(103) = 0
TAB = -2(103)N # m
And the free - body diagram of segment BC, Fig. b, ©Mx = 0;
Angle of Twist: The polar moment p J = A 0.042 B = 1.28(10 - 6)p m4. We have 2 fA = ©
1.28(10 - 6)p(26)(109)
0.6 m -
+
1 = -
of
inertia
of
the
shaft
is
LBC TiLi TABLAB TBC dx = + JiGi JGal JGal L0
-2(103)(0.6) =
TBC = - C 10(103)x + 2(103) D N # m
-TBC - 10(103)x - 2(103) = 0
1.28(10 - 6)p(26)(109)
L0
C 10(103)x + 2(103) D dx
1.28(10 - 6)p(26)(109)
b 1200 + C 5(103)x2 + 2(103)x D 2
0.6m 0
r
= -0.04017 rad = 2.30°
Ans.
263
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•5–73. The tapered shaft has a length L and a radius r at end A and 2r at end B. If it is fixed at end B and is subjected to a torque T, determine the angle of twist of end A. The shear modulus is G.
B 2r L
T
Geometry: r A
r rL + rx r(x) = r + x = L L p rL + rx 4 p r4 (L + x)4 a b = 2 L 2L4
J(x) = Angle of Twist:
L
T dx L0 J(x)G
f =
L
=
2TL4 dx p r4G L0 (L + x)4
=
L 1 2TL4 cd2 4 3 pr G 3(L + x) 0
=
7TL 12p r4G
Ans.
5–74. The rod ABC of radius c is embedded into a medium where the distributed torque reaction varies linearly from zero at C to t0 at B. If couple forces P are applied to the lever arm, determine the value of t0 for equilibrium. Also, find the angle of twist of end A. The rod is made from material having a shear modulus of G.
L 2 L 2
Equilibrium: Referring to the free-body diagram of the entire rod shown in Fig. a, 1 L ©Mx = 0; Pd - (t0)a b = 0 2 2 to =
Ans.
Internal Loading: The distributed torque expressed as a function of x, measured 4Pd>L to 8Pd from the left end, is t = ¢ ≤x = ¢ ≤ x = ¢ 2 ≤ x. Thus, the resultant L>2 L>2 L torque within region x of the shaft is
TR =
1 1 8Pd 4Pd 2 tx = B ¢ 2 ≤ x R x = x 2 2 L L2
Referring to the free - body diagram shown in Fig. b, ©Mx = 0; TBC -
4Pd 2 x = 0 L2
d 2
B P
4Pd L
TBC =
4Pd 2 x L2
Referring to the free - body diagram shown in Fig. c, ©Mx = 0; Pd - TAB = 0
TAB = Pd
264
d 2
t0
C
A
P
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5–74.
Continued
Angle of Twist: f = ©
LBC TAB LAB TBC dx TiLi = + JiGi JG JG L0
L>2
=
L0
4Pd 2 x dx L2
Pd(L>2) +
p 2
p 2
¢ c4 ≤ G
¢ c4 ≤ G L>2
8Pd x3 = £ ≥3 4 2 pc L G 3
+
PLd pc4G
0
=
4PLd 3pc4G
Ans.
265
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5–75. When drilling a well, the deep end of the drill pipe is assumed to encounter a torsional resistance TA . Furthermore, soil friction along the sides of the pipe creates a linear distribution of torque per unit length, varying from zero at the surface B to t0 at A. Determine the necessary torque TB that must be supplied by the drive unit to turn the pipe. Also, what is the relative angle of twist of one end of the pipe with respect to the other end at the instant the pipe is about to turn? The pipe has an outer radius ro and an inner radius ri . The shear modulus is G.
TB
B
L
t0 A
1 t L + TA - TB = 0 2 0 TB =
t0L + 2TA 2
T(x) +
t0 2 t0L + 2TA x = 0 2L 2
T(x) =
t0 2 t0 L + 2TA x 2 2L
f =
Ans.
T(x) dx L JG
=
L t0L + 2TA t0 2 1 ( x ) dx J G L0 2 2L
=
t0 3 L 1 t0 L + 2TA c x x dƒ JG 2 6L 0
=
t0 L2 + 3TAL 3JG
However, J =
f =
p (r 4 - ri 4) 2 o 2L(t0 L + 3TA)
Ans.
3p(ro 4 - ri 4)G
266
TA
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*5–76. A cylindrical spring consists of a rubber annulus bonded to a rigid ring and shaft. If the ring is held fixed and a torque T is applied to the rigid shaft, determine the angle of twist of the shaft. The shear modulus of the rubber is G. Hint: As shown in the figure, the deformation of the element at radius r can be determined from rdu = drg. Use this expression along with t = T>12pr2h2 from Prob. 5–26, to obtain the result.
ro r ri T h
gdr rdu
dr g du r
r du = g dr du =
gdr r
(1)
From Prob. 5-26, t =
T 2p r2h
g =
T 2p r2hG
and
g =
t G
From (1), du =
T dr 2p hG r3 r
u =
o dr T 1 ro T = cd| 3 2p hG Lri r 2p hG 2 r2 ri
=
1 1 T c- 2 + 2d 2p hG 2ro 2ri
=
1 1 T c - 2d 4p hG r2i ro
Ans.
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•5–77.
The A-36 steel shaft has a diameter of 50 mm and is fixed at its ends A and B. If it is subjected to the torque, determine the maximum shear stress in regions AC and CB of the shaft.
A
300 Nm
0.4 m C 0.8 m
Equilibrium: TA + TB - 300 = 0
[1]
Compatibility: fC>A = fC>B TA(0.4) TB(0.8) = JG JG TA = 2.00TB
[2]
Solving Eqs. [1] and [2] yields: TA = 200 N # m
TB = 100 N # m
Maximum Shear stress: (tAC)max =
200(0.025) TAc = 8.15 MPa = p 4 J 2 (0.025 )
Ans.
(tCB)max =
100(0.025) TBc = 4.07 MPa = p 4 J 2 (0.025 )
Ans.
268
B
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5–78. The A-36 steel shaft has a diameter of 60 mm and is fixed at its ends A and B.If it is subjected to the torques shown, determine the absolute maximum shear stress in the shaft.
200 N⭈m B 500 N⭈m
D 1.5 m
C A
Referring to the FBD of the shaft shown in Fig. a, TA + TB - 500 - 200 = 0
©Mx = 0;
(1)
Using the method of superposition, Fig. b fA = (fA)TA - (fA)T 0 =
500 (1.5) 700 (1) TA (3.5) - c + d JG JG JG TA = 414.29 N # m
Substitute this result into Eq (1), TB = 285.71 N # m Referring to the torque diagram shown in Fig. c, segment AC is subjected to maximum internal torque. Thus, the absolute maximum shear stress occurs here. tAbs =
414.29 (0.03) TAC c = = 9.77 MPa p J (0.03)4 2
Ans.
269
1m
1m
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5–79. The steel shaft is made from two segments: AC has a diameter of 0.5 in, and CB has a diameter of 1 in. If it is fixed at its ends A and B and subjected to a torque of determine the maximum shear stress in the shaft. Gst = 10.811032 ksi.
A
0.5 in. C D 500 lbft
5 in.
1 in.
8 in.
B 12 in.
Equilibrium: TA + TB - 500 = 0
(1)
Compatibility condition: fD>A = fD>B TA(5) p 2
4
(0.25 )G
TA(8) +
p 2
4
(0.5 )G
TB(12) =
p 2
(0.54)G
1408 TA = 192 TB
(2)
Solving Eqs. (1) and (2) yields TA = 60 lb # ft
TB = 440 lb # ft
tAC =
60(12)(0.25) TC = 29.3 ksi = p 4 J 2 (0.25 )
tDB =
440(12)(0.5) TC = 26.9 ksi = p 4 J 2 (0.5 )
Ans.
(max)
270
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*5–80. The shaft is made of A-36 steel, has a diameter of 80 mm, and is fixed at B while A is loose and can rotate 0.005 rad before becoming fixed. When the torques are applied to C and D, determine the maximum shear stress in regions AC and CD of the shaft.
2 kN⭈m B
4 kN⭈m D
600 mm C 600 mm A
Referring to the FBD of the shaft shown in Fig. a,
600 mm
TA + TB + 2 - 4 = 0
©Mx = 0;
(1)
Using the method of superposition, Fig. b, fA = (fA)T - (uA)TA 0.005 = B p 2
4(103)(0.6)
(0.04 ) C 75(10 ) D 4
9
2(103)(0.6)
+
p 2
(0.04 ) C 75(10 ) D 4
9
R -
TA (1.8)
p 2
(0.044) C 75(109) D
TA = 1162.24 N # m = 1.162 kN # m Substitute this result into Eq (1), TB = 0.838 kN # m Referring to the torque diagram shown in Fig. c, segment CD is subjected to a maximum internal torque. Thus, the absolute maximum shear stress occurs here. t$$$ =
2.838 (103)(0.04) TCD c = = 28.23 (106) Pa = 28.2 MPa p 4 J 2 (0.04)
271
Ans.
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•5–81. The shaft is made of A-36 steel and has a diameter of 80 mm. It is fixed at B and the support at A has a torsional stiffness of k = 0.5 MN # m>rad. If it is subjected to the gear torques shown, determine the absolute maximum shear stress in the shaft.
2 kN⭈m B
4 kN⭈m D
600 mm C 600 mm A 600 mm
Referring to the FBD of the shaft shown in Fig. a, TA + TB + 2 - 4 = 0
©Mx = 0;
(1)
Using the method of superposition, Fig. b, fA = (fA)T - (fA)TA TA 6
0.5(10 )
= D
4(103)(0.6)
p 2
(0.04 ) C 75(10 ) D 4
9
2(103)(0.6)
+
p 2
(0.04 ) C 75(10 ) D 4
9
T -
TA(1.8)
p 2
(0.044) C 75(109) D
TA = 1498.01 N # m = 1.498 kN # m Substituting this result into Eq (1), TB = 0.502 kN # m Referring to the torque diagram shown in Fig. c, segment CD subjected to maximum internal torque. Thus, the maximum shear stress occurs here. t$$$ =
2.502(103)(0.04) TCD C = = $$$ = 24.9 MPa p 4 J 2 (0.04)
272
Ans.
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5–82. The shaft is made from a solid steel section AB and a tubular portion made of steel and having a brass core. If it is fixed to a rigid support at A, and a torque of T = 50 lb # ft is applied to it at C, determine the angle of twist that occurs at C and compute the maximum shear stress and maximum shear strain in the brass and steel. Take Gst = 11.511032 ksi, Gbr = 5.611032 ksi.
3 ft
2 ft A 0.5 in. B 1 in.
Equilibrium: Tbr + Tst - 50 = 0
(1)
Both the steel tube and brass core undergo the same angle of twist fC>B fC>B =
TL = JG
Tst (2)(12)
Tst (2)(12) p 2
(0.54)(5.6)(104)
=
p 2
4
(1 - 0.54)(11.5)(106)
Tbr = 0.032464 Tst
(2)
Solving Eqs. (1) and (2) yields: Tst = 48.428 lb # ft; fC = ©
Tbr = 1.572 lb # ft
50(12)(3)(12) 1.572(12)(2)(12) TL + p 4 = p 4 6 6 JG 2 (0.5 )(5.6)(10 ) 2 (1 )(11.5)(10 ) = 0.002019 rad = 0.116°
Ans.
(tst)max AB =
50(12)(1) TABc = 382 psi = p 4 J 2 (1 )
(tst)max BC =
48.428(12)(1) Tst c = 394.63 psi = 395 psi (Max) = p 4 4 J 2 (1 - 0.5 )
Ans.
(gst)max =
(tst)max 394.63 = 343.(10 - 6) rad = G 11.5(106)
Ans.
(tbr)max =
1.572(12)(0.5) Tbr c = 96.07 psi = 96.1 psi (Max) = p 4 J 2 (0.5 )
Ans.
(gbr)max =
(tbr)max 96.07 = 17.2(10 - 6) rad = G 5.6(106)
Ans.
273
C
T 50 lbft
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5–83. The motor A develops a torque at gear B of 450 lb # ft, which is applied along the axis of the 2-in.-diameter steel shaft CD. This torque is to be transmitted to the pinion gears at E and F. If these gears are temporarily fixed, determine the maximum shear stress in segments CB and BD of the shaft. Also, what is the angle of twist of each of these segments? The bearings at C and D only exert force reactions on the shaft and do not resist torque. Gst = 1211032 ksi.
B E 4 ft
(1)
Compatibility condition: fB>C = fB>D TC(4) TD(3) = JG JG TC = 0.75 TD
(2)
Solving Eqs. (1) and (2), yields TD = 257.14 lb # ft TC = 192.86 lb # ft
(tBD)max =
f =
192.86(12)(1) p 2
(14)
257.14(12)(1) p 2
(14)
192.86(12)(4)(12) p 2
(14)(12)(106)
3 ft D A
TC + TD - 450 = 0
(tBC)max =
F
C
Equilibrium:
= 1.47 ksi
Ans.
= 1.96 ksi
Ans.
= 0.00589 rad = 0.338°
Ans.
274
450 lb⭈ft
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*5–84. A portion of the A-36 steel shaft is subjected to a linearly distributed torsional loading. If the shaft has the dimensions shown, determine the reactions at the fixed supports A and C. Segment AB has a diameter of 1.5 in. and segment BC has a diameter of 0.75 in.
300 lb⭈in./in. A
60 in.
B
C 48 in.
Equilibrium: TA + TC - 9000 = 0 TR = t x +
1 tx (300 - t)x = 150x + 2 2
300 t = ; 60 - x 60
But
(1)
TR = 150 x +
t = 5(60 - x)
1 [5(60 - x)]x 2
= (300x - 2.5x2) lb # in. Compatibility condition: fB>A = fB>C fB>A =
60 T(x) dx 1 = [TA - (300x - 2.5x2)] dx JG L0 L JG
=
60 1 [TAx - 150x2 + 0.8333x3] | JG 0
=
60TA - 360 000 JG TC(48)
60TA - 360 000 p 2
(0.754)G
=
p 2
(0.3754)G (2)
60TA - 768TC = 360 000 Solving Eqs. (1) and (2) yields: TC = 217.4 lb # in. = 18.1 lb # ft
Ans.
TA = 8782.6 lb # in. = 732 lb # ft
Ans.
275
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•5–85.
Determine the rotation of joint B and the absolute maximum shear stress in the shaft in Prob. 5–84.
300 lb⭈in./in. A
Equilibrium: TA + TC - 9000 = 0 TR = tx +
But
(1)
1 tx (300 - t)x = 150x + 2 2
300 t = ; 60 - x 60
C
t = 5(60 - x)
= (300x - 2.5x2) lb # in. Compatibility condition: fB>A = fB>C fB>A =
60 T(x) dx 1 = [TA - (300x - 2.5x2)] dx JG L0 L JG
=
60 1 [TAx - 150x2 + 0.8333x3] | JG 0
=
60TA - 360 000 JG TC(48)
60TA - 360 000 p 2
=
4
(0.75 )G
p 2
(0.3754)G
60TA - 768TC = 360 000
(2)
Solving Eqs. (1) and (2) yields: TC = 217.4 lb # in. = 18.1 lb # ft TA = 8782.6 lb # in. = 732 lb # ft For segment BC: fB = fB>C =
TCL = JG
217.4(48) p 2
(0.375)4(11.0)(106)
= 0.030540 rad
fB = 1.75° tmax =
Ans.
217.4(0.375) TC = p = 2.62 ksi 4 J 2 (0.375)
For segment AB, tmax =
B
48 in.
1 [5(60 - x)]x 2
TR = 150x +
60 in.
8782.6(0.75) TC = p = 13.3 ksi 4 J 2 (0.75)
abs = 13.3 ksi tmax
Ans.
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5–86. The two shafts are made of A-36 steel. Each has a diameter of 25 mm and they are connected using the gears fixed to their ends. Their other ends are attached to fixed supports at A and B. They are also supported by journal bearings at C and D, which allow free rotation of the shafts along their axes. If a torque of 500 N # m is applied to the gear at E as shown, determine the reactions at A and B.
B F
D
50 mm
0.75 m
100 mm 500 Nm E
C 1.5 m A
Equilibrium: TA + F(0.1) - 500 = 0
[1]
TB - F(0.05) = 0
[2]
TA + 2TB - 500 = 0
[3]
From Eqs. [1] and [2]
Compatibility: 0.1fE = 0.05fF fE = 0.5fF TA(1.5) TB(0.75) = 0.5 c d JG JG TA = 0.250TB
[4]
Solving Eqs. [3] and [4] yields: TB = 222 N # m
Ans.
TA = 55.6 N # m
Ans.
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5–87. Determine the rotation of the gear at E in Prob. 5–86.
B F
D
50 mm
0.75 m
100 mm 500 Nm E
C 1.5 m A
Equilibrium: TA + F(0.1) - 500 = 0
[1]
TB - F(0.05) = 0
[2]
TA + 2TB - 500 = 0
[3]
From Eqs. [1] and [2]
Compatibility: 0.1fE = 0.05fF fE = 0.5fF TA(1.5) TB(0.75) = 0.5 c d JG JG TA = 0.250TB
[4]
Solving Eqs. [3] and [4] yields: TB = 222.22 N # m
TA = 55.56 N # m
Angle of Twist: fE =
TAL = JG
55.56(1.5) p 2
(0.01254)(75.0)(109)
= 0.02897 rad = 1.66°
Ans.
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*5–88. The shafts are made of A-36 steel and have the same diameter of 4 in. If a torque of 15 kip # ft is applied to gear B, determine the absolute maximum shear stress developed in the shaft.
2.5 ft 2.5 ft
Equilibrium: Referring to the free - body diagrams of shafts ABC and DE shown in Figs. a and b, respectively, we have ©Mx = 0; TA + F(0.5) - 15 = 0
(1)
and
A B 6 in. 15 kip⭈ft
C
D
12 in.
©Mx = 0; F(1) - TE = 0
E
(2) 3 ft
Internal Loadings: The internal torques developed in segments AB and BC of shaft ABC and shaft DE are shown in Figs. c, d, and e, respectively. Compatibility Equation: fCrC = fDrD
¢
TABLAB TDE LDE TBCLBC + ≤ rC = ¢ ≤ rD JGst JGst JGst
C -TA(2.5) + F(0.5)(2.5) D (0.5) = -TE(3)(1) TA - 0.5F = 2.4TE
(3)
Solving Eqs. (1), (2), and (3), we have F = 4.412 kip
TE = 4.412 kip # ft
TA = 12.79 kip # ft
Maximum Shear Stress: By inspection, segment AB of shaft ABC is subjected to the greater torque.
A tmax B abs =
12.79(12)(2) TAB c = 12.2 ksi = Jst p 4 a2 b 2
Ans.
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•5–89.
The shafts are made of A-36 steel and have the same diameter of 4 in. If a torque of 15 kip # ft is applied to gear B, determine the angle of twist of gear B.
2.5 ft
Equilibrium: Referring to the free - body diagrams of shafts ABC and DE shown in Figs. a and b, respectively, ©Mx = 0; TA + F(0.5) - 15 = 0
2.5 ft A B
(1)
6 in. 15 kip⭈ft
and ©Mx = 0; F(1) - TE = 0
(2)
Internal Loadings: The internal torques developed in segments AB and BC of shaft ABC and shaft DE are shown in Figs. c, d, and e, respectively. Compatibility Equation: It is required that fCrC = fDrD
¢
TAB LAB TDE LDE TBC LBC + ≤ rC = ¢ ≤ rD JGst JGst JGst
C -TA(2.5) + F(0.5)(2.5) D (0.5) = -TE(3)(1) TA - 0.5F = 2.4TE
(3)
Solving Eqs. (1), (2), and (3), F = 4.412 kip
TE = 4.412 kip # ft
TA = 12.79 kip # ft
Angle of Twist: Here, TAB = -TA = -12.79 kip # ft fB =
-12.79(12)(2.5)(12) TAB LAB = JGst p 4 a 2 b(11.0)(103) 2
= -0.01666 rad = 0.955°
Ans.
280
C
D
12 in. E 3 ft
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5–90. The two 3-ft-long shafts are made of 2014-T6 aluminum. Each has a diameter of 1.5 in. and they are connected using the gears fixed to their ends. Their other ends are attached to fixed supports at A and B. They are also supported by bearings at C and D, which allow free rotation of the shafts along their axes. If a torque of 600 lb # ft is applied to the top gear as shown, determine the maximum shear stress in each shaft.
A B
E
2 in.
TA + F a
4 b - 600 = 0 12
(1)
TB - F a
2 b = 0 12
(2)
From Eqs. (1) and (2) TA + 2TB - 600 = 0
TAL TBL = 0.5 a b; JG JG
(3)
fE = 0.5fF TA = 0.5TB
(4)
Solving Eqs. (3) and (4) yields: TB = 240 lb # ft;
TA = 120 lb # ft
(tBD)max =
240(12)(0.75) TB c = 4.35 ksi = p 4 J 2 (0.75 )
Ans.
(tAC)max =
120(12)(0.75) TA c = 2.17 ksi = p 4 J 2 (0.75 )
Ans.
281
3 ft D
4 in.
4(fE) = 2(fF);
C
600 lbft
F
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5–91. The A-36 steel shaft is made from two segments: AC has a diameter of 0.5 in. and CB has a diameter of 1 in. If the shaft is fixed at its ends A and B and subjected to a uniform distributed torque of 60 lb # in.>in. along segment CB, determine the absolute maximum shear stress in the shaft.
A
0.5 in. C
5 in.
60 lbin./in. 1 in. 20 in.
Equilibrium: TA + TB - 60(20) = 0
(1)
Compatibility condition: fC>B = fC>A fC>B =
20 (TB - 60x) dx T(x) dx = p JG L L0 2 (0.54)(11.0)(106)
= 18.52(10-6)TB - 0.011112 18.52(10-6)TB - 0.011112 =
TA(5) p 4 6 2 (0.25 )(11.0)(10 )
18.52(10-6)TB - 74.08(10-6)TA = 0.011112 (2)
18.52TB - 74.08TA = 11112 Solving Eqs. (1) and (2) yields: TA = 120.0 lb # in. ;
TB = 1080 lb # in.
(tmax)BC =
1080(0.5) TB c = 5.50 ksi = p 4 J 2 (0.5 )
(tmax)AC =
120.0(0.25) TA c = 4.89 ksi = p 4 J 2 (0.25 )
abs = 5.50 ksi tmax
Ans.
282
B
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*5–92. If the shaft is subjected to a uniform distributed torque of t = 20 kN # m>m, determine the maximum shear stress developed in the shaft. The shaft is made of 2014-T6 aluminum alloy and is fixed at A and C.
400 mm
20 kN⭈m/m 600 mm a A 80 mm 60 mm
B a C
Equilibrium: Referring to the free - body diagram of the shaft shown in Fig. a, we have ©Mx = 0; TA + TC - 20(103)(0.4) = 0
(1)
Compatibility Equation: The resultant torque of the distributed torque within the region x of the shaft is TR = 20(103)x N # m. Thus, the internal torque developed in the shaft as a function of x when end C is free is T(x) = 20(103)x N # m, Fig. b. Using the method of superposition, Fig. c, fC = A fC B t - A fC B TC 0 =
0 =
0.4 m
T(x)dx TCL JG JG
0.4 m
20(103)xdx TC(1) JG JG
L0 L0
0 = 20(103) ¢
x2 2 0.4 m - TC ≤ 2 0
TC = 1600 N # m Substituting this result into Eq. (1), TA = 6400 N # m Maximum Shear Stress: By inspection, the maximum internal torque occurs at support A. Thus,
A tmax B abs =
6400(0.04) TA c = 93.1 MPa = J p 4 4 a0.04 - 0.03 b 2
Ans.
283
Section a–a
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•5–93.
The tapered shaft is confined by the fixed supports at A and B. If a torque T is applied at its mid-point, determine the reactions at the supports. 2c
T
A
B
Equilibrium:
c
TA + TB - T = 0
[1] L/2 L/ 2
Section Properties: r(x) = c +
J(x) =
c c x = (L + x) L L
4 p c pc4 c (L + x) d = (L + x)4 2 L 2L4
Angle of Twist: fT =
Tdx = Lp2 L J(x)G
L
Tdx pc4 2L4
(L + x)4 G L
=
dx 2TL4 pc4 G Lp2 (L + x)4
= -
=
fB =
L 1 2TL4 c d 2 4 3 p 3pc G (L + x) 2
37TL 324 pc4 G
Tdx = J(x)G L L0
L
TBdx pc4 2L4
(L + x)4G L
2TBL4 =
dx pc G L0 (L + x)4 4
2TBL4
= -
L 1 d 2 3 3pc G (L + x) 0 4
c
7TB L =
12pc4G
Compatibility: 0 = fT - fB 0 =
7TBL 37TL 4 324pc G 12pc4G
TB =
37 T 189
Ans.
Substituting the result into Eq. [1] yields: TA =
152 T 189
Ans.
284
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5–94. The shaft of radius c is subjected to a distributed torque t, measured as torque>length of shaft. Determine the reactions at the fixed supports A and B.
B t0
(
t t0 1
( Lx ) 2 )
x L
2t0
x
x2 x2 T(x) = t0 a1 + 2 b dx = t0 ax + b L 3L2 L0
(1)
By superposition: 0 = fB - fB L
0 =
L0 TB =
A
t0 a x +
x 3L2 b 3
2
dx -
JG
TB(L) 7t0L = - TB(L) JG 12
7t0 L 12
Ans.
From Eq. (1), TA = t0 a L + TA +
4t0 L L3 b = 2 3 3L
7t0 L 4t0 L = 0 12 3 TA =
3t0 L 4
Ans.
285
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5–95. Compare the values of the maximum elastic shear stress and the angle of twist developed in 304 stainless steel shafts having circular and square cross sections. Each shaft has the same cross-sectional area of 9 in2, length of 36 in., and is subjected to a torque of 4000 lb # in. r A
Maximum Shear Stress: For circular shaft 1
A = pc2 = 9;
(tc)max =
9 2 c = a b p
2(4000) Tc Tc 2T = = 525 psi = p 4 = 1 3 J pc p A 9x B 2 2 c
Ans.
For rectangular shaft A = a2 = 9 ; (tr)max =
a = 3 in.
4.81(4000) 4.81T = = 713 psi 3 a 33
Ans.
Angle of Twist: For circular shaft fc =
TL = JG
4000(36)
p 2
A B 11.0(106) 9 2 p
Ans.
= 0.001015 rad = 0.0582° For rectangular shaft fr =
7.10(4000)(36) 7.10 TL = 4 a4 G 3 (11.0)(106) Ans.
= 0.001147 rad = 0.0657° The rectangular shaft has a greater maximum shear stress and angle of twist.
286
a
A a
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*5–96. If a = 25 mm and b = 15 mm, determine the maximum shear stress in the circular and elliptical shafts when the applied torque is T = 80 N # m. By what percentage is the shaft of circular cross section more efficient at withstanding the torque than the shaft of elliptical cross section?
b
a
a
For the circular shaft: (tmax)c =
80(0.025) Tc = 3.26 MPa = p 4 J 2 (0.025 )
Ans.
For the elliptical shaft: (tmax)c =
2(80) 2T = 9.05 MPa = p a b2 p(0.025)(0.0152)
Ans.
(tmax)c - (tmax)c (100%) (tmax)c
% more efficient =
9.05 - 3.26 (100%) = 178 % 3.26
=
Ans.
•5–97.
It is intended to manufacture a circular bar to resist torque; however, the bar is made elliptical in the process of manufacturing, with one dimension smaller than the other by a factor k as shown. Determine the factor by which the maximum shear stress is increased.
kd
For the circular shaft: (tmax)c =
d
TA B Tc 16T = = 3 p J AdB4 p d d 2
2 2
For the elliptical shaft: (tmax)c =
d
2T 2T 16T = = 2 p a b2 p k2 d3 p A d2 B A kd B 2
Factor of increase in shear stress =
=
(tmax)c = (tmax)c
16T p k2 d3 16T p d3
1 k2
Ans.
287
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5–98. The shaft is made of red brass C83400 and has an elliptical cross section. If it is subjected to the torsional loading shown, determine the maximum shear stress within regions AC and BC, and the angle of twist f of end B relative to end A.
A 20 Nm
50 Nm
30 Nm 2m
Maximum Shear Stress:
C
(tBC)max =
2(30.0)
2TBC p a b2
=
B
p(0.05)(0.022) Ans.
= 0.955 MPa (tAC)max =
2(50.0)
2TAC 2
50 mm 20 mm
1.5 m
=
pab
p(0.05)(0.022) Ans.
= 1.59 MPa Angle of Twist: fB>A = a
(a2 + b2)T L p a3b3 G (0.052 + 0.022)
=
p(0.053)(0.023)(37.0)(109)
[(-30.0)(1.5) + (-50.0)(2)]
= -0.003618 rad = 0.207°
Ans.
5–99. Solve Prob. 5–98 for the maximum shear stress within regions AC and BC, and the angle of twist f of end B relative to C.
A 20 Nm
50 Nm
30 Nm 2m
Maximum Shear Stress:
C
(tBC)max =
2(30.0)
2TBC p a b2
=
B
p(0.05)(0.022)
= 0.955 MPa (tAC)max =
2
Ans.
2(50.0)
2TAC =
pab
p(0.05)(0.022)
= 1.59 MPa
Ans.
Angle of Twist: fB>C =
(a2 + b2) TBC L p a3 b3 G (0.052 + 0.022)(-30.0)(1.5)
=
50 mm 20 mm
1.5 m
p(0.053)(0.023)(37.0)(109)
= -0.001123 rad = | 0.0643°|
Ans.
288
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*5–100. Segments AB and BC of the shaft have circular and square cross sections, respectively. If end A is subjected to a torque of T = 2 kN # m, determine the absolute maximum shear stress developed in the shaft and the angle of twist of end A. The shaft is made from A-36 steel and is fixed at C.
600 mm C 600 mm
90 mm
B
30 mm
90 mm
Internal Loadings: The internal torques developed in segments AB and BC are shown in Figs. a, and b, respectively. Maximum Shear Stress: For segment AB,
A tmax B AB =
2(103)(0.03) TAB c = 47.2 MPa (max) J p a 0.034 b 2
For segment BC,
A tmax B BC =
4.81TBC 3
=
a
4.81 C 2(103) D (0.09)3
Ans.
= 13.20 MPa
Angle of Twist: fA =
7.10TBCLBC TABLAB + JG a4G 2(103)(0.6)
=
p a 0.034 b(75)(109) 2
7.10(2)(103)(0.6) +
(0.09)4(75)(109)
= 0.01431 rad = 0.820°
Ans.
289
A T
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•5–101. Segments AB and BC of the shaft have circular and square cross sections, respectively. The shaft is made from A-36 steel with an allowable shear stress of tallow = 75 MPa, and an angle of twist at end A which is not allowed to exceed 0.02 rad. Determine the maximum allowable torque T that can be applied at end A.The shaft is fixed at C.
600 mm C 600 mm
Internal Loadings: The internal torques developed in segments AB and BC are shown in Figs. a, and b, respectively.
90 mm
B
30 mm
90 mm
Allowable Shear Stress: For segment AB,
A T
tallow =
TAB c ; J
75(106) =
T(0.03) p a0.034 b 2
T = 3180.86 N # m For segment BC, tallow =
4.81TBC a3
75(106) =
;
4.81T (0.09)3 T = 11 366.94 N # m
Angle of Twist: fA =
TABLAB 7.10TBC LBC + JG a4G
0.02 =
T(0.6)
7.10T(0.6)
p a 0.034 b(75)(109) 2
+
(0.09)4 (75)(109)
T = 2795.90 N # m = 2.80 kN # m (controls)
Ans.
5–102. The aluminum strut is fixed between the two walls at A and B. If it has a 2 in. by 2 in. square cross section, and it is subjected to the torque of 80 lb # ft at C, determine the reactions at the fixed supports. Also, what is the angle of twist at C? Gal = 3.811032 ksi.
A C 2 ft 80 lb⭈ft
By superposition:
3 ft
0 = f - fB 0 =
7.10(TB)(5)
7.10(80)(2) 4
-
a G
a4 G
TB = 32 lb # ft
Ans.
TA + 32 - 80 = 0 TA = 48 lb # ft fC =
7.10(32)(12)(3)(12) (24)(3.8)(106)
Ans. = 0.00161 rad = 0.0925°
Ans.
290
B
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5–103. The square shaft is used at the end of a drive cable in order to register the rotation of the cable on a gauge. If it has the dimensions shown and is subjected to a torque of 8 N # m, determine the shear stress in the shaft at point A. Sketch the shear stress on a volume element located at this point.
5 mm A 5 mm
Maximum shear stress:
8 Nm
(tmax)A =
4.81(8) 4.81T = = 308 MPa a3 (0.005)3
Ans.
*5–104. The 6061-T6 aluminum bar has a square cross section of 25 mm by 25 mm. If it is 2 m long, determine the maximum shear stress in the bar and the rotation of one end relative to the other end.
C 1.5 m 20 N⭈m B
Maximum Shear Stress: tmax =
0.5 m A
4.81(80.0)
4.81Tmax a3
=
(0.0253)
Ans.
= 24.6 MPa
60 N·m 25 mm
Angle of Twist: 7.10(-20.0)(1.5) 7.10(-80.0)(0.5) 7.10TL fA>C = a 4 = + 4 9 aG (0.025 )(26.0)(10 ) (0.0254)(26.0)(109) = -0.04894 rad = | 2.80° |
Ans.
291
80 N⭈m 25 mm
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•5–105.
The steel shaft is 12 in. long and is screwed into the wall using a wrench. Determine the largest couple forces F that can be applied to the shaft without causing the steel to yield. tY = 8 ksi.
1 in. 12 in.
F(16) - T = 0 tmax = tY = 8(103) =
(1)
4.81T a3
F
8 in.
4.81T (1)3
1 in.
8 in.
T = 1663.2 lb # in.
F
From Eq. (1), F = 104 lb
Ans.
5–106. The steel shaft is 12 in. long and is screwed into the wall using a wrench. Determine the maximum shear stress in the shaft and the amount of displacement that each couple force undergoes if the couple forces have a magnitude of F = 30 lb, Gst = 10.811032 ksi.
1 in. 12 in.
T - 30(16) = 0 F
T = 480 lb # in. tmax =
4.81(480) 4.18T = a3 (1)3
1 in.
8 in.
Ans.
= 2.31 ksi f =
8 in.
7.10(480)(12) 7.10TL = = 0.00379 rad a4 G (1)4(10.8)(106)
dF = 8(0.00397) = 0.0303 in.
Ans.
292
F
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5–107. Determine the constant thickness of the rectangular tube if the average shear stress is not to exceed 12 ksi when a torque of T = 20 kip # in. is applied to the tube. Neglect stress concentrations at the corners. The mean dimensions of the tube are shown.
T 4 in.
Am = 2(4) = 8 in2 tavg =
12 =
T 2 t Am
2 in.
20 2 t (8)
t = 0.104 in.
Ans.
*5–108. Determine the torque T that can be applied to the rectangular tube if the average shear stress is not to exceed 12 ksi. Neglect stress concentrations at the corners. The mean dimensions of the tube are shown and the tube has a thickness of 0.125 in.
T 4 in.
Am = 2(4) = 8 in2 tavg =
T ; 2 t Am
12 =
2 in.
T 2(0.125)(8)
T = 24 kip # in. = 2 kip # ft
Ans.
293
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•5–109.
For a given maximum shear stress, determine the factor by which the torque carrying capacity is increased if the half-circular section is reversed from the dashed-line position to the section shown. The tube is 0.1 in. thick.
1.80 in. 0.6 in. 1.20 in. 0.5 in.
Am
p(0.552) = 1.4498 in2 = (1.10)(1.75) 2
Am ¿ = (1.10)(1.75) +
tmax =
p(0.552) = 2.4002 in2 2
T 2t Am
T = 2 t Am tmax Factor =
=
2t Am ¿ tmax 2t Am tmax Am ¿ 2.4002 = = 1.66 Am 1.4498
Ans.
5–110. For a given average shear stress, determine the factor by which the torque-carrying capacity is increased if the half-circular sections are reversed from the dashed-line positions to the section shown. The tube is 0.1 in. thick.
1.80 in. 0.6 in. 1.20 in. 0.5 in.
Section Properties: œ Am = (1.1)(1.8) - B
p (0.552) R (2) = 1.02967 in2 2
Am = (1.1)(1.8) + B
p (0.552) R (2) = 2.93033 in2 2
Average Shear Stress: tavg = Hence,
T ; 2 t Am
T = 2 t Am tavg œ tavg T¿ = 2 t Am
The factor of increase =
Am 2.93033 T = œ = T¿ Am 1.02967
= 2.85
Ans.
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5–111. A torque T is applied to two tubes having the cross sections shown. Compare the shear flow developed in each tube.
t t t
Circular tube:
a
T T 2T = = 2Am 2p (a>2)2 p a2
qct =
a
a
Square tube: qst =
T T = 2Am 2a2
qst T>(2a2) p = = qct 4 2T>(p a2) Thus; qst =
p q 4 ct
Ans.
*5–112. Due to a fabrication error the inner circle of the tube is eccentric with respect to the outer circle. By what percentage is the torsional strength reduced when the eccentricity e is one-fourth of the difference in the radii?
ab 2 a
Average Shear Stress:
e 2
For the aligned tube tavg =
T T = 2 t Am 2(a - b)(p) A a
T = tavg (2)(a - b)(p)a
B
+ b 2 2
a + b 2 b 2
For the eccentric tube tavg =
b
T¿ 2 t Am
t = a -
= a -
e e - a + bb = a - e - b 2 2 1 3 (a - b) - b = (a - b) 4 4
3 a + b 2 b T¿ = tavg (2)c (a - b) d(p)a 4 2 Factor =
tavg (2) C 34 (ab) D (p) A a T¿ = T tavg (2)(a - b)(p) A a
Percent reduction in strength = a1 -
B
+ b 2 2
B
+ b 2 2
=
3 4
3 b * 100 % = 25 % 4
295
Ans.
e 2
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•5–113.
The mean dimensions of the cross section of an airplane fuselage are shown. If the fuselage is made of 2014-T6 aluminum alloy having allowable shear stress of tallow = 18 ksi, and it is subjected to a torque of 6000 kip # ft, determine the required minimum thickness t of the cross section to the nearest 1>16 in. Also, find the corresponding angle of twist per foot length of the fuselage.
t 3 ft
4.5 ft
3 ft
Section Properties: Referring to the geometry shown in Fig. a, Am = p A 32 B + 4.5(6) = 55.2743 ft2 ¢ F
144 in2 ≤ = 7959.50 in2 1 ft2
ds = 2p(3) + 2(4.5) = 27.8496 fta
12 in. b = 334.19 in. 1 ft
Allowable Average Shear Stress:
A tavg B allow =
T ; 2tAm
18 =
6000(12) 2t(7959.50)
t = 0.2513 in. =
Angle of Twist: Using the result of t =
f = ©
Ans.
5 in, 16
ds TL 4Am 2G F t 6000(12)(1)(12)
=
5 in. 16
4(7959.502)(3.9)(103)
¢
334.19 ≤ 5>16
= 0.9349(10 - 3) rad = 0.0536°
Ans.
296
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5–114. The mean dimensions of the cross section of an airplane fuselage are shown. If the fuselage is made from 2014-T6 aluminum alloy having an allowable shear stress of tallow = 18 ksi and the angle of twist per foot length of fuselage is not allowed to exceed 0.001 rad>ft, determine the maximum allowable torque that can be sustained by the fuselage. The thickness of the wall is t = 0.25 in.
t 3 ft
4.5 ft
3 ft
Section Properties: Referring to the geometry shown in Fig. a, Am = p A 32 B + 4.5(6) = 55.2743 ft2 ¢ F
144 in2 ≤ = 7959.50 in2 1 ft2
ds = 2p(3) + 2(4.5) = 27.8496 fta
12 in. b = 334.19 in. 1 ft
Allowable Average Shear Stress:
A tavg B allow =
T ; 2tAm
18 =
T 2(0.25)(7959.50)
T = 71635.54 kip # ina
1ft b = 5970 kip # ft 12 in.
Angle of Twist: f =
ds TL 4Am 2G F t
0.001 =
T(1)(12) 4(7959.502)(3.9)(103)
T = 61610.65 kip # ina
a
334.19 b 0.25
1ft b = 5134 kip # ft (controls) 12 in.
Ans.
297
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5–115. The tube is subjected to a torque of 750 N # m. Determine the average shear stress in the tube at points A and B.
4 mm 6 mm
A
100 mm
Referring to the geometry shown in Fig. a,
6 mm 2
Am = 0.06 (0.1) = 0.006 m
B 750 N⭈m
Thus, (tavg)A
T 750 = = = 15.63(106)Pa = 15.6 MPa 2tA Am 2(0.004)(0.006)
Ans.
T 750 = = 10.42(106)Pa = 10.4 MPa 2tB Am 2(0.006)(0.006)
Ans.
(tavg)B =
4 mm 60 mm
*5–116. The tube is made of plastic, is 5 mm thick, and has the mean dimensions shown. Determine the average shear stress at points A and B if it is subjected to the torque of T = 5 N # m. Show the shear stress on volume elements located at these points. Am = (0.11)(0.08) +
tA = tB = tavg =
A
1 (0.08)(0.03) = 0.01 m2 2
B
50 mm 60 mm
T 5 = = 50 kPa 2tAm 2(0.005)(0.01)
T
Ans. 30 mm 40 mm
298
40 mm
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•5–117.
The mean dimensions of the cross section of the leading edge and torsion box of an airplane wing can be approximated as shown. If the wing is made of 2014-T6 aluminum alloy having an allowable shear stress of tallow = 125 MPa and the wall thickness is 10 mm, determine the maximum allowable torque and the corresponding angle of twist per meter length of the wing.
10 mm 0.5 m
10 mm
Section Properties: Referring to the geometry shown in Fig. a, Am =
F
p 1 a 0.52 b + A 1 + 0.5 B (2) = 1.8927 m2 2 2
ds = p(0.5) + 2 222 + 0.252 + 0.5 = 6.1019 m
Allowable Average Shear Stress:
A tavg B allow =
T ; 2tAm
125(106) =
T 2(0.01)(1.8927)
T = 4.7317(106)N # m = 4.73 MN # m
Ans.
Angle of Twist: f =
ds TL 2 4Am G F t 4.7317(106)(1)
=
4(1.89272)(27)(109)
¢
0.25 m 10 mm
6.1019 ≤ 0.01
= 7.463(10 - 3) rad = 0.428°>m
Ans.
299
2m
0.25 m
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5–118. The mean dimensions of the cross section of the leading edge and torsion box of an airplane wing can be approximated as shown. If the wing is subjected to a torque of 4.5 MN # m and the wall thickness is 10 mm, determine the average shear stress developed in the wing and the angle of twist per meter length of the wing. The wing is made of 2014-T6 aluminum alloy.
10 mm 0.5 m
10 mm
Section Properties: Referring to the geometry shown in Fig. a, Am =
F
p 1 a0.52 b + A 1 + 0.5 B (2) = 1.8927 m2 2 2
ds = p(0.5) + 2222 + 0.252 + 0.5 = 6.1019 m
Average Shear Stress: tavg =
4.5(106) T = = 119 MPa 2tAm 2(0.01)(1.8927)
Ans.
Angle of Twist: f =
ds TL 4Am 2G F t 4.5(106)(1)
=
4(1.89272)(27)(109)
¢
0.25 m 10 mm
6.1019 ≤ 0.01
= 7.0973(10 - 3) rad = 0.407°>m
Ans.
300
2m
0.25 m
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5–119. The symmetric tube is made from a high-strength steel, having the mean dimensions shown and a thickness of 5 mm. If it is subjected to a torque of T = 40 N # m, determine the average shear stress developed at points A and B. Indicate the shear stress on volume elements located at these points.
20 mm
30 mm
60 mm A B
40 Nm
Am = 4(0.04)(0.06) + (0.04)2 = 0.0112 m2 tavg =
T 2 t Am
(tavg)A = (tavg)B =
40 = 357 kPa 2(0.005)(0.0112)
Ans.
*5–120. The steel used for the shaft has an allowable shear stress of tallow = 8 MPa. If the members are connected with a fillet weld of radius r = 4 mm, determine the maximum torque T that can be applied.
50 mm
20 mm
T 2
Allowable Shear Stress: D 50 = = 2.5 d 20
and
4 r = = 0.20 d 20
From the text, K = 1.25 tmax = tallow = K
Tc J t 2 (0.01) 4 R 2 (0.01 )
8(10)4 = 1.25 B p T = 20.1 N # m
Ans.
301
T
20 mm
T 2
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•5–121.
The built-up shaft is to be designed to rotate at 720 rpm while transmitting 30 kW of power. Is this possible? The allowable shear stress is tallow = 12 MPa. v = 720
T =
75 mm
rev 2p rad 1 min a b = 24 p rad>s min 1 rev 60 s
60 mm
30(103) P = = 397.89 N # m v 24 p
tmax = K
Tc ; J
12(106) = Kc
397.89(0.03) p 4 2 (0.03 )
d;
K = 1.28
D 75 = = 1.25 d 60 From Fig. 5-32,
r = 0.133 d
r = 0.133 ; 60
r = 7.98 mm
Check: D - d 75 - 60 15 = = = 7.5 mm 6 7.98 mm 2 2 2 No, it is not possible.
Ans.
5–122. The built-up shaft is designed to rotate at 540 rpm. If the radius of the fillet weld connecting the shafts is r = 7.20 mm, and the allowable shear stress for the material is tallow = 55 MPa, determine the maximum power the shaft can transmit. D 75 = = 1.25; d 60
75 mm
60 mm
r 7.2 = = 0.12 d 60
From Fig. 5-32, K = 1.30 tmax = K
v = 540
Tc ; J
T(0.03) d; 55(106) = 1.30 c[ p 4 2 (0.03 )
T = 1794.33 N # m
rev 2p rad 1 min a b = 18 p rad>s min 1 rev 60 s
P = Tv = 1794.33(18p) = 101466 W = 101 kW
Ans.
302
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5–123. The steel shaft is made from two segments: AB and BC, which are connected using a fillet weld having a radius of 2.8 mm. Determine the maximum shear stress developed in the shaft.
C 50 mm
D
100 N⭈m
20 mm B
(tmax)CD =
100(0.025) TCDc = p 4 J 2 (0.025 )
40 N⭈m
A
= 4.07 MPa
60 N⭈m
For the fillet: D 50 = = 2.5; d 20
r 2.8 = = 0.14 d 20
From Fig. 5-32, K = 1.325 (tmax)f = K
60(0.01) TABc d = 1.325 c p 4 J 2 (0.01 ) = 50.6 MPa (max)
Ans.
*5–124. The steel used for the shaft has an allowable shear stress of tallow = 8 MPa. If the members are connected together with a fillet weld of radius r = 2.25 mm, determine the maximum torque T that can be applied.
30 mm
30 mm
15 mm
T T 2
Allowable Shear Stress: D 30 = = 2 d 15
r 2.25 = = 0.15 d 15
and
From the text, K = 1.30 tmax = tallow = K
Tc J
8(106) = 1.3 C
A 2r B (0.0075) p 4 2 (0.0075 )
S
T = 8.16 N # m
Ans.
303
T 2
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•5–125. The assembly is subjected to a torque of 710 lb # in. If the allowable shear stress for the material is tallow = 12 ksi, determine the radius of the smallest size fillet that can be used to transmit the torque.
tmax = tallow = K
0.75 in. A
Tc J
710 lb⭈in.
B 1.5 in.
3
12(10 ) =
K(710)(0.375) p 4 2 (0.375 )
C
K = 1.40 710 lb⭈ft
D 1.5 = = 2 d 0.75 From Fig. 5-32, r = 0.1; d
r = 0.1(0.75) = 0.075 in.
Ans.
Check: D - d 1.5 - 0.75 = = 0.375 7 0.075 in. 2 2
OK
5–126. A solid shaft is subjected to the torque T, which causes the material to yield. If the material is elastic plastic, show that the torque can be expressed in terms of the angle of twist f of the shaft as T = 43 TY11 - f3Y>4f32, where TY and fY are the torque and angle of twist when the material begins to yield. gY gL = L r rY
f =
rY =
gYL f
(1)
When rY = c, f = fY From Eq. (1), c =
gYL fY
(2)
Dividing Eq. (1) by Eq. (2) yields: rY fY = c f
(3)
Use Eq. 5-26 from the text. T =
r3Y p tY 2p tYc3 (4 c3 - r3Y) = a1 )b 6 3 4 c3
Use Eq. 5-24, TY =
T =
p t c3 from the text and Eq. (3) 2 Y
f3Y 4 TY a1 b 3 4 f3
QED
304
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5–127. A solid shaft having a diameter of 2 in. is made of elastic-plastic material having a yield stress of tY = 16 ksi and shear modulus of G = 1211032 ksi. Determine the torque required to develop an elastic core in the shaft having a diameter of 1 in. Also, what is the plastic torque?
Use Eq. 5-26 from the text: T =
p (16) p tY (4 c3 - rY 3) = [4(13) - 0.53] 6 6
= 32.46 kip # in. = 2.71 kip # ft
Ans.
Use Eq. 5-27 from the text: TP =
2p 2p t c3 = (16)(13) 3 Y 3
= 33.51 kip # in. = 2.79 kip # ft
Ans.
*5–128. Determine the torque needed to twist a short 3-mm-diameter steel wire through several revolutions if it is made from steel assumed to be elastic plastic and having a yield stress of tY = 80 MPa. Assume that the material becomes fully plastic.
When the material becomes fully plastic then, from Eq. 5-27 in the text, TP =
2 p (80)(106) 2 p tY 3 c = (0.00153) = 0.565 N # m 3 3
305
Ans.
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•5–129.
The solid shaft is made of an elastic-perfectly plastic material as shown. Determine the torque T needed to form an elastic core in the shaft having a radius of rY = 20 mm. If the shaft is 3 m long, through what angle does one end of the shaft twist with respect to the other end? When the torque is removed, determine the residual stress distribution in the shaft and the permanent angle of twist.
80 mm T
T
t (MPa) 160
Elastic-Plastic Torque: Applying Eq. 5-26 from the text T =
=
0.004
p tY A 4c3 - r3Y B 6 p(160)(106) C 4 A 0.043 B - 0.023 D 6
= 20776.40 N # m = 20.8 kN # m
Ans.
Angle of Twist: gY 0.004 L = a b(3) = 0.600 rad = 34.4° rY 0.02
f =
Ans.
When the reverse T = 20776.4 N # m is applied, G =
160(106) = 40 GPa 0.004
f¿ =
TL = JG
20776.4(3) p 4 9 2 (0.04 )(40)(10 )
= 0.3875 rad
The permanent angle of twist is, fr = f - f¿ = 0.600 - 0.3875 = 0.2125 rad = 12.2°
Ans.
Residual Shear Stress: (t¿)r = c =
20776.4(0.04) Tc = 206.67 MPa = p 4 J 2 (0.04 )
(t¿)r = 0.02 m =
20776.4(0.02) Tc = 103.33 MPa = p 4 J 2 (0.04 )
(tr)r = c = -160 + 206.67 = 46.7 MPa (tr)r = 0.02m = -160 + 103.33 = -56.7 MPa
306
g (rad)
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5–130. The shaft is subjected to a maximum shear strain of 0.0048 rad. Determine the torque applied to the shaft if the material has strain hardening as shown by the shear stress–strain diagram. 2 in.
From the shear - strain diagram, rY 2 = ; 0.0006 0.0048
T
rY = 0.25 in.
t (ksi) 12
From the shear stress–strain diagram, t1 =
6 r = 24r 0.25
6
t2 - 6 12 - 6 = ; r - 0.25 2 - 0.25
t2 = 3.4286 r + 5.1429
0.0006
0.0048
c
T = 2p
L0
t r2 dr 0.25
= 2p
2
24r3 dr + 2p
L0
= 2p[6r4] | + 2p c 0.25 0
L0.25
(3.4286r + 5.1429)r2 dr
3.4286r4 5.1429r3 2 + d | 4 3 0.25
= 172.30 kip # in. = 14.4 kip # ft
Ans.
5–131. An 80-mm diameter solid circular shaft is made of an elastic-perfectly plastic material having a yield shear stress of tY = 125 MPa. Determine (a) the maximum elastic torque TY; and (b) the plastic torque Tp. Maximum Elastic Torque. TY =
=
1 3 pc tY 2 1 pa 0.043 b A 125 B a 106 b 2
= 12 566.37 N # m = 12.6 kN # m
Ans.
Plastic Torque. TP =
=
2 3 pc tY 3 2 pa 0.043 b A 125 B a 106 b 3
= 16755.16 N # m = 16.8 kN # m
Ans. 307
g (rad)
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*5–132. The hollow shaft has the cross section shown and is made of an elastic-perfectly plastic material having a yield shear stress of tY. Determine the ratio of the plastic torque Tp to the maximum elastic torque TY.
c c 2
Maximum Elastic Torque. In this case, the torsion formula is still applicable. tY =
TY c J
TY =
J t c Y c 4 p 4 B c - a b R tY 2 2
=
=
c 15 3 pc tY 32
Plastic Torque. Using the general equation, with t = tY, c
TP = 2ptY
r2dr Lc>2 c
r3 = 2ptY ¢ ≤ ` 3 c>2 =
7 pc3tY 12
The ratio is 7 pc3tY TP 12 = = 1.24 TY 15 3 pc tY 32
Ans.
5–133. The shaft consists of two sections that are rigidly connected. If the material is elastic plastic as shown, determine the largest torque T that can be applied to the shaft. Also, draw the shear-stress distribution over a radial line for each section. Neglect the effect of stress concentration.
T 1 in.
0.75 in. T
0.75 in. diameter segment will be fully plastic. From Eq. 5-27 of the text: T = Tp =
2p tY 3 (c ) 3
t (ksi) 12
3
=
2p (12)(10 ) (0.3753) 3
= 1325.36 lb # in. = 110 lb # ft
Ans.
For 1 – in. diameter segment: tmax =
1325.36(0.5) Tc = p 4 J 2 (0.5)
= 6.75 ksi 6 tY
308
0.005
g (rad)
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5–134. The hollow shaft is made of an elastic-perfectly plastic material having a shear modulus of G and a yield shear stress of tY. Determine the applied torque Tp when the material of the inner surface is about to yield (plastic torque). Also, find the corresponding angle of twist and the maximum shear strain. The shaft has a length of L.
ci
Plastic Torque. Using the general equation with t = tY, co
TP = 2ptY
Lci
r2dr co
r3 = 2ptY ¢ ≤ ` 3 ci =
2 pt A c 3 - ci 3 B 3 Y o
Ans.
Angle of Twist. When the material is about to yield at the inner surface, g = gY at r = rY = ci. Also, Hooke’s Law is still valid at the inner surface. gY =
f =
tY G gY tY>G tYL L = L = rY ci ciG
Ans.
Shear Strain. Since the shear strain varies linearly along the radial line, Fig. a, gmax gY = co ci gmax = ¢
co co tY cotY ≤ gY = ¢ ≤ a b = ci ci G ciG
Ans.
309
c0
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5–135. The hollow shaft has inner and outer diameters of 60 mm and 80 mm, respectively. If it is made of an elasticperfectly plastic material, which has the t-g diagram shown, determine the reactions at the fixed supports A and C.
150 mm 450 mm B
C 15 kN⭈m
A
t (MPa)
120
Equation of Equilibrium. Refering to the free - body diagram of the shaft shown in Fig. a, ©Mx = 0; TA + TC - 15 A 103 B = 0
(1)
Elastic Analysis. It is required that fB>A = fB>C. Thus, the compatibility equation is fB>A = fB>C TCLBC TALAB = JG JG TA (0.45) = TC(0.15) TC = 3TA
(2)
Solving Eqs. (1) and (2), TA = 3750 N # m
TC = 11 250N # m
The maximum elastic torque and plastic torque in the shaft can be determined from p A 0.044 - 0.034 B J 2 T(120) A 106 B = 8246.68 N # m TY = tY = D c 0.04
co
TP = 2ptY
Lci
r2dr
= 2p(120) A 106 B ¢
g (rad) 0.0016
0.04 m
r3 = 9299.11 N # m ≤` 3 0.03 m
Since TC 7 TY, the results obtained using the elastic analysis are not valid. Plastic Analysis. Assuming that segment BC is fully plastic, TC = TP = 9299.11N # m = 9.3kN # m
Ans.
Substituting this result into Eq. (1), TA = 5700 N # m = 5.70 kN # m
Ans.
310
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5–135.
Continued
Since TA 6 TY, segment AB of the shaft is still linearly elastic. Here, 120 A 106 B = 75GPa. G = 0.0016 fB>C = fB>A =
fB>C =
gi L ; ci BC
5700.89(0.45) TALAB = 0.01244 rad = p JG A 0.044 - 0.034 B (75) A 109 B 2 0.01244 =
gi (0.15) 0.03
gi = 0.002489 rad Since gi 7 gY, segment BC of the shaft is indeed fully plastic.
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*5–136. The tubular shaft is made of a strain-hardening material having a t-g diagram as shown. Determine the torque T that must be applied to the shaft so that the maximum shear strain is 0.01 rad.
T 0.5 in. 0.75 in.
t (ksi) 15 10
0.005
From the shear–strain diagram, g 0.01 = ; 0.5 0.75
g = 0.006667 rad
From the shear stress–strain diagram, 15 - 10 t - 10 = ; t = 11.667 ksi 0.006667 - 0.005 0.01 - 0.005 15 - 11.667 t - 11.667 = ; r - 0.5 0.75 - 0.50
t = 13.333 r + 5
co
T = 2p
tr2 dr
Lci
0.75
= 2p
(13.333r + 5) r2 dr
L0.5 0.75
= 2p
L0.5
= 2p c
(13.333r3 + 5r2) dr
13.333r4 5r3 0.75 + d | 4 3 0.5
= 8.426 kip # in. = 702 lb # ft
Ans.
312
0.01
g (rad)
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•5–137.
The shear stress–strain diagram for a solid 50-mm-diameter shaft can be approximated as shown in the figure. Determine the torque T required to cause a maximum shear stress in the shaft of 125 MPa. If the shaft is 1.5 m long, what is the corresponding angle of twist?
T
1.5 m T t (MPa) 125 50
Strain Diagram: rg 0.0025
=
0.025 ; 0.01
0.0025
rg = 0.00625 m
Stress Diagram: t1 =
50(106) r = 8(109) r 0.00625
t2 - 50(106) 125(106) - 50(106) = r - 0.00625 0.025 - 0.00625 t2 = 4 A 109 B r + 25 A 106 B The Ultimate Torque: c
T = 2p
L0
t r2dr 0.00625 m
= 2p
L0
8 A 109 B r3 dr 0.025 m
+ 2p
L0.00625 m
9 6 C 4 A 10 B r + 25 A 10 B D r2dr
m = 2p C 2 A 109 B r4 D |0.00625 0
+ 2p B 1 A 109 B r4 +
25(106)r3 0.025 m R 2 3 0.00625 m
= 3269.30 N # m = 3.27 kN # m
Ans.
Angle of Twist: f =
gmax 0.01 L = a b (1.5) = 0.60 rad = 34.4° c 0.025
Ans.
313
0.010
g (rad)
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5–138. A tube is made of elastic-perfectly plastic material, which has the t-g diagram shown. If the radius of the elastic core is rY = 2.25 in., determine the applied torque T. Also, find the residual shear-stress distribution in the shaft and the permanent angle of twist of one end relative to the other when the torque is removed.
3 ft
3 in. T
T
6 in.
t (ksi)
Elastic - Plastic Torque. The shear stress distribution due to T is shown in Fig. a. The 10 r = 4.444r. Thus, linear portion of this distribution can be expressed as t = 2.25 tr = 1.5 in. = 4.444(1.5) = 6.667 ksi.
T = 2p
L
tr2dr
g (rad) 0.004
2.25 in.
= 2p
L1.5 in.
= 8.889p ¢
4.444r A r2dr B + 2p(10)
3 in.
L2.25 in.
r2dr
r4 2.25 in. r3 3 in. + 20p ¢ ≤ 2 ≤2 4 1.5 in. 3 2.25 in.
= 470.50 kip # in = 39.2 kip # ft
Ans.
Angle of Twist. f =
gY 0.004 L = (3)(12) = 0.064 rad rY 2.25
The process of removing torque T is equivalent to the application of T¿ , which is equal magnitude but opposite in sense to that of T. This process occurs in a linear 10 = 2.5 A 103 B ksi. manner and G = 0.004 f¿ =
10
T¿L = JG
470.50(3)(2)
p 2
A 34 - 1.54 B (2.5) A 103 B 470.50(3)
= 0.0568 rad
trœ = co =
T¿co = J
trœ = rY =
T¿rY 470.50(2.25) = = 8.875 ksi p 4 4 J 2 A 3 - 1.5 B
trœ = ci =
470.50(1.5) T¿ci = = 5.917 ksi p 4 4 J 2 A 3 - 1.5 B
p 2
A 34 - 1.54 B
= 11.83 ksi
Thus, the permanent angle of twist is fP = f - f¿ = 0.064 - 0.0568 = 0.0072 rad = 0.413°
Ans.
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5–138.
Continued
And the residual stresses are (tr)r = co = tr = c + trœ = c = -10 + 11.83 = 1.83 ksi (tr)r = rY = tr = rY + trœ = rY = -10 + 8.875 = -1.125 ksi (tr)r = ci = tr = ci + trœ = ci = -6.667 + 5.917 = -0.750 ksi The residual stress distribution is shown in Fig. a.
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5–139. The tube is made of elastic-perfectly plastic material, which has the t-g diagram shown. Determine the torque T that just causes the inner surface of the shaft to yield. Also, find the residual shear-stress distribution in the shaft when the torque is removed.
3 ft
3 in. T
T
6 in.
t (ksi)
Plastic Torque. When the inner surface of the shaft is about to yield, the shaft is about to become fully plastic. T = 2p
L
10
tr2dr 3 in.
= 2ptY
L1.5 in.
= 2p(10)a
g (rad)
r2dr
0.004
r3 3 in. b2 3 1.5 in.
= 494.80 kip # in. = 41.2 kip # ft
Ans.
Angle of Twist. f =
gY 0.004 (3)(12) = 0.096 rad L = rY 1.5
The process of removing torque T is equivalent to the application of T¿ , which is equal magnitude but opposite in sense to that of T. This process occurs in a linear 10 manner and G = = 2.5 A 103 B ksi. 0.004 f¿ =
494.80(3)(12) T¿L = = 0.05973 rad p 4 JG A 3 - 1.54 B (2.5) A 103 B 2
trœ = co =
trœ = ci =
494.80(3) T¿co = = 12.44 ksi p 4 J 4 3 1.5 A B 2 494.80(1.5) T¿ci = = 6.222 ksi p 4 J A 3 - 1.54 B 2
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5–139.
Continued
And the residual stresses are (tr)r = co = tr = c + trœ = c = -10 + 12.44 = 2.44 ksi
Ans.
(tr)r = ci = tr = ci + trœ = ci = -10 + 6.22 = -3.78 ksi
Ans.
The shear stress distribution due to T and T¿ and the residual stress distribution are shown in Fig. a.
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*5–140. The 2-m-long tube is made of an elastic-perfectly plastic material as shown. Determine the applied torque T that subjects the material at the tube’s outer edge to a shear strain of gmax = 0.006 rad. What would be the permanent angle of twist of the tube when this torque is removed? Sketch the residual stress distribution in the tube.
T 35 mm
30 mm t (MPa) 210
Plastic Torque: The tube is fully plastic if gi Ú gr = 0.003 rad. g 0.006 = ; 0.03 0.035
0.003
g = 0.005143 rad
Therefore the tube is fully plastic. co
TP = 2p
Lci
2p tg =
=
3
tg r2 dr
A c3o - c3i B
2p(210)(106) A 0.0353 - 0.033 B 3
= 6982.19 N # m = 6.98 kN # m
Ans.
Angle of Twist: fP =
gmax 0.006 L = a b(2) = 0.34286 rad co 0.035
When a reverse torque of TP = 6982.19 N # m is applied, G =
fPœ =
210(106) tY = = 70 GPa gY 0.003 TPL = JG
6982.19(2) p 4 2 (0.035
- 0.034)(70)(109)
= 0.18389 rad
Permanent angle of twist, fr = fP - fPœ = 0.34286 - 0.18389 = 0.1590 rad = 9.11°
Ans.
Residual Shear Stress: 6982.19(0.035)
tPœ o =
TP c = J
p 4 2 (0.035
tPœ i =
TP r = J
p 4 2 (0.035
- 0.034)
6982.19(0.03) - 0.034)
= 225.27 MPa
= 193.09 MPa
(tP)o = -tg + tPœ o = -210 + 225.27 = 15.3 MPa (tP)i = -tg + tPœ i = -210 + 193.09 = -16.9 MPa
318
g (rad)
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•5–141.
A steel alloy core is bonded firmly to the copper alloy tube to form the shaft shown. If the materials have the t-g diagrams shown, determine the torque resisted by the core and the tube.
450 mm A
100 mm 60 mm
B 15 kN⭈m
t (MPa)
Equation of Equilibrium. Refering to the free - body diagram of the cut part of the assembly shown in Fig. a, ©Mx = 0; Tc + Tt - 15 A 103 B = 0
180
(1)
Elastic Analysis. The shear modulus of steel and copper are Gst = 36 A 106 B and G q = = 18 GPa. Compatibility requires that 0.002
180 A 106 B 0.0024
g (rad) 0.0024
= 75 GPa
Steel Alloy t (MPa)
fC = ft 36
TcL TtL = JcGst JtG q
g (rad) 0.002
Tc
p 2
A 0.03 B (75) A 10 B 4
Tt
=
9
p 2
Copper Alloy
A 0.05 - 0.034 B (18) A 109 B 4
Tc = 0.6204Tt
(2)
Solving Eqs. (1) and (2), Tt = 9256.95 N # m
Tc = 5743.05 N # m
The maximum elastic torque and plastic torque of the core and the tube are (TY)c =
1 3 1 pc (tY)st = p A 0.033 B (180) A 106 B = 7634.07 N # m 2 2
(TP)c =
2 3 2 pc (tY)st = p A 0.033 B (180) A 106 B = 10 178.76 N # m 3 3
and p A 0.054 - 0.034 B J 2 T c(36) A 106 B d = 6152.49 N # m (TY)t = tY = D c 0.05
r2dr = 2p(36) A 106 B ¢
co
(TP)t = 2p(tY) q
Lci
r3 0.05 m = 7389.03 N # m ≤2 3 0.03 m
Since Tt 7 (TY)t, the results obtained using the elastic analysis are not valid.
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5–141.
Continued
Plastic Analysis. Assuming that the tube is fully plastic, Tt = (TP)t = 7389.03 N # m = 7.39 kN # m
Ans.
Substituting this result into Eq. (1), Tc = 7610.97 N # m = 7.61 kN # m
Ans.
Since Tc 6 (TY)c, the core is still linearly elastic. Thus, ft = ftc =
ft =
gi L; ci
TcL = JcGst
7610.97(0.45) p 4 9 2 (0.03 )(75)(10 )
0.3589 =
= 0.03589 rad
gi (0.45) 0.03
gi = 0.002393 rad Since gi 7 (gY) q = 0.002 rad, the tube is indeed fully plastic.
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5–142. A torque is applied to the shaft of radius r. If the material has a shear stress–strain relation of t = kg1>6, where k is a constant, determine the maximum shear stress in the shaft. r
r r g = gmax = gmax c r
T
1
gmax 6 1 b r6 t = kg = ka r 1 6
r
T = 2p
tr2 dr
L0
1
r
= 2p
L0
gmax = a
ka
1
1
gmax 6 13 gmax 6 6 12p kg6max r3 19 b r 6 dr = 2pk a b a b r6 = r r 19 19
6 19T b 3 12p kr
19T 12p r3
1
tmax = kg6max =
Ans.
5–143. Consider a thin-walled tube of mean radius r and thickness t. Show that the maximum shear stress in the tube due to an applied torque T approaches the average shear stress computed from Eq. 5–18 as r>t : q .
t
r
t 2r + t ; ro = r + = 2 2 J =
=
t 2r - t ri = r - = 2 2
2r - t 4 p 2r + t 4 ca b - a b d 2 2 2 p p [(2r + t)4 - (2r - t)4] = [64 r3 t + 16 r t3] 32 32
tmax =
Tc ; J
c = ro =
2r + t 2
T(2 r 2+ t) =
p 3 32 [64 r t 2r T(2r 2 +
=
+ 16 r t3]
2p r t[r2 + 14t2]
t 2 r2 )
2p r t c rr2 + 2
As
T(2 r 2+ t) =
1 t2 4 r2 d
t r : q , then : 0 r t
tmax =
=
T(1r + 0) 2p r t(1 + 0)
=
T 2p r2 t
T 2 t Am
QED
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*5–144. The 304 stainless steel shaft is 3 m long and has an outer diameter of 60 mm. When it is rotating at 60 rad>s, it transmits 30 kW of power from the engine E to the generator G. Determine the smallest thickness of the shaft if the allowable shear stress is tallow = 150 MPa and the shaft is restricted not to twist more than 0.08 rad.
E
Internal Torque: P = 30(103) W a T =
1 N # m>s b = 30(103) N # m>s W
30(103) P = = 500 N # m v 60
Allowable Shear Stress: Assume failure due to shear stress. tmax = tallow = 150(106) =
Tc J 500(0.03) p 4 2 (0.03
- r4i )
ri = 0.0293923 m = 29.3923 mm Angle of Twist: Assume failure due to angle of twist limitation. f =
0.08 =
TL JG 500(3)
p 2
A 0.03 - r4i B (75.0)(109) 4
ri = 0.0284033 m = 28.4033 mm Choose the smallest value of ri = 28.4033 mm t = ro - ri = 30 - 28.4033 = 1.60 mm
Ans.
322
G
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•5–145.
The A-36 steel circular tube is subjected to a torque of 10 kN # m. Determine the shear stress at the mean radius r = 60 mm and compute the angle of twist of the tube if it is 4 m long and fixed at its far end. Solve the problem using Eqs. 5–7 and 5–15 and by using Eqs. 5–18 and 5–20.
r 60 mm 4m
t 5 mm 10 kNm
Shear Stress: Applying Eq. 5-7, ro = 0.06 +
tr = 0.06 m =
0.005 = 0.0625 m 2
Tr = J
ri = 0.06 -
10(103)(0.06) p 4 2 (0.0625
- 0.05754)
0.005 = 0.0575 m 2
= 88.27 MPa
Ans.
Applying Eq. 5-18, tavg =
10(103) T = 88.42 MPa = 2 t Am 29(0.005)(p)(0.062)
Ans.
Angle of Twist: Applying Eq. 5-15, f =
TL JG 10(103)(4)
=
p 4 2 (0.0625
- 0.05754)(75.0)(109)
= 0.0785 rad = 4.495°
Ans.
Applying Eq. 5-20, f =
=
ds TL 4A2mG L t TL ds 4A2mG t L
Where
L
ds = 2pr
2pTLr =
4A2mG t 2p(10)(103)(4)(0.06)
=
4[(p)(0.062)]2 (75.0)(109)(0.005)
= 0.0786 rad = 4.503°
Ans.
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5–146. Rod AB is made of A-36 steel with an allowable shear stress of 1tallow2st = 75 MPa, and tube BC is made of AM1004-T61 magnesium alloy with an allowable shear stress of 1tallow2mg = 45 MPa. The angle of twist of end C is not allowed to exceed 0.05 rad. Determine the maximum allowable torque T that can be applied to the assembly.
0.3 m
0.4 m a A C 60 mm
Internal Loading: The internal torque developed in rod AB and tube BC are shown in Figs. a and b, respectively. Allowable Shear Stress: The polar moment of inertia of rod AB and tube p p a0.0154 b = 25.3125(10 - 9)p m4 and JBC = a0.034 - 0.0254 b BC are JAB = 2 2 = 0.2096875(10 - 6)p m4. We have
A tallow B st =
TAB cAB ; JAB
75(106) =
T(0.015) 25.3125(10 - 9)p
T = 397.61 N # m and
A tallow B mg =
TBC cBC ; JBC
45(106) =
T(0.03) 0.2096875(10 - 6)p
T = 988.13 N # m Angle of Twist: fB>A =
-T(0.7) TAB LAB = -0.11737(10 - 3)T = 0.11737(10 - 3)T = JAB Gst 25.3125(10 - 9)p(75)(109)
and fC>B =
T(0.4) TBC LBC = 0.03373(10 - 3)T = JBC Gmg 0.2096875(10 - 6)p(18)(109)
It is required that fC>A = 0.05 rad. Thus, fC>A = fB>A + fC>B 0.05 = 0.11737(10 - 3)T + 0.03373(10 - 3)T T = 331 N # m A controls B
Ans.
324
50 mm
30 mm Section a–a
T
a
B
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5–147. A shaft has the cross section shown and is made of 2014-T6 aluminum alloy having an allowable shear stress of tallow = 125 MPa. If the angle of twist per meter length is not allowed to exceed 0.03 rad, determine the required minimum wall thickness t to the nearest millimeter when the shaft is subjected to a torque of T = 15 kN # m.
30⬚ 30⬚
75 mm
Section Properties: Referring to the geometry shown in Fig. a, Am =
C
0.075 1 1 (0.15) ¢ ≤ + p A 0.0752 B = 0.01858 m2 2 tan 30° 2
ds = 2(0.15) + p(0.075) = 0.53562 m
Allowable Shear Stress:
A tavg B allow =
T ; 2tAm
125(106) =
15(103) 2t(0.01858)
t = 0.00323 m = 3.23 mm Angle of Twist: f =
ds TL 2 4Am G C t
0.03 =
15(103)(1) 4(0.018582)(27)(109)
a
0.53562 b t
t = 0.007184 m = 7.18 mm (controls) Use t = 8 mm
Ans.
325
t
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*5–148. The motor A develops a torque at gear B of 500 lb # ft, which is applied along the axis of the 2-in.diameter A-36 steel shaft CD. This torque is to be transmitted to the pinion gears at E and F. If these gears are temporarily fixed, determine the maximum shear stress in segments CB and BD of the shaft. Also, what is the angle of twist of each of these segments? The bearings at C and D only exert force reactions on the shaft.
B E
F 2 ft
1.5 ft
C
D A
Equilibrium: TC + TD - 500 = 0
[1]
Compatibility: fB>C = fB>D TC(2) TD(1.5) = JG JG TC = 0.75TD
[2]
Solving Eqs. [1] and [2] yields: TD = 285.71 lb # ft
TC = 214.29 lb # ft
Maximum Shear Stress: (tCB)max =
214.29(12)(1) TCc = 1.64 ksi = p 4 J 2 (1 )
Ans.
(tBD)max =
285.71(12)(1) TDc = 2.18 ksi = p 4 J 2 (1 )
Ans.
Angle of Twist: fCB = fBD =
TD LBD JG 285.71(12)(1.5)(12)
=
p 2
(14)(11.0)(106)
= 0.003572 rad = 0.205°
Ans.
326
500 lb·ft
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5–149. The coupling consists of two disks fixed to separate shafts, each 25 mm in diameter. The shafts are supported on journal bearings that allow free rotation. In order to limit the torque T that can be transmitted, a “shear pin” P is used to connect the disks together. If this pin can sustain an average shear force of 550 N before it fails, determine the maximum constant torque T that can be transmitted from one shaft to the other. Also, what is the maximum shear stress in each shaft when the “shear pin” is about to fail?
25 mm
P
130 mm 25 mm
T
Equilibrium: T - 550(0.13) = 0
©Mx = 0;
T = 71.5 N # m
Ans.
Maximum Shear Stress: tmax =
71.5(0.0125) Tc = 23.3 MPa = p 4 J 2 (0.0125 )
Ans.
5–150. The rotating flywheel and shaft is brought to a sudden stop at D when the bearing freezes. This causes the flywheel to oscillate clockwise–counterclockwise, so that a point A on the outer edge of the flywheel is displaced through a 10-mm arc in either direction. Determine the maximum shear stress developed in the tubular 304 stainless steel shaft due to this oscillation. The shaft has an inner diameter of 25 mm and an outer diameter of 35 mm. The journal bearings at B and C allow the shaft to rotate freely.
D 2m
B A
80 mm
Angle of Twist: f =
0.125 =
TL JG
Where
f =
10 = 0.125 rad 80
T(2) p 4 2 (0.0175
- 0.01254)(75.0)(109)
T = 510.82 N # m Maximum Shear Stress: tmax =
Tc = J
510.82(0.0175) p 4 2 (0.0175
- 0.01254)
Ans.
= 82.0 MPa
327
C
T
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5–151. If the solid shaft AB to which the valve handle is attached is made of C83400 red brass and has a diameter of 10 mm, determine the maximum couple forces F that can be applied to the handle just before the material starts to fail. Take tallow = 40 MPa. What is the angle of twist of the handle? The shaft is fixed at A.
B
A 150 mm 150 mm F 150 mm
tmax = tallow = 40(106) =
Tc J
F
0.3F(0.005) p 4 2 (0.005)
F = 26.18 N = 26.2 N
Ans.
T = 0.3F = 7.85 N # m f =
TL = JG
7.85(0.15) p 4 9 2 (0.005) (37)(10 )
= 0.03243 rad = 1.86°
Ans.
328
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6–1. Draw the shear and moment diagrams for the shaft. The bearings at A and B exert only vertical reactions on the shaft.
B
A
800 mm
250 mm
24 kN
6–2. Draw the shear and moment diagrams for the simply supported beam.
4 kN M 2 kNm A
B 2m
329
2m
2m
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6–3. The engine crane is used to support the engine, which has a weight of 1200 lb. Draw the shear and moment diagrams of the boom ABC when it is in the horizontal position shown. a + ©MA = 0;
4 F (3) - 1200(8) = 0; 5 A
+ c ©Fy = 0;
-Ay +
+ ©F = 0; ; x
Ax -
4 (4000) - 1200 = 0; 5
3 (4000) = 0; 5
A
3 ft
5 ft B
FA = 4000 lb
4 ft
Ay = 2000 lb
Ax = 2400 lb
*6–4. Draw the shear and moment diagrams for the cantilever beam.
2 kN/m
A
6 kNm 2m
The free-body diagram of the beam’s right segment sectioned through an arbitrary point shown in Fig. a will be used to write the shear and moment equations of the beam. + c ©Fy = 0;
C
V - 2(2 - x) = 0
V = {4 - 2x} kN‚
(1)
1 a + ©M = 0; -M - 2(2 - x)c (2 - x) d - 6 = 0 M = {-x2 + 4x - 10}kN # m‚(2) 2 The shear and moment diagrams shown in Figs. b and c are plotted using Eqs. (1) and (2), respectively. The value of the shear and moment at x = 0 is evaluated using Eqs. (1) and (2). Vx = 0 = 4 - 2(0) = 4 kN Mx = 0 = C -0 + 4(0) - 10 D = -10kN # m
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6–5. Draw the shear and moment diagrams for the beam.
10 kN
8 kN
15 kNm
2m
3m
6–6. Draw the shear and moment diagrams for the overhang beam.
8 kN/m
C
A B 2m
4m
6–7. Draw the shear and moment diagrams for the compound beam which is pin connected at B.
6 kip
8 kip
A C B 4 ft
331
6 ft
4 ft
4 ft
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*6–8. Draw the shear and moment diagrams for the simply supported beam.
150 lb/ft 300 lbft A
B 12 ft
The free-body diagram of the beam’s left segment sectioned through an arbitrary point shown in Fig. b will be used to write the shear and moment equations. The intensity of the triangular distributed load at the point of sectioning is w = 150 a
x b = 12.5x 12
Referring to Fig. b, + c ©Fy = 0; a + ©M = 0; M +
275 -
1 (12.5x)(x) - V = 0 2
V = {275 - 6.25x2}lb‚ (1)
x 1 (12.5x)(x)a b - 275x = 0 M = {275x - 2.083x3}lb # ft‚(2) 2 3
The shear and moment diagrams shown in Figs. c and d are plotted using Eqs. (1) and (2), respectively. The location where the shear is equal to zero can be obtained by setting V = 0 in Eq. (1). 0 = 275 - 6.25x2
x = 6.633 ft
The value of the moment at x = 6.633 ft (V = 0) is evaluated using Eq. (2). M x = 6.633 ft = 275(6.633) - 2.083(6.633)3 = 1216 lb # ft
332
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6–9. Draw the shear and moment diagrams for the beam. Hint: The 20-kip load must be replaced by equivalent loadings at point C on the axis of the beam.
15 kip 1 ft
A
C 4 ft
20 kip
B
4 ft
4 ft
6–10. Members ABC and BD of the counter chair are rigidly connected at B and the smooth collar at D is allowed to move freely along the vertical slot. Draw the shear and moment diagrams for member ABC. Equations of Equilibrium: Referring to the free-body diagram of the frame shown in Fig. a, + c ©Fy = 0;
P 150 lb
Ay - 150 = 0 C A
Ay = 150 lb a + ©MA = 0;
B
1.5 ft 1.5 ft
ND(1.5) - 150(3) = 0 D
ND = 300 lb Shear and Moment Diagram: The couple moment acting on B due to ND is MB = 300(1.5) = 450 lb # ft. The loading acting on member ABC is shown in Fig. b and the shear and moment diagrams are shown in Figs. c and d.
333
1.5 ft
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6–11. The overhanging beam has been fabricated with a projected arm BD on it. Draw the shear and moment diagrams for the beam ABC if it supports a load of 800 lb. Hint: The loading in the supporting strut DE must be replaced by equivalent loads at point B on the axis of the beam.
E
800 lb
B
Support Reactions: a + ©MC = 0;
5 ft
D 2 ft
C
A
800(10) -
3 4 FDE(4) - FDE(2) = 0 5 5
6 ft
4 ft
FDE = 2000 lb + c ©Fy = 0;
-800 +
+ ©F = 0; : x
-Cx +
3 (2000) - Cy = 0 5 4 (2000) = 0 5
Cy = 400 lb
Cx = 1600 lb
Shear and Moment Diagram:
*6–12. A reinforced concrete pier is used to support the stringers for a bridge deck. Draw the shear and moment diagrams for the pier when it is subjected to the stringer loads shown. Assume the columns at A and B exert only vertical reactions on the pier.
60 kN 60 kN 35 kN 35 kN 35 kN 1 m 1 m 1.5 m 1.5 m 1 m 1 m
A
334
B
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6–13. Draw the shear and moment diagrams for the compound beam. It is supported by a smooth plate at A which slides within the groove and so it cannot support a vertical force, although it can support a moment and axial load.
P
Support Reactions:
P
A
D
B C
From the FBD of segment BD a + ©MC = 0; + c ©Fy = 0; + ©F = 0; : x
By (a) - P(a) = 0 Cy - P - P = 0
By = P
a
a
a
a
Cy = 2P
Bx = 0
From the FBD of segment AB a + ©MA = 0; + c ©Fy = 0;
P(2a) - P(a) - MA = 0
MA = Pa
P - P = 0 (equilibrium is statisfied!)
6–14. The industrial robot is held in the stationary position shown. Draw the shear and moment diagrams of the arm ABC if it is pin connected at A and connected to a hydraulic cylinder (two-force member) BD. Assume the arm and grip have a uniform weight of 1.5 lbin. and support the load of 40 lb at C.
4 in. A
10 in. B
50 in.
120 D
335
C
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*6–16. Draw the shear and moment diagrams for the shaft and determine the shear and moment throughout the shaft as a function of x. The bearings at A and B exert only vertical reactions on the shaft.
500 lb 800 lb A
B x 3 ft
For 0 6 x 6 3 ft + c ©Fy = 0.
220 - V = 0
a + ©MNA = 0.
V = 220 lb‚
Ans.
M - 220x = 0 M = (220x) lb ft‚
Ans.
For 3 ft 6 x 6 5 ft + c ©Fy = 0;
220 - 800 - V = 0 V = -580 lb
a + ©MNA = 0;
Ans.
M + 800(x - 3) - 220x = 0 M = {-580x + 2400} lb ft‚
Ans.
For 5 ft 6 x … 6 ft + c ©Fy = 0; a + ©MNA = 0;
V - 500 = 0
V = 500 lb‚
Ans.
-M - 500(5.5 - x) - 250 = 0 M = (500x - 3000) lb ft
Ans.
336
2 ft
0.5 ft
0.5 ft
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•6–17.
Draw the shear and moment diagrams for the cantilevered beam.
300 lb
200 lb/ft
A 6 ft
The free-body diagram of the beam’s left segment sectioned through an arbitrary point shown in Fig. b will be used to write the shear and moment equations. The intensity of the triangular distributed load at the point of sectioning is x w = 200 a b = 33.33x 6 Referring to Fig. b, + c ©Fy = 0;
-300 -
a + ©M = 0; M +
1 (33.33x)(x) - V = 0 2
V = {-300 - 16.67x2} lb (1)
1 x (33.33x)(x)a b + 300x = 0 M = {-300x - 5.556x3} lb # ft (2) 2 3
The shear and moment diagrams shown in Figs. c and d are plotted using Eqs. (1) and (2), respectively.
337
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6–18. Draw the shear and moment diagrams for the beam, and determine the shear and moment throughout the beam as functions of x.
2 kip/ft
10 kip
8 kip 40 kip⭈ft
Support Reactions: As shown on FBD. Shear and Moment Function:
x 6 ft
For 0 … x 6 6 ft: + c ©Fy = 0;
4 ft
30.0 - 2x - V = 0 V = {30.0 - 2x} kip
Ans.
x a + ©MNA = 0; M + 216 + 2xa b - 30.0x = 0 2 M = {-x2 + 30.0x - 216} kip # ft
Ans.
For 6 ft 6 x … 10 ft: + c ©Fy = 0; a + ©MNA = 0;
V - 8 = 0
V = 8.00 kip
Ans.
-M - 8(10 - x) - 40 = 0 M = {8.00x - 120} kip # ft
Ans.
6–19. Draw the shear and moment diagrams for the beam.
2 kip/ ft 30 kip⭈ft
B A 5 ft
338
5 ft
5 ft
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*6–20. Draw the shear and moment diagrams for the simply supported beam.
10 kN 10 kN/m
A
B
3m
Since the area under the curved shear diagram can not be computed directly, the value of the moment at x = 3 m will be computed using the method of sections. By referring to the free-body diagram shown in Fig. b, a + ©M = 0; Mx= 3 m +
1 (10)(3)(1) - 20(3) = 0 2
Mx= 3m = 45 kN # m
339
Ans.
3m
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•6–21. The beam is subjected to the uniform distributed load shown. Draw the shear and moment diagrams for the beam.
2 kN/m
Equations of Equilibrium: Referring to the free-body diagram of the beam shown in Fig. a, a + ©MA = 0;
3 FBC a b (2) - 2(3)(1.5) = 0 5
B
A 1.5 m
FBC = 7.5 kN + c ©Fy = 0;
C
3 Ay + 7.5 a b - 2(3) = 0 5 Ay = 1.5 kN
3 Shear and Moment Diagram: The vertical component of FBC is A FBC B y = 7.5a b 5 = 4.5 kN. The shear and moment diagrams are shown in Figs. c and d.
340
2m
1m
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6–22. Draw the shear and moment diagrams for the overhang beam.
4 kN/m
A B 3m
Since the loading is discontinuous at support B, the shear and moment equations must be written for regions 0 … x 6 3 m and 3 m 6 x … 6 m of the beam. The free-body diagram of the beam’s segment sectioned through an arbitrary point within these two regions is shown in Figs. b and c. Region 0 … x 6 3 m, Fig. b + c ©Fy = 0;
-4 -
a + ©M = 0; M +
2 V = e - x2 - 4 f kN 3
1 4 a xb(x) - V = 0 2 3
1 4 x a xb(x)a b + 4x = 0 2 3 3
(1)
2 M = e - x3 - 4x f kN # m (2) 9
Region 3 m 6 x … 6 m, Fig. c + c ©Fy = 0;
V - 4(6 - x) = 0
1 a + ©M = 0; -M - 4(6 - x) c (6 - x) d = 0 2
V = {24 - 4x} kN
(3)
M = {-2(6 - x)2}kN # m
(4)
The shear diagram shown in Fig. d is plotted using Eqs. (1) and (3). The value of shear just to the left and just to the right of the support is evaluated using Eqs. (1) and (3), respectively. 2 Vx= 3 m - = - (32) - 4 = -10 kN 3 Vx=3 m + = 24 - 4(3) = 12 kN The moment diagram shown in Fig. e is plotted using Eqs. (2) and (4). The value of the moment at support B is evaluated using either Eq. (2) or Eq. (4). 2 Mx= 3 m = - (33) - 4(3) = -18 kN # m 9 or Mx= 3 m = -2(6 - 3)2 = -18 kN # m
341
3m
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6–23. Draw the shear and moment diagrams for the beam. It is supported by a smooth plate at A which slides within the groove and so it cannot support a vertical force, although it can support a moment and axial load.
w
B
A
L
*6–24. Determine the placement distance a of the roller support so that the largest absolute value of the moment is a minimum. Draw the shear and moment diagrams for this condition.
w
A
B a
wL2 wL - wx = 0 2a
+ c ©Fy = 0;
x = L -
L
L2 2a
x wL2 Mmax (+) + wxa b - a wL bx = 0 2 2a
a + ©M = 0;
Substitute x = L -
L2 ; 2a
Mmax (+) = a wL =
wL2 L2 w L2 2 b aL b aL b 2a 2a 2 2a
w L2 2 aL b 2 2a Mmax (-) - w(L - a)
©M = 0;
Mmax (-) =
(L - a) = 0 2
w(L - a)2 2
To get absolute minimum moment, Mmax (+) = Mmax (-) L2 2 w w (L ) = (L - a)2 2 2a 2 L a =
L2 = L - a 2a L 22
‚
Ans.
342
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6–25. The beam is subjected to the uniformly distributed moment m (moment>length). Draw the shear and moment diagrams for the beam.
m A L
Support Reactions: As shown on FBD. Shear and Moment Function: V = 0
+ c ©Fy = 0; a + ©MNA = 0;
M + mx - mL = 0
M = m(L - x)
Shear and Moment Diagram:
6–27. Draw the shear and moment diagrams for the beam.
+ c ©Fy = 0;
w0
w0L 1 w0x - a b(x) = 0 4 2 L B
x = 0.7071 L a + ©MNA = 0;
M +
w0L 1 w0x x L a b (x)a b ax - b = 0 2 L 3 4 3
Substitute x = 0.7071L, M = 0.0345 w0L2
343
L 3
A
2L 3
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*6–28. Draw the shear and moment diagrams for the beam.
w0
B
A L – 3
Support Reactions: As shown on FBD. Shear and Moment Diagram: Shear and moment at x = L>3 can be determined using the method of sections. + c ©Fy = 0;
w0 L w0 L - V = 0 3 6
a + ©MNA = 0;
M +
V =
w0 L 6
w0 L L w0 L L a b a b = 0 6 9 3 3 M =
5w0 L2 54
344
L – 3
L – 3
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•6–29.
Draw the shear and moment diagrams for the beam.
5 kN/m
5 kN/m
B
A 4.5 m
From FBD(a) + c ©Fy = 0; a + ©MNA = 0;
9.375 - 0.5556x2 = 0
x = 4.108 m
M + (0.5556) A 4.1082 B a
4.108 b - 9.375(4.108) = 0 3
M = 25.67 kN # m From FBD(b) a + ©MNA = 0;
M + 11.25(1.5) - 9.375(4.5) = 0 M = 25.31 kN # m
345
4.5 m
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6–30. Draw the shear and moment diagrams for the compound beam.
150 lb/ft
150 lb/ft
A 6 ft
Support Reactions: From the FBD of segment AB a + ©MB = 0;
450(4) - Ay (6) = 0
Ay = 300.0 lb
+ c ©Fy = 0;
By - 450 + 300.0 = 0
By = 150.0 lb
+ ©F = 0; : x
Bx = 0
From the FBD of segment BC a + ©MC = 0;
225(1) + 150.0(3) - MC = 0 MC = 675.0 lb # ft
+ c ©Fy = 0; + ©F = 0; : x
Cy - 150.0 - 225 = 0
Cy = 375.0 lb
Cx = 0
Shear and Moment Diagram: The maximum positive moment occurs when V = 0. + c ©Fy = 0; a + ©MNA = 0;
150.0 - 12.5x2 = 0
x = 3.464 ft
150(3.464) - 12.5 A 3.4642 B a
3.464 b - Mmax = 0 3
Mmax = 346.4 lb # ft
346
C
B 3 ft
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6–31. Draw the shear and moment diagrams for the beam and determine the shear and moment in the beam as functions of x.
w0
Support Reactions: As shown on FBD.
A
B x
Shear and Moment Functions:
L – 2
For 0 … x 6 L>2 + c ©Fy = 0;
3w0 L -w0x - V = 0 4 V =
a + ©MNA = 0;
w0 (3L - 4x) 4
Ans.
7w0 L2 3w0 L x x + w0 xa b + M = 0 24 4 2 M =
w0 A -12x2 + 18Lx - 7L2) 24
Ans.
For L>2 6 x … L + c ©Fy = 0;
V -
1 2w0 c (L - x) d(L - x) = 0 2 L V =
a + ©MNA = 0;
-M -
w0 (L - x)2 L
Ans.
1 2w0 L - x c (L - x) d(L - x)a b = 0 2 L 3
M = -
w0 (L - x)3 3L
Ans.
347
L – 2
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*6–32. The smooth pin is supported by two leaves A and B and subjected to a compressive load of 0.4 kNm caused by bar C. Determine the intensity of the distributed load w0 of the leaves on the pin and draw the shear and moment diagram for the pin.
0.4 kN/m C
A
+ c ©Fy = 0;
B
w0
1 2(w0)(20)a b - 60(0.4) = 0 2
20 mm 60 mm 20 mm
w0 = 1.2 kN>m
Ans.
•6–33.
The ski supports the 180-lb weight of the man. If the snow loading on its bottom surface is trapezoidal as shown, determine the intensity w, and then draw the shear and moment diagrams for the ski.
180 lb 3 ft
w 1.5 ft
Ski: + c ©Fy = 0;
1 1 w(1.5) + 3w + w(1.5) - 180 = 0 2 2 Ans.
w = 40.0 lb>ft Segment: + c ©Fy = 0;
30 - V = 0;
a + ©M = 0;
M - 30(0.5) = 0;
w0
V = 30.0 lb M = 15.0 lb # ft
348
w 3 ft
1.5 ft
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6–34. Draw the shear and moment diagrams for the compound beam.
5 kN 3 kN/m
A B 3m
6–35. Draw the shear and moment diagrams for the beam and determine the shear and moment as functions of x.
A x 3m
200 - V = 0
V = 200 N
Ans.
M - 200 x = 0 M = (200 x) N # m
Ans.
For 3 m 6 x … 6 m: 200 - 200(x - 3) V = e-
1 200 c (x - 3) d(x - 3) - V = 0 2 3
100 2 x + 500 f N 3
Ans.
Set V = 0, x = 3.873 m a + ©MNA = 0;
M +
1 200 x - 3 c (x - 3) d(x - 3)a b 2 3 3
+ 200(x - 3)a M = e-
1.5 m
B
For 0 … x 6 3 m:
+ c ©Fy = 0;
1.5 m
200 N/ m
Shear and Moment Functions:
a + ©MNA = 0;
3m
400 N/m
Support Reactions: As shown on FBD.
+ c ©Fy = 0;
D
C
x - 3 b - 200x = 0 2
100 3 x + 500x - 600 f N # m 9
Ans.
Substitute x = 3.87 m, M = 691 N # m
349
3m
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*6–36. Draw the shear and moment diagrams for the overhang beam.
18 kN 6 kN
A B 2m
6–37. Draw the shear and moment diagrams for the beam.
2m
M 10 kNm
2m
50 kN/m
50 kN/m
B A 4.5 m
350
4.5 m
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6–38. The dead-weight loading along the centerline of the airplane wing is shown. If the wing is fixed to the fuselage at A, determine the reactions at A, and then draw the shear and moment diagram for the wing.
3000 lb
400 lb/ft
250 lb/ft
A
8 ft
2 ft
Support Reactions:
3 ft
15 000 lb
-1.00 - 3 + 15 - 1.25 - 0.375 - Ay = 0
+ c ©Fy = 0; Ay = 9.375 kip
Ans.
a + ©MA = 0;
1.00(7.667) + 3(5) - 15(3) + 1.25(2.5) + 0.375(1.667) + MA = 0
MA = 18.583 kip # ft = 18.6 kip # ft
Ans.
+ ©F = 0; : x
Ans.
Ax = 0
Shear and Moment Diagram:
6–39. The compound beam consists of two segments that are pinned together at B. Draw the shear and moment diagrams if it supports the distributed loading shown. + c ©Fy = 0;
2wL 1w 2 x = 0 27 2L x =
a + ©M = 0;
w
B 2/3 L
4 L = 0.385 L A 27
M +
C
A
1w 1 2wL (0.385L)2 a b(0.385L) (0.385L) = 0 2L 3 27 M = 0.0190 wL2
351
1/3 L
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*6–40. Draw the shear and moment diagrams for the simply supported beam.
10 kN
10 kN
15 kNm A
B 2m
6–41. Draw the shear and moment diagrams for the compound beam. The three segments are connected by pins at B and E.
3 kN
2m
2m
3 kN
0.8 kN/m
B
E F
A C 2m
352
1m
1m
D 2m
1m
1m
2m
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6–42. Draw the shear and moment diagrams for the compound beam.
5 kN/m
Support Reactions:
A
From the FBD of segment AB a + ©MA = 0; + c ©Fy = 0;
B 2m
By (2) - 10.0(1) = 0
By = 5.00 kN
Ay - 10.0 + 5.00 = 0
Ay = 5.00 kN
C 1m
D
1m
From the FBD of segment BD a + ©MC = 0;
5.00(1) + 10.0(0) - Dy (1) = 0 Dy = 5.00 kN
+ c ©Fy = 0;
Cy - 5.00 - 5.00 - 10.0 = 0 Cy = 20.0 kN
+ ©F = 0; : x
Bx = 0
From the FBD of segment AB + ©F = 0; : x
Ax = 0
Shear and Moment Diagram:
6–43. Draw the shear and moment diagrams for the beam. The two segments are joined together at B.
8 kip
3 kip/ft
A
C B 3 ft
353
5 ft
8 ft
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*6–44. Draw the shear and moment diagrams for the beam.
w
8
FR =
x =
1 2 x dx = 21.33 kip 8 L0
1 8 3 8 10 x dx
21.33
8 kip/ft 1 w ⫽ x2 8
= 6.0 ft
x B
A 8 ft
•6–45.
Draw the shear and moment diagrams for the beam. L
FR =
dA =
LA
L0 w0
wdx =
w0
L
L L0 2
x2 dx =
w
w0 L 3
w
LA
2
x A
w0L w0x = 0 12 3L2
1 1>3 x = a b L = 0.630 L 4 w0L w0x3 1 a + ©M = 0; (x) a xb - M = 0 12 3L2 4
M =
B L
3
+ c ©Fy = 0;
w0
L
x3dx L L0 3L x = = = w0 L 4 dA 3 LA xdA
w0 2 x L2
w0Lx w0x4 12 12L2
Substitute x = 0.630L M = 0.0394 w0L2
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6–46. Draw the shear and moment diagrams for the beam. L
FR =
dA = w0
LA
L0
sina
w
2w0 L p xb dx = p L
w0
A L – 2
6–47. A member having the dimensions shown is used to resist an internal bending moment of M = 90 kN # m. Determine the maximum stress in the member if the moment is applied (a) about the z axis (as shown) (b) about the y axis. Sketch the stress distribution for each case.
200 mm y 150 mm
The moment of inertia of the cross-section about z and y axes are 1 (0.2)(0.153) = 56.25(10 - 6) m4 12
Iy =
1 (0.15)(0.23) = 0.1(10 - 3) m4 12
M z x
For the bending about z axis, c = 0.075 m. smax =
90(103) (0.075) Mc = 120(106)Pa = 120 MPa = Iz 56.25 (10 - 6)
Ans.
For the bending about y axis, C = 0.1 m. smax =
x
B L – 2
Iz =
p w w0 sin – x L
90(103) (0.1) Mc = 90 (106)Pa = 90 MPa = Iy 0.1 (10 - 3)
Ans.
The bending stress distribution for bending about z and y axes are shown in Fig. a and b respectively.
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*6–48. Determine the moment M that will produce a maximum stress of 10 ksi on the cross section.
0.5 in. A
3 in.
0.5 in.
0.5 in. B
C 3 in. M 10 in.
D 0.5 in.
Section Properties: y =
=
©yA ©A 0.25(4)(0.5) + 2[2(3)(0.5)] + 5.5(10)(0.5) = 3.40 in. 4(0.5) + 2[(3)(0.5)] + 10(0.5)
INA =
1 (4) A 0.53 B + 4(0.5)(3.40 - 0.25)2 12 + 2c
1 (0.5)(33) + 0.5(3)(3.40 - 2)2 d 12
+
1 (0.5) A 103 B + 0.5(10)(5.5 - 3.40)2 12
= 91.73 in4 Maximum Bending Stress: Applying the flexure formula smax = 10 =
Mc I M (10.5 - 3.4) 91.73
M = 129.2 kip # in = 10.8 kip # ft
Ans.
356
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•6–49.
Determine the maximum tensile and compressive bending stress in the beam if it is subjected to a moment of M = 4 kip # ft.
0.5 in. A 0.5 in.
3 in.
0.5 in. B
C 3 in. M 10 in.
D 0.5 in.
Section Properties: y =
=
©yA ©A 0.25(4)(0.5) + 2[2(3)(0.5)] + 5.5(10)(0.5) = 3.40 in. 4(0.5) + 2[(3)(0.5)] + 10(0.5)
INA =
1 (4) A 0.53 B + 4(0.5)(3.40 - 0.25)2 12 + 2c
1 (0.5)(33) + 0.5(3)(3.40 - 2)2 d 12
+
1 (0.5) A 103 B + 0.5(10)(5.5 - 3.40)2 12
= 91.73 in4 Maximum Bending Stress: Applying the flexure formula smax =
Mc I
(st)max =
4(103)(12)(10.5 - 3.40) = 3715.12 psi = 3.72 ksi 91.73
Ans.
(sc)max =
4(103)(12)(3.40) = 1779.07 psi = 1.78 ksi 91.73
Ans.
357
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6–50. The channel strut is used as a guide rail for a trolley. If the maximum moment in the strut is M = 30 N # m, determine the bending stress at points A, B, and C.
50 mm C 5 mm 5 mm
y =
B
2.5(50)(5) + 7.5(34)(5) + 2[20(5)(20)] + 2[(32.5)(12)(5)] 50(5) + 34(5) + 2[5(20)] + 2[(12)(5)]
30 mm
= 13.24 mm A
I = c
1 (50)(53) + 50(5)(13.24 - 2.5)2 d 12
+ c
5 mm
5 mm 5 mm 7 mm 10 mm 7 mm
1 (34)(53) + 34(5)(13.24 - 7.5)2 d 12
+ 2c
1 1 (5)(203) + 5(20)(20 - 13.24)2 d + 2c (12)(53) + 12(5)(32.5 - 13.24)2 d 12 12
= 0.095883(10 - 6) m4 30(35 - 13.24)(10 - 3)
sA =
0.095883(10 - 6) 30(13.24 - 10)(10 - 3)
sB =
0.095883(10 - 6)
= 6.81 MPa
Ans.
= 1.01 MPa
Ans.
6–51. The channel strut is used as a guide rail for a trolley. If the allowable bending stress for the material is sallow = 175 MPa, determine the maximum bending moment the strut will resist.
50 mm C 5 mm 5 mm B
-3
30(13.24)(10 )
sC =
-6
0.095883(10 )
= 4.14 MPa
©y2A 2.5(50)(5) + 7.5(34)(5) + 2[20(5)(20)] + 2[(32.5)(12)(5)] = = 13.24 mm y = ©A 50(5) + 34(5) + 2[5(20)] + 2[(12)(5)] I = c
1 1 (50)(53) + 50(5)(13.24 - 2.5)2 d + c (34)(53) + 34(5)(13.24 - 7.5)2 d 12 12
+ 2c
1 1 (5)(203) + 5(20)(20 - 13.24)2 d + 2c (12)(53) + 12(5)(32.5 - 13.24)2 d 12 12
= 0.095883(10 - 6) m4 s =
Mc ; I
175(106) =
30 mm
Ans.
M(35 - 13.24)(10 - 3) 0.095883(10 - 6)
M = 771 N # m
Ans.
358
A
5 mm
5 mm 5 mm 7 mm 10 mm 7 mm
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*6–52. The beam is subjected to a moment M. Determine the percentage of this moment that is resisted by the stresses acting on both the top and bottom boards, A and B, of the beam.
A 25 mm M
D
Section Property: I =
1 1 (0.2) A 0.23 B (0.15) A 0.153 B = 91.14583 A 10 - 6 B m4 12 12
150 mm 25 mm 25 mm
Bending Stress: Applying the flexure formula
B 150 mm
25 mm
My I
s =
sE =
sD =
M(0.1) 91.14583(10 - 6) M(0.075) 91.14583(10 - 6)
= 1097.143 M
= 822.857 M
Resultant Force and Moment: For board A or B F = 822.857M(0.025)(0.2) +
1 (1097.143M - 822.857M)(0.025)(0.2) 2
= 4.800 M M¿ = F(0.17619) = 4.80M(0.17619) = 0.8457 M sc a
M¿ b = 0.8457(100%) = 84.6 % M
Ans.
•6–53. Determine the moment M that should be applied to the beam in order to create a compressive stress at point D of sD = 30 MPa . Also sketch the stress distribution acting over the cross section and compute the maximum stress developed in the beam.
A 25 mm
Section Property:
150 mm
1 1 I = (0.2) A 0.23 B (0.15) A 0.153 B = 91.14583 A 10 - 6 B m4 12 12
25 mm 25 mm
Bending Stress: Applying the flexure formula s = 30 A 106 B =
My I M(0.075) 91.14583(10 - 6)
M = 36458 N # m = 36.5 kN # m smax =
M
D
Ans.
36458(0.1) Mc = 40.0 MPa = I 91.14583(10 - 6)
Ans.
359
B 150 mm
25 mm
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6–54. The beam is made from three boards nailed together as shown. If the moment acting on the cross section is M = 600 N # m, determine the maximum bending stress in the beam. Sketch a three-dimensional view of the stress distribution acting over the cross section.
25 mm
150 mm 20 mm
(0.0125)(0.24)(0.025) + 2 (0.1)(0.15)(0.2) = 0.05625 m y = 0.24 (0.025) + 2 (0.15)(0.02)
200 mm M 600 Nm
1 (0.24)(0.0253) + (0.24)(0.025)(0.043752) 12
I =
+ 2a
20 mm
1 b (0.02)(0.153) + 2(0.15)(0.02)(0.043752) 12
= 34.53125 (10 - 6) m4 smax = sB =
Mc I 600 (0.175 - 0.05625)
=
34.53125 (10 - 6)
= 2.06 MPa sC =
Ans.
My 600 (0.05625) = 0.977 MPa = I 34.53125 (10 - 6)
6–55. The beam is made from three boards nailed together as shown. If the moment acting on the cross section is M = 600 N # m, determine the resultant force the bending stress produces on the top board.
25 mm
150 mm
(0.0125)(0.24)(0.025) + 2 (0.15)(0.1)(0.02) = 0.05625 m 0.24 (0.025) + 2 (0.15)(0.02)
y =
20 mm 200 mm M 600 Nm
1 (0.24)(0.0253) + (0.24)(0.025)(0.043752) 12
I =
+ 2a
20 mm
1 b (0.02)(0.153) + 2(0.15)(0.02)(0.043752) 12
= 34.53125 (10 - 6) m4 s1 =
My 600(0.05625) = 0.9774 MPa = I 34.53125(10 - 6)
sb =
My 600(0.05625 - 0.025) = 0.5430 MPa = I 34.53125(10 - 6)
F =
1 (0.025)(0.9774 + 0.5430)(106)(0.240) = 4.56 kN 2
Ans.
360
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*6–56. The aluminum strut has a cross-sectional area in the form of a cross. If it is subjected to the moment M = 8 kN # m, determine the bending stress acting at points A and B, and show the results acting on volume elements located at these points.
A 100 mm 20 mm 100 mm
B
M ⫽ 8 kN⭈m
20 mm
50 mm 50 mm
Section Property: I =
1 1 (0.02) A 0.223 B + (0.1) A 0.023 B = 17.8133 A 10 - 6 B m4 12 12
Bending Stress: Applying the flexure formula s =
sA =
sB =
8(103)(0.11) 17.8133(10 - 6) 8(103)(0.01) 17.8133(10 - 6)
My I
= 49.4 MPa (C)
Ans.
= 4.49 MPa (T)
Ans.
•6–57. The aluminum strut has a cross-sectional area in the form of a cross. If it is subjected to the moment M = 8 kN # m, determine the maximum bending stress in the beam, and sketch a three-dimensional view of the stress distribution acting over the entire cross-sectional area.
A 100 mm 20 mm 100 mm
B 20 mm
M ⫽ 8 kN⭈m 50 mm
50 mm
Section Property: I =
1 1 (0.02) A 0.223 B + (0.1) A 0.023 B = 17.8133 A 10 - 6 B m4 12 12
Bending Stress: Applying the flexure formula smax =
smax =
8(103)(0.11) 17.8133(10 - 6)
sy = 0.01m =
My Mc and s = , I I
= 49.4 MPa
8(103)(0.01) 17.8133(10 - 6)
Ans.
= 4.49 MPa
361
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6–58. If the beam is subjected to an internal moment of M = 100 kip # ft, determine the maximum tensile and compressive bending stress in the beam.
3 in. 3 in. 6 in. M 2 in.
1.5 in.
Section Properties: The neutral axis passes through centroid C of the cross section as shown in Fig. a. The location of C is
©yA y = = ©A
4(8)(6) - 2 cp A 1.52 B d 8(6) - p A 1.52 B
= 4.3454 in.
Thus, the moment of inertia of the cross section about the neutral axis is I = ©I + Ad2
=
1 1 (6)a83 b + 6(8) A 4.3454 - 4 B 2 - B pa 1.54 b + pa 1.52 b A 4.3454 - 2 B 2 R 12 4
= 218.87 in4 Maximum Bending Stress: The maximum compressive and tensile bending stress occurs at the top and bottom edges of the cross section.
A smax B T =
100(12)(4.3454) Mc = = 23.8 ksi (T) I 218.87
Ans.
A smax B C =
My 100(12)(8 - 4.3454) = = 20.0 ksi (C) I 218.87
Ans.
362
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6–59. If the beam is made of material having an allowable tensile and compressive stress of (sallow)t = 24 ksi and (sallow)c = 22 ksi, respectively, determine the maximum allowable internal moment M that can be applied to the beam.
3 in. 3 in. 6 in. M 2 in.
1.5 in.
Section Properties: The neutral axis passes through centroid C of the cross section as shown in Fig. a. The location of C is
©yA y = = ©A
4(8)(6) - 2 cp A 1.52 B d 8(6) - p A 1.52 B
= 4.3454 in.
Thus, the moment of inertia of the cross section about the neutral axis is I = ©I + Ad2
=
1 1 (6) A 83 B + 6(8) A 4.3454 - 4 B 2 - B p A 1.54 B + p A 1.52 B A 4.3454 - 2 B 2 R 12 4
= 218.87 in4 Allowable Bending Stress: The maximum compressive and tensile bending stress occurs at the top and bottom edges of the cross section. For the top edge, (sallow)c =
My ; I
22 =
M(8 - 4.3454) 218.87 M = 1317.53 kip # ina
1 ft b = 109.79 kip # ft 12 in.
For the bottom edge,
A smax B t =
Mc ; I
24 =
M(4.3454) 218.87
M = 1208.82 kip # ina
1 ft b = 101 kip # ft (controls) 12 in.
363
Ans.
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*6–60. The beam is constructed from four boards as shown. If it is subjected to a moment of Mz = 16 kip # ft, determine the stress at points A and B. Sketch a three-dimensional view of the stress distribution.
y
A
C
1 in. 10 in. 1 in. 10 in.
2[5(10)(1)] + 10.5(16)(1) + 16(10)(1) y = 2(10)(1) + 16(1) + 10(1)
Mz 16 kipft z
= 9.3043 in.
14 in.
1 1 I = 2c (1)(103) + 1(10)(9.3043 - 5)2 d + (16)(13) + 16(1)(10.5 - 9.3043)2 12 12 +
B
1 in.
x
1 in.
1 (1)(103) + 1(10)(16 - 9.3043)2 = 1093.07 in4 12
sA =
16(12)(21 - 9.3043) Mc = = 2.05 ksi I 1093.07
Ans.
sB =
My 16(12)(9.3043) = = 1.63 ksi I 1093.07
Ans.
•6–61. The beam is constructed from four boards as shown. If it is subjected to a moment of Mz = 16 kip # ft, determine the resultant force the stress produces on the top board C.
y
A
C
1 in. 10 in. 1 in. 10 in.
y =
2[5(10)(1)] + 10.5(16)(1) + 16(10)(1) = 9.3043 in. 2(10)(1) + 16(1) + 10(1)
Mz 16 kipft z 14 in.
1 1 I = 2c (1)(103) + (10)(9.3043 - 5)2 d + (16)(13) + 16(1)(10.5 - 9.3043)2 12 12 +
1 (1)(103) + 1(10)(16 - 9.3043)2 = 1093.07 in4 12
sA =
16(12)(21 - 9.3043) Mc = = 2.0544 ksi I 1093.07
sD =
My 16(12)(11 - 9.3043) = = 0.2978 ksi I 1093.07
(FR)C =
1 (2.0544 + 0.2978)(10)(1) = 11.8 kip 2
Ans.
364
1 in.
B
1 in.
x
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6–62. A box beam is constructed from four pieces of wood, glued together as shown. If the moment acting on the cross section is 10 kN # m, determine the stress at points A and B and show the results acting on volume elements located at these points.
20 mm
160 mm
25 mm A 250 mm
25 mm
B
M 10 kNm
The moment of inertia of the cross-section about the neutral axis is I =
1 1 (0.2)(0.33) (0.16)(0.253) = 0.2417(10 - 3) m4. 12 12
For point A, yA = C = 0.15 m. sA =
10(103) (0.15) MyA = 6.207(106)Pa = 6.21 MPa (C) = I 0.2417(10 - 3)
Ans.
For point B, yB = 0.125 m. sB =
MyB 10(103)(0.125) = 5.172(106)Pa = 5.17 MPa (T) = I 0.2417(10 - 3)
Ans.
The state of stress at point A and B are represented by the volume element shown in Figs. a and b respectively.
365
20 mm
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6–63. Determine the dimension a of a beam having a square cross section in terms of the radius r of a beam with a circular cross section if both beams are subjected to the same internal moment which results in the same maximum bending stress.
a a
r
Section Properties: The moments of inertia of the square and circular cross sections about the neutral axis are 1 a4 a A a3 B = 12 12
IS =
IC =
1 4 pr 4
Maximum Bending Stress: For the square cross section, c = a>2.
A smax B S =
M(a>2) 6M Mc = 3 = 4 IS a >12 a
For the circular cross section, c = r.
A smax B c =
Mc Mr 4M = Ic 1 4 pr3 pr 4
It is required that
A smax B S = A smax B C 6M 4M = a3 pr3 a = 1.677r
Ans.
*6–64. The steel rod having a diameter of 1 in. is subjected to an internal moment of M = 300 lb # ft. Determine the stress created at points A and B. Also, sketch a three-dimensional view of the stress distribution acting over the cross section. I =
A B
p 4 p r = (0.54) = 0.0490874 in4 4 4
sA =
M ⫽ 300 lb⭈ft 45⬚
300(12)(0.5) Mc = = 36.7 ksi I 0.0490874
Ans. 0.5 in.
My 300(12)(0.5 sin 45°) sB = = = 25.9 ksi I 0.0490874
Ans.
366
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•6–65.
If the moment acting on the cross section of the beam is M = 4 kip # ft, determine the maximum bending stress in the beam. Sketch a three-dimensional view of the stress distribution acting over the cross section.
A
1.5 in. 12 in.
The moment of inertia of the cross-section about the neutral axis is
12 in.
1 1 (12)(153) (10.5)(123) = 1863 in4 I = 12 12
M 1.5 in.
1.5 in.
Along the top edge of the flange y = c = 7.5 in. Thus smax =
4(103)(12)(7.5) Mc = = 193 psi I 1863
Ans.
Along the bottom edge to the flange, y = 6 in. Thus s =
4(103)(12)(6) My = = 155 psi I 1863
6–66. If M = 4 kip # ft, determine the resultant force the bending stress produces on the top board A of the beam. A
1.5 in.
The moment of inertia of the cross-section about the neutral axis is
12 in.
1 1 (12)(153) (10.5)(123) = 1863 in4 12 12
I =
12 in. M
Along the top edge of the flange y = c = 7.5 in. Thus 1.5 in.
smax =
4(103)(12)(7.5) Mc = = 193.24 psi I 1863
Along the bottom edge of the flange, y = 6 in. Thus s =
4(103)(12)(6) My = = 154.59 psi I 1863
The resultant force acting on board A is equal to the volume of the trapezoidal stress block shown in Fig. a. FR =
1 (193.24 + 154.59)(1.5)(12) 2
= 3130.43 lb = 3.13 kip
Ans.
367
1.5 in.
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6–67. The rod is supported by smooth journal bearings at A and B that only exert vertical reactions on the shaft. If d = 90 mm, determine the absolute maximum bending stress in the beam, and sketch the stress distribution acting over the cross section.
12 kN/m d A
B 3m
Absolute Maximum Bending Stress: The maximum moment is Mmax = 11.34 kN # m as indicated on the moment diagram. Applying the flexure formula smax =
Mmax c I 11.34(103)(0.045)
=
p 4
(0.0454)
= 158 MPa
Ans.
368
1.5 m
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*6–68. The rod is supported by smooth journal bearings at A and B that only exert vertical reactions on the shaft. Determine its smallest diameter d if the allowable bending stress is sallow = 180 MPa.
12 kN/m d A
B 3m
1.5 m
Allowable Bending Stress: The maximum moment is Mmax = 11.34 kN # m as indicated on the moment diagram. Applying the flexure formula Mmax c I
smax = sallow =
11.34(103) A d2 B
180 A 106 B =
p 4
A d2 B 4
d = 0.08626 m = 86.3 mm
Ans.
•6–69.
Two designs for a beam are to be considered. Determine which one will support a moment of M = 150 kN # m with the least amount of bending stress. What is that stress?
200 mm
200 mm
30 mm
15 mm
300 mm 30 mm
Section Property:
300 mm 15 mm
For section (a) I =
1 1 (0.2) A 0.333 B (0.17)(0.3)3 = 0.21645(10 - 3) m4 12 12
15 mm (a)
For section (b) I =
1 1 (0.2) A 0.363 B (0.185) A 0.33 B = 0.36135(10 - 3) m4 12 12
Maximum Bending Stress: Applying the flexure formula smax =
Mc I
For section (a) smax =
150(103)(0.165) 0.21645(10 - 3)
= 114.3 MPa
For section (b) smax =
150(103)(0.18) 0.36135(10 - 3)
= 74.72 MPa = 74.7 MPa
Ans.
369
30 mm (b)
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6–70. The simply supported truss is subjected to the central distributed load. Neglect the effect of the diagonal lacing and determine the absolute maximum bending stress in the truss. The top member is a pipe having an outer diameter of 1 in. 3 and thickness of 16 in., and the bottom member is a solid rod having a diameter of 12 in.
y =
100 lb/ft
6 ft
5.75 in.
6 ft
6 ft
©yA 0 + (6.50)(0.4786) = = 4.6091 in. ©A 0.4786 + 0.19635
I = c
1 1 1 p(0.5)4 - p(0.3125)4 d + 0.4786(6.50 - 4.6091)2 + p(0.25)4 4 4 4
+ 0.19635(4.6091)2 = 5.9271 in4 Mmax = 300(9 - 1.5)(12) = 27 000 lb # in. smax =
27 000(4.6091 + 0.25) Mc = I 5.9271
= 22.1 ksi
Ans.
6–71. The axle of the freight car is subjected to wheel loadings of 20 kip. If it is supported by two journal bearings at C and D, determine the maximum bending stress developed at the center of the axle, where the diameter is 5.5 in. A
C
B
60 in. 10 in. 20 kip
smax =
200(2.75) Mc = 1 = 12.2 ksi 4 I 4 p(2.75)
Ans.
370
D
10 in. 20 kip
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*6–72. The steel beam has the cross-sectional area shown. Determine the largest intensity of distributed load w0 that it can support so that the maximum bending stress in the beam does not exceed smax = 22 ksi.
w0
12 ft
12 ft 8 in. 0.30 in. 10 in.
0.3 in.
Support Reactions: As shown on FBD.
0.30 in.
Internal Moment: The maximum moment occurs at mid span. The maximum moment is determined using the method of sections. Section Property: I =
1 1 (8) A 10.63 B (7.7) A 103 B = 152.344 in4 12 12
Absolute Maximum Bending Stress: The maximum moment is Mmax = 48.0w0 as indicated on the FBD. Applying the flexure formula smax = 22 =
Mmax c I 48.0w0 (12)(5.30) 152.344
w0 = 1.10 kip>ft
Ans.
•6–73.
The steel beam has the cross-sectional area shown. If w0 = 0.5 kip>ft, determine the maximum bending stress in the beam.
w0
12 ft
12 ft 8 in.
Support Reactions: As shown on FBD.
0.3 in.
0.30 in.
Internal Moment: The maximum moment occurs at mid span. The maximum moment is determined using the method of sections. Section Property: I =
1 1 (8) A 10.63 B (7.7) A 103 B = 152.344 in4 12 12
Absolute Maximum Bending Stress: The maximum moment is Mmax = 24.0 kip # ft as indicated on the FBD. Applying the flexure formula smax =
=
Mmax c I
24.0(12)(5.30) 152.344
= 10.0 ksi
Ans.
371
0.30 in. 10 in.
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6–74. The boat has a weight of 2300 lb and a center of gravity at G. If it rests on the trailer at the smooth contact A and can be considered pinned at B, determine the absolute maximum bending stress developed in the main strut of the trailer. Consider the strut to be a box-beam having the dimensions shown and pinned at C.
B 1 ft
G
C
A 3 ft
D
5 ft
4 ft
1.75 in.
1 ft
1.75 in.
3 in. 1.5 in.
Boat: + ©F = 0; : x a + ©MB = 0;
Bx = 0 -NA(9) + 2300(5) = 0 NA = 1277.78 lb
+ c ©Fy = 0;
1277.78 - 2300 + By = 0 By = 1022.22 lb
Assembly: a + ©MC = 0;
-ND(10) + 2300(9) = 0 ND = 2070 lb
+ c ©Fy = 0;
Cy + 2070 - 2300 = 0 Cy = 230 lb
I =
1 1 (1.75)(3)3 (1.5)(1.75)3 = 3.2676 in4 12 12
smax =
3833.3(12)(1.5) Mc = = 21.1 ksi I 3.2676
Ans.
372
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6–75. The shaft is supported by a smooth thrust bearing at A and smooth journal bearing at D. If the shaft has the cross section shown, determine the absolute maximum bending stress in the shaft.
40 mm A
B 0.75 m
D
C 1.5 m
25 mm
0.75 m
3 kN
3 kN
Shear and Moment Diagrams: As shown in Fig. a. Maximum Moment: Due to symmetry, the maximum moment occurs in region BC of the shaft. Referring to the free-body diagram of the segment shown in Fig. b. Section Properties: The moment of inertia of the cross section about the neutral axis is I =
p A 0.044 - 0.0254 B = 1.7038 A 10 - 6 B m4 4
Absolute Maximum Bending Stress:
sallow =
2.25 A 103 B (0.04) Mmaxc = = 52.8 MPa I 1.7038 A 10 - 6 B
Ans.
*6–76. Determine the moment M that must be applied to the beam in order to create a maximum stress of 80 MPa.Also sketch the stress distribution acting over the cross section.
300 mm
20 mm
The moment of inertia of the cross-section about the neutral axis is I =
M
1 1 (0.3)(0.33) (0.21)(0.263) = 0.36742(10 - 3) m4 12 12
260 mm
Thus, 20 mm 30 mm
smax
Mc = ; I
6
80(10 ) =
M(0.15) 0.36742(10 - 3)
M = 195.96 (103) N # m = 196 kN # m The bending stress distribution over the cross-section is shown in Fig. a.
373
Ans.
30 mm 30 mm
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•6–77.
The steel beam has the cross-sectional area shown. Determine the largest intensity of distributed load w that it can support so that the bending stress does not exceed smax = 22 ksi.
I = smax
w
1 1 (8)(10.6)3 (7.7)(103) = 152.344 in4 12 12
8 ft
w
8 ft
8 in.
Mc = I
22 =
8 ft
0.30 in. 10 in.
0.3 in.
0.30 in.
32w(12)(5.3) 152.344
w = 1.65 kip>ft
Ans.
6–78. The steel beam has the cross-sectional area shown. If w = 5 kip>ft, determine the absolute maximum bending stress in the beam.
w
8 ft
w
8 ft
8 ft 8 in. 0.3 in.
0.30 in. 10 in. 0.30 in.
From Prob. 6-78: M = 32w = 32(5)(12) = 1920 kip # in. I = 152.344 in4 smax =
1920(5.3) Mc = = 66.8 ksi I 152.344
Ans.
374
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6–79. If the beam ACB in Prob. 6–9 has a square cross section, 6 in. by 6 in., determine the absolute maximum bending stress in the beam.
15 kip 1 ft
A
20 kip
C 4 ft
B
4 ft
4 ft
Mmax = 46.7 kip # ft smax =
46.7(103)(12)(3) Mc = 15.6 ksi = 1 3 I 12 (6)(6 )
Ans.
*6–80. If the crane boom ABC in Prob. 6–3 has a rectangular cross section with a base of 2.5 in., determine its required height h to the nearest 14 in. if the allowable bending stress is sallow = 24 ksi.
A
a + ©MA = 0;
+ c ©Fy = 0;
-Ay +
+ ©F = 0; ; x
Ax -
smax =
4 (4000) - 1200 = 0; 5
3 (4000) = 0; 5
5 ft B
4 ft
4 F (3) - 1200(8) = 0; 5 B
3 ft
FB = 4000 lb
Ay = 2000 lb
Ax = 2400 lb
6000(12) A h2 B Mc = 24(10)3 = 1 3 I 12 (2.5)(h )
h = 2.68 in.
Ans.
Use h = 2.75 in.
Ans.
375
C
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•6–81.
If the reaction of the ballast on the railway tie can be assumed uniformly distributed over its length as shown, determine the maximum bending stress developed in the tie. The tie has the rectangular cross section with thickness t = 6 in.
15 kip 1.5 ft
15 kip 5 ft
1.5 ft
t
w
Support Reactions: Referring to the free - body diagram of the tie shown in Fig. a, we have + c ©Fy = 0;
w(8) - 2(15) = 0 w = 3.75 kip>ft
Maximum Moment: The shear and moment diagrams are shown in Figs. b and c. As indicated on the moment diagram, the maximum moment is Mmax = 7.5 kip # ft. Absolute Maximum Bending Stress: smax =
12 in.
7.5(12)(3) Mmaxc = 1.25 ksi = I 1 (12)(63) 12
Ans.
376
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6–82. The reaction of the ballast on the railway tie can be assumed uniformly distributed over its length as shown. If the wood has an allowable bending stress of sallow = 1.5 ksi, determine the required minimum thickness t of the rectangular cross sectional area of the tie to the nearest 18 in.
15 kip 1.5 ft
15 kip 5 ft
1.5 ft
t
w
Support Reactions: Referring to the free-body diagram of the tie shown in Fig. a, we have + c ©Fy = 0;
w(8) - 2(15) = 0 w = 3.75 kip>ft
Maximum Moment: The shear and moment diagrams are shown in Figs. b and c. As indicated on the moment diagram, the maximum moment is Mmax = 7.5 kip # ft. Absolute Maximum Bending Stress:
smax
t 7.5(12)a b 2 1.5 = 1 (12)t3 12
Mc = ; I
t = 5.48 in. Use
t = 5
12 in.
1 in. 2
Ans.
377
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6–83. Determine the absolute maximum bending stress in the tubular shaft if di = 160 mm and do = 200 mm.
15 kN/m 60 kN m d i do A
B 3m
Section Property: I =
p A 0.14 - 0.084 B = 46.370 A 10 - 6 B m4 4
Absolute Maximum Bending Stress: The maximum moment is Mmax = 60.0 kN # m as indicated on the moment diagram. Applying the flexure formula smax =
Mmaxc I
60.0(103)(0.1) =
46.370(10 - 6)
= 129 MPa
Ans.
378
1m
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*6–84. The tubular shaft is to have a cross section such that its inner diameter and outer diameter are related by di = 0.8do. Determine these required dimensions if the allowable bending stress is sallow = 155 MPa.
15 kN/m 60 kN m d i do A
B 3m
Section Property: I =
0.8do 4 do 4 dl 4 p do 4 p - a b R = 0.009225pd4o Ba b - a b R = B 4 2 2 4 16 2
Allowable Bending Stress: The maximum moment is Mmax = 60.0 kN # m as indicated on the moment diagram. Applying the flexure formula smax = sallow = 155 A 106 B =
Thus,
Mmax c I 60.0(103) A 2o B d
0.009225pd4o
do = 0.1883 m = 188 mm
Ans.
dl = 0.8do = 151 mm
Ans.
379
1m
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6–85. The wood beam has a rectangular cross section in the proportion shown. Determine its required dimension b if the allowable bending stress is sallow = 10 MPa.
500 N/m
1.5b A
B b 2m
Allowable Bending Stress: The maximum moment is Mmax = 562.5 N # m as indicated on the moment diagram. Applying the flexure formula smax = sallow = 10 A 106 B =
Mmax c I 562.5(0.75b) 1 12
(b)(1.5b)3
b = 0.05313 m = 53.1 mm
Ans.
380
2m
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6–86. Determine the absolute maximum bending stress in the 2-in.-diameter shaft which is subjected to the concentrated forces. The journal bearings at A and B only support vertical forces.
800 lb 600 lb
A
15 in.
B
15 in. 30 in.
The FBD of the shaft is shown in Fig. a. The shear and moment diagrams are shown in Fig. b and c, respectively. As indicated on the moment diagram, Mmax = 15000 lb # in. The moment of inertia of the cross-section about the neutral axis is I =
p 4 (1 ) = 0.25 p in4 4
Here, c = 1 in. Thus smax =
=
Mmax c I 15000(1) 0.25 p
= 19.10(103) psi Ans.
= 19.1 ksi
381
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6–87. Determine the smallest allowable diameter of the shaft which is subjected to the concentrated forces. The journal bearings at A and B only support vertical forces. The allowable bending stress is sallow = 22 ksi.
800 lb 600 lb
A
15 in.
B
15 in. 30 in.
The FBD of the shaft is shown in Fig. a The shear and moment diagrams are shown in Fig. b and c respectively. As indicated on the moment diagram, Mmax = 15,000 lb # in The moment of inertia of the cross-section about the neutral axis is I =
p 4 p d 4 a b = d 4 2 64
Here, c = d>2. Thus sallow =
Mmax c ; I
22(103) =
15000(d> 2) pd4>64
d = 1.908 in = 2 in.
Ans.
382
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*6–88. If the beam has a square cross section of 9 in. on each side, determine the absolute maximum bending stress in the beam.
1200 lb
800 lb/ft
B A 8 ft
Absolute Maximum Bending Stress: The maximum moment is Mmax = 44.8 kip # ft as indicated on moment diagram. Applying the flexure formula smax =
44.8(12)(4.5) Mmax c = = 4.42 ksi 1 3 I 12 (9)(9)
Ans.
•6–89.
If the compound beam in Prob. 6–42 has a square cross section, determine its dimension a if the allowable bending stress is sallow = 150 MPa.
Allowable Bending Stress: The maximum moments is Mmax = 7.50 kN # m as indicated on moment diagram. Applying the flexure formula smax = sallow = 150 A 106 B =
Mmax c I 7.50(103) A a2 B 1 12
a4
a = 0.06694 m = 66.9 mm
Ans.
383
8 ft
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6–90. If the beam in Prob. 6–28 has a rectangular cross section with a width b and a height h, determine the absolute maximum bending stress in the beam.
Absolute Maximum Bending Stress: The maximum moments is Mmax =
23w0 L2 216
as indicated on the moment diagram. Applying the flexure formula
smax
Mmax c = = I
A B
23w0 L2 h 2 216 1 3 12 bh
23w0 L2 =
Ans.
36bh2
384
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6–91. Determine the absolute maximum bending stress in the 80-mm-diameter shaft which is subjected to the concentrated forces. The journal bearings at A and B only support vertical forces.
A
0.5 m
B
0.4 m
0.6 m
12 kN 20 kN
The FBD of the shaft is shown in Fig. a The shear and moment diagrams are shown in Fig. b and c, respectively. As indicated on the moment diagram, Mmax = 6 kN # m. The moment of inertia of the cross-section about the neutral axis is I =
p (0.044) = 0.64(10 - 6)p m4 4
Here, c = 0.04 m. Thus smax =
6(103)(0.04) Mmax c = I 0.64(10 - 6)p = 119.37(106) Pa = 119 MPa
Ans.
385
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*6–92. Determine the smallest allowable diameter of the shaft which is subjected to the concentrated forces. The journal bearings at A and B only support vertical forces. The allowable bending stress is sallow = 150 MPa.
A
0.5 m
B
0.4 m
0.6 m
12 kN 20 kN
The FBD of the shaft is shown in Fig. a. The shear and moment diagrams are shown in Fig. b and c, respectively. As indicated on the moment diagram, Mmax = 6 kN # m. The moment of inertia of the cross-section about the neutral axis is I =
pd4 p d 4 a b = 4 2 64
Here, c = d>2. Thus sallow =
Mmax c ; I
150(106) =
6(103)(d> 2) pd4>64
d = 0.07413 m = 74.13 mm = 75 mm
386
Ans.
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•6–93. The man has a mass of 78 kg and stands motionless at the end of the diving board. If the board has the cross section shown, determine the maximum normal strain developed in the board. The modulus of elasticity for the material is E = 125 GPa. Assume A is a pin and B is a roller.
350 mm 30 mm A 1.5 m
Internal Moment: The maximum moment occurs at support B. The maximum moment is determined using the method of sections. Section Property: y =
=
I =
©yA ©A 0.01(0.35)(0.02) + 0.035(0.03)(0.03) = 0.012848 m 0.35(0.02) + 0.03(0.03) 1 (0.35) A 0.023 B + 0.35(0.02)(0.012848 - 0.01)2 12 +
1 (0.03) A 0.033 B + 0.03(0.03)(0.035 - 0.012848)2 12
= 0.79925 A 10 - 6 B m4 Absolute Maximum Bending Stress: The maximum moment is Mmax = 1912.95 N # m as indicated on the FBD. Applying the flexure formula smax =
Mmax c I 1912.95(0.05 - 0.012848)
=
0.79925(10 - 6)
= 88.92 MPa Absolute Maximum Normal Strain: Applying Hooke’s law, we have emax =
88.92(106) smax = 0.711 A 10 - 3 B mm>mm = E 125(109)
Ans.
387
B
2.5 m
C
20 mm 10 mm 10 mm 10 mm
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6–94. The two solid steel rods are bolted together along their length and support the loading shown. Assume the support at A is a pin and B is a roller. Determine the required diameter d of each of the rods if the allowable bending stress is sallow = 130 MPa.
20 kN/m
80 kN
A B
2m
Section Property: I = 2B
2m
p d 4 p d 2 5p 4 a b + d2 a b R = d 4 2 4 2 32
Allowable Bending Stress: The maximum moment is Mmax = 100 kN # m as indicated on moment diagram. Applying the flexure formula smax = sallow = 130 A 106 B =
Mmax c I 100(103)(d) 5p 32
d4
d = 0.1162 m = 116 mm
Ans.
6–95. Solve Prob. 6–94 if the rods are rotated 90° so that both rods rest on the supports at A (pin) and B (roller).
20 kN/m
Section Property: I = 2B
A
p d 4 p 4 a b R = d 4 2 32
smax = sallow =
2m
Mmax c I
100(103)(d) p 32
B
2m
Allowable Bending Stress: The maximum moment is Mmax = 100 kN # m as indicated on the moment diagram. Applying the flexure formula
130 A 106 B =
80 kN
d4
d = 0.1986 m = 199 mm
Ans.
388
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*6–96. The chair is supported by an arm that is hinged so it rotates about the vertical axis at A. If the load on the chair is 180 lb and the arm is a hollow tube section having the dimensions shown, determine the maximum bending stress at section a–a.
180 lb
1 in. a 3 in.
A a
0.5 in.
8 in.
c + ©M = 0;
M - 180(8) = 0 M = 1440 lb # in.
Ix =
1 1 (1)(33) (0.5)(2.53) = 1.59896 in4 12 12
smax =
1440 (1.5) Mc = = 1.35 ksi I 1.59896
Ans.
s (ksi)
•6–97.
A portion of the femur can be modeled as a tube having an inner diameter of 0.375 in. and an outer diameter of 1.25 in. Determine the maximum elastic static force P that can be applied to its center. Assume the bone to be roller supported at its ends. The s– P diagram for the bone mass is shown and is the same in tension as in compression.
P
2.30 1.25 4 in.
0.02
I =
1 p 4
0.375 4 4 4 C A 1.25 2 B - A 2 B D = 0.11887 in
Mmax =
P (4) = 2P 2
Require smax = 1.25 ksi smax =
Mc I
1.25 =
2P(1.25>2) 0.11887
P = 0.119 kip = 119 lb
Ans.
389
2.5 in.
0.05
P (in./ in.)
4 in.
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6–98. If the beam in Prob. 6–18 has a rectangular cross section with a width of 8 in. and a height of 16 in., determine the absolute maximum bending stress in the beam. 16 in.
Absolute Maximum Bending Stress: The maximum moment is Mmax = 216 kip # ft as indicated on moment diagram. Applying the flexure formula smax =
216(12)(8) Mmax c = 7.59 ksi = 1 3 I 12 (8)(16 )
8 in.
Ans.
6–99. If the beam has a square cross section of 6 in. on each side, determine the absolute maximum bending stress in the beam.
400 lb/ft
B A 6 ft
The maximum moment occurs at the fixed support A. Referring to the FBD shown in Fig. a, a + ©MA = 0;
Mmax - 400(6)(3) -
1 (400)(6)(8) = 0 2
Mmax = 16800 lb # ft The moment of inertia of the about the neutral axis is I =
smax =
1 (6)(63) = 108 in4. Thus, 12
16800(12)(3) Mc = I 108 = 5600 psi = 5.60 ksi
Ans.
390
6 ft
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*6–100. The steel beam has the cross-sectional area shown. Determine the largest intensity of the distributed load w0 that it can support so that the maximum bending stress in the beam does not exceed sallow = 22 ksi.
w0
9 ft
9 ft
9 in. 0.25 in.
0.25 in. 12 in. 0.25 in.
Support Reactions. The FBD of the beam is shown in Fig. a. The shear and moment diagrams are shown in Fig. a and b, respectively. As indicated on the moment diagram, Mmax = 27wo. The moment of inertia of the cross-section about the neutral axis is I =
1 1 (9)(12.53) (8.75)(123) 12 12
= 204.84375 in4 Here, ¢ = 6.25 in. Thus, sallow =
Mmax c ; I
22(103) =
(27wo)(12)(6.25) 204.84375
wo = 2 225.46 lb>ft = 2.23 kip>ft
Ans.
391
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•6–101.
The steel beam has the cross-sectional area shown. If w0 = 2 kip>ft, determine the maximum bending stress in the beam.
w0
9 ft
9 ft
9 in. 0.25 in.
0.25 in. 12 in. 0.25 in.
The FBD of the beam is shown in Fig. a The shear and moment diagrams are shown in Fig. b and c, respectively. As indicated on the moment diagram, Mmax = 54 kip # ft. The moment of inertia of the I cross-section about the bending axis is I =
1 1 (9) A 12.53 B (8.75) A 123 B 12 12
= 204.84375 in4 Here, c = 6.25 in. Thus smax =
=
Mmax c I 54 (12)(6.25) 204.84375
= 19.77 ksi = 19.8 ksi
Ans.
392
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6–102. The bolster or main supporting girder of a truck body is subjected to the uniform distributed load. Determine the bending stress at points A and B.
1.5 kip/ft
A 8 ft
B 12 ft F2
F1 0.75 in. 6 in.
12 in.
0.5 in. A B
0.75 in.
Support Reactions: As shown on FBD. Internal Moment: Using the method of sections. + ©MNA = 0;
M + 12.0(4) - 15.0(8) = 0 M = 72.0 kip # ft
Section Property: I =
1 1 (6) A 13.53 B (5.5) A 123 B = 438.1875 in4 12 12
Bending Stress: Applying the flexure formula s =
My I
sB =
72.0(12)(6.75) = 13.3 ksi 438.1875
Ans.
sA =
72.0(12)(6) = 11.8 ksi 438.1875
Ans.
393
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6–103. Determine the largest uniform distributed load w that can be supported so that the bending stress in the beam does not exceed s allow = 5 MPa .
w
The FBD of the beam is shown in Fig. a
0.5 m
The shear and moment diagrams are shown in Fig. b and c, respectively. As indicated on the moment diagram, |Mmax| = 0.125 w.
150 mm
The moment of inertia of the cross-section is, I =
1 (0.075) A 0.153 B = 21.09375 A 10 - 6 B m4 12
Here, c = 0.075 w. Thus, sallow = 5 A 106 B =
Mmax c ; I 0.125w(0.075)
21.09375 A 10 - 6 B
w = 11250 N>m = 11.25 kN>m
Ans.
394
1m 75 mm
0.5 m
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w
*6–104. If w = 10 kN>m, determine the maximum bending stress in the beam. Sketch the stress distribution acting over the cross section. Support Reactions. The FBD of the beam is shown in Fig. a
0.5 m
75 mm
The shear and moment diagrams are shown in Figs. b and c, respectively. As indicated on the moment diagram, |Mmax| = 1.25 kN # m.
150 mm
The moment of inertia of the cross-section is I =
1 (0.075) A 0.153 B = 21.09375 A 10 - 6 B m4 12
Here, c = 0.075 m. Thus smax =
=
Mmax c I 1.25 A 103 B (0.075) 21.09375 A 10 - 6 B
= 4.444 A 106 B Pa = 4.44 MPa
Ans.
The bending stress distribution over the cross section is shown in Fig. d
395
1m
0.5 m
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400 lb/ft
•6–105.
If the allowable bending stress for the wood beam is sallow = 150 psi, determine the required dimension b to the nearest 14 in. of its cross section. Assume the support at A is a pin and B is a roller.
B
A 3 ft
The FBD of the beam is shown in Fig. a The shear and moment diagrams are shown in Figs. b and c, respectively. As indicated on the moment diagram, Mmax = 3450 lb # ft.
2b b
The moment of inertia of the cross section is I =
2 1 (b)(2b)3 = b4 12 3
Here, c = 2b> 2 = b. Thus, sallow = 150 =
Mmax c ; I 3450(12)(b) > 3 b4
2
b = 7.453 in = 7
1 in. 2
Ans.
396
3 ft
3 ft
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400 lb/ft
6–106. The wood beam has a rectangular cross section in the proportion shown. If b 7.5 in., determine the absolute maximum bending stress in the beam.
B
A
The FBD of the beam is shown in Fig. a.
3 ft
The shear and moment diagrams are shown in Fig. b and c, respectively. As indicated on the moment diagram, Mmax = 3450 lb # ft.
2b b
The moment of inertia of the cross-section is I =
1 (7.5) A 153 B = 2109.375 in4 12
Here, c =
15 = 7.5 in. Thus 2
smax =
3450(12)(7.5) Mmax c = = 147 psi I 2109.375
Ans.
397
3 ft
3 ft
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6–107. A beam is made of a material that has a modulus of elasticity in compression different from that given for tension. Determine the location c of the neutral axis, and derive an expression for the maximum tensile stress in the beam having the dimensions shown if it is subjected to the bending moment M.
M h s
P
Ec(emax)t (h - c) c
Ec
Location of neutral axis: + ©F = 0; :
1 1 - (h - c)(smax)c (b) + (c)(smax)t (b) = 0 2 2
(h - c)(smax)c = c(smax)t (h - c)Ec (emax)t
[1]
(h - c) = cEt (emax)t ; c
Ec (h - c)2 = Etc2
Taking positive root: Ec c = h - c A Et Ec h A Et h2Ec c = = Ec 2Et + 2Ec 1 + A Et
[2] Ans.
©MNA = 0; 1 2 1 2 M = c (h - c)(smax)c (b) d a b (h - c) + c (c)(smax)t(b) d a b(c) 2 3 2 3 M =
1 1 (h - c)2 (b)(smax)c + c2b(smax)t 3 3
From Eq. [1]. (smax)c =
c (s ) h - c max t
M =
c 1 1 (h - c)2 (b)a b (smax)t + c2b(smax)t 3 h - c 3
M =
1 bc(smax)t (h - c + c) ; 3
(smax)t =
3M bhc
From Eq. [2]
(smax)t =
b
Et
(emax)t (h - c) (emax)c = c (smax)c = Ec(emax)c =
c
3M 2Et + 2Ec £ ≥ b h2 2Ec
Ans.
398
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*6–108. The beam has a rectangular cross section and is subjected to a bending moment M. If the material from which it is made has a different modulus of elasticity for tension and compression as shown, determine the location c of the neutral axis and the maximum compressive stress in the beam.
M h s
c b
Et
P Ec
See the solution to Prob. 6–107 c =
h2Ec
Ans.
2Et + 2Ec
Since (smax)c =
(smax)c =
c (s ) = h - c max t
2Ec 2Et
h2Ec ( 2Et + 2Ec)ch - a
h 1Ec 1Et + 1Ec
bd
(smax)t
(smax)t
(smax)c =
2Et + 2Ec 2Ec 3M ¢ 2≤¢ ≤ bh 2Et 2Ec
(smax)c =
3M 2Et + 2Ec ¢ ≤ bh2 2Et
Ans.
399
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•6–109.
The beam is subjected to a bending moment of M = 20 kip # ft directed as shown. Determine the maximum bending stress in the beam and the orientation of the neutral axis.
y 8 in. C
B
The y and z components of M are negative, Fig. a. Thus,
14 in. z
My = -20 sin 45° = -14.14 kip # ft
45 16 in.
Mz = -20 cos 45° = -14.14 kip # ft. The moments of inertia of the cross-section about the principal centroidal y and z axes are Iy =
1 1 (16) A 103 B (14) A 83 B = 736 in4 12 12
Iz =
1 1 (10) A 163 B (8) A 143 B = 1584 in4 12 12
My z
Mz y Iz
+
smax = sC = -
Iy -14.14(12)(8) -14.14(12)( - 5) + 1584 736
= 2.01 ksi smax = sA = -
(T)
Ans.
-14.14(12)(-8) -14.14(12)(5) + 1584 736
= -2.01 ksi = 2.01 ksi (C)
Ans.
Here, u = 180° + 45° = 225° tan a =
tan a =
Iz Iy
D 10 in. M
By inspection, the bending stress occurs at corners A and C are s = -
A
tan u
1584 tan 225° 736
a = 65.1°
Ans.
The orientation of neutral axis is shown in Fig. b.
400
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6–110. Determine the maximum magnitude of the bending moment M that can be applied to the beam so that the bending stress in the member does not exceed 12 ksi.
y 8 in. C
B
The y and z components of M are negative, Fig. a. Thus, 14 in.
My = -M sin 45° = -0.7071 M
z
45 16 in.
Mz = -M cos 45° = -0.7071 M The moments of inertia of the cross-section about principal centroidal y and z axes are Iy =
1 1 (16) A 103 B (14) A 83 B = 736 in4 12 12
Iz =
1 1 (10) A 163 B (8) A 143 B = 1584 in4 12 12
12 = -
Myzc
Mz yc Iz
+
D 10 in. M
By inspection, the maximum bending stress occurs at corners A and C. Here, we will consider corner C. sC = sallow = -
A
Iy
-0.7071 M(12)( -5) -0.7071 M (12)(8) + 1584 736
M = 119.40 kip # ft = 119 kip # ft
Ans.
401
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6–111. If the resultant internal moment acting on the cross section of the aluminum strut has a magnitude of M = 520 N # m and is directed as shown, determine the bending stress at points A and B. The location y of the centroid C of the strut’s cross-sectional area must be determined. Also, specify the orientation of the neutral axis.
y M 520 Nm 12
20 mm z
–y
5
13
B C
200 mm
20 mm
20 mm A 200 mm
Internal Moment Components: Mz = -
12 (520) = -480 N # m 13
My =
5 (520) = 200 N # m 13
Section Properties: 0.01(0.4)(0.02) + 2[(0.110)(0.18)(0.02)] ©yA = ©A 0.4(0.02) + 2(0.18)(0.02)
y =
= 0.057368 m = 57.4 mm Iz =
Ans.
1 (0.4) A 0.023 B + (0.4)(0.02)(0.057368 - 0.01)2 12 +
1 (0.04) A 0.183 B + 0.04(0.18)(0.110 - 0.057368)2 12
= 57.6014 A 10 - 6 B m4 Iy =
1 1 (0.2) A 0.43 B (0.18) A 0.363 B = 0.366827 A 10 - 3 B m4 12 12
Maximum Bending Stress: Applying the flexure formula for biaxial at points A and B s = -
Myz
Mzy +
Iz
Iy 200(-0.2)
-480(-0.142632) sA = -
+
-6
57.6014(10 )
0.366827(10 - 3)
= -1.298 MPa = 1.30 MPa (C) 200(0.2)
-480(0.057368) sB = -
Ans.
+
-6
57.6014(10 )
0.366827(10 - 3)
= 0.587 MPa (T)
Ans.
Orientation of Neutral Axis: tan a =
tan a =
Iz Iy
tan u
57.6014(10 - 6) 0.366827(10 - 3)
tan (-22.62°)
a = -3.74°
Ans.
402
200 mm
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*6–112. The resultant internal moment acting on the cross section of the aluminum strut has a magnitude of M = 520 N # m and is directed as shown. Determine maximum bending stress in the strut. The location y of the centroid C of the strut’s cross-sectional area must be determined. Also, specify the orientation of the neutral axis.
y M 520 Nm 12
20 mm z
–y
5
13
B C
200 mm
20 mm
20 mm A 200 mm
Internal Moment Components: Mz = -
12 (520) = -480 N # m 13
My =
5 (520) = 200 N # m 13
Section Properties: 0.01(0.4)(0.02) + 2[(0.110)(0.18)(0.02)] ©yA = ©A 0.4(0.02) + 2(0.18)(0.02)
y =
= 0.057368 m = 57.4 mm Iz =
Ans.
1 (0.4) A 0.023 B + (0.4)(0.02)(0.057368 - 0.01)2 12 1 (0.04) A 0.183 B + 0.04(0.18)(0.110 - 0.057368)2 12
+
= 57.6014 A 10 - 6 B m4 Iy =
1 1 (0.2) A 0.43 B (0.18) A 0.363 B = 0.366827 A 10 - 3 B m4 12 12
Maximum Bending Stress: By inspection, the maximum bending stress can occur at either point A or B. Applying the flexure formula for biaxial bending at points A and B s = -
My z
Mz y +
Iz
Iy 200(-0.2)
-480(-0.142632) sA = -
+
57.6014(10 - 6)
0.366827(10 - 3)
= -1.298 MPa = 1.30 MPa (C) (Max) 200(0.2)
-480(0.057368) sB = -
-6
57.6014(10 )
Ans.
+
0.366827(10 - 3)
= 0.587 MPa (T) Orientation of Neutral Axis: tan a =
tan a =
Iz Iy
tan u
57.6014(10 - 6) 0.366827(10 - 3)
tan (-22.62°)
a = -3.74°
Ans.
403
200 mm
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6–113. Consider the general case of a prismatic beam subjected to bending-moment components My and Mz, as shown, when the x, y, z axes pass through the centroid of the cross section. If the material is linear-elastic, the normal stress in the beam is a linear function of position such that s = a + by + cz. Using the equilibrium conditions 0 = 1A s dA, My = 1A zs dA, Mz = 1A - ys dA, determine the constants a, b, and c, and show that the normal stress can be determined from the equation s = [-1MzIy + MyIyz2y + 1MyIz + MzIyz2z]>1IyIz - Iyz22, where the moments and products of inertia are defined in Appendix A.
y z My dA sC y Mz z
Equilibrium Condition: sx = a + by + cz 0 =
LA
sx dA
0 =
LA
(a + by + cz) dA
0 = a
LA
dA + b
LA
y dA + c
My =
LA
z sx dA
=
LA
z(a + by + cz) dA
= a
Mz =
=
LA
= -a
LA
LA
z dA + b
LA
LA
z dA
yz dA + c
LA
[1]
z2 dA
[2]
-y sx dA
-y(a + by + cz) dA
LA
ydA - b
y2 dA - c
LA
LA
yz dA
[3]
Section Properties: The integrals are defined in Appendix A. Note that LA
y dA =
LA
z dA = 0.Thus,
From Eq. [1]
Aa = 0
From Eq. [2]
My = bIyz + cIy
From Eq. [3]
Mz = -bIz - cIyz
Solving for a, b, c: a = 0 (Since A Z 0) b = -¢
Thus,
MzIy + My Iyz
sx = - ¢
Iy Iz -
I2yz
≤
Mz Iy + My Iyz Iy Iz -
I2yz
c =
≤y + ¢
My Iz + Mz Iyz Iy Iz - I2yz My Iy + MzIyz Iy Iz - I2yz
≤z
(Q.E.D.)
404
x
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6–114. The cantilevered beam is made from the Z-section having the cross-section shown. If it supports the two loadings, determine the bending stress at the wall in the beam at point A. Use the result of Prob. 6–113.
50 lb 50 lb 3 ft
(My)max Iy =
= 50(3) + 50(5) = 400 lb # ft = 4.80(103)lb # in.
2 ft
0.25 in. 2 in.
1 1 (3.25)(0.25)3 + 2c (0.25)(2)3 + (0.25)(2)(1.125)2 d = 1.60319 in4 12 12
A
B 2.25 in.
1 1 Iz = (0.25)(3.25)3 + 2 c (2)(0.25)3 + (0.25)(2)(1.5)2 d = 2.970378 in4 12 12
0.25 in. 3 in.
0.25 in.
Iyz = 2[1.5(1.125)(2)(0.25)] = 1.6875 in4 Using the equation developed in Prob. 6-113. s = -a
sA =
Mz Iy + My Iyz Iy Iz -
I2yz
by + a
My Iz + Mz Iyz Iy Iz - I2yz
bz
{-[0 + (4.80)(103)(1.6875)](1.625) + [(4.80)(103)(2.970378) + 0](2.125)} [1.60319(2.970378) - (1.6875)2]
= 8.95 ksi
Ans.
6–115. The cantilevered beam is made from the Z-section having the cross-section shown. If it supports the two loadings, determine the bending stress at the wall in the beam at point B. Use the result of Prob. 6–113.
50 lb 50 lb 3 ft
3
(My)max = 50(3) + 50(5) = 400 lb # ft = 4.80(10 )lb # in. Iy =
1 1 (3.25)(0.25)3 + 2c (0.25)(2)3 + (0.25)(2)(1.125)2 d = 1.60319 in4 12 12
1 1 Iz = (0.25)(3.25)3 + 2 c (2)(0.25)3 + (0.25)(2)(1.5)2 d = 2.970378 in4 12 12
2.25 in.
sB =
Iy Iz -
I2yz
by + a
My Iz + Mz Iyz Iy Iz - I2yz
0.25 in. 3 in.
0.25 in.
Using the equation developed in Prob. 6-113. Mz Iy + My Iyz
A
B
Iyz = 2[1.5(1.125)(2)(0.25)] = 1.6875 in4
s = -a
2 ft
0.25 in. 2 in.
bz
-[0 + (4.80)(103)(1.6875)]( -1.625) + [(4.80)(103)(2.976378) + 0](0.125) [(1.60319)(2.970378) - (1.6875)2]
= 7.81 ksi
Ans.
405
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*6–116. The cantilevered wide-flange steel beam is subjected to the concentrated force P at its end. Determine the largest magnitude of this force so that the bending stress developed at A does not exceed sallow = 180 MPa.
200 mm 10 mm 150 mm 10 mm
Internal Moment Components: Using method of section
10 mm A
y
©Mz = 0;
Mz + P cos 30°(2) = 0
Mz = -1.732P
©My = 0;
My + P sin 30°(2) = 0
My = -1.00P
z
Section Properties:
x
2m
30
1 1 Iz = (0.2) A 0.173 B (0.19) A 0.153 B = 28.44583(10 - 6) m4 12 12 Iy = 2 c
P
1 1 (0.01) A 0.23 B d + (0.15) A 0.013 B = 13.34583(10 - 6) m4 12 12
Allowable Bending Stress: By inspection, maximum bending stress occurs at points A and B. Applying the flexure formula for biaxial bending at point A. sA = sallow = 180 A 106 B = -
Myz
Mzy Iz
+
Iy
(-1.732P)(0.085) 28.44583(10 - 6)
-1.00P(-0.1) +
13.34583(10 - 6)
P = 14208 N = 14.2 kN
Ans.
•6–117.
The cantilevered wide-flange steel beam is subjected to the concentrated force of P = 600 N at its end. Determine the maximum bending stress developed in the beam at section A.
200 mm 10 mm 150 mm 10 mm
Internal Moment Components: Using method of sections
A
y
©Mz = 0;
Mz + 600 cos 30°(2) = 0
Mz = -1039.23 N # m
©My = 0;
My + 600 sin 30°(2) = 0;
My = -600.0 N # m
z
Section Properties:
x
1 1 Iz = (0.2) A 0.173 B (0.19) A 0.153 B = 28.44583(10 - 6) m4 12 12 Iy = 2 c
Maximum Bending Stress: By inspection, maximum bending stress occurs at A and B. Applying the flexure formula for biaxial bending at point A s = -
Myz
Mzy Iz
+
Iy -600.0(-0.1)
-1039.32(0.085) sA = -
-6
28.44583(10 )
= 7.60 MPa (T)
+
13.34583(10 - 6)
(Max)
Ans.
406
2m
30 P
1 1 (0.01) A 0.23 B d + (0.15) A 0.013 B = 13.34583(10 - 6) m4 12 12
10 mm
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6–118. If the beam is subjected to the internal moment of M = 1200 kN # m, determine the maximum bending stress acting on the beam and the orientation of the neutral axis.
y 150 mm 150 mm
Internal Moment Components: The y component of M is positive since it is directed towards the positive sense of the y axis, whereas the z component of M, which is directed towards the negative sense of the z axis, is negative, Fig. a. Thus,
M 300 mm 30
My = 1200 sin 30° = 600 kN # m
150 mm
Mz = -1200 cos 30° = -1039.23 kN # m
z
x 150 mm
Section Properties: The location of the centroid of the cross-section is given by ©yA 0.3(0.6)(0.3) - 0.375(0.15)(0.15) = = 0.2893 m ©A 0.6(0.3) - 0.15(0.15)
y =
150 mm
The moments of inertia of the cross section about the principal centroidal y and z axes are Iy =
1 1 (0.6) A 0.33 B (0.15) A 0.153 B = 1.3078 A 10 - 3 B m4 12 12
Iz =
1 (0.3) A 0.63 B + 0.3(0.6)(0.3 - 0.2893)2 12 - c
1 (0.15) A 0.153 B + 0.15(0.15)(0.375 - 0.2893)2 d 12
= 5.2132 A 10 - 3 B m4 Bending Stress: By inspection, the maximum bending stress occurs at either corner A or B. s = -
Myz
Mzy
sA = -
+
Iz
Iy
c -1039.23 A 103 B d(0.2893) 5.2132 A 10 - 3 B
+
600 A 103 B (0.15) 1.3078 A 10 - 3 B
= 126 MPa (T)
sB = -
c -1039.23 A 103 B d(-0.3107) 5.2132 A 10 - 3 B
+
600 A 103 B ( -0.15) 1.3078 A 10 - 3 B
= -131 MPa = 131 MPa (C)(Max.)
Ans.
Orientation of Neutral Axis: Here, u = -30°. tan a =
tan a =
Iz Iy
tan u
5.2132 A 10 - 3 B
1.3078 A 10 - 3 B
tan(-30°)
a = -66.5°
Ans.
The orientation of the neutral axis is shown in Fig. b.
407
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6–119. If the beam is made from a material having an allowable tensile and compressive stress of (sallow)t = 125 MPa and (sallow)c = 150 MPa, respectively, determine the maximum allowable internal moment M that can be applied to the beam.
y 150 mm 150 mm M 300 mm
Internal Moment Components: The y component of M is positive since it is directed towards the positive sense of the y axis, whereas the z component of M, which is directed towards the negative sense of the z axis, is negative, Fig. a. Thus,
30 150 mm z
My = M sin 30° = 0.5M
x 150 mm
Mz = -M cos 30° = -0.8660M Section Properties: The location of the centroid of the cross section is y =
150 mm
0.3(0.6)(0.3) - 0.375(0.15)(0.15) ©yA = = 0.2893 m ©A 0.6(0.3) - 0.15(0.15)
The moments of inertia of the cross section about the principal centroidal y and z axes are Iy =
1 1 (0.6) A 0.33 B (0.15) A 0.153 B = 1.3078 A 10 - 3 B m4 12 12
Iz =
1 (0.3) A 0.63 B + 0.3(0.6)(0.3 - 0.2893)2 12 - c
1 (0.15) A 0.153 B + 0.15(0.15)(0.375 - 0.2893)2 d 12
= 5.2132 A 10 - 3 B m4 Bending Stress: By inspection, the maximum bending stress can occur at either corner A or B. For corner A which is in tension, sA = (sallow)t = 125 A 106 B = -
My zA
Mz yA Iz
+
Iy
(-0.8660M)(0.2893) 5.2132 A 10
-3
B
0.5M(0.15)
+
1.3078 A 10 - 3 B
M = 1185 906.82 N # m = 1186 kN # m (controls)
Ans.
For corner B which is in compression, sB = (sallow)c = -150 A 106 B = -
My zB
Mz yB Iz
+
Iy
(-0.8660M)(-0.3107) 5.2132 A 10 - 3 B
0.5M(-0.15)
+
1.3078 A 10 - 3 B
M = 1376 597.12 N # m = 1377 kN # m
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*6–120. The shaft is supported on two journal bearings at A and B which offer no resistance to axial loading. Determine the required diameter d of the shaft if the allowable bending stress for the material is sallow = 150 MPa .
z y
0.5 m
0.5 m
C
0.5 m 200 N
The FBD of the shaft is shown in Fig. a.
A
200 N 300 N
The shaft is subjected to two bending moment components Mz and My, Figs. b and c, respectively. Since all the axes through the centroid of the circular cross-section of the shaft are principal axes, then the resultant moment M = 2My 2 + Mz 2 can be used for design. The maximum moment occurs at D (x = 1m). Then, Mmax = 21502 + 1752 = 230.49 N # m Then, sallow =
Mmax C ; I
150(106) =
230.49(d>2) p 4
(d>2)4
d = 0.02501 m = 25 mm
Ans.
409
300 N
0.5 m D B E x 150 N 150 N
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•6–121.
The 30-mm-diameter shaft is subjected to the vertical and horizontal loadings of two pulleys as shown. It is supported on two journal bearings at A and B which offer no resistance to axial loading. Furthermore, the coupling to the motor at C can be assumed not to offer any support to the shaft. Determine the maximum bending stress developed in the shaft.
1m 1m 1m 1m A D
150 N 150 N
Support Reactions: As shown on FBD. Internal Moment Components: The shaft is subjected to two bending moment components My and Mz. The moment diagram for each component is drawn. Maximum Bending Stress: Since all the axes through the circle’s center for circular shaft are principal axis, then the resultant moment M = 2My 2 + Mz 2 can be used to determine the maximum bending stress. The maximum resultant moment occurs at E Mmax = 24002 + 1502 = 427.2 N # m. Applying the flexure formula Mmax c I 427.2(0.015) =
p 4
A 0.0154 B
= 161 MPa
Ans.
410
E
C
B 400 N
100 mm 400 N 60 mm
x
smax =
y
z
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6–122. Using the techniques outlined in Appendix A, Example A.5 or A.6, the Z section has principal moments of inertia of Iy = 0.060110-32 m4 and Iz = 0.471110-32 m4, computed about the principal axes of inertia y and z, respectively. If the section is subjected to an internal moment of M = 250 N # m directed horizontally as shown, determine the stress produced at point A. Solve the problem using Eq. 6–17.
50 mm y
A
200 mm
32.9
y¿ 250 Nm z
My = 250 cos 32.9° = 209.9 N # m
z¿ 300 mm
Mz = 250 sin 32.9° = 135.8 N # m
200 mm
50 mm B
50 mm
y = 0.15 cos 32.9° + 0.175 sin 32.9° = 0.2210 m z = -(0.175 cos 32.9° - 0.15 sin 32.9°) = -0.06546 m sA = -
Myz
Mzy +
Iz
Iy
209.9(-0.06546)
-135.8(0.2210) =
0.471(10 - 3)
+
60.0(10 - 6)
= -293 kPa = 293 kPa (C)
Ans.
6–123. Solve Prob. 6–122 using the equation developed in Prob. 6–113.
50 mm y
A
Internal Moment Components: My = 250 N # m
200 mm
Mz = 0
32.9
y¿
Section Properties: Iy =
250 Nm
1 1 (0.3) A 0.053 B + 2c (0.05) A 0.153 B + 0.05(0.15) A 0.12 B d 12 12
= 0.18125 A 10 Iz =
-3
z z¿ 300 mm
Bm
4
1 1 (0.05) A 0.33 B + 2c (0.15) A 0.053 B + 0.15(0.05) A 0.1252 B d 12 12
= 0.350(10 - 3) m4 Iyz = 0.15(0.05)(0.125)(-0.1) + 0.15(0.05)(-0.125)(0.1) = -0.1875 A 10 - 3 B m4 Bending Stress: Using formula developed in Prob. 6-113 s =
sA =
-(Mz Iy + My Iyz)y + (My Iz + MzIyz)z IyIz - I2yz -[0 + 250(-0.1875)(10 - 3)](0.15) + [250(0.350)(10 - 3) + 0](-0.175) 0.18125(10 - 3)(0.350)(10 - 3) - [0.1875(10 - 3)]2
= -293 kPa = 293 kPa (C)
Ans.
411
200 mm
50 mm B
50 mm
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*6–124. Using the techniques outlined in Appendix A, Example A.5 or A.6, the Z section has principal moments of inertia of Iy = 0.060110-32 m4 and Iz = 0.471110-32 m4, computed about the principal axes of inertia y and z, respectively. If the section is subjected to an internal moment of M = 250 N # m directed horizontally as shown, determine the stress produced at point B. Solve the problem using Eq. 6–17.
50 mm y
A
200 mm
32.9
y¿ 250 Nm z z¿ 300 mm
Internal Moment Components: My¿ = 250 cos 32.9° = 209.9 N # m Mz¿ = 250 sin 32.9° = 135.8 N # m Section Property: y¿ = 0.15 cos 32.9° + 0.175 sin 32.9° = 0.2210 m z¿ = 0.15 sin 32.9° - 0.175 cos 32.9° = -0.06546 m Bending Stress: Applying the flexure formula for biaxial bending s =
sB =
My¿z¿
Mz¿y¿ Iz¿
+
Iy¿ 209.9(-0.06546)
135.8(0.2210) 0.471(10 - 3)
-
0.060(10 - 3)
= 293 kPa = 293 kPa (T)
Ans.
412
200 mm
50 mm B
50 mm
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z
•6–125. Determine the bending stress at point A of the beam, and the orientation of the neutral axis. Using the method in Appendix A, the principal moments of inertia of the cross section are I¿z = 8.828 in4 and I¿y = 2.295 in4, where z¿ and y¿ are the principal axes. Solve the problem using Eq. 6–17.
1.183 in. 0.5 in.
z¿
A
4 in.
45 C y 1.183 in. 0.5 in.
M 3 kip ft
y′ 4 in.
Internal Moment Components: Referring to Fig. a, the y¿ and z¿ components of M are negative since they are directed towards the negative sense of their respective axes. Thus, Section Properties: Referring to the geometry shown in Fig. b, œ = 2.817 cos 45° - 1.183 sin 45° = 1.155 in. zA œ yA = -(2.817 sin 45° + 1.183 cos 45°) = -2.828 in.
Bending Stress: sA = -
= -
œ My¿zA
œ Mz¿yA
Iz¿
+
Iy¿
(-2.121)(12)(-2.828) (-2.121)(12)(1.155) + 8.828 2.295
= -20.97 ksi = 21.0 ksi (C)
Ans.
413
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z
6–126. Determine the bending stress at point A of the beam using the result obtained in Prob. 6–113. The moments of inertia of the cross sectional area about the z and y axes are Iz = Iy = 5.561 in4 and the product of inertia of the cross sectional area with respect to the z and y axes is Iyz 3.267 in4. (See Appendix A)
1.183 in. 0.5 in.
z¿
A
4 in.
45 C y 1.183 in. 0.5 in.
M 3 kip ft
y′ 4 in.
Internal Moment Components: Since M is directed towards the negative sense of the y axis, its y component is negative and it has no z component. Thus, My = -3 kip # ft
Mz = 0
Bending Stress:
sA =
=
- A MzIy + MyIyz B yA + A MyIz + MzIyz B zA IyIz - Iyz 2
- C 0(5.561) + (-3)(12)(-3.267) D (-1.183) + C -3(12)(5.561) + 0(-3.267) D (2.817) 5.561(5.561) - (-3.267)2
= -20.97 ksi = 21.0 ksi
Ans.
414
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6–127. The composite beam is made of 6061-T6 aluminum (A) and C83400 red brass (B). Determine the dimension h of the brass strip so that the neutral axis of the beam is located at the seam of the two metals. What maximum moment will this beam support if the allowable bending stress for the aluminum is 1sallow2al = 128 MPa and for the brass 1sallow2br = 35 MPa?
h B A 150 mm
Section Properties: n =
68.9(109) Eal = 0.68218 = Ebr 101(109)
bbr = nbal = 0.68218(0.15) = 0.10233 m y =
0.05 =
©yA ©A 0.025(0.10233)(0.05) + (0.05 + 0.5h)(0.15)h 0.10233(0.05) + (0.15)h
h = 0.04130 m = 41.3 mm INA =
Ans.
1 (0.10233) A 0.053 B + 0.10233(0.05)(0.05 - 0.025)2 12 +
1 (0.15) A 0.041303 B + 0.15(0.04130)(0.070649 - 0.05)2 12
= 7.7851 A 10 - 6 B m4 Allowable Bending Stress: Applying the flexure formula Assume failure of red brass (sallow)br = 35 A 106 B =
Mc INA M(0.04130) 7.7851(10 - 6)
M = 6598 N # m = 6.60 kN # m (controls!)
Ans.
Assume failure of aluminium (sallow)al = n
Mc INA
128 A 106 B = 0.68218c
M(0.05) 7.7851(10 - 6)
d
M = 29215 N # m = 29.2 kN # m
415
50 mm
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*6–128. The composite beam is made of 6061-T6 aluminum (A) and C83400 red brass (B). If the height h = 40 mm, determine the maximum moment that can be applied to the beam if the allowable bending stress for the aluminum is 1sallow2al = 128 MPa and for the brass 1sallow2br = 35 MPa.
h B A
Section Properties: For transformed section. 150 mm
68.9(109) Eal = 0.68218 = n = Ebr 101.0(109) bbr = nbal = 0.68218(0.15) = 0.10233 m y =
=
©yA ©A 0.025(0.10233)(0.05) + (0.07)(0.15)(0.04) 0.10233(0.05) + 0.15(0.04)
= 0.049289 m INA =
1 (0.10233) A 0.053 B + 0.10233(0.05)(0.049289 - 0.025)2 12 +
1 (0.15) A 0.043 B + 0.15(0.04)(0.07 - 0.049289)2 12
= 7.45799 A 10 - 6 B m4 Allowable Bending Stress: Applying the flexure formula Assume failure of red brass (sallow)br = 35 A 106 B =
Mc INA M(0.09 - 0.049289) 7.45799(10 - 6)
M = 6412 N # m = 6.41 kN # m (controls!)
Ans.
Assume failure of aluminium (sallow)al = n
Mc INA
128 A 106 B = 0.68218c
M(0.049289) 7.45799(10 - 6)
d
M = 28391 N # m = 28.4 kN # m
416
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•6–129. Segment A of the composite beam is made from 2014-T6 aluminum alloy and segment B is A-36 steel. If w = 0.9 kip>ft, determine the absolute maximum bending stress developed in the aluminum and steel. Sketch the stress distribution on the cross section.
w
15 ft A
3 in.
B
3 in. 3 in.
Maximum Moment: For the simply-supported beam subjected to the uniform 0.9 A 152 B wL2 = distributed load, the maximum moment in the beam is Mmax = 8 8 = 25.3125 kip # ft. Section Properties: The cross section will be transformed into that of steel as Eal 10.6 = = 0.3655. shown in Fig. a. Here, n = Est 29 Then bst = nbal = 0.3655(3) = 1.0965 in. The location of the centroid of the transformed section is y =
©yA 1.5(3)(3) + 4.5(3)(1.0965) = = 2.3030 in. ©A 3(3) + 3(1.0965)
The moment of inertia of the transformed section about the neutral axis is I = ©I + Ad2 =
1 (3) A 33 B + 3(3)(2.3030 - 1.5)2 12 +
1 (1.0965) A 33 B + 1.0965(3)(4.5 - 2.3030)2 12
= 30.8991 in4 Maximum Bending Stress: For the steel, (smax)st =
25.3125(12)(2.3030) Mmaxcst = = 22.6 ksi I 30.8991
Ans.
At the seam, ssty = 0.6970 in. =
Mmaxy 25.3125(12)(0.6970) = = 6.85 ksi I 30.8991
For the aluminium, (smax)al = n
25.3125(12)(6 - 2.3030) Mmaxcal = 0.3655c d = 13.3 ksi I 30.8991
Ans.
At the seam, saly = 0.6970 in. = n
Mmaxy 25.3125(12)(0.6970) = 0.3655c d = 2.50 ksi I 30.8991
The bending stress across the cross section of the composite beam is shown in Fig. b.
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6–130. Segment A of the composite beam is made from 2014-T6 aluminum alloy and segment B is A-36 steel. If the allowable bending stress for the aluminum and steel are (sallow)al = 15 ksi and (sallow)st = 22 ksi, determine the maximum allowable intensity w of the uniform distributed load.
w
15 ft A
3 in.
B
3 in. 3 in.
Maximum Moment: For the simply-supported beam subjected to the uniform distributed load, the maximum moment in the beam is w A 152 B wL2 = = 28.125w. Mmax = 8 8 Section Properties: The cross section will be transformed into that of steel as Eal 10.6 = = 0.3655. shown in Fig. a. Here, n = Est 29 Then bst = nbal = 0.3655(3) = 1.0965 in. The location of the centroid of the transformed section is y =
©yA 1.5(3)(3) + 4.5(3)(1.0965) = = 2.3030 in. ©A 3(3) + 3(1.0965)
The moment of inertia of the transformed section about the neutral axis is I = ©I + Ad2 =
1 1 (3) A 33 B + 3(3)(2.3030 - 1.5)2 + (1.0965) A 33 B 12 12 + 1.0965 A 33 B + 1.0965(3)(4.5 - 2.3030)2
= 30.8991 in4 Bending Stress: Assuming failure of steel, (sallow)st =
Mmax cst ; I
22 =
(28.125w)(12)(2.3030) 30.8991
w = 0.875 kip>ft (controls)
Ans.
Assuming failure of aluminium alloy, (sallow)al = n
Mmax cal ; I
15 = 0.3655c
(28.125w)(12)(6 - 2.3030) d 30.8991
w = 1.02 kip>ft
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6–131. The Douglas fir beam is reinforced with A-36 straps at its center and sides. Determine the maximum stress developed in the wood and steel if the beam is subjected to a bending moment of Mz = 7.50 kip # ft. Sketch the stress distribution acting over the cross section.
y
0.5 in.
0.5 in.
0.5 in.
z
6 in.
2 in.
Section Properties: For the transformed section. n =
1.90(103) Ew = 0.065517 = Est 29.0(103)
bst = nbw = 0.065517(4) = 0.26207 in. INA =
1 (1.5 + 0.26207) A 63 B = 31.7172 in4 12
Maximum Bending Stress: Applying the flexure formula (smax)st =
7.5(12)(3) Mc = = 8.51 ksi I 31.7172
(smax)w = n
Ans.
7.5(12)(3) Mc = 0.065517c d = 0.558 ksi I 31.7172
Ans.
419
2 in.
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*6–132. The top plate is made of 2014-T6 aluminum and is used to reinforce a Kevlar 49 plastic beam. Determine the maximum stress in the aluminum and in the Kevlar if the beam is subjected to a moment of M = 900 lb # ft.
6 in. 0.5 in. 0.5 in. 12 in. M
0.5 in. 0.5 in.
Section Properties: n =
10.6(103) Eal = 0.55789 = Ek 19.0(103)
bk = n bal = 0.55789(12) = 6.6947 in. y =
0.25(13)(0.5) + 2[(3.25)(5.5)(0.5)] + 5.75(6.6947)(0.5) ©yA = ©A 13(0.5) + 2(5.5)(0.5) + 6.6947(0.5) = 2.5247 in.
INA =
1 (13) A 0.53 B + 13(0.5)(2.5247 - 0.25)2 12 +
1 (1) A 5.53 B + 1(5.5)(3.25 - 2.5247)2 12 +
1 (6.6947) A 0.53 B + 6.6947(0.5)(5.75 - 2.5247)2 12
= 85.4170 in4 Maximum Bending Stress: Applying the flexure formula (smax)al = n
(smax)k =
900(12)(6 - 2.5247) Mc = 0.55789 c d = 245 psi I 85.4170
900(12)(6 - 2.5247) Mc = = 439 psi I 85.4168
Ans.
Ans.
420
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•6–133.
The top plate made of 2014-T6 aluminum is used to reinforce a Kevlar 49 plastic beam. If the allowable bending stress for the aluminum is (sallow)al = 40 ksi and for the Kevlar (sallow)k = 8 ksi, determine the maximum moment M that can be applied to the beam.
6 in. 0.5 in. 0.5 in.
Section Properties: n =
10.6(103) Eal = 0.55789 = Ek 19.0(103)
12 in.
bk = n bal = 0.55789(12) = 6.6947 in. y =
0.5 in.
© yA 0.25(13)(0.5) + 2[(3.25)(5.5)(0.5)] + 5.75(6.6947(0.5) = ©A 13(0.5) + 2(5.5)(0.5) + 6.6947(0.5) = 2.5247 in.
INA =
1 (13) A 0.53 B + 13(0.5)(2.5247 - 0.25)2 12 +
1 (1) A 5.53 B + 1(5.5)(3.25 - 2.5247)2 12 +
1 (6.6947) A 0.53 B + 6.6947(0.5)(5.75 - 2.5247)2 12
= 85.4170 in4 Maximum Bending Stress: Applying the flexure formula Assume failure of aluminium (sallow)al = n
Mc I
40 = 0.55789 c
M(6 - 2.5247) d 85.4170
M = 1762 kip # in = 146.9 kip # ft Assume failure of Kevlar 49 (sallow)k = 8 =
Mc I M(6 - 2.5247) 85.4170
M = 196.62 kip # in = 16.4 kip # ft
M
0.5 in.
(Controls!)
Ans.
421
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6–134. The member has a brass core bonded to a steel casing. If a couple moment of 8 kN # m is applied at its end, determine the maximum bending stress in the member. Ebr = 100 GPa, Est = 200 GPa.
8 kNm
3m 20 mm 100 mm 20 mm
n =
Ebr 100 = = 0.5 Est 200
I =
1 1 (0.14)(0.14)3 (0.05)(0.1)3 = 27.84667(10 - 6)m4 12 12
20 mm
100 mm
20 mm
Maximum stress in steel: (sst)max =
8(103)(0.07) Mc1 = 20.1 MPa = I 27.84667(10 - 6)
Ans.
(max)
Maximum stress in brass: (sbr)max =
0.5(8)(103)(0.05) nMc2 = 7.18 MPa = I 27.84667(10 - 6)
6–135. The steel channel is used to reinforce the wood beam. Determine the maximum stress in the steel and in the wood if the beam is subjected to a moment of M = 850 lb # ft. Est = 29(103) ksi, Ew = 1600 ksi.
y =
4 in.
0.5 in.
(0.5)(16)(0.25) + 2(3.5)(0.5)(2.25) + (0.8276)(3.5)(2.25) = 1.1386 in. 0.5(16) + 2(3.5)(0.5) + (0.8276)(3.5)
15 in. M 850 lbft
0.5 in.
1 1 I = (16)(0.53) + (16)(0.5)(0.88862) + 2 a b(0.5)(3.53) + 2(0.5)(3.5)(1.11142) 12 12 +
1 (0.8276)(3.53) + (0.8276)(3.5)(1.11142) = 20.914 in4 12
Maximum stress in steel: (sst) =
850(12)(4 - 1.1386) Mc = = 1395 psi = 1.40 ksi I 20.914
Ans.
Maximum stress in wood: (sw) = n(sst)max = 0.05517(1395) = 77.0 psi
Ans.
422
0.5 in.
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*6–136. A white spruce beam is reinforced with A-36 steel straps at its top and bottom as shown. Determine the bending moment M it can support if (sallow)st = 22 ksi and (sallow)w = 2.0 ksi.
y 0.5 in.
4 in.
M
0.5 in.
x z 3 in.
Section Properties: For the transformed section. n =
1.40(103) Ew = 0.048276 = Est 29.0(103)
bst = nbw = 0.048276(3) = 0.14483 in. INA =
1 1 (3) A 53 B (3 - 0.14483) A 43 B = 16.0224 in4 12 12
Allowable Bending Stress: Applying the flexure formula Assume failure of steel (sallow)st = 22 =
Mc I M(2.5) 16.0224
M = 141.0 kip # in = 11.7 kip # ft (Controls !)
Ans.
Assume failure of wood (sallow)w = n
My I
2.0 = 0.048276 c
M(2) d 16.0224
M = 331.9 kip # in = 27.7 kip # ft
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•6–137. If the beam is subjected to an internal moment of
M = 45 kN # m, determine the maximum bending stress developed in the A-36 steel section A and the 2014-T6 aluminum alloy section B.
A 50 mm
M 15 mm 150 mm
Section Properties: The cross section will be transformed into that of steel as shown in Fig. a. 73.1 A 109 B Eal = = 0.3655. Thus, bst = nbal = 0.3655(0.015) = 0.0054825 m. The Here, n = Est 200 A 109 B location of the transformed section is
©yA y = = ©A
0.075(0.15)(0.0054825) + 0.2cp A 0.052 B d 0.15(0.0054825) + p A 0.052 B
= 0.1882 m The moment of inertia of the transformed section about the neutral axis is I = ©I + Ad2 =
1 (0.0054825) A 0.153 B + 0.0054825(0.15)(0.1882 - 0.075)2 12 +
1 p A 0.054 B + p A 0.052 B (0.2 - 0.1882)2 4
= 18.08 A 10 - 6 B m4 Maximum Bending Stress: For the steel,
(smax)st =
45 A 103 B (0.06185) Mcst = = 154 MPa I 18.08 A 10 - 6 B
Ans.
For the aluminum alloy,
(smax)al = n
45 A 103 B (0.1882) Mcal = 0.3655 C S = 171 MPa I 18.08 A 10 - 6 B
424
Ans.
B
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6–138. The concrete beam is reinforced with three 20-mm diameter steel rods. Assume that the concrete cannot support tensile stress. If the allowable compressive stress for concrete is (sallow)con = 12.5 MPa and the allowable tensile stress for steel is (sallow)st = 220 MPa, determine the required dimension d so that both the concrete and steel achieve their allowable stress simultaneously. This condition is said to be ‘balanced’. Also, compute the corresponding maximum allowable internal moment M that can be applied to the beam. The moduli of elasticity for concrete and steel are Econ = 25 GPa and Est = 200 GPa, respectively.
200 mm M
Bending Stress: The cross section will be transformed into that of concrete as shown Est 200 = = 8. It is required that both concrete and steel in Fig. a. Here, n = Econ 25 achieve their allowable stress simultaneously. Thus, (sallow)con =
12.5 A 106 B =
Mccon ; I
Mccon I
M = 12.5 A 106 B ¢ (sallow)st = n
I ≤ ccon
220 A 106 B = 8 B
Mcst ; I
(1)
M(d - ccon) R I
M = 27.5 A 106 B ¢
I ≤ d - ccon
(2)
Equating Eqs. (1) and (2), 12.5 A 106 B ¢
I I ≤ = 27.5 A 106 B ¢ ≤ ccon d - ccon
ccon = 0.3125d (3) Section Properties: The area of the steel bars is Ast = 3c
(3) p A 0.022 B d = 0.3 A 10 - 3 B p m2. 4
Thus, the transformed area of concrete from steel is (Acon)t = nAs = 8 C 0.3 A 10 - 3 B p D
= 2.4 A 10 - 3 B p m2. Equating the first moment of the area of concrete above and below the neutral axis about the neutral axis, 0.2(ccon)(ccon>2) = 2.4 A 10 - 3 B p (d - ccon)
0.1ccon 2 = 2.4 A 10 - 3 B pd - 2.4 A 10 - 3 B pccon ccon 2 = 0.024pd - 0.024pccon
(4)
Solving Eqs. (3) and (4), d = 0.5308 m = 531 mm
Ans.
ccon = 0.1659 m Thus, the moment of inertia of the transformed section is I =
1 (0.2) A 0.16593 B + 2.4 A 10 - 3 B p(0.5308 - 0.1659)2 3
425
d
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6–138.
Continued
= 1.3084 A 10 - 3 B m4 Substituting this result into Eq. (1), M = 12.5 A 106 B C
1.3084 A 10 - 3 B 0.1659
S
= 98 594.98 N # m = 98.6 kN # m‚
Ans.
6–139. The beam is made from three types of plastic that are identified and have the moduli of elasticity shown in the figure. Determine the maximum bending stress in the PVC. (bbk)1 = n1 bEs =
160 (3) = 0.6 in. 800
(bbk)2 = n2 bpvc =
450 (3) = 1.6875 in. 800
500 lb
PVC EPVC 450 ksi Escon EE 160 ksi Bakelite EB 800 ksi 3 ft
y =
©yA (1)(3)(2) + 3(0.6)(2) + 4.5(1.6875)(1) = = 1.9346 in. ©A 3(2) + 0.6(2) + 1.6875(1)
I =
1 1 (3)(23) + 3(2)(0.93462) + (0.6)(23) + 0.6(2)(1.06542) 12 12 +
4 ft
1 in. 2 in. 2 in. 3 in.
1 (1.6875)(13) + 1.6875(1)(2.56542) = 20.2495 in4 12
(smax)pvc = n2
500 lb
450 1500(12)(3.0654) Mc = a b I 800 20.2495 = 1.53 ksi
Ans.
426
3 ft
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*6–140. The low strength concrete floor slab is integrated with a wide-flange A-36 steel beam using shear studs (not shown) to form the composite beam. If the allowable bending stress for the concrete is (sallow)con = 10 MPa, and allowable bending stress for steel is (sallow)st = 165 MPa, determine the maximum allowable internal moment M that can be applied to the beam.
1m
100 mm
15 mm 400 mm M 15 mm 15 mm
Section Properties: The beam cross section will be transformed into Econ 22.1 that of steel. Here, Thus, = = 0.1105. n = Est 200 bst = nbcon = 0.1105(1) = 0.1105 m. The location of the transformed section is y =
=
©yA ©A 0.0075(0.015)(0.2) + 0.2(0.37)(0.015) + 0.3925(0.015)(0.2) + 0.45(0.1)(0.1105) 0.015(0.2) + 0.37(0.015) + 0.015(0.2) + 0.1(0.1105)
= 0.3222 m The moment of inertia of the transformed section about the neutral axis is I = ©I + Ad2 =
1 (0.2) A 0.0153 B 12 + 0.2(0.015)(0.3222 - 0.0075)2 +
1 (0.015) A 0.373 B + 0.015(0.37)(0.3222 - 0.2)2 12
+
1 (0.2) A 0.0153 B + 0.2(0.015)(0.3925 - 0.3222)2 12
+
1 (0.1105) A 0.13 B + 0.1105(0.1)(0.45 - 0.3222)2 12 = 647.93 A 10 - 6 B m4
Bending Stress: Assuming failure of steel, (sallow)st =
M(0.3222) Mcst ; 165 A 106 B = I 647.93 A 10 - 6 B M = 331 770.52 N # m = 332 kN # m
Assuming failure of concrete,
(sallow)con = n
Mccon ; I
10 A 106 B = 0.1105C
M(0.5 - 0.3222) 647.93 A 10 - 6 B
S
M = 329 849.77 N # m = 330 kN # m (controls) Ans.
427
200 mm
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•6–141.
The reinforced concrete beam is used to support the loading shown. Determine the absolute maximum normal stress in each of the A-36 steel reinforcing rods and the absolute maximum compressive stress in the concrete. Assume the concrete has a high strength in compression and yet neglect its strength in supporting tension.
10 kip
8 in.
15 in. 4 ft
8 ft
Mmax = (10 kip)(4 ft) = 40 kip # ft Ast = 3(p)(0.5)2 = 2.3562 in2 Est = 29.0(103) ksi Econ = 4.20(103) ksi A¿ = nAst =
©yA = 0;
29.0(103) 4.20(103) 8(h¿)a
(2.3562) = 16.2690 in2
h¿ b - 16.2690(13 - h¿) = 0 2
h¿ 2 + 4.06724h - 52.8741 = 0 Solving for the positive root: h¿ = 5.517 in. I = c
1 (8)(5.517)3 + 8(5.517)(5.517>2)2 d + 16.2690(13 - 5.517)2 12
= 1358.781 in4 (scon)max =
My 40(12)(5.517) = = 1.95 ksi I 1358.781
(sst)max = na
10 kip
Ans.
My 29.0(103) 40(12)(13 - 5.517) ba b = a b = 18.3 ksi I 1358.781 4.20(103)
428
Ans.
4 ft
2 in. 1 in. diameter rods
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6–142. The reinforced concrete beam is made using two steel reinforcing rods. If the allowable tensile stress for the steel is (sst)allow = 40 ksi and the allowable compressive stress for the concrete is (sconc)allow = 3 ksi, determine the maximum moment M that can be applied to the section. Assume the concrete cannot support a tensile stress. Est = 29(103) ksi, Econc = 3.8(103) ksi.
8 in. 6 in. 4 in.
8 in.
M 18 in. 2 in. 1-in. diameter rods
Ast = 2(p)(0.5)2 = 1.5708 in2 A¿ = nAst = ©yA = 0;
29(103) 3.8(103)
(1.5708) = 11.9877 in2
22(4)(h¿ + 2) + h¿(6)(h¿>2) - 11.9877(16 - h¿) = 0 3h2 + 99.9877h¿ - 15.8032 = 0
Solving for the positive root: h¿ = 0.15731 in. I = c
1 1 (22)(4)3 + 22(4)(2.15731)2 d + c (6)(0.15731)3 + 6(0.15731)(0.15731>2)2 d 12 12 + 11.9877(16 - 0.15731)2 = 3535.69 in4
Assume concrete fails: (scon)allow =
My ; I
3 =
M(4.15731) 3535.69
M = 2551 kip # in. Assume steel fails: (sst)allow = na
My b; I
40 = ¢
29(103) 3
3.8(10 )
≤¢
M(16 - 0.15731) ≤ 3535.69
M = 1169.7 kip # in. = 97.5 kip # ft (controls) Ans.
429
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6–143. For the curved beam in Fig. 6–40a, show that when the radius of curvature approaches infinity, the curved-beam formula, Eq. 6–24, reduces to the flexure formula, Eq. 6–13. Normal Stress: Curved-beam formula M(R - r)
s =
where A¿ =
Ar(r - R)
dA LA r
and R =
A 1A
dA r
=
A A¿
M(A - rA¿)
s =
[1]
Ar(rA¿ - A)
r = r + y rA¿ = r
[2]
dA r = a - 1 + 1 b dA LA r + y LA r =
LA
a
= A -
r - r - y r + y y
+ 1b dA
dA
LA r + y
[3]
Denominator of Eq. [1] becomes, y
Ar(rA¿ - A) = Ar ¢ A -
LA r + y
dA - A ≤ = -Ar
y LA r + y
dA
Using Eq. [2], Ar(rA¿ - A) = -A
= A
=
¢
ry
LA r + y y2
LA r + y
+ y - y ≤ dA - Ay
LA r + y
dA - A 1A y dA - Ay
y LA r + y
as
y r
: 0
A I r
Then,
Ar(rA¿ - A) :
Eq. [1] becomes
s =
Mr (A - rA¿) AI
Using Eq. [2],
s =
Mr (A - rA¿ - yA¿) AI
Using Eq. [3],
s =
=
dA
dA
y2 y Ay A ¢ ¢ y ≤ dA - A 1A y dA r LA 1 + r r LA 1 + 1A y dA = 0,
But,
y
y Mr dA C A - ¢A dA ≤ - y S AI r + y r LA LA + y
y Mr dA C dA - y S AI LA r + y r LA + y
430
y≤ r
dA
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6–143. Continued y
=
y
As
r
Mr r C ¢ AI LA 1 +
y ≤ dA r
y -
r LA
¢
dA ≤S 1 + yr
=
: 0
¢
y r
LA 1 +
y≤ r
dA = 0
s =
Therefore,
and
y r LA
¢
y yA dA A dA = y≤ = 1 1 + r r r
yA My Mr b = aAI I r
(Q.E.D.)
*6–144. The member has an elliptical cross section. If it is subjected to a moment of M = 50 N # m, determine the stress at points A and B. Is the stress at point A¿ , which is located on the member near the wall, the same as that at A? Explain.
75 mm
150 mm A¿ 250 mm A
dA 2p b = (r - 2r2 - a2 ) a LA r
100 mm
2p(0.0375) = (0.175 - 20.1752 - 0.0752 ) = 0.053049301 m 0.075 A = p ab = p(0.075)(0.0375) = 2.8125(10 - 3)p R =
A 1A
dA r
=
B
2.8125(10 - 3)p = 0.166556941 0.053049301
r - R = 0.175 - 0.166556941 = 0.0084430586 sA =
sB =
M(R - rA)
50(0.166556941 - 0.1) =
2.8125(10 - 3)p (0.1)(0.0084430586)
=
2.8125(10 - 3)p (0.25)(0.0084430586)
ArA (r - R) M(R - rB) ArB (r - R)
50(0.166556941 - 0.25)
= 446k Pa (T)
= 224 kPa (C)
No, because of localized stress concentration at the wall.
Ans.
Ans. Ans.
431
M
Mr AI
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•6–145. The member has an elliptical cross section. If the allowable bending stress is sallow = 125 MPa determine the maximum moment M that can be applied to the member.
75 mm
150 mm A¿ 250 mm A 100 mm
B
b = 0.0375 m
a = 0.075 m;
A = p(0.075)(0.0375) = 0.0028125 p 2p(0.0375) dA 2pb (0.175 - 20.1752 - 0.0752) = (r - 2r2 - a2) = r a 0.075 LA = 0.053049301 m R =
A dA 1A r
=
0.0028125p = 0.166556941 m 0.053049301
r - R = 0.175 - 0.166556941 = 8.4430586(10 - 3) m s =
M(R - r) Ar(r - R)
Assume tension failure. 125(106) =
M(0.166556941 - 0.1) 0.0028125p(0.1)(8.4430586)(10 - 3)
M = 14.0 kN # m (controls)
Ans.
Assume compression failure: -125(106) =
M(0.166556941 - 0.25) 0.0028125p(0.25)(8.4430586)(10 - 3)
M = 27.9 kN # m
432
M
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6–146. Determine the greatest magnitude of the applied forces P if the allowable bending stress is (sallow)c = 50 MPa in compression and (sallow)t = 120 MPa in tension.
75 mm P
10 mm
10 mm 160 mm
10 mm
P 150 mm 250 mm
Internal Moment: M = 0.160P is positive since it tends to increase the beam’s radius of curvature. Section Properties: r =
=
©yA ©A 0.255(0.15)(0.01) + 0.335(0.15)(0.01) + 0.415(0.075)(0.01) 0.15(0.01) + 0.15(0.01) + 0.075(0.01)
= 0.3190 m A = 0.15(0.01) + 0.15(0.01) + 0.075(0.01) = 0.00375 m2 ©
dA 0.26 0.41 0.42 = 0.15 ln + 0.01 ln + 0.075 ln 0.25 0.26 0.41 LA r = 0.012245 m
R =
A © 1A dA r
=
0.00375 = 0.306243 m 0.012245
r - R = 0.319 - 0.306243 = 0.012757 m Allowable Normal Stress: Applying the curved-beam formula Assume tension failure (sallow)t = 120 A 106 B =
M(R - r) Ar(r - R) 0.16P(0.306243 - 0.25) 0.00375(0.25)(0.012757)
P = 159482 N = 159.5 kN Assume compression failure (sallow)t = -50 A 106 B =
M(R - r) Ar(r - R) 0.16P(0.306243 - 0.42) 0.00375(0.42)(0.012757)
P = 55195 N = 55.2 kN (Controls !)
Ans.
433
150 mm
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6–147. If P = 6 kN, determine the maximum tensile and compressive bending stresses in the beam.
75 mm P
10 mm
10 mm 160 mm
10 mm
P 150 mm 250 mm
Internal Moment: M = 0.160(6) = 0.960 kN # m is positive since it tends to increase the beam’s radius of curvature. Section Properties: r =
=
©yA ©A 0.255(0.15)(0.01) + 0.335(0.15)(0.01) + 0.415(0.075)(0.01) 0.15(0.01) + 0.15(0.01) + 0.075(0.01)
= 0.3190 m A = 0.15(0.01) + 0.15(0.01) + 0.075(0.01) = 0.00375 m2 ©
dA 0.41 0.42 0.26 = 0.15 ln + 0.01 ln + 0.075 ln 0.25 0.26 0.41 LA r = 0.012245 m
R =
A ©1A dA r
=
0.00375 = 0.306243 m 0.012245
r - R = 0.319 - 0.306243 = 0.012757 m Normal Stress: Applying the curved-beam formula (smax)t =
=
M(R - r) Ar(r - R) 0.960(103)(0.306243 - 0.25) 0.00375(0.25)(0.012757)
= 4.51 MPa (smax)c =
=
Ans.
M(R - r) Ar(r - R) 0.960(103)(0.306243 - 0.42) 0.00375(0.42)(0.012757)
= -5.44 MPa
Ans.
434
150 mm
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*6–148. The curved beam is subjected to a bending moment of M = 900 N # m as shown. Determine the stress at points A and B, and show the stress on a volume element located at each of these points.
A C B
100 mm C
A
30
20 mm
15 mm
150 mm
400 mm B M
Internal Moment: M = -900 N # m is negative since it tends to decrease the beam’s radius curvature. Section Properties: ©A = 0.15(0.015) + 0.1(0.02) = 0.00425 m2 ©rA = 0.475(0.15)(0.015) + 0.56(0.1)(0.02) = 2.18875(10 - 3) m3 r =
©
2.18875 (10 - 3) ©rA = = 0.5150 m ©A 0.00425
dA 0.57 0.55 = 0.015 ln + 0.1 ln = 8.348614(10 - 3) m 0.4 0.55 LA r
R =
A ©1A
dA r
=
0.00425 = 0.509067 m 8.348614(10 - 3)
r - R = 0.515 - 0.509067 = 5.933479(10 - 3) m Normal Stress: Applying the curved-beam formula sA =
M(R - rA)
-900(0.509067 - 0.57) =
ArA (r - R)
0.00425(0.57)(5.933479)(10 - 3) Ans.
= 3.82 MPa (T) sB =
M(R - rB)
-900(0.509067 - 0.4) =
ArB (r - R)
0.00425(0.4)(5.933479)(10 - 3)
= -9.73 MPa = 9.73 MPa (C)
Ans.
435
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•6–149.
The curved beam is subjected to a bending moment of M = 900 N # m. Determine the stress at point C. A C B
100 mm C
A
30
20 mm
15 mm
150 mm
400 mm B M
Internal Moment: M = -900 N # m is negative since it tends to decrease the beam’s radius of curvature. Section Properties: ©A = 0.15(0.015) + 0.1(0.02) = 0.00425 m2 ©rA = 0.475(0.15)(0.015) + 0.56(0.1)(0.02) = 2.18875(10 - 3) m r =
©
2.18875 (10 - 3) ©rA = = 0.5150 m ©A 0.00425
dA 0.57 0.55 = 0.015 ln + 0.1 ln = 8.348614(10 - 3) m 0.4 0.55 LA r
R =
A ©1A
dA r
=
0.00425 = 0.509067 m 8.348614(10 - 3)
r - R = 0.515 - 0.509067 = 5.933479(10 - 3) m Normal Stress: Applying the curved-beam formula sC =
M(R - rC)
-900(0.509067 - 0.55) =
ArC(r - R)
0.00425(0.55)(5.933479)(10 - 3)
= 2.66 MPa (T)
Ans.
436
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6–150. The elbow of the pipe has an outer radius of 0.75 in. and an inner radius of 0.63 in. If the assembly is subjected to the moments of M = 25 lb # in., determine the maximum stress developed at section a-a.
a 30
M 25 lbin.
1 in. a
dA = ©2p (r - 2r2 - c2) LA r = 2p(1.75 - 21.752 - 0.752) - 2p (1.75 - 21.752 - 0.632)
0.63 in. 0.75 in.
= 0.32375809 in. A = p(0.752) - p(0.632) = 0.1656 p R =
A dA 1A r
=
M = 25 lbin.
0.1656 p = 1.606902679 in. 0.32375809
r - R = 1.75 - 1.606902679 = 0.14309732 in. (smax)t =
M(R - rA) = ArA(r - R)
(smax)c = =
25(1.606902679 - 1) = 204 psi (T) 0.1656 p(1)(0.14309732)
M(R - rB) = ArB(r - R)
Ans.
25(1.606902679 - 2.5) = 120 psi (C) 0.1656p(2.5)(0.14309732)
Ans.
6–151. The curved member is symmetric and is subjected to a moment of M = 600 lb # ft. Determine the bending stress in the member at points A and B. Show the stress acting on volume elements located at these points.
0.5 in. B 2 in. A
1 A = 0.5(2) + (1)(2) = 2 in2 2 r =
1.5 in. 8 in.
9(0.5)(2) + 8.6667 A 12 B (1)(2) ©rA = = 8.83333 in. ©A 2
M
M
1(10) dA 10 10 = 0.5 ln + c cln d - 1 d = 0.22729 in. r 8 (10 - 8) 8 LA R =
A dA 1A r
=
2 = 8.7993 in. 0.22729
r - R = 8.83333 - 8.7993 = 0.03398 in. s =
M(R - r) Ar(r - R)
sA =
600(12)(8.7993 - 8) = 10.6 ksi (T) 2(8)(0.03398)
Ans.
sB =
600(12)(8.7993 - 10) = -12.7 ksi = 12.7 ksi (C) 2(10)(0.03398)
Ans.
437
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*6–152. The curved bar used on a machine has a rectangular cross section. If the bar is subjected to a couple as shown, determine the maximum tensile and compressive stress acting at section a-a. Sketch the stress distribution on the section in three dimensions.
a 75 mm a
50 mm
162.5 mm
250 N 60
150 mm
60 250 N 75 mm
a + ©MO = 0;
M - 250 cos 60° (0.075) - 250 sin 60° (0.15) = 0 M = 41.851 N # m
r2 dA 0.2375 = b ln = 0.05 ln = 0.018974481 m r r 0.1625 1 LA A = (0.075)(0.05) = 3.75(10 - 3) m2 R =
A 1A
dA r
=
3.75(10 - 3) = 0.197633863 m 0.018974481
r - R = 0.2 - 0.197633863 = 0.002366137 sA =
M(R - rA)
41.851(0.197633863 - 0.2375) =
ArA(r - R)
3.75(10 - 3)(0.2375)(0.002366137)
= -791.72 kPa Ans.
= 792 kPa (C) sB =
M(R - rB)
41.851 (0.197633863 - 0.1625) =
ArB(r - R)
3.75(10 - 3)(0.1625)(0.002366137)
= 1.02 MPa (T)
438
Ans.
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•6–153.
The ceiling-suspended C-arm is used to support the X-ray camera used in medical diagnoses. If the camera has a mass of 150 kg, with center of mass at G, determine the maximum bending stress at section A.
G
1.2 m A
200 mm 100 mm 20 mm 40 mm
Section Properties: r =
©
1.22(0.1)(0.04) + 1.25(0.2)(0.02) ©rA = = 1.235 m ©A 0.1(0.04) + 0.2(0.02)
dA 1.26 1.24 = 0.1 ln + 0.2 ln = 6.479051 A 10 - 3 B m r 1.20 1.24 LA
A = 0.1(0.04) + 0.2(0.02) = 0.008 m2 R =
A dA 1A r
=
0.008 = 1.234749 m 6.479051 (10 - 3)
r - R = 1.235 - 1.234749 = 0.251183 A 10 - 3 B m Internal Moment: The internal moment must be computed about the neutral axis as shown on FBD. M = -1816.93 N # m is negative since it tends to decrease the beam’s radius of curvature. Maximum Normal Stress: Applying the curved-beam formula sA =
M(R - rA) ArA (r - R) -1816.93(1.234749 - 1.26)
=
0.008(1.26)(0.251183)(10 - 3)
= 18.1 MPa (T) sB =
M(R - rB) ArB (r - R) -1816.93(1.234749 - 1.20)
=
0.008(1.20)(0.251183)(10 - 3)
= -26.2 MPa = 26.2 MPa (C)
Ans.
(Max)
439
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6–154. The circular spring clamp produces a compressive force of 3 N on the plates. Determine the maximum bending stress produced in the spring at A. The spring has a rectangular cross section as shown.
10 mm 20 mm
Internal Moment: As shown on FBD, M = 0.660 N # m is positive since it tends to increase the beam’s radius of curvature.
210 mm
200 mm A
Section Properties: 220 mm
0.200 + 0.210 r = = 0.205 m 2 r2 dA 0.21 = 0.02 ln = b ln = 0.97580328 A 10 - 3 B m r r 0.20 1 LA A = (0.01)(0.02) = 0.200 A 10 - 3 B m2 R =
0.200(10 - 3)
A 1A
dA r
=
0.97580328(10 - 3)
= 0.204959343 m
r - R = 0.205 - 0.204959343 = 0.040657 A 10 - 3 B m Maximum Normal Stress: Applying the curved-beam formula sC =
M(R - r2) Ar2(r - R) 0.660(0.204959343 - 0.21)
=
0.200(10 - 3)(0.21)(0.040657)(10 - 3)
= -1.95MPa = 1.95 MPa (C) st =
M(R - r1) Ar1 (r - R) 0.660(0.204959343 - 0.2)
=
0.200(10 - 3)(0.2)(0.040657)(10 - 3)
= 2.01 MPa (T)
(Max)
Ans.
440
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6–155. Determine the maximum compressive force the spring clamp can exert on the plates if the allowable bending stress for the clamp is sallow = 4 MPa.
10 mm 20 mm
210 mm
200 mm A
220 mm
Section Properties: r =
0.200 + 0.210 = 0.205 m 2
r2 dA 0.21 = b ln = 0.02 ln = 0.97580328 A 10 - 3 B m r1 0.20 LA r A = (0.01)(0.02) = 0.200 A 10 - 3 B m2 R =
0.200(10 - 3)
A 1A
dA r
=
0.97580328(10 - 3)
= 0.204959 m
r - R = 0.205 - 0.204959343 = 0.040657 A 10 - 3 B m Internal Moment: The internal moment must be computed about the neutral axis as shown on FBD. Mmax = 0.424959P is positive since it tends to increase the beam’s radius of curvature. Allowable Normal Stress: Applying the curved-beam formula Assume compression failure sc = sallow = -4 A 106 B =
M(R - r2) Ar2(r - R) 0.424959P(0.204959 - 0.21) 0.200(10 - 3)(0.21)(0.040657)(10 - 3)
P = 3.189 N Assume tension failure st = sallow = 4 A 106 B =
M(R - r1) Ar1 (r - R) 0.424959P(0.204959 - 0.2) 0.200(10 - 3)(0.2)(0.040657)(10 - 3)
P = 3.09 N (Controls !)
Ans.
441
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*6–156. While in flight, the curved rib on the jet plane is subjected to an anticipated moment of M = 16 N # m at the section. Determine the maximum bending stress in the rib at this section, and sketch a two-dimensional view of the stress distribution. 16 Nm 5 mm 20 mm 5 mm
0.6 m 5 mm
30 mm
LA
0.625 0.630 0.605 + (0.005)ln + (0.03)ln = 0.650625(10 - 3) in. 0.6 0.605 0.625
dA>r = (0.03)ln
A = 2(0.005)(0.03) + (0.02)(0.005) = 0.4(10 - 3) in2 R =
0.4(10 - 3)
A 1A dA>r
=
0.650625(10 - 3)
= 0.6147933
(sc)max =
M(R - rc) 16(0.6147933 - 0.630) = -4.67 MPa = ArA(r - R) 0.4(10 3)(0.630)(0.615 - 0.6147933)
(ss)max =
M(R - rs) 16(0.6147933 - 0.6) = 4.77 MPa = ArA(r - R) 0.4(10 - 3)(0.6)(0.615 - 0.6147933)
Ans.
If the radius of each notch on the plate is r = 0.5 in., determine the largest moment that can be applied. The allowable bending stress for the material is sallow = 18 ksi.
•6–157.
14.5 in.
M
b =
14.5 - 12.5 = 1.0 in. 2 r 0.5 = = 0.04 h 12.5
1 b = = 2.0 r 0.5 From Fig. 6-44: K = 2.60 smax = K
Mc I
18(103) = 2.60c
(M)(6.25) 1 3 12 (1)(12.5)
d
M = 180 288 lb # in. = 15.0 kip # ft
Ans.
442
1 in.
12.5 in.
M
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6–158. The symmetric notched plate is subjected to bending. If the radius of each notch is r = 0.5 in. and the applied moment is M = 10 kip # ft, determine the maximum bending stress in the plate.
14.5 in.
M
M
12.5 in.
r 0.5 = = 0.04 h 12.5
1 b = 2.0 = r 0.5
1 in.
From Fig. 6-44: K = 2.60 smax = K
(10)(12)(6.25) Mc = 2.60 c 1 d = 12.0 ksi 3 I 12 (1)(12.5)
Ans.
6–159. The bar is subjected to a moment of M = 40 N # m. Determine the smallest radius r of the fillets so that an allowable bending stress of sallow = 124 MPa is not exceeded.
80 mm 7 mm
20 mm r
M
M r
Allowable Bending Stress: sallow = K
Mc I
124 A 106 B = K B
40(0.01)
R 1 3 12 (0.007)(0.02 )
K = 1.45 Stress Concentration Factor: From the graph in the text w 80 r with = = 4 and K = 1.45, then = 0.25. h 20 h r = 0.25 20 r = 5.00 mm
Ans.
*6–160. The bar is subjected to a moment of M = 17.5 N # m. If r = 5 mm, determine the maximum bending stress in the material.
80 mm 7 mm
20 mm r
M
M
Stress Concentration Factor: From the graph in the text with r w 80 5 = = 4 and = = 0.25, then K = 1.45. h 20 h 20
r
Maximum Bending Stress: smax = K
Mc I
= 1.45 B
17.5(0.01)
R 1 3 12 (0.007)(0.02 )
= 54.4 MPa
Ans.
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•6–161. The simply supported notched bar is subjected to two forces P. Determine the largest magnitude of P that can be applied without causing the material to yield.The material is A-36 steel. Each notch has a radius of r = 0.125 in.
P
P 0.5 in. 1.75 in.
1.25 in.
20 in.
b =
20 in.
20 in.
20 in.
1.75 - 1.25 = 0.25 2
0.25 b = = 2; r 0.125
r 0.125 = = 0.1 h 1.25
From Fig. 6-44. K = 1.92 sY = K
Mc ; I
36 = 1.92 c
20P(0.625) 1 3 12 (0.5)(1.25)
d
P = 122 lb
Ans.
6–162. The simply supported notched bar is subjected to the two loads, each having a magnitude of P = 100 lb. Determine the maximum bending stress developed in the bar, and sketch the bending-stress distribution acting over the cross section at the center of the bar. Each notch has a radius of r = 0.125 in.
P
0.5 in.
1.75 - 1.25 = 0.25 2
b 0.25 = = 2; r 0.125
r 0.125 = = 0.1 h 1.25
From Fig. 6-44, K = 1.92 smax = K
1.75 in.
1.25 in.
20 in.
b =
P
2000(0.625) Mc = 1.92c 1 d = 29.5 ksi 3 I 12 (0.5)(1.25)
Ans.
444
20 in.
20 in.
20 in.
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6–163. Determine the length L of the center portion of the bar so that the maximum bending stress at A, B, and C is the same. The bar has a thickness of 10 mm.
7 mm
350 N 60 mm
A
r 7 = = 0.175 h 40
60 w = = 1.5 h 40
200 mm
40 mm 7 mm
C L 2
B L 2
200 mm
From Fig. 6-43, K = 1.5 (sA)max = K
(35)(0.02) MAc d = 19.6875 MPa = 1.5c 1 3 I 12 (0.01)(0.04 )
(sB)max = (sA)max = 19.6875(106) =
MB c I
175(0.2 + L2 )(0.03) 1 3 12 (0.01)(0.06 )
L = 0.95 m = 950 mm
Ans.
*6–164. The stepped bar has a thickness of 15 mm. Determine the maximum moment that can be applied to its ends if it is made of a material having an allowable bending stress of sallow = 200 MPa.
45 mm 30 mm 3 mm
M
M
Stress Concentration Factor: w 30 6 r = = 3 and = = 0.6, we have K = 1.2 h 10 h 10 obtained from the graph in the text. For the smaller section with
w 45 3 r = = 1.5 and = = 0.1, we have K = 1.75 h 30 h 30 obtained from the graph in the text. For the larger section with
Allowable Bending Stress: For the smaller section smax = sallow = K
Mc ; I
200 A 106 B = 1.2 B
M(0.005)
R 1 3 12 (0.015)(0.01 )
M = 41.7 N # m (Controls !)
Ans.
For the larger section smax = sallow = K
Mc ; I
200 A 106 B = 1.75 B
M(0.015)
R 1 3 12 (0.015)(0.03 )
M = 257 N # m
445
10 mm 6 mm
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•6–165.
The beam is made of an elastic plastic material for which sY = 250 MPa. Determine the residual stress in the beam at its top and bottom after the plastic moment Mp is applied and then released.
15 mm
1 1 (0.2)(0.23)3 (0.18)(0.2)3 = 82.78333(10 - 6)m4 12 12
20 mm 200 mm
Ix =
Mp
C1 = T1 = sY (0.2)(0.015) = 0.003sY
15 mm
C2 = T2 = sY (0.1)(0.02) = 0.002sY 200 mm
Mp = 0.003sY (0.215) + 0.002sY (0.1) = 0.000845 sY = 0.000845(250)(106) = 211.25 kN # m s =
Mp c
211.25(103)(0.115) =
I
82.78333(10 - 6)
y 0.115 = ; 250 293.5
= 293.5 MPa
y = 0.09796 m = 98.0 mm
stop = sbottom = 293.5 - 250 = 43.5 MPa
Ans.
6–166. The wide-flange member is made from an elasticplastic material. Determine the shape factor.
t
Plastic analysis: T1 = C1 = sY bt;
h
T2 = C2 = sY a
MP = sY bt(h - t) + sY a
h - 2t bt 2
t t
h - 2t h - 2t b(t) a b 2 2 b
t = sY c bt(h - t) + (h - 2t)2 d 4 Elastic analysis: I =
=
1 1 bh3 (b - t)(h - 2t)3 12 12 1 [bh3 - (b - t)(h - 2 t)3] 12
MY =
sy I c
=
=
1 sY A 12 B [bh3 - (b - t)(h - 2t)3] h 2
bh3 - (b - t)(h - 2t)3 sY 6h
Shape factor: k =
[bt(h - t) + 4t (h - 2t)2]sY MP = bh3 - (b - t)(h - 2t)3 MY s 6h
=
Y
3h 4bt(h - t) + t(h - 2t)2 c d 2 bh3 - (b - t)(h - 2t)3
Ans.
446
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6–167.
Determine the shape factor for the cross section.
Maximum Elastic Moment: The moment of inertia about neutral axis must be determined first.
a
1 1 (a)(3a)3 + (2a) A a3 B = 2.41667a4 12 12
INA =
a a
Applying the flexure formula with s = sY, we have sY =
MY c I
MY =
a
a
a
sY (2.41667a4) sYI = = 1.6111a3sY c 1.5a
Plastic Moment: MP = sY (a)(a)(2a) + sY (0.5a)(3a)(0.5a) = 2.75a3sY Shape Factor: k =
MP 2.75a3sY = = 1.71 MY 1.6111a3sY
Ans.
*6–168. The beam is made of elastic perfectly plastic material. Determine the maximum elastic moment and the plastic moment that can be applied to the cross section. Take a = 2 in. and sY = 36 ksi.
a a
Maximum Elastic Moment: The moment of inertia about neutral axis must be determined first. INA
a
1 1 (2) A 63 B + (4) A 23 B = 38.667 in4 = 12 12
Applying the flexure formula with s = sY, we have sY = = MY =
a
MY c I
36(38.667) sY I = c 3 = 464 kip # in = 38.7 kip # ft
Ans.
Plastic Moment: MP = 36(2)(2)(4) + 36(1)(6)(1) = 792 kip # in = 66.0 kip # ft
Ans.
447
a
a
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•6–169.
The box beam is made of an elastic perfectly plastic material for which sY = 250 MPa . Determine the residual stress in the top and bottom of the beam after the plastic moment Mp is applied and then released.
Plastic Moment: MP = 250 A 106 B (0.2)(0.025)(0.175) + 250 A 106 B (0.075)(0.05)(0.075)
25 mm
= 289062.5 N # m
150 mm
Modulus of Rupture: The modulus of rupture sr can be determined using the flexure formula with the application of reverse, plastic moment MP = 289062.5 N # m. I =
25 mm
25 mm 150 mm 25 mm
1 1 (0.2) A 0.23 B (0.15) A 0.153 B 12 12
= 91.14583 A 10 - 6 B m4 sr =
289062.5 (0.1) MP c = 317.41 MPa = I 91.14583 A 10 - 6 B
Residual Bending Stress: As shown on the diagram. œ œ = sbot = sr - sY stop
= 317.14 - 250 = 67.1 MPa
Ans.
6–170. Determine the shape factor for the wideflange beam.
15 mm
1 1 (0.2)(0.23)3 (0.18)(0.2)3 = 82.78333 A 10 - 6 B m4 12 12
Ix =
20 mm 200 mm
C1 = T1 = sY(0.2)(0.015) = 0.003sY
Mp
C2 = T2 = sY(0.1)(0.02) = 0.002sY Mp = 0.003sY(0.215) + 0.002sY(0.1) = 0.000845 sY
15 mm 200 mm
sY =
MY =
k =
MY c I sY A 82.78333)10 - 6 B 0.115
Mp MY
=
= 0.000719855 sY
0.000845sY = 1.17 0.000719855sY
Ans.
448
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6–171. Determine the shape factor of the beam’s cross section. 3 in.
Referring to Fig. a, the location of centroid of the cross-section is y =
7.5(3)(6) + 3(6)(3) ©yA = = 5.25 in. ©A 3(6) + 6(3)
6 in.
The moment of inertia of the cross-section about the neutral axis is I =
1 1 (3) A 63 B + 3(6)(5.25 - 3)2 + (6) A 33 B + 6(3)(7.5 - 5.25)2 12 12
1.5 in. 3 in. 1.5 in.
4
= 249.75 in
Here smax = sY and c = y = 5.25 in. Thus smax =
Mc ; I
sY =
MY (5.25) 249.75
MY = 47.571sY Referring to the stress block shown in Fig. b, sdA = 0; LA
T - C1 - C2 = 0
d(3)sY - (6 - d)(3)sY - 3(6)sY = 0 d = 6 in. Since d = 6 in., c1 = 0, Fig. c. Here T = C = 3(6) sY = 18 sY Thus, MP = T(4.5) = 18 sY (4.5) = 81 sY Thus, k =
MP 81 sY = = 1.70 MY 47.571 sY
Ans.
449
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*6–172. The beam is made of elastic-perfectly plastic material. Determine the maximum elastic moment and the plastic moment that can be applied to the cross section. Take sY = 36 ksi.
3 in.
Referring to Fig. a, the location of centroid of the cross-section is 6 in.
7.5(3)(6) + 3(6)(3) ©yA y = = = 5.25 in. ©A 3(6) + 6(3) The moment of inertia of the cross-section about the neutral axis is
1.5 in. 3 in. 1.5 in.
I =
1 1 (3)(63) + 3(6)(5.25 - 3)2 + (6)(33) + 6(3)(7.5 - 5.25)2 12 12
= 249.75 in4 Here, smax = sY = 36 ksi and ¢ = y = 5.25 in. Then smax =
Mc ; I
36 =
MY (5.25) 249.75
MY = 1712.57 kip # in = 143 kip # ft
Ans.
Referring to the stress block shown in Fig. b, sdA = 0; LA
T - C1 - C2 = 0
d(3) (36) - (6 - d)(3)(36) - 3(6) (36) = 0 d = 6 in. Since d = 6 in., c1 = 0, Here, T = C = 3(6)(36) = 648 kip Thus, MP = T(4.5) = 648(4.5) = 2916 kip # in = 243 kip # ft
Ans.
450
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•6–173.
Determine the shape factor for the cross section of the H-beam.
Ix =
1 1 (0.2)(0.023) + 2 a b(0.02)(0.23) = 26.8(10 - 6)m4 12 12
200 mm
C1 = T1 = sY(2)(0.09)(0.02) = 0.0036sy 20 mm
C2 = T2 = sY(0.01)(0.24) = 0.0024sy
Mp
20 mm
200 mm
Mp = 0.0036sY(0.11) + 0.0024sY(0.01) = 0.00042sY
20 mm
MYc sY = I MY =
k =
sY(26.8)(10 - 6) = 0.000268sY 0.1
Mp MY
=
0.00042sY = 1.57 0.000268sY
Ans.
6–174. The H-beam is made of an elastic-plastic material for which sY = 250 MPa. Determine the residual stress in the top and bottom of the beam after the plastic moment Mp is applied and then released. 200 mm
Ix =
1 1 (0.2)(0.023) + 2 a b (0.02)(0.23) = 26.8(10 - 6)m4 12 12 20 mm
C1 = T1 = sY(2)(0.09)(0.02) = 0.0036sy
200 mm
C2 = T2 = sY(0.01)(0.24) = 0.0024sy
20 mm
Mp = 0.0036sY(0.11) + 0.0024sY(0.01) = 0.00042sY Mp = 0.00042(250) A 106 B = 105 kN # m s¿ =
Mp c I
y 0.1 = ; 250 392
105(103)(0.1) =
26.8(10 - 6)
Mp
= 392 MPa
y = 0.0638 = 63.8 mm
sT = sB = 392 - 250 = 142 MPa
Ans.
451
20 mm
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6–175.
Determine the shape factor of the cross section. 3 in.
The moment of inertia of the cross-section about the neutral axis is I =
3 in.
1 1 (3)(93) + (6) (33) = 195.75 in4 12 12
3 in.
Here, smax = sY and c = 4.5 in. Then smax =
Mc ; I
sY =
MY(4.5) 195.75
3 in.
MY = 43.5 sY Referring to the stress block shown in Fig. a, T1 = C1 = 3(3)sY = 9 sY T2 = C2 = 1.5(9)sY = 13.5 sY Thus, MP = T1(6) + T2(1.5) = 9sY(6) + 13.5sY(1.5) = 74.25 sY k =
74.25 sY MP = = 1.71 MY 43.5 sY
Ans.
452
3 in.
3 in.
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*6–176. The beam is made of elastic-perfectly plastic material. Determine the maximum elastic moment and the plastic moment that can be applied to the cross section. Take sY = 36 ksi.
3 in. 3 in.
The moment of inertia of the cross-section about the neutral axis is I =
3 in.
1 1 (3)(93) + (6)(33) = 195.75 in4 12 12
Here, smax = sY = 36 ksi and c = 4.5 in. Then smax
Mc = ; I
3 in.
MY (4.5) 36 = 195.75 MY = 1566 kip # in = 130.5 kip # ft
Ans.
Referring to the stress block shown in Fig. a, T1 = C1 = 3(3)(36) = 324 kip T2 = C2 = 1.5(9)(36) = 486 kip Thus, MP = T1(6) + T2(1.5) = 324(6) + 486(1.5) = 2673 kip # in. = 222.75 kip # ft = 223 kip # ft
Ans.
453
3 in.
3 in.
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•6–177.
Determine the shape factor of the cross section for the tube.
The moment of inertia of the tube’s cross-section about the neutral axis is I =
5 in.
p 4 p A r - r4i B = A 64 - 54 B = 167.75 p in4 4 o 4
6 in.
Here, smax = sY and C = ro = 6 in, smax =
Mc ; I
sY =
MY (6) 167.75 p
MY = 87.83 sY The plastic Moment of the table’s cross-section can be determined by super posing the moment of the stress block of the solid circular cross-section with radius ro = 6 in and ri = 5 in. as shown in Figure a, Here, T1 = C1 =
1 p(62)sY = 18psY 2
T2 = C2 =
1 p(52)sY = 12.5p sY 2
Thus, MP = T1 b 2 c
4(6) 4(5) d r - T2 b 2 c dr 3p 3p
= (18psY)a
16 40 b - 12.5psY a b p 3p
= 121.33 sY k =
121.33 sY MP = = 1.38 MY 87.83 sY
Ans.
454
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6–178. The beam is made from elastic-perfectly plastic material. Determine the shape factor for the thick-walled tube. ro
Maximum Elastic Moment. The moment of inertia of the cross-section about the neutral axis is I =
p A r 4 - r4i B 4 o
With c = ro and smax = sY, smax =
Mc ; I
sY =
MY =
MY(ro) p A r 4 - ri 4 B 4 o p A r 4 - ri 4 B sY 4ro o
Plastic Moment. The plastic moment of the cross section can be determined by superimposing the moment of the stress block of the solid beam with radius r0 and ri as shown in Fig. a, Referring to the stress block shown in Fig. a, T1 = c1 =
p 2 r s 2 o Y
T2 = c2 =
p 2 r s 2 i Y
MP = T1 c2 a
4ro 4ri b d - T2 c2 a b d 3p 3p
=
8ro 8ri p 2 p r s a b - ri 2sY a b 2 o Y 3p 2 3p
=
4 A r 3 - ri 3 B sY 3 o
Shape Factor. 4 A r 3 - ri 3 B sY 16ro A ro 3 - ri 3 B MP 3 o k = = = p MY 3p A ro 4 - ri 4 B A ro 4 - ri 4 B sY 4ro
Ans.
455
ri
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6–179.
Determine the shape factor for the member.
Plastic analysis: T = C =
–h 2
h 1 bh (b)a bsY = s 2 2 4 Y –h 2
b h2 bh h MP = sY a b = s 4 3 12 Y Elastic analysis: I = 2c
1 h 3 b h3 (b)a b d = 12 2 48
b
sY A bh sYI 48 B b h2 = s = h c 24 Y 2 3
MY =
Shape factor: k =
Mp MY
=
bh2 12
sY
bh2 24
sY
= 2
Ans.
*6–180. The member is made from an elastic-plastic material. Determine the maximum elastic moment and the plastic moment that can be applied to the cross section. Take b = 4 in., h = 6 in., sY = 36 ksi.
–h 2
Elastic analysis: I = 2c
1 (4)(3)3 d = 18 in4 12
MY =
36(18) sYI = = 216 kip # in. = 18 kip # ft c 3
–h 2
Ans. b
Plastic analysis: T = C =
1 (4)(3)(36) = 216 kip 2
6 Mp = 2160 a b = 432 kip # in. = 36 kip # ft 3
Ans.
456
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•6–181.
The beam is made of a material that can be assumed perfectly plastic in tension and elastic perfectly plastic in compression. Determine the maximum bending moment M that can be supported by the beam so that the compressive material at the outer edge starts to yield.
h
sY
M
sdA = 0; LA
C - T = 0
sY
a
1 s (d)(a) - sY(h - d)a = 0 2 Y d =
M =
2 h 3
11 11a h2 2 1 sY a hb (a)a hb = sY 2 3 18 54
Ans.
6–182. The box beam is made from an elastic-plastic material for which sY = 25 ksi. Determine the intensity of the distributed load w0 that will cause the moment to be (a) the largest elastic moment and (b) the largest plastic moment.
w0
Elastic analysis: I =
9 ft
1 1 (8)(163) (6)(123) = 1866.67 in4 12 12
Mmax
sYI = ; c
9 ft
8 in.
25(1866.67) 27w0(12) = 8 Ans.
w0 = 18.0 kip>ft Plastic analysis:
16 in.
12 in.
6 in.
C1 = T1 = 25(8)(2) = 400 kip C2 = T2 = 25(6)(2) = 300 kip MP = 400(14) + 300(6) = 7400 kip # in. 27w0(12) = 7400 w0 = 22.8 kip>ft
Ans.
457
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6–183. The box beam is made from an elastic-plastic material for which sY = 36 ksi. Determine the magnitude of each concentrated force P that will cause the moment to be (a) the largest elastic moment and (b) the largest plastic moment.
P
From the moment diagram shown in Fig. a, Mmax = 6 P.
P
8 ft
6 ft
6 ft
The moment of inertia of the beam’s cross-section about the neutral axis is 6 in.
1 1 (6)(123) (5)(103) = 447.33 in4 I = 12 12
12 in.
10 in.
Here, smax = sY = 36 ksi and c = 6 in. smax =
Mc ; I
36 =
5 in.
MY (6) 447.33
MY = 2684 kip # in = 223.67 kip # ft It is required that Mmax = MY 6P = 223.67 P = 37.28 kip = 37.3 kip
Ans.
Referring to the stress block shown in Fig. b, T1 = C1 = 6(1)(36) = 216 kip T2 = C2 = 5(1)(36) = 180 kip Thus, MP = T1(11) + T2(5) = 216(11) + 180(5) = 3276 kip # in = 273 kip # ft It is required that Mmax = MP 6P = 273 P = 45.5 kip
Ans.
458
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*6–184. The beam is made of a polyester that has the stress–strain curve shown. If the curve can be represented by the equation s = [20 tan-1115P2] ksi, where tan-1115P2 is in radians, determine the magnitude of the force P that can be applied to the beam without causing the maximum strain in its fibers at the critical section to exceed Pmax = 0.003 in.>in.
P 2 in. 4 in.
8 ft s(ksi)
8 ft s 20 tan1(15 P)
P(in./in.)
Maximum Internal Moment: The maximum internal moment M = 4.00P occurs at the mid span as shown on FBD. Stress–Strain Relationship: Using the stress–strain relationship. the bending stress can be expressed in terms of y using e = 0.0015y. s = 20 tan - 1 (15e) = 20 tan - 1 [15(0.0015y)] = 20 tan - 1 (0.0225y) When emax = 0.003 in.>in., y = 2 in. and smax = 0.8994 ksi Resultant Internal Moment: The resultant internal moment M can be evaluated from the integal
M = 2
LA
ysdA.
ysdA
2in
= 2
LA
L0
y C 20 tan
-1
(0.0225y) D (2dy)
2in
= 80
L0
= 80 B
y tan - 1 (0.0225y) dy
1 + (0.0225)2y2 2(0.0225)2
tan - 1 (0.0225y) -
2in. y R2 2(0.0225) 0
= 4.798 kip # in Equating M = 4.00P(12) = 4.798 P = 0.100 kip = 100 lb
Ans.
459
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•6–185.
The plexiglass bar has a stress–strain curve that can be approximated by the straight-line segments shown. Determine the largest moment M that can be applied to the bar before it fails.
s (MPa)
20 mm M 20 mm
failure
60 40
tension
0.06 0.04
P (mm/mm) 0.02
compression 80 100
Ultimate Moment: LA
s dA = 0;
C - T2 - T1 = 0
1 1 d 1 d sc (0.02 - d)(0.02) d - 40 A 106 B c a b(0.02) d - (60 + 40) A 106 B c(0.02) d = 0 2 2 2 2 2 s - 50s d - 3500(106)d = 0 Assume.s = 74.833 MPa; d = 0.010334 m From the strain diagram, 0.04 e = 0.02 - 0.010334 0.010334
e = 0.037417 mm>mm
From the stress–strain diagram, 80 s = 0.037417 0.04
s = 74.833 MPa (OK! Close to assumed value)
Therefore, 1 C = 74.833 A 106 B c (0.02 - 0.010334)(0.02) d = 7233.59 N 2 T1 =
1 0.010334 (60 + 40) A 106 B c(0.02)a b d = 5166.85 N 2 2
1 0.010334 b d = 2066.74 N T2 = 40 A 106 B c (0.02)a 2 2
y1 =
2 (0.02 - 0.010334) = 0.0064442 m 3
y2 =
2 0.010334 a b = 0.0034445 m 3 2
y3 =
0.010334 1 2(40) + 60 0.010334 + c1 - a bda b = 0.0079225m 2 3 40 + 60 2
M = 7233.59(0.0064442) + 2066.74(0.0034445) + 5166.85(0.0079255) = 94.7 N # m
Ans.
460
0.04
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6–186. The stress–strain diagram for a titanium alloy can be approximated by the two straight lines. If a strut made of this material is subjected to bending, determine the moment resisted by the strut if the maximum stress reaches a value of (a) sA and (b) sB.
3 in. M
2 in. s (ksi) B
sB 180 sA 140
A
0.01
a) Maximum Elastic Moment : Since the stress is linearly related to strain up to point A, the flexure formula can be applied. sA =
Mc I M =
=
sA I c 1 140 C 12 (2)(33) D
1.5
= 420 kip # in = 35.0 kip # ft b)
Ans.
The Ultimate Moment : C1 = T1 =
1 (140 + 180)(1.125)(2) = 360 kip 2
C2 = T2 =
1 (140)(0.375)(2) = 52.5 kip 2
M = 360(1.921875) + 52.5(0.5) = 718.125 kip # in = 59.8 kip # ft
Ans.
Note: The centroid of a trapezodial area was used in calculation of moment.
461
0.04
P (in./in.)
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6–187. A beam is made from polypropylene plastic and has a stress–strain diagram that can be approximated by the curve shown. If the beam is subjected to a maximum tensile and compressive strain of P = 0.02 mm>mm, determine the maximum moment M.
M
s (Pa)
s 10(106)P1/ 4
emax = 0.02 smax = 10 A 106 B (0.02)1>4 = 3.761 MPa
M
100 mm
30 mm P (mm/mm)
e 0.02 = y 0.05 e = 0.4 y s = 10 A 106 B (0.4)1>4y1>4 y(7.9527) A 106 B y1>4(0.03)dy
0.05
M =
y s dA = 2
LA
M = 0.47716 A 106 B
L0
4 y5>4dy = 0.47716 A 106 B a b(0.05)9>4 5
0.05
L0
M = 251 N # m
Ans.
*6–188. The beam has a rectangular cross section and is made of an elastic-plastic material having a stress–strain diagram as shown. Determine the magnitude of the moment M that must be applied to the beam in order to create a maximum strain in its outer fibers of P max = 0.008.
400 mm M
200 mm
s(MPa)
200
0.004
C1 = T1 = 200 A 106 B (0.1)(0.2) = 4000 kN C2 = T2 =
1 (200) A 106 B (0.1)(0.2) = 2000 kN 2
M = 4000(0.3) + 2000(0.1333) = 1467 kN # m = 1.47 MN # m
Ans.
462
P (mm/mm)
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s(ksi) 90 80
•6–189.
The bar is made of an aluminum alloy having a stress–strain diagram that can be approximated by the straight line segments shown. Assuming that this diagram is the same for both tension and compression, determine the moment the bar will support if the maximum strain at the top and bottom fibers of the beam is P max = 0.03. 90 - 80 s - 80 = ; 0.03 - 0.025 0.05 - 0.025
60
4 in. M
s = 82 ksi
C1 = T1 =
1 (0.3333)(80 + 82)(3) = 81 kip 2
C2 = T2 =
1 (1.2666)(60 + 80)(3) = 266 kip 2
C3 = T3 =
1 (0.4)(60)(3) = 36 kip 2
0.006
0.025
0.05
P (in./ in.)
3 in.
M = 81(3.6680) + 266(2.1270) + 36(0.5333) = 882.09 kip # in. = 73.5 kip # ft
Ans.
Note: The centroid of a trapezodial area was used in calculation of moment areas.
6–190. The beam is made from three boards nailed together as shown. If the moment acting on the cross section is M = 650 N # m, determine the resultant force the bending stress produces on the top board. 15 mm
Section Properties: y =
0.0075(0.29)(0.015) + 2[0.0775(0.125)(0.02)] 0.29(0.015) + 2(0.125)(0.02)
M 650 Nm 20 mm
125 mm
= 0.044933 m INA
20 mm
1 = (0.29) A 0.0153 B + 0.29(0.015) (0.044933 - 0.0075)2 12 +
1 (0.04) A 0.1253 B + 0.04(0.125)(0.0775 - 0.044933)2 12
= 17.99037 A 10 - 6 B m4 Bending Stress: Applying the flexure formula s =
sB =
sA =
650(0.044933 - 0.015) 17.99037(10 - 6) 650(0.044933) 17.99037(10 - 6)
My I
= 1.0815 MPa
= 1.6234 MPa
Resultant Force: FR =
1 (1.0815 + 1.6234) A 106 B (0.015)(0.29) 2
= 5883 N = 5.88 kN
Ans.
463
250 mm
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6–191. The beam is made from three boards nailed together as shown. Determine the maximum tensile and compressive stresses in the beam. 15 mm M 650 Nm 20 mm
125 mm 20 mm
Section Properties: y =
0.0075(0.29)(0.015) + 2[0.0775(0.125)(0.02)] 0.29(0.015) + 2(0.125)(0.02)
= 0.044933 m INA =
1 (0.29) A 0.0153 B + 0.29(0.015)(0.044933 - 0.0075)2 12 +
1 (0.04) A 0.1253 B + 0.04(0.125)(0.0775 - 0.044933)2 12
= 17.99037 A 10 - 6 B m4 Maximum Bending Stress: Applying the flexure formula s =
(smax)t =
(smax)c =
650(0.14 - 0.044933) 17.99037(10 - 6) 650(0.044933) 17.99037(10 - 6)
My I Ans.
= 3.43 MPa (T)
= 1.62 MPa (C)
Ans.
464
250 mm
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*6–192. Determine the bending stress distribution in the beam at section a–a. Sketch the distribution in three dimensions acting over the cross section.
80 N
80 N
a
a 300 mm
400 mm
a + ©M = 0;
300 mm
400 mm
80 N
M - 80(0.4) = 0
80 N 15 mm
M = 32 N # m
100 mm
1 1 Iz = (0.075)(0.0153) + 2 a b (0.015)(0.13) = 2.52109(10 - 6)m4 12 12 smax =
32(0.05) Mc = 635 kPa = I 2.52109(10 - 6)
15 mm
•6–193. The composite beam consists of a wood core and two plates of steel. If the allowable bending stress for the wood is (sallow)w = 20 MPa, and for the steel (sallow)st = 130 MPa, determine the maximum moment that can be applied to the beam. Ew = 11 GPa, Est = 200 GPa.
n =
75 mm
Ans.
y
z 125 mm
200(109) Est = 18.182 = Ew 11(109) M
1 (0.80227)(0.1253) = 0.130578(10 - 3)m4 I = 12
x
75 mm
Failure of wood : (sw)max
20 mm
Mc = I
20(106) =
M(0.0625) 0.130578(10 - 3)
;
M = 41.8 kN # m
Failure of steel : (sst)max =
20 mm
nMc I 130(106) =
18.182(M)(0.0625) 0.130578(10 - 3)
M = 14.9 kN # m (controls)
Ans.
465
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6–194. Solve Prob. 6–193 if the moment is applied about the y axis instead of the z axis as shown.
y
z 125 mm
M
x
20 mm 75 mm 20 mm
n =
I =
11(109) 200(104)
= 0.055
1 1 (0.125)(0.1153) (0.118125)(0.0753) = 11.689616(10 - 6) 12 12
Failure of wood : (sw)max =
nMc2 I
20(106) =
0.055(M)(0.0375) 11.689616(10 - 6)
;
M = 113 kN # m
Failure of steel : (sst)max =
Mc1 I 130(106) =
M(0.0575) 11.689616(10 - 6)
M = 26.4 kN # m (controls)
Ans.
466
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6–195. A shaft is made of a polymer having a parabolic cross section. If it resists an internal moment of M = 125 N # m, determine the maximum bending stress developed in the material (a) using the flexure formula and (b) using integration. Sketch a three-dimensional view of the stress distribution acting over the cross-sectional area. Hint: The moment of inertia is determined using Eq. A–3 of Appendix A.
y
100 mm y 100 – z 2/ 25 M 125 N· m z
Maximum Bending Stress: The moment of inertia about y axis must be determined first in order to use Flexure Formula I =
LA
50 mm 50 mm
y2 dA 100mm
= 2
L0
y2 (2z) dy 100mm
= 20
L0
y2 2100 - y dy
100 mm 3 5 7 3 8 16 y (100 - y)2 (100 - y)2 R 2 = 20 B - y2 (100 - y)2 2 15 105 0
= 30.4762 A 10 - 6 B mm4 = 30.4762 A 10 - 6 B m4 Thus, smax =
125(0.1) Mc = 0.410 MPa = I 30.4762(10 - 6)
Ans.
Maximum Bending Stress: Using integration dM = 2[y(s dA)] = 2 b yc a
M =
smax by d(2z dy) r 100
smax 100mm 2 y 2100 - y dy 5 L0
125 A 103 B =
100 mm smax 3 5 7 3 8 16 y(100 - y)2 (100 - y)2 R 2 B - y2(100 - y)2 5 2 15 105 0
125 A 103 B =
smax (1.5238) A 106 B 5
smax = 0.410 N>mm2 = 0.410 MPa
Ans.
467
x
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*6–196. Determine the maximum bending stress in the handle of the cable cutter at section a–a. A force of 45 lb is applied to the handles. The cross-sectional area is shown in the figure.
20
45 lb
a
5 in. 4 in.
3 in. 0.75 in.
A a
0.50 in.
45 lb
a + ©M = 0;
M - 45(5 + 4 cos 20°) = 0 M = 394.14 lb # in.
394.14(0.375) Mc = 8.41 ksi = 1 3 I 12 (0.5)(0.75 )
smax =
Ans.
M 85 Nm
•6–197.
The curved beam is subjected to a bending moment of M = 85 N # m as shown. Determine the stress at points A and B and show the stress on a volume element located at these points.
100 mm
A
r2 0.57 0.59 dA 0.42 + 0.015 ln + 0.1 ln = b ln = 0.1 ln r1 0.40 0.42 0.57 LA r
400 mm
= 0.012908358 m
= LA
dA r
6.25(10 - 3) = 0.484182418 m 0.012908358
r - R = 0.495 - 0.484182418 = 0.010817581 m sA =
M(R - rA)
85(0.484182418 - 0.59) =
ArA(r - R)
6.25(10 - 3)(0.59)(0.010817581)
= -225.48 kPa
sA = 225 kPa (C) sB =
Ans.
M(R - rB)
85(0.484182418 - 0.40) =
ArB(r - R)
150 mm
6.25(10 - 3)(0.40)(0.010817581)
20 mm B
2
A = 2(0.1)(0.02) + (0.15)(0.015) = 6.25(10 ) m A
20 mm
30 -3
R =
A 15 mm
B
= 265 kPa (T)
468
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6–198. Draw the shear and moment diagrams for the beam and determine the shear and moment in the beam as functions of x, where 0 … x 6 6 ft.
8 kip
2 kip/ ft 50 kipft
x 6 ft
+ c ©Fy = 0;
20 - 2x - V = 0 V = 20 - 2x
c + ©MNA = 0;
4 ft
Ans.
x 20x - 166 - 2xa b - M = 0 2 M = -x2 + 20x - 166
Ans.
6–199. Draw the shear and moment diagrams for the shaft if it is subjected to the vertical loadings of the belt, gear, and flywheel. The bearings at A and B exert only vertical reactions on the shaft.
300 N 450 N
A
B
200 mm
400 mm
300 mm
200 mm 150 N
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*6–200. A member has the triangular cross section shown. Determine the largest internal moment M that can be applied to the cross section without exceeding allowable tensile and compressive stresses of (sallow)t = 22 ksi and (sallow)c = 15 ksi, respectively.
4 in. 4 in.
M 2 in.
2 in.
y (From base) = I =
1 242 - 22 = 1.1547 in. 3
1 (4)(242 - 22)3 = 4.6188 in4 36
Assume failure due to tensile stress : smax =
My ; I
22 =
M(1.1547) 4.6188
M = 88.0 kip # in. = 7.33 kip # ft Assume failure due to compressive stress: smax =
Mc ; I
15 =
M(3.4641 - 1.1547) 4.6188
M = 30.0 kip # in. = 2.50 kip # ft
(controls)
Ans.
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•6–201.
The strut has a square cross section a by a and is subjected to the bending moment M applied at an angle u as shown. Determine the maximum bending stress in terms of a, M, and u. What angle u will give the largest bending stress in the strut? Specify the orientation of the neutral axis for this case.
y
a
z
x
a M
Internal Moment Components: Mz = -M cos u
My = -M sin u
Section Property: Iy = Iz =
1 4 a 12
Maximum Bending Stress: By Inspection, Maximum bending stress occurs at A and B. Applying the flexure formula for biaxial bending at point A s = -
My z
Mzy +
Iz
Iy
-M cos u (a2) = -
=
1 12
a4
-Msin u ( - a2) +
1 12
a4
6M (cos u + sin u) a3
Ans.
6M ds = 3 (-sin u + cos u) = 0 du a cos u - sin u = 0 u = 45°
Ans.
Orientation of Neutral Axis: tan a =
Iz Iy
tan u
tan a = (1) tan(45°) a = 45°
Ans.
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•7–1.
If the wide-flange beam is subjected to a shear of V = 20 kN, determine the shear stress on the web at A. Indicate the shear-stress components on a volume element located at this point.
200 mm
A
20 mm
20 mm B
V 300 mm 200 mm
The moment of inertia of the cross-section about the neutral axis is I =
1 1 (0.2)(0.343) (0.18)(0.33) = 0.2501(10 - 3) m4 12 12
From Fig. a, QA = y¿A¿ = 0.16 (0.02)(0.2) = 0.64(10 - 3) m3 Applying the shear formula, VQA 20(103)[0.64(10 - 3)] = tA = It 0.2501(10 - 3)(0.02) = 2.559(106) Pa = 2.56 MPa
Ans.
The shear stress component at A is represented by the volume element shown in Fig. b.
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7–2. If the wide-flange beam is subjected to a shear of V = 20 kN, determine the maximum shear stress in the beam.
200 mm
A
20 mm
20 mm B
V 300 mm 200 mm
The moment of inertia of the cross-section about the neutral axis is I =
1 1 (0.2)(0.343) (0.18)(0.33) = 0.2501(10 - 3) m4 12 12
From Fig. a. Qmax = ©y¿A¿ = 0.16 (0.02)(0.2) + 0.075 (0.15)(0.02) = 0.865(10 - 3) m3 The maximum shear stress occurs at the points along neutral axis since Q is maximum and thicknest t is the smallest. tmax =
VQmax 20(103) [0.865(10 - 3)] = It 0.2501(10 - 3) (0.02) = 3.459(106) Pa = 3.46 MPa
Ans.
473
20 mm
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7–3. If the wide-flange beam is subjected to a shear of V = 20 kN, determine the shear force resisted by the web of the beam.
200 mm
A
20 mm
20 mm B
V 300 mm 200 mm
The moment of inertia of the cross-section about the neutral axis is I =
1 1 (0.2)(0.343) (0.18)(0.33) = 0.2501(10 - 3) m4 12 12
For 0 … y 6 0.15 m, Fig. a, Q as a function of y is Q = ©y¿A¿ = 0.16 (0.02)(0.2) +
1 (y + 0.15)(0.15 - y)(0.02) 2
= 0.865(10 - 3) - 0.01y2 For 0 … y 6 0.15 m, t = 0.02 m. Thus. t =
20(103) C 0.865(10 - 3) - 0.01y2 D VQ = It 0.2501(10 - 3) (0.02) =
E 3.459(106) - 39.99(106) y2 F Pa.
The sheer force resisted by the web is, 0.15 m
Vw = 2
L0
0.15 m
tdA = 2
L0
C 3.459(106) - 39.99(106) y2 D (0.02 dy)
= 18.95 (103) N = 19.0 kN
Ans.
474
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*7–4. If the T-beam is subjected to a vertical shear of V = 12 kip, determine the maximum shear stress in the beam. Also, compute the shear-stress jump at the flangeweb junction AB. Sketch the variation of the shear-stress intensity over the entire cross section.
4 in. 4 in.
3 in.
4 in. B
6 in.
A V ⫽ 12 kip
Section Properties: y =
INA =
1.5(12)(3) + 6(4)(6) ©yA = = 3.30 in. ©A 12(3) + 4(6)
1 1 (12) A 33 B + 12(3)(3.30 - 1.5)2 + (4) A 63 B + 4(6)(6 - 3.30)2 12 12
= 390.60 in4 Qmax = y1œ A¿ = 2.85(5.7)(4) = 64.98 in3 QAB = y2œ A¿ = 1.8(3)(12) = 64.8 in3 Shear Stress: Applying the shear formula t =
tmax =
VQ It
VQmax 12(64.98) = = 0.499 ksi It 390.60(4)
Ans.
(tAB)f =
VQAB 12(64.8) = = 0.166 ksi Itf 390.60(12)
Ans.
(tAB)W =
VQAB 12(64.8) = = 0.498 ksi I tW 390.60(4)
Ans.
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•7–5.
If the T-beam is subjected to a vertical shear of V = 12 kip, determine the vertical shear force resisted by the flange.
4 in. 4 in.
3 in.
4 in. B
6 in.
A V ⫽ 12 kip
Section Properties: y =
©yA 1.5(12)(3) + 6(4)(6) = = 3.30 in. ©A 12(3) + 4(6)
INA =
1 1 (12) A 33 B + 12(3)(3.30 - 1.5)2 + (4) A 63 B + 6(4)(6 - 3.30)2 12 12
= 390.60 in4 Q = y¿A¿ = (1.65 + 0.5y)(3.3 - y)(12) = 65.34 - 6y2 Shear Stress: Applying the shear formula t =
VQ 12(65.34 - 6y2) = It 390.60(12) = 0.16728 - 0.01536y2
Resultant Shear Force: For the flange Vf =
tdA LA 3.3 in
=
L0.3 in
A 0.16728 - 0.01536y2 B (12dy)
= 3.82 kip
Ans.
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7–6. If the beam is subjected to a shear of V = 15 kN, determine the web’s shear stress at A and B. Indicate the shear-stress components on a volume element located at these points. Show that the neutral axis is located at y = 0.1747 m from the bottom and INA = 0.2182110-32 m4.
200 mm
A
30 mm
25 mm V
(0.015)(0.125)(0.03) + (0.155)(0.025)(0.25) + (0.295)(0.2)(0.03) y = = 0.1747 m 0.125(0.03) + (0.025)(0.25) + (0.2)(0.03) I =
1 (0.125)(0.033) + 0.125(0.03)(0.1747 - 0.015)2 12
+
1 (0.025)(0.253) + 0.25(0.025)(0.1747 - 0.155)2 12
+
1 (0.2)(0.033) + 0.2(0.03)(0.295 - 0.1747)2 = 0.218182 (10 - 3) m4 12
B
250 mm
30 mm
125 mm
œ QA = yAA = (0.310 - 0.015 - 0.1747)(0.2)(0.03) = 0.7219 (10 - 3) m3
QB = yABœ = (0.1747 - 0.015)(0.125)(0.03) = 0.59883 (10 - 3) m3 tA =
15(103)(0.7219)(10 - 3) VQA = 1.99 MPa = It 0.218182(10 - 3)(0.025)
Ans.
tB =
VQB 15(103)(0.59883)(10 - 3) = 1.65 MPa = It 0.218182(10 - 3)0.025)
Ans.
7–7. If the wide-flange beam is subjected to a shear of V = 30 kN, determine the maximum shear stress in the beam.
200 mm
A
30 mm
25 mm V B 250 mm 30 mm
Section Properties: I =
1 1 (0.2)(0.310)3 (0.175)(0.250)3 = 268.652(10) - 6 m4 12 12
Qmax = © y¿A = 0.0625(0.125)(0.025) + 0.140(0.2)(0.030) = 1.0353(10) - 3 m3 tmax =
VQ 30(10)3(1.0353)(10) - 3 = 4.62 MPa = It 268.652(10) - 6 (0.025)
Ans.
477
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*7–8. If the wide-flange beam is subjected to a shear of V = 30 kN, determine the shear force resisted by the web of the beam.
200 mm
A
30 mm
1 1 (0.2)(0.310)3 (0.175)(0.250)3 = 268.652(10) - 6 m4 12 12
I =
Q = a
25 mm V B
0.155 + y b (0.155 - y)(0.2) = 0.1(0.024025 - y2) 2
250 mm
30(10)3(0.1)(0.024025 - y2)
tf =
268.652(10)
-6
30 mm
200 mm
(0.2) 0.155
Vf =
L
tf dA = 55.8343(10)6
L0.125
= 11.1669(10)6[ 0.024025y -
(0.024025 - y2)(0.2 dy)
1 3 0.155 y ] 2 0.125
Vf = 1.457 kN Vw = 30 - 2(1.457) = 27.1 kN
Ans.
•7–9. Determine the largest shear force V that the member can sustain if the allowable shear stress is tallow = 8 ksi.
3 in. 1 in. V 3 in. 1 in.
1 in.
y =
(0.5)(1)(5) + 2 [(2)(1)(2)] = 1.1667 in. 1 (5) + 2 (1)(2)
I =
1 (5)(13) + 5 (1)(1.1667 - 0.5)2 12
+ 2a
1 b (1)(23) + 2 (1)(2)(2 - 1.1667)2 = 6.75 in4 12
Qmax = ©y¿A¿ = 2 (0.91665)(1.8333)(1) = 3.3611 in3 tmax = tallow = 8 (103) = -
VQmax It
V (3.3611) 6.75 (2)(1)
V = 32132 lb = 32.1 kip
Ans.
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7–10. If the applied shear force V = 18 kip, determine the maximum shear stress in the member. 3 in. 1 in. V 3 in. 1 in.
1 in.
y =
(0.5)(1)(5) + 2 [(2)(1)(2)] = 1.1667 in. 1 (5) + 2 (1)(2)
I =
1 (5)(13) + 5 (1)(1.1667 - 0.5)2 12
+ 2a
1 b (1)(23) + 2 (1)(2)(2 - 1.1667) = 6.75 in4 12
Qmax = ©y¿A¿ = 2 (0.91665)(1.8333)(1) = 3.3611 in3 tmax =
18(3.3611) VQmax = = 4.48 ksi It 6.75 (2)(1)
Ans.
7–11. The wood beam has an allowable shear stress of tallow = 7 MPa. Determine the maximum shear force V that can be applied to the cross section.
50 mm
50 mm 100 mm
50 mm
200 mm V 50 mm
I =
1 1 (0.2)(0.2)3 (0.1)(0.1)3 = 125(10 - 6) m4 12 12
tallow = 7(106) =
VQmax It V[(0.075)(0.1)(0.05) + 2(0.05)(0.1)(0.05)] 125(10 - 6)(0.1)
V = 100 kN
Ans.
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*7–12. The beam has a rectangular cross section and is made of wood having an allowable shear stress of tallow = 200 psi. Determine the maximum shear force V that can be developed in the cross section of the beam. Also, plot the shear-stress variation over the cross section.
V 12 in.
8 in.
Section Properties The moment of inertia of the cross-section about the neutral axis is I =
1 (8) (123) = 1152 in4 12
Q as the function of y, Fig. a, Q =
1 (y + 6)(6 - y)(8) = 4 (36 - y2) 2
Qmax occurs when y = 0. Thus, Qmax = 4(36 - 02) = 144 in3 The maximum shear stress occurs of points along the neutral axis since Q is maximum and the thickness t = 8 in. is constant. tallow =
VQmax ; It
200 =
V(144) 1152(8)
V = 12800 16 = 12.8 kip
Ans.
Thus, the shear stress distribution as a function of y is t =
12.8(103) C 4(36 - y2) D VQ = It 1152 (8) =
E 5.56 (36 - y2) F psi
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7–13. Determine the maximum shear stress in the strut if it is subjected to a shear force of V = 20 kN.
12 mm
Section Properties: INA
60 mm
1 1 = (0.12) A 0.0843 B (0.04) A 0.063 B 12 12
V
= 5.20704 A 10 - 6 B m4
12 mm 80 mm
Qmax = ©y¿A¿
20 mm
20 mm
= 0.015(0.08)(0.03) + 0.036(0.012)(0.12) = 87.84 A 10 - 6 B m3 Maximum Shear Stress: Maximum shear stress occurs at the point where the neutral axis passes through the section. Applying the shear formula tmax =
VQmax It 20(103)(87.84)(10 - 6)
=
5.20704(10 - 6)(0.08)
= 4 22 MPa
Ans.
7–14. Determine the maximum shear force V that the strut can support if the allowable shear stress for the material is tallow = 40 MPa.
12 mm
60 mm
Section Properties: INA =
V
1 1 (0.12) A 0.0843 B (0.04) A 0.063 B 12 12
12 mm
= 5.20704 A 10 - 6 B m4
80 mm
Qmax = ©y¿A¿
20 mm
= 0.015(0.08)(0.03) + 0.036(0.012)(0.12) = 87.84 A 10 - 6 B m3 Allowable shear stress: Maximum shear stress occurs at the point where the neutral axis passes through the section. Applying the shear formula tmax = tallow = 40 A 106 B =
VQmax It V(87.84)(10 - 6) 5.20704(10 - 6)(0.08)
V = 189 692 N = 190 kN
Ans.
481
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7–15. Plot the shear-stress distribution over the cross section of a rod that has a radius c. By what factor is the maximum shear stress greater than the average shear stress acting over the cross section? c y V
x = 2c2 - y2 ;
p 4 c 4
I =
t = 2 x = 2 2c2 - y2 dA = 2 x dy = 22c2 - y2 dy dQ = ydA = 2y 2c2 - y2 dy x
Q =
Ly
2y2c2 - y2 dy = -
3 x 2 2 2 2 (c - y2)2 | y = (c2 - y2)3 3 3
3
V[23 (c2 - y2)2] VQ 4V 2 t = = = [c - y2) p 4 2 2 It 3pc4 ( 4 c )(2 2c - y ) The maximum shear stress occur when y = 0 tmax =
4V 3 p c2
tavg =
V V = A p c2
The faector =
tmax = tavg
4V 3 pc2 V pc2
=
4 3
Ans.
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*7–16. A member has a cross section in the form of an equilateral triangle. If it is subjected to a shear force V, determine the maximum average shear stress in the member using the shear formula. Should the shear formula actually be used to predict this value? Explain.
I =
V
1 (a)(h)3 36
y h ; = x a>2 Q =
a
LA¿
Q = a
y =
y dA = 2c a
2h x a
1 2 2 b (x)(y) a h - yb d 2 3 3
4h2 2x b (x2)a 1 b a 3a
t = 2x t =
t =
V(4h2>3a)(x2)(1 - 2x VQ a) = It ((1>36)(a)(h3))(2x) 24V(x - a2 x2) a2h
24V 4 dt = 2 2 a 1 - xb = 0 a dx ah At x =
y =
a 4 h 2h a a b = a 4 2
tmax =
24V a 2 a a b a1 - a b b a 4 a2h 4
tmax =
3V ah
Ans.
No, because the shear stress is not perpendicular to the boundary. See Sec. 7-3.
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•7–17.
Determine the maximum shear stress in the strut if it is subjected to a shear force of V = 600 kN.
30 mm
150 mm
V 100 mm 100 mm 100 mm
The moment of inertia of the cross-section about the neutral axis is I =
1 1 (0.3)(0.213) (0.2)(0.153) = 0.175275(10 - 3) m4 12 12
From Fig. a, Qmax = ©y¿A¿ = 0.09(0.03)(0.3) + 0.0375(0.075)(0.1) = 1.09125(10 - 3) m3 The maximum shear stress occurs at the points along the neutral axis since Q is maximum and thickness t = 0.1 m is the smallest. tmax =
VQmax 600(103)[1.09125(10 - 3)] = It 0.175275(10 - 3) (0.1) = 37.36(106) Pa = 37.4 MPa
Ans.
484
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7–18. Determine the maximum shear force V that the strut can support if the allowable shear stress for the material is tallow = 45 MPa.
30 mm
150 mm
V 100 mm 100 mm 100 mm
The moment of inertia of the cross-section about the neutral axis is I =
1 1 (0.3)(0.213) (0.2)(0.153) = 0.175275 (10 - 3) m4 12 12
From Fig. a Qmax = ©y¿A¿ = 0.09(0.03)(0.3) + 0.0375 (0.075)(0.1) = 1.09125 (10 - 3) m3 The maximum shear stress occeurs at the points along the neutral axis since Q is maximum and thickness t = 0.1 m is the smallest. tallow =
VQmax ; It
45(106) =
V C 1.09125(10 - 3) D
0.175275(10 - 3)(0.1)
V = 722.78(103) N = 723 kN
Ans.
485
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7–19. Plot the intensity of the shear stress distributed over the cross section of the strut if it is subjected to a shear force of V = 600 kN.
30 mm
The moment of inertia of the cross-section about the neutral axis is I =
1 1 (0.3)(0.213) (0.2)(0.153) = 0.175275 (10 - 3) m4 12 12
For 0.075 m 6 y … 0.105 m, Fig. a, Q as a function of y is Q = y¿A¿ =
1 (0.105 + y) (0.105 - y)(0.3) = 1.65375(10 - 3) - 0.15y2 2
For 0 … y 6 0.075 m, Fig. b, Q as a function of y is Q = ©y¿A¿ = 0.09 (0.03)(0.3) +
1 (0.075 + y)(0.075 - y)(0.1) = 1.09125(10 - 3) - 0.05 y2 2
For 0.075 m 6 y … 0.105 m, t = 0.3 m. Thus, t =
600 (103) C 1.65375(10 - 3) - 0.15y2 D VQ = (18.8703 - 1711.60y2) MPa = It 0.175275(10 - 3) (0.3)
At y = 0.075 m and y = 0.105 m, t|y = 0.015 m = 9.24 MPa
ty = 0.105 m = 0
For 0 … y 6 0.075 m, t = 0.1 m. Thus, t =
VQ 600 (103) [1.09125(10 - 3) - 0.05 y2] = (37.3556 - 1711.60 y2) MPa = It 0.175275(10 - 3) (0.1)
At y = 0 and y = 0.075 m, t|y = 0 = 37.4 MPa
ty = 0.075 m = 27.7 MPa
The plot shear stress distribution over the cross-section is shown in Fig. c.
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150 mm
V 100 mm 100 mm 100 mm
30 mm
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*7–20. The steel rod is subjected to a shear of 30 kip. Determine the maximum shear stress in the rod. The moment of inertia of the ciralor cross-section about the neutral axis (x axis) is p p I = r4 = (24) = 4 p in4 4 4
30 kip
dQ = ydA = y (2xdy) = 2xy dy 1
However, from the equation of the circle, x = (4 - y2)2 , Then 1
dQ = 2y(4 - y2)2 dy Thus, Q for the area above y is 2 in 1
2y (4 - y2)2 dy
Ly 3 2 in 2 = - (4 - y2)2 y 3 =
3 2 (4 - y2)2 3
1
Here, t = 2x = 2 (4 - y2)2 . Thus
30 C 23 (4 - y2)2 D VQ = t = 1 It 4p C 2(4 - y2)2 D 3
t =
5 (4 - y2) ksi 2p
By inspecting this equation, t = tmax at y = 0. Thus ¿= tmax
A 2 in.
Q for the differential area shown shaded in Fig. a is
Q =
1 in.
20 10 = 3.18 ksi = p 2p
Ans.
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•7–21.
The steel rod is subjected to a shear of 30 kip. Determine the shear stress at point A. Show the result on a volume element at this point. 1 in. A
The moment of inertia of the circular cross-section about the neutral axis (x axis) is I =
2 in.
p 4 p r = (24) = 4p in4 4 4
30 kip
Q for the differential area shown in Fig. a is dQ = ydA = y (2xdy) = 2xy dy 1
However, from the equation of the circle, x = (4 - y2)2 , Then 1
dQ = 2y (4 - y2)2 dy Thus, Q for the area above y is 2 in. 1
Q =
Ly
= -
2y (4 - y2)2 dy
2 in. 3 3 2 2 (4 - y2)2 ` = (4 - y2)2 3 3 y
1
Here t = 2x = 2 (4 - y2)2 . Thus,
30 C 23 (4 - y2)2 D VQ = t = 1 It 4p C 2(4 - y2)2 D 3
t =
5 (4 - y2) ksi 2p
For point A, y = 1 in. Thus tA =
5 (4 - 12) = 2.39 ksi 2p
Ans.
The state of shear stress at point A can be represented by the volume element shown in Fig. b.
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7–22. Determine the shear stress at point B on the web of the cantilevered strut at section a–a.
2 kN 250 mm
a
250 mm
4 kN 300 mm
a
20 mm 70 mm
(0.01)(0.05)(0.02) + (0.055)(0.07)(0.02) y = = 0.03625 m (0.05)(0.02) + (0.07)(0.02) I =
+
B
20 mm 50 mm
1 (0.05)(0.023) + (0.05)(0.02)(0.03625 - 0.01)2 12
1 (0.02)(0.073) + (0.02)(0.07)(0.055 - 0.03625)2 = 1.78625(10 - 6) m4 12
yBœ = 0.03625 - 0.01 = 0.02625 m QB = (0.02)(0.05)(0.02625) = 26.25(10 - 6) m3 tB =
6(103)(26.25)(10 - 6) VQB = It 1.78622(10 - 6)(0.02) = 4.41 MPa
Ans.
7–23. Determine the maximum shear stress acting at section a–a of the cantilevered strut.
2 kN 250 mm
a
250 mm
4 kN 300 mm
a
20 mm 70 mm
y =
(0.01)(0.05)(0.02) + (0.055)(0.07)(0.02) = 0.03625 m (0.05)(0.02) + (0.07)(0.02)
I =
1 (0.05)(0.023) + (0.05)(0.02)(0.03625 - 0.01)2 12
+
20 mm 50 mm
1 (0.02)(0.073) + (0.02)(0.07)(0.055 - 0.03625)2 = 1.78625(10 - 6) m4 12
Qmax = y¿A¿ = (0.026875)(0.05375)(0.02) = 28.8906(10 - 6) m3 tmax =
B
VQmax 6(103)(28.8906)(10 - 6) = It 1.78625(10 - 6)(0.02) = 4.85 MPa
Ans.
489
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*7–24. Determine the maximum shear stress in the T-beam at the critical section where the internal shear force is maximum.
10 kN/m
A 1.5 m
3m
The shear diagram is shown in Fig. b. As indicated, Vmax = 27.5 kN
150 mm
The neutral axis passes through centroid c of the cross-section, Fig. c. ' 0.075(0.15)(0.03) + 0.165(0.03)(0.15) © y A = y = ©A 0.15(0.03) + 0.03(0.15)
150 mm
1 (0.03)(0.153) + 0.03(0.15)(0.12 - 0.075)2 12 +
1 (0.15)(0.033) + 0.15(0.03)(0.165 - 0.12)2 12
= 27.0 (10 - 6) m4 From Fig. d, Qmax = y¿A¿ = 0.06(0.12)(0.03) = 0.216 (10 - 3) m3 The maximum shear stress occurs at points on the neutral axis since Q is maximum and thickness t = 0.03 m is the smallest. tmax =
27.5(103) C 0.216(10 - 3) D Vmax Qmax = It 27.0(10 - 6)(0.03) = 7.333(106) Pa = 7.33 MPa
Ans.
490
30 mm 30 mm
= 0.12 m I =
B
C
The FBD of the beam is shown in Fig. a,
1.5 m
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•7–25.
Determine the maximum shear stress in the T-beam at point C. Show the result on a volume element at this point.
10 kN/m
A
B
C
1.5 m
3m
150 mm
150 mm
30 mm
using the method of sections, + c ©Fy = 0;
VC + 17.5 -
1 (5)(1.5) = 0 2
VC = -13.75 kN The neutral axis passes through centroid C of the cross-section, 0.075 (0.15)(0.03) + 0.165(0.03)(0.15) ©yA = ©A 0.15(0.03) + 0.03(0.15)
y =
= 0.12 m I =
1 (0.03)(0.15) + 0.03(0.15)(0.12 - 0.075)2 12
+
1 (0.15)(0.033) + 0.15(0.03)(0.165 - 0.12)2 12
= 27.0 (10 - 6) m4 Qmax = y¿A¿ = 0.06 (0.12)(0.03) = 0.216 (10 - 3) m3 490 The maximum shear stress occurs at points on the neutral axis since Q is maximum and thickness t = 0.03 m is the smallest. tmax =
30 mm
13.75(103) C 0.216(10 - 3) D VC Qmax = It 27.0(10 - 6) (0.03)
= 3.667(106) Pa = 3.67 MPa
Ans.
491
1.5 m
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7–26. Determine the maximum shear stress acting in the fiberglass beam at the section where the internal shear force is maximum.
200 lb/ft
150 lb/ft
D
A 6 ft
6 ft
2 ft
4 in.
6 in.
0.5 in. 4 in.
Support Reactions: As shown on FBD. Internal Shear Force: As shown on shear diagram, Vmax = 878.57 lb. Section Properties: INA =
1 1 (4) A 7.53 B (3.5) A 63 B = 77.625 in4 12 12
Qmax = ©y¿A¿ = 3.375(4)(0.75) + 1.5(3)(0.5) = 12.375 in3 Maximum Shear Stress: Maximum shear stress occurs at the point where the neutral axis passes through the section. Applying the shear formula tmax =
=
VQmax It 878.57(12.375) = 280 psi 77.625(0.5)
Ans.
492
0.75 in.
0.75 in.
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7–27. Determine the shear stress at points C and D located on the web of the beam.
3 kip/ft
D
A
C
B 6 ft
6 ft
6 in.
0.75 in.
The FBD is shown in Fig. a. Using the method of sections, Fig. b, + c ©Fy = 0;
18 -
1 (3)(6) - V = 0 2
V = 9.00 kip. The moment of inertia of the beam’s cross section about the neutral axis is I =
1 1 (6)(103) (5.25)(83) = 276 in4 12 12
QC and QD can be computed by refering to Fig. c. QC = ©y¿A¿ = 4.5 (1)(6) + 2 (4)(0.75) = 33 in3 QD = y3œ A¿ = 4.5 (1)(6) = 27 in3 Shear Stress. since points C and D are on the web, t = 0.75 in. tC =
VQC 9.00 (33) = = 1.43 ksi It 276 (0.75)
Ans.
tD =
VQD 9.00 (27) = = 1.17 ksi It 276 (0.75)
Ans.
493
6 ft 1 in.
C D
4 in. 4 in.
6 in.
1 in.
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*7–28. Determine the maximum shear stress acting in the beam at the critical section where the internal shear force is maximum.
3 kip/ft
D
A
C
B 6 ft
6 ft
6 in.
The FBD is shown in Fig. a. The shear diagram is shown in Fig. b, Vmax = 18.0 kip.
0.75 in.
6 ft 1 in.
C D
4 in. 4 in.
6 in.
1 in.
The moment of inertia of the beam’s cross-section about the neutral axis is I =
1 1 (6)(103) (5.25)(83) 12 12
= 276 in4 From Fig. c Qmax = ©y¿A¿ = 4.5 (1)(6) + 2(4)(0.75) = 33 in3 The maximum shear stress occurs at points on the neutral axis since Q is the maximum and thickness t = 0.75 in is the smallest tmax =
Vmax Qmax 18.0 (33) = = 2.87 ksi It 276 (0.75)
Ans.
494
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7–30. The beam has a rectangular cross section and is subjected to a load P that is just large enough to develop a fully plastic moment Mp = PL at the fixed support. If the material is elastic-plastic, then at a distance x 6 L the moment M = Px creates a region of plastic yielding with an associated elastic core having a height 2y¿. This situation has been described by Eq. 6–30 and the moment M is distributed over the cross section as shown in Fig. 6–48e. Prove that the maximum shear stress developed in the beam is given by tmax = 321P>A¿2, where A¿ = 2y¿b, the crosssectional area of the elastic core.
P x Plastic region 2y¿
h
b Elastic region
Force Equilibrium: The shaded area indicares the plastic zone. Isolate an element in the plastic zone and write the equation of equilibrium. ; ©Fx = 0;
tlong A2 + sg A1 - sg A1 = 0 tlong = 0
This proves that the longitudinal shear stress. tlong, is equal to zero. Hence the corresponding transverse stress, tmax, is also equal to zero in the plastic zone. Therefore, the shear force V = P is carried by the malerial only in the elastic zone. Section Properties: INA =
1 2 (b)(2y¿)3 = b y¿ 3 12 3
Qmax = y¿ A¿ =
y¿ y¿ 2b (y¿)(b) = 2 2
Maximum Shear Stress: Applying the shear formula V A y¿2 b B 3
tmax However,
VQmax = = It
A¿ = 2by¿ tmax =
3P ‚ 2A¿
A by¿ B (b) 2 3
3
=
3P 4by¿
hence (Q.E.D.)
495
L
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7–31. The beam in Fig. 6–48f is subjected to a fully plastic moment Mp . Prove that the longitudinal and transverse shear stresses in the beam are zero. Hint: Consider an element of the beam as shown in Fig. 7–4c.
P x Plastic region 2y¿
h
b Elastic region
L
Force Equilibrium: If a fully plastic moment acts on the cross section, then an element of the material taken from the top or bottom of the cross section is subjected to the loading shown. For equilibrium ; ©Fx = 0;
sg A1 + tlong A2 - sg A1 = 0 tlong = 0
Thus no shear stress is developed on the longitudinal or transverse plane of the element. (Q. E. D.)
*7–32. The beam is constructed from two boards fastened together at the top and bottom with two rows of nails spaced every 6 in. If each nail can support a 500-lb shear force, determine the maximum shear force V that can be applied to the beam.
6 in. 6 in. 2 in. 2 in.
V
6 in.
Section Properties: I =
1 (6) A 43 B = 32.0 in4 12
Q = y¿A¿ = 1(6)(2) = 12.0 in4
Shear Flow: There are two rows of nails. Hence, the allowable shear flow 2(500) = 166.67 lb>in. q = 6 q =
166.67 =
VQ I V(12.0) 32.0
V = 444 lb
Ans.
496
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•7–33.
The beam is constructed from two boards fastened together at the top and bottom with two rows of nails spaced every 6 in. If an internal shear force of V = 600 lb is applied to the boards, determine the shear force resisted by each nail.
6 in. 6 in. 2 in. 2 in.
Section Properties: I =
1 (6) A 43 B = 32.0 in4 12
V
6 in.
Q = y¿A¿ = 1(6)(2) = 12.0 in4 Shear Flow: q =
VQ 600(12.0) = = 225 lb>in. I 32.0
There are two rows of nails. Hence, the shear force resisted by each nail is q 225 lb>in. F = a bs = a b(6 in.) = 675 lb 2 2
Ans.
7–34. The beam is constructed from two boards fastened together with three rows of nails spaced s = 2 in. apart. If each nail can support a 450-lb shear force, determine the maximum shear force V that can be applied to the beam. The allowable shear stress for the wood is tallow = 300 psi.
s s 1.5 in.
The moment of inertia of the cross-section about the neutral axis is I =
V
1 (6)(33) = 13.5 in4 12
6 in.
Refering to Fig. a, QA = Qmax = y¿A¿ = 0.75(1.5)(6) = 6.75 in3 The maximum shear stress occurs at the points on the neutral axis where Q is maximum and t = 6 in. tallow =
VQmax ; It
300 =
V(6.75) 13.5(6)
V = 3600 lb = 3.60 kips
Shear Flow: Since there are three rows of nails, F 450 b = 675 lb>in. qallow = 3 a b = 3 a s 2 VQA V(6.75) ; 675 = qallow = I 13.5 V = 1350 lb = 1.35 kip
497
Ans.
1.5 in.
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7–35. The beam is constructed from two boards fastened together with three rows of nails. If the allowable shear stress for the wood is tallow = 150 psi, determine the maximum shear force V that can be applied to the beam. Also, find the maximum spacing s of the nails if each nail can resist 650 lb in shear.
s s 1.5 in. V
6 in.
The moment of inertia of the cross-section about the neutral axis is I =
1 (6)(33) = 13.5 in4 12
Refering to Fig. a, QA = Qmax = y¿A¿ = 0.75(1.5)(6) = 6.75 in3 The maximum shear stress occurs at the points on the neutral axis where Q is maximum and t = 6 in. tallow =
VQmax ; It
150 =
V(6.75) 13.5(6)
V = 1800 lb = 1.80 kip Since there are three rows of nails, qallow = 3 a qallow =
VQA ; I
Ans.
F 650 1950 lb b = 3¢ b ≤ = a s s s in.
1800(6.75) 1950 = s 13.5
s = 2.167 in = 2
1 in 8
Ans.
498
1.5 in.
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*7–36. The beam is fabricated from two equivalent structural tees and two plates. Each plate has a height of 6 in. and a thickness of 0.5 in. If a shear of V = 50 kip is applied to the cross section, determine the maximum spacing of the bolts. Each bolt can resist a shear force of 15 kip.
0.5 in. s 3 in.
1 in. A
Section Properties: INA =
V
6 in.
1 1 (3) A 93 B (2.5) A 83 B 12 12
0.5 in.
N
1 1 (0.5) A 23 B + (1) A 63 B 12 12
3 in.
= 93.25 in4 Q = ©y¿A¿ = 2.5(3)(0.5) + 4.25(3)(0.5) = 10.125 in3 Shear Flow: Since there are two shear planes on the bolt, the allowable shear flow is 2(15) 30 . = q = s s VQ q = I 50(10.125) 30 = s 93.25 s = 5.53 in.
Ans.
•7–37.
The beam is fabricated from two equivalent structural tees and two plates. Each plate has a height of 6 in. and a thickness of 0.5 in. If the bolts are spaced at s = 8 in., determine the maximum shear force V that can be applied to the cross section. Each bolt can resist a shear force of 15 kip.
0.5 in. s 3 in.
1 in. A
Section Properties: INA
-
1 1 (0.5) A 23 B + (1) A 63 B 12 12
3 in.
Q = ©y¿A¿ = 2.5(3)(0.5) + 4.25(3)(0.5) = 10.125 in3 Shear Flow: Since there are two shear planes on the bolt, the allowable shear flow is 2(15) = 3.75 kip>in. q = 8
3.75 =
0.5 in.
N
= 93.25 in4
q =
V
6 in.
1 1 = (3) A 93 B (2.5) A 83 B 12 12
VQ I V(10.125) 93.25
y = 34.5 kip
Ans.
499
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7–38. The beam is subjected to a shear of V = 2 kN. Determine the average shear stress developed in each nail if the nails are spaced 75 mm apart on each side of the beam. Each nail has a diameter of 4 mm. The neutral axis passes through centroid C of the cross-section as shown in Fig. a. ' 0.175(0.05)(0.2) + 0.1(0.2)(0.05) © y A y = = = 0.1375 m ©A 0.05(0.2) + 0.2(0.05)
200 mm
25 mm
75 mm 50 mm 75 mm
V 200 mm
Thus, I =
1 (0.2)(0.053) + 0.2 (0.05)(0.175 - 0.1375)2 12 +
25 mm
1 (0.05)(0.23) + 0.05(0.2)(0.1375 - 0.1)2 12
= 63.5417(10 - 6) m4 Q for the shaded area shown in Fig. b is Q = y¿A¿ = 0.0375 (0.05)(0.2) = 0.375(10 - 3) m3 Since there are two rows of nails q = 2a
q =
VQ ; I
26.67 F =
F 2F b = = (26.67 F) N>m. s 0.075
2000 C 0.375 (10 - 3) D 63.5417 (10 - 6)
F = 442.62 N Thus, the shear stress developed in the nail is tn =
F 442.62 = = 35.22(106)Pa = 35.2 MPa p A 2 (0.004 ) 4
Ans.
500
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7–39. A beam is constructed from three boards bolted together as shown. Determine the shear force developed in each bolt if the bolts are spaced s = 250 mm apart and the applied shear is V = 35 kN.
25 mm 25 mm 100 mm 250 mm
2 (0.125)(0.25)(0.025) + 0.275 (0.35)(0.025) y = = 0.18676 m 2 (0.25)(0.025) + 0.35 (0.025) I = (2)a
+
1 b(0.025)(0.253) + 2 (0.025)(0.25)(0.18676 - 0.125)2 12
V
1 (0.025)(0.35)3 + (0.025)(0.35)(0.275 - 0.18676)2 12
350 mm
s = 250 mm
= 0.270236 (10 - 3) m4
25 mm -3
3
Q = y¿A¿ = 0.06176(0.025)(0.25) = 0.386(10 ) m q =
35 (0.386)(10 - 3) VQ = 49.997 kN>m = I 0.270236 (10 - 3)
F = q(s) = 49.997 (0.25) = 12.5 kN
Ans.
*7–40. The double-web girder is constructed from two plywood sheets that are secured to wood members at its top and bottom. If each fastener can support 600 lb in single shear, determine the required spacing s of the fasteners needed to support the loading P = 3000 lb. Assume A is pinned and B is a roller.
2 in. 2 in. s 10 in. A 4 ft 2 in. 2 in. 6 in. 0.5 in. 0.5 in.
Support Reactions: As shown on FBD. Internal Shear Force: As shown on shear diagram, Vmax = 1500 lb. Section Properties: INA =
P
1 1 (7) A 183 B (6) A 103 B = 2902 in4 12 12
Q = y¿A¿ = 7(4)(6) = 168 in3 Shear Flow: Since there are two shear planes on the bolt, the allowable shear flow is 2(600) 1200 = q = . s s VQ q = I 1500(168) 1200 = s 2902 s = 13.8 in.
Ans.
501
4 ft
B
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•7–41.
The double-web girder is constructed from two plywood sheets that are secured to wood members at its top and bottom. The allowable bending stress for the wood is sallow = 8 ksi and the allowable shear stress is tallow = 3 ksi. If the fasteners are spaced s = 6 in. and each fastener can support 600 lb in single shear, determine the maximum load P that can be applied to the beam.
2 in. 2 in. s 10 in. A 4 ft 2 in. 2 in. 6 in. 0.5 in. 0.5 in.
Support Reactions: As shown on FBD. Internal Shear Force and Moment: As shown on shear and moment diagram, Vmax = 0.500P and Mmax = 2.00P. Section Properties: INA =
P
1 1 (7) A 183 B (6) A 103 B = 2902 in4 12 12
Q = y2œ A¿ = 7(4)(6) = 168 in3 Qmax = ©y¿A¿ = 7(4)(6) + 4.5(9)(1) = 208.5 in3 Shear Flow: Assume bolt failure. Since there are two shear planes on the bolt, the 2(600) = 200 lb>in. allowable shear flow is q = 6 VQ q = I 0.500P(168) 200 = 2902 P = 6910 lb = 6.91 kip (Controls !)
Ans.
Shear Stress: Assume failure due to shear stress. VQmax It 0.500P(208.5) 3000 = 2902(1) tmax = tallow =
P = 22270 lb = 83.5 kip Bending Stress: Assume failure due to bending stress. Mc I 2.00P(12)(9) 8(103) = 2902
smax = sallow =
P = 107 ksi
502
4 ft
B
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7–42. The T-beam is nailed together as shown. If the nails can each support a shear force of 950 lb, determine the maximum shear force V that the beam can support and the corresponding maximum nail spacing s to the nearest 18 in. The allowable shear stress for the wood is tallow = 450 psi.
2 in.
s
The neutral axis passes through the centroid c of the cross-section as shown in Fig. a. ' 13(2)(12) + 6(12)(2) © y A y = = = 9.5 in. ©A 2(12) + 12(2) I =
1 (2)(123) + 2(12)(9.5 - 6)2 12 +
= 884 in4 Refering to Fig. a, Qmax and QA are Qmax = y1œ A1œ = 4.75(9.5)(2) = 90.25 in3 QA = y2œ A2œ = 3.5 (2)(12) = 84 in3 The maximum shear stress occurs at the points on the neutral axis where Q is maximum and t = 2 in. VQmax ; It
450 =
V (90.25) 884 (2)
V = 8815.51 lb = 8.82 kip Here, qallow =
F 950 = lb>in. Then s s VQA ; qallow = I
Ans.
8815.51(84) 950 = s 884 s = 1.134 in = 1
12 in. V
2 in.
1 (12)(23) + 12(2)(13 - 9.5)2 12
tallow =
s
12 in.
1 in 8
Ans.
503
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7–43. Determine the average shear stress developed in the nails within region AB of the beam. The nails are located on each side of the beam and are spaced 100 mm apart. Each nail has a diameter of 4 mm. Take P = 2 kN.
P 2 kN/m
A
B
C
1.5 m
The FBD is shown in Fig. a. As indicated in Fig. b, the internal shear force on the cross-section within region AB is constant that is VAB = 5 kN.
1.5 m
100 mm
The neutral axis passes through centroid C of the cross section as shown in Fig. c. ' 0.18(0.04)(0.2) + 0.1(0.2)(0.04) © y A = y = ©A 0.04(0.2) + 0.2(0.04)
40 mm
= 0.14 m
200 mm
1 I = (0.04)(0.23) + 0.04(0.2)(0.14 - 0.1)2 12 1 + (0.2)(0.043) + 0.2(0.04)(0.18 - 0.14)2 12
200 mm 20 mm 20 mm
= 53.333(10 - 6) m4 Q for the shaded area shown in Fig. d is Q = y¿A¿ = 0.04(0.04)(0.2) = 0.32(10 - 3) m3 Since there are two rows of nail, q = 2 a q =
VAB Q ; I
20F =
F F b = 2a b = 20F N>m. s 0.1
5(103) C 0.32(10 - 3) D 53.333(10 - 6)
F = 1500 N Thus, the average shear stress developed in each nail is
A tnail B avg =
F 1500 = = 119.37(106)Pa = 119 MPa p Anail 2 (0.004 ) 4
504
Ans.
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*7–44. The nails are on both sides of the beam and each can resist a shear of 2 kN. In addition to the distributed loading, determine the maximum load P that can be applied to the end of the beam. The nails are spaced 100 mm apart and the allowable shear stress for the wood is tallow = 3 MPa.
P 2 kN/m
A
B
C
1.5 m
1.5 m
100 mm
The FBD is shown in Fig. a. 40 mm
As indicated the shear diagram, Fig. b, the maximum shear occurs in region AB of Constant value, Vmax = (P + 3) kN. The neutral axis passes through Centroid C of the cross-section as shown in Fig. c. ' 0.18(0.04)(0.2) + 0.1(0.2)(0.04) © y A = y = ©A 0.04(0.2) + 0.2(0.04) = 0.14 m I =
1 (0.04)(0.23) + 0.04(0.2)(0.14 - 0.1)2 12 1 + (0.2)(0.043) + 0.2(0.04)(0.18 - 0.142) 12
Refering to Fig. d, Qmax = y1œ A1œ = 0.07(0.14)(0.04) = 0.392(10 - 3) m3 QA = y2œ A2œ = 0.04(0.04)(0.2) = 0.32(10 - 3) m3 The maximum shear stress occurs at the points on Neutral axis where Q is maximum and t = 0.04 m. Vmax Qmax ; It
3(106) =
(P + 3)(103) C 0.392(10 - 3) D 53.333(10 - 6)(0.04)
P = 13.33 kN Since there are two rows of nails qallow = 2 a qallow
Vmax QA = ; I
40 000 =
200 mm 20 mm 20 mm
= 53.333(10 - 6) m4
tallow =
200 mm
2(103) F d = 40 000 N>m. b = 2c s 0.1
(P + 3)(103) C 0.32(10 - 3) D 53.333(10 - 6)
P = 3.67 kN (Controls!)
Ans.
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7–44.
Continued
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•7–45.
The beam is constructed from four boards which are nailed together. If the nails are on both sides of the beam and each can resist a shear of 3 kN, determine the maximum load P that can be applied to the end of the beam.
3 kN
A
P
B
C
2m
2m
100 mm
30 mm
150 mm
30 mm
250 mm 30 mm 30 mm
Support Reactions: As shown on FBD. Internal Shear Force: As shown on shear diagram, VAB = (P + 3) kN. Section Properties: INA =
1 1 (0.31) A 0.153 B (0.25) A 0.093 B 12 12
= 72.0 A 10 - 6 B m4
Q = y¿A¿ = 0.06(0.25)(0.03) = 0.450 A 10 - 3 B m3 Shear Flow: There are two rows of nails. Hence the allowable shear flow is 3(2) = 60.0 kN>m. q = 0.1 VQ q = I (P + 3)(103)0.450(10 - 3) 60.0 A 103 B = 72.0(10 - 6) P = 6.60 kN
Ans.
507
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7–47. The beam is made from four boards nailed together as shown. If the nails can each support a shear force of 100 lb., determine their required spacing s and s if the beam is subjected to a shear of V = 700 lb.
D 1 in. 1 in. 2 in.
s¿ s¿ s
A
C s
10 in.
1 in.
10 in. V B 1.5 in.
Section Properties: y =
©yA 0.5(10)(1) + 1.5(2)(3) + 6(1.5)(10) = ©A 10(1) + 2(3) + 1.5(10) = 3.3548 in
INA =
1 (10) A 13 B + 10(1)(3.3548 - 0.5)2 12 1 + (2) A 33 B + 2(3)(3.3548 - 1.5)2 12
= 337.43 in4 QC = y1 ¿A¿ = 1.8548(3)(1) = 5.5645 in3 QD = y2 ¿A¿ = (3.3548 - 0.5)(10)(1) + 2 C (3.3548 - 1.5)(3)(1) D = 39.6774 in3 Shear Flow: The allowable shear flow at points C and D is qC = 100 , respectively. qB = s¿ VQC qC = I 700(5.5645) 100 = s 337.43 s = 8.66 in. VQD qD = I 700(39.6774) 100 = s¿ 337.43
100 and s
Ans.
s¿ = 1.21 in.
Ans.
508
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*7–48. The box beam is constructed from four boards that are fastened together using nails spaced along the beam every 2 in. If each nail can resist a shear of 50 lb, determine the greatest shear V that can be applied to the beam without causing failure of the nails.
1 in. 12 in. 5 in.
V 2 in.
1 in. 6 in.
1 in.
y =
©yA 0.5 (12)(1) + 2 (4)(6)(1) + (6.5)(6)(1) = = 3.1 in. ©A 12(1) + 2(6)(1) + (6)(1)
I =
1 (12)(13) + 12(1)(3.1 - 0.5)2 12 + 2a +
1 b (1)(63) + 2(1)(6)(4 - 3.1)2 12
1 (6)(13) + 6(1)(6.5 - 3.1)2 = 197.7 in4 12
QB = y1œ A¿ = 2.6(12)(1) = 31.2 in3
qB =
V(31.2) 1 VQB a b = = 0.0789 V 2 I 2(197.7)
qB s = 0.0789V(2) = 50 V = 317 lb (controls)
Ans.
QA = y2œ A¿ = 3.4(6)(1) = 20.4 in3
qA =
V(20.4) 1 VQA a b = = 0.0516 V 2 I 2(197.7)
qA s = 0.0516V(2) = 50 V = 485 lb
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7–50. A shear force of V = 300 kN is applied to the box girder. Determine the shear flow at points A and B. 90 mm
90 mm C
A
D
200 mm B
190 mm V
200 mm
10 mm 180 mm
10 mm
The moment of inertia of the cross-section about the neutral axis is I =
1 1 (0.2)(0.43) (0.18)(0.383) = 0.24359(10 - 3) m4 12 12
Refering to Fig. a Fig. b, QA = y1œ A1œ = 0.195 (0.01)(0.19) = 0.3705 (10 - 3) m3 QB = 2yzœ A2œ + y3œ A3œ = 2 [0.1(0.2)(0.01)] + 0.195(0.01)(0.18) = 0.751(10 - 3) m3 Due to symmety, the shear flow at points A and A¿ , Fig. a, and at points B and B¿ , Fig. b, are the same. Thus qA
3 -3 1 300(10 ) C 0.3705(10 ) D 1 VQA s b = c = a 2 I 2 0.24359(10 - 3)
= 228.15(103) N>m = 228 kN>m
qB =
Ans.
3 -3 1 VQB 1 300(10 ) C 0.751(10 ) D s a b = c 2 I 2 0.24359(10 - 3)
= 462.46(103) N>m = 462 kN>m
Ans.
510
100 mm
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7–51. A shear force of V = 450 kN is applied to the box girder. Determine the shear flow at points C and D. 90 mm
90 mm C
A
D
200 mm B
190 mm V
200 mm
10 mm 180 mm
10 mm
The moment of inertia of the cross-section about the neutral axis is I =
1 1 (0.2)(0.43) (0.18)(0.383) = 0.24359(10 - 3) m4 12 12
Refering to Fig. a, due to symmetry ACœ = 0. Thus QC = 0 Then refering to Fig. b, QD = y1œ A1œ + y2œ A2œ = 0.195 (0.01)(0.09) + 0.15(0.1)(0.01) = 0.3255(10 - 3) m3 Thus, qC =
qD =
VQC = 0 I
Ans.
450(103) C 0.3255(10 - 3) D VQD = I 0.24359(10 - 3)
= 601.33(103) N>m = 601 kN>m
Ans.
100 mm
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*7–52. A shear force of V = 18 kN is applied to the symmetric box girder. Determine the shear flow at A and B.
10 mm 30 mm 10 mm
A
100 mm C
B
100 mm
150 mm
10 mm 10 mm
V 150 mm 10 mm 125 mm 10 mm
Section Properties: INA =
1 1 (0.145) A 0.33 B (0.125) A 0.283 B 12 12 + 2c
1 (0.125) A 0.013 B + 0.125(0.01) A 0.1052 B d 12
= 125.17 A 10 - 6 B m4 QA = y2œ A¿ = 0.145(0.125)(0.01) = 0.18125 A 10 - 3 B m3 QB = y1œ A¿ = 0.105(0.125)(0.01) = 0.13125 A 10 - 3 B m3
Shear Flow: qA =
=
1 VQA c d 2 I 1 18(103)(0.18125)(10 - 3) d c 2 125.17(10 - 6)
= 13033 N>m = 13.0 kN>m qB =
=
Ans.
1 VQB c d 2 I 1 18(103)(0.13125)(10 - 3) d c 2 125.17(10 - 6)
= 9437 N>m = 9.44 kN>m
Ans.
512
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A shear force of V = 18 kN is applied to the box girder. Determine the shear flow at C.
•7–53.
10 mm 30 mm 10 mm
A
100 mm C
B
100 mm
150 mm
10 mm 10 mm
V 150 mm 10 mm 125 mm 10 mm
Section Properties: INA =
1 1 (0.145) A 0.33 B (0.125) A 0.283 B 12 12 +2 c
1 (0.125) A 0.013 B + 0.125(0.01) A 0.1052 B d 12
= 125.17 A 10 - 6 B m4 QC = ©y¿A¿ = 0.145(0.125)(0.01) + 0.105(0.125)(0.01) + 0.075(0.15)(0.02) = 0.5375 A 10 - 3 B m3 Shear Flow: qC =
=
1 VQC c d 2 I 1 18(103)(0.5375)(10 - 3) d c 2 125.17(10 - 4)
= 38648 N>m = 38.6 kN>m
Ans.
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7–54. The aluminum strut is 10 mm thick and has the cross section shown. If it is subjected to a shear of, V = 150 N, determine the shear flow at points A and B. 10 mm
2[0.005(0.03)(0.01)] + 2[0.03(0.06)(0.01)] + 0.055(0.04)(0.01) = 0.027727 m y = 2(0.03)(0.01) + 2(0.06)(0.01) + 0.04(0.01) I = 2c
1 (0.03)(0.01)3 + 0.03(0.01)(0.027727 - 0.005)2 d 12
+ 2c +
40 mm 10 mm 30 mm
1 (0.01)(0.06)3 + 0.01(0.06)(0.03 - 0.027727)2 d 12
B A V 40 mm 10 mm
30 mm 10 mm
1 (0.04)(0.01)3 + 0.04(0.01)(0.055 - 0.027727)2 = 0.98197(10 - 6) m4 12
yB ¿ = 0.055 - 0.027727 = 0.027272 m yA ¿ = 0.027727 - 0.005 = 0.022727 m QA = yA ¿A¿ = 0.022727(0.04)(0.01) = 9.0909(10 - 6) m3 QB = yB ¿A¿ = 0.027272(0.03)(0.01) = 8.1818(10 - 6) m3 qA =
VQA 150(9.0909)(10 - 6) = 1.39 kN>m = I 0.98197(10 - 6)
Ans.
qB =
VQB 150(8.1818)(10 - 6) = 1.25 kN>m = I 0.98197(10 - 6)
Ans.
7–55. The aluminum strut is 10 mm thick and has the cross section shown. If it is subjected to a shear of V = 150 N, determine the maximum shear flow in the strut.
y =
2[0.005(0.03)(0.01)] + 2[0.03(0.06)(0.01)] + 0.055(0.04)(0.01) 2(0.03)(0.01) + 2(0.06)(0.01) + 0.04(0.01)
10 mm 40 mm
B A
= 0.027727 m I = 2c
10 mm
1 (0.03)(0.01)3 + 0.03(0.01)(0.027727 - 0.005)2 d 12
30 mm
1 + 2 c (0.01)(0.06)3 + 0.01(0.06)(0.03 - 0.027727)2 d 12 +
10 mm
1 (0.04)(0.01)3 + 0.04(0.01)(0.055 - 0.027727)2 12
= 0.98197(10 - 6) m4
Qmax = (0.055 - 0.027727)(0.04)(0.01) + 2[(0.06 - 0.027727)(0.01)]a
0.06 - 0.0277 b 2
= 21.3(10 - 6) m3 qmax =
V 40 mm
1 150(21.3(10 - 6)) 1 VQmax b = 1.63 kN>m a b = a 2 I 2 0.98197(10 - 6)
Ans.
514
30 mm 10 mm
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*7–56. The beam is subjected to a shear force of V = 5 kip. Determine the shear flow at points A and B.
0.5 in. C
5 in. 5 in. 0.5 in.
0.5 in. 2 in.
0.25(11)(0.5) + 2[4.5(8)(0.5)] + 6.25(10)(0.5) ©yA y = = = 3.70946 in. ©A 11(0.5) + 2(8)(0.5) + 10(0.5)
A D
8 in.
1 1 (11)(0.53) + 11(0.5)(3.70946 - 0.25)2 + 2c (0.5)(83) + 0.5(8)(4.5 - 3.70946)2 d 12 12
I =
+
0.5 in.
V
B
1 (10)(0.53) + 10(0.5)(6.25 - 3.70946)2 12
= 145.98 in4 œ = 3.70946 - 0.25 = 3.45946 in. yA
yBœ = 6.25 - 3.70946 = 2.54054 in. œ QA = yA A¿ = 3.45946(11)(0.5) = 19.02703 in3
QB = yBœ A¿ = 2.54054(10)(0.5) = 12.7027 in3 qA =
1 VQA 1 5(103)(19.02703) a b = a b = 326 lb>in. 2 I 2 145.98
Ans.
qB =
1 VQB 1 5(103)(12.7027) a b = a b = 218 lb>in. 2 I 2 145.98
Ans.
•7–57.
The beam is constructed from four plates and is subjected to a shear force of V = 5 kip. Determine the maximum shear flow in the cross section.
0.5 in. C
5 in. 5 in. 0.5 in.
y =
©yA 0.25(11)(0.5) + 2[4.5(8)(0.5)] + 6.25(10)(0.5) = = 3.70946 in. ©A 11(0.5) + 2(8)(0.5) + 10(0.5)
I =
1 1 (11)(0.53) + 11(0.5)(3.45952) + 2 c (0.5)(83) + 0.5(8)(0.79052) d 12 12
0.5 in. 2 in.
A D
8 in. V
+
1 (10)(0.53) + 10(0.5)(2.54052) 12
= 145.98 in4 Qmax = 3.4594 (11)(0.5) + 2[(1.6047)(0.5)(3.7094 - 0.5)] = 24.177 in3 qmax =
1 VQmax 1 5(103)(24.177) a b = a b 2 I 2 145.98
= 414 lb>in.
Ans.
515
0.5 in.
B
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7–58. The channel is subjected to a shear of V = 75 kN. Determine the shear flow developed at point A.
30 mm 400 mm
A
200 mm
30 mm
V ⫽ 75 kN
30 mm
y =
©yA 0.015(0.4)(0.03) + 2[0.13(0.2)(0.03)] = = 0.0725 m ©A 0.4(0.03) + 2(0.2)(0.03)
I =
1 (0.4)(0.033) + 0.4(0.03)(0.0725 - 0.015)2 12 + 2c
1 (0.03)(0.23) + 0.03(0.2)(0.13 - 0.0725)2 d = 0.12025(10 - 3) m4 12
œ A¿ = 0.0575(0.2)(0.03) = 0.3450(10 - 3) m3 QA = yA
q =
qA =
VQ I 75(103)(0.3450)(10 - 3) 0.12025(10 - 3)
= 215 kN>m
Ans.
7–59. The channel is subjected to a shear of V = 75 kN. Determine the maximum shear flow in the channel.
30 mm 400 mm
A V ⫽ 75 kN
30 mm
y =
©yA 0.015(0.4)(0.03) + 2[0.13(0.2)(0.03)] = ©A 0.4(0.03) + 2(0.2)(0.03) = 0.0725 m
1 I = (0.4)(0.033) + 0.4(0.03)(0.0725 - 0.015)2 12 1 + 2c (0.03)(0.23) + 0.03(0.2)(0.13 - 0.0725)2 d 12 = 0.12025(10 - 3) m4 Qmax = y¿A¿ = 0.07875(0.1575)(0.03) = 0.37209(10 - 3) m3 qmax =
75(103)(0.37209)(10 - 3) 0.12025(10 - 3)
= 232 kN>m
Ans.
516
200 mm
30 mm
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*7–60. The angle is subjected to a shear of V = 2 kip. Sketch the distribution of shear flow along the leg AB. Indicate numerical values at all peaks.
A 5 in.
5 in. 45⬚ 45⬚
0.25 in.
Section Properties: b =
0.25 = 0.35355 in. sin 45°
h = 5 cos 45° = 3.53553 in. INA = 2c
1 (0.35355) A 3.535533 B d = 2.604167 in4 12
Q = y¿A¿ = [0.25(3.53553) + 0.5y]a2.5 -
y b(0.25) sin 45°
= 0.55243 - 0.17678y2 Shear Flow: VQ I 2(103)(0.55243 - 0.17678y2) = 2.604167
q =
= {424 - 136y2} lb>in. At y = 0,
Ans.
q = qmax = 424 lb>in.
Ans.
517
B V
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•7–61.
The assembly is subjected to a vertical shear of V = 7 kip. Determine the shear flow at points A and B and the maximum shear flow in the cross section.
A
0.5 in.
B V 2 in.
0.5 in.
0.5 in.
6 in.
6 in. 2 in. 0.5 in.
y =
©yA (0.25)(11)(0.5) + 2(3.25)(5.5)(0.5) + 6.25(7)(0.5) = = 2.8362 in. ©A 0.5(11) + 2(0.5)(5.5) + 7(0.5)
I =
1 1 (11)(0.53) + 11(0.5)(2.8362 - 0.25)2 + 2a b(0.5)(5.53) + 2(0.5)(5.5)(3.25 - 2.8362)2 12 12 +
1 (7)(0.53) + (0.5)(7)(6.25 - 2.8362)2 = 92.569 in4 12
QA = y1 ¿A1 ¿ = (2.5862)(2)(0.5) = 2.5862 in3 QB = y2 ¿A2 ¿ = (3.4138)(7)(0.5) = 11.9483 in3 Qmax = ©y¿A¿ = (3.4138)(7)(0.5) + 2(1.5819)(3.1638)(0.5) = 16.9531 in3 q =
VQ I
7(103)(2.5862) = 196 lb>in. 92.569 1 7(103)(11.9483) qB = a b = 452 lb>in. 2 92.569 1 7(103)(16.9531) b = 641 lb>in. qmax = a 2 92.569 qA =
Ans. Ans. Ans.
518
0.5 in.
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7–62. Determine the shear-stress variation over the cross section of the thin-walled tube as a function of elevation y and show that t max = 2V>A, where A = 2prt. Hint: Choose a differential area element dA = Rt du. Using dQ = y dA, formulate Q for a circular section from u to (p - u) and show that Q = 2R2t cos u, where cos u = 2R2 - y2>R.
ds du y
u t
dA = R t du dQ = y dA = yR t du Here y = R sin u Therefore dQ = R2 t sin u du p-u
Q =
p-u
R2 t sin u du = R2 t(-cos u) |
Lu
u
2
= R t [-cos (p - u) - (-cos u)] = 2R2 t cos u dI = y2 dA = y2 R t du = R3 t sin2 u du 2p
I =
L0
2p
R3 t sin2 u du = R3 t 2p
=
t =
sin 2u R3 t [u ] 2 2 0
R3 t [2p - 0] = pR3 t 2
VQ V(2R2t cos u) V cos u = = 3 It pR t pR t(2t)
Here cos u =
t =
=
L0
(1 - cos 2u) du 2
2R2 - y2 R
V 2R2 - y2 pR2t
Ans.
tmax occurs at y = 0; therefore tmax =
V pR t
A = 2pRt; therefore tmax =
2V A
QED
519
R
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7–63. Determine the location e of the shear center, point O, for the thin-walled member having the cross section shown where b2 7 b1. The member segments have the same thickness t.
t h
e
b2
Section Properties: I =
1 h 2 t h2 t h3 + 2c(b1 + b2)ta b d = C h + 6(b1 + b2) D 12 2 12 Q1 = y¿A¿ =
h ht (x )t = x 2 1 2 1
Q2 = y¿A¿ =
h ht (x )t = x 2 2 2 2
Shear Flow Resultant: VQ1 q1 = = I q2 =
VQ2 = I
P A ht2 x1 B P A ht2 x2 B
h C h + 6(b1 + b2) D h C h + 6(b1 + b2) D
6P
t h2 12
C h + 6(b1 + b2) D
=
t h2 12
C h + 6(b1 + b2) D
=
6P
b1
(Ff)1 =
L0
q1 dx1 =
6P
x1
x2
b1
h C h + 6(b1 + b2) D L0
x1 dx1
3Pb21
= b2
(Ff)2 =
L0
q2 dx2 =
h C h + 6(b1 + b2) D 6P
b2
h C h + 6(b1 + b2) D L0
x2 dx2
3Pb22
=
h C h + 6(b1 + b2) D
Shear Center: Summing moment about point A. Pe = A Ff B 2 h - A Ff B 1 h Pe =
e =
3Pb22
h C h + 6(b1 + b2) D
3(b22 - b21) h + 6(b1 + b2)
(h) -
3Pb21
h C h + 6(b1 + b2) D
(h)
Ans.
Note that if b2 = b1, e = 0 (I shape).
520
b1
O
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*7–64. Determine the location e of the shear center, point O, for the thin-walled member having the cross section shown. The member segments have the same thickness t.
b d 45⬚ O e
Section Properties: I =
=
t 1 a b(2d sin 45°)3 + 2 C bt(d sin 45°)2 D 12 sin 45° td2 (d + 3b) 3
Q = y¿A¿ = d sin 45° (xt) = (td sin 45°)x Shear Flow Resultant: qf =
P(td sin 45°)x VQ 3P sin 45° = = x td2 I d(d + 3b) (d + 3b) 3 b
Ff =
L0
b
qfdx =
2
3P sin 45° 3b sin 45° P xdx = d(d + 3b) L0 2d(d + 3b)
Shear Center: Summing moments about point A, Pe = Ff(2d sin 45°) Pe = c e =
3b2 sin 45° P d(2d sin 45°) 2d(d + 3b)
3b2 2(d + 3b)
Ans.
521
45⬚
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•7–65.
Determine the location e of the shear center, point O, for the thin-walled member having a slit along its side. Each element has a constant thickness t.
a e
a
t
a
Section Properties: I =
1 10 3 (2t)(2a)3 + 2 C at A a2 B D = a t 12 3
Q1 = y1œ A¿ =
y t (yt) = y2 2 2
Q2 = ©y¿A¿ =
a at (at) + a(xt) = (a + 2x) 2 2
Shear Flow Resultant: q1 =
P A 12 y2 B VQ1 3P 2 = 10 3 = y 3 I 20a a t 3
P C at2 (a + 2x) D VQ2 3P = = (a + 2x) q2 = 10 3 2 I 20a a t 3 a
(Fw)1 =
L0
a
q1 dy =
a
Ff =
L0
3P P y2 dy = 20 20a3 L0 a
q2 dx =
3P 3 (a + 2x)dx = P 2 10 20a L0
Shear Center: Summing moments about point A. Pe = 2(Fw)1 (a) + Ff(2a) Pe = 2 a e =
3 P b a + a Pb 2a 20 10
7 a 10
Ans.
522
O
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7–66. Determine the location e of the shear center, point O, for the thin-walled member having the cross section shown.
a 60⬚ O
a 60⬚ a e
Summing moments about A. Pe = F2 a I =
13 ab 2
t 1 1 1 (t)(a)3 + a b(a)3 = t a3 12 12 sin 30° 4
q1 =
V(a)(t)(a>4) 1 4
q2 = q1 +
F2 =
=
3
ta
V a
V(a>2)(t)(a>4) 1 4
ta
3
= q1 +
V 2a
V 4V 2 V (a) + a b (a) = a 3 2a 3 e =
223 a 3
Ans.
523
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7–67. Determine the location e of the shear center, point O, for the thin-walled member having the cross section shown. The member segments have the same thickness t.
b t h 2 O e h 2 b
Shear Flow Resultant: The shear force flows through as Indicated by F1, F2, and F3 on FBD (b). Hence, The horizontal force equilibrium is not satisfied (©Fx Z 0). In order to satisfy this equilibrium requirement. F1 and F2 must be equal to zero. Shear Center: Summing moments about point A. Pe = F2(0)
e = 0
Ans.
Also, The shear flows through the section as indicated by F1, F2, F3. + ©F Z 0 However, : x
To satisfy this equation, the section must tip so that the resultant of : : : : F1 + F2 + F3 = P Also, due to the geometry, for calculating F1 and F3, we require F1 = F3. Hence, e = 0
Ans.
524
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*7–68. Determine the location e of the shear center, point O, for the beam having the cross section shown. The thickness is t. 1 — r 2
e r
O
I = (2)c
1 r 2 (t)(r>2)3 + (r>2)(t)ar + b d + Isemi-circle 12 4
= 1.583333t r3 + Isemi-circle p>2
Isemi-circle =
p>2 2
L-p>2
(r sin u) t r du = t r3
L-p>2
sin2 u du
p Isemi-circle = t r3 a b 2 Thus, p I = 1.583333t r3 + t r3 a b = 3.15413t r3 2 r r Q = a b t a + rb + 2 4 Lu
p>2
r sin u (t r du)
Q = 0.625 t r2 + t r2 cos u q =
VQ P(0.625 + cos u)t r2 = I 3.15413 t r3
Summing moments about A: p>2
Pe =
L-p>2
(q r du)r p>2
Pe =
e =
Pr (0.625 + cos u)du 3.15413 L-p>2
r (1.9634 + 2) 3.15413
e = 1.26 r
Ans.
525
1 — r 2
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•7–69.
Determine the location e of the shear center, point O, for the thin-walled member having the cross section shown. The member segments have the same thickness t.
h1
h
O e h1 b
Summing moments about A. Pe = F(h) + 2V(b)
h 2 1 1 (t)(h3) + 2b(t)a b + (t)[h3 - (h - 2h1)3] 12 2 12
I =
=
(1)
t(h - 2h1)3 bth2 th3 + 6 2 12
Q1 = y¿A¿ =
t(hy - 2h1 y + y2) 1 (h - 2h1 + y)yt = 2 2
VQ Pt(hy - 2h1 y + y2) = I 2I
q1 =
V =
L
h1 Pt Pt hh1 2 2 (hy - 2h1 y + y2)dy = c - h31 d 2I L0 2I 2 3
q1 dy =
Q2 = ©y¿A¿ =
1 1 h (h - h1)h1 t + (x)(t) = t[h1 (h - h1) + hx] 2 2 2
VQ2 Pt = (h (h - h1) + hx) I 2I 1
q2 =
b
F =
L
q2 dx =
Pt Pt hb2 [h1 (h - h1) + hx]dx = ah1 hb - h21 b + b 2I L0 2I 2
From Eq, (1). Pe =
h2b2 4 Pt [h1 h2b - h21 hb + + hh21 b - h31 b] 2I 2 3
I =
t (2h3 + 6bh2 - (h - 2h1)3) 12
e =
b(6h1 h2 + 3h2b - 8h31) t (6h1 h2b + 3h2b2 - 8h1 3b) = 12I 2h3 + 6bh2 - (h - 2h1)3
526
Ans.
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7–70. Determine the location e of the shear center, point O, for the thin-walled member having the cross section shown.
t
r
a O
a e
Summing moments about A. Pe = r
dF L dA = t ds = t r du
(1)
y = r sin u dI = y2 dA = r2 sin2 u(t r du) = r3 t sin2 udu p+a
I = r3 t
L
sin2 u du = r3 t
Lp - a
1 - cos 2u du 2
=
sin 2u p + a r3 t (u ) 2 2 p - a
=
sin 2(p + a) sin 2(p - a) r3 t c ap + a b - ap - a bd 2 2 2
=
r3 t r3 t 2 (2a - 2 sin a cos a) = (2a - sin 2a) 2 2
dQ = y dA = r sin u(t r du) = r2 t sin u du u
Q = r2 t
q =
L
u
sin u du = r2 t (-cos u)|
Lp-a
= r2 t(-cos u - cos a) = -r2 t(cos u + cos a)
p-a
P(-r2t)(cos u + cos a) -2P(cos u + cos a) VQ = = r3t I r(2a - sin 2a) 2 (2a - sin 2a)
dF =
L
q ds =
L
q r du p+p
L =
dF =
2P r -2P (cos u + cos a) du = (2a cos a - 2 sin a) r(2a - sin 2a) Lp - a 2a - sin 2a
4P (sin a - a cos a) 2a - sin 2a
4P (sin a - a cos a) d 2a - sin 2a 4r (sin a - a cos a) e = 2a - sin 2a
From Eq. (1); P e = r c
Ans.
527
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7–71. Sketch the intensity of the shear-stress distribution acting over the beam’s cross-sectional area, and determine the resultant shear force acting on the segment AB. The shear acting at the section is V = 35 kip. Show that INA = 872.49 in4.
C
V 8 in.
B
A
6 in.
Section Properties: y =
4(8)(8) + 11(6)(2) ©yA = = 5.1053 in. ©A 8(8) + 6(2)
INA =
2 in.
1 (8) A 83 B + 8(8)(5.1053 - 4)2 12 +
1 (2) A 63 B + 2(6)(11 - 5.1053)2 12
= 872.49 in4 (Q.E.D) Q1 = y1œ A¿ = (2.55265 + 0.5y1)(5.1053 - y1)(8) = 104.25 - 4y21 Q2 = y2œ A¿ = (4.44735 + 0.5y2)(8.8947 - y2)(2) = 79.12 - y22 Shear Stress: Applying the shear formula t =
tCB =
VQ , It
VQ1 35(103)(104.25 - 4y21) = It 872.49(8) = {522.77 - 20.06y21} psi
At y1 = 0,
tCB = 523 psi
At y1 = -2.8947 in.
tCB = 355 psi
tAB =
VQ2 35(103)(79.12 - y22) = It 872.49(2)
= {1586.88 - 20.06y22} psi At y2 = 2.8947 in.
tAB = 1419 psi
Resultant Shear Force: For segment AB. VAB =
L
tAB dA 0.8947 in
=
L2.8947 in 0.8947 in
=
L2.8947 in
3 in. 3 in.
A 1586.88 - 20.06y22 B (2dy) A 3173.76 - 40.12y22 B dy
= 9957 lb = 9.96 kip
Ans.
528
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*7–72. The beam is fabricated from four boards nailed together as shown. Determine the shear force each nail along the sides C and the top D must resist if the nails are uniformly spaced at s = 3 in. The beam is subjected to a shear of V = 4.5 kip.
1 in. 1 in. 3 in.
10 in. A 1 in.
12 in. V
B
Section Properties: y =
0.5(10)(1) + 2(4)(2) + 7(12)(1) © yA = = 3.50 in. ©A 10(1) + 4(2) + 12(1)
INA =
1 (10) A 13 B + (10)(1)(3.50 - 0.5)2 12 +
1 (2) A 43 B + 2(4)(3.50 - 2)2 12 1 + (1) A 123 B + 1(12)(7 - 3.50)2 12
= 410.5 in4 QC = y1œ A¿ = 1.5(4)(1) = 6.00 in2 QD = y2œ A¿ = 3.50(12)(1) = 42.0 in2 Shear Flow: qC =
VQC 4.5(103)(6.00) = = 65.773 lb>in. I 410.5
qD =
VQD 4.5(103)(42.0) = = 460.41 lb>in. I 410.5
Hence, the shear force resisted by each nail is FC = qC s = (65.773 lb>in.)(3 in.) = 197 lb
Ans.
FD = qD s = (460.41 lb>in.)(3 in.) = 1.38 kip
Ans.
529
1 in.
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•7–73.
The member is subjected to a shear force of V = 2 kN. Determine the shear flow at points A, B, and C. The thickness of each thin-walled segment is 15 mm.
200 mm B 100 mm A C V ⫽ 2 kN
Section Properties: y =
=
© yA ©A 0.0075(0.2)(0.015) + 0.0575(0.115)(0.03) + 0.165(0.3)(0.015) 0.2(0.015) + 0.115(0.03) + 0.3(0.015)
= 0.08798 m 1 (0.2) A 0.0153 B + 0.2(0.015)(0.08798 - 0.0075)2 12 1 + (0.03) A 0.1153 B + 0.03(0.115)(0.08798 - 0.0575)2 12 1 + (0.015) A 0.33 B + 0.015(0.3)(0.165 - 0.08798)2 12
INA =
= 86.93913 A 10 - 6 B m4
QA = 0 ' QB = y 1œ A¿ = 0.03048(0.115)(0.015) = 52.57705 A 10 - 6 B m3
Ans.
QC = ©y¿A¿ = 0.03048(0.115)(0.015) + 0.08048(0.0925)(0.015) = 0.16424 A 10 - 3 B m3 Shear Flow: qA =
VQA = 0 I
Ans.
qB =
VQB 2(103)(52.57705)(10 - 6) = 1.21 kN>m = I 86.93913(10 - 6)
Ans.
qC =
VQC 2(103)(0.16424)(10 - 3) = 3.78 kN>m = I 86.93913(10 - 6)
Ans.
530
300 mm
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7–74. The beam is constructed from four boards glued together at their seams. If the glue can withstand 75 lb>in., what is the maximum vertical shear V that the beam can support? 3 in. 0.5 in.
Section Properties: INA =
1 1 (1) A 103 B + 2c (4) A 0.53 B + 4(0.5) A 1.752 B d 12 12
3 in. 0.5 in.
= 95.667 in4
V
Q = y¿A¿ = 1.75(4)(0.5) = 3.50 in3
4 in.
Shear Flow: There are two glue joints in this case, hence the allowable shear flow is 2(75) = 150 lb>in. q =
150 =
3 in.
0.5 in.
0.5 in.
VQ I V(3.50) 95.667
V = 4100 lb = 4.10 kip
Ans.
7–75. Solve Prob. 7–74 if the beam is rotated 90° from the position shown.
3 in. 0.5 in. 3 in. 0.5 in.
V
3 in.
4 in. 0.5 in.
Section Properties: INA =
1 1 (10) A 53 B (9) A 43 B = 56.167 in4 12 12
Q = y¿A¿ = 2.25(10)(0.5) = 11.25 in3 Shear Flow: There are two glue joints in this case, hence the allowable shear flow is 2(75) = 150 lb>in. q = 150 =
VQ I V(11.25) 56.167
V = 749 lb
Ans.
531
0.5 in.
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8–1. A spherical gas tank has an inner radius of r = 1.5 m. If it is subjected to an internal pressure of p = 300 kPa, determine its required thickness if the maximum normal stress is not to exceed 12 MPa. pr ; 2t
sallow =
12(106) =
300(103)(1.5) 2t
t = 0.0188 m = 18.8 mm
Ans.
8–2. A pressurized spherical tank is to be made of 0.5-in.-thick steel. If it is subjected to an internal pressure of p = 200 psi, determine its outer radius if the maximum normal stress is not to exceed 15 ksi.
sallow =
pr ; 2t
15(103) =
200 ri 2(0.5)
ri = 75 in. ro = 75 in. + 0.5 in. = 75.5 in.
Ans.
8–3. The thin-walled cylinder can be supported in one of two ways as shown. Determine the state of stress in the wall of the cylinder for both cases if the piston P causes the internal pressure to be 65 psi. The wall has a thickness of 0.25 in. and the inner diameter of the cylinder is 8 in.
P
Case (a): s1 =
pr ; t
s1 =
65(4) = 1.04 ksi 0.25
Ans.
s2 = 0
Ans.
Case (b): s1 =
pr ; t
s1 =
65(4) = 1.04 ksi 0.25
Ans.
s2 =
pr ; 2t
s2 =
65(4) = 520 psi 2(0.25)
Ans.
532
P
8 in.
8 in.
(a)
(b)
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*8–4. The tank of the air compressor is subjected to an internal pressure of 90 psi. If the internal diameter of the tank is 22 in., and the wall thickness is 0.25 in., determine the stress components acting at point A. Draw a volume element of the material at this point, and show the results on the element.
Hoop Stress for Cylindrical Vessels: Since
A
11 r = = 44 7 10, then thin wall t 0.25
analysis can be used. Applying Eq. 8–1
s1 =
pr 90(11) = = 3960 psi = 3.96 ksi t 0.25
Ans.
Longitudinal Stress for Cylindrical Vessels: Applying Eq. 8–2
s2 =
pr 90(11) = = 1980 psi = 1.98 ksi 2t 2(0.25)
Ans.
•8–5.
The spherical gas tank is fabricated by bolting together two hemispherical thin shells of thickness 30 mm. If the gas contained in the tank is under a gauge pressure of 2 MPa, determine the normal stress developed in the wall of the tank and in each of the bolts.The tank has an inner diameter of 8 m and is sealed with 900 bolts each 25 mm in diameter.
Normal Stress: Since
4 r = = 133.33 7 10, thin-wall analysis is valid. For the t 0.03
spherical tank’s wall,
s =
Referring
pr 2(4) = = 133 MPa 2t 2(0.03)
to
the free-body diagram p 2 6 P = pA = 2 A 10 B c A 8 B d = 32p A 10 B N. Thus, 4
Ans.
shown
in
Fig.
a,
6
+ c ©Fy = 0;
32p A 106 B - 450Pb - 450Pb = 0
Pb = 35.56 A 103 B p N
The normal stress developed in each bolt is then sb =
35.56 A 103 B p Pb = = 228 MPa p Ab A 0.0252 B 4
Ans.
533
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8–6. The spherical gas tank is fabricated by bolting together two hemispherical thin shells. If the 8-m inner diameter tank is to be designed to withstand a gauge pressure of 2 MPa, determine the minimum wall thickness of the tank and the minimum number of 25-mm diameter bolts that must be used to seal it. The tank and the bolts are made from material having an allowable normal stress of 150 MPa and 250 MPa, respectively.
Normal Stress: For the spherical tank’s wall, sallow =
pr 2t
150 A 106 B =
2 A 106 B (4) 2t
t = 0.02667 m = 26.7 mm Since
Ans.
r 4 = = 150 7 10, thin-wall analysis is valid. t 0.02667
Referring
the free-body diagram p P = pA = 2 A 106 B c A 82 B d = 32p A 106 B N. Thus, 4 + c ©Fy = 0;
to
32p A 106 B n =
shown
in
Fig.
a,
n n (P ) - (Pb)allow = 0 2 b allow 2
32p A 106 B
(1)
(Pb)allow
The allowable tensile force for each bolt is (Pb)allow = sallowAb = 250 A 106 B c
p A 0.0252 B d = 39.0625 A 103 B pN 4
Substituting this result into Eq. (1), n =
32p A 106 B
39.0625p A 103 B
= 819.2 = 820
Ans.
534
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8–7. A boiler is constructed of 8-mm thick steel plates that are fastened together at their ends using a butt joint consisting of two 8-mm cover plates and rivets having a diameter of 10 mm and spaced 50 mm apart as shown. If the steam pressure in the boiler is 1.35 MPa, determine (a) the circumferential stress in the boiler’s plate apart from the seam, (b) the circumferential stress in the outer cover plate along the rivet line a–a, and (c) the shear stress in the rivets.
a
8 mm
50 mm
a)
s1 =
pr 1.35(106)(0.75) = = 126.56(106) = 127 MPa t 0.008
Ans.
126.56 (106)(0.05)(0.008) = s1 ¿(2)(0.04)(0.008)
b)
s1 ¿ = 79.1 MPa
Ans.
c) From FBD(a) + c ©Fy = 0;
Fb - 79.1(106)[(0.008)(0.04)] = 0 Fb = 25.3 kN
(tavg)b =
Fb 25312.5 - p = 322 MPa 2 A 4 (0.01)
Ans.
535
0.75 m a
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*8–8. The gas storage tank is fabricated by bolting together two half cylindrical thin shells and two hemispherical shells as shown. If the tank is designed to withstand a pressure of 3 MPa, determine the required minimum thickness of the cylindrical and hemispherical shells and the minimum required number of longitudinal bolts per meter length at each side of the cylindrical shell. The tank and the 25 mm diameter bolts are made from material having an allowable normal stress of 150 MPa and 250 MPa, respectively. The tank has an inner diameter of 4 m. Normal Stress: For the cylindrical portion of the tank, the hoop stress is twice as large as the longitudinal stress. sallow =
pr ; t
150 A 106 B =
3 A 106 B (2) tc
tc = 0.04 m = 40 mm For the hemispherical cap, sallow
pr = ; t
150 A 10
6
B =
Ans. 3 A 106 B (2) 2ts
ts = 0.02 m = 20 mm Since
Ans.
r 6 10, thin-wall analysis is valid. t
Referring to the free-body diagram of the per meter length of the cylindrical portion, Fig. a, where P = pA = 3 A 106 B [4(1)] = 12 A 106 B N, we have + c ©Fy = 0;
12 A 106 B - nc(Pb)allow - nc(Pb)allow = 0 nc =
6 A 106 B
(1)
(Pb)allow
The allowable tensile force for each bolt is (Pb)allow = sallowAb = 250 A 106 B c
p A 0.0252 B d = 122.72 A 103 B N 4
Substituting this result into Eq. (1), nc = 48.89 = 49 bolts>meter
Ans.
536
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•8–9.
The gas storage tank is fabricated by bolting together two half cylindrical thin shells and two hemispherical shells as shown. If the tank is designed to withstand a pressure of 3 MPa, determine the required minimum thickness of the cylindrical and hemispherical shells and the minimum required number of bolts for each hemispherical cap. The tank and the 25 mm diameter bolts are made from material having an allowable normal stress of 150 MPa and 250 MPa, respectively. The tank has an inner diameter of 4 m.
Normal Stress: For the cylindrical portion of the tank, the hoop stress is twice as large as the longitudinal stress. sallow =
pr ; t
150 A 106 B =
3 A 106 B (2) tc
tc = 0.04 m = 40 mm For the hemispherical cap, sallow =
pr ; t
150 A 106 B =
Ans. 3 A 106 B (2) 2ts
ts = 0.02 m = 20 mm Since
Ans.
r 6 10, thin-wall analysis is valid. t
The allowable tensile force for each bolt is (Pb)allow = sallowAb = 250 A 106 B c
p A 0.0252 B d = 122.72 A 103 B N 4
Referring to the free-body diagram of the hemispherical cap, Fig. b, where p P = pA = 3 A 106 B c A 42 B d = 12p A 106 B N, 4 + ©F = 0; : x
12p A 106 B ns =
ns ns (Pb)allow (Pb)allow = 0 2 2
12p A 106 B
(1)
(Pb)allow
Substituting this result into Eq. (1), ns = 307.2 = 308 bolts
Ans.
537
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8–10. A wood pipe having an inner diameter of 3 ft is bound together using steel hoops each having a crosssectional area of 0.2 in2. If the allowable stress for the hoops is sallow = 12 ksi, determine their maximum spacing s along the section of pipe so that the pipe can resist an internal gauge pressure of 4 psi. Assume each hoop supports the pressure loading acting along the length s of the pipe.
s
4 psi
4 psi
s
s
Equilibrium for the steel Hoop: From the FBD + ©F = 0; : x
P = 72.0s
2P - 4(36s) = 0
Hoop Stress for the Steel Hoop: s1 = sallow = 12(103) =
P A 72.0s 0.2
s = 33.3 in.
Ans.
8–11. The staves or vertical members of the wooden tank are held together using semicircular hoops having a thickness of 0.5 in. and a width of 2 in. Determine the normal stress in hoop AB if the tank is subjected to an internal gauge pressure of 2 psi and this loading is transmitted directly to the hoops. Also, if 0.25-in.-diameter bolts are used to connect each hoop together, determine the tensile stress in each bolt at A and B. Assume hoop AB supports the pressure loading within a 12-in. length of the tank as shown.
18 in.
6 in. 6 in.
FR = 2(36)(12) = 864 lb ©F = 0; 864 - 2F = 0; F = 432 lb sh =
sb =
F 432 = = 432 psi Ah 0.5(2)
Ans.
F 432 = = 8801 psi = 8.80 ksi p Ab (0.25)2 4
Ans.
538
12 in. A
B 12 in.
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*8–12. Two hemispheres having an inner radius of 2 ft and wall thickness of 0.25 in. are fitted together, and the inside gauge pressure is reduced to -10 psi. If the coefficient of static friction is ms = 0.5 between the hemispheres, determine (a) the torque T needed to initiate the rotation of the top hemisphere relative to the bottom one, (b) the vertical force needed to pull the top hemisphere off the bottom one, and (c) the horizontal force needed to slide the top hemisphere off the bottom one.
0.25 in. 2 ft
Normal Pressure: Vertical force equilibrium for FBD(a). + c ©Fy = 0;
10 C p(242) D - N = 0
N = 5760p lb
The Friction Force: Applying friction formula Ff = ms N = 0.5(5760p) = 2880p lb a) The Required Torque: In order to initiate rotation of the two hemispheres relative to each other, the torque must overcome the moment produced by the friction force about the center of the sphere. T = Ffr = 2880p(2 + 0.125>12) = 18190 lb # ft = 18.2 kip # ft
Ans.
b) The Required Vertical Force: In order to just pull the two hemispheres apart, the vertical force P must overcome the normal force. P = N = 5760p = 18096 lb = 18.1 kip
Ans.
c) The Required Horizontal Force: In order to just cause the two hemispheres to slide relative to each other, the horizontal force F must overcome the friction force. F = Ff = 2880p = 9048 lb = 9.05 kip
Ans.
•8–13. The 304 stainless steel band initially fits snugly around the smooth rigid cylinder. If the band is then subjected to a nonlinear temperature drop of ¢T = 20 sin2 u °F, where u is in radians, determine the circumferential stress in the band.
1 64
10 in.
Compatibility: Since the band is fixed to a rigid cylinder (it does not deform under load), then dF - dT = 0 2p P(2pr) a¢Trdu = 0 AE L0
2pr P a b = 20ar E A L0 2p s = 10a E c L0
2p
sin2 udu
however,
P = sc A
2p
(1 - cos 2u)du
sc = 10aE = 10(9.60) A 10 - 6 B 28.0 A 103 B = 2.69 ksi
Ans.
539
u
in.
1 in.
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8–14. The ring, having the dimensions shown, is placed over a flexible membrane which is pumped up with a pressure p. Determine the change in the internal radius of the ring after this pressure is applied. The modulus of elasticity for the ring is E.
ro ri w p
Equilibrium for the Ring: Form the FBD + ©F = 0; : x
2P - 2pri w = 0
P = pri w
Hoop Stress and Strain for the Ring: s1 =
pri w pri P = = rs - ri A (rs - ri)w
Using Hooke’s Law e1 =
However,
e1 =
pri s1 = E E(rs - ri)
[1]
2p(ri)1 - 2pri (ri)1 - ri dri = = . ri ri 2pr
Then, from Eq. [1] pri dri = ri E(rs - ri) dri =
pr2i E(rs - ri)
Ans.
540
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8–15. The inner ring A has an inner radius r1 and outer radius r2. Before heating, the outer ring B has an inner radius r3 and an outer radius r4, and r2 7 r3. If the outer ring is heated and then fitted over the inner ring, determine the pressure between the two rings when ring B reaches the temperature of the inner ring. The material has a modulus of elasticity of E and a coefficient of thermal expansion of a.
r1 A
Equilibrium for the Ring: From the FBD + ©F = 0; : x
P = priw
2P - 2priw = 0
Hoop Stress and Strain for the Ring: s1 =
priw pri P = = ro - ri A (ro - ri)w
Using Hooke’s law e1 =
However,
e1 =
pri s1 = E E(ro - ri)
[1]
2p(ri)1 - 2pri (ri)1 - ri dri = = . ri ri 2pr
Then, from Eq. [1] pri dri = ri E(ro - ri) dri =
pr2i E(ro - ri)
Compatibility: The pressure between the rings requires dr2 + dr3 = r2 - r3
[2]
From the result obtained above dr2 =
pr22 E(r2 - r1)
dr3 =
pr23 E(r4 - r3)
Substitute into Eq. [2] pr22 pr23 + = r2 - r3 E(r2 - r1) E(r4 - r3) p =
r4
r2
E(r2 - r3)
Ans.
r22 r23 + r2 - r1 r4 - r3
541
r3 B
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*8–16. The cylindrical tank is fabricated by welding a strip of thin plate helically, making an angle u with the longitudinal axis of the tank. If the strip has a width w and thickness t, and the gas within the tank of diameter d is pressured to p, show that the normal stress developed along the strip is given by su = (pd>8t)(3 - cos 2u).
w u
Normal Stress: sh = s1 =
pr p(d>2) pd = = t t 2t
sl = s2 =
p(d>2) pd pr = = 2t 2t 4t
Equilibrium: We will consider the triangular element cut from the strip shown in Fig. a. Here, Ah = (w sin u)t and Thus, Al = (w cos u)t. pd pwd and (w sin u)t = sin u Fh = shAh = 2t 2 pwd pd (w cos u)t = cos u. 4t 4
Fl = slAl =
Writing the force equation of equilibrium along the x¿ axis, ©Fx¿ = 0;
c
pwd pwd sin u d sin u + c cos u d cos u - Nu = u 2 4 Nu =
pwd A 2 sin2 u + cos2 u B 4
However, sin2 u + cos2 u = 1. This equation becomes Nu = Also, sin2 u =
pwd A sin2 u + 1 B 4
1 (1 - cos 2u), so that 2 pwd Nu = (3 - cos 2u) 8
Since Au = wt, then Nu = su = Au su =
pwd (3 - cos 2u) 8 wt
pd (3 - cos 2u) 8t
(Q.E.D.)
542
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8–17. In order to increase the strength of the pressure vessel, filament winding of the same material is wrapped around the circumference of the vessel as shown. If the pretension in the filament is T and the vessel is subjected to an internal pressure p, determine the hoop stresses in the filament and in the wall of the vessel. Use the free-body diagram shown, and assume the filament winding has a thickness t and width w for a corresponding length of the vessel.
L w s1
t¿
T
p
t s1 T
Normal Stress in the Wall and Filament Before the Internal Pressure is Applied: The entire length w of wall is subjected to pretension filament force T. Hence, from equilibrium, the normal stress in the wall at this state is 2T - (sl ¿)w (2wt) = 0
(sl ¿)w =
T wt
and for the filament the normal stress is (sl ¿)fil =
T wt¿
Normal Stress in the Wall and Filament After the Internal Pressure is Applied: The stress in the filament becomes sfil = sl + (sl ¿)fil =
pr T + (t + t¿) wt¿
Ans.
sw = sl - (sl ¿)w =
pr T (t + t¿) wt
Ans.
And for the wall,
8–18. The vertical force P acts on the bottom of the plate having a negligible weight. Determine the shortest distance d to the edge of the plate at which it can be applied so that it produces no compressive stresses on the plate at section a–a. The plate has a thickness of 10 mm and P acts along the center line of this thickness.
300 mm a
a
200 mm
500 mm
sA = 0 = sa - sb 0 =
0 =
P Mc A I P (0.2)(0.01)
d
P(0.1 - d)(0.1) 1 12
P
(0.01)(0.23)
P(-1000 + 15000 d) = 0 d = 0.0667 m = 66.7 mm
Ans.
543
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8–19. Determine the maximum and minimum normal stress in the bracket at section a–a when the load is applied at x = 0.
100 kN 15 mm x 15 mm 200 mm 150 mm
a
Consider the equilibrium of the FBD of the top cut segment in Fig. a, + c ©Fy = 0; a + ©MC = 0;
N - 100 = 0
N = 100 kN
100(0.1) - M = 0
A = 0.2(0.03) = 0.006 m2
I =
M = 10 kN # m
1 (0.03)(0.23) = 20.0(10 - 6) m4 12
The normal stress developed is the combination of axial and bending stress. Thus, s =
My N ; A I
For the left edge fiber, y = C = 0.1 m. Then sL = -
100(103) 10(103)(0.1) 0.006 20.0(10 - 6)
= -66.67(106) Pa = 66.7 MPa (C) (Max)
Ans.
For the right edge fiber, y = 0.1 m. Then sR = -
100 (103) 10(103)(0.1) = 33.3 MPa (T) + 0.006 20.0(10 - 6)
Ans.
544
a
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*8–20. Determine the maximum and minimum normal stress in the bracket at section a–a when the load is applied at x = 300 mm.
100 kN 15 mm x 15 mm 200 mm 150 mm
a
Consider the equilibrium of the FBD of the top cut segment in Fig. a, + c ©Fy = 0; a + ©MC = 0;
N - 100 = 0
N = 100 kN
M - 100(0.2) = 0
A = 0.2 (0.03) = 0.006 m2
I =
M = 20 kN # m
1 (0.03)(0.23) = 20.0(10 - 6) m4 12
The normal stress developed is the combination of axial and bending stress. Thus, s =
My N ; A I
For the left edge fiber, y = C = 0.1 m. Then sC = -
100(103) 20.0(103)(0.1) + 0.006 20.0(10 - 6)
= 83.33(106) Pa = 83.3 MPa (T)(Min)
Ans.
For the right edge fiber, y = C = 0.1 m. Thus sR = -
100(103) 20.0(103)(0.1) 0.006 20.0(10 - 6)
= 117 MPa
Ans.
545
a
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•8–21.
The coping saw has an adjustable blade that is tightened with a tension of 40 N. Determine the state of stress in the frame at points A and B.
8 mm 75 mm
A
3 mm 8 mm 3 mm B
100 mm
50 mm
sA = -
sB =
P Mc 40 + = + A I (0.008)(0.003)
Mc = I
2(0.004) 1 12
(0.003)(0.008)3
4(0.004) 1 3 12 (0.003)(0.008)
= 123 MPa
Ans.
Ans.
= 62.5 MPa
8–22. The clamp is made from members AB and AC, which are pin connected at A. If it exerts a compressive force at C and B of 180 N, determine the maximum compressive stress in the clamp at section a–a. The screw EF is subjected only to a tensile force along its axis.
30 mm
40 mm
F C
180 N
15 mm 15 mm Section a – a
a
a
B
A E
There is no moment in this problem. Therefore, the compressive stress is produced by axial force only. smax =
P 240 = = 1.07 MPa A (0.015)(0.015)
Ans.
546
180 N
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8–23. The clamp is made from members AB and AC, which are pin connected at A. If it exerts a compressive force at C and B of 180 N, sketch the stress distribution acting over section a–a. The screw EF is subjected only to a tensile force along its axis.
30 mm
40 mm
F C
180 N
15 mm 15 mm Section a – a
a
a
180 N
B
A E
There is moment in this problem. Therefore, the compressive stress is produced by axial force only. smax =
240 P = = 1.07 MPa A (0.015)(0.015)
*8–24. The bearing pin supports the load of 700 lb. Determine the stress components in the support member at point A. The support is 0.5 in. thick.
0.75 in. A 2 in. 30⬚
A B
3 in.
©Fx = 0;
N - 700 cos 30° = 0;
N = 606.218 lb
©Fy = 0;
V - 700 sin 30° = 0;
V = 350 lb
a + ©M = 0;
M - 700(1.25 - 2 sin 30°) = 0; sA =
1.25 in. 700 lb
M = 175 lb # in.
(175)(0.375) N Mc 606.218 = - 1 3 A I (0.75)(0.5) 12 (0.5)(0.75)
sA = -2.12 ksi
Ans.
tA = 0
Ans.
(since QA = 0)
547
B
0.5 in.
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•8–25.
The bearing pin supports the load of 700 lb. Determine the stress components in the support member at point B. The support is 0.5 in. thick.
0.75 in. A 2 in. 30⬚
A B
B
0.5 in.
3 in.
©Fx = 0;
N - 700 cos 30° = 0;
N = 606.218 lb
©Fy = 0;
V - 700 sin 30° = 0;
V = 350 lb
a + ©M = 0;
M - 700(1.25 - 2 sin 30°) = 0; sB =
N Mc 606.218 + = + A I (0.75)(0.5)
1.25 in. 700 lb
M = 175 lb # in.
175(0.375) 1 12
(0.5)(0.75)3
sB = 5.35 ksi
Ans.
tB = 0
Ans.
(since QB = 0)
8–26. The offset link supports the loading of P = 30 kN. Determine its required width w if the allowable normal stress is sallow = 73 MPa. The link has a thickness of 40 mm.
P
s due to axial force: sa =
30(103) 750(103) P = = A (w)(0.04) w
sb =
w
50 mm
s due to bending: 30(103)(0.05 + w2)(w2) Mc = 1 3 I 12 (0.04)(w) 4500 (103)(0.05 + w2) =
w2 P
smax = sallow = sa + sb 73(106) =
4500(103)(0.05 + w2) 750(103) + w w2
73 w2 = 0.75 w + 0.225 + 2.25 w 73 w2 - 3 w - 0.225 = 0 w = 0.0797 m = 79.7 mm
Ans.
548
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8–27. The offset link has a width of w = 200 mm and a thickness of 40 mm. If the allowable normal stress is sallow = 75 MPa, determine the maximum load P that can be applied to the cables.
P
A = 0.2(0.04) = 0.008 m2 I =
s =
1 (0.04)(0.2)3 = 26.6667(10 - 6) m4 12
w
50 mm
P Mc + A I
75(106) =
0.150 P(0.1) P + 0.008 26.6667(10 - 6)
P = 109 kN
Ans.
P
*8–28. The joint is subjected to a force of P 80 lb and F 0. Sketch the normal-stress distribution acting over section a–a if the member has a rectangular cross-sectional area of width 2 in. and thickness 0.5 in. a B
s due to axial force: s =
0.5 in.
P 80 = = 80 psi A (0.5)(2)
2 in.
F
s due to bending: s =
A a
100(0.25) Mc = 1 = 1200 psi 3 I 12 (2)(0.5)
1.25 in. P
(smax)t = 80 + 1200 = 1280 psi = 1.28 ksi
Ans.
(smax)c = 1200 - 80 = 1120 psi = 1.12 ksi
Ans.
y (0.5 - y) = 1.28 1.12 y = 0.267 in.
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The joint is subjected to a force of P = 200 lb and F = 150 lb. Determine the state of stress at points A and B and sketch the results on differential elements located at these points. The member has a rectangular cross-sectional area of width 0.75 in. and thickness 0.5 in. •8–29.
a B
A a
0.5 in. 2 in.
F 1.25 in. P
A = 0.5(0.75) = 0.375 in2 œ QA = yA A¿ = 0.125(0.75)(0.25) = 0.0234375 in3 ;
I =
QB = 0
1 (0.75)(0.53) = 0.0078125 in4 12
Normal Stress: My N ; A I
s =
sA =
200 + 0 = 533 psi (T) 0.375
Ans.
sB =
50(0.25) 200 = -1067 psi = 1067 psi (C) 0.375 0.0078125
Ans.
Shear stress: t =
VQ It
tA =
150(0.0234375) = 600 psi (0.0078125)(0.75)
Ans.
tB = 0
Ans.
550
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8–30. If the 75-kg man stands in the position shown, determine the state of stress at point A on the cross section of the plank at section a–a. The center of gravity of the man is at G. Assume that the contact point at C is smooth.
C
G
600 mm A
a
50 mm
1.5 m B
12.5 mm
30 a 600 mm
300 mm
Support Reactions: Referring to the free-body diagram of the entire plank, Fig. a, a + ©MB = 0;
FC sin 30°(2.4) - 75(9.81) cos 30°(0.9) = 0 FC = 477.88 N
©Fx¿ = 0; Bx¿ - 75(9.81) sin 30° - 477.88 cos 30° = 0 Bx¿ = 781.73 N ©Fy¿ = 0; By¿ + 477.88 sin 30° - 75(9.81) cos 30° = 0 By¿ = 398.24 N Internal Loadings: Consider the equilibrium of the free-body diagram of the plank’s lower segment, Fig. b, ©Fx¿ = 0; 781.73 - N = 0
N = 781.73 N
©Fy¿ = 0; 398.24 - V = 0
V = 398.24 N
a + ©MO = 0;
M - 398.24(0.6) = 0
M = 238.94 N # m
Section Properties: The cross-sectional area and the moment of inertia about the centroidal axis of the plank’s cross section are A = 0.6(0.05) = 0.03 m2 I =
1 (0.6) A 0.053 B = 6.25 A 10 - 6 B m4 12
Referring to Fig. c, QA is QA = y¿A¿ = 0.01875(0.0125)(0.6) = 0.140625 A 10 - 3 B m3 Normal Stress: The normal stress is the combination of axial and bending stress. Thus, s =
My N ; A I
For point A, y = 0.0125 m. Then sA =
238.94(0.0125) -781.73 0.03 6.25 A 10 - 6 B
= -503.94 kPa = 504 kPa (C)
Ans.
551
Section a – a and b – b
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8–30.
Continued
Shear Stress: The shear stress is contributed by transverse shear stress. Thus,
tA
VQA = = It
398.24 c0.140625 A 10 - 3 B d 6.25 A 10 - 6 B (0.6)
Ans.
= 14.9 kPa
The state of stress at point A is represented on the element shown in Fig. d.
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8–31. Determine the smallest distance d to the edge of the plate at which the force P can be applied so that it produces no compressive stresses in the plate at section a–a. The plate has a thickness of 20 mm and P acts along the centerline of this thickness.
a
P
200 mm d 300 mm a
Consider the equilibrium of the FBD of the left cut segment in Fig. a, + : ©Fx = 0;
N - P = 0
a + ©MC = 0;
N = P
M - P(0.1 - d) = 0
A = 0.2 (0.02) = 0.004 m4
I =
M = P(0.1 - d)
1 (0.02)(0.23) = 13.3333(10 - 6) m4 12
The normal stress developed is the combination of axial and bending stress. Thus s =
My N ; A I
Since no compressive stress is desired, the normal stress at the top edge fiber must be equal to zero. Thus,
0 =
P(0.1 - d)(0.1) P ; 0.004 13.3333 (10 - 6)
0 = 250 P - 7500 P (0.1 - d) d = 0.06667 m = 66.7 mm
Ans.
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*8–32. The horizontal force of P = 80 kN acts at the end of the plate. The plate has a thickness of 10 mm and P acts along the centerline of this thickness such that d = 50 mm. Plot the distribution of normal stress acting along section a–a.
a
P
200 mm d 300 mm
Consider the equilibrium of the FBD of the left cut segment in Fig. a, + : ©Fx = 0; a + ©MC = 0;
N - 80 = 0
N = 80 kN
M - 80(0.05) = 0
A = 0.01(0.2) = 0.002 m2
I =
M = 4.00 kN # m
1 (0.01)(0.23) = 6.667(10 - 6) m4 12
The normal stress developed is the combination of axial and bending stress. Thus,
s =
My N ; A I
At point A, y = 0.1 m. Then sA =
80(103) 4.00(103)(0.1) 0.002 6.667(10 - 6)
= -20.0(106) Pa = 20.0 Mpa (C) At point B, y = 0.1 m. Then sB =
80(103) 4.00(103)(0.1) + 0.002 6.667(10 - 6)
= 100 (106) Pa = 100 MPa (T) The location of neutral axis can be determined using the similar triangles.
554
a
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•8–33.
The pliers are made from two steel parts pinned together at A. If a smooth bolt is held in the jaws and a gripping force of 10 lb is applied at the handles, determine the state of stress developed in the pliers at points B and C. Here the cross section is rectangular, having the dimensions shown in the figure.
0.18 in.
10 lb
D 0.2 in.
0.1 in. D
3 in.
30
E A B
0.2 in. 0.2 in.
B E
C C
0.2 in.
1.75 in. 2.5 in.
Q ©Fx = 0;
N - 10 sin 30° = 0;
N = 5.0 lb
a+ ©Fy = 0;
V - 10 cos 30° = 0;
V = 8.660 lb
+
a + ©MC = 0;
10 lb
M = 30 lb # in.
M - 10(3) = 0
A = 0.2(0.4) = 0.08 in2 I =
1 (0.2)(0.43) = 1.0667(10 - 3) in4 12
QB = 0 QC = y¿A¿ = 0.1(0.2)(0.2) = 4(10 - 3) in3 Point B: sB =
My 30(0.2) N -5.0 = 5.56 ksi(T) + = + A I 0.08 1.0667(10 - 3)
Ans.
VQ = 0 It
Ans.
My N -5.0 + = + 0 = -62.5 psi = 62.5 psi(C) A I 0.08
Ans.
VQ 8.660(4)(10 - 3) = 162 psi = It 1.0667(10 - 3)(0.2)
Ans.
tB = Point C: sC = Shear Stress : tC =
4 in.
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8–34.
Solve Prob. 8–33 for points D and E.
0.18 in.
10 lb
D 0.2 in.
0.1 in. D
3 in.
30
E A B
0.2 in. 0.2 in.
B E
C C
0.2 in.
1.75 in. 2.5 in.
a + ©MA = 0;
-F(2.5) + 4(10) = 0;
F = 16 lb
10 lb
Point D: sD = 0
tD =
Ans.
16(0.05)(0.1)(0.18) VQ = 667 psi = 1 It [12 (0.18)(0.2)3](0.18)
Ans.
Point E: sE =
My = I
28(0.1) 1 12
(0.18)(0.2)3
4 in.
Ans.
= 23.3 ksi (T)
tE = 0
Ans.
556
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8–35. The wide-flange beam is subjected to the loading shown. Determine the stress components at points A and B and show the results on a volume element at each of these points. Use the shear formula to compute the shear stress.
500 lb
3000 lb
2500 lb A B
2 ft
2 ft
2 ft
4 ft
6 ft
A 0.5 in.
1 1 I = (4)(73) (3.5)(63) = 51.33 in4 12 12
B
QB = ©y¿A¿ = 3.25(4)(0.5) + 2(2)(0.5) = 8.5 in3 QA = 0 -11500 (12)(3.5) -Mc = = -9.41 ksi I 51.33
Ans.
tA = 0 sB = tB =
4 in. 2 in. 4 in. 0.5 in.
A = 2(0.5)(4) + 6(0.5) = 7 in2
sA =
0.5 in.
Ans.
My 11500(12)(1) = = 2.69 ksi I 51.33
Ans.
VQB 2625(8.5) = = 0.869 ksi It 51.33(0.5)
Ans.
557
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*8–36. The drill is jammed in the wall and is subjected to the torque and force shown. Determine the state of stress at point A on the cross section of drill bit at section a–a.
y 400 mm a 20 N ·m x a 125 mm y A z
5 mm
B Section a – a
Internal Loadings: Consider the equilibrium of the free-body diagram of the drill’s right cut segment, Fig. a, 4 ©Fx = 0; N - 150 a b = 0 5
N = 120 N
3 ©Fy = 0; 150 a b - Vy = 0 5
Vy = 90 N T = 20 N # m
©Mx = 0; 20 - T = 0 4 3 ©Mz = 0; -150 a b (0.4) + 150 a b(0.125) + Mz = 0 5 5 Mz = 21 N # m
Section Properties: The cross-sectional area, the moment of inertia about the z axis, and the polar moment of inertia of the drill’s cross section are A = p A 0.0052 B = 25p A 10 - 6 B m2 Iz =
p A 0.0054 B = 0.15625p A 10 - 9 B m4 4
J =
p A 0.0054 B = 0.3125p A 10 - 9 B m4 2
Referring to Fig. b, QA is QA = 0 Normal Stress: The normal stress is a combination of axial and bending stress. Thus, s =
Mzy N A Iz
For point A, y = 0.005 m. Then sA =
-120
25p A 10
-6
B
21(0.005)
-
0.15625p A 10 - 9 B
= -215.43 MPa = 215 MPa (C)
558
Ans.
3
5 4
150 N
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8–36. Continued
Shear Stress: The transverse shear stress developed at point A is c A txy B V d
VyQA = A
Izt
Ans.
= 0
The torsional shear stress developed at point A is
C (txz)T D A =
20(0.005) Tc = 101.86 MPa = J 0.3125p A 10 - 9 B
Thus,
A txy B A = 0
Ans.
A txz B A = c A txz B T d = 102 MPa
Ans.
A
The state of stress at point A is represented on the element shown in Fig c.
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•8–37.
The drill is jammed in the wall and is subjected to the torque and force shown. Determine the state of stress at point B on the cross section of drill bit at section a–a.
y 400 mm a 20 N ·m x a 125 mm y
Internal Loadings: Consider the equilibrium of the free-body diagram of the drill’s right cut segment, Fig. a, 4 ©Fx = 0; N - 150 a b = 0 5
N = 120 N
3 ©Fy = 0; 150 a b - Vy = 0 5
Vy = 90 N
z
Section a – a
4 3 ©Mz = 0; -150 a b (0.4) + 150 a b (0.125) + Mz = 0 5 5 Mz = 21 N # m Section Properties: The cross-sectional area, the moment of inertia about the z axis, and the polar moment of inertia of the drill’s cross section are A = p A 0.0052 B = 25p A 10 - 6 B m2 Iz =
p A 0.0054 B = 0.15625p A 10 - 9 B m4 4
J =
p A 0.0054 B = 0.3125p A 10 - 9 B m4 2
Referring to Fig. b, QB is
QB = y¿A¿ =
4(0.005) p c A 0.0052 B d = 83.333 A 10 - 9 B m3 3p 2
Normal Stress: The normal stress is a combination of axial and bending stress. Thus, Mzy N A Iz
For point B, y = 0. Then
sB =
-120
25p A 10 - 6 B
- 0 = -1.528 MPa = 1.53 MPa(C)
560
5 mm
B
T = 20 N # m
©Mx = 0; 20 - T = 0
s =
A
Ans.
3
5 4
150 N
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8–37. Continued Shear Stress: The transverse shear stress developed at point B is c A txy B V d
90 c83.333 A 10 - 9 B d
VyQB = B
Izt
=
0.15625p A 10 - 9 B (0.01)
= 1.528 MPa
Ans.
The torsional shear stress developed at point B is c A txy B T d
= B
20(0.005) TC = 101.86 MPa = J 0.3125p A 10 - 9 B
Thus,
A tC B B = 0
Ans.
A txy B B = c A txy B T d - c A txy B V d B
B
Ans.
= 101.86 - 1.528 = 100.33 MPa = 100 MPa The state of stress at point B is represented on the element shown in Fig. d.
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8–38. Since concrete can support little or no tension, this problem can be avoided by using wires or rods to prestress the concrete once it is formed. Consider the simply supported beam shown, which has a rectangular cross section of 18 in. by 12 in. If concrete has a specific weight of 150 lb>ft3, determine the required tension in rod AB, which runs through the beam so that no tensile stress is developed in the concrete at its center section a–a. Neglect the size of the rod and any deflection of the beam.
a 16 in. B 2 in.
A a 4 ft
Support Reactions: As shown on FBD. Internal Force and Moment: + : ©Fx = 0; a + ©Mo = 0;
T - N = 0
N = T
M + T(7) - 900(24) = 0 M = 21600 - 7T
Section Properties: A = 18(12) = 216 in2 I =
1 (12) A 183 B = 5832 in4 12
Normal Stress: Requires sA = 0 sA = 0 = 0 =
N Mc + A I (21600 - 7T)(9) -T + 216 5832
T = 2160 lb = 2.16 kip
Ans.
562
4 ft
18 in. 6 in. 6 in.
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8–39. Solve Prob. 8–38 if the rod has a diameter of 0.5 in. Use the transformed area method discussed in Sec. 6.6. Est = 2911032 ksi, Ec = 3.6011032 ksi.
a 16 in. B 2 in.
A a 4 ft
Support Reactions: As shown on FBD. Section Properties: n =
29(103) Est = 8.0556 = Econ 3.6(103)
p Acon = (n - 1)Aat = (8.0556 - 1) a b A 0.52 B = 1.3854 in2 4 A = 18(12) + 1.3854 = 217.3854 in2 y =
©yA 9(18)(12) + 16(1.3854) = = 9.04461 in. ©A 217.3854
I =
1 (12) A 183 B + 12(18)(9.04461 - 9)2 + 1.3854(16 - 9.04461)2 12
= 5899.45 in4 Internal Force and Moment: + : ©Fx = 0;
T - N = 0
a + ©Mo = 0;
M + T(6.9554) - 900(24) = 0
N = T
M = 21600 - 6.9554T Normal Stress: Requires sA = 0 sA = 0 =
0 =
N Mc + A I (21600 - 6.9554T)(8.9554) -T + 217.3854 5899.45
T = 2163.08 lb = 2.16 kip
Ans.
563
4 ft
18 in. 6 in. 6 in.
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*8–40. Determine the state of stress at point A when the beam is subjected to the cable force of 4 kN. Indicate the result as a differential volume element.
4 kN
250 mm G 375 mm
D 2m
0.75 m
100 mm
B 1m
a + ©MD = 0;
A
150 mm
Cy = 0.6667 kN Cx - 4 = 0
Cx = 4.00 kN
Internal Forces and Moment: + : ©Fx = 0;
4.00 - N = 0
+ c ©Fy = 0;
V - 0.6667 = 0
a + ©Mo = 0;
N = 4.00 kN V = 0.6667 kN M = 0.6667 kN # m
M - 0.6667(1) = 0
Section Properties: A = 0.24(0.15) - 0.2(0.135) = 9.00 A 10 - 3 B m2 I =
1 1 (0.15) A 0.243 B (0.135) A 0.23 B = 82.8 A 10 - 6 B m4 12 12
QA = ©y¿A¿ = 0.11(0.15)(0.02) + 0.05(0.1)(0.015) = 0.405 A 10 - 3 B m3 Normal Stress: s = sA =
My N ; A I 4.00(103) -3
0.6667(103)(0) +
9.00(10 )
82.8(10 - 6)
= 0.444 MPa (T)
Ans.
Shear Stress: Applying shear formula.
tA =
=
200 mm
B
4(0.625) - Cy (3.75) = 0
+ : ©Fx = 0;
C
20 mm
15 mm
Support Reactions:
A
VQA It
0.6667(103) C 0.405(10 - 3) D 82.8(10 - 6)(0.015)
= 0.217 MPa
Ans.
564
20 mm
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•8–41.
Determine the state of stress at point B when the beam is subjected to the cable force of 4 kN. Indicate the result as a differential volume element.
4 kN
250 mm G 375 mm
D 2m
0.75 m
100 mm
B 1m
a + ©MD = 0;
A
150 mm
4(0.625) - Cy (3.75) = 0
Cx - 4 = 0
Cx = 4.00 kN
Internal Forces and Moment: + : ©Fx = 0;
4.00 - N = 0
+ c ©Fy = 0;
V - 0.6667 = 0
a + ©Mo = 0;
N = 4.00 kN V = 0.6667 kN M = 0.6667 kN # m
M - 0.6667(1) = 0
Section Properties: A = 0.24(0.15) - 0.2(0.135) = 9.00 A 10 - 3 B m2 I =
200 mm
B
Cy = 0.6667 kN + : ©Fx = 0;
C
20 mm
15 mm
Support Reactions:
A
1 1 (0.15) A 0.243 B (0.135) A 0.23 B = 82.8 A 10 - 6 B m 12 12
QB = 0 Normal Stress: s = sB =
My N ; A I 4.00(103) 9.00(10 - 3)
0.6667(103)(0.12) -
82.8(10 - 6)
= -0.522 MPa = 0.522 MPa (C)
Ans.
Shear Stress: Since QB = 0, then tB = 0
Ans.
565
20 mm
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8–42. The bar has a diameter of 80 mm. Determine the stress components that act at point A and show the results on a volume element located at this point. 200 mm 300 mm B A
©Fy = 0;
3 Vy - 5a b = 0 5
Vy = 3 kN
©Fz = 0;
4 Vz + 5 a b = 0 5
Vz = -4 kN
4
5 kN
©My = 0;
4 My + 5a b(0.3) = 0 5
My = -1.2 kN # m
©Mz = 0;
3 Mz + 5 a b(0.3) = 0 5
Mz = -0.9 kN # m
Iy = It =
p (0.044) = 0.64(10 - 6)p m4 4
Referring to Fig. b, (Qy)A = 0
(Qz)A = z¿A¿ =
4(0.04) p c (0.042) d = 42.67(10 - 6) m3 3p 2
The normal stress is contributed by bending stress only. Thus,
s = -
Myz
Mzy Iz
+
Iy
For point A, y = -0.04 m and z = 0. Then
s = -
-0.9(103)(-0.04) 0.64(10 - 6)p
+ 0 = -17.90(106)Pa = 17.9 MPa (C)
Ans.
The transverse shear stress developed at point A is
A txy B v =
Vy(Qy)A
A txz B v =
Vz(Qz)A
Iz t
Iy t
= 0
=
5
3
Consider the equilibrium of the FBD of bar’s left cut segment shown in Fig. a,
Ans.
4(103) C 42.67(10 - 6) D 0.64(10 - 6)p (0.08)
= 1.061(106) Pa = 1.06 MPa
Ans.
The state of stress for point A can be represented by the volume element shown in Fig. c, 566
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8–42. Continued
567
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8–43. The bar has a diameter of 80 mm. Determine the stress components that act at point B and show the results on a volume element located at this point. 200 mm 300 mm B A
Consider the equilibrium of the FBD of bar’s left cut segment shown in Fig. a,
5
3 4
©Fy = 0;
3 Vy - 5a b = 0 5
Vy = 3 kN
©Fz = 0;
4 Vz + 5 a b = 0 5
Vz = -4 kN
5 kN
©My = 0;
4 My + 5a b(0.3) = 0 5
My = -1.2 kN # m
©Mz = 0;
3 Mz + 5 a b (0.3) = 0 5
Mz = -0.9 kN # m
Iy = Iz =
p (0.044) = 0.64(10 - 6)p m4 4
Referring to Fig. b,
A Qy B B = y¿A¿ = c
4(0.04) p d c (0.042) d = 42.67(10 - 6) m3 3p 2
A Qz B B = 0 The normal stress is contributed by bending stress only. Thus, s = -
Myz
Mzy Iz
+
Iy
For point B, y = 0 and z = 0.04 m. Then
s = -0 +
-1.2(103)(0.04) 0.64(10 - 6)p
= -23.87(106) Pa = 23.9 MPa (C)
Ans.
The transverse shear stress developed at point B is
A txy B v =
Vy(Qy)B Iz t
=
3(103) C 42.67(10 - 6) D 0.64(10 - 6)p (0.08)
= 0.7958(106) MPa = 0.796 MPa
568
Ans.
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8–43. Continued
A txz B v =
Vz (Qz)B Iy t
= 0
Ans.
The state of stress for point B can be represented by the volume element shown in Fig. c
569
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*8–44. Determine the normal stress developed at points A and B. Neglect the weight of the block.
6 kip 3 in.
Referring to Fig. a,
12 kip
6 in.
©Fx = (FR)x;
-6 - 12 = F
©My = (MR)y;
6(1.5) - 12(1.5) = My
©Mz = (MR)z;
12(3) - 6(3) = Mz
a
F = -18.0 kip
A
My = -9.00 kip # in Mz = 18.0 kip # in
The cross-sectional area and moment of inertia about the y and z axes of the crosssection are A = 6(3) = 18 in2 Iy =
1 (6)(3)3 = 13.5 in4 12
Iz =
1 (3)(63) = 54.0 in4 12
The normal stress developed is the combination of axial and bending stress. Thus, s =
My z Mz y F + A Iz Iy
For point A, y = 3 in. and z = -1.5 in. sA =
18.0(3) -9.00(-1.5) -18.0 + 18.0 54.0 13.5 Ans.
= -1.00 ksi = 1.00 ksi (C) For point B, y = 3 in and z = 1.5 in.
sB =
18.0(3) -9.00(1.5) -18.0 + 18.0 54 13.5
= -3.00 ksi = 3.00 ksi (C)
Ans.
570
B a
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•8–45.
Sketch the normal stress distribution acting over the cross section at section a–a. Neglect the weight of the block.
6 kip 3 in.
12 kip
6 in. a A
B a
Referring to Fig. a, ©Fx = (FR)x;
-6 - 12 = F
F = -18.0 kip
©My = (MR)y;
6(1.5) - 12(1.5) = My
©Mz = (MR)z;
12(3) - 6(3) = Mz
My = -9.00 kip # in Mz = 18.0 kip # in
The cross-sectional area and the moment of inertia about the y and z axes of the cross-section are A = 3 (6) = 18.0 in2 Iy =
1 (6)(33) = 13.5 in4 12
Iz =
1 (3)(63) = 54.0 in4 12
The normal stress developed is the combination of axial and bending stress. Thus, s =
Myz Mzy F + A Iz Iy
For point A, y = 3 in. and z = -1.5 in. sA =
18.0(3) -9.00(-1.5) -18.0 + 18.0 54.0 13.5
= -1.00 ksi = 1.00 ksi (C) For point B, y = 3 in. and z = 1.5 in. sB =
18.0(3) -9.00(1.5) -18.0 + 18.0 54.0 13.5
= -3.00 ksi = 3.00 ksi (C)
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8–45. Continued
For point C, y = -3 in. and z = 1.5 in. sC =
18.0(-3) -9.00(1.5) -18.0 + 18.0 54.0 13.5
= -1.00 ksi = 1.00 ksi (C) For point D, y = -3 in. and z = -1.5 in. sD =
18.0(-3) -9.00(-1.5) -18.0 + 18.0 54.0 13.5
= 1.00 ksi (T) The normal stress distribution over the cross-section is shown in Fig. b
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8–46. The support is subjected to the compressive load P. Determine the absolute maximum and minimum normal stress acting in the material.
a — a 2 — 2 a a — 2 — 2
Section Properties: w = a +x A = a(a + x) I =
a 1 (a) (a + x)3 = (a + x)3 12 12
Internal Forces and Moment: As shown on FBD. Normal Stress: s =
=
N Mc ; A I
0.5Px C 12 (a + x) D -P ; a 3 a(a + x) 12 (a + x) =
P 3x -1 ; B R a a+x (a + x)2
P 1 3x + B R a a+x (a + x)2 P 4x + a = - B R a (a + x)2 P -1 3x sB = + B R a a+x (a + x)2
sA = -
=
[1]
P 2x - a B R a (a + x)2
In order to have maximum normal stress,
[2]
dsA = 0. dx
dsA P (a + x)2(4) - (4x + a)(2)(a + x)(1) = - B R = 0 a dx (a + x)4
-
Since
P (2a - 4x) = 0 a(a + x)3
P Z 0, then a(a + x)3 2a - 4x = 0
x = 0.500a
573
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8–46. Continued
Substituting the result into Eq. [1] yields smax = -
= -
P 4(0.500a) + a B R a (a + 0.5a)2 1.33P 1.33P = (C) 2 a a2
In order to have minimum normal stress,
Ans.
dsB = 0. dx
dsB P (a + x)2 (2) - (2x - a)(2)(a + x)(1) = B R = 0 a dx (a + x)4 P (4a - 2x) = 0 a(a + x)3 Since
P Z 0, then a(a + x)3 4a - 2x = 0
x = 2a
Substituting the result into Eq. [2] yields smin =
P 2(2a) - a P B R = 2 (T) a (a + 2a)2 3a
Ans.
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8–47. The support is subjected to the compressive load P. Determine the maximum and minimum normal stress acting in the material. All horizontal cross sections are circular.
P r
Section Properties: d¿ = 2r + x A = p(r + 0.5x)2 p (r + 0.5x)4 4
I =
Internal Force and Moment: As shown on FBD. Normal Stress: s =
Mc N ; A I
=
0.5Px(r + 0.5x) –P ; p 2 4 p(r + 0.5x) 4 (r + 0.5)
=
–1 P 2x ; B R p (r + 0.5x)2 (r + 0.5x)3
sA = -
= -
sB =
=
P 1 2x + B R p (r + 0.5x)2 (r + 0.5x)3 P r + 2.5x B R p (r + 0.5x)3
[1]
P –1 2x + B R p (r + 0.5x)2 (r + 0.5x)3 P 1.5x - r B R p (r + 0.5x)3
In order to have maximum normal stress,
[2] dsA = 0. dx
dsA P (r + 0.5x)3 (2.5) - (r + 2.5x)(3)(r + 0.5x)2 (0.5) = B R = 0 p dx (r + 0.5x)6 -
Since
P (r - 2.5x) = 0 p(r + 0.5x)4
P Z 0, then p(r + 0.5x)4 r - 2.5x = 0
x = 0.400r
Substituting the result into Eq. [1] yields smax = -
= -
P r + 2.5(0.400r) B R p [r + 0.5(0.400r)]3 0.368P 0.368P = (C) r2 r2
Ans.
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8–47. Continued
In order to have minimum normal stress,
dsB = 0. dx
dsB P (r + 0.5x)3 (1.5) - (1.5x - r)(3)(r + 0.5x)2 (0.5) = B R = 0 p dx (r + 0.5x)6 P (3r - 1.5x) = 0 p(r + 0.5x)4 Since
P Z 0, then p(r + 0.5x)4 x = 2.00r
3r - 1.5x = 0 Substituting the result into Eq. [2] yields smin =
P 1.5(2.00r) - r 0.0796P (T) B R = p [r + 0.5(2.00r)]3 r2
Ans.
*8–48. The post has a circular cross section of radius c. Determine the maximum radius e at which the load can be applied so that no part of the post experiences a tensile stress. Neglect the weight of the post.
P c e
Require sA = 0 sA = 0 =
P Mc + ; A I e =
0 =
(Pe)c -P + p 4 2 pc 4c
c 4
Ans.
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•8–49.
If the baby has a mass of 5 kg and his center of mass is at G, determine the normal stress at points A and B on the cross section of the rod at section a–a. There are two rods, one on each side of the cradle.
500 mm 15⬚ G a
75 mm a
6 mm A
B
Section a–a
Section Properties: The location of the neutral surface from the center of curvature of the rod, Fig. a, can be determined from A
R = ©
dA LA r
where A = p A 0.0062 B = 36p A 10 - 6 B m2 ©
dA = 2p ¢ r - 2r2 - c2 ≤ = 2p ¢ 0.081 - 20.0812 - 0.0062 ≤ = 1.398184 A 10 - 3 B m LA r
Thus, R =
36p A 10 - 6 B
1.398184 A 10 - 3 B
= 0.080889 m
Then e = r - R = 0.081 - 0.080889 = 0.111264 A 10 - 3 B m Internal Loadings: Consider the equilibrium of the free-body diagram of the cradle’s upper segment, Fig. b, + c ©Fy = 0;
-5(9.81) - 2N = 0
N = -24.525 N
a + ©MO = 0;
5(9.81)(0.5+ 0.080889) - 2M = 0
M = 14.2463 N # m
Normal Stress: The normal stress is the combination of axial and bending stress. Thus, s =
M(R - r) N + A Aer
Here, M = -14.1747 (negative) since it tends to increase the curvature of the rod. For point A, r = rA = 0.075 m. Then, sA =
-24.525
36p A 10
-6
B
-14.2463(0.080889 - 0.075)
+
36p A 10 - 6 B (0.111264) A 10 - 3 B (0.075)
= -89.1 MPa = 89.1 MPa (C)
Ans.
For point B, r = rB = 0.087 m. Then, sB =
-24.525
36p A 10
-6
B
-14.2463(0.080889 - 0.087)
+
36p A 10 - 6 B (0.111264) A 10 - 3 B (0.087) Ans.
= 79.3 kPa (T) dA 5 = 0.25 ln = 0.055786 4 LA r
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8–50. The C-clamp applies a compressive stress on the cylindrical block of 80 psi. Determine the maximum normal stress developed in the clamp.
1 in.
4 in.
4.5 in.
0.75 in.
R =
A dA 1 r
=
1(0.25) = 4.48142 0.055786
P = sbA = 80p (0.375)2 = 35.3429 lb M = 35.3429(8.98142) = 317.4205 lb # in. s =
M(R - r) P + Ar(r - R) A
(st)max =
317.4205(4.48142 - 4) 35.3429 + = 8.37 ksi (1)(0.25)(4)(4.5 - 4.48142) (1)(0.25)
(sc)max =
317.4205(4.48142 - 5) 35.3429 + = -6.95 ksi 1(0.25)(5)(4.5 - 4.48142) (1)(0.25)
578
Ans.
0.25 in.
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8–51. A post having the dimensions shown is subjected to the bearing load P. Specify the region to which this load can be applied without causing tensile stress to be developed at points A, B, C, and D.
x z a
a A
B
Equivalent Force System: As shown on FBD. Section Properties: 1 A = 2a(2a) + 2 B (2a)a R = 6a2 2 Iz =
1 1 1 a 2 (2a)(2a)3 + 2 B (2a) a3 + (2a) aa a + b R 12 36 2 3
= 5a4 Iy =
=
1 1 1 a 2 (2a)(2a)3 + 2 B (2a) a3 + (2a) aa b R 12 36 2 3 5 4 a 3
Normal Stress: s =
My z Mzy N + A Iz Iy
=
Peyy Pez z -P + 5 2 4 4 6a 5a 3a
=
P A -5a2 - 6eyy + 18ez z B 30a4
At point A where y = -a and z = a, we require sA 6 0. 0 7
P C -5a2 - 6(-a) ey + 18(a) ez D 30a4
0 7 -5a + 6ey + 18ez 6ey + 18ez 6 5a When
ez = 0,
When
ey = 0,
Ans.
5 a 6 5 ez 6 a 18
ey 6
Repeat the same procedures for point B, C and D. The region where P can be applied without creating tensile stress at points A, B, C and D is shown shaded in the diagram.
579
a P
a
D ez
ey C
a a
y
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*8–52. The hook is used to lift the force of 600 lb. Determine the maximum tensile and compressive stresses at section a–a. The cross section is circular and has a diameter of 1 in. Use the curved-beam formula to compute the bending stress.
300 lb
a
300 lb
2.5 in. a 1.5 in.
Section Properties: r = 1.5 + 0.5 = 2.00 in. dA = 2p A r - 2r2 - c2 B LA r
600 lb
= 2p A 2.00 - 22.002 - 0.52 B = 0.399035 in. A = p A 0.52 B = 0.25p in2 A
R =
dA 1A r
=
0.25p = 1.968246 in. 0.399035
r - R = 2 - 1.968246 = 0.031754 in. Internal Force and Moment: As shown on FBD. The internal moment must be computed about the neutral axis. M = 1180.95 lb # in. is positive since it tends to increase the beam’s radius of curvature. Normal Stress: Applying the curved-beam formula. For tensile stress (st)max =
=
M(R - r1) N + A Ar1 (r - R) 1180.95(1.968246 - 1.5) 600 + 0.25p 0.25p(1.5)(0.031754)
= 15546 psi = 15.5 ksi (T)
Ans.
For compressive stress, (sc)max =
=
M(R - r2) N + A Ar2 (r - R) 1180.95(1.968246 - 2.5) 600 + 0.25p 0.25p(2.5)(0.031754)
= -9308 psi = 9.31 ksi (C)
Ans.
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•8–53.
The masonry pier is subjected to the 800-kN load. Determine the equation of the line y = f1x2 along which the load can be placed without causing a tensile stress in the pier. Neglect the weight of the pier.
800 kN 1.5 m y 1.5 m
2.25 m y 2.25 m
x
x
C A B
A = 3(4.5) = 13.5 m2 Ix =
1 (3)(4.53) = 22.78125 m4 12
Iy =
1 (4.5)(33) = 10.125 m4 12
Normal Stress: Require sA = 0 sA =
0 =
Myx Mxy P + + A Ix Iy
800(103)y(2.25) 800(103)x(1.5) -800(103) + + 13.5 22.78125 10.125
0 = 0.148x + 0.0988y - 0.0741 y = 0.75 - 1.5 x
Ans.
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8–54. The masonry pier is subjected to the 800-kN load. If x = 0.25 m and y = 0.5 m, determine the normal stress at each corner A, B, C, D (not shown) and plot the stress distribution over the cross section. Neglect the weight of the pier.
800 kN 1.5 m y 1.5 m
2.25 m y 2.25 m
x
x
A = 3(4.5) = 13.5 m2 Ix =
1 (3)(4.53) = 22.78125 m4 12 C
1 Iy = (4.5)(33) = 10.125 m4 12 s =
sA =
A B
Myx Mxy P + + A Ix Iy
-800(103) 400(103)(2.25) 200(103)(1.5) + + 13.5 22.78125 10.125 Ans.
= 9.88 kPa (T) 3
sB =
3
3
400(10 )(2.25) 200(10 )(1.5) -800(10 ) + 13.5 22.78125 10.125 Ans.
= -49.4 kPa = 49.4 kPa (C) sC =
-800(103) 400(103)(2.25) 200(103)(1.5) + 13.5 22.78125 10.125 Ans.
= -128 kPa = 128 kPa (C) sD =
-800(103) 400(103)(2.25) 200(103)(1.5) + 13.5 22.78125 10.125
= -69.1 kPa = 69.1 kPa (C)
Ans.
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8–55. The bar has a diameter of 40 mm. If it is subjected to the two force components at its end as shown, determine the state of stress at point A and show the results on a differential volume element located at this point.
x 100 mm 150 mm z A
B
500 N
y 300 N
Internal Forces and Moment: ©Fx = 0;
Nx = 0
©Fy = 0;
Vy + 300 = 0
Vy = -300 N
©Fz = 0;
Vz - 500 = 0
Vz = 500 N
©Mx = 0;
Tx = 0
©My = 0;
My - 500(0.15) = 0
My = 75.0 N # m
©Mz = 0;
Mz - 300(0.15) = 0
Mz = 45.0 N # m
Section Properties: A = p A 0.022 B = 0.400 A 10 - 3 B p m2 p A 0.024 B = 40.0 A 10 - 9 B p m4 4
Ix = Iy =
J =
p A 0.024 B = 80.0 A 10 - 9 B p m4 2
(QA)z = 0 4(0.02) 1 c p A 0.022 B d = 5.333 A 10 - 6 B m3 3p 2
(QA)y = Normal Stress: s =
Myz Mzy N + A Iz Iy
sA = 0 -
75.0(0.02)
45.0(0) -9
+
40.0(10 )p
40.0(10 - 9)p
= 11.9 MPa (T)
Ans.
Shear Stress: The tranverse shear stress in the z and y directions can be obtained VQ using the shear formula, tV = . It (txy)A = -tVy = -
300 C 5.333(10 - 6) D
40.0(10 - 9)p (0.04) Ans.
= -0.318 MPa (txz)A = tVz = 0
Ans.
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*8–56.
Solve Prob. 8–55 for point B. x 100 mm 150 mm z A
B
500 N
y 300 N
Internal Forces and Moment: ©Fx = 0;
Nx = 0
©Fy = 0;
Vy + 300 = 0
Vy = -300 N
©Fz = 0;
Vz - 500 = 0
Vz = 500 N
©Mx = 0;
Tx = 0
©My = 0;
My - 500(0.15) = 0
My = 75.0 N # m
©Mz = 0;
Mz - 300(0.15) = 0
Mz = 45.0 N # m
Section Properties: A = p A 0.022 B = 0.400 A 10 - 3 B p m2 Ix = Iy =
J =
p A 0.024 B = 40.0 A 10 - 9 B p m4 4
p A 0.024 B = 80.0 A 10 - 9 B p m4 2
(QB)y = 0 (QB)z =
4(0.02) 1 c p A 0.022 B d = 5.333 A 10 - 6 B m3 3p 2
Normal Strees: s =
Myz Mzy N + A Iz Iy
sB = 0 -
75.0(0)
45.0(0.02) -9
40.0(10 ) p
+
40.0(10 - 9) p
= -7.16 MPa = 7.16 MPa (C)
Ans.
Shear Stress: The tranverse shear stress in the z and y directions can be obtained VQ using the shear formula, tV = . It (txz)B = tVz =
500 C 5.333(10 - 6) D
40.0(10 - 9) p (0.04) Ans.
= 0.531 MPa (txy)B = tVy = 0
Ans.
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•8–57. The 2-in.-diameter rod is subjected to the loads shown. Determine the state of stress at point A, and show the results on a differential element located at this point.
z B x
A
8 in.
y
600 lb 12 in.
Consider the equilibrium of the FBD of the right cut segment, Fig. a, ©Fy = 0 ;
Ny + 800 = 0
©Fz = 0 ;
Vz + 600 = 0
Vz = -600 lb
©Fx = 0 ;
Vx - 500 = 0
Vx = 500 lb
©My = 0 ;
Ty - 600(12) = 0
©Mz = 0 ;
Mz + 800(12) + 500(8) = 0
©Mx = 0 ;
Mx + 600(8) = 0
J =
Ny = -800 lb
p 4 p (1 ) = in4 4 4
Ix = Iz =
500 lb 800 lb
Ty = 7200 lb # in Mz = -13600 lb # in
Mx = -4800 lb # in A = p(12) = p in2
p 4 p (1 ) = in4 2 2
Referring to Fig. b, (Qz)A = y¿A¿ =
(Qx)A = 0
4(1) p 2 c (1 ) d = 0.6667 in3 3p 2
The normal stress is contributed by axial and bending stress. Thus, s =
Mzx Mxz N + A Ix Iz
For point A, z = 0 and x = 1 in. 4800(0) -13600(1) 800 p p>4 p>4
s =
= 17.57(103) psi = 17.6 ksi (T)
Ans.
The torsional shear stress developed at point A is (tyz)T =
TyC J
=
7200(1) = 4.584(103) psi = 4.584 ksi T p>2
The transverse shear stress developed at point A is. (tyz)g =
(txy)g =
Vz(Qz)A Ixt
=
600(0.6667) = 254.64 psi = 0.2546 ksi T p (2) 4
Vx(Qx)A 500(0) = = 0 p Izt (2) 4 585
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8–57. Continued Combining these two shear stress components, tyz = A tyz B T + A tyz B g = 4.584 + 0.2546 Ans.
= 4.838 ksi = 4.84 ksi txy = 0
Ans.
The state of stress of point A can be represented by the volume element shown in Fig. c.
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8–58. The 2-in.-diameter rod is subjected to the loads shown. Determine the state of stress at point B, and show the results on a differential element located at this point.
z B x
A
8 in.
y
600 lb 12 in.
Consider the equilibrium of the FBD of the right cut segment, Fig. a, ©Fy = 0;
Ny + 800 = 0
Ny = -800 lb
©Fz = 0;
Vz + 600 = 0
Vz = -600 lb
©Fx = 0;
Vx - 500 = 0
Vx = 500 lb
©My = 0;
Ty - 600(12) = 0
©Mz = 0;
Mz + 800(12) + 500(8) = 0 Mz = -13600 lb # in
©Mx = 0;
Mx + 600(8) = 0
800 lb
Ty = 7200 lb # in
Mx = -4800 lb # in.
The cross-sectional area the moment of inertia about x and Z axes and polar moment of inertia of the rod are ¿
A = p(12) = p in2
Ix = Iz =
p 4 p (1 ) = in4 4 4
J =
p 4 p (1 ) = in4 2 2
Referring to Fig. b, (Qz)B = 0 (Qx)B = z¿A¿ =
4(1) p 2 c (1 ) d = 0.6667 in4 3p 2
The normal stress is contributed by axial and bending stress. Thus, s =
Mzx Mxz N + A Ix Iz
For point B, x = 0 and z = 1 in. s =
4800 (1) 13600 (0) 800 + p p>4 p>4
= 5.86 ksi (C) The torsional shear stress developed at point B is (txy)T =
TyC J
=
7200(1) = 4.584(103) psi = 4.584 ksi : p>2
The transverse shear stress developed at point B is. (txy)v =
(tyz)v =
Vx(Qx)B 500 (0.6667) = = 212.21 psi = 0.2122 ksi : p Izt (2) 4 Vz(Qz)B Ixt
=
500 lb
600 (0) = 0 p (2) 4
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8–58.
Continued
Combining these two shear stress components, txy = (txy)T + (txy)v = 4.584 + 0.2122 = 4.796 ksi = 4.80 ksi
Ans.
tyz = 0
Ans.
The state of stress of point B can be represented by the volume element shown in Fig. c.
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8–59. If P = 60 kN, determine the maximum normal stress developed on the cross section of the column.
2P
150 mm 150 mm
Equivalent Force System: Referring to Fig. a, + c ©Fx = A FR B x;
F = 180 kN
-60 - 120 = -F
15 mm
©My = (MR)y;
-60(0.075) = -My
My = 4.5kN # m
©Mz = (MR)z;
-120(0.25) = -Mz
Mz = 30kN # m
100 mm 100 mm
Section Properties: The cross-sectional area and the moment of inertia about the y and z axes of the cross section are A = 0.2(0.3) - 0.185(0.27) = 0.01005 m2 Iz =
1 1 (0.2) A 0.33 B (0.185) A 0.273 B = 0.14655 A 10 - 3 B m4 12 12
Iy = 2c
1 1 (0.015) A 0.23 B d + (0.27) A 0.0153 B = 20.0759 A 10 - 6 B m4 12 12
Normal Stress: The normal stress is the combination of axial and bending stress. Here, F is negative since it is a compressive force. Also, My and Mz are negative since they are directed towards the negative sense of their respective axes. By inspection, point A is subjected to a maximum normal stress. Thus, s =
Myz Mzy N + A Iz Iy
smax = sA =
-180 A 103 B 0.01005
-
C -30 A 103 B D ( -0.15) 0.14655 A 10 - 3 B
15 mm 15 mm P
+
C -4.5 A 103 B D (0.1) 20.0759 A 10 - 6 B
= -71.0 MPa = 71.0 MPa(C)
Ans.
589
75 mm
100 mm
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*8–60. Determine the maximum allowable force P, if the column is made from material having an allowable normal stress of sallow = 100 MPa .
2P
150 mm 150 mm
Equivalent Force System: Referring to Fig. a, + c ©Fx = (FR)x;
15 mm
-P - 2P = -F
100 mm 100 mm
F = 3P ©My = (MR)y;
-P(0.075) = -My My = 0.075 P
©Mz = (MR)z;
-2P(0.25) = -Mz Mz = 0.5P
Section Properties: The cross-sectional area and the moment of inertia about the y and z axes of the cross section are A = 0.2(0.3) - 0.185(0.27) = 0.01005 m2 Iz =
1 1 (0.2) A 0.33 B (0.185) A 0.273 B = 0.14655 A 10 - 3 B m4 12 12
Iy = 2 c
1 1 (0.15) A 0.23 B d + (0.27) A 0.0153 B = 20.0759 A 10 - 6 B m4 12 12
Normal Stress: The normal stress is the combination of axial and bending stress. Here, F is negative since it is a compressive force. Also, My and Mz are negative since they are directed towards the negative sense of their respective axes. By inspection, point A is subjected to a maximum normal stress, which is in compression. Thus, s =
Myz Mzy N + A Iz Iy
-100 A 106 B = -
15 mm 15 mm P
(-0.5P)(-0.15) -0.075P(0.1) 3P + 0.01005 0.14655 A 10 - 3 B 20.0759 A 10 - 6 B
P = 84470.40 N = 84.5 kN
Ans.
590
75 mm
100 mm
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•8–61.
The beveled gear is subjected to the loads shown. Determine the stress components acting on the shaft at point A, and show the results on a volume element located at this point. The shaft has a diameter of 1 in. and is fixed to the wall at C.
z y A 200 lb
C B
x 8 in.
©Fx = 0;
Vx - 125 = 0;
©Fy = 0;
75 - Ny = 0;
Ny = 75 lb
©Fz = 0;
Vz - 200 = 0;
Vz = 200 lb
3 in.
Vx = 125 lb
75 lb
Mx = 1600 lb # in.
©Mx = 0;
200(8) - Mx = 0;
©My = 0;
200(3) - Ty = 0;
©Mz = 0;
Mz + 75(3) - 125(8) = 0;
Ty = 600 lb # in. Mz = 775 lb # in.
A = p(0.52) = 0.7854 in2 J =
p (0.54) = 0.098175 in4 2
I =
p (0.54) = 0.049087 in4 4
(QA)x = 0 (QA)z =
4(0.5) 1 a b (p)(0.52) = 0.08333 in3 3p 2
(sA)y = -
= -
Ny A
+
Mx c I
1600(0.5) 75 + 0.7854 0.049087
= 16202 psi = 16.2 ksi (T)
Ans.
(tA)yx = (tA)V - (tA)twist Tyc
Vx(QA)z =
=
It
-
J
600(0.5) 125(0.08333) 0.049087 (1) 0.098175
= -2843 psi = -2.84 ksi (tA)yz =
Vz(QA)x It
Ans.
= 0
Ans.
591
125 lb
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8–62. The beveled gear is subjected to the loads shown. Determine the stress components acting on the shaft at point B, and show the results on a volume element located at this point. The shaft has a diameter of 1 in. and is fixed to the wall at C.
z y A 200 lb
C B
x 8 in. 3 in. 75 lb
©Fx = 0;
Vx - 125 = 0;
Vx = 125 lb
©Fy = 0;
75 - Ny = 0;
Ny = 75 lb
©Fz = 0;
Vz - 200 = 0;
Vz = 200 lb Mx = 1600 lb # in.
©Mx = 0;
200(8) - Mx = 0;
©My = 0;
200(3) - Ty = 0;
©Mz = 0;
Mz + 75(3) - 125(8) = 0;
Ty = 600 lb # in. Mz = 775 lb # in.
A = p(0.52) = 0.7854 in2 J =
p (0.54) = 0.098175 in4 2
I =
p (0.54) = 0.049087 in4 4
(QB)z = 0 (QB)x =
4(0.5) 1 a b(p)(0.52) = 0.08333 in3 3p 2
(sB)y = -
= -
Ny A
Mz c +
I
775(0.5) 75 + 0.7854 0.049087
= 7.80 ksi (T)
Ans.
(tB)yz = (tB)V + (tB)twist Ty c
Vz(QB)x =
=
It
+
J
600(0.5) 200(0.08333) + 0.049087 (1) 0.098175 Ans.
= 3395 psi = 3.40 ksi (tB)yx =
Vx (QB)z It
Ans.
= 0
592
125 lb
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8–63. The uniform sign has a weight of 1500 lb and is supported by the pipe AB, which has an inner radius of 2.75 in. and an outer radius of 3.00 in. If the face of the sign is subjected to a uniform wind pressure of p = 150 lb>ft2, determine the state of stress at points C and D. Show the results on a differential volume element located at each of these points. Neglect the thickness of the sign, and assume that it is supported along the outside edge of the pipe.
12 ft B 150 lb/ft2 6 ft
F
E
3 ft
D C
A z
y x
Section Properties: A = p A 32 - 2.752 B = 1.4375p in2 Iy = Iz =
p 4 A 3 - 2.754 B = 18.6992 in4 4
(QC)z = (QD)y = 0 4(3) 1 4(2.75) 1 c (p) A 32 B d c (p) A 2.752 B d 3p 2 3p 2
(QC)y = (QD)z =
= 4.13542 in3 J =
p 4 A 3 - 2.754 B = 37.3984 in4 2
Normal Stress: s =
sC =
My z Mz y N + A Iz Iy (-64.8)(12)(0) 9.00(12)(2.75) -1.50 + 1.4375p 18.6992 18.6992 Ans.
= 15.6 ksi (T) sD =
(-64.8)(12)(3) 9.00(12)(0) -1.50 + 1.4375p 18.6992 18.6992 Ans.
= 124 ksi (T)
Shear Stress: The tranverse shear stress in the z and y directions and the torsional shear stress can be obtained using the shear formula and the torsion formula, VQ Tr and ttwist = , respectively. tV = It J (txz)D = ttwist =
64.8(12)(3) = 62.4 ksi 37.3984
Ans.
(txy)D = tVy = 0
Ans.
593
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8–63.
Continued
(txy)C = tVy - ttwist =
64.8(12)(2.75) 10.8(4.13542) 18.6992(2)(0.25) 37.3984 Ans.
= -52.4 ksi (txz)C = tVz = 0
Ans.
Internal Forces and Moments: As shown on FBD. ©Fx = 0;
1.50 + Nx = 0
Nx = -15.0 kip
©Fy = 0;
Vy - 10.8 = 0
Vy = 10.8 kip
©Fz = 0;
Vz = 0
©Mx = 0;
Tx - 10.8(6) = 0
Tx = 64.8 kip # ft
©My = 0;
My - 1.50(6) = 0
My = 9.00 kip # ft
©Mz = 0;
10.8(6) + Mz = 0
Mz = -64.8 kip # ft
594
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*8–64.
Solve Prob. 8–63 for points E and F. 12 ft B 150 lb/ft2 6 ft
F
E
3 ft
D C
A z
y x
Internal Forces and Moments: As shown on FBD. ©Fx = 0;
1.50 + Nx = 0
Nx = -1.50 kip
©Fy = 0;
Vy - 10.8 = 0
Vy = 10.8 kip
©Fz = 0;
Vz = 0
©Mx = 0;
Tx - 10.8(6) = 0
Tx = 64.8 kip # ft
©My = 0;
My - 1.50(6) = 0
My = 9.00 kip # ft
©Mz = 0;
10.8(6) + Mz = 0
Mz = -64.8 kip # ft
Section Properties: A = p A 32 - 2.752 B = 1.4375p in2 Iy = Iz =
p 4 A 3 - 2.754 B = 18.6992 in4 4
(QC)z = (QD)y = 0 (QC)y = (QD)z =
4(3) 1 4(2.75) 1 c (p) A 32 B d c (p) A 2.752 B d 3p 2 3p 2
= 4.13542 in3 J =
p 4 A 3 - 2.754 B = 37.3984 in4 2
Normal Stress: s =
sF =
My z Mzy N + A Iz Iy (-64.8)(12)(0) 9.00(12)(-3) -1.50 + 1.4375p 18.6992 18.6992 Ans.
= -17.7 ksi = 17.7 ksi (C) sE =
(-64.8)(12)(-3) 9.00(12)(0) -1.50 + 1.4375p 18.6992 18.6992
= -125 ksi = 125 ksi (C)
Ans.
595
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8–64.
Continued
Shear Stress: The tranverse shear stress in the z and y directions and the torsional shear stress can be obtained using the shear formula and the torsion formula, VQ Tr and ttwist = , respectively. tV = It J (txz)E = -ttwist = -
64.8(12)(3) = -62.4 ksi 37.3984
Ans.
(txy)E = tVy = 0
Ans.
(txy)F = tVy + ttwist =
64.8(12)(3) 10.8(4.13542) + 18.6992(2)(0.25) 37.3984
= 67.2 ksi
Ans.
(txy)F = tVy = 0
Ans.
596
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•8–65.
Determine the state of stress at point A on the cross section of the pipe at section a–a.
A 0.75 in. B y
50 lb
1 in. Section a–a
x a
60°
z
a
10 in.
Internal Loadings: Referring to the free - body diagram of the pipe’s right segment, Fig. a, ©Fy = 0; Vy - 50 sin 60° = 0
Vy = 43.30 lb
©Fz = 0; Vz - 50 cos 60° = 0
Vz = 25 lb T = -519.62 lb # in
©Mx = 0; T + 50 sin 60°(12) = 0 ©My = 0; My - 50 cos 60°(10) = 0
My = 250 lb # in
©Mz = 0; Mz + 50 sin 60° (10) = 0
Mz = -433.01 lb # in
Section Properties: The moment of inertia about the y and z axes and the polar moment of inertia of the pipe are Iy = Iz =
J =
p 4 A 1 - 0.754 B = 0.53689 in4 4
p 4 A 1 - 0.754 B = 1.07379 in4 2
Referring to Fig. b,
A Qy B A = 0 A Qz B A = y1œ A1œ - y2œ A2œ =
4(1) p 2 4(0.75) p c A1 B d c A 0.752 B d = 0.38542 in3 3p 2 3p 2
Normal Stress: The normal stress is contributed by bending stress only. Thus, s = -
Myz
Mzy Iz
+
Iy
For point A, y = 0.75 in and z = 0. Then sA =
-433.01(0.75) + 0 = 604.89 psi = 605 psi (T) 0.53689
Shear Stress: The torsional shear stress developed at point A is c A txz B T d
= A
TrA 519.62(0.75) = = 362.93 psi J 1.07379
597
Ans.
12 in.
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8–65.
Continued
The transverse shear stress developed at point A is c A txy B V d c A txz B V d
= 0 A
= A
Vz A Qz B A Iy t
=
25(0.38542) = 35.89 psi 0.53689(2 - 1.5)
Combining these two shear stress components,
A txy B A = 0
Ans.
A txz B A = c A txz B T d - c A txz B V d A
A
= 362.93 - 35.89 = 327 psi
Ans.
598
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8–66. Determine the state of stress at point B on the cross section of the pipe at section a–a.
A 0.75 in. B y
50 lb
1 in. Section a–a
x a
60°
z
a
10 in.
Internal Loadings: Referring to the free - body diagram of the pipe’s right segment, Fig. a, ©Fy = 0; Vy - 50 sin 60° = 0
Vy = 43.30 lb
©Fz = 0; Vz - 50 cos 60° = 0
Vz = 25 lb T = -519.62 lb # in
©Mx = 0; T + 50 sin 60°(12) = 0 ©My = 0; My - 50 cos 60°(10) = 0
My = 250 lb # in
©Mz = 0; Mz + 50 sin 60°(10) = 0
Mz = -433.01 lb # in
Section Properties: The moment of inertia about the y and z axes and the polar moment of inertia of the pipe are p 4 A 1 - 0.754 B = 0.53689 in4 4
Iy = Iz =
J =
p 4 A 1 - 0.754 B = 1.07379 in4 2
Referring to Fig. b,
A Qz B B = 0 A Qy B B = y1œ A1œ - y2œ A2œ =
4(1) p 2 4(0.75) p c A1 B d c A 0.752 B d = 0.38542 in3 3p 2 3p 2
Normal Stress: The normal stress is contributed by bending stress only. Thus, s = -
Myz
Mzy Iz
+
Iy
For point B, y = 0 and z = -1. Then sB = -0 +
250(1) = -465.64 psi = 466 psi (C) 0.53689
Ans.
Shear Stress: The torsional shear stress developed at point B is c A txy B T d
= B
519.62(1) TrC = = 483.91 psi J 1.07379
599
12 in.
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8–66.
Continued
The transverse shear stress developed at point B is c A txz B V d c A txy B V d
= 0 B
= B
Vy A Qy B B Izt
=
43.30(0.38542) = 62.17 psi 0.53689(2 - 1.5)
Combining these two shear stress components,
A txy B B = c A txy B T d - c A txy B V d B
B
Ans.
= 483.91 - 62.17 = 422 psi
A txz B B = 0
Ans.
600
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•8–67. The eccentric force P is applied at a distance ey from the centroid on the concrete support shown. Determine the range along the y axis where P can be applied on the cross section so that no tensile stress is developed in the material.
x
z
P
b 2
ey b 2
2h 3 h 3
Internal Loadings: As shown on the free - body diagram, Fig. a. Section Properties: The cross-sectional area and moment of inertia about the z axis of the triangular concrete support are A =
1 bh 2
Iz =
1 bh3 36
Normal Stress: The normal stress is the combination of axial and bending stress. Thus, s =
Mzy N A Iz
A Pey B y -P 1 1 bh bh3 2 36
s =
2P 2 A h + 18eyy B bh3
s = -
Here, it is required that sA … 0 and sB … 0. For point A, y =
(1) h , Then. Eq. (1) gives 3
2P 2 h ch + 18ey a b d 3 3 bh
0 Ú -
0 … h2 + 6hey ey Ú -
For Point B, y = -
h 6
2 h. Then. Eq. (1) gives 3 0 Ú -
2P 2 2 ch + 18ey a - hb d 3 3 bh
0 … h2 - 12hey ey …
h 12
Thus, in order that no tensile stress be developed in the concrete support, ey must be in the range of -
h h … ey … 6 12
Ans.
601
y
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*8–68. The bar has a diameter of 40 mm. If it is subjected to a force of 800 N as shown, determine the stress components that act at point A and show the results on a volume element located at this point.
150 mm 200 mm
1 1 I = p r4 = (p)(0.024) = 0.1256637 (10 - 6) m4 4 4 A
A = p r2 = p(0.022) = 1.256637 (10 - 3) m2 QA = y¿A¿ = a
z
B
4 (0.02) p (0.02)2 ba b = 5.3333 (10 - 6) m3 3p 2
y
x 30⬚
800 N
sA
P Mz = + A I 400 + 0 = 0.318 MPa 1.256637 (10 - 3)
Ans.
692.82 (5.3333) (10 - 6) VQA = 0.735 MPa = It 0.1256637 (10 - 6)(0.04)
Ans.
=
tA =
•8–69.
Solve Prob. 8–68 for point B.
150 mm 200 mm
A
z
B y
x 30⬚
800 N
1 1 I = p r4 = (p)(0.024) = 0.1256637 (10 - 6) m4 4 4 A = p r2 = p(0.022) = 1.256637 (10 - 3) m2 QB = 0 sB =
138.56 (0.02) P Mc 400 = -21.7 MPa = -3 A I 1.256637 (10 ) 0.1256637 (10 - 6)
tB = 0
Ans. Ans.
602
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8–70. The 43-in.-diameter shaft is subjected to the loading shown. Determine the stress components at point A. Sketch the results on a volume element located at this point. The journal bearing at C can exert only force components Cy and Cz on the shaft, and the thrust bearing at D can exert force components Dx , Dy , and Dz on the shaft.
D
z 125 lb 2 in.
8 in. 125 lb 2 in.
A
20 in.
8 in. B
C
10 in.
y
20 in.
x
A =
p (0.752) = 0.44179 in2 4
I =
p (0.3754) = 0.015531 in4 4
QA = 0 tA = 0 sA =
Ans.
My c I
=
-1250(0.375) = -30.2 ksi = 30.2 ksi (C) 0.015531
Ans.
8–71. Solve Prob. 8–70 for the stress components at point B.
D
z 125 lb 2 in.
8 in. 125 lb 2 in.
A
8 in. B
C
10 in. x
p A = (0.752) = 0.44179 in2 4 I =
p (0.3754) = 0.015531 in4 4
QB = y¿A¿ =
4(0.375) 1 a b(p)(0.3752) = 0.035156 in3 3p 2
sB = 0 tB =
Ans.
VzQB It
=
125(0.035156) = 0.377 ksi 0.015531(0.75)
Ans.
603
20 in.
20 in.
y
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*8–72. The hook is subjected to the force of 80 lb. Determine the state of stress at point A at section a–a. The cross section is circular and has a diameter of 0.5 in. Use the curved-beam formula to compute the bending stress.
80 lb 1.5 in. 45⬚
The location of the neutral surface from the center of curvature of the hook, Fig. a, can be determined from
where A = p(0.252) = 0.0625p in2 dA = 2p A r - 2r2 - c2 B = 2p A 1.75 - 21.752 - 0.252 B = 0.11278 in. LA r
Thus, R =
0.0625p = 1.74103 in. 0.11278
Then e = r - R = 1.75 - 1.74103 = 0.0089746 in. Referring to Fig. b, I and QA are I =
p (0.254) = 0.9765625(10 - 3)p in4 4
QA = 0 Consider the equilibrium of the FBD of the hook’s cut segment, Fig. c, + ©F = 0; ; x
N - 80 cos 45° = 0
N = 56.57 lb
+ c ©Fy = 0;
80 sin 45° - V = 0
V = 56.57 lb
a + ©Mo = 0;
M - 80 cos 45°(1.74103) = 0
M = 98.49 lb # in
The normal stress developed is the combination of axial and bending stress. Thus, s =
M(R - r) N + A Ae r
Here, M = 98.49 lb # in since it tends to reduce the curvature of the hook. For point A, r = 1.5 in. Then s =
(98.49)(1.74103 - 1.5) 56.57 + 0.0625p 0.0625p(0.0089746)(1.5)
= 9.269(103) psi = 9.27 ksi (T)
Ans.
The shear stress in contributed by the transverse shear stress only. Thus t =
VQA = 0 It
Ans.
The state of strees of point A can be represented by the element shown in Fig. d.
604
A B
B a
A R = dA © LA r
©
a A
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•8–73.
The hook is subjected to the force of 80 lb. Determine the state of stress at point B at section a–a. The cross section has a diameter of 0.5 in. Use the curved-beam formula to compute the bending stress.
80 lb 1.5 in. 45⬚
The location of the neutral surface from the center of curvature of the the hook, Fig. a, can be determined from R =
dA © LA r
dA = 2p A r - 2r2 - c2 B = 2p A 1.75 - 21.752 - 0.252 B = 0.11278 in. LA r
Thus, R =
0.0625p = 1.74103 in 0.11278
Then e = r - R = 1.75 - 1.74103 = 0.0089746 in Referring to Fig. b, I and QB are computed as p (0.254) = 0.9765625(10 - 3)p in4 4
I =
QB = y¿A¿ =
4(0.25) p c (0.252) d = 0.0104167 in3 3p 2
Consider the equilibrium of the FBD of the hook’s cut segment, Fig. c, + ©F = 0; ; x
N - 80 cos 45° = 0
+ c ©Fy = 0;
80 sin 45° - V = 0
a + ©Mo = 0;
N = 56.57 lb V = 56.57 lb
M - 80 cos 45° (1.74103) = 0
M = 98.49 lb # in
The normal stress developed is the combination of axial and bending stress. Thus, s =
M(R - r) N + A Ae r
Here, M = 98.49 lb # in since it tends to reduce. the curvature of the hook. For point B, r = 1.75 in. Then s =
(98.49)(1.74103 - 1.75) 56.57 + 0.0625p 0.0625 p (0.0089746)(1.75)
= 1.62 psi (T)
Ans.
The shear stress is contributed by the transverse shear stress only. Thus, t =
56.57 (0.0104167) VQB = 3.84 psi = It 0.9765625(10 - 3)p (0.5)
Ans.
The state of stress of point B can be represented by the element shown in Fig. d.
605
A B
B a
A
Where A = p(0.252) = 0.0625p in2 ©
a A
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8–74. The block is subjected to the three axial loads shown. Determine the normal stress developed at points A and B. Neglect the weight of the block.
100 lb 250 lb
50 lb
2 in.4 in.
5 in.
2 in.
3 in. 5 in.
A B
Mx = -250(1.5) - 100(1.5) + 50(6.5) = -200 lb # in. My = 250(4) + 50(2) - 100(4) = 700 lb # in. Ix =
1 1 (4)(133) + 2 a b (2)(33) = 741.33 in4 12 12
Iy =
1 1 (3)(83) + 2 a b (5)(43) = 181.33 in4 12 12
A = 4(13) + 2(2)(3) = 64 in2 s =
My x Mx y P + A Iy Ix
sA = -
700(4) -200 (-1.5) 400 + 64 181.33 741.33
= -21.3 psi sB = -
Ans.
700(2) -200 (-6.5) 400 + 64 181.33 741.33
= -12.2 psi
Ans.
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8–75. The 20-kg drum is suspended from the hook mounted on the wooden frame. Determine the state of stress at point E on the cross section of the frame at section a–a. Indicate the results on an element.
50 mm
25 mm
E 75 mm
Section a – a 0.5 m 0.5 m
1m
a B C
a 1m 30⬚
Support Reactions: Referring to the free-body diagram of member BC shown in Fig. a, a + ©MB = 0;
F sin 45°(1) - 20(9.81)(2) = 0
+ ©F = 0; : x
554.94 cos 45° - Bx = 0
Bx = 392.4 N
+ c ©Fy = 0;
554.94 sin 45° - 20(9.81) - By = 0
By = 196.2 N
1m b
F = 554.94 N
b
75 mm
1m D
F A 25 mm
Internal Loadings: Consider the equilibrium of the free - body diagram of the right segment shown in Fig. b.
Section b – b
+ ©F = 0; : x
N - 392.4 = 0
N = 392.4 N
+ c ©Fy = 0;
V - 196.2 = 0
V = 196.2 N
a + ©MC = 0;
196.2(0.5) - M = 0
M = 98.1 N # m
Section Properties: The cross -sectional area and the moment of inertia of the cross section are A = 0.05(0.075) = 3.75 A 10 - 3 B m2 I =
1 (0.05) A 0.0753 B = 1.7578 A 10 - 6 B m4 12
Referring to Fig. c, QE is
QE = y¿A¿ = 0.025(0.025)(0.05) = 3.125 A 10 - 6 B m3
Normal Stress: The normal stress is the combination of axial and bending stress. Thus, s =
My N ; A I
For point A, y = 0.0375 - 0.025 = 0.0125 m. Then sE =
392.4
3.75 A 10
-3
B
98.1(0.0125)
+
1.7578 A 10 - 6 B
= 802 kPa
Ans.
Shear Stress: The shear stress is contributed by transverse shear stress only. Thus, tE =
196.2 C 31.25 A 10 - 6 B D VQA = = 69.8 kPa It 1.7578 A 10 - 6 B (0.05)
Ans.
The state of stress at point E is represented on the element shown in Fig. d. 607
75 mm
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8–75.
Continued
608
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*8–76. The 20-kg drum is suspended from the hook mounted on the wooden frame. Determine the state of stress at point F on the cross section of the frame at section b–b. Indicate the results on an element.
50 mm
25 mm
E 75 mm
Section a – a 0.5 m 0.5 m
1m
a B C
a 1m 30⬚
1m b
FBD sin 30°(3) - 20(9.81)(2) = 0
+ c ©Fy = 0;
Ay - 261.6 cos 30° - 20(9.81) = 0
Ay = 422.75 N
+ ©F = 0; : x
Ax - 261.6 sin 30° = 0
Ax = 130.8 N
75 mm
1m
Support Reactions: Referring to the free-body diagram of the entire frame shown in Fig. a, a + ©MA = 0;
b
D
F A
FBD = 261.6 N 25 mm Section b – b
Internal Loadings: Consider the equilibrium of the free - body diagram of the lower cut segment, Fig. b, + ©F = 0; : x
130.8 - V = 0
V = 130.8 N
+ c ©Fy = 0;
422.75 - N = 0
N = 422.75 N
a + ©MC = 0;
130.8(1) - M = 0
M = 130.8 N # m
Section Properties: The cross -sectional area and the moment of inertia about the centroidal axis of the cross section are A = 0.075(0.075) = 5.625 A 10 - 3 B m2 I =
1 (0.075) A 0.0753 B = 2.6367 A 10 - 6 B m4 12
Referring to Fig. c, QE is
QF = y¿A¿ = 0.025(0.025)(0.075) = 46.875 A 10 - 6 B m3
Normal Stress: The normal stress is the combination of axial and bending stress. Thus, s =
My N ; A I
For point F, y = 0.0375 - 0.025 = 0.0125 m. Then sF =
-422.75
5.625 A 10
-3
B
130.8(0.0125) -
2.6367 A 10 - 6 B
= -695.24 kPa = 695 kPa (C)
Ans.
609
75 mm
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8–76.
Continued
Shear Stress: The shear stress is contributed by transverse shear stress only. Thus,
tA
130.8 c46.875 A 10 - 6 B d VQA = = = 31.0 kPa It 2.6367 A 10 - 6 B (0.075)
Ans.
The state of stress at point A is represented on the element shown in Fig. d.
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•8–77. The eye is subjected to the force of 50 lb. Determine the maximum tensile and compressive stresses at section a-a. The cross section is circular and has a diameter of 0.25 in. Use the curved-beam formula to compute the bending stress.
50 lb
0.25 in. 1.25 in. a
Section Properties: r = 1.25 +
0.25 = 1.375 in. 2
dA = 2p A r - 2r2 - c2 B LA r = 2p A 1.375 - 21.3752 - 0.1252 B = 0.035774 in. A = p A 0.1252 B = 0.049087 in2 R =
A dA 1A r
=
0.049087 = 1.372153 in. 0.035774
r - R = 1.375 - 1.372153 = 0.002847 in. Internal Force and Moment: As shown on FBD. The internal moment must be computed about the neutral axis. M = 68.608 lb # in is positive since it tends to increase the beam’s radius of curvature. Normal Stress: Applying the curved - beam formula, For tensile stress (st)max =
=
M(Rr1) N + A Ar1(r - R) 68.608(1.372153 - 1.25) 50.0 + 0.049087 0.049087(1.25)(0.002847)
= 48996 psi = 49.0 ksi (T)
Ans.
For compressive stress (sc)max =
=
M(R - r2) N + A Ar2(r - R) 68.608(1.372153 - 1.50) 50.0 + 0.049087 0.049087(1.50)(0.002847)
= -40826 psi = 40.8 ksi (C)
Ans.
611
a
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8–78. Solve Prob. 8–77 if the cross section is square, having dimensions of 0.25 in. by 0.25 in.
50 lb
0.25 in. 1.25 in. a
Section Properties: r = 1.25 +
0.25 = 1.375 in. 2
r2 dA 1.5 = bln = 0.25 ln = 0.45580 in. r1 1.25 LA r A = 0.25(0.25) = 0.0625 in2 R =
0.0625 A = = 1.371204 in. dA 0.045580 1A r
r - R = 1.375 - 1.371204 = 0.003796 in. Internal Force and Moment: As shown on FBD. The internal moment must be computed about the neutral axis. M = 68.560 lb # in. is positive since it tends to increase the beam’s radius of curvature. Normal Stress: Applying the curved -beam formula, For tensile stress (st)max =
=
M(R - r1) N + A Ar1(r - R) 68.560(1.371204 - 1.25) 50.0 + 0.0625 0.0625(1.25)(0.003796)
= 28818 psi = 28.8 ksi (T)
Ans.
For Compressive stress (sc)max =
=
M(R - r2) N + A Ar2 (r - R) 68.560(1.371204 - 1.5) 50.0 + 0.0625 0.0625(1.5)(0.003796)
= -24011 psi = 24.0 ksi (C)
Ans.
612
a
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8–79. If the cross section of the femur at section a–a can be approximated as a circular tube as shown, determine the maximum normal stress developed on the cross section at section a–a due to the load of 75 lb.
2 in.
75 lb
a
a
0.5 in. 1 in. Section a – a
M F
Internal Loadings: Considering the equilibrium for the free-body diagram of the femur’s upper segment, Fig. a, + c ©Fy = 0;
N - 75 = 0
N = 75 lb
a + ©MO = 0;
M - 75(2) = 0
M = 150 lb # in
Section Properties: The cross-sectional area, the moment of inertia about the centroidal axis of the femur’s cross section are A = p A 12 - 0.52 B = 0.75p in2 I =
p 4 A 1 - 0.54 B = 0.234375p in4 4
Normal Stress: The normal stress is a combination of axial and bending stress. Thus, s =
My N + A I
By inspection, the maximum normal stress is in compression. smax =
150(1) -75 = -236 psi = 236 psi (C) 0.75p 0.234375p
613
Ans.
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*8–80. The hydraulic cylinder is required to support a force of P = 100 kN. If the cylinder has an inner diameter of 100 mm and is made from a material having an allowable normal stress of sallow = 150 MPa, determine the required minimum thickness t of the wall of the cylinder.
P
t
100 mm
Equation of Equilibrium: The absolute pressure developed in the hydraulic cylinder can be determined by considering the equilibrium of the free-body diagram of the piston shown in Fig. a. The resultant force of the pressure on the p piston is F = pA = pc A 0.12 B d = 0.0025pp. Thus, 4 ©Fx¿ = 0; 0.0025pp - 100 A 103 B = 0
p = 12.732 A 106 B Pa
Normal Stress: For the cylinder, the hoop stress is twice as large as the longitudinal stress, sallow =
pr ; t
150 A 106 B =
12.732 A 106 B (50) t
t = 4.24 mm Since
Ans.
r 50 = = 11.78 7 10, thin -wall analysis is valid. t 4.24
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•8–81. The hydraulic cylinder has an inner diameter of 100 mm and wall thickness of t = 4 mm. If it is made from a material having an allowable normal stress of sallow = 150 MPa, determine the maximum allowable force P.
P
t
100 mm
Normal Stress: For the hydraulic cylinder, the hoop stress is twice as large as the longitudinal stress. Since
50 r = = 12.5 7 10, thin-wall analysis can be used. t 4
sallow =
pr ; t
150 A 106 B =
p(50) 4
p = 12 A 106 B MPa
Ans.
Equation of Equilibrium: The resultant force on the piston is F = pA = 12 A 106 B c
p A 0.12 B d = 30 A 103 B p. Referring to the free-body diagram of 4 the piston shown in Fig. a, ©Fx¿ = 0; 30 A 103 B p - P = 0
P = 94.247 A 103 B N = 94.2 kN
Ans.
615
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8–82. The screw of the clamp exerts a compressive force of 500 lb on the wood blocks. Determine the maximum normal stress developed along section a-a. The cross section there is rectangular, 0.75 in. by 0.50 in. 4 in.
Internal Force and Moment: As shown on FBD.
a
Section Properties: A = 0.5(0.75) = 0.375 in2 I =
a
1 (0.5) A 0.753 B = 0.017578 in4 12
0.75 in.
Maximum Normal Stress: Maximum normal stress occurs at point A. smax = sA =
=
Mc N + A I 2000(0.375) 500 + 0.375 0.017578
= 44000 psi = 44.0 ksi (T)
Ans.
8–83. Air pressure in the cylinder is increased by exerting forces P = 2 kN on the two pistons, each having a radius of 45 mm. If the cylinder has a wall thickness of 2 mm, determine the state of stress in the wall of the cylinder.
p =
s1 =
P
47 mm
2(103) P = 314 380.13 Pa = A p(0.0452) pr 314 380.13(0.045) = = 7.07 MPa t 0.002
Ans.
s2 = 0
P
Ans.
The pressure P is supported by the surface of the pistons in the longitudinal direction.
*8–84. Determine the maximum force P that can be exerted on each of the two pistons so that the circumferential stress component in the cylinder does not exceed 3 MPa. Each piston has a radius of 45 mm and the cylinder has a wall thickness of 2 mm.
s =
pr ; t
3(106) =
P
47 mm
p(0.045) 0.002
P = 133.3 kPa
Ans.
P = pA = 133.3 A 103 B (p)(0.045)2 = 848 N
Ans.
616
P
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•8–85.
The cap on the cylindrical tank is bolted to the tank along the flanges. The tank has an inner diameter of 1.5 m and a wall thickness of 18 mm. If the largest normal stress is not to exceed 150 MPa, determine the maximum pressure the tank can sustain. Also, compute the number of bolts required to attach the cap to the tank if each bolt has a diameter of 20 mm. The allowable stress for the bolts is 1sallow2b = 180 MPa.
Hoop Stress for Cylindrical Tank: Since
750 r = = 41.7 7 10, then thin wall t 18
analysis can be used. Applying Eq. 8–1
s1 = sallow = 150 A 106 B =
pr t p(750) 18
p = 3.60 MPa
Ans.
Force Equilibrium for the Cap: + c ©Fy = 0;
3.60 A 106 B C p A 0.752 B D - Fb = 0 Fb = 6.3617 A 106 B N
Allowable Normal Stress for Bolts: (sallow)b = 180 A 106 B =
P A 6.3617(106)
n C p4 (0.022) D
n = 112.5 Use n = 113 bolts
Ans.
617
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8–86. The cap on the cylindrical tank is bolted to the tank along the flanges. The tank has an inner diameter of 1.5 m and a wall thickness of 18 mm. If the pressure in the tank is p = 1.20 MPa, determine the force in each of the 16 bolts that are used to attach the cap to the tank. Also, specify the state of stress in the wall of the tank.
Hoop Stress for Cylindrical Tank: Since
750 r = = 41.7 7 10, then thin wall t 18
analysis can be used. Applying Eq. 8–1
s1 =
pr 1.20(106)(750) = = 50.0 MPa t 18
Ans.
Longitudinal Stress for Cylindrical Tank: s2 =
pr 1.20(106)(750) = = 25.0 MPa 2t 2(18)
Ans.
Force Equilibrium for the Cap: + c ©Fy = 0;
1.20 A 106 B C p A 0.752 B D - 16Fb = 0 Fb = 132536 N = 133 kN
Ans.
618
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9–1. Prove that the sum of the normal stresses sx + sy = sx¿ + sy¿ is constant. See Figs. 9–2a and 9–2b.
Stress Transformation Equations: Applying Eqs. 9-1 and 9-3 of the text. sx¿ + sy¿ =
sx + sy 2
sx - sy +
2
cos 2u + txy sin 2u
sx + sy +
2
sx - sy -
2
cos 2u - txy sin 2u
sx¿ + sy¿ = sx + sy
(Q.E.D.)
619
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9–2. The state of stress at a point in a member is shown on the element. Determine the stress components acting on the inclined plane AB. Solve the problem using the method of equilibrium described in Sec. 9.1.
A
8 ksi
2 ksi 5 ksi 60⬚ B
Referring to Fig a, if we assume that the areas of the inclined plane AB is ¢A, then the area of the horizontal and vertical of the triangular element are ¢A cos 60° and ¢A sin 60° respectively. The forces act acting on these two faces indicated on the FBD of the triangular element, Fig. b. ¢Fx¿ + 2¢A sin 60° cos 60° + 5¢ A sin 60° sin 60°
+Q©Fx¿ = 0;
+ 2¢A cos 60° sin 60° - 8¢A cos 60° cos 60° = 0 ¢Fx¿ = -3.482 ¢A ¢Fy¿ + 2¢A sin 60° sin 60° - 5¢ A sin 60° cos 60°
+a©Fy¿ = 0;
- 8¢A cos 60° sin 60° - 2¢A cos 60° cos 60° = 0 ¢Fy¿ = 4.629 ¢A From the definition, sx¿ = lim¢A:0
¢Fx¿ = -3.48 ksi ¢A
tx¿y¿ = lim¢A:0
¢Fy¿
Ans.
Ans.
= 4.63 ksi
¢A
The negative sign indicates that sx¿, is a compressive stress.
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9–3. The state of stress at a point in a member is shown on the element. Determine the stress components acting on the inclined plane AB. Solve the problem using the method of equilibrium described in Sec. 9.1.
500 psi B 60⬚
A
Referring to Fig. a, if we assume that the area of the inclined plane AB is ¢A, then the areas of the horizontal and vertical surfaces of the triangular element are ¢A sin 60° and ¢A cos 60° respectively. The force acting on these two faces are indicated on the FBD of the triangular element, Fig. b +R©Fx¿ = 0;
¢Fx¿ + 500 ¢A sin 60° sin 60° + 350¢A sin 60° cos 60° +350¢A cos 60° sin 60° = 0 ¢Fx¿ = -678.11 ¢A
+Q©Fy¿ = 0;
¢Fy¿ + 350¢A sin 60° sin 60° - 500¢A sin 60° cos 60° -350¢A cos 60° cos 60° = 0 ¢Fy¿ = 41.51 ¢A
From the definition sx¿ = lim¢A:0
tx¿y¿ = lim¢A:0
¢Fx¿ = -6.78 psi ¢A ¢Fy¿
Ans.
Ans.
= 41.5 psi
¢A
The negative sign indicates that sx¿, is a compressive stress.
621
350 psi
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*9–4. The state of stress at a point in a member is shown on the element. Determine the stress components acting on the inclined plane AB. Solve the problem using the method of equilibrium described in Sec. 9.1.
A
650 psi
¢Fx¿ - 400(¢Acos 60°)cos 60° + 650(¢ A sin 60°)cos 30° = 0
Q+ ©Fx¿ = 0
400 psi
60⬚
¢Fx¿ = -387.5¢A ¢Fy¿ - 650(¢Asin 60°)sin 30° - 400(¢ A cos 60°)sin 60° = 0
a+ ©Fy¿ = 0
B
¢Fy¿ = 455 ¢A sx¿ = lim¢A:0
sx¿y¿ = lim¢A:0
¢Fx¿ = -388 psi ¢A ¢Fy¿
Ans.
Ans.
= 455 psi
¢A
The negative sign indicates that the sense of sx¿, is opposite to that shown on FBD.
•9–5.
Solve Prob. 9–4 using the stress-transformation equations developed in Sec. 9.2. sy = 400 psi
sx = -650 psi sx¿ =
=
sx + sy
sx - sy +
2
2
txy = 0
A
400 psi
u = 30° 650 psi
cos 2u + txy sin 2u
60⬚
-650 + 400 -650 - 400 + cos 60° + 0 = -388 psi 2 2
Ans. B
The negative sign indicates sx¿, is a compressive stress. tx¿y¿ = = -a
sx - sy 2
sin 2u + txy cos 2u
-650 - 400 bsin 60° = 455 psi 2
Ans.
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9–6. The state of stress at a point in a member is shown on the element. Determine the stress components acting on the inclined plane AB. Solve the problem using the method of equilibrium described in Sec. 9.1.
90 MPa
A
35 MPa 60⬚ 30⬚
R+ ©Fy¿ = 0
B 50 MPa
¢Fy¿ - 50¢A sin 30° cos 30° - 35¢A sin 30° cos 60° + 90¢A cos 30° sin 30° + 35¢A cos 30° sin 60° = 0 ¢Fy¿ = -34.82¢A
b+ ©Fx¿ = 0
¢Fx¿ - 50¢A sin 30° sin 30° + 35¢A sin 30° sin 60° -90¢A cos 30° cos 30° + 35¢A cos 30° cos 60° = 0 ¢Fx¿ = 49.69 ¢A
sx¿ = lim¢A:0
¢Fx¿ = 49.7 MPa ¢A
tx¿y¿ = lim¢A:0
¢Fy¿
Ans.
Ans.
= -34.8 MPa
¢A
The negative signs indicate that the sense of sx¿, and tx¿y¿ are opposite to the shown on FBD.
9–7. Solve Prob. 9–6 using the stress-transformation equations developed in Sec. 9.2. Show the result on a sketch.
90 MPa
A
35 MPa 60⬚ 30⬚
sy = 50 MPa
sx = 90 MPa sx¿ =
=
sx + sy
sx - sy +
2
2
txy = -35 MPa
u = -150°
cos 2u + txy sin 2u
90 - 50 90 + 50 + cos(-300°) + (-35) sin ( -300°) 2 2
= 49.7 MPa tx¿y¿ = -
sx - sy
= -a
2
Ans. sin 2u + txy cos 2u
90 - 50 bsin(-300°) + ( -35) cos ( -300°) = -34.8 MPa 2
The negative sign indicates tx¿y¿ acts in -y¿ direction.
623
Ans.
B 50 MPa
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*9–8. Determine the normal stress and shear stress acting on the inclined plane AB. Solve the problem using the method of equilibrium described in Sec. 9.1.
45 MPa
B
80 MPa 45⬚ A
Force Equllibrium: Referring to Fig. a, if we assume that the area of the inclined plane AB is ¢A, then the area of the vertical and horizontal faces of the triangular sectioned element are ¢A sin 45° and ¢A cos 45°, respectively. The forces acting on the free-body diagram of the triangular sectioned element, Fig. b, are ©Fx¿ = 0;
¢Fx¿ + c45 A 106 B ¢A sin 45° dcos 45° + c45 A 106 B ¢A cos 45° dsin 45° - c80 A 106 B ¢A sin 45° dcos 45° = 0 ¢Fx¿ = -5 A 106 B ¢A
©Fy¿ = 0;
¢Fy¿ + c45 A 106 B ¢A cos 45° dcos 45° - c45 A 106 B ¢A sin 45° dsin 45° - c80 A 106 B ¢ A sin 45° dsin 45° = 0 ¢Fy¿ = 40 A 106 B ¢A
Normal and Shear Stress: From the definition of normal and shear stress, sx¿ = lim¢A:0
¢Fx¿ = -5 MPa ¢A
tx¿y¿ = lim¢A:0
¢Fy¿
Ans.
Ans.
= 40 MPa
¢A
The negative sign indicates that sx¿ is a compressive stress.
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•9–9.
Determine the normal stress and shear stress acting on the inclined plane AB. Solve the problem using the stress transformation equations. Show the result on the sectioned element.
45 MPa
80 MPa 45⬚
Stress Transformation Equations: u = +135° (Fig. a)
sx = 80 MPa
sy = 0
txy = 45 MPa
we obtain, sx¿ =
=
sx + sy
sx - sy +
2
2
cos u + txysin 2u
80 - 0 80 + 0 + cos 270 + 45 sin 270° 2 2 Ans.
= -5 MPa
tx¿y¿ = -
= -
sx - sy 2
B
sinu + txy cos 2u
80 - 0 sin 270° + 45 cos 270° 2
= 40 MPa
Ans.
The negative sign indicates that sx¿ is a compressive stress. These results are indicated on the triangular element shown in Fig. b.
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9–10. The state of stress at a point in a member is shown on the element. Determine the stress components acting on the inclined plane AB. Solve the problem using the method of equilibrium described in Sec. 9.1.
2 ksi
A 3 ksi 30⬚
4 ksi B
Force Equllibrium: For the sectioned element, ¢Fy¿ - 3(¢A sin 30°) sin 60° + 4(¢ A sin 30°)sin 30°
a+ ©Fy¿ = 0;
-2(¢A cos 30°) sin 30° - 4(¢A cos 30°) sin 60° = 0 ¢Fy¿ = 4.165 ¢A ¢Fx¿ + 3(¢A sin 30°) cos 60° + 4(¢ A sin 30°)cos 30°
Q+ ©Fx¿ = 0;
-2(¢A cos 30°) cos 30° + 4(¢A cos 30°) cos 60° = 0 ¢Fx¿ = -2.714 ¢A Normal and Shear Stress: For the inclined plane. sx = lim¢A:0
tx¿y¿ = lim¢A:0
¢Fx¿ = -2.71 ksi ¢A ¢Fy¿
Ans.
Ans.
= 4.17 ksi
¢A
Negative sign indicates that the sense of sx¿, is opposite to that shown on FBD.
9–11. Solve Prob. 9–10 using the stress-transformation equations developed in Sec. 9.2. Show the result on a sketch.
2 ksi
Normal and Shear Stress: In accordance with the established sign convention, u = +60°
sx = -3 ksi
sy = 2 ksi
A
txy = -4 ksi
3 ksi 30⬚
Stress Transformation Equations: Applying Eqs. 9-1 and 9-2. sx¿ =
=
sx + sy
sx - sy +
2
2
B
cos 2u + txy sin 2u
-3 - 2 -3 + 2 + cos 120° + (-4 sin 120°) 2 2 Ans.
= -2.71 ksi tx¿y¿ = -
= -
sx - sy 2
4 ksi
sin 2u + txy cos 2u
-3 - 2 sin 120° + (-4 cos 120°) 2
= 4.17 ksi
Ans.
Negative sign indicates sx¿, is a compressive stress
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*9–12. Determine the equivalent state of stress on an element if it is oriented 50° counterclockwise from the element shown. Use the stress-transformation equations.
10 ksi
16 ksi
sy = 0
sx = -10 ksi
txy = -16 ksi
u = +50° sx¿ =
=
sx + sy
= -a
=
2
cos 2u + txy sin 2u
-10 - 0 -10 + 0 + cos 100° + ( -16)sin 100° = -19.9 ksi 2 2
tx¿y¿ = - a
sy¿ =
sx - sy +
2
sx - sy 2
b sin 2u + txy cos 2u
-10 - 0 b sin 100° + (-16)cos 100° = 7.70 ksi 2
sx + sy 2
sx - sy -
Ans.
2
Ans.
cos 2u - txy sin 2u
-10 + 0 -10 - 0 - a bcos 100° - (-16)sin 100° = 9.89 ksi 2 2
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Ans.
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•9–13.
Determine the equivalent state of stress on an element if the element is oriented 60° clockwise from the element shown. Show the result on a sketch.
350 psi
75 psi 200 psi
In accordance to the established sign covention, u = -60° (Fig. a)
sx = 200 psi
sy = -350 psi
txy = 75 psi
Applying Eqs 9-1, 9-2 and 9-3, sx¿ =
=
sx + sy
sx - sy +
2
2
cos 2u + txy sin 2u
200 - ( -350) 200 + (-350) + cos ( -120°) + 75 sin (-120°) 2 2
= -277.45 psi = -277 psi sy¿ =
=
sx + sy
sx - sy -
2
2
Ans.
cos 2u - txy sin 2u
200 - ( -350) 200 + (-350) cos ( -120°) - 75 sin ( -120°) 2 2
= 127.45 psi = 127 psi tx¿y¿ = -
= -
sx - sy 2
Ans.
sin 2u + txy cos 2u
200 - (-350) sin (-120°) + 75 cos (-120°) 2
= 200.66 psi = 201 psi
Ans.
Negative sign indicates that sx¿ is a compressive stress. These result, can be represented by the element shown in Fig. b.
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9–14. The state of stress at a point is shown on the element. Determine (a) the principal stress and (b) the maximum in-plane shear stress and average normal stress at the point. Specify the orientation of the element in each case. Show the results on each element.
30 ksi
12 ksi
sx = -30 ksi
sy = 0
txy = -12 ksi
a) sx + sy
s1, 2 =
;
2
C
a
sx - sy 2
2
b + txy 2 =
-30 + 0 -30 - 0 2 ; a b + (-12)2 2 C 2
s1 = 4.21 ksi
Ans.
s2 = -34.2 ksi
Ans.
Orientation of principal stress: txy
tan 2uP =
(sx - sy)>2
uP = 19.33° and
-12 = 0.8 (-30 -0)>2
=
-70.67°
Use Eq. 9-1 to determine the principal plane of s1 and s2. sx + sy
sx¿ =
sx - sy +
2
2
cos 2u + txy sin 2u
u = 19.33° sx¿ =
-30 + 0 -30 - 0 + cos 2(19.33°) + (-12)sin 2(19.33°) = -34.2 ksi 2 2
Therefore uP2 = 19.3°
Ans.
and uP1 = -70.7°
Ans.
b) tmaxin-plane = savg =
C
a
sx - sy 2
sx + sy 2
=
2
b + txy 2 =
-30 - 0 2 b + (-12)2 = 19.2 ksi C 2 a
-30 + 0 = -15 ksi 2
Ans.
Ans.
Orientation of max, in - plane shear stress: tan 2uP =
-(sx - sy)>2 =
txy
uP = -25.2°
and
-(-30 - 0)>2 = -1.25 -12 64.3°
Ans.
By observation, in order to preserve equllibrium along AB, tmax has to act in the direction shown in the figure.
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9–15. The state of stress at a point is shown on the element. Determine (a) the principal stress and (b) the maximum in-plane shear stress and average normal stress at the point. Specify the orientation of the element in each case. Show the results on each element.
80 MPa
50 MPa
60 MPa
In accordance to the established sign convention, sx = -60 MPa s1, 2 =
=
sy = -80 MPa
sx + sy ;
2
C
a
sx - sy 2
txy = 50 MPa
2
b + txy 2
-60 + (-80) -60 - (-80) 2 ; c d + 502 2 C 2
= -70 ; 22600 s2 = -121 MPa
s1 = -19.0 MPa txy
tan 2uP =
=
(sx - sy)>2
uP = 39.34°
Ans.
50 = 5 [-60 - (-80)]>2
and
-50.65°
Substitute u = 39.34° into Eq. 9-1, sx¿ =
=
sx + sy
sx - sy +
2
2
cos 2u + txy sin 2u
-60 + (-80) -60 - ( -80) + cos 78.69° + 50 sin 78.69° 2 2
= -19.0 MPa = s1 Thus, (uP)1 = 39.3°
Ans.
(uP)2 = -50.7°
The element that represents the state of principal stress is shown in Fig. a. t max
in-plane
=
C
a
sx - sy 2
tan 2uS =
2
b + txy 2 =
-(sx - sy)>2 =
txy
-60 - (-80) 2 d + 502 = 51.0 MPa C 2 c
-[-60 - (-80)]>2 = -0.2 50
uS = -5.65° and 84.3° By Inspection, t max
Ans.
Ans.
has to act in the sense shown in Fig. b to maintain
in-plane
equilibrium.
savg =
sx + sy 2
=
-60 + (-80) = -70 MPa 2
The element that represents the state of maximum in - plane shear stress is shown in Fig. c.
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9–15. Continued
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*9–16. The state of stress at a point is shown on the element. Determine (a) the principal stress and (b) the maximum in-plane shear stress and average normal stress at the point. Specify the orientation of the element in each case. Sketch the results on each element.
60 MPa
30 MPa
45 MPa
sx = 45 MPa
sy = -60 MPa
txy = 30 MPa
a) s1, 2 =
=
sx + sy ;
2
C
a
sx - sy 2
2
b + txy 2
45 - (-60) 2 45 - 60 a ; b + (30)2 2 C 2 s1 = 53.0 MPa
Ans.
s2 = -68.0 MPa
Ans.
Orientation of principal stress: tan 2uP =
txy (sx - sy)>2
uP = 14.87,
=
30 = 0.5714 (45 - (-60))>2
-75.13
Use Eq. 9-1 to determine the principal plane of s1 and s2: sx¿ =
=
sx + sy
sx - sy +
2
2
cos 2u + txy sin 2u,
where u = 14.87°
45 + (-60) 45 - (-60) + cos 29.74° + 30 sin 29.74° = 53.0 MPa 2 2
Therefore uP1 = 14.9°
Ans.
and uP2 = -75.1°
Ans.
b) tmaxin-plane = savg =
C
a
sx - sy 2
sx - sy 2
=
2
b + txy 2 =
45 - (-60) 2 b + 302 = 60.5 MPa C 2 a
45 + (-60) = -7.50 MPa 2
Ans.
Ans.
Orientation of maximum in - plane shear stress: tan 2uS =
-(sx - sy)>2 txy
=
-(45 - ( -60))>2 = -1.75 30
uS = -30.1°
Ans.
uS = 59.9°
Ans.
and
By observation, in order to preserve equilibrium along AB, tmax has to act in the direction shown.
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•9–17.
Determine the equivalent state of stress on an element at the same point which represents (a) the principal stress, and (b) the maximum in-plane shear stress and the associated average normal stress. Also, for each case, determine the corresponding orientation of the element with respect to the element shown. Sketch the results on each element.
75 MPa
125 MPa
50 MPa
Normal and Shear Stress: sx = 125 MPa
sy = -75 MPa
txy = -50 MPa
In - Plane Principal Stresses: s1,2 =
=
sx - sy ;
2
B
a
sx - sy 2
2
b + txy 2
125 + (-75) 125 - (-75) 2 a ; b + (-50)2 2 2 B
= 25; 212500 s2 = -86.8 MPa
s1 = 137 MPa
Ans.
Orientation of Principal Plane: tan 2uP =
txy
A sx - sy B >2
-50
=
A 125 -(-75) B >2
= -0.5
up = -13.28° and 76.72° Substitute u = -13.28° into sx¿ =
=
sx + sy
sx - sy +
2
2
cos 2u + txy sin 2u
125 + (-75) 125 - (-75) + cos(-26.57°)+(-50) sin(-26.57°) 2 2
= 137 MPa = s1 Thus,
A up B 1 = -13.3° and A up B 2 = 76.7°
Ans.
125 - (-75)>(-50) The element that represents the state of principal stress is shown in Fig. a. Maximum In - Plane Shear Stress: t max
in-plane
=
C
¢
sx - sy 2
2
≤ + txy 2 =
-100 - 0 2 b + 252 = 112 MPa 2 B a
Orientation of the Plane of Maximum In - Plane Shear Stress: tan 2us = -
A sx - sy B >2 txy
= -
A 125 - (-75) B >2
= 2
-50
us = 31.7° and 122°
633
Ans.
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9–17.
Continued
By inspection, t max
has to act in the same sense shown in Fig. b to maintain
in-plane
equilibrium. Average Normal Stress: savg =
sx + sy 2
=
125 + (-75) = 25 MPa 2
Ans.
The element that represents the state of maximum in - plane shear stress is shown in Fig. c.
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sy
9–18. A point on a thin plate is subjected to the two successive states of stress shown. Determine the resultant state of stress represented on the element oriented as shown on the right.
=
sx¿ + sy¿
sx¿ - sy¿ +
2
2
ⴙ
60⬚
Stress Transformation Equations: Applying Eqs. 9-1, 9-2, and 9-3 u = -30°, sx¿ = -200 MPa, to element (a) with sy¿ = -350 MPa and tx¿y¿ = 0. (sx)a =
350 MPa
cos 2u + tx¿y¿ sin 2u
-200 - (-350) -200 + (-350) + cos (-60°) + 0 2 2
= -237.5 MPa
A sy B a = =
sx¿ + sy¿
sx¿ - sy¿ -
2
2
cos 2u - tx¿y¿ sin 2u
-200 - (-350) -200 + (-350) cos (-60°) - 0 2 2
= -312.5 MPa
A txy B a = = -
sx¿ - sy¿ 2
sin 2u + tx¿y¿ cos 2u
-200 - (-350) sin (-60°) + 0 2
= 64.95 MPa For element (b), u = 25°, sx¿ = sy¿ = 0 and sx¿y¿ = 58 MPa. (sx)b =
sx¿ + sy¿
sx¿ - sy¿ +
2
2
cos 2u + tx¿y¿ sin 2u
= 0 + 0 + 58 sin 50° = 44.43 MPa
A sy B b =
sx¿ + sy¿
sx¿ - sy¿ -
2
2
cos 2u - tx¿y¿ sin 2u
= 0 - 0 - 58 sin 50° = -44.43 MPa
A txy B b = -
sx¿ - sy¿ 2
58 MPa
200 MPa
sin 2u + tx¿y¿ cos 2u
= -0 + 58 cos 50° = 37.28 MPa Combining the stress components of two elements yields ss = (sx)a + (sx)b = -237.5 + 44.43 = -193 MPa
Ans.
sy = A sy B a + A sy B b = -312.5 - 44.43 = -357 MPa
Ans.
txy = A txy B a + A txy B b = 64.95 + 37.28 = 102 MPa
Ans.
635
25⬚
ⴝ
txy sx
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9–19. The state of stress at a point is shown on the element. Determine (a) the principal stress and (b) the maximum in-plane shear stress and average normal stress at the point. Specify the orientation of the element in each case. Sketch the results on each element.
160 MPa
120 MPa
In accordance to the established sign Convention, sx = 0
sy = 160 MPa
s1, 2 =
=
sx + sy ;
2
B
a
txy = -120 MPa
sx - sy 2
2
b + t2xy
0 + 160 0 - 160 2 ; a b + (-120)2 2 2 B
= 80 ; 220800 s2 = -64.2 MPa
s1 = 224 MPa tan 2up =
txy (sx - sy)>2
up = 28.15°
=
Ans.
-120 = 1.5 (0 - 160)>2
and -61.85°
Substitute u = 28.15° into Eq. 9-1, sx¿ =
=
sx + sy
sx - sy +
2
2
cos 2u + txy sin 2u
0 + 160 0 - 160 + cos 56.31° + (-120) sin 56.31° 2 2
= -64.22 = s2 Thus, (up)1 = -61.8°
Ans.
(up)2 = 28.2°
The element that represents the state of principal stress is shown in Fig. a tmax
in-plane
=
B
a
sx - sy 2
tan 2us =
0 - 160 2 b + (-120)2 = 144 MPa 2 B a
-(sx - sy)>2
us = -16.8° By inspection, equilibrium.
2
b + t2xy =
tmax
in-plane
savg =
txy
Ans.
-(0 - 160)>2 = -0.6667 -120
=
Ans.
and 73.2°
has to act in the sense shown in Fig. b to maintain sx + sy 2
=
0 + 160 = 80 MPa 2
Ans.
The element that represents the state of Maximum in - plane shear stress is shown in Fig. (c)
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9–19. Continued
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*9–20. The stress acting on two planes at a point is indicated. Determine the normal stress sb and the principal stresses at the point.
a
4 ksi 60⬚ 45⬚
b
2 ksi sb a
Stress Transformation Equations: Applying Eqs. 9-2 and 9-1 with u = -135°, sy = 3.464 ksi, txy = 2.00 ksi, tx¿y¿ = -2 ksi, and sx¿ = sb¿., tx¿y¿ = -
-2 = -
sx - sy 2
sin 2u + txy cos 2u
sx - 3.464 sin (-270°) + 2cos ( -270°) 2
sx = 7.464 ksi sx¿ =
sy =
sx - sy
sx - sy +
2
2
cos 2u + txy sin 2u
7.464 - 3.464 7.464 + 3.464 + cos (-270°) + 2sin ( -270°) 2 2 Ans.
= 7.46 ksi In - Plane Principal Stress: Applying Eq. 9-5. s1, 2 =
=
sx + sy 2
;
B
a
sx - sy 2
2
b + t2xy
7.464 - 3.464 2 7.464 + 3.464 ; a b + 22 2 2 B
= 5.464 ; 2.828 s1 = 8.29 ksi
s2 = 2.64 ksi
Ans.
638
b
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•9–21. The stress acting on two planes at a point is indicated. Determine the shear stress on plane a–a and the principal stresses at the point.
b a
ta 45⬚
60 ksi 60⬚
txy = 60 cos 60° = 30 ksi sa =
80 =
sx + sy
sx - sy +
2
2
51.962 - sy
51.962 + sy +
2
cos 2u + txy sin 2u
2
cos (90°) + 30 sin (90°)
sy = 48.038 ksi ta = - a = -a
sx - sy 2
b sin 2u + txy cos u
51.962 - 48.038 bsin (90°) + 30 cos (90°) 2
ta = -1.96 ksi s1, 2 =
=
Ans.
sx + sy 2
;
C
a
sx - sy 2
2
b + t2xy
51.962 - 48.038 2 51.962 + 48.038 ; a b + (30)2 2 C 2
s1 = 80.1 ksi
Ans.
s2 = 19.9 ksi
Ans.
639
90⬚
a b
sx = 60 sin 60° = 51.962 ksi
80 ksi
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9–22. The T-beam is subjected to the distributed loading that is applied along its centerline. Determine the principal stress at point A and show the results on an element located at this point.
100 kN/m
A 1m
0.5 m 200 mm
75 mm
' ©yA 0.1(0.2)(0.02) + 0.21(0.02)(0.2) = = 0.155 m ©A 0.2(0.02) + 0.02(0.2) 1 (0.02)(0.2 3) + 0.02(0.2)(0.155 - 0.1)2 12
I =
+
1 (0.2)(0.023) + 0.2(0.02)(0.21 - 0.155)2 12
= 37.6667(10 - 6) m4 Referring to Fig. b, QA = y¿A¿ = 0.1175(0.075)(0.02) = 0.17625(10 - 3) m3 Using the method of sections and considering the FBD of the left cut segment of the beam, Fig. c, + c ©Fy = 0;
V - 100(1) = 0
a + ©MC = 0;
100(1)(0.5) - M = 0 M = 50 kN # m
V = 100 kN
The normal stress developed is contributed by bending stress only. For point A, y = 0.155 - 0.075 = 0.08 m. Thus s =
My 50(103) (0.08) = 106 MPa = I 37.6667(10 - 6)
The shear stress is contributed by the transverse shear stress only. Thus, t =
100(103)[0.17625(10 - 3)] VQA = 23.40(106)Pa = 23.40 MPa = It 37.6667(10 - 6) (0.02)
The state of stress of point A can be represented by the element shown in Fig. c. Here, sx = -106.19 MPa, sy = 0 and txy = 23.40 MPa. s1, 2 =
=
sx + sy 2
;
B
a
sx - sy 2
2
b + txy 2
-106.19 - 0 2 -106.19 + 0 ; b + 23.402 a 2 2 B
= -53.10 ; 58.02 s1 = 4.93 MPa
20 mm 200 mm
20 mm
The location of the centroid c of the T cross-section, Fig. a, is y =
A
s2 = -111 MPa
Ans.
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9–22.
Continued
tan 2up =
txy (sx - sy)>2
up = -11.89°
= ans
23.40 = -0.4406 ( -106.19 - 0)>2 78.11°
Substitute u = -11.89°, sx¿ =
=
sx + sy 2
sx - sy +
2
cos 2u + txy sin 2u
-106.19 + 0 -106.19 - 0 + cos (-23.78°) + 23.40 5m (-23.78°) 2 2
= -111.12 MPa = s2 Thus, (up)1 = 78.1°
Ans.
(up)2 = -11.9°
The state of principal stress can be represented by the element shown in Fig. e.
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•9–23.
The wood beam is subjected to a load of 12 kN. If a grain of wood in the beam at point A makes an angle of 25° with the horizontal as shown, determine the normal and shear stress that act perpendicular and parallel to the grain due to the loading.
I =
12 kN 1m
2m A 25⬚
300 mm 75 mm
1 (0.2)(0.3)3 = 0.45(10 - 3) m4 12
QA = yA¿ = 0.1125(0.2)(0.075) = 1.6875(10 - 3) m3 sA =
MyA 13.714(103)(0.075) = 2.2857 MPa (T) = I 0.45(10 - 3)
tA =
6.875(103)(1.6875)(10 - 3) VQA = 0.1286 MPa = It 0.45(10 - 3)(0.2)
sx = 2.2857 MPa sx¿ =
sx¿ =
sx + sy
sx - sy +
2
sy = 0
2
txy = -0.1286 MPa
u = 115°
cos 2u + txy sin 2u
2.2857 - 0 2.2857 + 0 + cos 230° + (-0.1286)sin 230° 2 2 Ans.
= 0.507 MPa tx¿y¿ = -
sx - sy
= -a
2
sin 2u + txy cos 2u
2.2857 - 0 b sin 230° + (-0.1286)cos 230° 2
= 0.958 MPa
Ans.
642
4m
200 mm
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*9–24. The wood beam is subjected to a load of 12 kN. Determine the principal stress at point A and specify the orientation of the element.
12 kN 1m
2m A 25⬚
I =
300 mm 75 mm
1 (0.2)(0.3)3 = 0.45(10 - 3) m4 12
QA = yA¿ = 0.1125(0.2)(0.075) = 1.6875(10 - 3) m3 sA =
13.714(103)(0.075) MyA = 2.2857 MPa (T) = I 0.45(10 - 3)
tA =
VQA 6.875(103)(1.6875)(10 - 3) = 0.1286 MPa = It 0.45(10 - 3)(0.2)
sx = 2.2857 MPa s1, 2 =
=
sy = 0
sx + sy ;
2
C
a
txy = -0.1286 MPa
sx - sy 2
2
b + t2xy
2.2857 - 0 2 2.2857 + 0 ; a b + (-0.1286)2 2 C 2
s1 = 2.29 MPa
Ans.
s2 = -7.20 kPa
Ans.
tan 2up =
txy (sx - sy)>2
=
-0.1286 (2.2857 - 0)>2
up = -3.21° Check direction of principal stress: sx¿ =
=
sx + sy 2
sx - sy +
2
cos 2u + txy sin 2u
2.2857 + 0 2.2857 - 0 + cos (-6.42°) - 0.1285 sin (-6.42) 2 2
= 2.29 MPa
643
4m
200 mm
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•9–25.
The bent rod has a diameter of 20 mm and is subjected to the force of 400 N. Determine the principal stress and the maximum in-plane shear stress that is developed at point A. Show the results on a properly oriented element located at this point.
100 mm
150 mm
400 N
400 N
250 mm
A
Using the method of sections and consider the FBD of the rod’s left cut segment, Fig. a. + ©F = 0; : x
N - 400 = 0 N = 400 N
a + ©MC = 0;
400(0.25) - M = 0 M = 100 N # m A = p(0.012) = 0.1(10 - 3) p m2 p (0.014) = 2.5(10 - 9)p m4 4
I =
The normal stress developed is the combination of axial and bending stress. Thus, My N ; A I
s = For point A, y = C = 0.01 m. s =
100(0.01) 400 -3 0.1(10 )p 2.5(10 - 9)p
= -126.05 (106)Pa = 126.05 MPa (C) Since no torque and transverse shear acting on the cross - section, t = 0 The state of stress at point A can be represented by the element shown in Fig. b Here, sx = -126.05 MPa, sy = 0 and txy = 0. Since no shear stress acting on the element s1 = sy = 0
s2 = sx = -126 MPa
Ans.
Thus, the state of principal stress can also be represented by the element shown in Fig. b. tmax
in-plane
=
B
a
sx - sy 2
tan 2us = -
2
b + t2xy =
(sx - sy)>2 = -
txy
us = 45° tx¿y¿ = -
-126.05 - 0 2 b + 02 = 63.0 MPa 2 B a
(-126.05 - 0)>2 = q 0
and -45°
sx - sy 2 = -
= 63.0 =
sin 2u + txy cos 2u
-126.05 - 0 sin 90° + 0 cos 90° 2 tmax
in-plane
644
Ans.
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9–25. Continued tmax This indicates that in-plane acts toward the positive sense of y¿ axis at the face of element defined by us = 45° savg =
sx + sy 2
=
-126.05 + 0 = -63.0 MPa 2
Ans.
The state of maximum In - plane shear stress can be represented by the element shown in Fig. c
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9–26. The bracket is subjected to the force of 3 kip. Determine the principal stress and maximum in-plane shear stress at point A on the cross section at section a–a. Specify the orientation of this state of stress and show the results on elements.
3 kip
3 kip
a 3 in.
a A
0.25 in.
2 in.
0.25 in.
Internal Loadings: Consider the equilibrium of the free - body diagram from the bracket’s left cut segment, Fig. a.
B 1 in.
+ ©F = 0; : x
N - 3 = 0
N = 3 kip Section a – a
M = 12 kip # in
©MO = 0; 3(4) - M = 0
Normal and Shear Stresses: The normal stress is the combination of axial and bending stress. Thus, s =
My N A I
The cross - sectional area and the moment of inertia about the z axis of the bracket’s cross section is A = 1(2) - 0.75(1.5) = 0.875 in2 I =
1 1 (1) A 23 B (0.75) A 1.53 B = 0.45573 in4 12 12
For point A, y = 1 in. Then sA =
(-12)(1) 3 = 29.76 ksi 0.875 0.45573
Since no shear force is acting on the section, tA = 0 The state of stress at point A can be represented on the element shown in Fig. b. In - Plane Principal Stress: sx = 29.76 ksi, sy = 0, and txy = 0. Since no shear stress acts on the element, s1 = sx = 29.8 ksi s2 = sy = 0
Ans.
The state of principal stresses can also be represented by the elements shown in Fig. b Maximum In - Plane Shear Stress: t max
in-plane
=
C
¢
sx - sy 2
2
≤ + txy 2 =
29.76 - 0 2 b + 02 = 14.9 ksi 2 B a
Ans.
Orientation of the Plane of Maximum In - Plane Shear Stress: tan 2us = -
A sx - sy B >2 txy
0.25 in.
= -
(29.76 - 0)>2 = -q 0
us = -45° and 45°
Ans.
646
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9–26.
Continued
Substituting u = -45° into tx¿y¿ = -
= -
sx - sy 2
sin 2u + txy cos 2u
29.76 - 0 sin(-90°) + 0 2
= 14.9 ksi = t max
in-plane
This indicates that t max
is directed in the positive sense of the y¿ axes on the ace
in-plane
of the element defined by us = -45°. Average Normal Stress: savg =
sx + sy 2
=
29.76 + 0 = 14.9 ksi 2
Ans.
The state of maximum in - plane shear stress is represented by the element shown in Fig. c.
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9–27. The bracket is subjected to the force of 3 kip. Determine the principal stress and maximum in-plane shear stress at point B on the cross section at section a–a. Specify the orientation of this state of stress and show the results on elements.
3 kip
3 kip
a 3 in.
a A
0.25 in.
2 in.
0.25 in.
Internal Loadings: Consider the equilibrium of the free - body diagram of the bracket’s left cut segment, Fig. a.
B 1 in.
+ ©F = 0; : x
N - 3 = 0
N = 3 kip Section a – a
M = 12 kip # in
©MO = 0; 3(4) - M = 0
Normal and Shear Stresses: The normal stress is the combination of axial and bending stress. Thus, s =
My N A I
The cross - sectional area and the moment of inertia about the z axis of the bracket’s cross section is A = 1(2) - 0.75(1.5) = 0.875 in2 I =
1 1 (1) A 23 B (0.75) A 1.53 B = 0.45573 in4 12 12
For point B, y = -1 in. Then sB =
(-12)(-1) 3 = -22.90 ksi 0.875 0.45573
Since no shear force is acting on the section, tB = 0 The state of stress at point A can be represented on the element shown in Fig. b. In - Plane Principal Stress: sx = -22.90 ksi, sy = 0, and txy = 0. Since no shear stress acts on the element, s1 = sy = 0
s2 = sx = -22.90 ksi
Ans.
The state of principal stresses can also be represented by the elements shown in Fig. b. Maximum In - Plane Shear Stress: t max
in-plane
=
C
¢
sx - sy 2
2
≤ + txy 2 =
-22.90 - 0 2 b + 02 = 11.5 ksi 2 B a
Ans.
Orientation of the Plane of Maximum In - Plane Shear Stress: tan 2us = -
A sx - sy B >2 txy
0.25 in.
= -
(-22.9 - 0)>2 = -q 0
us = 45° and 135°
Ans.
648
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9–27. Continued
Substituting u = 45° into tx¿y¿ = -
= -
sx - sy 2
sin 2u + txy cos 2u
-22.9 - 0 sin 90° + 0 2
= 11.5 ksi = t max
in-plane
is directed in the positive sense of the y¿ axes on the
This indicates that t max
in-plane
element defined by us = 45°. Average Normal Stress: savg =
sx + sy 2
=
-22.9 + 0 = -11.5 ksi 2
Ans.
The state of maximum in - plane shear stress is represented by the element shown in Fig. c.
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*9–28. The wide-flange beam is subjected to the loading shown. Determine the principal stress in the beam at point A and at point B. These points are located at the top and bottom of the web, respectively. Although it is not very accurate, use the shear formula to determine the shear stress.
8 kN/m A
B
1m
3m
B
Internal Forces and Moment: As shown on FBD(a).
200 mm
Section Properties: A = 0.2(0.22) - 0.19(0.2) = 6.00 A 10 - 3 B m2 1 1 (0.2) A 0.223 B (0.19) A 0.22 B = 50.8 A 10 - 6 B m4 12 12
I =
QA = QB = y¿A¿ = 0.105(0.01)(0.2) = 0.210 A 10 - 3 B m3 Normal Stress: s =
My N ; A I 21.65(103)
=
73.5(103)(0.1) ;
6.00(10 - 3)
50.8(10 - 6)
sA = 3.608 + 144.685 = 148.3 MPa sB = 3.608 - 144.685 = -141.1 MPa VQ . It
Shear Stress: Applying the shear formula t =
tA = tB =
36.5(103) C 0.210(10 - 3) D 50.8(10 - 6)(0.01)
= 15.09 MPa
In - Plane Principal Stress: sx = 148.3 MPa, sy = 0, and txy = -15.09 MPa for point A. Applying Eq. 9-5. s1, 2 =
=
sx + sy ;
2
C
sx - sy
a
2
2
b + t2xy
148.3 - 0 2 148.3 + 0 ; a b + (-15.09)2 2 C 2
= 74.147 ; 75.666 s1 = 150 MPa
s2 = -1.52 MPa
Ans.
sx = -141.1 MPa, sy = 0, and txy = -15.09 MPa for point B. Applying Eq. 9-5. s1, 2 =
=
sx + sy 2
;
C
a
sx - sy 2
2
b + t2xy
( -141.1) - 0 2 -141.1 + 0 ; a b + (-15.09)2 2 C 2
= -70.538 ; 72.134 s1 = 1.60 MPa
s2 = -143 MPa
Ans. 650
30⬚ 25 kN
A 10 mm
110 mm
10 mm 200 mm 10 mm
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•9–29.
The wide-flange beam is subjected to the loading shown. Determine the principal stress in the beam at point A, which is located at the top of the web. Although it is not very accurate, use the shear formula to determine the shear stress. Show the result on an element located at this point.
120 kN/m
30 kN
A 0.3 m
0.9 m
Using the method of sections and consider the FBD of the left cut segment of the bean, Fig. a V -
+ c ©Fy = 0;
1 2
a + ©MC = 0;
1 2
(90)(0.9)(0.3) + 30(0.9) - M = 0
M = 39.15 kN # m
150 mm
1 1 (0.15)(0.193) (0.13)(0.153) = 49.175(10 - 6) m4 12 12
Referring to Fig. b, QA = y¿A¿ = 0.085 (0.02)(0.15) = 0.255 (10 - 3) m3 The normal stress developed is contributed by bending stress only. For point A, y = 0.075 m. Thus, s =
My 39.15(103)(0.075) = 59.71(106)Pa = 59.71 MPa (T) = I 49.175(10 - 6)
The shear stress is contributed by the transverse shear stress only. Thus t =
70.5(103) C 0.255(10 - 3) D VQA = 18.28(106)Pa = 18.28 MPa = It 49.175(10 - 6) (0.02)
Here, sx = 59.71 MPa, sy = 0 and txy = 18.28 MPa. s1, 2 =
=
sx + sy ;
2
C
a
sx - sy 2
2
b + txy
59.71 - 0 2 59.71 + 0 a ; b + 18.282 2 C 2
= 29.86 ; 35.01 s2 = -5.15 MPa
s1 = 64.9 MPa tan 2uP =
txy (sx - sy)>2
uP = 15.74°
=
and
Ans.
18.28 = 0.6122 (59.71 - 0)>2 -74.26°
Substitute u = 15.74°, sx¿ =
=
sx + sy 2
sx - sy +
2
20 mm 150 mm 20 mm
V = 70.5 kN
(90)(0.9) - 30 = 0
The moment of inertia of the cross - section about the bending axis is I =
A 20 mm
cos 2u + txy sin 2u
59.71 - 0 59.71 + 0 + cos 31.48° + 18.28 sin 31.48° 2 2
= 64.9 MPa = s1
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9–29.
Continued
Thus,
A uP B 1 = 15.7°
A uP B 2 = -74.3°
Ans.
The state of principal stress can be represented by the element shown in Fig. d
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9–30. The cantilevered rectangular bar is subjected to the force of 5 kip. Determine the principal stress at points A and B. 1 I = (3)(63) = 54 in4 12
1.5 in. A 1.5 in.
A = (6)(3) = 18 in2
QA = 2.25(1.5)(3) = 10.125 in3
1.5 in. 1.5 in. B 1 in. 3 in.
QB = 2(2)(3) = 12 in3
1 in.
Point A:
15 in.
3 in.
3 5 4
5 kip
sA
45(1.5) Mxz 4 P = + = + = 1.472 ksi A I 18 54
tA =
Vz QA =
It
3(10.125) = 0.1875 ksi 54(3)
sx = 1.472 ksi s1, 2 =
=
sy = 0
sx + sy ;
2
C
a
sx - sy 2
txy = 0.1875 ksi 2
b + txy 2
1.472 - 0 2 1.472 + 0 ; a b + 0.18752 2 C 2
s1 = 1.50 ksi
Ans.
s2 = -0.0235 ksi
Ans.
Point B: sB =
tB =
45(1) Mxz 4 P = = -0.6111 ksi A I 18 54 Vz QB =
It
3(12) = 0.2222 ksi 54(3)
sx = -0.6111 ksi s1, 2 =
=
sy = 0
sx + sy 2
;
C
a
sx - sy 2
txy = 0.2222 ksi
2
b + txy 2
-0.6111 - 0 2 -0.611 + 0 ; a b + 0.2222 2 C 2
s1 = 0.0723 ksi
Ans.
s2 = -0.683 ksi
Ans.
653
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9–31. Determine the principal stress at point A on the cross section of the arm at section a–a. Specify the orientation of this state of stress and indicate the results on an element at the point.
7.5 mm A 50 mm
7.5 mm
Support Reactions: Referring to the free - body diagram of the entire arm shown in Fig. a, ©MB = 0; FCD sin 30°(0.3) - 500(0.65) = 0
FCD = 2166.67 N
+ ©F = 0; : x
Bx - 2166.67 cos 30° = 0
Bx = 1876.39 N
+ c ©Fy = 0;
2166.67 sin 30° - 500 - By = 0
By = 583.33 N
20 mm
Section a – a D
Internal Loadings: Consider the equilibrium of the free - body diagram of the arm’s left segment, Fig. b. + ©F = 0; : x
1876.39 - N = 0
N = 1876.39 N
+ c ©Fy = 0;
V - 583.33 = 0
V = 583.33 N
583.33(0.15) - M = 0
M = 87.5N # m
+ ©MO = 0;
0.15 m
Referring to Fig. b, QA = y¿A¿ = 0.02125(0.0075)(0.02) = 3.1875 A 10 - 6 B m3 Normal and Shear Stress: The normal stress is a combination of axial and bending stress. Thus,
-1876.39
0.5625 A 10
-3
B
MyA N + A I 87.5(0.0175)
+
0.16367 A 10 - 6 B
= 6.020 MPa
The shear stress is caused by transverse shear stress.
tA =
583.33 C 3.1875 A 10 - 6 B D VQA = = 1.515 MPa It 0.16367 A 10 - 6 B (0.0075)
The share of stress at point A can be represented on the element shown in Fig. d. In - Plane Principal Stress: sx = 6.020 MPa, sy = 0, and txy = 1.515 MPa. We have s1,2 =
=
sx + sy 2
;
C
¢
sx - sy 2
2
≤ + txy 2
6.020 - 0 2 6.020 + 0 ; a b + 1.5152 2 C 2
s1 = 6.38 MPa
s2 = -0.360 MPa
Ans.
654
C 0.15 m
0.35 m 500 N
1 1 (0.02) A 0.053 B (0.0125) A 0.0353 B = 0.16367 A 10 - 6 B m4 12 12
=
a
a
A = 0.02(0.05) - 0.0125(0.035) = 0.5625 A 10 - 3 B m2
sA =
60⬚
B
Section Properties: The cross - sectional area and the moment of inertia about the z axis of the arm’s cross section are
I =
7.5 mm
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9–31. Continued Orientation of the Principal Plane: tan 2uP =
txy
A sx - sy B >2
=
1.515 = 0.5032 (6.020 - 0)>2
up = 13.36° and 26.71° Substituting u = 13.36° into sx¿ =
=
sx + sy 2
sx - sy +
2
cos 2u + txy sin 2u
6.020 + 0 6.020 - 0 + cos 26.71° + 1.515 sin 26.71° 2 2
= 6.38 MPa = s1 Thus, A uP B 1 = 13.4 and A uP B 2 = 26.71°
Ans.
The state of principal stresses is represented by the element shown in Fig. e.
655
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*9–32. Determine the maximum in-plane shear stress developed at point A on the cross section of the arm at section a–a. Specify the orientation of this state of stress and indicate the results on an element at the point.
7.5 mm A 50 mm
7.5 mm
20 mm
7.5 mm
Section a – a
Support Reactions: Referring to the free - body diagram of the entire arm shown in Fig. a, ©MB = 0; FCD sin 30°(0.3) - 500(0.65) = 0
FCD = 2166.67 N
+ ©F = 0; : x
Bx - 2166.67 cos 30° = 0
Bx = 1876.39 N
+ c ©Fy = 0;
2166.67 sin 30° - 500 - By = 0
By = 583.33 N
D 60⬚
a
B
C
a 0.15 m
0.15 m
0.35 m 500 N
Internal Loadings: Considering the equilibrium of the free - body diagram of the arm’s left cut segment, Fig. b, + ©F = 0; : x
1876.39 - N = 0
N = 1876.39 N
+ c ©Fy = 0;
V - 583.33 = 0
V = 583.33 N
+ ©MO = 0;
583.33(0.15) - M = 0
M = 87.5 N # m
Section Properties: The cross - sectional area and the moment of inertia about the z axis of the arm’s cross section are A = 0.02(0.05) - 0.0125(0.035) = 0.5625 A 10 - 3 B m2 1 1 (0.02) A 0.053 B (0.0125) A 0.0353 B = 0.16367 A 10 - 6 B m4 12 12
I =
Referring to Fig. b, QA = y¿A¿ = 0.02125(0.0075)(0.02) = 3.1875 A 10 - 6 B m3 Normal and Shear Stress: The normal stress is a combination of axial and bending stress. Thus, sA =
MyA N + A I -1876.39
=
0.5625 A 10
-3
B
87.5(0.0175)
+
0.16367 A 10 - 6 B
= 6.020 MPa
The shear stress is contributed only by transverse shear stress. tA =
583.33 C 3.1875 A 10 - 6 B D VQA = = 1.515 MPa It 0.16367 A 10 - 6 B (0.0075)
Maximum In - Plane Shear Stress: sx = 6.020 MPa, sy = 0, and txy = 1.515 MPa. tmax
in-plane
=
C
¢
sx - sy 2
2
≤ + txy 2 =
6.020 - 0 2 b + 1.5152 = 3.37 MPa B 2 a
656
Ans.
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9–32.
Continued
Orientation of the Plane of Maximum In - Plane Shear Stress: tan 2us = -
A sx - sy B >2 txy
= -
(6.020 - 0)>2 = -1.9871 1.515
us = -31.6° and 58.4°
Ans.
Substituting u = -31.6° into tx¿y¿ = -
= -
sx - sy 2
sin 2u + txy cos 2u
6.020 - 0 sin(-63.29°) + 1.515 cos(-63.29°) 2
= 3.37 MPa = t max
in-plane
is directed in the positive sense of the y¿ axis on the face
This indicates that t max
in-plane
of the element defined by us = -31.6°. Average Normal Stress: savg =
sx + sy 2
=
6.020 + 0 = 3.01 MPa 2
Ans.
The state of maximum in - plane shear stress is represented on the element shown in Fig. e.
657
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•9–33. The clamp bears down on the smooth surface at E by tightening the bolt. If the tensile force in the bolt is 40 kN, determine the principal stress at points A and B and show the results on elements located at each of these points. The cross-sectional area at A and B is shown in the adjacent figure.
300 mm 50 mm 30 mm 100 mm
B A
Support Reactions: As shown on FBD(a). E
Internal Forces and Moment: As shown on FBD(b). Section Properties: I =
1 (0.03) A 0.053 B = 0.3125 A 10 - 6 B m4 12
QA = 0 QB = y¿A¿ = 0.0125(0.025)(0.03) = 9.375 A 10 - 6 B m3 Normal Stress: Applying the flexure formula s = -
sA = -
sB = -
2.40(103)(0.025)
= -192 MPa
0.3125(10 - 6) 2.40(103)(0) 0.3125(10 - 6)
= 0
Shear Stress: Applying the shear formula t =
tA =
tB =
My . I
24.0(103)(0) 0.3125(10 - 6)(0.03)
VQ It
= 0
24.0(103) C 9.375(10 - 6) D 0.3125(10 - 6)(0.03)
= 24.0 MPa
In - Plane Principal Stresses: sx = 0, sy = -192 MPa, and txy = 0 for point A. Since no shear stress acts on the element. s1 = sx = 0
Ans.
s2 = sy = -192 MPa
Ans.
sx = sy = 0 and txy = -24.0 MPa for point B. Applying Eq. 9-5 s1,2 =
sx + sy 2
;
C
a
sx - sy 2
2
b + t2xy
= 0 ; 20 + (-24.0)2 = 0 ; 24.0 s1 = 24.0
s2 = -24.0 MPa
Ans.
658
B
A
25 mm 100 mm 50 mm
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9–33. Continued Orientation of Principal Plane: Applying Eq. 9-4 for point B. tan 2up =
txy
A sx - sy B >2
up = -45.0°
=
and
-24.0 = -q 0 45.0°
Subsututing the results into Eq. 9-1 with u = -45.0° yields sx¿ =
sx + sy 2
sx - sy +
2
cos 2u + txy sin 2u
= 0 + 0 + [-24.0 sin (-90.0°)] = 24.0 MPa = s1 Hence, up1 = -45.0°
up2 = 45.0°
Ans.
659
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9–34. Determine the principal stress and the maximum inplane shear stress that are developed at point A in the 2-in.-diameter shaft. Show the results on an element located at this point. The bearings only support vertical reactions.
300 lb
Using the method of sections and consider the FBD of shaft’s left cut segment, Fig. a, + ©F = 0; : x
N - 3000 = 0
+ c ©Fy = 0;
75 - V = 0
a + ©MC = 0;
M - 75(24) = 0
N = 3000 lb V = 75 lb M = 1800 lb # in
A = p(12) = p in2
I =
p 4 p (1 ) = in4 4 4
Also, QA = 0 The normal stress developed is the combination of axial and bending stress. Thus My N ; A I
s = For point A, y = C = 1 in. Then s =
1800(1) 3000 p p>4
= -1.337 (103) psi = 1.337 ksi (c) The shear stress developed is due to transverse shear force. Thus, t =
VQA = 0 It
The state of stress at point A, can be represented by the element shown in Fig. b. Here, sx = -1.337 ksi, sy = 0 is txy = 0. Since no shear stress acting on the element, s1 = sy = 0
s2 = sx = -1.34 ksi
Ans.
Thus, the state of principal stress can also be represented by the element shown in Fig. b. t
max in-plane
=
C
a
sx - sy 2
tan 2us = us = 45°
2
b + t2xy =
-1.337 - 0 2 b + 02 = 0.668 ksi - 668 psi Ans. C 2 a
(sx - sy)>2 = -
txy and
(-1.337 - 0)>2 = q 0 Ans.
-45°
Substitute u = 45°, tx¿y¿ = -
= -
sx - sy 2
sin 2u + txy cos 2u
-1.337 - 0 sin 90° + 0 2
= 0.668 ksi = 668 psi =
A
3000 lb
tmax
in-plane
660
24 in.
3000 lb
12 in.
12 in.
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9–34. Continued tmax This indicates that in-plane acts toward the positive sense of y¿ axis at the face of the element defined by us = 45°. Average Normal Stress. The state of maximum in - plane shear stress can be represented by the element shown in Fig. c.
661
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9–35. The square steel plate has a thickness of 10 mm and is subjected to the edge loading shown. Determine the maximum in-plane shear stress and the average normal stress developed in the steel. sx = 5 kPa
sy = -5 kPa
tmax
sx - sy
in-plane
savg =
a
txy = 0
200 mm
b +
t2xy
C
=
5 + 5 2 b + 0 = 5 kPa C 2 a
sx + sy 3
50 N/m
2
=
2
50 N/m
Ans. 200 mm
5 - 5 = = 0 2
Ans.
Note: tan 2us =
tan 2us =
-(sx - sy)>2 txy -(5 + 5)>2 = q 0
us = 45°
*9–36. The square steel plate has a thickness of 0.5 in. and is subjected to the edge loading shown. Determine the principal stresses developed in the steel. sx = 0 s1,2 =
sy = 0
txy = 32 psi
sx + sy 2
;
16 lb/in.
C
a
sx - sy 2
2
b + t2xy
= 0 ; 20 + 322
4 in.
s1 = 32 psi
Ans.
s2 = -32 psi
Ans.
Note: tan 2up =
16 lb/in.
4 in.
txy (sx - sy)>2
=
32 = q 0
up = 45°
662
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•9–37.
The shaft has a diameter d and is subjected to the loadings shown. Determine the principal stress and the maximum in-plane shear stress that is developed at point A. The bearings only support vertical reactions.
P F
F A L 2
Support Reactions: As shown on FBD(a). Internal Forces and Moment: As shown on FBD(b). Section Properties: A =
p 2 d 4
p d 4 p 4 a b = d 4 2 64
I =
QA = 0
Normal Stress: N Mc ; A I
s =
-F ; d2
=
p 4
sA =
A B
pL d 4 2 p 4 d 64
4 2PL - Fb a 2 d pd
Shear Stress: Since QA = 0, tA = 0 In - Plane Principal Stress: sx =
4 2PL a - Fb. pd2 d
sy = 0 and txy = 0 for point A. Since no shear stress acts on the element, s1 = sx =
4 2PL a - Fb d pd2
Ans.
s2 = sy = 0
Ans.
Maximum In - Plane Shear Stress: Applying Eq. 9-7 for point A, t
max in-plane
=
=
=
Q
£
B
a
4 2 pd
sx - sy 2
2
b + t2xy
A 2PL d - FB - 0 2
2
≥ + 0
2PL 2 a - Fb d pd2
Ans.
663
L 2
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9–38. A paper tube is formed by rolling a paper strip in a spiral and then gluing the edges together as shown. Determine the shear stress acting along the seam, which is at 30° from the vertical, when the tube is subjected to an axial force of 10 N. The paper is 1 mm thick and the tube has an outer diameter of 30 mm. P = A
s =
p 4
= -
10 N
10 N 30 mm
10 = 109.76 kPa (0.032 - 0.0282)
sx = 109.76 kPa tx¿y¿ = -
30⬚
sx - sy 2
sy = 0
txy = 0
u = 30°
sin 2u + txy cos 2u
106.76 - 0 sin 60° + 0 = -47.5 kPa 2
Ans.
9–39. Solve Prob. 9–38 for the normal stress acting perpendicular to the seam.
30⬚
10 N
10 N 30 mm
s =
sn =
=
P = A
p 4
10 = 109.76 kPa (0.032 - 0.0282)
sx + sy 2
sx - sy +
2
cos 2u + txy sin 2u
109.76 - 0 109.76 + 0 + cos (60°) + 0 = 82.3 kPa 2 2
Ans.
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*9–40. Determine the principal stresses acting at point A of the supporting frame. Show the results on a properly oriented element located at this point.
800 mm B
A 300 mm
150 mm 12 mm
5
B 15 mm
130 mm A
y =
0.065(0.13)(0.015) + 0.136(0.15)(0.012) ©yA = = 0.0991 m ©A 0.13(0.015) + 0.15(0.012)
I =
1 (0.015)(0.133) + 0.015(0.13)(0.0991 - 0.065)2 12 1 (0.15)(0.012 3) + 0.15(0.012)(0.136 - 0.0991)2 = 7.4862(10 - 6) m4 12
+
QA = 0 A = 0.13(0.015) + 0.15(0.012) = 3.75(10 - 3) m2 Normal stress: s =
P Mc + A I
sA =
-3.6(103) -3
3.75(10 )
5.2767(103)(0.0991) -
7.4862(10 - 6)
= -70.80 MPa
Shear stress: tA = 0 Principal stress: s1 = 0
Ans.
s2 = -70.8 MPa
Ans.
665
4 3
6 kN
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•9–41. Determine the principal stress acting at point B, which is located just on the web, below the horizontal segment on the cross section. Show the results on a properly oriented element located at this point. Although it is not very accurate, use the shear formula to calculate the shear stress.
800 mm B
A 300 mm
150 mm 12 mm
y =
©yA 0.065(0.13)(0.015) + 0.136(0.15)(0.012) = = 0.0991 m ©A 0.13(0.015) + 0.15(0.012)
I =
1 (0.015)(0.133) + 0.015(0.13)(0.0991 - 0.065)2 12 +
130 mm A
1 (0.15)(0.0123) + 0.15(0.012)(0.136 - 0.0991)2 = 7.4862(10 - 6) m4 12
A = 0.13(0.015) + 0.15(0.012) = 3.75(10 - 3) m2 Normal stress: s =
Mc P + A I
sB = -
3.6(103) 3.75(10 - 3)
5.2767(103)(0.130 - 0.0991) +
7.4862(10 - 6)
= 20.834 MPa
Shear stress: tB =
VQ -4.8(103)(0.0369)(0.15)(0.012) = -2.84 MPa = It 7.4862(10 - 6)(0.015)
Principal stress: s1,2 = a
20.834 + 0 20.834 - 0 2 b ; a b + (-2.84)2 2 C 2
s1 = 21.2 MPa
Ans.
s2 = -0.380 MPa
Ans.
tan 2up =
A
-2.84 20.834 - 0 2
B
up = -7.63°
Ans.
666
5
B 15 mm
4 3
6 kN
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9–42. The drill pipe has an outer diameter of 3 in., a wall thickness of 0.25 in., and a weight of 50 lb>ft. If it is subjected to a torque and axial load as shown, determine (a) the principal stress and (b) the maximum in-plane shear stress at a point on its surface at section a.
1500 lb
800 lb⭈ft 20 ft
a 20 ft
Internal Forces and Torque: As shown on FBD(a). Section Properties: A =
p 2 A 3 - 2.52 B = 0.6875p in2 4
J =
p A 1.54 - 1.254 B = 4.1172 in4 2
s =
N -2500 = = -1157.5 psi A 0.6875p
Normal Stress:
Shear Stress: Applying the torsion formula. t =
800(12)(1.5) Tc = = 3497.5 psi J 4.1172
a) In - Plane Principal Stresses: sx = 0, sy = -1157.5 psi and txy = 3497.5 psi for any point on the shaft’s surface. Applying Eq. 9-5. s1,2 =
=
sx + sy 2
;
C
a
sx - sy 2
2
b + t2xy
0 - (-1157.5) 2 0 + (-1157.5) ; a b + (3497.5)2 2 C 2
= -578.75 ; 3545.08 s1 = 2966 psi = 2.97 ksi
Ans.
s2 = -4124 psi = -4.12 ksi
Ans.
b) Maximum In - Plane Shear Stress: Applying Eq. 9-7 t
max in-plane
a
sx - sy
2
b + t2xy
=
C
=
0 - (-1157.5) 2 ≤ + (3497.5)2 C 2
2
¢
= 3545 psi = 3.55 ksi
Ans.
667
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9–43. Determine the principal stress in the beam at point A.
60 kN 50 mm 150 kN
A
A 60 mm
0.5 m
Using the method of sections and consider the FBD of the beam’s left cut segment, Fig. a, + ©F = 0; : x
150 - N = 0
N = 150 kN
+ c ©Fy = 0;
V - 60 = 0
V = 60 kN
a + ©MC = 0;
M = 30 kN # m
60(0.5) - M = 0 A = 0.06(0.15) = 0.009 m2
1 (0.06)(0.153) = 16.875(10 - 6) m4 12
I = Referring to Fig. b,
QA = y¿A¿ = 0.05 (0.05)(0.06) = 0.15(10 - 3) m3 The normal stress developed is the combination of axial and bending stress. Thus My N ; A I
s =
For point A, y = 0.075 - 0.05 = 0.025 m. Then s =
30(103)(0.025) -150(103) 0.009 16.875(10 - 6)
= -61.11(106) Pa = 61.11 MPa (c) The shear stress developed is due to the transverse shear, Thus, t =
60(103) C 0.15(10 - 3) D VQA = 8.889 MPa = It 16.875(10 - 6) (0.06)
Here, sx = -61.11 MPa, sy = 0 and txy = 8.889 MPa, s1, 2 =
=
sx + sy ;
2
C
a
sx - sy 2
2
b + t2xy
-61.11 - 0 2 -61.11 + 0 ; a b + 8.8892 2 C 2
= -30.56 ; 31.82 s2 = -62.4 MPa
s1 = 1.27 MPa tan 2uP =
txy (sx - sy)>2
uP = -8.11°
=
and
Ans.
8.889 = -0.2909 (-61.11 - 0)>2 81.89°
668
0.25 m
150 mm
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9–43.
Continued
Substitute u = -8.11°, sx¿ =
=
sx + sy 2
sx - sy +
2
cos 2u + txy sin 2u
-61.11 + 0 -61.11 - 0 + cos (-16.22°) + 8.889 sin (-16.22°) 2 2
= -62.4 MPa = s2 Thus, (uP)1 = 81.9°
(uP)2 = -8.11°
The state of principal stresses can be represented by the elements shown in Fig. (c)
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*9–44. Determine the principal stress at point A which is located at the bottom of the web. Show the results on an element located at this point.
150 kN/m
Using the method of sections, consider the FBD of the bean’s left cut segment, Fig. a, V -
+ c ©Fy = 0;
I =
1 (100)(0.6) = 0 2
10 mm
M = 6 kN # m
A
1 1 (0.15)(0.223) (0.14)(0.23) = 39.7667(10 - 6) m4 12 12
150 mm
Referring to Fig. b QA = y¿A¿ = 0.105 (0.01)(0.15) = 0.1575(10 - 3) m3 The normal stress developed is due to bending only. For point A, y = 0.1 m. Then s =
My
6(103)(0.1) =
I
= 15.09(106)Pa = 15.09 MPa (c)
39.7667(10 - 6)
The shear stress developed is due to the transverse shear. Thus, t =
30(103) C 0.1575(10 - 3) D VQA = 11.88(106)Pa = 11.88 MPa = It 39.7667(10 - 6)(0.01)
Here, sx = -15.09 MPa, sy = 0 And txy = 11.88 MPa. s1, 2 =
=
sx + sy ;
2
C
a
sx - sy 2
2
b + t2xy
-15.09 - 0 2 -15.09 + 0 ; a b + 11.882 2 C 2
= -7.544 ; 14.074 s2 = -21.6 MPa
s1 = 6.53 MPa tan 2uP =
txy (sx - sy)>2
uP = -28.79°
=
Ans.
11.88 = -1.575 (-15.09 - 0)>2
and
61.21°
Substitute u = 61.21°, sx¿ =
=
sx + sy 2
sx - sy +
2
cos 2u + txy sin 2u
-15.09 + 0 -15.09 - 0 + cos 122.42° + 11.88 sin 122.42° 2 2
= 6.53 MPa = s1 Thus, (uP)1 = 61.2°
0.3 m
V = 30 kN
1 (100)(0.6)(0.2) - M = 0 2
a + ©MC = 0;
A 0.6 m
Ans.
(uP)2 = -28.8°
The state of principal stresses can be represented by the element shown in Fig. d.
670
10 mm 200 mm 10 mm
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9–44. Continued
671
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•9–45.
Determine the maximum in-plane shear stress in the box beam at point A. Show the results on an element located at this point.
10 kip 4 kip
A B
Using the method of section, consider the FBD, of bean’s left cut segment, Fig. a, 8 - 10 + V = 0
+ c ©Fy = 0; a + ©MC = 0;
M + 10(1.5) - 8(3.5) = 0
4 in. A
M = 13 kip # ft
4 in.
The moment of inertia of the cross - section about the neutral axis is
Referring to Fig. b, QA = 0 The normal stress developed is contributed by the bending stress only. For point A, y = C = 3 in. My =
I
13(12)(3) = 5.40 ksi (c) 86.6667
The shear stress is contributed by the transverse shear stress only. Thus t =
VQA = 0 It
The state of stress at point A can be represented by the element shown in Fig. c Here, sx = -5.40 ksi, sy = 0 and txy = 0. tmax
in-plane
=
C
a
sx - sy 2
tan 2us = -
2
b + txy 2 =
-5.40 - 0 2 b + 02 = 2.70 ksi C 2
(sx - sy)>2 = -
txy
us = 45°
and
a
Ans.
(-5.40 - 0)>2 = q 2
-45°
Substitute u = 45°, tx¿y¿ = -
sx - sy
= -
2
sin 2u + txy cos 2u
-5.40 - 0 sin 90° + 0 2
= 2.70 ksi =
tmax
in-plane
tmax This indicates that in-plane acts toward the positive sense of y¿ axis at the face of element defined by us = 45° savg =
sx + sy 2
=
-5.40 + 0 = -2.70 ksi 2
The state of maximum In - plane shear stress can be represented by the element shown in Fig. d. 672
B 6 in.
1 1 I = (6)(63) (4)(43) = 86.6667 in4 12 12
s =
1.5 ft
2 ft
V = 2 kip
2 ft 0.5 ft 3 in. 3 in.
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9–45. Continued
673
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9–46. Determine the principal stress in the box beam at point B. Show the results on an element located at this point.
10 kip 4 kip
A
Using the method of sections, consider the FBD of bean’s left cut segment, Fig. a, 8 - 10 + V = 0
+ c ©Fy = 0; a + ©MC = 0;
M = 13 kip # ft
M + 10(1.5) - 8(3.5) = 0 I =
B
V = 2 kip
4 in. A
1 1 (6)(63) (4)(43) = 86.6667 in4 12 12
4 in.
Referring to Fig. b, QB =
2y1œ A1œ
+
= 2 C 1(2)(1) D + 2.5(1)(6) = 19 in
The normal stress developed is contributed by the bending stress only. For point B, y = 0. My
s =
= 0
I
The shear stress is contributed by the transverse shear stress only. Thus 2(103)(19) VQB = = 219.23 psi It 86.6667(2)
t =
The state of stress at point B can be represented by the element shown in Fig. c Here, sx = sy = 0 and txy = 219.23 psi. s1, 2 =
sx + sy ;
2
C
a
sx - sy 2
2
b + txy 2
= 0 ; 20 + 219.232 s2 = -219 psi
s1 = 219 psi tan 2uP =
txy (sx - sy)>2
uP = 45°
=
and
Ans.
219.23 = q 0 -45°
Substitute u = 45°, sx¿ =
sx + sy 2
sx - sy +
2
cos 2u + txy sin 2u
= 0 + 0 + 219.23 sin 90° = 219 psi = s1 Thus, (uP)1 = 45°
Ans.
(uP)2 = -45°
The state of principal stress can be represented by the element shown in Fig. d.
674
B 6 in.
3
y2œ A2œ
1.5 ft
2 ft
2 ft 0.5 ft 3 in. 3 in.
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9–46. Continued
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9–47. The solid shaft is subjected to a torque, bending moment, and shear force as shown. Determine the principal stresses acting at point A. Ix = Iy =
J =
p (0.025)4 = 0.306796(10 - 6) m4 4
450 mm 300 N⭈m
p (0.025)4 = 0.613592(10 - 6) m4 2
25 mm
45 N⭈m
QA = 0 sA
800 N
60(0.025) Mx c = 4.889 MPa = = I 0.306796(10 - 6)
tA =
Ty c
45(0.025) =
J
0.613592(10 - 6)
sx = 4.889 MPa s1, 2 =
=
*9–48.
sy = 0
sx + sy ;
2
= 1.833 MPa
C
a
txy = -1.833 MPa
sx - sy 2
2
b + txy 2
4.889 - 0 2 4.889 + 0 ; b + (-1.833)2 a 2 C 2
s1 = 5.50 MPa
Ans.
s2 = -0.611 MPa
Ans.
Solve Prob. 9–47 for point B.
Ix = Iy =
p (0.025)4 = 0.306796(10 - 6) m4 4 450 mm
p J = (0.025)4 = 0.613592(10 - 6) m4 2 QB = yA¿ =
300 N⭈m
4(0.025) 1 a b p (0.0252) = 10.4167(10 - 6) m3 3p 2
800 N
Ty c
VzQB It
-
-6
800(10.4167)(10 ) =
J
0.306796(10 )(0.05)
sx = 0 s1, 2 =
sy = 0 sx + sy 2
;
C
a
sx - sy 2
45(0.025) -
-6
0.61359(10 - 6)
= -1.290 MPa
txy = -1.290 MPa 2
b + txy 2
= 0 ; 2(0)2 + (-1.290)2 s1 = 1.29 MPa
Ans.
s2 = -1.29 MPa
Ans.
676
A B 25 mm
45 N⭈m
sB = 0 tB =
A B
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•9–49.
The internal loadings at a section of the beam are shown. Determine the principal stress at point A. Also compute the maximum in-plane shear stress at this point.
50 mm A
200 mm 50 mm
50 mm y
200 mm
Section Properties:
z
40 kN⭈m 30 kN⭈m
A = 0.2(0.3) - 0.15(0.2) = 0.030 m4 800 kN
1 1 Iz = (0.2) A 0.33 B (0.15) A 0.23 B = 0.350 A 10 - 3 B m4 12 12 Iy =
1 1 (0.1) A 0.23 B + (0.2) A 0.053 B = 68.75 A 10 - 6 B m4 12 12
(QA)y = 0 Normal Stress: s =
sA =
Myz Mzy N + A Iz Iy -30(103)(0.1) -500(103) 40(103)(0.15) + 3 0.030 0.350(10 ) 68.75(10 - 6)
= -77.45 MPa tA = 0.
Shear Stress: Since (QA)y = 0,
In - Plane Principal Stresses: sx = -77.45 MPa. sy = 0. and txy = 0 for point A. Since no shear stress acts on the element. s1 = sy = 0
Ans.
s2 = sz = -77.4 MPa
Ans.
Maximum In-Plane Shear Stress: Applying Eq. 9–7. t
max in-plane
a
sx - sy
2
b + t2xy
=
C
=
-77.45 - 0 2 b + 0 C 2
2
a
= 38.7 MPa
Ans.
677
500 kN x
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9–50. The internal loadings at a section of the beam consist of an axial force of 500 N, a shear force of 800 N, and two moment components of 30 N # m and 40 N # m. Determine the principal stress at point A. Also calculate the maximum in-plane shear stress at this point. Ix =
A
1 (0.1)(0.2)3 = 66.67(10 - 6) in4 12
40 N⭈m
B C
50 mm
QA = 0 sA
200 mm
50 mm
100 mm
30(0.1) Mz P 500 = -20 kPa = = A Ix (0.1)(0.2) 66.67(10 - 6)
30 N⭈m
500 N 800 N
tA = 0 Here, the principal stresses are s1 = sy = 0
Ans.
s2 = sx = -20 kPa
Ans.
t
max in-plane=
=
C
a
sx - sy 2
2
b + txy 2
-20 - 0 2 b + 0 = 10 kPa C 2 a
Ans.
9–51. Solve Prob. 9–4 using Mohr’s circle.
A
400 psi
650 psi 60⬚
sx + sy 2
-650 + 400 = = -125 2
A(-650, 0)
B(400, 0)
C( -125, 0)
B
R = CA = = 650 - 125 = 525 sx¿ = -125 - 525 cos 60° = -388 psi
Ans.
tx¿y¿ = 525 sin 60° = 455 psi
Ans.
678
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*9–52.
Solve Prob. 9–6 using Mohr’s circle. 90 MPa
A
35 MPa 60⬚ 30⬚
sx = 90 MPa sx + sy 2
=
sy = 50 MPa
txy = -35 MPa
A(90, -35)
90 + 50 = 70 2
R = 2(90 - 70)2 + (35)2 = 40.311 Coordinates of point B: f = tan - 1 a
35 b = 60.255° 20
c = 300° - 180° - 60.255° = 59.745° sx¿ = 70 - 40.311 cos 59.745° = 49.7 MPa
Ans.
tx¿ = -40.311 sin 59.745° = -34.8 MPa
Ans.
679
B 50 MPa
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•9–53.
Solve Prob. 9–14 using Mohr’s circle. 30 ksi
12 ksi
sx + sy 2
=
-30 + 0 = -15 2
R = 2(30 - 15)2 + (12)2 = 19.21 ksi s1 = 19.21 - 15 = 4.21 ksi
Ans.
s2 = -19.21 - 15 = -34.2 ksi
Ans.
2uP2 = tan - 1 tmax
in-plane
12 ; (30 - 15)
uP2 = 19.3°
Ans.
= R = 19.2 ksi
Ans.
savg = -15 ksi 2uP2 = tan - 1
12 + 90°; (30 - 15)
Ans. us = 64.3°
Ans.
680
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9–54.
Solve Prob. 9–16 using Mohr’s circle.
350 psi
75 psi 200 psi
sx + sy 2
=
45 - 60 = -7.5 MPa 2
R = 2(45 + 7.5)2 + (30)2 = 60.467 MPa s1 = 60.467 - 7.5 = 53.0 MPa
Ans.
s2 = -60.467 - 7.5 = -68.0 MPa
Ans.
2uP1 = tan - 1 uP1 = 14.9° tmax
30 (45 + 7.5) Ans.
counterclockwise
= 60.5 MPa
Ans.
savg = -7.50 MPa
Ans.
in-plane
2uP1 = 90° - tan - 1 us1 = 30.1°
30 (45 + 7.5) Ans.
clockwise
681
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9–55.
Solve Prob. 9–12 using Mohr’s circle.
sx + sy 2
=
-10 + 0 = -5 ksi 2
10 ksi
16 ksi
R = 2(10 - 5)2 + (16)2 = 16.763 ksi f = tan - 1
16 = 72.646° (10 - 5)
a = 100 - 72.646 = 27.354° sx¿ = -5 - 16.763 cos 27.354° = -19.9 ksi
Ans.
tx¿y¿ = 16.763 sin 27.354° = 7.70 ksi
Ans.
sy¿ = 16.763 cos 27.354° - 5 = 9.89 ksi
*9–56. Solve Prob. 9–11 using Mohr’s circle.
2 ksi
Construction of the Circle: In accordance with the sign convention, sx = -3 ksi, sy = 2 ksi, and txy = -4 ksi. Hence, savg =
sx + sy 2
=
-3 + 2 = -0.500 ksi 2
3 ksi 30⬚
The coordinates for reference point A and C are A(-3, -4)
A
4 ksi B
C(-0.500, 0)
The radius of circle is R = 2(3 - 0.5)2 + 42 = 4.717 ksi Stress on the Inclined Plane: The normal and shear stress components A sx¿ and tx¿y¿ B are represented by the coordinates of point P on the circle. sx¿ = -0.500 - 4.717 cos 62.01° = -2.71 ksi
Ans.
tx¿y¿ = 4.717 sin 62.01° = 4.17 ksi
Ans.
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9–57. Mohr’s circle for the state of stress in Fig. 9–15a is shown in Fig. 9–15b. Show that finding the coordinates of point P1sx¿ , tx¿y¿2 on the circle gives the same value as the stress-transformation Eqs. 9–1 and 9–2.
A(sx, txy)
R =
sxœ =
C
Ca a
B(sy, -txy)
csx - a
sx + sy 2
sx + sy +
2
C
a
sx + sy 2
2
b d + t2xy =
sx - sy 2
C
a
b, 0b
sx - sy 2
2
b + t2xy
2
b + t2xy cos u¿
(1)
u¿ = 2uP - 2u (2)
cos (2uP - 2u) = cos 2uP cos 2u + sin 2up sin 2u From the circle: sx -
cos 2uP =
sin 2uP =
4A 4A
sx + sy 2
sx - sy 2
txy
sx - sy 2
B + 2
(3) t2xy (4)
B 2 + t2xy
Substitute Eq. (2), (3) and into Eq. (1) sx¿ =
tx¿y¿ =
sx + sy 2
C
a
sx - sy +
2
sx - sy 2
cos 2u + txy sin 2u
QED
2
b + t2xy sin u¿
(5)
sin u¿ = sin (2uP - 2u) (6)
= sin 2uP cos 2u - sin 2u cos 2uP Substitute Eq. (3), (4), (6) into Eq. (5), tx¿y¿ = -
sx - sy 2
sin 2u + txy cos 2u
QED
683
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9–58. Determine the equivalent state of stress if an element is oriented 25° counterclockwise from the element shown.
550 MPa
A(0, -550)
B(0, 550)
C(0, 0)
R = CA = CB = 550 sx¿ = -550 sin 50° = -421 MPa
Ans.
tx¿y¿ = -550 cos 50° = -354 MPa
Ans.
sy¿ = 550 sin 50° = 421 MPa
Ans.
9–59. Determine the equivalent state of stress if an element is oriented 20° clockwise from the element shown.
2 ksi
Construction of the Circle: In accordance with the sign convention, sx = 3 ksi, sy = -2 ksi, and tx¿y¿ = -4 ksi. Hence, savg =
sx + sy 2
=
3 + (-2) = 0.500 ksi 2
4 ksi
The coordinates for reference points A and C are A(3, -4)
3 ksi
C(0.500, 0)
The radius of the circle is R = 2(3 - 0.500)2 + 42 = 4.717 ksi Stress on the Rotated Element: The normal and shear stress components A sx¿ and tx¿y¿ B are represented by the coordinate of point P on the circle, sy¿, can be determined by calculating the coordinates of point Q on the circle. sx¿ = 0.500 + 4.717 cos 17.99° = 4.99 ksi
Ans.
tx¿y¿ = -4.717 sin 17.99° = -1.46 ksi
Ans.
sy¿ = 0.500 - 4.717 cos 17.99° = -3.99 ksi
Ans.
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*9–60. Determine the equivalent state of stress if an element is oriented 30° clockwise from the element shown. Show the result on the element.
9 ksi 4 ksi
In accordance to the established sign convention, sx = -6 ksi, sy = 9 ksi and txy = 4 ksi. Thus, savg =
sx + sy 2
=
-6 + 9 = 1.50 ksi 2
Then, the coordinates of reference point A and C are A(-6, 4)
C(1.5, 0)
The radius of the circle is R = CA = 2(-6 - 1.5)2 + 42 = 8.50 ksi Using these results, the circle shown in Fig. a can be constructed. Referring to the geometry of the circle, Fig. a, a = tan - 1 a
4 b = 28.07° 6 + 1.5
b = 60° - 28.07° = 31.93°
Then, sx¿ = 1.5 - 8.50 cos 31.93° = -5.71 ksi
Ans.
tx¿y¿ = -8.5 sin 31.95° = -4.50 ksi sy¿ = 8.71 ksi
Ans.
The results are shown in Fig. b.
685
6 ksi
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•9–61.
Determine the equivalent state of stress for an element oriented 60° counterclockwise from the element shown. Show the result on the element.
250 MPa 400 MPa
In accordance to the established sign convention, sx = -560 MPa, sy = 250 MPa and txy = -400 MPa. Thus, savg =
sx + sy 2
=
-560 + 250 = -155 MPa 2
Then, the coordinate of reference points A and C are A(-560, -400)
C(-155, 0)
The radius of the circle is R = CA = 3 C -560 - (-155) D 2 + (-400)2 = 569.23 MPa Using these results, the circle shown in Fig. a can be constructed. Referring to the geometry of the circle, Fig. a a = tan - 1 a
400 b = 44.64° 560 - 155
b = 120° - 44.64° = 75.36°
Then, sx¿ = -155 - 569.23 cos 75.36° = -299 MPa
Ans.
tx¿y¿ = 569.23 sin 75.36° = 551 MPa
Ans.
sy¿ = -155 + 569.23 cos 75.36° = -11.1 MPa
Ans.
The results are shown in Fig. b.
686
560 MPa
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9–62. Determine the equivalent state of stress for an element oriented 30° clockwise from the element shown. Show the result on the element.
5 ksi
In accordance to the established sign convention, sx = 2 ksi, sy = -5 ksi and txy = 0. Thus, savg =
sx + sy 2
=
2 + (-5) = -1.50 ksi 2
Then, the coordinate of reference points A and C are A(2, 0)
C(-1.5, 0)
The radius of the circle is R = CA = 3 C 2 - (-1.5) D 2 + 02 = 3.50 ksi Using these results, the circle shown in Fig. a can be constructed. Referring to the geometry of the circle, Fig. a, b = 60° Then, sx¿ = -1.50 + 3.50 cos 60° - 0.250 ksi
Ans.
tx¿y¿ = 3.50 sin 60° = 3.03 ksi
Ans.
sy¿ = -3.25 ksi
Ans.
The results are shown in Fig b.
687
2 ksi
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9–63. Determine the principal stress, the maximum in-plane shear stress, and average normal stress. Specify the orientation of the element in each case. 15 ksi 5 ksi
Construction of the Circle: In accordance with the sign convention, sx = 15 ksi, sy = 0 and txy = -5 ksi. Hence, sx + sy
savg =
=
2
15 + 0 = 7.50 ksi 2
Ans.
The coordinates for reference point A and C are A(15, -5)
C(7.50, 0)
The radius of the circle is R = 2(15 - 7.50)2 + 52 = 9.014 ksi a) In - Plane Principal Stress: The coordinates of points B and D represent s1 and s2, respectively. s1 = 7.50 + 9.014 = 16.5 ksi
Ans.
s2 = 7.50 - 9.014 = -1.51 ksi
Ans.
Orientation of Principal Plane: From the circle tan 2uP1 =
5 = 0.6667 15 - 7.50
uP1 = 16.8° (Clockwise)
Ans.
b) Maximum In - Plane Shear Stress: Represented by the coordinates of point E on the circle. tmax
in-plane
= -R = -9.01 ksi
Ans.
Orientation of the Plane for Maximum In - Plane Shear Stress: From the circle tan 2us =
15 - 7.50 = 1.500 5
us = 28.2° (Counterclockwise)
Ans.
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*9–64. Determine the principal stress, the maximum in-plane shear stress, and average normal stress. Specify the orientation of the element in each case.
20 MPa 80 MPa
30 MPa
In accordance to the established sign convention, sx = 30 MPa, sy = -20 MPa and txy = 80 MPa. Thus, savg =
sx + sy
30 + ( -20) = 5 MPa 2
=
2
Then, the coordinates of reference point A and the center C of the circle is A(30, 80)
C(5, 0)
Thus, the radius of circle is given by R = CA = 2(30 - 5)2 + (80 - 0)2 = 83.815 MPa Using these results, the circle shown in Fig. a, can be constructed. The coordinates of points B and D represent s1 and s2 respectively. Thus s1 = 5 + 83.815 = 88.8 MPa
Ans.
s2 = 5 - 83.815 = -78.8 MPa
Ans.
Referring to the geometry of the circle, Fig. a tan 2(uP)1 =
80 = 3.20 30 - 5
uP = 36.3° (Counterclockwise)
Ans.
The state of maximum in - plane shear stress is represented by the coordinate of point E. Thus tmax
in-plane
= R = 83.8 MPa
Ans.
From the geometry of the circle, Fig. a, tan 2us =
30 - 5 = 0.3125 80
us = 8.68° (Clockwise)
Ans.
The state of maximum in - plane shear stress is represented by the element in Fig. c
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9–64.
Continued
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•9–65. Determine the principal stress, the maximum inplane shear stress, and average normal stress. Specify the orientation of the element in each case.
120 psi
300 psi
A(300, 120)
B(0, -120)
C(150, 0)
R = 2(300 - 150)2 + 1202 = 192.094 s1 = 150 + 192.094 = 342 psi
Ans.
s2 = 150 - 192.094 = -42.1 psi
Ans.
tan 2uP =
120 = 0.8 300 - 150
uP1 = 19.3° Counterclockwise
Ans.
savg = 150 psi
Ans.
tmax
Ans.
in-plane
= 192 psi
tan 2us =
300 - 150 = 1.25 120
us = -25.7°
Ans.
9–66. Determine the principal stress, the maximum in-plane shear stress, and average normal stress. Specify the orientation of the element in each case. A(45, -50)
B(30, 50)
30 MPa
C(37.5, 0)
45 MPa
R = CA = CB = 27.52 + 502 = 50.56 50 MPa
a)
tan 2uP =
50 7.5
s1 = 37.5 + 50.56 = 88.1 MPa
Ans.
s2 = 37.5 - 50.56 = -13.1 MPa
Ans.
uP = -40.7°
2uP = 81.47°
b) t
max in-plane
= R = 50.6 MPa
Ans.
savg = 37.5 MPa
Ans.
2us = 90 - 2uP us = 4.27°
Ans.
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9–67. Determine the principal stress, the maximum in-plane shear stress, and average normal stress. Specify the orientation of the element in each case.
200 MPa 500 MPa
350 MPa
Construction of the Circle: In accordance with the sign convention, sx = 350 MPa, sy = -200 MPa, and txy = 500 MPa. Hence, savg =
sx + sy 2
=
350 + (-200) = 75.0 MPa 2
Ans.
The coordinates for reference point A and C are A(350, 500)
C(75.0, 0)
The radius of the circle is R = 2(350 - 75.0)2 + 5002 = 570.64 MPa a) In - Plane Principal Stresses: The coordinate of points B and D represent s1 and s2 respectively. s1 = 75.0 + 570.64 = 646 MPa
Ans.
s2 = 75.0 - 570.64 = -496 MPa
Ans.
Orientaion of Principal Plane: From the circle tan 2uP1 =
500 = 1.82 350 - 75.0
uP1 = 30.6° (Counterclockwise)
Ans.
b) Maximum In - Plane Shear Stress: Represented by the coordinates of point E on the circle. t
max in-plane
= R = 571 MPa
Ans.
Orientation of the Plane for Maximum In - Plane Shear Stress: From the circle tan 2us =
350 - 75.0 = 0.55 500
us = 14.4° (Clockwise)
Ans.
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*9–68. Draw Mohr’s circle that describes each of the following states of stress.
700 psi
4 ksi 40 MPa
600 psi
(a)
a) Here, sx = 600 psi, sy = 700 psi and txy = 0. Thus, savg =
sx + sy =
2
600 + 700 = 650 psi 2
Thus, the coordinate of reference point A and center of circle are A(600, 0)
C(650, 0)
Then the radius of the circle is R = CA = 650 - 600 = 50 psi The Mohr’s circle represents this state of stress is shown in Fig. a. b) Here, sx = 0, sy = 4 ksi and txy = 0. Thus, savg =
sx + sy =
2
0 + 4 = 2 ksi 2
Thus, the coordinate of reference point A and center of circle are A(0, 0)
C(2, 0)
Then the radius of the circle is R = CA = 2 - 0 = 2 psi c) Here, sx = sy = 0 and txy = -40 MPa. Thus, savg =
sx + sy 2
= 0
Thus, the coordinate of reference point A and the center of circle are A(0, -40)
C(0, 0)
Then, the radius of the circle is R = CA = 40 MPa The Mohr’s circle represents this state of stress shown in Fig. c
693
(b)
(c)
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9–68.
Continued
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9–69. The frame supports the distributed loading of 200 N兾m. Determine the normal and shear stresses at point D that act perpendicular and parallel, respectively, to the grain. The grain at this point makes an angle of 30° with the horizontal as shown.
200 N/ m B
30⬚ 1m
200 mm
75 mm D
1.5 m
C 100 mm
4m 60⬚
E
Support Reactions: As shown on FBD(a).
50 mm
30 mm
Internal Forces and Moment: As shown on FBD(b).
1.5 m 100 mm
Section Properties: I =
A
1 (0.1) A 0.23 B = 66.667 A 10 - 6 B m4 12
QD = y¿A¿ = 0.0625(0.075)(0.1) = 0.46875 A 10 - 3 B m3 Normal Stress: Applying the flexure formula. sD = -
My 150(-0.025) = 56.25 kPa = I 66.667(10 - 6)
Shear Stress: Applying the shear formula. tD =
50.0 C 0.46875(10 - 3) D VQD = 3.516 kPa = It 66.667(10 - 6)(0.1)
Construction of the Circle: In accordance to the established sign convention, sx = 56.25 kPa, sy = 0 and txy = -3.516 kPa. Hence. savg =
sx + sy 2
=
56.25 + 0 = 28.125 kPa 2
The coordinates for reference point A and C are A(56.25, -3.516)
C(28.125, 0)
The radius of the circle is R = 2(56.25 - 28.125)2 + 3.5162 = 28.3439 kPa Stresses on The Rotated Element: The normal and shear stress components A sx¿ and tx¿y¿ B are represented by the coordinates of point P on the circle. Here, u = 60°. sx¿ = 28.125 - 28.3439 cos 52.875° = 11.0 kPa
Ans.
tx¿y¿ = -28.3439 sin 52.875° = -22.6 kPa
Ans.
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9–70. The frame supports the distributed loading of 200 N兾m. Determine the normal and shear stresses at point E that act perpendicular and parallel, respectively, to the grain. The grain at this point makes an angle of 60° with the horizontal as shown.
200 N/ m B
30⬚ 1m
200 mm
75 mm D
1.5 m
C 100 mm
4m 60⬚
E
Support Reactions: As shown on FBD(a).
50 mm
30 mm
Internal Forces and Moment: As shown on FBD(b).
1.5 m 100 mm
Section Properties:
A
A = 0.1(0.05) = 5.00 A 10 - 3 B m2 Normal Stress: sE =
N -250 = -50.0 kPa = A 5.00(10 - 3)
Construction of the Circle: In accordance with the sign convention. sx = 0, sy = -50.0 kPa, and txy = 0. Hence. savg =
sx + sy 2
=
0 + (-50.0) = -25.0 kPa 2
The coordinates for reference points A and C are A(0, 0)
C(-25.0, 0)
The radius of circle is R = 25.0 - 0 = 25.0 kPa Stress on the Rotated Element: The normal and shear stress components A sx¿ and tx¿y¿ B are represented by coordinates of point P on the circle. Here, u = 150°. sx = -25.0 + 25.0 cos 60° = -12.5 kPa
Ans.
tx¿y¿ = 25.0 sin 60° = 21.7 kPa
Ans.
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9–71. The stair tread of the escalator is supported on two of its sides by the moving pin at A and the roller at B. If a man having a weight of 300 lb stands in the center of the tread, determine the principal stresses developed in the supporting truck on the cross section at point C. The stairs move at constant velocity.
1.25 ft 30⬚ A C 1.5 ft
30⬚
0.5 ft B 0.5 ft
2 in.
Support Reactions: As shown on FBD (a). Internal Forces and Moment: As shown on FBD (b). Section Properties: A = 2(0.5) = 1.00 in2 I =
1 (0.5) A 23 B = 0.3333 in4 12
QB = y¿A¿ = 0.5(1)(0.5) = 0.250 in3 Normal Stress: s =
sC =
My N ; A I 475.48(0) -137.26 + = -137.26 psi 1.00 0.3333
Shear Stress: Applying the shear formula t =
tC =
VQ . It
79.25(0.250) = 118.87 psi 0.3333(0.5)
Construction of the Circle: In accordance with the sign convention, sx = 0, sy = -137.26 psi, and txy = 118.87 psi. Hence, savg =
sx + sy 2
=
0 + (-137.26) = -68.63 psi 2
The coordinates for reference points A and C are A(0, 118.87)
1 in. C
C(-68.63, 0)
The radius of the circle is R = 2(68.63 - 0)2 + 118.872 = 137.26 psi In - Plane Principal Stress: The coordinates of point B and D represent s1 and s2, respectively. s1 = -68.63 + 137.26 = 68.6 psi
Ans.
s2 = -68.63 - 137.26 = -206 psi
Ans.
697
0.5 in.
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*9–72. The thin-walled pipe has an inner diameter of 0.5 in. and a thickness of 0.025 in. If it is subjected to an internal pressure of 500 psi and the axial tension and torsional loadings shown, determine the principal stress at a point on the surface of the pipe.
200 lb
200 lb 20 lb⭈ft
Section Properties: A = p A 0.2752 - 0.252 B = 0.013125p in2 J =
p A 0.2754 - 0.254 B = 2.84768 A 10 - 3 B in4 2
Normal Stress: Since
0.25 r = = 10, thin wall analysis is valid. t 0.025
slong =
pr 500(0.25) 200 N + = + = 7.350 ksi A 2t 0.013125p 2(0.025)
shoop =
pr 500(0.25) = = 5.00 ksi t 0.025
Shear Stress: Applying the torsion formula, t =
20(12)(0.275) Tc = 23.18 ksi = J 2.84768(10 - 3)
Construction of the Circle: In accordance with the sign convention sx = 7.350 ksi, sy = 5.00 ksi, and txy = -23.18 ksi. Hence, savg =
sx + sy 2
=
7.350 + 5.00 = 6.175 ksi 2
The coordinates for reference points A and C are A(7.350, -23.18)
C(6.175, 0)
The radius of the circle is R = 2(7.350 - 6.175)2 + 23.182 = 23.2065 ksi In - Plane Principal Stress: The coordinates of point B and D represent s1 and s2, respectively. s1 = 6.175 + 23.2065 = 29.4 ksi
Ans.
s2 = 6.175 - 23.2065 = -17.0 ksi
Ans.
698
20 lb⭈ft
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•9–73. The cantilevered rectangular bar is subjected to the force of 5 kip. Determine the principal stress at point A. 1.5 in. A 1.5 in.
1 in. B
1.5 in.
1.5 in.
3 in. 1 in.
15 in.
3 in.
3 5 4
5 kip
Internal Forces and Moment: As shown on FBD. Section Properties: A = 3(6) = 18.0 in2 I =
1 (3) A 63 B = 54.0 in4 12
QA = y¿A¿ = 2.25(1.5)(3) = 10.125 in3 Normal Stress: s =
sA =
My N ; A I 45.0(1.5) 4.00 + = 1.4722 ksi 18.0 54.0
Shear Stress: Applying the shear formula t =
tA =
VQ . It
3.00(10.125) = 0.1875 ksi 54.0(3)
Construction of the Circle: In accordance with the sign convention, sx = 1.4722 ksi, sy = 0, and txy = -0.1875 ksi. Hence, savg =
sx + sy 2
=
1.472 + 0 = 0.7361 ksi 2
The coordinates for reference points A and C are A(1.4722, -0.1875)
C(0.7361, 0)
The radius of the circle is R = 2(1.4722 - 0.7361)2 + 0.18752 = 0.7596 ksi In - Plane Principal Stress: The coordinates of point B and D represent s1 and s2, respectively. s1 = 0.7361 + 0.7596 = 1.50 ksi
Ans.
s2 = 0.7361 - 0.7596 = -0.0235 ksi
Ans.
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9–74. Solve Prob. 9–73 for the principal stress at point B. 1.5 in. A 1.5 in.
1 in. B
1.5 in.
1.5 in.
3 in. 1 in.
15 in.
3 in.
3 5 4
5 kip
Internal Forces and Moment: As shown on FBD. Section Properties: A = 3(6) = 18.0 in2 1 (3) A 63 B = 54.0 in4 12
I =
QB = y¿A¿ = 2(2)(3) = 12.0 in3 Normal Stress: My N ; A I
s =
45.0(1) 4.00 = -0.6111 ksi 18.0 54.0
sB =
Shear Stress: Applying the shear formula t =
tB =
VQ . It
3.00(12.0) = 0.2222 ksi 54.0(3)
Construction of the Circle: In accordance with the sign convention, sx = -0.6111 ksi, sy = 0, and txy = -0.2222 ksi. Hence. savg =
sx + sy 2
=
-0.6111 + 0 = -0.3055 ksi 2
The coordinates for reference points A and C are A(-0.6111, -0.2222)
C(-0.3055, 0)
The radius of the circle is R = 2(0.6111 - 0.3055)2 + 0.22222 = 0.3778 ksi In - Plane Principal Stress: The coordinates of point B and D represent s1 and s2, respectively. s1 = -0.3055 + 0.3778 = 0.0723 ksi
Ans.
s2 = -0.3055 - 0.3778 = -0.683 ksi
Ans.
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9–75. The 2-in.-diameter drive shaft AB on the helicopter is subjected to an axial tension of 10 000 lb and a torque of 300 lb # ft. Determine the principal stress and the maximum in-plane shear stress that act at a point on the surface of the shaft.
s =
10 000 P = = 3.183 ksi A p(1)2
t =
300(12)(1) Tc = = 2.292 ksi p 4 J 2 (1) s1, 2 =
=
t
sx + sy ;
2
A
(
sx - sy 2
B A
)2 + t2xy
3.183 - 0 2 3.183 + 0 ; ( ) + (2.292)2 2 A 2
s1 = 4.38 ksi
Ans.
s2 = -1.20 ksi
Ans.
max in-plane
=
A
(
=
A
(
sx - sy 2
)2 + t2xy
3.183 - 0 2 ) + (2.292)2 2
= 2.79 ksi
Ans.
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*9–76. The pedal crank for a bicycle has the cross section shown. If it is fixed to the gear at B and does not rotate while subjected to a force of 75 lb, determine the principal stress in the material on the cross section at point C.
75 lb B 3 in.
A 4 in.
C 0.4 in. 0.4 in.
0.2 in. 0.3 in.
Internal Forces and Moment: As shown on FBD Section Properties: I =
1 (0.3) A 0.83 B = 0.0128 in3 12
QC = y¿A¿ = 0.3(0.2)(0.3) = 0.0180 in3 Normal Stress: Applying the flexure formula. sC = -
My -300(0.2) = = 4687.5 psi = 4.6875 ksi I 0.0128
Shear Stress: Applying the shear formula. tC =
VQC 75.0(0.0180) = = 351.6 psi = 0.3516 ksi It 0.0128(0.3)
Construction of the Circle: In accordance with the sign convention, sx = 4.6875 ksi, sy = 0, and txy = 0.3516 ksi. Hence, savg =
sx + sy 2
=
4.6875 + 0 = 2.34375 ksi 2
The coordinates for reference points A and C are A(4.6875, 0.3516)
C(2.34375, 0)
The radius of the circle is R = 2(4.6875 - 2.34375)2 + 0.35162 = 2.3670 ksi In - Plane Principal Stress: The coordinates of point B and D represent s1 and s2, respectively. s1 = 2.34375 + 2.3670 = 4.71 ksi
Ans.
s2 = 2.34375 - 2.3670 = -0.0262 ksi
Ans.
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•9–77.
A spherical pressure vessel has an inner radius of 5 ft and a wall thickness of 0.5 in. Draw Mohr’s circle for the state of stress at a point on the vessel and explain the significance of the result. The vessel is subjected to an internal pressure of 80 psi.
Normal Stress: s1 = s2 =
pr 80(5)(12) = = 4.80 ksi 2t 2(0.5)
Mohr’s circle: A(4.80, 0)
B(4.80, 0)
C(4.80, 0)
Regardless of the orientation of the element, the shear stress is zero and the state of stress is represented by the same two normal stress components.
9–78. The cylindrical pressure vessel has an inner radius of 1.25 m and a wall thickness of 15 mm. It is made from steel plates that are welded along the 45° seam. Determine the normal and shear stress components along this seam if the vessel is subjected to an internal pressure of 8 MPa.
sx =
45⬚ 1.25 m
pr 8(1.25) = = 333.33 MPa 2t 2(0.015)
sy = 2sx = 666.67 MPa A(333.33, 0) sx¿ =
B(666.67, 0)
C(500, 0)
333.33 + 666.67 = 500 MPa 2
Ans.
tx¿y¿ = R = 666.67 - 500 = 167 MPa
Ans.
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•9–79.
Determine the normal and shear stresses at point D that act perpendicular and parallel, respectively, to the grains. The grains at this point make an angle of 30° with the horizontal as shown. Point D is located just to the left of the 10-kN force.
10 kN
A 100 mm
D
B 30⬚
1m
100 mm
D
100 mm
Using the method of section and consider the FBD of the left cut segment, Fig. a + c ©Fy = 0;
5 - V = 0
a + ©MC = 0;
V = 5 kN M = 5 kN # m
M - 5(1) = 0
The moment of inertia of the rectangular cross - section about the neutral axis is I =
1 (0.1)(0.33) = 0.225(10 - 3) m4 12
Referring to Fig. b, QD = y¿A¿ = 0.1(0.1)(0.1) = 0.001 m3 The normal stress developed is contributed by bending stress only. For point D, y = 0.05 m. Then s =
My 5(103)(0.05) = 1.111 MPa (T) = I 0.225(10 - 3)
The shear stress is contributed by the transverse shear stress only. Thus, t =
5(103)(0.001) VQD = 0.2222 MPa = It 0.225(10 - 3)(0.1)
The state of stress at point D can be represented by the element shown in Fig. c In accordance to the established sign convention, sx = 1.111 MPa, sy = 0 and txy = -0.2222 MPa, Thus. savg =
sx + sy 2
=
1.111 + 0 = 0.5556 MPa 2
Then, the coordinate of reference point A and the center C of the circle are A(1.111, -0.2222)
C(0.5556, 0)
Thus, the radius of the circle is given by R = 2(1.111 - 0.5556)2 + (-0.2222)2 = 0.5984 MPa Using these results, the circle shown in Fig. d can be constructed. Referring to the geometry of the circle, Fig. d, a = tan - 1 a
0.2222 b = 21.80° 1.111 - 0.5556
b = 180° - (120° - 21.80°) = 81.80°
704
1m 300 mm
2m C
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9–79. Continued Then sx¿ = 0.5556 - 0.5984 cos 81.80° = 0.4702 MPa = 470 kPa
Ans.
tx¿y¿ = 0.5984 sin 81.80° = 0.5922 MPa = 592 kPa
Ans.
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*9–80. Determine the principal stress at point D, which is located just to the left of the 10-kN force.
10 kN
A 100 mm
D
B 30⬚
1m
100 mm
D
100 mm
Using the method of section and consider the FBD of the left cut segment, Fig. a, + c ©Fy = 0;
5 - V = 0
a + ©MC = 0;
V = 5 kN M = 5 kN # m
M - 5(1) = 0 I =
1 (0.1)(0.33) = 0.225(10 - 3) m4 12
Referring to Fig. b, QD = y¿A¿ = 0.1(0.1)(0.1) = 0.001 m3 The normal stress developed is contributed by bending stress only. For point D, y = 0.05 m s =
My 5(103)(0.05) = 1.111 MPa (T) = I 0.225(10 - 3)
The shear stress is contributed by the transverse shear stress only. Thus, t =
5(103)(0.001) VQD = 0.2222 MPa = It 0.225(10 - 3)(0.1)
The state of stress at point D can be represented by the element shown in Fig. c. In accordance to the established sign convention, sx = 1.111 MPa, sy = 0, and txy = -0.2222 MPa. Thus, savg =
sx + sy 2
=
1.111 + 0 = 0.5556 MPa 2
Then, the coordinate of reference point A and center C of the circle are A(1.111, -0.2222)
C(0.5556, 0)
Thus, the radius of the circle is R = CA = 2(1.111 - 0.5556)2 + (-0.2222)2 = 0.5984 MPa Using these results, the circle shown in Fig. d. In-Plane Principal Stresses. The coordinates of points B and D represent s1 and s2, respectively. Thus, s1 = 0.5556 + 0.5984 = 1.15 MPa
Ans.
s2 = 0.5556 - 0.5984 = -0.0428 MPa
Ans.
706
1m 300 mm
2m C
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9–80. Continued Referring to the geometry of the circle, Fig. d, tan (2uP)1 =
0.2222 = 0.4 1.111 - 0.5556
(uP)1 = 10.9° (Clockwise)
Ans.
The state of principal stresses is represented by the element show in Fig. e.
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•9–81.
Determine the principal stress at point A on the cross section of the hanger at section a–a. Specify the orientation of this state of stress and indicate the result on an element at the point.
0.75 m
0.75 m a
250 mm a
900 N
Internal Loadings: Considering the equilibrium of the free - body diagram of the hanger’s left cut segment, Fig. a, + ©F = 0; : x
900 - N = 0
N = 900 N
+ c ©Fy = 0;
V - 900 = 0
V = 900 N
a + ©MO = 0;
900(1) - 900(0.25) - M = 0
M = 675 N # m
b
250 mm
Section Properties: The cross - sectional area and the moment of inertia about the centroidal axis of the hanger’s cross section are A = 0.05(0.1) - 0.04(0.09) = 1.4 A 10 - 3 B m2 1 1 (0.05) A 0.13 B (0.04) A 0.093 B = 1.7367 A 10 - 6 B m4 12 12
Referring to Fig. b, QA = 2y1œ A1œ + y2œ A2œ = 2[0.0375(0.025)(0.005)] + 0.0475(0.005)(0.04) = 18.875 A 10 - 6 B m3 Normal and Shear Stress: The normal stress is a combination of axial and bending stresses. Thus, sA =
675(0.025) MyA N 900 + = + = 9.074 MPa -3 A I 1.4 A 10 B 1.7367 A 10 - 6 B
The shear stress is caused by the transverse shear stress. tA =
900 C 18.875 A 10 - 6 B D VQA = = 0.9782 MPa It 1.7367 A 10 - 6 B (0.01)
The state of stress at point A is represented by the element shown in Fig. c. Construction of the Circle: sx = 9.074 MPa, sy = 0, and txy = 0.9782 MPa. Thus, savg =
sx + sy 2
=
9.074 + 0 = 4.537 MPa 2
The coordinates of reference points A and the center C of the circle are A(9.074, 0.9782)
C(4.537, 0)
Thus, the radius of the circle is R = CA = 2(9.074 - 4.537)2 + 0.97822 = 4.641 MPa Using these results, the circle is shown in Fig. d.
708
b
900 N 5 mm
25 mm A 100 mm
5 mm 50 mm
I =
0.5 m
5 mm
Sections a – a and b – b
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9–81. Continued In - Plane Principal Stress: The coordinates of point B and D represent s1 and s2, respectively. s1 = 4.537 + 4.641 = 9.18 MPa
Ans.
s2 = 4.537 - 4.641 = -0.104 MPa
Ans.
Orientaion of Principal Plane: Referring to the geometry of the circle, Fig. d, tan 2 A uP B 1 =
0.9782 = 0.2156 9.074 - 4.537
A uP B 1 = 6.08° (counterclockwise)
Ans.
The state of principal stresses is represented on the element shown in Fig. e.
709
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9–82. Determine the principal stress at point A on the cross section of the hanger at section b–b. Specify the orientation of the state of stress and indicate the results on an element at the point.
0.75 m
0.75 m a
250 mm a
900 N
Internal Loadings: Considering the equilibrium of the free - body diagram of the hanger’s left cut segment, Fig. a, V - 900 - 900 = 0
+ c ©Fy = 0; a + ©MO = 0;
900(2.25) + 900(0.25) - M = 0
Referring to Fig. b. QA = 2y1œ A1œ + y2œ A2œ = 2[0.0375(0.025)(0.005)] + 0.0475(0.005)(0.04) = 18.875 A 10 - 6 B m3 Normal and Shear Stress: The normal stress is contributed by the bending stress only. MyA 2250(0.025) = = 32.39 MPa I 1.7367 A 10 - 6 B
The shear stress is contributed by the transverse shear stress only. 1800 C 18.875 A 10 - 6 B D VQA = = 1.956 MPa It 1.7367 A 10 - 6 B (0.01)
The state stress at point A is represented by the element shown in Fig. c. Construction of the Circle: sx = 32.39 MPa, sy = 0, and txy = 1.956 MPa. Thus, savg =
sx + sy 2
=
32.39 + 0 = 16.19 MPa 2
The coordinates of reference point A and the center C of the circle are A(32.39, 1.956)
C(16.19, 0)
Thus, the radius of the circle is R = CA = 2(32.39 - 16.19)2 + 1.9562 = 16.313 MPa Using these results, the cricle is shown in Fig. d.
710
b
5 mm 25 mm A 100 mm
5 mm 50 mm
5 mm
Sections a – a and b – b
1 1 (0.05) A 0.13 B (0.04) A 0.093 B = 1.7367 A 10 - 6 B m4 12 12
tA =
250 mm
M = 2250 N # m
A = 0.05(0.1) - 0.04(0.09) = 1.4 A 10 - 3 B m2
sA =
b
900 N
V = 1800 N
Section Properties: The cross - sectional area and the moment of inertia about the centroidal axis of the hanger’s cross section are
I =
0.5 m
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9–82. Continued In - Plane Principal Stresses: The coordinates of reference point B and D represent s1 and s2, respectively. s1 = 16.19 + 16.313 = 32.5 MPa
Ans.
s2 = 16.19 - 16.313 = -0.118 MPa
Ans.
Orientaion of Principal Plane: Referring to the geometry of the circle, Fig. d, tan 2 A uP B 1 =
A uP B 1 = 3.44°
1.956 = 0.1208 32.39 - 16.19 Ans.
(counterclockwise)
The state of principal stresses is represented on the element shown in Fig. e.
711
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9–83. Determine the principal stresses and the maximum in-plane shear stress that are developed at point A. Show the results on an element located at this point. The rod has a diameter of 40 mm.
450 N
150 mm
Using the method of sections and consider the FBD of the member’s upper cut segment, Fig. a, + c ©Fy = 0;
450 - N = 0
a + ©MC = 0;
100 mm A
150 mm
N = 450 N
450(0.1) - M = 0
B
M = 45 N # m
A = p(0.022) = 0.4(10 - 3)p m2
I =
450 N
p (0.024) = 40(10 - 9)p m4 4
The normal stress is the combination of axial and bending stress. Thus, s =
My N + A I
For point A, y = C = 0.02 m. s =
45 (0.02) 450 + = 7.520 MPa 0.4(10 - 3)p 40(10 - 9)p
Since no transverse shear and torque is acting on the cross - section t = 0 The state of stress at point A can be represented by the element shown in Fig. b. In accordance to the established sign convention sx = 0, sy = 7.520 MPa and txy = 0. Thus savg =
sx + sy 2
=
0 + 7.520 = 3.760 MPa 2
Then, the coordinates of reference point A and the center C of the circle are A(0, 0)
C(3.760, 0)
Thus, the radius of the circle is R = CA = 3.760 MPa Using this results, the circle shown in Fig. c can be constructed. Since no shear stress acts on the element, s1 = sy = 7.52 MPa
s2 = sx = 0
Ans.
The state of principal stresses can also be represented by the element shown in Fig. b. The state of maximum in - plane shear stress is represented by point B on the circle, Fig. c. Thus. tmax
in-plane
= R = 3.76 MPa
Ans.
712
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9–83. Continued From the circle, 2us = 90° us = 45° (counter clockwise)
Ans.
The state of maximum In - Plane shear stress can be represented by the element shown in Fig. d.
713
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*9–84. Draw the three Mohr’s circles that describe each of the following states of stress.
5 ksi
(a) Here, smin = 0, sint = 3 ksi and smax = 5 ksi. The three Mohr’s circle of this state of stress are shown in Fig. a
3 ksi
(b) Here, smin = 0, sint = 140 MPa and smax = 180 MPa. The three Mohr’s circle of this state of stress are shown in Fig. b (a)
•9–85.
Draw the three Mohr’s circles that describe the following state of stress.
180 MPa
140 MPa (b)
300 psi
Here, smin = -300 psi, sint = 0 and smax = 400 psi. The three Mohr’s circle for this state of stress is shown in Fig. a. 400 psi
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z
9–86. The stress at a point is shown on the element. Determine the principal stress and the absolute maximum shear stress.
y
x 80 MPa
For y – z plane: A(0, -80)
B(90, 80)
C(45, 0)
R = 2452 + 802 = 91.79 s1 = 45 + 91.79 = 136.79 MPa s2 = 45 - 91.79 = -46.79 MPa Thus,
tabs
max
=
s1 = 0
Ans.
s2 = 137 MPa
Ans.
s3 = -46.8 MPa
Ans.
136.79 - (-46.79) smax - smin = = 91.8 MPa 2 2
Ans.
715
90 MPa
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z
9–87. The stress at a point is shown on the element. Determine the principal stress and the absolute maximum shear stress. Mohr’s circle for the element in y - 7 plane, Fig. a, will be drawn first. In accordance to the established sign convention, sy = 30 psi, sz = 120 psi and tyz = 70 psi. Thus savg =
sy + sz 2
=
x
y
120 psi 70 psi
30 + 120 = 75 psi 2 30 psi
Thus the coordinates of reference point A and the center C of the circle are A(30, 70)
C(75, 0)
Thus, the radius of the circle is R = CA = 2(75 - 30)2 + 702 = 83.217 psi Using these results, the circle shown in Fig. b. The coordinates of point B and D represent the principal stresses From the results, smax = 158 psi
smin = -8.22 psi
sint = 0 psi
Ans.
Using these results, the three Mohr’s circle are shown in Fig. c, From the geometry of the three circles, tabs
max
=
158.22 - ( -8.22) smax - smin = = 83.22 psi 2 2
716
Ans.
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z
*9–88. The stress at a point is shown on the element. Determine the principal stress and the absolute maximum shear stress. Mohr’s circle for the element in x - z plane, Fig. a, will be drawn first. In accordance to the established sign convention, sx = -2 ksi, sz = 0 and txz = 8 ksi. Thus savg =
sx + sz 2
=
-2 + 0 = -1 ksi 2
2 ksi 8 ksi
Thus, the coordinates of reference point A and the center C of the circle are A( -2, 8)
C(-1, 0)
Thus, the radius of the circle is R = CA = 2[-2 - (-1)]2 + 82 = 265 ksi Using these results, the circle in shown in Fig. b, The coordinates of points B and D represent s1 and s2, respectively. s = -1 + 265 = 7.062 ksi smax = 7.06 ksi sint = 0 smin = -9.06 ksi From the results obtained, sint = 0 ksi
smax = 7.06 ksi
smin = -9.06 ksi
Ans.
Using these results, the three Mohr’s circles are shown in Fig, c. From the geometry of the cricle, tabs
max
=
y
x
7.06 - (-9.06) smax - smin = = 8.06 ksi 2 2
Ans.
717
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•9–89.
The stress at a point is shown on the element. Determine the principal stress and the absolute maximum shear stress.
z
y
x
For x – y plane:
150 MPa
120 MPa
R = CA = 2(120 - 60)2 + 1502 = 161.55 s1 = 60 + 161.55 = 221.55 MPa s2 = 60 - 161.55 = -101.55 MPa s1 = 222 MPa tabs
max
=
s2 = 0 MPa
s3 = -102 MPa
Ans.
221.55 - (-101.55) smax - smin = = 162 MPa 2 2
Ans.
9–90. The state of stress at a point is shown on the element. Determine the principal stress and the absolute maximum shear stress.
z
x
For y - z plane: A(5, -4)
B(-2.5, 4)
4 ksi
s1 = 1.25 + 5.483 = 6.733 ksi
5 ksi
s2 = 1.25 - 5.483 = -4.233 ksi Thus,
tabs
max
=
2.5 ksi
C(1.25, 0)
R = 23.752 + 42 = 5.483
savg =
y
s1 = 6.73 ksi
Ans.
s2 = 0
Ans.
s3 = -4.23 ksi
Ans.
6.73 + (-4.23) = 1.25 ksi 2 6.73 - (-4.23) smax - smin = = 5.48 ksi 2 2
Ans.
718
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*9–92. The solid shaft is subjected to a torque, bending moment, and shear force as shown. Determine the principal stress acting at points A and B and the absolute maximum shear stress.
450 mm A B 300 N⭈m
25 mm
45 N⭈m 800 N
Internal Forces and Moment: As shown on FBD. Section Properties: Iz =
p A 0.0254 B = 0.306796 A 10 - 6 B m4 4
J =
p A 0.0254 B = 0.613592 A 10 - 6 B m4 2
(QA)x = 0 (QB)y = y¿A¿ =
4(0.025) 1 c (p) A 0.0252 B d = 10.417 A 10 - 6 B m3 3p 2
Normal stress: Applying the flexure formula. s = -
Mzy Iz -60.0(0.025)
sA = -
0.306796(10 - 6)
= 4.889 MPa
-60.0(0) sB = -
0.306796(10 - 6)
= 0
Shear Stress: Applying the torsion formula for point A, tA =
45.0(0.025) Tc = 1.833 MPa = J 0.613592(10 - 6)
The transverse shear stress in the y direction and the torsional shear stress can be VQ Tr obtained using shear formula and torsion formula. tv = and ttwist = , It J respectively. tB = (tv)y - ttwist =
800 C 10.417(10 - 6) D -6
0.306796(10 )(0.05)
45.0(0.025) -
0.613592(10 - 6)
= -1.290 MPa
Construction of the Circle: sx = 4.889 MPa, sz = 0, and txz = -1.833 MPa for point A. Hence, savg =
sx + sz 2
=
4.889 + 0 = 2.445 MPa 2
The coordinates for reference points A and C are A (4.889, –1.833) and C(2.445, 0).
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9–92.
Continued
The radius of the circle is R = 2(4.889 - 2.445)2 + 1.8332 = 3.056 MPa sx = sy = 0 and txy = -1.290 MPa for point B. Hence, savg =
sx + sz
= 0
2
The coordinates for reference points A and C are A(0. ‚–1.290) and C(0,0). The radius of the circle is R = 1.290 MPa In - Plane Principal Stresses: The coordinates of point B and D represent s1 and s2, respectively. For point A s1 = 2.445 + 3.056 = 5.50 MPa s2 = 2.445 - 3.506 = -0.611 MPa For point B s1 = 0 + 1.290 = 1.29 MPa s2 = 0 - 1.290 = -1.290 MPa Three Mohr’s Circles: From the results obtaired above, the principal stresses for point A are smax = 5.50 MPa
sint = 0
smin = -0.611 MPa
Ans.
sint = 0
smin = -1.29 MPa
Ans.
And for point B smax = 1.29 MPa
Absolute Maximum Shear Stress: For point A, tabs
max
=
5.50 - (-0.611) smax - smin = = 3.06 MPa 2 2
Ans.
1.29 - (-1.29) smax - smin = = 1.29 MPa 2 2
Ans.
For point B, tabs
max
=
720
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•9–93.
The propane gas tank has an inner diameter of 1500 mm and wall thickness of 15 mm. If the tank is pressurized to 2 MPa, determine the absolute maximum shear stress in the wall of the tank.
Normal Stress: Since
750 r = = 50 7 10, thin - wall analysis can be used. We have t 15 s1 =
2(750) pr = = 100 MPa t 15
s2 =
2(750) pr = = 50 MPa 2t 2(15)
The state of stress of any point on the wall of the tank can be represented on the element shown in Fig. a Construction of Three Mohr’s Circles: Referring to the element, smax = 100 MPa
sint = 50 MPa
smin = 0
Using these results, the three Mohr’s circles are shown in Fig. b. Absolute Maximum Shear Stress: From the geometry of three circles, tabs
max
=
smax - smin 100 - 0 = = 50 MPa 2 2
Ans.
721
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9–94. Determine the principal stress and absolute maximum shear stress developed at point A on the cross section of the bracket at section a–a.
12 in.
6 in.
5
3
a
4
a 0.5 in. B
0.25 in. A
0.25 in.
0.25 in. 1.5 in.1.5 in. Section a – a
Internal Loadings: Considering the equilibrium of the free - body diagram of the bracket’s upper cut segment, Fig. a, + c ©Fy = 0;
3 N - 500 a b = 0 5
N = 300 lb
+ ©F = 0; ; x
4 V - 500 a b = 0 5
V = 400 lb
3 4 ©MO = 0; M - 500 a b(12) - 500 a b(6) = 0 5 5
M = 6000 lb # in
Section Properties: The cross - sectional area and the moment of inertia of the bracket’s cross section are A = 0.5(3) - 0.25(2.5) = 0.875 in2 I =
1 1 (0.5) A 33 B (0.25) A 2.53 B = 0.79948 in4 12 12
Referring to Fig. b. QA = x1œ A1œ + x2œ A2œ = 0.625(1.25)(0.25) + 1.375(0.25)(0.5) = 0.3672 in3 Normal and Shear Stress: The normal stress is sA =
N 300 = = -342.86 psi A 0.875
The shear stress is contributed by the transverse shear stress. tA =
400(0.3672) VQA = = 734.85 psi It 0.79948(0.25)
The state of stress at point A is represented by the element shown in Fig. c. Construction of the Circle: sx = 0, sy = -342.86 psi, and txy = 734.85. Thus, savg =
sx + sy 2
=
0 + (-342.86) = -171.43 psi 2
The coordinates of reference point A and the center C of the circle are A(0, 734.85)
C(-171.43, 0)
Thus, the radius of the circle is R = CA = 2[0 - (-171.43)]2 + 734.852 = 754.58 psi 722
500 lb
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9–94. Continued Using these results, the cricle is shown in Fig. d. In - Plane Principal Stresses: The coordinates of reference point B and D represent s1 and s2, respectively. s1 = -171.43 + 754.58 = 583.2 psi s2 = -171.43 - 754.58 = -926.0 psi Three Mohr’s Circles: Using these results, smax = 583 psi
sint = 0 smin = -926 psi
Ans.
Absolute Maximum Shear Stress: tabs
max
=
583.2 - (-926.0) smax - smin = - 755 psi 2 2
Ans.
723
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9–95. Determine the principal stress and absolute maximum shear stress developed at point B on the cross section of the bracket at section a–a.
12 in.
Internal Loadings: Considering the equilibrium of the free - body diagram of the 6 in. bracket’s upper cut segment, Fig. a, a
+ c ©Fy = 0; + ©F = 0; ; x
3 N - 500 a b = 0 5
N = 300 lb
4 V - 500 a b = 0 5
V = 400 lb
1 1 (0.5) A 33 B (0.25) A 2.53 B = 0.79948 in4 12 12
Referring to Fig. b, QB = 0 Normal and Shear Stress: The normal stress is a combination of axial and bending stress. 6000(1.5) MxB N 300 + = + = 10.9 ksi A I 0.875 0.79948
Since QB = 0, tB = 0. The state of stress at point B is represented on the element shown in Fig. c. In - Plane Principal Stresses: Since no shear stress acts on the element, s2 = 0
Three Mohr’s Circles: Using these results, smax = 10.91 ksi
sint = smin = 0
Ans.
Absolute Maximum Shear Stress: tabs
max
=
0.25 in. 1.5 in.1.5 in. Section a – a
M = 6000 lb # in
smax - smin 10.91 - 0 = = 5.46 ksi 2 2
Ans.
724
500 lb
0.25 in. A
0.25 in.
A = 0.5(3) - 0.25(2.5) = 0.875 in2
s1 = 10.91 ksi
a 0.5 in.
Section Properties: The cross - sectional area and the moment of inertia about the centroidal axis of the bracket’s cross section are
sB =
4
B
4 3 ©MO = 0; M - 500 a b(12) - 500 a b(6) = 0 5 5
I =
5
3
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*9–96. The solid propeller shaft on a ship extends outward from the hull. During operation it turns at v = 15 rad>s when the engine develops 900 kW of power. This causes a thrust of F = 1.23 MN on the shaft. If the shaft has an outer diameter of 250 mm, determine the principal stresses at any point located on the surface of the shaft.
0.75 m A T
Power Transmission: Using the formula developed in Chapter 5, P = 900 kW = 0.900 A 106 B N # m>s 0.900(106) P = = 60.0 A 103 B N # m v 15
T0 =
Internal Torque and Force: As shown on FBD. Section Properties: A =
p A 0.252 B = 0.015625p m2 4
J =
p A 0.1254 B = 0.3835 A 10 - 3 B m4 2
Normal Stress: s =
-1.23(106) N = = -25.06 MPa A 0.015625p
Shear Stress: Applying the torsion formula, t =
60.0(103) (0.125) Tc = 19.56 MPa = J 0.3835(10 - 3)
In - Plane Principal Stresses: sx = -25.06 MPa, sy = 0 and txy = 19.56 MPa for any point on the shaft’s surface. Applying Eq. 9-5, s1,2 =
=
sx + sy 2
;
C
a
sx - sy 2
2
b + t2xy
-25.06 - 0 2 -25.06 + 0 ; a b + (19.56)2 2 C 2
= -12.53 ; 23.23 s1 = 10.7 MPa
s2 = -35.8 MPa
Ans.
725
F
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•9–97. The solid propeller shaft on a ship extends outward from the hull. During operation it turns at v = 15 rad>s when the engine develops 900 kW of power. This causes a thrust of F = 1.23 MN on the shaft. If the shaft has a diameter of 250 mm, determine the maximum in-plane shear stress at any point located on the surface of the shaft.
0.75 m A T
Power Transmission: Using the formula developed in Chapter 5, P = 900 kW = 0.900 A 106 B N # m>s T0 =
0.900(106) P = = 60.0 A 103 B N # m v 15
Internal Torque and Force: As shown on FBD. Section Properties: A =
p A 0.252 B = 0.015625p m2 4
J =
p A 0.1254 B = 0.3835 A 10 - 3 B m4 2
Normal Stress: s =
-1.23(106) N = = -25.06 MPa A 0.015625p
Shear Stress: Applying the torsion formula. t =
60.0(103) (0.125) Tc = 19.56 MPa = J 0.3835 (10 - 3)
Maximum In - Plane Principal Shear Stress: sx = -25.06 MPa, sy = 0, and txy = 19.56 MPa for any point on the shaft’s surface. Applying Eq. 9-7, t
max in-plane
a
sx - sy
2
b + t2xy
=
C
=
-25.06 - 0 2 b + (19.56)2 C 2
2
a
= 23.2 MPa
Ans.
726
F
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9–98. The steel pipe has an inner diameter of 2.75 in. and an outer diameter of 3 in. If it is fixed at C and subjected to the horizontal 20-lb force acting on the handle of the pipe wrench at its end, determine the principal stresses in the pipe at point A, which is located on the surface of the pipe.
20 lb 12 in.
10 in.
Internal Forces, Torque and Moment: As shown on FBD. A
Section Properties:
B
I =
p A 1.54 - 1.3754 B = 1.1687 in4 4
J =
p A 1.54 - 1.3754 B = 2.3374 in4 2
C y z
(QA)z = ©y¿A¿
x
4(1.5) 1 4(1.375) 1 = c p A 1.52 B d c p A 1.3752 B d 3p 2 3p 2 = 0.51693 in3 Normal Stress: Applying the flexure formula s =
sA =
My z Iy
,
200(0) = 0 1.1687
Shear Stress: The transverse shear stress in the z direction and the torsional shear VQ stress can be obtained using shear formula and torsion formula, tv = and It Tr ttwist = , respectively. J tA = (tv)z - ttwist =
20.0(0.51693) 240(1.5) 1.1687(2)(0.125) 2.3374
= -118.6 psi In - Plane Principal Stress: sx = 0, sz = 0 and txz = -118.6 psi for point A. Applying Eq. 9-5 s1,2 =
sx + sz 2
;
C
a
sx - sz 2
2
b + t2xz
= 0 ; 20 + (-118.6)2 s1 = 119 psi
s2 = -119 psi
Ans.
727
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9–99. Solve Prob. 9–98 for point B, which is located on the surface of the pipe.
20 lb 12 in.
10 in.
A B
Internal Forces, Torque and Moment: As shown on FBD. Section Properties:
C
I =
p A 1.54 - 1.3754 B = 1.1687 in4 4
y z x
p J = A 1.54 - 1.3754 B = 2.3374 in4 2 (QB)z = 0 Normal Stress: Applying the flexure formula s =
sB =
My z Iv
,
200(1.5) = 256.7 psi 1.1687
Shear Stress: Torsional shear stress can be obtained using torsion formula, Tr . ttwist = J tB = ttwist =
240(1.5) = 154.0 psi 2.3374
In - Plane Prinicipal Stress: sx = 256.7 psi, sy = 0, and txy = -154.0 psi for point B. Applying Eq. 9-5 s1,2 =
=
sx + sy 2
;
C
sx - sy
a
2
2
b + t2xy
256.7 - 0 2 256.7 + 0 ; a b + ( -154.0)2 2 C 2
= 128.35 ; 200.49 s1 = 329 psi
s2 = -72.1 psi
Ans.
728
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*9–100. The clamp exerts a force of 150 lb on the boards at G. Determine the axial force in each screw, AB and CD, and then compute the principal stresses at points E and F. Show the results on properly oriented elements located at these points. The section through EF is rectangular and is 1 in. wide.
A
150 lb
C
G 0.5 in. E
Support Reactions: FBD(a). a + ©MB = 0; + c ©Fy = 0;
F
FCD(3) - 150(7) = 0
FCD = 350 lb
Ans.
350 - 150 - FAB = 0
FAB = 200 lb
Ans.
B
1.5 in. 1.5 in.
Internal Forces and Moment: As shown on FBD(b). Section Properties: I =
1 (1) A 1.53 B = 0.28125 in4 12
QE = 0 QF = y¿A¿ = 0.5(0.5)(1) = 0.250 in3 Normal Stress: Applying the flexure formula s = -
My , I
sE = -
-300(0.75) = 800 psi 0.28125
sF = -
-300(0.25) = 266.67 psi 0.28125 VQ , It
Shear Stress: Applying the shear formula t =
tE =
200(0) = 0 0.28125(1)
tF =
200(0.250) = 177.78 psi 0.28125(1)
In - Plane Principal Stress: sx = 800 psi, sy = 0 and txy = 0 for point E. Since no shear stress acts upon the element. s1 = sx = 800 psi
Ans.
s2 = sy = 0
Ans.
sx = 266.67 psi, sy = 0, and txy = 177.78 psi for point F. Applying Eq. 9-5 s1,2 =
=
sx + sy 2
;
C
sx - sy
a
2
2
b + t2xy
266.67 - 0 2 266.67 + 0 ; a b + 177.782 2 C 2
= 133.33 ; 222.22 s1 = 356 psi
s2 = -88.9 psi
Ans.
729
150 lb
D 4 in.
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9–100.
Continued
Orientation of Principal Plane: Applying Eq. 9-4 for point F, tan 2up =
txy
A sx - sy B >2
up = 26.57°
=
and
177.78 = 1.3333 (266.67 - 0)>2 -63.43°
Substituting the results into Eq. 9-1 with u = 26.57° yields sx¿ =
=
sx + sy 2
sx - sy +
2
cos 2u + txy sin 2u
266.67 - 0 266.67 + 0 + cos 53.13° + 177.78 sin 53.13° 2 2
= 356 psi = s1 Hence, up1 = 26.6°
up2 = -63.4°
Ans.
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9–101. The shaft has a diameter d and is subjected to the loadings shown. Determine the principal stress and the maximum in-plane shear stress that is developed anywhere on the surface of the shaft.
F
T0
F T0
Internal Forces and Torque: As shown on FBD(b). Section Properties: A =
p 2 d 4
J =
p d 4 p 4 a b = d 2 2 32
Normal Stress: N -F 4F = p 2 = - 2 A pd 4 d
s =
Shear Stress: Applying the shear torsion formula, t =
T0 A d2 B 16T0 Tc = p 4 = J d pd3 32
16T0 4F , sy = 0, and txy = for any point on pd2 pd3 the shaft’s surface. Applying Eq. 9-5, In - Plane Principal Stress: sx = -
s1,2 =
sx + sy ;
2 - 4F2 pd
=
C
a
+ 0 ;
2
D
sx - sy 2
¢
- 4F2 pd
2
b + t2xy
- 0
2
2
≤ + a-
16T0 3
pd
b
2
=
64T20 2 -F ; F2 + ≤ 2 ¢ C pd d2
s1 =
64T20 2 -F + F2 + ≤ 2 ¢ C pd d2
Ans.
64T20 2 F + F2 + ≤ 2 ¢ C pd d2
Ans.
s2 = -
Maximum In - Plane Shear Stress: Applying Eq. 9-7, t
max in-plane
=
=
=
C
a
D
¢
sx - sy 2 - 4F2 pd 2
2
b + t2xy
- 0
2
≤ + a-
16T0 pd3
b
2
64T20 2 2 F + pd2 C d2
Ans.
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9–102. The state of stress at a point in a member is shown on the element. Determine the stress components acting on the plane AB.
A 50 MPa
30⬚
28 MPa
100 MPa B
Construction of the Circle: In accordance with the sign convention, sx = -50 MPa, sy = -100 MPa, and txy = -28 MPa. Hence, savg =
sx + sy 2
=
-50 + (-100) = -75.0 MPa 2
The coordinates for reference points A and C are A(–50, –28) and C(–75.0, 0). The radius of the circle is R = 2(75.0 - 50)2 + 282 = 37.54 MPa. Stress on the Rotated Element: The normal and shear stress components A sx¿ and tx¿y¿ B are represented by the coordinates of point P on the circle sx¿ = -75.0 + 37.54 cos 71.76° = -63.3 MPa
Ans.
tx¿y¿ = 37.54 sin 71.76° = 35.7 MPa
Ans.
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9–103. The propeller shaft of the tugboat is subjected to the compressive force and torque shown. If the shaft has an inner diameter of 100 mm and an outer diameter of 150 mm, determine the principal stress at a point A located on the outer surface. 10 kN
A 2 kN·m
Internal Loadings: Considering the equilibrium of the free - body diagram of the propeller shaft’s right segment, Fig. a, ©Fx = 0; 10 - N = 0
N = 10 kN
©Mx = 0; T - 2 = 0
T = 2 kN # m
Section Properties: The cross - sectional area and the polar moment of inertia of the propeller shaft’s cross section are A = p A 0.0752 - 0.052 B = 3.125p A 10 - 3 B m2 J =
p A 0.0754 - 0.054 B = 12.6953125p A 10 - 6 B m4 2
Normal and Shear Stress: The normal stress is a contributed by axial stress only. sA =
10 A 103 B N = = -1.019 MPa A 3.125p A 10 - 3 B
The shear stress is contributed by the torsional shear stress only. tA =
2 A 103 B (0.075) Tc = = 3.761 MPa J 12.6953125p A 10 - 6 B
The state of stress at point A is represented by the element shown in Fig. b. Construction of the Circle: sx = -1.019 MPa, sy = 0, and txy = -3.761 MPa. Thus, savg =
sx + sy 2
=
-1.019 + 0 = -0.5093 MPa 2
The coordinates of reference point A and the center C of the circle are A(-1.019, -3.761)
C(-0.5093, 0)
Thus, the radius of the circle is R = CA = 2[-1.019 - ( -0.5093)]2 + (-3.761)2 = 3.795 MPa Using these results, the circle is shown is Fig. c. In - Plane Principal Stress: The coordinates of reference points B and D represent s1 and s2, respectively. s1 = -0.5093 + 3.795 = 3.29 MPa
Ans.
s2 = -0.5093 - 3.795 = -4.30 MPa
Ans.
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9–103.
Continued
Orientation of the Principal Plane: Referring to the geometry of the circle, Fig. d, tan 2 A up B 2 =
3.761 = 7.3846 1.019 - 0.5093
A up B 2 = 41.1° (clockwise)
Ans.
The state of principal stresses is represented on the element shown in Fig. d.
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*9–104. The box beam is subjected to the loading shown. Determine the principal stress in the beam at points A and B.
6 in. A 6 in. B 8 in. 8 in.
Support Reactions: As shown on FBD(a). Internal Forces and Moment: As shown on FBD(b). Section Properties: I =
1 1 (8) A 83 B (6) A 63 B = 233.33 in4 12 12
QA = QB = 0 Normal Stress: Applying the flexure formula. s = -
My I
sA = -
-300(12)(4) = 61.71 psi 233.33
sB = -
-300(12)(-3) = -46.29 psi 233.33
1200 lb
800 lb
Shear Stress: Since QA = QB = 0, then tA = tB = 0. In - Plane Principal Stress: sx = 61.71 psi, sy = 0, and txy = 0 for point A. Since no shear stress acts on the element, s1 = sx = 61.7 psi
Ans.
s2 = sy = 0
Ans.
sx = -46.29 psi, sy = 0, and txy = 0 for point B. Since no shear stress acts on the element, s1 = sy = 0
Ans.
s2 = sx = -46.3 psi
Ans.
735
A B 3 ft
2.5 ft
2.5 ft
5 ft
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•9–105.
The wooden strut is subjected to the loading shown. Determine the principal stresses that act at point C and specify the orientation of the element at this point. The strut is supported by a bolt (pin) at B and smooth support at A.
50 N
50 N 60⬚ C
100 mm
40 N
40 N B
A 25 mm
50 mm 200 mm 200 mm 200 mm 200 mm 100 mm 100 mm
QC = y¿A¿ = 0.025(0.05)(0.025) = 31.25(10 - 6) m3 1 (0.025)(0.13) = 2.0833(10 - 6) m4 12
I =
Normal stress: sC = 0 Shear stress: VQC 44(31.25)(10 - 6) = 26.4 kPa = It 2.0833(10 - 6)(0.025)
t =
Principal stress: sx = sy = 0; s1,2 =
txy = -26.4 kPa
sx + sy ;
2
C
a
sx - sy 2
2
b + t2 xy
= 0 ; 20 + (26.4)2 s1 = 26.4 kPa
s2 = -26.4 kPa
;
Ans.
Orientation of principal stress: tan 2up =
txy (sx - sy)
= - q
2 up = +45° and -45° Use sx¿
Eq. 9-1 to determine the principal sx + sy sx - sy = + cos 2u + txy sin 2u 2 2
plane
of
s1
and
s2
u = up = -45° sx¿ = 0 + 0 + (-26.4) sin( -90°) = 26.4 kPa Therefore, up1 = -45°;
up2 = 45°
Ans.
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9–106. The wooden strut is subjected to the loading shown. If grains of wood in the strut at point C make an angle of 60° with the horizontal as shown, determine the normal and shear stresses that act perpendicular and parallel to the grains, respectively, due to the loading. The strut is supported by a bolt (pin) at B and smooth support at A.
50 N
50 N 60⬚ C
100 mm
40 N
40 N B
A 25 mm
50 mm 200 mm 200 mm 200 mm 200 mm 100 mm 100 mm
QC = y¿A¿ = 0.025(0.05)(0.025) = 31.25(10 - 6) m3 I =
1 (0.025)(0.13) = 2.0833(10 - 6) m4 12
Normal stress: sC = 0 Shear stress: t =
VQC 44(31.25)(10 - 6) = 26.4 kPa = It 2.0833(10 - 6)(0.025)
Stress transformation: sx = sy = 0; sx¿ =
sx + sy
sx - sy +
2
2
txy = -26.4 kPa;
u = 30°
cos 2u + txy sin 2u
= 0 + 0 + (-26.4) sin 60° = -22.9 kPa tx¿y¿ = -
sx - sy 2
Ans.
sin 2u + txy cos 2u
= -0 + (-26.4) cos 60° = -13.2 kPa
Ans.
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10–1. Prove that the sum of the normal strains in perpendicular directions is constant.
ex¿ =
ey¿ =
ex + ey 2
ex - ey +
ex + ey 2
2 ex - ey
-
2
cos 2u +
cos 2u -
gxy 2 gxy 2
sin 2u
(1)
sin 2u
(2)
Adding Eq. (1) and Eq. (2) yields: ex¿ + ey¿ = ex + ey = constant
QED
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10–2. The state of strain at the point has components of Px = 200 110-62, Py = -300 110-62, and gxy = 400(10-62. Use the strain-transformation equations to determine the equivalent in-plane strains on an element oriented at an angle of 30° counterclockwise from the original position. Sketch the deformed element due to these strains within the x–y plane.
y x
In accordance to the established sign convention, ex = 200(10 - 6), ex¿ =
ex + ey
ex - ey +
2
= c
ey = -300(10 - 6)
2
cos 2u +
gxy 2
gxy = 400(10 - 6)
u = 30°
sin 2u
200 - (-300) 200 + (-300) 400 + cos 60° + sin 60° d(10 - 6) 2 2 2
= 248 (10 - 6) gx¿y¿ 2
= -a
Ans.
ex - ey 2
b sin 2u +
gxy 2
cos 2u
gx¿y¿ = e - C 200 - ( -300) D sin 60° + 400 cos 60° f(10 - 6) = -233(10 - 6) ey¿ =
ex + ey
= c
2
Ans.
ex - ey -
2
cos 2u -
gxy 2
sin 2u
200 - ( -300) 200 + (-300) 400 cos 60° sin 60° d(10 - 6) 2 2 2
= -348(10 - 6)
Ans.
The deformed element of this equivalent state of strain is shown in Fig. a
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10–3. A strain gauge is mounted on the 1-in.-diameter A-36 steel shaft in the manner shown. When the shaft is rotating with an angular velocity of v = 1760 rev>min, the reading on the strain gauge is P = 800110-62. Determine the power output of the motor. Assume the shaft is only subjected to a torque.
v = (1760 rev>min)a
60⬚
2p rad 1 min ba b = 184.307 rad>s 60 sec 1 rev
ex = ey = 0 ex¿ =
ex + ey 2
ex - ey +
2
800(10 - 6) = 0 + 0 +
cos 2u +
gxy 2
gxy 2
sin 2u
sin 120°
gxy = 1.848(10 - 3) rad t = G gxy = 11(103)(1.848)(10 - 3) = 20.323 ksi
t =
Tc ; J
20.323 =
T(0.5) p 2
(0.5)4
;
T = 3.99 kip # in = 332.5 lb # ft P = Tv = 0.332.5 (184.307) = 61.3 kips # ft>s = 111 hp
Ans.
740
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*10–4. The state of strain at a point on a wrench has components Px = 120110-62, Py = -180110-62, gxy = 150110-62. Use the strain-transformation equations to determine (a) the in-plane principal strains and (b) the maximum in-plane shear strain and average normal strain. In each case specify the orientation of the element and show how the strains deform the element within the x–y plane. ex = 120(10 - 6) e1, 2 =
a)
ey = -180(10 - 6)
gxy = 150(10 - 6)
Ex - Ey 2 ex + ey gxy 2 ; a b + a b 2 A 2 2
120 + (-180) 120 - ( -180) 2 150 2 -6 ; a b + a b d 10 2 A 2 2 e1 = 138(10 - 6); e2 = -198(10 - 6) = c
Ans.
Orientation of e1 and e2 gxy 150 = = 0.5 tan 2up = ex - ey [120 - (-180)] up = 13.28° and -76.72° Use Eq. 10.5 to determine the direction of e1 and e2 ex¿ =
ex + ey
ex - ey +
2
2
cos 2u +
gxy 2
sin 2u
u = up = 13.28° ex¿ = c
120 + ( -180) 120 - ( -180) 150 + cos (26.56°) + sin 26.56° d 10 - 6 2 2 2
= 138 (10 - 6) = e1 Therefore up1 = 13.3° ; gmax
b)
=
2 in-plane
ex + ey 2
A
ex - ey
b + a 2
gxy
b
Ans. 2
2 2 150 2 120 - ( -180) 2 -6 -6 = 2c a b + a b d10 = 335 (10 ) 2 2 A
gmax
eavg =
a
in-plane
up2 = -76.7°
= c
120 + (-180) d 10 - 6 = -30.0(10 - 6) 2
Ans.
Ans.
Orientation of gmax tan 2us =
-(ex - ey) gxy
=
-[120 - ( -180)] = -2.0 150
us = -31.7° and 58.3°
Ans.
gmax Use Eq. 10–6 to determine the sign of in-plane gx¿y¿ ex - ey gxy = sin 2u + cos 2u 2 2 2 u = us = -31.7° gx¿y¿ = 2 c -
120 - (-180) 150 sin (-63.4°) + cos (-63.4°) d10 - 6 = 335(10 - 6) 2 2
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10–5. The state of strain at the point on the arm has components Px = 250110-62, Py = -450110-62, gxy = -825110-62. Use the strain-transformation equations to determine (a) the in-plane principal strains and (b) the maximum in-plane shear strain and average normal strain. In each case specify the orientation of the element and show how the strains deform the element within the x–y plane.
ex = 250(10 - 6)
ey = -450(10 - 6)
y
gxy = -825(10 - 6)
x
a) ex + ey
e1, 2 =
;
2
= c
A
ex - ey
a
2
2
b + a
gxy 2
b
2
250 - 450 250 - ( -450) 2 -825 2 -6 ; a b + a b d(10 ) 2 A 2 2
e1 = 441(10 - 6)
Ans.
e2 = -641(10 - 6)
Ans.
Orientation of e1 and e2 : gxy
tan 2up =
ex - ey
up = -24.84°
-825 250 - ( -450)
=
up = 65.16°
and
Use Eq. 10–5 to determine the direction of e1 and e2: ex¿ =
ex + ey
ex - ey +
2
2
cos 2u +
gxy 2
sin 2u
u = up = -24.84° ex¿ = c
250 - (-450) 250 - 450 -825 + cos (-49.69°) + sin (-49.69°) d(10 - 6) = 441(10 - 6) 2 2 2
Therefore, up1 = -24.8°
Ans.
up2 = 65.2°
Ans.
b) g
max in-plane
2 g
max in-plane
eavg =
=
A
= 2c
a
ex - ey 2
2
gxy 2
b
2
250 - (-450) 2 -825 2 -6 -3 b + a b d(10 ) = 1.08(10 ) A 2 2 a
ex + ey 2
b + a
= a
250 - 450 b (10 - 6) = -100(10 - 6) 2
Ans.
Ans.
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10–6. The state of strain at the point has components of Px = -100110-62, Py = 400110-62, and gxy = -300110-62. Use the strain-transformation equations to determine the equivalent in-plane strains on an element oriented at an angle of 60° counterclockwise from the original position. Sketch the deformed element due to these strains within the x–y plane.
y
x
In accordance to the established sign convention, ex = -100(10 - 6) ex¿ =
ex + ey
= c
ex - ey +
2
ey = 400(10 - 6)
2
gxy
cos 2u +
2
gxy = -300(10 - 6)
u = 60°
sin 2u
-100 - 400 -300 -100 + 400 + cos 120° + sin 120° d(10 - 6) 2 2 2
= 145(10 - 6) gx¿y¿ 2
= -a
Ans.
ex - ey 2
b sin 2u +
gxy 2
cos 2u
gx¿y¿ = c -(-100 - 400) sin 120° + (-300) cos 120° d(10 - 6) = 583(10 - 6) ey¿ =
ex + ey
= c
2
Ans.
ex - ey -
2
cos 2u -
gxy 2
sin 2u
-100 - 400 -300 -100 + 400 cos 120° sin 120° d(10 - 6) 2 2 2
= 155 (10 - 6)
Ans.
The deformed element of this equivalent state of strain is shown in Fig. a
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10–7. The state of strain at the point has components of Px = 100110-62, Py = 300110-62, and gxy = -150110-62. Use the strain-transformation equations to determine the equivalent in-plane strains on an element oriented u = 30° clockwise. Sketch the deformed element due to these strains within the x–y plane.
y
x
In accordance to the established sign convention, ex = 100(10 - 6) ex¿ =
ex + ey
ex - ey +
2
= c
ey = 300(10 - 6)
2
cos 2u +
gxy = -150(10 - 6) gxy 2
u = -30°
sin 2u
100 - 300 -150 100 + 300 + cos (-60°) + sin ( -60°) d (10 - 6) 2 2 2
= 215(10 - 6) gx¿y¿ 2
= -a
Ans.
ex - ey 2
b sin 2u +
gxy 2
cos 2u
gx¿y¿ = c -(100 - 300) sin ( -60°) + ( -150) cos ( -60°) d(10 - 6) = -248 (10 - 6) ey¿ =
ex + ey
= c
2
Ans.
ex - ey -
2
cos 2u -
gxy 2
sin 2u
100 - 300 -150 100 + 300 cos ( -60°) sin (-60°) d (10 - 6) 2 2 2
= 185(10 - 6)
Ans.
The deformed element of this equivalent state of strain is shown in Fig. a
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*10–8. The state of strain at the point on the bracket has components Px = -200110-62, Py = -650110-62, gxy ⫽ -175110-62. Use the strain-transformation equations to determine the equivalent in-plane strains on an element oriented at an angle of u = 20° counterclockwise from the original position. Sketch the deformed element due to these strains within the x–y plane.
ex = -200(10 - 6) ex¿ =
ex + ey
ex - ey +
2
= c
ey = -650(10 - 6)
2
cos 2u +
gxy 2
y
x
gxy = -175(10 - 6)
u = 20°
sin 2u
( -200) - (-650) (-175) -200 + (-650) + cos (40°) + sin (40°) d(10 - 6) 2 2 2
= -309(10 - 6) ey¿ =
ex + ey
ex - ey -
2
= c
Ans.
2
cos 2u -
gxy 2
sin 2u
-200 - ( -650) ( -175) -200 + (-650) cos (40°) sin (40°) d(10 - 6) 2 2 2
= -541(10 - 6) gx¿y¿ 2
ex - ey = -
2
Ans. sin 2u +
gxy 2
cos 2u
gx¿y¿ = [-(-200 - (-650)) sin (40°) + (-175) cos (40°)](10 - 6) = -423(10 - 6)
Ans.
745
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10–9. The state of strain at the point has components of Px = 180110-62, Py = -120110-62, and gxy = -100110-62. Use the strain-transformation equations to determine (a) the in-plane principal strains and (b) the maximum in-plane shear strain and average normal strain. In each case specify the orientation of the element and show how the strains deform the element within the x–y plane.
y
x
a)
In
accordance
to
the
established
sign
convention,
ex = 180(10 - 6),
ey = -120(10 - 6) and gxy = -100(10 - 6). ex + ey
e1, 2 =
;
2
= b
a
A
ex - ey 2
2
b + a
gxy 2
b
2
180 + (-120) 180 - ( -120) 2 -100 2 -6 ; c d + a b r (10 ) 2 A 2 2
= A 30 ; 158.11 B (10 - 6) e1 = 188(10 - 6) tan 2uP =
e2 = -128(10 - 6)
gxy
Ans.
-100(10 - 6)
ex - ey
C 180 - (-120) D (10 - 6)
=
uP = -9.217°
and
= -0.3333
80.78°
Substitute u = -9.217°, ex + ey
ex¿ =
2
= c
ex - ey +
2
cos 2u +
gxy 2
sin 2u
180 + ( -120) 180 - ( -120) -100 + cos (-18.43°) + sin (-18.43) d(10 - 6) 2 2 2
= 188(10 - 6) = e1 Thus, (uP)1 = -9.22°
(uP)2 = 80.8°
Ans.
The deformed element is shown in Fig (a). gmax ex - ey 2 gxy 2 in-plane = b) a b + a b 2 A 2 2 gmax
in-plane
tan 2us = - a
= b2
180 - (-120) 2 -100 2 -6 -6 d + a b r (10 ) = 316 A 10 B A 2 2
ex - ey gxy
c
b = -c
C 180 - (-120) D (10 - 6)
us = 35.78° = 35.8° and
-100(10 - 6)
Ans.
s = 3 Ans.
-54.22° = -54.2°
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10–9.
Continued
gmax The algebraic sign for in-plane when u = 35.78°. ex - ey gxy gx¿y¿ = -a b sin 2u + cos 2u 2 2 2 gx¿y¿ = e - C 180 - ( -120) D sin 71.56° + ( -100) cos 71.56° f(10 - 6)
eavg
= -316(10 - 6) ex + ey 180 + (-120) = = c d(10 - 6) = 30(10 - 6) 2 2
Ans.
The deformed element for the state of maximum In-plane shear strain is shown is shown in Fig. b
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10–10. The state of strain at the point on the bracket has components Px = 400110-62, Py = -250110-62, gxy ⫽ 310110-62. Use the strain-transformation equations to determine the equivalent in-plane strains on an element oriented at an angle of u = 30° clockwise from the original position. Sketch the deformed element due to these strains within the x–y plane.
ex = 400(10 - 6) ex¿ =
ex + ey
= c
ex - ey +
2
ey = -250(10 - 6)
2
cos 2u +
gxy 2
gxy = 310(10 - 6)
y
x
u = -30°
sin 2u
400 - ( -250) 400 + ( -250) 310 + cos (-60°) + a b sin (-60°) d(10 - 6) 2 2 2
= 103(10 - 6) ey¿ =
ex + ey
= c
ex - ey -
2
Ans.
2
cos 2u -
gxy 2
sin 2u
400 - (-250) 400 + (-250) 310 cos (60°) sin (-60°) d(10 - 6) 2 2 2
= 46.7(10 - 6) gx¿y¿ 2
ex - ey = -
2
Ans. sin 2u +
gxy 2
cos 2u
gx¿y¿ = [-(400 - (-250)) sin (-60°) + 310 cos ( -60°)](10 - 6) = 718(10 - 6)
748
Ans.
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10–11. The state of strain at the point has components of Px = -100110-62, Py = -200110-62, and gxy = 100110-62. Use the strain-transformation equations to determine (a) the in-plane principal strains and (b) the maximum in-plane shear strain and average normal strain. In each case specify the orientation of the element and show how the strains deform the element within the x–y plane. In accordance to the established ey = -200(10 - 6) and gxy = 100(10 - 6). ex + ey
e1, 2 =
;
2
= b
A
a
ex - ey 2
b + a 2
gxy 2
b
sign
y
x
convention,
ex = -100(10 - 6),
2
-100 + (-200) 100 2 -100 - (-200) 2 -6 ; c d + a b r (10 ) 2 A 2 2
A -150 ; 70.71 B (10 - 6)
=
e1 = -79.3(10 - 6) tan 2uP =
e2 = -221(10 - 6)
gxy
100(10 - 6)
C -100 - (-200) D (10 - 6)
=
ex - ey
uP = 22.5°
and
Ans. = 1
-67.5°
Substitute u = 22.5, ex + ey
ex¿ =
ex - ey
cos 2u +
gxy
sin 2u 2 2 2 -100 + (-200) -100 - (-200) 100 + cos 45° + sin 45° d(10 - 6) = c 2 2 2 +
= -79.3(10 - 6) = e1 Thus, (uP)1 = 22.5°
(uP)2 = -67.5°
Ans.
The deformed element of the state of principal strain is shown in Fig. a gmax ex - ey 2 gxy 2 in-plane = a b + a b 2 A 2 2 gmax
in-plane
= b2
tan 2us = - a
c
-100 - (-200) 2 100 2 -6 -6 d + a b r (10 ) = 141(10 ) A 2 2
ex - ey gxy
b = -c
us = -22.5° The algebraic sign for gx¿y¿ 2
= -a
ex - ey 2
C -100 - ( -200) D (10 - 6) 100(10 - 6)
and gmax
in-plane
b sin 2u +
Ans.
s = -1 Ans.
67.5° when u = -22.5°. gxy 2
cos 2u
gx¿y¿ = - C -100 - (-200) D sin ( -45°) + 100 cos (-45°)
eavg
= 141(10 - 6) ex + ey -100 + ( -200) = = c d(10 - 6) = -150(10 - 6) 2 2
Ans.
The deformed element for the state of maximum In-plane shear strain is shown in Fig. b. 749
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10–11.
Continued
*10–12. The state of plane strain on an element is given by Px = 500110-62, Py = 300110-62, and gxy = -200110-62. Determine the equivalent state of strain on an element at the same point oriented 45° clockwise with respect to the original element.
y
Pydy dy gxy 2
Strain Transformation Equations: ex = 500 A 10 - 6 B
ey = 300 A 10 - 6 B
gxy = -200 A 10 - 6 B
u = -45°
We obtain ex¿ =
ex + ey +
2
= c
ex - ey 2
cos 2u +
gxy 2
sin 2u
500 - 300 -200 500 + 300 + cos (-90°) + a b sin (-90°) d A 10 - 6 B 2 2 2
= 500 A 10 - 6 B gx¿y¿ 2
= -a
Ans.
ex - ey 2
b sin 2u +
gxy 2
cos 2u
gx¿y¿ = [-(500 - 300) sin ( -90°) + (-200) cos ( -90°)] A 10 - 6 B = 200 A 10 - 6 B
ey¿ =
ex + ey
= c
2
Ans.
ex - ey -
2
cos 2u -
gxy 2
sin 2u
500 + 300 500 - 300 -200 cos ( -90°) - a b sin (-90°) d A 10 - 6 B 2 2 2
= 300 A 10 - 6 B
Ans.
The deformed element for this state of strain is shown in Fig. a.
750
gxy 2 dx
x Pxdx
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10–13. The state of plane strain on an element is Px = -300110-62, Py = 0, and gxy = 150110-62. Determine the equivalent state of strain which represents (a) the principal strains, and (b) the maximum in-plane shear strain and the associated average normal strain. Specify the orientation of the corresponding elements for these states of strain with respect to the original element.
y
gxy dy 2 x
In-Plane Principal Strains: ex = -300 A 10 - 6 B , ey = 0, and gxy = 150 A 10 - 6 B . We obtain ex + ey
e1, 2 =
;
2
= C
C
¢
ex - ey 2
2
≤ + ¢
gxy 2
≤
2
-300 + 0 -300 - 0 2 150 2 ; ¢ ≤ + ¢ ≤ S A 10 - 6 B 2 C 2 2
= ( -150 ; 167.71) A 10 - 6 B
e1 = 17.7 A 10 - 6 B
e2 = -318 A 10 - 6 B
Ans.
Orientation of Principal Strain: tan 2up =
gxy ex - ey
=
150 A 10 - 6 B
(-300 - 0) A 10 - 6 B
= -0.5
uP = -13.28° and 76.72° Substituting u = -13.28° into Eq. 9-1, ex¿ =
ex + ey
= c
ex - ey +
2
2
cos 2u +
gxy 2
sin 2u
-300 + 0 -300 - 0 150 + cos (-26.57°) + sin (-26.57°) d A 10 - 6 B 2 2 2
= -318 A 10 - 6 B = e2 Thus,
A uP B 1 = 76.7° and A uP B 2 = -13.3°
Ans.
The deformed element of this state of strain is shown in Fig. a. Maximum In-Plane Shear Strain: gmax ex - ey 2 gxy 2 in-plane = ¢ ≤ + ¢ ≤ 2 C 2 2 gmax
in-plane
-300 - 0 2 150 2 -6 -6 b + a b R A 10 B = 335 A 10 B A 2 2
= B2
a
Ans.
Orientation of the Maximum In-Plane Shear Strain: tan 2us = - ¢
ex - ey gxy
≤ = -C
(-300 - 0) A 10 - 6 B 150 A 10 - 6 B
S = 2
us = 31.7° and 122°
Ans.
751
gxy 2 dx
Pxdx
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10–13.
Continued
The algebraic sign for gx¿y¿ 2
= -¢
ex - ey 2
gmax
in-plane
≤ sin 2u +
when u = us = 31.7° can be obtained using gxy 2
cos 2u
gx¿y¿ = [-(-300 - 0) sin 63.43° + 150 cos 63.43°] A 10 - 6 B = 335 A 10 - 6 B
Average Normal Strain: eavg =
ex + ey 2
= a
-300 + 0 b A 10 - 6 B = -150 A 10 - 6 B 2
Ans.
The deformed element for this state of strain is shown in Fig. b.
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10–14. The state of strain at the point on a boom of an hydraulic engine crane has components of Px = 250110-62, Py = 300110-62, and gxy = -180110-62. Use the straintransformation equations to determine (a) the in-plane principal strains and (b) the maximum in-plane shear strain and average normal strain. In each case, specify the orientation of the element and show how the strains deform the element within the x–y plane.
y
a) In-Plane Principal Strain: Applying Eq. 10–9, ex + ey
e1, 2 =
;
2
= B
a
A
ex - ey 2
b + a 2
gxy 2
b
2
250 - 300 2 250 + 300 -180 2 -6 ; a b + a b R A 10 B 2 A 2 2
= 275 ; 93.41 e1 = 368 A 10 - 6 B
e2 = 182 A 10 - 6 B
Ans.
Orientation of Principal Strain: Applying Eq. 10–8, gxy
tan 2uP =
-180(10 - 6)
ex - ey
=
(250 - 300)(10 - 6)
uP = 37.24°
and
= 3.600
-52.76°
Use Eq. 10–5 to determine which principal strain deforms the element in the x¿ direction with u = 37.24°. ex¿ =
ex + ey
= c
2
ex - ey +
2
cos 2u +
gxy 2
sin 2u
250 + 300 250 - 300 -180 + cos 74.48° + sin 74.48° d A 10 - 6 B 2 2 2
= 182 A 10 - 6 B = e2 Hence, uP1 = -52.8°
and
uP2 = 37.2°
Ans.
b) Maximum In-Plane Shear Strain: Applying Eq. 10–11, g max ex - ey 2 gxy 2 in-plane = a b + a b 2 A 2 2 g
max in-plane
= 2B
-180 2 250 - 300 2 -6 b + a b R A 10 B A 2 2 a
= 187 A 10 - 6 B
Ans.
753
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10–14.
Continued
Orientation of the Maximum In-Plane Shear Strain: Applying Eq. 10–10, tan 2us = -
ex - ey
us = -7.76°
and
The proper sign of gx¿y¿ 2
ex - ey = -
= -
gxy
2
g
max in-plane
250 - 300 = -0.2778 -180
82.2°
Ans.
can be determined by substituting u = -7.76° into Eq. 10–6.
sin 2u +
gxy 2
cos 2u
gx¿y¿ = {-[250 - 300] sin (-15.52°) + (-180) cos (-15.52°)} A 10 - 6 B = -187 A 10 - 6 B
Normal Strain and Shear strain: In accordance with the sign convention, ex = 250 A 10 - 6 B
ey = 300 A 10 - 6 B
gxy = -180 A 10 - 6 B
Average Normal Strain: Applying Eq. 10–12, eavg =
ex + ey 2
= c
250 + 300 d A 10 - 6 B = 275 A 10 - 6 B 2
Ans.
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*10–16. The state of strain at a point on a support has components of Px = 350110-62, Py = 400110-62, gxy = -675110-62. Use the strain-transformation equations to determine (a) the in-plane principal strains and (b) the maximum in-plane shear strain and average normal strain. In each case specify the orientation of the element and show how the strains deform the element within the x–y plane.
a)
e1, 2 =
=
ex + ey ;
2
B
a
ex -ey 2
b + a 2
gxy 2
b
2
350 - 400 2 -675 2 350 + 400 ; a b + a b 2 A 2 2
e1 = 713(10 - 6)
Ans.
e2 = 36.6(10 - 6)
Ans.
tan 2uP =
gxy ex - ey
=
-675 (350 - 400)
uP = 42.9°
Ans.
b) (gx¿y¿)max =
2 (gx¿y¿)max
=
2
A
a
ex - ey 2
b + a 2
gxy 2
b
2
a
350 - 400 2 -675 2 b + a b A 2 2
(gx¿y¿)max = 677(10 - 6) eavg =
ex + ey
tan 2us =
2
=
Ans.
350 + 400 = 375(10 - 6) 2
-(ex - ey) gxy
=
Ans.
350 - 400 675
us = -2.12°
Ans.
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•10–17.
Solve part (a) of Prob. 10–4 using Mohr’s circle.
ex = 120(10 - 6)
ey = -180(10 - 6)
gxy = 150(10 - 6)
A (120, 75)(10 - 6) C (-30, 0)(10 - 6) R = C 2[120 - (-30)]2 + (75)2 D (10 - 6) = 167.71 (10 - 6) e1 = (-30 + 167.71)(10 - 6) = 138(10 - 6)
Ans.
e2 = (-30 - 167.71)(10 - 6) = -198(10 - 6)
Ans.
75 tan 2uP = a b , uP = 13.3° 30 + 120
Ans.
756
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10–18.
Solve part (b) of Prob. 10–4 using Mohr’s circle.
ex = 120(10 - 6)
ey = -180(10 - 6)
gxy = 150(10 - 6)
A (120, 75)(10 - 6) C (-30, 0)(10 - 6) R = C 2[120 - (-30)]2 + (75)2 D (10 - 6) = 167.71 (10 - 6) gxy max 2 in-plane
gxy
= R = 167.7(10 - 6)
max in-plane
= 335(10 - 6)
Ans.
eavg = -30 (10 - 6) tan 2us =
120 + 30 75
Ans. us = -31.7°
Ans.
757
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10–19.
Solve Prob. 10–8 using Mohr’s circle.
ex = -200(10 - 6)
ey = -650(10 - 6)
gxy = -175(10 - 6)
gxy 2
= -87.5(10 - 6)
u = 20°, 2u = 40° A(-200, -87.5)(10 - 6)
C(-425, 0)(10 - 6)
R = [2(-200 - (-425))2 + 87.52 ](10 - 6) = 241.41(10 - 6) tan a =
87.5 ; -200 - (-425)
a = 21.25°
f = 40 + 21.25 = 61.25° ex¿ = (-425 + 241.41 cos 61.25°)(10 - 6) = -309(10 - 6)
Ans.
ey¿ = (-425 - 241.41 cos 61.25°)(10 - 6) = -541(10 - 6)
Ans.
-gx¿y¿ 2
= 241.41(10 - 6) sin 61.25°
gx¿y¿ = -423(10 - 6)
Ans.
758
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*10–20.
Solve Prob. 10–10 using Mohr’s circle.
ex = 400(10 - 6) A(400, 155)(10 - 6)
ey = -250(10 - 6)
gxy = 310(10 - 6)
gxy 2
= 155(10 - 6)
C(75, 0)(10 - 6)
R = [2(400 - 75)2 + 1552 ](10 - 6) = 360.1(10 - 6) tan a =
155 ; 400 - 75
a = 25.50°
f = 60 + 25.50 = 85.5° ex¿ = (75 + 360.1 cos 85.5°)(10 - 6) = 103(10 - 6)
Ans.
ey¿ = (75 - 360.1 cos 85.5°)(10 - 6) = 46.7(10 - 6)
Ans.
gx¿y¿ 2
= (360.1 sin 85.5°)(10 - 6)
gx¿y¿ = 718(10 - 6)
Ans.
759
u = 30°
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•10–21.
Solve Prob. 10–14 using Mohr’s circle.
Construction of the Circle: In accordance with the sign convention, ex = 250 A 10 - 6 B , gxy ey = 300 A 10 - 6 B , and = -90 A 10 - 6 B . Hence, 2 eavg =
ex + ey 2
= a
250 + 300 b A 10 - 6 B = 275 A 10 - 6 B 2
Ans.
The coordinates for reference points A and C are A(250, -90) A 10 - 6 B
C(275, 0) A 10 - 6 B
The radius of the circle is R = a 2(275 - 250)2 + 902 b A 10 - 6 B = 93.408 In-Plane Principal Strain: The coordinates of points B and D represent e1 and e2, respectively. e1 = (275 + 93.408) A 10 - 6 B = 368 A 10 - 6 B
Ans.
e2 = (275 - 93.408) A 10 - 6 B = 182 A 10 - 6 B
Ans.
Orientation of Principal Strain: From the circle, tan 2uP2 =
90 = 3.600 275 - 250
2uP2 = 74.48°
2uP1 = 180° - 2uP2 uP1 =
180° - 74.78° = 52.8° (Clockwise) 2
Ans.
Maximum In-Plane Shear Strain: Represented by the coordinates of point E on the circle. g max in-plane
2 g
= -R = -93.408 A 10 - 6 B
max in-plane
= -187 A 10 - 6 B
Ans.
Orientation of the Maximum In-Plane Shear Strain: From the circle, tan 2us =
275 - 250 = 0.2778 90
us = 7.76° (Clockwise)
Ans.
760
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10–22. The strain at point A on the bracket has components Px = 300110-62, Py = 550110-62, gxy = -650110-62. Determine (a) the principal strains at A in the x– y plane, (b) the maximum shear strain in the x–y plane, and (c) the absolute maximum shear strain. ex = 300(10 - 6)
ey = 550(10 - 6)
A(300, -325)10 - 6
gxy = -650(10 - 6)
y
gxy 2
= -325(10 - 6)
A
C(425, 0)10 - 6
R = C 2(425 - 300)2 + (-325)2 D 10 - 6 = 348.2(10 - 6) a) e1 = (425 + 348.2)(10 - 6) = 773(10 - 6)
Ans.
e2 = (425 - 348.2)(10 - 6) = 76.8(10 - 6)
Ans.
b) g
max in-plane
= 2R = 2(348.2)(10 - 6) = 696(10 - 6)
Ans.
773(10 - 6) ; 2
Ans.
c) gabs max
=
2
gabs max
= 773(10 - 6)
10–23. The strain at point A on the leg of the angle has components Px = -140110-62, Py = 180110-62, gxy = -125110-62. Determine (a) the principal strains at A in the x–y plane, (b) the maximum shear strain in the x–y plane, and (c) the absolute maximum shear strain. ex = -140(10 - 6) A( -140, -62.5)10 - 6
ey = 180(10 - 6)
gxy = -125(10 - 6)
A
gxy 2
= -62.5(10 - 6)
C(20, 0)10 - 6
A 2(20 - ( -140))2 + (-62.5)2 B 10 - 6 = 171.77(10 - 6)
R = a)
e1 = (20 + 171.77)(10 - 6) = 192(10 - 6)
Ans.
e2 = (20 - 171.77)(10 - 6) = -152(10 - 6)
Ans.
(b, c) gabs max
=
g
max in-plane
= 2R = 2(171.77)(10 - 6) = 344(10 - 6)
Ans.
761
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*10–24. The strain at point A on the pressure-vessel wall has components Px = 480110-62, Py = 720110-62, gxy = 650110-62. Determine (a) the principal strains at A, in the x– y plane, (b) the maximum shear strain in the x–y plane, and (c) the absolute maximum shear strain.
ex = 480(10 - 6)
ey = 720(10 - 6)
A(480, 325)10 - 6
C(600, 0)10 - 6
gxy = 650(10 - 6)
y A
gxy 2
= 325(10 - 6)
R = (2(600 - 480)2 + 3252 )10 - 6 = 346.44(10 - 6) a) e1 = (600 + 346.44)10 - 6 = 946(10 - 6)
Ans.
e2 = (600 - 346.44)10 - 6 = 254(10 - 6)
Ans.
b) g
max in-plane
= 2R = 2(346.44)10 - 6 = 693(10 - 6)
Ans.
946(10 - 6) ; 2
Ans.
c) gabs max
2
=
gabs max
= 946(10 - 6)
762
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•10–25.
The 60° strain rosette is mounted on the bracket. The following readings are obtained for each gauge: Pa = -100110-62, Pb = 250110-62, and Pc = 150110-62. Determine (a) the principal strains and (b) the maximum inplane shear strain and associated average normal strain. In each case show the deformed element due to these strains.
b c 60⬚ 60⬚
This is a 60° strain rosette Thus, ex = ea = -100(10 - 6) 1 A 2eb + 2ec - ea B 3
ey =
1 C 2(250) + 2(150) - (-100) D (10 - 6) 3
=
= 300(10 - 6) gxy =
2 23
(eb - ec) =
2 23
(250 - 150)(10 - 6) = 115.47(10 - 6)
In accordance to the established sign convention, ex = -100(10 - 6), ey = 300(10 - 6) gxy and = 57.74(10 - 6). 2 Thus, eavg =
ex + ey 2
= a
-100 + 300 b(10 - 6) = 100(10 - 6) 2
Ans.
Then, the coordinates of reference point A and Center C of the circle are A( -100, 57.74)(10 - 6)
C(100, 0)(10 - 6)
Thus, the radius of the circle is R = CA = a 2(-100 - 100)2 + 208.16b(10 - 6) = 208.17(10 - 6) Using these result, the circle is shown in Fig. a. The coordinates of points B and D represent e1 and e2 respectively. e1 = (100 + 208.17)(10 - 6) = 308(10 - 6)
Ans.
e2 = (100 - 208.17)(10 - 6) = -108(10 - 6)
Ans.
Referring to the geometry of the circle, tan 2(uP)2 =
57.74(10 - 6) (100 + 100)(10 - 6)
= 0.2887
A uP B 2 = 8.05° (Clockwise)
Ans.
The deformed element for the state of principal strain is shown in Fig. b.
763
a
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10–25.
Continued gmax
The coordinates for point E represent eavg and
in-plane
2
. Thus,
gmax
in-plane
2
= R = 208.17(10 - 6)
gmax
in-plane
= 416(10 - 6)
Ans.
Referring to the geometry of the circle, tan 2us =
100 + 100 57.74
us = 36.9° (Counter Clockwise)
Ans.
The deformed element for the state of maximum In-plane shear strain is shown in Fig. c.
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10–26. The 60° strain rosette is mounted on a beam. The following readings are obtained for each gauge: Pa = 200110-62, Pb = - 450110-62, and Pc = 250110-62. Determine (a) the in-plane principal strains and (b) the maximum in-plane shear strain and average normal strain. In each case show the deformed element due to these strains.
b a 30⬚ 30⬚ c
With ua = 60°, ub = 120° and uc = 180°, ea = ex cos2 ua + ey sin2 ua + gxy sin ua cos ua 200(10 - 6) = ex cos2 60° + ey sin2 60° + gxy sin 60° cos 60° 0.25ex + 0.75ey + 0.4330 gxy = 200(10 - 6)
(1)
eb = ex cos2 ub + ey sin2 ub + gxy sin ub cos ub -450(10 - 6) = ex cos2 120° + ey sin2 120° + gxy sin 120° cos 120° 0.25ex + 0.75ey - 0.4330 gxy = -450(10 - 6)
(2)
ec = ex cos2 uc + ey sin2 uc + gxy sin uc cos uc 250(10 - 6) = ex cos2 180° + ey sin2 180° + gxy sin 180° cos 180° ex = 250(10 - 6) Substitute this result into Eqs. (1) and (2) and solve them, ey = -250 (10 - 6)
gxy = 750.56 (10 - 6)
In accordance to the established sign convention, ex = 250(10 - 6), ey = -250(10 - 6), gxy and = 375.28(10 - 6), Thus, 2 eavg =
ex + ey 2
= c
250 + ( -250) d(10 - 6) = 0 2
Ans.
Then, the coordinates of the reference point A and center C of the circle are A(250, 375.28)(10 - 6)
C(0, 0)
Thus, the radius of the circle is R = CA =
A 2(250 - 0)2 + 375.282 B (10 - 6) = 450.92(10 - 6)
Using these results, the circle is shown in Fig. a. The coordinates for points B and D represent e1 and e2, respectively. Thus, e1 = 451(10 - 6)
e2 = -451(10 - 6)
Ans.
Referring to the geometry of the circle, tan 2(uP)1 =
375.28 = 1.5011 250
(uP)1 = 28.2° (Counter Clockwise)
Ans.
The deformed element for the state of principal strains is shown in Fig. b.
765
60⬚
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10–26.
Continued gmax
in-plane
The coordinates of point E represent eavg and . Thus, 2 gmax in-plane gmax = 902(10 - 6) = R = 450.92(10 - 6) in-plane 2
Ans.
Referring to the geometry of the circle, tan 2us =
250 = 0.6662 375.28
us = 16.8° (Clockwise)
Ans.
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10–27. The 45° strain rosette is mounted on a steel shaft. The following readings are obtained from each gauge: Pa = 300110-62, Pb = -250110-62, and Pc = -450110-62. Determine (a) the in-plane principal strains and (b) the maximum in-plane shear strain and average normal strain. In each case show the deformed element due to these strains.
b
c 45⬚
With ua = 45°, ub = 90° and uc = 135°, ea = ex cos2 ua + ey sin2 ua + gxy sin ua cos ua 300(10 - 6) = ex cos2 45° + ey sin2 45° + gxy sin 45° cos 45° ex + ey + gxy = 600(10 - 6)
(1)
eb = ex cos2 ub + ey sin2 ub + gxy sin ub cos ub -250(10 - 6) = ex cos2 90° + ey sin2 90° + gxy sin 90° cos 90° ey = -250(10 - 6) ec = ex cos2 uc + ey sin2 uc + gxy sin uc cos uc -450(10 - 6) = ex cos2 135° + ey sin2 135° + gxy sin 135° cos 135° ex + ey - gxy = -900(10 - 6)
(2)
Substitute the result of ey into Eq. (1) and (2) and solve them ex = 100(10 - 6)
gxy = 750(10 - 6)
In accordance to the established sign convention, ex = 100(10 - 6), ey = -250(10 - 6) gxy and = 375(10 - 6). Thus, 2 eavg =
ex + ey 2
= c
100 + (-250) d(10 - 6) = -75(10 - 6) 2
Ans.
Then, the coordinates of the reference point A and the center C of the circle are A(100, 375)(10 - 6)
C(-75, 0)(10 - 6)
Thus, the radius of the circle is R = CA = a 2 C 100 - (-75) D 2 + 3752 b(10 - 6) = 413.82(10 - 6) Using these results, the circle is shown in Fig. a. The Coordinates of points B and D represent e1 and e2, respectively. Thus, e1 = e2 =
A -75 + 413.82 B (10 - 6) = 339(10 - 6)
Ans.
A -75 - 413.82 B (10 - 6) = -489(10 - 6)
Ans.
Referring to the geometry of the circle tan 2(uP)1 =
375 = 2.1429 100 + 75
(uP)1 = 32.5° (Counter Clockwise)
Ans.
767
45⬚
a
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10–27.
Continued
The deformed element for the state of principal strains is shown in Fig. b. gmax The coordinates of point E represent eavg and gmax
in-plane
2
= R = 413.82(106)
in-plane
gmax
2
in-plane
. Thus
= 828(10 - 6)
Ans.
Referring to the geometry of the circle tan 2us =
-100 + 75 = 0.4667 375
us = 12.5° (Clockwise)
Ans.
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*10–28. The 45° strain rosette is mounted on the link of the backhoe. The following readings are obtained from each gauge: Pa = 650110-62, Pb = -300110-62, Pc = 480110-62. Determine (a) the in-plane principal strains and (b) the maximum in-plane shear strain and associated average normal strain.
a 45⬚ b
ea = 650(10 - 6); ua = 180°;
eb = -300(10 - 6);
ub = 225°
c
uc = 270°
Applying Eq. 10–16, e = ex cos2 u + ey sin2 u + gxy sin u cos u 650(10 - 6) = ex cos2 (180°) + ey sin2 (180°) + gxy sin (180°) cos (180°) ex = 650 (10 - 6) 480 (10 - 6) = ex cos2 (270°) + ey sin2 (270°) + gxy sin (270°) cos (270°) ey = 480 (10 - 6) -300 (10 - 6) = 650 (10 - 6) cos2 (225°) + 480 (10 - 6) sin2 (225°) + gxy sin (225°) cos (225°) gxy = -1730 (10 - 6) Therefore, ex = 650 (10 - 6) gxy 2
ey = 480 (10 - 6)
gxy = -1730 (10 - 6)
= -865 (10 - 6)
Mohr’s circle: A(650, -865) 10 - 6
C(565, 0) 10 - 6
R = CA = C 2(650 - 565)2 + 8652 D 10 - 6 = 869.17 (10 - 6) (a)
(b)
e1 = [565 + 869.17]10 - 6 = 1434 (10 - 6)
Ans.
e2 = [565 - 869.17]10 - 6 = -304 (10 - 6)
Ans.
gmax
in-plane
45⬚
ec = 480(10 - 6)
= 2 R = 2(869.17) (10 - 6) = 1738 (10 - 6)
Ans.
eavg = 565(10 - 6)
Ans.
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10–30. For the case of plane stress, show that Hooke’s law can be written as sx =
E E 1Px + nPy2, sy = 1Py + nPx2 11 - n22 11 - n22
Generalized Hooke’s Law: For plane stress, sz = 0. Applying Eq. 10–18, ex =
1 A s - v sy B E x
vEex = A sx - v sy B v vEex = v sx - v2 sy ey =
[1]
1 (s - v sx) E y
E ey = -v sx + sy
[2]
Adding Eq [1] and Eq.[2] yields. vE ex - E ey = sy - v2 sy sy =
E A vex + ey B 1 - v2
(Q.E.D.)
Substituting sy into Eq. [2] E ey = -vsx +
sx =
E A v ex + ey B 1 - v2
E A v ex + ey B v (1 - v2)
Eey -
v
E v ex + E ey - E ey + Eey v2 =
=
v(1 - v2) E (ex + v ey) 1 - v2
(Q.E.D.)
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10–31. Use Hooke’s law, Eq. 10–18, to develop the straintransformation equations, Eqs. 10–5 and 10–6, from the stress-transformation equations, Eqs. 9–1 and 9–2.
Stress transformation equations: sx + sy
sx¿ =
2
tx¿y¿ =
sy¿ =
sx - sy +
sx - sy
2
(1)
sin 2u + txy cos 2u
2 sx + sy
sx - sy -
2
cos 2u + txy sin 2u
2
(2)
cos 2u - txy sin 2u
(3)
Hooke’s Law: ex =
v sy sx E E
(4)
ey =
sy -v sx + E E
(5)
txy = G gxy G =
(6)
E 2 (1 + v)
(7)
From Eqs. (4) and (5) ex + ey =
ex - ey =
(1 - v)(sx + sy)
(8)
E (1 + v)(sx - sy)
(9)
E
From Eqs. (6) and (7) txy =
E g 2 (1 + v) xy
(10)
From Eq. (4) ex¿ =
v sy¿ sx¿ E E
(11)
Substitute Eqs. (1) and (3) into Eq. (11) ex¿ =
(1 - v)(sx - sy)
(1 + v)(sx - sy) +
2E
2E
cos 2u +
(1 + v)txy sin 2u E
(12)
By using Eqs. (8), (9) and (10) and substitute into Eq. (12), ex¿ =
ex + ey 2
ex - ey +
2
cos 2u +
gxy 2
QED
sin 2u
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10–31.
Continued
From Eq. (6). gx¿y¿ = G gx¿y¿ =
E g 2 (1 + v) x¿y¿
(13)
Substitute Eqs. (13), (6) and (9) into Eq. (2), E (ex - ey) E E gx¿y¿ = sin 2u + g cos 2u 2 (1 + v) 2 (1 + v) 2 (1 + v) xy gx¿y¿
(ex - ey) = -
2
2
sin 2u +
gxy 2
QED
cos 2u
*10–32. A bar of copper alloy is loaded in a tension machine and it is determined that Px = 940110-62 and sx = 14 ksi, sy = 0, sz = 0. Determine the modulus of elasticity, Ecu, and the dilatation, ecu, of the copper. ncu = 0.35.
ex =
1 [s - v(sy + sz)] E x
940(10 - 6) =
ecu =
1 [14(103) - 0.35(0 + 0)] Ecu
Ecu = 14.9(103) ksi
Ans.
1 - 2(0.35) 1 - 2v (14 + 0 + 0) = 0.282(10 - 3) (sx + sy + sz) = E 14.9(103)
Ans.
•10–33.
The principal strains at a point on the aluminum fuselage of a jet aircraft are P1 = 780110-62 and P2 = 400110-62. Determine the associated principal stresses at the point in the same plane. Eal = 1011032 ksi, nal = 0.33. Hint: See Prob. 10–30.
Plane stress, s3 = 0 See Prob 10-30, s1 =
E (e1 + ve2) 1 - v2 10(103)
=
s2 =
1 - 0.332
Ans.
E (e2 + ve1) 1 - v2 10(103)
=
(780(10 - 6) + 0.33(400)(10 - 6)) = 10.2 ksi
1 - 0.332
(400(10 - 6) + 0.33(780)(10 - 6)) = 7.38 ksi
Ans.
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10–34. The rod is made of aluminum 2014-T6. If it is subjected to the tensile load of 700 N and has a diameter of 20 mm, determine the absolute maximum shear strain in the rod at a point on its surface.
700 N
Normal Stress: For uniaxial loading, sy = sz = 0. sx =
P = A
p 4
700 = 2.228 MPa (0.022)
Normal Strain: Applying the generalized Hooke’s Law. ex =
=
1 C s - v A sy + sz B D E x 1 C 2.228 A 106 B - 0 D 73.1(109)
= 30.48 A 10 - 6 B ey =
=
1 C s - v(sx + sz) D E y 1 C 0 - 0.35 A 2.228 A 106 B + 0 B D 73.1(109)
= -10.67 A 10 - 6 B ez =
=
1 C s - v A sx + sy B D E z 1 C 0 - 0.35 A 2.228 A 106 B + 0 B D 73.1(109)
= -10.67 A 10 - 6 B Therefore. emax = 30.48 A 10 - 6 B
emin = -10.67 A 10 - 6 B
Absolute Maximum Shear Strain: gabs
max
= emax - emin = [30.48 - (-10.67)] A 10 - 6 B = 41.1 A 10 - 6 B
Ans.
773
700 N
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10–35. The rod is made of aluminum 2014-T6. If it is subjected to the tensile load of 700 N and has a diameter of 20 mm, determine the principal strains at a point on the surface of the rod.
700 N
Normal Stress: For uniaxial loading, sy = sz = 0. sx =
P = A
p 4
700 = 2.228 MPa (0.022)
Normal Strains: Applying the generalized Hooke’s Law. ex =
=
1 C s - v A sy + sz B D E x 1 C 2.228 A 106 B - 0 D 73.1(109)
= 30.48 A 10 - 6 B ey =
=
1 C s - v(sx + sz) D E y 1 C 0 - 0.35 A 2.228 A 106 B + 0 B D 73.1(109)
= -10.67 A 10 - 6 B ez =
=
1 C s - v A sx + sy B D E z 1 C 0 - 0.35 A 2.228 A 106 B + 0 B D 73.1(109)
= -10.67 A 10 - 6 B Principal Strains: From the results obtained above, emax = 30.5 A 10 - 6 B
eint = emin = -10.7 A 10 - 6 B
Ans.
774
700 N
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*10–36. The steel shaft has a radius of 15 mm. Determine the torque T in the shaft if the two strain gauges, attached to the surface of the shaft, report strains of Px¿ = -80110-62 and Py¿ = 80110-62. Also, compute the strains acting in the x and y directions. Est = 200 GPa, nst = 0.3. ex¿ = -80(10 - 6)
y T
ex = ey = 0
Ans.
ex¿ = ex cos2 u + ey sin2 u + gxy sin u cos u u = 45° -80(10 - 6) = 0 + 0 + gxy sin 45° cos 45° gxy = -160(10 - 6)
Ans.
Also, u = 135° 80(10 - 6) = 0 + 0 + g sin 135° cos 135° gxy = -160(10 - 6) 200(109) E = = 76.923(109) 2(1 + V) 2(1 + 0.3)
t = Gg = 76.923(109)(160)(10 - 6) = 12.308(106) Pa 12.308(106) A
p B (0.015)4 2 = 65.2 N # m 0.015
Ans.
10–37. Determine the bulk modulus for each of the following materials: (a) rubber, Er = 0.4 ksi, nr = 0.48, and (b) glass, Eg = 811032 ksi, ng = 0.24. a) For rubber: Kr =
Er 0.4 = = 3.33 ksi 3 (1 - 2 vr) 3[1 - 2(0.48)]
Ans.
b) For glass: Kg =
Eg 3 (1 - 2 vg)
=
x T
Pure shear
tJ T = = c
x¿ 45⬚
ey¿ = 80(10 - 6)
G =
y¿
8(103) = 5.13 (103) ksi 3[1 - 2(0.24)]
Ans.
775
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10–38. The principal stresses at a point are shown in the figure. If the material is A-36 steel, determine the principal strains.
12 ksi
e1 =
1 1 e 12 - 0.32 C 8 + (-20) D f = 546 (10-6) C s - v(s2 + s3) D = E 1 29.0(103)
e2 =
1 1 e 8 - 0.32 C 12 + (-20) D f = 364 (10-6) C s - v(s1 + s3) D = E 2 29.0(103)
e3 =
20 ksi
8 ksi
1 1 C s3 - v(s1 + s2) D = C -20 - 0.32(12 + 8) D = -910 (10-6) E 29.0(103)
emax = 546 (10 - 6)
eint = 346 (10 - 6)
emin = -910 (10 - 6)
Ans.
10–39. The spherical pressure vessel has an inner diameter of 2 m and a thickness of 10 mm. A strain gauge having a length of 20 mm is attached to it, and it is observed to increase in length by 0.012 mm when the vessel is pressurized. Determine the pressure causing this deformation, and find the maximum in-plane shear stress, and the absolute maximum shear stress at a point on the outer surface of the vessel. The material is steel, for which Est = 200 GPa and nst = 0.3.
20 mm
1000 r = = 100 7 10, the thin wall analysis is valid to t 10 determine the normal stress in the wall of the spherical vessel. This is a plane stress Normal Stresses: Since
problem where smin = 0 since there is no load acting on the outer surface of the wall.
smax = slat =
pr p(1000) = = 50.0p 2t 2(10)
[1]
Normal Strains: Applying the generalized Hooke’s Law with emax = elat =
0.012 = 0.600 A 10 - 3 B mm>mm 20
emax =
1 C s - V (slat + smin) D E max
0.600 A 10 - 3 B =
1 [50.0p - 0.3 (50.0p + 0)] 200(104)
p = 3.4286 MPa = 3.43 MPa
Ans.
From Eq.[1] smax = slat = 50.0(3.4286) = 171.43 MPa Maximum In-Plane Shear (Sphere’s Surface): Mohr’s circle is simply a dot. As the result, the state of stress is the same consisting of two normal stresses with zero shear stress regardless of the orientation of the element. t
max in-plane
= 0
Ans.
smax - smin 171.43 - 0 = = 85.7MPa 2 2
Ans.
Absolute Maximum Shear Stress: tabs
max
=
776
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*10–40. The strain in the x direction at point A on the steel beam is measured and found to be Px = -100110-62. Determine the applied load P. What is the shear strain gxy at point A? Est = 2911032 ksi, nst = 0.3.
P y 3 in. A 3 ft
1 1 (5.5)(83) = 129.833 in4 (6)(9)3 12 12
Ix =
QA = (4.25)(0.5)(6) + (2.75)(0.5)(2.5) = 16.1875 in3 s = Eex = 29(103)(100)(10 - 6) = 2.90 ksi My , I
s =
2.90 =
1.5P(12)(1.5) 129.833
P = 13.945 = 13.9 kip tA =
0.5(13.945)(16.1875) VQ = = 1.739 ksi It 129.833(0.5)
G =
29(103) E = = 11.154(103) ksi 2(1 + v) 2(1 + 0.3)
gxy =
txy G
=
Ans.
1.739 = 0.156(10 - 3) rad 11.154(103)
Ans.
777
x
4 ft
7 ft
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•10–41.
The cross section of the rectangular beam is subjected to the bending moment M. Determine an expression for the increase in length of lines AB and CD. The material has a modulus of elasticity E and Poisson’s ratio is n.
C D B h
For line AB, sz = -
ey = -
A M
12My My My = 1 = 3 I b h3 12 b h v sz =
E
E b h3
h 2
¢LAB =
=
b
12 v My
L0
h
ey dy =
2 12 v M y dy 3 E b h L0
3vM 2Ebh
Ans.
For line CD, sz = -
ex = -
M h2 Mc 6M = - 1 = - 2 3 I bh b h 12 v sz E
=
6vM E b h2
¢LCD = ex LCD =
=
6vM (b) E b h2
6vM E h2
Ans.
10–42. The principal stresses at a point are shown in the figure. If the material is aluminum for which Eal = 1011032 ksi and nal = 0.33, determine the principal strains.
26 ksi
ex =
1 1 (s - v(sy + sz)) = (10 - 0.33(-15 - 26)) = 2.35(10 - 3) E x 10(103)
ey =
1 1 (s - v(sx + sz)) = (-15 - 0.33)(10 - 26)) = -0.972(10 - 3)Ans. E y 10(103)
ez =
Ans.
1 1 (s - v(sx + sy)) = (-26 - 0.33(10 - 15)) = -2.44(10 - 3) Ans. E z 10(103)
778
15 ksi 10 ksi
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10–43. A single strain gauge, placed on the outer surface and at an angle of 30° to the axis of the pipe, gives a reading at point A of Pa = -200(10-6). Determine the horizontal force P if the pipe has an outer diameter of 2 in. and an inner diameter of 1 in. The pipe is made of A-36 steel.
1.5 ft
Using the method of section and consider the equilibrium of the FBD of the pipe’s upper segment, Fig. a, Vz - p = 0
©Fz = 0;
Vz = p
©Mx = 0;
Tx - p(1.5) = 0 Tx = 1.5p
©My = 0;
My - p(2.5) = 0 My = 2.5p
30⬚ A
The normal strees is due to bending only. For point A, z = 0. Thus sx =
My z Iy
= 0
The shear stress is the combination of torsional shear stress and transverse shear stress. Here, J = p2 (14 - 0.54) = 0.46875 p in4. Thus, for point A tt =
1.5p(12)(1) 38.4 p Txc = = p J 0.46875p
Referring to Fig. b, (QA)z = y1œ A1œ - y2œ A2œ =
4 (1) p 2 4(0.5) p c (1 ) d c (0.52) d 3p 2 3p 2
= 0.5833 in3 Iy =
p 4
(14 - 0.54) = 0.234375 p in4
Combine these two shear stress components, t = tt + tv =
P 2.5 ft
38.4P 2.4889P 40.8889P + = p p p
Since no normal stress acting on point A, it is subjected to pure shear which can be represented by the element shown in Fig. c. For pure shear, ex = ez = 0, ea = ex cos3 ua + ez sin2 ua + gxz sin ua cos ua -200(10 - 6) = 0 + 0 + gxz sin 150° cos 150° gxz = 461.88(10 - 6) Applying the Hooke’s Law for shear, txz = G gxz 40.8889P = 11.0(103) C 461.88(10 - 6) D p P = 0.3904 kip = 390 lb
Ans.
779
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*10–44. A single strain gauge, placed in the vertical plane on the outer surface and at an angle of 30° to the axis of the pipe, gives a reading at point A of Pa = -200(10-6). Determine the principal strains in the pipe at point A. The pipe has an outer diameter of 2 in. and an inner diameter of 1 in. and is made of A-36 steel.
1.5 ft
P 2.5 ft
Using the method of sections and consider the equilibrium of the FBD of the pipe’s upper segment, Fig. a, Vz - P = 0
©Fz = 0;
Vz = P
©Mx = 0;
Tx - P(1.5) = 0 Tx = 1.5P
©My = 0;
My - P(2.5) = 0 My = 2.5P
By observation, no normal stress acting on point A. Thus, this is a case of pure shear. For the case of pure shear, ex = ez = ey = 0 ea = ex cos2 ua + ez sin2 ua + gxz sin ua cos ua -200(10 - 6) = 0 + 0 + gxz sin 150° cos 150° gxz = 461.88(10 - 6) e1, 2 =
ex + ez
= B
2
+
A
a
ex - ez 2
b + a 2
gxz 2
b
2
0 - 0 2 461.88 2 0 + 0 -6 ; a b + a b R (10 ) 2 A 2 2
e1 = 231(10 - 6)
e2 = -231(10 - 6)
Ans.
780
30⬚ A
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10–45. The cylindrical pressure vessel is fabricated using hemispherical end caps in order to reduce the bending stress that would occur if flat ends were used. The bending stresses at the seam where the caps are attached can be eliminated by proper choice of the thickness th and tc of the caps and cylinder, respectively. This requires the radial expansion to be the same for both the hemispheres and cylinder. Show that this ratio is tc>th = 12 - n2>11 - n2.Assume that the vessel is made of the same material and both the cylinder and hemispheres have the same inner radius. If the cylinder is to have a thickness of 0.5 in., what is the required thickness of the hemispheres? Take n = 0.3.
tc th r
For cylindrical vessel: s1 =
pr ; tc
e1 =
1 [s - v (s2 + s3)] E 1
=
s2 =
pr 2 tc s3 = 0
vpr pr 1 1 pr a b = a1 - v b E tc 2 tc E tc 2
d r = e1 r =
p r2 1 a1 - v b E tc 2
(1)
For hemispherical end caps: s1 = s2 =
e1 =
=
pr 2 th
1 [s - v (s2 + s3)] ; E 1
s3 = 0
vpr pr 1 pr a b = (1 - v) E 2 th 2 th 2 E th
d r = e1 r =
p r2 (1 - v) 2 E th
(2)
Equate Eqs. (1) and (2): p r2 p r2 1 a1 - vb = (1 - v) E tc 2 2 E th 2 (1 - 12 v) tc 2 - v = = th 1 - v 1 - v th =
QED
(1 - v) tc (1 - 0.3) (0.5) = = 0.206 in. 2 - v 2 - 0.3
Ans.
781
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10–46. The principal strains in a plane, measured experimentally at a point on the aluminum fuselage of a jet aircraft, are P1 = 630(10-6) and P2 = 350(10-6). If this is a case of plane stress, determine the associated principal stresses at the point in the same plane. Eal = 10(103) ksi and nal = 0.33. Normal Stresses: For plane stress, s3 = 0. Normal Strains: Applying the generalized Hooke’s Law. e1 =
1 C s - v (s2 + s3) D E 1
630 A 10 - 6 B =
1 [s1 - 0.33(s2 + 0)] 10(103)
6.30 = s1 - 0.33s2 e2 =
[1]
1 C s - v (s1 + s3) D E 2
350 A 10 - 6 B =
1 C s2 - 0.33(s1 + 0) D 10(103)
3.50 = s2 - 0.33s1
[2]
Solving Eqs.[1] and [2] yields: s1 = 8.37 ksi
s2 = 6.26 ksi
Ans.
10–47. The principal stresses at a point are shown in the figure. If the material is aluminum for which Eal = 1011032 ksi and nal = 0.33, determine the principal strains.
3 ksi
e1 =
1 1 e 8 - 0.33 C 3 + (-4) D f = 833 (10 - 6) C s - v(s2 + s3) D = E 1 10(103)
e2 =
1 1 e 3 - 0.33 C 8 + (-4) D f = 168 (10 - 6) C s - v(s1 + s3) D = E 2 10(103)
e3 =
1 1 C s3 - v(s1 + s2) D = C -4 - 0.33(8 + 3) D = -763 (10 - 6) E 10(103)
Using these results, e1 = 833(10 - 6)
e2 = 168(10 - 6)
e3 = -763(10 - 6)
782
8 ksi
4 ksi
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*10–48. The 6061-T6 aluminum alloy plate fits snugly into the rigid constraint. Determine the normal stresses sx and sy developed in the plate if the temperature is increased by ¢T = 50°C. To solve, add the thermal strain a¢T to the equations for Hooke’s Law.
y
400 mm
300 mm
x
Generalized Hooke’s Law: Since the sides of the aluminum plate are confined in the rigid constraint along the x and y directions, ex = ey = 0. However, the plate is allowed to have free expansion along the z direction. Thus, sz = 0. With the additional thermal strain term, we have ex =
0 =
1 cs - v A sy + sz B d + a¢T E x 1
68.9 A 109 B
csx - 0.35 A sy + 0 B d + 24a 10 - 6 b(50)
sx - 0.35sy = -82.68 A 106 B ey =
0 =
(1)
1 C s - v A sx + sz B D + a¢T E y 1 68.9a 10 b 9
C sy - 0.35(sx + 0) D + 24 A 10 - 6 B (50)
sy - 0.35sx = -82.68 A 106 B
(2)
Solving Eqs. (1) and (2), sx = sy = -127.2 MPa = 127.2 MPa (C)
Ans.
Since sx = sy and sy 6 sY, the above results are valid.
783
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•10–49.
Initially, gaps between the A-36 steel plate and the rigid constraint are as shown. Determine the normal stresses sx and sy developed in the plate if the temperature is increased by ¢T = 100°F. To solve, add the thermal strain a¢T to the equations for Hooke’s Law.
y 0.0015 in.
6 in.
8 in.
0.0025 in. x
Generalized Hooke’s Law: Since there are gaps between the sides of the plate and the rigid constraint, the plate is allowed to expand before it comes in contact with the constraint. dy dx 0.0025 0.0015 ey = = = 0.3125 A 10 - 3 B and = = 0.25 A 10 - 3 B . Thus, ex = Lx 8 Ly 6 However, the plate is allowed to have free expansion along the z direction. Thus, sz = 0. With the additional thermal strain term, we have ex =
1 csx - v A sy + sz B d + a¢T E
0.3125 a 10 - 3 b =
1 29.0 a103 b
C sx - 0.32 A sy + 0 B D + 6.60 A 10 - 6 B (100)
sx - 0.32sy = -10.0775 ey =
(1)
1 C s - v A sx + sz B D + a¢T E y
0.25 A 10 - 3 B =
1
29.0 A 103 B
C sy - 0.32(sx + 0) D + 6.60 A 10 - 6 B (100)
sy - 0.32sx = -11.89
(2)
Solving Eqs. (1) and (2), sx = -15.5 ksi = 15.5 ksi (C)
Ans.
sy = -16.8 ksi = 16.8 ksi (C)
Ans.
Since sx 6 sY and sy 6 sY, the above results are valid.
784
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10–50. Two strain gauges a and b are attached to a plate made from a material having a modulus of elasticity of E = 70 GPa and Poisson’s ratio n = 0.35. If the gauges give a reading of Pa = 450110-62 and Pb = 100110-62, determine the intensities of the uniform distributed load wx and wy acting on the plate. The thickness of the plate is 25 mm.
wy
b 45⬚
y
a
Normal Strain: Since no shear force acts on the plane along the x and y axes, gxy = 0. With ua = 0 and ub = 45°, we have 2
2
ea = ex cos ua + ey sin ua + gxy sin ua cos ua 450 A 10 - 6 B = ex cos2 0° + ey sin2 0°+0 ex = 450 A 10 - 6 B
eb = ex cos2 ub + ey sin2 ub + gxy sin ub cos ub 100 A 10 - 6 B = 450 A 10 - 6 B cos2 45° + ey sin2 45° + 0
ey = -250 A 10 - 6 B
Generalized Hooke’s Law: This is a case of plane stress. Thus, sz = 0. ex =
1 C s - v A sy + sz B D E x
450 A 10 - 6 B =
1
70 A 109 B
C sy - 0.35 A sy + 0 B D
sx - 0.35sy = 31.5 A 106 B ey =
(1)
1 C s - v A sx + sz B D E y
-250 A 10-6 B =
1
70 A 109 B
C sy - 0.35 A sy + 0 B D
sy - 0.35sx = -17.5 A 106 B
(2)
Solving Eqs. (1) and (2), sy = -7.379 A 106 B N>m2
sx = 28.917 A 106 B N>m2
Then, wy = syt = -7.379 A 106 B (0.025) = -184 N>m
Ans.
wx = sxt = 28.917 A 106 B (0.025) = 723 N>m
Ans.
785
z
x
wx
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10–51. Two strain gauges a and b are attached to the surface of the plate which is subjected to the uniform distributed load wx = 700 kN>m and wy = -175 kN>m. If the gauges give a reading of Pa = 450110-62 and Pb = 100110-62, determine the modulus of elasticity E, shear modulus G, and Poisson’s ratio n for the material.
wy
b 45⬚
y
Normal Stress and Strain: The normal stresses along the x, y, and z axes are sx =
700 A 103 B 0.025
sy = -
a
= 28 A 10 B N>m 6
175 A 103 B 0.025
2
= -7 A 106 B N>m2
z
sz = 0 (plane stress) Since no shear force acts on the plane along the x and y axes, gxy = 0. With ua = 0° and ub = 45°, we have ea = ex cos2 ua + ey sin2 ua + gxy sin ua cos ua 450 A 10 - 6 B = ex cos2 0° + ey sin2 0° + 0
ex = 450 A 10 - 6 B
eb = ex cos2 ub + ey sin2 ub + gxy sin ub cos ub 100 A 10 - 6 B = 450 A 10 - 6 B cos2 45°+ ey sin2 45° + 0 ey = -250 A 10 - 6 B Generalized Hooke’s Law: ex =
1 C s - v A sy + sz B D E x
450 A 10 - 6 B =
1 B 28 A 106 B - v C -7 A 106 B + 0 D R E
450 A 10 - 6 B E - 7 A 106 B v = 28 A 106 B ey =
(1)
1 [s - v(sx + sz)] E y
-250 A 10 - 6 B =
1 b -7 A 106 B - v C 28 A 106 B + 0 D r E
250 A 10 - 6 B E - 28 A 106 B v = 7 A 106 B
(2)
Solving Eqs. (1) and (2), E = 67.74 A 109 B N>m2 = 67.7 GPa
Ans.
v = 0.3548 = 0.355
Ans.
Using the above results, G =
67.74 A 109 B E = 2(1 + v) 2(1 + 0.3548)
= 25.0 A 109 B N>m2 = 25.0 GPa
Ans.
786
x
wx
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*10–52. The block is fitted between the fixed supports. If the glued joint can resist a maximum shear stress of tallow = 2 ksi, determine the temperature rise that will cause the joint to fail. Take E = 10 (103) ksi, n = 0.2, and Hint: Use Eq. 10–18 with an additional strain term of a¢T (Eq. 4–4).
40⬚
Normal Strain: Since the aluminum is confined along the y direction by the rigid frame, then ey = 0 and sx = sz = 0. Applying the generalized Hooke’s Law with the additional thermal strain, ey =
0 =
1 C s - v(sx + sz) D + a¢T E y
1 C sy - 0.2(0 + 0) D + 6.0 A 10 - 6 B (¢T) 10.0(103)
sy = -0.06¢T Construction of the Circle: In accordance with the sign convention. sx = 0, sy = -0.06¢T and txy = 0. Hence. savg =
sx + sy 2
=
0 + ( -0.06¢T) = -0.03¢T 2
The coordinates for reference points A and C are A (0, 0) and C( -0.03¢T, 0). The radius of the circle is R = 2(0 - 0.03¢T)2 + 0 = 0.03¢T Stress on The inclined plane: The shear stress components tx¿y¿, are represented by the coordinates of point P on the circle. tx¿y¿ = 0.03¢T sin 80° = 0.02954¢T Allowable Shear Stress: tallow = tx¿y¿ 2 = 0.02954¢T ¢T = 67.7 °F
Ans.
787
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z
•10–53.
The smooth rigid-body cavity is filled with liquid 6061-T6 aluminum. When cooled it is 0.012 in. from the top of the cavity. If the top of the cavity is covered and the temperature is increased by 200°F, determine the stress components sx , sy , and sz in the aluminum. Hint: Use Eqs. 10–18 with an additional strain term of a¢T (Eq. 4–4).
0.012 in.
4 in. 4 in.
6 in.
y
Normal Strains: Since the aluminum is confined at its sides by a rigid container and 0.012 allowed to expand in the z direction, ex = ey = 0; whereas ez = = 0.002. 6 Applying the generalized Hooke’s Law with the additional thermal strain,
ex =
0 =
1 C s - v(sy + sz) D + a¢T E x 1 C sx - 0.35 A sy + sz B D + 13.1 A 10 - 6 B (200) 10.0(103)
0 = sx - 0.35sy - 0.35sz + 26.2 ey =
0 =
[1]
1 C s - v(sx + sz) + a¢T E y 1 C sy - 0.35(sx + sz) D + 13.1 A 10 - 6 B (200) 10.0(103)
0 = sy - 0.35sx - 0.35sz + 26.2 ez =
0.002 =
[2]
1 C s - v A sx + sy B D + a¢T E z 1 C sz - 0.35 A sx + sy B D + 13.1 A 10 - 6 B (200) 10.0(103)
0 = sz - 0.35sx - 0.35sy + 6.20
[3]
Solving Eqs.[1], [2] and [3] yields: sx = sy = -70.0 ksi
sz = -55.2 ksi
Ans.
788
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z
10–54. The smooth rigid-body cavity is filled with liquid 6061-T6 aluminum. When cooled it is 0.012 in. from the top of the cavity. If the top of the cavity is not covered and the temperature is increased by 200°F, determine the strain components Px , Py , and Pz in the aluminum. Hint: Use Eqs. 10–18 with an additional strain term of a¢T (Eq. 4–4).
0.012 in.
4 in. 4 in.
6 in.
y
Normal Strains: Since the aluminum is confined at its sides by a rigid container, then ex = ey = 0
Ans.
and since it is not restrained in z direction, sz = 0. Applying the generalized Hooke’s Law with the additional thermal strain, ex =
0 =
1 C s - v A sy + sz B D + a¢T E x 1 C sx - 0.35 A sy + 0 B D + 13.1 A 10 - 6 B (200) 10.0(103)
0 = sx - 0.35sy + 26.2 ey =
0 =
[1]
1 C s - v(sx + sz) D + a¢T E y 1 C sy - 0.35(sx + 0) D + 13.1 A 10 - 6 B (200) 10.0(103)
0 = sy - 0.35sx + 26.2
[2]
Solving Eqs. [1] and [2] yields: sx = sy = -40.31 ksi ez =
=
1 C s - v A sx + sy B D + a¢T E z 1 {0 - 0.35[-40.31 + (-40.31)]} + 13.1 A 10 - 6 B (200) 10.0(103)
= 5.44 A 10 - 3 B
Ans.
789
x
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10–55. A thin-walled spherical pressure vessel having an inner radius r and thickness t is subjected to an internal pressure p. Show that the increase in the volume within the vessel is ¢V = 12ppr4>Et211 - n2. Use a small-strain analysis.
pr 2t
s1 = s2 = s3 = 0 e1 = e2 =
1 (s - vs2) E 1
e1 = e2 =
pr (1 - v) 2t E
e3 =
1 (-v(s1 + s2)) E e3 = -
V =
v pr tE
4pr3 3
V + ¢V =
4p 4pr3 ¢r 3 (r + ¢r)3 = (1 + ) r 3 3
where ¢V V V, ¢r V r V + ¢V -
eVol =
¢r 4p r3 a1 + 3 b r 3
¢V ¢r = 3a b V r
Since e1 = e2 =
eVol = 3e1 =
2p(r + ¢r) - 2p r ¢r = r 2p r
3pr (1 - v) 2t E
¢V = VeVol =
2pp r4 (1 - v) Et
QED
790
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*10–56. A thin-walled cylindrical pressure vessel has an inner radius r, thickness t, and length L. If it is subjected to an internal pressure p, show that the increase in its inner radius is dr = rP1 = pr211 - 12 n2>Et and the increase in its length is ¢L = pLr112 - n2>Et. Using these results, show that the change in internal volume becomes dV = pr211 + P12211 + P22L - pr2L. Since P1 and P2 are small quantities, show further that the change in volume per unit volume, called volumetric strain, can be written as dV>V = pr12.5 - 2n2>Et. Normal stress: pr ; t
s1 =
s2 =
pr 2t
Normal strain: Applying Hooke’s law e1 =
=
1 [s - v (s2 + s3)], E 1
vpr pr 1 1 pr a b = a1 - vb E t 2t Et 2
d r = et r =
e2 =
=
s3 = 0
p r2 1 a1 - v b Et 2
1 [s - v (s1 + s3)], E 2
QED
s3 = 0
vpr pr 1 1 pr a b = a - vb E 2t t Et 2
¢L = e2 L =
pLr 1 a - vb Et 2
V¿ = p(r + e1 r)2 (L + e2L) ;
QED V = p r2 L
dV = V¿ - V = pr2 (1 + e1)2 (1 + e2)L - pr2 L
QED
(1 + e1)2 = 1 + 2 e1 neglect e21 term (1 + e1)2 (1 + e2) = (1 + 2 e1)(1 + e2) = 1 + e2 + 2 e1 neglect e1 e2 term dV = 1 + e2 + 2 e1 - 1 = e2 + 2 e1 V =
2pr pr 1 1 a - vb + a1 - v b Et 2 Et 2
=
pr (2.5 - 2 v) Et
QED
791
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10–57. The rubber block is confined in the U-shape smooth rigid block. If the rubber has a modulus of elasticity E and Poisson’s ratio n, determine the effective modulus of elasticity of the rubber under the confined condition.
P
Generalized Hooke’s Law: Under this confined condition, ex = 0 and sy = 0. We have ex =
0 =
1 C s - v A sy + sz B D E x 1 (s - vsz) E x
sx = vsz
(1)
ez =
1 C s - v A sx + sy B D E z
ez =
1 [s - v(sx + 0)] E z
ez =
1 (s - vsx) E z
(2)
Substituting Eq. (1) into Eq. (2), ez =
sz E
A 1 - v2 B
The effective modulus of elasticity of the rubber block under the confined condition can be determined by considering the rubber block as unconfined but rather undergoing the same normal strain of ez when it is subjected to the same normal stress sz, Thus, sz = Eeff ez Eeff =
sz ez
sz =
sz E
A 1 - v2 B
=
E 1 - v2
Ans.
792
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z
10–58. A soft material is placed within the confines of a rigid cylinder which rests on a rigid support. Assuming that Px = 0 and Py = 0, determine the factor by which the modulus of elasticity will be increased when a load is applied if n = 0.3 for the material.
P
x
Normal Strain: Since the material is confined in a rigid cylinder. ex = ey = 0. Applying the generalized Hooke’s Law, ex =
1 C s - v(sy + sx) D E z
0 = sx - v(sy + sz) ey =
[1]
1 C s - v(sx + sz) D E y
0 = sy - v(sx + sz)
[2]
Solving Eqs.[1] and [2] yields: sx = sy =
v s 1 - v z
Thus, ez =
=
1 C s - v(sx + sy) D E z v v 1 csz - v a sz + s bd E 1 - v 1 - v z sz
=
E
c1 -
2v2 d 1 - v
=
sz 1 - v - 2v2 c d E 1 - v
=
sz (1 + v)(1 - 2v c d E 1 - v
Thus, when the material is not being confined and undergoes the same normal strain of ez, then the requtred modulus of elasticity is E¿ =
sz ez
=
The increased factor is k =
1 - v E (1 - 2v)(1 + v) E¿ 1 - v = E (1 - 2v)(1 + v) =
1 - 0.3 [1 - 2(0.3)](1 + 0.3)
= 1.35
Ans.
793
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10–59. A material is subjected to plane stress. Express the distortion-energy theory of failure in terms of sx , sy , and txy . Maximum distortion energy theory: (s21 - s1 s2 + s22) = s2Y s1,2 =
sx + sy ;
2
Let a =
sx + sy 2
s1 = a + b;
A
a
(1)
sx - sy 2
and b =
A
a
2
2 b + txy
sx - sy 2
2
2 b + txy
s2 = a - b
s21 = a2 + b2 + 2 a b;
s22 = a2 + b2 - 2 a b
s1 s2 = a2 - b2 From Eq. (1) (a2 + b2 + 2 a b - a2 + b2 + a2 + b2 - 2 a b) = s2y (a2 + 3 b2) = s2Y (sx + sy)2 4
+ 3
(sx - sy)2 4
+ 3 t2xy = s2Y
s2x + s2y - sxsy + 3 t2xy = s2Y
Ans.
*10–60. A material is subjected to plane stress. Express the maximum-shear-stress theory of failure in terms of sx , sy , and txy . Assume that the principal stresses are of different algebraic signs. Maximum shear stress theory: |s1 - s2| = sY s1,2 =
(1)
sx + sy ;
2
` s1 - s2 ` = 2
A
a
A
a
sx - sy 2
sx - sy 2
2
2 b + txy
2 b + txy 2
From Eq. (1) 4 ca
sx - sy 2
2
b + t2xy d = s2Y
2
(sx - sy) + 4 t2xy = s2Y
Ans.
794
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•10–61.
An aluminum alloy 6061-T6 is to be used for a solid drive shaft such that it transmits 40 hp at 2400 rev>min. Using a factor of safety of 2 with respect to yielding, determine the smallest-diameter shaft that can be selected based on the maximum-shear-stress theory. v = a2400 T =
2p rad 1 min rev ba ba b = 80 p rad>s min rev 60s
40 (550) (12) P 3300 # = = lb in. p v 80 p Tc J
Applying t =
t =
A 3300 p B c p 2
=
c4
6600 p3 c3
The principal stresses: s1 = t =
6600 ; p2 c3
s2 = -t =
6600 p2 c3
Maximum shear stress theory: Both principal stresses have opposite sign, hence,
` s1 - s2 ` =
2a
sY ; F.S.
37 (103) 6600 b = ` ` 2 3 2 pc
c = 0.4166 in. d = 0.833 in.
Ans.
10–62. Solve Prob. 10–61 using the maximum-distortionenergy theory. v = a2400 T =
2p rad 1 min rev ba ba b = 80 p rad>s min rev 60s
40 (550) (12) P 3300 = = lb.in. p v 80 p
Applying t =
t =
A 3300 p B c p 2
c4
=
Tc J 6600 p2 c3
The principal stresses: s1 = t =
6600 ; p2 c3
s2 = - t = -
6600 p2 c3
The maximum distortion-energy theory: s21 - s1 s2 + s22 = a 3B
sY 2 b F.S.
37(103) 2 6600 2 = a b R 2 p2 c3
c = 0.3971 in. d = 0.794 in.
Ans. 795
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10–63. An aluminum alloy is to be used for a drive shaft such that it transmits 25 hp at 1500 rev>min. Using a factor of safety of 2.5 with respect to yielding, determine the smallest-diameter shaft that can be selected based on the maximum-distortion-energy theory. sY = 3.5 ksi. 1500(2p) = 50p 60
T =
P v
T =
25(550)(12) 3300 = p 50p
t =
Tc , J
t =
3300 p c p 4 2c
s1 =
v =
J =
=
p 4 c 2
6600 p2c3
6600 p2c3
s2 =
s21 - s1 s2 + s22 = a 3a
-6600 p2c3
sY 2 b F.S.
3.5(103) 2 6600 2 b b = a 2.5 p2c3
c = 0.9388 in. d = 1.88 in.
Ans.
*10–64. A bar with a square cross-sectional area is made of a material having a yield stress of sY = 120 ksi. If the bar is subjected to a bending moment of 75 kip # in., determine the required size of the bar according to the maximumdistortion-energy theory. Use a factor of safety of 1.5 with respect to yielding. Normal and Shear Stress: Applying the flexure formula, s =
75 A a2 B Mc 450 = 1 4 = 3 I a 12 a
In-Plane Principal Stress: Since no shear stress acts on the element s1 = sx =
450 a3
s2 = sy = 0
Maximum Distortion Energy Theory: s21 - s1 s2 + s22 = s2allow a
120 2 450 2 b - 0 + 0 = a b 3 1.5 a a = 1.78 in.
Ans.
796
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•10–65.
Solve Prob. 10–64 using the maximum-shearstress theory.
Normal and Shear Stress: Applying the flexure formula, s =
75 A a2 B Mc 450 = 1 4 = 3 I a a 12
In-Plane Principal Stress: Since no shear stress acts on the element. s1 = sx =
450 a3
s2 = sx = 0
Maximum Shear Stress Theory: |s2| = 0 6 sallow =
120 = 80.0 ksi 1.5
(O.K!)
|s1| = sallow 120 450 = 1.5 a3 a = 1.78 in.
Ans.
10–66. Derive an expression for an equivalent torque Te that, if applied alone to a solid bar with a circular cross section, would cause the same energy of distortion as the combination of an applied bending moment M and torque T.
t =
Te c J
Principal stress: s1 = tx ¿ ud =
s2 = -t
1 + v 2 (s1 - s1 s2 + s22) 3E
(ud)1 =
1 + v 1 + v 3 T2x c2 ( 3 t2) = a b 3E 3E J2
Bending moment and torsion: s =
Mc ; I
t =
Tc J
Principal stress: s1, 2 =
s1 =
s + 0 s - 0 2 2 ; a b + t 2 A 2
s s2 + + t2 ; 2 A4
s2 =
s s2 + t2 2 A4
797
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10–66.
Continued
Let a =
s 2
b =
s2 + t2 A4
s21 = a2 + b2 + 2 a b s1 s2 = a2 - b2 s22 = a2 + b2 - 2 a b s21 - s1 s2 + s22 = 3 b2 + a2 1 + v 2 (s1 - s1 s2 + s22) 3E
ud =
(ud)2 =
=
1 + v 1 + v 3 s2 s2 (3 b2 + a2) = a + 3t2 + b 3E 3E 4 4 c2(1 + v) M2 1 + v 2 3 T2 b (s + 3 t2) = a 2 + 3E 3E I J2
(ud)1 = (ud)2 c3(1 + v) 3 Tx 2 c2(1 + v) M2 3 T2 = a + b 3E 3E J2 I2 J2 For circular shaft J = I
p 3 p 4
c4 c4
=2
Te =
J2 M2 + T2 A I2 3
Te =
4 2 M + T2 A3
Ans.
10–67. Derive an expression for an equivalent bending moment Me that, if applied alone to a solid bar with a circular cross section, would cause the same energy of distortion as the combination of an applied bending moment M and torque T. Principal stresses: s1 =
Me c ; I
ud =
1 + v 2 (s1 - s1 s2 + s22) 3E
(ud)1 =
s2 = 0
1 + v M2e c2 a 2 b 3E I
(1)
798
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10–67.
Continued
Principal stress: s + 0 s - 0 2 3 ; a b + t 2 A 2
s1, 2 =
s s2 + + t2; 2 A4
s1 =
s2 =
s s2 + t2 2 A4
Distortion Energy: s s2 ,b = + t2 2 A4
Let a =
s21 = a2 + b2 + 2 a b s1 s2 = a2 - b2 s22 = a2 + b2 - 2 a b s22 - s1 s2 + s22 = 3 b2 + a2 Apply s =
(ud)2 =
=
Mc ; I
t =
Tc J
3s2 1 + v 1 + v s2 (3 b2 + a2) = a + + 3 t2 b 3E 3E 4 4 1 + v 2 1 + v M2 c2 3 T2 c2 (s + 3 t2) = a 2 + b 3E 3E I J2
(2)
Equating Eq. (1) and (2) yields: (1 + v) Me c2 1 + v M2 c2 3T2 c2 a 2 b = a 2 + b 3E 3E I I J2 M2e 2
=
I
M1 3 T2 + 2 I J2
M2e = M1 + 3 T2 a
I 2 b J
For circular shaft I = J
p 4 p 2
c4 c4
=
1 2
1 2 Hence, M2e = M2 + 3 T2 a b 2 Me =
A
M2 +
3 2 T 4
Ans.
799
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*10–68. The short concrete cylinder having a diameter of 50 mm is subjected to a torque of 500 N # m and an axial compressive force of 2 kN. Determine if it fails according to the maximum-normal-stress theory. The ultimate stress of the concrete is sult = 28 MPa. A =
p (0.05)2 = 1.9635(10 - 3) m2 4
J =
p (0.025)4 = 0.61359(10 - 4) m4 2
2 kN
500 N⭈m
500 N⭈m
2 kN
3
s =
2(10 ) P = 1.019 MPa = A 1.9635(10 - 3)
t =
500(0.025) Tc = 20.372 MPa = J 0.61359(10 - 6)
sx = 0
sy = -1.019 MPa sx + sy
s1, 2 =
s1,2 =
2
;
A
a
sx - sy 2
txy = 20.372 MPa 2 b + txy 2
0 - 1.018 0 - (-1.019) 2 2 ; a b + 20.372 2 A 2
s1 = 19.87 MPa
s2 = -20.89 MPa
Failure criteria: |s1| 6 salt = 28 MPa
OK
|s2| 6 salt = 28 MPa
OK
No.
Ans.
•10–69.
Cast iron when tested in tension and compression has an ultimate strength of 1sult2t = 280 MPa and 1sult2c = 420 MPa, respectively. Also, when subjected to pure torsion it can sustain an ultimate shear stress of tult = 168 MPa. Plot the Mohr’s circles for each case and establish the failure envelope. If a part made of this material is subjected to the state of plane stress shown, determine if it fails according to Mohr’s failure criterion.
120 MPa
100 MPa
220 MPa
s1 = 50 + 197.23 = 247 MPa s2 = 50 - 197.23 = -147 MPa The principal stress coordinate is located at point A which is outside the shaded region. Therefore the material fails according to Mohr’s failure criterion. Yes.
Ans.
800
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10–69.
Continued
10–70. Derive an expression for an equivalent bending moment Me that, if applied alone to a solid bar with a circular cross section, would cause the same maximum shear stress as the combination of an applied moment M and torque T. Assume that the principal stresses are of opposite algebraic signs. Bending and Torsion: Mc Mc 4M = p 4 = ; I c p c3 4
s =
t =
Tc Tc 2T = p 4 = J c p c3 2
The principal stresses:
s1, 2 =
= tabs max
sx + sy 2
;
A
a
sx - sy 2
2
2 b + txy =
4M pc3
4M
+ 0 2
3
;
Q
¢pc
- 0 2
2
≤ + a
2T 3
pc
b
2
2 2M ; 2M2 + T2 p c3 p c3
= s1 - s2 = 2 c
2 2M2 + T2 d p c3
(1)
Pure bending: s1 = tabs max
Me c 4 Me Mc = p 4 = ; I c p c3 4
= s1 - s2 =
s2 = 0
4 Me
(2)
p c3
Equating Eq. (1) and (2) yields: 4 Me 4 2M2 + T2 = p c3 p c3 Me = 2M2 + T2
Ans.
801
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10–71. The components of plane stress at a critical point on an A-36 steel shell are shown. Determine if failure (yielding) has occurred on the basis of the maximum-shearstress theory.
60 MPa
40 MPa
In accordance to the established sign convention, sx = 70 MPa, sy = -60 MPa and txy = 40 MPa. s1, 2 =
=
sx + sy 2
;
A
a
sx - sy 2
70 MPa
2 b + txy 2
70 + (-60) 70 - (-60) 2 2 ; c d + 40 2 A 2
= 5 ; 25825 s1 = 81.32 MPa
s2 = -71.32 MPa
In this case, s1 and s2 have opposite sign. Thus, |s1 - s2| = |81.32 - (-71.32)| = 152.64 MPa 6 sy = 250 MPa Based on this result, the steel shell does not yield according to the maximum shear stress theory.
*10–72. The components of plane stress at a critical point on an A-36 steel shell are shown. Determine if failure (yielding) has occurred on the basis of the maximumdistortion-energy theory.
60 MPa
40 MPa
In accordance to the established sign convention, sx = 70 MPa, sy = -60 MPa and txy = 40 MPa. s1, 2 =
=
sx + sy 2
;
A
a
sx - sy 2
2 b + txy 2
70 + (-60) 70 - ( -60) 2 2 ; c d + 40 2 A 2
= 5 ; 25825 s1 = 81.32 MPa
s2 = -71.32 MPa
s1 2 - s1 s2 + s2 2 = 81.322 - 81.32(-71.32) + (-71.32)2 = 17,500 6 sy 2 = 62500 Based on this result, the steel shell does not yield according to the maximum distortion energy theory.
802
70 MPa
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•10–73.
If the 2-in. diameter shaft is made from brittle material having an ultimate strength of sult = 50 ksi for both tension and compression, determine if the shaft fails according to the maximum-normal-stress theory. Use a factor of safety of 1.5 against rupture.
30 kip 4 kip · ft
Normal Stress and Shear Stresses. The cross-sectional area and polar moment of inertia of the shaft’s cross-section are A = p A 12 B = pin2
J =
p 4 p A 1 B = in4 2 2
The normal stress is caused by axial stress. s =
N 30 = = -9.549 ksi p A
The shear stress is contributed by torsional shear stress. t =
4(12)(1) Tc = = 30.56 ksi p J 2
The state of stress at the points on the surface of the shaft is represented on the element shown in Fig. a. In-Plane Principal Stress. sx = -9.549 ksi, sy = 0 and txy = -30.56 ksi. We have s1, 2 =
=
sx + sy 2
;
A
a
sx - sy 2
2 b + txy 2
-9.549 - 0 2 -9.549 + 0 2 ; a b + ( -30.56) 2 A 2
= (-4.775 ; 30.929) ksi s1 = 26.15 ksi
s2 = -35.70 ksi
Maximum Normal-Stress Theory. sallow =
sult 50 = = 33.33 ksi F.S. 1.5
|s1| = 26.15 ksi 6 sallow = 33.33 ksi
(O.K.)
|s2| = 35.70 ksi 7 sallow = 33.33 ksi
(N.G.)
Based on these results, the material fails according to the maximum normal-stress theory.
803
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10–74. If the 2-in. diameter shaft is made from cast iron having tensile and compressive ultimate strengths of 1sult2t = 50 ksi and 1sult2c = 75 ksi, respectively, determine if the shaft fails in accordance with Mohr’s failure criterion. Normal Stress and Shear Stresses. The cross-sectional area and polar moment of inertia of the shaft’s cross-section are A = p A 12 B = p in2
J =
p 4 p A 1 B = in4 2 2
The normal stress is contributed by axial stress. s =
N 30 = = -9.549 ksi p A
The shear stress is contributed by torsional shear stress. t =
4(12)(1) Tc = = 30.56 ksi p J 2
The state of stress at the points on the surface of the shaft is represented on the element shown in Fig. a. In-Plane Principal Stress. sx = -9.549 ksi, sy = 0, and txy = -30.56 ksi. We have s1, 2 =
=
sx + sy 2
;
A
a
sx - sy 2
2 b + txy 2
-9.549 - 0 2 -9.549 + 0 2 ; a b + ( -30.56) 2 A 2
= ( -4.775 ; 30.929) ksi s1 = 26.15 ksi
s2 = -35.70 ksi
Mohr’s Failure Criteria. As shown in Fig. b, the coordinates of point A, which represent the principal stresses, are located inside the shaded region. Therefore, the material does not fail according to Mohr’s failure criteria.
804
30 kip 4 kip · ft
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10–75. If the A-36 steel pipe has outer and inner diameters of 30 mm and 20 mm, respectively, determine the factor of safety against yielding of the material at point A according to the maximum-shear-stress theory.
900 N
150 mm A
100 mm
200 mm
Internal Loadings. Considering the equilibrium of the free - body diagram of the post’s right cut segment Fig. a, ©Fy = 0; Vy + 900 - 900 = 0
Vy = 0 T = -360 N # m
©Mx = 0; T + 900(0.4) = 0
©Mz = 0; Mz + 900(0.15) - 900(0.25) = 0 Mz = 90 N # m Section Properties. The moment of inertia about the z axis and the polar moment of inertia of the pipe’s cross section are Iz =
p A 0.0154 - 0.014 B = 10.15625p A 10 - 9 B m4 4
J =
p A 0.0154 - 0.014 B = 20.3125p A 10 - 9 B m4 2
Normal Stress and Shear Stress. The normal stress is contributed by bending stress. Thus, sY = -
MyA 90(0.015) = = -42.31MPa Iz 10.15625p A 10 - 9 B
The shear stress is contributed by torsional shear stress. t =
360(0.015) Tc = = 84.62 MPa J 20.3125p A 10 - 9 B
The state of stress at point A is represented by the two - dimensional element shown in Fig. b. In - Plane Principal Stress. sx = -42.31 MPa, sz = 0 and txz = 84.62 MPa. We have s1, 2 =
=
sx + sz 2
;
A
a
sx - sz 2
2 b + txz 2
-42.31 - 0 2 -42.31 + 0 2 ; a b + 84.62 2 A 2
= (-21.16 ; 87.23) MPa s1 = 66.07 MPa
s2 = -108.38 MPa
805
200 mm
900 N
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10–75.
Continued
Maximum Shear Stress Theory. s1 and s2 have opposite signs. This requires |s1 - s2| = sallow 66.07 - (-108.38) = sallow sallow = 174.45 MPa The factor of safety is F.S. =
sY 250 = = 1.43 sallow 174.45
Ans.
806
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*10–76. If the A-36 steel pipe has an outer and inner diameter of 30 mm and 20 mm, respectively, determine the factor of safety against yielding of the material at point A according to the maximum-distortion-energy theory.
900 N
150 mm A
100 mm
200 mm
Internal Loadings: Considering the equilibrium of the free - body diagram of the pipe’s right cut segment Fig. a, ©Fy = 0; Vy + 900 - 900 = 0
Vy = 0 T = -360 N # m
©Mx = 0; T + 900(0.4) = 0
©Mz = 0; Mz + 900(0.15) - 900(0.25) = 0 Mz = 90 N # m Section Properties. The moment of inertia about the z axis and the polar moment of inertia of the pipe’s cross section are Iz =
p A 0.0154 - 0.014 B = 10.15625p A 10 - 9 B m4 4
J =
p A 0.0154 - 0.014 B = 20.3125p A 10 - 9 B m4 2
Normal Stress and Shear Stress. The normal stress is caused by bending stress. Thus, sY = -
MyA 90(0.015) = = -42.31MPa Iz 10.15625p A 10 - 9 B
The shear stress is caused by torsional stress. t =
360(0.015) Tc = = 84.62 MPa J 20.3125p A 10 - 9 B
The state of stress at point A is represented by the two -dimensional element shown in Fig. b. In - Plane Principal Stress. sx = -42.31 MPa, sz = 0 and txz = 84.62 MPa. We have s1, 2 =
=
sx + sz 2
;
A
a
sx - sz 2
2 b + txz 2
-42.31 - 0 2 -42.31 + 0 2 ; a b + 84.62 2 A 2
= (-21.16 ; 87.23) MPa s1 = 66.07 MPa
s2 = -108.38 MPa
807
200 mm
900 N
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10–76.
Continued
Maximum Distortion Energy Theory. s1 2 - s1s2 + s2 2 = sallow 2 66.072 - 66.07(-108.38) + (-108.38)2 = sallow 2 sallow = 152.55 MPa Thus, the factor of safety is F.S. =
sY 250 = = 1.64 sallow 152.55
Ans.
808
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•10–77.
The element is subjected to the stresses shown. If sY = 36 ksi, determine the factor of safety for the loading based on the maximum-shear-stress theory. sx = 4 ksi s1, 2 =
=
sy = -12 ksi
sx + sy ;
2
A
a
sx - sy 2
txy = -8 ksi 4 ksi
2 b + txy 2
8 ksi
4 - (-12) 2 4 - 12 2 ; a b + (-8) 2 A 2 s2 = -15.314 ksi
s1 = 7.314 ksi
tabsmax =
7.314 - (-15.314) s1 - s2 = = 11.314 ksi 2 2
tallow =
sY 36 = = 18 ksi 2 2
F.S. =
12 ksi
tallow 18 = = 1.59 abs tmax 11.314
Ans.
10–78. Solve Prob. 10–77 using the maximum-distortionenergy theory.
sx = 4 ksi s1, 2 =
=
sy = -12 ksi
sx + sy 2
;
A
a
sx - sy
txy = -8 ksi 4 ksi
2 b + txy 2
8 ksi
4 - (-12) 2 4 - 12 2 ; a b + (-8) 2 A 2
s1 = 7.314 ksi
s2 = -15.314 ksi
s1 2 - s1 s2 + s2 2 = a F.S. =
2
12 ksi
sY 2 b F.S.
362 = 1.80 A (7.314)2 - (7.314)(-15.314) + (-15.314)2
Ans.
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10–79. The yield stress for heat-treated beryllium copper is sY = 130 ksi. If this material is subjected to plane stress and elastic failure occurs when one principal stress is 145 ksi, what is the smallest magnitude of the other principal stress? Use the maximum-distortion-energy theory. Maximum Distortion Energy Theory : With s1 = 145 ksi, s21 - s1 s2 + s22 = s2Y 1452 - 145s2 + s22 = 1302 s22 - 145s2 + 4125 = 0 s2 =
-(-145) ; 2( -145)2 - 4(1)(4125) 2(1)
= 72.5 ; 33.634 Choose the smaller root, s2 = 38.9 ksi
Ans.
sy ⫽ 0.5sx
*10–80. The plate is made of hard copper, which yields at sY = 105 ksi. Using the maximum-shear-stress theory, determine the tensile stress sx that can be applied to the plate if a tensile stress sy = 0.5sx is also applied.
s1 = sx
sx
1 s2 = sx 2
|s1| = sY sx = 105 ksi
Ans.
sy ⫽ 0.5sx
•10–81.
Solve Prob. 10–80 using the maximum-distortionenergy theory. s1 = sx s2 =
sx 2
sx
s21 - s1 s2 + s22 = s2Y s2x -
s2x s2x + = (105)2 2 4
sx = 121 ksi
Ans.
810
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10–82. The state of stress acting at a critical point on the seat frame of an automobile during a crash is shown in the figure. Determine the smallest yield stress for a steel that can be selected for the member, based on the maximumshear-stress theory. Normal and Shear Stress: In accordance with the sign convention. sx = 80 ksi
sy = 0
25 ksi
txy = 25 ksi
80 ksi
In - Plane Principal Stress: Applying Eq. 9-5. s1,2 =
=
sx + sy ;
2
A
a
sx - sy 2
2 b + txy 2
80 - 0 2 80 + 0 2 ; a b + 25 2 A 2
= 40 ; 47.170 s1 = 87.170 ksi
s2 = -7.170 ksi
Maximum Shear Stress Theory: s1 and s2 have opposite signs so |s1 - s2| = sY |87.170 - (-7.170)| = sY sY = 94.3 ksi
Ans.
10–83. Solve Prob. 10–82 using the maximum-distortionenergy theory. Normal and Shear Stress: In accordance with the sign convention. sx = 80 ksi
sy = 0
txy = 25 ksi
In - Plane Principal Stress: Applying Eq. 9-5. s1,2 =
=
sx + sy ;
2
A
a
25 ksi
sx - s 2 b + txy 2 2
80 ksi
80 - 0 2 80 + 0 2 ; a b + 25 2 A 2
= 40 ; 47.170 s1 = 87.170 ksi
s2 = -7.170 ksi
Maximum Distortion Energy Theory: s21 - s1s2 + s22 = s2Y 87.1702 - 87.170(-7.170) + (-7.170)2 = s2Y sY = 91.0 ksi
Ans.
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*10–84. A bar with a circular cross-sectional area is made of SAE 1045 carbon steel having a yield stress of sY = 150 ksi. If the bar is subjected to a torque of 30 kip # in. and a bending moment of 56 kip # in., determine the required diameter of the bar according to the maximum-distortion-energy theory. Use a factor of safety of 2 with respect to yielding.
Normal and Shear Stresses: Applying the flexure and torsion formulas. 56 A d2 B Mc 1792 = = p d 4 I pd3 A B
s =
4
t =
Tc = J
2
30 A d2 B
A B
d 4 2
p 2
=
480 pd3
The critical state of stress is shown in Fig. (a) or (b), where sx =
1792 pd3
sy = 0
txy =
480 pd3
In - Plane Principal Stresses : Applying Eq. 9-5, s1,2 =
sx + sy 2 1792 3 pd
=
=
s1 =
;
A
+ 0 2
;
D
a
¢
sx - sy 2 1792 3 pd
- 0 2
2 b + txy 2
2
≤ + a
480 2 b pd3
896 1016.47 ; pd3 pd3
1912.47 pd3
s2 = -
120.47 pd3
Maximum Distortion Energy Theory : s21 - s1s2 + s22 = s2allow a
1912.47 2 1912.47 120.47 120.47 2 150 2 b - a bab + ab = a b 3 3 3 3 2 pd pd pd pd d = 2.30 in.
Ans.
812
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•10–85.
The state of stress acting at a critical point on a machine element is shown in the figure. Determine the smallest yield stress for a steel that might be selected for the part, based on the maximum-shear-stress theory.
10 ksi
The Principal stresses: s1,2 =
=
sx + sy ;
2
A
4 ksi
a
sx - sy 2
2 b + txy 2
8 ksi
8 - (-10) 2 8 - 10 2 ; a b + 4 2 A 2
s1 = 8.8489 ksi
s2 = -10.8489 ksi
Maximum shear stress theory: Both principal stresses have opposite sign. hence, |s1 - s2| = sY
8.8489 - (-10.8489) = sY
sY = 19.7 ksi
Ans.
10–86. The principal stresses acting at a point on a thinwalled cylindrical pressure vessel are s1 = pr>t, s2 = pr>2t, and s3 = 0. If the yield stress is sY, determine the maximum value of p based on (a) the maximum-shear-stress theory and (b) the maximum-distortion-energy theory.
a) Maximum Shear Stress Theory: s1 and s2 have the same signs, then |s2| = sg
2
pr 2 = sg 2t
p =
2t s r g
|s1| = sg
2
pr 2 = sg t
p =
t s (Controls!) r g
Ans.
b) Maximum Distortion Energy Theory : s21 - s1s2 + s22 = s2g a
pr 2 pr pr pr 2 b - a b a b + a b = s2g t t 2t 2t p =
2t 23r
sg
Ans.
813
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10–87. If a solid shaft having a diameter d is subjected to a torque T and moment M, show that by the maximumshear-stress theory the maximum allowable shear stress is tallow = 116>pd322M2 + T2. Assume the principal stresses to be of opposite algebraic signs.
T
T
M
M
Section properties : I =
p d 4 pd4 a b = ; 4 2 64
J =
p d 4 pd4 a b = 2 2 32
Thus, M(d2 ) Mc 32 M = = p d4 I pd3
s =
64
T (d2 ) Tc 16 T t = = = p d4 J pd3 32
The principal stresses : s1,2 =
=
sx + sy ;
2
A
a
sx - sy 2
2 b + txy 2
16 M 16 M 2 16 T 2 16 M 16 ; ; 2M2 + T2 a b + a b = 3 3 A pd pd p d3 pd3 p d3
Assume s1 and s2 have opposite sign, hence, tallow
2 C 163 2M2 + T2 D s1 - s2 16 pd = = = 2M2 + T2 2 2 pd3
QED
*10–88. If a solid shaft having a diameter d is subjected to a torque T and moment M, show that by the maximum-normalstress theory the maximum allowable principal stress is sallow = 116>pd321M + 2M2 + T22.
T M
M
Section properties : I =
p d4 ; 64
p d4 32
J =
Stress components : s =
M (d2 ) Mc 32 M = p 4 = ; I d p d3 64
t =
T(d2 ) Tc 16 T = p 4 = J d p d3 32
The principal stresses : s1,2 =
=
sx + sy 2
;
A
a
sx - sy 2
2 b + txy = 2
32 M 3 pd
32 M
+ 0
2
3
;
D
¢ pd
- 0
2
2
≤ + a
16 T 2 b p d3
16 M 16 ; 2M2 + T2 p d3 p d3
Maximum normal stress theory. Assume s1 7 s2 sallow = s1 =
=
16 M 16 + 2M2 + T2 p d3 p d3
16 [M + 2M2 + T2] p d3
QED
814
T
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•10–89.
The shaft consists of a solid segment AB and a hollow segment BC, which are rigidly joined by the coupling at B. If the shaft is made from A-36 steel, determine the maximum torque T that can be applied according to the maximum-shear-stress theory. Use a factor of safety of 1.5 against yielding.
A
B T C
Shear Stress: This is a case of pure shear, and the shear stress is contributed by p torsion. For the hollow segment, Jh = A 0.054 - 0.044 B = 1.845p A 10 - 6 B m4. Thus, 2 (tmax)h =
T(0.05) Tch = = 8626.28T Jh 1.845p A 10 - 6 B
For the solid segment, Js =
(tmax)s =
p A 0.044 B = 1.28p A 10 - 6 B m4. Thus, 2
T(0.04) Tcs = = 9947.18T Js 1.28p A 10 - 6 B
By comparision, the points on the surface of the solid segment are critical and their state of stress is represented on the element shown in Fig. a. In - Plane Principal Stress. sx = sy = 0 and txy = 9947.18T. We have s1,2 =
=
sx + sy 2
;
C
¢
sx - sy 2
2
≤ + t2xy
0 - 0 2 0 + 0 ; ¢ ≤ + (9947.18T)2 2 C 2
s1 = 9947.18T
s2 = -9947.18T
Maximum Shear Stress Theory. sallow =
80 mm
sY 250 = = 166.67 MPa F.S. 1.5
Since s1 and s2 have opposite sings, |s1 - s2| = sallow 9947.18T - (-9947.18T) = 166.67 A 106 B T = 8377.58 N # m = 8.38 kN # m
Ans.
815
80 mm 100 mm
T
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10–90. The shaft consists of a solid segment AB and a hollow segment BC, which are rigidly joined by the coupling at B. If the shaft is made from A-36 steel, determine the maximum torque T that can be applied according to the maximum-distortion-energy theory. Use a factor of safety of 1.5 against yielding.
A
B T C
Shear Stress. This is a case of pure shear, and the shear stress is contributed by p torsion. For the hollow segment, Jh = A 0.054 - 0.044 B = 1.845p A 10 - 6 B m4. Thus, 2 (tmax)h =
T(0.05) Tch = = 8626.28T Jh 1.845p A 10 - 6 B
For the solid segment, Js =
(tmax)s =
p A 0.044 B = 1.28p A 10 - 6 B m4. Thus, 2
T(0.04) Tcs = = 9947.18T Js 1.28p A 10 - 6 B
By comparision, the points on the surface of the solid segment are critical and their state of stress is represented on the element shown in Fig. a. In - Plane Principal Stress. sx = sy = 0 and txy = 9947.18T. We have s1,2 =
=
sx + sy 2
;
C
¢
sx - sy 2
2
≤ + t2xy
0 - 0 2 0 + 0 ; ¢ ≤ + (9947.18T)2 2 C 2
s1 = 9947.18T
s2 = -9947.18T
Maximum Distortion Energy Theory. sallow =
80 mm
sY 250 = = 166.67 MPa F.S. 1.5
Then, s1 2 - s1s2 + s2 2 = sallow 2 (9947.18T)2 - (9947.18T)( -9947.18T) + (-9947.18T)2 = C 166.67 A 106 B D 2 T = 9673.60 N # m = 9.67 kN # m
Ans.
816
80 mm 100 mm
T
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10–91. The internal loadings at a critical section along the steel drive shaft of a ship are calculated to be a torque of 2300 lb # ft, a bending moment of 1500 lb # ft, and an axial thrust of 2500 lb. If the yield points for tension and shear are sY = 100 ksi and tY = 50 ksi, respectively, determine the required diameter of the shaft using the maximum-shearstress theory. 2300 lb⭈ft 2500 lb
p I = c4 4
A = p c2
sA =
p J = c4 2
1500(12)(c) P Mc 2500 2500 72 000 + b = -a + b + = -a pc4 A I p c2 p c2 p c3 4
tA =
2300(12)(c) Tc 55 200 = = p c4 J p c3 2
s1,2 =
sx + sy
= -a
2
;
A
a
sx - sy 2
2 b + txy 2
2500 c + 72 000 2500c + 72 000 2 55200 2 b ; a b + a b 3 3 A 2p c 2p c p c3
(1)
Assume s1 and s2 have opposite signs: |s1 - s2| = sg 2500c + 72 000 2 55 200 2 3 b + a b = 100(10 ) 3 A 2p c p c3
2
a
(2500c + 72000)2 + 1104002 = 10 000(106)p2 c6 6.25c2 + 360c + 17372.16 - 10 000p2 c6 = 0 By trial and error: c = 0.750 57 in. Substitute c into Eq. (1): s1 = 22 193 psi
s2 = -77 807 psi
s1 and s2 are of opposite signs
OK
Therefore, d = 1.50 in.
Ans.
817
1500 lb⭈ft
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*10–92. The gas tank has an inner diameter of 1.50 m and a wall thickness of 25 mm. If it is made from A-36 steel and the tank is pressured to 5 MPa, determine the factor of safety against yielding using (a) the maximum-shear-stress theory, and (b) the maximum-distortion-energy theory.
(a) Normal Stress. Since
0.75 r = = 30 7 10, thin - wall analysis can be used.We have t 0.025
s1 = sh =
5(0.75) pr = = 150 MPa t 0.025
s2 = slong =
pr 5(0.75) = = 75 MPa 2t 2(0.025)
Maximum Shear Stress Theory. s1 and s2 have the sign. Thus, |s1| = sallow sallow = 150 MPa The factor of safety is F.S. =
sY 250 = = 1.67 sallow 150
Ans.
(b) Maximum Distortion Energy Theory. s1 2 - s1s2 + s2 2 = sallow 2 1502 - 150(75) + 752 = sallow 2 sallow = 129.90 MPa The factor of safety is F.S. =
sY 250 = = 1.92 sallow 129.90
Ans.
818
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•10–93.
The gas tank is made from A-36 steel and has an inner diameter of 1.50 m. If the tank is designed to withstand a pressure of 5 MPa, determine the required minimum wall thickness to the nearest millimeter using (a) the maximum-shear-stress theory, and (b) maximumdistortion-energy theory. Apply a factor of safety of 1.5 against yielding.
(a) Normal Stress. Assuming that thin - wall analysis is valid, we have s1 = sh =
5 A 106 B (0.75) 3.75 A 106 B pr = = t t t
s2 = slong =
5 A 106 B (0.75) 1.875 A 106 B pr = = 2t 2t t
Maximum Shear Stress Theory. sallow =
250 A 106 B sY = = 166.67 A 106 B Pa FS. 1.5
s1 and s2 have the same sign. Thus, |s1| = sallow 3.75 A 106 B
= 166.67 A 106 B
t
t = 0.0225 m = 22.5 mm
Since
Ans.
0.75 r = = 33.3 7 10, thin - wall analysis is valid. t 0.0225
(b) Maximum Distortion Energy Theory. sallow =
250 A 106 B sY = = 166.67 A 106 B Pa F.S. 1.5
Thus, s1 2 - s1s2 + s2 2 = sallow 2
C
3.75 A 106 B t
3.2476 A 106 B t
2
S - C
3.75 A 106 B t
SC
1.875 A 106 B t
S + C
1.875 A 106 B t
2
S = c166.67 A 106 B d
= 166.67 A 106 B
t = 0.01949 m = 19.5 mm
Since
2
Ans.
0.75 r = = 38.5 7 10, thin - wall analysis is valid. t 0.01949
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10–94. A thin-walled spherical pressure vessel has an inner radius r, thickness t, and is subjected to an internal pressure p. If the material constants are E and n, determine the strain in the circumferential direction in terms of the stated parameters. s1 = s2 =
pr 2t
e1 = e2 = e =
e =
1 (s - vs) E
pr 1 - v pr 1 - v s = a b = (1 - v) E E 2t 2Et
Ans.
10–95. The strain at point A on the shell has components Px = 250(10 - 6), Py = 400(10 - 6), gxy = 275(10 - 6), Pz = 0. Determine (a) the principal strains at A, (b) the maximum shear strain in the x–y plane, and (c) the absolute maximum shear strain. ex = 250(10 - 6) A(250, 137.5)10 - 6
ey = 400(10 - 6)
gxy = 275(10 - 6)
y A
gxy 2
= 137.5(10 - 6)
C(325, 0)10 - 6
R = a 2(325 - 250)2 + (137.5)2 b10 - 6 = 156.62(10 - 6) a) e1 = (325 + 156.62)10 - 6 = 482(10 - 6)
Ans.
e2 = (325 - 156.62)10 - 6 = 168(10 - 6)
Ans.
b) g
max in-plane
= 2R = 2(156.62)(10 - 6) = 313(10 - 6)
Ans.
c) gabs
max
2 gabs
max
=
482(10 - 6) 2
= 482(10 - 6)
Ans.
820
x
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*10–96. The principal plane stresses acting at a point are shown in the figure. If the material is machine steel having a yield stress of sY = 500 MPa, determine the factor of safety with respect to yielding if the maximum-shear-stress theory is considered.
100 MPa
150 MPa
Have, the in plane principal stresses are s1 = sy = 100 MPa
s2 = sx = -150 MPa
Since s1 and s2 have same sign, F.S =
sy =
|s1 - s2|
500 = 2 |100 - (-150)|
Ans.
•10–97.
The components of plane stress at a critical point on a thin steel shell are shown. Determine if failure (yielding) has occurred on the basis of the maximumdistortion-energy theory. The yield stress for the steel is sY = 650 MPa.
340 MPa
65 MPa
55 MPa
sx = -55 MPa s1, 2 =
=
sy = 340 MPa
sx + sy 2
;
A
a
sx - sy 2
txy = 65 MPa
2 b + txy 2
-55 - 340 2 -55 + 340 2 ; a b + 65 2 A 2
s1 = 350.42 MPa
s2 = -65.42 MPa
(s1 2 - s1s2 + s2 ) = [350.422 - 350.42(-65.42) + ( -65.42)2] = 150 000 6 s2Y = 422 500
OK
No.
Ans.
821
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10–98. The 60° strain rosette is mounted on a beam. The following readings are obtained for each gauge: Pa = 600110-62, Pb = -700110-62, and Pc = 350110-62. Determine (a) the in-plane principal strains and (b) the maximum in-plane shear strain and average normal strain. In each case show the deformed element due to these strains.
a 60⬚ 60⬚ b
Strain Rosettes (60º): Applying Eq. 10-15 with ex = 600 A 10 - 6 B , eb = -700 A 10
-6
c
B , ec = 350 A 10 B , ua = 150°, ub = -150° and uc = -90°, -6
350 A 10 - 6 B = ex cos2 (-90°) + ey sin2( -90°) + gxy sin (-90°) cos ( -90°) ey = 350 A 10 - 6 B
600 A 10 - 6 B = ex cos2 150° + 350 A 10 - 6 B sin2 150° + gxy sin 150° cos 150° 512.5 A 10 - 6 B = 0.75 ex - 0.4330 gxy
[1]
-787.5 A 10 - 6 B = 0.75ex + 0.4330 gxy
[2]
-700 A 10 - 6 B = ex cos2 ( -150°) + 350 A 10 - 6 B sin2(-150°) + gxy sin (-150°) cos (-150°)
Solving Eq. [1] and [2] yields ex = -183.33 A 10 - 6 B
gxy = -1501.11 A 10 - 6 B
Construction of she Circle: With ex = -183.33 A 10 - 6 B , ey = 350 A 10 - 6 B , and gxy = -750.56 A 10 - 6 B . 2 eavg =
ex + ey 2
= a
-183.33 + 350 b A 10 - 6 B = 83.3 A 10 - 6 B 2
Ans.
The coordinates for reference points A and C are A( -183.33, -750.56) A 10 - 6 B
C(83.33, 0) A 10 - 6 B
The radius of the circle is R = a 2(183.33 + 83.33)2 + 750.562 b A 10 - 6 B = 796.52 A 10 - 6 B a) In-plane Principal Strain: The coordinates of points B and D represent e1 and e2, respectively. e1 = (83.33 + 796.52) A 10 - 6 B = 880 A 10 - 6 B
Ans.
e2 = (83.33 - 796.52) A 10 - 6 B = -713 A 10 - 6 B
Ans.
Orientation of Principal Strain: From the circle, tan 2uP1 =
750.56 = 2.8145 183.33 + 83.33
2uP2 = 70.44°
2uP1 = 180° - 2uP2 uP =
180° - 70.44° = 54.8° (Clockwise) 2
Ans.
822
60⬚
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10–98.
Continued
b) Maximum In - Plane Shear Strain: Represented by the coordinates of point E on the circle. g max in-plane
2 g
max in-plane
= -R = -796.52 A 10 - 6 B = -1593 A 10 - 6 B
Ans.
Orientation of Maximum In-Plane Shear Strain: From the circle. tan 2uP =
183.33 + 83.33 = 0.3553 750.56
uP = 9.78° (Clockwise)
Ans.
10–99. A strain gauge forms an angle of 45° with the axis of the 50-mm diameter shaft. If it gives a reading of P = -200110-62 when the torque T is applied to the shaft, determine the magnitude of T. The shaft is made from A-36 steel.
T 45⬚
Shear Stress. This is a case of pure shear, and the shear stress developed is p contributed by torsional shear stress. Here, J = A 0.0254 B = 0.1953125p A 10 - 6 B m4. 2 Then 0.128 A 106 B T T(0.025) Tc = = t = p J 0.1953125p A 10 - 6 B
T
The state of stress at points on the surface of the shaft can be represented by the element shown in Fig. a. Shear Strain: For pure shear ex = ey = 0. We obtain, ea = ex cos2ua + ey sin2ua + gxysin ua cos ua -200 A 10 - 6 B = 0 + 0 + gxy sin 45° cos 45°
gxy = -400 A 10 - 6 B
Shear Stress and Strain Relation: Applying Hooke’s Law for shear, txy = Ggxy -
0.128 A 106 B T p
= 75 A 109 B C -400 A 10 - 6 B D
T = 736 N # m
Ans.
823
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*10–100. The A-36 steel post is subjected to the forces shown. If the strain gauges a and b at point A give readings of Pa = 300110-62 and Pb = 175110-62, determine the magnitudes of P1 and P2.
P1
P2 a
A
Internal Loadings: Considering the equilibrium of the free - body diagram of the 1 in. post’s segment, Fig. a, P2 - V = 0
V = P2
+ c ©Fy = 0;
N - P1 = 0
N = P1
a + ©MO = 0;
M + P2(2) = 0
M = 2P2
Section Properties: The cross - sectional area and the moment of inertia about the bending axis of the post’s cross - section are A = 4(2) = 8 in2 I =
1 (2) A 43 B = 10.667 in4 12
Referring to Fig. b,
A Qy B A = x¿A¿ = 1.5(1)(2) = 3 in3 Normal and Shear Stress: The normal stress is a combination of axial and bending stress. sA =
2P2(12)(1) MxA P1 N + = + = 2.25P2 - 0.125P1 A I 8 10.667
The shear stress is caused by transverse shear stress. tA =
VQA P2(3) = = 0.140625P2 It 10.667(2)
Thus, the state of stress at point A is represented on the element shown in Fig. c. Normal and Shear Strain: With ua = 90° and ub = 45°, we have ea = ex cos2ua + ey sin2ua + gxysin ua cos ua 300 A 10 - 6 B = ex cos2 90° + ey sin2 90° + gxysin 90° cos 90°
ey = 300 A 10 - 6 B eb = ex cos2ub + ey sin2 ub + gxysin ub cos ub
175 A 10 - 6 B = ex cos2 45° + 300 A 10 - 6 B sin2 45° + gxy sin 45°cos 45° ex + gxy = 50 A 10 - 6 B
(1)
824
A 1 in.
b 45⬚ c
+ ©F = 0; : x
2 in.
2 ft
A
4 in.
c
Section c– c
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10–100.
Continued
Since sy = sz = 0, ex = -vey = -0.32(300) A 10 - 6 B = -96 A 10 - 6 B Then Eq. (1) gives gxy = 146 A 10 - 6 B Stress and Strain Relation: Hooke’s Law for shear gives tx = Ggxy 0.140625P2 = 11.0 A 103 B C 146 A 10 - 6 B D P2 = 11.42 kip = 11.4 kip
Ans.
Since sy = sz = 0, Hooke’s Law gives sy = Eey 2.25(11.42) - 0.125P1 = 29.0 A 103 B C 300 A 10 - 6 B D P1 = 136 kip
Ans.
825
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10–101. A differential element is subjected to plane strain that has the following components: Px = 950110-62, Py = 420110-62, gxy = -325110-62. Use the strain-transformation equations and determine (a) the principal strains and (b) the maximum in-plane shear strain and the associated average strain. In each case specify the orientation of the element and show how the strains deform the element.
e1, 2 =
ex + ey ;
2
= c
A
a
ex - ey 2
2 b + gxy 2
950 - 420 2 -325 2 950 + 420 -6 ; a b + a b d(10 ) 2 A 2 2 e1 = 996(10 - 6)
Ans.
e2 = 374(10 - 6)
Ans.
Orientation of e1 and e2 : gxy
tan 2uP =
ex - ey
-325 950 - 420
=
uP = -15.76°, 74.24° Use Eq. 10.5 to determine the direction of e1 and e2. ex¿ =
ex + ey
ex - ey +
2
2
cos 2u +
gxy 2
sin 2u
u = uP = -15.76° ex¿ = b
( -325) 950 - 420 950 + 420 + cos (-31.52°) + sin (-31.52°) r (10 - 6) = 996(10 - 6) 2 2 2
uP1 = -15.8°
Ans.
uP2 = 74.2°
Ans.
b) gmax
in-plane
2 gmax
in-plane
eavg =
=
A
= 2c
a
ex - ey
A
2 a
ex + ey 2
b + a 2
gxy 2
b
2
950 - 420 2 -325 2 -6 -6 b + a b d(10 ) = 622(10 ) 2 2
Ans.
= a
Ans.
950 + 420 b (10 - 6) = 685(10 - 6) 2
826
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10–101.
Continued
Orientation of gmax : -(ex - ey)
tan 2uP =
-(950 - 420) -325
=
gxy
uP = 29.2° and uP = 119°
Ans.
Use Eq. 10.6 to determine the sign of ex - ey
gx¿y¿ = -
2
sin 2u +
2
gxy 2
g
max in-plane
:
cos 2u
u = uP = 29.2° gx¿y¿ = 2 c
-(950 - 420) -325 sin (58.4°) + cos (58.4°) d(10 - 6) 2 2
gxy = -622(10 - 6)
10–102. The state of plane strain on an element is Px = 400110-62, Py = 200110-62, and gxy = -300110-62. Determine the equivalent state of strain on an element at the same point oriented 30° clockwise with respect to the original element. Sketch the results on the element.
y
Pydy
dy
Stress Transformation Equations: ex = 400 A 10 - 6 B
ey = 200 A 10 - 6 B
gxy = -300 A 10 - 6 B
u = -30° gxy 2 dx
We obtain, ex¿ =
ex + ey +
2
= B
ex - ey 2
cos 2u +
gxy 2
sin 2u
400 + 200 400 - 200 -300 + cos (-60°) + a b sin (-60°) R A 10 - 6 B 2 2 2
= 480 A 10 - 6 B gx¿y¿ 2
= -¢
Ans.
ex - ey 2
≤ sin 2u +
gxy 2
cos 2u
gx¿y¿ = [-(400 - 200) sin ( -60°) + (-300) cos ( -60°)] A 10 - 6 B = 23.2 A 10 - 6 B
ey¿ =
ex + ey
= B
2
Ans.
ex - ey -
2
cos 2u -
gxy 2
sin 2u
400 + 200 400 - 200 -300 cos ( -60°) - a b sin (-60°) R A 10 - 6 B 2 2 2
= 120 A 10 - 6 B
Ans.
827
gxy 2 x Pxdx
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10–103. The state of plane strain on an element is Px = 400110-62, Py = 200110-62, and gxy = -300110-62. Determine the equivalent state of strain, which represents (a) the principal strains, and (b) the maximum in-plane shear strain and the associated average normal strain. Specify the orientation of the corresponding element at the point with respect to the original element. Sketch the results on the element.
y
Pydy
dy
Construction of the Circle: ex = 400 A 10 - 6 B , ey = 200 A 10 - 6 B , and
gxy 2
= -150 A 10 - 6 B .
Thus, eavg =
ex + ey 2
= a
400 + 200 b A 10 - 6 B = 300 A 10 - 6 B 2
Ans.
The coordinates for reference points A and the center C of the circle are A(400, -150) A 10 - 6 B
C(300, 0) A 10 - 6 B
The radius of the circle is R = CA = 2(400 - 300)2 + (-150)2 = 180.28 A 10 - 6 B Using these results, the circle is shown in Fig. a. In - Plane Principal Stresses: The coordinates of points B and D represent e1 and e2, respectively. Thus, e1 = (300 + 180.28) A 10 - 6 B = 480 A 10 - 6 B
Ans.
e2 = (300 - 180.28) A 10 - 6 B = 120 A 10 - 6 B
Ans.
Orientation of Principal Plane: Referring to the geometry of the circle, tan 2 A up B 1 =
150 = 1.5 400 - 300
A up B 1 = 28.2° (clockwise)
Ans.
The deformed element for the state of principal strains is shown in Fig. b. Maximum In - Plane Shear Stress: The coordinates of point E represent eavg and gmax . Thus in-plane gmax = -R = -180.28 A 10 - 6 B
in-plane
2 gmax
in-plane
= -361 A 10 - 6 B
Ans.
Orientation of the Plane of Maximum In - Plane Shear Strain: Referring to the geometry of the circle, tan 2us =
400 - 300 = 0.6667 150
uS = 16.8° (counterclockwise)
Ans.
The deformed element for the state of maximum in - plane shear strain is shown in Fig. c.
828
gxy 2
gxy 2 dx
x Pxdx
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10–103.
Continued
829
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11–1. The simply supported beam is made of timber that has an allowable bending stress of sallow = 6.5 MPa and an allowable shear stress of tallow = 500 kPa. Determine its dimensions if it is to be rectangular and have a height-towidth ratio of 1.25.
8 kN/m
2m
Ix =
1 (b)(1.25b)3 = 0.16276b4 12
Qmax = y¿A¿ = (0.3125b)(0.625b)(b) = 0.1953125b3 Assume bending moment controls: Mmax = 16 kN # m sallow =
Mmax c I
6.5(106) =
16(103)(0.625b) 0.16276b4
b = 0.21143 m = 211 mm
Ans.
h = 1.25b = 264 mm
Ans.
Check shear: Qmax = 1.846159(10 - 3) m3
I = 0.325248(10 - 3) m4 tmax =
VQmax 16(103)(1.846159)(10 - 3) = 429 kPa 6 500 kPa‚ OK = It 0.325248(10 - 3)(0.21143)
830
4m
2m
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11–2. The brick wall exerts a uniform distributed load of 1.20 kip>ft on the beam. If the allowable bending stress is sallow = 22 ksi and the allowable shear stress is tallow = 12 ksi, select the lightest wide-flange section with the shortest depth from Appendix B that will safely support the load.
1.20 kip/
4 ft
10 ft
ft
6 ft b
Bending Stress: From the moment diagram, Mmax = 44.55 kip # ft. Assuming bending controls the design and applying the flexure formula. Sreq d =
=
44.55 (12) = 24.3 in3 22
W12 * 22
A Sx = 25.4 in3, d = 12.31 in., tw = 0.260 in. B V for the W12 * 22 wide tw d = 6.60 kip.
Shear Stress: Provide a shear stress check using t = flange section. From the shear diagram, Vmax tmax =
=
Vmax tw d 6.60 0.260(12.31)
= 2.06 ksi 6 tallow = 12 ksi (O.K!) Hence,
Use
9 in. 0.5 in.
Mmax sallow
Two choices of wide flange section having the weight 22 lb>ft can be made. They are W12 * 22 and W14 * 22. However, W12 * 22 is the shortest. Select
0.5 in.
0.5 in.
Ans.
W12 * 22
831
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11–3. The brick wall exerts a uniform distributed load of 1.20 kip>ft on the beam. If the allowable bending stress is sallow = 22 ksi, determine the required width b of the flange to the nearest 14 in.
1.20 kip/
4 ft
10 ft
ft
6 ft b 0.5 in.
0.5 in. 9 in. 0.5 in.
Section Property: I =
1 1 (b) A 103 B (b - 0.5) A 93 B = 22.583b + 30.375 12 12
Bending Stress: From the moment diagram, Mmax = 44.55 kip # ft. sallow = 22 =
Mmax c I 44.55(12)(5) 22.583b + 30.375
b = 4.04 in. Use
b = 4.25 in.
Ans.
832
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*11–4. Draw the shear and moment diagrams for the shaft, and determine its required diameter to the nearest 1 4 in. if sallow = 7 ksi and tallow = 3 ksi. The bearings at A and D exert only vertical reactions on the shaft. The loading is applied to the pulleys at B, C, and E.
14 in.
20 in.
15 in.
12 in.
E A
C
B
D 35 lb
80 lb 110 lb
sallow = 7(103) =
Mmax c I 1196 c p 4 ; 4 c
c = 0.601 in.
d = 2c = 1.20 in. Use d = 1.25 in.
Ans.
Check shear: 2
tmax =
0.625 108(4(0.625) Vmax Q 3p )(p)( 2 ) = 117 psi 6 3 ksi OK = p 4 It 4 (0.625) (1.25)
•11–5.
Select the lightest-weight steel wide-flange beam from Appendix B that will safely support the machine loading shown. The allowable bending stress is sallow = 24 ksi and the allowable shear stress is tallow = 14 ksi.
2 ft
Bending Stress: From the moment diagram, Mmax = 30.0 kip # ft. Assume bending controls the design. Applying the flexure formula. Sreq¿d =
= Select
W12 * 16
Mmax sallow 30.0(12) = 15.0 in3 24
A Sx = 17.1 in3, d = 11.99 in., tw = 0.220 in. B V for the W12 * 16 wide tw d = 10.0 kip
Shear Stress: Provide a shear stress check using t = flange section. From the shear diagram, Vmax tmax =
=
Vmax tw d 10.0 0.220(11.99)
= 3.79 ksi 6 tallow = 14 ksi (O.K!) Hence, Use
5 kip
5 kip
Ans.
W12 * 16
833
2 ft
5 kip
2 ft
5 kip
2 ft
2 ft
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11–6. The compound beam is made from two sections, which are pinned together at B. Use Appendix B and select the lightest-weight wide-flange beam that would be safe for each section if the allowable bending stress is sallow = 24 ksi and the allowable shear stress is tallow = 14 ksi. The beam supports a pipe loading of 1200 lb and 1800 lb as shown.
C
A B 6 ft
Bending Stress: From the moment diagram, Mmax = 19.2 kip # ft for member AB. Assuming bending controls the design, applying the flexure formula. Sreq¿d =
= Select
Mmax sallow 19.2(12) = 9.60 in3 24
A Sx = 10.9 in3, d = 9.87 in., tw = 0.19 in. B
W10 * 12
For member BC, Mmax = 8.00 kip # ft. Sreq¿d =
= Select
Mmax sallow 8.00(12) = 4.00 in3 24
A Sx = 5.56 in3, d = 5.90 in., tw = 0.17 in. B
W6 * 9
V for the W10 * 12 widetw d flange section for member AB. From the shear diagram, Vmax = 2.20 kip. Shear Stress: Provide a shear stress check using t =
tmax =
=
Vmax tw d 2.20 0.19(9.87)
= 1.17 ksi 6 tallow = 14 ksi (O.K!) Use
Ans.
W10 * 12
For member BC (W6 * 9), Vmax = 1.00 kip. tmax =
=
Vmax tw d 1.00 0.17(5.90)
= 0.997 ksi 6 tallow = 14 ksi (O.K!) Hence, Use
1800 lb
1200 lb
W6 * 9
Ans.
834
6 ft
8 ft
10 ft
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11–7. If the bearing pads at A and B support only vertical forces, determine the greatest magnitude of the uniform distributed loading w that can be applied to the beam. sallow = 15 MPa, tallow = 1.5 MPa.
w
A
B 1m
1m 150 mm 25 mm 150 mm 25 mm
The location of c, Fig. b, is y =
0.1625(0.025)(0.15) + 0.075(0.15)(0.025) ©yA = ©A 0.025(0.15) + 0.15(0.025) = 0.11875 m
I =
+
1 (0.025)(0.153) + (0.025)(0.15)(0.04375)2 12 1 (0.15)(0.0253) + 0.15(0.025)(0.04375)2 12
= 21.58203125(10 - 6) m4 Referring to Fig. b, Qmax = y¿A¿ = 0.059375 (0.11875)(0.025) = 0.176295313(10 - 4) m3 Referring to the moment diagram, Mmax = 0.28125 w. Applying the Flexure formula with C = y = 0.11875 m, sallow =
Mmax c ; I
15(106) =
0.28125w(0.11875) 21.582(10 - 6)
W = 9.693(103) N>m Referring to shear diagram, Fig. a, Vmax = 0.75 w. tallow =
Vallow Qmax ; It
1.5(106) =
0.75w C 0.17627(10 - 3) D 21.582(10 - 6)(0.025)
W = 6.122(103) N>m = 6.12 kN>m (Control!)
Ans.
835
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*11–8. The simply supported beam is made of timber that has an allowable bending stress of sallow = 1.20 ksi and an allowable shear stress of tallow = 100 psi. Determine its smallest dimensions to the nearest 18 in. if it is rectangular and has a height-to-width ratio of 1.5.
12 kip/ft
B
A 3 ft
3 ft
1.5 b b
The moment of inertia of the beam’s cross-section about the neutral axis is 1 (b)(1.5b)3 = 0.28125b4. Referring to the moment diagram, I = 12 Mmax = 45.375 kip # ft. sallow =
Mmax c ; I
1.2 =
45.375(12)(0.75b) 0.28125b4
b = 10.66 in Referring to Fig. b, Qmax = y¿A¿ = 0.375b (0.75b)(b) = 0.28125b3. Referring to the shear diagram, Fig. a, Vmax = 33 kip. tmax =
Vmax Qmax ; It
100 =
33(103)(0.28125b3) 0.28125b4(b)
b = 18.17 in (Control!) Thus, use b = 18
1 in 4
Ans.
836
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•11–9.
Select the lightest-weight W12 steel wide-flange beam from Appendix B that will safely support the loading shown, where P = 6 kip. The allowable bending stress is sallow = 22 ksi and the allowable shear stress is tallow = 12 ksi.
P
P
9 ft
From the Moment Diagram, Fig. a, Mmax = 54 kip # ft. Mmax sallow
Sreq¿d =
54(12) 22
=
= 29.45 in3 Select W12 * 26
C Sx = 33.4 in3, d = 12.22 in and tw = 0.230 in. D
From the shear diagram, Fig. a, Vmax = 7.5 kip. Provide the shear-stress check for W 12 * 26, tmax =
=
Vmax tw d 7.5 0.230(12.22)
= 2.67 ksi 6 tallow = 12 ksi (O.K!) Hence Use
Ans.
W12 * 26
837
6 ft
6 ft
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11–10. Select the lightest-weight W14 steel wide-flange beam having the shortest height from Appendix B that will safely support the loading shown, where P = 12 kip. The allowable bending stress is sallow = 22 ksi and the allowable shear stress is tallow = 12 ksi.
P
P
9 ft
From the moment diagram, Fig. a, Mmax = 108 kip # ft. Mmax sallow
Sreq¿d =
108(12) 22
=
= 58.91 in3 Select W14 * 43
C Sx = 62.7 in3, d = 13.66 in and tw = 0.305 in. D
From the shear diagram, Fig. a, Vmax = 15 kip . Provide the shear-stress check for W14 * 43 , tmax =
=
Vmax tw d 15 0.305(13.66)
= 3.60 ksi 6 tallow = 12 ksi‚ (O.K!) Hence, Use
Ans.
W14 * 43
838
6 ft
6 ft
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11–11. The timber beam is to be loaded as shown. If the ends support only vertical forces, determine the greatest magnitude of P that can be applied. sallow = 25 MPa, tallow = 700 kPa.
150 mm 30 mm 120 mm
40 mm P 4m
A
y =
(0.015)(0.150)(0.03) + (0.09)(0.04)(0.120) = 0.05371 m (0.150)(0.03) + (0.04)(0.120)
I =
1 1 (0.150)(0.03)3 + (0.15)(0.03)(0.05371 - 0.015)2 + (0.04)(0.120)3 + 12 12
B
(0.04)(0.120)(0.09 - 0.05371)2 = 19.162(10 - 6) m4 Maximum moment at center of beam: Mmax =
P (4) = 2P 2
Mc ; I
s =
25(106) =
(2P)(0.15 - 0.05371) 19.162(10 - 6)
P = 2.49 kN Maximum shear at end of beam: Vmax =
P 2
VQ ; t = It
700(103) =
P 1 C (0.15 - 0.05371)(0.04)(0.15 - 0.05371) D 2 2 19.162(10 - 6)(0.04)
P = 5.79 kN Thus, P = 2.49 kN
Ans.
839
4m
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*11–12. Determine the minimum width of the beam to the nearest 14 in. that will safely support the loading of P = 8 kip. The allowable bending stress is sallow = 24 ksi and the allowable shear stress is tallow = 15 ksi.
P 6 ft
6 ft
6 in.
B A
Beam design: Assume moment controls. sallow =
Mc ; I
24 =
48.0(12)(3) 1 3 12 (b)(6 )
b = 4 in.
Ans.
Check shear: 8(1.5)(3)(4) VQ = 0.5 ksi 6 15 ksi OK = 1 3 It 12 (4)(6 )(4)
tmax =
•11–13.
Select the shortest and lightest-weight steel wideflange beam from Appendix B that will safely support the loading shown.The allowable bending stress is sallow = 22 ksi and the allowable shear stress is tallow = 12 ksi.
10 kip
6 kip
4 kip
A
B 4 ft
Beam design: Assume bending moment controls. Sreq¿d =
60.0(12) Mmax = = 32.73 in3 sallow 22
Select a W 12 * 26 Sx = 33.4 in3, d = 12.22 in., tw = 0.230 in. Check shear: tavg =
V 10.5 = = 3.74 ksi 6 12 ksi Aweb (12.22)(0.230)
Use W 12 * 26
Ans.
840
4 ft
4 ft
4 ft
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11–14. The beam is used in a railroad yard for loading and unloading cars. If the maximum anticipated hoist load is 12 kip, select the lightest-weight steel wide-flange section from Appendix B that will safely support the loading. The hoist travels along the bottom flange of the beam, 1 ft … x … 25 ft, and has negligible size. Assume the beam is pinned to the column at B and roller supported at A. The allowable bending stress is sallow = 24 ksi and the allowable shear stress is tallow = 12 ksi.
x
27 ft
A
B
12 kip
15 ft
C
Maximum moment occurs when load is in the center of beam. Mmax = (6 kip)(13.5 ft) = 81 lb # ft sallow =
M ; S
24 =
81(12) Sreq¿d Sreq¿d = 40.5 in3
Select a W 14 * 30, Sx = 42.0 in3, d = 13.84 in, tw = 0.270 in. At x = 1 ft, V = 11.56 kip t =
11.36 V = = 3.09 ksi 6 12 ksi Aweb (13.84)(0.270)
Use W14 * 30
Ans.
841
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11–15. The simply supported beam is made of timber that has an allowable bending stress of sallow = 960 psi and an allowable shear stress of tallow = 75 psi. Determine its dimensions if it is to be rectangular and have a heightto-width ratio of 1.25.
5 kip/ft
6 ft
1 I = (b)(1.25b)3 = 0.16276b4 12 Sreq¿d
b
Assume bending moment controls: Mmax = 60 kip # ft
960 =
Mmax Sreq¿d
60(103)(12) 0.26042 b3
b = 14.2 in. Check shear: tmax =
1.5(15)(103) 1.5V = = 88.9 psi 7 75 psi NO A (14.2)(1.25)(14.2)
Shear controls: tallow =
6 ft
1.25 b
I 0.16276b4 = = = 0.26042b3 c 0.625b
sallow =
B
A
1.5(15)(103) 1.5V = A (b)(1.25b)
b = 15.5 in.
Ans.
842
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*11–16. The simply supported beam is composed of two W12 * 22 sections built up as shown. Determine the maximum uniform loading w the beam will support if the allowable bending stress is sallow = 22 ksi and the allowable shear stress is tallow = 14 ksi.
w
Section properties: 24 ft
For W12 * 22 (d = 12.31 in. Ix = 156 in4 tw = 0.260 in. A = 6.48 in2) I = 2c 156 + 6.48a
S =
12.31 2 b d = 802.98 in4 2
I 802.98 = = 65.23 in3 c 12.31
Maximum Loading: Assume moment controls. M = sallowS(72 w)(12) = 22(65.23) w = 1.66 kip>ft Check Shear: tmax =
Ans. (Neglect area of flanges.)
12(1.66) Vmax = 3.11 ksi 6 tallow = 14 ksi OK = Aw 2(12.31)(0.26)
•11–17.
The simply supported beam is composed of two W12 * 22 sections built up as shown. Determine if the beam will safely support a loading of w = 2 kip>ft. The allowable bending stress is sallow = 22 ksi and the allowable shear stress is tallow = 14 ksi.
w
24 ft
Section properties: For W 12 * 22 (d = 12.31 in.
Ix = 156 in4
tw = 0.260 in.
A = 6.48 in2)
I = 2[156 + 6.48(6.1552)] = 802.98 in4 S =
802.98 I = = 65.23 in3 c 12.31
Bending stress: smax =
144 (12) Mallow = = 26.5 ksi 7 sallow = 22 ksi S 65.23
No, the beam falls due to bending stress criteria. Check shear: tmax =
Ans.
(Neglect area of flanges.)
Vmax 24 = = 3.75 ksi 6 tallow = 14 ksi OK Aw 2(12.31)(0.26)
843
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11–18. Determine the smallest diameter rod that will safely support the loading shown. The allowable bending stress is sallow = 167 MPa and the allowable shear stress is tallow = 97 MPa.
25 N/m 15 N/m
15 N/m
1.5 m
Bending Stress: From the moment diagram, Mmax = 24.375 N # m. Assume bending controls the design. Applying the flexure formula. sallow = 167 A 10
6
B =
Mmax c I 24.375 p 4
A d2 B
A d2 B 4
d = 0.01141 m = 11.4 mm
Ans.
Shear Stress: Provide a shear stress check using the shear formula with I =
p A 0.0057074 B = 0.8329 A 10 - 9 B m4 4
Qmax =
4(0.005707) 1 c (p) A 0.0057062 B d = 0.1239 A 10 - 6 B m3 3p 2
From the shear diagram, Vmax = 30.0 N. tmax =
=
Vmax Qmax It 30.0 C 0.1239(10 - 6) D
0.8329 (10 - 9)(0.01141)
= 0.391 MPa 6 tallow = 97 MPa (O.K!)
844
1.5 m
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11–19. The pipe has an outer diameter of 15 mm. Determine the smallest inner diameter so that it will safely support the loading shown. The allowable bending stress is sallow = 167 MPa and the allowable shear stress is tallow = 97 MPa.
25 N/m 15 N/m
15 N/m
1.5 m
Bending Stress: From the moment diagram, Mmax = 24.375 N # m. Q. Assume bending controls the design. Applying the flexure formula. sallow = 167 A 106 B =
Mmax c I 24.375(0.0075) p 4
C 0.00754 - A 2i B 4 D d
di = 0.01297 m = 13.0 mm
Ans.
Shear Stress: Provide a shear stress check using the shear formula with I =
p A 0.00754 - 0.0064864 B = 1.0947 A 10 - 9 B m4 4
Qmax =
4(0.0075) 1 4(0.006486) 1 c (p) A 0.00752 B d c (p) A 0.0064862 B d 3p 2 3p 2
= 99.306 A 10 - 9 B m3 From the shear diagram, Vmax = 30.0 N. Q tmax =
=
Vmax Qmax It 30.0 C 99.306(10 - 9) D
1.0947(10 - 9)(0.015 - 0.01297)
= 1.34 MPa 6 tallow = 97 MPa (O.K!)
845
1.5 m
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*11–20. Determine the maximum uniform loading w the W12 * 14 beam will support if the allowable bending stress is sallow = 22 ksi and the allowable shear stress is tallow = 12 ksi.
w 10 ft 10 ft
From the moment diagram, Fig. a, Mmax = 28.125 w. For W12 * 14, Sx = 14.9 in3, d = 11.91 in and tw = 0.200 in. sallow = 22 =
Mmax S 28.125 w (12) 14.9 Ans.
w = 0.9712 kip>ft = 971 lb>ft
From the shear diagram, Fig. a, Vmax = 7.5(0.9712) = 7.284 kip. Provide a shear stress check on W12 * 14, tmax =
=
Vmax tw d 7.284 0.200(11.91)
= 3.06 ksi 6 tallow = 12 ksi (O.K)
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•11–21.
Determine if the W14 * 22 beam will safely support a loading of w = 1.5 kip>ft. The allowable bending stress is sallow = 22 ksi and the allowable shear stress is tallow = 12 ksi.
w 10 ft 10 ft
For W14 * 22, Sx = 29.0 in3, d = 13.74 in and tw = 0.23 in. From the moment diagram, Fig. a, Mmax = 42.1875 kip # ft. smax =
=
Mmax S 42.1875(12) 29.0
= 17.46 ksi 6 sallow = 22 ksi (O.K!) From the shear diagram, Fig. a, Vmax = 11.25 kip. tmax =
=
Vmax tw d
11.25 0.23(13.74)
= 3.56 ksi 6 tallow = 12 ksi (O.K!) Based on the investigated results, we conclude that W14 * 22 can safely support the loading.
847
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11–22. Determine the minimum depth h of the beam to the nearest 18 in. that will safely support the loading shown. The allowable bending stress is sallow = 21 ksi and the allowable shear stress is tallow = 10 ksi. The beam has a uniform thickness of 3 in.
4 kip/ft
h
A B 12 ft
The section modulus of the rectangular cross-section is S =
I = C
1 12
(3)(h3) h>2
= 0.5 h2
From the moment diagram, Mmax = 72 kip # ft. Sreq¿d =
Mmax sallow
0.5h2 =
72(12) 21
h = 9.07 in Use
h = 9 18 in
Ans.
From the shear diagram, Fig. a, Vmax = 24 kip . Referring to Fig. b, 9.125 9.125 ba b (3) = 31.22 in3 and Qmax = y¿A¿ = a 4 2 1 I = (3) A 9.1253 B = 189.95 in4 . Provide the shear stress check by applying 12 shear formula, tmax =
=
Vmax Qmax It 24(31.22) 189.95(3)
= 1.315 ksi 6 tallow = 10 ksi (O.K!)
848
6 ft
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11–23. The box beam has an allowable bending stress of sallow = 10 MPa and an allowable shear stress of tallow = 775 kPa. Determine the maximum intensity w of the distributed loading that it can safely support. Also, determine the maximum safe nail spacing for each third of the length of the beam. Each nail can resist a shear force of 200 N.
w 30 mm 250 mm 30 mm 150 mm 30 mm
Section Properties: I =
1 1 (0.21) A 0.253 B (0.15) A 0.193 B = 0.1877 A 10 - 3 B m4 12 12
QA = y1 ¿A¿ = 0.11(0.03)(0.15) = 0.495 A 10 - 3 B m3 Qmax = ©y¿A¿ = 0.11(0.03)(0.15) + 0.0625(0.125)(0.06) = 0.96375 A 10 - 3 B m3 Bending Stress: From the moment diagram, Mmax = 4.50w. Assume bending controls the design. Applying the flexure formula. sallow = 10 A 106 B =
Mmax c I 4.50w (0.125) 0.1877 (10 - 3)
w = 3336.9 N>m Shear Stress: Provide a shear stress check using the shear formula. From the shear diagram, Vmax = 3.00w = 10.01 kN. tmax =
=
Vmax Qmax It 10.01(103) C 0.96375(10 - 3) D 0.1877(10 - 3)(0.06)
= 857 kPa 7 tallow = 775 kPa (No Good!) Hence, shear stress controls. tallow = 775 A 103 B =
Vmax Qmax It 3.00w C 0.96375(10 - 3) D 0.1877(10 - 3)(0.06)
w = 3018.8 N>m = 3.02 kN>m
Ans.
Shear Flow: Since there are two rows of nails, the allowable shear flow is 2(200) 400 = q = . s s
849
6m
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11–23.
Continued
For 0 … x 6 2 m and 4 m 6 x … 6 m, the design shear force is V = 3.00w = 9056.3 N. q =
VQA I
9056.3 C 0.495(10 - 3) D 400 = s 0.1877(10 - 3) s = 0.01675 m = 16.7 mm
Ans.
For 2 m 6 x 6 4 m, the design shear force is V = w = 3018.8 N. q =
VQA I
3018.8 C 0.495(10 - 3) D 400 = s 0.1877(10 - 3) s = 0.05024 m = 50.2 mm
Ans.
850
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*11–24. The simply supported joist is used in the construction of a floor for a building. In order to keep the floor low with respect to the sill beams C and D, the ends of the joists are notched as shown. If the allowable shear stress for the wood is tallow = 350 psi and the allowable bending stress is sallow = 1500 psi, determine the height h that will cause the beam to reach both allowable stresses at the same time. Also, what load P causes this to happen? Neglect the stress concentration at the notch.
P
2 in.
15 ft B
h
15 ft D
A 10 in. C
Bending Stress: From the moment diagram, Mmax = 7.50P. Applying the flexure formula. Mmax c I
salllow =
7.50P(12)(5)
1500 =
1 12
(2)(103)
P = 555.56 lb = 556 lb
Ans.
Shear Stress: From the shear diagram, Vmax = 0.500P = 277.78 lb. The notch is the critical section. Using the shear formula for a rectangular section. tallow = 350 =
3Vmax 2A 3(277.78) 2(2) h
h = 0.595 in.
Ans.
11–25. The simply supported joist is used in the construction of a floor for a building. In order to keep the floor low with respect to the sill beams C and D, the ends of the joists are notched as shown. If the allowable shear stress for the wood is tallow = 350 psi and the allowable bending stress is sallow = 1700 psi, determine the smallest height h so that the beam will support a load of P = 600 lb. Also, will the entire joist safely support the load? Neglect the stress concentration at the notch.
P
B
tallow =
1.5V ; A
350 =
D
A 10 in.
600 = 300 lb 2
1.5(300) (2)(h)
h = 0.643 in. smax =
Ans.
4500(12)(5) Mmax c = 1620 psi 6 1700 psi OK = 1 3 I 12 (2)(10)
Yes, the joist will safely support the load.
Ans.
851
h
15 ft
C
The reaction at the support is
2 in.
15 ft
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11–26. Select the lightest-weight steel wide-flange beam from Appendix B that will safely support the loading shown. The allowable bending stress is sallow = 22 ksi and the allowable shear stress is tallow = 12 ksi.
5 kip 18 kip ft B A 6 ft
From the moment diagram, Fig. a, Mmax = 48 kip # ft. Sreq¿d =
=
Mmax sallow 48(12) 22
= 26.18 in3 Select W 14 * 22 C Sx = 29.0 in3, d = 13.74 in. and tw = 0.230 in. D From the shear diagram, Fig. a, Vmax = 5 kip. Provide the shear stress check for W 14 * 22, tmax =
=
Vmax twd
5 0.230(13.74)
= 1.58 ksi 6 tallow = 12 ksi‚ (O.K!) Use
Ans.
W14 * 22
W12 * 22 would work also.
852
12 ft
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11–27. The T-beam is made from two plates welded together as shown. Determine the maximum uniform distributed load w that can be safely supported on the beam if the allowable bending stress is sallow = 150 MPa and the allowable shear stress is tallow = 70 MPa.
w
A 1.5 m
1.5 m 200 mm 20 mm 200 mm 20 mm
The neutral axis passes through centroid c of the beam’s cross-section. The location of c, Fig. b, is y =
0.21(0.02)(0.2) + 0.1(0.2)(0.02) ©yA = ©A 0.02(0.2) + 0.2(0.02) = 0.155 m
I =
1 (0.2)(0.023) + 0.2(0.02)(0.055)2 12
+
1 (0.02)(0.23) + 0.02(0.2)(0.055)2 12
= 37.667 (10 - 6) m4 Referring to Fig. b, Qmax = y¿A¿ = 0.0775(0.155)(0.02) = 0.24025(10 - 3) m3 Referring to the moment diagram, Mmax = -3.375 w. Applying the flexure formula with C = y = 0.155 m, sallow =
Mmax c ; I
150(106) =
3.375 w (0.155) 37.667(10 - 6)
w = 10.80(103) N>m = 10.8 kN>m (Control!)
Ans.
Referring to the shear diagram, Vmax = 1.5w. tallow =
Vmax Qmax ; It
70(106) =
1.5 w C 0.24025(10 - 3) D 37.667(10 - 6)(0.02)
w = 146.33(103) N>m = 146 kN>m
853
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*11–28. The beam is made of a ceramic material having an allowable bending stress of sallow = 735 psi and an allowable shear stress of tallow = 400 psi. Determine the width b of the beam if the height h = 2b.
15 lb 10 lb
6 lb/in.
2 in.
6 in.
2 in.
h b
Bending Stress: From the moment diagram, Mmax = 30.0 lb # in. Assume bending controls the design. Applying the flexure formula. sallow =
Mmax c I 30.0
735 =
1 12
A 2b2 B
(b) (2b)3
b = 0.3941 in. = 0.394 in.
Ans.
Shear Stress: Provide a shear stress check using the shear formula for a rectangular section. From the shear diagram, Vmax = 19.67 lb. tmax =
=
3Vmax 2A 3(19.67) 2(0.3941)(2)(0.3941)
= 94.95 psi 6 tallow = 400 psi (O.K!)
854
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•11–29.
The wood beam has a rectangular cross section. Determine its height h so that it simultaneously reaches its allowable bending stress of sallow = 1.50 ksi and an allowable shear stress of tallow = 150 psi. Also, what is the maximum load P that the beam can then support?
P
P
B
A 1.5 ft
3 ft
1.5 ft h
6 in.
The section modulus of the rectangular cross-section about the neutral axis is S =
I = C
1 12
(6) h3 h>2
= h2
From the moment diagram, Fig. a, Mmax = 1.5P. Mmax = sallow S 1.5P(12) = 1.50(103) h2 P = 83.33h2
(1)
h h 1 a b (6) = 0.75 h2 and I = (6) h3 = 0.5h3. 4 2 12 From the shear diagram, Fig. a, Vmax = P. Referring to Fig. b, Qmax = y¿A¿ =
tmax =
150 =
Vmax Qmax It P (0.75 h2) 0.5 h3 (6)
P = 600 h
(2)
Solving Eq (1) and (2) h = 7.20 in
P = 4320 lb = 4.32 kip
Ans.
855
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11–30. The beam is constructed from three boards as shown. If each nail can support a shear force of 300 lb, determine the maximum allowable spacing of the nails, s, s¿, s– , for regions AB, BC, and CD respectively. Also, if the allowable bending stress is sallow = 1.5 ksi and the allowable shear stress is tallow = 150 psi, determine if it can safely support the load.
1500 lb
500 lb s¿
s
A
s¿¿
C
B 6 ft
6 ft
6 ft
10 in. 4 in. 10 in.
The neutral axis passes through centroid c of the beam’s cross-section. The location of c, Fig. b, is y =
12(4)(10) + 2 C 5(10)(2) D ©yA = ©A 4(10) + 2(10)(2) = 8.50 in
The moment of inertia of the beam’s cross-section about the neutral axis is I = 2c +
1 (2)(103) + 2(10)(3.50)2 d 12 1 (10)(43) + 10(4)(3.50)2 12
= 1366.67 in4 Referring to Fig. b, Qmax = 2y2œ A2œ = 2 C 4.25(8.50)(2) D = 144.5 in3 QA = y1œ A1œ = 3.50(4)(10) = 140 in3 Referring to the moment diagram, Fig. a, Mmax = 9000 lb # ft. Applying flexure formula with C = y = 8.50 in, smax =
=
Mmax c I
9000(12)(8.50) 1366.67
= 671.70 psi 6 sallow = 1.50 ksi (O.K!) Referring to shear diagram, Fig. a, Vmax = 1500 lb. tmax =
=
Vmax Qmax It 1500 (144.5) = 39.65 psi 6 tallow = 150 psi (O.K!) 1366.67 (4)
1 S– = 11 in. Yes, it can support the load. 2
Ans.
856
2 in.
D
2 in.
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11–30.
Continued
Since there are two rows of nails, the allowable shear flow is 2(300) 600 2F qallow = = = . For region AB, V = 1500 lb. Thus S S S qallow =
VQA ; I
1500 (140) 600 = S 1366.67 Use
S = 3.904 in
S = 3 34 in
Ans.
For region BC, V = 1000 lb. Thus qallow =
VQA ; I
1000(140) 600 = S¿ 1366.67 Use
S¿ = 5.85 in
S¿ = 5 34 in
Ans.
For region CD, V = 500 lb. Thus qallow =
VQA ; I
500 (140) 600 = S– 1366.67 Use
S– = 11.71 in
S– = 1112 in
Ans.
857
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11–31. The tapered beam supports a concentrated force P at its center. If it is made from a plate that has a constant width b, determine the absolute maximum bending stress in the beam.
2h0
h0 L 2
L 2 P
Section Properties: h - h0 h0 = L x 2 I =
S =
h =
h0 (2x + L) L
h30 1 (b) a 3 b(2x + L)3 12 L 1 12
(b) A h3 2L
h30 3
L
B (2x + L)3
bh20 =
(2x + L)
6L2
(2x + L)2
Bending Stress: Applying the flexure formula. s =
M = S
Px 2 bh20 2
6L
= (2x + L)2
bh20
3PL2x (2x + L)2
In order to have the absolute maximum bending stress,
[1]
ds = 0. dx
3PL2 (2x + L)2(1) - x(2)(2x + L)(2) ds = c d = 0 dx bh20 (2x + L)4 x =
Substituting x =
L 2
L into Eq. [1] yields 2 smax =
3PL 8bh20
Ans.
858
h0
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*11–32. The beam is made from a plate that has a constant thickness b. If it is simply supported and carries a uniform load w, determine the variation of its depth as a function of x so that it maintains a constant maximum bending stress sallow throughout its length.
w
h0
y
x L –– 2
Moment Function: As shown on FBD(b). Section Properties: I =
1 3 by 12
S =
I = c
1 3 12 by y 2
=
1 2 by 6
Bending Stress: Applying the flexure formula. M = S
sallow =
w 2 2 8 (L - 4x ) 1 2 6 by
3w (L2 - 4x2)
sallow =
[1]
4by2
At x = 0, y = h0. From Eq. [1], sallow =
3wL2 4bh20
[2]
Equating Eq. [1] and [2] yields y2 =
h20 L2
y2 h20
+
A L2 - 4x2 B 4x2 = 1 L2
Ans.
The beam has a semi-elliptical shape.
859
L –– 2
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•11–33.
The beam is made from a plate having a constant thickness t and a width that varies as shown. If it supports a concentrated force P at its center, determine the absolute maximum bending stress in the beam and specify its location x, 0 6 x 6 L>2.
P P — 2
b0
L — 2
x
b L — 2
t P — 2
Section properties: b x = L; b0 2
2b0 x L
b =
I =
b0 t3 1 2b0 a xb t3 = x 12 L 6L
S =
I = c
b0 t
6L x t 2
=
b0 t2 x 3L
Bending stress: s =
M = S
P 2x b 0 t2
3L x
=
3PL 2b0t2
Ans.
The bending stress is independent of x. Therefore, the stress is constant throughout the span. Ans.
860
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11–34. The beam is made from a plate that has a constant thickness b. If it is simply supported and carries the distributed loading shown, determine the variation of its depth as a function of x so that it maintains a constant maximum bending stress sallow throughout its length.
w0
A
L –– 2
Moment Function: The distributed load as a function of x is w0 w = x L>2
w =
2w0 x L
The free-body diagram of the beam’s left cut segment is shown in Fig. b. Considering the moment equilibrium of this free-body diagram, d+ ©MO = 0;
M +
1 2w0 x 1 x R x ¢ ≤ - w0Lx = 0 B 2 L 3 4
M =
w0 A 3L2x - 4x3 B 12L
Section Properties: At position x, the height of the beam’s cross section is h. Thus 1 bh3 12
I = Then
1 bh3 I 12 1 S = = = bh2 c h>2 6 Bending Stress: The maximum bending stress smax as a function of x can be obtained by applying the flexure formula.
smax
At x =
w0 A 3L2x - 4x3 B w0 M 12L = = = A 3L2x - 4x3 B ‚ S 1 2 2bh2L bh 6
(1)
L , h = h0. From Eq. (1), 2 smax =
w0L2
(2)
2bh0 2
Equating Eqs. (1) and (2), w0 2
2bh L h =
A 3L2x - 4x3 B =
h0 L3>2
w0L2 2bh0 2
A 3L2x - 4x3 B 1>2
Ans.
861
h0 B
x
Support Reactions: As shown on the free-body diagram of the entire beam, Fig. a.
C
h
L –– 2
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11–35. The beam is made from a plate that has a constant thickness b. If it is simply supported and carries the distributed loading shown, determine the maximum bending stress in the beam.
w0
h0
h0 2h0 L – 2
Support Reactions: As shown on the free - body diagram of the entire beam, Fig. a. Moment Function: The distributed load as a function of x is w0 w = ; x L>2
w =
2w0 x L
The free - body diagram of the beam’s left cut segment is shown in Fig. b. Considering the moment equilibrium of this free - body diagram, d+ ©MO = 0;
M +
w0L x 1 2w0 a xbxa b x = 0 2 L 3 4
M =
w0 A 3L2x - 4x3 B 12L
Section Properties: Referring to the geometry shown in Fig. c, h - h0 h0 = ; x L>2
h =
h0 (2x + L) L
At position x, the height of the beam’s cross section is h. Thus I =
1 bh3 12
Then 1 bh3 bh0 2 12 I 1 = S = = bh2 = (2x + L)2 c h>2 6 6L2 Bending Stress: Applying the flexure formula,
smax
w0 A 3L2x - 4x3 B M 12L = = S bh0 2 (2x + L)2 6L2
smax =
w0L 2bh0 2
B
3L2x - 4x3 R (2x + L)2
In order to have absolute maximum bending stress,
(1) dsmax = 0. dx
2 2 2 2 3 dsmax w0L (2x + L) A 3L - 12x B - A 3L x - 4x B (2)(2x + L)(2) = C S = 0 dx 2bh0 2 (2x + L)4
w0L 2bh0
2
B
3L3 - 8x3 - 6L2x - 12Lx2 R = 0 (2x + L)3
862
L – 2
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11–35.
Since
Continued
w0L 2bh0 2
Z 0, then
3L3 - 8x3 - 6L2x - 12Lx2 = 0 Solving by trial and error, x = 0.2937L = 0.294L Substituting this result into Eq. (1), sabs = max
0.155w0L2
Ans.
bh0 2
863
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*11–36. Determine the variation of the radius r of the cantilevered beam that supports the uniform distributed load so that it has a constant maximum bending stress smax throughout its length.
w r0 r
Moment Function: As shown on FBD. Section Properties: I =
p 4 r 4
I = c
S =
p 4
r r
x
L
4
=
p 3 r 4
Bending Stress: Applying the flexure formula. smax =
wx2 2 p 3 4r
M = S
smax =
2wx2 pr3
[1]
At x = L, r = r0. From Eq. [1], smax =
2wL2 pr30
[2]
Equating Eq. [1] and [2] yields r3 =
r30 L2
x2
Ans.
•11–37.
Determine the variation in the depth d of a cantilevered beam that supports a concentrated force P at its end so that it has a constant maximum bending stress sallow throughout its length. The beam has a constant width b0 .
P d0
d
L
Section properties: I =
1 (b )(d3) 12 0
sallow =
S =
I = c
1 12
(b0)(d3) d>2
=
b0d2 6
M Px = S b0d2>6
(1)
PL b0d0 2>6
(2)
At x = L sallow =
Equate Eqs. (1) and (2): PL Px = b0d2>6 b0 d0 2>6 d2 = a
d0 2 bx ; L
x d = d0 AL
Ans.
864
x
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11–38. Determine the variation in the width b as a function of x for the cantilevered beam that supports a uniform distributed load along its centerline so that it has the same maximum bending stress sallow throughout its length. The beam has a constant depth t.
b —0 2 b —0 2 b — 2 w L
x t
Section properties: I =
1 b t3 12
S =
I = c
1 12
b t3 t 2
=
t2 b 6
Bending stress: sallow
M = = S
w x2 2 2
t 6b
=
3wx2 t2b
(1)
At x = L, b = b0 sallow =
3wL2 t2b0
(2)
Equating Eqs. (1) and (2) yields: 3wL2 3wx2 = 2 2 t b t b0 b =
b0 L2
x2
Ans.
865
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z
11–39. The shaft is supported on journal bearings that do not offer resistance to axial load. If the allowable normal stress for the shaft is sallow = 80 MPa, determine to the nearest millimeter the smallest diameter of the shaft that will support the loading. Use the maximum-distortionenergy theory of failure.
A
150 mm D
30 250 mm 50 N
x
30
C 30 150 N 100 mm 500 mm
100 N 30
250 N 250 mm
Torque and Moment Diagrams: As shown. In-Plane Principal Stresses: Applying Eq. 9–5 with sy = 0, sx = txy =
Mc 4M , and = I pc3
2T Tc . = J pc3 s1, 2 =
sx + sy ;
2
Aa
sx - sy 2
2
2 b + txy
=
2M 2M 2 2T 2 ; A a pc3 b + a pc3 b pc3
=
2M 2 ; 2M2 + T2 pc3 pc3
2 2M 2M2 + T2, then and b = pc3 pc3 s21 = a2 + b2 + 2ab, s1s2 = a2 - b2, s22 = a2 + b2 - 2ab, and s21 - s1 s2 + s22 = 3b2 + a2.
Maximum Distortion Energy Theory: Let a =
s21 - s1 s2 + s22 = s2allow 3a
2 2 2M 2 2 2 2M + T b + a b = s2allow pc3 pc3 6 4 A 4M2 + 3T2 B R p2s2allow 1
c = B
Shaft Design: By observation, the critical section is located just to the left of gear C, where M = 239.06252 + 46.012 = 60.354 N # m and T = 15.0 N # m. Using the maximum distortion energy theory, 6 4 A 4M2 + 3T2 B R p2s2allow 1
c = B
= b
4 p2 [80(106)]2
C 4(60.354)2 + 3(15.0)2 D r
1 6
= 0.009942 m d = 2c = 2(0.009942) = 0.01988 m = 19.88 mm Use
d = 20 mm
Ans.
866
B y
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z
*11–40. The shaft is supported on journal bearings that do not offer resistance to axial load. If the allowable shear stress for the shaft is tallow = 35 MPa, determine to the nearest millimeter the smallest diameter of the shaft that will support the loading. Use the maximum-shear-stress theory of failure.
A
x
150 mm D
30 250 mm 50 N
30
C 30 150 N 100 mm 500 mm
100 N 30
250 N 250 mm
Shaft Design: By observation, the critical section is located just to the left of gear C, where M = 239.06252 + 46.012 = 60.354 N # m and T = 15.0 N # m. Using the maximum shear stress theory. c = a = B
1
3 2 2M2 + T2 b ptallow
2 p(35)(106)
260.3542 + 15.02 R
1 3
= 0.01042 m d = 2c = 2(0.01042) = 0.02084 m = 20.84 mm Use
d = 21 mm
Ans.
867
B y
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z
•11–41.
The end gear connected to the shaft is subjected to the loading shown. If the bearings at A and B exert only y and z components of force on the shaft, determine the equilibrium torque T at gear C and then determine the smallest diameter of the shaft to the nearest millimeter that will support the loading. Use the maximum-shear-stress theory of failure with tallow = 60 MPa.
100 mm T 250 mm C 50 mm 150 mm
A
x
100 mm Fz 1.5 kN
From the free - body diagrams: T = 100 N # m
Ans.
Critical section is at support A. 1
1
3 3 2 2 c = c 22252 + 1502 d 2M2 + T2 d = c p tallow p(60)(106)
= 0.01421 m d = 2c = 0.0284 m = 28.4 mm Use d = 29 mm
Ans.
868
B
75 mm y
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z
11–42. The end gear connected to the shaft is subjected to the loading shown. If the bearings at A and B exert only y and z components of force on the shaft, determine the equilibrium torque T at gear C and then determine the smallest diameter of the shaft to the nearest millimeter that will support the loading. Use the maximum-distortionenergy theory of failure with sallow = 80 MPa.
100 mm T 250 mm C 50 mm 150 mm
A
x
T = 100 N # m
Ans.
Critical section is at support A. s1, 2 =
sx s2x 2 ; 2 A 4 + txy sx s2x 2 ,b = A 4 + txy 2
Let a =
s1 = a + b, s2 = a - b Require, s21 - s1 s2 + s22 = s2allowa2 + 2ab + b2 - [a2 - b2] + a2 - 2ab + b2 = s2allow a2 + 3b2 = s2allow s2x s2x + 3a + t2xy b = s2allow 4 4 s2x + 3t2xy = s2allow Mt 2 Tc 2 a p 4 b + 3a p 4 b = s2allow 4 c 2 c 4M 2 2T 2 1 ca b + 3a b d = s2allow 4 p p c c4 =
16 s2allow
c = a = c
2
p
M2 +
4 s2allow p2
12T2 p2
s2allow
(4M + 3T ) b 2
4 (80(106))2(p)2
2
100 mm Fz 1.5 kN
From the free-body diagrams:
1 2
(4(225) + 3(150) ) d 2
2
1 2
= 0.01605 m d = 2c = 0.0321 m = 32.1 mm Use d = 33 mm
Ans.
869
B
75 mm y
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11–43. The shaft is supported by bearings at A and B that exert force components only in the x and z directions on the shaft. If the allowable normal stress for the shaft is sallow = 15 ksi, determine to the nearest 18 in. the smallest diameter of the shaft that will support the loading. Use the maximum-distortion-energy theory of failure.
z C F¿x 100 lb
6 in.
A x
8 in. 12 in.
Critical moment is just to the right of D.
T = 1200 lb # in. Both states of stress will yield the same result.
Let
s s 2 2 ; a 2 A 2b + t
2 s = A and s + t2 = B 2 A4
s2a = (A + B)2, s2b = (A - B)2 sa sb = (A + B)(A - B) = A2 - B2 s2a - sa sb + s2b = A2 + B2 + 2AB - A2 + B2 + A2 + B2 - 2AB = A2 + 3B2 =
s2 s2 + 3a + t2 b = s2 + 3t2 4 4
s2a - sa sb + s2b = s2allow s2 + 3t2 = s2allow‚ s =
Mc Mc 4M = p 4 = I c pc3 4
t =
Tc Tc 2T = p 4 = J c p c3 2
(1)
From Eq. (1) 16M2 2
6
p c
c = a
12T2 +
p2 c6
= s2allow
16(2396)2 + 12(12002) 1>6 16M2 + 12T2 1>6 b = c d = 0.605 in. p2s2allow p2((15)(103))2
d = 2c = 1.210 in. Use d = 1
2 in. Fz 300 lb 10 in.
6 in.
M = 220572 + 12292 = 2396 lb # in.
sa, b =
F y 300 lb
D
1 in. 4
Ans.
870
4 in.
E
B y
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*11–44. The shaft is supported by bearings at A and B that exert force components only in the x and z directions on the shaft. If the allowable normal stress for the shaft is sallow = 15 ksi, determine to the nearest 18 in. the smallest diameter of the shaft that will support the loading. Use the maximum-shear-stress theory of failure. Take tallow = 6 ksi.
z C F¿x 100 lb
6 in.
A x
8 in. 12 in.
Critical moment is just to the right of D.
T = 1200 lb # in. Use Eq. 11-2, 1>3 2 2M2 + T2 b p tallow
c = a
1>3 2 2(2396)2 + (1200)2 b = 0.6576 in. 3 p(6)(10 )
2 in. Fz 300 lb 10 in.
6 in.
M = 2(2057)2 + (1229)2 = 2396 lb # in.
c = a
F y 300 lb
D
dreq¿d = 2c = 1.315 in. 3 Use d = 1 in. 8
Ans.
871
4 in.
E
B y
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z
•11–45. The bearings at A and D exert only y and z components of force on the shaft. If tallow = 60 MPa, determine to the nearest millimeter the smallest-diameter shaft that will support the loading. Use the maximum-shearstress theory of failure.
350 mm D 400 mm 200 mm B A
Critical moment is at point B: M = 2(473.7)2 + (147.4)2 = 496.1 N # m
x
T = 150 N # m
c = a
1>3 1>3 2 2 2 2 2496.1 2M2 + T2 b = a + 150 b = 0.0176 m p tallow p(60)(106)
c = 0.0176 m = 17.6 mm d = 2c = 35.3 mm Use d = 36 mm
Ans.
872
y
C 75 mm Fy ⫽ 3 kN 50 mm
Fz ⫽ 2 kN
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z
11–46. The bearings at A and D exert only y and z components of force on the shaft. If tallow = 60 MPa, determine to the nearest millimeter the smallest-diameter shaft that will support the loading. Use the maximumdistortion-energy theory of failure. sallow = 130 MPa.
350 mm D 400 mm 200 mm B A
The critical moment is at B. M = 2(473.7)2 + (147.4)2 = 496.1 N # m
x
T = 150 N # m Since, sa, b =
Let
s s 2 2 ; a 2 A 2b + t
s = A 2
and
s 2 2 = B a A 2b + t
s2a = (A + B)2
s2b = (A - B)2
sa sb = (A + B)(A - B) s2a - sa sb + s2b = A2 + B2 + 2AB - A2 + B2 + A2 + B2 - 2AB = A2 + 3B2 =
s2 s2 + 3a + t2 b 4 4
= s2 + 3t2 s2a - sasb + s2b = s2allow s2 + 3t2 = s2allow
(1)
s =
Mc Mc 4M = p 4 = I c pc3 4
t =
Tc Tc 2T = p 4 = J pc3 2 c
From Eq (1) 12T2 16M2 + 2 4 = s2allow 2 4 pc pc c = a = a
16M2 + 12T2 1>6 b p2s2allow 16(496.1)2 + 12(150)2 2
4
2
p ((130)(10 ))
b
1>4
= 0.01712 m
d = 2c = 34.3 mm
Ans.
873
y
C 75 mm Fy ⫽ 3 kN 50 mm
Fz ⫽ 2 kN
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11–47. Draw the shear and moment diagrams for the shaft, and then determine its required diameter to the nearest millimeter if sallow = 140 MPa and tallow = 80 MPa. The bearings at A and B exert only vertical reactions on the shaft.
1500 N 800 N A
B
600 mm 125 mm
Bending Stress: From the moment diagram, Mmax = 111 N # m. Assume bending controls the design. Applying the flexure formula. sallow = 140 A 106 B =
Mmax c I 111 A d2 B p 4
A d2 B 4
d = 0.02008 m = 20.1 mm d = 21 mm
Use
Ans.
Shear Stress: Provide a shear stress check using the shear formula with I =
p A 0.01054 B = 9.5466 A 10 - 9 B m4 4
Qmax =
4(0.0105) 1 c (p)(0.0105)2 d = 0.77175 A 10 - 6 B m3 3p 2
From the shear diagram, Vmax = 1484 N. tmax =
=
Vmax Qmax It
1484 C 0.77175(10 - 6) D 9.5466(10 - 9)(0.021)
= 5.71 MPa 6 tallow = 80 MPa (O.K!)
874
75 mm
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*11–48. The overhang beam is constructed using two 2-in. by 4-in. pieces of wood braced as shown. If the allowable bending stress is sallow = 600 psi, determine the largest load P that can be applied. Also, determine the associated maximum spacing of nails, s, along the beam section AC if each nail can resist a shear force of 800 lb. Assume the beam is pin-connected at A, B, and D. Neglect the axial force developed in the beam along DA.
D
2 ft 3 ft A
2 ft
Section properties: I =
1 (4)(4)3 = 21.33 in4 12
S =
21.33 I = = 10.67 in3 c 2
Mmax = sallow S 3P(12) = 600(10.67) P = 177.78 = 178 lb
Ans.
Nail Spacing: V = P = 177.78 lb Q = (4)(2)(1) = 8 in3 q =
177.78(8) VQ = = 66.67 lb>in. I 21.33
S =
800 lb = 12.0 in. 66.67 lb>in.
Ans.
875
2 in. 2 in.
s B
MA = Mmax = 3P
P
C
4 in.
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z
•11–49. The bearings at A and B exert only x and z components of force on the steel shaft. Determine the shaft’s diameter to the nearest millimeter so that it can resist the loadings of the gears without exceeding an allowable shear stress of tallow = 80 MPa. Use the maximum-shear-stress theory of failure.
Fx 5 kN A 75 mm
x
50 mm
150 mm 350 mm
B Fz 7.5 kN 250 mm
Maximum resultant moment M = 212502 + 2502 = 1274.75 N # m 1
1
3 3 2 2 21274.752 + 3752 d = 0.0219 m 2M2 + T2 d = c c = c 6 p tallow p(80)(10 )
d = 2c = 0.0439 m = 43.9 mm Use d = 44 mm
Ans.
876
y
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z
11–50. The bearings at A and B exert only x and z components of force on the steel shaft. Determine the shaft’s diameter to the nearest millimeter so that it can resist the loadings of the gears without exceeding an allowable shear stress of tallow = 80 MPa. Use the maximumdistortion-energy theory of failure with sallow = 200 MPa.
Fx 5 kN A 75 mm
x
50 mm
150 mm 350 mm
Maximum resultant moment M = 212502 + 2502 = 1274.75 N # m s1, 2 =
sx s2x 2 ; 2 A 4 + txy sx s2x 2 ,b = 2 A 4 + txy
Let a =
s1 = a + b,
s2 = a - b
Require, s21 - s1 s2 + s22 = s2allow a2 + 2ab + b2 - [a2 - b2] + a2 - 2ab + b2 = s2allow a2 + 3b2 = s2allow s2x s2x + 3a + t2xy b = s2allow 4 4 s2x + 3t2xy = s2allow Mc 2 Tc 2 a p 4 b + 3a p 4 b = s2allow 4 c 2 c 1 6
c
Ba
c6 =
4M 2 2T 2 b + 3a b R = s2allow p p 16
s2allow p2
c = B
= B
M2 +
4 s2allow p2
12T2 s2allow p2
(4M2 + 3 T2) R
4
1 4
(4(1274.75)2 + 3(375)2) R (200(106))2(p)2
1 4
= 0.0203 m = 20.3 mm d = 40.6 mm
Ans.
Use d = 41 mm
Ans.
877
B Fz 7.5 kN 250 mm
y
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11–51. Draw the shear and moment diagrams for the beam. Then select the lightest-weight steel wide-flange beam from Appendix B that will safely support the loading. Take sallow = 22 ksi, and tallow = 12 ksi.
3 kip/ft 1.5 kip ft A B 12 ft
Bending Stress: From the moment diagram, Mmax = 18.0 kip # ft. Assume bending controls the design. Applying the flexure formula. Sreq¿d =
= Select
Mmax sallow 18.0(12) = 9.82 in3 22
A Sx = 10.9 in3, d = 9.87 in., tw = 0.19 in. B
W10 * 12
V for the W10 * 12 wide twd = 9.00 kip
Shear Stress: Provide a shear stress check using t = flange section. From the shear diagram, Vmax tmax =
=
Vmax tw d 9.00 0.19(9.87)
= 4.80 ksi 6 tallow = 12 ksi (O.K!) Hence,
Use
W10 * 12
Ans.
878
6 ft
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*11–52. The beam is made of cypress having an allowable bending stress of sallow = 850 psi and an allowable shear stress of tallow = 80 psi. Determine the width b of the beam if the height h = 1.5b.
300 lb
75 lb/ft
B
A 5 ft
5 ft
h 1.5b b
Ix =
1 (b)(1.5b)3 = 0.28125 b4 12
Qmax = y¿A¿ = (0.375b) (0.75b)(b) = 0.28125 b3 Assume bending controls. Mmax = 527.34 lb # ft sallow =
Mmax c ; I
850 =
527.34(12)(0.75 b) 0.28125 b4
b = 2.71 in.
Ans.
Check shear: I = 15.12 in4 tmax =
Qmax = 5.584 in3
VQmax 281.25(5.584) = It 15.12(2.71)
= 38.36 psi 6 80 psi
OK
879
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•11–53.
The tapered beam supports a uniform distributed load w. If it is made from a plate and has a constant width b, determine the absolute maximum bending stress in the beam.
w h0 L –– 2
Support Reactions: As shown on FBD(a). Moment Function: As shown on FBD(b). Section Properties: h - h0 h0 = L x 2 I =
S =
h =
h0 (2x + L) L
h30 1 (b) a 3 b(2x + L)3 12 L 1 12
(b) A h0 2L
B (2x + L)3
h30 3
L
bh20 =
(2x + L)
6L2
(2x + L)2
Bending Stress: Applying the flexure formula. s =
M = S
w 2
(Lx - x2)
bh20 2
6L
3wL2 (Lx - x2) =
(2x + L)2
[1]
bh20 (2x + L)2
In order to have the absolute maximum bending stress,
ds = 0. dx
3wL2 (2x + L)2(L - 2x) - (Lx - x2)(2)(2x + L)(2) ds = c d = 0 dx bh20 (2x + L)4 x =
Substituting x =
L 4
L into Eq. [1] yields 4 smax =
wL2 4bh20
Ans.
880
h0
2 h0 L –– 2
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11–54. The tubular shaft has an inner diameter of 15 mm. Determine to the nearest millimeter its outer diameter if it is subjected to the gear loading. The bearings at A and B exert force components only in the y and z directions on the shaft. Use an allowable shear stress of tallow = 70 MPa, and base the design on the maximum-shear-stress theory of failure.
z
100 mm B 500 N 150 mm
A 200 mm 150 mm
x
I =
p 4 p (c - 0.00754) and J = (c4 - 0.00754) 4 2
tallow =
Aa
sx - sy
tallow =
Aa
Mc 2 Tc 2 b + a b 2I J
t2allow =
M2 c2 T2 c2 + 2 4I J2
¢
2
100 mm
2
b + t2xy
c4 - 0.00754 2 4M2 4T2 ≤ = 2 + 2 c p p
c4 - 0.00754 2 = 2M2 + T2 c p tallow c4 - 0.00754 2 2752 + 502 = c p(70)(106) c4 - 0.00754 = 0.8198(10 - 6)c Solving, c = 0.0103976 m d = 2c = 0.0207952 m = 20.8 mm Use d = 21 mm
Ans.
881
500 N
y
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11–55. Determine to the nearest millimeter the diameter of the solid shaft if it is subjected to the gear loading. The bearings at A and B exert force components only in the y and z directions on the shaft. Base the design on the maximum-distortion-energy theory of failure with sallow = 150 MPa.
z
100 mm B 500 N 150 mm
A 200 mm 150 mm
x
s1, 2 =
100 mm
sx 2 ; 2 A 4 + txy s2x
sx s2x 2 ,b = A 4 + txy 2
Let a =
s1 = a + b, s2 = a - b Require, s21 - s1 s2 + s21 = s2allow a2 + 2ab + b2 - [a2 - b2] + a2 - 2ab + b2 = sallow a2 + 3b2 = s2allow s2x s2x + 3a + t2xy b = s2allow 4 4 s2x + 3t2xy = s2allow Mc 2 Tc 2 a p 4 b + 3a p 4 b = s2allow 4 c 2 c 1 6
c
ca
c6 =
4M 2 2T 2 b + 3a b d = s2allow p p 16 s2allow p2
c = a = c
M2 +
4 s2allow p2
12T2 s2allow p2
(4M2 + 3T2) b
4 (150(106))2(p)2
1 4
1 4
(4(75) + 3(50) ) d = 0.009025 m 2
2
d = 2c = 0.0181 m Use d = 19 mm
Ans.
882
500 N
y
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•12–1. An A-36 steel strap having a thickness of 10 mm and a width of 20 mm is bent into a circular arc of radius r = 10 m. Determine the maximum bending stress in the strap.
Moment-Curvature Relationship: M 1 = r EI
however,
M =
I s c
1 1 c s = r EI
s =
0.005 c E = a b C 200 A 109 B D = 100 MPa r 10
12–2. A picture is taken of a man performing a pole vault, and the minimum radius of curvature of the pole is estimated by measurement to be 4.5 m. If the pole is 40 mm in diameter and it is made of a glass-reinforced plastic for which Eg = 131 GPa, determine the maximum bending stress in the pole.
r ⫽ 4.5 m
Moment-Curvature Relationship: M 1 = r EI
however,
M =
I s c
I 1 c s = r EI
s =
0.02 c E = a b C 131 A 109 B D = 582 MPa r 4.5
Ans.
883
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12–3. When the diver stands at end C of the diving board, it deflects downward 3.5 in. Determine the weight of the diver. The board is made of material having a modulus of elasticity of E = 1.5(103) ksi.
B
A
3.5 in.
2 in.
C 9 ft
3 ft
Support Reactions and Elastic Curve. As shown in Fig. a. Moment Functions. Referring to the free-body diagrams of the diving board’s cut segments, Fig. b, M A x1 B is a + ©MO = 0; and M A x2 B is a + ©MO = 0;
M A x1 B + 3Wx1 = 0
M A x1 B = -3Wx1
-M A x2 B - Wx2 = 0
M A x2 B = -Wx2
Equations of Slope and Elastic Curve. EI
d2v = M(x) dx2
For coordinate x1, EI
d2v1 dx1 2
= -3Wx1
d2v1 3 = - Wx1 2 + C1 dx1 2
(1)
1 EIv1 = - Wx1 3 + C1x1 + C2 2
(2)
EI
For coordinate x2 EI
EI
d2v2 dx2 2
= -Wx2
dv2 1 = - Wx2 2 + C3 dx2 2
EIv2 = -
(3)
1 Wx2 3 + C3x2 + C4 6
(4)
Boundary Conditions. At x1 = 0, v1 = 0. Then, Eq. (2) gives 1 EI(0) = - W A 03 B + C1(0) + C2 2
C2 = 0
At x1 = 3 ft, v1 = 0. Then, Eq. (2) gives 1 EI(0) = - W A 33 B + C1(3) + 0 2
C1 = 4.5W
884
18 in.
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12-3.
Continued
At x2 = 9 ft, v2 = 0. Then, Eq. (4) gives 1 EI(0) = - W A 93 B + C3(9) + C4 6 9C3 + C4 = 121.5W Continuity Conditions. At x1 = 3 ft and x2 = 9 ft,
(5) dv2 dv1 . Thus, Eqs. (1) and = dx1 dx2
(3) give 1 3 - W A 32 B + 4.5W = - c - W A 92 B + C3 d 2 2
C3 = 49.5W
Substituting the value of C3 into Eq. (5), C4 = -324W Substituting the values of C3 and C4 into Eq. (4), v2 =
1 1 a - Wx2 3 + 49.5Wx2 - 324Wb EI 6
At x2 = 0, v2 = -3.5 in. Then, -324W(1728) -3.5 =
1.5 A 106 B c
1 (18) A 2 3 B d 12
W = 112.53 lb = 113 lb
Ans.
885
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*12–4. Determine the equations of the elastic curve using the x1 and x2 coordinates. EI is constant.
P A
EI
d2v1 dx1 2
= M1 (x)
M1(x) = 0;
EI
EI
d v1 dx1
2
L
= 0
x3
dv1 = C1 dx1
(1)
EI v1 = C1x1 + C2
(2)
M2(x) = Px2 - P(L - a) EI
EI
d2 v2 dx2 2
= Px2 - P(L - a)
dv2 P 2 = x - P(L - a)x2 + C3 dx2 2 2
EI v2 =
(3)
P(L - a)x22 P 3 x2 + C3x2 + C4 6 2
(4)
Boundary conditions: At x2 = 0,
dv2 = 0 dx2
From Eq. (3), 0 = C3 At x2 = 0, v2 = 0 0 = C4 Continuity condition: At x1 = a, x2 = L - a;
dv1 dv2 = dx1 dx2
From Eqs. (1) and (3), C1 = - c
P(L - a)2 - P(L - a)2 d ; 2
C1 =
P(L - a)2 2
At x1 = a, x2 = L - a, v1 = v2 From Eqs. (2) and (4), a
P(L - a)3 P(L - a)3 P(L - a)2 b a + C2 = 2 6 2
C2 = -
Pa(L - a)2 P(L - a)3 2 3
From Eq. (2), v1 =
P [3(L - a)2x1 - 3a(L - a)2 - 2(L - a)3] 6EI
Ans.
For Eq. (4), v2 =
B
x1 2
P [x2 - 3(L - a)x33] 6EI 2
Ans.
886
L 2
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•12–5.
Determine the equations of the elastic curve for the beam using the x1 and x2 coordinates. EI is constant.
P A
B
x1 L
Moment Functions. Referring to the FBDs of the beam’s cut segments shown in Fig. b and c, 1 P(x1) = 0 2
a + ©MO = 0;
M(x1) +
a + ©MO = 0;
-Px2 - M(x2) = 0
M(x1) = -
P x 2 1
And
EI
M(x2) = -Px2
d2v = M(x) dx2
For coordinate x1, EI
EI
d2v1 dx1 2
= -
P x 2 1
dv1 P = - x1 2 + C1 dx1 4
EI v1 = -
(1)
P 3 x + C1x + C2 12 1
(2)
For coordinate x2, EI
EI
d2v2 dx2 2
= -Px2
dv2 P = - x2 2 + C3 dx2 2
EI v2 = -
(3)
P 3 x + C3x2 + C4 6 2
(4)
At x1 = 0, v1 = 0. Then, Eq (2) gives EI(0) = -
P (0) + C1(0) + C2 12
C2 = 0
At x1 = L, v1 = 0. Then, Eq (2) gives EI(0) = At x2 =
P (L3) + C1L + 0 12
C1 =
PL2 12
L , v2 = 0. Then Eq (4) gives 2 EI(0) = -
P L 3 L a b + C3 a b + C4 6 2 2
C3L + 2C4 =
PL3 24
(5)
887
x2 L 2
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•12–5.
Continued
At x1 = L and x2 =
-
dv2 L dv1 , = . Thus, Eqs. (1) and (3) gives 2 dx1 dx2
P 2 P L 2 PL2 = - c - a b + C3 d AL B + 4 12 2 2 C3 =
7PL2 24
Substitute the result of C3 into Eq. (5) C4 = -
PL3 8
Substitute the values of C1 and C2 into Eq. (2) and C3 and C4 into Eq (4), v1 =
P A -x1 3 + L2x1 B 12EI
Ans.
v2 =
P A -4x2 3 + 7L2x2 - 3L3 B 24EI
Ans.
888
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12–6. Determine the equations of the elastic curve for the beam using the x1 and x3 coordinates. Specify the beam’s maximum deflection. EI is constant.
P A
Support Reactions and Elastic Curve: As shown on FBD(a). L
Slope and Elastic Curve:
For M(x1) = -
x3
d2v = M(x) dx2
EI
P x. 2 1 EI
d2y1 dx21
EI y1 = For M(x3) = Px3 -
= -
P x 2 1
dy1 P = - x21 + C1 dx1 4
EI
[1]
P 3 x + C1x1 + C2 12 1
[2]
3PL . 2 EI
d2y3 dx23
= Px3 -
3PL 2
dy3 P 2 3PL = x3 x3 + C3 dx3 2 2
EI
EI y3 =
[3]
P 3 3PL 3 x x3 + C3x3 + C4 6 3 4
[4]
Boundary Conditions: y1 = 0 at x1 = 0. From Eq. [2], C2 = 0 y1 = 0 at x1 = L. From Eq. [2]. 0 = -
PL3 + C1L 12
C1 =
PL2 12
y3 = 0 at x3 = L. From Eq. [4]. 0 =
PL3 3PL3 + C3L + C4 6 4
0 = -
7PL3 + C3L + C4 12
[5]
Continuity Condition: At x1 = x3 = L,
-
dy1 dy3 . From Eqs. [1] and [3], = dx1 dx3
PL2 PL2 PL2 3PL2 + = + C3 4 12 2 2
From Eq. [5], C4 = -
B
x1
Moment Function: As shown on FBD(b) and (c).
C3 =
5PL2 6
PL3 4
889
L 2
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12–6.
Continued
The Slope: Substitute the value of C1 into Eq. [1], dy1 P = A L2 - 3x21 B dx1 12EI dy1 P = 0 = A L2 - 3x21 B dx1 12EI
x1 =
L 23
The Elastic Curve: Substitute the values of C1, C2, C3, and C4 into Eqs. [2] and [4], respectively. y1 =
Px1 A -x21 + L2 B 12EI
yO = y1 |x1 =
y3 =
L 23
=
PA
L 23
B
12EI
Ans. a-
0.0321PL3 L3 + L2 b = 3 EI
P A 2x33 - 9Lx23 + 10L2x3 - 3L3 B 12EI
Ans.
yC = y3 |x3 = 32 L =
2 P 3 3 3 3 c2 a L b - 9La Lb + 10L2 a L b - 3L3 d 12EI 2 2 2
= -
PL3 8EI
Hence, ymax =
PL3 8EI
Ans.
890
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12–7. The beam is made of two rods and is subjected to the concentrated load P. Determine the maximum deflection of the beam if the moments of inertia of the rods are IAB and IBC , and the modulus of elasticity is E. EI
P B A
C l
d2y = M(x) dx2
L
M1(x) = - Px1 EIBC
EIBC
d2y1 dx1 2
= - Px1
dy1 Px21 = + C1 dx1 2
EIBC y1 = -
(1)
Px31 + C1x1 + C2 6
(2)
M2(x) = - Px2 EIAB
EIAB
d2y2 dx2 2
= - Px2
dy2 P = - x2 2 + C3 dx2 2
EIAB y2 = -
(3)
P 3 x + C3x2 + C4 2 2
(4)
Boundary conditions: At x2 = L,
0 = -
dy2 = 0 dx2
PL2 + C3; 2
C3 =
PL2 2
At x2 = L, y = 0 0 = -
PL3 PL3 + + C4; 6 2
C4 = -
PL3 3
Continuity Conditions: At x1 = x2 = l,
dy1 dy2 = dx1 dx2
891
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12–7.
Continued
From Eqs. (1) and (3), 1 PI 2 1 PI 2 PL2 cc+ C1 d = + d EIBC 2 EIAB 2 2 C1 =
IBC PL2 Pl2 Pl2 c+ d + IAB 2 2 2
At x1 = x2 = l, y1 = y2 From Eqs. (2) and (4), IBC PL2 Pl2 1 Pl3 Pl2 e+ c a+ b + dl + C2 f EIBC 6 IAB 2 2 2 =
1 PL2l PL3 Pl3 c+ d EIAB 6 2 3
C2 =
IBC PL3 IBC Pl3 Pl3 IAB 3 IAB 3 3
Therefore, y1 =
Px1 3 IBC 1 Pl2 PL2 Pl2 e+ c a+ b + dx1 EIBC 6 IAB 2 2 2 +
IBC PL3 IBC Pl3 Pl3 f IAB 3 IAB 3 3
At x1 = 0, y1 |x = 0 = ymax ymax =
=
IBC Pl3 IBC PL3 IAB 3 I Pl3 P e f = e l3 - L3 - a bl f EIBC IAB 3 IAB 3 3 3EIAB IBC IAB 3 P e a1 b l - L3 f 3EIAB IBC
Ans.
892
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*12–8. Determine the equations of the elastic curve for the beam using the x1 and x2 coordinates. EI is constant.
P
Referring to the FBDs of the beam’s cut segments shown in Fig. b and c, a + ©MO = 0;
M(x1) +
PL - Px1 = 0 2
M(x1) = Px1 -
PL 2
x1 x2
And a + ©MO = 0; EI
L 2
M(x2) = 0
d2v = M(x) dx2
For coordinate x1, EI
EI
d2v1 dx21
= Px1 -
PL 2
dv1 P 2 PL = x x + C1 dx1 2 1 2 1
EI v1 =
(1)
P 3 PL 2 x x + C1x1 + C2 6 1 4 1
(2)
For coordinate x2, EI
EI
d2v2 dx22
= 0
dv2 = C3 dx2
(3)
EI v2 = C3x2 = C4 At x1 = 0,
(4)
dv1 = 0. Then, Eq.(1) gives dx1
EI(0) =
PL P 2 (0 ) (0) + C1 2 2
C1 = 0
At x1 = 0, v1 = 0. Then, Eq(2) gives EI(0) = At x1 = x2 =
PL 2 P 3 (0 ) (0 ) + 0 + C2 6 4
C2 = 0
dv2 L dv1 = , . Thus, Eqs.(1) and (3) gives 2 dx1 dx2
P L 2 PL L a b a b = C3 2 2 2 2 Also, at x1 = x2 =
C3 = -
PL2 8
L , v = v2. Thus, Eqs, (2) and (4) gives 2 1
PL L 2 PL2 L P L 3 a b a b = ab a b + C4 6 2 4 2 8 2
C4 =
PL3 48
Substitute the values of C1 and C2 into Eq. (2) and C3 and C4 into Eq (4), v1 =
P A 2x31 - 3Lx21 B 12EI
Ans.
v2 =
PL2 (-6x2 + L) 48EI
Ans.
893
L 2
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•12–9.
Determine the equations of the elastic curve using the x1 and x2 coordinates. EI is constant. EI
d2y = M(x) dx2
M1 =
EI
EI
P
A
x1
Pb x L 1
d2y1 dx21
=
a
EI y3 =
L
(1)
Pb 3 x + C3x1 + C2 6L 1
(2)
Pb x - P(x2 - a) L 2
But b = L - a. Thus M2 = Paa1 -
EI
EI
d2y2 dx2 2
x2 b L
= Pa a1 -
x2 b L
dy2 x22 = Paax2 b + C3 dx2 2L
EI y2 = Pa a
(3)
x22 x22 b + C3x2 + C4 2 6L
(4)
Applying the boundary conditions: y1 = 0 at x1 = 0 Therefore,C2 = 0, y2 = 0 at x2 = L 0 =
b
x2
Pb x L 1
dy1 Pb 2 = x + C1 dx1 2L 1
M2 =
B
Pa L2 + C3L + C4 3
(5)
Applying the continuity conditions: y1 |x1 = a = y2 |x2 = a Pb 3 a2 a3 a + C1a = Pa a b + C3a + C4 6L 2 6L
(6)
dy1 dy2 2 2 = dx1 x1 = a dx2 x2 = a a2 Pb 2 a + C1 = Pa a a b + C3 2L 2L
(7)
894
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•12–9.
Continued
Solving Eqs. (5), (6) and (7) simultaneously yields, C1 = C4 =
Pb 2 A L - b2 B ; 6L
C3 = -
Pa A 2L2 + a2 B 6L
Pa3 6
Thus, EIy1 =
Pb 3 Pb 2 x A L - b 2 B x1 6L 1 6L
or v1 =
Pb A x3 - A L2 - b2 B x1 B 6EIL 1
Ans.
and EIy2 = Pa a y2 =
x22 x32 Pa Pa3 b A 2L2 + a2 B x2 + 2 6L 6L 6
Pa C 3x22 L - x32 - A 2L2 + a2 B x2 + a2L D 6EIL
Ans.
895
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12–10. Determine the maximum slope and maximum deflection of the simply supported beam which is subjected to the couple moment M0 . EI is constant.
M0
A
B L
Support Reactions and Elastic Curve: As shown on FBD(a). Moment Function: As shown on FBD(b). Slope and Elastic Curve: EI
d2y = M(x) dx2
EI
M0 d2y = x 2 L dx
EI
M0 2 dy = x + C1 dx 2L
EI y =
[1]
M0 3 x + C1x + C2 6L
[2]
Boundary Conditions: y = 0 at x = 0. From Eq. [2]. 0 = 0 + 0 + C2
C2 = 0
y = 0 at x = L. From Eq. [2]. 0 =
M0 3 A L B + C1 (L) 6L
C1 = -
M0L 6
The Slope: Substitute the value of C1 into Eq. [1], M0 dy = A 3x2 - L2 B dx 6LEI M0 dy = 0 = A 3x2 - L2 B dx 6LEI uB =
x =
23 L 3
M0L dy 2 = dx x = 0 6EI
umax = uA =
M0L dy 2 = dx x = L 3EI
Ans.
The Elastic Curve: Substituting the values of C1 and C2 into Eq. [2], y =
ymax occurs at x =
M0 A x3 - L2x B 6LEI
23 L, 3 ymax = -
23M0L2 Ans 27EI
The negative sign indicates downward displacement.
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12–11. Determine the equations of the elastic curve for the beam using the x1 and x2 coordinates. Specify the beam’s maximum deflection. EI is constant.
P
Referring to the FBDs of the beam’s cut segments shown in Fig. b and c, 2 M(x1) - Px1 = 0 3
a + ©M0 = 0;
A B
x1
2P x M(x1) = 3 1
a
2a x2
And 1 P P(3a - x2) - M(x2) = 0 M(x2) = Pa x 3 3 2
a + ©M0 = 0; EI
d2y = M(x) dx2
For coordinate x1, EI
EI
d2y1 dx21
=
2P x 3 1
dy1 P 2 = x + C1 dx1 3 1
EI y1 =
(1)
P 3 x = C1x1 + C2 9 1
(2)
For coordinate x2, EI
EI
d2y2 dx2
2
= Pa -
P x 3 2
dy2 P 2 = Pax2 x + C3 dx2 6 2
EI y2 =
(3)
Pa 2 P 3 x x + C3x2 + C4 2 2 18 2
(4)
At x1 = 0, y1 = 0. Then, Eq (2) gives EI(0) =
P 3 A 0 B + C1(0) + C2 9
C2 = 0
At x2 = 3a, y2 = 0. Then Eq (4) gives EI(0) =
Pa P (3a)2 (3a)3 + C3(3a) + C4 2 18 C3(3a) + C4 = -3Pa3
At x1 = x2 = a,
(5)
dy1 dy2 = . Thus, Eq. (1) and (3) gives dx1 dx2
P 2 P 2 a + C1 = Pa(a) a + C3 3 6 C1 - C3 =
Pa2 2
(6)
897
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12–11.
Continued
Also, At x1 = x2 = a, y1 = y2. Thus, Eqs, (2) and (4) gives. P 3 Pa 2 P 3 a + C1a = (a ) a + C3a + C4 9 2 18 C1a - C3a - C4 =
Pa3 3
(7)
Solving Eqs. (5), (6) and (7), C4 =
Pa3 6
C3 = -
19 Pa2 18
C1 = -
5Pa2 9
Substitute the values of C1 into Eq. (1) and C3 into Eq. (3), dy1 P = A 3x1 2 - 5a2 B dx1 9EI dy1 P = 0 = A 3x1 2 - 5a2 B dx1 9EI
x1 =
5 a 7 a (Not Valid) A3
And dy2 P = A 18ax2 - 3x2 2 - 19a2 B dx2 18EI dy2 P = 0 = A 18ax2 - 3x2 2 - 19a2) dx2 18EI x2 = 4.633a 7 3a (Not Valid)
x2 = 1.367a
Substitute the values of C1 and C2 into Eq. (2) and C3 and C4 into Eq.(4), y1 =
P A x 3 - 5a2x1 B 9EI 1
Ans.
y2 =
P A -x2 3 + 9ax2 2 - 19a2x2 + 3a3 B 18EI
Ans.
Vmax occurs at x2 = 1.367a. Thus. ymax = -
0.484 Pa3 0.484 Pa3 = T EI EI
Ans.
898
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*12–12. Determine the equations of the elastic curve for the beam using the x1 and x2 coordinates. Specify the slope at A and the maximum displacement of the shaft. EI is constant.
P
P
a
a
A
B
Referring to the FBDs of the beam’s cut segments shown in Fig. b and c, a + ©M0 = 0;
M(x1) - Px1 = 0
x1
M(x1) = Px1
x2
And
L
a + ©M0 = 0;
M(x2) - Pa = 0 EI
M(x2) = Pa
d2y = M(x) dx2
For coordinate x1, EI
EI
d2y1
= Px1
dx21
dy1 P 2 = x + C1 dx1 2 1
(1)
P 3 x + C1x1 + C2 6 1
EI y1 =
(2)
For coordinate x2, EI
EI
d2y2 dx2 2
= Pa
dy2 = Pax2 + C3 dx2
(3)
Pa 2 x + C3x2 + C4 2 2
EI y2 =
(4)
At x1 = 0, y1 = 0. Then, Eq. (2) gives EI (0) = Due to symmetry, at x2 =
P 3 (0 ) + C1(0) + C2 6
L dv2 = 0. Then, Eq. (3) gives , 2 dx2
EI (0) = Pa a At x1 = x2 = a,
C2 = 0
L b + C3 2
C3 = -
PaL 2
dy1 dy2 = . Thus, Eqs(1) and (3) give dx1 dx2 P 2 PaL a + C1 = Pa (a) + a b 2 2 C1 =
Pa2 PaL 2 2
899
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*12–12.
Continued
Also, at x1 = x2 = a, y1 = y2. Thus, Eq. (2) and (4) give P 3 Pa2 PaL Pa 2 PaL a + a ba = (a ) + a ba + C4 6 2 2 2 2 C4 =
Pa3 6
Substituting the value of C1 and C2 into Eq. (2) and C3 and C4 into Eq.(4), y1 =
P C x 3 + a(3a - 3L)x1 D 6EI 1
Ans.
y2 =
Pa A 3x2 2 - 3Lx2 + a2 B 6EI
Ans.
Due to symmetry, ymax occurs at x2 = ymax =
L . Thus 2
Pa Pa A 4a2 - 3L2 B = A 3L2 - 4a2 B T 24EI 24EI
Ans.
Substitute the value C1 into Eq (1), dy1 P = A x 2 + a2 - aL B dx1 2EI 1 At point A, x1 = 0. Then uA =
dy1 Pa Pa 2 = (a - L) = (L - a) T dx1 x1 = 0 2EI 2EI
900
Ans.
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12–13. The bar is supported by a roller constraint at B, which allows vertical displacement but resists axial load and moment. If the bar is subjected to the loading shown, determine the slope at A and the deflection at C. EI is constant d2y1
EI
EI
dx21
P C A
L 2
= M1 = Px1
dy1 Px21 = + C1 dx1 2
EI y1 =
Px31 + C1x1 + C2 6
EI
d2y2 PL = M2 = dx2 2
EI
dy2 PL = x + C3 dx2 2 2
EI y2 =
PL 2 x + C3x2 + C4 4 2
Boundary conditions: At x1 = 0, y1 = 0 0 = 0 + 0 + C2 ; At x2 = 0,
At x1 = P(L2 )3
C3 = 0
P A L2 B 2 2
dy1 dy2 L L = , x = , y1 = y2, 2 2 2 dx1 dx2
+ C1 a
6
PL(L2 )2 L b = + C4 2 4
+ C1 = -
C4 = -
C2 = 0
dy2 = 0 dx2
0 + C3 = 0 ;
PL A L2 B 2
;
3 C1 = - PL2 8
11 PL3 48
At x1 = 0 dy1 3 PL2 = uA = dx1 8 EI At x1 =
yC =
yC =
Ans.
L 2
P A L2 B 3 6EI
- a
B
3 PL2 L ba b + 0 8 EI 2
-PL2 6EI
Ans.
901
L 2
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12–14. The simply supported shaft has a moment of inertia of 2I for region BC and a moment of inertia I for regions AB and CD. Determine the maximum deflection of the beam due to the load P. M1 (x) =
P x 2 1
M2(x) =
P x 2 2
P B
A
L – 4
Elastic curve and slope: EI
EI
EI
d2v = M(x) dx2 d2v1 dx1
=
2
P x 2 1
dv1 Px21 = + C1 dx1 4
EIv1 =
2EI
2EI
(1)
Px31 + C1x1 + C2 12
d2v2
=
dx2 2
(2)
P x 2 2
dv2 Px22 = + C3 dx1 4
2EIv2 =
(3)
Px32 + C3x2 + C4 12
(4)
Boundary Conditions: v1 = 0 at x1 = 0 From Eq. (2), C2 = 0 dv2 L = 0 at x2 = dx2 2 From Eq. (3), 0 =
PL2 + C3 16
C3 =
PL2 16
Continuity conditions: dv1 dv2 L = at x1 = x2 = dx1 dx2 4
902
C
L – 4
L – 4
D
L – 4
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12–14.
Continued
From Eqs. (1) and (3), PL2 1 PL2 PL2 + C1 = - a b 64 128 2 16 C1 =
-5PL2 128
v1 = v2 at x1 = x2 =
L 4
From Eqs. (2) and (4) PL3 5PL2 L PL3 1 PL2 L 1 a b = - a b a b + C4 768 128 4 1536 2 16 4 2 C4 = v2 =
-PL3 384 P A 32x32 - 24L2 x2 - L3 B 768EI
vmax = v2 2
= x2 = L2
-3PL3 3PL3 = T 256EI 256EI
Ans.
903
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12–15. Determine the equations of the elastic curve for the shaft using the x1 and x3 coordinates. Specify the slope at A and the deflection at the center of the shaft. EI is constant.
P
P
A
Support Reactions and Elastic Curve: As shown on FBD(a).
x1
Moment Function: As shown on FBD(b) and (c).
x3 a
Slope and Elastic Curve: EI
d2 y = M(x) dx2
For M(x1) = -Px1, EI
EI
d2y1 dx21
= -Px1
dy1 P = - x21 + C1 dx1 2
EI y1 = -
[1]
P 3 x + C1x1 + C2 6 1
[2]
For M(x3) = -Pa, EI
EI
d2y3 dx23
= -Pa
dy3 = -Pax3 + C3 dx3
EI y3 = -
[3]
Pa 2 x + C3x3 + C4 2 3
[4]
Boundary Conditions: y1 = 0 at x1 = a. From Eq. [2], 0 = Due to symmetry,
Pa3 + C1a + C2 6
[5]
dy3 b = 0 at x3 = . From Eq. [3] dx3 2
b 0 = -Paa b + C3 2
C3 =
Pab 2
y3 = 0 at x3 = 0 From Eq.[4]. C4 = 0 Continuity Condition: At x1 = a and x3 = 0,
-
From Eq. [5]
dy1 dy3 = . From Eqs. [1] and [3], dx1 dx3
Pa2 Pab + C1 = 2 2 C2 = -
C1 =
B
Pa (a + b) 2
Pa2 (2a + 3b) 6
904
b
a
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12–15. Continued The Slope: Either Eq. [1] or [3] can be used. Substitute the value of C1 into Eq. [1], dy1 P = C -x21 + a(a + b) D dx1 2EI uA =
dy1 P Pab 2 = C -a2 + a(a + b) D = dx1 x1 = a 2EI 2EI
Ans.
The Elastic Curve: Substitute the values of C1, C2, C3, and C4 into Eqs. [2] and [4], respectively, y1 =
P C -x31 + 3a(a + b)x1 - a2(2b + 3b) D 6EI
Ans.
y3 =
Pax3 ( -x3 + b) 2EI
Ans.
yC = y3 |x3 = b2
=
=
Pa A b2 B 2EI
a-
b + bb 2
Pab2 8EI
Ans.
905
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*12–16. The fence board weaves between the three smooth fixed posts. If the posts remain along the same line, determine the maximum bending stress in the board. The board has a width of 6 in. and a thickness of 0.5 in. E = 1.60(103) ksi. Assume the displacement of each end of the board relative to its center is 3 in.
4 ft
3 in. A
Support Reactions and Elastic Curve: As shown on FBD(a). Moment Function: As shown on FBD(b). Slope and Elastic Curve: EI
d2y = M(x) dx2
EI EI
d2y P = x 2 dx2
dy P 2 = x + C1 dx 4
EI y =
[1]
P 3 x + C1x + C2 12
Boundary Conditions: Due to symmetry,
[2]
L dy = 0 at x = . dx 2
Also, y = 0 at x = 0. From Eq. [1] 0 =
P L 2 a b + C1 4 2
From Eq. [2] 0 = 0 + 0 + C2
C1 = -
PL2 16
C2 = 0
The Elastic Curve: Substitute the values of C1 and C2 into Eq. [2], y =
Px A 4x2 - 3L2 B 48EI
[1]
Require at x = 48 in., y = -3 in. From Eq.[1], P(48)
-3 =
1 48(1.60) A 106 B A 12 B (6) A 0.53 B
C 4 A 482 B - 3 A 962 B D
P = 16.28 lb Maximum Bending Stress: From the moment diagram, the maximum moment is Mmax = 390.625 lb # in. Applying the flexure formula, smax =
4 ft
390.625(0.25) Mc = = 1562.5 psi = 1.56 ksi 1 3 I 12 (6) A 0.5 B
Ans.
906
B
C
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•12–17.
Determine the equations of the elastic curve for the shaft using the x1 and x2 coordinates. Specify the slope at A and the deflection at C. EI is constant.
a + ©MO = 0;
a + ©MO = 0;
MO - M(x2) = 0
M(x2) = MO
d2y = M(x) dx2
For coordinate x1, EI
EI
d2y1 dx1 2
=
MO x L 1
dy1 MO 2 = x + C1 dx1 2L 1
EI y1 =
(1)
MO 3 x + C1x1 + C2 6L 1
(2)
For coordinate x2, EI
EI
d2y2 dx2 2
= MO
dy2 = MOx2 + C3 dx2
EI y2 =
(3)
MO 2 x + C3x2 + C4 2 2
(4)
At x1 = 0, y1 = 0. Then, Eq. (2) gives EI (0) =
MO 3 (0 ) + C1(0) + C2 6L
C2 = 0
At x1 = L, y1 = 0. Then, Eq. (2) gives C1 = Also, at x2 =
-ML 6
L , y = 0. Then Eq. (4) gives. 2 2 EI(0) =
MO L 2 L a b + C3 a b + C4 2 2 2
C3L + 2C4 = -
MOL2 4
(5)
907
C
x2 L
MO M(x1) = x L 1
And
EI
B
x1
Referring to the FBDs of the shaft’s cut segments shown in Fig. b and c, MO M(x1) x = 0 L 1
M0
A
L 2
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•12–17.
Continued
At x1 = L and x2 =
dy2 L dy1 , = . Then, Eq. (1) and (3) give 2 dx1 dx2 MO 2 MOL L = - cMO a b + C3 d AL B 2L 6 2
C3 = -
5MOL 6
Substitute the result of C3 into Eq. (5), C4 =
7M0L2 24
Substitute the value of C1 into Eq. (1), dy1 MO = A 3x1 2 - L2 B dx1 6LEI At A, x1 = 0. Thus uA =
MOL dy1 MO 2 = = dx1 x1 = 0 6EI 6EI
Ans.
Substitute the values of C1 and C2 into Eq (2) and C3 and C4 into Eq. (4), y1 =
MO A x 3 - L2x1 B 6EIL 1
Ans.
y2 =
MO A 12x2 2 - 20 Lx2 + 7L2 B 24EI
Ans.
At C, x2 = 0. Thus yC = y2 2
= x2 = 0
7MOL2 24EI
Ans.
c
908
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12–18. Determine the equation of the elastic curve for the beam using the x coordinate. Specify the slope at A and the maximum deflection. EI is constant.
M0
A
M0
x
B L
Referring to the FBD of the beam’s cut segment shown in Fig. b, a + ©MO = 0;
M(x) +
2MO x - MO = 0 L
EI
d2y = M(x) dx2
EI
2MO d2y = MO x L dx2
EI
MO 2 dy = MOx x + C1 dx L
M(x) = MO -
2MO x L
(1)
MO 3 MO 2 x x + C1x + C2 2 3L
EI y =
(2)
At x = 0, y = 0. Then Eq (2) gives EI(0) =
MO 3 MO 2 A0 B A 0 B + C1(0) + C2 2 3L
C2 = 0
Also, at x = L, y = 0. Then Eq (2) gives EI(0) =
MO 2 MO 3 AL B A L B + C1L + 0 2 3L
C1 = -
MOL 6
Substitute the value of C1 into Eq (1), MO dy = A 6Lx - 6x2 - L2 B dx 6EIL MO dy = 0 = A 6Lx - 6x2 - L2 B dx 6EIL x = 0.2113 L
and
0.7887 L
At A, x = 0. Thus uA = -
MOL 6EI
Ans.
Substitute the values of C1 and C2 into Eq (2) y =
MO A 3Lx2 - 2x3 - L2x B 6EIL
Ans.
909
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12–18.
Continued
vmax occurs at x = 0.21132 L ymax =
or
0.7887 L. Thus,
MO c3L(0.2113L)2 - 2(0.2113L)3 - L2(0.2113L) d 6EIL
= -
0.0160 MOL2 0.0160 MOL2 = EI EI
Ans.
T
and ymax =
=
MO c3L(0.7887L)2 - 2(0.7887L)3 - L2(0.7887L) d 6EIL 0.0160 MOL2 EI
Ans.
c
910
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12–19. Determine the deflection at the center of the beam and the slope at B. EI is constant.
M0
A
M0
x
B L
Referring to FBD of the beam’s cut segment shown in Fig. b, a + ¢Mo = 0;
M(x) +
2Mo x - Mo = 0 L
EI
d2v = M(x) dx2
EI
2Mo d2v = Mo x L dx2
EI
Mo 2 dv = Mox x + C1 dx L
EI v =
M(x) = Mo -
2Mo x L
(1)
Mo 3 Mo 2 x x + C1x + C2 2 3L
(2)
At x = 0, v = 0. Then Eq. (2) gives EI(0) =
Mo 3 Mo 2 (0 ) A 0 B + C1(0) + C2 2 3L
C2 = 0
Also, at x = L, v = 0. Then Eq. (2) gives EI (0) =
Mo 2 Mo 3 AL B A L B + C1L + 0 2 3L
C1 = -
MoL 6
Substitute the value of C1 into Eq. (1), Mo dv = A 6Lx - 6x2 - L2 B dx 6EIL At B, x = L. Thus uB =
MoL MoL dv 2 = = dx x = L 6EI 6EI
Ans.
Substitute the values of C1 and C2 into Eq. (2), v =
Mo A 3Lx2 - 2x3 - L2x B 6EIL
At the center of the beam, x =
L . Thus 2
v冷x = L2 = 0
Ans.
911
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*12–20. Determine the equations of the elastic curve using the x1 and x2 coordinates, and specify the slope at A and the deflection at C. EI is constant.
8 kip
A
C
B
x1
x2 20 ft
Referring to the FBDs of the beam’s cut segments shown in Fig. b, and c, M(x1) = (-5x1) kip # ft
a + ©Mo = 0;
M(x1) + 5x1 = 0
a + ©Mo = 0;
-M(x2) - 8x2 - 20 = 0 M(x2) = ( -8x2 - 20) kip # ft
And
EI
d2v = M(x) dx2
For coordinate x1, EI
EI
d2v1 dx21
= (-5x1) kip # ft
dv1 5 = a - x21 + C1 b kip # ft2 dx1 2
(1)
5 EI v1 = a - x1 3 + C1x1 + C2 b kip # ft3 6
(2)
For coordinate x2, EI
EI
d2v2 dx2 2
= (-8x2 - 20) kip # ft
dv2 = dx2
A -4x2 2 - 20x2 + C3 B kip # ft2
(3)
4 EI v2 = a - x2 3 - 10x2 2 + C3x2 + C4 b kip # ft3 3
(4)
At x1 = 0, v1 = 0. Then, Eq (2) gives EI(0) = -
5 3 A 0 B + C1(0) + C2 6
C2 = 0
Also, at x1 = 20 ft, v1 = 0. Then, Eq (2) gives EI(0) = -
5 A 203 B + C1 (20) + 0 6
C1 = 333.33 kip # ft2
Also, at x2 = 10 ft, v2 = 0. Then, Eq. (4) gives EI(0) = -
4 A 103 B - 10 A 102 B + C3(10) + C4 3
10C3 + C4 = 2333.33
(5)
912
10 ft
20 kip⭈ft
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*12–20.
Continued
At x1 = 20 ft and x2 = 10 ft,
-
dv1 dv2 = . Then Eq. (1) and (3) gives dx1 dx2
5 A 202 B + 333.33 = - C -4 A 102 B - 20(10) + C3 D 2 C3 = 1266.67 kip # ft2
Substitute the value of C3 into Eq (5), C4 = -10333.33 kip # ft3 Substitute the value of C1 into Eq. (1), dv1 5 1 a - x1 2 + 333.33b kip # ft2 = dx1 EI 2 At A, x1 = 0. Thus, uA =
333 kip # ft2 dv1 2 = dx1 x1 = 0 EI
uA
Ans.
Substitute the values of C1 and C2 into Eq. (2) and C3 and C4 into Eq (4), v1 =
1 5 a - x1 3 + 333 x1 b kip # ft3 EI 6
Ans.
v2 =
4 1 a - x2 3 - 10x2 2 + 1267x2 - 10333b kip # ft3 EI 3
Ans.
At C, x2 = 0. Thus vC = v2 冷x2 = 0 = -
10 333 kip # ft3 10 333 kip # ft3 = T EI EI
913
Ans.
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•12–21.
Determine the elastic curve in terms of the and coordinates and the deflection of end C of the overhang beam. EI is constant.
w
A
C
Support Reactions and Elastic Curve. As shown in Fig. a.
B
Moment Functions. Referring to the free-body diagrams of the beam’s cut segments, Fig. b, M(x1) is a + ©MO = 0;
M(x1) +
wL x = 0 8 1
M(x1) = -
-M(x2) - wx2 a
x2 b = 0 2
M(x2) = -
w 2 x 2 2
Equations of Slope and Elastic Curve. EI
d2v = M(x) dx2
For coordinate x1, EI
EI
d2v1 dx1 2
= -
wL x 8 1
dv1 wL 2 = x + C1 dx1 16 1
EIv1 = -
(1)
wL 3 x + C1x1 + C2 48 1
(2)
For coordinate x2, EI
EI
d2v2 dx2 2
= -
w 2 x 2 2
dv2 w = - x2 3 + C3 dx2 6
EIv2 = -
(3)
w 4 x + C3x2 + C4 24 2
(4)
Boundary Conditions. At x1 = 0, v1 = 0. Then, Eq. (2) gives EI(0) = -
wL 3 A 0 B + C1(0) + C2 48
C2 = 0
At x1 = L, v1 = 0. Then, Eq. (2) gives EI(0) = At x2 =
wL 3 A L B + C1L + 0 48
C1 =
wL3 48
L , v = 0. Then, Eq. (4) gives 2 2
EI(0) = -
w L 4 L a b + C3 a b + C4 24 2 2
L wL4 C3 + C4 = 2 384
(5)
914
x2 L
wL x 8 1
and M(x2) is a + ©MO = 0;
x1
L 2
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•12–21.
Continued
Continuity Conditions. At x1 = L Land x2 =
dv2 L dv1 , = . Thus, Eqs. (1) and 2 dx1 dx2
(3) give
-
w L 3 wL3 wL 2 = - C - a b + C3 S AL B + 16 48 6 2
C3 =
wL3 16
Substituting the value of C3 into Eq. (5), C4 = -
11wL4 384
Substituting the values of C3 and C4 into Eq. (4), v2 =
w A -16x2 4 + 24L3x2 - 11L4 B 384EI
At C, x2 = 0. Thus, vC = v2冷x2 = 0 = -
11wL4 11wL4 = T 384EI 384EI
Ans.
915
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12–22. Determine the elastic curve for the cantilevered beam using the x coordinate. Specify the maximum slope and maximum deflection. E = 29(103) ksi.
3 kip/ft
A
Referring to the FBD of the beam’s cut segment shown in Fig. b, a + ©Mo = 0; M(x) + 81 +
1 2
x 3
9 ft
M(x) = A 13.5x - 0.05556x3 - 81 B kip # ft. EI
d2v = M(x) dx2
EI
d2v = A 13.5x - 0.05556x3 - 81 B kip # ft dx2
EI
dv = A 6.75x2 - 0.01389x4 - 81x + C1 B kip # ft2 dx
(1)
EI v = A 2.25x3 - 0.002778x5 - 40.5x2 + C1x + C2 B kip # ft3 (2) At x = 0,
dv = 0. Then, Eq (1) gives dx
EI(0) = 6.75 A 02 B - 0.01388 A 04 B - 81(0) + C1
C1 = 0
Also, at x = 0, v = 0. Then Eq. (2) gives EI(0) = 2.25 A 03 B - 0.002778 A 05 B - 40.5 A 02 B + 0 + C2
C2 = 0
Substitute the value of C1 into Eq (1) gives. 1 dv = A 6.75x2 - 0.01389x4 - 81x B kip # ft2 dx EI The Maximum Slope occurs at x = 9 ft. Thus, umax =
273.375 kip # ft2 dv 2 = dx x = 9ft EI =
273.375 kip # ft2 EI
umax
For W14 * 30, I = 291 in4. Thus u = 273.375 A 12 2 B = 0.00466 rad
Ans.
Substitute the values of C1 and C2 into Eq (2), v =
1 A 2.25x3 - 0.002778x5 - 40.5x2 B kip # ft3 EI
The maximum deflection occurs at x = 9 ft, Thus, vmax = v 冷x = 9 ft = -
=
=
B
x
A x B (x) A B - 13.5x = 0 1 3
1804.275 kip # ft3 EI
1804.275 kip # ft3 T EI 1804.275 A 12 3 B
29.0 A 103 B (291)
= 0.369 in T
Ans.
916
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12–23. The beam is subjected to the linearly varying distributed load. Determine the maximum slope of the beam. EI is constant.
w0
A
B
x L
EI
d2y = M(x) dx2
EI
w0 d2y = A L2x - x3 B 2 6L dx
EI
w0 L2x2 x4 dy = a b + C1 dx 6L 2 4
EI y =
(1)
w0 L2x3 x5 a b + C1x + C2 6L 6 20
(2)
Boundary conditions: At x = 0, y = 0. From Eq. (2), C2 = 0 At x = L, y = 0 From Eq. (2), 0 =
w0 L5 L5 a b + C1L ; 6L 6 20
C1 = -
7w0L3 360
The slope: From Eq.(1), w0 dy L2x2 x4 7L4 = a b dx 6EIL 2 4 60 umax =
w0L3 dy 2 = dx x = L 45EI
Ans.
917
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*12–24. The beam is subjected to the linearly varying distributed load. Determine the maximum deflection of the beam. EI is constant.
w0
A
B
x L
EI
d2y = M(x) dx2
EI
w0 d2y = A L2x - x3 B 2 6L dx
EI
w0 L2x2 x4 dy = a b + C1 dx 6L 2 4
EI y =
(1)
w0 L2x3 x5 a b + C1x + C2 6L 6 20
(2)
Boundary conditions: y = 0 at x = 0. From Eq. (2), C2 = 0 y = 0 at x = L. From Eq. (2), 0 =
w0 L2 L5 a b + C1L; 6L 6 20
C1 = -
7w0L3 360
w0 dy L2x2 x4 7L4 = a b dx 6EIL 2 4 60 dy L2x2 x4 7L4 = 0 = a b dx 2 4 60 15x4 - 30L2x2 + 7L4 = 0; y =
x = 0.5193L
w0x A 10L2x2 - 3x4 - 7L4 B 360EIL
Substitute x = 0.5193L into y, ymax = -
0.00652w0L4 EI
Ans.
918
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•12–25.
Determine the equation of the elastic curve for the simply supported beam using the x coordinate. Determine the slope at A and the maximum deflection. EI is constant.
12 kN/m
A
B
x 6m
Referring to the FBD of the beam’s cut segment shown in Fig. b, a + ©Mo = 0; M(x) +
1 1 (2x)(x) A x3 B - 36x = 0 M(x) = a36x - x3 bkN # m 2 3
EI
d2v = M(x) dx2
EI
d2v 1 = a36x - x3 b kN # m 2 3 dx dv 1 4 = a18x2 x + C1 b kN # m2 dx 12
EI
EI v = a6x3 Due to the Symmetry,
(1)
1 5 x + C1x + C2 b kN # m3 60
(2)
dv = 0 at x = 6 m. Then, Eq (1) gives dx
EI(0) = 18 A 62 B -
1 A 64 B + C1 12
C1 = -540 kN # m2
Also, at x = 0, v = 0. Then, Eq (2) gives EI(0) = 6 A 03 B -
1 A 05 B + C1(0) + C2 60
C2 = 0
Substitute the value of C1 into Eq. (1), dv 1 1 4 = a18x2 x - 540b kN # m2 dx EI 12 At A, x = 0. Then uA =
dv 540 kN # m2 540kN # m2 2 = = dx x = 0 EI EI
Ans.
Substitute the values of C1 and C2 into Eq (2) v =
1 1 5 a6x3 x - 540xb kN # m3 EI 60
Ans.
Due to Symmetry, vmax occurs at mind span x = 6 m. Thus, vmax =
1 1 c6 A 63 B A 65 B - 540(6) d EI 60
= -
2074 kN # m3 2073.6 kN # m3 = EI EI
Ans.
T
919
6m
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12–26. Determine the equations of the elastic curve using the coordinates x1 and x2 , and specify the slope and deflection at B. EI is constant.
w
C
A
B
x1 a x2 L
EI
d2y = M(x) dx2
For M1(x) = -
EI
EI
d2y1 dx21
= -
w 2 wa2 x1 + wax1 2 2
w 2 wa2 x1 + wax1 2 2
dy1 w wa 2 wa2 = - x21 + x1 x + C1 dx1 6 2 2 1
EI y1 = -
w 4 wa 3 wa2 2 x1 + x1 x1 + C1x1 + C2 24 6 4
For M2(x) = 0 ;
EI
(1)
EI
d2y2 dx2 2
(2)
= 0
dy2 = C3 dx2
(3)
EI y2 = C3x2 + C4
(4)
Boundary conditions: At x1 = 0.
dy1 = 0 dx1
From Eq. (1), C1 = 0 At x1 = 0. y1 = 0 From Eq. (2): C2 = 0 Continuity conditions: At x1 = a,
dy1 dy2 = dx1 dx2
x2 = a ;
From Eqs. (1) and (3), -
wa3 wa3 wa3 + = C3; 6 2 2
C3 = -
wa3 6
From Eqs. (2) and (4), At x1 = a, x2 = a -
y1 = y2
wa4 wa4 wa4 wa4 + = + C4 ; 24 6 4 6
C4 =
wa4 24
920
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12–26.
Continued
The slope, from Eq. (3). uB =
dy2 wa3 = dx2 6EI
Ans.
The elastic curve: y1 =
w a -x41 + 4ax31 - 6a2 x21 b 24EI
Ans.
y2 =
wa3 a -4x2 + ab 24EI
Ans.
yB = y2 2
= x3 = L
wa3 a -4L + ab 24EI
Ans.
921
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12–27. Wooden posts used for a retaining wall have a diameter of 3 in. If the soil pressure along a post varies uniformly from zero at the top A to a maximum of 300 lb>ft at the bottom B, determine the slope and displacement at the top of the post. Ew = 1.6(103) ksi.
A
6 ft
Moment Function: As shown on FBD. Slope and Elastic Curve: B
d2y EI 2 = M(y) dy EI
EI
d2y = -8.333y3 dy2
dy = -2.0833y4 + C1 dy
[1]
EI y = -0.4167y5 + C1y + C2 Boundary Conditions:
[2]
dy = 0 at y = 6 ft and y = 0 at y = 6 ft dy
From Eq. [1], 0 = -2.0833 A 64 B + C1
C1 = 2700
From Eq. [2], 0 = -0.4167 A 65 B + 2700(6) + C2
C2 = -12960
The Slope: Substituting the value of C1 into Eq. [1], 1 dy = b A -2.0833y4 + 2700 B r lb # ft2 dy EI uA =
dy 2 2700 lb # ft2 = dy y = 0 EI 2700(144)
=
1.6 A 106 B A p4 B A 1.54 B
= 0.0611 rad
Ans.
The Elastic Curve: Substituting the values of C1 C2 into Eq. [2], y =
1 EI
E A -0.4167y5 + 2700y - 12960 B F lb # ft3
yA = y|y = 0 = -
12960 lb # ft3 EI 12960(1728)
= -
1.6 A 106 B A p4 B A 1.54 B
= -3.52 in.
Ans.
The negative sign indicates leftward displacement.
922
300 lb/ft
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*12–28. Determine the slope at end B and the maximum deflection of the cantilevered triangular plate of constant thickness t. The plate is made of material having a modulus of elasticity E.
b 2 b 2 L
Section Properties. Referring to the geometry shown in Fig. a, b(x) b = ; x L
b(x) =
A
b x L t
Thus, the moment of the plate as a function of x is I(x) =
1 bt3 x C b(x) D t3 = 12 12L
x
Moment Functions. Referring to the free-body diagram of the plate’s cut segments, Fig. b, + ©MO = 0;
x -M(x) - w(x) a b = 0 2
M(x) = -
w 2 x 2
Equations of Slope and Elastic Curve. E
M(x) d 2v = I(x) dx2
d 2v E 2 = dx
E
w 2 x 6wL 2 = - 3 x 3 bt bt x 12L
-
dv 3wL = - 3 x2 + C1 dx bt
Ev = -
(1)
wL 3 x + C1x + C2 bt3
Boundary Conditions. At x = L, E(0) = -
(2)
dv = 0. Then Eq. (1) gives dx
3wL 2 A L B + C1 bt3
C1 =
3wL3 bt3
At x = L, v = 0. Then Eq. (2) gives E(0) = -
wL 3 A L B + C1(L) + C2 bt3
C2 = -
2wL4 bt3
Substituting the value of C1 into Eq. (1), dv 3wL = A -x2 + L2 B dx Ebt3 At B, x = 0. Thus, uB =
dv 2 3wL3 = dx x = 0 Ebt3
Substituting the values of C1 and C2 into Eq. (2), v =
wL A -x3 + 3L2x - 2L3 B Ebt3
vmax occurs at x = 0. Thus, vmax = v冷x = 0 = -
2wL4 2wL4 = 3 Ebt Ebt3
w
Ans.
T
923
B
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•12–29.
The beam is made of a material having a specific weight g. Determine the displacement and slope at its end A due to its weight. The modulus of elasticity for the material is E.
L
Section Properties: h
bh 2 1 h a xb(x)(b) = x 2 L 2L
h(x) =
h x L
I(x) =
3 h bh3 3 1 (b)a xb = x 12 L 12L3
V(x) =
Moment Function: As shown on FBD. Slope and Elastic Curve: E
E
M(x) d2y = I(x) dx2
d2y = dx2
E
bhg 6L
x3
2gL2 = -
3
bh 3 3x 12L
h2
2gL2 dy = - 2 x + C1 dx h
Ey = -
gL2
x2 + C1 x + C2
h2
From Eq. [2], 0 = -
[2]
dy = 0 at x = L and y = 0 at x = L. dx
Boundary Conditions: From Eq. [1], 0 = -
[1]
2gL2 2
h
gL2 2
h
(L) + C1
A L2 B +
C1 =
2gL3 2
h
2gL3
(L) + C2
h3 C2 = -
gL4 h2
The Slope: Substituting the value of C1 into Eq. [1], 2gL2 dy = 2 ( -x + L) dx hE uA =
2gL3 dy 2 = 2 dx x = 0 hE
Ans.
The Elastic Curve: Substituting the values of C1 and C2 into Eq. [2], y =
gL2 h2E
A -x2 + 2Lx - L2 B
yA |x = 0 = -
gL4
Ans.
h2E
The negative sign indicates downward displacement.
924
A
b
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12–30. The beam is made of a material having a specific weight of g. Determine the displacement and slope at its end A due to its weight. The modulus of elasticity for the material is E.
r
Section Properties: r(x) =
r x L 4
I(x) =
A
2 p r pr2 3 a xb x = x 3 L 3L2
V(x) =
L
4
pr 4 p r a xb = x 4 L 4L4
Moment Function: As shown on FBD. Slope and Elastic Curve: M(x) d2y = I(x) dx2
E
2
pr g2 4 gL2 d 2y 12L x E 2 = - 4 = - 2 pr 4 4 dx 3r x 4L
E
gL2 dy = - 2 x + C1 dx 3r
Ey = Boundary Conditions: From Eq. [1], 0 = -
From Eq. [2], 0 = -
6r2
x2 + C1x + C2
[2]
dy = 0 at x = L and y = 0 at x = L. dx
gL2 3r
gL2
[1]
2
(L) + C1
L2 B + ¢ 2 A
C1 =
gL2
gL3
6r
3r
gL3 3r2
L + C2 2 ≤
C2 = -
gL4 6r2
The Slope: Substituting the value of C1 into Eq. [1], gL2 dy = 2 (-x + L) dx 3r E uA =
gL3 dy 2 = 2 dx x = 0 3r E
Ans.
The Elastic Curve: Substituting the values of C1 and C2 into Eq. [2], y =
gL2 6r2E
A -x2 + 2Lx - L2 B
yA |x = 0 = -
gL4
Ans.
6r2E
The negative sign indicates downward displacement.
925
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12–31. The tapered beam has a rectangular cross section. Determine the deflection of its free end in terms of the load P, length L, modulus of elasticity E, and the moment of inertia I0 of its fixed end.
b
A
P
Moment function: M(x) = -Px
L
Moment of inertia: w =
b x; L
I =
l0 1 b 1 x a xb t3 = b t3 a b = x 12 L 12 L L
Slope and elastic curve: EI(x)
d2y = M(x) dx2
Ea
l0 d2y bx 2 = -Px ; L dx
El0
dy = -PLx + C1 dx
El0 y =
El0
d2y = -PL dx2 (1)
-PL 2 x + C1x + C2 2
(2)
Boundary conditions: dy = 0, x = L dx From Eq. (1), 0 = -PL2 + C1 ;
C1 = PL2
y = 0, x = L From Eq. (2), 0 = y =
PL3 + PL3 + C2 ; 2
C2 = -
PL3 2
PL ( -x2 + Lx - L2) 2El0
x = 0,
ymax = y 2
= u=0
PL3 2El0
Ans.
The negative sign indicates downward displacement.
926
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*12–32. The beam is made from a plate that has a constant thickness t and a width that varies linearly. The plate is cut into strips to form a series of leaves that are stacked to make a leaf spring consisting of n leaves. Determine the deflection at its end when loaded. Neglect friction between the leaves.
P
b
L
Use the triangular plate for the calculation. M = Px I =
1 b a xb(t)3 12 L
d2v M Px = = 1 EI dx2 E A 12 B A Lb B x(t)3 d2v 12PL = 2 dx Ebt3 dv 12PL = x + C1 dx Ebt3 v =
6PL 2 x + C1x + C2 Ebt3
dv = 0 at x = L dx C1 =
-12PL2 Ebt3
v = 0 at x = L C2 =
6PL3 Ebt3
When x = 0 vmax =
6PL3 Ebt3
Ans.
927
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•12–33.
The tapered beam has a rectangular cross section. Determine the deflection of its center in terms of the load P, length L, modulus of elasticity E, and the moment of inertia Ic of its center.
P
b
L — 2
Moment of inertia: L — 2
2b w = x L I =
2IC 1 2b 1 2x a xb (t3) = (b) A t3 B a b = a bx 12 L 12 L L
Elastic curve and slope: EI(x)
Ea
d2v = M(x) dx2
2IC d2v P b(x) 3 = x L 2 dx
EIC
dv PL = x + C1 dx 4
(1)
PL 2 x + C1x + C2 8
EICv1 =
(2)
Boundary condition: Due to symmetry: dv = 0 dx
x =
at
L 2
From Eq. (1), 0 =
PL2 + C1 8
C1 = -
PL2 8
v = 0 at x = 0 C2 = 0 v =
PLx (x - L) 8EIC
vC = v `
= x = L2
PL3 32EIC
Ans.
The negative sign indicates downward displacement.
928
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12–34. The leaf spring assembly is designed so that it is subjected to the same maximum stress throughout its length. If the plates of each leaf have a thickness t and can slide freely between each other, show that the spring must be in the form of a circular arc in order that the entire spring becomes flat when a large enough load P is applied. What is the maximum normal stress in the spring? Consider the spring to be made by cutting the n strips from the diamond-shaped plate of thickness t and width b.The modulus of elasticity for the material is E. Hint: Show that the radius of curvature of the spring is constant.
nb
b x P
L 2
Section Properties: Since the plates can slide freely relative to each other, the plates resist the moment individually. At an arbitrary distance x from the support, the 2nx nx numbers, of plates is L = . Hence, L 2 I(x) =
1 2nx nbt3 a b (b) A t3 B = x 12 L 6L
Moment Function: As shown on FBD. Bending Stress: Applying the flexure formula,
smax
M(x) c = = I(x)
Px 2
A 2t B
nbt3 6L x
=
3PL 2nbt2
Ans.
Moment - Curvature Relationship: Px
M(x) 1 3PL 2 = Constant (Q.E.D.) = = = 3 nbt3 r EI(x) nbt E E A 6L x B
929
x L 2
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12–35. The shaft is made of steel and has a diameter of 15 mm. Determine its maximum deflection. The bearings at A and B exert only vertical reactions on the shaft. Est = 200 GPa.
15 mm A
B
200 mm
300 mm 250 N
M = -(-201.43) 6 x - 0 7 -250 6 x - 0.2 7 - 80 6 x - 0.5 7 M = 201.43x - 250 6 x - 0.2 7 - 80 6 x - 0.5 7 Elastic curve and slope: EI
d2v = M = 201.43x - 250 6 x - 0.2 7 - 80 6 x - 0.5 7 dx3
EI
dv = 100.71x2 - 125 6 x - 0.2 7 2 - 40 6 x - 0.5 7 2 + C1 dx
EIv = 33.57x3 - 41.67 6 x - 0.2 7 3 - 13.33 6 x - 0.5 7 3 + C1x + C2 (1) Boundary conditions: v = 0
at
x = 0
From Eq. (1) C2 = 0 v = 0
at
x = 0.7 m
0 = 11.515 - 5.2083 - 0.1067 + 0.7C1 C1 = -8.857 dv 1 = C 100.71x2 - 125 6 x - 0.2 7 2 - 40 6 x - 0.5 7 2 - 8.857 D dx EI Assume vmax occurs at 0.2 m 6 x 6 0.5 m dv 1 = 0 = C 100.71x2 - 125(x - 0.2)2 - 8.857 D dx EI 24.28x2 - 50x + 13.857 = 0 x = 0.3300 m v =
O.K.
1 C 33.57x3 - 41.67 6 x - 0.2 7 3 - 13.33 6 x - 0.5 7 3 - 8.857x D EI
Substitute x = 0.3300 m into the elastic curve: vmax = -
1.808N # m3 1.808 = = -0.00364 = -3.64 mm Ans. 9 p EI 200 A 10 B 4 (0.0075)4
The negative sign indicates downward displacement.
930
200 mm 80 N
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*12–36. The beam is subjected to the loads shown. Determine the equation of the elastic curve. EI is constant.
4 kip
2 kip
4 kip⭈ft
A B x
8 ft
M = -(-2.5) 6 x - 0 7 - 2 6 x - 8 7 - 4 6 x - 16 7 M = 2.5x - 2 6 x - 8 7 - 4 6 x - 16 7 Elastic curve and slope: EI
d2v = M = 2.5x - 2 6 x - 8 7 - 4 6 x - 16 7 dx2
EI
dv = 1.25x2 - 6 x - 8 7 2 - 2 6 x - 16 7 2 + C1 dx
EIv = 0.417x3 - 0.333 6 x - 8 7 3 - 0.667 6 x - 16 7 3 + C1x + C2
(1)
Boundary conditions: v = 0
at
x = 0
From Eq. (1), C2 = 0 v = 0
at
x = 24 ft
0 = 5760 - 1365.33 - 341.33 + 24C1 C1 = -169 v =
1 C 0.417x3 - 0.333 6 x - 8 7 3 - 0.667 6 x - 16 7 3 - 169x D kip # ft3 Ans. EI
931
8 ft
8 ft
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•12–37.
Determine the deflection at each of the pulleys C, D, and E. The shaft is made of steel and has a diameter of 30 mm. The bearings at A and B exert only vertical reactions on the shaft. Est = 200 GPa.
C
E
D
A
B
250 mm
250 mm
250 mm
250 mm
M = -(-180) 6 x - 0 7 - 150 6 x - 0.25 7 150 N
- 60 6 x - 0.5 7 -150 6 x - 0.75 7 M = 180x - 150 6 x - 0.25 7 - 60 6 x - 0.5 7 -150 6 x - 0.75 7 Elastic curve and slope: EI
d2v = M = 180x - 150 6 x - 0.25 7 - 60 6 x - 0.5 7 dx2 -150 6 x - 0.75 7
EI
dv = 90x2 - 75 6 x - 0.25 7 2 - 30 6 x - 0.50 7 2 dx - 75 6 x - 0.75 7 2 + C1
(1)
EIv = 30x3 - 25 6 x - 0.25 7 3 - 10 6 x - 0.50 7 3 - 25 6 x - 0.75 7 3 + C1x + x2
(2)
Boundary conditions: v = 0
at
x = 0
From Eq. (2) C2 = 0 v = 0
x = 1.0 m
at
0 = 30 - 10.55 - 1.25 - 0.39 + C1 C1 = -17.8125 dv 1 = C 90x2 - 75 6 x - 0.25 7 2 - 30 6 x - 0.5 7 2 dx EI - 75 6 x - 0.75 7 2 - 17.8125 D v =
(3)
1 C 30x3 - 25 6 x - 0.25 7 3 - 10 6 x - 0.5 7 3 EI - 25 6 x - 0.75 7 3 - 17.8125x D
vC = v `
= x = 0.25m
-3.984 -3.984 = = -0.000501 m EI 200 A 109 B p4 (0.015)4 Ans.
= -0.501 mm vD = v `
x = 0.5m
vE = v `
x = 0.75 m
-5.547
=
=
200 A 109 B p4 (0.015)4
= -0.000698 m = -0.698 mm
-3.984 = -0.501 mm EI
Ans.
Ans. (symmetry check !)
The negative signs indicate downward displacement.
932
60 N
150 N
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12–38. The shaft supports the two pulley loads shown. Determine the equation of the elastic curve.The bearings at A and B exert only vertical reactions on the shaft. EI is constant.
A
B
x 20 in.
20 in. 40 lb
M = -10 6 x - 0 7 -40 6 x - 20 7 -(-110) 6 x - 40 7 M = -10x - 40 6 x - 20 7 + 110 6 x - 40 7 Elastic curve and slope: EI
d2v = M dx2
EI
d2v = -10x - 40 6 x - 20 7 + 110 6 x - 40 7 dx2
EI
dv = -5x2 - 20 6 x - 20 7 2 + 55 6 x - 40 7 2 + C1 dx
EIv = -1.667x3 - 6.667 6 x - 20 7 3 + 18.33 6 x - 40 7 3 + C1x + C2 (1) Boundary conditions: v = 0 at x = 0 From Eq. (1): C2 = 0 v = 0 at x = 40 in. 0 = -106,666.67 - 53,333.33 + 0 + 40C1. C1 = 4000 v =
1 C -1.67x3 - 6.67 6 x - 20 7 3 + 18.3 6 x - 40 7 3 + 4000x D lb # in3 EI
933
Ans.
20 in. 60 lb
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12–39. Determine the maximum deflection of the simply supported beam. E = 200 GPa and I = 65.0(106) mm4.
30 kN 15 kN
Support Reactions and Elastic Curve. As shown in Fig. a. Moment Function. From Fig. a, we obtain
A
M = -(-25)(x - 0) - 30(x - 2) - 15(x - 4)
2m
= 25x - 30(x - 2) - 15(x - 4) Equations of Slope and Elastic Curve. EI
d2v = M dx2
EI
d2v = 25x - 30(x - 2) - 15(x - 4) dx2
EI
dv = 12.5x2 - 15(x - 2)2 - 7.5(x - 4)2 + C1 dx
(1)
EIv = 4.1667x3 - 5(x - 2)3 - 2.5(x - 4)3 + C1x + C2
(2)
Boundary Conditions. At x = 0, v = 0. Then, Eq. (2) gives 0 = 0 - 0 - 0 + C1(0) + C2
C2 = 0
At x = 6 m, v = 0. Then Eq. (2) gives 0 = 4.1667 A 63 B - 5(6 - 2)3 - 2.5(6 - 4)3 + C1(6) + C2 C1 = -93.333 kN # m3 Substituting the value of C1 into Eq. (1), dv 1 = c12.5x2 - 15(x - 2)2 - 7.5(x - 4)2 - 93.333 d dx EI Assuming that
B
dv = 0 occurs in the region 2 m 6 x 6 4 m. Then dx
dv 1 = 0 = c12.5x2 - 15(x - 2)2 - 93.333 d dx EI 12.5x2 - 15(x - 2)2 - 93.333 = 0 2.5x2 - 60x + 153.333 = 0 Solving for the root 2 m 6 x 6 4 m, x = 2.9079 ft O.K.
934
2m
2m
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12–39. Continued Substituting the values of C1 and C2 into Eq. (2), v =
1 c4.1667x3 - 5(x - 2)3 - 2.5(x - 4)3 - 93.333x d EI
vmax occurs at x = 2.9079 m, where
Ans.
dv = 0. Thus, dx
vmax = v|x = 2.9079 ft =
1 c4.1667 A 2.90793 B - 5(2.9079 - 2)3 - 0 - 93.333(2.9079) d EI
= -
172.69 A 103 B 172.69kN # m3 = EI 200 A 109 B C 65.0 A 10 - 6 B D
= -0.01328 m = 13.3 mm T
Ans.
935
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*12–40. Determine the eqution of the elastic curve, the slope at A, and the deflection at B of the simply supported beam. EI is constant.
M0
M0
A B L 3
Support Reactions and Elastic Curve. As shown in Fig. a. Moment Function. M = -(-MO)a x = MO a x -
L 0 2 0 b - MO ax - L b 3 3
L 0 2 0 b - MO a x - L b 3 3
Equations of Slope and Elastic Curve. EI
d2v = M dx2
EI
0 d2v L 0 2 = M a x b M a x Lb O O 3 3 dx2
EI
dv L 2 = MO ax - b - MO a x - Lb + C1 dx 3 3
EIv
(1)
2 MO MO L 2 2 ax - b a x - Lb + C1x + C2 2 3 2 3
Boundary Conditions. Due to symmetry, EI(0) = MO a
L L - b - 0 + C1 2 3
(2)
dv L = 0 at x = . Then Eq. (1) gives dx 2 C1 = -
MOL 6
At x = 0, v = 0. Then, Eq. (2) gives EI(0) = 0 - 0 + C1(0) + C2
C2 = 0
Substituting the value of C1 into Eq. (1), 2 MO L 2 dv = B 6 ax - b - 6 ax - Lb - L R dx 6EI 3 3
At A, x = 0. Thus, uA =
MO MO L MOL dv = = C 6(0) - 6(0) - L D = ` dx x = 0 6EI 6EI 6EI
Ans.
Substituting the values of C1 and C2 into Eq. (2), v =
MO L 2 2 2 B 3 a x - b - 3 ax - L b - Lx R 6EI 3 3
At B, x =
Ans.
L . Thus, 3
vB = v|x = L3 =
= -
MO L B 3(0) - 3(0) - L a b R 6EI 3 MOL2 MOL2 = 18EI 18EI
Ans.
T
936
D
C L 3
L 3
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•12–41.
Determine the equation of the elastic curve and the maximum deflection of the simply supported beam. EI is constant.
M0
M0
A B
Support Reactions and Elastic Curve. As shown in Fig. a.
L 3
Moment Function. M = -(-MO)a x -
L 0 2 b - MO ax - L b 3 3
L 0 2 b - MO a x - L b 3 3
= MO a x -
Equations of Slope and Elastic Curve. EI
d2v = M dx2
EI
d2v L 0 2 = M a x b - MO a x - Lb O 3 3 dx2
EI
dv L 2 = MO ax - b - MO a x - Lb + C1 dx 3 3
EIv
(1)
2 MO MO L 2 2 ax - b a x - Lb + C1x + C2 2 3 2 3
Boundary Conditions. Due to symmetry, EI(0) = MO a
(2)
dv L = 0 at x = . Then Eq. (1) gives dx 2
L L - b - 0 + C1 2 3
C1 = -
MOL 6
At x = 0, v = 0. Then, Eq. (2) gives EI(0) = 0 - 0 + C1(0) + C2
C2 = 0
Substituting the values of C1 and C2 into Eq. (2), v =
2 MO L 2 2 B 3 a x - b - 3 ax - Lb - Lx R 6EI 3 3
vmax occurs at x =
L dv = 0. Then, , where 2 dx
vmax = v|x = L2 =
= -
Ans.
MO L L 2 L B3a - b - 0 - La b R 6EI 2 3 2
5MOL2 5MOL2 = T 72EI 72EI
Ans.
937
D
C L 3
L 3
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12–42. Determine the equation of the elastic curve, the slope at A, and the maximum deflection of the simply supported beam. EI is constant.
P
Moment Function.
= Px - Pax -
L 3
L 2 b - Pa x - Lb 3 3
L 2 b - Pax - Lb 3 3
Equations of Slope and Elastic Curve. EI
d2v = M dx2
EI
2 d2v L = Px - Pax - b - Pax - Lb 3 3 dx2
EI
2 P dv P L 2 P 2 = x2 ax - b a x - Lb + C1 dx 2 2 3 2 3
EIv =
(1)
3 P 3 P L 3 P 2 x ax - b a x - Lb + C1x + C2 6 6 3 6 3
Boundary Conditions. Due to symmetry,
EI(0) =
(2)
dv L = 0 at x = . Then Eq. (1) gives dx 2
P L 2 P L L 2 a b a - b - 0 + C1 2 2 2 2 3
C1 = -
PL2 9
At x = 0, v = 0. Then, Eq. (2) gives EI(0) = 0 - 0 - 0 + C1(0) + C2
C2 = 0
Substituting the value of C1 into Eq. (1), 2 P L 2 2 dv = B 9x2 - 9a x - b - 9ax - Lb - 2L2 R dx 18EI 3 3
At A, x = 0. Thus, uA =
dv P PL2 PL2 = = C 0 - 0 - 0 - 2L2 D = ` dx x = 0 18EI 9EI 9EI
Ans.
SubStituting the values of C1 and C2 into Eq. (2), v =
3 P L 3 2 B 3x3 - 3ax - b - 3ax - Lb - 2L2x R 18EI 3 3
vmax occurs at x =
= -
Ans.
L dv , where = 0. Then, 2 dx
vmax = v|x = L2 =
B
A
Support Reactions and Elastic Curve. As shown in Fig. a.
M = -(-P)(x - 0) - Pax -
P
P L 3 L L 3 L B 3 a b - 3a - b - 0 - 2L2 a b R 18EI 2 2 3 2
23PL3 23PL3 = T 648EI 648EI
Ans.
938
L 3
L 3
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12–43. Determine the maximum deflection of the cantilevered beam. The beam is made of material having an E = 200 GPa and I = 65.0(106) mm6.
15kN
30 kN/m
A
Support Reactions and Elastic Curve. As shown in Fig. a. Moment Function. From Fig. b, we obtain M = -(-37.5)(x - 0) - 67.5(x - 0)0 - a-
1.5 m
20 (x - 0)3 6
20 30 b (x - 1.5)3 - a - b(x - 1.5)2 6 2
= 37.5x - 67.5 -
10 3 10 x + (x - 1.5)3 + 15(x - 1.5)2 3 3
Equations of Slope and Elastic Curve. EI
d2v = M dx2
EI
d2v 10 3 10 = 37.5x - 67.5 x + (x - 1.5)3 + 15(x - 1.5)2 3 3 dx2
EI
dv 5 5 = 18.75x2 - 67.5x - x4 + (x - 1.5)4 + 5(x - 1.5)3 + C1 dx 6 6
EIv = 6.25x3 - 33.75x2 -
(1)
1 5 1 5 x + (x - 1.5)5 + (x - 1.5)4 + C1x + C2 (2) 6 6 4
Boundary Conditions. At x = 0,
dv = 0 Then Eq. (1) gives dx
0 = 0 - 0 - 0 + 0 + 0 + C1
C1 = 0
At x = 0, v = 0. Then Eq. (2) gives 0 = 0 - 0 - 0 + 0 + 0 + 0 + C2
C2 = 0
Substituting the values of C1 and C2 into Eq. (2), v =
1 1 1 5 c6.25x3 - 33.75x2 - x5 + (x - 1.5)5 + (x - 1.5)4 d EI 6 6 4
Ans.
vmax occurs at x = 3 m Thus vmax = v|x = 3 m =
1 1 1 5 c6.25 A 33 B - 33.75 A 32 B - A 35 B + (3 - 1.5)5 + (3 - 1.5)4 d EI 6 6 4
= -
167.91kN # m3 = EI
167.91 A 103 B
200 A 109 B c65.0 A 10 - 6 B d
= -0.01292 m = 12.9 mm T
Ans.
939
1.5 m
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*12–44. The beam is subjected to the load shown. Determine the equation of the elastic curve. EI is constant.
50 kN 3 kN/m
Support Reactions and Elastic Curve: As shown on FBD. Moment Function: Using discontinuity function,
B
A x
M = 24.6 6 x - 0 7 - 1.5 6 x - 0 7 2 - (-1.5) 6 x - 4 7 2
4m
- 50 6 x - 7 7 = 24.6x - 1.5x2 + 1.5 6 x - 47 2 - 50 6 x - 7 7 Slope and Elastic Curve: EI
EI EI
d2 y = M dx2
d2 y = 24.6x - 1.5x2 + 1.5 6 x - 47 2 - 50 6 x - 7 7 dx2
dy = 12.3x2 - 0.5x3 + 0.5 6 x - 4 7 3 - 25 6 x - 7 7 2 + C1 dx
[1]
EI y = 4.10x3 - 0.125x4 + 0.125 6 x - 4 7 4 - 8.333 6 x - 7 7 3 + C1x + C2
[2]
Boundary Conditions: y = 0 at x = 0. From Eq.[2], C2 = 0 y = 0 at x = 10 m. From Eq.[2], 0 = 4.10 A 103 B - 0.125 A 104 B + 0.125(10 - 4)4 - 8.333(10 - 7)3 + C1 (10) C1 = -278.7 The Elastic Curve: Substituting the values of C1 and C2 into Eq.[2], y =
1 {4.10x3 - 0.125x4 + 0.125 6 x - 4 7 4 EI - 8.33 6 x - 7 7 3 - 279x} kN # m3
Ans.
940
3m
3m
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•12–45.
The beam is subjected to the load shown. Determine the displacement at x = 7 m and the slope at A. EI is constant.
50 kN 3 kN/m
Support Reactions and Elastic Curve: As shown on FBD.
B
A
Moment Function: Using the discontinuity function.
x
2
M = 24.6 6 x - 0 7 - 1.5 6 x - 0 7 - (-1.5) 6 x - 4 7
4m
2
- 50 6 x - 7 7 = 24.6x - 1.5x2 + 1.5 6 x - 47 2 - 50 6 x - 7 7 Slope and Elastic Curve: EI
EI EI
d2 y = M dx2
d2 y = 24.6x - 1.5x2 + 1.5 6 x - 47 2 - 50 6 x - 7 7 dx2
dy = 12.3x2 - 0.5x3 + 0.5 6 x - 4 7 3 - 25 6 x - 7 7 2 + C1 dx
[1]
EI y = 4.10x3 - 0.125x4 + 0.125 6 x - 4 7 4 - 8.333 6 x - 7 7 3 + C1x + C2
[2]
Boundary Conditions: y = 0 at x = 0. From Eq.[2], C2 = 0 y = 0 at x = 10 m. From Eq.[2], 0 = 4.10 A 103 B - 0.125 A 104 B + 0.125(10 - 4)4 - 8.333(10 - 7)3 + C1 (10) C1 = -278.7 The Slope: Substituting the value of C1 into Eq.[1], 1 dy E 12.3x2 - 0.5x3 + 0.5 6 x - 4 7 3 - 25 6 x - 7 7 2 - 278.7 F kN # m2 = dx EI uA =
279kN # m2 dy 1 {0 - 0 + 0 - 0 - 278.7} = = ` dx x = 0 EI EI
Ans.
The Elastic Curve: Substituting the values of C1 and C2 into Eq.[2], y =
1 E 4.10x3 - 0.125x4 + 0.125 6 x - 4 7 4 - 8.33 6 x - 7 7 3 EI - 278.7x F kN # m3
y |x = 7 m =
1 E 4.10 A 73 B - 0.125 A 74 B + 0.125(7 - 4)4 - 0 - 278.7(7) F kN # m3 EI
= -
835 kN # m3 EI
Ans.
941
3m
3m
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12–46. Determine the maximum deflection of the simply supported beam. E = 200 GPa and I = 65.0(106) mm4.
20 kN
15 kN/m
Support Reactions and Elastic Curve. As shown in Fig. a. A
Moment Function. From Fig. b, we obtain M = -(-22.5)(x - 0) - 20(x - 1.5) -
15 5 (x - 3)2 - a - b(x - 3)3 2 6
= 22.5x - 20(x - 1.5) - 7.5(x - 3)2 +
1.5 m
5 (x - 3)3 6
Equations of Slope and Elastic Curve. EI
d2v = M dx2
EI
d2v 5 = 22.5x - 20(x - 1.5) - 7.5(x - 3)2 + (x - 3)3 2 6 dx
EI
dv 5 (x - 3)4 + C1 = 11.25x2 - 10(x - 1.5)2 - 2.5(x - 3)3 + dx 24
EIv = 3.75x3 -
(1)
10 1 (x - 1.5)3 - 0.625(x - 3)4 + (x - 3)5 + C1x + C2 (2) 3 24
Boundary Conditions. At x = 0, v = 0. Then, Eq. (2) gives 0 = 0 - 0 - 0 + 0 + C1(0) + C2
C2 = 0
At x = 6 m, v = 0. Then Eq. (2) gives 0 = 3.75 A 63 B -
10 1 (6 - 1.5)3 - 0.625(6 - 3)4 + (6 - 3)5 + C1(6) + C2 3 24
C1 = -77.625 kN # m2 Substituting the value of C1 into Eq. (1), dv 1 5 = c11.25x2 - 10(x - 1.5)2 - 2.5(x - 3)3 + (x - 3)4 - 77.625 d dx EI 24 Assuming that
dv = 0 occurs in the region 1.5 m 6 x 6 3 m, then dx
dv 1 = 0 = c11.25x2 - 10(x - 1.5)2 - 0 + 0 - 77.625 d dx EI Solving for the root 1.5 m 6 x 6 3 m, x = 2.970 m O.K. Substituting the values of C1 and C2 into Eq. (2), v =
1 10 1 c3.75x3 (x - 1.5)3 - 0.625(x - 3)4 + (x - 3)5 - 77.625x d Ans. EI 3 24
vmax occurs at x = 2.970 m, where
dv = 0. Thus, dx
942
1.5 m
3m
B
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12–46.
Continued
vmax = v ƒ x = 2.970 m =
1 10 c3.75 A 2.9703 B (2.970 - 1.5)3 - 0 + 0 - 77.625(2.970) d EI 3
= -
142.89kN # m3 = EI
142.89 A 103 B
200 A 109 B c65.0 A 10 - 6 B d
= -0.01099 m = 11.0 mm T
Ans.
943
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12–47. The wooden beam is subjected to the load shown. Determine the equation of the elastic curve. If Ew = 12 GPa, determine the deflection and the slope at end B. M = -63 6 x - 0 7 0 - (-16) 6 x - 0 7 -
6 kN
4 kN
2 kN/m A
B x
2 6 x - 0 72 2
3m
1.5 m
1.5 m
2 - a - b 6 x - 3 7 2 - 4 6 x - 4.5 7 2
400 mm
M = -63 + 16x - x2 + 6 x - 3 7 2 - 4 6 x - 4.5 7
200 mm
Elastic curve and Slope: EI
d2v = M = -63 + 16x - x2 + 6 x - 3 7 2 - 4 6 x - 4.5 7 dx2
EI
1 dv x3 = -63x + 8x2 + 6 x - 3 7 3 - 2 6 x - 4.5 7 2 + C1 dx 3 3
EIv = -31.5x2 +
(1)
1 8 3 x4 2 x + 6 x - 3 7 4 - 6 x - 4.5 7 3 3 12 12 3
+ C1x + C2
(2)
Boundary condition: dv = 0 dx
at
x = 0
From Eq. (1), C1 = 0 v = 0
x = 0
at
From Eq. (2), C2 = 0 1 x3 1 dv = c -63x + 8x2 + 6 x - 3 7 3 - 2 6 x - 4.5 7 2 d dx EI 3 3 v =
(3)
1 8 x4 1 c -31.5x2 + x3 + 6 x - 3 74 EI 3 12 12 -
2 6 x - 4.5 7 3 d kN # m3 (4) 3
Ans.
1 (0.20)(0.40)3 = 1.067 A 10 - 3 B m4 12
I =
At point B, x = 6m
uB =
-157.5 A 103 B dv -157.5 = = = -0.0123 rad = -0.705° Ans. ` dx x = 6m EI 12 A 103 B (1.067) A 10 - 3 B
The negative sign indicates clockwise rotation.
vB =
-661.5 A 103 B -661.5 = = -0.0517m = -51.7 mm EI 12 A 103 B (1.067) A 10 - 3 B
944
Ans.
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*12–48. The beam is subjected to the load shown. Determine the slopes at A and B and the displacement at C. EI is constant.
30 kN
A
The negative sign indicates downward displacement.
C
Moment Function: Using the discontinuity function,
3m
M = 66.75 6 x - 0 7 -6 6 x - 0 7 2 - 30 6 x - 3 7 = 66.75x - 6x2 - 30 6 x - 3 7 Slope and Elastic Curve:
EI EI
d2y = M dx2
d2y = 66.75x - 6x2 - 30 6 x - 3 7 dx2
dy = 33.375x2 - 2x3 - 15 6 x - 3 7 2 + C1 dx
[1]
EI y = 11.125x3 - 0.5x4 - 5 6 x - 3 7 3 + C1x + C2
[2]
Boundary Conditions: y = 0 at x = 0. From Eq.[2], C2 = 0 y = 0 at x = 8 m. From Eq.[2], 0 = 11.125 A 83 B - 0.5 A 84 B - 5(8 - 3)3 + C1 (8) C1 = -377.875 The Slope: Substituting the value of C1 into Eq.[1], 1 dy = E 33.375x2 - 2x3 - 15 6 x - 3 7 2 - 377.875 F kN # m2 dx EI uA =
dy 1 378 kN # m2 = {0 - 0 - 0 - 377.875} = ` dx x = 0 EI EI
uB =
dy ` dx x = 8 m
=
1 E 33.375 A 82 B - 2 A 83 B - 15(8 - 3)2 - 377.875 F EI
=
359 kN # m2 EI
Ans.
Ans.
The Elastic Curve: Substituting the values of C1 and C2 into Eq.[2], y =
1 E 11.125x3 - 0.5x4 - 5 6 x - 3 7 3 - 377.875 x F kN # m3 EI
yC = y|x = 3 m =
1 E 11.125 A 33 B - 0.5 A 34 B - 0 - 377.875(3) F EI
= -
B
x
Support Reactions and Elastic Curve: As shown on FBD.
EI
12 kN/m
874 kN # m3 EI
Ans.
945
5m
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•12–49.
Determine the equation of the elastic curve of the simply supported beam and then find the maximum deflection. The beam is made of wood having a modulus of elasticity E = 1.5(103) ksi.
600 lb 500 lb/ft 3 in.
A
B 6 ft
Support Reactions and Elastic Curve. As shown in Fig. a. Moment Function. From Fig. b, we obtain M = -(-2400)(x - 0) - 600(x - 9) -
500 500 (x - 0)2 - ¢ ≤ (x - 6) 2 2
= 2400x - 600(x - 9) - 250x2 + 250(x - 6)2 Equations of Slope and Elastic Curve. EI
d2v = M dx2
EI
d2v = 2400x - 600(x - 9) - 250x2 + 250(x - 6)2 dx2
EI
dv 250 3 250 = 1200x2 - 300(x - 9)2 x + (x - 6)3 + C1 dx 3 3
EIv = 400x3 - 100(x - 9)3 -
125 4 125 x + (x - 6)4 + C1x + C2 6 6
Boundary Conditions. At x = 0, v = 0. Then Eq.(2) gives C2 = 0 At x = 12ft, v = 0. Then Eq.(2) gives 0 = 400 A 12 3 B - 100(12 - 9)3 -
125 125 (12)4 + (12 - 6)4 + C1(12) 6 6
C1 = -23625 lb # ft2 Substituting the value of C1 into Eq.(1), dv 1 250 3 250 = x + (x - 6)3 - 23625 R B 1200x2 - 300(x - 9)2 dx EI 3 3 Assuming that
dv = 0 occurs in the region 0 6 x 6 6 ft. Then dx
dv 1 250 3 = 0 = c1200x2 x - 23625 d dx EI 3 1200x2 -
250 3 x - 23625 = 0 3
Solving x = 5.7126 ft O.K.
946
(1) (2)
3 ft
3 ft
6 in.
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•12–49.
Continued
Substituting the values of C1 and C2 into Eq.(2), v =
1 125 4 125 c400x3 - 100(x - 9)3 x + (x - 6)4 - 23625x d EI 6 6
vmax occurs at x = 5.7126 ft, where
Ans.
dv = 0. Thus, dx
vmax = v|x = 5.7126 ft =
1 125 c400 A 5.71263 B - 0 A 5.71264 B + 0 - 23625(5.7126) d EI 6
= -
82.577.41lb # ft3 = EI
82577.41 A 12 3 B
1.5 A 106 B c
1 (3) A 63 B d 12
= -1.76 in = 1.76 in T
Ans.
947
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12–50. The beam is subjected to the load shown. Determine the equations of the slope and elastic curve. EI is constant.
2 kN/m
8 kN⭈m
A B
Support Reactions and Elastic Curve: As shown on FBD. x
Moment Function: Using the discontinuity function,
5m
1 1 M = 0.200 6 x - 0 7 - (2) 6 x - 0 7 2 - ( -2) 6 x - 5 7 2 2 2 - (-17.8) 6 x - 5 7 = 0.200x - x2 + 6 x - 5 7 2 + 17.8 6 x - 5 7 Slope and Elastic Curve: EI
EI EI
d2y = M dx2
d2y = 0.200x - x2 + 6 x - 5 7 2 + 17.8 6 x - 5 7 dx2
dy = 0.100x2 - 0.3333x3 + 0.3333 6 x - 5 7 3 + 8.90 6 x - 5 7 2 + C1 [1] dx
EI y = 0.03333x3 - 0.08333x4 + 0.08333 6 x - 5 7 4 + 2.9667 6 x - 5 7 3 + C1x + C2
[2]
Boundary Conditions: y = 0 at x = 0. From Eq.[2], C2 = 0 y = 0 at x = 5 m. From Eq.[2], 0 = 0.03333 A 53 B - 0.08333 A 54 B + 0 + 0 + C1 (5) C1 = 9.5833 The Slope: Substituting the value of C1 into Eq.[1], dy 1 = E 0.100x2 - 0.333x3 + 0.333 6 x - 5 7 3 + 8.90 6 x - 5 7 2 dx EI + 9.58 F kN # m2
Ans.
948
3m
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12–51. The beam is subjected to the load shown. Determine the equation of the elastic curve. EI is constant. The Elastic Curve: Substituting the values of C1 and C2 into Eq.[2], y =
+ 9.58x F kN # m3
1.5 m
Ans.
6 6 6 x - 0 7 2 -(-1.25) 6 x - 1.5 7 -(- ) 6 x - 1.5 7 2 2 2
- (-27.75) 6 x - 4.5 7 M = -3x2 + 1.25 6 x - 1.5 7 +3 6 x - 1.5 7 2 + 27.75 6 x - 4.5 7 Elastic curve and slope: EI
d2v = M dx2 = -3x2 + 1.25 6 x - 1.5 7 + 3 6 x - 1.5 7 2 + 27.75 6 x - 4.5 7
EI
dv = -x3 + 0.625 6 x - 1.5 7 2 + 6 x - 1.5 7 3 dx + 13.875 6 x - 4.5 7 2 + C1
EIv = -0.25x4 + 0.208 6 x - 1.5 7 3 + 0.25 6 x - 1.5 7 4 + 4.625 6 x - 4.5 7 3 + C1x + C2
(1)
Boundary conditions: v = 0
at
x = 1.5 m
From Eq.(1) 0 = -1.266 + 1.5C1 + C2 1.5C1 + C2 = 1.266 v = 0
at
(2)
x = 4.5 m
From Eq.(1) 0 = -102.516 + 5.625 + 20.25 + 4.5C1 + C2 4.5C1 + C2 = 76.641
(3)
Solving Eqs. (2) and (3) yields: C1 = 25.12 C2 = -36.42 v =
Ans.
1 C -0.25x4 + 0.208 6 x - 1.5 7 3 + 0.25 6 x - 1.5 7 4 EI + 4.625 6 x - 4.5 7 3 + 25.1x - 36.4 D kN # m3
949
B
A
1 E 0.0333x3 - 0.0833x4 + 0.0833 6 x - 5 7 4 + 2.97 6 x - 5 7 3 EI
M = -
20 kN
6 kN/m
3m
1.5 m
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*12–52. The wooden beam is subjected to the load shown. Determine the equation of the elastic curve. Specify the deflection at the end C. Ew = 1.6(103) ksi.
0.8 kip/ft
1.5 kip
A
12 in.
C
B x 9 ft
M = -0.3 6 x - 0 7 -a-
1 1.6 a b 6 x - 0 7 3 -(-5.4) 6 x - 9 7 6 18
0.8 0.8 1 b 6 x - 9 72 - a b 6 x - 9 73 2 6 9
M = -0.3x - 0.0148x3 + 5.4 6 x - 9 7 +0.4 6 x - 9 7 2 + 0.0148 6 x - 9 7 3 Elastic curve and slope: EI
d2v = M = -0.3x - 0.0148x3 + 5.4 6 x - 9 7 +0.4 6 x - 9 7 2 dx2 +0.0148 6 x - 9 7 3
EI
dv = -0.15x2 - 0.003704x4 + 2.7 6 x - 9 7 2 + 0.1333 6 x - 9 7 3 dx + 0.003704 6 x - 9 7 4 + C1
EIv = -0.05x3 + 0.0007407x5 + 0.9 6 x - 9 7 3 + 0.03333 6 x - 9 7 4 + 0.0007407 6 x - 9 7 5 + C1x + C2
(1)
Boundary conditions: v = 0
at
x = 0
at
x = 9 ft
From Eq.(1) C2 = 0 v = 0 From Eq.(1) 0 = -36.45 - 43.74 + 0 + 0 + 0 + 9C1 C1 = 8.91 v =
1 C -0.05x3 - 0.000741x5 + 0.9 6 x - 9 7 3 + 0.0333 6 x - 9 7 4 EI + 0.000741 6 x - 9 7 5 + 8.91x D kip # ft3
Ans.
At point C, x = 18 ft
vC =
-612.29 A 12 3 B -612.29kip # ft3 = = -0.765 in. 1 EI 1.6 A 103 B A 12 B (6) A 12 3 B
Ans.
The negative sign indicates downward displacement.
950
9 ft
6 in.
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12–53. Determine the displacement at C and the slope at A of the beam.
8 kip/ft
Support Reactions and Elastic Curve: As shown on FBD.
C
Moment Function: Using the discontinuity function, M = -
x
1 1 8 (8) 6 x - 0 7 2 - a - b 6 x - 6 7 3 - (-88) 6 x - 6 7 2 6 9
= -4x2 +
6 ft
4 6 x - 6 7 3 + 88 6 x - 6 7 27
Slope and Elastic Curve: EI
EI EI
d2y = M dx2
d 2y 4 = -4x2 + 6 x - 6 7 3 + 88 6 x - 6 7 27 dx2
dy 4 1 = - x3 + 6 x - 6 7 4 + 44 6 x - 6 7 2 + C1 dx 3 27
1 1 44 EI y = - x4 + 6 x - 6 75 + 6 x - 6 7 3 + C1x + C2 3 135 3
[1] [2]
Boundary Conditions: y = 0 at x = 6 ft. From Eq.[2], 0 = -
1 4 A 6 B + 0 + 0 + C1 (6) + C2 3 432 = 6C1 + C2
[3]
y = 0 at x = 15 ft. From Eq.[2], 0 = -
1 1 44 (15 - 6)3 + (15 - 6)3 + C1 (15) + C2 A 154 B + 3 135 3 5745.6 = 15C1 + C2
[4]
Solving Eqs. [3] and [4] yields, C1 = 590.4
C2 = -3110.4
The Slope: Substitute the value of C1 into Eq.[1], dy 1 1 4 = 6 x - 6 7 4 + 44 6 x - 6 7 2 + 590.4 r kip # ft2 b - x3 + dx EI 3 27
uA =
302 kip # ft2 dy 1 4 = b - A 63 B + 0 + 0 + 590.4 r = ` dx x = 6 ft EI 3 EI
Ans.
The Elastic Curve: Substitute the values of C1 and C2 into Eq. [2], y =
1 1 1 44 6 x - 6 75 + 6 x - 6 73 b - x4 + EI 3 135 3 + 590.4x - 3110.4 r kip # ft3
yC = y |x = 0 =
B A
3110kip # ft3 1 {-0 + 0 + 0 + 0 - 3110.4} kip # ft3 = EI EI
951
Ans.
9 ft
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12–54. The beam is subjected to the load shown. Determine the equation of the elastic curve. EI is constant.
6 kip/ft B A x 9 ft
M = -
1 16 1 10 a b 6 x - 0 7 3 -(-77.4) 6 x - 9 7 - a - b a b 6 x - 9 7 3 6 24 6 15
M = -0.1111x3 + 77.4 6 x - 9 7 +0.1111 6 x - 9 7 3 Elastic curve and slope: EI
d2v = M = -0.1111x3 + 77.4 6 x - 9 7 +0.1111 6 x - 9 7 3 dx2
EI
dv = -0.02778x4 + 38.7 6 x - 9 7 2 + 0.02778 6 x - 9 7 4 + C1 dx
EIv = -0.005556x5 + 12.9 6 x - 9 7 3 + 0.005556 6 x - 9 7 5 + C1x + C2
(1)
Boundary conditions: v = 0
at
x = 9 ft
From Eq.(1) 0 = -328.05 + 0 + 0 + 9C1 + C2 9C1 + C2 = 328.05 v = 0
at
(2)
x = 24 ft
0 = -44236.8 + 43537.5 + 4218.75 + 24C1 + C2 24C1 + C2 = -3519.45
(3)
Solving Eqs. (2) and (3) yields, C1 = -256.5 C2 = 2637 v =
1 C -0.00556x5 + 12.9 6 x - 9 7 3 + 0.00556 6 x - 9 7 5 EI - 265.5x + 2637 D kip # ft3
Ans.
952
15 ft
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12–55. Determine the slope and deflection at C. EI is constant.
15 kip
A C B 30 ft
|tB>A|
uA =
30 -33 750 1 -225 a b(30)(10) = 2 EI EI
tB>A =
1125 EI
uA = uC>A =
1 -225 -5062.5 5062.5 1 -225 a b(30) + a b(15) = = 2 EI 2 EI EI EI
uC = uC>A + uA uC =
5062.5 1125 3937.5 = EI EI EI
¢ C = |tC>A| tC>A =
¢C =
Ans.
45 |t | 30 B>A
1 -225 1 225 101 250 ab(30)(25) + a b(15)(10) = 2 EI 2 EI EI 45 33 750 50 625 101.250 b = a EI 30 EI EI
Ans.
953
15 ft
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*12–56. Determine the slope and deflection at C. EI is constant.
10 kN
A C B 6m
Referring to Fig. b, |uC>A| =
1 30 135 kN # m2 a b A9B = 2 EI EI
|tB>A| =
6 1 30 180 kN # m3 c a b A6B d = 3 2 EI EI
|tC>A| = a =
6 1 30 2 1 30 + 3b c a b A 6 B d + c (3) d c a b A3B d 3 2 EI 3 2 EI
540 kN # m3 EI
From the geometry shown in Fig. b, uA =
|tB>A| =
6
180>EI 30 kN # m2 = 6 EI
Here, + buC = uA + uC>A uC = uC =
135 30 + EI EI
105 kN # m2 EI
uC
Ans.
9 yC = 2 tC>A 2 - 2 tB>A 2 a b 6 =
540 180 9 a b EI EI 6
=
270 kN # m3 T EI
Ans.
954
3m
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•12–57. Determine the deflection of end B of the cantilever beam. E is constant.
P
P
B
A L 2
Support Reactions and
M Diagram. As shown in Fig. a. EI
Moment Area Theorem. Since A is a fixed support, uA = 0. Referring to the geometry of the elastic curve, Fig. b, uB = |uB>A| =
=
1 3PL PL L 1 PL L + B R¢ ≤ + B R¢ ≤ 2 2EI 2EI 2 2 2EI 2
5PL2 8 EI
Ans.
¢ B = |tB>A| - ¢
=
7PL3 16EI
3L PL L L 5L 1 PL L L 1 PL ≤¢ ≤¢ ≤ + B ¢ ≤¢ ≤R + B ¢ ≤¢ ≤R 4 2EI 2 6 2 EI 2 3 2 2EI 2 Ans.
T
955
L 2
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12–58. Determine the slope at A and the maximum deflection. EI is constant.
20 kip⭈ft
20 kip⭈ft
A 6 ft
Point D is located at the mid span of the beam. Due to symmetry, the slope at D is zero. Referring to Fig. b, |uD>A| = a
120 kip # ft2 20 b(6) = EI EI
|tD>A| = 3 a
360 kip # ft3 20 b A6B = EI EI
|tC>D| = 6a
1440 kip # ft3 20 b A 12 B = EI EI
From the geomtry shown in Fig. b uA = |uD>A| =
120 kip # ft2 EI
uA
Ans.
y D = uA(6) - |tD>A| =
360 120 (6) EI EI
=
360 kip # ft3 EI
c
yC = |tC>D| - 4D =
1440 360 EI EI
=
1080 kip # ft EI
T (max)
Ans.
956
C
B 12 ft
6 ft
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12–59. Determine the slope and deflection at C. EI is constant.
20 kip⭈ft
20 kip⭈ft
A
Referring to Fig. b, |uC>A| = a
360 kip # ft2 20 b A 18 B = EI EI
|uB>A| = 6a |tC>A| = 9 a
6 ft
b
1440 kip # ft3 20 b A 12 B = = EI EI
3240 kip # ft3 20 b A 18 B = EI EI
From the geometry shown in Fig. b uA =
|tB>A| 12
=
1440>EI 12
=
120 kip # ft2 EI
uA
Here, + b uC = uA + uC>A uC = -
uC =
360 120 + EI EI
240 kip # ft2 EI
yC = |tC>A| - |tB>A| a
uC
Ans.
18 b 12
=
1440 18 3240 a b EI EI 12
=
1080 kip # ft3 T EI
Ans.
957
C
B 12 ft
6 ft
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*12–60. If the bearings at A and B exert only vertical reactions on the shaft, determine the slope at A and the maximum deflection of the shaft. EI is constant.
50 lb⭈ft
B C 2 ft
Point E is located at the mid span of the shaft. Due to symmetry, the slope at E is zero. Referring to Fig. b, |uE>A| =
100 lb # ft2 50 (2) = EI EI
|tE>A| = (1)a
100 lb # ft3 50 b (2) = EI EI
Here, uA = |uE>A| =
100 lb # ft2 EI
uA
Ans.
ymax = uA (4) - |tE>A| =
100 100 (4) EI EI
=
300 lb # ft3 EI
50 lb⭈ft
A
Ans.
c
958
D 4 ft
2 ft
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•12–61.
Determine the maximum slope and the maximum deflection of the beam. EI is constant.
M0
B
A
M0L M0 L = a b = EI 2 2EI
uC>A
M0
L
uC = uC>A + uA 0 =
M0 L + uA 2EI
umax = uA =
M0 L -M0 L = 2EI 2EI
¢ max = |tB>C| =
Ans.
M0 L2 M0 L L a ba b = EI 2 4 8EI
Ans.
12–62. Determine the deflection and slope at C. EI is constant.
A
B
C M0
tB>A
M0L2 1 -M0 1 = a b(L)a b(L) = 2 EI 3 6EI
L
¢ C = |tC>A| - 2|tB>A| -M0 7M0L2 L L 1 -M0 a b(L)aL + b + a b (L) a b = 2 EI 3 EI 2 6EI
tC>A =
¢C =
uA =
7M0 L2 M0L2 5M0L2 - (2) a b = 6EI 6EI 6EI |tB>A| L
uC>A =
=
Ans.
M0L 6EI
M0 M0 3M0L 3M0L 1 ab(L) + a b(L) = = 2 EI EI 2EI 2EI
uC = uC>A + uA uC =
3M0L M0L 4M0L = 2EI 6EI 3EI
Ans.
959
L
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12–63. Determine the slope at A of the overhang beam. E = 200 GPa and I = 45.5(106) mm4.
30 kN 30 kN⭈m
M Support Reactions and Diagram. As shown in Fig. a. EI
A
Moment Area Theorem. Referring to Fig. b,
4m
1 30 30 1 1 |tB>A| = c (4) d B ¢ (4) R ≤ (4) R + c (4) d B 3 2 EI 2 EI
=
320 kN # m3 EI
From the geometry of the elastic curve, Fig. b, uA =
=
|tB>A| LAB
=
320>EI 80 kN # m2 = 4 EI
80 A 103 B
200 A 109 B C 45.5 A 10 - 6 B D
C B
Ans.
= 0.00879 rad
960
2m
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*12–64. Determine the deflection at C of the overhang beam. E = 200 GPa and I = 45.5(106) mm4.
30 kN 30 kN⭈m
1 1 30 1 30 |tB>A| = c (4) d B ¢ (4) R ≤ (4) R + c (4) d B 3 2 EI 2 EI
=
A
320 kN # m3 EI
4m
1 30 30 1 1 |tC>A| = c (4) + 2 d B ¢ (4) R ≤ (4) R + c (4) + 2 d B 3 2 EI 2 EI 1 60 2 + c (2) d B ¢ ≤ (2) R 3 2 EI
=
760 kN # m3 EI
¢ C = |tC>A| - |tB>A ¢
C B
L ≤ LAB
=
760 320 6 ¢ ≤ EI EI 4
=
280 A 103 B 280 kN # m3 = EI 200 A 109 B C 45.5 A 10 - 6 B D
= 0.03077 m = 30.8 mm T
Ans.
961
2m
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•12–65. Determine the position a of roller support B in terms of L so that the deflection at end C is the same as the maximum deflection of region AB of the overhang beam. EI is constant.
P L C
A B
M Support Reactions and Diagram. As shown in Fig. a. EI Moment Area Theorem. Referring to Fig. b, |tB>A| =
Pa2(L - a) a 1 P(L - a) B ¢ ≤ (a) R = 3 2 EI 6EI
|tC>A| = aL -
=
a
2(L - a) 1 P(L - a) 1 P(L - a) 2 ab B ¢ ≤ (a) R + B ¢ ≤ (L - a) R 3 2 EI 3 2 EI
P(L - a) A 2L2 - aL B 6EI
From the geometry shown in Fig. b, ¢ C = |tC>A| -
=
=
uA =
|tB>A| a
L
PL(L - a) A 2L - a B 6EI
-
Pa2 (L - a) L ¢ ≤ a 6EI
PL(L - a)2 3EI |tB>A| a
Pa2(L - a) Pa(L - a) 6EI = = a 6EI
The maximum deflection in region AB occurs at point D, where the slope of the elastic curve is zero (uD = 0). Thus, |uD>A| = uA Pa(L - a) 1 P(L - a) x R (x) = B 2 EIa 6EI x =
23 a 3
Also, ¢ D = |t4>D| = a
23Pa2(L - a) 2 23 1 P(L - a) 23 23 ab B c a ab d R a ab = 9 2 EIa 3 3 27EI
It is required that ¢C = ¢D PL(L - a)2 23Pa2(L - a) = 3EI 27EI 23 2 a + La - L2 = 0 9 Solving for the positive root, a = 0.858L
Ans.
962
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12–66. Determine the slope at A of the simply supported beam. EI is constant. Support Reactions and
P
M Diagram. As shown in Fig. a. EI
A
Moment Area Theorem.
2L 3
1 2PL 2 2 1 2PL L 5 tB>A = a Lb c a b a L b d + Lc a ba bd 9 2 9EI 3 9 2 9EI 3 =
4PL3 81EI
Referring to the geometry of the elastic curve, Fig. b,
uA
B
4PL3 |tB>A| 81EI 4PL2 = = = L L 81EI
Ans.
963
L 3
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12–67. The beam is subjected to the load P as shown. Determine the magnitude of force F that must be applied at the end of the overhang C so that the deflection at C is zero. EI is constant.
F P B
A
C 3
3
tB>A =
1 Fa 2 2Fa 1 Pa Pa a b(2a)(a) + a b(2a)a ab = 2 2EI 2 EI 3 2EI 3EI
tC>A =
1 -Fa 2a 1 -Fa 2a 1 Pa a b(2a)(2a) + a b(2a)a a + b + a b(a) a b 2 2EI 2 EI 3 2 EI 3
=
a
a
a
2Fa3 Pa3 EI EI
¢ C = tC>A -
3 t = 0 2 B>A
Pa3 2Fa3 3 Pa3 2Fa3 - a b = 0 EI EI 2 2EI 3EI F =
P 4
Ans.
M0 ⫽ Pa
*12–68. If the bearings at A and B exert only vertical reactions on the shaft, determine the slope at A and the maximum deflection.
tB>A
uA =
A C a
a Pa 17Pa3 1 Pa b(a)a3a + b + a b (2a)(a + a) = = a 2 EI 3 EI 3EI |tB>A| 4a
=
B
Ans.
Assume ¢ max is at point E located at 0 6 x 6 2a uE>A =
1 Pa Pa Pa2 Pax a b(a) + a b(x) = + 2 EI EI 2EI EI
uE = 0 = uE>A + uA 0 =
Pax -17Pa2 Pa2 + + a b 2EI EI 12EI
x =
11 a 12
¢ max = |tB>E| = a
(2a Pa 11 b a 2a ab c EI 12 2
11 12 a)
+ ad =
2a P
17Pa2 12EI
481Pa3 288EI
964
Ans.
D a
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•12–69.
The beam is subjected to the loading shown. Determine the slope at A and the displacement at C. Assume the support at A is a pin and B is a roller. EI is constant.
P
Support Reactions and Elastic Curve: As shown.
P
A
C a
M/EI Diagram: As shown.
P
a
B a
a
Moment - Area Theorems: Due to symmetry, the slope at midspan (point C) is zero. Hence the slope at A is uA = uA>C =
=
1 3Pa 3Pa 1 Pa a b(a) + a b(a) + a b(a) 2 2EI 2EI 2 2EI 5Pa2 2EI
Ans.
The displacement at C is ¢ C = tA>C =
=
2a 3Pa a 1 Pa 2a 1 3Pa a b (a)a b + a b aa + b + a b(a)a a + b 2 2EI 3 2EI 2 2 2EI 3 19Pa3 T 6EI
Ans.
12–70. The shaft supports the gear at its end C. Determine the deflection at C and the slopes at the bearings A and B. EI is constant.
A
L –– 2
1 -PL L L -PL3 = a ba ba b = 2 2EI 2 6 48EI
tB>A
L -PL3 1 -PL a b(L)a b = 2 2EI 2 8EI
tC>A =
L ¢ C = |tC>A| - a L b|tB>A| 2
=
uA =
PL3 PL3 PL3 - 2a b = 8EI 48EI 12EI |tB>A| L 2
uB>A =
=
PL3 48 EI L 2
=
Ans.
PL2 24EI
Ans.
1 -PL L -PL2 PL2 a ba b = = 2 2EI 2 8EI 8EI
uB = uB>A + uA uB =
B
PL2 PL2 PL2 = 8EI 24EI 12EI
Ans.
965
C L –– 2
P
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12–71. The shaft supports the gear at its end C. Determine its maximum deflection within region AB. EI is constant. The bearings exert only vertical reactions on the shaft.
A
B
L –– 2
uD>A =
C L –– 2
P
tB>A
A L2 B
1 Px a bx = 2 EI
PL A B A 2EI B A 13 B A L2 B
1 L 2 2
A L2 B
;
x = 0.288675 L
¢ max =
2 1 P(0.288675 L) a b (0.288675 L)a b(0.288675 L) 2 EI 3
¢ max =
0.00802PL3 EI
Ans.
*12–72. Determine the value of a so that the displacement at C is equal to zero. EI is constant.
P
P A
C
B
Moment-Area Theorems: (¢ C)1 = (tA>C)1 =
(tB>A)2 =
a
1 Pa 2 PaL2 ab(L)a Lb = 2 EI 3 3EI
(tC>A)2 = a -
(¢ C)2 =
1 PL L L PL3 a ba ba b = 2 4EI 2 3 48EI
Pa Pa L L 1 L L 5PaL2 b¢ ≤¢ ≤ + ab¢ ≤¢ ≤ = 2EI 2 4 2 2EI 2 3 48EI
PaL2 1 1 PaL2 5PaL2 |(tB>A)2| - |(tC>A)2| = ¢ = ≤ 2 2 3EI 48EI 16EI
Require, ¢ C = 0 = (¢ C)1 - (¢ C)2 0 =
PaL2 PL3 48EI 16EI
a =
L 3
Ans.
966
L 2
L 2
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•12–73.
The shaft is subjected to the loading shown. If the bearings at A and B only exert vertical reactions on the shaft, determine the slope at A and the displacement at C. EI is constant.
A
B C a
M/EI Diagram: As shown. Moment-Area Theorems: M0 M0 1 a 1 a ab(a)a b + a b(a)aa + b 2 EI 3 2 EI 3
tB>A =
= -
5M0 a2 6EI
M0 M0 a2 a 1 ab(a)a b = 2 EI 3 6EI
tC>A =
The slope at A is
uA =
5M0a 2
|tB>A| L
=
6EI
2a
=
5M0 a 12EI
Ans.
The displacement at C is, ¢C = `
1 t ` - |tC>A| 2 B>A
=
M0 a2 1 5M0 a2 ¢ ≤ 2 6EI 6EI
=
M0 a2 4EI
M0
M0
Ans.
c
967
a
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12–74. Determine the slope at A and the maximum deflection in the beam. EI is constant.
12 kip 24 kip⭈ft
A
B
6 ft
Here, tB>A = 20 c
=
1 12 36 1 24 1 60 a b(6) d + 12 c (12) d + 10 c a b(12) d + 4c a b(6) d 2 EI EI 2 EI 2 EI
8064 kip # ft3 EI
From the geometry of the elastic curve diagram, Fig. b, uA =
tB>A =
L
8064>EI 336 kip # ft2 = uA 24 EI
Ans.
Assuming that the zero slope of the elastic curve occurs in the region 6ft 6 x = 18ft such as point C where the maximum deflection occurs, then uC>A = uA 1 12 36 1 2x 336 a b (6) + a bx + a b(x) = 2 EI EI 2 EI EI x2 + 36x - 300 = 0 Solving for the root 0 6 x 6 12 ft, x = 6.980 ft O.K. Thus, ymax = tA>C = 4 c
=
1 12 36 1 a b(6) d + 9.490 c (6.980) d + 10.653 c (13.960)(6.980) d 2 EI EI 2
3048 kip # ft3 T EI
Ans.
968
12 ft
6 ft
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12–75. The beam is made of a ceramic material. In order to obtain its modulus of elasticity, it is subjected to the elastic loading shown. If the moment of inertia is I and the beam has a measured maximum deflection ¢ , determine E. The supports at A and D exert only vertical reactions on the beam.
P
P B
C
A
D a
a L
Moment-Area Theorems: Due to symmetry, the slope at midspan (point E) is zero. Hence the maximum displacement is, ¢ max = tA>E = a =
L - 2a 1 Pa 2 Pa L - 2a ba b aa + b + a b(a)a ab EI 2 4 2 EI 3
Pa A 3L2 - 4a2 B 24EI
Require, ¢ max = ¢ , then, ¢ =
Pa A 3L2 - 4a2 B 24EI
E =
Pa A 3L2 - 4a2 B 24¢I
Ans.
*12–76. The bar is supported by a roller constraint at B, which allows vertical displacement but resists axial load and moment. If the bar is subjected to the loading shown, determine the slope at A and the deflection at C. EI is constant.
P C A
B
L — 2
uA>B =
1 PL L PL L 3PL2 a ba b + a b = 2 2EI 2 2EI 2 8EI
uA = uA>B uA =
3PL2 8EI
Ans.
tA>B =
1 PL L L PL L L L 11PL3 a ba ba b + a ba + b = 2 2EI 2 3 2EI 2 2 4 48EI
tC>B =
PL L L PL3 a ba b = 2EI 2 4 16EI
¢ C = tA>B - tC>B =
11PL3 PL3 PL3 = 48EI 16EI 6EI
Ans.
969
L — 2
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•12–77. The bar is supported by the roller constraint at C, which allows vertical displacement but resists axial load and moment. If the bar is subjected to the loading shown, determine the slope and displacement at A. EI is constant.
P
B C
A
Support Reactions and Elastic Curve: As shown. a
M/EI Diagram: As shown.
2a
Moment-Area Theorems: uA>C = a -
Pa 1 Pa 5Pa2 b (2a) + a b(a) = EI 2 EI 2EI
tB>C = a -
Pa 2Pa3 b(2a)(a) = EI EI
tA>C = a -
1 2 Pa Pa 13Pa3 b(2a)(2a) + a b (a)a ab = EI 2 EI 3 3EI
Due to the moment constraint, the slope at support C is zero. Hence, the slope at A is uA = |uA>C| =
5Pa2 2EI
Ans.
and the displacement at A is ¢ A = |tA>C| - |tB>C| 2Pa3 7Pa3 13Pa3 = 3EI EI 3EI
=
Ans.
T
12–78. The rod is constructed from two shafts for which the moment of inertia of AB is I and of BC is 2I. Determine the maximum slope and deflection of the rod due to the loading. The modulus of elasticity is E.
P
L 2
uA>C =
1 -PL L -PL L -5PL2 5PL2 1 -PL L a ba b + a ba b + a ba b = = 2 2EI 2 2 4EI 2 4EI 2 16EI 16EI
uA = uA>C + uC umax = uA =
5PL2 5PL2 + 0 = 16EI 16EI
Ans.
¢ max = ¢ A = |tA>C| = `
1 -PL L L 1 -PL L L L a ba ba b + a ba ba + b 2 2 EI 2 3 2 4EI 2 2 3 + a
=
-PL L L L ba ba + b ` 4EI 2 2 4
3PL3 16EI
Ans.
970
C
B
A
L 2
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12–79. Determine the slope at point D and the deflection at point C of the simply supported beam. The beam is made of material having a modulus of elasticity E. The moment of inertia of segments AB and CD of the beam is I, while the moment of inertia of segment BC of the beam is 2I.
A L 4
Moment Area Theorem. Referring to Fig. b,
=
tC>D =
=
L L PL L 5L 1 PL L L 1 PL c a ba bd + c a bd + c a ba bd 6 2 4EI 4 2 8EI 2 6 2 4EI 4 PL3 16EI L 1 PL L c a ba bd 12 2 4EI 4 PL3 384EI
From the geometry of Fig. b,
uD =
|tA>D| L
PL3 PL2 18EI = = L 16EI
¢ C + tC>D =
Ans.
tA>D 4
PL3 PL 16EI = = 384EI 4 3
¢C
¢C =
D C
B
M Support Reactions and Diagram. As shown in Fig. a. EI
tA>D =
P
P
5PL3 384EI
Ans.
971
L 2
L 4
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*12–80. Determine the slope at point A and the maximum deflection of the simply supported beam. The beam is made of material having a modulus of elasticity E. The moment of inertia of segments AB and CD of the beam is I, while the moment of inertia of segment BC is 2I.
A L 4
Moment Area Theorem. Due to symmetry, the slope at the midspan of the beam, i.e., point E, is zero (uE = 0). Thus the maximum deflection occurs here. Referring to the geometry of the elastic curve, Fig. b,
=
1 PL L PL L a ba b + a b 2 4EI 4 8EI 4
PL2 16EI
¢ max = ¢ E = |tA>E| =
=
D C
B
M Support Reactions and Diagram. As shown in Fig. a. EI
uA = |uA>E| =
P
P
Ans. L 1 PL L 3 PL L Lc a bd + c a ba bd 8 8EI 4 6 2 4EI 4
13PL3 T 768EI
Ans.
972
L 2
L 4
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•12–81. Determine the position a of roller support B in terms of L so that deflection at end C is the same as the maximum deflection of region AB of the simply supported overhang beam. EI is constant.
Support Reactions and
A B a L
M Diagram. As shown in Fig. a. EI
Moment Area Theorem. Referring to Fig. b, |tB>A| =
MOa2 a 1 MO c a b (a) d = 3 2 EI 6EI
|tC>A| = aL =
2 1 MO L - a MO ab c a b (a) d + a bc (La) d 3 2 EI 2 EI
MO 2 A a + 3L2 - 3La B 6EI
From the geometry shown in Fig. b, ¢ C = |tC>A| -
uA
|tB>A| a
L
=
MO 2 MOa2 L a b A a + 3L2 - 3La B 6EI 6EI a
=
MO 2 A a + 3L2 - 4La B 6EI
MOa2 |tB>A| MO a 6EI = = = a a 6EI
The maximum deflection in region AB occurs at point D, where the slope of the elastic curve is zero (uD = 0). Thus, |uD>A| = uA MOa 1 MO a b (x)2 = 2 EIa 6EI x =
23 a 3
Also, ¢ D = |tA>D| =
23MOa2 2 23 1 MO 23 23 a≤ B a b¢ a≤ R ¢ a≤ = ¢ 3 3 2 EIa 3 3 27EI
It is required that ¢C = ¢D MO 2 23MO a2 A a + 3L2 - 4La B = 6EI 27EI 0.6151a2 - 4La + 3L2 = 0 Solving for the root 6 L, a = 0.865L
Ans.
973
C
M0
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12–82. The W10 * 15 cantilevered beam is made of A-36 steel and is subjected to the loading shown. Determine the slope and displacement at its end B.
3 kip/ft
B A
Here, uB = 冷uB>A冷 =
1 54 a b (6) 3 EI
=
108 kip # ft2 EI
6 ft
uC
For W 10 * 15 I = 68.9 in4, and for A36 steel E = 29.0 A 103 B ksi. Thus uB =
108 A 12 2 B
29 A 103 B (68.9)
= 0.00778 rad uB
Ans.
1 54 3 b (6) d yB = 冷 tB>A冷 = c (6) + 6 d c a 4 3 EI =
=
1134 kip # ft3 EI 1134 A 12 3 B
29 A 103 B (68.9)
= 0.981 in.
Ans.
T
974
6 ft
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12–83. The cantilevered beam is subjected to the loading shown. Determine the slope and displacement at C. Assume the support at A is fixed. EI is constant.
P w
Support Reactions and Elastic Curve: As shown.
A
C
B a
M/EI Diagrams: The M/EI diagrams for the uniform distributed load and concentrated load are drawn separately as shown.
a
Moment-Area Theorems: The slope at support A is zero. The slope at C is 2Pa 1 wa 2 1 ab (2a) + a b(a) 2 EI 3 2EI
uC = 冷uC>A冷 =
a2 (12P + wa) 6EI
=
Ans.
The displacement at C is ¢ C = 冷tC>A 冷 =
=
2Pa 4 1 wa 2 3 1 ab(2a) a ab + a b(a)a a + a b 2 EI 3 3 2EI 4 a3 (64P + 7wa) 24EI
Ans.
T
*12–84. Determine the slope at C and deflection at B. EI is constant.
w C A
B
Support Reactions and Elastic Curve: As shown.
a
M/EI Diagram: As shown. Moment-Area Theorems: The slope at support A is zero. The slope at C is uC = 冷uC>A冷 =
=
1 wa2 wa2 ab(a) + a b (a) 2 EI 2EI wa3 EI
Ans.
The displacement at B is ¢ B = 冷tB>A冷 =
1 wa2 2 wa2 a 1 wa2 3 ab(a)aa + ab + a b(a)a a + b + a b(a)a ab 2 EI 3 2EI 2 3 2EI 4
=
41wa4 24EI
Ans.
T
975
a
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•12–85.
Determine the slope at B and the displacement at C. The member is an A-36 steel structural tee for which I = 76.8 in4.
5 kip 1.5 kip/ft
B
A
C 3 ft
3 ft
Support Reactions and Elastic Curve: As shown. M/EI Diagrams: The M/EI diagrams for the uniform distributed load and concentrated load are drawn separately as shown. Moment-Area Theorems: Due to symmetry, the slope at midspan C is zero. Hence the slope at B is uB = 冷uB>C冷 =
=
2 6.75 1 7.50 a b (3) + a b (3) 2 EI 3 EI 24.75 kip # ft2 EI 24.75(144)
=
29.0 A 103 B (76.8)
= 0.00160 rad
Ans.
The dispacement at C is ¢ C = 冷tA>C冷 =
=
2 2 6.75 5 1 7.50 a b (3)a b (3) + a b (3) a b(3) 2 EI 3 3 EI 8 47.8125 kip # ft3 EI 47.8125(1728)
=
29.0 A 103 B (76.8)
= 0.0371 in.
Ans.
T
12–86. The A-36 steel shaft is used to support a rotor that exerts a uniform load of 5 kN兾m within the region CD of the shaft. Determine the slope of the shaft at the bearings A and B. The bearings exert only vertical reactions on the shaft.
uE>A =
5 kN/m A
1 75 4.6875 2 3.5156 4.805 a b(0.1) + a b(0.15) + a b(0.15) = 2 EI EI 3 EI EI
uA = uE>A =
4.805 = EI
4.805 = 0.00306rad = 0.175° 200 (109)(0.01)4
976
B
C 20 mm 100 mm
Ans.
40 mm 300 mm
D 20 mm 100 mm
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12–87. The W12 * 45 simply supported beam is made of A-36 steel and is subjected to the loading shown. Determine the deflection at its center C.
12 kip 50 kip⭈ft B
A C 12 ft
A ¢C B 1 = ¢ 2 (x) =
12 A 24 3 B 3456 PL3 = = T 48EI 48EI EI Mx A L2 - x2 B 6LEI
At point C, x =
A ¢C B 2 = =
L 2
MN A L2 B 6LEI
A L2 - A L2 B 2 B
50 A 24 2 B 1800 ML2 = = T 16EI 16EI EI
¢C = A ¢C B 1 + A ¢C B 2 = 5256(1728)
=
29 A 103 B (350)
3456 1800 5256 + = EI EI EI
= 0.895 in. T
Ans.
977
12 ft
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*12–88. The W10 * 15 cantilevered beam is made of A-36 steel and is subjected to the loading shown. Determine the displacement at B and the slope at B.
6 kip
4 kip
A
B 6 ft
Using the table in appendix, the required slopes and deflections for each load case are computed as follow: (¢ B)1 =
(uB)1 =
(¢ B)2 =
(uB)2 =
5(4) A 12 3 B 720 kip # in.3 5PL3 = = T 48EI 48EI EI 4 A 12 2 B 72 kip # in.2 PL2 = = 8EI 8EI EI
(uB)1
6 A 12 3 B 3456 kip # in.3 PL3 = = T 3EI 3EI EI 6 A 12 2 B 432 kip # in.2 PL2 = = 2EI 2EI EI
(uB)2
Then the slope and deflection at B are uB = (uB)1 + (uB)2 =
432 72 + EI EI
=
504 kip # ft2 EI
¢ B = (¢ B)1 + (¢ B)2 =
3456 720 + EI EI
=
4176 kip # in.3 EI
For A36 steel W10 * 15, I = 68.9 in4 And E = 29.0 A 103 B ksi uB =
504
29.0 A 103 B (68.9)
= 0.252(10-3) rad
Ans.
4176
¢B =
29.0 A 103 B (68.9)
= 0.00209 in
Ans.
978
6 ft
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•12–89.
Determine the slope and deflection at end C of the overhang beam. EI is constant.
w
A
C D a
Elastic Curves. The uniform distributed load on the beam is equivalent to the sum of the seperate loadings shown in Fig.a. The elastic curve for each seperate loading is shown Fig. a. Method of Superposition. Using the table in the appendix, the required slopes and deflections are (uC)1 = (uB)1 =
w(2a)3 wa3 wL3 = = 24EI 24EI 3EI
(¢ C)1 = (uB)1(a) =
wa3 wa4 (a) = 3EI 3EI
(uC)2 =
wa3 wL3 = 6EI 6EI
(¢ C)2 =
wL4 wa4 = 8EI 8EI
T
MOL (uC)3 = (uB)3 = = 3EI (¢ C)3 = (uB)3 (a) =
c
¢
wa2 2
≤ (2a)
3EI
wa3 wa4 (a) = 3EI 3EI
=
wa3 3EI
T
Then the slope and deflection of C are uC = (uC)1 + (uC)2 + (uC)3 = -
=
wa3 wa 3 wa3 + + 3EI 6EI 3EI
wa3 6EI
Ans.
¢ C = (¢ C)1 + (¢ C)2 + (¢ C)3 = -
=
wa4 wa 4 wa4 + + 3EI 8EI 3EI
wa4 T 8EI
Ans.
979
B a
a
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12–90. Determine the slope at A and the deflection at point D of the overhang beam. EI is constant.
w
A
C D a
Elastic Curves. The uniform distributed load on the deformation of span AB is equivalent to the sum of the seperate loadings shown in Fig. a. The elastic curve for each seperate loading is shown in Fig. a. Method of Superposition. Using the table in the appendix, the required slope and deflections are (uA)1 =
w(2a)3 wL3 wa3 = = 24EI 24EI 3EI
(¢ D)1 =
5w(2a)4 5wa4 5wL4 = = 384EI 384EI 24EI
T
wa2 (2a) MOL 2 wa3 (uA)2 = = = 6EI 6EI 6EI
¢
(¢ D)2 =
=
wa2 ≤ (a) 2
MOx A L2 - x2 B = C (2a)2 - a2 D 6EIL 6EI(2a) wa4 8EI
c
Then the slope and deflection of point D are uA = (uA)1 + (uA)2 =
wa3 wa3 wa3 = 3EI 6EI 6EI
Ans.
¢ D = (¢ D)1 + (¢ D)2 =
5wa4 wa4 wa4 = T 24EI 8EI 12EI
Ans.
980
B a
a
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12–91. Determine the slope at B and the deflection at point C of the simply supported beam. E = 200 GPa and I = 45.5(106) mm4.
9 kN/m
10 kN
A
B
C 3m
Elastic Curves. The loading system on the beam is equivalent to the sum of the seperate loadings shown in Fig. a. The elastic curves for each loading are shown in Fig. a. Method of Superposition. Using the table in the appendix, the required slope and deflections are
(uB)1 = (¢ C)1 =
9 A 63 B wOL3 43.2kN # m2 = = 45EI 45EI EI wOx A 3x4 - 10L2x2 + 7L4 B 360EIL
=
9(3) C 3 A 34 B - 10 A 62 B A 32 B + 7 A 64 B D 360EI(6)
=
75.9375kN # m3 EI
(uB)2 =
(¢ C)2 =
T
10 A 62 B PL2 22.5kN # m2 = = 16EI 16EI EI 10 A 63 B PL3 45kN # m3 = = 48EI 48EI EI
T
Then the slope at B and deflection at C are uB = (uB)1 + (uB)2
=
65.7 A 103 B 43.2 22.5 65.7kN # m2 + = = = 0.00722 rad EI EI EI 200 A 109 B C 45.5 A 10 - 6 B D
Ans.
¢ C = (¢ C)1 + (¢ C)2 120.9375 A 10 B 75.9375 45 120.9375 kN # m3 + = = EI EI EI 200 A 109 B C 45.5 A 10 - 6 B D 3
=
= 0.01329 m = 13.3 mm T
Ans.
981
3m
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*12–92. Determine the slope at A and the deflection at point C of the simply supported beam. The modulus of elasticity of the wood is E = 10 GPa .
3 kN
100 m C
A
1.5 m
Method of Superposition. Using the table in the appendix, the required slopes and deflections are (uA)1 =
Pab(L + b) 3(1.5)(4.5)(6 + 4.5) 5.90625kN # m2 = = 6EIL 6EI(6) EI
(¢ C)1 =
3(4.5)(1.5) 2 Pbx A L2 - b2 - x2 B = A 6 - 4.52 - 1.52 B 6EIL 6EI(6)
(uA)2 = (¢ C)2 =
7.594kN # m3 EI
T
3 A 62 B PL2 6.75 kN # m2 = = 16EL 16EI EI 3(1.5) Px 9.281 a3(6)2 - 4(1.5)2 b = A 3L2 - 4x2 B = 48EI 48EI EI
Then the slope and deflection at C are uA = (uA)1 + (uA)2 =
6.75 5.90625 + EI EI
12.65625kN # m2 = = EI
12.6525 A 103 B
10 A 109 B c
1 (0.1) A 0.2 3 B d 12
Ans.
= 0.0190 rad
and ¢ C = (¢ C)1 + (¢ C)2
=
9.281 7.594 + = EI EI
16.88 A 103 B
10 A 109 B c
1 (0.1) A 0.2 3 B d 12
= 0.0253 m = 25.3 mm
982
B
1.5 m
Elastic Curves. The two concentrated forces P are applied seperately on the beam and the resulting elastic curves are shown in Fig. a.
=
3 kN
Ans.
3m
200 m
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•12–93.
The W8 * 24 simply supported beam is made of A-36 steel and is subjected to the loading shown. Determine the deflection at its center C.
6 kip/ft 5 kip⭈ft A
B
I = 82.8 in4
C 8 ft
5(6) A 164 B 2560 5wL4 (¢ C)1 = = = T 768EI 768EI EI ¢ 2 (x) =
Mx A L2 - x2 B 6LEI
At point C, x =
(¢ C)2 =
=
M A L2 B
6LEI
L 2
A L2 - A L2 B 2 B
5 A 162 B 80 ML2 = = 16EI 16EI EI
¢ C = (¢ C)1 + (¢ C)2 = 2640(1728)
=
8 ft
29 A 103 B (82.8)
T
80 2640 2560 + = EI EI EI Ans.
= 1.90 in.
12–94. Determine the vertical deflection and slope at the end A of the bracket. Assume that the bracket is fixed supported at its base, and neglect the axial deformation of segment AB. EI is constant.
3 in. B
6 in.
A 3
¢A =
uA =
3
8(3) 72 PL = = 3EI 3EI EI
Ans.
8 A 32 B PL2 36 = = 2EI 2EI EI
Ans.
983
8 kip
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12–95. The simply supported beam is made of A-36 steel and is subjected to the loading shown. Determine the deflection at its center C. I = 0.1457(10-3) m4.
20 kN 4 kN/m
A 5m
Using the table in appendix, the required deflections for each load case are computed as follow: (yC)1 =
5(4) A 104 B 5wL4 = 768EI 768 EI =
(yC)2 =
260.42 kN # m3 EI
T
20N A 103 B PL3 416.67 kN # m3 = = T 48EI 48EI EI
Then the deflection of point C is yC = (yC)1 + (yC)2 =
260.42 416.67 + EI EI
=
677.08 kN # m3 EI
T
= 0.1457 A 10 - 3 B m4 and E = 200GPa
¢C =
677.08 A 103 B
200 A 109 B C 0.1457 A 10 - 3 B D
Ans.
= 0.0232 m = 23.2 m T
984
B
C 5m
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*12–96. Determine the deflection at end E of beam CDE. The beams are made of wood having a modulus of elasticity of E = 10 GPa .
2m
1.5 m A
C
1m
3 kN
Method of Superposition. Referring to the table in the appendix, the deflection of point D is 4.5 A 33 B 2.53125 kN # m3 PL3 = = 48EI 48EI EI
T
Subsequently, 3 2.53125 3 3.796875 kN # m3 (¢ E)1 = ¢ D a b = a b = T 2 EI 2 EI Also,
(¢ E)2 = (uD)3 =
3 A 13 B 1 kN # m3 PL3 = = T 3EI 3EI EI 3(2) MOL 2 kN # m2 = = 3EI 3EI EI
(¢ E)3 = (uD)3L =
2 2 kN # m3 (1) = T EI EI
Thus, the deflection of end E is ¢ E = (¢ E)1 + (¢ E)2 + (¢ E)3
=
3.796875 1 2 6.796875kN # m3 + + = = EI EI EI EI
6.796875 A 103 B
10 A 109 B c
1 (0.075) A 0.153 B d 12
= 0.03222 m = 32.2 mm T
Ans.
985
a a
E
75 mm
D
a
¢D =
1.5 m
a
150 mm Section a – a B
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•12–97.
The pipe assembly consists of three equal-sized pipes with flexibility stiffness EI and torsional stiffness GJ. Determine the vertical deflection at point A.
L – 2
C
¢B =
P A L2 B 3 3EI
(¢ A)1 =
=
P A L2 B 3 3EI
PL3 24EI
=
L – 2
P
PL3 24EI
L – 2
A
B
(PL>2) A L2 B PL2 TL u = = = JG JG 4JG (¢ A)2 = u a
L PL3 b = 2 8JG
¢ A = ¢ B + (¢ A)1 + (¢ A)2 =
PL3 PL3 PL3 + + 24EI 24EI 8JG
= PL3 a
1 1 + b 12EI 8JG
Ans.
12–98. Determine the vertical deflection at the end A of the bracket. Assume that the bracket is fixed supported at its base B and neglect axial deflection. EI is constant. u =
a
A
ML Pab = EI EI
b
Pa2b (¢ A)1 = u(a) = EI (¢ A)2 =
B
PL3 Pa3 = 3EI 3EI
¢ A = (¢ A)1 + (¢ A)2 =
P
Pa2 (3b + a) Pa3 Pa2b + = EI 3EI 3EI
Ans.
986
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12–99. Determine the vertical deflection and slope at the end A of the bracket. Assume that the bracket is fixed supported at its base, and neglect the axial deformation of segment AB. EI is constant.
20 lb/in. 80 lb
B 4 in. 3 in.
C
Elastic Curve: The elastic curves for the concentrated load, uniform distibuted load, and couple moment are drawn separately as shown. Method of Superposition: Using the table in Appendix C, the required slopes and displacements are
(uA)1 =
20 A 4 3 B wL3AB 213.33 lb # in2 = = 6EI 6EI EI
(uA)2 = (uB)2 =
(uA)3 = (uB)3 =
(¢ A)v1 =
160(3) M0 LBC 480 lb # in2 = = EI EI EI 80 A 32 B PL2BC 360 lb # in2 = = 2EI 2EI EI
20 A 4 4 B wL4AB 640 lb # in3 = = 8EI 8EI EI
T
(¢ A)v2 = (uB)2 (LAB) =
480 1920lb # in3 (4) = EI EI
T
(¢ A)v3 = (uB)3 (LAB) =
1440lb # in3 360 (4) = EI EI
T
The slope at A is uA = (uA)1 + (uA)2 + (uA)3 =
213.33 480 360 + + EI EI EI
=
1053 lb # in2 EI
Ans.
The vertical displacement at A is (¢ A)v = (¢ A)v1 + (¢ A)v2 (¢ A)v3 =
640 1920 1440 + + EI EI EI
=
4000 lb # in3 EI
Ans.
T
987
A
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*12–100. The framework consists of two A-36 steel cantilevered beams CD and BA and a simply supported beam CB. If each beam is made of steel and has a moment of inertia about its principal axis of Ix = 118 in4, determine the deflection at the center G of beam CB.
A
15 kip B
D C
G 8 ft
16 ft
¢C =
8 ft
7.5 A 163 B PL3 10.240 = = T 3EI 3EI EI 15 A 163 B PL3 1.280 = = T 48EI 48EI EI
¢ ¿G =
¢G = ¢C + ¢ ¿G =
1,280 11,520 10,240 + = EI EI EI 11,520(1728)
=
29 A 103 B (118)
Ans.
= 5.82 in. T
•12–101.
The wide-flange beam acts as a cantilever. Due to an error it is installed at an angle u with the vertical. Determine the ratio of its deflection in the x direction to its deflection in the y direction at A when a load P is applied at this point. The moments of inertia are Ix and Iy. For the solution, resolve P into components and use the method of superposition. Note: The result indicates that large lateral deflections (x direction) can occur in narrow beams, Iy V Ix, when they are improperly installed in this manner. To show this numerically, compute the deflections in the x and y directions for an A-36 steel W10 * 15, with P = 1.5 kip, u = 10°, and L = 12 ft.
ymax =
xmax = ymax
P cos L3 ; 3EIx P sin u L3 3 EIy Pcosu L3 3 EIx
W 10 * 15 ymax =
xmax =
=
xmax =
u Vertical P y u
L A x
P sinu L3 3EIy
Ix tan u Iy
Ans.
Ix = 68.9 in4
1.5( cos 10°)(144)3 3(29) A 103 B (68.9)
1.5( sin 10°)(144)3 3(29) A 103 B (2.89)
Iy = 2.89 in4
= 0.736 in.
Ans.
= 3.09 in.
Ans.
988
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12–102. The simply supported beam carries a uniform load of 2 kip>ft. Code restrictions, due to a plaster ceiling, require the maximum deflection not to exceed 1>360 of the span length. Select the lightest-weight A-36 steel wide-flange beam from Appendix B that will satisfy this requirement and safely support the load. The allowable bending stress is sallow = 24 ksi and the allowable shear stress is tallow = 14 ksi. Assume A is a pin and B a roller support.
8 kip
A
Strength criterion:
24 =
M Sreq’d
96(12) Sreq’d
Sreq’d = 48 in3 Choose W14 * 34, S = 48.6 in3, tw = 0.285 in., d = 13.98 in., I = 340 in4. tallow = 14 Ú
V A web
24 = 6.02 ksi O.K. (13.98)(0.285)
Deflection criterion: Maximum is at center. vmax =
P(4)(8) 5wL4 + (2) C (16)2 - (4)2 - (8)2) D (12)3 384EI 6EI(16)
= c
117.33(8) 5(2)(16)4 + d(12)3 384EI EI 4.571(106)
=
29(106)(340)
= 0.000464 in. 6
B 4 ft
Mmax = 96 kip # ft
sallow =
8 kip 2 kip/ft
1 (16)(12) = 0.533 in. O.K. 360
Use W14 * 34
Ans.
989
8 ft
4 ft
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12–103. Determine the reactions at the supports A and B, then draw the moment diagram. EI is constant.
M0 A B L
Support Reactions: FBD(a). + ©F = 0; : x
Ax = 0 A y - By = 0
[1]
M0 - A y L + MB = 0
[2]
+ c ©Fy = 0; a + ©MB = 0;
Ans.
Moment Function: FBD(b) a + ©MNA = 0;
M(x) + M0 - A y x = 0 M(x) = A y x - M0
Slope and Elastic Curve: EI
EI
EI
d2y = M(x) dx2
d 2y = A y x - M0 dx2
Ay dy = x2 - M0x + C1 dx 2
EI y =
Ay 6
x3 -
[3]
M0 2 x + C1x + C2 2
[4]
Boundary Conditions: At x = 0, y = 0. From Eq.[4], C2 = 0 At x = L,
0 =
dy = 0. From Eq. [3], dx
A y L2
- M0 L + C1
2
[5]
At x = L, y = 0. From Eq. [4], 0 =
A y L3 -
6
M0 L2 + C1 L 2
[6]
Solving Eqs. [5] and [6] yields, Ay =
3M0 2L
C1 =
Ans.
M0 L 4
Substituting Ay, into Eqs. [1] and [2] yields: By =
3M0 2L
MB =
M0 2
Ans.
990
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*12–104. Determine the value of a for which the maximum positive moment has the same magnitude as the maximum negative moment. EI is constant.
P
a L
A y + By - P = 0
+ c ©Fy = 0; a + ©MA = 0;
[1]
MA + By L - Pa = 0
[2]
Moment Functions: FBD(b) and (c). M(x1) = Byx1 M(x2) = Byx2 - Px2 + PL - Pa Slope and Elastic Curve: EI
d2y = M(x) dx2
For M(x1) = Byx1, EI
EI
d2y1 dx21
= Byx1
By dy1 = x2 + C1 dx1 2 1
EI y1 =
By 6
[3]
x31 + C1x1 + C2
[4]
For M(x2) = Byx2 - Px2 + PL - Pa, EI
EI
d2y2 dx22
= Byx2 - Px2 + PL - Pa
By dy2 P = x2 - x22 + PLx2 - Pax2 + C3 dx2 2 2 2
EI y2 =
By 6
x32 -
[5]
P 3 PL 2 Pa 2 x + x x + C3x2 + C4 6 2 2 2 2 2
[6]
Boundary Conditions: y1 = 0 at x1 = 0. From Eq.[4], C2 = 0 dy2 dx2
= 0 at x2 = L. From Eq.[5]
0 =
ByL2 2
-
C3 = -
PL2 + PL2 - PaL + C3 2 By L2 2
-
PL2 + PaL 2
991
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*12–104.
Continued
y2 = 0 at x2 = L. From Eq.[6], 0 =
By L3 6
By L2 PL3 PL3 PaL2 PL2 + + a+ PaL bL + C4 6 2 2 2 2 C4 =
By L3 3
+
PL3 PaL2 6 2
Continuity Conditions: At x1 = x2 = L - a, By 2
dy1 dy2 . From Eqs.[3] and [5], = dx1 dx2
(L - a)2 + C1 =
By 2
(L - a)2 -
P (L - a)2 + PL(L - a) 2
- Pa(L - a) + a C1 =
By L2 2
-
PL2 + PaL b 2
By L2 Pa2 2 2
At x1 = x2 = L - a, y1 = y2. From Eqs.[4] and [6], By L2 Pa2 (L - a) + a b(L - a) 6 2 2
By
3
By =
6
(L - a)3 -
+ a-
By L2 2
-
P PL Pa (L - a)3 + (L - a)2 (L - a)2 6 2 2 By L3 PL2 PL3 PaL2 + PaL b(L - a) + + 2 3 6 2
By L3 Pa3 Pa2L + = 0 6 2 3 By =
3Pa2 Pa3 Pa2 = (3L - a) 2 3 2L 2L 2L3
Substituting By into Eqs.[1] and [2], we have Ay =
P A 2L3 - 3a2L + a3 B 2L3
MA =
Pa A -3aL + a2 + 2L2 B 2L2
Require |Mmax( + )| = |Mmax( - )|. From the moment diagram, Pa2 Pa (3L - a)(L - a) = A -3aL + a2 + 2L2 B 2L3 2L2 a2 - 4aL + 2L2 = 0 a = A 2 - 22 B L
Ans.
992
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•12–105. Determine the reactions at the supports A, B, and C; then draw the shear and moment diagrams. EI is constant.
P
A L 2
Support Reactions: FBD(a). + ©Fx = 0; :
Ax = 0
+ c ©Fy = 0;
A y + By + Cy - 2P = 0
a + ©MA = 0;
By L + Cy (2L) - Pa
Ans. [1]
3L L b - Pa b = 0 2 2
[2]
Moment Function: FBD(b) and (c). M(x1) = Cy x1 M(x2) = Cy x2 - Px2 +
PL 2
Slope and Elastic Curve: d2y = M(x) dx2
EI
For M(x1) = Cy x1, EI
d2y1
= Cyx1
dx21
Cy dy1 = x2 + C1 dx1 2 1
EI
EI y1 =
Cy 6
x31 + C1x1 + C2
For M(x2) = Cyx2 - Px2 +
EI
EI
[3]
d2y2
PL , 2
= Cyx2 - Px2 +
dx22
[4]
PL 2
Cy dy2 P PL = x22 - x22 + x + C3 dx2 2 2 2 2
EI y2 =
Cy 6
x32 -
[5]
P 3 PL 2 x + x + C3x2 + C4 6 2 4 2
[6]
Boundary Conditions: y1 = 0 at x1 = 0. From Eq.[4], C2 = 0 Due to symmetry,
0 =
Cy L2 2
-
dy2 = 0 at x2 = L. From Eq.[5], dx2 PL2 PL2 + + C3 2 2
C3 = -
Cy L2 2
993
P
B L 2
C L 2
L 2
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•12–105.
Continued
y2 = 0 at x2 = L. From Eq. [6], 0 =
Cy L3
2
-
6
Cy L PL3 PL3 + + abL + C4 6 4 2 Cy L3
C4 =
-
3
PL3 12
Continuity Conditions: At x1 = x2 =
dy2 L dy1 , . From Eqs.[3] and [5], = 2 dx1 dx2 Cy 2
a
Cy L 2 Cy L2 L 2 P L 2 PL L b + C1 = a b - a b + a b 2 2 2 2 2 2 2 2 C1 =
At x1 = x2 = Cy
a
6
Cy =
6
Cy L2 PL2 8 2
L , y = y2. From Eqs.[4] and [6], 2 1
Cy L2 L PL2 L 3 b + a ba b 2 8 2 2 a
Cy L2 L Cy L3 L 3 P L 3 PL L 2 PL3 b a b + a b + aba b + 2 6 2 4 2 2 2 3 12 Cy =
5 P 16
Ans.
Substituting Cy into Eqs.[1] and [2], By =
11 P 8
Ay =
5 P 16
Ans.
994
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12–106. Determine the reactions at the supports, then draw the shear and moment diagram. EI is constant.
P
A
B L
Support Reactions: FBD(a). + ©F = 0; : x
Ax = 0
Ans.
By - A y - P = 0
+ c ©Fy = 0; a + ©MB = 0;
[1]
A y L - MA - PL = 0
[2]
Moment Functions: FBD(b) and (c). M(x1) = -Px1 M(x2) = MA - A yx2 Slope and Elastic Curve: EI
d2y = M(x) dx2
For M(x1) = -Px1, EI
EI
d2y1 dx21
= -Px1
dy1 P = - x21 + C1 dx1 2
[3]
P EI y1 = - x31 + C1x1 + C2 6
[4]
For M(x2) = MA - A yx2, EI
EI
EI y2 =
d2y2 dx22
= MA - A yx2
Ay dy2 = MAx2 x2 + C3 dx2 2 2
[5]
Ay MA 2 x2 x3 + C3x2 + C4 2 6 2
[6]
995
L
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12–106.
Continued
Boundary Conditions: y2 = 0 at x2 = 0. From Eq.[6], C4 = 0 dy2 = 0 at x2 = 0. From Eq.[5], C3 = 0 dx2 y2 = 0 at x2 = L. From Eq. [6], 0 =
A y L3 MA L2 2 6
[7]
Solving Eqs.[2] and [7] yields, MA =
PL 2
Ay =
3P 2
Ans.
Substituting the value of Ay into Eq.[1], By =
5P 2
Ans.
Note: The other boundary and continuity conditions can be used to determine the constants C1 and C2 which are not needed here.
996
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12–107. Determine the moment reactions at the supports A and B. EI is constant.
P
P
A
B a
Support Reactions: FBD(a).
a L
a + ©MB = 0;
Pa + P(L - a) + MA - A y L - MB = 0 PL + MA - A y L - MB = 0
[1]
Moment Functions: FBD(b) and (c). M(x1) = A y x1 - MA M(x2) = A y x2 - Px2 + Pa - MA Slope and Elastic Curve: d2y = M(x) dx2
EI
For M(x1) = A y x1 - MA, EI
EI
d2y1 dx21
= A y x1 - MA
Ay dy1 = x2 - MA x1 + C1 dx1 2 1
EI y1 =
Ay 6
x31 -
[2]
MA 2 x + C1x1 + C2 2 1
[3]
For M(x2) = A y x2 - Px2 + Pa - Ma, EI
EI
EI y2 =
d2y = A y x2 - Px2 + Pa - MA dx22
Ay dy2 P = x2 - x22 + Pax2 - MA x2 + C3 dx2 2 2 2 Ay 6
x32 -
[4]
MA 2 P 3 Pa 2 x + x x + C3 x2 + C4 6 2 2 2 2 2
[5]
Boundary Conditions: dy1 = 0 at x1 = 0. From Eq.[2], C1 = 0 dx1 y1 = 0 at x1 = 0. From Eq.[3], C2 = 0 Due to symmetry,
0 =
Ay 2
a
dy2 dx2
= 0 at x2 =
L . From Eq.[4], 2
L 2 P L 2 L L b a b + Paa b - MA a b + C3 2 2 2 2 2
C3 = -
A y L2 8
+
MA L PL2 PaL + 8 2 2
997
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12–107.
Continued
Due to symmetry, A y a2
dy1 dy2 = at x1 = a and x2 = L - a. From Eqs.[2] and [4], dx1 dx2
- MA a = -
2
Ay 2
(L - a)2 +
+ MA (L - a) +
-A y a2 -
3A y L2 8
+ A y aL +
P (L - a)2 - Pa(L - a) 2
A y L2 8
-
MA L PL2 PaL + 8 2 2
MAL 3PaL 3Pa2 3PL2 + + = 0 8 2 2 2
[6]
Continuity Conditions: At x1 = x2 = a, A y a2 2
dy1 dy2 . From Eqs.[2] and [4], = dx1 dx2
- MA a
A y a2 =
2
A y L2 MA L Pa2 PL2 PaL + Pa2 - MA a + + 2 8 8 2 2
-
A y L2 MA L Pa2 PL2 PaL + + = 0 2 8 8 2 2
[7]
Solving Eqs.[6] and [7] yields, MA =
Pa (L - a) L
Ans.
Ay = P Substitute the value of MA and Ay obtained into Eqs.[1], MB =
Pa (L - a) L
Ans.
998
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*12–108. Determine the reactions at roller support A and fixed support B.
w
Equations of Equilibrium. Referring to the free-body diagram of the entire beam, Fig. a, A
+ ©F = 0; : x
Bx = 0
+ c ©Fy = 0;
A y + By - wL = 0
a+ ©MB = 0;
wL a
(1)
L 2 b - A y a Lb - MB = 0 2 3
MB =
wL2 2 - A yL 2 3
(2)
Moment Functions. Referring to the free-body diagram of the beam’s segment, Fig. b, x L L M(x) + wxa b + wa b ¢ x + ≤ - A yx = 0 2 3 6
a+ ©MO = 0;
M(x) = A yx -
w 2 wL wL2 x x 2 3 18
Equations of Slope and Elastic Curves. EI
d2v = M(x) dx2
EI
wL2 d2v w 2 wL x x = A x y 2 3 18 dx2
EI
Ay dv w 3 wL 2 wL2 = x2 x x x + C1 dx 2 6 6 18
EIv =
Ay 6
x3 -
(3)
w 4 wL 3 wL2 2 x x x + C1x + C2 24 18 36
(4)
Boundary Conditions. At x = 0, v = 0. Then Eq. (4) gives 0 = 0 - 0 - 0 - 0 + 0 + C2 0 =
At x =
0 =
C2 = 0At x =
2 dv L, = 0. Then Eq. (3) gives 3 dx
Ay
2 2 w 2 3 wL 2 2 wL2 2 a Lb a Lb a Lb a L b + C1 2 3 6 3 6 3 18 3
C1 =
2A yL2 13wL3 81 9
(5)
2 L, v = 0. Then Eq. (4) gives 3
Ay
C1 =
B
Ans.
2 3 w 2 4 wL 2 3 wL2 2 2 2 a Lb a Lb a Lb a Lb + C1 a Lb 6 3 24 3 18 3 36 3 3 2A yL2 wL3 18 27
(6)
Solving Eqs. (5) and (6), Ay =
17wL wL Ans.C1 = Substituting the result of Ay into Eqs. (1) and (2), 24 324
By =
7wL 24
MB =
wL2 36
Ans.
The shear and moment diagrams are shown in Figs. c and d, respectively. 999
L 3
2L 3
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•12–109.
Use discontinuity functions and determine the reactions at the supports, then draw the shear and moment diagrams. EI is constant.
3 kip/ft C A
B 8 ft
+ ©F = 0 : x
Cx = 0
+ c ©Fy = 0
A y + By + Cy - 24 = 0
a + ©MA = 0
Ans. (1)
18 Cy + 8By - 24(4) = 0
(2)
Bending Moment M(x): M(x) = -(-Cy) 6 x - 0 7 -( -By) 6 x - 10 7 = Cyx + By 6 x - 10 7 -
3 6 x - 10 7 2 2
3 6 x - 10 7 2 2
Elastic curve and slope: EI
d2v 3 = M(x) = Cyx + By 6 x - 10 7 - 6 x - 10 7 2 2 dx2
EI
Cyx2 By dv 1 = + 6 x - 10 7 2 - 6 x - 10 7 3 + C1 dx 2 2 2
EIv =
Cyx3 6
By +
6
6 x - 10 7 3 -
1 6 x - 10 7 4 + C1x + C2 8
(3)
(4)
Boundary conditions: v = 0
at
x = 0
From Eq. (4) C2 = 0 v = 0
at
x = 10 ft
From Eq. (4) 0 = 166.67 Cy + 10C1 v = 0
at
(5)
x = 18 ft
0 = 972Cy + 85.33By - 512 + 18C1
(6)
Solving Eqs. (2),(5) and (6) yields: By = 14.4 kip
Ans.
Cy = -1.07 kip = 1.07 kip T
Ans.
C1 = 17.78 From Eq. (1): A y = 10.7 kip
Ans.
1000
10 ft
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12–110. Determine the reactions at the supports, then draw the shear and moment diagrams. EI is constant.
w0
A
C
B L
Support Reaction: FBD(b). + ©F = 0; : x
Ax = 0
Ans.
A y + By + Cy - w0L = 0
+ c ©Fy = 0; a + ©MA = 0;
[1]
By L + Cy (2L) - w0 L(L) = 0
[2]
Moment Function: FBD(b). a + ©MNA = 0;
-M(x) -
x 1 w0 a xbx a b + Cyx = 0 2 L 3
M(x) = Cyx -
w0 3 x 6L
Slope and Elastic Curve: EI
EI
EI
d 2y = M(x) dx2
w0 3 d 2y = Cyx x 2 6L dx
Cy w0 4 dy = x2 x + C1 dx 2 24L
EI y =
Cy 6
x3 -
[3]
w0 5 x + C1x + C2 120L
[4]
Boundary Conditions: At x = 0, y = 0. From Eq.[4], C2 = 0 Due to symmetry,
0 =
Cy L2 -
2
C1 = -
dy = 0 at x = L. From Eq. [3], dx w0L3 + C1 24
CyL2 2
+
w0L3 24
At x = L, y = 0. From Eq. [4], 0 =
Cy L3 6
2
-
Cy L w0L4 w0L3 + a+ bL 120 2 24 Cy =
w0L 10
Ans.
1001
L
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12–110.
Continued
Substituting Cy into Eqs. [1] and [2] yields: By =
4w0L 5
Ay =
w0L 10
Ans.
Shear and Moment diagrams: The maximum span (positive) moment occurs when the shear force V = 0. From FBD(c), + c ©Fy = 0;
w0L 1 w0 - a xbx = 0 10 2 L x =
+ ©MNA = 0;
M +
25 L 5
w0L 1 w0 x a xb (x) a b (x) = 0 2 L 3 10 M =
At x =
25 L, 5
At x = L,
M =
w0L w0 3 x x 10 6L 25w0L2 75
M = -
w0L2 15
1002
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12–111. Determine the reactions at pin support A and roller supports B and C. EI is constant.
w
Equations of Equilibrium. Referring to the free-body diagram of the entire beam, Fig. a, + ©F = 0; : x
Ax = 0
+ c ©Fy = 0;
A y + By + Cy - wL = 0
a + ©MB = 0;
Ans.
Cy (L) + wLa A y - Cy =
(1)
L b - A y(L) = 0 2
wL 2
(2)
Moment Functions. Referring to the free-body diagram of the beam’s segment, Fig. b, M(x1) is M(x1) + wx1 ¢
a + ©MO = 0;
x1 ≤ - A yx1 = 0 2
M(x1) = A yx1 -
w 2 x 2 1
and M(x2) is given by a + ©MO = 0;
Cyx2 - M(x2) = 0 M(x2) = Cyx2
Equations of Slope and Elastic Curves. EI
d2v = M(x) dx2
For coordinate x1, EI
d2v w 2 = A yx1 x 2 1 dx21
EI
Ay dv w 3 x 2 x + C1 = dx1 2 1 6 1
EIv =
Ay 6
x1 3 -
(3)
w 4 x + C1x1 + C2 24 1
(4)
For coordinate x2, EI
d2v = Cyx2 dx22
EI
Cy dv = x 2 + C3 dx2 2 2
EIv =
Cy 6
(5)
x2 3 + C3x2 + C4
(6)
1003
A
C
B L
L
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12–111.
Continued
Boundary Conditions. At x1 = 0, v1 = 0. Then Eq.(4) gives 0 = 0 - 0 + 0 + C2
C2 = 0
At x1 = L, v1 = 0. Then Eq. (4) gives 0 =
Ay 6
A L3 B -
w A L4 B + C1L 24
C1 =
A yL2 wL3 24 6
At x2 = 0, v2 = 0. Then Eq. (6) gives 0 = 0 + 0 + C4
C4 = 0
At x2 = L, v2 = 0. Then Eq. (6) gives 0 =
Cy 6
A L B + C3L 3
Continuity Conditions. At x1 = x2 = L, Ay 2
A L2 B -
A y + Cy =
C3 = -
CyL2 6
dv1 dv2 . Then Eqs.(3) and (5) give = dx1 dx2
Cy A yL2 CyL2 wL3 w 3 ≤ = - B A L2 B R AL B + ¢ 6 24 6 2 6 3wL 8
(7)
Solving Eqs. (2) and (7), Ay =
7wL 16
Cy = -
wL 16
Ans.
The negative sign indicates that Cy acts in the opposite sense to that shown on freebody diagram. Substituting these results into Eq. (1), By =
5wL 8
Ans.
The shear and moment diagrams are shown in Figs. c and d, respectively.
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*12–112. Determine the moment reactions at fixed supports A and B. EI is constant.
w0
Equations of Equilibrium. Due to symmetry, A y = By = R and MA = MB = M. Referring to the free-body diagram of the entire beam, Fig. a, + c ©Fy = 0;
2R R =
A
1 wL = 0 2 0
L 2
w0L 4
Moment Function. Referring to the free-body diagram of the beam’s segment, Fig. b, M(x) + c
w0L x 1 2w0 a xb (x) d a b + M x = 0 2 L 3 4 M(x) =
w0L w0 3 x x - M 4 3L
Equations of Slope and Elastic Curves. EI
EI
EI
d2v = M(x) dx2 d2v dx 2
=
w0L w0 3 x x - M 4 3L
w0L 2 w0 4 dv = x x - Mx + C1 dx 8 12L
EIv =
Due to symmetry,
(2)
dv = 0. Then Eq. (1) gives dx
0 = 0 - 0 - 0 + C1
0 =
(1)
w0L 3 w0 5 M 2 x x x + C1x + C2 24 60L 2
Boundary Conditions. At x = 0,
C1 = 0
dv L = 0 at x = . Then Eq. (1) gives dx 2
w0 L 4 w0L L 2 L a b a b - Ma b 8 2 12L 2 2
MA = MB = M =
B
5w0L2 96
Note. The boundary condition v = 0 at x = 0 can be used to determine C2 using Eq.(2).
1005
L 2
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The beam has a constant E1I1 and is supported by the fixed wall at B and the rod AC. If the rod has a cross-sectional area A2 and the material has a modulus of elasticity E2 , determine the force in the rod.
•12–113.
C w
L2
B
A L1
TAC + By - wL1 = 0
+ c ©Fy = 0 c + ©MB = 0
TAC(L1) + MB MB =
wL1 2 = 0 2
(1)
wL1 2 - TACL1 2
(2)
Bending Moment M(x): wx2 2
M(x) = TACx -
Elastic curve and slope: EI
d2v wx2 = M(x) = TACx 2 2 dx
EI
TACx2 dv wx3 = + C1 dx 2 6
EIv =
(3)
TACx3 wx4 + C1x + C2 6 24
(4)
Boundary conditions: v =
TACL2 A 2E2
x = 0
From Eq. (4) -E2I1 a
TACL2 b = 0 - 0 + 0 + C2 A 2E2
C2 = a v = 0
-E1I1L2 b TAC A 2E2
at
x = L1
From Eq. (4) 0 =
TACL1 3 wL1 4 E1I1L2 + C1L1 T 6 24 A 2E2 AC dv = 0 dx
at
(5)
x = L1
From Eq. (3) 0 =
TACL1 2 wL1 3 + C1 2 6
(6)
Solving Eqs. (5) and (6) yields: TAC =
3A 2E2wL1 4
8 A A 2E2L1 3 + 3E1I1L2 B
Ans.
1006
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12–114. The beam is supported by a pin at A, a roller at B, and a post having a diameter of 50 mm at C. Determine the support reactions at A, B, and C. The post and the beam are made of the same material having a modulus of elasticity E = 200 GPa, and the beam has a constant moment of inertia I = 255(106) mm4.
15 kN/m
A 1m 6m
Equations of Equilibrium. Referring to the free-body diagram of the entire beam, Fig. a, + ©F = 0; : x
Ax = 0
+ c ©Fy = 0;
A y + By + FC - 15(12) = 0
a+ ©MB = 0;
15(12)(6) - FC(6) - A y(12) = 0
Ans. (1)
2A y + FC = 180
(2)
Moment Functions. Referring to the free-body diagram of the beam’s segment, Fig. b, x M(x) + 15xa b - A yx = 0 2
a+ ©MO = 0;
M(x) = A yx - 7.5x2 Equations of Slope and Elastic Curves. EI
d2v = M(x) dx2
EI
d2v = A yx - 7.5x2 dx2
EI
Ay dv = x2 - 2.5x3 + C1 dx 2
EIv =
Ay 6
(3)
x3 - 0.625x4 + C1x + C2
(4)
Boundary Conditions. At x = 0, v = 0. Then Eq. (4) gives C2 = 0
0 = 0 - 0 + 0 + C2 At x = 6 m, v = - ¢ C = -
E C 255 A 10 - 6 B D a -
FC(1) FCLC 1600FC . Then Eq. (4) gives = = p A CE pE 2 0.05 E A B 4
Ay 1600FC b = A 63 B - 0.625 A 64 B + C1(6) pE 6
C1 = 135 - 6A y - 0.02165FC Due to symmetry,
0 =
Ay 2
dv = 0 at x = 6 m. Then Eq. (3) gives dx
A 62 B - 2.5 A 63 B + 135 - 6A y - 0.02165FC (5)
12A y - 0.02165FC = 405 Solving Eqs. (2) and (5), FC = 112.096 kN = 112 kN
A y = 33.95 kN = 34.0 kN
Ans.
Substituting these results into Eq. (1), By = 33.95 kN = 34.0 kN
Ans.
1007
B
C 6m
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12–115. Determine the moment reactions at the supports A and B, then draw the shear and moment diagrams. EI is constant.
A B L
Support Reaction: FBD(a). + ©F = 0; : x + c ©Fy = 0; a + ©MA = 0;
Ax = 0
Ans.
By - A y = 0
[1]
ByL - MA - M0 = 0
[2]
Elastic Curve: As shown. M/EI Diagrams: M/EI diagrams for By and M0 acting on a cantilever beam are shown. Moment-Area Theorems: From the elastic curve, tB>A = 0. tB>A = 0 =
M0 2 L 1 By L a b(L)a Lb + a b (L)a b 2 EI 3 EI 2 By =
3M0 2L
Ans.
Substituting the value of By into Eqs.[1] and [2] yields, Ay =
3M0 2L
MA =
M0 2
Ans.
1008
M0
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*12–116. The rod is fixed at A, and the connection at B consists of a roller constraint which allows vertical displacement but resists axial load and moment. Determine the moment reactions at these supports. EI is constant.
w
B
A L
Support Reaction: FBD(a). a + ©MA = 0;
MB + MA - wLa
L b = 0 2
[1]
Elastic Curve: As shown. M/EI Diagrams: M/EI diagrams for MB and the uniform distributed load acting on a cantilever beam are shown. Moment-Area Theorems: Since both tangents at A and B are horizontal (parallel), uB>A = 0. uB>A = 0 = a
MB 1 wL2 b(L) + a b(L) EI 3 2EI
MB =
wL2 6
Ans.
Substituting MB into Eq.[1], MA =
wL2 3
Ans.
1009
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•12–117. Determine the value of a for which the maximum positive moment has the same magnitude as the maximum negative moment. EI is constant.
P a
L
(tA>B)1 =
2(L - a) -P(L - a)2(2L + a) 1 -P(L - a) a b (L - a)a a + b = 2 EI 3 6EI
(tA>B)2 =
A yL3 2L 1 A yL a b(L)a b = 2 EI 3 3EI
tA>B = 0 = (tA>B)1 + (tA>B)2 0 =
A yL3 -P(L - a)2(2L + a) + 6EI 3EI Ay =
P(L - a)2(2L + a) 2L3
Require: |M1| = |M2| Pa(L - a)2(2L + a) 3
2L
Pa(L - a)(L + a) =
2L2
a2 + 2La - L2 = 0 a = 0.414L
Ans.
1010
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12–118. Determine the reactions at the supports, then draw the shear and moment diagrams. EI is constant.
M0
M0
A L
Require: tA>B = 0 = a 0 =
M0 L 1 -A y L 2L b(L)a b + a b(L)a b EI 2 2 EI 3
A y L3 M0L2 ; 2EI 3EI
Ay =
3M0 2L
Ans.
Equilibrium: a+ ©MB = 0;
3M0 (L) - Cy (L) = 0 2L Cy =
+ c ©Fy = 0;
By -
Ans.
3M0 3M0 = 0 2L 2L
By = + ©F = 0; : x
3M0 2L
3M0 L
Ans.
Cx = 0
Ans.
1011
C
B L
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12–119. Determine the reactions at the supports, then draw the shear and moment diagrams. EI is constant. Support B is a thrust bearing.
P A
B
L
Support Reactions: FBD(a). + ©F = 0; : x + c ©Fy = 0; a+ ©MA = 0;
Bx = 0
Ans.
-A y + By + Cy - P = 0 By (L) + Cy (2L) - Pa
[1]
3L b = 0 2
[2]
Elastic Curve: As shown. M/EI Diagrams: M/EI diagrams for P and By acting on a simply supported beam are drawn separately. Moment-Area Theorems: (tA>C)1 =
=
1 3PL L 3L L 1 3PL 3L 2 3L a ba ba ba b + a ba ba + b 2 8EI 2 3 2 2 8EI 2 2 6 7PL3 16EI
(tA>C)2 =
By L By L3 1 ab (2L)(L) = 2 2EI 2EI
(tB>C)1 =
1 PL L 2 L PL L L a ba ba ba b + a ba ba b 2 8EI 2 3 2 4EI 2 4 +
=
(tB>C)2 =
1 3PL L L L a ba ba + b 2 8EI 2 2 6
5PL3 48EI By L By L3 1 L ab(L)a b = 2 2EI 3 12EI
tA>C = (tA>C)1 + (tA>C)2 =
By L3 7PL3 16EI 2EI
tB>C = (tB>C)1 + (tB>C)2 =
By L3 5PL3 48EI 12EI
From the elastic curve, tA>C = 2tB>C By L3 By L3 5PL3 7PL3 = 2a b 16EI 2EI 48EI 12EI By =
11P 16
Ans.
Substituting By into Eqs. [1] and [2] yields, Cy =
13P 32
Ay =
3P 32
Ans.
1012
C
L 2
L 2
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*12–120. Determine the moment reactions at the supports A and B. EI is constant.
w
B
A L – 2
-MA 1 -wL2 L 1 Ay L a b(L) + a b (L) + a ba b 2 EI EI 3 8EI 2
uB>A = 0 =
Ay L
0 =
tB>A = 0 =
0 =
2
- MA -
wL2 48
(1)
-MA L L 1 -wL2 L L 1 Ay L a b(L)a b + a b(L)a b + a ba ba b 2 EI 3 EI 2 3 8EI 2 8 Ay L 6
-
MA wL2 2 384
(2)
Solving Eqs. (1) and (2) yields: Ay =
3wL 32
MA =
5wL2 192
c + ©MB = 0;
Ans.
MB +
3wL 5wL2 wL L (L) a b = 0 32 192 2 4
MB =
11wL2 192
Ans.
1013
L – 2
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•12–121.
Determine the reactions at the bearing supports A, B, and C of the shaft, then draw the shear and moment diagrams. EI is constant. Each bearing exerts only vertical reactions on the shaft.
A
1m
1m 400 N
Support Reactions: FBD(a). + c ©Fy = 0; a + ©MA = 0;
A y + By + Cy - 800 = 0
[1]
By (2) + Cy (4) - 400(1) - 400(3) = 0
[2]
Method of superposition: Using the table in Appendix C, the required displacements are yB œ =
yB
fl
Pbx A L2 - b2 - x2 B 6EIL
=
400(1)(2) 2 A 4 - 12 - 2 2 B 6EI(4)
=
366.67 N # m3 EI
T
By A 4 3 B 1.3333By m3 PL3 = = = 48EI 48EI EI
c
The compatibility condition requires (+ T)
0 = 2yB ¿ + yB – 0 = 2a
1.3333By 366.67 b + ab EI EI
By = 550 N
Ans.
Substituting By into Eqs. [1] and [2] yields, A y = 125 N
Cy = 125 N
Ans.
1014
C
B
1m
1m 400 N
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12–122. Determine the reactions at the supports A and B. EI is constant.
P
A
B L
Referring to the FBD of the beam, Fig. a + ©F = 0; : x
Ax = 0
Ans.
By - P - A y = 0
+ c ©Fy = 0;
A y = By - P
(1)
3 a+ ©MA = 0; -MA + By L - Pa Lb = 0 2 MA = By L -
3 PL 2
(2)
Referring to Fig. b and the table in appendix, the necessary deflections are computed as follow: yP =
Px2 (3LAC - x) 6EI
=
P(L2) 3 c 3a L b - L d 6EI 2
=
7PL3 12EI
yBy =
T
By L3 PL3AB c = 3EI 3EI
The compatibility condition at support B requires that (+ T)
0 = vP + vBy 0 =
-By L3 7PL3 + a b 12EI 3EI
By =
7P 4
Ans.
Substitute this result into Eq (1) and (2) Ay =
3P 4
MA =
PL 4
Ans.
1015
L 2
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12–123. Determine the reactions at the supports A, B, and C, then draw the shear and moment diagrams. EI is constant.
12 kip
A
Support Reaction: FBD(b).
+ c ©Fy = 0; a + ©MA = 0;
Cx = 0
Ans.
A y + By + Cy - 12 - 36.0 = 0
[1]
By (12) + Cy (24) - 12(6) - 36.0(18) = 0
[2]
Method of superposition: Using the table in Appendix C, the required displacements are
yB ¿ = yB – =
=
yB –¿ =
5(3) A 24 4 B 6480 kip # ft3 5wL4 = = 768EI 768EI EI
T
Pbx A L2 - b2 - x2 B 6EIL 2376 kip # ft3 12(6)(12) A 24 2 - 62 - 12 2 B = 6EI(24) EI By A 24 3 B 288By ft3 PL3 = = 48EI 48EI EI
T
c
The compatibility condition requires (+ T)
0 = yB ¿ + yB – + yB –¿ 0 =
288By 2376 6480 + + ab EI EI EI
By = 30.75 kip
Ans.
Substituting By into Eqs.[1] and [2] yields, A y = 2.625 kip
Cy = 14.625 kip
Ans.
1016
C
B 6 ft
+ ©F = 0; : x
3 kip/ft
6 ft
12 ft
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*12–124. The assembly consists of a steel and an aluminum bar, each of which is 1 in. thick, fixed at its ends A and B, and pin connected to the rigid short link CD. If a horizontal force of 80 lb is applied to the link as shown, determine the moments created at A and B. Est = 2911032 ksi, Eal = 1011032 ksi.
C
80 lb
D
1 in.
Steel
30 in. Aluminum
0.5 in.
A
; ©Fx = 0
Pal + Pst - 80 = 0
(1)
Compatibility condition: ¢ st = ¢ al PstL3 Pal L3 = 3EstIst 3EalIal Pst = a
1 (29) A 103 B A 12 B (1) A 0.53 B EstIst b (Pal) = Pal 1 EalIal (10) A 103 B A 12 B (1) A 13 B (N)
Pst = 0.3625 Pal
(2)
Solving Eqs. (1) and (2) yields: Pal = 58.72 lb
Pst = 21.28 lb
MA = Pst (30) = 639 lb # in. = 0.639 kip # in.
Ans.
MB = Pal (30) = 1761 lb # in. = 1.76 kip # in.
Ans.
1017
B
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•12–125. Determine the reactions at the supports A, B, and C, then draw the shear and moment diagrams. EI is constant. A 3m
Cx = 0
Ans.
a+ ©MC = 0; A y(12) + By(16) - 10(3) - 10(9) = 0 2A y + By = 20 + c ©Fy = 0;
(1)
A y + By + Cy - 10 - 10 = 0 A y + By + Cy = 20
(2)
Referring to Fig. b and table in appendix, the necessary deflections are: (vP)1 = (vP)2 =
Pbx A L2 - b2 - x 2 B 6EILAC AC
=
10(3)(6) A 12 2 - 32 - 62 B 6EI(12)
=
247.5 kN # m3 EI
T
By(12 3) 36 By PL3AC c (vB)y = = = 48EI 48EI EI The compatibility condition at support B requires that (+ T) 0 = (vP)1 + (vP)2 + (vB)y 0 =
36 By 247.5 247.5 + + ab EI EI EI
By = 13.75 kN
Ans.
Substitute this result into Eq. (1) and (2) and solve, A y = Cy = 3.125 kN
Ans.
The shear And moment diagrams are shown in Fig. b and c respectively.
1018
C
B
Referring to the FBD of the beam, Fig. a, + ©F = 0; ; x
10 kN
10 kN
3m
3m
3m
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12–126. Determine the reactions at the supports A and B. EI is constant.
M0 A B L
Referring to the FBD of the beam, Fig. a, + ©F = 0; : x
Ax = 0
+ c ©Fy = 0;
By - A y = 0
Ans. (1)
By(L) - Mo - MA = 0
a+ ©MA = 0;
MA = ByL - Mo
(2)
Referring to Fig. b and the table in the appendix, the necessary deflections are: vMo =
vBy
MoL2 2EI
T
ByL3 PL3 = = 3EI 3EI
c
Compatibility condition at roller support B requires (+ T)
0 = vMo + (vB)y 0 =
By =
ByL3 MoL2 + ab 2EI 3EI 3Mo 2L
Ans.
Substitute this result into Eq. (1) and (2) Ay =
3Mo 2L
MA =
Mo 2
Ans.
1019
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12–127. Determine the reactions at support C. EI is constant for both beams.
P D
B
A
C L 2
Support Reactions: FBD (a). + ©F = 0; : x
Cx = 0
a + ©MA = 0;
Cy(L) - By a
Ans. L b = 0 2
[1]
Method of superposition: Using the table in Appendix C, the required displacements are yB =
yB ¿ =
yB – =
By L3 PL3 = 48EI 48EI
T
P A L2 B 3 PL3BD PL3 = = 3EI 3EI 24EI By L3 PL3BD = 3EI 24EI
T
c
The compatibility condition requires yB = yB ¿ + yB –
(+ T)
By L3 48EI
=
By =
By L3 PL3 + ab 24EI 24EI 2P 3
Substituting By into Eq. [1] yields, Cy =
P 3
Ans.
1020
L 2
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*12–128. The compound beam segments meet in the center using a smooth contact (roller). Determine the reactions at the fixed supports A and B when the load P is applied. EI is constant.
P
A C
B L
L
(P - R)L3 RL3 = 3EI 3EI
¢C =
R =
P 2
Member AC: ©Fy = 0;
Ay =
P 2
Ans.
©Fx = 0;
Ax = 0
Ans.
© MA = 0;
MA =
PL 2
Ans.
Member BC: ©Fy = 0;
By =
P 2
Ans.
©Fx = 0;
Bx = 0
Ans.
©MB = 0;
MB =
PL 2
Ans.
The beam has a constant E1 I1 and is supported by the fixed wall at B and the rod AC. If the rod has a crosssectional area A2 and the material has a modulus of elasticity E2, determine the force in the rod.
•12–129.
(¢ A)¿ =
dA =
wL41 ; 8E1I1
¢A =
C L2
TACL2 A 2E2
L1
By superposition: ¢ A = (¢ A)¿ - dA
TACL2 wL41 TACL31 = A 2E2 8E1I1 3E1I1 TAC a
L2 L31 wL41 + b = A 2E2 3E1I1 8E1I1
TAC =
B
A
TAC L31 3E1I1
(+ T)
w
3wA 2E2L41
8 C 3E1I1L2 + A 2E2L31 D
Ans.
1021
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12–130. Determine the reactions at A and B. Assume the support at A only exerts a moment on the beam. EI is constant.
P
A 2
(uA)1 =
PL ; 8EI
(uA)2 =
B
MAL EI
L – 2
By superposition:
L – 2
0 = (uA)1 - (uA)2 0 =
MAL PL2 8EI EI
MA =
PL 8
Ans.
Equilibrium: a + ©MB = 0; MB =
-
PL PL + - MB = 0 8 2
3PL 8
Ans.
+ ©F = 0 ; : x
Bx = 0
Ans.
+ c ©Fy = 0 ;
By = P
Ans.
12–131. The beam is supported by the bolted supports at its ends. When loaded these supports do not provide an actual fixed connection, but instead allow a slight rotation a before becoming fixed. Determine the moment at the connections and the maximum deflection of the beam.
P
u - u¿ = a L — 2
ML ML PL2 = a 16EI 3EI 6EI ML = a M = a
PL2 - a b(2EI) 16EI
2EI PL ab 8 L
¢ max = ¢ - ¢ ¿ =
Ans.
M(L2 ) PL3 - 2c C L2 - (L>2)2 D d 48EI 6EIL
¢ max =
PL3 L2 PL 2EIa a b 48EI 8EI 8 L
¢ max =
aL PL3 + 192EI 4
Ans.
1022
L — 2
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*12–132. The beam is supported by a pin at A, a spring having a stiffness k at B, and a roller at C. Determine the force the spring exerts on the beam. EI is constant.
w
A B
Method of Superposition: Using the table in appendix C, the required displacements are 5w(2L)4 5wL4AC 5wL4 = = 384EI 384EI 24EI
T
Fsp (2L)3 Fsp L3 PL3AC = = yB – = 48EI 48EI 6EI
c
yB ¿ =
Using the spring formula, ysp =
Fsp k
L
C k L
.
The compatibility condition requires ysp = yB ¿ + yB –
(+ T)
3
Fsp k
=
Fsp =
Fsp L 5wL4 + ab 24EI 6EI 5wkL4
4 A 6EI + kL3 B
Ans.
•12–133.
The beam is made from a soft linear elastic material having a constant EI. If it is originally a distance ¢ from the surface of its end support, determine the distance a at which it rests on this support when it is subjected to the uniform load w0 , which is great enough to cause this to happen.
w0 ⌬
a L
The curvature of the beam in region BC is zero, therefore there is no bending moment in the region BC, The reaction F is at B where it touches the support. The slope is zero at this point and the deflection is ¢ where ¢ =
R(L - a)3 w0(L - a)4 8EI 3EI
u1
=
w0(L - a)3 R(L - a)2 6EI 2EI
Thus, 1
8¢EI 4 R = a b 9w30
Ans. 1
72¢EI 4 L - a = a b w0 1
72¢EI 4 a = L - a b w0
Ans.
1023
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12–134. Before the uniform distributed load is applied on the beam, there is a small gap of 0.2 mm between the beam and the post at B. Determine the support reactions at A, B, and C. The post at B has a diameter of 40 mm, and the moment of inertia of the beam is I = 875(106) mm4. The post and the beam are made of material having a modulus of elasticity of E = 200 GPa .
30 kN/m
A
6m
Equations of Equilibrium. Referring to the free-body diagram of the beam, Fig. a, + ©F = 0; : x
Ax = 0
+ c ©Fy = 0;
A y + FB + Cy - 30(12) = 0
a + ©MA = 0;
Ans. (1)
FB(6) + Cy(12) - 30(12)(6) = 0
(2)
Method of superposition: Referring to Fig. b and the table in the Appendix, the necessary deflections are
(vB)1 =
(vB)2 =
5(30) A 12 4 B 8100kN # m3 5wL4 = = T 384EI 384EI EI FB A 12 3 B 36FB PL3 = = 48EI 48EI EI
c
The deflection of point B is vB = 0.2 A 10 - 3 B +
FB(a) FBLB = 0.2 A 10 - 3 B + AE AE
T
The compatibility condition at support B requires
A+TB
vB = (vB)1 + (vB)2 0.2 A 10 - 3 B +
FB (1) 36FB 8100 = + ab AE EI EI
0.2 A 10 - 3 B E + FB
p A 0.04 2 B 4
+
FB 36FB 8100 = A I I 36FB
875 A 10 - 6 B
8100
=
875 A 10 - 6 B
-
C 1m
0.2 A 10 - 3 B C 200 A 109 B D 1000
FB = 219.78 kN = 220 kN
Ans.
Substituting the result of FB into Eqs. (1) and (2), A y = Cy = 70.11 kN = 70.1 kN
Ans.
1024
B
0.2 mm 6m
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12–135. The 1-in.-diameter A-36 steel shaft is supported by unyielding bearings at A and C. The bearing at B rests on a simply supported steel wide-flange beam having a moment of inertia of I = 500 in4. If the belt loads on the pulley are 400 lb each, determine the vertical reactions at A, B, and C.
3 ft 5 ft
A
2 ft 5 ft
B
For the shaft: (¢ b)1 =
(¢ b)2 =
400 lb
800(3)(5) 13200 A -52 - 32 + 102 B = 6EIs(10) EIs By A 103 B 48EIs
C
5 ft
20.833By =
EIs
For the beam:
¢b =
By A 103 B 48EIb
20.833By =
EIb
Compatibility condition: + T ¢ b = (¢ b)1 - (¢ b)2 20.833By EIb Is =
=
20.833By 13200 EIs EIs
p (0.5)4 = 0.04909 in4 4
20.833By (0.04909) 500
400 lb
= 13200 - 20.833By
By = 634 lb
Ans.
Form the free-body digram, A y = 243 lb
Ans.
Cy = 76.8 lb
Ans.
1025
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*12–136. If the temperature of the 75-mm-diameter post CD is increased by 60°C, determine the force developed in the post. The post and the beam are made of A-36 steel, and the moment of inertia of the beam is I = 255(106) mm4.
3m
3m
A B
C 3m
D
Method of Superposition. Referring to Fig. a and the table in the Appendix, the necessary deflections are (vC)1 =
FCD A 33 B 9FCD PLBC 3 c = = 3EI 3EI EI
(vC)2 = (uB)2LBC =
3FCD (3) 9FCD MOLAB c (LBC) = (3) = 3EI 3EI EI
The compatibility condition at end C requires
A+cB
vC = (vC)1 + (vC)2 =
9FCD 9FCD 18FCD c + = EI EI EI
Referring to Fig. b, the compatibility condition of post CD requires that dFCD + vC = dT dFCD =
(1)
FCD (3) FCD LCD = AE AE
dT = a¢TL = 12 A 10 - 6 B (60)(3) = 2.16 A 10 - 3 B m Thus, Eq. (1) becomes 3FCD 18FCD + = 2.16 A 10 - 3 B AE EI 3FCD
p A 0.0752 B 4
+
18FCD
255 A 10 - 6 B
= 2.16 A 10 - 3 B C 200 A 109 B D
FCD = 6061.69N = 6.06 kN
Ans.
1026
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•12–137.
The shaft supports the two pulley loads shown. Using discontinuity functions, determine the equation of the elastic curve. The bearings at A and B exert only vertical reactions on the shaft. EI is constant.
x A
12 in.
B
12 in. 70 lb
180 lb
M = -180 6 x - 0 7 - (-277.5) 6 x - 12 7 - 70 6 x - 24 7 M = -180x + 277.5 6 x - 12 7 - 70 6 x - 24 7 Elastic curve and slope: EI
d2v = M = -180x + 277.5 6 x - 12 7 - 70 6 x - 24 7 dx2
EI
dv = -90x2 + 138.75 6 x - 12 7 dx
EIv = -30x3 + 46.25 6 x - 12 7
3
2
- 35(x - 24 7
2
- 11.67 6 x - 24 7
+ C1 3
+ C1x + C2 (1)
Boundary conditions: v = 0
at
x = 12 in,
From Eq. (1) 0 = -51,840 + 12C1 + C2 12C1 + C2 = 51 840 v = 0
at
(2)
x = 60 in.
From Eq.(1) 0 = -6 480 000 + 5 114 880 - 544 320 + 60C1 + C2 60C1 + C2 = 1909440
(3)
Solving Eqs. (2) and (3) yields: C1 = 38 700 v =
C2 = -412 560
1 [-30x3 + 46.25 6 x - 12 7 EI
3
- 11.7 6 x - 24 7
3
+ 38 700x - 412 560]
Ans.
1027
36 in.
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12–138. The shaft is supported by a journal bearing at A, which exerts only vertical reactions on the shaft, and by a thrust bearing at B, which exerts both horizontal and vertical reactions on the shaft. Draw the bending-moment diagram for the shaft and then, from this diagram, sketch the deflection or elastic curve for the shaft’s centerline. Determine the equations of the elastic curve using the coordinates x1 and x2 . EI is constant.
80 lb A
x1
EI
d2v1 dx21
= 26.67x1
dv1 = 13.33x21 + C1 dx1
(1)
EIv1 = 4.44x31 + C1x1 + C2
(2)
EI
For M2 (x) = -26.67x2 EI
d2v2 dx22
= -26.67x2
dv2 = -13.33x22 + C3 dx2
(3)
EIv2 = -4.44x32 + C3x2 + C4
(4)
EI
Boundary conditions: v1 = 0
at
x1 = 0
at
x2 = 0
From Eq.(2) C2 = 0 v2 = 0 C4 = 0 Continuity conditions: dv1 dv2 = dx1 dx2
at
x1 = x2 = 12
From Eqs. (1) and (3) 1920 + C1 = -( -1920 + C3) C1 = -C3 v1 = v2
(5)
x1 = x2 = 12
at
7680 + 12C1 = -7680 + 12C3 C3 - C1 = 1280
(6)
Solving Eqs. (5) and (6) yields: C3 = 640
80 lb 12 in.
For M1 (x) = 26.67 x1
C1 = -640
v1 =
1 A 4.44x31 - 640x1 B lb # in3 EI
Ans
v2 =
1 A -4.44x32 + 640x2 B lb # in3 EI
Ans.
1028
B
4 in. 4 in. x2 12 in.
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12–139. The W8 * 24 simply supported beam is subjected to the loading shown. Using the method of superposition, determine the deflection at its center C. The beam is made of A-36 steel.
6 kip/ft 5 kip⭈ft A
B C 8 ft
Elastic Curves: The elastic curves for the uniform distributed load and couple moment are drawn separately as shown. Method of superposition: Using the table in Appendix C, the required displacements are
(¢ C)1 =
-5(6) A 164 B 2560 kip # ft3 -5wL4 = = T 768EI 768EI EI
(¢ C)2 = -
= -
=
M0x A L2 - x2 B 6EIL 5(8) C (16)2 - (8)2 D 6EI(16)
80 kip # ft3 EI
T
The displacement at C is ¢ C = (¢ C)1 + (¢ C)2 =
80 2560 + EI EI
=
2640 kip # ft3 EI 2640(1728)
=
29 A 103 B (82.8)
= 1.90 in.
Ans.
T
1029
8 ft
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*12–140. Using the moment-area method, determine the slope and deflection at end C of the shaft. The 75-mmdiameter shaft is made of material having E = 200 GPa. Support Reactions and
B
A
M Diagram. As shown in Fig. a. EI
1m
1m
15 kN
= (1) B
=
冷tC>A 冷
1 7.5 1 3 1 a b (2) R + c (2) d B a b(2) R 2 EI 3 2 EI
5.5 kN # m3 EI
= (1 + 1) B
1 7.5 1 1 3 a b(2) R + c (2) + 1 d B a b(2) R 2 EI 3 2 EI 2 3 1 + c (1) d B a b(1) R 3 2 EI
冷uC>A 冷
=
9 kN # m3 EI
=
1 3 1 7.5 a b(2) + a b(3) 2 EI 2 EI
=
3 kN # m3 EI
Referring to the geometry of the elastic curve, Fig. b,
uA =
冷tB>A 冷 LAB
5.5 EI 2.75kN # m2 = = 2 EI
uC = uC>A - uA =
=
3 2.75 EI EI
0.25 kN # m2 = EI
0.25 A 103 B
200 A 109 B c
p A 0.03754 B d 4
= 0.805 A 10 - 3 B rad
Ans.
and ¢ C = 冷 tC>A冷 - 冷 tB>A ¢
=
9 5.5 3 a b EI EI 2
=
0.75 kN # m3 = EI
LAC ≤ LAB
0.75 A 103 B
p 200 A 10 B c A 0.03754 B d 4
1m 3 kN
Moment Area Theorem. Referring to Fig. b,
冷tB>A 冷
C
= 0.002414 m = 2.41 mm c Ans.
9
1030
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•12–141. Determine the reactions at the supports. EI is constant. Use the method of superposition.
w A
wL C L3 - 2(3L)L2 + (3L)3 D 24EI
¢B = ¢C =
D B L
4
11wL 12EI
=
Due to symmetry, By = Cy By (L)(2L) ¢ BB = ¢ CC =
6EI(3L)
C (3L)2 - (2L)2 - L2 D
4By L3 =
9EI By (L)(L)
¢ BC = ¢ CB =
6EI(3L)
C -L2 - L2 + (3L)2 D
7By L3 =
18EI
By superposition: +T
0 =
0 = ¢ B - ¢ BB - ¢ BC 4By L3 7By L3 11wL4 12EI 9EI 18EI
By = Cy =
11wL 10
Ans.
Equilibrium: a+ ©MD = 0; Ay = c + ©Fy = 0; Dy = + ©F = 0; ; x
3wLa
3L 11wL 11wL b (L) (2L) - A y (3L) = 0 2 10 10
2wL 5
Ans.
2wL 11wL 11wL + + + Dy - 3wL = 0 5 10 10 2wL 5
Ans.
Dx = 0
Ans.
1031
C L
L
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12–142. Determine the moment reactions at the supports A and B. Use the method of integration. EI is constant.
w0
A
B L
Support Reactions: FBD(a). A y + By -
+ c ©Fy = 0; a + ©MA = 0;
w0L = 0 2
[1]
ByL + MA - MB -
w0L L a b = 0 2 3
[2]
Moment Function: FBD(b). a+ ©MNA = 0;
-M(x) -
x 1 w0 a xb x a b - MB + Byx = 0 2 L 3
M(x) = Byx -
w0 3 x - MB 6L
Slope and Elastic Curve: EI
EI
EI
EI y =
d2y = M(x) dx2
w0 3 d2y = Byx x - MB 6L dx2
By w0 4 dy = x2 x - MBx + C1 dx 2 24L By 6
x3 -
[3]
w0 5 MB 2 x x + C1x + C2 120L 2
[4]
Boundary Conditions: At x = 0,
dy = 0 dx
From Eq.[3],
At x = 0, y = 0. At x = L,
0 =
From Eq.[4],
dy = 0. dx
By L2 2
-
C1 = 0 C2 = 0
From Eq. [3].
w0L3 - MBL 24
0 = 12By L - w0 L2 - 24MB At x = L, y = 0. 0 =
By L3 6
-
[5]
From Eq. [4], w0 L4 MB L2 120 2
0 = 20By L - w0 L2 - 60MB
[6]
1032
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12–142.
Continued
Solving Eqs. [5] and [6] yields, MB = By =
w0 L2 30
Ans.
3w0L 20
Substituting By and MB into Eqs. [1] and [2] yields, MA =
w0L2 20
Ay =
7w0 L 20
Ans.
1033
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12–143. If the cantilever beam has a constant thickness t, determine the deflection at end A. The beam is made of material having a modulus of elasticity E.
L w0 x
Section Properties: Referring to the geometry shown in Fig. a, A
h(x) h0 = ; x L
h0 x h(x) = L
h0 B
Thus, the moment of inertia of the tapered beam as a function of x is I(x) =
3 h0 th0 3 3 1 1 t C h(x) D 3 = t¢ x≤ = x 12 12 L 12L3
Moment Function. Referring to the free-body diagram of the beam’s segment, Fig. b, M(x) + B
a + ©MO = 0;
1 w0 x a xbx R a b = 0 2 L 3
M(x) = -
w0 3 x 6L
Equations of slope and Elastic Curve. E
M(x) d2v = 2 I(x) dx
w0 3 x 2w0L2 dv 6L E 2 = = dx th0 3 3 th0 3 x 3 12L -
2
E
2w0L2 dv = x + C1 dx th0 3
Ev = -
w0L2 th0 3
(1)
x2 + C1x + C2
Boundary conditions. At x = L,
0 = -
2w0L2 th0 3
(2)
dv = 0. Then Eq. (1) gives dx
(L) + C1
C1 =
2w0L3 th0 3
At x = L, v = 0. Then Eq. (2) gives 0 = -
w0L2 th0
3
A L2 B +
2w0L3 th0
3
(L) + C2
C2 = -
w0L4 th0 3
Substituting the results of C1 and C2 into Eq. (2), v =
w0L2 Eth0 3
A -x2 + 2Lx - L2 B
At A, x = 0. Then vA = v冷x = 0 = -
w0L4 Eth0 3
=
w0L4 Eth0 3
Ans.
T
1034
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*12–144. Beam ABC is supported by beam DBE and fixed at C. Determine the reactions at B and C. The beams are made of the same material having a modulus of elasticity E = 200 GPa, and the moment of inertia of both beams is I = 25.0(106) mm4.
100 lb/ft
a
A
B
C
D
E a 4 ft
4 ft 6 ft
6 in. Section a – a
+ ©F = 0; : x
Cx = 0
+ c ©Fy = 0;
By + Cy - 9(6) = 0
a+ ©MC = 0;
9(6)(3) - By(4) - MC = 0
Ans. (1)
MC = 162 - 4By
(2)
Method of superposition: Referring to Fig. b and the table in the appendix, the deflections are
vB =
By A 63 B 4.5By PLDE 3 = = T 48EI 48EI EI
(vB)1 =
=
(vB)2 =
9 A 42 B wx2 A x2 - 4Lx + 6L2 B = C 4 2 - 4(6)(4) + 6 A 62 B D 24EI 24EI 816 kN # m3 T EI By A 4 3 B 21.3333By PLBC 3 c = = 3EI 3EI EI
The compatibility condition at support B requires that
A+TB
vB = (vB)1 + (vB)2 4.5By EI
=
21.3333By 816 + ab EI EI
By = 31.59 kN = 31.6 kN
Ans.
Substituting the result of By into Eqs. (1) and (2), MC = 35.65 kN # m = 35.7 kN # m
Ans.
Cy = 22.41 kN = 22.4 kN
Ans.
1035
a
6 ft
3 in.
Equation of Equilibrium. Referring to the free-body diagram of the beam, Fig. a,
a
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•12–145.
Using the method of superposition, determine the deflection at C of beam AB. The beams are made of wood having a modulus of elasticity of E = 1.5(103) ksi.
100 lb/ft
a
A
B
C
D
E a 4 ft
4 ft 6 ft
6 in. Section a – a
Method of superposition. Referring to Fig. b and the table in the appendix, the deflection of point B is
¢B =
600 A 83 B PLDE 3 6400 lb # ft3 = = T 48EI 48EI EI
Subsequently, referring to Fig. c, (¢ C)1 = ¢ B a
(¢ C)2 =
6 6400 6 3200 lb # ft3 b = a b = T 12 EI 12 EI
5(100) A 12 4 B 5wL4 27000 lb # ft3 = = T 384EI 384EI EI
Thus, the deflection of point C is
A+TB
¢ C = (¢ C)1 + (¢ C)2 =
3200 27000 + EI EI
30200 lb # ft3 = = EI
30200 A 12 3 B
1.5 A 106 B c
1 (3) A 63 B d 12
= 0.644 in T
Ans.
1036
a
6 ft
3 in.
Support Reactions: The reaction at B is shown on the free-body diagram of beam AB, Fig. a.
a
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12–146. The rim on the flywheel has a thickness t, width b, and specific weight g. If the flywheel is rotating at a constant rate of v, determine the maximum moment developed in the rim. Assume that the spokes do not deform. Hint: Due to symmetry of the loading, the slope of the rim at each spoke is zero. Consider the radius to be sufficiently large so that the segment AB can be considered as a straight beam fixed at both ends and loaded with a uniform centrifugal force per unit length. Show that this force is w ⫽ btgv2r>g.
A
B
v r
Centrifugal Force: The centrifugal force action on a unit length of the rim rotating at a constant rate of v is g btgv2r w = mv2 r = bta bv2r = g g
(Q.E.D.)
Elastic Curve: Member AB of the rim is modeled as a straight beam with both of its ends fixed and subjected to a uniform centrifigal force w. Method of Superposition: Using the table in Appendix C, the required displacements are uB ¿ =
wL3 6EI
yB ¿ =
wL4 c 8EI
uB – =
yB – =
MBL EI
uB ¿– =
MBL2 c 2EI
yB –¿ =
ByL2 2EI
ByL3 3EI
T
Computibility requires, 0 = uB ¿ + uB – + uB ¿– 2
0 =
By L MBL wL3 + + ab 6EI EI 2EI
0 = wL2 + 6MB - 3By L
[1]
0 = yB ¿ + yB – + yB –¿
(+ c )
3
0 =
By L MB L2 wL4 + + ab 8EI 2EI 3EI
0 = 3wL2 + 12MB - 8By L
[2]
Solving Eqs. [1] and [2] yields, By =
wL 2
MB =
Due to symmetry, A y =
wL 2
wL2 12 MA =
wL2 12
Maximum Moment: From the moment diagram, the maximum moment occurs at btgv2r pr the two fixed end supports. With w = and L = ru = . g 3
Mmax
wL2 = = 12
A B
btgv2r pr 2 g 3
12
=
t
p2btgv2r3 108g
Ans.
1037
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•13–1.
Determine the critical buckling load for the column. The material can be assumed rigid.
P
L 2 k
Equilibrium: The disturbing force F can be determined by summing moments about point A. a + ©MA = 0;
P(Lu) - F a
L b = 0 2
A
F = 2Pu Spring Formula: The restoring spring force F1 can be determine using spring formula Fs = kx. Fs = ka
L kLu ub = 2 2
Critical Buckling Load: For the mechanism to be on the verge of buckling, the disturbing force F must be equal to the restoring spring force F1. 2Pcr u =
Pcr =
L 2
kLu 2 kL 4
Ans.
1038
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13–2. Determine the critical load Pcr for the rigid bar and spring system. Each spring has a stiffness k.
P
Equilibrium: The disturbing forces F1 and F2 can be related to P by writing the moment equation of equlibrium about point A. Using small angle ananlysis, where cos u ⬵ 1 and sin u = u, + ©MA = 0;
F2 a
L 3 k
L 2 b + F1 a Lb - PLu = 01 3 3
L 3
F2 + 2F1 = 3Pu
k
(1)
Spring Force. The restoring spring force A Fsp B 1 and A Fsp B 2 can be determined using the spring formula, 2 1 Lu and x2 = Lu, Fig. b. Thus, 3 3 2 2 = kx1 = ka Lu b = kLu 3 3
L 3 A
Fsp = kx, where x1 =
A Fsp B 1
A Fsp B 2 = kx2 = ka Lu b = 1 3
Critical Buckling Load. When the mechanism is on the verge of buckling the disturbing force F must be equal to the restoring force of the spring Fsp. Thus, F1 = A Fsp B 1 =
2 kLu 3
F2 = A Fsp B 2 =
1 kLu 3
Substituting this result into Eq. (1), 2 1 kLu + 2 a kLu b = 3Pcr u 3 3 Pcr =
5 kL 9
Ans.
1039
1 kLu 3
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13–3. The leg in (a) acts as a column and can be modeled (b) by the two pin-connected members that are attached to a torsional spring having a stiffness k (torque兾rad). Determine the critical buckling load. Assume the bone material is rigid.
a + ©MA = 0;
-P(u)a
P
L b + 2ku = 0 2
L — 2
Require:
k
Pcr =
4k L
Ans.
L — 2
(a)
*13–4. Rigid bars AB and BC are pin connected at B. If the spring at D has a stiffness k, determine the critical load Pcr for the system.
(b)
P A
Equilibrium. The disturbing force F can be related P by considering the equilibrium of joint A and then the equilibrium of member BC,
a B
Joint A (Fig. b) + c ©Fy = 0;
FAB cos f - P = 0
FAB =
a
P cos f
k D
Member BC (Fig. c)
a
©MC = 0; F(a cos u) -
P P cos f (2a sin u) sin f(2a cos u) = 0 cos f cos f
C
F = 2P(tan u + tan f) Since u and f are small, tan u ⬵ u and tan f ⬵ f. Thus, F = 2P(u + f)
(1)
Also, from the geometry shown in Fig. a, 2au = af
f = 2u
Thus Eq. (1) becomes F = 2P(u + 2u) = 6Pu Spring Force. The restoring spring force Fsp can be determined using the spring formula, Fsp = kx, where x = au, Fig. a. Thus, Fsp = kx = kau
1040
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13–4.
Continued
Critical Buckling Load. When the mechanism is on the verge of buckling the disturbing force F must be equal to the restoring spring force Fsp. F = Fsp
6Pcru = kau
Pcr =
ka 6
Ans.
1041
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•13–5.
An A-36 steel column has a length of 4 m and is pinned at both ends. If the cross sectional area has the dimensions shown, determine the critical load. 25 mm
Section Properties: A = 0.01(0.06) + 0..05(0.01) = 1.10 A 10 - 3 B m2 Ix = Iy =
10 mm
1 1 (0.01) A 0.063 B + (0.05) A 0.013 B = 0.184167 A 10 - 6 B m4 12 12
25 mm
Critical Buckling Load: K = 1 for pin supported ends column. Applying Euler’s formula, Pcr =
25 mm
25 mm 10 mm
p2EI (KL)2 p2 (200)(109)(0.184167)(10 - 6)
=
[1(4)]2 Ans.
= 22720.65 N = 22.7 kN Critical Stress: Euler’s formula is only valid if scr 6 sg. scr =
Pcr 22720.65 = 20.66 MPa 6 sg = 250 MPa = A 1.10(10 - 3)
O.K.
13–6. Solve Prob. 13–5 if the column is fixed at its bottom and pinned at its top. 25 mm
Section Properties: A = 0.01(0.06) + 0.05(0.01) = 1.10 A 10 - 3 B m2
10 mm
1 1 Ix = Iy = (0.01) A 0.063 B + (0.05) A 0.013 B = 0.184167 A 10 - 6 B m4 12 12
25 mm
Critical Buckling Load: K = 0.7 for one end fixed and the other end pinned column. Applying Euler’s formula, Pcr =
p EI (EL)2 p2 (200)(109)(0.184167)(10 - 6)
=
[0.7(4)]2 Ans.
= 46368.68 N = 46.4 kN Critical Stress: Euler’s formula is only valid if scr 6 sg. scr =
25 mm
25 mm 10 mm
2
Pcr 46368.68 = 42.15 MPa 6 sg = 250 MPa = A 1.10(10 - 3)
1042
O.K.
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13–7. A column is made of A-36 steel, has a length of 20 ft, and is pinned at both ends. If the cross-sectional area has the dimensions shown, determine the critical load.
6 in. 0.25 in.
The cross sectional area and moment of inertia of the square tube is
5.5 in.
A = 6(6) - 5.5(5.5) = 5.75 in2 I =
0.25 in.
1 1 (6)(63) (5.5)(5.53) = 31.74 in4 12 12
0.25 in.
0.25 in.
The column is pinned at both of its end, k = 1. For A36 steel, E = 29.0(103) ksi and sg = 36 ksi (table in appendix). Applying Euler’s formula, Pcr =
p2 C 29.0(103) D (31.74) p2EI = (KL)2 C 1(20)(12) D 2
Ans.
= 157.74 kip = 158 Critical Stress. Euler’s formula is valid only if scr 6 sg. scr =
Pcr 157.74 = = 27.4 ksi 6 sg = 36 ksi A 5.75
O.K.
*13–8. A column is made of 2014-T6 aluminum, has a length of 30 ft, and is fixed at its bottom and pinned at its top. If the cross-sectional area has the dimensions shown, determine the critical load.
6 in. 0.25 in.
5.5 in.
The cross-sectional area and moment of inertia of the square tube is 0.25 in.
A = 6(6) - 5.5(5.5) = 5.75 in2
0.25 in.
1 1 I = (6)(63) (5.5)(5.53) = 31.74 in4 12 12 The column is fixed at one end, K = 0.7. For 2014–76 aluminium, E = 10.6(103) ksi and sg = 60 ksi (table in appendix). Applying Euler’s formula, Pcr =
p2 C 10.6(103) D (31.74) p2EI = (KL)2 C 0.7(30)(12) D 2
Ans.
= 52.29 kip = 52.3 kip Critical Stress. Euler’s formula is valid only if scr 6 sg. scr =
Pcr 52.3 = = 9.10 ksi 6 sg = 60 ksi A 5.75
O.K.
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•13–9.
The W14 * 38 column is made of A-36 steel and is fixed supported at its base. If it is subjected to an axial load of P = 15 kip, determine the factor of safety with respect to buckling.
P
From the table in appendix, the cross-sectional area and moment of inertia about weak axis (y-axis) for W14 * 38 are 20 ft
A = 11.2 in2
Iy = 26.7 in4
The column is fixed at its base and free at top, k = 2. Here, the column will buckle about the weak axis (y axis). For A36 steel, E = 29.0(103) ksi and sy = 36 ksi. Applying Euler’s formula, p2 C 29.0(103) D (26.7)
p2EIy
Pcr =
C 2 (20)(12) D 2
=
(KL)2
= 33.17 kip
Thus, the factor of safety with respect to buckling is F.S =
Pcr 33.17 = = 2.21 P 15
Ans.
The Euler’s formula is valid only if scr 6 sg. scr =
Pcr 33.17 = = 2.96 ksi 6 sg = 36 ksi A 11.2
O.K.
13–10. The W14 * 38 column is made of A-36 steel. Determine the critical load if its bottom end is fixed supported and its top is free to move about the strong axis and is pinned about the weak axis.
P
From the table in appendix, the cross-sectional area and moment of inertia about weak axis (y-axis) for W14 * 38 are A = 11.2 in2
Ix = 385 in4
Iy = 26.7 in4
The column is fixed at its base and free at top about strong axis. Thus, kx = 2. For A36 steel, E = 29.0(103) ksi and sg = 36 ksi. Pcr =
p2EIx (KxLx)
2
=
p2 C 29.0(103) D (385)
C 2 (20)(12) D 2
= 478.28 kip
The column is fixed at its base and pinned at top about weak axis. Thus, ky = 0.7. Pcr =
p2EIy 2
(KyLy)
=
p2 C 29.0(103) D (26.7)
C 0.7(20)(12) D 2
Ans.
= 270.76 kip = 271 kip (Control) The Euler’s formula is valid only if scr 6 sg. scr =
Pcr 270.76 = = 24.17 ksi 6 sg = 36 ksi A 11.2
O.K.
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13–11. The A-36 steel angle has a cross-sectional area of A = 2.48 in2 and a radius of gyration about the x axis of rx = 1.26 in. and about the y axis of ry = 0.879 in. The smallest radius of gyration occurs about the z axis and is rz = 0.644 in. If the angle is to be used as a pin-connected 10-ft-long column, determine the largest axial load that can be applied through its centroid C without causing it to buckle.
y z C
x
x z
y
The least radius of gyration: r2 = 0.644 in. scr =
p2E
2 A KL r B
controls. K = 1.0
;
p2 (29)(103) (120) 2 C 1.00.644 D
=
= 8.243 ksi 6 sg
O.K.
Pcr = scr A = 8.243 (2.48) = 20.4 kip
Ans.
*13–12. An A-36 steel column has a length of 15 ft and is pinned at both ends. If the cross-sectional area has the dimensions shown, determine the critical load.
8 in. 0.5 in.
0.5 in. 6 in. 0.5 in.
Ix =
1 1 (8)(73) (7.5)(63) = 93.67 in4 12 12
Iy = 2 a Pcr =
1 1 b(0.5)(83) + (6)(0.53) = 42.729 in4 (controls) 12 12
p2(29)(103)(42.729) p2EI = 2 (EL) [(1.0)(15)(12)]2 = 377 kip
Ans.
Check: A = (2)(8)(0.5) + 6(0.5) = 11 in2 scr =
Pcr 377 = = 34.3 ksi 6 sg A 11
Therefore, Euler’s formula is valid
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•13–13.
An A-36 steel column has a length of 5 m and is fixed at both ends. If the cross-sectional area has the dimensions shown, determine the critical load. I =
10 mm 10 mm
50 mm
1 1 (0.1)(0.053) (0.08)(0.033) = 0.86167 (10 - 6) m4 12 12
Pcr =
100 mm
p2(200)(109)(0.86167)(10 - 6) p2EI = 2 (KL) [(0.5)(5)]2 = 272 138 N = 272 kN
scr =
=
Pcr ; A
Ans.
A = (0.1)(0.05) - (0.08)(0.03) = 2.6(10 - 3) m2
272 138 = 105 MPa 6 sg 2.6 (10 - 3)
Therefore, Euler’s formula is valid.
13–14. The two steel channels are to be laced together to form a 30-ft-long bridge column assumed to be pin connected at its ends. Each channel has a cross-sectional area of A = 3.10 in2 and moments of inertia Ix = 55.4 in4, Iy = 0.382 in4. The centroid C of its area is located in the figure. Determine the proper distance d between the centroids of the channels so that buckling occurs about the x–x and y¿ – y¿ axes due to the same load. What is the value of this critical load? Neglect the effect of the lacing. Est = 2911032 ksi, sY = 50 ksi.
y 0.269 in.
C d
y
In order for the column to buckle about x - x and y - y at the same time, Iy must be equal to Ix Iy = Ix 0.764 + 1.55 d2 = 110.8 d = 8.43 in.
Ans.
Check: d 7 2(1.231) = 2.462 in.
O.K. 3
p (29)(10 )(110.8) p2 EI = 2 (KL) [1.0(360)]2
= 245 kip
Ans.
Check stress: scr =
x C
d 2 Iy = 2(0.382) + 2 (3.10)a b = 0.764 + 1.55 d2 2
Pcr =
1.231 in.
x
Ix = 2(55.4) = 110.8 in.4
2
y¿
Pcr 245 = = 39.5 ksi 6 sg A 2(3.10)
Therefore, Euler’s formula is valid. 1046
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13–15. An A-36-steel W8 * 24 column is fixed at one end and free at its other end. If it is subjected to an axial load of 20 kip, determine the maximum allowable length of the column if F.S. = 2 against buckling is desired. Section Properties. From the table listed in the appendix, the cross-sectional area and moment of inertia about the y axis for a W8 * 24 are A = 7.08 in2
Iy = 18.3 in4
Critical Buckling Load. The critical buckling load is Pcr = Pallow (F.S) = 20(2) = 40 kip Applying Euler’s formula, p2 EIy
Pcr =
40 =
(KL)2
p2 C 29 A 103 B D (18.3) (2L)2
L = 180.93 in = 15.08 ft = 15.1 ft
Ans.
Critical Stress. Euler’s formula is valid only if scr 6 sY. scr =
Pcr 40 = = 5.65 ksi 6 sY = 36 ksi A 7.08
O.K.
*13–16. An A-36-steel W8 * 24 column is fixed at one end and pinned at the other end. If it is subjected to an axial load of 60 kip, determine the maximum allowable length of the column if F.S. = 2 against buckling is desired. Section Properties. From the table listed in the appendix, the cross-sectional area and moment of inertia about the y axis for a W8 * 24 are A = 7.08 in2
Iy = 18.3 in4
Critical Buckling Load. The critical buckling load is Pcr = Pallow (F.S.) = 60(2) = 120 kip Applying Euler’s formula, Pcr =
120 =
p2EIy (KL)2
p2 C 24 A 103 B D (18.3) (0.7L)2
L = 298.46 in = 24.87 ft = 24.9 ft
Ans.
Critical Stress. Euler’s formula is valid only if scr 6 sY. scr =
Pcr 120 = = 16.95 ksi 6 sY = 36 ksi A 7.08
O.K.
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•13–17.
The 10-ft wooden rectangular column has the dimensions shown. Determine the critical load if the ends are assumed to be pin connected. Ew = 1.611032 ksi, sY = 5 ksi.
Section Properties: 10 ft
A = 4(2) = 8.00 in2
4 in.
Ix =
1 (2) A 43 B = 10.667 in4 12
Iy =
1 (4) A 23 B = 2.6667 in4 (Controls !) 12
2 in.
Critical Buckling Load: K = 1 for pin supported ends column. Applying Euler’s formula,. Pcr =
p2EI (KL)2 p2(1.6)(103)(2.6667)
=
[1(10)(12)]2 Ans.
= 2.924 kip = 2.92 kip Critical Stress: Euler’s formula is only valid if scr 6 sg. scr =
Pcr 2.924 = = 0.3655 ksi 6 sg = 5 ksi A 8.00
O.K.
13–18. The 10-ft column has the dimensions shown. Determine the critical load if the bottom is fixed and the top is pinned. Ew = 1.611032 ksi, sY = 5 ksi. Section Properties: A = 4(2) = 8.00 in2 10 ft
1 (2) A 43 B = 10.667 in4 Ix = 12
4 in. 2 in.
1 Iy = (4) A 23 B = 2.6667 in4 (Controls!) 12 Critical Buckling Load: K = 0.7 for column with one end fixed and the other end pinned. Applying Euler’s formula. Pcr =
p2EI (KL)2 p2 (1.6)(103)(2.6667)
=
[0.7(10)(12)]2 Ans.
= 5.968 kip = 5.97 kip Critical Stress: Euler’s formula is only valid if scr 6 sg. scr =
Pcr 5.968 = = 0.7460 ksi 6 sg = 5 ksi A 8.00
O.K.
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13–19. Determine the maximum force P that can be applied to the handle so that the A-36 steel control rod BC does not buckle. The rod has a diameter of 25 mm.
P 350 mm A 250 mm
45⬚
Support Reactions: a + ©MA = 0;
P(0.35) - FBC sin 45°(0.25) = 0 FBC = 1.9799P
Section Properties: A =
p A 0.0252 B = 0.15625 A 10 - 3 B p m2 4
I =
p A 0.01254 B = 19.17476 A 10 - 9 B m4 4
Critical Buckling Load: K = 1 for a column with both ends pinned. Appyling Euler’s formula, Pcr = FBC =
1.9799P =
p2EI (KLBC)2
p2(200)(109) C 19.17476(10 - 9) D [1(0.8)]2
P = 29 870 N = 29.9 kN
Ans.
Critical Stress: Euler’s formula is only valid if scr 6 sg. scr =
1.9799(29 870) Pcr = 120.5 MPa 6 sg = 250 MPa = A 0.15625(10 - 3)p
1049
O.K.
C
B
800 mm
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*13–20. The W10 * 45 is made of A-36 steel and is used as a column that has a length of 15 ft. If its ends are assumed pin supported, and it is subjected to an axial load of 100 kip, determine the factor of safety with respect to buckling.
P
Critical Buckling Load: Iy = 53.4 in4 for a W10 * 45 wide flange section and K = 1 for pin supported ends column. Applying Euler’s formula, Pcr =
15 ft
p2EI (KL)2 p2 (29)(103)(53.4)
=
[1(15)(12)]2 P
= 471.73 kip Critical Stress: Euler’s formula is only valid if scr 6 sg. A = 13.3 in2 for the W10 * 45 wide-flange section. scr =
Pcr 471.73 = = 35.47 ksi 6 sg = 36 ksi A 13.3
O.K.
Pcr 471.73 = = 4.72 P 100
Ans.
Factor of Safety: F.S =
The W10 * 45 is made of A-36 steel and is used as a column that has a length of 15 ft. If the ends of the column are fixed supported, can the column support the critical load without yielding?
•13–21.
P
Critical Buckling Load: Iy = 53.4 in4 for W10 * 45 wide flange section and K = 0.5 for fixed ends support column. Applying Euler’s formula, Pcr =
15 ft
p2EI (KL)2 p2 (29)(103)(53.4)
=
[0.5(15)(12)]2 P
= 1886.92 kip Critical Stress: Euler’s formula is only valid if scr 6 sg. A = 13.3 in2 for W10 * 45 wide flange section. scr =
Pcr 1886.92 = = 141.87 ksi 7 sg = 36 ksi (No!) A 13.3
Ans.
The column will yield before the axial force achieves the critical load Pcr and so Euler’s formula is not valid.
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13–22. The W12 * 87 structural A-36 steel column has a length of 12 ft. If its bottom end is fixed supported while its top is free, and it is subjected to an axial load of P = 380 kip, determine the factor of safety with respect to buckling.
W 12 * 87
A = 25.6 in2
Ix = 740 in4
P
Iy = 241 in4 (controls) 12 ft
K = 2.0 Pcr =
p2(29)(103)(241) p2EI = = 831.63 kip 2 (KL) [(2.0)(12)(12)]2 Pcr 831.63 = = 2.19 P 380
F.S. =
Ans.
Check: scr =
=
Pcr A 831.63 = 32.5 ksi 6 sg 25.6
O.K.
13–23. The W12 * 87 structural A-36 steel column has a length of 12 ft. If its bottom end is fixed supported while its top is free, determine the largest axial load it can support. Use a factor of safety with respect to buckling of 1.75.
W 12 * 87
A = 25.6 in2
Ix = 740 in4
P
Iy = 241 in4
(controls)
K = 2.0 12 ft
Pcr
p2(29)(103)(241) p2EI = = = 831.63 kip 2 (KL) (2.0(12)(12))2
P =
Pcr 831.63 = = 475 ksi F.S 1.75
Ans.
Check: scr =
P 831.63 = = 32.5 ksi 6 sg A 25.6
O.K.
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*13–24. An L-2 tool steel link in a forging machine is pin connected to the forks at its ends as shown. Determine the maximum load P it can carry without buckling. Use a factor of safety with respect to buckling of F.S. = 1.75. Note from the figure on the left that the ends are pinned for buckling, whereas from the figure on the right the ends are fixed.
P
P
1.5 in.
0.5 in.
24 in.
Section Properties: A = 1.5(0.5) = 0.750 in2 Ix =
1 (0.5) A 1.53 B = 0.140625 in4 12
Iy =
1 (1.5) A 0.53 B = 0.015625 in4 12
P
Critical Buckling Load: With respect to the x - x axis, K = 1 (column with both ends pinned). Applying Euler’s formula, Pcr =
p2EI (KL)2 p2(29.0)(103)(0.140625)
=
[1(24)]2
= 69.88 kip With respect to the y - y axis, K = 0.5 (column with both ends fixed). Pcr =
p2EI (KL)2 p2(29.0)(103)(0.015625)
=
[0.5(24)]2
= 31.06 kip
(Controls!)
Critical Stress: Euler’s formula is only valid if scr 6 sg. scr =
Pcr 31.06 = = 41.41 ksi 6 sg = 102 ksi A 0.75
O.K.
Factor of Safety: F.S =
1.75 =
Pcr P 31.06 P
P = 17.7 kip
Ans.
1052
P
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The W14 * 30 is used as a structural A-36 steel column that can be assumed pinned at both of its ends. Determine the largest axial force P that can be applied without causing it to buckle. •13–25.
P
From the table in appendix, the cross-sectional area and the moment of inertia about weak axis (y-axis) for W14 * 30 are A = 8.85 in2
Iy = 19.6 in4
25 ft
Critical Buckling Load: Since the column is pinned at its base and top, K = 1. For A36 steel, E = 29.0(103) ksi and sg = 36 ksi. Here, the buckling occurs about the weak axis (y-axis). P = Pcr =
p2EIy (KL)2
=
p2 C 29.0(103) D (19.6)
C 1(25)(12) D 2
Ans.
= 62.33 kip = 62.3 kip Euler’s formula is valid only if scr 6 sg. scr =
Pcr 62.33 = = 7.04 ksi 6 sg = 36 ksi A 8.85
O.K.
13–26. The A-36 steel bar AB has a square cross section. If it is pin connected at its ends, determine the maximum allowable load P that can be applied to the frame. Use a factor of safety with respect to buckling of 2. a + ©MA = 0;
C
FBC sin 30°(10) - P(10) = 0 FBC = 2 P
+ : ©Fx = 0;
A
1.5 in.
30⬚ B
1.5 in.
FA - 2P cos 30° = 0 1.5 in.
10 ft
FA = 1.732 P
P
Buckling load: Pcr = FA(F.S.) = 1.732 P(2) = 3.464 P L = 10(12) = 120 in. I =
1 (1.5)(1.5)3 = 0.421875 in4 12
Pcr =
p2 EI (KL)2
3.464 P =
p2 (29)(103)(0.421875) [(1.0)(120)]2
P = 2.42 kip
Ans.
Pcr = FA(F.S.) = 1.732(2.42)(2) = 8.38 kip Check: scr =
Pcr 8.38 = = 3.72 ksi 6 sg A 1.5 (1.5)
O.K.
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13–27. Determine the maximum allowable intensity w of the distributed load that can be applied to member BC without causing member AB to buckle. Assume that AB is made of steel and is pinned at its ends for x–x axis buckling and fixed at its ends for y–y axis buckling. Use a factor of safety with respect to buckling of 3. Est = 200 GPa, sY = 360 MPa.
w
C
1.5 m
B 0.5 m
2m 30 mm x
Ix =
1 (0.02)(0.033) = 45.0(10 - 9)m4 12
Iy =
1 (0.03)(0.023) = 20(10 - 9) m4 12
x
x-x axis: Pcr = FAB (F.S.) = 1.333w(3) = 4.0 w K = 1.0, Pcr =
L = 2m
p2EI (KL)2
4.0w =
p2(200)(109)(45.0)(10 - 9) [(1.0)(2)]2
w = 5552 N>m = 5.55 kN>m
Ans.
(controls)
y-y axis K = 0.5, 4.0w =
L = 2m
p2 (200)(109)(20)(10 - 9) [(0.5)(2)]2
w = 9870 N>m = 9.87 kN>m Check: scr =
20 mm y
y
Moment of inertia:
4(5552) Pcr = = 37.0 MPa 6 sg A (0.02)(0.03)
O.K.
1054
30 mm
A
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*13–28. Determine if the frame can support a load of w = 6 kN>m if the factor of safety with respect to buckling of member AB is 3. Assume that AB is made of steel and is pinned at its ends for x–x axis buckling and fixed at its ends for y–y axis buckling. Est = 200 GPa, sY = 360 MPa.
w
C
B
1.5 m
0.5 m
Check x-x axis buckling: Ix =
1 (0.02)(0.03)3 = 45.0(10 - 9) m4 12
K = 1.0 Pcr
2m 30 mm x 20 mm y
y
L = 2m
p2(200)(109)(45.0)(10 - 9) p2EI = = 2 (KL) ((1.0)(2))2
x
A
30 mm
Pcr = 22.2 kN a + ©MC = 0;
FAB(1.5) - 6(2)(1) = 0 FAB = 8 kN
Preq’d = 8(3) = 24 kN 7 22.2 kN No, AB will fail.
Ans.
The beam supports the load of P = 6 kip. As a result, the A-36 steel member BC is subjected to a compressive load. Due to the forked ends on the member, consider the supports at B and C to act as pins for x–x axis buckling and as fixed supports for y–y axis buckling. Determine the factor of safety with respect to buckling about each of these axes. •13–29.
a + ©MA = 0;
P 4 ft
A 3 ft
1 )(1)(3)3 p2(29)(103)(12 p2EI = = 178.9 kip 2 (KL) (1.0(5)(12))2
178.9 = 8.94 20
Ans.
y-y axis buckling: Pcr =
F.S. =
3 in.
y
x-x axis buckling:
F.S. =
B
C x
3 FBC a b(4) - 6000(8) = 0 5 FBC = 20 kip
Pcr =
4 ft
1 )(3)(1)3 p2 (29)(103)(12 p2EI = = 79.51 2 (KL) (0.5(5)(12))2
79.51 = 3.98 20
Ans.
1055
1 in.
x
y
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13–30. Determine the greatest load P the frame will support without causing the A-36 steel member BC to buckle. Due to the forked ends on the member, consider the supports at B and C to act as pins for x–x axis buckling and as fixed supports for y–y axis buckling.
P 4 ft
A 3 ft
3 FBC a b(4) - P(8) = 0 5
a + ©MA = 0;
4 ft
B
y
3 in. C x
1 in.
FBC = 3.33 P y
x-x axis buckling: Pcr =
x
1 )(1)(3)3 p2(29)(103)(12 p2EI = = 178.9 kip (KL)2 (1.0(5)(12))2
y -y axis buckling: Pcr =
1 )(3)(1)3 p2(29)(103)(12 p2EI = = 79.51 kip (KL)2 (0.5(5)(12))2
Thus, 3.33 P = 79.51 P = 23.9 kip
Ans.
13–31. Determine the maximum distributed load that can be applied to the bar so that the A-36 steel strut AB does not buckle. The strut has a diameter of 2 in. It is pin connected at its ends.
w
C
A
2 ft
The compressive force developed in member AB can be determined by writing the moment equation of equilibrium about C. a + ©MC = 0;
FAB(2) - w(2)(3) = 0
A = p(12) = p in2
I =
FAB = 3w
4 ft
p 4 p (1 ) = in4 4 4
Since member AB is pinned at both ends, K = 1. For A36 steel, E = 29.0(103) ksi and sg = 36 ksi. Pcr =
p EI ; (KL)2
p C 29.0(10 ) D (p>4) 2
2
3w =
3
C 1(4)(12) D 2
Ans.
w = 32.52 kip>ft = 32.5 kip>ft The Euler’s formula is valid only if scr 6 sg. scr =
3(32.52) Pcr = = 31.06 ksi 6 sg = 36 ksi p A
O.K.
1056
B
2 ft
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*13–32. The members of the truss are assumed to be pin connected. If member AC is an A-36 steel rod of 2 in. diameter, determine the maximum load P that can be supported by the truss without causing the member to buckle.
P
C B
4 ft
D
A 3 ft
Section the truss through a-a, the FBD of the top cut segment is shown in Fig. a. The compressive force developed in member AC can be determined directly by writing the force equation of equilibrium along x axis. + : ©Fx = 0;
3 FAC a b - P = 0 5
A = p(12) = p in2
I =
FAC =
5 P (C) 3
p 4 p (1 ) = in4 4 4
Since both ends of member AC are pinned, K = 1. For A-36 steel, E = 29.0(103) ksi and sg = 36 ksi. The length of member AC is LAC = 232 + 42 = 5 ft. Pcr =
p2EI ; (KL)2
p2 C 29.0(103) D (p>4) 5 P = 3 C 1(5)(12) D 2
P = 37.47 kip = 37.5 kip
Ans.
Euler’s formula is valid only if scr 6 sg.
scr
5 (37.47) Pcr 3 = = = 19.88 ksi 6 sg = 36 ksi p A
O.K.
1057
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•13–33.
The steel bar AB of the frame is assumed to be pin connected at its ends for y–y axis buckling. If w = 3 kN>m, determine the factor of safety with respect to buckling about the y–y axis due to the applied loading. Est = 200 GPa, sY = 360 MPa.
6m w B C 40 mm 40 mm
3m
40 mm
y
x A
4m
The force with reference to the FBD shown in Fig. a. a + ©MC = 0;
3 3(6)(3) - FAB a b(6) = 0 5
A = 0.04(0.08) = 3.2(10 - 3) m2
Iy =
FAB = 15 kN
1 (0.08)(0.043) = 0.4267(10 - 6)m4 12
The length of member AB is L = 232 + 42 = 5m. Here, buckling will occur about the weak axis, (y-axis). Since both ends of the member are pinned, Ky = 1. Pcr =
p2EIy (KyLy)2
=
p2 C 200(109) D C 0.4267(10 - 6) D
C 1.0(5) D 2
= 33.69 kN
Euler’s formula is valid only if scr 6 sg. scr =
33.69(103) Pcr = 10.53(106)Pa = 10.53 MPa 6 sg = 360 MPa = A 3.2(10 - 3)
O.K.
Thus, the factor of safety against buckling is F.S =
Pcr 33.69 = = 2.25 FAB 15
Ans.
1058
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13–34. The members of the truss are assumed to be pin connected. If member AB is an A-36 steel rod of 40 mm diameter, determine the maximum force P that can be supported by the truss without causing the member to buckle.
2m
C
E
D 1.5 m B A 2m P
By inspecting the equilibrium of joint E, FAB = 0. Then, the compressive force developed in member AB can be determined by analysing the equilibrium of joint A, Fig. a. + c ©Fy = 0;
3 FAC a b - P = 0 5
+ : ©Fx = 0;
5 4 P a b - FAB = 0 3 5
A = p(0.022) = 0.4(10 - 3)p m2
I =
FAC =
5 P (T) 3
FAB =
4 P(c) 3
p (0.024) = 40(10 - 9) p m4 4
Since both ends of member AB are pinned, K = 1. For A36 steel, E = 200 GPa and sg = 250 MPa. Pcr =
p2EI ; (KL)2
p2 C 200(109) D C 40(10 - 9)p D 4 P = 3 C 1(2) D 2 P = 46.51(103) N = 46.5 kN
Ans.
The Euler’s formula is valid only if scr 6 sg.
scr
4 (46.51)(103) Pcr 3 = 49.35(106) Pa = 49.35 MPa 6 sg = 250 MPa O.K. = = A 0.4(10 - 3)p
1059
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13–35. The members of the truss are assumed to be pin connected. If member CB is an A-36 steel rod of 40 mm diameter, determine the maximum load P that can be supported by the truss without causing the member to buckle.
2m
C
E
D 1.5 m B A 2m P
Section the truss through a–a, the FBD of the left cut segment is shown in Fig. a. The compressive force developed in member CB can be obtained directly by writing the force equation of equilibrium along y axis. + c ©Fy = 0;
FCB - P = 0
A = p(0.022) = 0.4(10 - 3)p m2
FCB = P (C) I =
p (0.024) = 40(10 - 9)p m4 4
Since both ends of member CB are pinned, K = 1. For A36 steel, E = 200 GPa and sg = 250 MPa. Pcr =
p2EI ; (KL)2
P =
p2 C 200(109) D C 40(10 - 9)p D
C 1(1.5) D 2
= 110.24(103) N = 110 kN
Ans.
The Euler’s formula is valid only if scr 6 sg. scr =
110.24(103) Pcr = 87.73(106) Pa = 87.73 MPa 6 sg = 250 MPa = A 0.4(10 - 3)p
1060
O.K.
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*13–36. If load C has a mass of 500 kg, determine the required minimum diameter of the solid L2-steel rod AB to the nearest mm so that it will not buckle. Use F.S. = 2 against buckling.
A
45°
4m
D
Equilibriun. The compressive force developed in rod AB can be determined by analyzing the equilibrium of joint A, Fig. a. ©Fy¿ = 0; FAB sin 15° - 500(9.81) cos 45° = 0
FAB = 13 400.71 N
Section Properties. The cross-sectional area and moment of inertia of the solid rod are A =
p 2 d 4
I =
p d 4 p 4 a b = d 4 2 64
Critical Buckling Load. Since the rod is pinned at both of its ends, K = 1. Here, Pcr = FAB (F.S.) = 13400.71(2) = 26801.42 N. Applying Euler’s formula,
Pcr =
p2EIy (KL)2
26801.42 =
p2 C 200 A 109 B D c
p 4 d d 64
[1(4)]2
d = 0.04587 m = 45.87 mm Use d = 46 mm
Ans.
Critical Stress. Euler’s formula is valid only if scr 6 sY. scr =
Pcr 26801.42 = = 16.13 MPa 6 sY = 703 MPa p A 2 A 0.046 B 4
1061
O.K.
60° B
C
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•13–37. If the diameter of the solid L2-steel rod AB is 50 mm, determine the maximum mass C that the rod can support without buckling. Use F.S. = 2 against buckling.
A
45°
4m
D
Equilibrium. The compressive force developed in rod AB can be determined by analyzing the equilibrium of joint A, Fig. a. ©Fy¿ = 0; FAB sin 15° - m(9.81) cos 45° = 0
FAB = 26.8014m
B
Section Properties. The cross-sectional area and moment of inertia of the rod are A =
I =
p A 0.052 B = 0.625 A 10 - 3 B pm2 4 p A 0.0254 B = 97.65625 A 10 - 9 B pm4 4
Critical Buckling Load. Since the rod is pinned at both of its ends, K = 1. Here, Pcr = FAB (F.S.) = 26.8014m(2) = 53.6028m. Applying Euler’s formula, Pcr =
p2EIy (KL)2
53.6028m =
p2 c200 A 109 B d c97.65625 A 10 - 9 B p d [1(4)]2
m = 706.11 kg = 7.06 kg
Ans.
Critical Stress. Euler’s formula is valid only if scr 6 sY. scr =
53.6028(706.11) Pcr = = 19.28 MPa 6 sY = 703 MPa A p 0.625 A 10 - 3 B
1062
60°
O.K.
C
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13–38. The members of the truss are assumed to be pin connected. If member GF is an A-36 steel rod having a diameter of 2 in., determine the greatest magnitude of load P that can be supported by the truss without causing this member to buckle.
H
16 ft P
Support Reactions: As shown on FBD(a). Member Forces: Use the method of sections [FBD(b)]. FGF = 1.3333P (C)
Section Properties: A =
I =
p 2 A 2 B = p in2 4 p 4 A 1 B = 0.250p in4 4
Critical Buckling Load: K = 1 for a column with both ends pinned. Applying Euler’s formula, Pcr = FGF =
1.3333P =
p2EI (KLGF)2 p2 (29)(103)(0.250p) [1(16)(12)]2
P = 4.573 kip = 4.57 kip
Ans.
Critical Stress: Euler’s formula is only valid if scr 6 sg. scr =
1.3333(4.573) Pcr = = 1.94 ksi 6 sg = 36 ksi p A
1063
O.K.
D
C
B 16 ft
FGF (12) - P(16) = 0
E
12 ft
A
+ ©MB = 0;
F
G
16 ft P
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13–39. The members of the truss are assumed to be pin connected. If member AG is an A-36 steel rod having a diameter of 2 in., determine the greatest magnitude of load P that can be supported by the truss without causing this member to buckle.
H
16 ft P
Support Reactions: As shown on FBD(a). Member Forces: Use the method of joints [FBD(b)]. 3 = 0 F 5 AG
FAG = 1.6667P (C)
Section Properties: LAG = 2162 + 122 = 20.0 ft
A =
I =
p 2 A 2 B = p in2 4 p 4 A 1 B = 0.250p in4 4
Critical Buckling Load: K = 1 for a column with both ends pinned. Applying Euler’s formula, Pcr = FGF =
1.6667P =
p2EI (KLGF)2 p2 (29)(103)(0.250p) [1(20)(12)]2
P = 2.342 kip = 2.34 kip
Ans.
Critical Stress: Euler’s formula is only valid if scr = sg. scr =
1.6667(2.342) Pcr = = 1.24 ksi 6 sg = 36 ksi p A
1064
O.K.
D
C
B 16 ft
P -
E
12 ft
A
+ c ©Fy = 0;
F
G
16 ft P
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*13–40. The column is supported at B by a support that does not permit rotation but allows vertical deflection. Determine the critical load Pcr . EI is constant.
L
B
Pcr A
Elastic curve: EI
d2y = M = -P y dx2
P d2y + y = 0 EI dx2 y = C1 sin c
P P x d + C2 cos c xd A EI A EI
Boundry conditions: At x = 0; 0 = 0 + C2; At x = L;
y = 0 C2 = 0 dv = 0 dx
P P dv = C1 cos c L] d = 0; dx A EI A EI cos c
P L d = 0; A EI
For n = 1 ;
Pcr =
C1
P p L = na b A EI 2
P Z 0 A EI n = 1, 3, 5
p2 P = EI 4L2
p2EI 4L2
Ans.
1065
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w
The ideal column has a weight w (force兾length) and rests in the horizontal position when it is subjected to the axial load P. Determine the maximum moment in the column at midspan. EI is constant. Hint: Establish the differential equation for deflection, Eq. 13–1, with the origin at the mid span. The general solution is v = C1 sin kx + C2 cos kx + 1w>12P22x2 - 1wL>12P22x - 1wEI>P22 where k2 = P>EI. •13–41.
P
L
Moment Functions: FBD(b). a + ©Mo = 0; M(x) =
wL x wx a b - M(x) - a bx - Pv = 0 2 2
w 2 A x - Lx B - Pv 2
[1]
Differential Equation of The Elastic Curve: EI
d2y = M(x) dx2
EI
d2y w 2 = A x - Lx B - Py 2 dx2
w d2y P y = + A x2 - Lx B EI 2EI dx2 The solution of the above differential equation is of the form v = C1 sin a
P P w 2 wL wEI xb + C2 cos ¢ xb + x x A EI A EI 2P 2P P2
[2]
dv P P P P w wL = C1 cos ¢ x ≤ - C2 sin ¢ x≤ + xdx A EI A EI A EI A EI P 2P
[3]
and
The integration constants can be determined from the boundary conditions. Boundary Condition: At x = 0, y = 0. From Eq. [2], 0 = C2 -
wEI P2
C2 =
wEI P2
At x =
L dy = 0. From Eq.[3], , 2 dx
0 = C1
P wEI P w L P L P L wL cos ¢ sin ¢ ≤ ≤ + a b A EI A EI 2 A EI 2 P 2 2P P2 A EI C1 =
wEI P L tan ¢ ≤ A EI 2 P2
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13–41.
Continued
Elastic Curve: y =
w EI P L P EI P x2 L EI x≤ + x≤ + tan ¢ cos ¢ - x B ≤ sin ¢ R P P A EI 2 A EI P A EI 2 2 P
However, y = ymax at x =
ymax =
=
L . Then, 2
P L P L P L EI w EI EI L2 tan ¢ cos ¢ B ≤ sin ¢ ≤ + ≤ R P P A EI 2 A EI 2 P A EI 2 8 P wEI P L PL2 sec - 1R B ¢ ≤ A EI 2 8EI P2
Maximum Moment: The maximum moment occurs at x =
Mmax =
L . From, Eq.[1], 2
w L2 L - L a b R - Pymax B 2 4 2
= -
wL2 wEI P L PL2 - P b 2 B sec ¢ - 1R r ≤ 8 A EI 2 8EI P
= -
PL wEI B sec ¢ ≤ - 1R P A EI 2
Ans.
13–42. The ideal column is subjected to the force F at its midpoint and the axial load P. Determine the maximum moment in the column at midspan. EI is constant. Hint: Establish the differential equation for deflection, Eq. 13–1. The general solution is v = C1 sin kx + C2 cos kx - c2x>k2, where c2 = F>2EI, k2 = P>EI.
F P
L 2
Moment Functions: FBD(b). a + ©Mo = 0;
M(x) +
F x + P(v) = 0 2 M(x) = -
F x - Pv 2
[1]
Differential Equation of The Elastic Curve: EI
d2y = M(x) dx2
EI
F d2y = - x - Py 2 2 dx
d2y F P y = x + EI 2EI dx2 The solution of the above differential equation is of the form, v = C1 sin a
P P F xb + C2 cos ¢ xb x A EI A EI 2P
1067
[2]
L 2
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13–42.
Continued
and dv P P P P F = C1 cos ¢ x ≤ - C2 sin ¢ x≤ dx A EI A EI A EI A EI 2P
[3]
The integration constants can be determined from the boundary conditions. Boundary Conditions: At x = 0, y = 0. From Eq.[2], C2 = 0 At x =
L dy = 0. From Eq.[3], , 2 dx 0 = C1
C1 =
P L P F cos ¢ ≤ A EI A EI 2 2P F EI P L sec ¢ ≤ 2P A P A EI 2
Elastic Curve: y =
F EI P L P F sec ¢ x≤ x ≤ sin ¢ 2P A P A EI 2 A EI 2P
=
F EI P L P sec ¢ x≤ - xR B ≤ sin ¢ 2P A P A EI 2 A EI
However, y = ymax at x =
ymax =
=
L . Then, 2
F EI P L P L L sec ¢ B ≤ sin ¢ ≤ - R 2P A P A EI 2 A EI 2 2 F EI P L L tan ¢ B ≤ - R 2P A P A EI 2 2
Maximum Moment: The maximum moment occurs at x =
Mmax = -
L . From Eq.[1], 2
F L a b - Pymax 2 2
= -
FL F EI P L L - Pb tan ¢ B ≤ - Rr 4 2P A P A EI 2 2
= -
P L F EI tan ¢ ≤ 2 AP A EI 2
Ans.
1068
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13–43. The column with constant EI has the end constraints shown. Determine the critical load for the column.
P
L
Moment Function. Referring to the free-body diagram of the upper part of the deflected column, Fig. a, a + ©MO = 0;
M + Pv = 0
M = -Pv
Differential Equation of the Elastic Curve. EI
d2v = M dx2
EI
d2v = -Pv dx2
d2v P + v = 0 2 EI dx The solution is in the form of v = C1 sin a
P P xb + C2 cos ¢ xb A EI A EI
(1)
dv P P P P = C1 cos ¢ x ≤ - C2 sin ¢ x≤ dx A EI A EI A EI A EI
(2)
Boundary Conditions. At x = 0, v = 0. Then Eq. (1) gives 0 = 0 + C2 At x = L,
C2 = 0
dv = 0. Then Eq. (2) gives dx 0 = C1
P P cos ¢ L≤ A EI A EI
C1 = 0 is the trivial solution, where v = 0. This means that the column will remain straight and buckling will not occur regardless of the load P. Another possible solution is cos ¢
P L≤ = 0 A EI
P np L = A EI 2
n = 1, 3, 5
The smallest critical load occurs when n = 1, then p Pcr L = A EI 2 Pcr =
p2EI 4L2
Ans.
1069
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*13–44. Consider an ideal column as in Fig. 13–10c, having both ends fixed. Show that the critical load on the column is given by Pcr = 4p2EI>L2. Hint: Due to the vertical deflection of the top of the column, a constant moment M¿ will be developed at the supports. Show that d2v>dx2 + 1P>EI2v = M¿>EI. The solution is of the form v = C1 sin11P>EIx2 + C2 cos11P>EIx2 + M¿>P. Moment Functions: M(x) = M¿ - Py Differential Equation of The Elastic Curve: EI
d2y = M(x) dx2
d2y = M¿ - Py dx2
EI
M¿ d2y P + y = 2 EI EI dx
(Q.E.D.)
The solution of the above differential equation is of the form v = C1 sin a
P P M¿ xb + C2 cos ¢ xb + A EI P A EI
[1]
and dv P P P P = C1 cos ¢ x ≤ - C2 sin ¢ x≤ dx A EI A EI A EI A EI
[2]
The integration constants can be determined from the boundary conditions. Boundary Conditions: At x = 0, y = 0. From Eq.[1], C2 = At x = 0,
M¿ P
dy = 0. From Eq.[2], C1 = 0 dx
Elastic Curve: y =
M¿ P x≤ R B 1 - cos ¢ P A EI
and M¿ P P dy = sin ¢ x≤ dx P A EI A EI However, due to symmetry sin B
L dy = 0 at x = . Then, dx 2
P L a bR = 0 A EI 2
or
P L a b = np A EI 2
where n = 1, 2, 3,...
The smallest critical load occurs when n = 1. Pce =
4p2EI L2
(Q.E.D.)
1070
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•13–45. Consider an ideal column as in Fig. 13–10d, having one end fixed and the other pinned. Show that the critical load on the column is given by Pcr = 20.19EI>L2. Hint: Due to the vertical deflection at the top of the column,a constant moment M¿ will be developed at the fixed support and horizontal reactive forces R¿ will be developed at both supports. Show that d2v>dx2 + 1P>EI2v = 1R¿>EI21L - x2. The solution is of the form v = C1 sin 11P>EIx2 + C2 cos 11P>EIx2 + 1R¿>P21L - x2. After application of the boundary conditions show that tan 11P>EIL2 = 1P>EI L. Solve by trial and error for the smallest nonzero root.
Equilibrium. FBD(a). Moment Functions: FBD(b). M(x) = R¿(L - x) - Py Differential Equation of The Elastic Curve: EI
d2y = M(x) dx2
EI
d2y = R¿(L - x) - Py dx2
d2y P R¿ + y = (L - x) 2 EI EI dx
(Q.E.D.)
The solution of the above differential equation is of the form v = C1 sin a
P P R¿ (L - x) xb + C2 cos ¢ xb + A EI P A EI
[1]
and dv P P P P R¿ = C1 cos ¢ x ≤ - C2 sin ¢ x≤ dx A EI A EI A EI A EI P The integration constants can be determined from the boundary conditions. Boundary Conditions: At x = 0, y = 0. From Eq.[1], C2 = -
At x = 0,
R¿L P
dy R¿ EI = 0. From Eq.[2], C1 = dx P AP
Elastic Curve: y =
=
R¿ EI P R¿L P R¿ sin ¢ x≤ cos ¢ x≤ + (L - x) P AP A EI P A EI P EI P P R¿ sin ¢ x ≤ - L cos ¢ x ≤ + (Lx) R B P AP A EI A EI
1071
[2]
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13–45.
Continued
However, y = 0 at x = L. Then, 0 =
P P EI sin ¢ L ≤ - L cos ¢ L≤ A EI A EI AP tan ¢
P P L≤ = L A EI A EI
(Q.E.D.)
By trial and error and choosing the smallest root, we have P L = 4.49341 A EI Then, Pcr =
20.19EI L2
(Q.E.D.)
13–46. Determine the load P required to cause the A-36 steel W8 * 15 column to fail either by buckling or by yielding. The column is fixed at its base and free at its top.
1 in.
P
Section properties for W8 * 15: A = 4.44 in2
Ix = 48.0 in4
rx = 3.29 in.
d = 8.11 in.
Iy = 3.41 in4
8 ft
Buckling about y-y axis: K = 2.0 P = Pcr =
L = 8(12) = 96 in. p2EIy (KL)2
Check: scr =
p2(29)(103)(3.41) =
[(2.0)(96)]2
= 26.5 kip
(controls)
Pcr 26.5 = = 5.96 ksi 6 sg A 4.44
Ans.
O.K.
Check yielding about x-x axis: smax =
P ec KL P c1 + 2 sec a bd A 2r A EA r
26.5 P = = 5.963 ksi A 4.44
(1) A 8.11 ec 2 B = = 0.37463 2 r (3.29)2
2.0(96) P 26.5 KL = = 0.4184 2r A EA 2(3.29) A 29(103)(4.44) smax = 5.963[1 + 0.37463 sec (0.4184)] = 8.41 ksi 6 sg = 36 ksi
1072
O.K.
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13–47. The hollow red brass C83400 copper alloy shaft is fixed at one end but free at the other end. Determine the maximum eccentric force P the shaft can support without causing it to buckle or yield. Also, find the corresponding maximum deflection of the shaft.
2m a
a
P 150 mm
30 mm
Section Properties. A = p A 0.032 - 0.022 B = 0.5 A 10 - 3 B p m2 I =
20 mm Section a – a
p A 0.034 - 0.024 B = 0.1625 A 10 - 6 B p m4 4
0.1625 A 10 B p I = = 0.01803 m C 0.5 A 10 - 3 B p AA -6
r =
e = 0.15 m
c = 0.03 m
For a column that is fixed at one end and free at the other, K = 2. Thus, KL = 2(2) = 4 m Yielding. In this case, yielding will occur before buckling. Applying the secant formula, smax =
P ec KL P B 1 + 2 sec ¢ ≤R A 2rx A EA rx
70.0 A 106 B = 70.0 A 106 B =
P
0.5 A 10
-3
Bp
P
0.5 A 10 - 3 B p
D1 +
0.15(0.03) 0.018032
secC
P 4 ST 2(0.01803)A 101 A 109 B C 0.5 A 10 - 3 B p D
a1 + 13.846 sec 8.8078 A 10 - 3 B 2Pb
Solving by trial and error, P = 5.8697 kN = 5.87 kN
Ans.
Maximum Deflection. vmax = e B sec ¢
P KL ≤ - 1R A EI 2
= 0.15 D sec C
5.8697 A 103 B
4 a b S - 1T
C 101 A 109 B C 0.1625 A 10 - 6 B p D 2
= 0.04210 m = 42.1 mm
Ans.
1073
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*13–48. The hollow red brass C83400 copper alloy shaft is fixed at one end but free at the other end. If the eccentric force P = 5 kN is applied to the shaft as shown, determine the maximum normal stress and the maximum deflection.
2m a
a
P 150 mm
30 mm
Section Properties. A = p A 0.032 - 0.022 B = 0.5 A 10 - 3 B p m2 I =
20 mm Section a – a
p A 0.034 - 0.024 B = 0.1625 A 10 - 6 B p m4 4
0.1625 A 10 - 6 B p I = 0.01803 m = r = C 0.5 A 10 - 3 B p AA e = 0.15 m
c = 0.03 m
For a column that is fixed at one end and free at the other, K = 2. Thus, KL = 2(2) = 4 m Yielding. Applying the secant formula, smax =
P ec KL P B 1 + 2 sec ¢ ≤R A 2r A EA r 5 A 103 B
5 A 103 B 4 ST secC D1 + = 2(0.01803) C 101 A 109 B C 0.5 A 10 - 3 B p D 0.018032 0.5 A 10 - 3 B p 0.15(0.03)
Ans.
= 57.44 MPa = 57.4 MPa Since smax 6 sY = 70 MPa, the shaft does not yield. Maximum Deflection. vmax = e B sec ¢
P KL ≤ - 1R A EI 2
= 0.15 Dsec C
5 A 103 B
4 a b S - 1T
C 101 A 109 B C 0.1625 A 10 - 6 B p D 2
= 0.03467 m = 34.7 mm
Ans.
1074
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•13–49.
The tube is made of copper and has an outer diameter of 35 mm and a wall thickness of 7 mm. Using a factor of safety with respect to buckling and yielding of F.S. = 2.5, determine the allowable eccentric load P. The tube is pin supported at its ends. Ecu = 120 GPa, sY = 750 MPa.
2m
P 14 mm
Section Properties: A =
p (0.0352 - 0.0212) = 0.61575(10 - 3) m2 4
I =
p (0.01754 - 0.01054) = 64.1152(10 - 9) m4 4
r =
I 64.1152(10 - 9) = = 0.010204 m AA A 0.61575(10 - 3)
For a column pinned at both ends, K = 1. Then KL = 1(2) = 2 m. Buckling: Applying Euler’s formula, Pmax = Pcr =
p2 (120)(109) C 64.1152(10 - 9) D p2EI = = 18983.7 N = 18.98 kN (KL)2 22
Critical Stress: Euler’s formula is only valid if scr 6 sg. scr =
Pcr 18983.7 = 30.83 MPa 6 sg = 750 MPa = A 0.61575(10 - 3)
O.K.
Yielding: Applying the secant formula, smax =
(KL) Pmax Pmax ec B 1 + 2 sec ¢ ≤R A 2r A EA r
750 A 106 B =
0.61575(10 - 3)
750 A 106 B =
0.61575(10 - 3)
Pmax
Pmax
B1 +
0.014(0.0175) 0.0102042
sec ¢
Pmax 2 ≤R 2(0.010204)A 120(109)[0.61575(10 - 3)]
A 1 + 2.35294 sec 0.01140062Pmax B
Solving by trial and error, Pmax = 16 885 N = 16.885 kN (Controls!) Factor of Safety: P =
Pmax 16.885 = = 6.75 kN F.S. 2.5
Ans.
1075
P
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13–50. The tube is made of copper and has an outer diameter of 35 mm and a wall thickness of 7 mm. Using a factor of safety with respect to buckling and yielding of F.S. = 2.5, determine the allowable eccentric load P that it can support without failure. The tube is fixed supported at its ends. Ecu = 120 GPa, sY = 750 MPa.
2m
P 14 mm
Section Properties: A =
p A 0.0352 - 0.0212 B = 0.61575 A 10 - 3 B m2 4
I =
p A 0.01754 - 0.01054 B = 64.1152 A 10 - 9 B m4 4
r =
I 64.1152(10 - 9) = = 0.010204 ms AA A 0.61575(10 - 3)
For a column fixed at both ends, K = 0.5. Then KL = 0.5(2) = 1 m. Buckling: Applying Euler’s formula, Pmax = Pcr =
p2(120)(109) C 64.1152(10 - 9) D p2EI = = 75 935.0 N = 75.93 kN (KL)2 12
Critical Stress: Euler’s formula is only valid if scr 6 sg. scr =
Pcr 75 935.0 = 123.3 MPa 6 sg = 750 MPa = A 0.61575(10 - 3)
O. K.
Yielding: Applying the secant formula, smax =
(KL) Pmax Pmax ec B 1 + 2 sec ¢ ≤R A 2r A EA r
750 A 106 B =
0.61575(10 - 3)
750 A 106 B =
0.61575(10 - 3)
Pmax
Pmax
B1 +
0.014(0.0175) 0.0102042
sec ¢
2 Pmax ≤R 2(0.010204)A 120(109)[0.61575(10 - 3)]
A 1 + 2.35294 sec 5.70032 A 10 - 3 B 2P B
Solving by trial and error, Pmax = 50 325 N = 50.325 kN (Controls!) Factor of Safety: P =
Pmax 50.325 = = 20.1 kN F.S. 2.5
Ans.
1076
P
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13–51. The wood column is fixed at its base and can be assumed pin connected at its top. Determine the maximum eccentric load P that can be applied without causing the column to buckle or yield. Ew = 1.811032 ksi, sY = 8 ksi.
P
y 4 in. x
x P y 10 in.
10 ft
Section Properties: A = 10(4) = 40 in2
ry =
Iy =
1 (4)(103) = 333.33 in4 12
Ix =
1 (10)(43) = 53.33 in4 12
Ix 333.33 = = 2.8868 in. AA A 40
Buckling about x-x axis: P = Pcr =
p2(1.8)(103)(53.33) p2EI = = 134 kip (KL)2 [(0.7)(10)(12)]2
Check: scr =
Pcr 134 = = 3.36 ksi 6 sg A 40
O.K.
Yielding about y -y axis: smax =
P ec KL P a1 + 2 seca b b A 2r A EA r
5(5) ec = = 3.0 r2 2.88682 a
0.7(10)(12) P P KL b = = 0.0542212P 2r A EA 2(2.8868) A 1.8(103)(40)
8(40) = P[1 + 3.0 sec (0.0542212P)] By trial and error: P = 73.5 kip
Ans.
(controls)
1077
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*13–52. The wood column is fixed at its base and can be assumed fixed connected at its top. Determine the maximum eccentric load P that can be applied without causing the column to buckle or yield. Ew = 1.811032 ksi, sY = 8 ksi.
P
y 4 in. x
x P y 10 in.
10 ft
Section Properties: A = 10(4) = 40 in2
ry =
Iy =
1 (4)(103) = 333.33 in4 12
Ix =
1 (10)(43) = 53.33 in4 12
Iy 333.33 = = 2.8868 in. AA A 40
Buckling about x-x axis: P = Pcr =
p2(1.8)(103)(53.33) p2EI = = 263 kip 2 (KL) [(0.5)(10)(12)]2
Check: scr =
Pcr 263 = = 6.58 ksi 6 sg A 40
O.K.
Yielding about y-y axis: smax =
P ec KL P a1 + 2 seca bb A 2r A EA r
5(5) ec = = 3.0 r2 2.88682 a
0.5(10)(12) P P KL b = = 0.0387292P 2r A EA 2(2.8868) A 1.8(103)(40)
8(40) = P[1 + 3.0 sec (0.0387292P)] By trial and error: P = 76.5 kip
Ans.
(controls)
1078
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The W200 * 22 A-36-steel column is fixed at its base. Its top is constrained to rotate about the y–y axis and free to move along the y–y axis. Also, the column is braced along the x–x axis at its mid-height. Determine the allowable eccentric force P that can be applied without causing the column either to buckle or yield. Use F.S. = 2 against buckling and F.S. = 1.5 against yielding.
•13–53.
100 mm
y
A = 2860 mm2 = 2.86 A 10 - 3 B m2
ry = 22.3 mm = 0.0223 m
Ix = 20.0 A 106 B mm4 = 20.0 A 10 - 6 B m4
c =
bf 2
=
102 = 51 mm = 0.051 m 2
e = 0.1m Buckling About the Strong Axis. Since the column is fixed at the base and free at the top, Kx = 2. Applying Euler’s formula,
Pcr =
p2EIx (KL)x 2
=
p2 c200 A 109 B d c20.0 A 10 - 6 B d [2(10)]2
= 98.70kN
Euler’s formula is valid if scr 6 sY. scr =
98.70 A 103 B Pcr = = 34.51 MPa 6 sY = 250MPa A 2.86 A 10 - 3 B
O.K.
Then, Pallow =
Pcr 98.70 = = 49.35 kN F.S. 2
Yielding About Weak Axis. Since the support provided by the bracing can be considered a pin connection, the upper portion of the column is pinned at both of its ends. Then Ky = 1 and L = 5 m. Applying the secant formula, smax =
A KL B y Pmax Pmax ec C 1 + 2 sec B RS A 2ry A EA ry
250 A 106 B =
250 A 106 B =
Pmax
2.86 A 10
-3
Pmax
B
2.86 A 10 - 3 B
D1 +
0.1(0.051) 0.02232
secC
1(5) Pmax ST 2(0.0223)A 200 A 109 B C 2.86 A 10 - 3 B D
c1 + 10.2556 sec 4.6875 A 10 - 3 B 2Pmax d
Solving by trial and error, Pmax = 39.376 kN Then, Pallow =
Pmax 39.376 = = 26.3 kN (controls) 1.5 1.5
Ans.
1079
5m
y
x
5m
Section Properties. From the table listed in the appendix, the necessary section properties for a W200 * 22 are
P
x
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13–54. The W200 * 22 A-36-steel column is fixed at its base. Its top is constrained to rotate about the y–y axis and free to move along the y–y axis. Also, the column is braced along the x–x axis at its mid-height. If P = 25 kN, determine the maximum normal stress developed in the column.
100 mm
y
Section Properties. From the table listed in the appendix, necessary section properties for a W200 * 22 are A = 2860 mm2 = 2.86 A 10 - 3 B m2
ry = 22.3 mm = 0.0223 m
Ix = 20.0 A 106 B mm4 = 20.0 A 10 - 6 B m4
c =
bf 2
=
102 = 51 mm = 0.051 m 2
e = 0.1m Buckling About the Strong Axis. Since the column is fixed at the base and free at the top, Kx = 2. Applying Euler’s formula,
Pcr =
p2EIx (KL)x 2
=
p2 c200 A 109 B d c20.0 A 10 - 6 B d [2(10)]2
= 98.70kN
Euler’s formula is valid only if scr 6 sY. scr =
98.70 A 103 B Pcr = = 34.51 MPa 6 sY = 250 MPa A 2.86 A 10 - 3 B
O.K.
Since P = 25 kN 6 Pcr, the column does not buckle. Yielding About Weak Axis. Since the support provided by the bracing can be considered a pin connection, the upper portion of the column is pinned at both of its ends. Then Ky = 1 and L = 5 m. Applying the secant formula, smax =
=
(KL) P P ec C 1 + 2 sec B RS A 2ry A EA ry 2.5 A 103 B
2.86 A 10 - 3 B
D1 +
0.1(0.051) 0.02232
secC
25 A 103 B 1(5) ST 2(0.0223) C 200 A 109 B C 2.86 A 10 - 3 B D
= 130.26 MPa = 130 MPa
Ans.
Since smax 6 sY = 250 MPa, the column does not yield.
1080
y
x
5m
5m
P
x
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13–55. The wood column is fixed at its base, and its top can be considered pinned. If the eccentric force P = 10 kN is applied to the column, investigate whether the column is adequate to support this loading without buckling or yielding. Take E = 10 GPa and sY = 15 MPa.
x
P 150 mm
25 mm yx 25 mm 75 mm
5m
Section Properties. A = 0.05(0.15) = 7.5 A 10 - 3 B m2 Ix =
rx =
1 (0.05) A 0.153 B = 14.0625 A 10 - 6 B m4 12
14.0625 A 10 - 6 B Ix = 0.04330 m = AA C 7.5 A 10 - 3 B
1 (0.15) A 0.053 B = 1.5625 A 10 - 6 B m4 12 e = 0.15 m c = 0.075 m
Iy =
For a column that is fixed at one end and pinned at the other K = 0.7. Then, (KL)x = (KL)y = 0.7(5) = 3.5 m Buckling About the Weak Axis. Applying Euler’s formula, Pcr =
p2EIy (KL)y 2
=
p2 C 10 A 109 B D C 1.5625 A 10 - 6 B D 3.52
= 12.59 kN
Euler’s formula is valid if scr 6 sY. scr =
12.59 A 103 B Pcr = = 1.68 MPa 6 sY = 15 MPa A 7.5 A 10 - 3 B
O.K.
Since Pcr 7 P = 10 kN, the column will not buckle. Yielding About Strong Axis. Applying the secant formula. smax =
=
(KL)x P ec P C 1 + 2 sec B RS A 2rx A EA rx 10 A 103 B
7.5 A 10 - 3 B
D1 +
0.15(0.075) 2
0.04330
secC
10 A 103 B 3.5 ST 2(0.04330) C 10 A 109 B C 7.5 A 10 - 3 B D
= 10.29 MPa Since smax 6 sY = 15 MPa , the column will not yield.
Ans.
1081
75 mm
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*13–56. The wood column is fixed at its base, and its top can be considered pinned. Determine the maximum eccentric force P the column can support without causing it to either buckle or yield. Take E = 10 GPa and sY = 15 MPa .
x
P 150 mm
25 mm yx 25 mm 75 mm
5m
Section Properties. A = 0.05(0.15) = 7.5 A 10 - 3 B m2 Ix =
1 (0.05) A 0.153 B = 14.0625 A 10 - 6 B m4 12
14.0625 A 10 Ix = C 7.5 A 10 - 3 B AA
-6
rx =
B
= 0.04330 m
1 (0.15) A 0.053 B = 1.5625 A 10 - 6 B m4 12 e = 0.15 m c = 0.075 m
Iy =
For a column that is fixed at one end and pinned at the other K = 0.7. Then, (KL)x = (KL)y = 0.7(5) = 3.5 m Buckling About the Weak Axis. Applying Euler’s formula, Pcr =
p2EIy (KL)y
2
=
p2 C 10 A 109 B D C 1.5625 A 10 - 6 B D 3.52
= 12.59 kN = 12.6 kN
Ans.
Euler’s formula is valid if scr 6 sY. scr
12.59 A 103 B Pcr = = = 1.68 MPa 6 sY = 15 MPa A 7.5 A 10 - 3 B
O.K.
Yielding About Strong Axis. Applying the secant formula with P = Pcr = 12.59 kN, smax =
=
(KL)x P P ec C B 1 + 2 sec B RS A 2rx A EA rx 12.59 A 103 B 7.5 A 10 - 3 B
D1 +
0.15(0.075) 0.043302
secC
12.59 A 103 B B 3.5 ST 2(0.04330) C 10 A 109 B C 7.5 A 10 - 3 B D
= 13.31 MPa 6 sY = 15 MPa
O.K.
1082
75 mm
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The W250 * 28 A-36-steel column is fixed at its base. Its top is constrained to rotate about the y–y axis and free to move along the y–y axis. If e = 350 mm, determine the allowable eccentric force P that can be applied without causing the column either to buckle or yield. Use F.S. = 2 against buckling and F.S. = 1.5 against yielding.
•13–57.
P e x
x y
6m
Section Properties. From the table listed in the appendix, necessary section properties for a W250 * 28 are A = 3620 mm2 = 3.62 A 10 - 3 B m2
rx = 105 mm = 0.105 m
Iy = 1.78 A 106 B mm4 = 1.78 A 10 - 6 B m4
c =
260 d = = 130 mm = 0.13 m 2 2
e = 0.35 m Buckling About the Strong Axis. Since the column is fixed at the base and pinned at the top, Kx = 0.7. Applying Euler’s formula, Pcr =
p2EIy (KL)y 2
=
p2 C 200 A 109 B D C 1.78 A 10 - 6 B D [0.7(6)]2
= 199.18 kN
Euler’s formula is valid only if scr 6 sY. scr =
199.18 A 103 B Pcr = = 55.02 MPa 6 sY = 250 MPa A 3.62 A 10 - 3 B
O.K.
Thus, Pallow =
Pcr 199.18 = = 99.59 kN F.S. 2
Yielding About Strong Axis. Since the column is fixed at its base and free at its top, Kx = 2. Applying the secant formula, smax =
(KL)x Pmax Pmax ec C 1 + 2 sec B RS A 2rx A EA rx
250 A 106 B =
250 A 106 B =
Pmax
3.62 A 10
-3
Pmax
B
3.62 A 10 - 3 B
D1 +
0.35(0.13) 0.1052
secC
2(6) Pmax ST 2(0.105)A 200 A 109 B C 3.62 A 10 - 3 B D
A 1 + 4.1270 sec (0.0021237)2Pmax B
Solving by trial and error, Pmax = 133.45 kN Then, Pallow =
y
Pmax 133.45 = = 88.97 kN = 89.0 kN (controls) 1.5 1.5
Ans.
1083
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13–58. The W250 * 28 A-36-steel column is fixed at its base. Its top is constrained to rotate about the y–y axis and free to move along the y–y axis. Determine the force P and its eccentricity e so that the column will yield and buckle simultaneously.
P e x
x y
6m
Section Properties. From the table listed in the appendix, the necessary section properties for a W250 * 28 are A = 3620 mm2 = 3.62 A 10 - 3 B m2
rx = 105 mm = 0.105 m
Iy = 1.78 A 106 B mm4 = 1.78 A 10 - 6 B m4
c =
d 260 = = 130 mm = 0.13 m 2 2
Buckling About the Weak Axis. Since the column is fixed at the base and pinned at its top, Kx = 0.7. Applying Euler’s formula, Pcr =
p2EIy (KL)y 2
=
p2 C 200 A 109 B D C 1.78 A 10 - 6 B D [0.7(6)]2
= 199.18 kN = 199 kN
Ans.
Euler’s formula is valid only if scr 6 sY. scr =
199.18 A 103 B Pcr = = 55.02 MPa 6 sY = 250 MPa A 3.62 A 10 - 3 B
O.K.
Yielding About Strong Axis. Since the column is fixed at its base and free at its top, Kx = 2. Applying the secant formula with P = Pcr = 199.18 kN, smax =
(KL)x P P ec C 1 + 2 sec B RS A 2rx A EA rx
250 A 106 B =
199.18 A 103 B 3.62 A 10 - 3 B
D1 +
e(0.13) 0.1052
secC
y
199.18 A 103 B 2(6) ST 2(0.105) C 200 A 109 B C 3.62 A 10 - 3 B D
e = 0.1753 m = 175 mm
Ans.
1084
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13–59. The steel column supports the two eccentric loadings. If it is assumed to be pinned at its top, fixed at the bottom, and fully braced against buckling about the y–y axis, determine the maximum deflection of the column and the maximum stress in the column. Est = 200 GPa, sY = 360 MPa.
130 kN 50 kN 80 mm 120 mm
100 mm 10 mm 6m
A = 0.12(0.1) - (0.1)(0.09) = 3.00 A 10 - 3 B m2 Ix =
1 1 (0.1) A 0.123 B (0.09) A 0.13 B = 6.90 A 10 - 6 B m4 12 12
rx =
Ix 6.90(10 - 6) = = 0.047958 m AA A 3.00(10 - 3)
For a column fixed at one end and pinned at the other end, K = 0.7. (KL)x = 0.7(6) = 4.2 m The eccentricity of the two applied loads is, e =
130(0.12) - 50(0.08) = 0.06444 m 180
Yielding About x–x Axis: Applying the secant formula, (KL)x P P ec B 1 + 2 sec ¢ ≤R A 2rx A EA rx 180(103) =
-3
3.00(10 )
y
B1 +
0.06444(0.06) 0.047958
2
sec ¢
180(103) 4.2 ≤R 2(0.047958)A 200(109)(3.00)(10 - 3)
= 199 MPa
Ans.
Since smax 6 sg = 360 MPa, the column does not yield. Maximum Displacement: ymax = e B sec ¢
P KL ≤ - 1R A EI 2
= 0.06444 B sec ¢
4.2 180(103) a b ≤ - 1R A 200(109)[6.90(10 - 6)] 2
= 0.02433 m = 24.3 mm
Ans.
1085
10 mm 100 mm
x
Section Properties:
smax =
x
10 mm
y
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*13–60. The steel column supports the two eccentric loadings. If it is assumed to be fixed at its top and bottom, and braced against buckling about the y–y axis, determine the maximum deflection of the column and the maximum stress in the column. Est = 200 GPa, sY = 360 MPa.
130 kN 50 kN 80 mm 120 mm
100 mm 10 mm 6m
A = 0.12(0.1) - (0.1)(0.09) = 3.00 A 10 - 3 B m2 Ix =
1 1 (0.1) A 0.123 B (0.09) A 0.013 B = 6.90 A 10 - 6 B m4 12 12
rx =
6.90(10 - 6) Ix = = 0.047958 m AA A 3.00(10 - 3)
For a column fixed at both ends, K = 0.5. (KL)x = 0.5(6) = 3.00 m The eccentricity of the two applied loads is, e =
130(0.12) - 50(0.08) = 0.06444 m 180
Yielding About x–x Axis: Applying the secant formula, (KL)x P P ec B 1 + 2 sec ¢ ≤R A 2rx A EA rx 180(103) =
-3
3.00(10 )
y
B1 +
0.06444(0.06) 0.047958
2
sec ¢
180(103) 3.00 ≤R 2(0.047958)A 200(109)(3.00)(10 - 3)
= 178 MPa
Ans.
Since smax 6 sg = 360 MPa, the column does not yield. Maximum Displacement: ymax = e B sec ¢
P KL ≤ - 1R A EI 2
= 0.06444 B sec ¢
3 180(103) a b ≤ - 1R 9 6 A 200(10 )[6.90(10 )] 2
= 0.01077 m = 10.8 mm
Ans.
1086
10 mm 100 mm
x
Section Properties:
smax =
x
10 mm
y
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13–61. The W250 * 45 A-36-steel column is pinned at its top and fixed at its base. Also, the column is braced along its weak axis at mid-height. If P = 250 kN, investigate whether the column is adequate to support this loading. Use F.S. = 2 against buckling and F.S. = 1.5 against yielding.
P 4 250 mm
P 250 mm
4m
Section Properties. From the table listed in the appendix, the necessary section properties for a W250 * 45 are A = 5700 mm = 5.70 A 10 2
-3
Bm
2
rx = 112 mm = 0.112 m
Iy = 7.03 A 106 B mm4 = 7.03 A 10 - 6 B m4
c =
266 d = = 133 mm = 0.133 m 2 2
The eccentricity of the equivalent force P¿ = 250 +
250 = 312.5 kN is 4
250 (0.25) 4 = 0.15 m 250 250 + 4
250(0.25) e =
Buckling About the Weak Axis. The column is braced along the weak axis at midheight and the support provided by the bracing can be considered as a pin. The top portion of the column is critical is since the top is pinned so Ky = 1 and L = 4 m Applying Euler’s formula, Pcr =
p2EIy (KL)y 2
=
p2 C 200 A 109 B D C 7.03 A 10 - 6 B D [1(4)]2
= 867.29 kN
Euler’s equation is valid only if scr 6 sY. scr =
867.29 A 103 B Pcr = = 152.16 MPa 6 sY = 250 MPa A 5.70 A 10 - 3 B
O.K.
Then, œ Pallow =
Pcr 867.29 = = 433.65 kN F.S. 2
œ Since Pallow 7 P¿ , the column does not buckle.
Yielding About Strong Axis. Since the column is fixed at its base and pinned at its top, Kx = 0.7 and L = 8 m. Applying the secant formula with œ = P¿(F.S.) = 312.5(1.5) = 468.75 kN Pmax smax =
=
œ œ (KL)x Pmax Pmax ec C 1 + 2 sec B RS A 2rx A EA rx
468.75 A 103 B 5.70 A 10 - 3 B
C1 +
0.15(0.133) 0.112
2
sec B
468.75 A 103 B 0.7(8) RS 2(0.112) C 200 A 109 B C 5.70 A 10 - 3 B D
= 231.84 MPa Since smax 6 sY = 250 MPa, the column does not yield.
1087
4m
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The W250 * 45 A-36-steel column is pinned at its top and fixed at its base. Also, the column is braced along its weak axis at mid-height. Determine the allowable force P that the column can support without causing it either to buckle or yield. Use F.S. = 2 against buckling and F.S. = 1.5 against yielding.
•13–62.
P 4 250 mm
P 250 mm
4m
Section Properties. From the table listed in the appendix, the necessary section properties for a W250 * 45 are A = 5700 mm2 = 5.70 A 10 - 3 B m2
rx = 112 mm = 0.112 m
Iy = 7.03 A 106 B mm4 = 7.03 A 10 - 6 B m4
c =
The eccentricity of the equivalent force P¿ = P +
d 266 = = 133 mm = 0.133 m 2 2 P = 1.25P is 4
P (0.25) 4 = 0.15 m P P + 4
P(0.25) e =
Buckling About the Weak Axis. The column is braced along the weak axis at midheight and the support provided by the bracing can be considered as a pin. The top portion of the column is critical is since the top is pinned so Ky = 1 and L = 4 m. Applying Euler’s formula, Pcr =
p2EIy (KL)y
2
=
p2 C 200 A 109 B D C 7.03 A 10 - 6 B D [1(4)]2
= 867.29 kN
Euler’s equation is valid only if scr 6 sY. scr
867.29 A 103 B Pcr = = = 152.16 MPa 6 sY = 250 MPa A 5.70 A 10 - 3 B
O.K.
Then, œ = Pallow
Pcr F.S.
867.29 2 = 346.92 kN
1.25Pallow = Pallow
Yielding About Strong Axis. Since the column is fixed at its base and pinned at its top, Kx = 0.7 and L = 8 m. Applying the secant formula, smax =
œ œ (KL)x Pmax Pmax ec C 1 + 2 sec B RS A 2rx A EA rx
250 A 106 B = 250 A 106 B =
1.25Pmax
5.70 A 10
-3
1.25Pmax
B
5.70 A 10 - 3 B
C1 +
0.15(0.133) 0.1122
sec B
0.7(8) 1.25Pmax RS 2(0.112) A 200 A 109 B C 5.70 A 10 - 3 B D
A 1 + 1.5904 sec (0.00082783)2Pmax B
Solving by trial and error, Pmax = 401.75 kN Then, Pallow =
401.75 = 267.83 kN = 268 kN (controls) 1.5
Ans.
1088
4m
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13–63. The W14 * 26 structural A-36 steel member is used as a 20-ft-long column that is assumed to be fixed at its top and fixed at its bottom. If the 15-kip load is applied at an eccentric distance of 10 in., determine the maximum stress in the column.
15 kip
Section Properties for W 14 * 26 A = 7.69 in2
d = 13.91 in.
10 in.
20 ft
rx = 5.65 in.
Yielding about x-x axis: smax =
P ec KL P b d; c1 + 2 sec a A 2 r AE A r
15 P = = 1.9506 ksi; A 7.69
K = 0.5
10 A 13.91 ec 2 B = = 2.178714 r2 (5.65)2
0.5 (20)(12) 15 KL P = = 0.087094 2 r A EA 2(5.65) A 29 (103)(7.69) smax = 1.9506[1 + 2.178714 sec (0.087094)] = 6.22 ksi 6 sg = 36 ksi
O.K.
Ans.
*13–64. The W14 * 26 structural A-36 steel member is used as a column that is assumed to be fixed at its top and pinned at its bottom. If the 15-kip load is applied at an eccentric distance of 10 in., determine the maximum stress in the column.
15 kip
Section Properties for W 14 * 26 A = 7.69 in2
d = 13.91 in.
20 ft
rx = 5.65 in.
Yielding about x-x axis: smax =
P ec KL P c1 + 2 sec a b d; A 2 r AE A r
15 P = = 1.9506 ksi ; A 7.69
K = 0.7
10 A 13.91 ec 2 B = = 2.178714 2 r (5.65)2
0.7 (20)(12) 15 KL P = = 0.121931 2 r A EA 2(5.65) A 29 (103)(7.69) smax = 1.9506[1 + 2.178714 sec (0.121931)] = 6.24 ksi 6 sg = 36 ksi
10 in.
O.K.
Ans.
1089
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•13–65. Determine the maximum eccentric load P the 2014-T6-aluminum-alloy strut can support without causing it either to buckle or yield. The ends of the strut are pin-connected.
P 150 mm
100 mm
a a 3m 50 mm
100 mm Section a – a
Section Properties. The necessary section properties are A = 0.05(0.1) = 5 A 10 - 3 B m2 Iy =
1 (0.1) A 0.053 B = 1.04167 A 10 - 6 B m4 12
4.1667 A 10 Ix = C 5 A 10 - 3 B AA
-6
rx =
B
= 0.02887 m
For a column that is pinned at both of its ends K = 1. Thus, (KL)x = (KL)y = 1(3) = 3 m Buckling About the Weak Axis. Applying Euler’s formula, Pcr =
p2EIy (KL)y 2
=
p2 C 73.1 A 109 B D C 1.04167 A 10 - 6 B D 32
= 83.50 kN = 83.5 kN
Ans.
Critical Stress: Euler’s formula is valid only if scr 6 sY. scr =
83.50 A 103 B Pcr = = 16.70 MPa 6 sY = 414 MPa A 5 A 10 - 3 B
O.K.
Yielding About Strong Axis. Applying the secant formula, smax =
=
(KL)x P P ec C 1 + 2 sec B RS A 2rx A EA rx
83.50 A 103 B 5 A 10 - 3 B
D1 +
0.15(0.05) 0.028872
83.50 A 10 B 3 ST 2(0.02887) C 73.1 A 109 B C 5 A 10 - 3 B D 3
secC
= 229.27 MPa 6 sY = 414 MPa
O.K.
1090
P 150 mm
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13–66. The W8 * 48 structural A-36 steel column is fixed at its bottom and free at its top. If it is subjected to the eccentric load of 75 kip, determine the factor of safety with respect to either the initiation of buckling or yielding.
75 kip
y
12 ft
Section Properties: For a wide flange section W8 * 48, A = 14.1 in2
rx = 3.61 in.
Iy = 60.9 in4
d = 8.50 in.
For a column fixed at one end and free at the other and, K = 2. (KL)y = (KL)x = 2(12)(12) = 288 in. Buckling About y–y Axis: Applying Euler’s formula, P = Pcr =
p2EIy (KL)2y p2 (29.0)(103)(60.9)
=
2882
= 210.15 kip Critical Stress: Euler’s formula is only valid if scr 6 sg. scr =
Pcr 210.15 = = 14.90 ksi 6 sg = 36 ksi A 14.1
O. K.
Yielding About x–x Axis: Applying the secant formula, smax =
36 =
(KL)x Pmax Pmax ec B 1 + 2 sec ¢ ≤R A 2rx A EA rx
8 A 1.50 Pmax Pmax 288 2 B sec ¢ B1 + ≤R 14.1 2(3.61)A 29.0(103)(14.1) 3.612
36(14.1) = Pmax A 1 + 2.608943 sec 0.06238022Pmax B Solving by trial and error, Pmax = 117.0 kip (Controls!) Factor of Safety: F.S. =
Pmax 117.0 = = 1.56 P 75
Ans.
1091
8 in. y x
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13–67. The W8 * 48 structural A-36 steel column is fixed at its bottom and pinned at its top. If it is subjected to the eccentric load of 75 kip, determine if the column fails by yielding. The column is braced so that it does not buckle about the y–y axis.
75 kip
y
12 ft
Section Properties: For a wide flange section W8 * 48, A = 14.1 in2
rx = 3.61 in.
d = 8.50 in.
For a column fixed at one end and pinned at the other end, K = 0.7. (KL)x = 0.7(12)(12) = 100.8 in. Yielding About x–x Axis: Applying the secant formula, smax =
=
(KL)x P P ec B 1 + 2 sec ¢ ≤R A 2rx A EA rx
8 A 1.50 75 100.8 75 2 B sec ¢ B1 + ≤R 2 14.1 2(3.61)A 3.61 29.0(103)(14.1)
= 19.45 ksi 6 sg = 36 ksi
O.K.
Hence, the column does not fail by yielding.
Ans.
1092
8 in. y x
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*13–68. Determine the load P required to cause the steel W12 * 50 structural A-36 steel column to fail either by buckling or by yielding. The column is fixed at its bottom and the cables at its top act as a pin to hold it.
2 in.
P
25 ft
Section Properties: For a wide flange section W12 * 50, A = 14.7 in2
rx = 5.18 in.
Iy = 56.3 in4
d = 12.19 in.
For a column fixed at one end and pinned at the other end, K = 0.7. (KL)y = (KL)x = 0.7(25)(12) = 210 in. Buckling About y–y Axis: Applying Euler’s formula, P = Pcr =
p2EIy (KL)2y p2 (29.0)(103)(56.3)
=
2102
= 365.40 kip Critical Stress: Euler’s formula is only valid if scr 6 sg. scr =
Pcr 365.4 = = 24.86 ksi 6 sg = 36 ksi A 14.7
O. K.
Yielding About x–x Axis: Applying the secant formula, smax =
36 =
(KL)x P P ec B 1 + 2 sec ¢ ≤R A 2rx A EA rx
2 A 12.19 P 210 P 2 B sec ¢ B1 + ≤R 14.7 2(5.18)A 29.0(103)(14.7) 5.182
36(14.7) = P A 1 + 0.454302 sec 0.03104572P B Solving by trial and error, Pmax = 343.3 kip = 343 kip (Controls!)
Ans.
1093
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•13–69.
Solve Prob. 13–68 if the column is an A-36 steel W12 * 16 section.
2 in.
P
25 ft
Section Properties: For a wide flange section W12 * 16, A = 4.71 in2
rx = 4.67 in.
Iy = 2.82 in4
d = 11.99 in.
For a column fixed at one end and pinned at the other end, K = 0.7. (KL)y = (KL)x = 0.7(25)(12) = 210 in. Buckling About y–y Axis: Applying Euler’s formula, P = Pcr =
p 2EIy (KL)2y p2 (29.0)(103)(2.82)
=
2102
= 18.30 kip = 18.3 kip‚
Ans.
Critical Stress: Euler’s formula is only valid if scr 6 sg. scr =
Pcr 18.30 = = 3.89 ksi 6 sg = 36 ksi A 4.71
Yielding About x–x Axis: Applying the secant formula, smax =
=
(KL)x P P ec B 1 + 2 sec ¢ ≤R A 2rx A EA rx
2 A 11.99 18.30 210 18.30 2 B seca bR B1 + 4.71 2(4.67)A 29.0(103)(4.71) 4.672
= 6.10 ksi 6 sg = 36 ksi
O.K.
1094
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13–70. A column of intermediate length buckles when the compressive stress is 40 ksi. If the slenderness ratio is 60, determine the tangent modulus. p2 Et
A KrL B 2
scr =
40 =
;
a
KL b = 60 r
p2 Et (60)2
Et = 14590 ksi = 14.6 (103) ksi ‚
Ans
s(ksi)
13–71. The 6-ft-long column has the cross section shown and is made of material which has a stress-strain diagram that can be approximated as shown. If the column is pinned at both ends, determine the critical load Pcr for the column.
0.5 in. 55 0.5 in.
5 in.
25
0.5 in. 3 in.
Section Properties: The neccessary section properties are A = 2[0.5(3)] + 5(0.5) = 5.5 in2
P (in./in.) 0.001
I = 2B ry =
1 1 (0.5) A 33 B R + (5) A 0.53 B = 2.3021 in4 12 12
2.3021 Iy = = 0.6470 in. AA A 5.5
For the column pinned at both of its ends, K = 1. Thus, 1(6)(12) KL = 111.29 = ry 0.6470 Critical Stress. Applying Engesser’s equation,
scr =
p2Et a
KL
r
b
=
p2Et 111.292
= 0.7969 A 10 - 3 B Et
(1)
From the stress - strain diagram, the tangent moduli are (Et)1 =
25 ksi = 25 A 103 B ksi 0.001
(Et)2 =
(55 - 25) ksi = 10 A 103 B (ksi) 0.004 - 0.001
0 … s 6 25 ksi
25ksi 6 s … 40 ksi
Substituting (Et)1 = 25 A 103 B into Eq. (1),
scr = 0.7969 A 10 - 3 B c25 A 103 B d = 19.92 ksi
Since scr 6 sY = 25 ksi, elastic buckling occurs. Thus, Pcr = scr A = 19.92(5.5) = 109.57 kip = 110 kip‚
Ans. 1095
0.004
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s(ksi)
*13–72. The 6-ft-long column has the cross section shown and is made of material which has a stress-strain diagram that can be approximated as shown. If the column is fixed at both ends, determine the critical load Pcr for the column.
0.5 in. 55 0.5 in.
5 in.
25
0.5 in. 3 in.
Section Properties. The neccessary section properties are A = 2[0.5(3)] + 5(0.5) = 5.5 in2
P (in./in.) 0.001
I = 2B ry =
1 1 (0.5) A 33 B R + (5) A 0.53 B = 2.3021 in4 12 12
Iy 2.3021 = = 0.6470 in. AA A 5.5
For the column fixed at its ends, K = 0.5. Thus, 0.5(6)(12) KL = 55.64 = ry 0.6470 Critical Stress. Applying Engesser’s equation, From the stress - strain diagram, the tangent moduli are (Et)1 =
25 ksi = 25 A 103 B ksi 0.001
(Et)2 =
(55 - 25)ksi = 10 A 103 B ksi 0.004 - 0.001
0 … s 6 25 ksi
25 ksi 6 s … 40 ksi
Substituting (Et)1 = 25 A 103 B ksi into Eq. (1),
scr = 3.1875 A 10 - 3 B c25 A 103 B d = 79.69 ksi
Since scr 7 sY = 25 ksi, the inelastic buckling occurs. Substituting (Et)2 into Eq. (1), scr = 3.1875 A 10 - 3 B c10 A 103 B d = 31.88 ksi Since 25 ksi 6 scr 6 55 ksi, this result can be used to calculate the critical load. Pcr = scr A = 31.88(5.5) = 175.31 kip = 175 kip
Ans.
1096
0.004
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s (MPa)
•13–73.
The stress-strain diagram of the material of a column can be approximated as shown. Plot P兾A vs. KL兾r for the column.
350
200
Tangent Moduli. From the stress - strain diagram, (Et)1 =
(Et)2 =
200 A 106 B 0.001
0 … s 6 200MPa
= 200 GPa
(350 - 200) A 106 B 0.004 - 0.001
= 50 GPa
0
P (in./in.) 0.001
0.004
200MPa 6 s … 350 MPa
Critical Stress. Applying Engesser’s equation, scr
P = A
p2Et
(1)
KL 2 a b r
If Et = (Et)1 = 200 GPa, Eq. (1) becomes p2 C 200 A 109 B D 1.974 A 106 B P = = MPa A KL 2 KL 2 a b a b r r
when scr =
P = sY = 200 MPa, this equation becomes A
200 A 106 B =
p2 C 200 A 109 B D a
KL 2 b r
KL = 99.346 = 99.3 r
If Et = (Et)2 = 50 GPa, Eq. (1) becomes P = A
p2 c50 A 109 B d
0.4935 A 106 B
MPa KL 2 KL 2 ¢ ≤ ¢ ≤ r r P when scr = = sY = 200 MPa, this A equation gives 200 A 106 B =
=
p2 C 50 A 109 B D a
KL 2 b r
KL = 49.67 = 49.7 r Using these results, the graphs of
P KL vs. is shown in Fig. a can be plotted. r A 1097
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s (MPa)
13–74. Construct the buckling curve, P兾A versus L兾r, for a column that has a bilinear stress–strain curve in compression as shown. The column is pinned at its ends.
260
140
0.001
Tangent modulus: From the stress–strain diagram, (Et)1 =
140(106) = 140 GPa 0.001
(Et)2 =
(260 - 140)(106) = 40 GPa 0.004 - 0.001
Critical Stress: Applying Engesser’s equation, scr
p2Et P = A L 2 a b r
[1]
Substituting (Et)1 = 140 GPa into Eq. [1], we have p2 C 140(109) D P = A ALB2 r
P = A When
1.38(106)
A Lr B 2
MPa
L P = 140 MPa, = 99.3 r A
Substitute (Et)2 = 40 GPa into Eq. [1], we have p2 C 40(109) D P = A ALB2 r
P = A When
0.395(106)
A Lr B 2
MPa
L P = 140 MPa, = 53.1 r A
1098
0.004
P (mm/mm)
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13–75. The stress-strain diagram for a material can be approximated by the two line segments shown. If a bar having a diameter of 80 mm and a length of 1.5 m is made from this material, determine the critical load provided the ends are pinned. Assume that the load acts through the axis of the bar. Use Engesser’s equation.
s (MPa) 1100
200 0.001
E1 =
200 (106) = 200 GPa 0.001
E2 =
1100 (106) - 200 (106) = 150 GPa 0.007 - 0.001
Section properties: I =
p 4 c; 4
r =
p 4 I 0.04 c 4 c = = = = 0.02 m AA C p c2 2 2
A = pc2
Engesser’s equation: 1.0(1.5) KL = = 75 r 0.02 scr =
p2 Et
A
B
KL 2 r
=
p2 Et (75)2
= 1.7546(10 - 3) Et
Assume Et = E1 = 200 GPa scr = 1.7546 (10 - 3)(200)(109) = 351 MPa 7 200 MPa Therefore, inelastic buckling occurs: Assume Et = E2 = 150 GPa scr = 1.7546 (10 - 3)(150)(109) = 263.2 MPa 200 MPa 6 scr 6 1100 MPa
O.K.
Critical load: Pcr = scr A = 263.2 (106)(p)(0.042) = 1323 kN‚
Ans
1099
0.007
P (mm/mm)
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*13–76. The stress-strain diagram for a material can be approximated by the two line segments shown. If a bar having a diameter of 80 mm and a length of 1.5 m is made from this material, determine the critical load provided the ends are fixed. Assume that the load acts through the axis of the bar. Use Engesser’s equation.
s (MPa) 1100
200 0.001
E1 =
200 (106) = 200 GPa 0.001
E2 =
1100 (106) - 200 (106) = 150 GPa 0.007 - 0.001
Section properties: I =
p 4 c; 4
r =
p 4 I 0.04 c 4 c = = = = 0.02 m AA C p c2 2 2
A = pc2
Engesser’s equation: 0.5 (1.5) KL = = 37.5 r 0.02 scr =
p2 Et
A
B
KL 2 r
=
p2 Et (37.5)2
= 7.018385(10 - 3) Et
Assume Et = E1 = 200 GPa scr = 7.018385 (10 - 3)(200)(109) = 1403.7 MPa 7 200 MPa
NG
Assume Et = E2 = 150 GPa scr = 7.018385 (10 - 3)(150)(109) = 1052.8 MPa 200 MPa 6 scr 6 1100 MPa
O.K.
Critical load: Pcr = scr A = 1052.8 (106)(p)(0.042) = 5292 kN
Ans.
1100
0.007
P (mm/mm)
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•13–77.
The stress-strain diagram for a material can be approximated by the two line segments shown. If a bar having a diameter of 80 mm and length of 1.5 m is made from this material, determine the critical load provided one end is pinned and the other is fixed. Assume that the load acts through the axis of the bar. Use Engesser’s equation.
s (MPa) 1100
200 0.001
E1 =
200 (106) = 200 GPa 0.001
E2 =
1100 (106) - 200 (106) = 150 GPa 0.007 - 0.001
Section properties: I =
p 4 c ; 4
r =
p 4 I 0.04 c 4 c = = = = 0.02 m AA C pc2 2 2
A = pc2
Engesser’s equation: 0.7 (1.5) KL = = 52.5 r 0.02 scr =
p2 Et
A
B
KL 2 r
=
p2Et (52.5)2
= 3.58081 (10 - 3) Et
Assume Et = E1 = 200 GPa scr = 3.58081 (10 - 3)(200)(109) = 716.2 MPa 7 200 MPa
NG
Assume Et = E2 = 150 GPa scr = 3.58081 (10 - 3)(150)(109) = 537.1 MPa 200 MPa 6 scr 6 1100 MPa
O.K.
Critical load: Pcr = scr A = 537.1 (106)(p)(0.042) = 2700 kN
Ans.
1101
0.007
P (mm/mm)
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13–78. Determine the largest length of a structural A-36 steel rod if it is fixed supported and subjected to an axial load of 100 kN. The rod has a diameter of 50 mm. Use the AISC equations.
Section Properties: A = p A 0.0252 B = 0.625 A 10 - 3 B p m2 I =
p A 0.0254 B = 97.65625 A 10 - 9 B p m4 4
r =
97.65625(10 - 9)p I = = 0.0125 m AA A 0.625(10 - 3)p
Slenderness Ratio: For a column fixed at both ends, K = 0.5. Thus, 0.5L KL = = 40.0L r 0.0125 AISC Column Formula: Assume a long column. sallow = 100(103) 0.625(10 - 3)p
=
12p2E
2 23 A KL r B
12p2 C 200(109) D 23(40.0L)3
L = 3.555 m KL KL 2p2E = 40.0(3.555) = 142.2 and for A–36 steel, a b = r r e A sg KL KL 2p2[200(109)] = = 125.7. Since a b … … 200, the assumption is correct. r e r A 250(106)
Here,
Thus, L = 3.56 m
Ans.
1102
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13–79. Determine the largest length of a W10 * 45 structural steel column if it is pin supported and subjected to an axial load of 290 kip. Est = 29(103) ksi, sY = 50 ksi. Use the AISC equations.
Section Properties: For a W10 * 45 wide flange section, A = 13.3 in2
ry = 2.01 in
Slenderness Ratio: For a column pinned at both ends, K = 1. Thus, a
1(L) KL b = = 0.49751L r y 2.01
AISC Column Formula: Assume a long column, sallow =
12p2E
2 23 A KL r B
12p2 C 29(103) D 290 = 13.3 23(0.49751L)2 L = 166.3 in. KL KL 2p2E = 0.49751 (166.3) = 82.76 and for grade 50 steel, a b = r r c A sg KL KL 2p2[29(103)] 6 a b , the assumption is not correct. = = 107.0. Since r r c 50 A Thus, the column is an intermediate column. Here,
Applying Eq. 13–23,
B1 sallow =
(KL>r)2 2(KL>r)2c
R sg
3(KL>r) (KL>r)3 5 + 3 8(KL>r)c 8(KL>r)3c
B1 -
(0.49751L)2
R (50) 2(107.02) 290 = 13.3 3(0.49751L) (0.49751L)3 5 + 3 8(107.0) 8(107.03)
0 = 12.565658 A 10 - 9 B L3 - 24.788132 A 10 - 6 B L2 - 1.743638 A 10 - 3 B L + 0.626437
Solving by trial and error, L = 131.12 in. = 10.9 ft
Ans.
1103
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*13–80. Determine the largest length of a W10 * 12 structural A-36 steel section if it is pin supported and is subjected to an axial load of 28 kip. Use the AISC equations. For a W 10 * 12,
A = 3.54 in2
ry = 0.785 in. s =
28 P = = 7.91 ksi A 3.54
Assume a long column: sallow = a a
12p2E 23(KL>r)2
KL 12p2(29)(103) 12p2E = = 137.4 b = r A 23(7.91) A 23sallow
KL 2p2(29)(103) 2p2E = = 126.1, b = r c A sg A 36
KL KL 7 a b r r c
Long column. KL = 137.4 r L = 137.4 ¢
r 0.785 ≤ = 137.4 ¢ ≤ = 107.86 in. K 1
= 8.99 ft
Ans.
•13–81.
Using the AISC equations, select from Appendix B the lightest-weight structural A-36 steel column that is 14 ft long and supports an axial load of 40 kip.The ends are pinned. Take sY = 50 ksi.
Try, W6 * 15 (A = 4.43 in2
ry = 1.46 in.)
a
KL 2p2(29)(103) 2p2E b = = = 107 r c A sY 50 A
a
(1.0)(14)(12) KL b = = 115.1, ry 1.46
a
KL KL b 7 a b ry r c
Long column sallow =
12p2(29)(103) 12 p2E = = 11.28 ksi 2 23(KL>r) 23(115.1)2
Pallow = sallowA = 11.28(4.43) = 50.0 kip 7 40 kip
O.K.
Use W6 * 15
Ans.
1104
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13–82. Using the AISC equations, select from Appendix B the lightest-weight structural A-36 steel column that is 12 ft long and supports an axial load of 40 kip. The ends are fixed. Take sY = 50 ksi.
A = 2.68 in2
Try W6 * 9 a
ry = 0.905 in.
KL 2p2(29)(103) 2p2E b = = = 107 r c A sY A 50
0.5(12)(12) KL = = 79.56 ry 0.905 KL KL 6 a b ry r c Intermediate column sallow =
KL>r 2 C 1 - 12 A (KL>r)c B D sg
KL>r KL>r 3 C 53 + 38 A (KL>r)c B - 18 A (KL>r)c B D
=
2 C 1 - 12 A 79.56 126.1 B D 36 ksi
1 79.56 3 C 53 + 38 A 79.56 126.1 B - 8 A 12.61 B D
= 15.40 ksi
Pallow = sallowA = 15.40(2.68) = 41.3 kip 7 40 kip
O.K.
Use W6 * 9
Ans.
1105
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13–83. Using the AISC equations, select from Appendix B the lightest-weight structural A-36 steel column that is 24 ft long and supports an axial load of 100 kip.The ends are fixed.
Section Properties: Try a W8 * 24 wide flange section, A = 7.08 in2
ry = 1.61 in
Slenderness Ratio: For a column fixed at both ends, K = 0.5. Thus, a AISC
Column
0.5(24)(12) KL b = = 89.44 r y 1.61 Formula:
For
A–36
steel,
a
KL 2p2E b = r c A sg
KL KL 2p2[29(103)] 6 a b , the column is an intermediate = 126.1. Since r r c A 36 column. Applying Eq. 13–23, =
(KL>r)2
B1 sallow =
R sg
3(KL>r) (KL>r)3 5 + 3 8(KL>r)c 8(KL>r)3c
B1 =
2(KL>r)2c
(89.442) 2(126.12)
R (36)
3(89.44) (89.443) 5 + 3 8(126.1) 8(126.13)
= 14.271 ksi The allowable load is Pallow = sallowA = 14.271(7.08) = 101 kip 7 P = 100 kip Thus, Use
O.K. Ans.
W8 * 24
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*13–84. Using the AISC equations, select from Appendix B the lightest-weight structural A-36 steel column that is 30 ft long and supports an axial load of 200 kip.The ends are fixed.
a
A = 14.1 in2
ry = 2.08 in.
Try W8 * 48
KL 2 p2 (29)(103) 2 p2E b = = = 126.1 r c A sg A 36
0.5 (30)(12) KL = = 86.54 ry 2.08 a
KL KL b 6 a b intermediate column. ry r c
b 1 - 12 B sallow =
b 53 + 38 B
KL r
A KL r Bc
e1 =
e 53
+
3 8
C
1 2
KL r
A KL r Bc
2
R r sg
R - 18 B
KL r
A KL r Bc
2 C 86.54 126.1 D f36
86.54 126.1
D - C
D
1 86.54 3 f 8 126.1
3
R r
= 14.611 ksi
Pallow = sallow A = 14.611 (14.1) = 206 kip 7 P = 200 kip Use
O.K.
W 8 * 48
Ans.
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•13–85.
A W8 * 24 A-36-steel column of 30-ft length is pinned at both ends and braced against its weak axis at midheight. Determine the allowable axial force P that can be safely supported by the column. Use the AISC column design formulas.
Section Properties. From the table listed in the appendix, the necessary section properties for a W8 * 24 are A = 7.08 in2
rx = 3.42 in.
ry = 1.61 in.
Slenderness Ratio. For a column pinned at both of its ends, K = 1. Here, Lx = 30(12) = 360 in. and Ly = 15(12) = 180 in. Thus,
¢
1(360) KL = 105.26 ≤ = r x 3.42
¢
1(180) KL = 111.80 (controls) ≤ = r y 1.61
AISC
=
C
Column 2p2 C 29 A 103 B D 36
Formulas.
For
= 126.10. Since
¢
A-36
steel
¢
KL KL ≤ 6 ¢ ≤ , the r y r c
KL 2p2E ≤ = r c A sY column
is
an
intermediate column. (KL>r)2
B1 sallow =
R sY
3(KL>r) (KL>r)3 5 + 3 8(KL>r)c 8(KL>r)c 3 C1 -
=
2(KL>r)c 2
111.802
2 A 126.102 B
S(36)
3(111.80) 5 111.803 + 3 8(126.10) 8 A 126.103 B
= 11.428 ksi Thus, the allowable force is Pallow = sallowA = 11.428(7.08) = 80.91 kip = 80.9 kip
1108
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13–86. Check if a W10 * 39 column can safely support an axial force of P = 250 kip. The column is 20 ft long and is pinned at both ends and braced against its weak axis at mid-height. It is made of steel having E = 29(103) ksi and sY = 50 ksi. Use the AISC column design formulas.
Section Properties. From the table listed in the appendix, the necessary section properties for a W10 * 39 are A = 11.5 in2
rx = 4.27 in.
ry = 1.98 in.
Slenderness Ratio. For a column pinned at both of its ends, K = 1. Here, Lx = 20(12) = 240 in. and Ly = 10(12) = 120 in. Thus,
¢
1(240) KL = 56.21 ≤ = r x 4.27
¢
1(120) KL = 60.606 (controls) ≤ = r y 1.98
AISC
=
Column 2p2 c29 A 103 B d
S
50
Formulas.
For
A-36
= 107.00 . Since ¢
steel
¢
KL 2p2E ≤ = r c A sY
KL KL ≤ 6 ¢ ≤ , the column is an r y r c
intermediate column.
B1 sallow =
2(KL>r)c 2
R sY
3(KL>r) (KL>r)3 5 + 3 8(KL>r)c 8(KL>r)c 3 C1 -
=
(KL>r)2
60.6062
2 A 107.002 B
S(50)
3(60.606) 5 60.6063 + 3 8(107.00) 8 A 107.003 B = 22.614 ksi
Thus, the allowable force is Pallow = sallowA = 22.614(11.5) = 260.06 kip 7 P = 250 kip Thus, a W10 * 39 column is adequate.
1109
O.K.
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13–87. A 5-ft-long rod is used in a machine to transmit an axial compressive load of 3 kip. Determine its smallest diameter if it is pin connected at its ends and is made of a 2014-T6 aluminum alloy.
Section properties: A =
p 2 d ; 4
I =
p d 4 pd4 a b = 4 2 64
pd4
r =
I d 64 = = AA C p4 d2 4
sallow =
P = A
p 4
3 3.820 = d2 d2
Assume long column: 1.0 (5)(12) 240 KL = = d r d 4
sallow =
54 000
A
B
KL 2 r
;
3.820 54000 = d2 C 240 D 2 d
d = 1.42 in.
Ans.
KL 240 = = 169 7 55 r 1.42
O.K.
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*13–88. Check if a W10 * 45 column can safely support an axial force of P = 200 kip. The column is 15 ft long and is pinned at both of its ends. It is made of steel having E = 29(103) ksi and sY = 50 ksi. Use the AISC column design formulas.
Section Properties. Try W10 * 45. From the table listed in the appendix, the necessary section properties are A = 13.3 in2
ry = 2.01 in.
Slenderness Ratio. For a column pinned at both of its ends, K = 1. Thus, a
1(15)(12) KL b = = 89.552 r y 2.01
KL 2p2E AISC Column Formulas. Here, a b = = r c A sY S KL KL Since a b 6 a b , the r y r c
2p2 c29 A 103 B d 50
= 107.00.
column is an intermediate column. (KL>r)2
B1 sallow =
R sY
(KL>r)3 3(KL>r) 5 + 3 8(KL>r)c 8(KL>r)c 3 C1 -
=
2(KL>r)c 2
89.5522
2 A 107.002 B
S(50)
3(89.552) 5 89.5523 + 3 8(107.00) 8 A 107.003 B
= 17.034 ksi Thus, the allowable force is Pallow = sallowA = 17.034(13.3) = 226.55 kip 7 P = 200 kip
O.K.
Thus, A W10 * 45 can be used
Ans.
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•13–89.
Using the AISC equations, check if a column having the cross section shown can support an axial force of 1500 kN. The column has a length of 4 m, is made from A-36 steel, and its ends are pinned.
20 mm
350 mm
300 mm 10 mm
Section Properties: A = 0.3(0.35) - 0.29(0.31) = 0.0151 m2 Iy =
1 1 (0.04) A 0.33 B + (0.31) A 0.013 B = 90.025833 A 10 - 6 B m4 12 12
ry =
Iy 90.02583(10 - 6) = = 0.077214 m AA A 0.0151
Slenderness Ratio: For a column pinned at both ends, K = 1. Thus, a
AISC =
Column 2p2[200(109)]
1(4) KL b = = 51.80 r y 0.077214
Formula: = 125.7. Since
A 250(10 ) column. Applying Eq. 13–23, 6
For
a
KL 2p2E b = r c A sg
(KL>r)2 2(KL>r)2c
R sg
3(KL>r) (KL>r)3 5 + 3 8(KL>r)c 8(KL>r)3c
B1 =
steel,
KL KL 6 a b , the column is an intermediate r r c
B1 sallow =
A–36
(51.802) 2(125.72)
20 mm
R (250)(106)
3(51.80) (51.803) 5 + 3 8(125.7) 8(125.73)
= 126.2 MPa The allowable load is Pallow = sallowA = 126.2 A 106 B (0.0151) = 1906 kN 7 P = 1500 kN
O.K.
Thus, the column is adequate.
Ans.
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13–90. The A-36-steel tube is pinned at both ends. If it is subjected to an axial force of 150 kN, determine the maximum length that the tube can safely support using the AISC column design formulas.
100 mm
80 mm
Section Properties. A = p A 0.052 - 0.042 B = 0.9 A 10 - 3 B p m2 I =
r =
p A 0.054 - 0.044 B = 0.9225 A 10 - 6 B p m4 4
0.9225 A 10 - 6 B p I = = 0.03202 m AA C 0.9 A 10 - 3 B p
Slenderness Ratio. For a column pinned at both of its ends, K = 1. Thus, 1(L) KL = = 31.23L r 0.03202 AISC Column Formulas. sallow =
12p2E 23(KL>r)2
150 A 103 B
.9 A 10 - 3 B p
=
12p2 C 200 A 109 B D 23(31.23L)2
L = 4.4607 m = 4.46 m
Here,
KL = 31.23(4.4607) = 139.33. r 2p2 C 200 A 109 B D
= 125.66. Since a
250 A 10 B long column is correct. =
C
6
Ans.
For
A-36
steel
a
KL 2p2E b = r c A sY
KL KL b 6 6 200, the assumption of a r c r
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13–91. The bar is made of a 2014-T6 aluminum alloy. Determine its smallest thickness b if its width is 5b. Assume that it is pin connected at its ends.
600 lb b 5b
8 ft
Section Properties: A = b(5b) = 5b2
Iy =
1 5 4 (5b) A b3 B = b 12 12
ry =
5 4 Iy 23 12 b = = b AA C 5b2 6
600 lb
Slenderness Ratio: For a column pinned at both ends, K = 1. Thus, a
1(8)(12) 332.55 KL = b = 23 b r y 6 b
Aluminum (2014 - T6 alloy) Column Formulas: Assume a long column and apply Eq. 13–26. sallow =
54 000 (KL>r)2
0.600 54 000 = 2 5b A 332.55 B 2 b
b = 0.7041 in. Here,
KL KL 332.55 = 7 55, the assumption is correct. Thus, = 472.3. Since r r 0.7041 b = 0.704 in.
Ans.
1114
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*13–92. The bar is made of a 2014-T6 aluminum alloy. Determine its smallest thickness b if its width is 5b. Assume that it is fixed connected at its ends.
600 lb b 5b
8 ft
Section Properties: A = b(5b) = 5b2 Iy =
1 5 4 (5b) A b3 B = b 12 12
ry =
Iy 23 12 b = = b AA C 5b2 6
5
600 lb
4
Slenderness Ratio: For a column fixed at both ends, K = 0.5. Thus, a
0.5(8)(12) 166.28 KL = b = r y b 23 6 b
Aluminum (2014 - T6 alloy) Column Formulas: Assume a long column and apply Eq. 13–26. sallow =
54 000 (KL>r)2
0.600 54 000 = 5b2 A 166.28 B 2 b
b = 0.4979 in. Here,
KL KL 166.28 = 334.0. Since = 7 55, the assumption is correct. r r 0.4979
Thus, b = 0.498 in.
Ans.
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•13–93.
The 2014-T6 aluminum column of 3-m length has the cross section shown. If the column is pinned at both ends and braced against the weak axis at its mid-height, determine the allowable axial force P that can be safely supported by the column.
15 mm
170 mm
15 mm
15 mm 100 mm
Section Properties. A = 0.1(0.2) - 0.085(0.17) = 5.55 A 10 - 3 B m2 Ix =
1 1 (0.1) A 0.23 B (0.085) A 0.173 B = 31.86625 A 10 - 6 B m4 12 12
Iy = 2 c rx =
ry =
1 1 (0.015) A 0.13 B d + (0.17) A 0.0153 B = 2.5478 A 10 - 6 B m4 12 12
31.86625 A 10 - 6 B Ix = = 0.07577 AA C 5.55 A 10 - 3 B
2.5478 A 10 - 6 B Iy = = 0.02143 m AA C 5.55 A 10 - 3 B
Slenderness Ratio. For a column pinned at both of its ends, K = 1. Here, Lx = 3 m and Ly = 1.5 m. Thus, a
(1)(3) KL b = = 39.592 r x 0.07577
a
(1)(1.5) KL b = = 70.009 (controls) r y 0.02143
2014-T6 Alumimum Alloy Column Formulas. Since a
KL b 7 55, the column can r y
be classified a long column, sallow = D
373 A 103 B
T Mpa
= C
373 A 103 B
S MPa
a
KL 2 b r
70.0092
= 76.103 MPa Thus, the allowed force is Pallow = sallowA = 76.103 A 106 B C 5.55 A 10 - 3 B D = 422.37 kN = 422 kN
1116
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13–94. The 2014-T6 aluminum column has the cross section shown. If the column is pinned at both ends and subjected to an axial force P = 100 kN, determine the maximum length the column can have to safely support the loading.
15 mm
170 mm
15 mm
15 mm 100 mm
Section Properties. A = 0.1(0.2) - 0.085(0.17) = 5.55 A 10 - 3 B m2 Iy = 2c
ry =
1 1 (0.015) A 0.13 B d + (0.17) A 0.0153 B = 2.5478 A 10 - 6 B m4 12 12
2.5478 A 10 - 6 B Iy = = 0.02143 m AA C 5.55 A 10 - 3 B
Slenderness Ratio. For a column pinned at both of its ends, K = 1. Then, a
1(L) KL b = = 46.6727L r y 0.02143
2014-T6 Alumimum Alloy Column Formulas. Assuming a long column, sallow = D 100 A 103 B
373 A 103 B
5.55 A 10 - 3 B
a
KL 2 b r
= C
T MPa
373 A 103 B
(46.672L)2
S A 106 B Pa
L = 3.083 m = 3.08 m Since a
Ans.
KL b = 46.6727(3.083) = 143.88 7 55, the assumption is correct. r y
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13–95. The 2014-T6 aluminum hollow section has the cross section shown. If the column is 10 ft long and is fixed at both ends, determine the allowable axial force P that can be safely supported by the column.
4 in.
3 in.
Section Properties. A = p A 22 - 1.52 B = 1.75p in2 r =
I =
p 4 A 2 - 1.54 B = 2.734375p in4 4
I 2.734375p = = 1.25 in. AA A 1.75p
Slenderness Ratio. For a column fixed at both of its ends, K = 0.5. Thus, 0.5(10)(12) KL = = 48 r 1.25 2014-T6 Aluminum Alloy Column Formulas. Since 12 6
KL 6 55, the column can r
be classified as an intermediate column. sallow = c30.7 - 0.23 a
KL b d ksi r
= [30.7 - 0.23(48)] ksi = 19.66 ksi Thus, the allowable load is Pallow = sallowA = 19.66 A 106 B (1.75p) = 108.09 kip = 108 kip
1118
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*13–96. The 2014-T6 aluminum hollow section has the cross section shown. If the column is fixed at its base and pinned at its top, and is subjected to the axial force P = 100 kip, determine the maximum length of the column for it to safely support the load.
4 in.
3 in.
Section Properties. A = p A 22 - 1.52 B = 1.75p in2 r =
I =
p 4 A 2 - 1.54 B = 2.734375p in4 4
I 2.734375p = = 1.25 in. AA A 1.75p
Slenderness Ratio. For a column fixed at its base and pinned at its top, K = 0.7. Thus, 0.7(L) KL = = 0.56L r 1.25 2014-T6 Aluminum Alloy Column Formulas. Assuming an intermediate column, sallow = c30.7 - 0.23a
KL b d ksi r
100 = 30.7 - 0.23(0.56L) 1.75p L = 97.13 in. = 8.09 ft
Ans.
KL = 0.56(97.13) = 54.39 6 55, the assumption of an intermediate column r is correct. Since
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•13–97.
The tube is 0.25 in. thick, is made of a 2014-T6 aluminum alloy, and is fixed at its bottom and pinned at its top. Determine the largest axial load that it can support.
P x
y 6 in. x
6 in. y
10 ft
Section Properties: P
A = 6(6) - 5.5(5.5) = 5.75 in2 I =
1 1 (6) A 63 B (5.5) A 5.53 B = 31.7448 in4 12 12
r =
I 31.7448 = = 2.3496 in. A 5.75 AA
Slenderness Ratio: For a column fixed at one end and pinned at the other end, K = 0.7. Thus, 0.7(10)(12) KL = = 35.75 r 2.3496 Aluminium (2014 –∑ T6 alloy) Column Formulas: Since 12 6
KL 6 55, the r
column is classified as an intermediate column. Applying Eq. 13–25, sallow = c30.7 - 0.23 a
KL b d ksi r
= [30.7 - 0.23(33.75)] = 24.48 ksi The allowable load is Pallow = sallowA = 22.48(5.75) = 129 kip
Ans.
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13–98. The tube is 0.25 in. thick, is made of a 2014-T6 aluminum alloy, and is fixed connected at its ends. Determine the largest axial load that it can support.
P x
y 6 in. x
6 in. y
10 ft
Section Properties: P
A = 6(6) - 5.5(5.5) = 5.75 in2 I =
1 1 (6) A 63 B (5.5) A 5.53 B = 31.7448 in4 12 12
r =
I 31.7448 = = 2.3496 in. AA A 5.75
Slenderness Ratio: For column fixed at both ends, K = 0.5. Thus, 0.5(10)(12) KL = = 25.54 r 2.3496 Aluminium (2014 – T6 alloy) Column Formulas: Since 12 6
KL 6 55, the r
column is classified as an intermediate column. Applying Eq. 13–25, sallow = c30.7 - 0.23 a
KL b d ksi r
= [30.7 - 0.23(25.54)] = 24.83 ksi The allowable load is Pallow = sallowA = 24.83(5.75) = 143 kip
Ans.
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13–99. The tube is 0.25 in. thick, is made of 2014-T6 aluminum alloy and is pin connected at its ends. Determine the largest axial load it can support.
P x
y 6 in. x
6 in. y
Section Properties: A = 6(6) - 5.5(5.5) = 5.75 in2 I =
1 1 (6) A 63 B (5.5) A 5.53 B = 31.7448 in4 12 12
r =
I 31.7448 = = 2.3496 in. AA A 5.75
10 ft
P
Slenderness Ratio: For a column pinned as both ends, K = 1. Thus, 1(10)(12) KL = = 51.07 r 2.3496 Aluminum (2014 – T6 alloy) Column Formulas: Since 12 6
KL 6 55, the r
column is classified as an intermediate column. Applying Eq. 13–25, sallow = c30.7 - 0.23 a
KL b d ksi r
= [30.7 - 0.23(51.07)] = 18.95 ksi The allowable load is Pallow = sallowA = 18.95(5.75) = 109 kip
Ans.
*13–100. A rectangular wooden column has the cross section shown. If the column is 6 ft long and subjected to an axial force of P = 15 kip, determine the required minimum 1 dimension a of its cross-sectional area to the nearest 16 in. so that the column can safely support the loading. The column is pinned at both ends.
a
2a
Slenderness Ratio. For a column pinned at both of its ends, K = 1. Then, (1)(6)(12) KL 72 = = a a d NFPA Timber Column Formula. Assuming an intermediate column, sallow = 1.20 c1 -
1 KL>d 2 a b d ksi 3 26.0
15 1 72>a 2 = 1.20 c1 - a b d 2a(a) 3 26.0 a = 2.968 in. Use a = 3 in.
Ans.
KL 72 KL = = 24. Since 11 6 6 26, the assumption is correct. d 3 d
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•13–101.
A rectangular wooden column has the cross section shown. If a = 3 in. and the column is 12 ft long, determine the allowable axial force P that can be safely supported by the column if it is pinned at its top and fixed at its base.
a
2a
Slenderness Ratio. For a column fixed at its base and pinned at its top K = 0.7. Then, 0.7(12)(12) KL = = 33.6 d 3 NFPA Timer Column Formula. Since 26 6
KL 6 50, the column can be classified d
as a long column.
sallow =
540 ksi 540 = = 0.4783 ksi 2 (KL>d) 33.62
The allowable force is Pallow = sallowA = 0.4783(3)(6) = 8.61 kip
Ans.
13–102. A rectangular wooden column has the cross section shown. If a = 3 in. and the column is subjected to an axial force of P = 15 kip, determine the maximum length the column can have to safely support the load. The column is pinned at its top and fixed at its base.
a
2a
Slenderness Ratio. For a column fixed at its base and pinned at its top, K = 0.7. Then, KL 0.7L = = 0.2333L d 3 NFPA Timber Column Formula. Assuming an intermediate column, sallow = 1.20 c1 -
1 KL>d 2 a b d ksi 3 26.0
15 1 0.2333L 2 = 1.20c1 - a b d 3(6) 3 26.0 L = 106.68 in. = 8.89 ft
Ans.
KL KL = 0.2333(106.68) = 24.89. Since 11 6 6 26, the assumption is d d correct. Here,
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13–103. The timber column has a square cross section and is assumed to be pin connected at its top and bottom. If it supports an axial load of 50 kip, determine its smallest side dimension a to the nearest 12 in. Use the NFPA formulas.
14 ft a
Section properties: A = a2
sallow = s =
50 P = 2 A a
Assume long column: sallow =
50 = a2
C
540
2 A KL d B
540 (1.0)(14)(12) a
D
2
a = 7.15 in. (1.0)(14)(12) KL KL = = 23.5, 6 26 d 7.15 d
Assumption NG
Assume intermediate column: sallow = 1.20 B 1 -
1 KL>d 2 a b R 3 26.0
2 50 1 a b R = 1.20 B 1 - a 2 a 26.0 3 1.0(14)(12)
a = 7.46 in. 1.0(14)(12) KL KL = = 22.53, 11 6 6 26 d 7.46 d
Assumption O.K.
1 Use a = 7 in. 2
Ans.
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*13–104. The wooden column shown is formed by gluing together the 6 in. * 0.5 in. boards. If the column is pinned at both ends and is subjected to an axial load P = 20 kip, determine the required number of boards needed to form the column in order to safely support the loading.
P 6 in.
0.5 in.
9 ft
Slenderness Ratio. For a column pinned at both of its ends, K = 1. If the number of the boards required is n and assuming that n(0.5) 6 6 in. Then, d = n(0.5). Thus, (1)(9)(12) KL 216 = = n d n(0.5)
P
NFPA Timber Column Formula. Assuming an intermediate column, sallow = 1.20 B 1 -
1 KL>d 2 a b R ksi 3 26.0
1 216>n 2 20 = 1.20 B 1 - a b R [n(0.5)](6) 3 26.0 n2 - 5.5556n - 23.01 = 0 Solving for the positive root, n = 8.32 Use n = 9
Ans.
KL KL 216 = = 24. Since n(0.5) = 9(0.5) = 4.5 in. 6 6 in. and 11 6 6 26, d 9 d the assumptions made are correct. Here,
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•13–105.
The column is made of wood. It is fixed at its bottom and free at its top. Use the NFPA formulas to determine its greatest allowable length if it supports an axial load of P = 2 kip.
P 2 in. y
y
x
x 4 in.
Slenderness Ratio: For a column fixed at one end and free at the other end, K = 2. Thus, L
2(L) KL = = 1.00L d 2 NFPA Timber Column Formulas: Assume a long column. Apply Eq. 13–29, sallow =
540 ksi (KL>d)2
2 540 = 2(4) (1.00L)2 L = 46.48 in Here,
KL KL = 1.00(46.48) = 46.48. Since 26 6 6 50, the assumption is correct. d d
Thus, L = 46.48 in. = 3.87 ft
Ans.
13–106. The column is made of wood. It is fixed at its bottom and free at its top. Use the NFPA formulas to determine the largest allowable axial load P that it can support if it has a length L = 4 ft.
P 2 in. y
x
y x 4 in.
Slenderness Ratio: For a column fixed at one end and free at the other end, K = 2. Thus, L
2(4)(12) KL = = 48.0 d 2 NFPA Timber Column Formulas: Since 26 6
KL 6 50, it is a long column. Apply d
Eq. 13–29,
sallow =
=
540 ksi (KL>d)2 540 48.02
= 0.234375 ksi The allowable axial force is Pallow = sallowA = 0.234375[2(4)] = 1.875 kip
Ans.
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13–107. The W14 * 53 structural A-36 steel column supports an axial load of 80 kip in addition to an eccentric load P. Determine the maximum allowable value of P based on the AISC equations of Sec. 13.6 and Eq. 13–30. Assume the column is fixed at its base, and at its top it is free to sway in the x–z plane while it is pinned in the y–z plane.
z 80 kip x y
12 ft
Section Properties: For a W14 * 53 wide flange section. A = 15.6 in2
Ix = 541 in4
d = 13.92 in.
rx = 5.89 in.
ry = 1.92 in. Slenderness Ratio: By observation, the largest slenderness ratio is about y-y axis. For a column fixed at one end and free at the other end, K = 2. Thus, a
2(12)(12) KL b = = 150 r y 1.92
Allowable Stress: The allowable stress can be determined using AISC Column 2p2[29(103)] 2p2E KL b = = = 126.1. Since Formulas. For A–36 steel, a r c B sY B 36 KL KL b … … 200, the column is a long column. Applying Eq. 13–21, a r c r sallow =
12p2E 23(KL>r)2 12p2(29.0)(103)
=
23(1502)
= 6.637 ksi Maximum Stress: Bending is about x-x axis. Applying we have smax = sallow =
6.637 =
Mc P + A I
P(10) A 13.92 P + 80 2 B + 15.6 541
P = 7.83 kip
Ans.
1127
P
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*13–108. The W12 * 45 structural A-36 steel column supports an axial load of 80 kip in addition to an eccentric load of P = 60 kip. Determine if the column fails based on the AISC equations of Sec. 13.6 and Eq. 13–30. Assume that the column is fixed at its base, and at its top it is free to sway in the x–z plane while it is pinned in the y–z plane.
z 80 kip x y
12 ft
Section Properties: For a W12 * 45 wide flange section, A = 13.2 in2
d2 = 12.06 in.
Ix = 350 in4
rx = 5.15 in.
ry = 1.94 in. Slenderness Ratio: By observation, the largest slenderness ratio is about y -y axis. For a column fixed at one end and free at the other end, K = 2. Thus, a
2(12)(12) KL b = = 148.45 r y 1.94
Allowable Stress: The allowable stress can be determined using AISC Column 2p2[29(103)] 2p2E KL b = = = 126.1. Since Formulas. For A–36 steel, a r c B sY B 36 KL KL b … … 200, the column is a long column. Applying Eq. 13–21, a r c r sallow =
12p2E 23(KL>r)2 12p2(29.0)(103)
=
23(148.452)
= 6.776 ksi Maximum Stress: Bending is about x-x axis. Applying Eq. 1 we have smax =
=
Mc P + A I
60(10) A 12.06 140 2 B + 13.2 350
= 20.94 ksi Since smax 7 sallow, the column is not adequate.
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•13–109.
The W14 * 22 structural A-36 steel column is fixed at its top and bottom. If a horizontal load (not shown) causes it to support end moments of M = 10 kip # ft, determine the maximum allowable axial force P that can be applied. Bending is about the x–x axis. Use the AISC equations of Sec. 13.6 and Eq. 13–30.
P x
M y
y
x
12 ft
Section properties for W14 * 22: A = 6.49 in2
d = 13.74 in2
Ix = 199 in4
ry = 1.04 in. M
Allowable stress method:
P
0.5(12)(12) KL = = 69.231 ry 1.04 a
KL KL 2p2E KL 2p2(29)(103) b = = = 126.1, 6 a b r c r r c B sY B 36 y
Hence,
B1 (sa)allow =
B 53
+
3 8
smax = (sa)allow =
16.510 =
1 2
¢ KL r
2 A KL r B 2 A KL r Bc
A KL r Bc
-
c1 -
≤ R sY
1 8
3 A KL r B 3 A KL r Bc
=
R
c 53 +
3 8
1 2
2 A 69.231 126.1 B d36
1 69.231 3 A 69.231 126.1 B - 8 A 126.1 B d
= 16.510 ksi
My c P + A Ix
10(12)(13.74 P 2 ) + 6.49 199
P = 80.3 kip
Ans.
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13–110. The W14 * 22 column is fixed at its top and bottom. If a horizontal load (not shown) causes it to support end moments of M = 15 kip # ft, determine the maximum allowable axial force P that can be applied. Bending is about the x–x axis. Use the interaction formula with 1sb2allow = 24 ksi.
P x
M y
y
x
12 ft
Section Properties for W 14 * 22: A = 6.49 in2
d = 13.74 in2
Ix = 199 in4
ry = 1.04 in. M P
Interaction method: 0.5(12)(12) KL = = 69.231 ry 1.04 a
KL 2p2E KL KL 2p2(29)(103) b = = = 126.1, 6 a b r c ry r c B sY B 36
Hence,
B1 (sa)allow =
B 53 sa =
+
3 8
1 2
¢ KL r
2 A KL r B 2 A KL r Bc
A KL r Bc
-
c1 -
≤ R sY
1 8
P P = = 0.15408 P A 6.49
3 A KL r B 3 A KL r Bc
=
R
c 53 +
3 8
A
1 2
2 A 69.231 126.1 B d36
69.231 126.1
B
-
1 8
A
B
69.231 3 d 126.1
= 16.510 ksi
15(12) A 13.74 Mxc 2 B = = 6.214 ksi sb = Ix 199 sb sa + = 1.0 (sa)allow (sb)allow 0.15408 P 6.2141 + = 1.0 16.510 24 P = 79.4 kip
Ans.
1130
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13–111. The W14 * 43 structural A-36 steel column is fixed at its bottom and free at its top. Determine the greatest eccentric load P that can be applied using Eq. 13–30 and the AISC equations of Sec. 13.6.
40 kip
16 in.
10 ft
Section properties for W14 * 43: A = 12.6 in2
d = 13.66 in.
Iy = 45.2 in4
ry = 1.89 in.
b = 7.995 Allowable stress method: 2(10)(12) KL = = 126.98 ry 1.89 a
2p2E KL KL KL 2p2 (29)(103) b = = = 126.1, 200 7 7 a b r c ry r c B sY B 36
(sa)allow =
12p2(29)(103) 12p2E = = 9.26 ksi 2 23(KL>r) 23(126.98)2
smax = (sa)allow =
P
My c P + A Iy
P(16) A 7.995 P + 40 A B 9.26 = + 12.6 45.2 P = 4.07 kip
Ans.
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*13–112. The W10 * 45 structural A-36 steel column is fixed at its bottom and free at its top. If it is subjected to a load of P = 2 kip, determine if it is safe based on the AISC equations of Sec. 13.6 and Eq. 13–30.
40 kip
16 in.
10 ft
Section Properties for W10 * 45: A = 13.3 in2
d = 10.10 in.
Iy = 53.4 in4
ry = 2.01 in.
b = 8.020 in. Allowable stress method: 2.0(10)(12) KL = = 119.4 ry 2.01 a
KL 2p2E 2p2(29)(103) b = = = 126.1 r c B sY B 36
KL KL 6 a b r r c
(sa)allow Ú
(sa)allow =
10.37 Ú
c1 5 3
+
A
1 (KL>r) 2 (KL>r)3 dsY
3 KL>r 8 KL>rc
My c P + A Iy
2 C 1 - 12 A 119.4 126.1 B D 36
2
B -
c
3
1 (KL>r) 8 (KL>rc)3
= 5 3
+
3 8
1 119.4 3 A 119.4 136.1 B - 8 A 126.1 B
P
= 10.37 ksi
2(16) A 8.020 42 2 B + 13.3 53.4 O.K.
10.37 Ú 5.56 Column is safe. Yes.
Ans.
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•13–113.
The A-36-steel W10 * 45 column is fixed at its base. Its top is constrained to move along the x–x axis but free to rotate about and move along the y–y axis. Determine the maximum eccentric force P that can be safely supported by the column using the allowable stress method.
12 in.
y
Section Properties. From the table listed in the appendix, the section properties for a W10 * 45 are bf = 8.02 in.
rx = 4.32 in.
Iy = 53.4 in4
ry = 2.01 in.
Slenderness Ratio. Here, Lx = 24(12) = 288 in. and for a column fixed at its base and free at its top, Kx = 2. Thus, a
2(288) KL b = = 133.33 (controls) r x 4.32
Since the column is fixed at its base and pinned at its top, Ky = 0.7 and Ly = 24(12) = 288 in. Then, a
0.7(288) KL b = = 100.30 r y 2.01
Allowable Stress. The allowable stress will be determined using the AISC column formulas. For A-36 steel, 2p2 C 29 A 103 B D KL 2p2E KL KL b = = = 126.10. Since a b 6 a b 6 200, r c r c r x B sY C 36 the column is classified as a long column. a
sallow =
=
12p2E 23(KL>r)2
12p2 C 29 A 103 B D 23(133.332)
= 8.400 ksi
Maximum Stress. Bending is about the weak axis. Since M = P(12) and bf 8.02 = = 4.01 in, c = 2 2 sallow =
Mc P + A I
8.400 =
[P(12)](4.01) P + 13.3 53.4
y
x
24 ft
A = 13.3 in2
P
P = 8.604 kip = 8.60 kip
Ans.
1133
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13–114. The A-36-steel W10 * 45 column is fixed at its base. Its top is constrained to move along the x–x axis but free to rotate about and move along the y–y axis. Determine the maximum eccentric force P that can be safely supported by the column using an interaction formula. The allowable bending stress is (sb)allow = 15 ksi.
12 in.
bf = 8.02 in.
rx = 4.32 in.
y 24 ft
Iy = 53.4 in4
ry = 2.01 in.
Slenderness Ratio. Here, Lx = 24(12) = 288 in and for a column fixed at its base and free at its top, Kx = 2. Thus, a
2(288) KL b = = 133.33 (controls) r x 4.32
Since the column is fixed at its base and pinned at its top, Ky = 0.7 and Ly = 24(12) = 288 in. Then, a
0.7(288) KL b = = 100.30 r y 2.01
Allowable Stress. The allowable stress will be determined using the AISC column 2p2 C 29 A 103 B D 2p2E KL = 126.10. Since formulas. For A-36 steel, a b = = C r c B sY 36 a
KL KL b 6 a b 6 200, the column is classified as a long column. r c r x sallow =
=
12p2E 23(KL>r)2
12p2 C 29 A 103 B D 23 A 133.332 B
= 8.400 ksi
Interaction Formula. Bending is about the weak axis. Here, M = P(12) and bf 8.02 = = 4.01 in. c = 2 2 P>A Mc>Ar2 + = 1 (sa)allow (sb)allow P>13.3 + 8.400
P(12)(4.01) n C 13.3 A 2.012 B D 15
y
x
Section Properties. From the table listed in the appendix, the section properties for a W10 * 45 are A = 13.3 in2
P
= 1
P = 14.57 kip = 14.6 kip
Ans.
14.57>13.3 sa = = 0.1304 6 0.15 (sa)allow 8.400
O.K.
1134
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13–115. The A-36-steel W12 * 50 column is fixed at its base. Its top is constrained to move along the x–x axis but free to rotate about and move along the y–y axis. If the eccentric force P = 15 kip is applied to the column, investigate if the column is adequate to support the loading. Use the allowable stress method.
12 in.
bf = 8.08 in.
rx = 5.18 in.
y 24 ft
Iy = 56.3 in4
ry = 1.96 in. Slenderness Ratio. Here, Lx = 24(12) = 288 in. and for a column fixed at its base and pinned at its top, Kx = 2. Thus, a
2(288) KL b = = 111.20 (controls) r x 5.18
Since the column is fixed at its base and pinned at its top, Ky = 0.7 and Ly = 24(12) = 288 in. Thus, a
0.7(288) KL b = = 102.86 r y 1.96
Allowable Stress. The allowable stress will be determined using the AISC column formulas. For A-36 steel, a a
2p2 C 29 A 103 B D KL 2p2E = 126.10. Since b = = C r c A sY 36
KL KL b 6 a b , the column can be classified as an intermediate column. r x r c
B1 sallow =
2(KL>r)C 2
R sY
(KL>r)3 3(KL>r) 5 + 3 8(KL>r)C 8(KL>r)C 3 C1 -
=
(KL>r)2
111.202
2 A 126.102 B
S(36)
3(111.20) 5 111.203 + 3 8(126.10) 8 A 126.103 B
= 11.51 ksi Maximum Stress. Bending is about the weak axis. Since, M = 15(12) = 180 kip # in. bf 8.08 and c = = = 4.04 in., 2 2 smax =
y
x
Section Properties. From the table listed in the appendix, the section properties for a W12 * 50 are A = 14.7 in2
P
180(4.04) P Mc 15 + = + = 13.94 ksi A I 14.7 56.3
Since smax 7 sallow, the W12 * 50 column is inadequate according to the allowable stress method.
1135
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*13–116. The A-36-steel W12 * 50 column is fixed at its base. Its top is constrained to move along the x–x axis but free to rotate about and move along the y–y axis. If the eccentric force P = 15 kip is applied to the column, investigate if the column is adequate to support the loading. Use the interaction formula. The allowable bending stress is (sb)allow = 15 ksi.
12 in.
bf = 8.08 in.
rx = 5.18 in.
y 24 ft
ry = 1.96 in.
Slenderness Ratio. Here, Lx = 24(12) = 288 in and for a column fixed at its base and pinned at its top, Kx = 2. Thus, a
2(288) KL b = = 111.20 (controls) r x 5.18
Since the column is fixed at its base and pinned at its top, Ky = 0.7 and Ly = 24(12) = 288 in. Then, a Allowable =
C
Axial
2p2 C 29 A 103 B D 36
0.7(288) KL b = = 102.86 r y 1.96 Stress.
For
= 126.10. Since a
A-36
steel,
a
KL 2p2E b = r c A sY
KL KL b 6 a b , the column can be r x r c
classified as an intermediate column. C1 sallow =
2(KL>r)c 2
SsY
3(KL>r) (KL>r)3 5 + 3 8(KL>r)c 8(KL>r)c 3 C1 -
=
(KL>r)2
111.202
2 A 126.102 B
S(36)
3(111.20) 5 111.203 + 3 8(126.10) 8 A 126.103 B
= 11.51 ksi Interaction Formula. Bending is about the weak axis. Here, M = 15(12) bf 8.08 = 180 kip # in. and c = = = 4.04 in. 2 2
P>A Mc>Ar2 15>14.7 + + = (sa)allow (sb)allow 11.51
180(4.04) n C 14.7 A 1.962 B D 15
y
x
Section Properties. From the table listed in the appendix, the section properties for a W12 * 50 are A = 14.7 in2
P
= 0.9471 6 1
P = 14.57 kip = 14.6 kip
Ans.
15>14.7 sa = = 0.089 6 0.15 (sa)allow 11.51
O.K.
Thus, a W12 * 50 column is adequate according to the interaction formula.
1136
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•13–117.
A 16-ft-long column is made of aluminum alloy 2014-T6. If it is fixed at its top and bottom, and a compressive load P is applied at point A, determine the maximum allowable magnitude of P using the equations of Sec. 13.6 and Eq. 13–30.
P A 4.25 in. x 0.5 in. y 8 in.
Section properties: A = 2(0.5)(8) + 8(0.5) = 12 in2 Ix =
1 1 (8)(93) (7.5)(83) = 166 in4 12 12
Iy = 2 a ry =
1 1 b(0.5)(83) + (8)(0.53) = 42.75 in4 12 12
Iy 42.75 = = 1.8875 in. AA A 12
Allowable stress method: 0.5(16)(12) KL KL = 50.86, 12 6 = 6 55 ry ry 1.8875 sallow = c30.7 - 0.23a
KL bd r
= [30.7 - 0.23(50.86)] = 19.00 ksi smax = sallow =
19.00 =
Mx c P + A Ix
P(4.25)(4.5) P + 12 166
P = 95.7 kip
Ans.
1137
y x 8 in. 0.5 in.
0.5 in.
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13–118. A 16-ft-long column is made of aluminum alloy 2014-T6. If it is fixed at its top and bottom, and a compressive load P is applied at point A, determine the maximum allowable magnitude of P using the equations of Sec. 13.6 and the interaction formula with 1sb2allow = 20 ksi.
P A 4.25 in. x 0.5 in. y 8 in.
Section Properties: A = 2(0.5)(8) + 8(0.5) = 12 in2 Ix =
1 1 (8)(93) (7.5)(83) = 166 in4 12 12
Iy = 2 a ry =
1 1 b(0.5)(83) + (8)(0.53) = 42.75 in4 12 12
Iy 42.75 = = 1.8875 in. AA A 12
Interaction method: 0.5(16)(12) KL KL = 50.86, 12 6 = 6 55 ry ry 1.8875 sallow = c30.7 - 0.23a
KL bd r
= [30.7 - 0.23(50.86)] = 19.00 ksi sa =
P P = = 0.08333P A 12
sb =
P(4.25)(4.50) Mc = = 0.1152P Ix 166 sb sa + = 1.0 (sa)allow (sb)allow 0.08333P 0.1152P + = 1 19.00 20
P = 98.6 kip
Ans.
1138
y x 8 in. 0.5 in.
0.5 in.
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13–119. The 2014-T6 hollow column is fixed at its base and free at its top. Determine the maximum eccentric force P that can be safely supported by the column. Use the allowable stress method. The thickness of the wall for the section is t = 0.5 in.
6 in.
P
3 in. 6 in.
8 ft
Section Properties. A = 6(3) - 5(2) = 8 in2 Ix =
1 1 (3) A 63 B (2) A 53 B = 33.1667 in4 12 12
rx =
33.1667 Ix = = 2.036 in. AA A 8
Iy =
1 1 (6) A 33 B (5) A 23 B = 10.1667 in4 12 12
ry =
Iy 10.1667 = = 1.127 in. AA A 8
Slenderness Ratio. For a column fixed at its base and free at its top, K = 2. Thus, a
2(8)(12) KL b = = 170.32 r y 1.127
Allowable Stress. Since a
KL b 7 55, the column can be classified as a long r y
column.
sallow =
54 000 ksi 54 000 ksi = = 1.862 ksi (KL>r)2 170.312
Maximum Stress. Bending occurs about the strong axis so that M = P(6) and 6 c = = 3 in. 2 sallow =
1.862 =
Mc P + A I
C P(6) D (3) P + 8 33.1667
P = 2.788 kip = 2.79 kip
Ans.
1139
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*13–120. The 2014-T6 hollow column is fixed at its base and free at its top. Determine the maximum eccentric force P that can be safely supported by the column. Use the interaction formula. The allowable bending stress is (sb)allow = 30 ksi. The thickness of the wall for the section is t = 0.5 in.
6 in.
P
3 in. 6 in.
8 ft
Section Properties. A = 6(3) - 5(2) = 8 in2 Ix =
1 1 (3) A 63 B (2) A 53 B = 33.1667 in4 12 12
rx =
33.1667 Ix = = 2.036 in. AA A 8
Iy =
1 1 (6) A 33 B (5) A 23 B = 10.1667 in4 12 12
ry =
Iy 10.1667 = = 1.127 in. AA A 8
Slenderness Ratio. For a column fixed at its base and pinned at its top, K = 2. Thus, a
2(8)(12) KL b = = 170.32 r y 1.127
KL b 7 55, the column can be classified as the column is r y classified as a long column. Allowable Stress. Since a
sallow =
54000 ksi 54000 ksi = = 1.862 ksi (KL>r)2 170.312
Interaction Formula. Bending is about the strong axis. Since M = P(6) and 6 c = = 3 in, 2 P>A Mc>Ar2 + = 1 (sa)allow (sb)allow P>8 + 1.862
[P(6)](3) n C 8 A 2.0362 B D 30
= 1
P = 11.73 kip = 11.7 kip
Ans.
1140
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•13–121. The 10-ft-long bar is made of aluminum alloy 2014-T6. If it is fixed at its bottom and pinned at the top, determine the maximum allowable eccentric load P that can be applied using the formulas in Sec. 13.6 and Eq. 13–30.
P x 1.5 in. 1.5 in. x
Section Properties: A = 6(4) = 24.0 in2 Ix =
1 (4) A 63 B = 72.0 in4 12
Iy =
1 (6) A 43 B = 32.0 in4 12
ry =
Iy 32.0 = = 1.155 in. AA A 24
Slenderness Ratio: The largest slenderness ratio is about y-y axis. For a column pinned at one end fixed at the other end, K = 0.7. Thus, a
0.7(10)(12) KL b = = 72.75 r y 1.155
Allowable Stress: The allowable stress can be determined using aluminum KL (2014 –T6 alloy) column formulas. Since 7 55, the column is classified as a long r column. Applying Eq. 13–26, sallow = c =
54 000 d ksi (KL>r)2
54 000 72.752
= 10.204 ksi Maximum Stress: Bending is about x-x axis. Applying Eq. 13–30, we have smax = sallow =
10.204 =
P Mc + A I P(1.5)(3) P + 24.0 72.0
P = 98.0 kip
Ans.
1141
y 3 in.
2 in. y 2 in.
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13–122. The 10-ft-long bar is made of aluminum alloy 2014-T6. If it is fixed at its bottom and pinned at the top, determine the maximum allowable eccentric load P that can be applied using the equations of Sec. 13.6 and the interaction formula with 1sb2allow = 18 ksi.
P x 1.5 in. 1.5 in. x
Section Properties: A = 6(4) = 24.0 in2 Ix =
1 (4) A 63 B = 72.0 in4 12
Iy =
1 (6) A 43 B = 32.0 in4 12
rx =
Ix 72.0 = = 1.732 in. AA A 24.0
ry =
Iy 32.0 = = 1.155 in. AA A 24.0
Slenderness Ratio: The largest slenderness radio is about y-y axis. For a column pinned at one end and fixed at the other end, K = 0.7. Thus a
0.7(10)(12) KL b = = 72.75 r y 1.155
Allowable Stress: The allowable stress can be determined using aluminum KL (2014 –T6 alloy) column formulas. Since 7 55, the column is classified as a long r column. Applying Eq. 13–26, (sa)allow = c =
54 000 d ksi (KL>r)2
54 000 72.752
= 10.204 ksi Interaction Formula: Bending is about x-x axis. Applying Eq. 13–31, we have Mc>Ar2 P>A + = 1 (sa)allow (sb)allow P(1.5)(3)>24.0(1.7322) P>24.0 + = 1 10.204 18 P = 132 kip
Ans.
1142
y 3 in.
2 in. y 2 in.
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13–123. The rectangular wooden column can be considered fixed at its base and pinned at its top. Also, the column is braced at its mid-height against the weak axis. Determine the maximum eccentric force P that can be safely supported by the column using the allowable stress method.
6 in.
6 in.
Section Properties. Ix =
dx = 6 in.
5 ft
1 (3) A 63 B = 54 in4 12
dy = 3 in.
Slenderness Ratio. Here, Lx = 10(12) = 120 in. and for a column fixed at its base and pinned at its top, K = 0.7. Thus, a
0.7(120) KL b = = 14 d x 6
Since the bracing provides support equivalent to a pin, Ky = 1 and Ly = 5(12) = 60 in. Then a
1(60) KL b = = 20 (controls) d y 3
KL 6 26, the column can be classified as the column d is classified as an intermediate column. Allowable Stress. Since 11 6
sallow = 1.20c 1 = 1.20 c 1 -
1 KL>d 2 a b d ksi 3 26.0 1 20 2 a b d ksi = 0.9633 ksi 3 26.0
Maximum Stress. Bending occurs about the strong axis. Here, M = P(6) and 6 c = = 3 in. 2 sallow =
0.9633 =
6 in. 3 in.
5 ft
A = 6(3) = 18 in2
P
P Mc + A I [P(6)](3) P + 18 54
P = 2.477 kip = 2.48 kip
Ans.
1143
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*13–124. The rectangular wooden column can be considered fixed at its base and pinned at its top. Also, the column is braced at its mid-height against the weak axis. Determine the maximum eccentric force P that can be safely supported by the column using the interaction formula. The allowable bending stress is (sb)allow = 1.5 ksi.
6 in.
6 in.
5 ft
Section Properties.
rx =
Ix =
1 (3) A 63 B = 54 in4 12
Ix 54 = = 1.732 in. AA A 18
dx = 6 in.
dy = 3 in.
Slenderness Ratio. Here, Lx = 10(12) = 120 in. and for a column fixed at its base pinned at its top, K = 0.7. Thus, a
0.7(120) KL b = = 14 d x 6
Since the bracing provides support equivalent to a pin, Ky = 1 and Ly = 5(12) = 60 in. Then a
1(60) KL b = = 20 (controls) d y 3
KL 6 26, the column can be classified as the d column is classified as an intermediate column. Allowable Axial Stress. Since 11 6
sallow = 1.20c 1 = 1.20 c 1 -
1 KL>d 2 a b d ksi 3 26.0 1 20 2 a b d ksi = 0.9633 ksi 3 26.0
Interaction Formula. Bending occurs about the strong axis. Since M = P(6) and 6 c = = 3 in. 2 P>A Mc>Ar2 + = 1 (sa)allow (sb)allow P>18 + 0.9633
[P(6)](3) n C 18 A 1.7322 B D 1.5
6 in. 3 in.
5 ft
A = 6(3) = 18 in2
P
= 1
P = 3.573 kip = 3.57 kip
Ans.
1144
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•13–125.
The 10-in.-diameter utility pole supports the transformer that has a weight of 600 lb and center of gravity at G. If the pole is fixed to the ground and free at its top, determine if it is adequate according to the NFPA equations of Sec. 13.6 and Eq. 13–30.
G
15 in.
18 ft
2(18)(12) KL = = 43.2 in. d 10 26 6 43.2 … 50 Use Eq. 13–29, sallow =
540 540 = = 0.2894 ksi (KL>d) (43.2)2
smax =
Mc P + A I
smax =
(600)(15)(5) 600 + 2 p (5) A p4 B (5)4
smax = 99.31 psi 6 0.289 ksi
O.K.
Yes.
Ans.
1145
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13–126. Using the NFPA equations of Sec. 13.6 and Eq. 13–30, determine the maximum allowable eccentric load P that can be applied to the wood column. Assume that the column is pinned at both its top and bottom.
P 0.75 in. 6 in. 3 in.
12 ft
Section Properties: A = 6(3) = 18.0 in2
Iy =
1 (6) A 33 B = 13.5 in4 12
Slenderness Ratio: For a column pinned at both ends, K = 1.0. Thus, a
1.0(12)(12) KL b = = 48.0 d y 3
Allowable Stress: The allowable stress can be determined using NFPA timber KL column formulas. Since 26 6 6 50, it is a long column. Applying Eq. 13–29, d sallow =
=
540 ksi (KL>d)2 540 = 0.234375 ksi 48.02
Maximum Stress: Bending is about y -y axis. Applying Eq. 13–30, we have smax = sallow =
0.234375 =
P Mc + A I P(0.75)(1.5) P + 18.0 13.5
P = 1.69 kip
Ans.
1146
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13–127. Using the NFPA equations of Sec. 13.6 and Eq. 13–30, determine the maximum allowable eccentric load P that can be applied to the wood column. Assume that the column is pinned at the top and fixed at the bottom.
P 0.75 in. 6 in. 3 in.
12 ft
Section Properties: A = 6(3) = 18.0 in2 Iy =
1 (6) A 33 B = 13.5 in4 12
Slenderness Ratio: For a column pinned at one end and fixed at the other end, K = 0.7. Thus, a
0.7(12)(12) KL b = = 33.6 d y 3
Allowable Stress: The allowable stress can be determined using NFPA timber KL column formulas. Since 26 6 6 50, it is a long column. Applying Eq. 13–29, d sallow =
=
540 ksi (KL>d)2 540 = 0.4783 ksi 33.62
Maximum Stress: Bending is about y -y axis. Applying Eq. 13–30, we have smax = sallow =
0.4783 =
Mc P + A I P(0.75)(1.5) P + 18.0 13.5
P = 3.44 kip
Ans.
1147
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*13–128. The wood column is 4 m long and is required to support the axial load of 25 kN. If the cross section is square, determine the dimension a of each of its sides using a factor of safety against buckling of F.S. = 2.5. The column is assumed to be pinned at its top and bottom. Use the Euler equation. Ew = 11 GPa, and sY = 10 MPa.
25 kN
a
4m a
1 a4 (a) A a3 B = , P = (2.5)25 = 62.5 kN and K = 1 12 12 cr for pin supported ends column. Applying Euler’s formula, Critical Buckling Load: I =
Pcr = 62.5 A 10
3
B =
p2EI (KL)2
a p2(11)(109) A 12 B 4
[1(4)]2
a = 0.1025 m = 103 mm
Ans.
Critical Stress: Euler’s formula is only valid if scr 6 sY. scr =
62.5(103) Pcr = = 5.94 MPa 6 s Y = 10 MPa A 0.1025(0.1025)
1148
O.K.
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•13–129. If the torsional springs attached to ends A and C of the rigid members AB and BC have a stiffness k, determine the critical load Pcr.
P
k
A
L 2 B
Equilibrium. When the system is given a slight lateral disturbance, the configuration shown in Fig. a is formed. The couple moment M can be related to P by considering the equilibrium of members AB and BC. Member AB + c ©Fy = 0; a + ©MA = 0;
By - P = 0
(1)
By a
(2)
L L sin u b + Bx a cos ub - M = 0 2 2
Member BC a + ©MC = 0; -By a
L L sin u b + Bx a cos ub + M = 0 2 2
(3)
Solving Eqs. (1), (2), and (3), we obtain Bx = 0
By =
2M L sin u
M =
PL sin u 2
Since u is very small, the small angle analysis gives sin u ⬵ u. Thus, M =
PL u 2
(4)
Torslonal Spring Moment. The restoring couple moment Msp can be determined using the torsional spring formula, M = ku. Thus, Msp = ku Critical Buckling Load. When the mechanism is on the verge of bucklling M must equal Msp. M = Msp Pcr L u = ku 2 Pcr =
2k L
Ans.
1149
L 2 k
C
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13–129.
Continued
1150
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13–130. Determine the maximum intensity w of the uniform distributed load that can be applied on the beam without causing the compressive members of the supporting truss to buckle. The members of the truss are made from A-36-steel rods having a 60-mm diameter. Use F.S. = 2 against buckling.
w
B A
C 2m
Equilibrium. The force developed in member BC can be determined by considering the equilibrium of the free-body diagram of the beam AB, Fig. a. 3 w(5.6)(2.8) - FBC a b(5.6) = 0 FBC = 4.6667w 5
a + ©MA = 0;
The Force developed in member CD can be obtained by analyzing the equilibrium of joint C, Fig. b, + c ©Fy = 0;
FAC a
+ : ©Fx = 0;
4 12 4.6667wa b + 7.28a b w-FCD = 0 5 13
5 3 b - 4.6667wa b = 0 13 5
FAC = 7.28w (T)
FCD = 10.4533w (C)
Section Properties. The cross-sectional area and moment of inertia of the solid circular rod CD are A = p A 0.032 B = 0.9 A 10 - 3 B p m2
I =
p A 0.034 B = 0.2025 A 10 - 6 B p m4 4
Critical Buckling Load. Since both ends of member CD are pinned, K = 1. The critical buckling load is Pcr = FCD (F.S.) = 10.4533w(2) = 20.9067w Applying Euler’s formula, Pcr =
p2EI (KL)2
20.9067w =
p2 C 200 A 109 B D C 0.2025 A 10 - 6 B p D [1(3.6)]2
Ans.
w = 4634.63 N>m = 4.63 kN>m Critical Stress: Euler’s formula is valid only if scr 6 sY. scr =
20.907(4634.63) Pcr = = 34.27 MPa 6 sY = 250 MPa A 0.9 A 10 - 3 B p
1151
O.K.
3.6 m
D
1.5 m
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13–131. The W10 * 45 steel column supports an axial load of 60 kip in addition to an eccentric load P. Determine the maximum allowable value of P based on the AISC equations of Sec. 13.6 and Eq. 13–30. Assume that in the x–z plane Kx = 1.0 and in the y–z plane Ky = 2.0. Est = 2911032 ksi, sY = 50 ksi.
z P 60 kip
x y
10 ft
Section properties for W 10 * 45: A = 13.3 in2
d = 10.10 in.
rx = 4.32 in.
ry = 2.01 in.
Ix = 248 in4
Allowable stress method: a
1.0(10)(12) KL b = = 27.8 r x 4.32
a
2.0(10)(12) KL b = = 119.4 r y 2.01
a
2p2(29)(103) KL 2p2E b = = = 107 r c B sg B 50
(controls)
KL KL 7 a b r r c (sa)allow =
12p2(29)(103) 12p2E = = 10.47 ksi 2 23(KL>r) 23(119.4)4
smax = (sa)allow =
Mc P + A I
P(8) A 10.10 P + 60 2 B + 10.47 = 13.3 248 P = 25.0 kip
Ans.
1152
y x 8 in.
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*13–132. The A-36-steel column can be considered pinned at its top and fixed at its base. Also, it is braced at its mid-height along the weak axis. Investigate whether a W250 * 45 section can safely support the loading shown. Use the allowable stress method.
600 mm 10 kN
4.5 m
Section Properties. From the table listed in the appendix, the necessary section properties for a W250 * 45 are A = 5700 mm2 = 5.70 A 10 - 3 B m2
d = 266 mm = 0.266 m
Ix = 71.1 A 106 B mm4 = 71.1 A 10 - 6 B m4
rx = 112 mm = 0.112 m
ry = 35.1 mm = 0.0351 mm
Slenderness Ratio. Here, Lx = 9 m and for a column fixed at its base and pinned at its top, Kx = 0.7. Thus,
¢
0.7(9) KL = 56.25 ≤ = r x 0.112
Since the bracing provides support equivalent to a pin, Ky = 1 and Ly = 4.5 m. Then,
¢
1(4.5) KL = 128.21 (controls) ≤ = r y 0.0351
Allowable Stress. For A-36 steel, a Since a
2p2 C 200 A 109 B D KL 2p2E b = = = 125.66. r c C 250 A 106 B B sY
KL KL b 6 a b 6 200, the column can be classified as a long column. r c r y sallow =
12p2 C 200 A 109 B D 12p2E = = 62.657 MPa 2 23(KL>r) 23(128.21)2
Maximum Stress. Bending occurs about the strong axis. Here, P = 10 + 40 0.266 d = 50 kN, M = 40(0.6) = 24 kN # m and c = = = 0.133 m, 2 2 smax =
50 A 103 B 24 A 103 B (0.133) P Mc + = + = 53.67 MPa A I 5.70 A 10 - 3 B 71.1 A 10 - 6 B
Since smax 6 sallow, the column is adequate according to the allowable stress method.
1153
4.5 m
40 kN
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•13–133.
The A-36-steel column can be considered pinned at its top and fixed at its base. Also, it is braced at its mid-height along the weak axis. Investigate whether a W250 * 45 section can safely support the loading shown. Use the interaction formula. The allowable bending stress is (sb)allow = 100 MPa.
600 mm 10 kN
4.5 m
Section Properties. From the table listed in the appendix, the necessary section properties for a W250 * 45 are A = 5700 mm2 = 5.70 A 10 - 3 B m2 Ix = 71.1 A 10 B mm = 71.1 A 10 6
4
-6
d = 266 mm = 0.266 m 4.5 m
Bm
4
rx = 112 mm = 0.112 m
ry = 35.1 mm = 0.0351 mm
Slenderness Ratio. Here, Lx = 9 m and for a column fixed at its base and pinned at its top, Kx = 0.7. Thus, a
0.7(9) KL b = = 56.25 r x 0.112
Since the bracing provides support equivalent to a pin, Ky = 1 and Ly = 4..5 m. Then, a
Allowable
=
C
1(4.5) KL b = = 128.21 (controls) r y 0.0351
Axial
2p2 C 200 A 109 B D 250 A 10
6
B
Stress.
For
= 125.66. Since a
A–36
steel,
a
KL 2p2E b = r c B sY
KL KL b 6 a b 6 200, the column can be r c r y
classified as a long column.
sallow =
12p2 C 200 A 109 B D 12p2E = = 62.657 MPa 23(KL>r)2 23(128.21)2
Interaction Formula. Bending is about the strong axis. Here, P = 10 + 40 = 50 kN, d 0.266 M = 40(0.6) = 24 kN # m and c = = = 0.133 m, 2 2 Mc>Ar2 P>A + = (sa)allow (sb)allow
50 A 103 B n 5.70 A 10 - 3 B 62.657 A 106 B
+
24 A 103 B (0.133) n C 5.70 A 10 - 3 B A 0.1122 B D 100 A 106 B
= 0.5864 6 1 sa = (sa)allow
50 A 103 B n 5.7 A 10 - 3 B 62.657 A 106 B
O.K.
= 0.140 6 0.15
O.K.
Thus, a W250 * 45 column is adequate according to the interaction formula.
1154
40 kN
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13–134. The member has a symmetric cross section. If it is pin connected at its ends, determine the largest force it can support. It is made of 2014-T6 aluminum alloy.
0.5 in. 2 in.
P
5 ft
Section properties: A = 4.5(0.5) + 4(0.5) = 4.25 in2 P
1 1 (0.5)(4.53) + (4)(0.5)3 = 3.839 in4 I = 12 12 r =
I 3.839 = = 0.9504 in. AA A 4.25
Allowable stress: 1.0(5)(12) KL = = 63.13 r 0.9504 KL 7 55 r Long column sallow =
54000 54000 = = 13.55 ksi (KL>r)2 63.132
Pallow = sallowA = 13.55(4.25) = 57.6 kip
Ans.
1155
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13–135. The W200 * 46 A-36-steel column can be considered pinned at its top and fixed at its base. Also, the column is braced at its mid-height against the weak axis. Determine the maximum axial load the column can support without causing it to buckle. 6m
Section Properties. From the table listed in the appendix, the section properties for a W200 * 46 are A = 5890 mm2 = 5.89 A 10 - 3 B m2
Iy = 15.3 A 106 B mm4 = 15.3 A 10 - 6 B m4
Ix = 45.5 A 106 B mm4 = 45.5 A 10 - 6 B m4
Critical Buckling Load. For buckling about the strong axis, Kx = 0.7 and Lx = 12 m. Since the column is fixed at its base and pinned at its top,
Pcr =
p2EIx (KL)x 2
=
p2 c200 A 109 B d c45.5 A 10 - 6 B d [0.7(12)]2
= 1.273 A 106 B N = 1.27 MN
For buckling about the weak axis, Ky = 1 and Ly = 6 m since the bracing provides a support equivalent to a pin. Applying Euler’s formula,
Pcr =
p2EIy (KL)y 2
=
p2 c200 A 109 B d c15.3 A 10 - 6 B d [1(6)]2
= 838.92 kN = 839 kN (controls)Ans.
Critical Stress. Euler’s formula is valid only if scr 6 sY. scr =
838.92 A 103 B Pcr = = 142.43 MPa 6 sY = 250 MPa A 5.89 A 10 - 3 B
1156
O.K.
6m
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*13–136. The structural A-36 steel column has the cross section shown. If it is fixed at the bottom and free at the top, determine the maximum force P that can be applied at A without causing it to buckle or yield. Use a factor of safety of 3 with respect to buckling and yielding.
P 20 mm A 10 mm
Section properties:
100 mm -3
2
4m
©A = 0.2(0.01) + 0.15 (0.01) + 0.1(0.01) = 4.5(10 ) m
150 mm A
100 mm 10 mm
0.005 (0.2)(0.01) + 0.085 (0.15)(0.01) + 0.165 (0.1)(0.01) ©xA x = = ©A 4.5(10 - 3) = 0.06722 m 1 (0.2)(0.013) + 0.2 (0.01)(0.06722 - 0.005)2 12
Iy =
+
1 (0.01)(0.153) + 0.01 (0.15)(0.085 - 0.06722)2 12
+
1 (0.1)(0.013) + 0.1 (0.01)(0.165 - 0.06722)2 12
= 20.615278 (10 - 6) m4 1 1 1 (0.01)(0.23) + (0.15)(0.013) + (0.01)(0.13) 12 12 12
Ix =
= 7.5125 (10 - 6) m4 ry =
Iy 20.615278(10 - 6) = = 0.0676844 AA A 4.5 (10 - 3)
Buckling about x-x axis: Pcr =
p2(200)(109)(7.5125)(10 - 6) p2 EI = 2 (KL) [2.0(4)]2
= 231.70 kN scr =
(controls)
231.7 (103) Pcr = 51.5 MPa 6 sg = 250 MPa = A 4.5 (10 - 3)
Yielding about y-y axis: smax =
P ec KL P c1 + 2 sec a b d; A 2r A EA r
e = 0.06722 - 0.02 = 0.04722 m
0.04722 (0.06722) ec = = 0.692919 0.0676844 r2 2.0 (4) P P KL = = 1.96992 P (10 - 3) 2P 2r A EA 2(0.0676844) A 200 (109)(4.5)(10 - 3) 250(106)(4.5)(10 - 3) = P[1 + 0.692919 sec (1.96992P (10 - 3)2P)] By trial and error: P = 378.45 kN Hence, Pallow =
231.70 = 77.2 kN 3
Ans.
1157
10 mm 100 mm
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•13–137.
The structural A-36 steel column has the cross section shown. If it is fixed at the bottom and free at the top, determine if the column will buckle or yield when the load P = 10 kN. Use a factor of safety of 3 with respect to buckling and yielding.
P 20 mm A 10 mm 100 mm
Section properties: 4m
©A = 0.2 (0.01) + 0.15 (0.01) + 0.1 (0.01) = 4.5 (10 - 3) m2 0.005 (0.2)(0.01) + 0.085 (0.15)(0.01) + 0.165 (0.1)(0.01) ©xA = 0.06722 m = ©A 4.5 (10 - 3)
Iy =
1 (0.2)(0.013) + 0.2 (0.01)(0.06722 - 0.005)2 12
ry =
+
1 (0.01)(0.153) + 0.01 (0.15)(0.085 - 0.06722)2 12
+
1 (0.1)(0.013) + 0.1 (0.01)(0.165 - 0.06722)2 = 20.615278 (10 - 6) m4 12
1 1 1 (0.01)(0.23) + (0.15)(0.013) + (0.01)(0.13) = 7.5125 (10 - 6) m4 12 12 12 Iy BA
20.615278 (10 - 6) =
B
4.5 (10 - 3)
= 0.067843648 m
Buckling about x-x axis: Pcr =
p2(200)(109)(7.5125)(10 - 6) p2 EI = = 231.70 kN 2 (KL) [2.0(4)]2
scr =
231.7 (103) Pcr = 51.5 MPa 6 sg = 250 MPa = A 4.5 (10 - 3)
Pallow =
O.K.
Pcr 231.7 = = 77.2 kN 7 P = 10 kN FS 3
Hence the column does not buckle. Yielding about y-y axis: smax =
P =
P KL ec P bd c1 + 2 sec a A 2r A EA r
A
100 mm 10 mm
x =
Ix =
150 mm
e = 0.06722 - 0.02 = 0.04722 m
10 = 3.333 kN 3
3.333 (103) P = 0.7407 MPa = A 4.5 (10 - 3) 0.04722 (0.06722) ec = 0.689815 = (0.067844) r2 2.0 (4) P KL 3.333 (103) = = 0.1134788 2 r AE A 2(0.06783648) A 200 (109)(4.5)(10 - 3) smax = 0.7407 [1 + 0.692919 sec (0.1134788)] = 1.25 MPa 6 sg = 250 MPa Hence the column does not yield! No.
Ans. 1158
10 mm 100 mm
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sy
14–1. A material is subjected to a general state of plane stress. Express the strain energy density in terms of the elastic constants E, G, and n and the stress components sx , sy , and txy .
txy
sx
‘Strain Energy Due to Normal Stresses: We will consider the application of normal stresses on the element in two successive stages. For the first stage, we apply only sx on the element. Since sx is a constant, from Eq. 14-8, we have s2x s2x V dV = 2E Lv 2E
(Ui)1 =
When sy is applied in the second stage, the normal strain ex will be strained by ex ¿ = -vey = -
vsy E
. Therefore, the strain energy for the second stage is
(Ui)2 =
=
s2y
¢
Lv 2E
B
s2y
Lv 2E
+ sx ex ¿ ≤ dV + sx a -
vsy E
b R dV
Since sx and sy are constants, (Ui)2 =
V (s2 - 2vsx sy) 2E y
Strain Energy Due to Shear Stresses: The application of txy does not strain the element in normal direction. Thus, from Eq. 14–11, we have (Ui)3 =
t2xy Lv 2G
dV =
t2xy V 2G
The total strain energy is Ui = (Ui)1 + (Ui)2 + (Ui)3 =
t2xy V s2x V V + (s2y - 2vsx sy) + 2E 2E 2G
=
t2xy V V (s2x + s2y - 2vsx sy) + 2E 2G
and the strain energy density is t2xy Ui 1 = (s2x + s2y - 2vsx sy) + V 2E 2G
Ans.
1159
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14–2. The strain-energy density must be the same whether the state of stress is represented by sx , sy , and txy , or by the principal stresses s1 and s2 . This being the case, equate the strain–energy expressions for each of these two cases and show that G = E>[211 + n2].
U =
1 v 1 2 (s2x + s2y) - sxsy + t R dV E 2 G xy Lv 2 E
U =
1 v (s21 + s22) s s R dV B E 1 2 Lv 2 E
B
Equating the above two equations yields. v 1 2 1 v 1 (s2x + s2y) sxsy + txy = (s21 + s22) s s 2E E 2G 2E E 1 2 However, s1, 2 =
sx + sy 2
;
A
a
sx - sy 2
(1)
2 b + txy 2
Thus, A s21 + s22 B = s2x + s2y + 2 t2xy s1 s2 = sxsy - t2xy Substitute into Eq. (1) v 1 2 1 v v 2 1 t = (s2 + s2y + 2t2xy) ss + t A s2 + s2y B - sxsy + 2E x E 2 G xy 2E x E x y E xy t2xy v 2 1 2 txy = + t 2G E E xy 1 v 1 = + 2G E E 1 1 = (1 + v) 2G E G =
E 2(1 + v)
QED
1160
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14–3. Determine the strain energy in the stepped rod assembly. Portion AB is steel and BC is brass. Ebr = 101 GPa, Est = 200 GPa, (sY)br = 410 MPa, (sY)st = 250 MPa.
100 mm A
B
30 kN
30 kN 1.5 m
Referring to the FBDs of cut segments in Fig. a and b, + ©F = 0; : x
NBC - 20 = 0
+ ©F = 0; : x
NAB - 30 - 30 - 20 = 0
NBC = 20 kN NAB = 80 kN
p The cross-sectional area of segments AB and BC are AAB = (0.12) = 2.5(10 - 3)p m2 and 4 p ABC = (0.0752) = 1.40625(10 - 3)p m2. 4 (Ui)a = ©
NAB 2LAB NBC 2LBC N2L = + 2AE 2AAB Est 2ABC Ebr =
C 80(103) D 2 (1.5)
2 C 2.5(10 - 3)p D C 200(109) D
+
C 20(103) D 2(0.5)
2 C 1.40625(10 - 3) p D C 101(109) D
= 3.28 J
Ans.
This result is valid only if s 6 sy. sAB =
80(103) NAB = 10.19(106)Pa = 10.19 MPa 6 (sy)st = 250 MPa = AAB 2.5(10 - 3)p
O.K.
sBC =
20 (103) NBC = 4.527(106)Pa = 4.527 MPa 6 (sy)br = 410 MPa = ABC 1.40625(10 - 3) p
O.K.
1161
0.5 m
75 mm C 20 kN
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*14–4. Determine the torsional strain energy in the A-36 steel shaft. The shaft has a diameter of 40 mm.
900 N⭈m 200 N⭈m
0.5 m
Referring to the FBDs of the cut segments shown in Fig. a, b and c,
300 N⭈m
0.5 m
TAB = 300 N # m
©Mx = 0;
TAB - 300 = 0
©Mx = 0;
TBC - 200 - 300 = 0
©Mx = 0;
TCD - 200 - 300 + 900 = 0 TCD = -400 N # m
0.5 m
TBC = 500 N # m
The shaft has a constant circular cross-section and its polar moment of inertia is p J = (0.024) = 80(10 - 9)p m4. 2 (Ui)t = ©
TAB 2 LAB TBC 2LBC TCD LCD T2L = + + 2GJ 2GJ 2GJ 2GJ 1
=
2 C 75(10 ) 80 (10 - 9)p D 9
c3002 (0.5) + 5002 (0.5) + (-400)2 (0.5) d
= 6.63 J
Ans.
•14–5.
Determine the strain energy in the rod assembly. Portion AB is steel, BC is brass, and CD is aluminum. Est = 200 GPa, Ebr = 101 GPa, and Eal = 73.1 GPa.
15 mm A
20 mm 2 kN B
25 mm D 5 kN C
2 kN
5 kN
3 kN
300 mm
N2 L Ui = © 2AE [3 (103) ]2 (0.3) =
2 (p4 )(0.0152)(200)(109)
[7 (103) ]2 (0.4) +
2(p4 )(0.022)(101)(109) [-3 (103) ]2 (0.2) + 2
(p4 )(0.0252)(73.1)(109)
= 0.372 N # m = 0.372 J
Ans.
1162
400 mm
200 mm
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14–6. If P = 60 kN, determine the total strain energy stored in the truss. Each member has a cross-sectional area of 2.511032 mm2 and is made of A-36 steel.
2m B C
Normal Forces. The normal force developed in each member of the truss can be determined using the method of joints.
1.5 m
Joint A (Fig. a) D
+ ©F = 0; : x
FAD = 0
+ c ©Fy = 0;
FAB - 60 = 0
FAB = 60 kN (T)
P
Joint B (Fig. b) + c ©Fy = 0;
3 FBD a b - 60 = 0 5
FBD = 100 kN (C)
+ ©F = 0; : x
4 100 a b - FBC = 0 5
FBC = 80 kN (T)
Axial Strain Energy. LBD = 222 + 1.52 = 2.5 m (Ui)a = ©
=
A = 2.5 A 103 B mm2 = 2.5 A 10 - 3 B m2
and
N2L 2AE
2 C 2.5 A 10
1
-3
B D C 200 A 109 B D
c C 60 A 103 B D 2 (1.5) + C 100 A 103 B D 2 (2.5) + C 80 A 103 B D 2 (2) d
= 43.2 J
Ans.
This result is only valid if s 6 sY. We only need to check member BD since it is subjected to the greatest normal force sBD =
A
100 A 103 B FBD = = 40 MPa 6 sY = 250 MPa A 2.5 A 10 - 3 B
O.K.
1163
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14–7. Determine the maximum force P and the corresponding maximum total strain energy stored in the truss without causing any of the members to have permanent deformation. Each member has the crosssectional area of 2.511032 mm2 and is made of A-36 steel.
2m B C
1.5 m
D
Normal Forces. The normal force developed in each member of the truss can be determined using the method of joints. Joint A (Fig. a) + ©F = 0; : x
FAD = 0
+ c ©Fy = 0;
FAB - P = 0
FAB = P (T)
Joint B (Fig. b) + c ©Fy = 0;
3 FBD a b - P = 0 5
+ ©F = 0; : x
4 1.6667Pa b - FBC = 0 5
FBD = 1.6667P (C)
FBC = 1.3333P(T)
Axial Strain Energy. A = 2.5 A 103 B mm2 = 2.5 A 10 - 3 B m2. Member BD is critical since it is subjected to the greatest force. Thus, sY =
FBD A
250 A 106 B =
1.6667P
2.5 A 10 - 3 B
P = 375 kN
Ans.
Using the result of P FAB = 375 kN
FBD = 625 kN
FBC = 500 kN
Here, LBD = 21.52 + 22 = 2.5 m. (Ui)a = ©
N2L = 2AE 1
=
2 C 2.5 A 10 - 3 B D C 200 A 109 B D
c C 375 A 103 B D 2 (1.5) + C 625 A 103 B D 2 (2.5) + C 500 A 103 B D 2 (2) d
= 1687.5 J = 1.6875 kJ
Ans.
1164
A P
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*14–8. Determine the torsional strain energy in the A-36 steel shaft. The shaft has a radius of 30 mm.
4 kN⭈m 3 kN⭈m
0.5 m
T2L 1 = [02(0.5) + ((3)(103))2(0.5) + ((1)(103))2(0.5)] Ui = © 2JG 2JG =
0.5 m
2.5(106) JG 2.5(106)
=
0.5 m
75(109)(p2 )(0.03)4
= 26.2 N # m = 26.2 J
Ans.
•14–9.
Determine the torsional strain energy in the A-36 steel shaft. The shaft has a radius of 40 mm.
12 kN⭈m 6 kN⭈m
0.5 m
Internal Torsional Moment: As shown on FBD. 8 kN⭈m
Torsional Strain Energy: With polar moment p J = A 0.044 B = 1.28 A 10 - 6 B p m4. Applying Eq. 14–22 gives 2
of
inertia
T2L Ui = a 2GJ =
1 C 80002 (0.6) + 20002 (0.4) + 2GJ
=
45.0(106) N2 # m3 GJ
A -100002 B (0.5) D
45.0(106) =
9
75(10 )[1.28(10 - 6) p]
= 149 J
Ans.
1165
0.4 m 0.6 m
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14–10. Determine the torsional strain energy stored in the tapered rod when it is subjected to the torque T. The rod is made of material having a modulus of rigidity of G. L
2r0
Internal Torque. The internal torque in the shaft is constant throughout its length as shown in the free-body diagram of its cut segment, Fig. a, Torsional Strain Energy. Referring to the geometry shown in Fig. b, r = r0 +
T
r0 r0 (x) = (L + x) L L
The polar moment of inertia of the bar in terms of x is J(x) =
4 pr0 4 p 4 p r0 (L + x)4 r = c (L + x) d = 2 2 L 2L4
We obtain, L
(Ui)t =
T2dx dx = L0 2GJ L0
L
T2 dx 2G B
pr0 4 2L4
(L + x)4 R
L
=
T2L4 dx pr0 4G L0 (L + x)4
=
L T2L4 1 B R ` pr0 4G 3(L + x)3 0
=
7 T2L 24pr0 4 G
Ans.
1166
r0
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14–11. The shaft assembly is fixed at C. The hollow segment BC has an inner radius of 20 mm and outer radius of 40 mm, while the solid segment AB has a radius of 20 mm. Determine the torsional strain energy stored in the shaft. The shaft is made of 2014-T6 aluminum alloy. The coupling at B is rigid.
600 mm 20 mm
600 mm C 40 mm B 60 N⭈m
Internal Torque. Referring to the free-body diagram of segment AB, Fig. a, TAB = -30 N # m
©Mx = 0; TAB + 30 = 0
Referring to the free-body diagram of segment BC, Fig. b, ©Mx = 0; TBC + 30 + 60 = 0
TAB = -90 N # m
p Torsional Strain Energy. Here, JAB = A 0.024 B = 80 A 10 - 9 B p m4 2 p JBC = A 0.044 - 0.024 B = 1200 A 10 - 9 B p m4, 2 (Ui)t = ©
TAB 2LAB TBC 2LBC T2L = + 2GJ 2GJAB 2GJBC (-30)2(0.6)
=
and
2 C 27 A 109 B D C 80 A 10 - 9 B p D
(-90)2(0.6)
+
2 C 27 A 109 B D C 1200 A 10 - 9 B p D
= 0.06379 J
Ans.
1167
A 20 mm 30 N⭈m
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*14–12. Consider the thin-walled tube of Fig. 5–28. Use the formula for shear stress, tavg = T>2tAm, Eq. 5–18, and the general equation of shear strain energy, Eq. 14–11, to show that the twist of the tube is given by Eq. 5–20, Hint: Equate the work done by the torque T to the strain energy in the tube, determined from integrating the strain energy for a differential element, Fig. 14–4, over the volume of material.
Ui =
t2 dV Lv 2 G
but t =
T 2 t Am
Thus, Ui =
T2 dV 2 2 Lv 8 t AmG L
=
2
T2 dV T2 dA TL dA = dx = 2 2 2 2 2 2 8 A m G Lv t 8 A m G LA t L0 8 A mG LA t
However, dA = t ds. Thus, Ui =
ds T2L 2 8 AmG L t
Ue =
1 Tf 2
Ue = Ui ds T2L 1 Tf = 2 8 A2mG L t f =
ds TL 4 A2mG L t
QED
1168
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•14–13.
Determine the ratio of shearing strain energy to bending strain energy for the rectangular cantilever beam when it is subjected to the loading shown. The beam is made of material having a modulus of elasticity of E and Poisson’s ratio of n.
P L a
a
b h Section a – a
Internal Moment. Referring to the free-body diagram of the left beam’s cut segment, Fig. a, + c ©Fy = 0;
-V - P = 0
V = -P
+ ©MO = 0;
M + Px = 0
M = -Px
Shearing Strain Energy. For the rectangular cross section, the form factor is fs =
6 5
6 2 2 L (-P) dx L fsV dx 5 3P2 3P2L (Ui)v = = = dx = 2GA 2G(bh) 5bhG L0 5bhG L0 L0 L
However, G =
E , then 2(1 + v) (Ui)v =
6(1 + v)P2L 5bhE
Bending Strain Energy. L
(Ui)b =
M2dx = L0 2EI L0
L
(-Px)2dx2 2Ea
1 bh3 b 12
6P2 6P2 x3 L 2P2L3 2 x dx = = ¢ ≤ ` bh3E L0 bh3E 3 0 bh3E L
=
Then, the ratio is 6(1 + v)P2L 3(1 + v) h 2 (Ui)v 5bhE a b = = 2 3 (Ui)b 5 L 2P L bh3E
Ans.
From this result, we can conclude that the proportion of the shearing strain energy stored in the beam increases if the depth h of the beam’s cross section increases but (Ui)v 1 = 0.009. the decreases if L increases. Suppose that v = and L = 10h, then 2 (Ui)b shearing strain energy is only 0.09% of the bending strain energy. Therefore, the effect of the shearing strain energy is usually neglected if L 7 10h.
1169
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14–14. Determine the bending strain-energy in the beam due to the loading shown. EI is constant.
M0 A
L
Ui =
L0
C B
2
M dx 2EI
=
1 c 2EI L0
=
M20L 24EI
L>2
L — 2
a
L — 2
L>2 2 2 -M0 M0 x1 b dx1 + a x2 b dx2] L L L0
Ans.
Note: Strain energy is always positive regardless of the sign of the moment function.
14–15. Determine the bending strain energy in the beam. EI is constant.
P
P
Referring to the FBD of the entire beam, Fig. a, a + ©MB = 0;
Pa
L 3L b + Pa b - Ay (L) = 0 4 4
L 4
Ay = P
Using the coordinates, x1 and x2, the FBDs of the beam’s cut segments in Figs. b and c are drawn. For coordinate x1, a + ©Mc = 0;
M(x1) - Px1 = 0
M(x1) = Px1
For coordinate x2 coordinate, a + ©Mc = 0;
M(x2) - Pa
L b = 0 4
L
(Ui)b = ©
1 M2dx = c2 2EI 2EI L0 L0
M(x2) =
PL 4
L>4
L>2
(Px1)2dx1 +
L0
a
=
P2 3 P2L2 2 1 c2 a x1 b ` + x ` d 2EI 3 16 2 0 0
=
P2L3 48EI
L 4
PL 2 b dx2 d 4
L
Ans.
1170
L 2
L 4
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*14–16. Determine the bending strain energy in the A-36 structural steel W10 * 12 beam. Obtain the answer using the coordinates 1a2 x1 and x4, and 1b2 x2 and x3.
6 kip
x1
x4 x2 12 ft
Support Reactions: As shown on FBD(a). Internal Moment Function: As shown on FBD(b), (c), (d) and (e). Bending Strain Energy: Using coordinates x1 and x4 and applying Eq. 14–17 gives L
Ui =
M2dx L0 2EI 1 c 2EI L0
12ft
=
6ft
1 c 2EI L0
12ft
=
=
3888 kip2 # ft3 EI
(-3.00x1)2 dx1 +
L0
(-6.00x4)2 dx4 d
6ft
9.00x21dx1 +
L0
36.0x24 dx4 d
For W10 * 12 wide flange section, I = 53.8 in4. Ui =
3888(123) 29.0(103)(53.8)
= 4.306 in # kip = 359 ft # lb
Ans.
b) Using coordinates x2 and x3 and applying Eq. 14–17 gives L
Ui =
M2dx L0 2EI 1 c 2EI L0
12ft
=
6ft
1 c 2EI L0
12ft
=
=
3888 kip2 # ft3 EI
(3.00x2 - 36.0)2dx2 +
L0
(6.00x3 - 36.0)2 dx3 d
A 9.00x22 - 216x + 1296 B dx2 +
6ft
L0
A 36.0x23 - 432x + 1296 B dx3 d
For W 10 * 12 wide flange section, I = 53.8 in4. Ui =
3888(123) 29.0(103)(53.8)
= 4.306 in # kip = 359 ft # lb
Ans.
1171
x3 6 ft
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•14–17.
Determine the bending strain energy in the A-36 steel beam. I = 99.2 (106) mm4.
9 kN/m
6m
Referring to the FBD of the entire beam, Fig. a, 1 (9)(6)(2) - Ay (6) = 0 2
a + ©MB = 0;
Ay = 9 kN
Referring to the FBD of the beam’s left cut segment, Fig. b,
a + ©M0 = 0;
M(x) +
1 3 a xb (x) (x>3) - 9x = 0 2 2
M(x) = a 9x L
(Ui)b =
For
A
36
1 3 x b 4
kN # m
M2 dx 1 = 2EI L0 L0 2EI
6m
a 9x -
1 2EI L0
6m
=
a 81x2 +
=
1 7 9 5 2 6m 1 c a27x3 + x x b d 2EI 112 10 0
=
666.51 kN2 # m3 EI
steel,
E = 200 GPa.
Here,
= 99.2(10 - 6) m4. Then
(Ui)b =
666.51 (10002)
200(109) C 99.2(10 - 6) D
1 3 2 x b dx 4 1 6 9 x - x4 bdx 16 2
I = C 99.2 (106) mm4 D a
4 1m b 1000 mm
Ans.
= 33.6 J
1172
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14–18. Determine the bending strain energy in the A-36 steel beam due to the distributed load. I = 122 (106) mm4.
15 kN/m
A
B 3m
Referring to the FBD of the entire beam, Fig. a, 1 (15)(3) = 0 2
Ay -
+ c ©Fy = 0; a + ©MA = 0;
MA -
Ay = 22.5 kN
1 (15)(3)(2) = 0 2
MA = 45 kN # m
Referring to the FBD of the beam’s left cut segment, Fig. b, a + ©M0 = 0;
M(x) +
1 (5x)(x)(x>3) - 22.5x + 45 = 0 2
M(x) = (22.5x - 0.8333x3 - 45) kN # m L
(Ui)b =
=
1 M2dx = c 2EI 2EI L0 L0 1 c 2EI L0
3m
(22.5x - 0.8333x3 - 45)2 dx
3m
0.6944x6 - 37.5x4 + 75x3 + 506.25x2 - 2025x + 2025)dx d
=
1 a0.09921x7 - 7.5x5 + 18.75x4 + 168.75x3 2EI - 1012.5x2 + 2025xb 2
3m 0
715.98 kN2 # m2 = EI For
A
36
steel,
E = 200 GPa.
Here,
I = c122(106) mm4 d a
4 1m b 1000 mm
= 122(10 - 6) m4. Thus,
(Ui)b =
715.98 (10002)
200(109) C 122 (10 - 6) D
Ans.
= 29.3 J
1173
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14–19. Determine the strain energy in the horizontal curved bar due to torsion. There is a vertical force P acting at its end. JG is constant.
r 90⬚
P
T = Pr(1 - cos u) Strain energy: L
Ui =
T2 ds L0 2JG
However, ds = rdu
s = ru; u
Ui =
T2rdu r = 2JG L0 L0 2JG P2r3 2JG L0
p>2
=
P2r3 2JG L0
p>2
=
P2r3 2JG L0
p>2
=
=
P2r3 3p a - 1b JG 8
p>2
[Pr(1 - cos u)]2du
(1 - cos u)2 du (1 + cos2 u - 2 cos u)du
(1 +
cos 2u + 1 - 2 cos u) du 2 Ans.
1174
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*14–20. Determine the bending strain energy in the beam and the axial strain energy in each of the two rods. The beam is made of 2014-T6 aluminum and has a square cross section 50 mm by 50 mm. The rods are made of A-36 steel and have a circular cross section with a 20-mm diameter.
2m 8 kN
Support Reactions: As shown on FBD(a).
8 kN
Internal Moment Function: As shown on FBD(b) and (c). Axial Strain Energy: Applying Eq. 14–16 gives
1m
N2L (Ui)a = 2AE
C 8.00(103) D 2 (2)
=
2AE
64.0(106) N2 # m AE
=
64.0(106) =
p 4
(0.022) [200(109)] Ans.
= 1.02 J Bending Strain Energy: Applying Eq. 14–17 gives L
(Ui) b =
M2dx L0 2EI
=
1 B2 2EI L0
=
85.333 kN2 # m3 EI
1m
(8.00x1)2 dx 1 +
2m
L0
8.002 dx2 R
85.333(106)
=
1 73.1(109) C 12 (0.05) (0.053) D
= 2241.3 N # m = 2.24 kJ
Ans.
1175
2m
1m
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•14–21.
The pipe lies in the horizontal plane. If it is subjected to a vertical force P at its end, determine the strain energy due to bending and torsion. Express the results in terms of the cross-sectional properties I and J, and the material properties E and G.
z
L
C x
B L — 2
A
P
Ui =
M2 dx T2 dx + L 2E I L 2JG L 2
=
L L PL 2 ( 2 ) dx (P x)2 dx (P x)2 dx + + 2 E I 2 EI L0 L0 2 J G L0
=
L 31 P2 L3 P2L2 P2 a b + + (L) 2EI 2 3 2EI 3 8JG
=
P2 L3 3 P2 L3 + 16 E I 8JG
= P2 L3 c
3 1 + d 16 E I 8JG
Ans.
1176
y
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14–22. The beam shown is tapered along its width. If a force P is applied to its end, determine the strain energy in the beam and compare this result with that of a beam that has a constant rectangular cross section of width b and height h.
b
h
P L
Moment of Inertia: For the beam with the uniform section, I =
bh3 = I0 12
For the beam with the tapered section, I =
I0 bh3 1 b a xb A h3 B = x = x 12 L 12L L
Internal Moment Function: As shown on FBD. Bending Strain Energy: For the beam with the tapered section, applying Eq. 14–17 gives L
UI =
M2 dx L0 2EI
=
L (-Px)2 1 dx I0 2E L0 L x
=
P2L xdx 2EI0 L0
=
P2L3 3P2 L3 = 4EI0 bh3 E
L
Ans.
For the beam with the uniform section, L
Ui =
M2dx L0 2EI L
=
1 (-Px)2 dx 2EI0 L0
=
P3 L3 6EI0
The strain energy in the capered beam is 1.5 times as great as that in the beam having a uniform cross section. Ans.
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14–23. Determine the bending strain energy in the cantilevered beam due to a uniform load w. Solve the problem two ways. (a) Apply Eq. 14–17. (b) The load w dx acting on a segment dx of the beam is displaced a distance y, where y = w1-x4 + 4L3x - 3L42>124EI2, the equation of the elastic curve. Hence the internal strain energy in the differential segment dx of the beam is equal to the external work, i.e., dUi = 121w dx21-y2. Integrate this equation to obtain the total strain energy in the beam. EI is constant.
w dx w
dx
Internal Moment Function: As shown on FBD. Bending Strain Energy: a) Applying Eq. 14–17 gives L
Ui =
M2dx L0 2EI L
=
1 w 2 c - x2 d dx R B 2EI L0 2
=
w2 x4 dx R B 8EI L0
=
w2 L5 40EI
L
b) Integrating dUi =
1 w (wdx) B A -x4 + 4L3x - 3L4 B R 2 24EI
dUi =
w2 A x4 -4L3x + 3L4 B dx 48EI
Ui =
=
Ans.
1 (wdx)( -y) 2
dUi =
w2 48EI L0
L
x L
A x4 - 4L3x + 3L4 B dx
w2L5 40EI
Ans.
1178
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*14–24. Determine the bending strain energy in the simply supported beam due to a uniform load w. Solve the problem two ways. (a) Apply Eq. 14–17. (b) The load w dx acting on the segment dx of the beam is displaced a distance y, where y = w1-x4 + 2Lx3 - L3x2>124EI2, the equation of the elastic curve. Hence the internal strain energy in the differential segment dx of the beam is equal to the external work, i.e., dUi = 121w dx21-y2. Integrate this equation to obtain the total strain energy in the beam. EI is constant.
w dx w
x
Support Reactions: As shown on FBD(a). Internal Moment Function: As shown on FBD(b). Bending Strain Energy: a) Applying Eq. 14–17 gives L
Ui =
M2dx L0 2EI L
=
2 1 w 2 c (Lx - x ) d dx R B 2EI L0 2
=
w2 (L2x2 + x4 - 2Lx3) dx R B 8EI L0
=
w2L5 240EI
L
b) Integrating dUi =
Ans.
1 (wdx) (-y) 2
dUi =
1 w (wdx) B (-x4 + 2Lx3 - L3x) R 2 24EI
dUi =
w2 (x4 - 2Lx3 + L3x) dx 48EI L
Ui =
=
w2 (x4 - 2Lx3 + L3x) dx 48EI L0 w2L5 240EI
Ans.
1179
dx L
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•14–25. Determine the horizontal displacement of joint A. Each bar is made of A-36 steel and has a cross-sectional area of 1.5 in2.
2 kip
A
3 ft D 3 ft C
B 4 ft
Member Forces: Applying the method of joints to joint at A, we have + ©F = 0; : x
4 - 2 = 0 F 5 AD
+ c ©Fy = 0;
FAB -
FAD = 2.50 kip (T)
3 (2.50) = 0 5
FAB = 1.50 kip (C)
At joint D + ©F = 0; : x
4 4 F - (2.50) = 0 5 DB 5
+ c ©Fy = 0;
3 3 ( 2.50) + (2.50) - FDC = 0 5 5
FDB = 2.50 kip (C)
FDC = 3.00 kip (T) Axial Strain Energy: Applying Eq. 14–16, we have N2L Ui = a 2AE =
1 [2.502 (5) + (-1.50)2 (6) + (-2.50)2 (5) + 3.002(3)] 2AE
=
51.5 kip2 # ft AE 51.5(12)
=
1.5[29.0(103)]
= 0.014207 in # kip
External Work: The external work done by 2 kip force is Ue =
1 (2) (¢ A)h = (¢ A)h 2
Conservation of Energy: Ue = Ui (¢ A)h = 0.014207 = 0.0142 in.
Ans.
1180
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14–26. Determine the horizontal displacement of joint C. AE is constant.
C P
L
L
A
B L
Member Forces: Applying the method of joints to C, we have + c ©Fy = 0;
FBC cos 30° - FAC cos 30° = 0
+ ©F = 0; : x
P - 2F sin 30° = 0
Hence,
FBC = P (C)
FBC = FAC = F
F = P
FAC = P (T)
Axial Strain Energy: Applying Eq. 14–16, we have N2L Ui = a 2AE =
1 C P2L + (-P)2 L D 2AE
=
P2L AE
External Work: The external work done by force P is Ui =
1 P(¢ c)k 2
Conservation of Energy: Ue = Ui P2L 1 P(¢ C)k = 2 AE (¢ C)k =
2PL AE
Ans.
1181
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14–27. Determine the vertical displacement of joint C. AE is constant.
A
L
L
C
L
B
Joint C: + ©F = 0 : x
FCB cos 30° - FCA cos 30° = 0 FCB = FCA
+ c © Fy = 0
FCA sin 30° + FCB sin 30° - P = 0 FCB = FCA = P
Conservation of energy: Ue = Ui N2L 1 P¢ C = © 2 2EA L 1 2 P¢ C = [F 2 + FCA ] 2 2EA CB P¢ C =
¢C =
L (P2 + P2) EA 2PL AE
Ans.
1182
P
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*14–28. Determine the horizontal displacement of joint D. AE is constant.
D
C
P
Joint B:
L
+ c ©Fy = 0;
FBC = 0.75P
+ ©F = 0; ; x
FBA = P
0.6 L
B
A
Joint D:
0.8 L
+ T ©Fy = 0;
FDA = 0
+ ©Fx = 0; :
FDC = P
Joint A: + T ©Fy = 0;
3 F - 0.75P = 0 5 AC FAC = 1.25P
Conservation of energy: Ue = Ui N2L 1 P¢ D = © 2 2AE 1 1 P¢ D = [(0.75P)2(0.6L) + (P)2(0.8L) + (02)(0.6L) 2 2AE + (P2)(0.8L) + (1.25P)2(L)] ¢D =
3.50PL AE
Ans.
•14–29.
The cantilevered beam is subjected to a couple moment M0 applied at its end. Determine the slope of the beam at B. EI is constant.
M0
A B
Ui =
2 L M20L M dx 1 = M20 dx = 2EI L0 2EI L0 2EI
Ue =
1 (M0 uB) 2
L
L
Conservation of energy: Ue = Ui M0 2L 1 M0 uB = 2 2EI uB =
M0L EI
Ans.
1183
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14–30. Determine the vertical displacement of point C of the simply supported 6061-T6 aluminum beam. Consider both shearing and bending strain energy.
100 kip a B
A C
1.5 ft
Support Reactions. Referring to the free-body diagram of the entire beam, Fig. a, a + ©MB = 0;
100(1.5) - Ay(3) = 0
Ay = 50 kip
50 - V = 0
+ c ©Fy = 0; a + ©MO = 0;
V = 50 kip
M - 50x = 0
M = 50x
Shearing Strain Energy. For the rectangular beam, the form factor is fs =
6 5
6 A 502 B dx 18 in. fsV2dx 5 (Ul)v = = 2 = 0.3041 in # kip L0 2GA L0 2 C 3.7 A 103 B D [4(12)] L
Bending Strain Energy. I =
1 (4) A 123 B = 576 in4. We obtain 12
L
(Ui)b =
L (50x)2 dx M2dx = 2 L0 2EI L0 2 C 10.0 A 103 B D (576)
= 0.4340 A 10-3 B
18 in.
L0
= 0.4340 A 10-3 B ¢
x2 dx
x3 18in. ≤` 3 0
= 0.84375 in # kip Thus, the total strain energy stored in the beam is Ui = (Ui)v + (Ui)b = 0.3041 + 0.84375 = 1.1478 in # kip
Ans.
External Work. The external work done by the external force (100 kip) is Ue =
1 1 P¢ = (100)¢ C = 50¢ C 2 2
Conservation of Energy. Ue = Ui 50¢ C = 1.1478 Ans.
¢ C = 0.02296 in. = 0.0230 in. 1184
1.5 ft 4 in.
12 in.
Internal Loading. Referring to the free-body diagram of the beam’s left cut segment, Fig. b,
a
Section a – a
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14–31. Determine the slope at the end B of the A-36 steel beam. I = 8011062 mm4.
6 kN⭈m A
B
M = -750x 8m L
2
M dx 1 M uB = 2 L0 2EI B ( -750x)2dx 1 (6(103)) uB = 2 2EI L0
uB =
16000 200 (109)(80)(10-6)
= 1 (10-3) rad
Ans.
*14–32. Determine the deflection of the beam at its center caused by shear. The shear modulus is G.
P b
Support Reactions: As shown on FBD(a).
h
Shear Functions: As shown on FBD(b). Shear Strain Energy: Applying 14–19 with fe =
L 2
6 for a rectangular section, we have 5
L
Ui =
feV2dx L0 2GA L
2 6 P 2 1 a b a b dx R = B2 2bhG 2 L0 5
=
3P2L 20bhG
External Work: The external work done by force P is Ue =
1 (P) ¢ 2
Conservation of Energy: Ue = Ui 3P2L 1 (P)¢ = 2 20bhG ¢ =
3PL 10bhG
Ans.
1185
L 2
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•14–33. The A-36 steel bars are pin connected at B and C. If they each have a diameter of 30 mm, determine the slope at E. 3
2
C D
A E 2m
3m L
300 N⭈m
B
2m
3m
2
Ui =
M dx 1 1 65625 (75x1)2dx1 + (2) (-75x2)2dx2 = = (2) 2EI 2EI 2EI EI L0 L0 L0
Ue =
1 1 (M¿)u = (300) uE = 150 uE 2 2
Conservation of energy: Ue = Ui
150 uE =
uE =
65625 EI
473.5 473.5 = 0.0550 rad = 3.15° = EI (200)(109)(p4 )(0.0154)
14–34. The A-36 steel bars are pin connected at B. If each has a square cross section, determine the vertical displacement at B.
Ans.
800 lb 2 in. A
B
C
D 2 in.
Support Reactions: As shown on FBD(a). 8 ft
Moment Functions: As shown on FBD(b) and (c). Bending Strain Energy: Applying 14–17, we have L
Ui =
M2dx L0 2EI 4ft
10ft
=
1 B 2EI L0
=
23.8933(106) lb2 # ft3 EI
( -800x1)2 dx1 +
23.8933(106)(123)
=
1 29.0(106) C 12 (2)(23) D
L0
( -320x2)2 dx2 R
= 1067.78 in # lb
External Work: The external work done by 800 lb force is Ue =
1 (800)(¢ B) = 400¢ B 2
Conservation of Energy: Ue = Ui 400¢ B = 1067.78 ¢ B = 2.67 in.
Ans.
1186
4 ft
10 ft
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14–35. Determine the displacement of point B on the A-36 steel beam. I = 8011062 mm4.
20 kN
A
C B 3m
L
3
5m
5
1 M2dx = [(12.5)(103)(x1)]2dx1 + [(7.5)(103)(x2)]2 dx2 R Ui = B 2EI L0 L0 2EI L0 = Ue =
1.875(109) EI
1 1 P¢ = (20)(103)¢ B = 10(103)¢ B 2 2
Conservation of energy: Ue = Ui 10(103)¢ B =
¢B =
1.875(109) EI
187500 187500 = 0.0117 m = 11.7 mm = EI 200(109)(80)(10 - 6)
Ans.
*14–36. The rod has a circular cross section with a moment of inertia I. If a vertical force P is applied at A, determine the vertical displacement at this point. Only consider the strain energy due to bending. The modulus of elasticity is E.
r A
Moment function: P
a + ©MB = 0;
P[r(1 - cos u)] - M = 0;
M = P r (1 - cos u)
Bending strain energy: s
Ui =
M2 ds L0 2 E I
ds = r du
u
p
=
r M2 r du = [P r (1 - cos u) ]2 du 2 E I L0 L0 2 E I
=
P2 r3 (1 + cos2 u - 2 cos u)du 2 E I L0
=
1 cos 2u P2 r3 a1 + + - 2 cos ub du 2 E I L0 2 2
=
P2 r3 3 cos 2u P2 r3 3 3 p P2 r3 a + - 2 cos u b du = a pb = 2 E I L0 2 2 2 EI 2 4 EI
p
p
p
Conservation of energy: Ue = Ui ;
¢A =
3 p P2 r3 1 P ¢A = 2 4 EI
3 p P r3 2 EI
Ans.
1187
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•14–37.
The load P causes the open coils of the spring to make an angle u with the horizontal when the spring is stretched. Show that for this position this causes a torque T = PR cos u and a bending moment M = PR sin u at the cross section. Use these results to determine the maximum normal stress in the material.
P
R
d
u
T = P R cos u;
M = P R sin u
Bending: smax =
P R sin u d Mc = d4 I 2 (p4 )(16 )
tmax =
P R cos u d2 Tc = p d4 J ( )
P
2 16
smax =
sx + sy 2
;
C
a
sx - sy 2
2
b + t2xy
=
16 P R sin u 2 16 P R cos u 2 16 P R sin u ; a b + a b C p d3 pd3 p d3
=
16 P R [sin u + 1] p d3
Ans.
14–38. The coiled spring has n coils and is made from a material having a shear modulus G. Determine the stretch of the spring when it is subjected to the load P. Assume that the coils are close to each other so that u L 0° and the deflection is caused entirely by the torsional stress in the coil.
P
R u
Bending Strain Energy: Applying 14–22, we have Ui =
P2R2L T2L 16P2R2L = = p 2GJ pd4G 2G C 32 (d4) D P
However, L = n(2pR) = 2npR. Then Ui =
32nP2R3 d4G
External Work: The external work done by force P is Ue =
1 P¢ 2
Conservation of Energy: Ue = Ui 1 32nP2R3 P¢ = 2 d4G ¢ =
64nPR3 d4G
Ans.
1188
d
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14–39. The pipe assembly is fixed at A. Determine the vertical displacement of end C of the assembly. The pipe has an inner diameter of 40 mm and outer diameter of 60 mm and is made of A-36 steel. Neglect the shearing strain energy.
A
800 mm
600 N
B
400 mm
Internal Loading: Referring to the free-body diagram of the cut segment BC, Fig. a, ©My = 0; My + 600x = 0
My = -600x
Referring to the free-body diagram of the cut segment AB, Fig. b, ©Mx = 0; Mx - 600y = 0
Mx = 600y
©My = 0; 600(0.4) - Ty = 0
Ty = 240 N # m
p A 0.034 - 0.024 B = 0.325 A 10 - 6 B pm4. We obtain 2
Torsional Strain Energy. J = L
(Ui)t =
T2dx = L0 2GJ L0
Bending Strain Energy. I =
0.8 m
2 C 75 A 109 B D C 0.325 A 10 - 6 B p D
1 M2dx = B 2EI 2EI L0 L0
0.4 m
0.8 m
(-600x)2 dx +
=
0.4 m 0.8 m 1 + 120 A 103 B y3 2 B 120 A 103 B x3 2 R 2EI 0 0
=
34 560 N2 # m3 EI 34 560
=
200 A 10 B c0.1625 A 10 9
= 0.3009 J
p A 0.034 - 0.024 B = 0.1625 A 10 - 6 B pm4. We obtain 4
L
(Ui)b =
2402 dx
-6
Bp d
L0
A 600y)2 dy R
= 0.3385 J
Thus, the strain energy stored in the pipe is Ui = (Ui)t + (Ui)b = 0.3009 + 0.3385 = 0.6394 J External Work. The work done by the external force P = 600 N is Ue =
1 1 P¢ = (600)¢ C = 300¢ C 2 2
Conservation of Energy. Ue = Ut 300¢ C = 0.6394 ¢ C = 2.1312 A 10 - 3 B = 2.13 mm
Ans.
1189
C
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*14–40. The rod has a circular cross section with a polar moment of inertia J and moment of inertia I. If a vertical force P is applied at A, determine the vertical displacement at this point. Consider the strain energy due to bending and torsion. The material constants are E and G.
z
y r P
T = Pr(1 - cos u);
M = Pr sin u
Torsion strain energy:
x s
Ui =
u
T2 ds T2 rdu = L0 2GJ L0 2GJ p
=
r [Pr(1 - cos u)]2 du 2GJ L0
=
P2 r3 (1 + cos2 u - 2 cos u)du 2GJ L0
=
cos 2u + 1 P2 r3 a1 + - 2 cos u bdu 2GJ L0 2
=
3P2r3 p 4GJ
p
p
Bending strain energy: s
Ui =
M2ds L0 2EI u
p
=
M2r du r = [Pr sin u]2 du 2EI 2EI L0 L0
=
P2 r3 1 - cos 2u P2 r3 p a bdu = 2EI L0 2 4EI
p
Conservation of energy: Ue = Ui 3P2 r3 p P2 r3 p 1 P¢ = + 2 4GJ 4EI ¢ =
1 Pr3 p 3 a + b 2 GJ EI
Ans.
1190
A
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•14–41. Determine the vertical displacement of end B of the frame. Consider only bending strain energy. The frame is made using two A-36 steel W460 * 68 wide-flange sections.
3m B
Internal Loading. Using the coordinates x1 and x2, the free-body diagrams of the frame’s segments in Figs. a and b are drawn. For coordinate x1, -M1 - 20 A 103 B x1 = 0
+ ©MO = 0; For coordinate x2, + ©MO = 0;
M1 = -20 A 103 B x1
M2 - 20 A 103 B (3) = 0
M2 = 60 A 103 B N # m
20 kN
A
Bending Strain Energy. L
(Ub)i =
M2dx 1 = B 2EI L0 L0 2EI
400 A 10 1 D£ = 2EI 3
6
B
3m
c -20 A 103 B x1 d dx1 +
3m
x1 ≥ 3 3
4m
2
+ 3.6 A 109 B x 2
4m
L0
c60 A 103 B d dx2 R 2
T
0
0
=
9 A 109 B N2 # m2 EI
For a W460 * 68, I = 297 A 106 B mm4 = 297 A 10 - 6 B m4. Then (Ub)i =
9 A 109 B
200 A 109 B (297) A 10 - 6 B
= 151.52 J
External Work. The work done by the external force P = 20 kN is Ue =
4m
1 1 P¢ = c20 A 103 B d ¢ B = 10 A 103 B ¢ B 2 2
Conservation of Energy. Ue = Ui 10 A 103 B ¢ B = 151.52 ¢ B = 0.01515 m = 15.2 mm
Ans.
1191
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14–42. A bar is 4 m long and has a diameter of 30 mm. If it is to be used to absorb energy in tension from an impact loading, determine the total amount of elastic energy that it can absorb if (a) it is made of steel for which Est = 200 GPa, sY = 800 MPa, and (b) it is made from an aluminum alloy for which Eal = 70 GPa, sY = 405 MPa.
a) eg =
800(106) sY = 4(10 - 3) m>m = E 200(109)
ur =
1 1 (s )(e ) = (800)(106)(N>m2)(4)(10 - 3)m>m = 1.6 MJ>m3 2 Y g 2
V =
p (0.03)2(4) = 0.9(10 - 3)p m2 4
ui = 1.6(106)(0.9)(10 - 3)p = 4.52 kJ
Ans.
b)
eg =
405(106) sY = 5.786(10 - 3) m>m = E 70(109)
ur =
1 1 (s )(e ) = (405)(106)(N>m2)(5.786)(10 - 3)m>m = 1.172 MJ>m3 2 Y g 2
V =
p (0.03)2 (4) = 0.9(10 - 3)p m3 4
ui = 1.172(106)(0.9)(10 - 3)p = 3.31 kJ
Ans.
14–43. Determine the diameter of a red brass C83400 bar that is 8 ft long if it is to be used to absorb 800 ft # lb of energy in tension from an impact loading. No yielding occurs. Elastic Strain Energy: The yielding axial force is PY = sgA. Applying Eq. 14–16, we have Ui =
(sgA)2L s2gAL N2L = = 2AE 2AE 2E
Substituting, we have Ui =
0.8(12) =
s2gAL 2E
11.42 C p4 (d2) D (8)(12) 2[14.6(103)]
d = 5.35 in.
Ans.
1192
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*14–44. A steel cable having a diameter of 0.4 in. wraps over a drum and is used to lower an elevator having a weight of 800 lb. The elevator is 150 ft below the drum and is descending at the constant rate of 2 ft兾s when the drum suddenly stops. Determine the maximum stress developed in the cable when this occurs. Est = 2911032 ksi, sY = 50 ksi. k =
AE = L
p 4
(0.42)(29)(103) 150 (12)
150 ft
= 2.0246 kip>in.
Ue = Ui 1 1 mv2 + W ¢ max = k¢ 2max 2 2 800 1 1 c d[(12) (2)]2 + 800 ¢ max = (2.0246)(103)¢ 2max 2 32.2 (12) 2 596.27 + 800 ¢ max = 1012.29 ¢ 2max ¢ max = 1.2584 in. Pmax = k¢ max = 2.0246 (1.2584) = 2.5477 kip smax =
Pmax 2.5477 = p = 20.3 ksi 6 sg 2 A 4 (0.4)
O.K.
Ans.
•14–45.
The composite aluminum bar is made from two segments having diameters of 5 mm and 10 mm. Determine the maximum axial stress developed in the bar if the 5-kg collar is dropped from a height of h = 100 mm. Eal = 70 GPa, sY = 410 MPa.
5 mm
200 mm
300 mm
¢ st = ©
WL = AE
Pmax = W B 1 +
(0.0052)(70)(109)
C
1 + 2a
= 5(9.81) B 1 + smax =
Pmax = A
5(9.81)(0.3)
5(9.81)(0.2) p 4
p 4
C
+
p 4
(0.012)(70)(109)
h
= 9.8139(10 - 4) m
10 mm
h bR ¢ st
1 + 2a
0.1 9.8139(10 - 6)
7051 = 359 MPa 6 sy (0.0052)
b R = 7051 N O.K.
1193
Ans.
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14–46. The composite aluminum bar is made from two segments having diameters of 5 mm and 10 mm. Determine the maximum height h from which the 5-kg collar should be dropped so that it produces a maximum axial stress in the bar of s max = 300 MPa, Eal = 70 GPa, sY = 410 MPa.
5 mm
200 mm
300 mm h 10 mm
¢ st = ©
WL = AE
Pmax = W B 1 +
5(9.81)(0.3)
5(9.81)(0.2) p 2 9 4 (0.005 )(70)(10 )
C
1 + 2a
+
p 4
(0.012)(70)(109)
= 9.8139(10 - 6) m
h bR ¢ st
p h bR 300(106)a b (0.0052) = 5(9.81) B 1 + 1 + 2a 4 C 9.8139(10 - 6) 120.1 = 1 + 21 + 203791.6 h h = 0.0696 m = 69.6 mm
Ans.
1194
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14–47. The 5-kg block is traveling with the speed of v = 4 m>s just before it strikes the 6061-T6 aluminum stepped cylinder. Determine the maximum normal stress developed in the cylinder.
C
B 40 mm
Equilibrium. The equivalent spring constant for segments AB and BC are
kAB
kBC
p A 0.022 B c68.9 A 109 B d AAB E 4 = = = 72.152 A 106 B N>m LAB 0.3 p A 0.042 B c68.9 A 109 B d ABC E 4 = = = 288.608 A 106 B N>m LBC 0.3
Equilibrium requires FAB = FBC kAB ¢ AB = kBC ¢ BC 72.152 A 106 B ¢ AB = 288.608 A 106 B ¢ BC ¢ BC =
1 ¢ 4 AB
(1)
Conservation of Energy. Ue = Ui 1 1 1 mv2 = kAB ¢ AB 2 + kBC ¢ BC 2 2 2 2
(2)
Substituting Eq. (1) into Eq. (2), 2 1 1 1 1 mv2 = kAB ¢ AB 2 + kBC a ¢ AB b 2 2 2 4
1 1 1 mv2 = kAB ¢ AB 2 + k ¢ 2 2 2 32 BC AB 1 1 1 (5) A 42 B = c 72.152 A 106 B d ¢ AB 2 + c288.608 A 106 B d ¢ AB 2 2 2 32 ¢ AB = 0.9418 A 10 - 3 B m Maximum Stress. The force developed in segment AB is FAB = kAB ¢ AB = 72.152 A 106 B c0.9418 A 10 - 3 B d = 67.954 A 103 B N.
Thus, smax = sAB =
300 mm
300 mm
67.954 A 103 B FAB = = 216.30 MPa = 216 MPa p AAB 2 0.02 A B 4
Since smax 6 sY = 255 MPa, this result is valid.
1195
Ans.
v A
20 mm
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*14–48. Determine the maximum speed v of the 5-kg block without causing the 6061-T6 aluminum stepped cylinder to yield after it is struck by the block.
C
B 40 mm
Equilibrium. The equivalent spring constant for segments AB and BC are
kAB
kBC
p A 0.022 B C 68.9 A 109 B D AAB E 4 = = = 72.152 A 106 B N>m LAB 0.3 p A 0.042 B C 68.9 A 109 B D ABC E 4 = = = 288.608 A 106 B N>m LBC 0.3
Equilibrium requires FAB = FBC kAB ¢ AB = kBC ¢ BC 72.152 A 106 B ¢ AB = 288.608 A 106 B ¢ BC ¢ BC =
1 ¢ 4 AB
(1)
Conservation of Energy. Ue = Ui 1 1 1 mv2 = kAB ¢ AB 2 + kBC ¢ BC 2 2 2 2
(2)
Substituting Eq. (1) into Eq. (2), 2 1 1 1 1 mv2 = kAB ¢ AB 2 + kBC a ¢ AB b 2 2 2 4
1 1 1 mv2 = kAB ¢ AB 2 + k ¢ 2 2 2 32 BC AB 1 1 1 (5)v2 = c72.152 A 106 B d ¢ AB 2 + c288.608 A 106 B d ¢ AB 2 2 2 32 ¢ AB = 0.23545 A 10 - 3 B v Maximum Stress. The force developed in segment AB is FAB = kAB ¢ AB = 72.152 A 106 B C 0.23545 A 10 - 3 B v D = 16988.46v. Thus, smax = sAB = 255 A 106 B =
300 mm
300 mm
FAB AAB
16988.46v p A 0.022 B 4
v = 4.716 m>s = 4.72 m>s
Ans.
1196
v A
20 mm
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•14–49. The steel beam AB acts to stop the oncoming railroad car, which has a mass of 10 Mg and is coasting towards it at v = 0.5 m>s . Determine the maximum stress developed in the beam if it is struck at its center by the car. The beam is simply supported and only horizontal forces occur at A and B. Assume that the railroad car and the supporting framework for the beam remains rigid. Also, compute the maximum deflection of the beam. Est = 200 GPa,sY = 250 MPa.
v ⫽ 0.5 m/s
k =
1m B
10(103)(9.81)(23) PL3 = 0.613125(10 - 3) m = 1 48EI 48(200)(104)(12 )(0.2)(0.23)
10(103)(9.81) W = 160(106) N>m = ¢ st 0.613125(10 - 3)
¢ max =
0.613125(10 - 3)(0.52) ¢ st v2 = = 3.953(10 - 3) m = 3.95 mm C g C 9.81
Ans.
W¿ = k¢ max = 160(106)(3.953)(10 - 3) = 632455.53 N M¿ =
632455.53(2) w¿L = = 316228 N # m 4 4
smax =
316228(0.1) M¿c = 237 MPa 6 sg = 1 3 I 12 (0.2)(0.2 )
A 1m
From Appendix C: ¢ st =
200 mm 200 mm
O.K.
Ans.
1197
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14–50. The aluminum bar assembly is made from two segments having diameters of 40 mm and 20 mm. Determine the maximum axial stress developed in the bar if the 10-kg collar is dropped from a height of h = 150 mm. Take Eal = 70 GPa, sY = 410 MPa.
C
1.2 m 40 mm B 0.6 m
kAB =
kBC =
AAB E = LAB
p (0.012) C 70(109) D 0.6
= 11.667(106) p N>m
A
p (0.022) C 70(109) D ABC E = = 23.333 (106) p N>m LBC 1.2
Equilibrium requires FAB = FBC kAB ¢ AB = kBC ¢ BC 11.667(106) p ¢ AB = 23.333(106) p ¢ BC ¢ BC = 0.5 ¢ AB
(1)
Ue = Ui mg (h + ¢ AB + ¢ BC) =
1 1 k ¢ 2 + kBC ¢ 2BC 2 AB AB 2
(2)
Substitute Eq. (1) into (2), mg (h + ¢ AB + 0.5 ¢ AB) =
mg (h + 1.5¢ AB) =
1 1 k ¢ 2 + kBC (0.5¢ AB)2 2 AB AB 2 1 k ¢ 2 + 0.125 kBC ¢ 2AB 2 AB AB
10(9.81)(0.15 + 1.5¢ AB) =
1 C 11.667(106)p D ¢ 2AB + 0.125 C 23.333(106)p D ¢ 2AB 2
27.4889 (106)¢ 2AB - 147.15 ¢ AB - 14.715 = 0 ¢ AB = 0.7343(10 - 3) m The force developed in segment AB C 11.667(106)p D C 0.7343(10 - 3) D = 26.915(103) N. Thus smax = sAB =
h
20 mm
is
FAB = kAB ¢ AB =
26.915(103) FAB = 85.67(106)Pa = 85.7 MPa = AAB p (0.012)
Since smax 6 sy = 410 MPa, this result is valid
1198
Ans.
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14–51. The aluminum bar assembly is made from two segments having diameters of 40 mm and 20 mm. Determine the maximum height h from which the 60-kg collar can be dropped so that it will not cause the bar to yield. Take Eal = 70 GPa, sY = 410 MPa.
C
1.2 m 40 mm B 0.6 m
kAB =
kBC =
p(0.012) C 70(109) D AAB E = = 11.667(106) p N>m LAB 0.6
A
p(0.022) C 70(109) D ABC E = = 23.333(106) p N>m LBC 1.2
Here, FAB = kAB ¢ AB = C 11.667(106)p D ¢ AB. It is required that smax = sAB = sy. sy =
FAB ; AAB
410(106) =
C 11.667(106)p D ¢ AB p(0.012)
¢ AB = 0.003514 m Equilibrium requires that FAB = FBC kAB ¢ AB = kBC ¢ BC 11.6667(106)p ¢ AB = 23.333(106)p ¢ BC ¢ BC = 0.5 ¢ B = 0.5(0.003514) = 0.001757 m Ue = Ui mg(h + ¢ AB + ¢ BC) =
1 1 k ¢ 2 + kBC ¢ 2BC 2 AB AB 2
60(9.81)(h + 0.003514 + 0.001757) =
1 C 11.667(106)p D (0.0035142) 2 +
h
20 mm
The equivalent spring constants for segment AB and BC are
1 C 23.333(106)p D (0.0017572) 2
h = 0.571 m
Ans.
1199
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*14–52. The 50-lb weight is falling at 3 ft>s at the instant it is 2 ft above the spring and post assembly. Determine the maximum stress in the post if the spring has a stiffness of k = 200 kip>in. The post has a diameter of 3 in. and a modulus of elasticity of E = 6.8011032 ksi. Assume the material will not yield.
3 ft/s
2 ft
k
2 ft
Equilibrium: This requires Fsp = FP. Hence ksp ¢ sp = kP ¢ P
and
¢ sp = =
P
ksp
[1]
¢P
Conservation of Energy: The equivalent spring constant for the post is kp =
AE = L
p 4
(32) C 6.80(103) D 2(12)
= 2.003 A 106 B lb>in. Ue = Ui
1 1 1 my2 + W(h + ¢ max) = kP ¢ 2P + ksp ¢ 2sp 2 2 2
[2]
However, ¢ max = ¢ P + ¢ sp. Then, Eq. [2] becomes 1 1 1 my2 + W A h + ¢ P + ¢ sp B = kP ¢ 2P + ksp ¢ 2sp 2 2 2
[3]
Substituting Eq. [1] into [3] yields kp 1 1 1 k2P 2 my2 + W ¢ h + ¢ P + ¢ P ≤ = kP¢ 2P + ¢ ¢ ≤ 2 ksp 2 2 ksp P 2.003(106) 1 50 ¢P R ¢ ≤ A 32 B (12) + 50 B 24 + ¢ p + 2 32.2 200(103) =
1 [2.003(106)]2 1 ≤ ¢ 2P C 2.003 A 106 B D ¢ 2P + ¢ 2 2 200(103)
11.029 A 106 B ¢ 2P - 550.69¢ P - 1283.85 = 0 Solving for positive root, we have ¢ P = 0.010814 in. Maximum Stress: The maximum axial force for the post is Pmax = kp ¢ p = 2.003 A 106 B (0.010814) = 21.658 kip. smax =
Pmax 21.658 = p 2 = 3.06 ksi A 4 (3 )
Ans.
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The 50-kg block is dropped from h = 600 mm onto the bronze C86100 tube. Determine the minimum length L the tube can have without causing the tube to yield.
•14–53.
A 30 mm 20 mm
Maximum Stress. A = p A 0.03 - 0.02 2
h ⫽ 600 mm
B = 0.5 A 10 B p
2
-3
¢ st =
50(9.81)L WL = 3.0317 A 10 - 6 B L = AE C 0.5 A 10 - 3 B p D C 103 A 109 B D
sst =
50(9.81) W = = 0.3123 MPa A 0.5 A 10 - 3 B p
Section a – a
a
a L
B
Using these results, n = 1 +
C
1 + 2a
h 0.6 395 821.46 b = 1 + 1 + 2B R = 1 + 1 + -6 ¢ st C C L 3.0317 A 10 B L
Then, smax = sY = nsst 345 = ¢ 1 +
A
1 +
395 821.46 ≤ (0.3123) L
L = 0.3248 m = 325 mm
Ans.
14–54. The 50-kg block is dropped from h = 600 mm onto the bronze C86100 tube. If L = 900 mm, determine the maximum normal stress developed in the tube.
A 30 mm 20 mm
Maximum Stress. A = p A 0.03 - 0.02 2
2
B = 0.5 A 10 B p
¢ st =
50(9.81)(0.9) WL = = 2.7285 A 10 - 6 B AE C 0.5 A 10 - 3 B p D C 103 A 109 B D
sst =
50(9.81) W = = 0.3123 MPa A 0.5 A 10 - 3 B p
Section a – a
C
1 + 2a
B
h 0.6 b = 1 + 1 + 2B R = 664.18 ¢ st C 2.7285 A 10 - 6 B
Thus, smax = nsst = 664.18(0.3123) = 207.40 MPa = 207 MPa Since smax 6 sY = 345 MPa, this result is valid.
1201
a
a L
Using these results, n = 1 +
h ⫽ 600 mm
-3
Ans.
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14–55. The steel chisel has a diameter of 0.5 in. and a length of 10 in. It is struck by a hammer that weighs 3 lb, and at the instant of impact it is moving at 12 ft兾s. Determine the maximum compressive stress in the chisel, assuming that 80% of the impacting energy goes into the chisel. Est = 2911032 ksi, sY = 100 ksi. k =
AE = L
p 4
(0.52)(29)(103) 10
= 569.41 kip>in.
10 in.
0.8 Ue = Ui 0.8 c
3 1 1 a b((12)(12))2 + 3¢ max d = (569.41)(103)¢ 2max 2 (32.2)(12) 2
¢ max = 0.015044 in. P = k¢ max = 569.41(0.015044) = 8.566 kip smax =
Pmax 8.566 = 43.6 ksi 6 sg = p 2 A 4 (0.5)
O.K.
Ans.
*14–56. The sack of cement has a weight of 90 lb. If it is dropped from rest at a height of h = 4 ft onto the center of the W10 * 39 structural steel A-36 beam, determine the maximum bending stress developed in the beam due to the impact. Also, what is the impact factor?
h
12 ft
Impact Factor: From the table listed in Appendix C, ¢ st =
90[24(12)]3 PL3 = 7.3898 A 10 - 3 B in. = 48EI 48[29.0(106)](209) n = 1 +
= 1 +
C
1 + 2a
C
1 + 2a
h b ¢ st 4(12) 7.3898(10 - 3)
b
= 114.98 = 115
Ans.
Maximum Bending Stress: The maximum moment occurs at mid-span where 90(24)(12) PL = = 6480 lb # in. Mmax = 4 4 sst =
6480(9.92>2) Mmax c = = 153.78 psi I 209
Thus, smax = nsst = 114.98(153.78) = 17.7 ksi
Ans.
Since smax 6 sg = 36 ksi, the above analysis is valid.
1202
12 ft
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•14–57.
The sack of cement has a weight of 90 lb. Determine the maximum height h from which it can be dropped from rest onto the center of the W10 * 39 structural steel A-36 beam so that the maximum bending stress due to impact does not exceed 30 ksi.
h
12 ft
Maximum Bending Stress: The maximum moment occurs at mid-span where 90(24)(12) PL = = 6480 lb # in. Mmax = 4 4 sst =
6480(9.92>2) Mmax c = = 153.78 psi I 209
However, smax = nsst 30 A 103 B = n(153.78) n = 195.08 Impact Factor: From the table listed in Appendix C, ¢ st =
90[24(12)]3 PL3 = 7.3898 A 10 - 3 B in. = 48EI 48[29.0(106)](209) n = 1 +
C
1 + 2a
h b ¢ st
195.08 = 1 +
C
1 + 2a
h b 7.3898(10 - 3)
h = 139.17 in. = 11.6 ft
Ans.
1203
12 ft
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14–58. The tugboat has a weight of 120 000 lb and is traveling forward at 2 ft兾s when it strikes the 12-in.-diameter fender post AB used to protect a bridge pier. If the post is made from treated white spruce and is assumed fixed at the river bed, determine the maximum horizontal distance the top of the post will move due to the impact. Assume the tugboat is rigid and neglect the effect of the water.
3 ft
A C
12 ft
B
From Appendix C: Pmax =
3EI(¢ C)max (LBC)3
Conservation of energy: 1 1 mv2 = Pmax (¢ C)max 2 2 1 1 3EI(¢ C)2max mv2 = a b 2 2 (LBC)3 (¢ C)max =
(¢ C)max =
Pmax =
uC =
mv2L3BC C 3EI (120 000>32.2)(2)2(12)3
C (3)(1.40)(106)(144)(p4 )(0.5)4
3[1.40(106)](p4 )(6)4(11.177) (144)3
= 0.9315 ft = 11.177 in.
= 16.00 kip
16.00(103)(144)2 PmaxL2BC = = 0.11644 rad 2EI 2(1.40)(106)(p4 )(6)4
(¢ A)max = (¢ C)max + uC(LCA) (¢ A)max = 11.177 + 0.11644(36) = 15.4 in.
Ans.
1204
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14–59. The wide-flange beam has a length of 2L, a depth 2c, and a constant EI. Determine the maximum height h at which a weight W can be dropped on its end without exceeding a maximum elastic stress s max in the beam.
W
h
A
2c B L
L
1 1 (-Px)2 dx P¢ C = 2a b 2 2EI L0 ¢C =
2PL3 3EI
¢ st =
2WL3 3EI
n = 1 +
C
1 + 2a
h b ¢ st
smax = n(sst)max smax = B 1 + a
C
1 + 2a
(sst)max =
WLc I
h WLc b R ¢ st I
2 smax I 2h - 1b = 1 + WLc ¢ st
h =
=
2 ¢ st smax I - 1b - 1R Ba 2 WLc
smax I 2 2smaxI smax L2 smax I WL3 b - 2R Ba R = B 3EI WLc WLc 3Ec WLc
Ans.
1205
L
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*14–60. The 50-kg block C is dropped from h = 1.5 m onto the simply supported beam. If the beam is an A-36 steel W250 * 45 wide-flange section, determine the maximum bending stress developed in the beam.
C h
Equilibrium. Referring to the free-body diagram of the beam under static A condition, Fig. a a + ©MA = 0;
By(6) - P(4) = 0
By =
2 P 3
4m
Then, the maximum moment in the beam occurs at the position where P is applied. Thus, Mmax = By(2) =
2 4 P(2) = P 3 3
Impact Factor. From the table listed in the appendix, the deflection of the beam at Pba the point of application of P is ¢ = A L2 - b2 - a2 B , where P = 50(9.81) 6EIL = 490.5 N, L = 6 m, a = 4 m, and b = 2 m. From the table listed in the appendix, the necessary section properties for a W250 * 45 are d = 266 mm = 0.266 m and Ix = 71.1 A 106 B mm4 = 71.1 A 10 - 6 B m4. Then 490.5(2)(4) ¢ st =
6c 200 A 10 B d c 71.1 A 10 9
-6
B d(6)
A 62 - 22 - 42 B = 0.1226 A 10 - 3 B m
We have,
n = 1 +
Maximum
C
1 + 2¢
Stress.
h 1.5 1 + 2C S = 157.40 ≤ = 1 + ¢ st Q 0.1226 A 10 - 3 B
The
maximum
= 157.40(490.5) = 77.21 A 10 B N. 3
= 102.94 A 10
smax =
3
force Then,
on
B N # m. Applying the flexure formula,
beam is Pmax = nP 4 4 = Pmax = C 77.21 A 103 B D 3 3
the
Mmax
102.94 A 103 B (0.266>2) Mmaxc = = 192.56 MPa = 193 MPa I 71.1 A 10 - 6 B
Since smax 6 sY = 250 MPa, this result is valid.
1206
B
Ans.
2m
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•14–61.
Determine the maximum height h from which the 50-kg block C can be dropped without causing yielding in the A-36 steel W310 * 39 wide flange section when the block strikes the beam.
C h A
B 4m
Equilibrium. Referring to the free-body diagram of the beam under static condition, Fig. a
a + ¢MA = 0;
By(6) - P(4) = 0
By =
2 P 3
Then, the maximum moment in the beam occurs at the position where P is applied. Thus, Mmax = By(2) =
4 2 P(2) = P 3 3
Maximum Stress. Since P = 50(9.81) = 490.5 N. Then the maximum force on the 4 4 beam is Pmax = nP = 490.5n and Mmax = P = (490.5n) = 654n. From the 3 3 table listed in the appendix, the necessary section properties for a W310 * 39 are d = 310 mm = 0.31 m and Ix = 84.8 A 106 B mm4 = 84.8 A 10 - 6 B m4. Applying the
flexure formula,
smax =
Mmax c I
250 A 106 B =
654n(0.31>2) 84.8 A 10 - 6 B
n = 209.13 Impact Factor. From the table listed in the appendix, the deflection of the beam at Pba the point of where P is applied is ¢ = A L2 - b2 - a2 B , where L = 6 m, 6EIL a = 4 m, and b = 2 m. Then 490.5(2)(4)
¢ st =
6 C 200 A 109 B D C 84.8 A 10 - 6 B D (6)
A 62 - 22 - 42 B = 0.1028 A 10 - 3 B m
We have, n = 1 +
C
1 + 2¢
209.13 = 1 +
S
h ≤ ¢ st
1 + 2C
h
0.1028 A 10 - 3 B
S
h = 2.227 m = 2.23 m
Ans.
Since smax 6 sY = 250 MPa, this result is valid.
1207
2m
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14–62. The diver weighs 150 lb and, while holding himself rigid, strikes the end of a wooden diving board 1h = 02 with a downward velocity of 4 ft>s. Determine the maximum bending stress developed in the board. The board has a thickness of 1.5 in. and width of 1.5 ft. Ew = 1.811032 ksi, sY = 8 ksi.
v
h 4 ft
Static Displacement: The static displacement at the end of the diving board can be determined using the conservation of energy. L
M2 dx 1 P¢ = 2 L0 2EI 1 1 (150)¢ st = c 2 2EI L0 ¢ st =
4 ft
10 ft
(-375x1)2 dx1 +
L0
(-150x2) dx2 d
70.0(103) lb # ft3 EI 70.0(103)(123)
=
1 1.8(106) C 12 (18)(1.53) D
= 13.274 in. Conservation of Energy: The equivalent spring constant for the board is W 150 k = = = 11.30 lb>in., ¢ st 13.274 Ue = Ui 1 1 my2 + W¢ max = k¢ 2max 2 2 c
1 150 1 a b A 42 B d (12) + 150¢ max = (11.30)¢ 2max 2 32.2 2
Solving for the positive root, we have ¢ max = 29.2538 in. Maximum Stress: The maximum force on to the beam is Pmax = k¢ max = 11.30(29.2538) = 330.57 lb. The maximum moment occurs at the middle support Mmax = 330.57(10)(12) = 39668.90 lb # in. smax =
39668.90(0.75) Mmax c = 5877 psi = 5.88 ksi = 1 3 I 12 (18)(1.5 )
Ans.
Note: The result will be somewhat inaccurate since the static displacement is so large.
1208
10 ft
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14–63. The diver weighs 150 lb and, while holding himself rigid, strikes the end of the wooden diving board. Determine the maximum height h from which he can jump onto the board so that the maximum bending stress in the wood does not exceed 6 ksi. The board has a thickness of 1.5 in. and width of 1.5 ft. Ew = 1.811032 ksi.
v
Static Displacement: The static displacement at the end of the diving board can be determined using the conservation of energy.
h
L
M2 dx 1 P¢ = 2 L0 2EI 1 1 (150)¢ st = c 2 2EI L0 ¢ st =
4 ft
4 ft
10 ft
(-375x1)2 dx1 +
L0
(-150x2) dx2 d
70.0(103) lb # ft3 EI 70.0(103)(123)
=
1 1.8(106) C 12 (18)(1.53) D
= 13.274 in. Maximum Stress: The maximum force on the beam is Pmax. The maximum moment occurs at the middle support Mmax = Pmax (10)(12) = 120Pmax. smax = 6 A 103 B =
Mmax c I 120Pmax (0.75) 1 12
(18)(1.53)
Pmax = 337.5 lb Conservation of Energy: The equivalent spring constant for the board is 150 W = 11.30 lb>in.. The maximum displacement at the end of the = k = ¢ st 13.274 Pmax 337.5 board is ¢ max = = = 29.687 in. k 11.30 Ue = Ui W(h + ¢ max) =
150(h + 29.867) =
1 k¢ 2max 2
1 (11.30) A 29.8672 B 2
h = 3.73 in.
Ans.
Note: The result will be somewhat at inaccurate since the static displacement is so large.
1209
10 ft
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*14–64. The weight of 175 lb is dropped from a height of 4 ft from the top of the A-36 steel beam. Determine the maximum deflection and maximum stress in the beam if the supporting springs at A and B each have a stiffness of k = 500 lb>in. The beam is 3 in. thick and 4 in. wide.
4 ft A
k 8 ft
8 ft
From Appendix C: ¢ beam =
PL3 48EI
kbeam =
1 )(4)(33) 48(29)(103)(12 48EI = = 1.7700 kip>in. 3 L (16(12))3
From equilibrium (equivalent system): 2Fsp = Fbeam 2ksp ¢ sp = kbeam ¢ beam ¢ sp =
1.7700(103) ¢ beam 2(500)
¢ sp = 1.7700¢ beam
(1)
Conservation of energy: Ue = Ui W(h + ¢ sp + ¢ beam) =
1 1 k ¢2 + 2a bksp ¢ 2sp 2 beam beam 2
From Eq. (1): 175[(4)(12) + 1.770¢ beam + ¢ beam] =
1 (1.7700)(103)¢ 2beam + 500(1.7700¢ beam)2 2
2451.5¢ 2beam - 484.75¢ beam - 8400 = 0 ¢ beam = 1.9526 in. From Eq. (1): ¢ sp = 3.4561 in. ¢ max = ¢ sp + ¢ beam = 3.4561 + 1.9526 = 5.41 in.
Ans.
Fbeam = kbeam ¢ beam = 1.7700(1.9526) = 3.4561 kip Mmax =
smax =
3.4561(16)(12) Fbeam L = = 165.893 kip # in. 4 4 165.893(1.5) Mmax c = 27.6 ksi 6 sg = 1 3 I 12 (4)(3 )
O.K.
Ans.
1210
3 in.
B k
4 in.
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•14–65.
The weight of 175 lb, is dropped from a height of 4 ft from the top of the A-36 steel beam. Determine the load factor n if the supporting springs at A and B each have a stiffness of k = 300 lb/in. The beam is 3 in. thick and 4 in. wide.
4 ft A
k 8 ft
8 ft
From Appendix C: ¢ beam =
PL3 48EI
kbeam =
1 )(4)(33) 48(29)(103)(12 48EI = = 1.7700 kip>in. L3 (16(12))3
From equilibrium (equivalent system): 2Fsp = Fbeam 2ksp ¢ sp = kbeam ¢ beam ¢ sp =
1.7700(103) ¢ beam 2(300) (1)
¢ sp = 2.95¢ beam Conservation of energy: Ue = Ui W(h + ¢ beam + ¢ sp) =
1 1 k ¢2 + 2a bksp ¢ 2sp 2 beam beam 2
From Eq. (1): 175[(4)(12) + ¢ beam + 2.95¢ beam] =
1 (1.7700)(103)¢ 2beam + 300(2.95¢ beam)2 2
3495.75¢ 2beam - 691.25¢ beam - 8400 = 0 ¢ beam = 1.6521 in. Fbeam = kbeam ¢ beam = 1.7700(1.6521) = 2.924 kip n =
2.924(103) = 16.7 175
smax = n(sst)max = na
M =
Ans. Mc b I
175(16)(12) = 8.40 kip # in. 4
smax = 16.7 a
8.40(1.5) 1 12
(4)(33)
b = 23.4 ksi 6 sg
O.K.
1211
3 in.
B k
4 in.
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14–66. Block C of mass 50 kg is dropped from height h = 0.9 m onto the spring of stiffness k = 150 kN>m mounted on the end B of the 6061-T6 aluminum cantilever beam. Determine the maximum bending stress developed in the beam.
C h a
k
100 mm B
Conservation of Energy. From the table listed in the appendix, the
A
a
200 mm
3
PL . Thus, the = 3EI
displacement of end B under static conditions is ¢ st
equivalent spring constant for the beam is kb = I =
1 (0.1) A 0.23 B = 66.6667 A 10 - 6 B m4, 12
L = 3 m,
3m
3EI , where L3
and
E = Eal
= 68.9 GPa. Thus,
3EI kb = = L3
3 c68.9 A 109 B d c66.6667 A 10 - 6 B d 33
= 510.37 A 103 B N>m
Equilibrium requires, Fsp = P ksp ¢ sp = kb ¢ b 150 A 103 B ¢ sp = 510.37 A 103 B ¢ b ¢ sp = 3.4025¢ b
(1)
We have, Ue = Ui mg A h + ¢ sp + ¢ b B =
1 1 k ¢ 2 + ksp ¢ sp 2 2 b b 2
Substituting Eq. (1) into this equation, 50(9.81)(0.9 + 3.4025¢ b + ¢ b) =
1 1 c510.37 A 103 B d ¢ b 2 + c150 A 103 B d(3.4025¢ b)2 2 2
1123444.90¢ b 2 - 2159.41¢ b - 441.45 = 0 Solving for the positive root, ¢ b = 0.020807 m Maximum
Stress.
The
maximum
force
Pmax = kb ¢ b = 510.37 A 10 B (0.020807) = 10.619 A 10 3
occurs
at
= 31.858 A 103 B smax =
3
on
the
beam
is
B N. The maximum moment
Mmax = Pmax L = 10.619 A 103 B (3) 0.2 N # m. Applying the flexure formula with c = = 0.1 m, 2
fixed
support
A,
where
31.858 A 103 B (0.1) Mmax c = = 47.79 MPa = 47.8 MPa I 66.6667 A 10 - 6 B
Since smax 6 sY = 255 MPa, this result is valid. 1212
Ans.
Section a – a
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14–67. Determine the maximum height h from which 200-kg block C can be dropped without causing the 6061-T6 aluminum cantilever beam to yield. The spring mounted on the end B of the beam has a stiffness of k = 150 kN>m.
C h a
k
100 mm B
A
a 3m
Maximum Stress. From the table listed in the appendix, the displacement of end B PL3 under static conditions is ¢ st = . Thus, the equivalent spring constant for the 3EI 3EI 1 beam is kb = , where I = (0.1) A 0.23 B = 66.6667 A 10 - 6 B m4, L = 3 m, and 12 L3 E = Eal = 68.9 GPa. Thus, 3EI kb = = L3
3 c68.9 A 109 B d c66.6667 A 10 - 6 B d 33
= 510.37 A 103 B N>m
The maximum force on the beam is Pmax = kb ¢ b = 510.37 A 103 B ¢ b. The maximum moment occurs at the fixed support A, where Mmax = Pmax L = 510.37 A 103 B ¢ b(3)
= 1.5311 A 106 B ¢ b. Applying the flexure formula with smax = sY = 255 MPa and 0.2 = 0.1 m, c = 2 Mmax c I
smax = sY = 255 A 106 B =
1.5311 A 106 B ¢ b(0.1) 66.6667 A 10 - 6 B
¢ b = 0.11103 m Equilibrium requires, Fsp = P ksp ¢ sp = kb ¢ b 150 A 103 B ¢ sp = 510.37 A 103 B (0.11103) ¢ sp = 0.37778 m Conservation of Energy. Ue = Ui mg A h + ¢ sp + ¢ b B =
1 1 kb ¢ b 2 + ksp ¢ sp 2 2 2
200(9.81)(h + 0.37778 + 0.11103) =
1 1 c510.37 A 103 B d(0.11103)2 + c150 A 103 B d(0.37778)2 2 2
h = 6.57 m
Ans.
1213
200 mm Section a – a
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*14–68. The 2014-T6 aluminum bar AB can slide freely along the guides mounted on the rigid crash barrier. If the railcar of mass 10 Mg is traveling with a speed of v = 1.5 m>s, determine the maximum bending stress developed in the bar. The springs at A and B have a stiffness of k = 15 MN>m.
300 mm k
A
v
2m
2m a
a
k
B
Equilibrium. Referring to the free-body diagram of the bar for static conditions, Fig. a, + ©F = 0; : x
2Fsp - P = 0
Fsp =
P 2
(1)
300 mm 400 mm
Referring to the table listed in the appendix, the displacement of the bar at the PL3 position where P is applied under static conditions is ¢ st = . Thus, the 48EI 48EI 1 equivalent spring constant for the bar is kb = , where I = (0.4) A 0.33 B 12 L3 and Thus, = 0.9 A 10 - 3 B m4, L = 4 m, E = Eal = 73.1 GPa. kb =
48 c73.1 A 109 B d c0.9 A 10 - 3 B d 43
= 49.3425 A 106 B N>m
Using Eq. (1) Fsp =
P 2
ksp ¢ sp =
1 k ¢ 2 b b
1 kb 1 49.3425 A 10 ¢ ≤ ¢b = C 2 ksp 2 15 A 106 B
6
¢ sp =
B
S ¢ b = 1.64475¢ b
Conservation of Energy. 1 1 1 mv2 = kb ¢ b 2 + 2c ksp ¢ sp 2 d 2 2 2 Substituting Eq. (2) into this equation, 1 1 mv2 = kb ¢ b 2 + ksp (1.64475¢ b)2 2 2 1 1 mv2 = kb ¢ b 2 + 2.7052ksp ¢ b 2 2 2 1 1 c10 A 103 B d A 1.52 B = c49.3425 A 106 B d ¢ b 2 + 2.7052c15 A 106 B d ¢ b 2 2 2 ¢ b = 0.01313 m
1214
(2)
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14–68.
Continued
Maximum Stress. The maximum force on the bar is (Pb)max = kb ¢ b = 49.3425 A 106 B (0.01313) = 647.90 A 103 B N. The maximum moment occurs at the midspan of the bar, where Mmax = = 647.90 A 103 B N # m. Applying the flexure formula, smax =
647.90 A 103 B (4) (Pb)max L = 4 4
647.90 A 103 B (0.15) Mmax c = = 107.98 MPa = 108 MPa I 0.9 A 10 - 3 B
Since smax 6 sY = 414 MPa, this result is valid.
1215
Ans.
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•14–69.
The 2014-T6 aluminum bar AB can slide freely along the guides mounted on the rigid crash barrier. Determine the maximum speed v the 10-Mg railcar without causing the bar to yield when it is struck by the railcar. The springs at A and B have a stiffness of k = 15 MN>m.
300 mm k
A
v
2m
2m a
a
k
B
Equilibrium. Referring to the free-body diagram of the bar for static conditions, Fig. a, + ©F = 0; : x
2Fsp - P = 0
Fsp
P = 2
(1)
300 mm 400 mm
Referring to the table listed in the appendix, the displacement of the bar at the PL3 position where P is applied under static conditions is ¢ st = . Thus, the 48EI 48EI 1 equivalent spring constant for the bar is kb = , where I = (0.4) A 0.33 B 12 L3 and Thus, = 0.9 A 10 - 3 B m4, L = 4 m, E = Eal = 73.1 GPa.
kb =
48c73.1 A 109 B d c 0.9 A 10 - 3 B d 43
= 49.3425 A 106 B N>m
Using Eq. (1) Fsp =
P 2
ksp ¢ sp =
1 k ¢ 2 b b
1 kb 1 49.3425 A 10 ¢ ≤ ¢b = C 2 ksp 2 15 A 106 B
6
B
S ¢ b = 1.64475¢ b
(2)
Maximum Stress. The maximum force on the bar is
(Pb)max = kb ¢ b
¢ sp =
= 49.3425 A 106 B ¢ b. The maximum moment occurs at the midspan of the bar, where 49.3425 A 106 B ¢ b(4) (Pb)max L = = 49.3425 A 106 B ¢ b. Applying the flexure Mmax = 4 4 formula with smax = sY = 414 MPa,
smax =
Mmax c I
414 A 106 B =
49.3425 A 106 B ¢ b (0.15) 0.9 A 10 - 3 B
¢ b = 0.050342 m
1216
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14–69. Continued
Substituting this result into Eq. (2), ¢ sp = 0.0828 m Conservation of Energy. 1 1 1 mv2 = kb ¢ b 2 + 2 B ksp ¢ sp 2 R 2 2 2 1 1 1 c 10 A 103 B dv2 = c49.3425 A 106 B d A 0.0503422 B + 2 B c15 A 106 B d A 0.08282 B R 2 2 2 v = 5.75 m>s
Ans.
1217
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14–70. The simply supported W10 * 15 structural A-36 steel beam lies in the horizontal plane and acts as a shock absorber for the 500-lb block which is traveling toward it at 5 ft兾s. Determine the maximum deflection of the beam and the maximum stress in the beam during the impact. The spring has a stiffness of k = 1000 lb>in. For W 10 * 15: I = 68.9 in4
12 ft v ⫽ 5 ft/s
d = 9.99 in. k
From Appendix C: ¢ beam =
PL3 48EI
kbeam =
48(29)(103)(68.9) 48EI = = 4.015 kip>in. L3 (24(12))3
12 ft
Equilibrium (equivalent system): Fsp = Fbeam ksp ¢ sp = kbeam ¢ beam ¢ sp =
4.015(103) ¢ beam 1000
¢ sp = 4.015¢ beam
(1)
Conservation of energy: Ue = Ui 1 1 1 mv2 = kbeam ¢ 2beam + ksp ¢ 2sp 2 2 2 From Eq. (1): 500 1 1 1 a b (5(12))2 = (4.015)(103)¢ 2beam + (1000)(4.015¢ beam)2 2 32.2(12) 2 2 10067.6¢ 2beam = 2329.2 ¢ beam = 0.481 in.
Ans.
Fbeam = kbeam ¢ beam = 4.015(0.481) = 1.931 kip Mmax = a smax =
1.931 b (12) (12) = 139.05 kip # in. 2
139.05(9.99 Mmax c 2 ) = = 10.1 ksi 6 sg I 68.9
O.K.
Ans.
1218
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14–71. The car bumper is made of polycarbonatepolybutylene terephthalate. If E = 2.0 GPa, determine the maximum deflection and maximum stress in the bumper if it strikes the rigid post when the car is coasting at v = 0.75 m/s. The car has a mass of 1.80 Mg, and the bumper can be considered simply supported on two spring supports connected to the rigid frame of the car. For the bumper take I = 30011062 mm4, c = 75 mm, sY = 30 MPa and k = 1.5 MN>m.
0.9 m
k
0.9 m
k
v ⫽ 0.75 m/s
Equilibrium: This requires Fsp =
ksp ¢ sp =
k¢ beam 2
Pbeam . Then 2 or
¢ sp =
k ¢ 2ksp beam
[1]
Conservation of Energy: The equivalent spring constant for the beam can be determined using the deflection table listed in the Appendix C. k =
48 C 2(109) D C 300(10 - 6) D 48EI = = 4 938 271.6 N>m L3 1.83
Thus, Ue = Ui 1 1 1 my2 = k¢ 2beam + 2a ksp ¢ 2sp b 2 2 2
[2]
Substitute Eq. [1] into [2] yields 1 1 k2 mv2 = k¢ 2beam + ¢2 2 2 4ksp beam (4 93 8271.6)2 2 1 1 ¢ beam (1800) A 0.752 B = (493 8271.6) ¢ 2beam + 2 2 4[1.5(106)] ¢ beam = 8.8025 A 10 - 3 B m Maximum Displacement: From Eq. [1] ¢ sp =
4 938 271.6 2[1.5(106)]
C 8.8025 A 10 - 3 B D =
0.014490 m. ¢ max = ¢ sp + ¢ beam = 0.014490 + 8.8025 A 10 - 3 B = 0.02329 m = 23.3 mm
Ans.
Maximum Stress: The maximum force on the beam is Pbeam = k¢ beam = 4 938 271.6 C 8.8025 A 10 - 3 B D = 43 469.3 N. The maximum moment
occurs at mid-span. Mmax =
smax =
43 469.3(1.8) Pbeam L = = 19 561.2 N # m. 4 4
19 561.2(0.075) Mmax c = 4.89 MPa = I 300(10 - 6)
Ans.
Since smax 6 sg = 30 MPa, the above analysis is valid.
1219
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*14–72. Determine the horizontal displacement of joint B on the two-member frame. Each A-36 steel member has a cross-sectional area of 2 in2.
800 lb B
30⬚
60⬚
A
Member
n
N
L
nNL
AB
1.1547
800
120
11085.25
BC
–0.5774
0
60
0
5 ft
C
© = 110851.25 1 # ¢ Bh = ©
¢ Bh =
nNL AE
110851.25 110851.25 = 0.00191 in. = AE 29(106)(2)
Ans.
•14–73. Determine the horizontal displacement of point B. Each A-36 steel member has a cross-sectional area of 2 in2.
B
200 lb
Member Real Forces N: As shown on figure(a). 8 ft
Member Virtual Forces n: As shown on figure(b).
6 ft
nNL 1#¢ = a AE 1 lb # (¢ B)h =
1 [0.8333(166.67)(10)(12) AE +( -0.8333)(-166.67)(10)(12) +0.500(100)(12)(12)]
1 lb # (¢ B)h =
(¢ B)h =
40533.33 lb2 # in AE 40533.33 2[29.0(106)]
A
C
Virtual-Work Equation:
= 0.699 A 10 - 3 B in. :
Ans.
1220
6 ft
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14–74. Determine the vertical displacement of point B. Each A-36 steel member has a cross-sectional area of 2 in2.
B
200 lb
Member Real Forces N: As shown on figure(a). 8 ft
Member Virtual Forces n: As shown on figure(b). A
C
Virtual-Work Equation: nNL 1#¢ = a AE 1 lb # (¢ B)v =
6 ft
6 ft
1 [(-0.625)(166.67)(10)(12) AE +(-0.625)(-166.67)(10)(12) +0.375(100)(12)(12)]
1 lb # (¢ B)v =
(¢ B)v =
5400 lb2 # in AE 5400 2[29.0(106)]
= 0.0931 A 10 - 3 B in. T
Ans.
14–75. Determine the vertical displacement of joint C on the truss. Each A-36 steel member has a cross-sectional area of A = 300 mm2. Member
n
N
L
20 kN 3m
C A
1.50
45.0
3
202.5
AD
0
18.03
213
0
BC
1.50
45.0
3
202.5
BD
0
–20.0
2
0
CD
–1.803
–54.08
213
351.56
DE
–1.803
–72.11
213
468.77
4m D
E
© = 1225.33 nNL AE 1225.33(103) ¢ Cv =
300(10 - 6)(200)(109)
3m B
nNL
AB
1 # ¢ Cv = ©
30 kN
Ans.
= 0.0204 m = 20.4 mm
1221
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*14–76. Determine the vertical displacement of joint D on the truss. Each A-36 steel member has a cross-sectional area of A = 300 mm2.
20 kN
30 kN
3m
3m B C
A
Member
n
N
L
nNL
4m
AB
0
45.0
3
0
AD
0.9014
18.03
213
58.60
BC
0
45.0
3
0
BD
0
–20.0
2
0
CD
0
–54.08
213
0
DE
–0.9014
–72.11
213
234.36
D
E
© = 292.96 1 # ¢ Dv = ©
nNL AE 292.96(103)
¢ Dv =
300(10 - 6)(200)(109)
= 4.88(10 - 3) m = 4.88 mm
Ans.
•14–77. Determine the vertical displacement of point B. Each A-36 steel member has a cross-sectional area of 4.5 in2.
F
E
D
Virtual-Work Equation: Applying Eq. 14–39, we have Member
n
N
AB
0.6667
BC
0.6667
6 ft
L
nNL
3.333
96
213.33
3.333
96
213.33
A
C
B
CD
0
0
72
0
DE
0
0
96
0
EF
0
0
96
0
AF
0
0
72
0
AE
–0.8333
–4.167
120
416.67
CE
–0.8333
–4.167
120
416.67
BE
1.00
5.00
72
360.00
8 ft
©1620 kip2 # in. nNL 1#¢ = a AE 1 kip # (¢ B)v =
(¢ B)v =
8 ft 5 kip
1620 kip2 # in. AE 1620 = 0.0124 in. T 4.5[29.0(103)]
Ans.
1222
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14–78. Determine the vertical displacement of point E. Each A-36 steel member has a cross-sectional area of 4.5 in2.
F
E
D
6 ft
Virtual-Work Equation: Applying Eq. 14–39, we have Member
n
N
L
nNL
AB
0.6667
3.333
96
213.33
BC
0.6667
3.333
96
213.33
CD
0
0
72
0
DE
0
0
96
0
EF
0
0
96
0
AF
0
0
72
0
AE
–0.8333
–4.167
120
416.67
CE
–0.8333
–4.167
120
416.67
BE
0
5.00
72
0
A
C
B 8 ft
8 ft 5 kip
©1260 kip2 # in. nNL 1#¢ = a AE 1 kip # (¢ E)v =
(¢ E)v =
1260 kip2 # in. AE 1260 = 0.00966 in. T 4.5[29.0(103)]
Ans.
14–79. Determine the horizontal displacement of joint B of the truss. Each A-36 steel member has a cross-sectional area of 400 mm2.
5 kN 4 kN
2m C
B
Member
n
N
L
nNL
AB
0
0
1.5
0
AC
–1.25
–5.00
2.5
15.625
AD
1.00
4.00
2.0
8.000
BC
1.00
4.00
2.0
8.000
CD
0.75
–2.00
1.5
–2.25
1.5 m
D A
© = 29.375 1 # ¢ Bh = ©
nNL AE 29.375(103)
¢ Bh =
400(10 - 6)(200)(109)
= 0.3672(10 - 3)m = 0.367 mm
Ans.
1223
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*14–80. Determine the vertical displacement of joint C of the truss. Each A-36 steel member has a cross-sectional area of 400 mm2.
5 kN 2m
4 kN
C
B
Member
n
N
L
nNL
AB
0
0
1.5
0
AC
0
–5.00
2.5
0
AD
0
4.00
2.0
0
BC
0
4.00
2.0
0
CD
–1.00
–2.00
1.5
3.00
1.5 m
D A
© = 3.00 1 # ¢ Cv = ©
nNL AE 3.00 (103)
¢ Cv =
400(10 - 6)(200)(109)
= 37.5(10 - 6)m = 0.0375 mm
Ans.
•14–81. Determine the vertical displacement of point A. Each A-36 steel member has a cross-sectional area of 400 mm2.
E
D
Virtual-Work Equation: Member
n
AB
–0.750
BC
–0.750
AE
1.25
CE
–1.25
BE
0
DE
2m
N
1.50
–22.5 A 103 B –22.5 A 103 B 37.5 A 10
3
–62.5 A 10
3
60.0 A 10
3
B B
22.0 A 103 B
B
L 1.5 1.5 2.5 2.5 2 1.5
nNL
25.3125 A 103 B
C A
25.3125 A 103 B
117.1875 A 10
3
195.3125 A 10
3
135.00 A 10
3
B 1.5 m
B B
30 kN
0
B
© 498.125 A 103 B N2 # m
nNL 1#¢ = a AE 1 N # (¢ A)v =
(¢ A)v =
498.125(103) N2 # m AE 498.125(103) 0.400(10 - 3)[200(109)]
= 6.227 A 10 - 3 B m = 6.23 mm T
Ans.
1224
1.5 m 20 kN
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14–82. Determine the vertical displacement of point B. Each A-36 steel member has a cross-sectional area of 400 mm2.
E
D
2m C A B 1.5 m
Virtual-Work Equation: Member AB
n
N
0
BC
0
AE
0
CE
–1.25
BE
1.00
DE
0.750
–22.5 A 103 B –22.5 A 10
3
–62.5 A 10
3
B
37.5 A 103 B
B
22.0 A 103 B 60.0 A 103 B
L
30 kN
nNL
1.5
0
1.5
0
2.5 2.5 2 1.5
195.3125 A 10
0 3
B
40.0 A 103 B 67.5 A 103 B
© 302.8125 A 103 B N2 # m
nNL 1#¢ = a AE 1 N # (¢ B)v =
(¢ B)v =
302.8125(103) N2 # m AE 302.8125(103) 0.400(10 - 3)[200(109)]
= 3.785 A 10 - 3 B m = 3.79 mm T
Ans.
1225
1.5 m 20 kN
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14–83. Determine the vertical displacement of joint C. Each A-36 steel member has a cross-sectional area of 4.5 in2. 1 # ¢ Cv = ©
¢ Cv =
J
A
nNL AE
B 12 ft
Ans.
*14–84. Determine the vertical displacement of joint H. Each A-36 steel member has a cross-sectional area of 4.5 in2.
¢ Hv =
H
G
F 9 ft
21 232 = 0.163 in. 4.5 (29(103))
1 # ¢ Nv = ©
I
C 12 ft
12 ft
12 ft
6 kip
8 kip
6 kip
I
H
G
J
E
D
F 9 ft
A
nNL AE
12 ft
20 368 = 0.156 in. 4.5 (29(103))
Ans.
1226
C
B 12 ft 6 kip
12 ft 8 kip
E
D 12 ft 6 kip
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•14–85.
Determine the vertical displacement of joint C. The truss is made from A-36 steel bars having a cross- sectional area of 150 mm2.
G 2m H
Member Real Forces N. As indicated in Fig. a.
F 2m
Member Virtual Forces n. As indicated in Fig. b. 18 A 103 B
0.15 A 10 - 3 B
Virtual Work Equation. Since smax = Member AB
n(N)
N(N)
9 A 10
0.375
DE
0.375
BC
0.375
CD
0.375
AH
–0.625
EF
–0.625
BH
0
DF
0
CH
0
CF
0
GH
–0.625
FG
–0.625
CG
1
9 A 10
3 3
B B
9 A 103 B 9 A 103 B
–15 A 103 B –15 A 103 B 6 A 103 B 6 A 10
3
–3.75 A 10
3
B
–3.75 A 103 B
B
–11.25 A 103 B –11.25 A 103 B 18 A 103 B
nNL(N
L(m)
2
# m)
5.0625 A 10
1.5
5.0625 A 10
1.5
3 3
B B
5.0625 A 103 B
1.5
24.4375 A 103 B
2.5 2.5
24.4375 A 103 B
2
0
2
0
2.5
0
2.5
0
17.578125 A 103 B
2.5
17.578125 A 103 B
2.5
72 A 103 B
©174.28125 A 103 B
Then 1#¢ = ©
nNL AE
1N # (¢ C)v =
174.28125 A 103 B
0.15 A 10 - 3 B C 200 A 109 B D
(¢ C)v = 5.809 A 10 - 3 B m = 5.81 mm T
Ans.
1227
B 1.5 m
1.5 m 6 kN
5.0625 A 103 B
1.5
4
E
A
= 120 MPa 6 sY = 250 MPa,
C 1.5 m 12 kN
D 1.5 m 6 kN
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14–86. Determine the vertical displacement of joint G. The truss is made from A-36 steel bars having a cross-sectional area of 150 mm2.
G 2m H
Member Real Forces N. As indicated in Fig. a.
F 2m
Member Virtual Forces n. As indicated in Fig. b. E
A
Virtual
Work
Equation.
Since
18 A 10 B FCG = = 120 MPa 6 sY = 250 MPa, = A 0.15 A 10 - 3 B 3
smax
Member
n(N)
AB
0.375
DE
0.375
BC
0.375
CD
0.375
AH
–0.625
EF
–0.625
BH
0
DF
0
CH
0
CF
0
GH FG CG
–0.625 –0.625
N(N)
9 A 103 B 9 A 103 B 9 A 103 B 9 A 103 B
–15 A 103 B –15 A 103 B 6 A 103 B 6 A 103 B
–3.75 A 10
3
–11.25 A 10
3
B
–3.75 A 103 B
–11.25 A 10 18 A 10
0
3 3
B B B
5.0625 A 103 B
1.5
5.0625 A 103 B
1.5
5.0625 A 103 B
1.5
5.0625 A 103 B
1.5
24.4375 A 103 B
2.5 2.5
24.4375 A 103 B
2
0
2
0
2.5
0
2.5
17.578125 A 10
2.5
17.578125 A 10
2.5 4
0 3 3
B B
0
©102.28125 A 103 B
Then 1#¢ = ©
nNL AE
1N # (¢ G)v =
1.5 m
1.5 m 6 kN
nNL(N 2 # m)
L(m)
B
102.28125 A 103 B
0.15 A 10 - 3 B C 200 A 109 B D
(¢ G)v = 3.409 A 10 - 3 B m = 3.41 mm T
Ans.
1228
C 1.5 m 12 kN
D 1.5 m 6 kN
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14–87. Determine the displacement at point C. EI is constant.
P
P
A
B
a
a– 2
C a– 2
a
L
1 # ¢C =
mM dx L0 EI
¢C = 2 a =
a
1 1 bc a x1 b(Px1)dx1 + EI L0 2 L0
a>2
1 (a + x2)(Pa)dx2 d 2
23Pa3 24EI
Ans.
*14–88. The beam is made of southern pine for which Ep = 13 GPa. Determine the displacement at A.
15 kN 4 kN/m
A
L
1 # ¢A =
mM L0 EI
B 1.5 m 1.5
¢A =
=
C 3m
3
1 c (x1)(15x1)dx1 + (0.5x2)(2x22 + 1.5x2)dx2 d EI L0 L0
43.875(103) 43.875 kN # m3 = 0.0579 m = 57.9 mm = 1 9 EI 13(10 )(12 )(0.12)(0.18)3
1229
180 mm
Ans.
120 mm
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•14–89. Determine the displacement at C of the A-36 steel beam. I = 7011062 mm4.
2 kN/m
Real Moment Function M(x): As shown on figure(a).
C B
A
Virtual Moment Functions m(x): As shown on figure(b).
10 m
5m
Virtual Work Equation: For the displacement at point C. L
1#¢ = 1 kN # ¢ C =
¢C =
mM dx L0 EI 1 EI L0
10 m
0.500x1 (2.50x1)dx1 +
1 EI L0
5m
x2 A x22 B dx2
572.92 kN # m3 EI 572.92(1000)
=
200(109)[70(10 - 6)]
= 0.04092 m = 40.9 mm T
Ans.
14–90. Determine the slope at A of the A-36 steel beam. I = 7011062 mm4.
2 kN/m
C
Real Moment Function M(x): As shown on figure(a). B
A
Virtual Moment Functions mu (x): As shown on figure(b).
10 m
Virtual Work Equation: For the slope at point A. L
1#u = 1 kN # m # uA =
uA =
muM dx L0 EI 1 EI L0
10 m
(1 - 0.100x1)(2.50x1) dx1 +
5m
0 A 1.00x22 B dx2
41.667 kN # m2 EI 41.667(1000)
=
1 EI L0
200(109)[70(10 - 6)]
= 0.00298 rad
Ans.
1230
5m
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14–91. Determine the slope at B of the A-36 steel beam. I = 7011062 mm4.
2 kN/m
C
Real Moment Function M(x): As shown on figure(a). B
A
Virtual Moment Functions mU (x): As shown on figure(b).
10 m
5m
Virtual Work Equation: For the slope at point B. L
muM dx L0 EI
1#u = 1 kN # m # uB =
uB =
1 EI L0
10 m
0.100x1(2.50x1) dx1 +
0 A 1.00x22 B dx2
5m
83.333 kN # m2 EI 83.333(1000)
=
1 EI L0
200(109)[70(10 - 6)]
Ans.
= 0.00595 rad
*14–92. Determine the displacement at B of the 1.5-indiameter A-36 steel shaft.
2 ft 2 ft
A
3 ft
D B
L
1 # ¢B =
mM dx L0 EI
140 lb
2
¢B
1.5 ft
140 lb C
2
1 = c (0.5294x1)(327.06x1)dx1 + 0.5294(2 + x2)(654.12 + 47.06x2)dx2 EI L0 L0 1.5
+
=
L0
3
(0.4706x3)(592.94x3)dx3 +
L0
0.4706(x4 + 1.5)(889.41 - 47.06x4)dx4 d
6437.67(123) 6437.67 lb # ft3 = = 1.54 in. EI 29(106) p4 (0.75)4
320 lb 320 lb
Ans.
1231
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•14–93.
Determine the slope of the 1.5-in-diameter A-36 steel shaft at the bearing support A.
2 ft
L
1 # uA =
muM dx L0 EI 2
uA =
1 (1 - 0.1176x1)(327.06x1)dx1 + B EI L0 L0 L0
3 ft
D
1.5
(0.1176x3)(592.94x3)dx3 140 lb
0.1176(x4 + 1.5)(889.41 - 47.06x4)dx4 R
2387.53(122) 2387.53 lb # ft2 = = 0.0477 rad = 2.73° EI 29(106) A p4 B (0.754)
1.5 ft
B
5
+
=
2 ft
A
140 lb C
Ans. 320 lb 320 lb
14–94. The beam is made of Douglas fir. Determine the slope at C.
8 kN
Virtual Work Equation: For the slope at point C. A
L
1#u =
muM dx L0 EI
1 kN # m # uC = 0 +
1 EI L0 +
uC =
B 1.5 m
1.5 m
1.5 m
1.5 m
1 EI L0
(0.3333x2)(4.00x2) dx2 180 mm 1.5 m
(1 - 0.3333x3)(4.00x3)dx3
120 mm
4.50 kN # m3 EI 4.50(1000)
= -
C
1 13.1(10 ) C 12 (0.12)(0.183) D 9
= 5.89 A 10 - 3 B rad
Ans.
1232
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14–95. The beam is made of oak, for which Eo = 11 GPa. Determine the slope and displacement at A.
200 mm 400 mm
4 kN/m
B
A 3m
Virtual Work Equation: For the displacement at point A, L
1#¢ = 1 kN # ¢ A =
+
¢A =
L0
mM dx EI
1 EI L0
1 EI L0
3m
3m
2 x1 a x31 bdx1 9
(x2 + 3) A 2.00x22 + 6.00x2 + 6.00 B dx2
321.3 kN # m3 EI 321.3(103)
=
1 11(109) C 12 (0.2)(0.43) D
Ans.
= 0.02738 m = 27.4 mm T For the slope at A. L
muM dx L0 EI
1#u = 1 kN # m # uA =
1 EI L0
3m
3m
+
uA = =
L0
2 1.00a x31 bdx1 9
1.00 A 2.00x22 + 6.00x2 + 6.00 B dx2
67.5 kN # m2 EI 67.5(1000)
1 11(109) C 12 (0.2)(0.43) D
= 5.75 A 10 - 3 B rad
Ans.
1233
3m
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*14–96. Determine the displacement at point C. EI is constant.
P
A
C B a
a
L
1 # ¢C =
mM dx L0 EI a
¢C =
=
•14–97.
a
1 c (x1)(Px1)dx1 + (x2)(Px2)dx2 d EI L0 L0 2Pa3 3EI
Ans.
Determine the slope at point C. EI is constant.
P
A
C B a
L
1 # uC =
L0 a
uC =
=
L0
muMdx EI
A a1 B Px1 dx1 x
EI
a
+
(1)Px2dx2 EI L0
Pa2 5Pa2 Pa2 + = 3EI 2EI 6EI
Ans.
1234
a
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14–98.
Determine the slope at point A. EI is constant.
P
A
L
1 # uA =
uA
muM dx L0 EI
C B a
a a x1 1 Pa2 = a1 (0)(Px2)dx2 R = b (Px1)dx1 + B a EI L0 6EI L0
14–99. Determine the slope at point A of the simply supported Douglas fir beam.
a
Ans.
3 kN 0.6 kN⭈m
a
A
B
C
Real Moment Function M. As indicated in Fig. a. 1.5 m
a 0.5 m
Virtual Moment Functions m. As indicated in Fig. b. 75 mm
Virtual Work Equation. 150 mm
L
1#u =
mu M dx EI L0
1kN # m # uA =
Section a – a
1 B EI L0
2m
(1 - 0.3333x1)(0.8x1 + 0.6)dx1 1m
+
uA =
=
=
1 B EI L0
2m
L0
(0.3333x2)(2.2x2)dx2 R
A -0.2667x1 2 + 0.6x1 + 0.6 B dx1 +
1m
L0
0.7333x2 2dx2 R
1.9333 kN # m2 EI
1.9333 A 103 B
1 13.1 A 109 B c (0.075) A 0.153 B d 2
= 0.00700 rad
Ans.
1235
1m
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14–99.
Continued
1236
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*14–100. Determine the displacement at C of the simply supported Douglas fir beam.
3 kN 0.6 kN⭈m
a
A
Real Moment Function M. As indicated in Fig. a.
B
C
Virtual Moment Functions m. As indicated in Fig. b.
1.5 m
Virtual Work Equation. L
1#¢ =
L0
1 kN # ¢ C =
150 mm
1 B EI L0
1.5 m
1m
(0.5x1)(0.8x1 + 0.6)dx1 +
L0
Section a – a
(0.5x2)(2.2x2)dx2
0.5 m
¢C =
1 B EI L0
1.5 m
(0.5x3 + 0.5)(2.2 - 0.8x3)dx3 R
L0
A 0.4x1 2 + 0.3x1 B dx1 + 0.5 m
+
=
=
L0
1m
L0
1.1x2 2 dx2
A -0.4x3 2 + 0.7x3 + 1.1 B dx3 R
1.775kN # m3 EI
1.775 A 103 B
13.1 A 109 B c
0.5 m 75 mm
mM dx EI
+
a
1 (0.075) A 0.153 B d 12
= 6.424 A 10 - 3 B m = 6.42 mm T
Ans.
1237
1m
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•14–101.
Determine the slope of end C of the overhang beam. EI is constant.
w
C
A
Real Moment Function M. As indicated in Fig. a.
B
D L 2
Virtual Moment Function mu. As indicated in Fig. b. Virtual Work Equation. L
1#u = 1 # uC =
uC =
mu M dx L0 EI L L>2 x1 w w 1 (1) ¢ x 3 ≤ dx2 R B ¢ - ≤ c A 11Lx1 - 12x1 2 B ddx1 + EI L0 L 24 3L 2 L0
1 w B EI 24L L0
uC = -
L
A 12x1 3 - 11Lx1 2 B dx1 +
w 3L L0
L>2
x2 3 dx2 R
13wL3 13wL3 = 576EI 576EI
Ans.
1238
L 2
L 2
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14–102. Determine the displacement of point D of the overhang beam. EI is constant.
w
L 2
Virtual Work Equation. L
1 # ¢D =
L0
mM dx EI
1 B EI L0
L>2
¢
x1 w ≤ c A 11Lx1 - 12x1 2 B ddx1 2 24
L>2
+
L0
¢D =
w B 48EI L0
¢D =
wL4 T 96EI
L>2
¢
x2 w ≤ c A 13Lx2 - 12x2 2 - L2 B ddx2 R 2 24
A 11Lx1 2 - 12x1 3 B dx1 +
L>2
L0
B
D
Virtual Moment Function m. As indicated in Fig. b.
1#¢ =
C
A
Real Moment Function M. As indicated in Fig. a.
A 13Lx2 2 - 12x2 3 - L2x2 B dx2 R Ans.
1239
L 2
L 2
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14–103. Determine the displacement of end C of the overhang Douglas fir beam.
400 lb a
Real Moment Functions M. As indicated in Fig. a.
400 lb⭈ft
A B
a
Virtual Moment Functions m. As indicated in Fig. b. 8 ft
Virtual Work Equation.
4 ft 3 in.
L
mM 1#¢ = dx L0 EI 1 lb # ¢ C =
¢C =
=
=
1 B EI L0
1 B EI L0
6 in.
8 ft
8 ft
4 ft
125x1 2dx1 +
Section a – a
4 ft
(0.5x1)(250x1)dx1 +
L0
L0
x2(400x2 + 400)dx2 R
A 400x2 2 + 400x2 B dx2 R
33066.67 lb # ft3 EI 33066.67 A 12 3 B
1.90 A 106 B c
1 (3) A 63 B d 12
= 0.5569 in. = 0.557 in. T
Ans.
1240
C
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*14–104. Determine the slope at A of the overhang white spruce beam.
400 lb a 400 lb⭈ft
A
8 ft
Virtual Moment Functions m. indicated in Fig. b.
6 in.
L
muM 1#u = dx L0 EI
uA =
=
=
1 B EI L0
1 B EI L0
8 ft
Section a – a 8 ft
4 ft
(1 - 0.125x1)(250x1)dx1 +
L0
0(400x2 + 400)dx2 R
A 250x1 - 31.25x1 2 B dx1 + 0 R
2666.67 lb # ft2 EI 2666.67 A 12 2 B
1.940 A 106 B c
1 (3) A 63 B d 12
= 0.00508 rad = 0.00508 rad
Ans.
1241
C 4 ft
3 in.
Virtual Work Equation.
1 lb # ft # uA =
B
a
Real Moment Functions M. As indicated in Fig. a.
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•14–105. Determine the displacement at point B. The moment of inertia of the center portion DG of the shaft is 2I, whereas the end segments AD and GC have a moment of inertia I. The modulus of elasticity for the material is E.
w A
C D a
Real Moment Function M(x): As shown on Fig. a. Virtual Moment Functions m(x): As shown on Fig. b. Virtual Work Equation: For the slope at point B, apply Eq. 14–42. L
1#¢ =
mM dx L0 EI
1 # ¢B = 2 B
a x1 1 a b(w ax1)dx1 R EI L0 2
a
+ 2B
¢B =
1 1 w 2 (x + a) B wa(a + x2) x R dx2 R 2EI L0 2 2 2 2
65wa4 48EI
Ans.
T
1242
B a
G a
a
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14–106. Determine the displacement of the shaft at C. EI is constant.
w0
A
B C L – 2
L – 2
L
1 # ¢C =
mM dx L0 E I L
¢C
2 w0 L w0 3 1 1 = 2a b a x1 b a x1 x bdx1 E I L0 2 4 3L 1
=
w0 L4 120 E I
Ans.
14–107. Determine the slope of the shaft at the bearing support A. EI is constant.
w0
A
B C L – 2
L
1 # uA =
mu M dx L0 E I L
uA
2 w0 L w0 3 1 1 = C a1 x ba x1 x bdx1 S E I L0 L 1 4 3L 1
L 2
+
=
L0
a
w0L w0 3 1 x ba x x bdx2 L 2 4 2 3L 2
5 w0L3 192 E I
Ans.
1243
L – 2
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*14–108. Determine the slope and displacement of end C of the cantilevered beam. The beam is made of a material having a modulus of elasticity of E. The moments of inertia for segments AB and BC of the beam are 2I and I, respectively.
P
L 2
Real Moment Function M. As indicated in Fig. a. Virtual Moment Functions mu and M. As indicated in Figs. b and c. Virtual Work Equation. For the slope at C, L
1#u =
L0
1 # uC =
uC =
mu M dx EI
1 EI L0
L>2
1(Px1)dx1 +
1 2 EI L0
L>2
1 B Pa x2 +
L b R dx2 2
5PL2 16 EI
Ans.
For the displacement at C, L
1#¢ = 1 # ¢C =
¢C =
mM dx L0 EI 1 EI L0
L>2
x1(Px1)dx1 +
1 2EI L0
L>2
¢ x2 +
L L ≤ B P ¢ x2 + ≤ R dx2 2 2
3PL T 16EI
Ans.
1244
C
B
A
L 2
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•14–109.
Determine the slope at A of the A-36 steel W200 * 46 simply supported beam.
12 kN/m 6 kN/m
Real Moment Function M. As indicated in Fig. a. A 3m
Virtual Work Equation. L
1#u =
mu M dx EI L0
1kN # m # uA =
1 B EI L0
3m
(1 - 0.1667x1) A 31.5x1 - 6x1 2 B dx1
3m
+
uA =
=
=
1 B EI L0
3m
L0
(0.1667x2) A 22.5x2 - 3x2 2 B dx2 R
A x1 3 - 11.25x1 2 + 31.5x1 B dx1 +
3m
L0
B
C
Virtual Moment Functions m. As indicated in Fig. b.
A 3.75x2 2 - 0.5x2 3 B dx2 R
84.375 kN # m2 EI
84.375 A 103 B
200 A 109 B c45.5 A 10 - 6 B d
= 0.009272 rad = 0.00927 rad
Ans.
1245
3m
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14–110. Determine the displacement at point C of the A-36 steel W200 * 46 simply supported beam.
12 kN/m 6 kN/m
Real Moment Functions M. As indicated in Fig. a. A 3m
Virtual Work Equation. L
1#¢ =
mM dx L0 EI
1kN # ¢ C =
1 B EI L0
(0.5x1) A 31.5x1 - 6x1 2 B dx1
3m
3m
+
¢C =
=
=
1 B EI L0
3m
L0
(0.5x2) A 22.5x2 - 3x2 2 B dx2 R
A 15.75x1 2 - 3x1 3 B dx1 +
3m
L0
B
C
Virtual Moment Functions m. As indicated in Figs. b.
A 11.25x2 2 - 1.5x2 3 B dx2 R
151.875 kN # m3 EI
151.875 A 103 B
200 A 109 B c45.5 A 10 - 6 B d
= 0.01669 m = 16.7 mm T
Ans.
1246
3m
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14–111. The simply supported beam having a square cross section is subjected to a uniform load w. Determine the maximum deflection of the beam caused only by bending, and caused by bending and shear. Take E = 3G.
w
a L
For bending and shear, L
1#¢ =
L fsvV mM dx + dx EI L0 GA L0 L>2
¢ = 2
L0
x A 12 x B A wL 2 x - w 2 B dx 2
EI
L>2
+ 2
L0
A 65 B A 12 B A wL 2 - wx B dx GA
A B wL wx2 2 L>2 wx4 2 L>2 1 wL 3 a x b + a x b EI 6 8 GA 2 2 0 0 6 5
=
=
5wL4 3wL2 + 384EI 20 GA 5wL4
¢ =
=
1 384(3G) A 12 B a4
+
3wL2 20(G)a2
20wL4 3wL2 + 4 384Ga 20Ga2
= a
L 2 L 2 w 5 3 ba b Ba ba b + R a a G 96 20
Ans.
For bending only, ¢ =
5w L 4 a b 96G a
Ans.
1247
a
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*14–112. The frame is made from two segments, each of length L and flexural stiffness EI. If it is subjected to the uniform distributed load determine the vertical displacement of point C. Consider only the effect of bending.
w B C L
L
A
Real Moment Function M(x): As shown on Fig. a. Virtual Moment Functions m(x): As shown on Fig. b. Virtual Work Equation: For the vertical displacement at point C, L
1#¢ =
L0
mM dx EI L
1 # (¢ C)v =
(¢ C)v =
L
1 w 2 1 wL2 (1.00x1)a x1 b dx1 + (1.00L)a b dx2 EI L0 2 EI L0 2 5wL4 8EI
Ans.
T
1248
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•14–113.
The frame is made from two segments, each of length L and flexural stiffness EI. If it is subjected to the uniform distributed load, determine the horizontal displacement of point B. Consider only the effect of bending.
w B C L
Real Moment Function M(x): As shown on Fig. a. L
Virtual Moment Functions m(x): As shown on Fig. b. Virtual Work Equation: For the horizontal displacement at point B,
A
L
1#¢
mM = dx L0 EI L
1 # (¢ B)h =
(¢ B)h =
L
1 w 2 1 wL2 (0)a x1 b dx1 + (1.00L - 1.00x2)a b dx2 EI L0 2 EI L0 2 wL4 : 4EI
Ans.
14–114. Determine the vertical displacement of point A on the angle bracket due to the concentrated force P. The bracket is fixed connected to its support. EI is constant. Consider only the effect of bending.
P L A
L
1 # ¢ Av =
mM dx L0 EI L L
¢ Av =
=
L
1 C (x1)(Px1)dx1 + (1L)(PL)dx2 S EI L0 L0 4PL3 3EI
Ans.
1249
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14–115. Beam AB has a square cross section of 100 mm by 100 mm. Bar CD has a diameter of 10 mm. If both members are made of A-36 steel, determine the vertical displacement of point B due to the loading of 10 kN.
C 10 kN
2m D
A 3m
Real Moment Function M(x): As shown on Fig. a. Virtual Moment Functions m(x): As shown on Fig. b. Virtual Work Equation: For the displacement at point B, L
1#¢ =
1 kN # ¢ B =
L0
nNL mM dx + EI AE
1 EI L0
+
3m
(0.6667x1)(6.667x1)dx1
1 EI L0
+
¢B =
2m
(1.00x2)(10.0x2)dx2
1.667(16.667)(2) AE
66.667 kN # m3 55.556 kN # m + EI AE 66.667(1000)
=
200(10 ) C 9
1 12
(0.1) A 0.1
3
BD
= 0.04354 m = 43.5 mm
55.556(1000)
+
C A 0.012 B D C 200 A 109 B D p 4
Ans.
T
1250
B 2m
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*14–116. Beam AB has a square cross section of 100 mm by 100 mm. Bar CD has a diameter of 10 mm. If both members are made of A-36 steel, determine the slope at A due to the loading of 10 kN.
C 10 kN
2m
Real Moment Function M(x): As shown on Fig. a.
D
A
Virtual Moment Functions mu(x): As shown on Fig. b.
B
3m
2m
Virtual Work Equation: For the slope at point A, L
1#u =
1 kN # m # uA =
muM nNL dx + EI AE L0 3m
1 EI L0
(1 - 0.3333x1)(6.667x1)dx1
+
uA =
1 EI L0
2m
0(10.0x2)dx2 +
10.0 kN # m2 11.111 kN EI AE 11.111(1000)
10.0(1000)
=
(-0.3333)(16.667)(2) AE
200 A 10
9
B C (0.1) A 0.1 B D 1 12
3
-
C A 0.012 B D C 200 A 109 B D p 4
= 0.00529 rad
Ans.
14–117. Bar ABC has a rectangular cross section of 300 mm by 100 mm. Attached rod DB has a diameter of 20 mm. If both members are made of A-36 steel, determine the vertical displacement of point C due to the loading. Consider only the effect of bending in ABC and axial force in DB.
D
4m 20 kN
300 mm 100 mm A
Real Moment Function M(x): As shown on Fig. a.
3m
Virtual Moment Functions m(x): As shown on Fig. b. Virtual Work Equation: For the displacement at point C, L
1#¢ =
L0
mM nNL dx + EI AE
1 kN # ¢ C = 2c
¢C =
1 EI L0
3m
(1.00x)(20.0x) dx d +
2.50(50.0) (5) AE
360 kN # m3 625 kN # m + EI AE 625(1000)
360(1000)
=
200 A 10
9
B C (0.1) A 0.3 B D 1 12
+
3
B
C A 0.02 2 B D C 200 A 109 B D p 4
= 0.017947 m = 17.9 mm T
Ans.
1251
3m
C
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14–118. Bar ABC has a rectangular cross section of 300 mm by 100 mm. Attached rod DB has a diameter of 20 mm. If both members are made of A-36 steel, determine the slope at A due to the loading. Consider only the effect of bending in ABC and axial force in DB.
D
4m 20 kN
300 mm 100 mm A
B 3m
Real Moment Function M(x): As shown on Fig. a. Virtual Moment Functions mu (x): As shown on Fig. b. Virtual Work Equation: For the slope at point A, L
1#u =
muM nNL dx + AE L0 EI
1 kN # m # uA =
uA =
1 EI L0
3m
(1 - 0.3333x)(20.0x)dx +
30.0 kN # m2 104.167 kN EI AE 104.167(1000)
30.0(1000)
=
( -0.41667)(50.0)(5) AE
200 A 10
9
B C (0.1) A 0.3 B D 1 12
3
-
C A 0.02 2 B D C 200 A 109 B D p 4
= -0.991 A 10 - 3 B rad = 0.991 A 10 - 3 B rad
Ans.
1252
3m
C
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14–119. Determine the vertical displacement of point C. The frame is made using A-36 steel W250 * 45 members. Consider only the effects of bending.
15 kN/m 15 kN
D B C 2.5 m
5m
A
Real Moment Functions M. As indicated in Fig. a. Virtual Moment Functions m. As indicated in Fig. b. Virtual Work Equation. L
1#¢ =
L0
mM dx EI
1 kN # (¢ C)v =
1 B EI L0
(0.5x1) A 52.5x1 - 7.5x1 2 B dx1
2.5 m
5m
+
0(15x2)dx2
L0
2.5 m
+
(¢ C)v =
1 B EI L0
2.5 m
L0
A 26.25x1 2 - 3.75x1 3 B dx1 + 0
2.5 m
+
=
=
(0.5x3) A 75 + 22.5x3 - 7.5x3 2 B dx2 R
L0
A 37.5x3 + 11.25x3 2 - 3.75x3 3 B dx3 R
239.26 kN # m3 EI
239.26 A 103 B
200 A 109 B c71.1 A 10 - 6 B d
= 0.01683 m = 16.8 mm T
Ans.
1253
2.5 m
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*14–120. Determine the horizontal displacement of end B. The frame is made using A-36 steel W250 * 45 members. Consider only the effects of bending.
15 kN/m 15 kN
D B C 2.5 m
5m
Real Moment Functions M. As indicated in Fig. a. A
Virtual Moment Functions m. As indicated in Fig. b. Virtual Work Equation. L
1#¢ =
L0
mM dx EI
1 kN # (¢ B)h =
(¢ B)h =
=
=
1 B EI L0
1 B EI L0
5m
x1 A 52.5x1 - 7.5x1 2 B dx1 +
5m
A 52.5x1 2 - 7.5x1 3 B dx1 +
5m
L0
x2(15x2)dx2 R
5m
L0
15x2 2dx2 R
1640.625 kN # m3 EI
1640.625 A 103 B
200 A 109 B c71.1 A 10 - 6 B d
= 0.1154 m = 115 mm :
Ans.
1254
2.5 m
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•14–121. Determine the displacement at point C. EI is constant.
A
B
a
C
M0
C
M0
a
L
mM dx L0 E I
1 # ¢C =
a
¢C =
L0
(1x) A
EI
5 M0 a 6EI
=
14–122.
M0 a
xB
a
dx +
(1x) M0 dx EI L0
2
Ans.
Determine the slope at B. EI is constant.
A
B
a L
mu M dx L0 E I
1 # uB =
a
uB =
=
L0
A xa B A Ma0 x B EI
dx
M0 a 3EI
14–123.
Ans.
Solve Prob. 14–72 using Castigliano’s theorem.
Member
N
0N>0P
N(P = 0)
L
N(0N>0P)L
AB
1.1547P + 800
1.1547
800
120
110851.25
BC
–0.5774P
–0.5774
0
60
0 © = 110851.25
¢ Bb = ©N a
0N L 110851.25 110851.25 = 0.00191 in. b = = 0P AE AE (2)(29)(106)
1255
Ans.
a
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*14–124.
Solve Prob. 14–73 using Castigliano’s theorem.
Member Force N: Member forces due to external force P and external applied forces are shown on the figure. Castigliano’s Second Theorem: N(P = 200 lb)
L
–0.8333P
0N 0P –0.8333
–166.67
10.0
0N bL 0P 1388.89
BC
0.8333P
0.8333
166.67
10.0
1388.89
AC
0.500P
0.500
100.00
12
600.00
Member
N
AB
Na
© 3377.78 lb # ft 0N L b ¢ = a Na 0P AE (¢ B)h =
3377.78 lb # ft AE 3377.78(12)
=
•14–125.
2 C 29.0 A 106 B D
= 0.699 A 10 - 3 B in. :
Ans.
Solve Prob. 14–75 using Castigliano’s theorem.
Member
N
N(P = 30)
0N>0P
L
N(0N>0P)L
AB
1.50P
1.50
45.00
3.0
202.50
AD
5 213
0
5213
213
0
BD
–20
0
–20
2.0
0
BC
1.5P
1.5
45.00
3.0
202.50
CD
–0.5 213P
–0.5 213
-15 213
213
351.54
–0.5 213
-20 213
213
468.72
DE
– A 0.5 213P + 5 213 B
1225.26 A 10 B 0N L b = 0P AE 300 A 10 - 6 B (200) A 109 B
© = 1225.26
3
¢ Cv = ©Na
= 0.02.04 m = 20.4 mm
Ans.
1256
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14–126. Solve Prob. 14–76 using Castigliano’s theorem.
Member
N
N(P = 0)
0N>0P
N(0N>0P)L
L
AB
45
0
45.00
3
0
AD
0.25 213P + 5 213
0.25213
5213
213
58.59
BC
45
0
45
3
0
BD
–20
0
–20
2
0
CD
–15 213
0
-15 213
213
0
–0.25213
-20 213
213
234.36
– A 0.25 213P + 20 213 B
DE
¢ Dv = ©Na
© = 292.95
292.95 A 103 B 292.95 0N L b = = 0P AE AE 300 A 10 - 6 B (200) A 109 B
= 4.88 A 10 - 3 B m = 4.88 mm
14–127.
Ans.
Solve Prob. 14–77 using Castigliano’s theorem.
Member Forces N: Member forces due to external force P and external applied forces are shown on the figure. Castigliano’s Second Theorem:
3.333
96
0N bL 0P 213.33
0.6667
3.333
96
213.33
0
0
0
72
0
DE
0
0
0
96
0
EF
0
0
0
96
0
AF
0
0
0
72
0
AE
– 0.8333P
– 0.8333
–4.167
120
416.67
CE
– 0.8333P
– 0.8333
–4.167
120
416.67
BE
1.00P
1.00
5.00
72
360.00
Member
N
AB
0.6667P
0N 0P 0.6667
BC
0.6667P
CD
N(P = 5 kip)
L
Na
©1620 kip # in 0N L b ¢ = a Na 0P AE (¢ B)v =
1620 kip # in. AE 1620
=
4.5 C 29.0 A 103 B D
= 0.0124 in. T
Ans.
1257
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*14–128.
Solve Prob. 14–78 using Castigliano’s theorem.
Member Forces N: Member forces due to external force P and external applied forces are shown on the figure. Castigliano’s Second Theorem: N(P = 0)
L
3.333
96
0N bL 0P 213.33
AB
0.6667P+3.333
0N 0P 0.6667
BC
0.6667P+3.333
0.6667
3.333
96
213.33
CD
0
0
0
72
0
DE
0
0
0
96
0
EF
0
0
0
96
0
AF
0
0
0
72
0
AE
–(0.8333P + 4.167)
– 0.8333
–4.167
120
416.67
CE
–(0.8333P + 4.167)
– 0.8333
–4.167
120
416.67
BE
5.0
0
5.00
72
0
Member
N
Na
©1260 kip # in 0N L b ¢ = a Na 0P AE (¢ E)v =
1260 kip # in. AE 1260
=
•14–129.
4.5 C 29.0 A 103 B D
= 0.00966 in. T
Ans.
Solve Prob. 14–79 using Castigliano’s theorem.
Member
N
0N>0P
N(P = 4)
L
N(0N>0P)L
AB
0
0
0
1.5
0
AC
–1.25P
–1.25
–5
2.5
15.625
AD
P
1
4
2.0
8.00
BC
P
1
4
2.0
8.00
CD
–(5 -0.75P)
0.75
–2
1.5
–2.25 © = 29.375
¢ Bh = ©Na
29.375 A 103 B 0N L ba b = = 0.367 A 10 - 3 B m 0P AE 400 A 10 - 6 B (200) A 109 B
= 0.367 mm
Ans.
1258
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14–130.
Solve Prob. 14–80 using Castigliano’s theorem.
Member
N
0N>0P
N(P = 5)
L
N(0N>0P)L
AB
0
0
0
1.5
0
AC
–5
0
–5
2.5
0
AD
4
0
4
2.0
0
BC
4
0
4
2.0
0
CD
–(P - 3)
–1
–2
1.5
3 © = 3
3 A 10 B 3 0N L b = = 0P AE AE 400 A 10 6 B (200) A 109 B 3
¢ Cv = ©Na
= 37.5 A 10 - 6 B m = 0.0375 mm
14–131.
Ans.
Solve Prob. 14–81 using Castigliano’s theorem.
Member Forces N: Member forces due to external force P and external applied forces are shown on the figure. Castigliano’s Second Theorem: N(P = 30 kN)
L
–22.5
1.5
0N bL 0P 25.3125
AB
– 0.750P
0N 0P – 0.750
BC
– 0.750P
– 0.750
–22.5
1.5
25.3125
AE
1.25P
1.25
37.5
2.5
117.1875
CE
–(1.25P + 25.0)
– 1.25
–62.5
2.5
195.3125
BE
20.0
0
20.0
2
0
DE
1.50P+15.0
1.50
60.0
1.5
135.00
Member
N
Na
# a 498.125 kN m 0N L b ¢ = a Na 0P AE (¢ A)v =
=
498.125 kN # m AE
498.125 A 103 B
0.400 A 10 - 3 B C 200 A 109 B D
= 6.227 A 10 - 3 B m = 6.23 mm T
Ans.
1259
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*14–132.
Solve Prob. 14–82 using Castigliano’s theorem.
Member Forces N: Member forces due to external force P and external applied forces are shown on the figure. Castigliano’s Second Theorem: 0N bL 0P 0
AB
– 22.5
0N 0P 0
BC
– 22.5
0
–22.5
1.5
0
AE
37.5
0
37.5
2.5
0
CE
–(1.25P + 37.5)
– 1.25
–62.5
2.5
195.3125
BE
1.00P
1.00
20.0
2
40.0
DE
0.750P + 45
0.750
60.0
1.5
67.50
Member
N
N(P = 20 kN)
L
–22.5
1.5
Na
# a 302.8125 kN m 0N L b ¢ = a Na 0P AE (¢ B)v =
=
302.8125 kN # m AE
302.8125 A 103 B
0.400 A 10 - 3 B C 200 A 109 B D
= 3.785 A 10 - 3 B m = 3.79 mm T
•14–133.
Ans.
Solve Prob. 14–83 using Castigliano’s theorem.
¢ Cv = ©N a
21232 21232 0N L = 0.163 in. b = = 0P AE AE 4.5 (29)(103)
Ans.
1260
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14–134.
Solve Prob. 14–84 using Castigliano’s theorem.
¢ Hv = ©N a
20368 20368 0N L b = = 0P AE AE 4.5 (29) A 103 B
= 0.156 in.
14–135.
Ans.
Solve Prob. 14–87 using Castigliano’s theorem.
0M1 x1 = 0P¿ 2
0M2 x2 a = + 0P¿ 2 2
Set P¿ = 0 M1 = Px1
M2 = Pa
a
¢C =
L0
Ma
0M dx b 0P EI a
= (2)
=
1 1 (Px1) a x1 b dx + B EI L0 2 L0
a>2
(Pa) a
1 a + x2 bdx2 R 2 2
23Pa3 24 EI
Ans.
1261
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*14–136.
Solve Prob. 14–88 using Castigliano’s theorem.
0M1 = x1 0P
0M2 = -0.5 x2 0P
Set P = 15 kN M2 = -1.5x2 - 2x22
M1 = 15x1 L
¢A =
Ma
L0
0M dx b 0P EI
1.5
3
A -1.5x2 - 2x22 B (-0.5x2)dx2 R
=
1 B EI L0
=
43.875(103) 43.875 kN # m3 = = 0.0579 m 1 9 EI 13(10 ) 12 (0.12)(0.18)3
(15x1)(x1)dx +
L0
= 57.9 mm
•14–137.
Ans.
Solve Prob. 14–90 using Castigliano’s theorem.
Internal Moment Function M(x): The internal moment function in terms of the couple moment M¿ and the applied load are shown on the figure. Castigliano’s Second Theorem: The slope at A can be determined with 0M(x1) 0M(x2) = 1 - 0.100x1, = 0 and setting M¿ = 0. 0M¿ 0M¿ L
u =
uA =
=
=
L0
Ma
1 EI L0
0M dx b 0M¿ EI
10 m
(2.50x1)(1 - 0.100x1)dx1 +
1 EI L0
5m
A 1.00x22 B (0)dx2
41.667 kN # m2 EI
41.667 A 103 B
200 A 109 B C 70 A 10 - 6 B D
Ans.
= 0.00298 rad
1262
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14–138.
Solve Prob. 14–92 using Castigliano’s theorem.
0M1 = 0.5294x1 0P
0M2 = 0.5294x2 + 1.0588 0P
0M3 = 0.4706x3 0P
0M4 = 0.4706x4 + 0.7059 0P
Set P = 0 M1 = 327.06x1
M2 = 47.06x2 + 654.12
M3 = 592.94x3
M4 = 889.41 - 47.06x4
L
¢B =
L0
Ma
0M dx b 0P EI 2
1 (327.06x1)(0.5294x1)dx1 c EI L0
=
2
+
(47.06x2 + 654.12)(0.5294x2 + 1.0588)dx2
L0
1.5
+
(592.94x3)(0.4706x3)dx3
L0 3
+
=
14–139.
L0
(889.41 - 47.06x4)(0.4706x4 + 0.7059)dx4 d
6437.69 A 12 3 B 6437.69 lb # ft3 = = 1.54 in. EI 29 A 106 B A p4 B A 0.754 B
Ans.
Solve Prob. 14–93 using Castigliano’s theorem.
0M1 = 1 - 0.1176 x1 0M¿
0M2 = 0.1176 x2 0M¿
0M3 = 0.1176x3 + 0.1764 0M¿
Set M¿ = 0 M1 = 327.06x1
uA =
L
Ma
M2 = 592.94x2
M3 = 889.41 - 47.06x3
2
0M dx 1 b = c (327.06x1)(1 - 0.1176x1)dx1 0M¿ EI EI L0
1.5
+
(592.94x2)(0.1176x2)dx2 +
L0 5
+
=
L0
(889.41 - 47.06x3)(0.1176x3 + 0.1764)dx3 d
2387.54(12 2) 2387.54 lb # ft2 = 0.0477 rad = 2.73° = EI 29(106)(p4 )(0.754)
1263
Ans.
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*14–140.
Solve Prob. 14–96 using Castigliano’s theorem.
0M1 = x1 0P¿
0M2 = x2 0P¿
Set P = P¿ M1 = Px1
M2 = Px2
L
¢C =
=
a
Ma
L0
a
1 0M bdx = (Px1)(x1)dx1 + (Px2)(x2)dx2 R B 0P¿ EI L0 L0
2Pa3 3EI
14–141.
Ans.
Solve Prob. 14–89 using Castigliano’s theorem.
Set M¿ = 0 L
uC =
Ma
L0
0M dx b 0M¿ EI
1 (Px1)a x1 b dx1 a (Px2)(1)dx2 a + = EI EI L0 L0 a
=
Pa2 5Pa2 Pa2 + = 3EI 2EI 6EI
14–142.
Ans.
Solve Prob. 14–98 using Castigliano’s theorem.
0M1 x1 = 1 a 0M¿
0M2 = 0 0M¿
Set M¿ = 0 M1 = -Px1 L
uA =
L0
Ma
M2 = Px2 0M dx b 0M¿ EI
=
a a x1 -Pa2 1 bdx1 + ( -Px1)a 1 (Px2)(0)dx2 R = B a EI L0 6EI L0
=
Pa2 6EI
Ans.
1264
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14–143.
Solve Prob. 14–112 using Castigliano’s theorem.
Internal Moment Function M(x): The internal moment function in terms of the load P and external applied load are shown on the figure. Castigliano’s Second Theorem: The vertical displacement at C can be determined 0M(x1) 0M(x2) with = 1.00x1, = 1.00L and setting P = 0. 0P 0P L
¢ =
Ma
L0
0M dx b 0P EI
L
(¢ C)v =
=
*14–144.
L
2
1 w 2 1 wL a x1 b(1.00x1) dx1 + a b(1.00L) dx2 EI L0 2 EI L0 2 5wL4 8EI
Ans.
T
Solve Prob. 14–114 using Castigliano’s theorem.
Castigliano’s Second Theorem: The horizontal displacement at A can be determined 0M(x1) 0M(x2) using = 1.00x1, = 1.00L and setting P¿ = P. 0P¿ 0P¿ L
¢ =
L0
Ma L
(¢ A)h =
=
0M dx b 0P EI L
1 1 (Px1)(1.00x1) dx1 + (PL)(1.00L) dx2 EI L0 EI L0 4PL3 3EI
Ans.
1265
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•14–145.
Solve Prob. 14–121 using Castigliano’s theorem.
L
0M dx b = = Ma 0P E I L0 L0
¢C
=
A
a M0 a
a M0 (1x) x B (1x) dx + dx EI EI L0
5 M0 a2 6EI
Ans.
14–146. Determine the bending strain energy in the beam due to the loading shown. EI is constant.
P
a
L
Ui =
=
a
a
1 M 2 dx = c2 (Px1)2 dx1 + (Pa)2 dx2 d 2EI 2EI L0 L0 L0 5P2a3 6EI
Ans.
1266
P
a
a
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14–147. The 200-kg block D is dropped from a height h = 1 m onto end C of the A-36 steel W200 * 36 overhang beam. If the spring at B has a stiffness k = 200 kN>m, determine the maximum bending stress developed in the beam.
D h A B 4m
Equilibrium. The support reactions and the moment functions for regions AB and BC of the beam under static conditions are indicated on the free-body diagram of the beam, Fig. a, Ue = Ui L
1 M 2dx P¢ st = © 2 L0 2EI 1 1 P¢ st = B 2 2EI L0
¢ st =
4m
a
2 P x2 b dx + 2 L0
2m
(Px1)2 dx R
8P EI
Here, I = 34.4 A 106 B mm4 = 34.4 A 10 - 6 B m4 (see the appendix) and E = Est = 200 GPa. Then, the equivalent spring constant can be determined from P = kb ¢ st P = kb a
8P b EI
EI = kb = 8
200 A 109 B c34.4 A 10 - 6 B d 8
= 860 A 103 B N>m
From the free-body diagram, Fsp =
3 P 2
ksp ¢ sp =
¢ sp =
3 (k ¢ ) 2 b b
3 3 kb 3 860 A 10 B ≥ ¢ b = 6.45 ¢ b ¢ ≤ ¢b = £ 2 ksp 2 200 A 103 B
(1)
Conservation of Energy. mga h + ¢ b +
3 1 1 ¢ b = ksp ¢ sp 2 + kb ¢ b 2 2 sp 2 2
1267
C
k 2m
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14–147.
Continued
Substiuting Eq. (1) into this equation. 200(9.81)c1 + ¢ b +
3 1 1 (6.45¢ b) d = c200 A 103 B d(6.45¢ b)2 + c860 A 103 B d ¢ b 2 2 2 2
4590.25 A 103 B ¢ b 2 - 20944.35¢ b - 1962 = 0 Solving for the positive root ¢ b = 0.02308 m Maximum
Stress.
The
maximum
force
on
the
beam
is
Pmax = kb ¢ b
= 860 A 103 B (0.02308) = 19.85 A 103 B N. The maximum moment occurs at the
Mmax = Pmax L = 19.85 A 103 B (2) = 39.70 A 103 B N # m. 0.201 d Applying the flexure formula with c = = = 0.1005 m. 2 2 supporting
smax =
spring,
where
39.70 A 103 B (0.1005) Mmax c = = 115.98 MPa = 116 MPa I 34.4 A 10 - 6 B
Since smax 6 sY = 250 MPa, this result is valid.
1268
Ans.
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*14–148. Determine the maximum height h from which the 200-kg block D can be dropped without causing the A-36 steel W200 * 36 overhang beam to yield. The spring at B has a stiffness k = 200 kN>m.
D h A B 4m
Equilibrium. The support reactions and the moment functions for regions AB and BC of the beam under static conditions are indicated on the free-body diagram of the beam, Fig. a, Ue = Ui L
1 M 2dx P¢ st = © 2 L0 2EI 1 1 P¢ st = B 2 2EI L0 ¢ st =
4m
a
2 P x2 b dx + 2 L0
2m
(Px1)2 dx R
8P EI
Here, I = 34.4 A 106 B mm4 = 34.4 A 10 - 6 B m4 (see the appendix) and E = Est = 200 GPa. Then, the equivalent spring constant can be determined from P = kb ¢ st P = kb a
8P b EI
EI kb = = 8
200 A 109 B c34.4 A 10 - 6 B d 8
= 860 A 103 B N>m
From the free-body diagram, Fsp =
3 P 2
ksp ¢ sp =
¢ sp =
3 (k ¢ ) 2 b b
3 3 kb 3 860 A 10 B ≥ ¢ b = 6.45¢ b ¢ ≤ ¢b = £ 2 ksp 2 200 A 103 B
(1)
1269
C
k 2m
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*14–148.
Continued
Maximum Stress. The maximum force on the beam is Pmax = kb ¢ b = 860 A 103 B ¢ b. The maximum moment occurs at the supporting spring, where Mmax = Pmax L = 860 A 103 B ¢ b(2) = 1720 A 103 B ¢ b. 0.201 d = = 0.1005 m, c = 2 2 smax =
Applying
the
flexure
formula
with
Mmaxc I
250 A 106 B =
1720 A 103 B ¢ b(0.1005) 34.4 A 10 - 6 B
¢ b = 0.04975 m Substituting this result into Eq. (1), ¢ sp = 0.3209 m Conservation of Energy. mga h + ¢ b +
3 1 1 ¢ b = ksp ¢ sp 2 + kb ¢ b 2 2 sp 2 2
200(9.81)ch + 0.04975 +
3 1 (0.3209) d = c200 A 103 B d(0.3209)2 2 2 +
1 c860 A 103 B d(0.04975)2 2
h = 5.26 m
Ans.
1270
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•14–149.
The L2 steel bolt has a diameter of 0.25 in., and the link AB has a rectangular cross section that is 0.5 in. wide by 0.2 in. thick. Determine the strain energy in the link AB due to bending, and in the bolt due to axial force. The bolt is tightened so that it has a tension of 350 lb. Neglect the hole in the link.
6 in.
4 in.
0.2 in. A
B
8 in.
Bending strain energy: L
(Ub)i =
=
6
4
1 M 2 dx = c (140x1)2dx1 + (210x2)2dx2 d 2EI 2EI L0 L0 L0 1.176(106) 1.176(106) = 122 in # lb = 10.1 ft # lb = 1 EI 29(106)(12 )(0.5)(0.2 3)
Ans.
Axial force strain energy: L
(Ua)i =
(350)2(8) N 2 dx N 2L = 0.344 in # lb = = 2AE 2(29)(106)(p4 )(0.252) L0 2EA
Ans.
14–150. Determine the vertical displacement of joint A. Each bar is made of A-36 steel and has a cross-sectional area of 600 mm2. Use the conservation of energy.
B
C
Joint A:
2m
+ c ©Fy = 0;
4 F - 5 = 0 5 AB
FAB = 6.25 kN
D A
+ ©F = 0; ; x
FAD -
3 (6.25) = 0 5
1.5 m
FAD = 3.75 kN 5 kN
Joint B: + c ©Fy = 0;
4 4 F - (6.25) = 0 5 BD 5
+ ©F = 0; : x
3 FBC - 2a b (6.25) = 0 5
FBD = 6.25 kN
FBC = 7.5 kN
Conservation of energy: Ue = Ui N 2L 1 P¢ = © 2 2AE 1 1 (5) A 103 B ¢ A c A 6.25 A 103 B B 2(2.5) + A 3.75 A 103 B B 2(3) 2 2AE + A 6.25 A 103 B B 2(2.5) + A 7.5 A 103 B B 2(1.5) d ¢A =
64 375 64 375 = = 0.5364 A 10 - 3 B m = 0.536 mm -6 AE 600 A 10 B (200) A 109 B
1271
Ans.
1.5 m
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14–151. Determine the total strain energy in the A-36 steel assembly. Consider the axial strain energy in the two 0.5-in.-diameter rods and the bending strain energy in the beam for which I = 43.4 in4. 3 ft
500 lb
Support Reactions: As shown FBD(a). Internal Moment Function: As shown on FBD(b). Total Strain Energy: L
(Ui)T =
4 ft
2
NL M dx + 2EI 2AE L0
= 2B
1 2EI L0
4 ft
(250x)2 dx R + 2 B
=
1.3333 A 106 B lb2 # ft3
=
1.3333 A 106 B A 12 3 B
+
EI
29.0 A 106 B (43.4)
+
p 4
2502(3) R 2AE
0.1875 A 106 B lb2 # ft AE
0.1875 A 106 B (12)
A 0.52 B C 29.0 A 106 B D
= 2.23 in # lb
Ans.
*14–152. Determine the vertical displacement of joint E. For each member A = 400 mm2, E = 200 GPa. Use the method of virtual work. Member
n
N
L
F
0
1.5
0
AE
– 0.8333
– 37.5
2.5
78.125
AB
0.6667
30.0
2.0
40.00
EF
0
0
2.0
0
EB
– 0.50
22.5
1.5
–16.875
ED
– 0.6667
– 30.0
2.0
40.00
BC
0
0
2.0
0
BD
0.8333
37.5
2.5
78.125
CD
– 0.5
– 22.5
1.5
16.875
A
2m
nNL AE
400 A 10 - 6 B (200) A 109 B
C
B 45 kN
© = 236.25
236.25 A 103 B
D
1.5 m
0
¢ Bv =
E
nNL
AF
1 # ¢ Bv = ©
4 ft
2
= 2.95 A 10 - 3 B = 2.95 mm
Ans.
1272
2m
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•14–153.
Solve Prob. 14–152 using Castigliano’s theorem.
F
E
D
1.5 m
A
C
B 45 kN 2m
Member
N
0N>0P
N(P = 45)
L
2m
N(0N>0P)L
AF
0
0
0
1.5
0
AE
–(0.8333P + 37.5)
– 0.8333
–37.5
2.5
78.125
AB
0.6667P + 30
0.6667
30.0
2.0
40.00
BE
22.5–0.5P
– 0.5
22.5
1.5
–16.875
BD
0.8333P + 37.5
0.8333
37.5
2.5
78.125
BC
0
0
0
2.0
0
CD
–(0.5P + 22.5)
– 0.5
–22.5
1.5
16.875
DE
–(0.6667P + 30)
– 0.6667
–30.0
2.0
40.00
EF
0
0
0
2.0
0 © = 236.25
¢ Bv = ©N
=
236.25 0N L = 0P AE AE
236.25 A 103 B
400 A 10 - 6 B (200) A 109 B
= 2.95 A 10 - 3 B m = 2.95 mm
Ans.
14–154. The cantilevered beam is subjected to a couple moment M0 applied at its end. Determine the slope of the beam at B. EI is constant. Use the method of virtual work.
M0
A B L
L
uB =
=
L (1) M0 muM dx = dx EI L0 EI L0
M0L EI
Ans.
1273
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14–155. Solve Prob. 14–154 using Castigliano’s theorem. M0
A L
uB =
=
L0
ma
B
L
M0(1) dm dy b = dx dm¿ EI EI L0
L
M0L EI
Ans.
*14–156. Determine the displacement of point B on the 3 aluminum beam. Eal = 10.6110 2 ksi. Use the conser
3 in.
3 kip
C B 12 ft
L
12(12)
M 2 dx 1 = (2) 2EI 2EI L0 L0
Ue =
1 1 P¢ = (3)¢ B = 1.5¢ B 2 2
(1.5x)2 dx =
2239488 EI
Conservation of energy: Ue = Ui
1.5¢ B =
2 239 488 EI
¢B =
1 492 992 EI
y =
0.5(7)(1) + (4)(6)(1) = 2.1154 in. 7(1) + 6(1)
I =
1 1 (7) A 13 B + (7)(1)(2.1154 - 0.5)2 + (1) A 63 B + (1)(6)(4 - 2.1154)2 12 12
= 58.16 in4 ¢B =
3 in. 1 in. 6 in.
A
Ui =
1 in.
1 492 992 = 2.42 in. (10.6)(103)(58.16)
Ans.
1274
12 ft
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14–157. A 20-lb weight is dropped from a height of 4 ft onto the end of a cantilevered A-36 steel beam. If the beam is a W12 * 50, determine the maximum stress developed in the beam. 4 ft
12 ft
From Appendix C: ¢ st =
20(12(12))3 PL3 = = 1.742216 A 10 - 3 B in. 3EI 3(29) A 106 B (394)
n = 1 +
A
1 + 2a
h 4(12) b = 235.74 b = 1 + 1 + 2a ¢ st A 1.742216(10 - 3)
smax = nsst = 235.74 £
20(12)(12) A 12.19 2 B 394
≥ = 10503 psi = 10.5 ksi 6 sg O.K. Ans.
1275
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1–1. Determine the resultant internal normal force acting on the cross section through point A in each column. In (a), segment BC weighs 180 lb>ft and segment CD weighs 250 lb>ft. In (b), the column has a mass of 200 kg>m. (a) + c ©Fy = 0;
5 kip
8 kN
B
200 mm
200 mm
6 kN
6 kN
FA - 1.0 - 3 - 3 - 1.8 - 5 = 0 10 ft
FA = 13.8 kip
8 in.
Ans.
3m
8 in. 200 mm
(b) + c ©Fy = 0;
3 kip
FA - 4.5 - 4.5 - 5.89 - 6 - 6 - 8 = 0 FA = 34.9 kN
3 kip
200 mm
4.5 kN
4.5 kN
C
Ans. 4 ft
A
A
1m
4 ft D (a)
1–2. Determine the resultant internal torque acting on the cross sections through points C and D. The support bearings at A and B allow free turning of the shaft. ©Mx = 0;
A 250 Nm 300 mm
TC - 250 = 0 TC = 250 N # m
©Mx = 0;
(b)
400 Nm
200 mm
Ans.
TD = 0
150 Nm
C
150 mm
Ans.
200 mm
B
D
250 mm 150 mm
1–3. Determine the resultant internal torque acting on the cross sections through points B and C. A
©Mx = 0;
TB = 150 lb # ft ©Mx = 0;
600 lbft B 350 lbft
TB + 350 - 500 = 0 Ans.
TC - 500 = 0 TC =
3 ft
C 500 lbft
1 ft
500 lb # ft
Ans.
2 ft 2 ft
1
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*1–4. A force of 80 N is supported by the bracket as shown. Determine the resultant internal loadings acting on the section through point A.
0.3 m A 30 0.1 m
80 N
Equations of Equilibrium: +
Q©Fx¿ = 0;
NA - 80 cos 15° = 0 NA = 77.3 N
a+ ©Fy¿ = 0;
Ans.
VA - 80 sin 15° = 0 VA = 20.7 N
a+
©MA = 0;
Ans.
MA + 80 cos 45°(0.3 cos 30°) - 80 sin 45°(0.1 + 0.3 sin 30°) = 0 MA = - 0.555 N # m
Ans.
or a+
©MA = 0;
MA + 80 sin 15°(0.3 + 0.1 sin 30°) -80 cos 15°(0.1 cos 30°) = 0 MA = - 0.555 N # m
Ans.
Negative sign indicates that MA acts in the opposite direction to that shown on FBD.
2
45
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•1–5.
Determine the resultant internal loadings in the beam at cross sections through points D and E. Point E is just to the right of the 3-kip load.
3 kip 1.5 kip/ ft
A D 6 ft
Support Reactions: For member AB a + ©MB = 0; + : ©Fx = 0; + c ©Fy = 0;
9.00(4) - A y(12) = 0
A y = 3.00 kip
Bx = 0 By + 3.00 - 9.00 = 0
By = 6.00 kip
Equations of Equilibrium: For point D + : ©Fx = 0; + c ©Fy = 0;
ND = 0
Ans.
3.00 - 2.25 - VD = 0 VD = 0.750 kip
a + ©MD = 0;
Ans.
MD + 2.25(2) - 3.00(6) = 0 MD = 13.5 kip # ft
Ans.
Equations of Equilibrium: For point E + : ©Fx = 0; + c ©Fy = 0;
NE = 0
Ans.
- 6.00 - 3 - VE = 0 VE = - 9.00 kip
a + ©ME = 0;
Ans.
ME + 6.00(4) = 0 ME = - 24.0 kip # ft
Ans.
Negative signs indicate that ME and VE act in the opposite direction to that shown on FBD.
3
E
B 6 ft
4 ft
4 ft
C
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1–6. Determine the normal force, shear force, and moment at a section through point C. Take P = 8 kN.
B
0.1 m
Support Reactions: a + ©MA = 0;
8(2.25) - T(0.6) = 0
0.5 m C
T = 30.0 kN
+ : ©Fx = 0;
30.0 - A x = 0
A x = 30.0 kN
+ c ©Fy = 0;
Ay - 8 = 0
A y = 8.00 kN
0.75 m
0.75 m
A 0.75 m
P
Equations of Equilibrium: For point C + : ©Fx = 0;
- NC - 30.0 = 0 NC = - 30.0 kN
+ c ©Fy = 0;
Ans.
VC + 8.00 = 0 VC = - 8.00 kN
a + ©MC = 0;
Ans.
8.00(0.75) - MC = 0 MC = 6.00 kN # m
Ans.
Negative signs indicate that NC and VC act in the opposite direction to that shown on FBD.
1–7. The cable will fail when subjected to a tension of 2 kN. Determine the largest vertical load P the frame will support and calculate the internal normal force, shear force, and moment at the cross section through point C for this loading.
B
0.1 m
0.5 m C
Support Reactions: a + ©MA = 0;
0.75 m
P(2.25) - 2(0.6) = 0 P
P = 0.5333 kN = 0.533 kN + : ©Fx = 0;
2 - Ax = 0
+ c ©Fy = 0;
A y - 0.5333 = 0
Ans.
A x = 2.00 kN A y = 0.5333 kN
Equations of Equilibrium: For point C + : ©Fx = 0;
- NC - 2.00 = 0 NC = - 2.00 kN
+ c ©Fy = 0;
Ans.
VC + 0.5333 = 0 VC = - 0.533 kN
a + ©MC = 0;
Ans.
0.5333(0.75) - MC = 0 MC = 0.400 kN # m
Ans.
Negative signs indicate that NC and VC act in the opposite direction to that shown on FBD.
4
0.75 m
A 0.75 m
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*1–8. Determine the resultant internal loadings on the cross section through point C. Assume the reactions at the supports A and B are vertical.
6 kN 3 kN/m
Referring to the FBD of the entire beam, Fig. a, a + ©MB = 0;
1 - A y(4) + 6(3.5) + (3)(3)(2) = 0 2
+ c ©Fy = 0; a + ©MC = 0;
C
A y = 7.50 kN
NC = 0
D 1.5 m
0.5 m 0.5 m
Referring to the FBD of this segment, Fig. b, + : ©Fx = 0;
B
A
1.5 m
Ans.
7.50 - 6 - VC = 0
VC = 1.50 kN
MC + 6(0.5) - 7.5(1) = 0
Ans.
MC = 4.50 kN # m
Ans.
•1–9.
Determine the resultant internal loadings on the cross section through point D. Assume the reactions at the supports A and B are vertical.
6 kN 3 kN/m
Referring to the FBD of the entire beam, Fig. a,
B
A
a + ©MA = 0;
By(4) - 6(0.5) -
1 (3)(3)(2) = 0 2
By = 3.00 kN 0.5 m 0.5 m
Referring to the FBD of this segment, Fig. b, + : ©Fx = 0; + c ©Fy = 0;
ND = 0 VD -
1 (1.5)(1.5) + 3.00 = 0 2
a + ©MD = 0; 3.00(1.5) -
C
Ans. VD = - 1.875 kN
Ans.
1 (1.5)(1.5)(0.5) - MD = 0 MD = 3.9375 kN # m 2 = 3.94 kN # m
5
Ans.
D 1.5 m
1.5 m
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1–10. The boom DF of the jib crane and the column DE have a uniform weight of 50 lb/ft. If the hoist and load weigh 300 lb, determine the resultant internal loadings in the crane on cross sections through points A, B, and C.
D 2 ft
F
A
B 8 ft
3 ft
5 ft C 300 lb 7 ft
E
Equations of Equilibrium: For point A + ; © Fx = 0; + c © Fy = 0;
NA = 0
Ans.
VA - 150 - 300 = 0 VA = 450 lb
a + ©MA = 0;
Ans.
- MA - 150(1.5) - 300(3) = 0 MA = - 1125 lb # ft = - 1.125 kip # ft
Ans.
Negative sign indicates that MA acts in the opposite direction to that shown on FBD. Equations of Equilibrium: For point B + ; © Fx = 0;
NB = 0
+ c © Fy = 0;
VB - 550 - 300 = 0
Ans.
VB = 850 lb a + © MB = 0;
Ans.
- MB - 550(5.5) - 300(11) = 0 MB = - 6325 lb # ft = - 6.325 kip # ft
Ans.
Negative sign indicates that MB acts in the opposite direction to that shown on FBD. Equations of Equilibrium: For point C + ; © Fx = 0; + c © Fy = 0;
VC = 0
Ans.
- NC - 250 - 650 - 300 = 0 NC = - 1200 lb = - 1.20 kip
a + ©MC = 0;
Ans.
- MC - 650(6.5) - 300(13) = 0 MC = - 8125 lb # ft = - 8.125 kip # ft
Ans.
Negative signs indicate that NC and MC act in the opposite direction to that shown on FBD.
6
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1–11. The force F = 80 lb acts on the gear tooth. Determine the resultant internal loadings on the root of the tooth, i.e., at the centroid point A of section a–a.
a
F 80 lb 30
Equations of Equilibrium: For section a–a +
Q©Fx¿ = 0;
VA - 80 cos 15° = 0
0.23 in.
VA = 77.3 lb a+ ©Fy¿ = 0;
Ans. A
NA - 80 sin 15° = 0
0.16 in.
NA = 20.7 lb a + ©MA = 0;
Ans.
- MA - 80 sin 15°(0.16) + 80 cos 15°(0.23) = 0 MA = 14.5 lb # in.
45
Ans.
*1–12. The sky hook is used to support the cable of a scaffold over the side of a building. If it consists of a smooth rod that contacts the parapet of a wall at points A, B, and C, determine the normal force, shear force, and moment on the cross section at points D and E.
a
0.2 m
0.2 m B
0.2 m
0.2 m D
E
Support Reactions: + c ©Fy = 0;
NB - 18 = 0
d+ ©MC = 0;
18(0.7) - 18.0(0.2) - NA(0.1) = 0
0.2 m
0.3 m
NB = 18.0 kN
A
NA = 90.0 kN + : ©Fx = 0;
NC - 90.0 = 0
0.3 m
NC = 90.0 kN
Equations of Equilibrium: For point D + : © Fx = 0;
18 kN
VD - 90.0 = 0 VD = 90.0 kN
+ c © Fy = 0;
Ans.
ND - 18 = 0 ND = 18.0 kN
d+ © MD = 0;
Ans.
MD + 18(0.3) - 90.0(0.3) = 0 MD = 21.6 kN # m
Ans.
Equations of Equilibrium: For point E + : © Fx = 0; 90.0 - VE = 0 VE = 90.0 kN + c © Fy = 0; d + © ME = 0;
Ans.
NE = 0
Ans.
90.0(0.2) - ME = 0 ME = 18.0 kN # m
Ans.
7
C
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•1–13.
The 800-lb load is being hoisted at a constant speed using the motor M, which has a weight of 90 lb. Determine the resultant internal loadings acting on the cross section through point B in the beam. The beam has a weight of 40 lb>ft and is fixed to the wall at A.
M 1.5 ft A D 4 ft
+ : ©Fx = 0;
3 ft
3 ft
4 ft
0.25 ft
Ans.
VB - 0.8 - 0.16 = 0 VB = 0.960 kip
a + ©MB = 0;
B
- NB - 0.4 = 0 NB = - 0.4 kip
+ c ©Fy = 0;
4 ft
C
Ans.
- MB - 0.16(2) - 0.8(4.25) + 0.4(1.5) = 0 MB = - 3.12 kip # ft
Ans.
1–14. Determine the resultant internal loadings acting on the cross section through points C and D of the beam in Prob. 1–13.
M 1.5 ft A D 4 ft
4 ft
C
B
3 ft
3 ft
4 ft
For point C: + ; ©Fx = 0; + c ©Fy = 0; a + ©MC = 0;
NC + 0.4 = 0;
NC = - 0.4kip
VC - 0.8 - 0.04 (7) = 0;
Ans.
VC = 1.08 kip
Ans.
- MC - 0.8(7.25) - 0.04(7)(3.5) + 0.4(1.5) = 0 MC = - 6.18 kip # ft
Ans.
ND = 0
Ans.
For point D: + ; ©Fx = 0; + c ©Fy = 0; a + ©MD = 0;
VD - 0.09 - 0.04(14) - 0.8 = 0;
VD = 1.45 kip
Ans.
- MD - 0.09(4) - 0.04(14)(7) - 0.8(14.25) = 0 MD = - 15.7 kip # ft
Ans.
8
0.25 ft
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1–15. Determine the resultant internal loading on the cross section through point C of the pliers. There is a pin at A, and the jaws at B are smooth.
20 N 40 mm
120 mm
15 mm C
+ c ©Fy = 0; + : ©Fx = 0; +d ©MC = 0;
- VC + 60 = 0;
VC = 60 N
Ans.
NC = 0 - MC + 60(0.015) = 0;
MC = 0.9 N.m
D
Ans. 80 mm 20 N
*1–16. Determine the resultant internal loading on the cross section through point D of the pliers.
B
A
Ans.
30
20 N 40 mm
120 mm
15 mm
R+ ©Fy = 0;
VD - 20 cos 30° = 0;
VD = 17.3 N
Ans.
+b©Fx = 0;
ND - 20 sin 30° = 0;
ND = 10 N
Ans.
+d ©MD = 0;
MD - 20(0.08) = 0;
MD = 1.60 N.m
Ans.
C A D 80 mm 20 N
9
30
B
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•1–17.
Determine resultant internal loadings acting on section a–a and section b–b. Each section passes through the centerline at point C.
5 kN
B
b a
Referring to the FBD of the entire beam, Fig. a, a + ©MA = 0;
NB sin 45°(6) - 5(4.5) = 0
1.5 m
NB = 5.303 kN
C
Referring to the FBD of this segment (section a–a), Fig. b, b
+b©Fx¿ = 0;
Na - a + 5.303 cos 45° = 0
Na - a = - 3.75 kN Va - a = 1.25 kN
Ans.
+a ©Fy¿ = 0;
Va - a + 5.303 sin 45° - 5 = 0
a + ©MC = 0;
5.303 sin 45°(3) - 5(1.5) - Ma - a = 0 Ma - a = 3.75 kN # m Ans.
Ans.
Referring to the FBD (section b–b) in Fig. c, + ; ©Fx = 0;
Nb - b - 5 cos 45° + 5.303 = 0 Nb - b = - 1.768 kN = - 1.77 kN
+ c ©Fy = 0; a + ©MC = 0;
Vb - b - 5 sin 45° = 0
Vb - b = 3.536 kN = 3.54 kN
Ans. Ans.
5.303 sin 45° (3) - 5(1.5) - Mb - b = 0 Mb - b = 3.75 kN # m
10
Ans.
A
45
3m
45
1.5 m a
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1–18. The bolt shank is subjected to a tension of 80 lb. Determine the resultant internal loadings acting on the cross section at point C.
C 6 in. 90
A
B
Segment AC: + : ©Fx = 0;
NC + 80 = 0;
+ c ©Fy = 0;
VC = 0
NC = - 80 lb
Ans. Ans.
MC = - 480 lb # in.
Ans.
1–19. Determine the resultant internal loadings acting on the cross section through point C. Assume the reactions at the supports A and B are vertical.
6 kip/ft
a + ©MC = 0;
MC + 80(6) = 0;
6 kip/ft
A
C 3 ft
Referring to the FBD of the entire beam, Fig. a, a + ©MB = 0;
1 1 (6)(6)(2) + (6)(6)(10) - A y(12) = 0 A y = 18.0 kip 2 2
Referring to the FBD of this segment, Fig. b, + : ©Fx = 0;
NC = 0
+ c ©Fy = 0;
18.0 -
a + ©MC = 0;
Ans. 1 (3)(3) - (3)(3) - VC = 0 2
MC + (3)(3)(1.5) +
VC = 4.50 kip
Ans.
1 (3)(3)(2) - 18.0(3) = 0 2
MC = 31.5 kip # ft
Ans.
11
B
D 3 ft
6 ft
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*1–20. Determine the resultant internal loadings acting on the cross section through point D. Assume the reactions at the supports A and B are vertical.
6 kip/ft
Referring to the FBD of the entire beam, Fig. a,
A
a + ©MB = 0;
+ c ©Fy = 0; a + ©MA = 0;
C
1 1 (6)(6)(2) + (6)(6)(10) - A y(12) = 0 A y = 18.0 kip 2 2
Referring to the FBD of this segment, Fig. b, + : ©Fx = 0; ND = 0 18.0 -
1 (6)(6) - VD = 0 2
MD - 18.0 (2) = 0
6 kip/ft
3 ft
B
D 3 ft
6 ft
Ans. VD = 0
Ans.
MD = 36.0 kip # ft
Ans.
The forged steel clamp exerts a force of F = 900 N on the wooden block. Determine the resultant internal loadings acting on section a–a passing through point A.
•1–21.
200 mm F 900 N
Internal Loadings: Referring to the free-body diagram of the section of the clamp shown in Fig. a, ©Fy¿ = 0;
900 cos 30° - Na - a = 0
Na - a = 779 N
Ans.
©Fx¿ = 0;
Va - a - 900 sin 30° = 0
Va - a = 450 N
Ans.
a + ©MA = 0;
900(0.2) - Ma - a = 0
Ma - a = 180 N # m
12
Ans.
a
30 a
A
F 900 N
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1–22. The floor crane is used to lift a 600-kg concrete pipe. Determine the resultant internal loadings acting on the cross section at G.
0.2 m 0.2 m
0.4 m
0.6 m
G E B
F 0.3 m
H
C 0.5 m 75
A
Support Reactions: We will only need to compute FEF by writing the moment equation of equilibrium about D with reference to the free-body diagram of the hook, Fig. a. a + ©MD = 0;
FEF(0.3) - 600(9.81)(0.5) = 0
FEF = 9810 N
Internal Loadings: Using the result for FEF, section FG of member EF will be considered. Referring to the free-body diagram, Fig. b, + : ©Fx = 0; + c ©Fy = 0; a + ©MG = 0;
9810 - NG = 0
NG = 9810 N = 9.81 kN
VG = 0
Ans. Ans.
MG = 0
Ans.
13
D
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1–23. The floor crane is used to lift a 600-kg concrete pipe. Determine the resultant internal loadings acting on the cross section at H.
0.2 m 0.2 m
0.4 m
0.6 m
G E B
F 0.3 m
H
C 0.5 m 75
A
Support Reactions: Referring to the free-body diagram of the hook, Fig. a. a + ©MF = 0;
Dx(0.3) - 600(9.81)(0.5) = 0
Dx = 9810 N
+ c ©Fy = 0;
Dy - 600(9.81) = 0
Dy = 5886 N
Subsequently, referring to the free-body diagram of member BCD, Fig. b, a + ©MB = 0;
FAC sin 75°(0.4) - 5886(1.8) = 0
+ : ©Fx = 0;
Bx + 27 421.36 cos 75° - 9810 = 0 Bx = 2712.83 N
+ c ©Fy = 0;
27 421.36 sin 75° - 5886 - By = 0
FAC = 27 421.36 N
By = 20 601 N
Internal Loadings: Using the results of Bx and By, section BH of member BCD will be considered. Referring to the free-body diagram of this part shown in Fig. c, + : ©Fx = 0;
NH + 2712.83 = 0
NH = - 2712.83 N = - 2.71 kN
Ans.
+ c ©Fy = 0;
- VH - 2060 = 0
VH = - 20601 N = - 20.6 kN
Ans.
a + ©MD = 0;
MH + 20601(0.2) = 0
MH = - 4120.2 N # m = - 4.12 kN # m Ans.
The negative signs indicates that NH, VH, and MH act in the opposite sense to that shown on the free-body diagram.
14
D
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*1–24. The machine is moving with a constant velocity. It has a total mass of 20 Mg, and its center of mass is located at G, excluding the front roller. If the front roller has a mass of 5 Mg, determine the resultant internal loadings acting on point C of each of the two side members that support the roller. Neglect the mass of the side members. The front roller is free to roll.
2m
G
C
B
A
1.5 m
Support Reactions: We will only need to compute NA by writing the moment equation of equilibrium about B with reference to the free-body diagram of the steamroller, Fig. a. a + ©MB = 0; NA (5.5) - 20(103)(9.81)(1.5) = 0
NA = 53.51(103) N
Internal Loadings: Using the result for NA, the free-body diagram of the front roller shown in Fig. b will be considered. + ; ©Fx = 0; 2NC = 0
NC = 0
+ c ©Fy = 0; 2VC + 53.51(103) - 5(103)(9.81) = 0
VC = - 2229.55 N = - 2.23 kN
Ans.
Ans.
a + ©MC = 0; 53.51(103)(2) - 5(103)(9.81)(2) - 2MC = 0 MC = 4459.10 N # m = 4.46 kN # m Ans.
15
4m
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z
•1–25.
Determine the resultant internal loadings acting on the cross section through point B of the signpost. The post is fixed to the ground and a uniform pressure of 7 lb>ft2 acts perpendicular to the face of the sign.
3 ft 2 ft
©Fx = 0;
(VB)x - 105 = 0;
(VB)x = 105 lb
©Fy = 0;
(VB)y = 0
Ans.
©Fz = 0;
(NB)z = 0
Ans.
Ans. 3 ft 7 lb/ft2
Ans.
©Mx = 0;
(MB)x = 0
©My = 0;
(MB)y - 105(7.5) = 0;
©Mz = 0;
(TB)z - 105(0.5) = 0;
(MB)y = 788 lb # ft
6 ft
Ans.
(TB)z = 52.5 lb # ft
B
Ans. A
4 ft
y
x
1–26. The shaft is supported at its ends by two bearings A and B and is subjected to the forces applied to the pulleys fixed to the shaft. Determine the resultant internal loadings acting on the cross section located at point C. The 300-N forces act in the z direction and the 500-N forces act in the x direction. The journal bearings at A and B exert only x and z components of force on the shaft.
z
A
400 mm 150 mm 200 mm C 250 mm
x
300 N
300 N B
500 N 500 N y
©Fx = 0;
(VC)x + 1000 - 750 = 0;
©Fy = 0;
(NC)y = 0
©Fz = 0;
(VC)z + 240 = 0;
(VC)x = - 250 N
Ans. Ans. Ans.
(VC)z = - 240 N (MC)x = - 108 N # m
©Mx = 0;
(MC)x + 240(0.45) = 0;
©My = 0;
(TC)y = 0
©Mz = 0;
(MC)z - 1000(0.2) + 750(0.45) = 0;
Ans. Ans.
(MC)z = - 138 N # m Ans.
16
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1–27. The pipe has a mass of 12 kg>m. If it is fixed to the wall at A, determine the resultant internal loadings acting on the cross section at B. Neglect the weight of the wrench CD.
z
A
300 mm
200 mm B 60 N
x
D 400 mm
60 N 150 mm C
©Fx = 0;
(NB)x = 0
Ans.
©Fy = 0;
(VB)y = 0
Ans.
©Fz = 0;
(VB)z - 60 + 60 - (0.2)(12)(9.81) - (0.4)(12)(9.81) = 0 Ans.
(VB)z = 70.6 N ©Mx = 0;
(TB)x + 60(0.4) - 60(0.4) - (0.4)(12)(9.81)(0.2) = 0 (TB)x = 9.42 N # m
©My = 0;
Ans.
(MB)y + (0.2)(12)(9.81)(0.1) + (0.4)(12)(9.81)(0.2) - 60(0.3) = 0 (MB)y = 6.23 N # m
©Mz = 0;
Ans. Ans.
(MB)z = 0
17
150 mm
y
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z
*1–28. The brace and drill bit is used to drill a hole at O. If the drill bit jams when the brace is subjected to the forces shown, determine the resultant internal loadings acting on the cross section of the drill bit at A.
O
x
Internal Loading: Referring to the free-body diagram of the section of the drill and brace shown in Fig. a, ©Fx = 0; ©Fy = 0; ©Fz = 0; ©Mx = 0; ©My = 0; ©Mz = 0;
A VA B x - 30 = 0
A NA B y - 50 = 0 A VA B z - 10 = 0
A MA B x - 10(2.25) = 0
A TA B y - 30(0.75) = 0
A MA B z + 30(1.25) = 0
A VA B x = 30 lb
Ans.
A NA B y = 50 lb
Ans.
A VA B z = 10 lb
Ans.
A MA B x = 22.5 lb # ft A TA B y = 22.5 lb # ft
A MA B z = - 37.5 lb # ft
Ans. Ans. Ans.
The negative sign indicates that (MA)Z acts in the opposite sense to that shown on the free-body diagram.
18
3 in. 9 in.
Fx 30 lb
A
Fz 10 lb 9 in.
6 in.
6 in.
6 in.
Fy 50 lb y
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•1–29. The curved rod has a radius r and is fixed to the wall at B. Determine the resultant internal loadings acting on the cross section through A which is located at an angle u from the horizontal.
B
A r
U
Equations of Equilibrium: For point A R+ ©Fx = 0;
P
P cos u - NA = 0 NA = P cos u
Q+ ©Fy = 0;
Ans.
VA - P sin u = 0 VA = P sin u
d+ ©MA = 0;
Ans.
MA - P[r(1 - cos u)] = 0 MA = Pr(1 - cos u)
Ans.
19
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1–30. A differential element taken from a curved bar is shown in the figure. Show that dN>du = V, dV>du = - N, dM>du = - T, and dT>du = M.
M dM V dV
©Fx = 0; (1)
N
©Fy = 0; N sin
du du du du - V cos + (N + dN) sin + (V + dV) cos = 0 2 2 2 2
(2)
©Mx = 0; T cos
du du du du + M sin - (T + dT) cos + (M + dM) sin = 0 2 2 2 2
(3)
©My = 0; du du du du - M cos + (T + dT) sin + (M + dM) cos = 0 2 2 2 2 du du du du Since is can add, then sin , cos = = 1 2 2 2 2 T sin
Eq. (1) becomes Vdu - dN +
dVdu = 0 2
Neglecting the second order term, Vdu - dN = 0 dN = V du Eq. (2) becomes Ndu + dV +
QED dNdu = 0 2
Neglecting the second order term, Ndu + dV = 0 dV = -N du Eq. (3) becomes Mdu - dT +
QED dMdu = 0 2
Neglecting the second order term, Mdu - dT = 0 dT = M du Eq. (4) becomes Tdu + dM +
QED dTdu = 0 2
Neglecting the second order term, Tdu + dM = 0 dM = -T du
N dN
M V
du du du du + V sin - (N + dN) cos + (V + dV) sin = 0 N cos 2 2 2 2
QED
20
(4)
T
T dT
du
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1–31. The column is subjected to an axial force of 8 kN, which is applied through the centroid of the cross-sectional area. Determine the average normal stress acting at section a–a. Show this distribution of stress acting over the area’s cross section.
8 kN 75 mm 75 mm
10 mm
10 mm
70 mm
10 mm
70 mm a a
A = (2)(150)(10) + (140)(10) = 4400 mm2 = 4.4 (10-3) m2 s =
8 (103) P = 1.82 MPa = A 4.4 (10 - 3)
Ans.
*1–32. The lever is held to the fixed shaft using a tapered pin AB, which has a mean diameter of 6 mm. If a couple is applied to the lever, determine the average shear stress in the pin between the pin and lever.
B 12 mm A 250 mm 20 N
a + ©MO = 0; tavg =
- F(12) + 20(500) = 0;
F = 833.33 N
V 833.33 = p 6 = 29.5 MPa 2 A 4 (1000 )
Ans.
21
250 mm 20 N
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•1–33.
The bar has a cross-sectional area A and is subjected to the axial load P. Determine the average normal and average shear stresses acting over the shaded section, which is oriented at u from the horizontal. Plot the variation of these stresses as a function of u 10 … u … 90°2.
P
P u A
Equations of Equilibrium: R+ ©Fx = 0;
V - P cos u = 0
V = P cos u
Q+ ©Fy = 0;
N - P sin u = 0
N = P sin u
Average Normal Stress and Shear Stress: Area at u plane, A¿ =
s =
A . sin u
P sin u N P = = sin2 u A A¿ A sin u
tavg =
Ans.
V P cos u = A A¿ sin u =
P P sin u cos u = sin 2u A 2A
Ans.
1–34. The built-up shaft consists of a pipe AB and solid rod BC. The pipe has an inner diameter of 20 mm and outer diameter of 28 mm. The rod has a diameter of 12 mm. Determine the average normal stress at points D and E and represent the stress on a volume element located at each of these points.
4 kN
D
At D: sD =
4(103)
P = A
p 2 4 (0.028
P = A
p 4
- 0.02 2)
= 13.3 MPa (C)
Ans.
At E: sE =
8(103) (0.012 2)
B
A
= 70.7 MPa (T)
Ans.
22
6 kN 6 kN E
C
8 kN
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1–35. The bars of the truss each have a cross-sectional area of 1.25 in2. Determine the average normal stress in each member due to the loading P = 8 kip. State whether the stress is tensile or compressive.
B
3 ft
A 4 ft
P
Joint A: sAB =
FAB 13.33 = = 10.7 ksi A AB 1.25
(T)
Ans.
sAE =
FAE 10.67 = = 8.53 ksi A AE 1.25
(C)
Ans.
(C)
Ans.
Joint E: sED =
FED 10.67 = = 8.53 ksi A ED 1.25
sEB =
FEB 6.0 = = 4.80 ksi A EB 1.25
C
Ans.
(T)
Joint B: sBC =
FBC 29.33 = 23.5 ksi = A BC 1.25
(T)
Ans.
sBD =
FBD 23.33 = = 18.7 ksi A BD 1.25
(C)
Ans.
23
E 0.75 P
4 ft
D
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*1–36. The bars of the truss each have a cross-sectional area of 1.25 in2. If the maximum average normal stress in any bar is not to exceed 20 ksi, determine the maximum magnitude P of the loads that can be applied to the truss.
B
3 ft
A 4 ft
P
Joint A: + c ©Fy = 0;
3 - P + a b FAB = 0 5
FAB = (1.667)P + : ©Fx = 0;
4 - FAE + (1.667)Pa b = 0 5 FAE = (1.333)P
Joint E: + c ©Fy = 0;
FEB - (0.75)P = 0 FEB = (0.75)P
+ : ©Fx = 0;
(1.333)P - FED = 0 FED = (1.333)P
Joint B: + c ©Fy = 0;
3 3 a b FBD - (0.75)P - (1.667)Pa b = 0 5 5 FBD = (2.9167)P
+ : ©Fx = 0;
4 4 FBC - (2.9167)Pa b - (1.667)P a b = 0 5 5
FBC = (3.67)P The highest stressed member is BC:
sBC =
C
(3.67)P = 20 1.25
P = 6.82 kip
Ans.
24
E 0.75 P
4 ft
D
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•1–37.
The plate has a width of 0.5 m. If the stress distribution at the support varies as shown, determine the force P applied to the plate and the distance d to where it is applied.
4m P d x
s (15x1/2) MPa
The resultant force dF of the bearing pressure acting on the plate of area dA = b dx = 0.5 dx, Fig. a, 1
1
dF = sb dA = (15x2)(106)(0.5dx) = 7.5(106)x2 dx + c ©Fy = 0;
L
dF - P = 0 4m 1
L0
7.5(106)x2 dx - P = 0
P = 40(106) N = 40 MN
Ans.
Equilibrium requires a + ©MO = 0;
L
xdF - Pd = 0
4m 1
L0
x[7.5(106)x2 dx] - 40(106) d = 0 d = 2.40 m
Ans.
25
30 MPa
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1–38. The two members used in the construction of an aircraft fuselage are joined together using a 30° fish-mouth weld. Determine the average normal and average shear stress on the plane of each weld. Assume each inclined plane supports a horizontal force of 400 lb. N - 400 sin 30° = 0;
N = 200 lb
400 cos 30° - V = 0;
V = 346.41 lb
A¿ =
1.5 in.
30
1 in. 1 in.
800 lb
800 lb 30
1.5(1) = 3 in2 sin 30°
s =
N 200 = = 66.7 psi A¿ 3
Ans.
t =
V 346.41 = = 115 psi A¿ 3
Ans.
1–39. If the block is subjected to the centrally applied force of 600 kN, determine the average normal stress in the material. Show the stress acting on a differential volume element of the material.
150 mm 600 kN
150 mm 150 mm 150 mm
The cross-sectional area of the block is A = 0.6(0.3) - 0.3(0.2) = 0.12 m2. savg =
600(103) P = = 5(106) Pa = 5 MPa A 0.12
Ans.
The average normal stress distribution over the cross-section of the block and the state of stress of a point in the block represented by a differential volume element are shown in Fig. a
26
50 mm 100 mm 100 mm 50 mm
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*1–40. The pins on the frame at B and C each have a diameter of 0.25 in. If these pins are subjected to double shear, determine the average shear stress in each pin.
3 ft 500 lb
3 ft A 3 ft
Support Reactions: FBD(a) a + ©Mg = 0;
500(6) + 300(3) - Dy (6) = 0 Dy = 650 lb
+ ; ©Fx = 0;
500 - Ex = 0
Ex = 500 lb
+ c ©Fy = 0;
650 - 300 - Ey = 0
Ey = 350 lb
a + ©MB = 0;
Cy (3) - 300(1.5) = 0
Cy = 150 lb
+ c ©Fy = 0;
By + 150 - 300 = 0
By = 150 lb
1.5 ft
From FBD (b) 150(1.5) + Bx(3) - 650(3) = 0 Bx = 575 lb From FBD (c), + : ©Fx = 0;
Cx - 575 = 0
Cx = 575 lb
Hence, FB = FC = 2 5752 + 1502 = 594.24 lb Average shear stress: Pins B and C are subjected to double shear as shown on FBD (d) (tB)avg = (tC)avg =
1.5 ft
300 lb D
From FBD (c),
a + ©MA = 0;
C
B
V 297.12 = p 2 A 4 (0.25 ) = 6053 psi = 6.05 ksi
Ans.
27
3 ft E
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•1–41.
Solve Prob. 1–40 assuming that pins B and C are subjected to single shear.
3 ft 500 lb
3 ft A 3 ft C
B
Support Reactions: FBD(a) a + ©Mg = 0;
1.5 ft
500(6) + 300(3) - Dy (6) = 0
300 lb D
Dy = 650 lb + ; ©Fx = 0;
500 - Ex = 0
Ex = 500 lb
+ c ©Fy = 0;
650 - 300 - Ey = 0
Ey = 350 lb
From FBD (c), a + ©MB = 0; + c ©Fy = 0;
Cy (3) - 300(1.5) = 0 By + 150 - 300 = 0
Cy = 150 lb By = 150 lb
From FBD (b) d+ ©MA = 0;
150(1.5) + Bx(3) - 650(3) = 0 Bx = 575 lb
From FBD (c), + : ©Fx = 0;
Cx - 575 = 0
Cx = 575 lb
Hence, FB = FC = 2 5752 + 1502 = 594.24 lb Average shear stress: Pins B and C are subjected to single shear as shown on FBD (d) (tB)avg = (tC)avg =
1.5 ft
594.24 V = p 2 A 4 (0.25 ) = 12106 psi = 12.1 ksi
Ans.
28
3 ft E
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1–42. The pins on the frame at D and E each have a diameter of 0.25 in. If these pins are subjected to double shear, determine the average shear stress in each pin.
3 ft 500 lb
3 ft A 3 ft C
B 1.5 ft
1.5 ft
Support Reactions: FBD(a) a + ©ME = 0;
300 lb
500(6) + 300(3) - Dy(6) = 0
D
Dy = 650 lb + ; ©Fx = 0;
500 - Ex = 0
Ex = 500 lb
+ c ©Fy = 0;
650 - 300 - Ey = 0
Ey = 350 lb
Average shear stress: Pins D and E are subjected to double shear as shown on FBD (b) and (c). For Pin D, FD = Dy = 650 lb then VD = (pD)avg =
VD = AD
FD z
= 325 lb
325 p 2 4 (0.25)
Ans.
= 6621 psi = 6.62 ksi For Pin E, FE = 2 5002 + 3502 = 610.32 lb then VE = (tE)avg =
Fg z
= 305.16 lb
VE 305.16 = p 2 AE 4 (0.25 ) = 6217 psi = 6.22 ksi
Ans.
29
3 ft E
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1–43. Solve Prob. 1–42 assuming that pins D and E are subjected to single shear.
3 ft 500 lb
3 ft A
Support Reactions: FBD(a) a + ©ME = 0;
3 ft
500(6) + 300(3) - Dy(6) = 0
+ ; ©Fx = 0;
500 - Ex = 0 650 - 300 - Ey = 0
+ c ©Fy = 0;
C
B
Dy = 650 lb
1.5 ft
Ex = 500 lb Ey = 350 lb
1.5 ft
300 lb D
Average shear stress: Pins D and E are subjected to single shear as shown on FBD (b) and (c). For Pin D, VD = FD = Dy = 650 lb (tD)avg =
VD = AD
650 p 2 4 (0.25 )
Ans.
= 13242 psi = 13.2 ksi For Pin E, VE = FE = 2 5002 + 3502 = 610.32 lb (tE)avg =
VE 610.32 = p 2 AE 4 (0.25 ) = 12433 psi = 12.4 ksi
Ans.
*1–44. A 175-lb woman stands on a vinyl floor wearing stiletto high-heel shoes. If the heel has the dimensions shown, determine the average normal stress she exerts on the floor and compare it with the average normal stress developed when a man having the same weight is wearing flat-heeled shoes. Assume the load is applied slowly, so that dynamic effects can be ignored. Also, assume the entire weight is supported only by the heel of one shoe. Stiletto shoes: A =
1.2 in.
1 (p)(0.3)2 + (0.6)(0.1) = 0.2014 in2 2
0.3 in. 0.1 in. 0.5 in.
P 175 lb = s = = 869 psi A 0.2014 in2
Ans.
Flat-heeled shoes: A =
1 (p)(1.2)2 + 2.4(0.5) = 3.462 in2 2
s =
P 175 lb = = 50.5 psi A 3.462 in2
Ans.
30
3 ft E
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•1–45.
The truss is made from three pin-connected members having the cross-sectional areas shown in the figure. Determine the average normal stress developed in each member when the truss is subjected to the load shown. State whether the stress is tensile or compressive.
500 lb
C
sBC =
FBC 375 = = 469 psi A BC 0.8
(T)
Ans.
(T)
Ans.
4 ft
Joint A: FAC 500 = = 833 psi A AC 0.6
.2 in
A
1–46. Determine the average normal stress developed in links AB and CD of the smooth two-tine grapple that supports the log having a mass of 3 Mg. The cross-sectional area of each link is 400 mm2. + c ©Fy = 0;
1.5
Ans.
(C)
AB
FAB 625 = = 417 psi A AB 1.5
A
sAB =
AAC 0.6 in.2
ABC 0.8 in.2
Joint B:
œ sAC =
3 ft
A
C
20 B
E
D
2(F sin 30°) - 29.43 = 0 0.2 m
F = 29.43 kN a + ©ME = 0; P cos 20°(0.2) - (29.43 cos 30°)(1.2) + (29.43 sin 30°)(0.4 cos 30°)
1.2 m
= 0 P = 135.61 kN
30
30 0.4 m
s =
135.61(103) P = 339 MPa = A 400(10 - 6)
Ans.
31
B
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1–47. Determine the average shear stress developed in pins A and B of the smooth two-tine grapple that supports the log having a mass of 3 Mg. Each pin has a diameter of 25 mm and is subjected to double shear. + c ©Fy = 0;
A
C
20 E
B
D
2(F sin 30°) - 29.43 = 0 0.2 m
F = 29.43 kN 1.2 m
a + ©ME = 0; P cos 20°(0.2) - (29.43 cos 30°)(1.2) + (29.43 sin 30°)(0.4 cos 30°) = 0 P = 135.61 kN tA = tB =
V = A
135.61(103) 2 p 2 (0.025) 4
30
30 0.4 m
= 138 MPa
Ans.
*1–48. The beam is supported by a pin at A and a short link BC. If P = 15 kN, determine the average shear stress developed in the pins at A, B, and C. All pins are in double shear as shown, and each has a diameter of 18 mm.
P
4P
4P
2P
0.5m
0.5 m 1m
1.5 m
1.5 m
C 30 B
For pins B and C: 82.5 (103) V tB = tC = = p 18 2 = 324 MPa A 4 (1000 )
A
Ans.
For pin A: FA = 2 (82.5)2 + (142.9)2 = 165 kN tA =
82.5 (103) V = p 18 2 = 324 MPa A 4 (1000 )
Ans.
32
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•1–49. The beam is supported by a pin at A and a short link BC. Determine the maximum magnitude P of the loads the beam will support if the average shear stress in each pin is not to exceed 80 MPa. All pins are in double shear as shown, and each has a diameter of 18 mm.
P
4P
4P
2P
0.5m
0.5 m 1m
1.5 m
1.5 m
C 30 B
a + ©MA = 0;
A
2P(0.5) + 4P(2) + 4P(3.5) + P(4.5) - (TCB sin 30°)(5) = 0
TCB = 11P + : ©Fx = 0;
A x - 11P cos 30° = 0
A x = 9.5263P + c ©Fy = 0;
A y - 11P + 11P sin 30° = 0
A y = 5.5P FA = 2 (9.5263P)2 + (5.5P)2 = 11P Require; t =
V ; A
80(106) =
11P>2 p 2 4 (0.018)
P = 3.70 kN
Ans.
33
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1–50. The block is subjected to a compressive force of 2 kN. Determine the average normal and average shear stress developed in the wood fibers that are oriented along section a–a at 30° with the axis of the block.
50 mm a
150 mm
2 kN
2 kN 30
a
Force equilibrium equations written perpendicular and parallel to section a–a gives +Q©Fx¿ = 0;
Va - a - 2 cos 30° = 0
Va - a = 1.732 kN
+a©Fy¿ = 0;
2 sin 30° - Na - a = 0
Na - a = 1.00 kN
The cross sectional area of section a–a is A = a
0.15 b (0.05) = 0.015 m2. Thus sin 30°
(sa - a)avg =
1.00(103) Na - a = = 66.67(103)Pa = 66.7 kPa A 0.015
Ans.
(ta - a)avg =
1.732(103) Va - a = = 115.47(103)Pa = 115 kPa A 0.015
Ans.
34
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1–51. During the tension test, the wooden specimen is subjected to an average normal stress of 2 ksi. Determine the axial force P applied to the specimen. Also, find the average shear stress developed along section a–a of the specimen.
P
a
4 in.
a 2 in.
1 in.
Internal Loading: The normal force developed on the cross section of the middle portion of the specimen can be obtained by considering the free-body diagram shown in Fig. a. + c ©Fy = 0;
P P + - N = 0 2 2
4 in.
N = P
Referring to the free-body diagram shown in fig. b, the shear force developed in the shear plane a–a is + c ©Fy = 0;
P - Va - a = 0 2
Va - a =
P 2
Average Normal Stress and Shear Stress: The cross-sectional area of the specimen is A = 1(2) = 2 in2. We have savg =
N ; A
2(103) =
P 2
P = 4(103)lb = 4 kip
Ans.
4(103) P = = 2(103) lb. The area of the shear plane is 2 2 = 2(4) = 8 in2. We obtain
Using the result of P, Va - a = Aa - a
A ta - a B avg =
2(103) Va - a = = 250 psi Aa - a 8
Ans.
35
P
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*1–52. If the joint is subjected to an axial force of P = 9 kN, determine the average shear stress developed in each of the 6-mm diameter bolts between the plates and the members and along each of the four shaded shear planes.
P
P 100 mm
Internal Loadings: The shear force developed on each shear plane of the bolt and the member can be determined by writing the force equation of equilibrium along the member’s axis with reference to the free-body diagrams shown in Figs. a. and b, respectively. ©Fy = 0; 4Vb - 9 = 0
Vb = 2.25 kN
©Fy = 0; 4Vp - 9 = 0
Vp = 2.25 kN
Average Shear Stress: The areas of each shear plane of the bolt and the member p are A b = (0.0062) = 28.274(10 - 6)m2 and A p = 0.1(0.1) = 0.01 m2, respectively. 4 We obtain
A tavg B b =
2.25(103) Vb = 79.6 MPa = Ab 28.274(10 - 6)
A tavg B p =
Vp Ap
=
Ans.
2.25(103) = 225 kPa 0.01
Ans.
36
100 mm
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•1–53.
The average shear stress in each of the 6-mm diameter bolts and along each of the four shaded shear planes is not allowed to exceed 80 MPa and 500 kPa, respectively. Determine the maximum axial force P that can be applied to the joint.
P
P 100 mm 100 mm
Internal Loadings: The shear force developed on each shear plane of the bolt and the member can be determined by writing the force equation of equilibrium along the member’s axis with reference to the free-body diagrams shown in Figs. a. and b, respectively. ©Fy = 0;
4Vb - P = 0
Vb = P>4
©Fy = 0;
4Vp - P = 0
Vp = P>4
Average Shear Stress: The areas of each shear plane of the bolts and the members p are A b = (0.0062) = 28.274(10 - 6)m2 and A p = 0.1(0.1) = 0.01m2, respectively. 4 We obtain
A tallow B b =
Vb ; Ab
80(106) =
P>4 28.274(10 - 6)
P = 9047 N = 9.05 kN (controls)
A tallow B p =
Vp Ap
;
500(103) =
Ans.
P>4 0.01
P = 20 000 N = 20 kN
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1–54. The shaft is subjected to the axial force of 40 kN. Determine the average bearing stress acting on the collar C and the normal stress in the shaft.
40 kN
30 mm
C
40 mm
Referring to the FBDs in Fig. a, + c ©Fy = 0;
Ns - 40 = 0
Ns = 40 kN
+ c ©Fy = 0;
Nb - 40 = 0
Nb = 40 kN
Here, the cross-sectional area of the shaft and the bearing area of the collar are p p A s = (0.032) = 0.225(10 - 3)p m2 and A b = (0.04 2) = 0.4(10 - 3)p m2. Thus, 4 4
A savg B s =
40(103) Ns = 56.59(106) Pa = 56.6 MPa = As 0.225(10 - 3)p
Ans.
A savg B b =
40(103) Nb = 31.83(106)Pa = 31.8 MPa = Ab 0.4(10 - 3)p
Ans.
38
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1–55. Rods AB and BC each have a diameter of 5 mm. If the load of P = 2 kN is applied to the ring, determine the average normal stress in each rod if u = 60°.
A u P B
Consider the equilibrium of joint B, Fig. a, + : ©Fx = 0;
2 - FAB sin 60° = 0
+ c ©Fy = 0;
2.309 cos 60° - FBC = 0
FAB = 2.309 kN
C
FBC = 1.155 kN
The cross-sectional area of wires AB and BC are A AB = A BC =
p (0.0052) 4
= 6.25(10 - 6)p m2. Thus,
A savg B AB =
2.309(103) FAB = 117.62(106) Pa = 118 MPa = A AB 6.25(10 - 6)p
Ans.
A savg B BC =
1.155(103) FBC = 58.81(106) Pa = 58.8 MPa = A BC 6.25(10 - 6)p
Ans.
39
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*1–56. Rods AB and BC each have a diameter of 5 mm. Determine the angle u of rod BC so that the average normal stress in rod AB is 1.5 times that in rod BC. What is the load P that will cause this to happen if the average normal stress in each rod is not allowed to exceed 100 MPa?
A u P B
Consider the equilibrium of joint B, Fig. a, FAB cos u - FBC = 0
+ c ©Fy = 0; + : ©Fx = 0;
(1)
P - FAB sin u = 0
(2)
p (0.0052) 4 = 6.25(10 - 6)p m2. Since the average normal stress in rod AB is required to be
The cross-sectional area of rods AB and BC are A AB = A BC =
1.5 times to that of rod BC, then
A savg B AB = 1.5 A savg B BC FAB FBC = 1.5 a b A AB A BC FAB 6.25(10 - 6)p
= 1.5 c
FBC 6.25(10 - 6)p
d
FAB = 1.5 FBC
(3)
Solving Eqs (1) and (3), u = 48.19° = 48.2°
Ans.
Since wire AB will achieve the average normal stress of 100 MPa first when P increases, then FAB = sallow A AB = C 100(106) D C 6.25(10 - 6)p D = 1963.50 N Substitute the result of FAB and u into Eq (2), P = 1.46 kN
Ans.
40
C
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•1–57.
The specimen failed in a tension test at an angle of 52° when the axial load was 19.80 kip. If the diameter of the specimen is 0.5 in., determine the average normal and average shear stress acting on the area of the inclined failure plane. Also, what is the average normal stress acting on the cross section when failure occurs? +
b © Fx = 0;
52 0.5 in.
V - 19.80 cos 52° = 0 V = 12.19 kip
+a © Fy = 0;
N - 19.80 sin 52° = 0 N = 15.603 kip
Inclined plane: s¿ =
P ; A
œ tavg =
s¿ =
V ; A
15.603 p(0.25)2 sin 52°
œ tavg =
Ans.
= 62.6 ksi
12.19 p(0.25)2 sin 52°
Ans.
= 48.9 ksi
Cross section: s =
P ; A
tavg =
s =
V ; A
19.80 = 101 ksi p(0.25)2
Ans.
tavg = 0
Ans.
1–58. The anchor bolt was pulled out of the concrete wall and the failure surface formed part of a frustum and cylinder. This indicates a shear failure occurred along the cylinder BC and tension failure along the frustum AB. If the shear and normal stresses along these surfaces have the magnitudes shown, determine the force P that must have been applied to the bolt.
P
A 45
45 50 mm
Average Normal Stress: For the frustum, A = 2pxL = 2p(0.025 + 0.025) A 2 0.05 + 0.05 2
2
B
3 MPa
3 MPa B
= 0.02221 m2 s =
P ; A
3 A 106 B =
4.5 MPa
F1 0.02221
C 25 mm 25 mm
F1 = 66.64 kN Average Shear Stress: For the cylinder, A = p(0.05)(0.03) = 0.004712 m2 F2 V ; 4.5 A 106 B = tavg = A 0.004712 F2 = 21.21 kN Equation of Equilibrium: + c ©Fy = 0;
P - 21.21 - 66.64 sin 45° = 0 P = 68.3 kN
Ans.
41
30 mm
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1–59. The open square butt joint is used to transmit a force of 50 kip from one plate to the other. Determine the average normal and average shear stress components that this loading creates on the face of the weld, section AB.
50 kip 30 30 2 in.
Equations of Equilibrium: a+ © Fy = 0; +
Q© Fx = 0;
N - 50 cos 30° = 0 - V + 50 sin 30° = 0
B
N = 43.30 kip
A 6 in.
V = 25.0 kip
50 kip
Average Normal and Shear Stress: A¿ = a s = tavg
2 b (6) = 13.86 in2 sin 60°
N 43.30 = = 3.125 ksi A¿ 13.86 V 25.0 = = = 1.80 ksi A¿ 13.86
Ans. Ans.
*1–60. If P = 20 kN, determine the average shear stress developed in the pins at A and C. The pins are subjected to double shear as shown, and each has a diameter of 18 mm.
C
Referring to the FBD of member AB, Fig. a a + ©MA = 0;
30
FBC sin 30° (6) - 20(2) - 20(4) = 0
FBC = 40 kN A
+ : ©Fx = 0;
A x - 40 cos 30° = 0
+ c ©Fy = 0;
A y - 20 - 20 + 40 sin 30°
A x = 34.64 kN A y = 20 kN
2m
P
Thus, the force acting on pin A is FA = 2 A x 2 + A y 2 = 2 34.64 2 + 202 = 40 kN Pins A and C are subjected to double shear. Referring to their FBDs in Figs. b and c, FA FBC 40 40 VA = VC = = = 20 kN = = 20 kN 2 2 2 2 p The cross-sectional area of Pins A and C are A A = A C = (0.0182) 4 = 81(10 - 6)p m2. Thus tA =
20(103) VA = 78.59(106) Pa = 78.6 MPa = AA 81(10 - 6)p
Ans.
tC =
20(103) VC = 78.59(106) Pa = 78.6 MPa = AC 81(10 - 6)p
Ans.
42
B 2m
2m P
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•1–61. Determine the maximum magnitude P of the load the beam will support if the average shear stress in each pin is not to allowed to exceed 60 MPa. All pins are subjected to double shear as shown, and each has a diameter of 18 mm.
C 30
Referring to the FBD of member AB, Fig. a, a + ©MA = 0; + : ©Fx = 0;
FBC sin 30°(6) - P(2) - P(4) = 0 A x - 2P cos 30° = 0
A x = 1.732P
A y - P - P + 2P sin 30° = 0
+ c ©Fy = 0;
A
FBC = 2P
FA = 2 A x 2 + A y 2 = 2 (1.732P)2 + P2 = 2P All pins are subjected to same force and double shear. Referring to the FBD of the pin, Fig. b, 2P F = = P 2 2
The cross-sectional area of the pin is A = tallow =
V ; A
60(106) =
2m
P
Ay = P
Thus, the force acting on pin A is
V =
B 2m
p (0.0182) = 81.0(10 - 6)p m2. Thus, 4
P 81.0(10 - 6)p
P = 15 268 N = 15.3 kN
Ans.
43
2m P
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1–62. The crimping tool is used to crimp the end of the wire E. If a force of 20 lb is applied to the handles, determine the average shear stress in the pin at A. The pin is subjected to double shear and has a diameter of 0.2 in. Only a vertical force is exerted on the wire.
20 lb
C
E A
Support Reactions:
5 in.
From FBD(a) a + ©MD = 0;
B D
1.5 in. 2 in. 1 in.
20(5) - By (1) = 0
+ : ©Fx = 0;
20 lb
By = 100 lb
Bx = 0
From FBD(b) + : ©Fx = 0; a + ©ME = 0;
Ax = 0 A y (1.5) - 100(3.5) = 0 A y = 233.33 lb
Average Shear Stress: Pin A is subjected to double shear. Hence, Ay FA VA = = = 116.67 lb 2 2 (tA)avg =
VA 116.67 = p 2 AA 4 (0.2 )
= 3714 psi = 3.71 ksi
Ans.
1–63. Solve Prob. 1–62 for pin B. The pin is subjected to double shear and has a diameter of 0.2 in.
20 lb
Support Reactions:
a + ©MD = 0; + : ©Fx = 0;
C
E
From FBD(a)
A
20(5) - By (1) = 0
By = 100 lb
5 in.
Bx = 0
1.5 in. 2 in. 1 in.
Average Shear Stress: Pin B is subjected to double shear. Hence, By FB VB = = = 50.0 lb 2 2 (tB)avg =
B D
VB = AB
p 4
50.0 (0.2 2)
= 1592 psi = 1.59 ksi
Ans.
44
20 lb
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*1–64. The triangular blocks are glued along each side of the joint. A C-clamp placed between two of the blocks is used to draw the joint tight. If the glue can withstand a maximum average shear stress of 800 kPa, determine the maximum allowable clamping force F.
50 mm
F glue
45
25 mm
Internal Loadings: The shear force developed on the glued shear plane can be obtained by writing the force equation of equilibrium along the x axis with reference to the free-body diagram of the triangular block, Fig. a. + : ©Fx = 0;
F cos 45° - V = 0
V =
F
2 2 F 2
Average Normal and Shear Stress: The area of the glued shear plane is A = 0.05(0.025) = 1.25(10 - 3)m2. We obtain
tavg =
V ; A
2 2 F 2 800(103) = 1.25(10 - 3) F = 1414 N = 1.41 kN
Ans.
•1–65.
The triangular blocks are glued along each side of the joint. A C-clamp placed between two of the blocks is used to draw the joint tight. If the clamping force is F = 900 N, determine the average shear stress developed in the glued shear plane.
50 mm 45
F glue 25 mm
Internal Loadings: The shear force developed on the glued shear plane can be obtained by writing the force equation of equilibrium along the x axis with reference to the free-body diagram of the triangular block, Fig. a. + : ©Fx = 0;
900 cos 45° - V = 0
V = 636.40 N
Average Normal and Shear Stress: The area of the glued shear plane is A = 0.05(0.025) = 1.25(10 - 3)m2. We obtain tavg =
V 636.40 = 509 kPa = A 1.25(10 - 3)
Ans.
45
F
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1–66. Determine the largest load P that can be a applied to the frame without causing either the average normal stress or the average shear stress at section a–a to exceed s = 150 MPa and t = 60 MPa, respectively. Member CB has a square cross section of 25 mm on each side.
B
Analyse the equilibrium of joint C using the FBD Shown in Fig. a, + c ©Fy = 0;
4 FBC a b - P = 0 5
FBC = 1.25P
2m
a
Referring to the FBD of the cut segment of member BC Fig. b. + : ©Fx = 0;
+ c ©Fy = 0; The
3 Na - a - 1.25Pa b = 0 5 4 1.25Pa b - Va - a = 0 5
cross-sectional
area
of
section
Na - a = 0.75P a A
Va - a = P a–a
is
A a - a = (0.025) a
0.025 b 3>5
= 1.0417(10 - 3)m2. For Normal stress, sallow =
Na - a ; Aa - a
150(106) =
0.75P 1.0417(10 - 3)
P = 208.33(103) N = 208.33 kN For Shear Stress Va - a ; tallow = Aa - a
60(106) =
P 1.0417(10 - 3)
P = 62.5(103) N = 62.5 kN (Controls!)
Ans.
46
C 1.5 m P
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1–67. The prismatic bar has a cross-sectional area A. If it is subjected to a distributed axial loading that increases linearly from w = 0 at x = 0 to w = w0 at x = a, and then decreases linearly to w = 0 at x = 2a, determine the average normal stress in the bar as a function of x for 0 … x 6 a.
w0
x a
a
Equation of Equilibrium: + : ©Fx = 0;
1 w0 1 a x + w0 b(a - x) + w0a = 0 2 a 2
-N +
N =
w0 A 2a2 - x2 B 2a
Average Normal Stress: N s = = A
w0 2a
(2a2 - x2) =
A
w0 A 2a2 - x2 B 2aA
Ans.
*1–68. The prismatic bar has a cross-sectional area A. If it is subjected to a distributed axial loading that increases linearly from w = 0 at x = 0 to w = w0 at x = a, and then decreases linearly to w = 0 at x = 2a, determine the average normal stress in the bar as a function of x for a 6 x … 2a.
w0
x a
a
Equation of Equilibrium: + : ©Fx = 0;
-N +
1 w0 c (2a - x) d(2a - x) = 0 2 a N =
w0 (2a - x)2 2a
Average Normal Stress: s =
N = A
w0 2a
(2a - x)2 =
A
w0 (2a - x)2 2aA
Ans.
The tapered rod has a radius of r = (2 - x>6) in. and is subjected to the distributed loading of w = (60 + 40x) lb>in. Determine the average normal stress at the center of the rod, B.
•1–69.
A = pa 2 -
r w (60 40x) lb/ in. x r = (2 — ) in. 6
3 2 b = 7.069 in2 6
x B
6
© Fx = 0;
N -
L3
(60 + 40x) dx = 0;
3 in.
N = 720 lb
720 s = = 102 psi 7.069
Ans.
47
3 in.
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1–70. The pedestal supports a load P at its center. If the material has a mass density r, determine the radial dimension r as a function of z so that the average normal stress in the pedestal remains constant. The cross section is circular.
P r1
z r
Require: P + W1 P + W1 + dW s = = A A + dA P dA + W1 dA = A dW P + W1 dW = = s dA A
(1)
dA = p(r + dr)2 - pr2 = 2p r dr dW = pr2(rg) dt From Eq. (1) pr2(rg) dz = s 2p r dr r rg dz = s 2 dr z r rg dr dz = 2s L0 Lr1 r
rg z r = ln ; r1 2s
p
r = r1 e(2a)z
However, s =
P p r21
r = r1 e
(p r12rg )z
Ans.
2P
48
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1–71. Determine the average normal stress at section a–a and the average shear stress at section b–b in member AB. The cross section is square, 0.5 in. on each side.
150 lb/ft
Consider the FBD of member BC, Fig. a, a + ©MC = 0;
B
4 ft
C
60 a
FAB sin 60°(4) - 150(4)(2) = 0
FAB = 346.41 lb a
Referring to the FBD in Fig. b, +
b©Fx¿ = 0;
Na - a + 346.41 = 0
b
Na - a = - 346.41 lb
Referring to the FBD in Fig. c. + c ©Fy = 0;
Vb - b - 346.41 sin 60° = 0
b
Vb - b = 300 lb
The cross-sectional areas of section a–a and b–b are A a - a = 0.5(0.5) = 0.25 in2 and 0.5 b = 0.5 in2. Thus A b - b = 0.5 a cos 60° Na - a 346.41 Ans. = = 1385.64 psi = 1.39 ksi sa - a = Aa - a 0.25 tb - b =
Vb - b 300 = = 600 psi Ab - b 0.5
Ans.
49
A
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•1–73. Member B is subjected to a compressive force of 800 lb. If A and B are both made of wood and are 38 in. thick, determine to the nearest 14 in. the smallest dimension h of the horizontal segment so that it does not fail in shear. The average shear stress for the segment is tallow = 300 psi.
tallow = 300 =
B
307.7
13
(32)
12
h h
3 in. 4
Ans.
1–74. The lever is attached to the shaft A using a key that has a width d and length of 25 mm. If the shaft is fixed and a vertical force of 200 N is applied perpendicular to the handle, determine the dimension d if the allowable shear stress for the key is tallow = 35 MPa. a + ©MA = 0;
5
A
h = 2.74 in. Use h = 2
800 lb
a A
d a 20 mm
500 mm 200 N
Fa - a (20) - 200(500) = 0 Fa - a = 5000 N
tallow =
Fa - a ; Aa - a
35(106) =
5000 d(0.025)
d = 0.00571 m = 5.71 mm
Ans.
1–75. The joint is fastened together using two bolts. Determine the required diameter of the bolts if the failure shear stress for the bolts is tfail = 350 MPa. Use a factor of safety for shear of F.S. = 2.5.
30 mm
30 mm
350(106) = 140(105) 2.5
40 kN 3
tallow = 140(106) =
20(10 ) p 4
80 kN
40 kN
d2
d = 0.0135 m = 13.5 mm
Ans.
50
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*1–76. The lapbelt assembly is to be subjected to a force of 800 N. Determine (a) the required thickness t of the belt if the allowable tensile stress for the material is (st)allow = 10 MPa, (b) the required lap length dl if the glue can sustain an allowable shear stress of (tallow)g = 0.75 MPa, and (c) the required diameter dr of the pin if the allowable shear stress for the pin is (tallow)p = 30 MPa.
800 N 45 mm
t dl
dr 800 N
Allowable Normal Stress: Design of belt thickness. 10 A 106 B =
P ; A
(st)allow =
800 (0.045)t
t = 0.001778 m = 1.78 mm
Ans.
Allowable Shear Stress: Design of lap length. VA 400 ; 0.750 A 106 B = (tallow)g = A (0.045) dt dt = 0.01185 m = 11.9 mm
Ans.
Allowable Shear Stress: Design of pin size. VB 400 ; 30 A 106 B = p 2 (tallow)P = A 4 dr dr = 0.004120 m = 4.12 mm
Ans.
•1–77.
The wood specimen is subjected to the pull of 10 kN in a tension testing machine. If the allowable normal stress for the wood is (st)allow = 12 MPa and the allowable shear stress is tallow = 1.2 MPa, determine the required dimensions b and t so that the specimen reaches these stresses simultaneously. The specimen has a width of 25 mm.
10 kN
Allowable Shear Stress: Shear limitation tallow =
V ; A
1.2 A 106 B =
t
A
b
3
5.00(10 ) (0.025) t
t = 0.1667 m = 167 mm
Ans.
Allowable Normal Stress: Tension limitation sallow =
P ; A
12.0 A 106 B =
10 kN
10(103) (0.025) b
b = 0.03333 m = 33.3 mm
Ans.
51
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1–78. Member B is subjected to a compressive force of 600 lb. If A and B are both made of wood and are 1.5 in. thick, determine to the nearest 1>8 in. the smallest dimension a of the support so that the average shear stress along the blue line does not exceed tallow = 50 psi. Neglect friction.
600 lb
3
5 4
B A
Consider the equilibrium of the FBD of member B, Fig. a, + : ©Fx = 0;
4 600 a b - Fh = 0 5
Fh = 480 lb
Referring to the FBD of the wood segment sectioned through glue line, Fig. b + : ©Fx = 0;
480 - V = 0
V = 480 lb
The area of shear plane is A = 1.5(a). Thus, tallow =
V ; A
50 =
480 1.5a
a = 6.40 in Use a = 612 in.
Ans.
52
a
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1–79. The joint is used to transmit a torque of T = 3 kN # m. Determine the required minimum diameter of the shear pin A if it is made from a material having a shear failure stress of tfail = 150 MPa. Apply a factor of safety of 3 against failure.
100 mm T
A
T
Internal Loadings: The shear force developed on the shear plane of pin A can be determined by writing the moment equation of equilibrium along the y axis with reference to the free-body diagram of the shaft, Fig. a. ©My = 0; V(0.1) - 3(103) = 0
V = 30(103)N
Allowable Shear Stress: tfail 150 tallow = = = 50 MPa F.S. 3 Using this result, tallow =
V ; A
50(106) =
30(103) p d 2 4 A
dA = 0.02764 m = 27.6 mm
Ans.
*1–80. Determine the maximum allowable torque T that can be transmitted by the joint. The shear pin A has a diameter of 25 mm, and it is made from a material having a failure shear stress of tfail = 150 MPa. Apply a factor of safety of 3 against failure.
100 mm T
Internal Loadings: The shear force developed on the shear plane of pin A can be determined by writing the moment equation of equilibrium along the y axis with reference to the free-body diagram of the shaft, Fig. a. ©My = 0; V(0.1) - T = 0
T
V = 10T
Allowable Shear Stress: tfail 150 tallow = = = 50 MPa F.S. 3 The area of the shear plane for pin A is A A =
p (0.0252) = 0.4909(10 - 3)m2. Using 4
these results, tallow =
V ; AA
50(106) =
10T 0.4909(10 - 3)
T = 2454.37 N # m = 2.45 kN # m
Ans.
53
A
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•1–81.
The tension member is fastened together using two bolts, one on each side of the member as shown. Each bolt has a diameter of 0.3 in. Determine the maximum load P that can be applied to the member if the allowable shear stress for the bolts is tallow = 12 ksi and the allowable average normal stress is sallow = 20 ksi.
60 P
P
N - P sin 60° = 0
a+ ©Fy = 0;
P = 1.1547 N
(1)
V - P cos 60° = 0
b+ ©Fx = 0;
P = 2V
(2)
Assume failure due to shear: tallow = 12 =
V (2) (0.3)2 p 4
V = 1.696 kip From Eq. (2), P = 3.39 kip Assume failure due to normal force: sallow = 20 =
N (2) p4 (0.3)2
N = 2.827 kip From Eq. (1), P = 3.26 kip
Ans.
(controls)
54
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1–82. The three steel wires are used to support the load. If the wires have an allowable tensile stress of sallow = 165 MPa, determine the required diameter of each wire if the applied load is P = 6 kN.
A C
The force in wire BD is equal to the applied load; ie, FBD = P = 6 kN. Analysing the equilibrium of joint B by referring to its FBD, Fig. a, + : ©Fx = 0;
FBC cos 30° - FAB cos 45° = 0
(1)
+ c ©Fy = 0;
FBC sin 30° + FAB sin 45° - 6 = 0
(2)
Solving Eqs. (1) and (2), FAB = 5.379 kN For wire BD, FBD ; sallow = A BD
FBC = 4.392 kN
165(106) =
6(103) p 2 4 dBD
dBD = 0.006804 m = 6.804 mm Use dBD = 7.00 mm For wire AB, FAB ; sallow = A AB
165(106) =
Ans.
5.379(103) p 4
dAB 2
dAB = 0.006443 m = 6.443 mm Use dAB = 6.50 mm For wire BC, FBC ; sallow = A BC
165(106) =
Ans.
4.392(103) p 4
dBC 2
dBC = 0.005822 m = 5.822 mm dBC = 6.00 mm
Ans.
55
45
B
30
D P
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1–83. The three steel wires are used to support the load. If the wires have an allowable tensile stress of sallow = 165 MPa, and wire AB has a diameter of 6 mm, BC has a diameter of 5 mm, and BD has a diameter of 7 mm, determine the greatest force P that can be applied before one of the wires fails.
A C
45
The force in wire BD is equal to the applied load; ie, FBD = P. Analysing the equilibrium of joint B by referring to its FBD, Fig. a, + : ©Fx = 0;
FBC cos 30° - FAB cos 45° = 0
(1)
+ c ©Fy = 0;
FBC sin 30° + FAB sin 45° - P = 0
(2)
Solving Eqs. (1) and (2), FAB = 0.8966 P For wire BD, FBD sallow = ; A BD
FBC = 0.7321 P
165(106) =
p 4
P (0.0072)
P = 6349.94 N = 6.350 kN For wire AB, FAB ; sallow = A AB
165(106) =
0.8966 P (0.0062)
p 4
P = 5203.42 N = 5.203 kN For wire BC, FBC sallow = ; A BC
165(106) =
0.7321 P (0.0052)
p 4
P = 4425.60 N = 4.43 kN (Controls!)
Ans.
56
B
30
D P
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*1–84. The assembly consists of three disks A, B, and C that are used to support the load of 140 kN. Determine the smallest diameter d1 of the top disk, the diameter d2 within the support space, and the diameter d3 of the hole in the bottom disk. The allowable bearing stress for the material is 1sallow2b = 350 MPa and allowable shear stress is tallow = 125 MPa.
140 kN d1
20 mm A
10 mm
B C d3 d2
Solution Allowable Bearing Stress: Assume bearing failure for disk B. (sb)allow =
P ; A
350 A 106 B =
140(103) p 4
d21
d1 = 0.02257 m = 22.6 mm Allowable Shear Stress: Assume shear failure for disk C. tallow =
125 A 106 B =
V ; A
140(103) pd2 (0.01)
d2 = 0.03565 m = 35.7 mm
Ans.
Allowable Bearing Stress: Assume bearing failure for disk C. 140(103) P ; 350 A 106 B = (sb)allow = p A A 0.035652 - d23 B 4
d3 = 0.02760 m = 27.6 mm
Ans.
Since d3 = 27.6 mm 7 d1 = 22.6 mm, disk B might fail due to shear. t =
140(103) V = = 98.7 MPa 6 tallow = 125 MPa (O. K !) A p(0.02257)(0.02) d1 = 22.6 mm
Therefore,
Ans.
•1–85.
The boom is supported by the winch cable that has a diameter of 0.25 in. and an allowable normal stress of sallow = 24 ksi. Determine the greatest load that can be supported without causing the cable to fail when u = 30° and f = 45°. Neglect the size of the winch.
B
u 20 ft
s =
P ; A
24(103) =
p 4
T ; (0.25)2
A
T = 1178.10 lb d
+ : ©Fx = 0;
- 1178.10 cos 30° + FAB sin 45° = 0
+ c ©Fy = 0;
- W + FAB cos 45° - 1178.10 sin 30° = 0 W = 431 lb
Ans.
FAB = 1442.9 lb
57
f
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1–86. The boom is supported by the winch cable that has an allowable normal stress of sallow = 24 ksi. If it is required that it be able to slowly lift 5000 lb, from u = 20° to u = 50°, determine the smallest diameter of the cable to 1 the nearest 16 in. The boom AB has a length of 20 ft. Neglect the size of the winch. Set d = 12 ft.
B
u 20 ft f
A
Maximum tension in cable occurs when u = 20°. sin c sin 20° = 20 12
d
c = 11.842° + : © Fx = 0;
- T cos 20° + FAB cos 31.842° = 0
+ c ©Fy = 0;
FAB sin 31.842° - T sin 20° - 5000 = 0 T = 20 698.3 lb FAB = 22 896 lb
P ; s = A
Use
20 698.3 p 2 4 (d) d = 1.048 in.
24(103) =
d = 1
1 in. 16
Ans.
1–87. The 60 mm * 60 mm oak post is supported on the pine block. If the allowable bearing stresses for these materials are soak = 43 MPa and spine = 25 MPa, determine the greatest load P that can be supported. If a rigid bearing plate is used between these materials, determine its required area so that the maximum load P can be supported. What is this load?
P
For failure of pine block: s =
P ; A
25(106) =
P (0.06)(0.06)
P = 90 kN
Ans.
For failure of oak post: s =
P ; A
43(106) =
P (0.06)(0.06)
P = 154.8 kN Area of plate based on strength of pine block: 154.8(10)3 P s = ; 25(106) = A A A = 6.19(10 - 3)m2
Ans.
Pmax = 155 kN
Ans.
58
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*1–88. The frame is subjected to the load of 4 kN which acts on member ABD at D. Determine the required diameter of the pins at D and C if the allowable shear stress for the material is tallow = 40 MPa. Pin C is subjected to double shear, whereas pin D is subjected to single shear.
4 kN 1m E
1.5 m C
45 D 1.5 m
Referring to the FBD of member DCE, Fig. a, a + ©ME = 0;
Dy(2.5) - FBC sin 45° (1) = 0
(1)
+ : ©Fx = 0
FBC cos 45° - Dx = 0
(2)
B 1.5 m
Referring to the FBD of member ABD, Fig. b, a + ©MA = 0;
4 cos 45° (3) + FBC sin 45° (1.5) - Dx (3) = 0
(3)
Solving Eqs (2) and (3), FBC = 8.00 kN
Dx = 5.657 kN
Substitute the result of FBC into (1) Dy = 2.263 kN Thus, the force acting on pin D is FD = 2 Dx 2 + Dy 2 = 2 5.6572 + 2.2632 = 6.093 kN Pin C is subjected to double shear white pin D is subjected to single shear. Referring to the FBDs of pins C, and D in Fig c and d, respectively, FBC 8.00 = = 4.00 kN VC = VD = FD = 6.093 kN 2 2 For pin C, tallow =
VC ; AC
40(106) =
4.00(103) p 4
dC 2
dC = 0.01128 m = 11.28 mm Use dC = 12 mm For pin D, VD ; tallow = AD
40(106) =
Ans.
6.093(103) p 4
dD 2
dD = 0.01393 m = 13.93 mm Use
dD = 14 mm
Ans.
59
A
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•1–89. The eye bolt is used to support the load of 5 kip. Determine its diameter d to the nearest 18 in. and the required thickness h to the nearest 18 in. of the support so that the washer will not penetrate or shear through it. The allowable normal stress for the bolt is sallow = 21 ksi and the allowable shear stress for the supporting material is tallow = 5 ksi.
1 in. h
d
5 kip
Allowable Normal Stress: Design of bolt size 5(103) P sallow = ; 21.0(103) = p 2 Ab 4 d d = 0.5506 in. Use d =
5 in. 8
Ans.
Allowable Shear Stress: Design of support thickness 5(103) V ; 5(103) = tallow = A p(1)(h) Use h =
3 in. 8
Ans.
60
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1–90. The soft-ride suspension system of the mountain bike is pinned at C and supported by the shock absorber BD. If it is designed to support a load P = 1500 N, determine the required minimum diameter of pins B and C. Use a factor of safety of 2 against failure. The pins are made of material having a failure shear stress of tfail = 150 MPa, and each pin is subjected to double shear.
P A
Internal Loadings: The forces acting on pins B and C can be determined by considering the equilibrium of the free-body diagram of the soft-ride suspension system shown in Fig. a. a + ©MC = 0;
FBD = 5905.36 N + : ©Fx = 0;
Cx - 5905.36 cos 60° = 0
+ c ©Fy = 0;
5905.36 sin 60° - 1500 - Cy = 0 Cy = 3614.20 N
Cx = 2952.68 N
Thus, FC = 2 Cx 2 + Cy 2 = 2 2952.682 + 3614.202 = 4666.98 N Since both pins are in double shear, FB 5905.36 VB = = = 2952.68 N 2 2
VC =
FC 4666.98 = = 2333.49 N 2 2
Allowable Shear Stress: tfail 150 tallow = = = 75 MPa F.S. 2 Using this result, VB tallow = ; AB
75(106) =
2952.68 p 2 d 4 B
dB = 0.007080 m = 7.08 mm tallow =
VC ; AC
75(106) =
Ans.
2333.49 p 2 d 4 C
dC = 0.006294 m = 6.29 mm
Ans.
61
30 mm
B 60
1500(0.4) - FBD sin 60°(0.1) - FBD cos 60°(0.03) = 0
FB = FBD = 5905.36 N
100 mm 300 mm
C
D
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1–91. The soft-ride suspension system of the mountain bike is pinned at C and supported by the shock absorber BD. If it is designed to support a load of P = 1500 N, determine the factor of safety of pins B and C against failure if they are made of a material having a shear failure stress of tfail = 150 MPa. Pin B has a diameter of 7.5 mm, and pin C has a diameter of 6.5 mm. Both pins are subjected to double shear.
P A
Internal Loadings: The forces acting on pins B and C can be determined by considerning the equilibrium of the free-body diagram of the soft-ride suspension system shown in Fig. a. + ©MC = 0;
1500(0.4) - FBD sin 60°(0.1) - FBD cos 60°(0.03) = 0 FBD = 5905.36 N
+ : ©Fx = 0;
Cx - 5905.36 cos 60° = 0
+ c ©Fy = 0;
5905.36 sin 60° - 1500 - Cy = 0 Cy = 3614.20 N
Cx = 2952.68 N
Thus, FB = FBD = 5905.36 N
FC = 2 Cx 2 + Cy 2 = 2 2952.682 + 3614.202 = 4666.98 N
Since both pins are in double shear, VB =
FB 5905.36 = = 2952.68N 2 2
VC =
FC 4666.98 = = 2333.49 N 2 2
Allowable Shear Stress: The areas of the shear plane for pins B and C are p p A B = (0.00752) = 44.179(10 - 6)m2 and A C = (0.00652) = 33.183(10 - 6)m2. 4 4 We obtain
A tavg B B =
VB 2952.68 = 66.84 MPa = AB 44.179(10 - 6)
A tavg B C =
VC 2333.49 = 70.32 MPa = AC 33.183(10 - 6)
Using these results, tfail = (F.S.)B = A tavg B B tfail (F.S.)C = = A tavg B C
100 mm 300 mm
150 = 2.24 66.84
Ans.
150 = 2.13 70.32
Ans.
62
30 mm
B 60
C
D
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*1–92. The compound wooden beam is connected together by a bolt at B. Assuming that the connections at A, B, C, and D exert only vertical forces on the beam, determine the required diameter of the bolt at B and the required outer diameter of its washers if the allowable tensile stress for the bolt is 1st2allow = 150 MPa and the allowable bearing stress for the wood is 1sb2allow = 28 MPa. Assume that the hole in the washers has the same diameter as the bolt.
2 kN 1.5 kN 1.5 m 1.5 m 1.5 m
3 kN 2m
2m
1.5 m C
D
A B
From FBD (a): a + ©MD = 0;
FB(4.5) + 1.5(3) + 2(1.5) - FC(6) = 0 4.5 FB - 6 FC = - 7.5
(1)
From FBD (b): a + ©MD = 0;
FB(5.5) - FC(4) - 3(2) = 0 5.5 FB - 4 FC = 6
(2)
Solving Eqs. (1) and (2) yields FB = 4.40 kN;
FC = 4.55 kN
For bolt: sallow = 150(106) =
4.40(103) p 2 4 (dB)
dB = 0.00611 m = 6.11 mm
Ans.
For washer: sallow = 28 (104) =
4.40(103) p 2 4 (d w
- 0.006112)
dw = 0.0154 m = 15.4 mm
Ans.
•1–93.
The assembly is used to support the distributed loading of w = 500 lb>ft. Determine the factor of safety with respect to yielding for the steel rod BC and the pins at B and C if the yield stress for the steel in tension is sy = 36 ksi and in shear ty = 18 ksi. The rod has a diameter of 0.40 in., and the pins each have a diameter of 0.30 in.
C
4 ft
For rod BC: s =
A
1.667 P = 13.26 ksi = p 2 A 4 (0.4 )
F. S. =
sy s
=
36 = 2.71 13.26
Ans. 3 ft
For pins B and C:
w
0.8333 V = 11.79 ksi = p t = 2 A 4 (0.3 ) F. S. =
ty t
=
B
1 ft
18 = 1.53 11.79
Ans.
63
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1–94. If the allowable shear stress for each of the 0.30in.-diameter steel pins at A, B, and C is tallow = 12.5 ksi, and the allowable normal stress for the 0.40-in.-diameter rod is sallow = 22 ksi, determine the largest intensity w of the uniform distributed load that can be suspended from the beam.
C
4 ft
Assume failure of pins B and C: tallow = 12.5 =
1.667w
A
p 2 4 (0.3 )
w = 0.530 kip>ft
Ans.
(controls)
B
Assume failure of pins A: 3 ft
FA = 2 (2w)2 + (1.333w)2 = 2.404 w tallow
w
1.202w = 12.5 = p 2 4 (0.3 )
1 ft
w = 0.735 kip>ft Assume failure of rod BC: sallow = 22 =
3.333w p 2 4 (0.4 )
w = 0.829 kip>ft
1–95. If the allowable bearing stress for the material under the supports at A and B is 1sb2allow = 1.5 MPa, determine the size of square bearing plates A¿ and B¿ required to support the load. Dimension the plates to the nearest mm. The reactions at the supports are vertical. Take P = 100 kN.
40 kN/m
Referring to the FBD of the bean, Fig. a
1.5 m
A
a + ©MA = 0;
NB(3) + 40(1.5)(0.75) - 100(4.5) = 0
NB = 135 kN
a + ©MB = 0;
40(1.5)(3.75) - 100(1.5) - NA(3) = 0
NA = 25.0 kN
For plate A¿ , NA (sb)allow = ; A A¿
1.5(106) =
25.0(103) a2A¿
aA¿ = 0.1291 m = 130 mm
Ans.
For plate B¿ , sallow =
NB ; A B¿
1.5(106) =
135(103) a2B¿
aB¿ = 0.300 m = 300 mm
Ans.
64
P
A¿
B¿ 3m
B
1.5 m
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40 kN/m
*1–96. If the allowable bearing stress for the material under the supports at A and B is 1sb2allow = 1.5 MPa, determine the maximum load P that can be applied to the beam. The bearing plates A¿ and B¿ have square cross sections of 150 mm * 150 mm and 250 mm * 250 mm, respectively.
A
P
A¿
B¿ 3m
1.5 m
B
1.5 m
Referring to the FBD of the beam, Fig. a, a + ©MA = 0;
NB(3) + 40(1.5)(0.75) - P(4.5) = 0
NB = 1.5P - 15
a + ©MB = 0;
40(1.5)(3.75) - P(1.5) - NA(3) = 0
NA = 75 - 0.5P
For plate A¿ , NA (sb)allow = ; A A¿
1.5(106) =
(75 - 0.5P)(103) 0.15(0.15)
P = 82.5 kN For plate B¿ , NB ; (sb)allow = A B¿
1.5(106) =
(1.5P - 15)(103) 0.25(0.25)
P = 72.5 kN (Controls!)
Ans.
•1–97.
The rods AB and CD are made of steel having a failure tensile stress of sfail = 510 MPa. Using a factor of safety of F.S. = 1.75 for tension, determine their smallest diameter so that they can support the load shown. The beam is assumed to be pin connected at A and C.
B
D 6 kN 5 kN
4 kN
Support Reactions: a + ©MA = 0;
FCD(10) - 5(7) - 6(4) - 4(2) = 0 A
FCD = 6.70 kN a + ©MC = 0;
4(8) + 6(6) + 5(3) - FAB(10) = 0 FAB = 8.30 kN
Allowable Normal Stress: Design of rod sizes For rod AB sallow =
sfail FAB = ; F.S A AB
510(106) 8.30(103) = p 2 1.75 4 d AB dAB = 0.006022 m = 6.02 mm
Ans.
For rod CD sallow =
FCD sfail = ; F.S A CD
C 2m
510(106) 6.70(103) = p 2 1.75 4 d CD
dCD = 0.005410 m = 5.41 mm
Ans.
65
2m
3m
3m
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1–98. The aluminum bracket A is used to support the centrally applied load of 8 kip. If it has a constant thickness of 0.5 in., determine the smallest height h in order to prevent a shear failure. The failure shear stress is tfail = 23 ksi. Use a factor of safety for shear of F.S. = 2.5.
A
Equation of Equilibrium: + c ©Fy = 0;
V - 8 = 0
V = 8.00 kip
Allowable Shear Stress: Design of the support size tallow =
tfail V = ; F.S A
23(103) 8.00(103) = 2.5 h(0.5)
8 kip
h = 1.74 in.
Ans.
1–99. The hanger is supported using the rectangular pin. Determine the magnitude of the allowable suspended load P if the allowable bearing stress is (sb)allow = 220 MPa, the allowable tensile stress is (st)allow = 150 MPa, and the allowable shear stress is tallow = 130 MPa. Take t = 6 mm, a = 5 mm, and b = 25 mm.
20 mm 75 mm 10 mm a
a
b
Allowable Normal Stress: For the hanger (st)allow =
P ; A
150 A 106 B =
P (0.075)(0.006)
P
P = 67.5 kN
37.5 mm
Allowable Shear Stress: The pin is subjected to double shear. Therefore, V = tallow =
130 A 106 B =
V ; A
P 2
P>2 (0.01)(0.025)
P = 65.0 kN Allowable Bearing Stress: For the bearing area (sb)allow =
P ; A
220 A 106 B =
37.5 mm
t
P>2 (0.005)(0.025)
P = 55.0 kN (Controls!)
Ans.
66
h
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*1–100. The hanger is supported using the rectangular pin. Determine the required thickness t of the hanger, and dimensions a and b if the suspended load is P = 60 kN. The allowable tensile stress is (st)allow = 150 MPa, the allowable bearing stress is (sb)allow = 290 MPa, and the allowable shear stress is tallow = 125 MPa.
20 mm 75 mm 10 mm a
a
37.5 mm
t P 37.5 mm
Allowable Normal Stress: For the hanger (st)allow =
P ; A
150 A 106 B =
60(103) (0.075)t
t = 0.005333 m = 5.33 mm
Ans.
Allowable Shear Stress: For the pin tallow =
125 A 106 B =
V ; A
30(103) (0.01)b
b = 0.0240 m = 24.0 mm
Ans.
Allowable Bearing Stress: For the bearing area (sb)allow =
P ; A
290 A 106 B =
30(103) (0.0240) a
a = 0.00431 m = 4.31 mm
Ans.
67
b
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•1–101. The 200-mm-diameter aluminum cylinder supports a compressive load of 300 kN. Determine the average normal and shear stress acting on section a–a. Show the results on a differential element located on the section.
300 kN
a
Referring to the FBD of the upper segment of the cylinder sectional through a–a shown in Fig. a, +Q©Fx¿ = 0;
Na - a - 300 cos 30° = 0
+a©Fy¿ = 0;
Va - a - 300 sin 30° = 0
30
Na - a = 259.81 kN
a
Va - a = 150 kN
0.1 Section a–a of the cylinder is an ellipse with a = 0.1 m and b = m. Thus, cos 30° 0.1 b = 0.03628 m2. A a - a = pab = p(0.1)a cos 30°
A sa - a B avg =
259.81(103) Na - a = = 7.162(106) Pa = 7.16 MPa Aa - a 0.03628
Ans.
A ta - a B avg =
150(103) Va - a = = 4.135(106) Pa = 4.13 MPa Aa - a 0.03628
Ans.
d
The differential element representing the state of stress of a point on section a–a is shown in Fig. b
1–102. The long bolt passes through the 30-mm-thick plate. If the force in the bolt shank is 8 kN, determine the average normal stress in the shank, the average shear stress along the cylindrical area of the plate defined by the section lines a–a, and the average shear stress in the bolt head along the cylindrical area defined by the section lines b–b.
8 mm
a 7 mm
b
8 kN
18 mm b a
P = ss = A
8 (103)
= 208 MPa
Ans.
(tavg)a =
8 (103) V = = 4.72 MPa A p (0.018)(0.030)
Ans.
(tavg)b =
8 (103) V = = 45.5 MPa A p (0.007)(0.008)
Ans.
p 4
(0.007)2
68
30 mm
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1–103. Determine the required thickness of member BC and the diameter of the pins at A and B if the allowable normal stress for member BC is sallow = 29 ksi and the allowable shear stress for the pins is tallow = 10 ksi.
C
1.5 in.
Referring to the FBD of member AB, Fig. a, a + ©MA = 0;
2(8)(4) - FBC sin 60° (8) = 0
+ : ©Fx = 0;
9.238 cos 60° - A x = 0
+ c ©Fy = 0;
9.238 sin 60° - 2(8) + A y = 0
FBC = 9.238 kip
60 B
A x = 4.619 kip
8 ft
A y = 8.00 kip 2 kip/ft
Thus, the force acting on pin A is FA = 2 A 2x + A 2y = 2 4.6192 + 8.002 = 9.238 kip Pin A is subjected to single shear, Fig. c, while pin B is subjected to double shear, Fig. b. FBC 9.238 VA = FA = 9.238 kip VB = = = 4.619 kip 2 2 For member BC FBC ; sallow = A BC
29 =
9.238 1.5(t)
t = 0.2124 in.
Use t = For pin A, VA ; tallow = AA
10 =
9.238 p 2 4 dA
1 in. 4
Ans.
dA = 1.085 in. 1 Use dA = 1 in 8
For pin B, VB ; tallow = AB
10 =
4.619 p 2 4 dB
Ans.
dB = 0.7669 in Use dB =
13 in 16
Ans.
69
A
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*1–104. Determine the resultant internal loadings acting on the cross sections located through points D and E of the frame.
150 lb/ft
Segment AD:
1.5 ft
+
: ©Fx = 0;
ND - 1.2 = 0;
+ T ©Fy = 0;
VD + 0.225 + 0.4 = 0;
a + ©MD = 0;
ND = 1.20 kip
Ans.
VD = - 0.625 kip
B
E
Ans. 2.5 ft
MD + 0.225(0.75) + 0.4(1.5) = 0 MD =
A 4 ft
D
C
- 0.769 kip # ft
Ans.
3 ft
5 ft
Segment CE: Q+ ©Fx = 0;
NE + 2.0 = 0;
R+ ©Fy = 0;
VE = 0
a + ©ME = 0;
NE = - 2.00 kip
Ans. Ans.
ME = 0
Ans.
•1–105.
The pulley is held fixed to the 20-mm-diameter shaft using a key that fits within a groove cut into the pulley and shaft. If the suspended load has a mass of 50 kg, determine the average shear stress in the key along section a–a. The key is 5 mm by 5 mm square and 12 mm long.
a + ©MO = 0;
a
75 mm
F (10) - 490.5 (75) = 0
F = 3678.75 N tavg =
a
3678.75 V = = 61.3 MPa A (0.005)(0.012)
Ans.
70
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1–106. The bearing pad consists of a 150 mm by 150 mm block of aluminum that supports a compressive load of 6 kN. Determine the average normal and shear stress acting on the plane through section a–a. Show the results on a differential volume element located on the plane.
6 kN
a
Equation of Equilibrium: +Q©Fx = 0;
Va - a - 6 cos 60° = 0
Va - a = 3.00 kN
a+ ©Fy = 0;
Na - a - 6 sin 60° = 0
Na - a = 5.196 kN
30 a
150 mm
Averge Normal Stress And Shear Stress: The cross sectional Area at section a–a is 0.15 b (0.15) = 0.02598 m2. A = a sin 60° sa - a =
5.196(103) Na - a = = 200 kPa A 0.02598
Ans.
ta - a =
3.00(103) Va - a = = 115 kPa A 0.02598
Ans.
1–107. The yoke-and-rod connection is subjected to a tensile force of 5 kN. Determine the average normal stress in each rod and the average shear stress in the pin A between the members.
5 kN
40 mm
For the 40 – mm – dia rod: s40
30 mm
5 (103) P = p = = 3.98 MPa 2 A 4 (0.04)
Ans.
For the 30 – mm – dia rod:
5 kN
3
s30 =
5 (10 ) V = p = 7.07 MPa 2 A 4 (0.03)
Ans.
Average shear stress for pin A: tavg =
A 25 mm
2.5 (103) P = p = 5.09 MPa 2 A 4 (0.025)
Ans.
71
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*1–108. The cable has a specific weight g (weight>volume) and cross-sectional area A. If the sag s is small, so that its length is approximately L and its weight can be distributed uniformly along the horizontal axis, determine the average normal stress in the cable at its lowest point C.
A
s C L/2
Equation of Equilibrium: a + ©MA = 0;
Ts -
gAL L a b = 0 2 4 T =
B
gAL2 8s
Average Normal Stress: gAL2
gL2 T 8s s = = = A A 8s
Ans.
72
L/2
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2–1. An air-filled rubber ball has a diameter of 6 in. If the air pressure within it is increased until the ball’s diameter becomes 7 in., determine the average normal strain in the rubber. d0 = 6 in. d = 7 in. e =
pd - pd0 7 - 6 = = 0.167 in./in. pd0 6
Ans.
2–2. A thin strip of rubber has an unstretched length of 15 in. If it is stretched around a pipe having an outer diameter of 5 in., determine the average normal strain in the strip. L0 = 15 in. L = p(5 in.) e =
L - L0 5p - 15 = = 0.0472 in.>in. L0 15
Ans.
2–3. The rigid beam is supported by a pin at A and wires BD and CE. If the load P on the beam causes the end C to be displaced 10 mm downward, determine the normal strain developed in wires CE and BD.
D
E
4m
P
¢LBD ¢LCE = 3 7
A
3 (10) = 4.286 mm 7 ¢LCE 10 = = = 0.00250 mm>mm L 4000
3m
¢LBD = eCE
eBD =
B
Ans.
¢LBD 4.286 = = 0.00107 mm>mm L 4000
Ans.
1
C
2m
2m
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*2–4. The two wires are connected together at A. If the force P causes point A to be displaced horizontally 2 mm, determine the normal strain developed in each wire.
C 300
œ = 23002 + 22 - 2(300)(2) cos 150° = 301.734 mm LAC
eAC = eAB
œ - LAC LAC 301.734 - 300 = = = 0.00578 mm>mm LAC 300
mm
30⬚
Ans.
30⬚
300
A
P
mm
B
•2–5. The rigid beam is supported by a pin at A and wires BD and CE. If the distributed load causes the end C to be displaced 10 mm downward, determine the normal strain developed in wires CE and BD.
E D 2m 1.5 m 3m
2m A
B
C
w
Since the vertical displacement of end C is small compared to the length of member AC, the vertical displacement dB of point B, can be approximated by referring to the similar triangle shown in Fig. a dB 10 = ; dB = 4 mm 2 5 The unstretched lengths of wires BD and CE are LBD = 1500 mm and LCE = 2000 mm. dB 4 Ans. = = 0.00267 mm>mm A eavg B BD = LBD 1500
A eavg B CE =
dC 10 = = 0.005 mm>mm LCE 2000
Ans.
2
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2–6. Nylon strips are fused to glass plates. When moderately heated the nylon will become soft while the glass stays approximately rigid. Determine the average shear strain in the nylon due to the load P when the assembly deforms as indicated.
y 2 mm P
3 mm 5 mm 3 mm 5 mm 3 mm
g = tan - 1 a
2 b = 11.31° = 0.197 rad 10
x
Ans.
2–7. If the unstretched length of the bowstring is 35.5 in., determine the average normal strain in the string when it is stretched to the position shown. 18 in.
6 in. 18 in.
Geometry: Referring to Fig. a, the stretched length of the string is L = 2L¿ = 2 2182 + 62 = 37.947 in. Average Normal Strain: eavg =
L - L0 37.947 - 35.5 = = 0.0689 in.>in. L0 35.5
Ans.
3
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u
*2–8. Part of a control linkage for an airplane consists of a rigid member CBD and a flexible cable AB. If a force is applied to the end D of the member and causes it to rotate by u = 0.3°, determine the normal strain in the cable. Originally the cable is unstretched.
D
P 300 mm
B
AB = 24002 + 3002 = 500 mm
300 mm
AB¿ = 2400 + 300 - 2(400)(300) cos 90.3° 2
2
A
C
= 501.255 mm eAB =
AB¿ - AB 501.255 - 500 = AB 500
400 mm
= 0.00251 mm>mm
Ans.
•2–9.
Part of a control linkage for an airplane consists of a rigid member CBD and a flexible cable AB. If a force is applied to the end D of the member and causes a normal strain in the cable of 0.0035 mm>mm, determine the displacement of point D. Originally the cable is unstretched.
u D
300 mm
B
AB = 23002 + 4002 = 500 mm
300 mm
AB¿ = AB + eABAB A
= 500 + 0.0035(500) = 501.75 mm
C
501.752 = 3002 + 4002 - 2(300)(400) cos a a = 90.4185°
400 mm
p (0.4185) rad u = 90.4185° - 90° = 0.4185° = 180° ¢ D = 600(u) = 600(
P
p )(0.4185) = 4.38 mm 180°
Ans.
4
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2–10. The corners B and D of the square plate are given the displacements indicated. Determine the shear strains at A and B.
y A
16 mm D
B
3 mm 3 mm 16 mm
16 mm
Applying trigonometry to Fig. a f = tan - 1 a
13 p rad b = 39.09° a b = 0.6823 rad 16 180°
a = tan - 1 a
16 p rad b = 50.91° a b = 0.8885 rad 13 180°
By the definition of shear strain,
A gxy B A =
p p - 2f = - 2(0.6823) = 0.206 rad 2 2
Ans.
A gxy B B =
p p - 2a = - 2(0.8885) = -0.206 rad 2 2
Ans.
5
C
16 mm
x
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2–11. The corners B and D of the square plate are given the displacements indicated. Determine the average normal strains along side AB and diagonal DB.
y A
16 mm D
B
3 mm 3 mm 16 mm
16 mm
Referring to Fig. a, LAB = 2162 + 162 = 2512 mm LAB¿ = 2162 + 132 = 2425 mm LBD = 16 + 16 = 32 mm LB¿D¿ = 13 + 13 = 26 mm Thus,
A eavg B AB =
LAB¿ - LAB 2425 - 2512 = = -0.0889 mm>mm LAB 2512
Ans.
A eavg B BD =
LB¿D¿ - LBD 26 - 32 = = -0.1875 mm>mm LBD 32
Ans.
6
C
16 mm
x
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*2–12. The piece of rubber is originally rectangular. Determine the average shear strain gxy at A if the corners B and D are subjected to the displacements that cause the rubber to distort as shown by the dashed lines.
y 3 mm
C D
2 = 0.006667 rad 300 3 u2 = tan u2 = = 0.0075 rad 400 u1 = tan u1 =
400 mm
gxy = u1 + u2
A
= 0.006667 + 0.0075 = 0.0142 rad
Ans.
•2–13.
The piece of rubber is originally rectangular and subjected to the deformation shown by the dashed lines. Determine the average normal strain along the diagonal DB and side AD.
f = tan
B 2 mm
3 mm
C D
400 mm
3 b = 0.42971° a 400
AB¿ = 2(300)2 + (2)2 = 300.00667 w = tan - 1 a
x
y
AD¿ = 2(400)2 + (3)2 = 400.01125 mm -1
300 mm
A
2 b = 0.381966° 300
a = 90° - 0.42971° - 0.381966° = 89.18832° D¿B¿ = 2(400.01125)2 + (300.00667)2 - 2(400.01125)(300.00667) cos (89.18832°) D¿B¿ = 496.6014 mm DB = 2(300)2 + (400)2 = 500 mm 496.6014 - 500 = -0.00680 mm>mm 500 400.01125 - 400 = = 0.0281(10 - 3) mm>mm 400
eDB =
Ans.
eAD
Ans.
7
300 mm
B 2 mm
x
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2–14. Two bars are used to support a load. When unloaded, AB is 5 in. long, AC is 8 in. long, and the ring at A has coordinates (0, 0). If a load P acts on the ring at A, the normal strain in AB becomes PAB = 0.02 in.>in., and the normal strain in AC becomes PAC = 0.035 in.>in. Determine the coordinate position of the ring due to the load.
y
B
C
60⬚
5 in.
8 in.
A
x
P
Average Normal Strain: œ = LAB + eAB LAB = 5 + (0.02)(5) = 5.10 in. LAB œ = LAC + eACLAC = 8 + (0.035)(8) = 8.28 in. LAC
Geometry: a = 282 - 4.33012 = 6.7268 in. 5.102 = 9.22682 + 8.282 - 2(9.2268)(8.28) cos u u = 33.317° x¿ = 8.28 cos 33.317° = 6.9191 in. y¿ = 8.28 sin 33.317° = 4.5480 in. x = -(x¿ - a) = -(6.9191 - 6.7268) = -0.192 in.
Ans.
y = -(y¿ - 4.3301) = -(4.5480 - 4.3301) = -0.218 in.
Ans.
8
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2–15. Two bars are used to support a load P. When unloaded, AB is 5 in. long, AC is 8 in. long, and the ring at A has coordinates (0, 0). If a load is applied to the ring at A, so that it moves it to the coordinate position (0.25 in., -0.73 in.), determine the normal strain in each bar.
y
B
C
60⬚
5 in.
8 in.
A
x
P
Geometry: a = 282 - 4.33012 = 6.7268 in. LA¿B = 2(2.5 + 0.25)2 + (4.3301 + 0.73)2 = 5.7591 in. LA¿C = 2(6.7268 - 0.25)2 + (4.3301 + 0.73)2 = 8.2191 in. Average Normal Strain: eAB =
=
eAC =
=
LA¿B - LAB LAB 5.7591 - 5 = 0.152 in.>in. 5
Ans.
LA¿C - LAC LAC 8.2191 - 8 = 0.0274 in.>in. 8
Ans.
9
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*2–16. The square deforms into the position shown by the dashed lines. Determine the average normal strain along each diagonal, AB and CD. Side D¿B¿ remains horizontal.
y 3 mm D¿
B¿
B
D
Geometry: AB = CD = 2502 + 502 = 70.7107 mm 53 mm
C¿D¿ = 2532 + 582 - 2(53)(58) cos 91.5°
50 mm 91.5⬚
= 79.5860 mm C
B¿D¿ = 50 + 53 sin 1.5° - 3 = 48.3874 mm
A
x
C¿
AB¿ = 2532 + 48.38742 - 2(53)(48.3874) cos 88.5°
50 mm 8 mm
= 70.8243 mm Average Normal Strain:
eAB =
=
eCD =
=
AB¿ - AB AB 70.8243 - 70.7107 = 1.61 A 10 - 3 B mm>mm 70.7107
Ans.
C¿D¿ - CD CD 79.5860 - 70.7107 = 126 A 10 - 3 B mm>mm 70.7107
Ans.
•2–17.
The three cords are attached to the ring at B. When a force is applied to the ring it moves it to point B¿ , such that the normal strain in AB is PAB and the normal strain in CB is PCB. Provided these strains are small, determine the normal strain in DB. Note that AB and CB remain horizontal and vertical, respectively, due to the roller guides at A and C.
A¿
B¿
A
B
L
Coordinates of B (L cos u, L sin u)
u
Coordinates of B¿ (L cos u + eAB L cos u, L sin u + eCB L sin u)
C¿ D
LDB¿ = 2(L cos u + eAB L cos u) + (L sin u + eCB L sin u) 2
2
LDB¿ = L 2cos2 u(1 + 2eAB + e2AB) + sin2 u(1 + 2eCB + e2CB) Since eAB and eCB are small, LDB¿ = L 21 + (2 eAB cos2 u + 2eCB sin2 u) Use the binomial theorem, LDB¿ = L ( 1 +
1 (2 eAB cos2 u + 2eCB sin2 u)) 2
= L ( 1 + eAB cos2 u + eCB sin2 u) Thus, eDB =
L( 1 + eAB cos2 u + eCB sin2 u) - L L
eDB = eAB cos2 u + eCB sin2 u
Ans.
10
C
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2–18. The piece of plastic is originally rectangular. Determine the shear strain gxy at corners A and B if the plastic distorts as shown by the dashed lines.
y 5 mm 2 mm 2 mm
B
4 mm
C 300 mm
Geometry: For small angles, 2 mm D
2 a = c = = 0.00662252 rad 302 b = u =
A
x
400 mm 3 mm
2 = 0.00496278 rad 403
Shear Strain: (gB)xy = a + b = 0.0116 rad = 11.6 A 10 - 3 B rad
Ans.
(gA)xy = -(u + c) = -0.0116 rad = -11.6 A 10 - 3 B rad
Ans.
2–19. The piece of plastic is originally rectangular. Determine the shear strain gxy at corners D and C if the plastic distorts as shown by the dashed lines.
y 5 mm 2 mm 2 mm
B
4 mm
C 300 mm 2 mm D
A 400 mm 3 mm
Geometry: For small angles, 2 = 0.00496278 rad 403 2 = 0.00662252 rad b = u = 302 Shear Strain: a = c =
(gC)xy = -(a + b) = -0.0116 rad = -11.6 A 10 - 3 B rad
Ans.
(gD)xy = u + c = 0.0116 rad = 11.6 A 10 - 3 B rad
Ans.
11
x
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*2–20. The piece of plastic is originally rectangular. Determine the average normal strain that occurs along the diagonals AC and DB.
y 5 mm 2 mm 2 mm
Geometry:
B
4 mm
C
AC = DB = 24002 + 3002 = 500 mm
300 mm
DB¿ = 24052 + 3042 = 506.4 mm
2 mm D
A¿C¿ = 2401 + 300 = 500.8 mm 2
2
x
A 400 mm 3 mm
Average Normal Strain: eAC =
A¿C¿ - AC 500.8 - 500 = AC 500
= 0.00160 mm>mm = 1.60 A 10 - 3 B mm>mm eDB =
Ans.
DB¿ - DB 506.4 - 500 = DB 500
= 0.0128 mm>mm = 12.8 A 10 - 3 B mm>mm
Ans.
•2–21.
The force applied to the handle of the rigid lever arm causes the arm to rotate clockwise through an angle of 3° about pin A. Determine the average normal strain developed in the wire. Originally, the wire is unstretched.
D
600 mm
Geometry: Referring to Fig. a, the stretched length of LB¿D can be determined using the consine law, A
LB¿D = 2(0.6 cos 45°)2 + (0.6 sin 45°)2 - 2(0.6 cos 45°)(0.6 sin 45°) cos 93°
B
= 0.6155 m Average Normal Strain: The unstretched length of wire BD is LBD = 0.6 m. We obtain eavg =
C
45⬚
LB¿D - LBD 0.6155 - 0.6 = 0.0258 m>m = LBD 0.6
Ans.
12
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2–22. A square piece of material is deformed into the dashed position. Determine the shear strain gxy at A.
y 15.18 mm B
Shear Strain: (gA)xy =
89.7° p - ¢ ≤p 2 180°
C
15.24 mm
15 mm
= 5.24 A 10 - 3 B rad
Ans.
89.7⬚ A
2–23. A square piece of material is deformed into the dashed parallelogram. Determine the average normal strain that occurs along the diagonals AC and BD.
15 mm 15.18 mm
x
D
y 15.18 mm B
C
15.24 mm
15 mm 89.7⬚ A
Geometry: AC = BD = 2152 + 152 = 21.2132 mm AC¿ = 215.182 + 15.242 - 2(15.18)(15.24) cos 90.3° = 21.5665 mm B¿D¿ = 215.182 + 15.242 - 2(15.18)(15.24) cos 89.7° = 21.4538 mm Average Normal Strain: eAC =
eBD
AC¿ - AC 21.5665 - 21.2132 = AC 21.2132
= 0.01665 mm>mm = 16.7 A 10 - 3 B mm>mm
Ans.
= 0.01134 mm>mm = 11.3 A 10 - 3 B mm>mm
Ans.
B¿D¿ - BD 21.4538 - 21.2132 = = BD 21.2132
13
15 mm 15.18 mm
D
x
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*2–24. A square piece of material is deformed into the dashed position. Determine the shear strain gxy at C.
y 15.18 mm B
C
15.24 mm
15 mm 89.7⬚ A
(gC)xy =
15 mm 15.18 mm
x
D
p 89.7° - ¢ ≤p 2 180° = 5.24 A 10 - 3 B rad
Ans.
u ⫽ 2⬚
•2–25.
The guy wire AB of a building frame is originally unstretched. Due to an earthquake, the two columns of the frame tilt u = 2°. Determine the approximate normal strain in the wire when the frame is in this position. Assume the columns are rigid and rotate about their lower supports.
u ⫽ 2⬚
B
Geometry: The vertical displacement is negligible 3m
xA
2° = (1) ¢ ≤ p = 0.03491 m 180° A
2° xB = (4) ¢ ≤ p = 0.13963 m 180°
1m
x = 4 + xB - xA = 4.10472 m A¿B¿ = 232 + 4.104722 = 5.08416 m AB = 232 + 42 = 5.00 m Average Normal Strain: eAB =
=
A¿B¿ - AB AB 5.08416 - 5 = 16.8 A 10 - 3 B m>m 5
Ans.
14
4m
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2–26. The material distorts into the dashed position shown. Determine (a) the average normal strains along sides AC and CD and the shear strain gxy at F, and (b) the average normal strain along line BE.
y 15 mm C
25 mm D
10 mm B E
75 mm
90 mm
A
Referring to Fig. a, LBE = 2(90 - 75)2 + 802 = 26625 mm LAC¿ = 21002 + 152 = 210225 mm LC¿D¿ = 80 - 15 + 25 = 90 mm f = tan-1 ¢
25 p rad ≤ = 14.04° ¢ ≤ = 0.2450 rad. 100 180°
When the plate deforms, the vertical position of point B and E do not change. LBB¿ 15 = ; LBB¿ = 13.5 mm 90 100 LEE¿ 25 = ; 75 100
LEE¿ = 18.75 mm
LB¿E¿ = 2(90 - 75)2 + (80 - 13.5 + 18.75)2 = 27492.5625 mm Thus,
A eavg B AC =
LAC¿ - LAC 210225 - 100 = = 0.0112 mm>mm LAC 100
Ans.
A eavg B CD =
LC¿D¿ - LCD 90 -80 = = 0.125 mm>mm LCD 80
Ans.
A eavg B BE =
LB¿E¿ - LBE 27492.5625 - 26625 = = 0.0635 mm>mm LBE 26625
Ans.
Referring to Fig. a, the angle at corner F becomes larger than 90° after the plate deforms. Thus, the shear strain is negative. 0.245 rad
Ans.
15
80 mm
F
x
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2–27. The material distorts into the dashed position shown. Determine the average normal strain that occurs along the diagonals AD and CF.
y 15 mm
25 mm D
C 10 mm
The undeformed length of diagonals AD and CF are
B E
LAD = LCF = 280 + 100 = 216400 mm 2
2
The deformed length of diagonals AD and CF are
75 mm
90 mm
LAD¿ = 2(80 + 25) + 100 = 221025 mm 2
2
LC¿F = 2(80 - 15)2 + 1002 = 214225 mm A
Thus,
A eavg B AD =
LAD¿ - LAD 221025 - 216400 = = 0.132 mm>mm LAD 216400
Ans.
A eavg B CF =
LC¿F - LCF 214225 - 216400 = = -0.0687 mm>mm LCF 216400
Ans.
*2–28. The wire is subjected to a normal strain that is 2 defined by P = xe - x , where x is in millimeters. If the wire has an initial length L, determine the increase in its length.
80 mm
P ⫽ xe⫺x
L
2
dL = e dx = x e-x dx L 2
L0
x e-x dx
L 1 1 1 2 2 = - c e-x d 冷 = - c e-L - d 2 2 2 0
=
x
2
x x
¢L =
F
1 2 [1 - e-L ] 2
Ans.
16
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•2–29. The curved pipe has an original radius of 2 ft. If it is heated nonuniformly, so that the normal strain along its length is P = 0.05 cos u, determine the increase in length of the pipe.
e = 0.05 cos u ¢L =
L
2 ft
e dL
=
A
u
90°
(0.05 cos u)(2 du)
L0
90°
= 0.1
90°
cos u du = [0.1[sin u] 0冷 ] = 0.100 ft
L0
Ans.
Solve Prob. 2–29 if P = 0.08 sin u.
2–30.
dL = 2 due = 0.08 sin u ¢L =
e dL
L
90°
=
2 ft
(0.08 sin u)(2 du)
L0
= 0.16
L0
90°
sin u du = 0.16[-cos u] 0冷 = 0.16 ft
Ans.
2–31. The rubber band AB has an unstretched length of 1 ft. If it is fixed at B and attached to the surface at point A¿, determine the average normal strain in the band. The surface is defined by the function y = (x2) ft, where x is in feet.
y y ⫽ x2
A¿
Geometry: 1 ft
L =
L0
A
1 + a
dy 2 b dx dx
However y = x2 then
1 ft
dy = 2x dx
B
1 ft
L =
=
L0
A
u 90°
21 + 4 x2 dx
1 1 ft C 2x21 + 4 x2 + ln A 2x + 21 + 4x2 B D 冷0 4
= 1.47894 ft Average Normal Strain: L - L0 1.47894 - 1 eavg = = = 0.479 ft>ft L0 1
Ans.
17
A 1 ft
x
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*2–32. The bar is originally 300 mm long when it is flat. If it is subjected to a shear strain defined by gxy = 0.02x, where x is in meters, determine the displacement ¢y at the end of its bottom edge. It is distorted into the shape shown, where no elongation of the bar occurs in the x direction.
y
⌬y x 300 mm
Shear Strain: dy = tan gxy ; dx
dy = tan (0.02 x) dx 300 mm
¢y
dy =
L0
L0
tan (0.02 x)dx
¢y = -50[ln cos (0.02x)]|0300 mm = 2.03 mm
Ans.
The fiber AB has a length L and orientation u. If its ends A and B undergo very small displacements uA and vB , respectively, determine the normal strain in the fiber when it is in position A¿B¿.
•2–33.
y B¿ vB B L
Geometry: LA¿B¿ = 2(L cos u - uA) + (L sin u + yB) 2
2
A
= 2L3 + u2A + y2B + 2L(yB sin u - uA cos u) Average Normal Strain: LA¿B¿ - L eAB = L =
A
1 +
2(yB sin u - uA cos u) u2A + y2B + - 1 L L2
Neglecting higher terms u2A and y2B 1
eAB
2(yB sin u - uA cos u) 2 = B1 + R - 1 L
Using the binomial theorem: eAB = 1 +
=
2uA cos u 1 2yB sin u ¢ ≤ + ... - 1 2 L L
yB sin u uA cos u L L
Ans.
18
uA A¿
u x
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2–34. If the normal strain is defined in reference to the final length, that is, Pnœ = lim a p : p¿
¢s¿ - ¢s b ¢s¿
instead of in reference to the original length, Eq. 2–2, show that the difference in these strains is represented as a second-order term, namely, Pn - Pnœ = PnPnœ .
eB =
¢S¿ - ¢S ¢S
œ = eB - eA
¢S¿ - ¢S ¢S¿ - ¢S ¢S ¢S¿
¢S¿ 2 - ¢S¢S¿ - ¢S¿¢S + ¢S2 ¢S¢S¿ 2 2 ¢S¿ + ¢S - 2¢S¿¢S = ¢S¢S¿ =
=
(¢S¿ - ¢S)2 ¢S¿ - ¢S ¢S¿ - ¢S = ¢ ≤¢ ≤ ¢S¢S¿ ¢S ¢S¿
= eA eBœ (Q.E.D)
19
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•3–1.
A concrete cylinder having a diameter of 6.00 in. and gauge length of 12 in. is tested in compression. The results of the test are reported in the table as load versus contraction. Draw the stress–strain diagram using scales of 1 in. = 0.5 ksi and 1 in. = 0.2110-32 in.>in. From the diagram, determine approximately the modulus of elasticity.
Stress and Strain: s =
P (ksi) A
e =
dL (in./in.) L
0
0
0.177
0.00005
0.336
0.00010
0.584
0.000167
0.725
0.000217
0.902
0.000283
1.061
0.000333
1.220
0.000375
1.362
0.000417
1.645
0.000517
1.768
0.000583
1.874
0.000625
Modulus of Elasticity: From the stress–strain diagram Eapprox =
1.31 - 0 = 3.275 A 103 B ksi 0.0004 - 0
Ans.
1
Load (kip)
Contraction (in.)
0 5.0 9.5 16.5 20.5 25.5 30.0 34.5 38.5 46.5 50.0 53.0
0 0.0006 0.0012 0.0020 0.0026 0.0034 0.0040 0.0045 0.0050 0.0062 0.0070 0.0075
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3–2. Data taken from a stress–strain test for a ceramic are given in the table. The curve is linear between the origin and the first point. Plot the diagram, and determine the modulus of elasticity and the modulus of resilience. Modulus of Elasticity: From the stress–strain diagram E =
33.2 - 0 = 55.3 A 103 B ksi 0.0006 - 0
S (ksi)
P (in./in.)
0 33.2 45.5 49.4 51.5 53.4
0 0.0006 0.0010 0.0014 0.0018 0.0022
S (ksi)
P (in./in.)
0 33.2 45.5 49.4 51.5 53.4
0 0.0006 0.0010 0.0014 0.0018 0.0022
Ans.
Modulus of Resilience: The modulus of resilience is equal to the area under the linear portion of the stress–strain diagram (shown shaded). ut =
1 lb in. in # lb (33.2) A 103 B ¢ 2 ≤ ¢ 0.0006 ≤ = 9.96 2 in. in in3
Ans.
3–3. Data taken from a stress–strain test for a ceramic are given in the table. The curve is linear between the origin and the first point. Plot the diagram, and determine approximately the modulus of toughness. The rupture stress is sr = 53.4 ksi. Modulus of Toughness: The modulus of toughness is equal to the area under the stress–strain diagram (shown shaded). (ut)approx =
lb in. 1 (33.2) A 103 B ¢ 2 ≤ (0.0004 + 0.0010) ¢ ≤ 2 in. in + 45.5 A 103 B ¢ +
1 lb in. (7.90) A 103 B ¢ 2 ≤ (0.0012) ¢ ≤ 2 in. in +
= 85.0
lb in. ≤ (0.0012) ¢ ≤ in. in2
1 lb in. (12.3) A 103 B ¢ 2 ≤ (0.0004) ¢ ≤ 2 in. in
in # lb in3
Ans.
2
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*3–4. A tension test was performed on a specimen having an original diameter of 12.5 mm and a gauge length of 50 mm. The data are listed in the table. Plot the stress–strain diagram, and determine approximately the modulus of elasticity, the ultimate stress, and the fracture stress. Use a scale of 20 mm = 50 MPa and 20 mm = 0.05 mm>mm. Redraw the linear-elastic region, using the same stress scale but a strain scale of 20 mm = 0.001 mm>mm. Stress and Strain: s =
dL P (MPa) e = (mm/mm) A L 0
0
90.45
0.00035
259.9
0.00120
308.0
0.00204
333.3
0.00330
355.3
0.00498
435.1
0.02032
507.7
0.06096
525.6
0.12700
507.7
0.17780
479.1
0.23876
Modulus of Elasticity: From the stress–strain diagram (E)approx =
228.75(106) - 0 = 229 GPa 0.001 - 0
Ans.
Ultimate and Fracture Stress: From the stress–strain diagram (sm)approx = 528 MPa
Ans.
(sf)approx = 479 MPa
Ans.
3
Load (kN)
Elongation (mm)
0 11.1 31.9 37.8 40.9 43.6 53.4 62.3 64.5 62.3 58.8
0 0.0175 0.0600 0.1020 0.1650 0.2490 1.0160 3.0480 6.3500 8.8900 11.9380
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3–5. A tension test was performed on a steel specimen having an original diameter of 12.5 mm and gauge length of 50 mm. Using the data listed in the table, plot the stress–strain diagram, and determine approximately the modulus of toughness. Use a scale of 20 mm = 50 MPa and 20 mm = 0.05 mm>mm. Stress and Strain: s =
P dL (MPa) e = (mm/mm) A L 0
0
90.45
0.00035
259.9
0.00120
308.0
0.00204
333.3
0.00330
355.3
0.00498
435.1
0.02032
507.7
0.06096
525.6
0.12700
507.7
0.17780
479.1
0.23876
Modulus of Toughness: The modulus of toughness is equal to the total area under the stress–strain diagram and can be approximated by counting the number of squares. The total number of squares is 187. (ut)approx = 187(25) A 106 B ¢
N m ≤ a 0.025 b = 117 MJ>m3 m m2
Ans.
4
Load (kN)
Elongation (mm)
0 11.1 31.9 37.8 40.9 43.6 53.4 62.3 64.5 62.3 58.8
0 0.0175 0.0600 0.1020 0.1650 0.2490 1.0160 3.0480 6.3500 8.8900 11.9380
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3–6. A specimen is originally 1 ft long, has a diameter of 0.5 in., and is subjected to a force of 500 lb. When the force is increased from 500 lb to 1800 lb, the specimen elongates 0.009 in. Determine the modulus of elasticity for the material if it remains linear elastic. Normal Stress and Strain: Applying s =
s1 =
s2 =
¢e =
0.500 p 2 4 (0.5 )
1.80 p 2 4 (0.5 )
dL P and e = . A L
= 2.546 ksi
= 9.167 ksi
0.009 = 0.000750 in.>in. 12
Modulus of Elasticity: E =
¢s 9.167 - 2.546 = = 8.83 A 103 B ksi ¢e 0.000750
Ans.
3–7. A structural member in a nuclear reactor is made of a zirconium alloy. If an axial load of 4 kip is to be supported by the member, determine its required cross-sectional area. Use a factor of safety of 3 relative to yielding. What is the load on the member if it is 3 ft long and its elongation is 0.02 in.? Ezr = 14(103) ksi, sY = 57.5 ksi. The material has elastic behavior. Allowable Normal Stress: F.S. =
3 =
sy sallow 57.5 sallow
sallow = 19.17 ksi sallow =
P A
19.17 =
4 A
A = 0.2087 in2 = 0.209 in2
Ans.
Stress–Strain Relationship: Applying Hooke’s law with e =
0.02 d = = 0.000555 in.>in. L 3 (12) s = Ee = 14 A 103 B (0.000555) = 7.778 ksi
Normal Force: Applying equation s =
P . A
P = sA = 7.778 (0.2087) = 1.62 kip
Ans.
5
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*3–8. The strut is supported by a pin at C and an A-36 steel guy wire AB. If the wire has a diameter of 0.2 in., determine how much it stretches when the distributed load acts on the strut.
A
60⬚ 200 lb/ft
a + ©MC = 0;
1 FAB cos 60°(9) - (200)(9)(3) = 0 2
9 ft
FAB = 600 lb
The normal stress the wire is sAB =
FAB = AAB
p 4
600 = 19.10(103) psi = 19.10 ksi (0.22)
Since sAB 6 sy = 36 ksi, Hooke’s Law can be applied to determine the strain in wire. sAB = EPAB;
19.10 = 29.0(103)PAB PAB = 0.6586(10 - 3) in>in
9(12) The unstretched length of the wire is LAB = = 124.71 in. Thus, the wire sin 60° stretches dAB = PAB LAB = 0.6586(10 - 3)(124.71) = 0.0821 in.
Ans.
6
B
C
Here, we are only interested in determining the force in wire AB.
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The s –P diagram for a collagen fiber bundle from which a human tendon is composed is shown. If a segment of the Achilles tendon at A has a length of 6.5 in. and an approximate cross-sectional area of 0.229 in2, determine its elongation if the foot supports a load of 125 lb, which causes a tension in the tendon of 343.75 lb.
•3–9.
s =
s (ksi) 4.50 A
3.75 3.00 2.25 1.50
P 343.75 = = 1.50 ksi A 0.229
125 lb
0.75 0.05
From the graph e = 0.035 in.>in. d = eL = 0.035(6.5) = 0.228 in.
0.10
P (in./in.)
Ans.
s (ksi)
3–10. The stress–strain diagram for a metal alloy having an original diameter of 0.5 in. and a gauge length of 2 in. is given in the figure. Determine approximately the modulus of elasticity for the material, the load on the specimen that causes yielding, and the ultimate load the specimen will support.
105 90 75 60
From the stress–strain diagram, Fig. a,
45
60 ksi - 0 E = ; 1 0.002 - 0 sy = 60 ksi
E = 30.0(103) ksi
Ans.
30 15
su>t = 100 ksi
0
Thus, PY = sYA = 60 C p4 (0.52) D = 11.78 kip = 11.8 kip
Ans.
Pu>t = su>t A = 100 C p4 (0.52) D = 19.63 kip = 19.6 kip
Ans.
7
0 0
0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.001 0.002 0.003 0.004 0.005 0.006 0.007
P (in./in.)
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s (ksi)
3–11. The stress–strain diagram for a steel alloy having an original diameter of 0.5 in. and a gauge length of 2 in. is given in the figure. If the specimen is loaded until it is stressed to 90 ksi, determine the approximate amount of elastic recovery and the increase in the gauge length after it is unloaded.
105 90 75 60 45 30 15 0
From the stress–strain diagram Fig. a, the modulus of elasticity for the steel alloy is E 60 ksi - 0 = ; 1 0.002 - 0
E = 30.0(103) ksi
when the specimen is unloaded, its normal strain recovered along line AB, Fig. a, which has a gradient of E. Thus Elastic Recovery =
90 90 ksi = 0.003 in>in = E 30.0(103) ksi
Ans.
Thus, the permanent set is PP = 0.05 - 0.003 = 0.047 in>in Then, the increase in gauge length is ¢L = PPL = 0.047(2) = 0.094 in
Ans.
8
0 0
0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.001 0.002 0.003 0.004 0.005 0.006 0.007
P (in./in.)
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*3–12. The stress–strain diagram for a steel alloy having an original diameter of 0.5 in. and a gauge length of 2 in. is given in the figure. Determine approximately the modulus of resilience and the modulus of toughness for the material. The Modulus of resilience is equal to the area under the stress–strain diagram up to the proportional limit. sPL = 60 ksi
PPL = 0.002 in>in.
Thus, (ui)r =
1 1 in # lb sPLPPL = C 60(103) D (0.002) = 60.0 2 2 in3
Ans.
The modulus of toughness is equal to the area under the entire stress–strain diagram. This area can be approximated by counting the number of squares. The total number is 38. Thus,
C (ui)t D approx = 38 c 15(103)
lb in in # lb d a0.05 b = 28.5(103) 2 in in in3
s (ksi) 105 90 75 60 45 30 15 0
0 0
0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.001 0.002 0.003 0.004 0.005 0.006 0.007
P (in./in.)
9
Ans.
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•3–13.
A bar having a length of 5 in. and cross-sectional area of 0.7 in2 is subjected to an axial force of 8000 lb. If the bar stretches 0.002 in., determine the modulus of elasticity of the material. The material has linear-elastic behavior.
8000 lb
8000 lb 5 in.
Normal Stress and Strain: 8.00 P = = 11.43 ksi A 0.7
s =
e =
dL 0.002 = = 0.000400 in.>in. L 5
Modulus of Elasticity: E =
s 11.43 = = 28.6(103) ksi e 0.000400
Ans.
3–14. The rigid pipe is supported by a pin at A and an A-36 steel guy wire BD. If the wire has a diameter of 0.25 in., determine how much it stretches when a load of P = 600 lb acts on the pipe.
B
Here, we are only interested in determining the force in wire BD. Referring 4 ft to the FBD in Fig. a a + ©MA = 0;
FBD A 45 B (3) - 600(6) = 0
FBD = 1500 lb
A
sBD
3 ft
1500 = 30.56(103) psi = 30.56 ksi p 2 (0.25 ) 4
Since sBD 6 sy = 36 ksi, Hooke’s Law can be applied to determine the strain in the wire. sBD = EPBD;
D C
The normal stress developed in the wire is FBD = = ABD
P
30.56 = 29.0(103)PBD PBD = 1.054(10 - 3) in.>in.
The unstretched length of the wire is LBD = 232 + 42 = 5ft = 60 in. Thus, the wire stretches dBD = PBD LBD = 1.054(10 - 3)(60) = 0.0632 in
Ans.
10
3 ft
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3–15. The rigid pipe is supported by a pin at A and an A-36 guy wire BD. If the wire has a diameter of 0.25 in., determine the load P if the end C is displaced 0.075 in. downward.
B
4 ft
P
A
D C 3 ft
Here, we are only interested in determining the force in wire BD. Referring to the FBD in Fig. a FBD A 45 B (3) - P(6) = 0
a + ©MA = 0;
FBD = 2.50 P
The unstretched length for wire BD is LBD = 232 + 42 = 5 ft = 60 in. From the geometry shown in Fig. b, the stretched length of wire BD is LBD¿ = 2602 + 0.0752 - 2(60)(0.075) cos 143.13° = 60.060017 Thus, the normal strain is PBD =
LBD¿ - LBD 60.060017 - 60 = = 1.0003(10 - 3) in.>in. LBD 60
Then, the normal stress can be obtain by applying Hooke’s Law. sBD = EPBD = 29(103) C 1.0003(10 - 3) D = 29.01 ksi Since sBD 6 sy = 36 ksi, the result is valid. sBD =
FBD ; ABD
29.01(103) =
2.50 P (0.252)
p 4
P = 569.57 lb = 570 lb
Ans.
11
3 ft
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s (MPa)
*3–16. Determine the elongation of the square hollow bar when it is subjected to the axial force P = 100 kN. If this axial force is increased to P = 360 kN and released, find the permanent elongation of the bar. The bar is made of a metal alloy having a stress–strain diagram which can be approximated as shown.
500 600 mm P
250
50 mm 5 mm
0.00125
Normal Stress and Strain: The cross-sectional area of the hollow bar is A = 0.052 - 0.042 = 0.9(10 - 3)m2. When P = 100 kN, s1 =
100(103) P = 111.11 MPa = A 0.9(10 - 3)
From the stress–strain diagram shown in Fig. a, the slope of the straight line OA which represents the modulus of elasticity of the metal alloy is E =
250(106) - 0 = 200 GPa 0.00125 - 0
Since s1 6 250 MPa, Hooke’s Law can be applied. Thus s1 = Ee1; 111.11(106) = 200(109)e1 e1 = 0.5556(10 - 3) mm>mm Thus, the elongation of the bar is d1 = e1L = 0.5556(10 - 3)(600) = 0.333 mm
Ans.
When P = 360 kN, s2 =
360(103) P = 400 MPa = A 0.9(10 - 3)
From the geometry of the stress–strain diagram, Fig. a, e2 - 0.00125 0.05 - 0.00125 = 400 - 250 500 - 250
e2 = 0.0305 mm>mm
When P = 360 kN is removed, the strain recovers linearly along line BC, Fig. a, parallel to OA. Thus, the elastic recovery of strain is given by s2 = Eer;
400(106) = 200(109)er er = 0.002 mm>mm
The permanent set is eP = e2 - er = 0.0305 - 0.002 = 0.0285 mm>mm Thus, the permanent elongation of the bar is dP = ePL = 0.0285(600) = 17.1 mm
Ans.
12
0.05
P (mm/mm) 50 mm
P 5 mm
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3–16. Continued
13
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s (ksi)
3–17. A tension test was performed on an aluminum 2014-T6 alloy specimen. The resulting stress–strain diagram is shown in the figure. Estimate (a) the proportional limit, (b) the modulus of elasticity, and (c) the yield strength based on a 0.2% strain offset method.
70 60 50 40 30 20 10 0
0.02 0.002
Proportional Limit and Yield Strength: From the stress–strain diagram, Fig. a, spl = 44 ksi
Ans.
sY = 60 ksi
Ans.
Modulus of Elasticity: From the stress–strain diagram, the corresponding strain for sPL = 44 ksi is epl = 0.004 in.>in. Thus, E =
44 - 0 = 11.0(103) ksi 0.004 - 0
Ans.
Modulus of Resilience: The modulus of resilience is equal to the area under the
14
0.04 0.004
0.06 0.006
0.08 0.008
0.10 0.010
P (in./in.)
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s (ksi)
3–18. A tension test was performed on an aluminum 2014-T6 alloy specimen. The resulting stress–strain diagram is shown in the figure. Estimate (a) the modulus of resilience; and (b) modulus of toughness.
70 60 50 40 30 20 10 0
0.02 0.002
0.04 0.004
0.06 0.006
0.08 0.008
0.10 0.010
P (in./in.)
stress–strain diagram up to the proportional limit. From the stress–strain diagram, spl = 44 ksi
epl = 0.004 in.>in.
Thus,
A Ui B r = splepl = (44)(103)(0.004) = 88 1 2
1 2
in # lb in3
Ans.
Modulus of Toughness: The modulus of toughness is equal to the area under the entire stress–strain diagram. This area can be approximated by counting the number of squares. The total number of squares is 65. Thus,
C A Ui B t D approx = 65 B 10(103)
lb in. in # lb c0.01 d = 6.50(103) 2R in. in in3
Ans.
The stress–strain diagram for a bone is shown, and can be described by the equation
3–19. The stress–strain diagram for a bone is shown, and can be described by the equation P = 0.45110-62 s ⫹ 0.36110-122 s3, where s is in kPa. Determine the yield strength assuming a 0.3% offset.
P
s
P ⫽ 0.45(10⫺6)s + 0.36(10⫺12)s3 P
e = 0.45(10-6)s + 0.36(10-12)s3, dP = A 0.45(10-6) + 1.08(10-12) s2 B ds E =
ds 1 2 = = 2.22 MPa dP 0.45(10 - 6)
Ans.
s=0
15
P
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*3–20. The stress–strain diagram for a bone is shown and can be described by the equation P = 0.45110-62 s ⫹ 0.36110-122 s3, where s is in kPa. Determine the modulus of toughness and the amount of elongation of a 200-mmlong region just before it fractures if failure occurs at P = 0.12 mm>mm.
P
s
P ⫽ 0.45(10⫺6)s + 0.36(10⫺12)s3 P
When e = 0.12
120(103) = 0.45 s + 0.36(10-6)s3 Solving for the real root: s = 6873.52 kPa 6873.52
ut =
LA
dA =
L0
(0.12 - e)ds
6873.52
ut =
L0
(0.12 - 0.45(10-6)s - 0.36(10-12)s3)ds 6873.52
= 0.12 s - 0.225(10-6)s2 - 0.09(10-12)s4|0 = 613 kJ>m3
Ans.
d = eL = 0.12(200) = 24 mm
Ans.
16
P
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•3–21.
The stress–strain diagram for a polyester resin is given in the figure. If the rigid beam is supported by a strut AB and post CD, both made from this material, and subjected to a load of P = 80 kN, determine the angle of tilt of the beam when the load is applied. The diameter of the strut is 40 mm and the diameter of the post is 80 mm.
B
2m P
A
C 0.75 m 0.75 m
D
0.5 m
From the stress–strain diagram, E =
32.2(10)6 = 3.22(109) Pa 0.01
s (MPa) 100 95
Thus,
70 60
40(10 ) FAB = p = 31.83 MPa 2 AAB 4 (0.04)
sAB =
eAB
50
31.83(106) sAB = 0.009885 mm>mm = = E 3.22(109)
20 0
7.958(106) sCD = 0.002471 mm>mm = E 3.22(109)
dAB = eABLAB = 0.009885(2000) = 19.771 mm dCD = eCDLCD = 0.002471(500) = 1.236 mm Angle of tilt a: tan a =
18.535 ; 1500
tension
40 32.2
40(103) FCD = p = 7.958 MPa 2 ACD 4 (0.08)
sCD =
eCD =
compression
80 3
a = 0.708°
Ans.
17
0
0.01 0.02 0.03 0.04
P (mm/mm)
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3–22. The stress–strain diagram for a polyester resin is given in the figure. If the rigid beam is supported by a strut AB and post CD made from this material, determine the largest load P that can be applied to the beam before it ruptures. The diameter of the strut is 12 mm and the diameter of the post is 40 mm.
B
2m P
Rupture of strut AB: sR =
FAB ; AAB
50(106) =
P>2
A
; p 2 4 (0.012)
0.75 m 0.75 m
P = 11.3 kN (controls)
D
0.5 m
Ans. s (MPa)
Rupture of post CD: FCD ; sR = ACD
C
95(10 ) =
100 95
P>2
6
p 2 4 (0.04)
compression
80 70 60
P = 239 kN
50 tension
40 32.2 20 0
0
0.01 0.02 0.03 0.04
P (mm/mm)
s (ksi)
3–23. By adding plasticizers to polyvinyl chloride, it is possible to reduce its stiffness. The stress–strain diagrams for three types of this material showing this effect are given below. Specify the type that should be used in the manufacture of a rod having a length of 5 in. and a diameter of 2 in., that is required to support at least an axial load of 20 kip and also be able to stretch at most 14 in.
15 P unplasticized 10
copolymer
flexible
5
(plasticized)
Normal Stress:
P
P s = = A
20 p 2 = 6.366 ksi (2 ) 4
0
0
Normal Strain: e =
0.25 = 0.0500 in.>in. 5
From the stress–strain diagram, the copolymer will satisfy both stress and strain requirements. Ans.
18
0.10
0.20
0.30
P (in./in.)
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*3–24. The stress–strain diagram for many metal alloys can be described analytically using the Ramberg-Osgood three parameter equation P = s>E + ksn, where E, k, and n are determined from measurements taken from the diagram. Using the stress–strain diagram shown in the figure, take E = 3011032 ksi and determine the other two parameters k and n and thereby obtain an analytical expression for the curve.
s (ksi) 80 60 40 20 0.1
0.2
0.3
0.4
0.5
P (10–6)
Choose, s = 40 ksi,
e = 0.1
s = 60 ksi,
e = 0.3
0.1 =
40 + k(40)n 30(103)
0.3 =
60 + k(60)n 30(103)
0.098667 = k(40)n 0.29800 = k(60)n 0.3310962 = (0.6667)n ln (0.3310962) = n ln (0.6667) n = 2.73
Ans.
k = 4.23(10-6)
Ans.
•3–25.
The acrylic plastic rod is 200 mm long and 15 mm in diameter. If an axial load of 300 N is applied to it, determine the change in its length and the change in its diameter. Ep = 2.70 GPa, np = 0.4. s =
P = A
elong =
300 p 2 4 (0.015)
300 N
300 N 200 mm
= 1.697 MPa
1.697(106) s = 0.0006288 = E 2.70(109)
d = elong L = 0.0006288 (200) = 0.126 mm
Ans.
elat = -Velong = -0.4(0.0006288) = -0.0002515 ¢d = elatd = -0.0002515 (15) = -0.00377 mm
Ans.
19
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3–26. The short cylindrical block of 2014-T6 aluminum, having an original diameter of 0.5 in. and a length of 1.5 in., is placed in the smooth jaws of a vise and squeezed until the axial load applied is 800 lb. Determine (a) the decrease in its length and (b) its new diameter.
800 lb
800 lb
a) s =
P = A
elong =
p 4
800 = 4074.37 psi (0.5)2
s -4074.37 = -0.0003844 = E 10.6(106)
d = elong L = -0.0003844 (1.5) = -0.577 (10 - 3) in.
Ans.
b) V =
-elat = 0.35 elong
elat = -0.35 (-0.0003844) = 0.00013453 ¢d = elat d = 0.00013453 (0.5) = 0.00006727 d¿ = d + ¢d = 0.5000673 in.
Ans.
s(MPa)
3–27. The elastic portion of the stress–strain diagram for a steel alloy is shown in the figure. The specimen from which it was obtained had an original diameter of 13 mm and a gauge length of 50 mm. When the applied load on the specimen is 50 kN, the diameter is 12.99265 mm. Determine Poisson’s ratio for the material.
400
Normal Stress: s =
P = A
50(103) p 4
(0.0132)
= 376.70 Mpa 0.002
Normal Strain: From the stress–strain diagram, the modulus of elasticity 400(106) = 200 GPa. Applying Hooke’s law E = 0.002 elong =
elat =
376.70(106) s = 1.8835 A 10 - 3 B mm>mm = E 200(104)
d - d0 12.99265 - 13 = = -0.56538 A 10 - 3 B mm>mm d0 13
Poisson’s Ratio: The lateral and longitudinal strain can be related using Poisson’s ratio. V = -
-0.56538(10 - 3) elat = 0.300 = elong 1.8835(10 - 3)
Ans.
20
P(mm/mm)
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s(MPa)
*3–28. The elastic portion of the stress–strain diagram for a steel alloy is shown in the figure. The specimen from which it was obtained had an original diameter of 13 mm and a gauge length of 50 mm. If a load of P = 20 kN is applied to the specimen, determine its diameter and gauge length. Take n = 0.4.
400
Normal Stress: s =
P = A
20(103) p 4
(0.0132)
= 150.68Mpa 0.002
P(mm/mm)
Normal Strain: From the Stress–Strain diagram, the modulus of elasticity 400(106) E = = 200 GPa. Applying Hooke’s Law 0.002 elong =
150.68(106) s = 0.7534 A 10 - 3 B mm>mm = E 200(109)
Thus, dL = elong L0 = 0.7534 A 10 - 3 B (50) = 0.03767 mm L = L0 + dL = 50 + 0.03767 = 50.0377 mm
Ans.
Poisson’s Ratio: The lateral and longitudinal can be related using poisson’s ratio. elat = -velong = -0.4(0.7534) A 10 - 3 B
= -0.3014 A 10 - 3 B mm>mm
dd = elat d = -0.3014 A 10 - 3 B (13) = -0.003918 mm d = d0 + dd = 13 + ( -0.003918) = 12.99608 mm
Ans.
•3–29.
The aluminum block has a rectangular cross section and is subjected to an axial compressive force of 8 kip. If the 1.5-in. side changed its length to 1.500132 in., determine Poisson’s ratio and the new length of the 2-in. side. Eal ⫽ 10(103) ksi. s =
elat =
2 in.
8 kip
8 kip 3 in.
P 8 = = 2.667 ksi A (2)(1.5)
elong =
v =
1.5 in.
s -2.667 = -0.0002667 = E 10(103)
1.500132 - 1.5 = 0.0000880 1.5
-0.0000880 = 0.330 -0.0002667
Ans.
h¿ = 2 + 0.0000880(2) = 2.000176 in.
Ans.
21
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3–30. The block is made of titanium Ti-6A1-4V and is subjected to a compression of 0.06 in. along the y axis, and its shape is given a tilt of u = 89.7°. Determine Px, Py, and gxy.
y
Normal Strain: ey =
4 in. u
dLy Ly
=
-0.06 = -0.0150 in.>in. 4
Ans.
Poisson’s Ratio: The lateral and longitudinal strain can be related using Poisson’s ratio.
x
5 in.
ex = -vey = -0.36(-0.0150) = 0.00540 in. >in.
Ans.
Shear Strain: b = 180° - 89.7° = 90.3° = 1.576032 rad gxy =
p p - b = - 1.576032 = -0.00524 rad 2 2
Ans.
3–31. The shear stress–strain diagram for a steel alloy is shown in the figure. If a bolt having a diameter of 0.75 in. is made of this material and used in the double lap joint, determine the modulus of elasticity E and the force P required to cause the material to yield. Take n = 0.3.
P/2 P/2
P
t(ksi) 60
The shear force developed on the shear planes of the bolt can be determined by considering the equilibrium of the FBD shown in Fig. a + ©F = 0; : x
V + V - P = 0
V = =
g(rad)
P
0.00545
2
From the shear stress–strain diagram, the yield stress is ty = 60 ksi. Thus, ty =
Vy A
;
60 =
P>2
p 4
A 0.752 B
P = 53.01 kip = 53.0 kip
Ans.
From the shear stress–strain diagram, the shear modulus is G =
60 ksi = 11.01(103) ksi 0.00545
Thus, the modulus of elasticity is G =
E ; 2(1 + y)
11.01(103) =
E 2(1 + 0.3)
E = 28.6(103) ksi
Ans.
22
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*3–32. A shear spring is made by bonding the rubber annulus to a rigid fixed ring and a plug. When an axial load P is placed on the plug, show that the slope at point y in the rubber is dy>dr = -tan g = -tan1P>12phGr22. For small angles we can write dy>dr = -P>12phGr2. Integrate this expression and evaluate the constant of integration using the condition that y = 0 at r = ro. From the result compute the deflection y = d of the plug.
P
h
ro
y
d
ri r y
Shear Stress–Strain Relationship: Applying Hooke’s law with tA =
g =
P . 2p r h
tA P = G 2p h G r
dy P = -tan g = -tan a b dr 2p h G r
(Q.E.D)
If g is small, then tan g = g. Therefore, dy P = dr 2p h G r
At r = ro,
y = -
dr P 2p h G L r
y = -
P ln r + C 2p h G
0 = -
P ln ro + C 2p h G
y = 0
C =
Then, y =
ro P ln r 2p h G
At r = ri,
y = d d =
P ln ro 2p h G
ro P ln ri 2p h G
Ans.
23
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•3–33.
The support consists of three rigid plates, which are connected together using two symmetrically placed rubber pads. If a vertical force of 5 N is applied to plate A, determine the approximate vertical displacement of this plate due to shear strains in the rubber. Each pad has cross-sectional dimensions of 30 mm and 20 mm. Gr = 0.20 MPa.
C
B
40 mm
40 mm
A
tavg =
g =
V 2.5 = = 4166.7 Pa A (0.03)(0.02)
5N
t 4166.7 = 0.02083 rad = G 0.2(106)
d = 40(0.02083) = 0.833 mm
Ans.
3–34. A shear spring is made from two blocks of rubber, each having a height h, width b, and thickness a. The blocks are bonded to three plates as shown. If the plates are rigid and the shear modulus of the rubber is G, determine the displacement of plate A if a vertical load P is applied to this plate. Assume that the displacement is small so that d = a tan g L ag.
P
d A
h
Average Shear Stress: The rubber block is subjected to a shear force of V =
P . 2
P
t =
V P 2 = = A bh 2bh
Shear Strain: Applying Hooke’s law for shear P
g =
t P 2bh = = G G 2bhG
Thus, d = ag = =
Pa 2bhG
Ans.
24
a
a
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s (ksi)
3–35. The elastic portion of the tension stress–strain diagram for an aluminum alloy is shown in the figure. The specimen used for the test has a gauge length of 2 in. and a diameter of 0.5 in. When the applied load is 9 kip, the new diameter of the specimen is 0.49935 in. Compute the shear modulus Gal for the aluminum.
70
0.00614
From the stress–strain diagram,
P (in./in.)
70 s = = 11400.65 ksi e 0.00614
Eal =
When specimen is loaded with a 9 - kip load, s =
P = A
p 4
9 = 45.84 ksi (0.5)2
s 45.84 = = 0.0040208 in.>in. E 11400.65
elong =
0.49935 - 0.5 d¿ - d = = - 0.0013 in.>in. d 0.5
elat =
V = -
Gal =
elat -0.0013 = 0.32332 = elong 0.0040208
11.4(103) Eat = = 4.31(103) ksi 2(1 + v) 2(1 + 0.32332)
Ans.
s (ksi)
*3–36. The elastic portion of the tension stress–strain diagram for an aluminum alloy is shown in the figure. The specimen used for the test has a gauge length of 2 in. and a diameter of 0.5 in. If the applied load is 10 kip, determine the new diameter of the specimen. The shear modulus is Gal = 3.811032 ksi. P s = = A
70
0.00614
10 = 50.9296 ksi p 2 (0.5) 4
From the stress–strain diagram E =
70 = 11400.65 ksi 0.00614
elong =
G =
s 50.9296 = = 0.0044673 in.>in. E 11400.65
E ; 2(1 + v)
3.8(103) =
11400.65 ; 2(1 + v)
v = 0.500
elat = - velong = - 0.500(0.0044673) = - 0.002234 in.>in. ¢d = elat d = - 0.002234(0.5) = - 0.001117 in. d¿ = d + ¢d = 0.5 - 0.001117 = 0.4989 in.
Ans.
25
P (in./in.)
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s(psi)
3–37. The s – P diagram for elastic fibers that make up human skin and muscle is shown. Determine the modulus of elasticity of the fibers and estimate their modulus of toughness and modulus of resilience.
55
11 1
E =
11 = 5.5 psi 2
Ans.
ut =
1 1 (2)(11) + (55 + 11)(2.25 - 2) = 19.25 psi 2 2
Ans.
ut =
1 (2)(11) = 11 psi 2
Ans.
3–38. A short cylindrical block of 6061-T6 aluminum, having an original diameter of 20 mm and a length of 75 mm, is placed in a compression machine and squeezed until the axial load applied is 5 kN. Determine (a) the decrease in its length and (b) its new diameter.
a)
s =
-5(103) P = p = - 15.915 MPa 2 A 4 (0.02)
s = E elong ;
- 15.915(106) = 68.9(109) elong elong = - 0.0002310 mm>mm
d = elong L = - 0.0002310(75) = - 0.0173 mm b)
v = -
elat ; elong
0.35 = -
Ans.
elat -0.0002310
elat = 0.00008085 mm>mm ¢d = elat d = 0.00008085(20) = 0.0016 mm d¿ = d + ¢d = 20 + 0.0016 = 20.0016 mm
Ans.
26
2 2.25
P(in./in.)
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3–39. The rigid beam rests in the horizontal position on two 2014-T6 aluminum cylinders having the unloaded lengths shown. If each cylinder has a diameter of 30 mm, determine the placement x of the applied 80-kN load so that the beam remains horizontal. What is the new diameter of cylinder A after the load is applied? nal = 0.35. a + ©MA = 0;
FB(3) - 80(x) = 0;
a + ©MB = 0;
-FA(3) + 80(3 - x) = 0;
FB =
80 kN x
A
80x 3 FA =
B
210 mm
220 mm
(1) 80(3 - x) 3
3m
(2)
Since the beam is held horizontally, dA = dB s =
P ; A
d = eL = a
P
e = P A
E
dA = dB ;
s A = E E
bL =
PL AE
80(3 - x) (220) 3
80x 3 (210)
=
AE
AE
80(3 - x)(220) = 80x(210) x = 1.53 m
Ans.
From Eq. (2), FA = 39.07 kN sA =
39.07(103) FA = 55.27 MPa = p 2 A 4 (0.03 )
elong =
55.27(106) sA = -0.000756 = E 73.1(109)
elat = -velong = -0.35(-0.000756) = 0.0002646 œ dA = dA + d elat = 30 + 30(0.0002646) = 30.008 mm
Ans.
*3–40. The head H is connected to the cylinder of a compressor using six steel bolts. If the clamping force in each bolt is 800 lb, determine the normal strain in the 3 bolts. Each bolt has a diameter of 16 in. If sY = 40 ksi and 3 Est = 29110 2 ksi, what is the strain in each bolt when the nut is unscrewed so that the clamping force is released?
C L
H
Normal Stress: s =
P = A
800
A B
p 3 2 4 16
= 28.97 ksi 6 sg = 40 ksi
Normal Strain: Since s 6 sg, Hooke’s law is still valid. e =
28.97 s = 0.000999 in.>in. = E 29(103)
Ans.
If the nut is unscrewed, the load is zero. Therefore, the strain e = 0
27
Ans.
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•3–41. The stone has a mass of 800 kg and center of gravity at G. It rests on a pad at A and a roller at B. The pad is fixed to the ground and has a compressed height of 30 mm, a width of 140 mm, and a length of 150 mm. If the coefficient of static friction between the pad and the stone is ms = 0.8, determine the approximate horizontal displacement of the stone, caused by the shear strains in the pad, before the stone begins to slip. Assume the normal force at A acts 1.5 m from G as shown. The pad is made from a material having E = 4 MPa and n = 0.35.
0.4 m
B
Equations of Equilibrium: a + ©MB = 0; + ©F = 0; : x
FA(2.75) - 7848(1.25) - P(0.3) = 0
[1]
P - F = 0
[2]
Note: The normal force at A does not act exactly at A. It has to shift due to friction. Friction Equation: F = ms FA = 0.8 FA
[3]
Solving Eqs. [1], [2] and [3] yields: FA = 3908.37 N
F = P = 3126.69 N
Average Shear Stress: The pad is subjected to a shear force of V = F = 3126.69 N. t =
V 3126.69 = = 148.89 kPa A (0.14)(0.15)
Modulus of Rigidity: G =
E 4 = = 1.481 MPa 2(1 + v) 2(1 + 0.35)
Shear Strain: Applying Hooke’s law for shear g =
148.89(103) t = 0.1005 rad = G 1.481(106)
Thus, dh = hg = 30(0.1005) = 3.02 mm
Ans.
28
P
G
1.25 m
0.3 m 1.5 m
A
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3–42. The bar DA is rigid and is originally held in the horizontal position when the weight W is supported from C. If the weight causes B to be displaced downward 0.025 in., determine the strain in wires DE and BC. Also, if the wires are made of A-36 steel and have a cross-sectional area of 0.002 in2, determine the weight W.
E 3 ft 2 ft D
3 ft B
5 3 = 0.025 d
A
4 ft
d = 0.0417 in eDE =
C
0.0417 d = = 0.00116 in.>in. L 3(12)
Ans. W
3
sDE = EeDE = 29(10 )(0.00116) = 33.56 ksi FDE = sDEADE = 33.56 (0.002) = 0.0672 kip a + ©MA = 0;
-(0.0672) (5) + 3(W) = 0
W = 0.112 kip = 112 lb
Ans.
sBC =
W 0.112 = = 55.94 ksi ABC 0.002
eBC =
sBC 55.94 = 0.00193 in.>in. = E 29 (103)
Ans.
3–43. The 8-mm-diameter bolt is made of an aluminum alloy. It fits through a magnesium sleeve that has an inner diameter of 12 mm and an outer diameter of 20 mm. If the original lengths of the bolt and sleeve are 80 mm and 50 mm, respectively, determine the strains in the sleeve and the bolt if the nut on the bolt is tightened so that the tension in the bolt is 8 kN. Assume the material at A is rigid. Eal = 70 GPa, Emg = 45 GPa.
50 mm A
30 mm
Normal Stress: 8(103)
sb =
P = Ab
p 2 4 (0.008 )
ss =
P = As
p 2 4 (0.02
= 159.15 MPa
8(103) - 0.0122)
= 39.79 MPa
Normal Strain: Applying Hooke’s Law eb =
159.15(106) sb = 0.00227 mm>mm = Eal 70(109)
Ans.
es =
39.79(106) ss = 0.000884 mm>mm = Emg 45(109)
Ans.
29
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*3–44. The A-36 steel wire AB has a cross-sectional area of 10 mm2 and is unstretched when u = 45.0°. Determine the applied load P needed to cause u = 44.9°.
A
400 mm
u 400 m
m
B P
¿ LAB 400 = sin 90.2° sin 44.9° ¿ = 566.67 mm LAB
LAB =
e =
400 = 565.69 sin 45°
¿ - LAB LAB 566.67 - 565.69 = = 0.001744 LAB 565.69
s = Ee = 200(109) (0.001744) = 348.76 MPa a + ©MA = 0 P(400 cos 0.2°) - FAB sin 44.9° (400) = 0
(1)
However, FAB = sA = 348.76(106)(10)(10 - 6) = 3.488 kN From Eq. (1), P = 2.46 kN
Ans.
30
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•4–1. The ship is pushed through the water using an A-36 steel propeller shaft that is 8 m long, measured from the propeller to the thrust bearing D at the engine. If it has an outer diameter of 400 mm and a wall thickness of 50 mm, determine the amount of axial contraction of the shaft when the propeller exerts a force on the shaft of 5 kN. The bearings at B and C are journal bearings. A
Internal Force: As shown on FBD.
B
C
D
5 kN
Displacement:
8m
dA =
PL = AE
-5.00 (103)(8) p 4
(0.42 - 0.32) 200(109)
= -3.638(10 - 6) m = -3.64 A 10 - 3 B mm
Ans.
Negative sign indicates that end A moves towards end D.
4–2. The copper shaft is subjected to the axial loads shown. Determine the displacement of end A with respect to end D. The diameters of each segment are dAB = 3 in., dBC = 2 in., and dCD = 1 in. Take Ecu = 1811032 ksi.
50 in.
A
p The cross-sectional area of segment AB, BC and CD are AAB = (32) = 2.25p in2, 4 p p ABC = (22) = p in2 and ACD = (12) = 0.25p in2. 4 4 Thus, PCD LCD PiLi PAB LAB PBC LBC = + + AiEi AAB ECu ABC ECu ACD ECu 2.00 (75)
6.00 (50)
=
(2.25p) C 18(10 ) D 3
+
p C 18(10 ) D 3
-1.00 (60)
+
(0.25p) C 18(103) D
= 0.766(10 - 3) in.
Ans.
The positive sign indicates that end A moves away from D.
122
60 in.
2 kip
6 kip
The normal forces developed in segment AB, BC and CD are shown in the FBDS of each segment in Fig. a, b and c respectively.
dA>D = ©
75 in.
B 2 kip
1 kip C
3 kip
D
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4–3. The A-36 steel rod is subjected to the loading shown. If the cross-sectional area of the rod is 50 mm2, determine the displacement of its end D. Neglect the size of the couplings at B, C, and D.
1m
A 9 kN B
The normal forces developed in segments AB, BC and CD are shown in the FBDS of each segment in Fig. a, b and c, respectively. The
cross-sectional areas of all 2 1 m b = 50.0(10 - 6) m2. A = A 50 mm2 B a 1000 mm dD = ©
the
segments
are
PiLi 1 = a PAB LAB + PBC LBC + PCD LCD b AiEi A ESC 1
=
50.0(10 ) C 200(109) D -6
c -3.00(103)(1) + 6.00(103)(1.5) + 2.00(103)(1.25) d
= 0.850(10 - 3) m = 0.850 mm
Ans.
The positive sign indicates that end D moves away from the fixed support.
123
1.5 m
1.25 m
C
4 kN
D
2 kN
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*4–4. The A-36 steel rod is subjected to the loading shown. If the cross-sectional area of the rod is 50 mm2, determine the displacement of C. Neglect the size of the couplings at B, C, and D.
1m
1.5 m
1.25 m
C
A 9 kN B
4 kN
D
2 kN
The normal forces developed in segments AB and BC are shown the FBDS of each segment in Fig. a and b, respectively. The cross-sectional area of these two segments 2 1m are A = A 50 mm2 B a b = 50.0 (10 - 6) m2. Thus, 10.00 mm dC = ©
PiLi 1 = A P L + PBC LBC B AiEi A ESC AB AB 1
=
50.0(10 - 6) C 200(109) D
c -3.00(103)(1) + 6.00(103)(1.5) d
= 0.600 (10 - 3) m = 0.600 mm
Ans.
The positive sign indicates that coupling C moves away from the fixed support.
4–5. The assembly consists of a steel rod CB and an aluminum rod BA, each having a diameter of 12 mm. If the rod is subjected to the axial loadings at A and at the coupling B, determine the displacement of the coupling B and the end A. The unstretched length of each segment is shown in the figure. Neglect the size of the connections at B and C, and assume that they are rigid. Est = 200 GPa, Eal = 70 GPa.
dB =
PL = AE
dA = ©
12(103)(3) p 4
12(103)(3) p 4
2
A
B
18 kN 6 kN 3m
= 0.00159 m = 1.59 mm
(0.012)2(200)(109)
PL = AE
C
2m
Ans.
18(103)(2) 9
(0.012) (200)(10 )
+
p 2 9 4 (0.012) (70)(10 )
= 0.00614 m = 6.14 mm
Ans.
4–6. The bar has a cross-sectional area of 3 in2, and E = 3511032 ksi. Determine the displacement of its end A when it is subjected to the distributed loading.
x
w ⫽ 500x1/3 lb/in.
A
4 ft x
P(x) =
L0
w dx = 500
x
L0
1
x3 dx =
1500 43 x 4
L
dA =
4(12) P(x) dx 1 3 1 1500 4 1500 b a b(48)3 = x3 dx = a AE 4 (3)(35)(108)(4) 7 (3)(35)(106) L0 L0
dA = 0.0128 in.
Ans.
124
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4–7. The load of 800 lb is supported by the four 304 stainless steel wires that are connected to the rigid members AB and DC. Determine the vertical displacement of the load if the members were horizontal before the load was applied. Each wire has a cross-sectional area of 0.05 in2.
E
F
4 ft H
D
C 2 ft
Referring to the FBD of member AB, Fig. a
5 ft
4.5 ft
a + ©MA = 0;
FBC (5) - 800(1) = 0
FBC = 160 lb
a + ©MB = 0;
800(4) - FAH (5) = 0
FAH = 640 lb
800 lb A
B 1 ft
Using the results of FBC and FAH, and referring to the FBD of member DC, Fig. b a + ©MD = 0;
FCF (7) - 160(7) - 640(2) = 0
a + ©MC = 0;
640(5) - FDE(7) = 0
FCF = 342.86 lb FDE = 457.14 lb
Since E and F are fixed, dD =
457.14(4)(2) FDE LDE = = 0.01567 in T A Est 0.05 C 28.0 (106) D
dC =
342.86 (4)(12) FCF LCF = = 0.01176 in T A Est 0.05 C 28.0 (106) D
From the geometry shown in Fig. c, dH = 0.01176 +
5 (0.01567 - 0.01176) = 0.01455 in T 7
Subsequently, dA>H =
640(4.5)(12) FAH LAH = = 0.02469 in T A Est 0.05 C 28.0(106) D
dB>C =
160(4.5)(12) FBC LBC = = 0.006171 in T A Est 0.05 C 28.0(106) D
Thus,
A + T B dA = dH + dA>H = 0.01455 + 0.02469 = 0.03924 in T
A + T B dB = dC + dB>C = 0.01176 + 0.006171 = 0.01793 in T From the geometry shown in Fig. d, dP = 0.01793 +
4 (0.03924 - 0.01793) = 0.0350 in T 5
125
Ans.
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*4–8. The load of 800 lb is supported by the four 304 stainless steel wires that are connected to the rigid members AB and DC. Determine the angle of tilt of each member after the load is applied.The members were originally horizontal, and each wire has a cross-sectional area of 0.05 in2.
E
F
4 ft H
D
C 2 ft
Referring to the FBD of member AB, Fig. a,
5 ft
4.5 ft
a + ©MA = 0;
FBC (5) - 800(1) = 0
FBC = 160 lb
a + ©MB = 0;
800(4) - FAH (5) = 0
FAH = 640 lb
800 lb A
B 1 ft
Using the results of FBC and FAH and referring to the FBD of member DC, Fig. b, a + ©MD = 0;
FCF (7) - 160(7) - 640(2) = 0
FCF = 342.86 lb
a + ©MC = 0;
640(5) - FDE (7) = 0
FDE = 457.14 lb
Since E and F are fixed, dD =
457.14 (4)(12) FDE LDE = = 0.01567 in T A Est 0.05 C 28.0(106) D
dC =
342.86 (4)(12) FCF LCF = = 0.01176 in T A Est 0.05 C 28.0(106) D
From the geometry shown in Fig. c dH = 0.01176 +
u =
5 (0.01567 - 0.01176) = 0.01455 in T 7
0.01567 - 0.01176 = 46.6(10 - 6) rad 7(12)
Ans.
Subsequently, dA>H =
640 (4.5)(12) FAH LAH = = 0.02469 in T A Est 0.05 C 28.0(106) D
dB>C =
160 (4.5)(12) FBC LBC = = 0.006171 in T A Est 0.05 C 28.0(106) D
Thus,
A + T B dA = dH + dA>H = 0.01455 + 0.02469 = 0.03924 in T
A + T B dB = dC + dB>C = 0.01176 + 0.006171 = 0.01793 in T From the geometry shown in Fig. d f =
0.03924 - 0.01793 = 0.355(10 - 3) rad 5(12)
Ans.
126
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4–8. Continued
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•4–9.
The assembly consists of three titanium (Ti-6A1-4V) rods and a rigid bar AC. The cross-sectional area of each rod is given in the figure. If a force of 6 kip is applied to the ring F, determine the horizontal displacement of point F.
D
4 ft
C
ACD ⫽ 1 in2 2 ft E
Internal Force in the Rods: a + ©MA = 0; + ©F = 0; : x
FCD(3) - 6(1) = 0
FCD = 2.00 kip
6 - 2.00 - FAB = 0
FAB = 4.00 kip
AAB ⫽ 1.5 in2 6 ft
B
1 ft F
6 kip 2 1 ft AEF ⫽ 2 in
A
Displacement: dC =
2.00(4)(12) FCD LCD = 0.0055172 in. = ACD E (1)(17.4)(103)
dA =
4.00(6)(12) FAB LAB = 0.0110344 in. = AAB E (1.5)(17.4)(103)
dF>E =
6.00(1)(12) FEF LEF = 0.0020690 in. = AEF E (2)(17.4)(103)
œ dE 0.0055172 = ; 2 3
œ dE = 0.0036782 in.
œ dE = dC + dE = 0.0055172 + 0.0036782 = 0.0091954 in.
dF = dE + dF>E = 0.0091954 + 0.0020690 = 0.0113 in.
Ans.
4–10. The assembly consists of three titanium (Ti-6A1-4V) rods and a rigid bar AC. The cross-sectional area of each rod is given in the figure. If a force of 6 kip is applied to the ring F, determine the angle of tilt of bar AC.
D
C
4 ft ACD ⫽ 1 in
2
2 ft E
Internal Force in the Rods: a + ©MA = 0;
FCD(3) - 6(1) = 0
FCD = 2.00 kip
+ ©F = 0; : x
6 - 2.00 - FAB = 0
FAB = 4.00 kip
AAB ⫽ 1.5 in2 B
Displacement: dC =
2.00(4)(12) FCD LCD = 0.0055172 in. = ACD E (1)(17.4)(103)
dA =
4.00(6)(12) FAB LAB = 0.0110344 in. = AAB E (1.5)(17.4)(103)
u = tan - 1
dA - dC 0.0110344 - 0.0055172 = tan - 1 3(12) 3(12) = 0.00878°
Ans.
128
6 ft
A
1 ft F
6 kip 2 1 ft AEF ⫽ 2 in
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4–11. The load is supported by the four 304 stainless steel wires that are connected to the rigid members AB and DC. Determine the vertical displacement of the 500-lb load if the members were originally horizontal when the load was applied. Each wire has a cross-sectional area of 0.025 in2.
E
F
G
3 ft 5 ft
H
D
C 1 ft
2 ft
1.8 ft I
Internal Forces in the wires: A
FBD (b) FBC(4) - 500(3) = 0
+ c ©Fy = 0;
FAH + 375.0 - 500 = 0
FAH = 125.0 lb
a + ©MD = 0;
FCF(3) - 125.0(1) = 0
FCF = 41.67 lb
+ c ©Fy = 0;
FDE + 41.67 - 125.0 = 0
FBC = 375.0 lb
FBD (a)
FDE = 83.33 lb
Displacement: dD =
83.33(3)(12) FDELDE = 0.0042857 in. = ADEE 0.025(28.0)(106)
dC =
41.67(3)(12) FCFLCF = 0.0021429 in. = ACFE 0.025(28.0)(106) œ dH = 0.0014286 in.
dH = 0.0014286 + 0.0021429 = 0.0035714 in. dA>H =
125.0(1.8)(12) FAHLAH = 0.0038571 in. = AAHE 0.025(28.0)(106)
dA = dH + dA>H = 0.0035714 + 0.0038571 = 0.0074286 in. dB =
375.0(5)(12) FBGLBG = 0.0321428 in. = ABGE 0.025(28.0)(106)
dlœ 0.0247143 = ; 3 4
1 ft 500 lb
a + ©MA = 0;
œ dH 0.0021429 = ; 2 3
B 3 ft
dlœ = 0.0185357 in.
dl = 0.0074286 + 0.0185357 = 0.0260 in.
Ans.
129
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*4–12. The load is supported by the four 304 stainless steel wires that are connected to the rigid members AB and DC. Determine the angle of tilt of each member after the 500-lb load is applied. The members were originally horizontal, and each wire has a cross-sectional area of 0.025 in2.
E
F
G
3 ft 5 ft
H
D
C 1 ft
2 ft
1.8 ft I
Internal Forces in the wires:
A
FBD (b) FBG(4) - 500(3) = 0
+ c ©Fy = 0;
FAH + 375.0 - 500 = 0
FAH = 125.0 lb
a + ©MD = 0;
FCF(3) - 125.0(1) = 0
FCF = 41.67 lb
+ c ©Fy = 0;
FDE + 41.67 - 125.0 = 0
FBG = 375.0 lb
FBD (a)
FDE = 83.33 lb
Displacement: dD =
83.33(3)(12) FDELDE = 0.0042857 in. = ADEE 0.025(28.0)(106)
dC =
41.67(3)(12) FCFLCF = 0.0021429 in. = ACFE 0.025(28.0)(106)
œ dH 0.0021429 = ; 2 3
œ dH = 0.0014286 in.
œ + dC = 0.0014286 + 0.0021429 = 0.0035714 in. dH = dH
tan a =
0.0021429 ; 36
dA>H =
125.0(1.8)(12) FAHLAH = 0.0038571 in. = AAHE 0.025(28.0)(106)
a = 0.00341°
Ans.
dA = dH + dA>H = 0.0035714 + 0.0038571 = 0.0074286 in. 375.0(5)(12) FBGLBG = 0.0321428 in. = ABGE 0.025(28.0)(106)
tan b =
1 ft 500 lb
a + ©MA = 0;
dB =
B 3 ft
0.0247143 ; 48
b = 0.0295°
Ans.
130
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•4–13. The bar has a length L and cross-sectional area A. Determine its elongation due to the force P and its own weight.The material has a specific weight g (weight>volume) and a modulus of elasticity E.
d =
=
L P(x) dx 1 = (gAx + P) dx AE L0 L A(x) E
L
gAL2 gL2 1 PL a + PLb = + AE 2 2E AE
Ans.
P
4–14. The post is made of Douglas fir and has a diameter of 60 mm. If it is subjected to the load of 20 kN and the soil provides a frictional resistance that is uniformly distributed along its sides of w = 4 kN>m, determine the force F at its bottom needed for equilibrium.Also, what is the displacement of the top of the post A with respect to its bottom B? Neglect the weight of the post.
20 kN A y w
2m
Equation of Equilibrium: For entire post [FBD (a)] + c ©Fy = 0;
F + 8.00 - 20 = 0
B
F = 12.0 kN
Ans.
Internal Force: FBD (b) + c ©Fy = 0;
-F(y) + 4y - 20 = 0 F(y) = {4y - 20} kN
Displacement: L
dA>B =
2m F(y)dy 1 = (4y - 20)dy AE L0 L0 A(y)E
=
1 A 2y2 - 20y B AE
= -
冷
2m 0
32.0 kN # m AE 32.0(103)
= -
p 2 4 (0.06 )
13.1 (109)
= - 0.8639 A 10 - 3 B m Ans.
= - 0.864 mm Negative sign indicates that end A moves toward end B.
131
F
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4–15. The post is made of Douglas fir and has a diameter of 60 mm. If it is subjected to the load of 20 kN and the soil provides a frictional resistance that is distributed along its length and varies linearly from w = 0 at y = 0 to w = 3 kN>m at y = 2 m, determine the force F at its bottom needed for equilibrium. Also, what is the displacement of the top of the post A with respect to its bottom B? Neglect the weight of the post.
20 kN A y w
2m
B F
Equation of Equilibrium: For entire post [FBD (a)] + c ©Fy = 0;
F + 3.00 - 20 = 0
F = 17.0 kN
Ans.
Internal Force: FBD (b) + c ©Fy = 0;
-F(y) +
1 3y a b y - 20 = 0 2 2
3 F(y) = e y2 - 20 f kN 4 Displacement: L
dA>B =
2m F(y) dy 1 3 = a y2 - 20b dy AE L0 4 L0 A(y)E
=
2m y3 1 a - 20y b 2 AE 4 0
= -
38.0 kN # m AE 38.0(103)
= -
p 2 4 (0.06 )
13.1 (109)
= -1.026 A 10 - 3 B m Ans.
= -1.03 mm Negative sign indicates that end A moves toward end B.
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*4–16. The linkage is made of two pin-connected A-36 steel members, each having a cross-sectional area of 1.5 in2. If a vertical force of P = 50 kip is applied to point A, determine its vertical displacement at A.
P A
2 ft B
C 1.5 ft
Analysing the equilibrium of Joint A by referring to its FBD, Fig. a, + ©F = 0 ; : x
+ c ©Fy = 0
The
3 3 FAC a b - FAB a b = 0 5 5 4 -2Fa b - 50 = 0 5
initial
FAC = FAB = F
F = -31.25 kip
length
of members AB and AC is 12 in b = 30 in. The axial deformation of members L = 21.52 + 22 = (2.50 ft)a 1 ft AB and AC is
d =
(-31.25)(30) FL = = -0.02155 in. AE (1.5) C 29.0(103) D
The negative sign indicates that end A moves toward B and C. From the geometry 1.5 shown in Fig. b, u = tan - 1 a b = 36.87°. Thus, 2
A dA B g =
d 0.02155 = = 0.0269 in. T cos u cos 36.87°
Ans.
133
1.5 ft
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•4–17.
The linkage is made of two pin-connected A-36 steel members, each having a cross-sectional area of 1.5 in2. Determine the magnitude of the force P needed to displace point A 0.025 in. downward.
P A
2 ft
Analysing the equilibrium of joint A by referring to its FBD, Fig. a + ©F = 0; : x
3 3 FAC a b - FAB a b = 0 5 5
+ c ©Fy = 0;
4 -2Fa b - P = 0 5
The
initial
B
1.5 ft
F = -0.625 P
of members AB and AC are 12 in b = 30 in. The axial deformation of members L = 21.5 + 2 = (2.50 ft)a 1 ft AB and AC is 2
length
2
d =
C
FAC = FAB = F
-0.625P(30) FL = = -0.4310(10 - 3) P AE (1.5) C 29.0(103) D
The negative sign indicates that end A moves toward B and C. From the geometry 1.5 shown in Fig. b, we obtain u = tan - 1 a b = 36.87°. Thus 2 (dA)g =
d cos u
0.025 =
0.4310(10 - 3) P cos 36.87°
P = 46.4 kips
Ans.
134
1.5 ft
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4–18. The assembly consists of two A-36 steel rods and a rigid bar BD. Each rod has a diameter of 0.75 in. If a force of 10 kip is applied to the bar as shown, determine the vertical displacement of the load.
C
A 3 ft 2 ft
B
Here, FEF = 10 kip. Referring to the FBD shown in Fig. a,
1.25 ft
a + ©MB = 0;
FCD (2) - 10(1.25) = 0
FCD = 6.25 kip
a + ©MD = 0;
10(0.75) - FAB(2) = 0
FAB = 3.75 kip
The cross-sectional area of the rods is A = A and C are fixed,
3.75 (2)(12) FAB LAB = = 0.007025 in. T A Est 0.140625p C 29.0(103) D
dD =
6.25(3)(12) FCD LCD = = 0.01756 in T A Est 0.140625p C 29.0(103) D
From the geometry shown in Fig. b dE = 0.007025 +
1.25 (0.01756 - 0.00725) = 0.01361 in. T 2
Here, dF>E =
10 (1) (12) FEF LEF = = 0.009366 in T A Est 0.140625p C 29.0(103) D
Thus,
A + T B dF = dE + dF>E = 0.01361 + 0.009366 = 0.0230 in T
Ans.
135
D
0.75 ft F
p (0.752) = 0.140625p in2. Since points 4
dB =
E
10 kip
1 ft
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4–19. The assembly consists of two A-36 steel rods and a rigid bar BD. Each rod has a diameter of 0.75 in. If a force of 10 kip is applied to the bar, determine the angle of tilt of the bar.
C
A
Here, FEF = 10 kip. Referring to the FBD shown in Fig. a, 3 ft
a + ©MB = 0;
FCD (2) - 10(1.25) = 0
FCD = 6.25 kip
a + ©MD = 0;
10(0.75) - FAB (2) = 0
FAB = 3.75 kip
2 ft
B
p The cross-sectional area of the rods is A = (0.752) = 0.140625p in2. Since points 4 A and C are fixed then,
dB =
3.75 (2)(12) FAB LAB = = 0.007025 in T A Est 0.140625p C 29.0(103) D
dD =
6.25 (3)(12) FCD LCD = = 0.01756 in T A Est 0.140625p C 29.0(103) D
0.01756 - 0.007025 = 0.439(10 - 3) rad 2(12)
Ans.
136
D
0.75 ft F
10 kip
From the geometry shown in Fig. b, u =
1.25 ft
E
1 ft
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*4–20. The rigid bar is supported by the pin-connected rod CB that has a cross-sectional area of 500 mm2 and is made of A-36 steel. Determine the vertical displacement of the bar at B when the load is applied.
C
3m
45 kN/m
Force In The Rod. Referring to the FBD of member AB, Fig. a a + ©MA = 0;
3 1 1 FBC a b (4) - (45)(4) c (4) d = 0 5 2 3
4m
Displacement. The initial length of rod BC is LBC = 232 + 42 = 5 m. The axial deformation of this rod is dBC =
50.0(103)(5) FBC LBC = = 2.50 (10 - 3) m ABC Est 0.5(10 - 3) C 200(109) D
3 From the geometry shown in Fig. b, u = tan - 1 a b = 36.87°. Thus, 4 (dB)g =
2.50(10 - 3) dBC = = 4.167 (10 - 3) m = 4.17 mm sin u sin 36.87°
137
B
A
FBC = 50.0 kN
Ans.
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•4–21.
A spring-supported pipe hanger consists of two springs which are originally unstretched and have a stiffness of k = 60 kN>m, three 304 stainless steel rods, AB and CD, which have a diameter of 5 mm, and EF, which has a diameter of 12 mm, and a rigid beam GH. If the pipe and the fluid it carries have a total weight of 4 kN, determine the displacement of the pipe when it is attached to the support.
F B
D
k G 0.75 m
0.75 m k H
E
A
C
Internal Force in the Rods: 0.25 m 0.25 m
FBD (a) a + ©MA = 0;
FCD (0.5) - 4(0.25) = 0 FAB + 2.00 - 4 = 0
+ c ©Fy = 0;
FCD = 2.00 kN
FAB = 2.00 kN
FBD (b) FEF - 2.00 - 2.00 = 0
+ c ©Fy = 0;
FEF = 4.00 kN
Displacement: dD = dE =
FEFLEF = AEFE
dA>B = dC>D =
4.00(103)(750) p 4
(0.012)2(193)(109)
PCDLCD = ACDE
= 0.1374 mm
2(103)(750) p 4
(0.005)2(193)(109)
= 0.3958 mm
dC = dD + dC>D = 0.1374 + 0.3958 = 0.5332 mm Displacement of the spring dsp =
Fsp k
=
2.00 = 0.0333333 m = 33.3333 mm 60
dlat = dC + dsp = 0.5332 + 33.3333 = 33.9 mm
Ans.
138
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4–22. A spring-supported pipe hanger consists of two springs, which are originally unstretched and have a stiffness of k = 60 kN>m, three 304 stainless steel rods, AB and CD, which have a diameter of 5 mm, and EF, which has a diameter of 12 mm, and a rigid beam GH. If the pipe is displaced 82 mm when it is filled with fluid, determine the weight of the fluid.
F B
D
k G 0.75 m
0.75 m k H
E
Internal Force in the Rods: A
C
FBD (a) a + ©MA = 0;
FCD(0.5) - W(0.25) = 0
FCD =
W - W = 0 2
W 2
FAB +
+ c ©Fy = 0;
FAB =
0.25 m 0.25 m
W 2
FBD (b) FEF -
+ c ©Fy = 0;
W W = 0 2 2
FEF = W
Displacement: dD = dE =
FEFLEF = AEFE
W(750) p 2 9 4 (0.012) (193)(10 )
= 34.35988(10 - 6) W dA>B = dC>D =
FCDLCD = ACDE
W 2
(750)
p 2 9 4 (0.005) (193)(10 )
= 98.95644(10 - 6) W dC = dD + dC>D = 34.35988(10 - 6) W + 98.95644(10 - 6) W = 0.133316(10 - 3) W Displacement of the spring dsp =
W 2
Fsp k
=
60(103)
(1000) = 0.008333 W
dlat = dC + dsp 82 = 0.133316(10 - 3) W + 0.008333W W = 9685 N = 9.69 kN
Ans.
139
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4–23. The rod has a slight taper and length L. It is suspended from the ceiling and supports a load P at its end. Show that the displacement of its end due to this load is d = PL>1pEr2r12. Neglect the weight of the material. The modulus of elasticity is E.
r(x) = r1 +
A(x) =
r2
r1L + (r2 - r1)x r2 - r1 x = L L
L
p (r1L + (r2 - r1)x)2 L2 r1
L
PL2 Pdx dx d = = 2 A(x)E pE [r L + (r L0 L 1 2 - r1)x] = -
L 1 PL2 c dƒ p E (r2 - r2)(r1L + (r2 - r1)x) 0
= -
=
= -
P
1 1 PL2 c d p E(r2 - r1) r1L + (r2 - r1)L r1L
r1 - r2 PL2 1 1 PL2 c d = c d p E(r2 - r1) r2L r1L p E(r2 - r1) r2r1L
r2 - r1 PL2 PL c d = p E(r2 - r1) r2r1L p E r2r1
QED
*4–24. Determine the relative displacement of one end of the tapered plate with respect to the other end when it is subjected to an axial load P.
P d2 t
w = d1 +
d1 h + (d2 - d1)x d2 - d1 x = h h
h P(x) dx P = d = E L0 [d1h L A(x)E
h
dx + ( d 2 - d1 )x ] t h
h
=
Ph dx E t L0 d1 h + (d2 - d1)x
d1 P
h
dx Ph = E t d1 h L0 1 + d2 -
h d1 h d2 - d1 Ph d1 = a b cln a1 + xb d ƒ E t d1 h d2 - d1 d1 h 0 d1 h x
=
d2 - d1 d1 + d2 - d1 Ph Ph cln a1 + bd = cln a bd E t(d2 - d1) d1 E t(d2 - d1) d1
=
d2 Ph cln d E t(d2 - d1) d1
Ans.
140
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4–25. Determine the elongation of the A-36 steel member when it is subjected to an axial force of 30 kN. The member is 10 mm thick. Use the result of Prob. 4–24.
20 mm 30 kN
30 kN 75 mm 0.5 m
Using the result of prob. 4-24 by substituting d1 = 0.02 m, d2 = 0.075 m t = 0.01 m and L = 0.5 m. d = 2c = 2c
d2 PL ln d Est t(d2 - d1) d1 30(103) (0.5) 9
200(10 )(0.01)(0.075 - 0.02)
ln a
0.075 bd 0.02
= 0.360(10 - 3) m = 0.360 mm
Ans.
4–26. The casting is made of a material that has a specific weight g and modulus of elasticity E. If it is formed into a pyramid having the dimensions shown, determine how far its end is displaced due to gravity when it is suspended in the vertical position.
b0
b0
L
Internal Forces: + c ©Fz = 0;
P(z) -
1 gAz = 0 3
P(z) =
1 gAz 3
Displacement: L
d =
P(z) dz L0 A(z) E L1 3
=
gAz
L0 A E
dz
=
L g z dz 3E L0
=
gL2 6E
Ans.
141
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4–27. The circular bar has a variable radius of r = r0eax and is made of a material having a modulus of elasticity of E. Determine the displacement of end A when it is subjected to the axial force P.
L
Displacements: The cross-sectional area of the bar as a function of x is A(x) = pr2 = pr0 2e2ax. We have
x
B
L
d =
L P(x)dx P dx = 2 A(x)E pr0 E L0 e2ax L0
r0 r ⫽ r0 eax
L 1 P 2 c d = pr0 2E 2ae2ax 0
= -
A
P a1 - e - 2aL b 2apr0 2E
P
Ans.
*4–28. The pedestal is made in a shape that has a radius defined by the function r = 2>12 + y1>22 ft, where y is in feet. If the modulus of elasticity for the material is E = 1411032 psi, determine the displacement of its top when it supports the 500-lb load.
y
500 lb 0.5 ft
r⫽
2 (2 ⫹ y 1/ 2)
4 ft
d =
=
P(y) dy L A(y) E y
4 dy 500 3 14(10 )(144) L0 p(2 + y2-1
2
)
2
1 ft
4 -3
= 0.01974(10 )
L0
r
1 2
(4 + 4y + y) dy
4 2 3 1 = 0.01974(10 - 3)c4y + 4 a y2 b + y2 d 3 2 0
= 0.01974(10 - 3)(45.33) = 0.8947(10 - 3) ft = 0.0107 in.
Ans.
142
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•4–29.
The support is made by cutting off the two opposite sides of a sphere that has a radius r0 . If the original height of the support is r0>2, determine how far it shortens when it supports a load P. The modulus of elasticity is E.
P
r0
Geometry: A = p r2 = p(r0 cos u)2 = p r20 cos2 u y = r0 sin u;
dy = r0 cos u du
Displacement: L
P(y) dy L0 A(y) E
d =
= 2B
When y =
u u r0 cos u du P P du = 2 R B R E L0 p r20 cos2 u p r0 E L0 cos u
=
u 2P [ln (sec u + tan u)] 2 p r0 E 0
=
2P [ln (sec u + tan u)] p r0 E
r0 ; 4
u = 14.48°
d =
=
2P [ln (sec 14.48° + tan 14.48°)] p r0 E 0.511P p r0 E
Ans.
Also, Geometry: A (y) = p x2 = p (r20 - y2) Displacement: L
d =
P(y) dy L0 A(y) E 0
0
r0 + y p 2P 1 2P p dy = ln = B R 2 2 2 pE L0 r0 - y p E 2r0 r0 - y 0 =
P [ln 1.667 - ln 1] p r0 E
=
0.511 P p r0 E
Ans.
143
r0 2
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4–30. The weight of the kentledge exerts an axial force of P ⫽ 1500 kN on the 300-mm diameter high strength concrete bore pile. If the distribution of the resisting skin friction developed from the interaction between the soil and the surface of the pile is approximated as shown, and the resisting bearing force F is required to be zero, determine the maximum intensity p0 kN>m for equilibrium. Also, find the corresponding elastic shortening of the pile. Neglect the weight of the pile.
P p0
12 m
Internal Loading: By considering the equilibrium of the pile with reference to its entire free-body diagram shown in Fig. a. We have 1 p (12) - 1500 = 0 2 0
+ c ©Fy = 0;
p0 = 250 kN>m
Ans.
Thus, p(y) =
250 y = 20.83y kN>m 12
The normal force developed in the pile as a function of y can be determined by considering the equilibrium of a section of the pile shown in Fig. b. 1 (20.83y)y - P(y) = 0 2
+ c ©Fy = 0;
P(y) = 10.42y2 kN
Displacement: The cross-sectional area of the pile is A =
p (0.32) = 0.0225p m2. 4
We have L
d =
12 m P(y)dy 10.42(103)y2dy = 0.0225p(29.0)(109) L0 L0 A(y)E 12 m
=
L0
5.0816(10 - 6)y2dy
= 1.6939(10 - 6)y3 冷 0
12 m
= 2.9270(10 - 3)m = 2.93 mm
Ans.
144
F
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4–31. The column is constructed from high-strength concrete and six A-36 steel reinforcing rods. If it is subjected to an axial force of 30 kip, determine the average normal stress in the concrete and in each rod. Each rod has a diameter of 0.75 in.
4 in. 30 kip
Equations of Equilibrium: 6Pst + Pcon - 30 = 0
+ c ©Fy = 0;
3 ft
[1]
Compatibility: dst = dcon Pcon(3)(12)
Pst(3)(12) p 4
(0.752)(29.0)(103)
=
[p4 (82) - 6(p4 )(0.75)2](4.20)(103)
Pst = 0.064065 Pcon
[2]
Solving Eqs. [1] and [2] yields: Pst = 1.388 kip
Pcon = 21.670 kip
Average Normal Stress: sst =
scon =
Pst = Ast
Pcon = Acon
1.388 p 2 4 (0.75 )
= 3.14 ksi
21.670
p 2 4 (8 )
- 6 A p4 B (0.752)
Ans.
Ans.
= 0.455 ksi
*4–32. The column is constructed from high-strength concrete and six A-36 steel reinforcing rods. If it is subjected to an axial force of 30 kip, determine the required diameter of each rod so that one-fourth of the load is carried by the concrete and three-fourths by the steel.
4 in. 30 kip
Equilibrium: The force of 30 kip is required to distribute in such a manner that 3/4 of the force is carried by steel and 1/4 of the force is carried by concrete. Hence Pst =
3 (30) = 22.5 kip 4
Pcon =
1 (30) = 7.50 kip 4 3 ft
Compatibility: dst = dcon PstL Pcon L = AstEst Acon Econ Ast =
22.5Acon Econ 7.50 Est
3 C p4 (82) - 6 A p4 B d2 D (4.20)(103) p 6 a bd2 = 4 29.0(103) d = 1.80 in.
Ans. 145
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•4–33.
The steel pipe is filled with concrete and subjected to a compressive force of 80 kN. Determine the average normal stress in the concrete and the steel due to this loading. The pipe has an outer diameter of 80 mm and an inner diameter of 70 mm. Est = 200 GPa, Ec = 24 GPa.
80 kN
Pst + Pcon - 80 = 0
+ c ©Fy = 0;
(1)
500 mm
dst = dcon Pcon L p 2 (0.07 ) (24) 4
Pst L p 2 4 (0.08
- 0.072) (200) (109)
=
(109)
Pst = 2.5510 Pcon
(2)
Solving Eqs. (1) and (2) yields Pst = 57.47 kN sst =
Pst = Ast
scon =
Pcon = 22.53 kN 57.47 (103)
p 4
(0.082 - 0.072)
Ans.
= 48.8 MPa
22.53 (103) Pcon = 5.85 MPa = p 2 Acon 4 (0.07 )
Ans.
4–34. The 304 stainless steel post A has a diameter of d = 2 in. and is surrounded by a red brass C83400 tube B. Both rest on the rigid surface. If a force of 5 kip is applied to the rigid cap, determine the average normal stress developed in the post and the tube.
5 kip
B B A 8 in.
Equations of Equilibrium: + c ©Fy = 0;
3 in.
Pst + Pbr - 5 = 0[1]
Compatibility:
d
dst = dbr Pst(8) p 2 3 4 (2 )(28.0)(10 )
Pbr(8) =
p 2 4 (6
- 52)(14.6)(103)
Pst = 0.69738 Pbr
[2]
Solving Eqs. [1] and [2] yields: Pbr = 2.9457 kip
Pst = 2.0543 kip
Average Normal Stress: sbr =
sst =
Pbr = Abr
2.9457 = 0.341 ksi - 52)
Ans.
p 2 4 (6
Pst 2.0543 = p 2 = 0.654 ksi Ast 4 (2 )
Ans.
146
0.5 in.
A
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4–35. The 304 stainless steel post A is surrounded by a red brass C83400 tube B. Both rest on the rigid surface. If a force of 5 kip is applied to the rigid cap, determine the required diameter d of the steel post so that the load is shared equally between the post and tube.
5 kip
B B
A
A 8 in.
Equilibrium: The force of 5 kip is shared equally by the brass and steel. Hence
3 in.
Pst = Pbr = P = 2.50 kip Compatibility:
d
0.5 in.
dst = dbr PL PL = AstEst AbrEbr Ast = p a b d2 = 4
AbrEbr Est p 4
(62 - 52)(14.6)(103) 28.0(103)
d = 2.39 in.
Ans.
*4–36. The composite bar consists of a 20-mm-diameter A-36 steel segment AB and 50-mm-diameter red brass C83400 end segments DA and CB. Determine the average normal stress in each segment due to the applied load. + ©F = 0; ; x
250 mm
D
FC - FD + 75 + 75 - 100 - 100 = 0 FC - FD - 50 = 0
+ ;
(1)
0 = ¢ D - dD 0 =
50(0.25)
150(0.5) p 2 9 4 (0.02) (200)(10 )
-
-
FD(0.5) p 2 9 4 (0.05 )(101)(10 )
p 2 9 4 (0.05 )(101)(10 )
-
500 mm
50 mm
FD(0.5) p 2 9 4 (0.02 )(200)(10 )
FD = 107.89 kN From Eq. (1), FC = 157.89 kN sAD =
107.89(103) PAD = 55.0 MPa = p 2 AAD 4 (0.05 )
Ans.
sAB =
42.11(103) PAB = 134 MPa = p 2 AAB 4 (0.02 )
Ans.
sBC =
157.89(103) PBC = 80.4 MPa = p 2 ABC 4 (0.05 )
Ans.
147
250 mm
20 mm 75 kN 100 kN A
75 kN
100 kN B
C
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•4–37.
The composite bar consists of a 20-mm-diameter A-36 steel segment AB and 50-mm-diameter red brass C83400 end segments DA and CB. Determine the displacement of A with respect to B due to the applied load.
250 mm
D
+ ; 0 =
-
500 mm
50 mm
250 mm
20 mm 75 kN 100 kN A
75 kN
100 kN B
C
0 = ¢ D - dD 150(103)(500)
50(103)(250)
p 2 9 4 (0.02 )(200)(10 )
FD(500) p 2 9 4 (0.05 )(101)(10 )
-
-
p 2 9 4 (0.05 )(101)(10 )
FD(500) p 2 9 4 (0.02) (200)(10 )
FD = 107.89 kN Displacement: dA>B =
42.11(103)(500) PABLAB = p 2 9 AABEst 4 (0.02 )200(10 )
= 0.335 mm
Ans.
4–38. The A-36 steel column, having a cross-sectional area of 18 in2, is encased in high-strength concrete as shown. If an axial force of 60 kip is applied to the column, determine the average compressive stress in the concrete and in the steel. How far does the column shorten? It has an original length of 8 ft.
16 in.
Pst + Pcon - 60 = 0
+ c ©Fy = 0; dst = dcon ;
60 kip
Pst(8)(12) 18(29)(103)
(1)
Pcon(8)(12) =
[(9)(16) - 18](4.20)(103)
Pst = 0.98639 Pcon
(2)
Solving Eqs. (1) and (2) yields Pst = 29.795 kip; sst =
Pcon = 30.205 kip
Pst 29.795 = = 1.66 ksi Ast 18
scon =
Ans.
Pcon 30.205 = = 0.240 ksi Acon 9(16) - 18
Ans.
Either the concrete or steel can be used for the deflection calculation. d =
29.795(8)(12) PstL = 0.0055 in. = AstE 18(29)(103)
Ans.
148
9 in.
8 ft
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4–39. The A-36 steel column is encased in high-strength concrete as shown. If an axial force of 60 kip is applied to the column, determine the required area of the steel so that the force is shared equally between the steel and concrete. How far does the column shorten? It has an original length of 8 ft.
60 kip 16 in.
9 in.
8 ft
The force of 60 kip is shared equally by the concrete and steel. Hence Pst = Pcon = P = 30 kip dcon = dst;
Ast =
PL PL = Acon Econ Ast Est
[9(16) - Ast] 4.20(103) AconEcon = Est 29(103) = 18.2 in2
d =
Ans.
30(8)(12) PstL = 0.00545 in. = AstEst 18.2(29)(103)
Ans.
*4–40. The rigid member is held in the position shown by three A-36 steel tie rods. Each rod has an unstretched length of 0.75 m and a cross-sectional area of 125 mm2. Determine the forces in the rods if a turnbuckle on rod EF undergoes one full turn. The lead of the screw is 1.5 mm. Neglect the size of the turnbuckle and assume that it is rigid. Note: The lead would cause the rod, when unloaded, to shorten 1.5 mm when the turnbuckle is rotated one revolution.
B
D
0.75 m E A
0.5 m
0.5 m
C 0.75 m
a + ©ME = 0;
-TAB(0.5) + TCD(0.5) = 0
F
TAB = TCD = T + T ©Fy = 0;
(1)
TEF - 2T = 0 TEF = 2T
(2)
Rod EF shortens 1.5mm causing AB (and DC) to elongate. Thus: 0.0015 = dA>B + dE>F 0.0015 =
T(0.75) -6
2T(0.75) 9
(125)(10 )(200)(10 )
+
(125)(10 - 6)(200)(109)
2.25T = 37500 T = 16666.67 N TAB = TCD = 16.7 kN
Ans.
TEF = 33.3 kN
Ans.
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•4–41.
The concrete post is reinforced using six steel reinforcing rods, each having a diameter of 20 mm. Determine the stress in the concrete and the steel if the post is subjected to an axial load of 900 kN. Est = 200 GPa, Ec = 25 GPa.
900 kN 250 mm
375 mm
Referring to the FBD of the upper portion of the cut concrete post shown in Fig. a Pcon + 6Pst - 900 = 0
+ c ©Fy = 0;
(1)
Since the steel rods and the concrete are firmly bonded, their deformation must be the same. Thus 0 con = dst Pcon L Pst L = Acon Econ Ast Est
C 0.25(0.375) -
Pcon L 6(p4 )(0.022)
D C 25(10 ) D
Pst L =
9
(p4 )(0.022)
C 200(109) D
Pcon = 36.552 Pst
(2)
Solving Eqs (1) and (2) yields Pst = 21.15 kN
Pcon = 773.10 kN
Thus, scon =
sst =
773.10(103) Pcon = 8.42 MPa = Acon 0.15(0.375) - 6(p4 )(0.022) 21.15(103) Pst = 67.3 MPa = p 2 Ast 4 (0.02 )
Ans.
Ans.
150
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4–42. The post is constructed from concrete and six A-36 steel reinforcing rods. If it is subjected to an axial force of 900 kN, determine the required diameter of each rod so that one-fifth of the load is carried by the steel and four-fifths by the concrete. Est = 200 GPa, Ec = 25 GPa.
900 kN 250 mm
375 mm
The normal force in each steel rod is Pst =
1 5
(900) 6
= 30 kN
The normal force in concrete is Pcon =
4 (900) = 720 kN 5
Since the steel rods and the concrete are firmly bonded, their deformation must be the same. Thus dcon = dst Pcon L Pst L = Acon Econ Ast Est 720(103) L
30(103)L
C 0.25(0.375) - 6(p4 d2) D C 25(109) D
=
49.5p d2 = 0.09375
p 4
d2 C 200(109) D
d = 0.02455 m = 24.6 mm
Ans.
4–43. The assembly consists of two red brass C83400 copper alloy rods AB and CD of diameter 30 mm, a stainless 304 steel alloy rod EF of diameter 40 mm, and a rigid cap G. If the supports at A, C and F are rigid, determine the average normal stress developed in rods AB, CD and EF.
300 mm
450 mm 40 kN
A
B
E
30 mm
F 40 mm
C
Equation of Equilibrium: Due to symmetry, FAB = FCD = F. Referring to the freebody diagram of the assembly shown in Fig. a, + ©F = 0; : x
2F + FEF - 2 C 40(103) D = 0
(1)
Compatibility Equation: Using the method of superposition, Fig. b, + B 0 = -d + d A: P EF 0 = -
40(103)(300) p 2 9 4 (0.03 )(101)(10 )
+ cp
FEF (450) 2
9
4 (0.04 )(193)(10 )
+
A
B
FEF>2 (300) d p 2 9 4 (0.03 )(101)(10 )
FEF = 42 483.23 N Substituting this result into Eq. (1), F = 18 758.38 N
151
30 mm
40 kN
D G
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4–43.
Continued
Normal Stress: We have, sAB = sCD =
sEF =
F 18 758.38 = 26.5 MPa = p 2 ACD 4 (0.03 )
Ans.
FEF 42 483.23 = 33.8 MPa = p 2 AEF 4 (0.04 )
Ans.
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*4–44. The two pipes are made of the same material and are connected as shown. If the cross-sectional area of BC is A and that of CD is 2A, determine the reactions at B and D when a force P is applied at the junction C.
B L – 2
Equations of Equilibrium: + ©F = 0; ; x
FB + FD - P = 0
[1]
Compatibility: + B A:
0 = dP - dB 0 =
0 =
P A L2 B
2AE
- C
FB
A L2 B
AE
FB +
A L2 B
2AE
S
3FBL PL 4AE 4AE
FB =
P 3
Ans.
From Eq. [1] FD =
C
2 P 3
Ans.
153
D P L – 2
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•4–45.
The bolt has a diameter of 20 mm and passes through a tube that has an inner diameter of 50 mm and an outer diameter of 60 mm. If the bolt and tube are made of A-36 steel, determine the normal stress in the tube and bolt when a force of 40 kN is applied to the bolt. Assume the end caps are rigid.
160 mm
40 kN
Referring to the FBD of left portion of the cut assembly, Fig. a + ©F = 0; : x
40(103) - Fb - Ft = 0
(1)
Here, it is required that the bolt and the tube have the same deformation. Thus dt = db Ft(150)
p 2 4 (0.06
- 0.05 ) C 200(10 ) D 2
Fb(160) =
9
p 2 4 (0.02 )
C 200(109) D
Ft = 2.9333 Fb
(2)
Solving Eqs (1) and (2) yields Fb = 10.17 (103) N
Ft = 29.83 (103) N
Thus, sb =
10.17(103) Fb = 32.4 MPa = p 2 Ab 4 (0.02 )
st =
Ft = At
29.83 (103) p 2 4 (0.06
- 0.052)
40 kN 150 mm
Ans.
= 34.5 MPa
Ans.
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4–46. If the gap between C and the rigid wall at D is initially 0.15 mm, determine the support reactions at A and D when the force P = 200 kN is applied. The assembly is made of A36 steel.
Equation of Equilibrium: Referring to the free-body diagram of the assembly shown in Fig. a, 200(103) - FD - FA = 0
(1)
Compatibility Equation: Using the method of superposition, Fig. b, + B A:
d = dP - dFD 0.15 =
200(103)(600) p 2 9 4 (0.05 )(200)(10 )
- Cp
FD (600)
2 9 4 (0.05 )(200)(10 )
0.15 mm
P A
+ ©F = 0; : x
600 mm
600 mm
+
FD (600) S p 2 (0.025 )(200)(109) 4
FD = 20 365.05 N = 20.4 kN
Ans.
Substituting this result into Eq. (1), FA = 179 634.95 N = 180 kN
Ans.
155
50 mm
D B
25 mm
C
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4–47. Two A-36 steel wires are used to support the 650-lb engine. Originally, AB is 32 in. long and A¿B¿ is 32.008 in. long. Determine the force supported by each wire when the engine is suspended from them. Each wire has a crosssectional area of 0.01 in2.
B¿ B
A¿ A
TA¿B¿ + TAB - 650 = 0
+ c ©Fy = 0;
(1)
dAB = dA¿B¿ + 0.008 TA¿B¿ (32.008)
TAB (32) (0.01)(29)(106)
=
(0.01)(29)(106)
+ 0.008
32TAB - 32.008TA¿B¿ = 2320 TAB = 361 lb
Ans.
TA¿B¿ = 289 lb
Ans.
*4–48. Rod AB has a diameter d and fits snugly between the rigid supports at A and B when it is unloaded. The modulus of elasticity is E. Determine the support reactions at A and B if the rod is subjected to the linearly distributed axial load.
p⫽
A
1 p L - FA - FB = 0 2 0
+ ©F = 0; : x
(1)
Compatibility Equation: Using the method of superposition, Fig. b, + B A:
0 = dP - dFA L
0 =
L0
FA (L) P(x)dx AE AE
L
0 =
L0
B x
Equation of Equilibrium: Referring to the free-body diagram of rod AB shown in Fig. a,
P(x)dx - FAL
156
p0
p0 x L
L
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4–48.
Continued
Here, P(x) =
p0 2 1 p0 a xb x = x . Thus, 2 L 2L 0 =
L p0 x2 dx - FAL 2L L0
0 =
p0 x3 L ¢ ≤ ` - FAL 2L 3 0
FA =
p0L 6
Ans.
Substituting this result into Eq. (1), FB =
p0L 3
Ans.
157
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•4–49.
The tapered member is fixed connected at its ends A and B and is subjected to a load P = 7 kip at x = 30 in. Determine the reactions at the supports. The material is 2 in. thick and is made from 2014-T6 aluminum.
A
B 3 in.
P
6 in.
x 60 in.
y 1.5 = 120 - x 60 y = 3 - 0.025 x + ©F = 0; : x
FA + FB - 7 = 0
(1)
dA>B = 0 30
-
L0
60 FA dx FBdx + = 0 2(3 - 0.025 x)(2)(E) L30 2(3 - 0.025 x)(2)(E) 30
-FA
L0
60
dx dx + FB = 0 (3 - 0.025 x) L30 (3 - 0.025x)
60 40 FA ln(3 - 0.025 x)|30 0 - 40 FB ln(3 - 0.025x)|30 = 0
-FA(0.2876) + 0.40547 FB = 0 FA = 1.40942 FB Thus, from Eq. (1). FA = 4.09 kip
Ans.
FB = 2.91 kip
Ans.
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4–50. The tapered member is fixed connected at its ends A and B and is subjected to a load P. Determine the location x of the load and its greatest magnitude so that the average normal stress in the bar does not exceed sallow = 4 ksi. The member is 2 in. thick.
A
B 3 in.
P
6 in.
x 60 in.
y 1.5 = 120 - x 60 y = 3 - 0.025 x + ©F = 0; : x
FA + FB - P = 0
dA>B = 0 x
-
60 FA dx FBdx + = 0 Lx 2(3 - 0.025 x)(2)(E) L0 2(3 - 0.025 x)(2)(E) x
-FA
60
dx dx + FB = 0 L0 (3 - 0.025 x) Lx (3 - 0.025 x)
FA(40) ln (3 - 0.025 x)|x0 - FB(40) ln (3 - 0.025x)|60 x = 0 FA ln (1 -
0.025 x 0.025x ) = -FB ln (2 ) 3 1.5
For greatest magnitude of P require, 4 =
FA ; 2(3 - 0.025 x)(2)
4 =
FB ; 2(3)
FA = 48 - 0.4 x
FB = 24 kip
Thus, (48 - 0.4 x) ln a 1 -
0.025 x 0.025 x b = -24 ln a2 b 3 1.5
Solving by trial and error, x = 28.9 in.
Ans.
Therefore, FA = 36.4 kip P = 60.4 kip
Ans.
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4–51. The rigid bar supports the uniform distributed load of 6 kip>ft. Determine the force in each cable if each cable has a cross-sectional area of 0.05 in2, and E = 3111032 ksi.
C
6 ft
6 kip/ft A D
B 3 ft
a + ©MA = 0; u = tan - 1
TCB a
2 25
b(3) - 54(4.5) + TCD a
2 25
b9 = 0
(1)
6 = 45° 6
L2B¿C¿ = (3)2 + (8.4853)2 - 2(3)(8.4853) cos u¿ Also, L2D¿C¿ = (9)2 + (8.4853)2 - 2(9)(8.4853) cos u¿
(2)
Thus, eliminating cos u¿ . -L2B¿C¿(0.019642) + 1.5910 = -L2D¿C¿(0.0065473) + 1.001735 L2B¿C¿(0.019642) = 0.0065473 L2D¿C¿ + 0.589256 L2B¿C¿ = 0.333 L2D¿C¿ + 30 But, LB¿C = 245 + dBC¿ ,
LD¿C = 245 + dDC¿
Neglect squares or d¿ B since small strain occurs. L2D¿C = (245 + dBC)2 = 45 + 2 245 dBC L2D¿C = (245 + dDC)2 = 45 + 2 245 dDC 45 + 2245 dBC = 0.333(45 + 2245 dDC) + 30 2 245 dBC = 0.333(2245 dDC) dDC = 3dBC Thus, TCD 245 TCB 245 = 3 AE AE TCD = 3 TCB From Eq. (1). TCD = 27.1682 kip = 27.2 kip
Ans.
TCB = 9.06 kip
Ans.
160
3 ft
3 ft
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*4–52. The rigid bar is originally horizontal and is supported by two cables each having a cross-sectional area of 0.05 in2, and E = 3111032 ksi. Determine the slight rotation of the bar when the uniform load is applied.
C
See solution of Prob. 4-51.
6 ft
TCD = 27.1682 kip dDC =
TCD 245 0.05(31)(103)
27.1682245 = 0.1175806 ft 0.05(31)(103)
=
6 kip/ft A D
B
Using Eq. (2) of Prob. 4-51, 3 ft
3 ft
3 ft
(245 + 0.1175806)2 = (9)2 + (8.4852)2 - 2(9)(8.4852) cos u¿ u¿ = 45.838° Thus, ¢u = 45.838° - 45° = 0.838°
Ans.
•4–53.
The press consists of two rigid heads that are held together by the two A-36 steel 12-in.-diameter rods. A 6061T6-solid-aluminum cylinder is placed in the press and the screw is adjusted so that it just presses up against the cylinder. If it is then tightened one-half turn, determine the average normal stress in the rods and in the cylinder. The single-threaded screw on the bolt has a lead of 0.01 in. Note: The lead represents the distance the screw advances along its axis for one complete turn of the screw.
12 in.
2 in.
10 in.
+ ©F = 0; : x
2Fst - Fal = 0 dst = 0.005 - dal
Fst(12) p ( 4 )(0.5)2(29)(103)
= 0.005 -
Fal(10) p(1)2(10)(103)
Solving, Fst = 1.822 kip Fal = 3.644 kip srod =
Fst 1.822 = p = 9.28 ksi Ast ( 4 )(0.5)2
Ans.
scyl =
Fal 3.644 = = 1.16 ksi Aal p(1)2
Ans.
161
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4–54. The press consists of two rigid heads that are held together by the two A-36 steel 12-in.-diameter rods. A 6061T6-solid-aluminum cylinder is placed in the press and the screw is adjusted so that it just presses up against the cylinder. Determine the angle through which the screw can be turned before the rods or the specimen begin to yield. The single-threaded screw on the bolt has a lead of 0.01 in. Note: The lead represents the distance the screw advances along its axis for one complete turn of the screw.
12 in.
2 in.
10 in.
+ ©F = 0; : x
2Fst - Fal = 0 dst = d - dal Fst(12)
(p4 )(0.5)2(29)(103)
= d-
Fal(10)
(1)
p(1)2(10)(103)
Assume steel yields first, sY = 36 =
Fst (p4 )(0.5)2
;
Fst = 7.068 kip
Then Fal = 14.137 kip; sal =
14.137 = 4.50 ksi p(1)2
4.50 ksi 6 37 ksi steel yields first as assumed. From Eq. (1), d = 0.01940 in. Thus, 0.01940 u = 360° 0.01 u = 698°
Ans.
162
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4–55. The three suspender bars are made of A-36 steel and have equal cross-sectional areas of 450 mm2. Determine the average normal stress in each bar if the rigid beam is subjected to the loading shown.
A 2m
1m
+ c ©Fy = 0; a + ©MD = 0;
FAD + FBE + FCF - 50(103) - 80(103) = 0 FBE(2) + FCF(4) - 50(103)(1) - 80(103)(3) = 0
(1) (2)
Referring to the geometry shown in Fig. b, dBE = dAD + a dBE =
dCF - dAD b(2) 4
1 A d + dCF B 2 AD
FBE L FCF L 1 FADL = a + b AE 2 AE AE FAD + FCF = 2 FBE
(3)
Solving Eqs. (1), (2) and (3) yields FBE = 43.33(103) N
FAD = 35.83(103) N
FCF = 50.83(103) N
Thus, sBE =
43.33(103) FBE = 96.3 MPa = A 0.45(10 - 3)
Ans.
sAD =
35.83(103) FAD = 79.6 MPa = A 0.45(10 - 3)
Ans.
sCF = 113 MPa
Ans.
163
80 kN
50 kN E
D
Referring to the FBD of the rigid beam, Fig. a,
C
B
1m
1m
F 1m
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*4–56. The rigid bar supports the 800-lb load. Determine the normal stress in each A-36 steel cable if each cable has a cross-sectional area of 0.04 in2.
C
12 ft
800 lb B A
5 ft
Referring to the FBD of the rigid bar, Fig. a, FBC a
a + ©MA = 0;
12 3 b(5) + FCD a b (16) - 800(10) = 0 13 5
(1)
The unstretched length of wires BC and CD are LBC = 2122 + 52 = 13 ft and LCD = 2122 + 162 = 20 ft. The stretches of wires BC and CD are dBC =
FBC (13) FBC LBC = AE AE
dCD =
FCD(20) FCD LCD = AE AE
Referring to the geometry shown in Fig. b, the vertical displacement of the points on d 12 3 the rigid bar is dg = . For points B and D, cos uB = and cos uD = . Thus, cos u 13 5 the vertical displacement of points B and D are
A dB B g =
FBC (13)>AE dBC 169 FBC = = cos uB 12>13 12AE
A dD B g =
FCD (20)>AE dCD 100 FCD = = cos uD 3>5 3 AE
The similar triangles shown in Fig. c give
A dB B g 5
=
A dD B g 16
1 169 FBC 1 100 FCD b = b a a 5 12 AE 16 3AE FBC =
125 F 169 CD
(2)
Solving Eqs. (1) and (2), yields FCD = 614.73 lb
FBC = 454.69 lb
Thus, sCD =
FCD 614.73 = 15.37(103) psi = 15.4 ksi = ACD 0.04
Ans.
sBC =
FBC 454.69 = = 11.37(103) psi = 11.4 ksi ABC 0.04
Ans.
164
D 5 ft
6 ft
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4–56.
Continued
165
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•4–57.
The rigid bar is originally horizontal and is supported by two A-36 steel cables each having a crosssectional area of 0.04 in2. Determine the rotation of the bar when the 800-lb load is applied.
C
12 ft
800 lb B A
Referring to the FBD of the rigid bar Fig. a, a + ©MA = 0;
FBC a
12 3 b(5) + FCD a b (16) - 800(10) = 0 13 5
5 ft
(1)
The unstretched length of wires BC and CD are LBC = 2122 + 52 = 13 ft and LCD = 2122 + 162 = 20 ft. The stretch of wires BC and CD are dBC =
FBC (13) FBC LBC = AE AE
dCD =
FCD(20) FCD LCD = AE AE
Referring to the geometry shown in Fig. b, the vertical displacement of the points on d 12 3 the rigid bar is dg = . For points B and D, cos uB = and cos uD = . Thus, cos u 13 5 the vertical displacement of points B and D are
A dB B g =
FBC (13)>AE dBC 169 FBC = = cos uB 12>13 12AE
A dD B g =
FCD (20)>AE dCD 100 FCD = = cos uD 3>5 3 AE
The similar triangles shown in Fig. c gives
A dB B g 5
=
A dD B g 16
1 169 FBC 1 100 FCD a b = a b 5 12 AE 16 3 AE FBC =
125 F 169 CD
(2)
Solving Eqs (1) and (2), yields FCD = 614.73 lb
FBC = 454.69 lb
Thus,
A dD B g =
100(614.73)
3(0.04) C 29.0 (106) D
= 0.01766 ft
Then u = a
0.01766 ft 180° ba b = 0.0633° p 16 ft
Ans.
166
D 5 ft
6 ft
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4–58. The horizontal beam is assumed to be rigid and supports the distributed load shown. Determine the vertical reactions at the supports. Each support consists of a wooden post having a diameter of 120 mm and an unloaded (original) length of 1.40 m. Take Ew = 12 GPa.
18 kN/m
A
B
C 1.40 m
2m
a + ©MB = 0; + c ©Fy = 0;
FC(1) - FA(2) = 0
(1)
FA + FB + FC - 27 = 0
dB - dA dC - dA = ; 2 3
1m
(2)
3dB - dA = 2dC
3FBL FAL 2FCL = ; AE AE AE
3FB - FA = 2FC
(3)
Solving Eqs. (1)–(3) yields : FA = 5.79 kN
Ans.
FB = 9.64 kN
Ans.
FC = 11.6 kN
Ans.
4–59. The horizontal beam is assumed to be rigid and supports the distributed load shown. Determine the angle of tilt of the beam after the load is applied. Each support consists of a wooden post having a diameter of 120 mm and an unloaded (original) length of 1.40 m. Take Ew = 12 GPa. a + ©MB = 0; c + ©Fy = 0;
18 kN/m
A
FC(1) - FA(2) = 0
2m
3FB - FA = 2FC
(3)
Solving Eqs. (1)–(3) yields : FA = 5.7857 kN;
FB = 9.6428 kN;
FC = 11.5714 kN
3
dA =
5.7857(10 )(1.40) FAL = 0.0597(10 - 3) m = p 2 9 AE (0.12 )12(10 ) 4
dC =
11.5714(103)(1.40) FCL = 0.1194(10 - 3) m = p 2 9 AE (0.12 )12(10 ) 4
tan u =
1.40 m
(2)
3dB - dA = 2dC
3FBL FAL 2FCL = ; AE AE AE
C
(1)
FA + FB + FC - 27 = 0
dB - dA dC - dA = ; 2 3
B
0.1194 - 0.0597 (10 - 3) 3
u = 0.0199(10 - 3) rad = 1.14(10 - 3)°
Ans.
167
1m
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*4–60. The assembly consists of two posts AD and CF made of A-36 steel and having a cross-sectional area of 1000 mm2, and a 2014-T6 aluminum post BE having a crosssectional area of 1500 mm2. If a central load of 400 kN is applied to the rigid cap, determine the normal stress in each post. There is a small gap of 0.1 mm between the post BE and the rigid member ABC.
400 kN 0.5 m
A
Equation of Equilibrium. Due to symmetry, FAD = FCF = F. Referring to the FBD of the rigid cap, Fig. a, FBE + 2F - 400(103) = 0
(1)
Compatibility Equation. Referring to the initial and final position of rods AD (CF) and BE, Fig. b, d = 0.1 + dBE F(400)
1(10 ) C 200(10 ) D -3
9
= 0.1 +
FBE (399.9)
1.5(10 - 3) C 73.1(109) D
F = 1.8235 FBE + 50(103)
(2)
Solving Eqs (1) and (2) yield FBE = 64.56(103) N
F = 167.72(103) N
Normal Stress. sAD = sCF =
sBE =
B
C 0.4 m
D
+ c ©Fy = 0;
0.5 m
167.72(103) F = 168 MPa = Ast 1(10 - 3)
Ans.
64.56(103) FBE = 43.0 MPa = Aal 1.5(10 - 3)
Ans.
168
E
F
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•4–61.
The distributed loading is supported by the three suspender bars. AB and EF are made of aluminum and CD is made of steel. If each bar has a cross-sectional area of 450 mm2, determine the maximum intensity w of the distributed loading so that an allowable stress of 1sallow2st = 180 MPa in the steel and 1sallow2al = 94 MPa in the aluminum is not exceeded. Est = 200 GPa, Eal = 70 GPa. Assume ACE is rigid.
1.5 m
1.5 m
B al
D st
A
F al
C
2m
E
w
a + ©MC = 0;
FEF(1.5) - FAB(1.5) = 0 FEF = FAB = F
+ c ©Fy = 0;
2F + FCD - 3w = 0
(1)
Compatibility condition : dA = dC FCDL FL = ; 9 A(70)(10 ) A(200)(109)
F = 0.35 FCD
(2)
Assume failure of AB and EF: F = (sallow)al A = 94(106)(450)(10 - 6) = 42300 N From Eq. (2) FCD = 120857.14 N From Eq. (1) w = 68.5 kN>m Assume failure of CD: FCD = (sallow)st A = 180(106)(450)(10 - 6) = 81000 N From Eq. (2) F = 28350 N From Eq. (1) w = 45.9 kN>m
(controls)
Ans.
169
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4–62. The rigid link is supported by a pin at A, a steel wire BC having an unstretched length of 200 mm and crosssectional area of 22.5 mm2, and a short aluminum block having an unloaded length of 50 mm and cross-sectional area of 40 mm2. If the link is subjected to the vertical load shown, determine the average normal stress in the wire and the block. Est = 200 GPa, Eal = 70 GPa.
C 200 mm
B 100 mm
D
450(250) - FBC(150) - FD(150) = 0
50 mm
750 - FBC - FD = 0
[1]
Compatibility: dBC = dD FD(50)
FBC(200) 22.5(10 - 6)200(109)
=
40(10 - 6)70(109)
FBC = 0.40179 FD
[2]
Solving Eqs. [1] and [2] yields: FD = 535.03 N
FBC = 214.97 N
Average Normal Stress: sD =
sBC =
150 mm
450 N
Equations of Equilibrium: a + ©MA = 0;
A 150 mm
FD 535.03 = 13.4 MPa = AD 40(10 - 6)
Ans.
FBC 214.97 = 9.55 MPa = ABC 22.5(10 - 6)
Ans.
170
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4–63. The rigid link is supported by a pin at A, a steel wire BC having an unstretched length of 200 mm and crosssectional area of 22.5 mm2, and a short aluminum block having an unloaded length of 50 mm and cross-sectional area of 40 mm2. If the link is subjected to the vertical load shown, determine the rotation of the link about the pin A. Report the answer in radians. Est = 200 GPa, Eal = 70 GPa.
C 200 mm
B 100 mm
D
450(250) - FBC(150) - FD(150) = 0
50 mm
750 - FBC - FD = 0
[1]
Compatibility: dBC = dD FD(50)
FBC(200) -6
9
22.5(10 )200(10 )
=
40(10 - 6)70(109)
FBC = 0.40179 FD
[2]
Solving Eqs. [1] and [2] yields : FD = 535.03 N
FBC = 214.97 N
Displacement: dD =
535.03(50) FDLD = 0.009554 mm = ADEal 40(10 - 6)(70)(109)
tan u =
150 mm
450 N
Equations of Equilibrium: a + ©MA = 0;
A 150 mm
dD 0.009554 = 150 150
u = 63.7(10 - 6) rad = 0.00365°
Ans.
171
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*4–64. The center post B of the assembly has an original length of 124.7 mm, whereas posts A and C have a length of 125 mm. If the caps on the top and bottom can be considered rigid, determine the average normal stress in each post. The posts are made of aluminum and have a cross-sectional area of 400 mm2. Eal = 70 GPa.
800 kN/m
A
a + ©MB = 0;
100 mm
B
C
-FA(100) + FC(100) = 0 (1) 2F + FB - 160 = 0
(2)
dA = dB + 0.0003 F (0.125)
FB (0.1247)
400 (10 - 6)(70)(106)
=
400 (10 - 6)(70)(106)
+ 0.0003
0.125 F - 0.1247FB = 8.4
(3)
Solving Eqs. (2) and (3) F = 75.762 kN FB = 8.547 kN sA = sC =
sB =
75.726 (103) 400(10 - 6)
8.547 (103) 400 (10 - 6)
= 189 MPa
Ans.
= 21.4 MPa
Ans.
•4–65.
The assembly consists of an A-36 steel bolt and a C83400 red brass tube. If the nut is drawn up snug against the tube so that L = 75 mm, then turned an additional amount so that it advances 0.02 mm on the bolt, determine the force in the bolt and the tube. The bolt has a diameter of 7 mm and the tube has a cross-sectional area of 100 mm2.
L
Equilibrium: Since no external load is applied, the force acting on the tube and the bolt is the same. Compatibility: 0.02 = dt + db 0.02 =
P(75)
P(75) -6
9
100(10 )(101)(10 )
+
125 mm
800 kN/m
FA = FC = F + c ©Fy = 0;
100 mm
p 2 9 4 (0.007 )(200)(10 )
P = 1164.83 N = 1.16 kN
Ans.
172
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4–66. The assembly consists of an A-36 steel bolt and a C83400 red brass tube. The nut is drawn up snug against the tube so that L = 75 mm. Determine the maximum additional amount of advance of the nut on the bolt so that none of the material will yield. The bolt has a diameter of 7 mm and the tube has a cross-sectional area of 100 mm2.
L
Allowable Normal Stress: (sg)st = 250 A 106 B =
Pst p 2 4 (0.007)
Pst = 9.621 kN (sg)br = 70.0 A 106 B =
Pbr 100(10 - 6)
Pbr = 7.00 kN Since Pst 7 Pbr, by comparison he brass will yield first. Compatibility: a = dt + db 7.00(103)(75) =
100(10 - 6)(101)(109)
7.00(103)(75) +
p 2 9 4 (0.007) (200)(10 )
= 0.120 mm
Ans.
173
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4–67. The three suspender bars are made of the same material and have equal cross-sectional areas A. Determine the average normal stress in each bar if the rigid beam ACE is subjected to the force P. a + ©MA = 0;
D
d FCD(d) + FEF(2d) - Pa b = 0 2 FCD + 2FEF =
+ c ©Fy = 0;
B
F
L
P
P 2
(1)
FAB + FCD + FEF - P = 0
A
C d 2
(2)
d 2
E d
dC - dE dA - dE = d 2d 2dC = dA + dE 2FCDL FABL FEFL = + AE AE AE 2FCD - FAB - FEF = 0
(3)
Solving Eqs. (1), (2) and (3) yields P 3
P 12
FAB =
7P 12
sAB =
7P 12A
Ans.
sCD =
P 3A
Ans.
sEF =
P 12A
Ans.
FCD =
FEF =
*4–68. A steel surveyor’s tape is to be used to measure the length of a line. The tape has a rectangular cross section of 0.05 in. by 0.2 in. and a length of 100 ft when T1 = 60°F and the tension or pull on the tape is 20 lb. Determine the true length of the line if the tape shows the reading to be 463.25 ft when used with a pull of 35 lb at T2 = 90°F. The ground on which it is placed is flat. ast = 9.60110-62>°F, Est = 2911032 ksi.
P
P 0.2 in. 0.05 in.
dT = a¢TL = 9.6(10 - 6)(90 - 60)(463.25) = 0.133416 ft d =
(35 - 20)(463.25) PL = 0.023961 ft = AE (0.2)(0.05)(29)(106)
L = 463.25 + 0.133416 + 0.023961 = 463.41 ft
Ans.
174
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•4–69.
Three bars each made of different materials are connected together and placed between two walls when the temperature is T1 = 12°C. Determine the force exerted on the (rigid) supports when the temperature becomes T2 = 18°C. The material properties and cross-sectional area of each bar are given in the figure.
Copper Steel Brass Est ⫽ 200 GPa Ebr ⫽ 100 GPa Ecu ⫽ 120 GPa ast ⫽ 12(10⫺6)/⬚C abr ⫽ 21(10⫺6)/°C acu ⫽ 17(10⫺6)/⬚C Ast ⫽ 200 mm2
300 mm
+ ) (;
Acu ⫽ 515 mm2
Abr ⫽ 450 mm2
200 mm
100 mm
0 = ¢T - d
0 = 12(10 - 6)(6)(0.3) + 21 (10 - 6)(6)(0.2) + 17 (10 - 6)(6)(0.1) F(0.3) -
-6
F(0.2) 9
200(10 )(200)(10 )
-
-6
F(0.1) 9
-
450(10 )(100)(10 )
515(10 - 6)(120)(109)
F = 4203 N = 4.20 kN
Ans.
k ⫽ 1000 lb/in.
4–70. The rod is made of A-36 steel and has a diameter of 0.25 in. If the rod is 4 ft long when the springs are compressed 0.5 in. and the temperature of the rod is T = 40°F, determine the force in the rod when its temperature is T = 160°F.
k ⫽ 1000 lb/ in.
4 ft
Compatibility: + B A:
x = dT - dF x = 6.60(10 - 6)(160 - 40)(2)(12) -
1.00(0.5)(2)(12) p 2 3 4 (0.25 )(29.0)(10 )
x = 0.01869 in. F = 1.00(0.01869 + 0.5) = 0.519 kip
Ans.
4–71. A 6-ft-long steam pipe is made of A-36 steel with sY = 40 ksi. It is connected directly to two turbines A and B as shown. The pipe has an outer diameter of 4 in. and a wall thickness of 0.25 in. The connection was made at T1 = 70°F. If the turbines’ points of attachment are assumed rigid, determine the force the pipe exerts on the turbines when the steam and thus the pipe reach a temperature of T2 = 275°F.
6 ft A
Compatibility: + B A:
0 = dT - dF 0 = 6.60(10 - 6)(275 - 70)(6)(12) -
F(6)(12) p 2 4 (4
- 3.52)(29.0)(103)
F = 116 kip
Ans. 175
B
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*4–72. A 6-ft-long steam pipe is made of A-36 steel with sY = 40 ksi. It is connected directly to two turbines A and B as shown. The pipe has an outer diameter of 4 in. and a wall thickness of 0.25 in. The connection was made at T1 = 70°F. If the turbines’ points of attachment are assumed to have a stiffness of k = 8011032 kip>in., determine the force the pipe exerts on the turbines when the steam and thus the pipe reach a temperature of T2 = 275°F.
6 ft A
B
Compatibility: x = dT - dF x = 6.60(10 - 6)(275 - 70)(3)(12) -
80(103)(x)(3)(12) p 2 4 (4
- 3.52)(29.0)(103)
x = 0.001403 in. F = k x = 80(103)(0.001403) = 112 kip
Ans.
•4–73.
The pipe is made of A-36 steel and is connected to the collars at A and B. When the temperature is 60° F, there is no axial load in the pipe. If hot gas traveling through the pipe causes its temperature to rise by ¢T = 140 + 15x2°F, where x is in feet, determine the average normal stress in the pipe. The inner diameter is 2 in., the wall thickness is 0.15 in.
A
Compatibility: L
0 = dT - dF 0 = 6.60 A 10 - 6 B
Where
dT =
L0
8ft
L0
(40 + 15 x) dx -
0 = 6.60 A 10 - 6 B B 40(8) +
a ¢T dx F(8) A(29.0)(103)
15(8)2 F(8) R 2 A(29.0)(103)
F = 19.14 A Average Normal Stress: s =
B 8 ft
19.14 A = 19.1 ksi A
Ans.
176
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4–74. The bronze C86100 pipe has an inner radius of 0.5 in. and a wall thickness of 0.2 in. If the gas flowing through it changes the temperature of the pipe uniformly from TA = 200°F at A to TB = 60°F at B, determine the axial force it exerts on the walls. The pipe was fitted between the walls when T = 60°F.
A
B 8 ft
Temperature Gradient: T(x) = 60 + a
8 - x b140 = 200 - 17.5x 8
Compatibility: 0 = dT - dF 0 = 9.60 A 10 - 6 B
Where
dT = 1 a¢Tdx
2ft
0 = 9.60 A 10 - 6 B
L0
[(200 - 17.5x) - 60] dx 2ft
L0
(140 - 17.5x) dx -
F(8) p 2 4 (1.4
- 12)15.0(103)
F(8) p 2 4 (1.4
- 12) 15.0(103)
F = 7.60 kip
Ans.
4–75. The 40-ft-long A-36 steel rails on a train track are laid with a small gap between them to allow for thermal expansion. Determine the required gap d so that the rails just touch one another when the temperature is increased from T1 = -20°F to T2 = 90°F. Using this gap, what would be the axial force in the rails if the temperature were to rise to T3 = 110°F? The cross-sectional area of each rail is 5.10 in2.
d
40 ft
Thermal Expansion: Note that since adjacent rails expand, each rail will be d required to expand on each end, or d for the entine rail. 2 d = a¢TL = 6.60(10 - 6)[90 - (-20)](40)(12) Ans.
= 0.34848 in. = 0.348 in. Compatibility: + B A:
0.34848 = dT - dF 0.34848 = 6.60(10 - 6)[110 - (-20)](40)(12) -
d
F(40)(12) 5.10(29.0)(103)
F = 19.5 kip
Ans.
177
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*4–76. The device is used to measure a change in temperature. Bars AB and CD are made of A-36 steel and 2014-T6 aluminum alloy respectively. When the temperature is at 75°F, ACE is in the horizontal position. Determine the vertical displacement of the pointer at E when the temperature rises to 150°F.
0.25 in.
A
3 in.
C
E
1.5 in.
Thermal Expansion:
A dT B CD = aal ¢TLCD = 12.8(10 - 6)(150 - 75)(1.5) = 1.44(10 - 3) in.
B
D
A dT B AB = ast ¢TLAB = 6.60(10 - 6)(150 - 75)(1.5) = 0.7425(10 - 3) in. From the geometry of the deflected bar AE shown Fig. b, dE = A dT B AB + C = 0.7425(10 - 3) + B
A dT B CD - A dT B AB 0.25
S(3.25)
1.44(10 - 3) - 0.7425(10 - 3) R (3.25) 0.25
= 0.00981 in.
Ans.
•4–77. The bar has a cross-sectional area A, length L, modulus of elasticity E, and coefficient of thermal expansion a. The temperature of the bar changes uniformly along its length from TA at A to TB at B so that at any point x along the bar T = TA + x1TB - TA2>L. Determine the force the bar exerts on the rigid walls. Initially no axial force is in the bar and the bar has a temperature of TA.
+ :
x A TA
0 = ¢ T - dF
(1)
However, d¢ T = a¢ T dx = a(TA +
TB - TA x - TA)dx L
L
¢T = a
= ac
L
TB - TA TB - TA 2 x dx = ac x d冷 L 2L L0 0 TB - TA aL Ld = (TB - TA) 2 2
From Eq.(1). 0 =
FL aL (TB - TA) 2 AE
F =
a AE (TB - TA) 2
B
Ans.
178
TB
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4–78. The A-36 steel rod has a diameter of 50 mm and is lightly attached to the rigid supports at A and B when T1 = 80°C. If the temperature becomes T2 = 20°C and an axial force of P = 200 kN is applied to its center, determine the reactions at A and B.
0.5 m
FB - FA + 200(103) = 0
(1)
When the rod is unconstrained at B, it has a free contraction of dT = ast ¢ TL = 12(10 - 6)(80 - 20)(1000) = 0.72 mm. Also, under force P and FB with unconstrained at B, the deformation of the rod are dP =
dFB =
PLAC = AE
FB LAB = AE
200(103)(500) p 2 4 (0.05 )
C 200(109) D
FB (1000) p 2 4 (0.05 )
C 200(109) D
= 0.2546 mm = 2.5465(10 - 6) FB
Using the method of super position, Fig. b, + B A:
B
P
Referring to the FBD of the rod, Fig. a + ©F = 0; : x
C
A
0 = -dT + dP + dFB 0 = -0.72 + 0.2546 + 2.5465(10 - 6) FB FB = 182.74(103) N = 183 kN
Ans.
Substitute the result of FB into Eq (1), FA = 382.74(103) N = 383 kN
Ans.
179
0.5 m
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4–79. The A-36 steel rod has a diameter of 50 mm and is lightly attached to the rigid supports at A and B when T1 = 50°C. Determine the force P that must be applied to the collar at its midpoint so that, when T2 = 30°C, the reaction at B is zero.
C
A
B
P 0.5 m
0.5 m
When the rod is unconstrained at B, it has a free contraction of dT = ast ¢TL = 12(10 - 6)(50 - 30)(1000) = 0.24 mm. Also, under force P with unconstrained at B, the deformation of the rod is dP =
PLAC = AE
P(500) p 2 4 (0.05 )
C 200(109) D
= 1.2732(10 - 6) P
Since FB is required to be zero, the method of superposition, Fig. b, gives + B A:
0 = -dT + dP 0 = -0.24 + 1.2732(10 - 6)P P = 188.50(103) N = 188 kN
Ans.
*4–80. The rigid block has a weight of 80 kip and is to be supported by posts A and B, which are made of A-36 steel, and the post C, which is made of C83400 red brass. If all the posts have the same original length before they are loaded, determine the average normal stress developed in each post when post C is heated so that its temperature is increased by 20°F. Each post has a cross-sectional area of 8 in2.
A
Equations of Equilibrium: a + ©MC = 0; + c ©Fy = 0;
FB(3) - FA(3) = 0
FA = FB = F
2F + FC - 80 = 0
[1]
Compatibility: (dC)F - (dC)T = dF
(+ T) FCL 8(14.6)(103)
- 9.80 A 10 - 5 B (20)L =
FL 8(29.0)(103)
8.5616 FC - 4.3103 F = 196
[2]
Solving Eqs. [1] and [2] yields: F = 22.81 kip
FC = 34.38 kip
average Normal Sress: sA = sB =
sC =
F 22.81 = = 2.85 ksi A 8
Ans.
FC 34.38 = = 4.30 ksi A 8
Ans.
180
C
B
3 ft
3 ft
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•4–81.
The three bars are made of A-36 steel and form a pin-connected truss. If the truss is constructed when T1 = 50°F, determine the force in each bar when T2 = 110°F. Each bar has a cross-sectional area of 2 in2.
t
5f
5f
t
A
4 ft
B
D 3 ft
(dT ¿)AB - (dF ¿)AB = (dT)AD + (dF)AD
(1)
œ cos u; However, dAB = dAB
œ dAB =
dAB 5 = dAB cos u 4
Substitute into Eq. (1) 5 5 (dT)AB - (dF)AB = (dT)AD + (dF)AD 4 4 FAB(5)(12) 5 d c6.60(10 - 6)(110° - 50°)(5)(12) 4 2(29)(103) = 6.60(10 - 6)(110° - 50°)(4)(12) +
FAD(4)(12) 2(29)(103)
620.136 = 75FAB + 48FAD + ©F = 0; : x
3 3 F - FAB = 0; 5 AC 5
+ c ©Fy = 0;
4 FAD - 2a FAB b = 0 5
(2) FAC = FAB
(3)
Solving Eqs. (2) and (3) yields : FAD = 6.54 kip
Ans.
FAC = FAB = 4.09 kip
Ans.
181
C 3 ft
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4–82. The three bars are made of A-36 steel and form a pinconnected truss. If the truss is constructed when T1 = 50°F, determine the vertical displacement of joint A when T2 = 150°F. Each bar has a cross-sectional area of 2 in2.
t
5f
5f
t
A
4 ft
(dT ¿)AB - (dF ¿)AB = (dT)AD + (dF)AD
(1)
œ However, dAB = dAB cos u;
œ dAB =
B
dAB 5 = dAB cos u 4
3 ft
Substitute into Eq. (1) 5 5 (d ) - (dT)AB = (dT)AD + (dF)AD 4 T AB 4 FAB(5)(12) 5 d c6.60(10 - 6)(150° - 50°)(5)(12) 4 2(29)(103) = 6.60(10 - 6)(150° - 50°)(4)(12) +
FAD(4)(12) 2(29)(103)
239.25 - 6.25FAB = 153.12 + 4 FAD 4 FAD + 6.25FAB = 86.13 + © F = 0; : x
3 3 F - FAB = 0; 5 AC 5
+ c © Fy = 0;
4 FAD - 2 a FAB b = 0; 5
(2) FAC = FAB
FAD = 1.6FAB
(3)
Solving Eqs. (2) and (3) yields: FAB = 6.8086 kip:
FAD = 10.8939 kip
(dA)r = (dT)AD + (dT)AD = 6.60(10 - 6)(150° - 50°)(4)(12) +
D
10.8939(4)(12) 2(29)(103)
= 0.0407 in. c
Ans.
182
C 3 ft
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4–83. The wires AB and AC are made of steel, and wire AD is made of copper. Before the 150-lb force is applied, AB and AC are each 60 in. long and AD is 40 in. long. If the temperature is increased by 80°F, determine the force in each wire needed to support the load. Take Est = 29(103) ksi, Ecu = 17(103) ksi, ast = 8(10-6)>°F, acu = 9.60(10-6)>°F. Each wire has a cross-sectional area of 0.0123 in2.
40 in. 60 in.
45⬚
45⬚
A 150 lb
Equations of Equilibrium: + ©F = 0; : x
FAC cos 45° - FAB cos 45° = 0 FAC = FAB = F 2F sin 45° + FAD - 150 = 0
+ c ©Fy = 0;
[1]
Compatibility: (dAC)T = 8.0 A 10 - 6 B (80)(60) = 0.03840 in. (dAC)Tr =
(dAC)T 0.03840 = = 0.05431 in. cos 45° cos 45°
(dAD)T = 9.60 A 10 - 6 B (80)(40) = 0.03072 in. d0 = (dAC)Tr - (dAD)T = 0.05431 - 0.03072 = 0.02359 in. (dAD)F = (dAC)Fr + d0 F(60)
FAD (40) 6
0.0123(17.0)(10 )
=
0.0123(29.0)(106) cos 45°
C
D
B
+ 0.02359
0.1913FAD - 0.2379F = 23.5858
[2]
Solving Eq. [1] and [2] yields: FAC = FAB = F = 10.0 lb
Ans.
FAD = 136 lb
Ans.
183
60 in.
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*4–84. The AM1004-T61 magnesium alloy tube AB is capped with a rigid plate E. The gap between E and end C of the 6061-T6 aluminum alloy solid circular rod CD is 0.2 mm when the temperature is at 30° C. Determine the normal stress developed in the tube and the rod if the temperature rises to 80° C. Neglect the thickness of the rigid cap.
25 mm a
Section a-a
E
B
A
20 mm
C
25 mm
a 0.2 mm 300 mm
Compatibility Equation: If tube AB and rod CD are unconstrained, they will have a free expansion of A dT B AB = amg ¢TLAB = 26(10 - 6)(80 - 30)(300) = 0.39 mm and
A dT)CD = aal ¢TLCD = 24(10 - 6)(80 - 30)(450) = 0.54 mm. Referring deformation diagram of the tube and the rod shown in Fig. a, d =
to
the
C A dT B AB - A dF B AB D + C A dT B CD - A dF B CD D
0.2 = C 0.39 -
F(300)
p A 0.025 - 0.02 B (44.7)(10 ) 2
2
9
S + C 0.54 -
F(450)
p 4
A 0.0252 B (68.9)(109)
S
F = 32 017.60 N Normal Stress: sAB =
F 32 017.60 = = 45.3 MPa AAB p A 0.0252 - 0.022 B
sCD =
F 32 017.60 = = 65.2 MPa p 2 ACD 4 A 0.025 B
Ans.
Ans.
F = 107 442.47 N
184
450 mm
D
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•4–85. The AM1004-T61 magnesium alloy tube AB is capped with a rigid plate. The gap between E and end C of the 6061-T6 aluminum alloy solid circular rod CD is 0.2 mm when the temperature is at 30° C. Determine the highest temperature to which it can be raised without causing yielding either in the tube or the rod. Neglect the thickness of the rigid cap.
25 mm a
Section a-a
E
B
A
20 mm
C
25 mm
a 0.2 mm 300 mm
Then sCD =
F 107 442.47 = = 218.88MPa 6 (sY)al p 2 ACD 4 A 0.025 B
(O.K.!)
Compatibility Equation: If tube AB and rod CD are unconstrained, they will have a free expansion of A dT B AB = amg ¢TLAB = 26(10 - 6)(T - 30)(300) = 7.8(10 - 6) (T - 30) and
A dT B CD = aal ¢TLCD = 24(10 - 6)(T - 30)(450) = 0.0108(T - 30).
Referring to the deformation diagram of the tube and the rod shown in Fig. a, d =
C A dT B AB - A dF B AB D + C A dT B CD - A dF B CD D
0.2 = C 7.8(10 - 3)(T - 30) -
+ C 0.0108(T - 30) -
107 442.47(300)
p A 0.0252 - 0.022 B (44.7)(109)
107 442.47(450)
p 4
A 0.0252 B (68.9)(109)
S
S
T = 172° C
Ans.
185
450 mm
D
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4–86. The steel bolt has a diameter of 7 mm and fits through an aluminum sleeve as shown. The sleeve has an inner diameter of 8 mm and an outer diameter of 10 mm. The nut at A is adjusted so that it just presses up against the sleeve. If the assembly is originally at a temperature of T1 = 20°C and then is heated to a temperature of T2 = 100°C, determine the average normal stress in the bolt and the sleeve. Est = 200 GPa, Eal = 70 GPa, ast = 14(10-6)>°C, aal = 23(10-6)>°C.
A
Compatibility: (ds)T - (db)T = (ds)F + (db)F 23(10 - 6)(100 - 20)L - 14(10 - 6)(100 - 20)L =
p 2 4 (0.01
FL + - 0.0082)70(109)
FL p 2 9 4 (0.007 )200(10 )
F = 1133.54 N Average Normal Stress: ss =
F = As
sb =
F 1133.54 = 29.5 MPa = p 2 Ab 4 (0.007 )
1133.54 = 40.1 MPa - 0.0082)
Ans.
p 2 4 (0.01
Ans.
4–87. Determine the maximum normal stress developed in the bar when it is subjected to a tension of P = 8 kN.
5 mm 40 mm
20 mm P
P
For the fillet:
r ⫽ 10 mm 20 mm
r 10 = = 0.5 h 20
w 40 = = 2 h 20 From Fig. 10-24. K = 1.4 smax = Ksavg = 1.4 a
8 (103) b 0.02 (0.005)
= 112 MPa For the hole: r 10 = = 0.25 w 40 From Fig. 4-25. K = 2.375 smax = Ksavg = 2.375 a
8 (103) b (0.04 - 0.02)(0.005)
= 190 MPa
Ans.
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*4–88. If the allowable normal stress for the bar is sallow = 120 MPa, determine the maximum axial force P that can be applied to the bar.
5 mm 40 mm
20 mm P
P
Assume failure of the fillet.
r ⫽ 10 mm 20 mm
r 10 = = 0.5 h 20
w 40 = = 2; h 20 From Fig. 4-24. K = 1.4
sallow = smax = Ksavg 120 (106) = 1.4 a
P b 0.02 (0.005)
P = 8.57 kN Assume failure of the hole. r 10 = = 0.25 w 20 From Fig. 4-25. K = 2.375 sallow = smax = Ksavg 120 (104) = 2.375 a
P b (0.04 - 0.02) (0.005)
P = 5.05 kN (controls)
Ans.
•4–89.
The member is to be made from a steel plate that is 0.25 in. thick. If a 1-in. hole is drilled through its center, determine the approximate width w of the plate so that it can support an axial force of 3350 lb. The allowable stress is sallow = 22 ksi.
0.25 in. w
3350 lb
sallow = smax = Ksavg
1 in.
3.35 d 22 = K c (w - 1)(0.25) w =
3.35K + 5.5 5.5
By trial and error, from Fig. 4-25, choose
w =
r = 0.2; w
K = 2.45
3.35(2.45) + 5.5 = 2.49 in. 5.5
Since
r 0.5 = = 0.2 w 2.49
3350 lb
Ans.
OK
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4–90. The A-36 steel plate has a thickness of 12 mm. If there are shoulder fillets at B and C, and sallow = 150 MPa, determine the maximum axial load P that it can support. Calculate its elongation, neglecting the effect of the fillets.
r = 30 mm 120 mm r = 30 mm 60 mm P A
w 120 = = 2 h 60
and
60 mm P D
B
Maximum Normal Stress at fillet: r 30 = = 0.5 h 60
C
800 mm
200 mm
200 mm
From the text, K = 1.4 smax = sallow = Ksavg 150(106) = 1.4 B
P R 0.06(0.012)
P = 77142.86 N = 77.1 kN
Ans.
Displacement: d = ©
PL AE 77142.86(800)
77142.86(400) =
9
+
(0.06)(0.012)(200)(10 )
(0.12)(0.012)(200)(109)
= 0.429 mm
Ans.
4–91. Determine the maximum axial force P that can be applied to the bar. The bar is made from steel and has an allowable stress of sallow = 21 ksi.
0.125 in. 1.25 in.
1.875 in.
P
Assume failure of the fillet. r 0.25 = = 0.2 h 1.25
P
w 1.875 = = 1.5 h 1.25 0.75 in.
From Fig. 4-24, K = 1.73 sallow = smax = Ksavg 21 = 1.73 a
P b 1.25 (0.125)
P = 1.897 kip Assume failure of the hole. r 0.375 = = 0.20 w 1.875 From Fig. 4-25, K = 2.45 sallow = smax = Ksavg 21 = 2.45 a
P b (1.875 - 0.75)(0.125)
P = 1.21 kip (controls)
Ans.
188
r ⫽ 0.25 in.
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*4–92. Determine the maximum normal stress developed in the bar when it is subjected to a tension of P = 2 kip.
0.125 in. 1.25 in.
1.875 in.
At fillet: P
r 0.25 = = 0.2 h 1.25
P
w 1.875 = = 1.5 h 1.25 0.75 in.
From Fig. 4-24, K = 1.73 smax = Ka
r ⫽ 0.25 in.
P 2 d = 22.1 ksi b = 1.73 c A 1.25(0.125)
At hole: r 0.375 = = 0.20 w 1.875 From Fig. 4-25, K = 2.45 smax = 2.45 c
2 d = 34.8 ksi (1.875 - 0.75)(0.125)
(Controls)
Ans.
•4–93.
Determine the maximum normal stress developed in the bar when it is subjected to a tension of P = 8 kN.
5 mm 60 mm P
Maximum Normal Stress at fillet: r 15 = = 0.5 h 30
P ht
= 1.4 B
8(103) R = 74.7 MPa (0.03)(0.005)
Maximum Normal Stress at the hole: r 6 = = 0.1 w 60 From the text, K = 2.65 smax = K savg = K
P (w - 2r) t
= 2.65 B
8(103) R (0.06 - 0.012)(0.005)
= 88.3 MPa
P r = 15 mm 12 mm
w 60 = = 2 h 30
and
From the text, K = 1.4 smax = Ksavg = K
30 mm
(Controls)
Ans.
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4–94. The resulting stress distribution along section AB for the bar is shown. From this distribution, determine the approximate resultant axial force P applied to the bar. Also, what is the stress-concentration factor for this geometry?
0.5 in. A P
4 in. 1 in.
B 12 ksi
P =
L
3 ksi
sdA = Volume under curve
Number of squares = 10 P = 10(3)(1)(0.5) = 15 kip savg =
K =
Ans.
15 kip P = = 7.5 ksi A (4 in.)(0.5 in.)
smax 12 ksi = = 1.60 savg 7.5 ksi
Ans.
4–95. The resulting stress distribution along section AB for the bar is shown. From this distribution, determine the approximate resultant axial force P applied to the bar. Also, what is the stress-concentration factor for this geometry?
0.5 in. A 0.6 in. 0.8 in.
Number of squares = 28 P = 28(6)(0.2)(0.5) = 16.8 kip savg
P 16.8 = = = 28 ksi A 2(0.6)(0.5)
K =
smax 36 = = 1.29 savg 28
0.6 in.
Ans. B
6 ksi
36 ksi
Ans.
*4–96. The resulting stress distribution along section AB for the bar is shown. From this distribution, determine the approximate resultant axial force P applied to the bar. Also, what is the stress-concentration factor for this geometry?
10 mm A 20 mm 80 mm B 5 MPa
Number of squares = 19
30 MPa
6
P = 19(5)(10 )(0.02)(0.01) = 19 kN savg =
K =
P
0.2 in.
Ans.
19(103) P = = 23.75 MPa A 0.08(0.01)
smax 30 MPa = = 1.26 savg 23.75 MPa
Ans.
190
P
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•4–97.
The 300-kip weight is slowly set on the top of a post made of 2014-T6 aluminum with an A-36 steel core. If both materials can be considered elastic perfectly plastic, determine the stress in each material. Aluminum 1 in. 2 in. Steel
Equations of Equilibrium: + c ©Fy = 0;
Pst + Pal - 300 = 0
[1]
Elastic Analysis: Assume both materials still behave elastically under the load. dst = dal Pst L p 2 (2) (29)(103) 4
Pal L =
p 2 4 (4
- 22)(10.6)(103)
Pst = 0.9119 Pal Solving Eqs. [1] and [2] yields: Pal = 156.91 kip
Pst = 143.09 kip
Average Normal Stress: sal =
Pal = Aal
156.91 - 22)
p 2 4 (4
(OK!)
= 16.65 ksi 6 (sg)al = 60.0 ksi sst =
Pst 143.09 = p 2 Ast 4 (2 ) = 45.55 ksi 7 (sg)st = 36.0 ksi
Therefore, the steel core yields and so the elastic analysis is invalid. The stress in the steel is sst = (sy)st = 36.0 ksi
Ans.
p Pst = (sg)stAst = 36.0a b A 22 B = 113.10 kip 4 From Eq. [1] Pal = 186.90 kip sal =
Pal = Aal
186.90 = 19.83 ksi 6 (sg)al = 60.0 ksi - 22)
p 2 4 (4
Then sal = 19.8 ksi
Ans.
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4–98. The bar has a cross-sectional area of 0.5 in2 and is made of a material that has a stress–strain diagram that can be approximated by the two line segments shown. Determine the elongation of the bar due to the applied loading.
A
B 5 ft
8 kip
C 5 kip
2 ft
s(ksi) 40
20
Average Normal Stress and Strain: For segment BC sBC =
0.001
PBC 5 = = 10.0 ksi ABC 0.5
10.0 20 = ; eBC 0.001
eBC =
0.001 (10.0) = 0.00050 in.>in. 20
Average Normal Stress and Strain: For segment AB sAB =
PAB 13 = = 26.0 ksi AAB 0.5
40 - 20 26.0 - 20 = eAB - 0.001 0.021 - 0.001 eAB = 0.0070 in.>in. Elongation: dBC = eBCLBC = 0.00050(2)(12) = 0.0120 in. dAB = eAB LAB = 0.0070(5)(12) = 0.420 in. dTot = dBC + dAB = 0.0120 + 0.420 = 0.432 in.
Ans.
192
0.021
P (in./in.)
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4–99. The rigid bar is supported by a pin at A and two steel wires, each having a diameter of 4 mm. If the yield stress for the wires is sY = 530 MPa, and Est = 200 GPa, determine the intensity of the distributed load w that can be placed on the beam and will just cause wire EB to yield. What is the displacement of point G for this case? For the calculation, assume that the steel is elastic perfectly plastic.
E
D
800 mm A
B
C G w
400 mm
Equations of Equilibrium: a + ©MA = 0;
FBE(0.4) + FCD(0.65) - 0.8w (0.4) = 0 0.4 FBE + 0.65FCD = 0.32w
[1]
Plastic Analysis: Wire CD will yield first followed by wire BE. When both wires yield FBE = FCD = (sg)A p = 530 A 106 B a b A 0.0042 B = 6.660 kN 4 Substituting the results into Eq. [1] yields: w = 21.9 kN>m
Ans.
Displacement: When wire BE achieves yield stress, the corresponding yield strain is eg =
sg E
530(106) =
200(109)
= 0.002650 mm>mm
dBE = eg LBE = 0.002650(800) = 2.120 mm From the geometry dBE dG = 0.8 0.4 dG = 2dBE = 2(2.120) = 4.24 mm
Ans.
193
250 mm
150 mm
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*4–100. The rigid bar is supported by a pin at A and two steel wires, each having a diameter of 4 mm. If the yield stress for the wires is sY = 530 MPa, and Est = 200 GPa, determine (a) the intensity of the distributed load w that can be placed on the beam that will cause only one of the wires to start to yield and (b) the smallest intensity of the distributed load that will cause both wires to yield. For the calculation, assume that the steel is elastic perfectly plastic.
E
D
800 mm A
B
C G w
400 mm
Equations of Equilibrium: a + ©MA = 0;
FBE(0.4) + FCD(0.65) - 0.8w (0.4) = 0 0.4 FBE + 0.65 FCD = 0.32w
[1]
(a) By observation, wire CD will yield first. p Then FCD = sg A = 530 A 106 B a b A 0.0042 B = 6.660 kN. 4 From the geometry dCD dBE = ; 0.4 0.65
dCD = 1.625dBE FBEL FCDL = 1.625 AE AE FCD = 1.625 FBE
[2]
Using FCD = 6.660 kN and solving Eqs. [1] and [2] yields: FBE = 4.099 kN w = 18.7 kN>m
Ans.
(b) When both wires yield FBE = FCD = (sg)A p = 530 A 106 B a b A 0.0042 B = 6.660 kN 4 Substituting the results into Eq. [1] yields: w = 21.9 kN>m
Ans.
194
250 mm
150 mm
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•4–101.
The rigid lever arm is supported by two A-36 steel wires having the same diameter of 4 mm. If a force of P = 3 kN is applied to the handle, determine the force developed in both wires and their corresponding elongations. Consider A-36 steel as an elastic-perfectly plastic material.
P
450 mm 150 mm 150 mm 30⬚ A
C
300 mm B
Equation of Equilibrium. Refering to the free-body diagram of the lever shown in Fig. a, FAB (300) + FCD (150) - 3 A 103 B (450) = 0
a + ©ME = 0;
2FAB + FCD = 9 A 103 B
(1)
Elastic Analysis. Assuming that both wires AB and CD behave as linearly elastic, the compatibility equation can be written by referring to the geometry of Fig. b. dAB = a
300 bd 150 CD
dAB = 2dCD
(2)
FAB L FCD L = 2a b AE AE FAB = 2FCD
(3)
Solving Eqs. (1) and (3), FCD = 1800 N
FAB = 3600 N
Normal Stress. sCD =
FCD = ACD
sAB =
FAB = AAB
1800
p 4
A 0.0042 B
p 4
A 0.0042 B
3600
= 143.24 MPa 6 (sY)st
(O.K.)
= 286.48 MPa 7 (sY)st
(N.G.)
Since wire AB yields, the elastic analysis is not valid. The solution must be reworked using FAB = (sY)st AAB = 250 A 106 B c
p A 0.0042 B d 4 Ans.
= 3141.59 N = 3.14 kN Substituting this result into Eq. (1), FCD = 2716.81 N = 2.72 kN sCD =
Ans.
FCD 2716.81 = = 216.20 MPa 6 (sY)st p 2 ACD 4 A 0.004 B
(O.K.)
195
D
E
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4–101.
Continued
Since wire CD is linearly elastic, its elongation can be determined by dCD =
FCD LCD = ACD Est
2716.81(300)
p 4
A 0.0042 B (200) A 109 B Ans.
= 0.3243 mm = 0.324 mm From Eq. (2), dAB = 2dCD = 2(0.3243) = 0.649 mm
Ans.
196
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4–102. The rigid lever arm is supported by two A-36 steel wires having the same diameter of 4 mm. Determine the smallest force P that will cause (a) only one of the wires to yield; (b) both wires to yield. Consider A-36 steel as an elastic-perfectly plastic material.
P
450 mm 150 mm 150 mm 30⬚ A
C
300 mm B
Equation of Equilibrium. Refering to the free-body diagram of the lever arm shown in Fig. a, a + ©ME = 0;
FAB (300) + FCD (150) - P(450) = 0 2FAB + FCD = 3P
(1)
Elastic Analysis. The compatibility equation can be written by referring to the geometry of Fig. b. dAB = a
300 bd 150 CD
dAB = 2dCD FAB L FCD L = 2a b AE AE FCD =
1 F 2 AB
(2)
Assuming that wire AB is about to yield first, FAB = (sY)st AAB = 250 A 106 B c
p A 0.0042 B d = 3141.59 N 4
From Eq. (2), FCD =
1 (3141.59) = 1570.80 N 2
Substituting the result of FAB and FCD into Eq. (1), P = 2618.00 N = 2.62 kN
Ans.
Plastic Analysis. Since both wires AB and CD are required to yield, FAB = FCD = (sY)st A = 250 A 106 B c
p A 0.0042 B d = 3141.59 N 4
Substituting this result into Eq. (1),
197
D
E
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4–103. The three bars are pinned together and subjected to the load P. If each bar has a cross-sectional area A, length L, and is made from an elastic perfectly plastic material, for which the yield stress is sY, determine the largest load (ultimate load) that can be supported by the bars, i.e., the load P that causes all the bars to yield. Also, what is the horizontal displacement of point A when the load reaches its ultimate value? The modulus of elasticity is E.
B L u
C L
P
u L
D
P = 3141.59 N = 3.14 kN
A
Ans.
When all bars yield, the force in each bar is, FY = sYA + ©F = 0; : x
P - 2sYA cos u - sYA = 0
P = sYA(2 cos u + 1)
Ans.
Bar AC will yield first followed by bars AB and AD. dAB = dAD =
dA =
FY(L) sYAL sYL = = AE AE E
dAB sYL = cos u E cos u
Ans.
*4–104. The rigid beam is supported by three 25-mm diameter A-36 steel rods. If the beam supports the force of P = 230 kN, determine the force developed in each rod. Consider the steel to be an elastic perfectly-plastic material.
D
F
E
600 mm P
Equation of Equilibrium. Referring to the free-body diagram of the beam shown in Fig. a, + c ©Fy = 0;
FAD + FBE + FCF - 230 A 103 B = 0
(1)
FBE + 3FCF = 460 A 103 B
(2)
FBE(400) + FCF(1200) - 230 A 103 B (800) = 0
a + ©MA = 0;
400 mm
Elastic Analysis. Referring to the deflection diagram of the beam shown in Fig. b, the compatibility equation can be written as dBE = dAD + a dBE =
dCF - dAD b(400) 1200
2 1 d + dCF 3 AD 3
FBEL 2 FCDL 1 FCF L = a b + a b AE 3 AE 3 AE FBE =
2 1 F + FCF 3 AD 3
(3)
Solving Eqs. (1), (2), and (3) FCF = 131 428.57 N
FBE = 65 714.29 N FAD = 32 857.14 N
198
A
B
400 mm
C
400 mm
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4–104.
Continued
Normal Stress. sCF =
FCF 131428.57 = = 267.74 MPa 7 (sY)st p 2 ACF 4 A 0.025 B
(N.G.)
sBE =
FBE 65714.29 = = 133.87 MPa 6 (sY)st p 2 ABE 4 A 0.025 B
(O.K.)
sAD =
FAD 32857.14 = = 66.94 MPa 6 (sY)st p 2 AAD 4 A 0.025 B
(O.K.)
Since rod CF yields, the elastic analysis is not valid. The solution must be reworked using FCF = (sY)st ACF = 250 A 106 B c
p A 0.0252 B d = 122 718.46 N = 123 kN 4
Ans.
Substituting this result into Eq. (2), FBE = 91844.61 N = 91.8 kN
Ans.
Substituting the result for FCF and FBE into Eq. (1), FAD = 15436.93 N = 15.4 kN
Ans.
sBE =
FBE 91844.61 = = 187.10 MPa 6 (sY)st p 2 ABE 4 A 0.025 B
(O.K.)
sAD =
FAD 15436.93 = = 31.45 MPa 6 (sY)st p 2 AAD 4 A 0.025 B
(O.K.)
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•4–105.
The rigid beam is supported by three 25-mm diameter A-36 steel rods. If the force of P = 230 kN is applied on the beam and removed, determine the residual stresses in each rod. Consider the steel to be an elastic perfectly-plastic material.
D 600 mm
P A
400 mm
Equation of Equilibrium. Referring to the free-body diagram of the beam shown in Fig. a, + c ©Fy = 0;
FAD + FBE + FCF - 230 A 103 B = 0
(1)
FBE + 3FCF = 460 A 103 B
(2)
FBE(400) + FCF(1200) - 230 A 103 B (800) = 0
a + ©MA = 0;
Elastic Analysis. Referring to the deflection diagram of the beam shown in Fig. b, the compatibility equation can be written as dBE = dAD + a dBE =
dCF - dAD b(400) 1200
2 1 d + dCF 3 AD 3
(3)
FBE L 2 FCD L 1 FCF L = a b + a b AE 3 AE 3 AE FBE =
2 1 F + FCF 3 AD 3
(4)
Solving Eqs. (1), (2), and (4) FCF = 131428.57 N
FBE = 65714.29 N
FAD = 32857.14 N
Normal Stress. sCF =
FCF 131428.57 = = 267.74 MPa (T) 7 (sY)st p 2 ACF 4 A 0.025 B
(N.G.)
sBE =
FBE 65714.29 = = 133.87 MPa (T) 6 (sY)st p 2 ABE 4 A 0.025 B
(O.K.)
sAD =
FAD 32857.14 = = 66.94 MPa (T) 6 (sY)st p 2 AAD 4 A 0.025 B
(O.K.)
Since rod CF yields, the elastic analysis is not valid. The solution must be reworked using sCF = (sY)st = 250 MPa (T) FCF = sCF ACF = 250 A 106 B c
F
E
p A 0.0252 B d = 122718.46 N 4
200
B
400 mm
C
400 mm
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4–105.
Continued
Substituting this result into Eq. (2), FBE = 91844.61 N Substituting the result for FCF and FBE into Eq. (1), FAD = 15436.93N sBE =
FBE 91844.61 = = 187.10 MPa (T) 6 (sY)st p 2 ABE 4 A 0.025 B
(O.K.)
sAD =
FAD 15436.93 = = 31.45 MPa (T) 6 (sY)st p 2 AAD 4 A 0.025 B
(O.K.)
Residual Stresses. The process of removing P can be represented by applying the force P¿ , which has a magnitude equal to that of P but is opposite in sense, Fig. c. Since the process occurs in a linear manner, the corresponding normal stress must have the same magnitude but opposite sense to that obtained from the elastic analysis. Thus, œ sCF = 267.74 MPa (C)
œ sBE = 133.87 MPa (C)
œ sAD = 66.94 MPa (C)
Considering the tensile stress as positive and the compressive stress as negative, œ = 250 + (-267.74) = -17.7 MPa = 17.7 MPa (C) (sCF)r = sCF + sCF
Ans.
œ = 187.10 + (-133.87) = 53.2 MPa (T) (sBE)r = sBE + sBE
Ans.
œ (sAD)r = sAD + sAD = 31.45 + (-66.94) = -35.5 MPa = 35.5 MPa (C)
Ans.
201
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4–106. The distributed loading is applied to the rigid beam, which is supported by the three bars. Each bar has a cross-sectional area of 1.25 in2 and is made from a s(ksi) material having a stress–strain diagram that can be approximated by the two line segments shown. If a load of 60 w = 25 kip>ft is applied to the beam, determine the stress in each bar and the vertical displacement of the beam.
4 ft A
B
5 ft
36
A
0.0012
a + ©MB = 0;
0.2
FC(4) - FA(4) = 0; FA = FC = F
+ c ©Fy = 0;
2F + FB - 200 = 0
(1)
Since the loading and geometry are symmetrical, the bar will remain horizontal. Therefore, the displacement of the bars is the same and hence, the force in each bar is the same. From Eq. (1). F = FB = 66.67 kip Thus, sA = sB = sC =
66.67 = 53.33 ksi 1.25
Ans.
From the stress-strain diagram: 60 - 36 53.33 - 36 = : e - 0.0012 0.2 - 0.0012
e = 0.14477 in.>in.
d = eL = 0.14477(5)(12) = 8.69 in.
Ans.
202
4 ft
∋ (in./in.)
B
C
w
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4–107. The distributed loading is applied to the rigid beam, which is supported by the three bars. Each bar has a cross-sectional area of 0.75 in2 and is s(ksi) made from a material having a stress–strain diagram that can be approximated by the two line segments 60 shown. Determine the intensity of the distributed loading w needed to cause the beam to be displaced 36 downward 1.5 in.
0.0012
a + ©MB = 0; + c ©Fy = 0;
FC(4) - FA(4) = 0;
4 ft A
A
0.2
(1)
Since the system and the loading are symmetrical, the bar will remain horizontal. Hence the displacement of the bars is the same and the force supported by each bar is the same. From Eq. (1), FB = F = 2.6667 w
(2)
From the stress-strain diagram: e =
1.5 = 0.025 in.>in. 5 (12)
60 - 36 s - 36 = ; 0.025 - 0.0012 0.2 - 0.0012
s = 38.87 ksi
Hence F = sA = 38.87 (0.75) = 29.15 kip From Eq. (2), w = 10.9 kip>ft
Ans.
203
B
5 ft
FA = FC = F
2F + FB - 8 w = 0
4 ft
∋ (in./in.)
B
C
w
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*4–108. The rigid beam is supported by the three posts A, B, and C of equal length. Posts A and C have a diameter of 75 mm and are made of aluminum, for which Eal = 70 GPa and 1sY2al = 20 MPa. Post B has a diameter of 20 mm and is made of brass, for which Ebr = 100 GPa and 1sY2br = 590 MPa. Determine the smallest magnitude of P so that (a) only rods A and C yield and (b) all the posts yield.
P
A
B
FA = FC = Fal Fat + 2Fat - 2P = 0
+ c ©Fy = 0;
(1)
(a) Post A and C will yield, Fal = (st)alA = 20(104)(pa )(0.075)2 = 88.36 kN (Eal)r =
(sr)al 20(104) = 0.0002857 = Eal 70(104)
Compatibility condition: dbr = dal = 0.0002857(L) Fbr (L) p 2 (0.02) (100)(104) 4
= 0.0002857 L
Fbr = 8.976 kN sbr =
8.976(103) p 3 4 (0.02 )
= 28.6 MPa 6 sr
OK.
From Eq. (1), 8.976 + 2(88.36) - 2P = 0 P = 92.8 kN
Ans.
(b) All the posts yield: Fbr = (sr)brA = (590)(104)(p4 )(0.022) = 185.35 kN Fal = 88.36 kN From Eq. (1); 185.35 + 2(88.36) - 2P = 0 P = 181 kN
Ans.
204
C br
al 2m
©MB = 0;
P
2m
2m
al 2m
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•4–109.
The rigid beam is supported by the three posts A, B, and C. Posts A and C have a diameter of 60 mm and are made of aluminum, for which Eal = 70 GPa and 1sY2al = 20 MPa. Post B is made of brass, for which Ebr = 100 GPa and 1sY2br = 590 MPa. If P = 130 kN, determine the largest diameter of post B so that all the posts yield at the same time.
P
A
B
2(Fg)al + Fbr - 260 = 0
(1)
(Fal)g = (sg)al A = 20(106)(p4 )(0.06)2 = 56.55 kN From Eq. (1), 2(56.55) + Fbr - 260 = 0 Fbr = 146.9 kN (sg)br = 590(106) =
146.9(103) p 3 4 (dB)
dB = 0.01779 m = 17.8 mm
Ans.
205
C br
al 2m
+ c ©Fy = 0;
P
2m
2m
al 2m
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4–110. The wire BC has a diameter of 0.125 in. and the material has the stress–strain characteristics shown in the figure. Determine the vertical displacement of the handle at D if the pull at the grip is slowly increased and reaches a magnitude of (a) P = 450 lb, (b) P = 600 lb.
C 40 in.
A
D
B 50 in.
30 in. P
s (ksi)
Equations of Equilibrium: a + ©MA = 0;
FBC(50) - P(80) = 0
(a) From Eq. [1] when P = 450 lb,
[1] 80 70
FBC = 720 lb
Average Normal Stress and Strain: sBC =
FBC = ABC
720 p 2 4 (0.125 )
P (in./in.)
= 58.67 ksi
0.007
From the Stress–Strain diagram 58.67 70 = ; eBC 0.007
eBC = 0.005867 in.>in.
Displacement: dBC = eBCLBC = 0.005867(40) = 0.2347 in. dBC dD = ; 80 50
dD =
8 (0.2347) = 0.375 in. 5
(b) From Eq. [1] when P = 600 lb,
Ans.
FBC = 960 lb
Average Normal Stress and Strain: sBC =
FBC = ABC
960 p 2 4 (0.125)
= 78.23 ksi
From Stress–Strain diagram 78.23 - 70 80 - 70 = eBC - 0.007 0.12 - 0.007
eBC = 0.09997 in.>in.
Displacement: dBC = eBCLBC = 0.09997(40) = 3.9990 in. dD dBC = ; 80 50
dD =
8 (3.9990) = 6.40 in. 5
Ans.
206
0.12
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4–111. The bar having a diameter of 2 in. is fixed connected at its ends and supports the axial load P. If the material is elastic perfectly plastic as shown by the stress–strain diagram, determine the smallest load P needed to cause segment CB to yield. If this load is released, determine the permanent displacement of point C.
P A
C
2 ft
B
3 ft
s (ksi)
20
0.001
When P is increased, region AC will become plastic first, then CB will become plastic. Thus, FA = FB = sA = 20(p)(1)2 = 62.832 kip + ©F = 0; : x
FA + FB - P = 0
(1)
P = 2(62.832) = 125.66 kip P = 126 kip
Ans.
The deflection of point C is, dC = eL = (0.001)(3)(12) = 0.036 in. ; Consider the reverse of P on the bar. FB ¿(3) FA ¿(2) = AE AE FA ¿ = 1.5 FB ¿ So that from Eq. (1) FB ¿ = 0.4P FA ¿ = 0.6P dC ¿ =
0.4(P)(3)(12) 0.4(125.66)(3)(12) FB ¿L = 0.02880 in. : = = AE AE p(1)2(20>0.001)
¢d = 0.036 - 0.0288 = 0.00720 in. ;
Ans.
207
P (in./in.)
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*4–112. Determine the elongation of the bar in Prob. 4–111 when both the load P and the supports are removed.
P A
C
2 ft
B
3 ft
s (ksi)
20
0.001
When P is increased, region AC will become plastic first, then CB will become plastic. Thus, FA = FB = sA = 20(p)(1)2 = 62.832 kip + ©F = 0; : x
FA + FB - P = 0
(1)
P = 2(62.832) = 125.66 kip P = 126 kip
Ans.
The deflection of point C is, dC = eL = (0.001)(3)(12) = 0.036 in. ; Consider the reverse of P on the bar. FB ¿(3) FA ¿(2) = AE AE FA ¿ = 1.5 FB ¿ So that from Eq. (1) FB ¿ = 0.4P FA ¿ = 0.6P The resultant reactions are FA ¿¿ = FB ¿¿ = -62.832 + 0.6(125.66) = 62.832 - 0.4(125.66) = 12.568 kip When the supports are removed the elongation will be, d =
12.568(5)(12) PL = 0.0120 in. = AE p(1)2(20>0.001)
Ans.
208
P (in./in.)
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s
•4–113.
A material has a stress–strain diagram that can be described by the curve s = cP1>2. Determine the deflection d of the end of a rod made from this material if it has a length L, cross-sectional area A, and a specific weight g. L 1 2
A
s2 = c2 e
s = ce ;
s2(x) = c2e(x)
P
(1) d P(x) ; A
However s(x) =
e(x) =
dd dx
From Eq. (1), P2(x)
= c2
A2
P2(x) dd = dx A2c2
dd ; dx
L
d =
1 1 P2(x) dx = 2 2 (gAx)2 dx 2 2 Ac L A c L0 g2
=
d =
L
c2 L0
x2 dx =
g2 x3 L 冷 c2 3 0
g3L3
Ans.
3c2
4–114. The 2014-T6 aluminum rod has a diameter of 0.5 in. and is lightly attached to the rigid supports at A and B when T1 = 70°F. If the temperature becomes T2 = -10°F, and an axial force of P = 16 lb is applied to the rigid collar as shown, determine the reactions at A and B.
A
B P/2 P/2 5 in.
8 in.
+ 0 = ¢ - ¢ + d : B T B 0 =
0.016(5) p 2 3 4 (0.5 )(10.6)(10 )
- 12.8(10 - 6)[70° - (-10°)](13) +
FB(13) p 2 (0.5 )(10.6)(103) 4
FB = 2.1251 kip = 2.13 kip + ©F = 0; : x
Ans.
2(0.008) + 2.1251 - FA = 0 FA = 2.14 kip
Ans.
4–115. The 2014-T6 aluminum rod has a diameter of 0.5 in. and is lightly attached to the rigid supports at A and B when T1 = 70°F. Determine the force P that must be applied to the collar so that, when T = 0°F, the reaction at B is zero. + :
A P/2 5 in.
0 = ¢ B - ¢ T + dB 0 =
P(5) p 2 3 4 (0.5 )(10.6)(10 )
B P/2
- 12.8(10 - 6)[(70)(13)] + 0
P = 4.85 kip
Ans.
209
8 in.
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*4–116. The rods each have the same 25-mm diameter and 600-mm length. If they are made of A-36 steel, determine the forces developed in each rod when the temperature increases to 50° C.
C
600 mm 60⬚ B
A
60⬚
600 mm
Equation of Equilibrium: Referring to the free-body diagram of joint A shown in Fig. a, FAD sin 60° - FAC sin 60° = 0
+ c ©Fx = 0; + ©F = 0; : x
FAC = FAD = F
FAB - 2F cos 60° = 0 FAB = F
(1)
Compatibility Equation: If AB and AC are unconstrained, they will have a free expansion of A dT B AB = A dT B AC = ast ¢TL = 12(10 - 6)(50)(600) = 0.36 mm. Referring to the initial and final position of joint A, dFAB - A dT B AB = a dT ¿ b
AC
- dFAC ¿
Due to symmetry, joint A will displace horizontally, and dAC ¿ = a dT ¿ b
AC
dAC = 2dAC. Thus, cos 60°
= 2(dT)AC and dFAC ¿ = 2dFAC. Thus, this equation becomes
dFAB - A dT B AB = 2 A dT B AC - 2dAC FAB (600)
p 4
A 0.025 B (200)(10 ) 2
9
- 0.36 = 2(0.36) - 2 C
F(600)
p 4
A 0.0252 B (200)(109)
FAB + 2F = 176 714.59
S (2)
Solving Eqs. (1) and (2), FAB = FAC = FAD = 58 904.86 N = 58.9 kN
Ans.
210
D
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•4–117.
Two A-36 steel pipes, each having a crosssectional area of 0.32 in2, are screwed together using a union at B as shown. Originally the assembly is adjusted so that no load is on the pipe. If the union is then tightened so that its screw, having a lead of 0.15 in., undergoes two full turns, determine the average normal stress developed in the pipe. Assume that the union at B and couplings at A and C are rigid. Neglect the size of the union. Note: The lead would cause the pipe, when unloaded, to shorten 0.15 in. when the union is rotated one revolution.
B
A 3 ft
C 2 ft
The loads acting on both segments AB and BC are the same since no external load acts on the system. 0.3 = dB>A + dB>C 0.3 =
P(2)(12)
P(3)(12) 3
0.32(29)(10 )
+
0.32(29)(103)
P = 46.4 kip P 46.4 = = 145 ksi A 0.32
sAB = sBC =
Ans.
4–118. The brass plug is force-fitted into the rigid casting. The uniform normal bearing pressure on the plug is estimated to be 15 MPa. If the coefficient of static friction between the plug and casting is ms = 0.3, determine the axial force P needed to pull the plug out. Also, calculate the displacement of end B relative to end A just before the plug starts to slip out. Ebr = 98 GPa.
100 mm
B
15 MPa
P - 4.50(106)(2)(p)(0.02)(0.1) = 0 P = 56.549 kN = 56.5 kN
Ans.
Displacement: PL dB>A = a AE 0.1 m
56.549(103)(0.15) =
2
9
p(0.02 )(98)(10 )
+
L0
P
A
Equations of Equilibrium: + ©F = 0; : x
150 mm
0.56549(106) x dx p(0.022)(98)(109)
= 0.00009184 m = 0.0918 mm
Ans.
211
20 mm
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4–119. The assembly consists of two bars AB and CD of the same material having a modulus of elasticity E1 and coefficient of thermal expansion a1, and a bar EF having a modulus of elasticity E2 and coefficient of thermal expansion a2. All the bars have the same length L and cross-sectional area A. If the rigid beam is originally horizontal at temperature T1, determine the angle it makes with the horizontal when the temperature is increased to T2.
D
B
L
A
C
d
Equations of Equilibrium: a + ©MC = 0; + c ©Fy = 0;
FAB = FEF = F FCD - 2F = 0
[1]
Compatibility: dAB = (dAB)T - (dAB)F
dCD = (dCD)T + (dCD)F
dEF = (dEF)T - (dEF)F From the geometry dCD - dAB dEF - dAB = d 2d 2dCD = dEF + dAB 2 C (dCD)T + (dCD)F D = (dEF)T - (dEF)F + (dAB)T - (dAB)F 2 B a1 (T2 - T1) L + = a2 (T2 - T1) L -
FCD (L) R AE1 F(L) F(L) + a1 (T2 - T1) L AE2 AE1
[2]
Substitute Eq. [1] into [2]. 2a1 (T2 - T1) L +
4FL FL FL = a2 (T2 - T1)L + a1 (T2 - T1)L AE1 AE2 AE1
F 5F + = a2 (T2 - T1) - a1 (T2 - T1) AE1 AE2 F¢
5E2 + E1 b = (T2 - T1)(a2 - a1) ; AE1E2
F =
AE1E2 (T2 - T1)(a2 - a1) 5E2 + E1
(dEF)T = a2 (T2 - T1) L (dEF)F =
AE1E2 (T2 - T1)(a2 - a1)(L) E1 (T2 - T1)(a2 - a1)(L) = AE2 (5E2 + E1) 5E2 + E1
dEF = (dEF)T - (dEF)F =
a2 L(T2 - T1)(5E2 - E1) - E1L(T2 - T1)(a2 - a1) 5E2 + E1
(dAB)T = a1 (T2 - T1) L (dAB)F =
AE1E2 (T2 - T1)(a2 - a1)(L) E2 (T2 - T1)(a2 - a1)(L) = AE1 (5E2 + E1) 5E2 + E1
dAB = (dAB)T - (dAB)F =
F
a1 L(5E2 + E1)(T2 - T1) - E2 L(T2 - T1)(a2 - a1) 5E2 + E1
212
E
d
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4–119.
Continued
dEF - dAB =
L(T2 - T1) [a2 (5E2 + E1) - E1 (a2 - a1) - a1 (5E2 + E1) 5E2 + E1 + E2 (a2 - a1)]
=
L(T2 - T1) C (5E2 + E1)(a2 - a1) + (a2 - a1)(E2 - E1) D 5E2 + E1
=
L(T2 - T1)(a2 - a1) (5E2 + E1 + E2 - E1) 5E2 + E1
=
L(T2 - T1)(a2 - a1)(6E2) 5E2 + E1
u =
3E2L(T2 - T1)(a2 - a1) dEF - dAB = 2d d(5E2 + E1)
Ans.
*4–120. The rigid link is supported by a pin at A and two A-36 steel wires, each having an unstretched length of 12 in. and cross-sectional area of 0.0125 in2. Determine the force developed in the wires when the link supports the vertical load of 350 lb.
12 in. C 5 in. B
Equations of Equilibrium: a + ©MA = 0;
4 in. A
-FC(9) - FB (4) + 350(6) = 0
[1]
Compatibility:
6 in.
dC dB = 4 9
350 lb
FC(L) FB (L) = 4AE 9AE 9FB - 4FC = 0‚
[2]
Solving Eqs. [1] and [2] yields: FB = 86.6 lb
Ans.
FC = 195 lb
Ans.
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•5–1.
A shaft is made of a steel alloy having an allowable shear stress of tallow = 12 ksi. If the diameter of the shaft is 1.5 in., determine the maximum torque T that can be transmitted. What would be the maximum torque T¿ if a 1-in.-diameter hole is bored through the shaft? Sketch the shear-stress distribution along a radial line in each case.
T T¿
Allowable Shear Stress: Applying the torsion formula tmax = tallow =
12 =
Tc J T (0.75) p 2
(0.754)
T = 7.95 kip # in.
Ans.
Allowable Shear Stress: Applying the torsion formula tmax = tallow =
12 =
T¿c J T¿ (0.75) p 2
(0.754 - 0.54)
T¿ = 6.381 kip # in. = 6.38 kip # in. tr = 0.5 in =
T¿r = J
6.381(0.5) p 2
(0.754 - 0.54)
Ans.
= 8.00 ksi
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5–2. The solid shaft of radius r is subjected to a torque T. Determine the radius r¿ of the inner core of the shaft that resists one-half of the applied torque 1T>22. Solve the problem two ways: (a) by using the torsion formula, (b) by finding the resultant of the shear-stress distribution.
r¿ r
T
a)
tmax =
t =
Since t =
r¿ =
Tc Tr 2T = p 4 = J p r3 2 r
(T2 )r¿ p 2
(r¿)4
=
T p(r¿)3 T r¿ 2T = a b r pr3 p(r¿)3
r¿ t ; r max r 1
= 0.841 r
Ans.
24 r 2
b)
r¿
dT = 2p
L0 r 2
r¿
dT = 2p
L0 r 2
L0
tr2 dr
L0
r tmax r2 dr L0 r r¿
dT = 2p
r 2T 2 a 3 br dr L0 r pr
r¿
4T T = 4 r3 dr 2 r L0 r¿ =
r 1
Ans.
= 0.841r
24
215
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5–3. The solid shaft is fixed to the support at C and subjected to the torsional loadings shown. Determine the shear stress at points A and B and sketch the shear stress on volume elements located at these points.
10 kN⭈m C A 50 mm
The internal torques developed at Cross-sections pass through point B and A are shown in Fig. a and b, respectively. The polar moment of inertia of the shaft is J =
p (0.0754) = 49.70(10 - 6) m4. For 2
point B, rB = C = 0.075 Thus,
tB =
4(103)(0.075) TB c = 6.036(106) Pa = 6.04 MPa = J 49.70(10 - 6)
Ans.
From point A, rA = 0.05 m. tA =
TArA 6(103)(0.05) = 6.036(106) Pa = 6.04 MPa. = J 49.70 (10 - 6)
216
Ans.
B
75 mm 4 kN⭈m 75 mm
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*5–4. The tube is subjected to a torque of 750 N # m. Determine the amount of this torque that is resisted by the gray shaded section. Solve the problem two ways: (a) by using the torsion formula, (b) by finding the resultant of the shear-stress distribution.
75 mm
100 mm 750 Nm 25 mm
a) Applying Torsion Formula: tmax =
Tc = J
750(0.1) p 2
(0.14 - 0.0254)
tmax = 0.4793 A 106 B =
= 0.4793 MPa
T¿(0.1) p 2
(0.14 - 0.0754)
T¿ = 515 N # m
Ans.
b) Integration Method: r t = a b tmax c
dA = 2pr dr
and
dT¿ = rt dA = rt(2pr dr) = 2ptr2 dr 0.1m
T¿ =
L
2ptr2 dr = 2p
r tmax a br2 dr c L0.075m
=
0.1m 2ptmax r3 dr c L0.075m
=
2p(0.4793)(106) r4 0.1 m c d2 0.1 4 0.075 m
= 515 N # m
Ans.
5–5. The copper pipe has an outer diameter of 40 mm and an inner diameter of 37 mm. If it is tightly secured to the wall at A and three torques are applied to it as shown, determine the absolute maximum shear stress developed in the pipe.
tmax =
Tmax c = J
A
30 N⭈m
90(0.02) p 2
4
4
(0.02 - 0.0185 )
20 N⭈m
= 26.7 MPa
Ans.. 80 N⭈m
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5–6. The solid shaft has a diameter of 0.75 in. If it is subjected to the torques shown, determine the maximum shear stress developed in regions BC and DE of the shaft. The bearings at A and F allow free rotation of the shaft.
F E D C B
(tBC)max =
35(12)(0.375) TBC c = 5070 psi = 5.07 ksi = p 4 J 2 (0.375)
Ans.
(tDE)max =
25(12)(0.375) TDE c = 3621 psi = 3.62 ksi = p 4 J 2 (0.375)
Ans.
A
35 lb⭈ft
5–7. The solid shaft has a diameter of 0.75 in. If it is subjected to the torques shown, determine the maximum shear stress developed in regions CD and EF of the shaft. The bearings at A and F allow free rotation of the shaft.
F E D C B
(tEF)max =
TEF c = 0 J (tCD)max =
25 lb⭈ft 40 lb⭈ft 20 lb⭈ft
Ans.
A
25 lb⭈ft 40 lb⭈ft 20 lb⭈ft
35 lb⭈ft
15(12)(0.375) TCD c = p 4 J 2 (0.375)
= 2173 psi = 2.17 ksi
Ans.
218
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300 Nm
*5–8. The solid 30-mm-diameter shaft is used to transmit the torques applied to the gears. Determine the absolute maximum shear stress on the shaft.
500 Nm
A 200 Nm
Internal Torque: As shown on torque diagram. C
Maximum Shear Stress: From the torque diagram Tmax = 400 N # m. Then, applying torsion Formula.
400 Nm 300 mm
abs = tmax
Tmax c J 400(0.015)
=
p 2
(0.0154)
Ans.
= 75.5 MPa
The shaft consists of three concentric tubes, each made from the same material and having the inner and outer radii shown. If a torque of T = 800 N # m is applied to the rigid disk fixed to its end, determine the maximum shear stress in the shaft.
500 mm
T 800 Nm
ri 20 mm ro 25 mm
2m
p p p ((0.038)4 - (0.032)4) + ((0.030)4 - (0.026)4) + ((0.025)4 - (0.020)4) 2 2 2 -6
ri 26 mm ro 30 mm
4
J = 2.545(10 ) m tmax =
B
400 mm
•5–9.
J =
D
800(0.038) Tc = 11.9 MPa = J 2.545(10 - 6)
Ans.
219
ri 32 mm ro 38 mm
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5–10. The coupling is used to connect the two shafts together. Assuming that the shear stress in the bolts is uniform, determine the number of bolts necessary to make the maximum shear stress in the shaft equal to the shear stress in the bolts. Each bolt has a diameter d.
T R
r
n is the number of bolts and F is the shear force in each bolt. T
T F = nR
T - nFR = 0; T
tavg =
F 4T nR = p 2 = A ( 4 )d nRpd2
Maximum shear stress for the shaft: tmax =
Tc Tr 2T = p 4 = J pr3 2r 4T 2T = nRpd2 p r3
tavg = tmax ;
n =
2 r3 Rd2
Ans.
5–11. The assembly consists of two sections of galvanized steel pipe connected together using a reducing coupling at B. The smaller pipe has an outer diameter of 0.75 in. and an inner diameter of 0.68 in., whereas the larger pipe has an outer diameter of 1 in. and an inner diameter of 0.86 in. If the pipe is tightly secured to the wall at C, determine the maximum shear stress developed in each section of the pipe when the couple shown is applied to the handles of the wrench.
C
B
A
15 lb 6 in.
tAB =
tBC
Tc = J
Tc = = J
8 in.
210(0.375) p 2
(0.3754 - 0.344)
Ans. 15 lb
210(0.5) p 2
= 7.82 ksi
(0.54 - 0.434)
= 2.36 ksi
Ans.
220
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*5–12. The motor delivers a torque of 50 N # m to the shaft AB. This torque is transmitted to shaft CD using the gears at E and F. Determine the equilibrium torque Tⴕ on shaft CD and the maximum shear stress in each shaft. The bearings B, C, and D allow free rotation of the shafts.
A 50 mm 30 mm
Equilibrium:
B
a + ©ME = 0; a + ©MF = 0;
50 - F(0.05) = 0
F = 1000 N
35 mm T¿
T¿ - 1000(0.125) = 0 T¿ = 125 N # m
C
E
125 mm D
F
Ans.
Internal Torque: As shown on FBD. Maximum Shear Stress: Applying torsion Formula. (tAB)max =
50.0(0.015) TAB c = 9.43 MPa = p 4 J 2 (0.015 )
Ans.
(tCD)max =
125(0.0175) TCDc = 14.8 MPa = p 4 J 2 (0.0175 )
Ans.
•5–13. If the applied torque on shaft CD is T¿ = 75 N # m, determine the absolute maximum shear stress in each shaft. The bearings B, C, and D allow free rotation of the shafts, and the motor holds the shafts fixed from rotating.
A 50 mm
Equilibrium:
30 mm
a + ©MF = 0;
75 - F(0.125) = 0;
a + ©ME = 0;
600(0.05) - TA = 0
B
F = 600 N
35 mm T¿
TA = 30.0 N # m Internal Torque: As shown on FBD. Maximum Shear Stress: Applying the torsion formula (tEA)max =
30.0(0.015) TEA c = 5.66 MPa = p 4 J 2 (0.015 )
Ans.
(tCD)max =
75.0(0.0175) TCDc = 8.91 MPa = p 4 J 2 (0.0175 )
Ans.
221
C
E
125 mm D
F
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250 N⭈m
5–14. The solid 50-mm-diameter shaft is used to transmit the torques applied to the gears. Determine the absolute maximum shear stress in the shaft.
75 N⭈m
A
325 N⭈m 150 N⭈ m
B 500 mm
The internal torque developed in segments AB , BC and CD of the shaft are shown in Figs. a, b and c.
C D
400 mm 500 mm
The maximum torque occurs in segment AB. Thus, the absolute maximum shear stress occurs in this segment. The polar moment of inertia of the shaft is p J = (0.0254) = 0.1953p(10 - 6)m4. Thus, 2
A tmax B abs =
250(0.025) TAB c = 10.19(106)Pa = 10.2 MPa = J 0.1953p(10 - 6)
222
Ans.
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5–15. The solid shaft is made of material that has an allowable shear stress of tallow = 10 MPa. Determine the required diameter of the shaft to the nearest mm.
15 N⭈m 25 N⭈m A
30 N⭈m B
60 N⭈m C
70 N⭈m D E
The internal torques developed in each segment of the shaft are shown in the torque diagram, Fig. a. Segment DE is critical since it is subjected to the greatest internal torque. The polar p d 4 p 4 moment of inertia of the shaft is J = a b = d . Thus, 2 2 32
tallow
TDE c = ; J
d 70a b 2 10(106) = p 4 d 32 d = 0.03291 m = 32.91 mm = 33 mm
223
Ans.
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*5–16. The solid shaft has a diameter of 40 mm. Determine the absolute maximum shear stress in the shaft and sketch the shear-stress distribution along a radial line of the shaft where the shear stress is maximum.
15 N⭈m 25 N⭈m A
30 N⭈m B
The internal torque developed in each segment of the shaft are shown in the torque diagram, Fig. a.
60 N⭈m C
70 N⭈m D E
Since segment DE subjected to the greatest torque, the absolute maximum shear p stress occurs here. The polar moment of inertia of the shaft is J = (0.024) 2 = 80(10 - 9)p m4. Thus,
tmax =
70(0.02) TDE c = 5.57(106) Pa = 5.57 MPa = J 80(10 - 9)p
Ans.
The shear stress distribution along the radial line is shown in Fig. b.
•5–17.
The rod has a diameter of 1 in. and a weight of 10 lb/ft. Determine the maximum torsional stress in the rod at a section located at A due to the rod’s weight.
4.5 ft B
Here, we are only interested in the internal torque. Thus, other components of the internal loading are not indicated in the FBD of the cut segment of the rod, Fig. a. ©Mx = 0;
TA - 10(4)(2) = 0
TA = 80 lb # ft a
The polar moment of inertia of the cross section at A is J =
12in b = 960 lb # in. 1ft
p (0.54) = 0.03125p in4. 2
Thus
tmax =
960 (0.5) TA c = = 4889.24 psi = 4.89 ksi J 0.03125p
Ans.
224
4 ft
A 1.5 ft
1.5 ft
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5–18. The rod has a diameter of 1 in. and a weight of 15 lb/ft. Determine the maximum torsional stress in the rod at a section located at B due to the rod’s weight.
4.5 ft B
4 ft
Here, we are only interested in the internal torque. Thus, other components of the internal loading are not indicated in the FBD of the cut segment of the rod, Fig. a. ©Mx = 0;
TB - 15(4)(2) = 0
TB = 120 lb # ft a
12 in b = 1440 lb # in. 1ft
p The polar moment of inertia of the cross-section at B is J = (0.54) 2 = 0.03125p in4. Thus,
tmax =
1440(0.5) TB c = = 7333.86 psi = 7.33 ksi J 0.03125p
Ans.
225
A 1.5 ft
1.5 ft
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5–19. Two wrenches are used to tighten the pipe. If P = 300 N is applied to each wrench, determine the maximum torsional shear stress developed within regions AB and BC. The pipe has an outer diameter of 25 mm and inner diameter of 20 mm. Sketch the shear stress distribution for both cases.
P
B
Internal Loadings: The internal torque developed in segments AB and BC of the pipe can be determined by writing the moment equation of equilibrium about the x axis by referring to their respective free - body diagrams shown in Figs. a and b. ©Mx = 0; TAB - 300(0.25) = 0
TAB = 75 N # m
TBC = 150 N # m
Allowable Shear Stress: The polar moment of inertia of the pipe is p J = A 0.01254 - 0.014 B = 22.642(10 - 9)m4. 2
A tmax B AB =
75(0.0125) TAB c = 41.4 MPa = J 22.642(10 - 9)
A tAB B r = 0.01 m = A tmax B BC =
Ans.
TAB r 75(0.01) = 33.1 MPa = J 22.642(10 - 9)
150(0.0125) TBC c = 82.8 MPa = J 22.642(10 - 9)
A tBC B r = 0.01 m =
Ans.
TBC r 150(0.01) = 66.2 MPa = J 22.642(10 - 9)
The shear stress distribution along the radial line of segments AB and BC of the pipe is shown in Figs. c and d, respectively.
226
A 250 mm P
And ©Mx = 0; TBC - 300(0.25) - 300(0.25) = 0
250 mm
C
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*5–20. Two wrenches are used to tighten the pipe. If the pipe is made from a material having an allowable shear stress of tallow = 85 MPa, determine the allowable maximum force P that can be applied to each wrench. The pipe has an outer diameter of 25 mm and inner diameter of 20 mm.
P 250 mm
C B
A 250 mm
Internal Loading: By observation, segment BC of the pipe is critical since it is subjected to a greater internal torque than segment AB. Writing the moment equation of equilibrium about the x axis by referring to the free-body diagram shown in Fig. a, we have ©Mx = 0; TBC - P(0.25) - P(0.25) = 0
TBC = 0.5P
Allowable Shear Stress: The polar moment of inertia of the pipe is p J = A 0.01254 - 0.014 B = 22.642(10 - 9)m4 2 tallow =
TBC c ; J
85(106) =
0.5P(0.0125) 22.642(10 - 9)
P = 307.93N = 308 N
Ans.
227
P
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•5–21.
The 60-mm-diameter solid shaft is subjected to the distributed and concentrated torsional loadings shown. Determine the absolute maximum and minimum shear stresses on the outer surface of the shaft and specify their locations, measured from the fixed end A.
A
2 kN⭈m/m
1.5 m 1200 N⭈m C
The internal torque for segment BC is Constant TBC = 1200 N # m, Fig. a. However, the internal for segment AB varies with x, Fig. b. TAB - 2000x + 1200 = 0
TAB = (2000x - 1200) N # m
The minimum shear stress occurs when the internal torque is zero in segment AB. By setting TAB = 0, 0 = 2000x - 1200
x = 0.6 m
Ans.
And d = 1.5 m - 0.6 m = 0.9 m
Ans.
tmin = 0
Ans.
The maximum shear stress occurs when the internal torque is the greatest. This occurs at fixed support A where d = 0
Ans.
At this location, (TAB)max = 2000(1.5) - 1200 = 1800 N # m The polar moment of inertia of the rod is J =
tmax =
p (0.034) = 0.405(10 - 6)p. Thus, 2
(TAB)max c 1800(0.03) = 42.44(106)Pa = 42.4 MPa = J 0.405(10 - 6)p
228
Ans.
B 0.8 m
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5–22. The solid shaft is subjected to the distributed and concentrated torsional loadings shown. Determine the required diameter d of the shaft to the nearest mm if the allowable shear stress for the material is tallow = 50 MPa.
A
2 kN⭈m/m
1.5 m 1200 N⭈m C
The internal torque for segment BC is constant TBC = 1200 N # m, Fig. a. However, the internal torque for segment AB varies with x, Fig. b. TAB - 2000x + 1200 = 0 TAB = (2000x - 1200) N # m For segment AB, the maximum internal torque occurs at fixed support A where x = 1.5 m. Thus,
A TAB B max = 2000(1.5) - 1200 = 1800 N # m Since A TAB B max 7 TBC, the critical cross-section is at A. The polar moment of inertia p d 4 pd4 of the rod is J = . Thus, a b = 2 2 32 tallow =
Tc ; J
50(106) =
1800(d>2) pd4>32
d = 0.05681 m = 56.81 mm = 57 mm
229
Ans.
B 0.8 m
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*5–24. The copper pipe has an outer diameter of 2.50 in. and an inner diameter of 2.30 in. If it is tightly secured to the wall at C and a uniformly distributed torque is applied to it as shown, determine the shear stress developed at points A and B. These points lie on the pipe’s outer surface. Sketch the shear stress on volume elements located at A and B.
B A
C
125 lbft/ft
4 in. 9 in. 12 in.
Internal Torque: As shown on FBD. Maximum Shear Stress: Applying the torsion formula tA =
TA c J 125.0(12)(1.25)
=
tB =
p 2
(1.254 - 1.154)
Ans.
= 3.02 ksi
Ans.
TB c J 218.75(12)(1.25)
=
= 1.72 ksi
p 2
(1.254 - 1.154)
•5–25.
The copper pipe has an outer diameter of 2.50 in. and an inner diameter of 2.30 in. If it is tightly secured to the wall at C and it is subjected to the uniformly distributed torque along its entire length, determine the absolute maximum shear stress in the pipe. Discuss the validity of this result.
B A
C
125 lbft/ft
4 in. 9 in.
Internal Torque: The maximum torque occurs at the support C. Tmax = (125 lb # ft>ft)a
12 in.
25 in. b = 260.42 lb # ft 12 in.>ft
Maximum Shear Stress: Applying the torsion formula abs = tmax
Tmax c J 260.42(12)(1.25)
=
p 2
(1.254 - 1.154)
Ans.
= 3.59 ksi
According to Saint-Venant’s principle, application of the torsion formula should be as points sufficiently removed from the supports or points of concentrated loading.
230
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5–26. A cylindrical spring consists of a rubber annulus bonded to a rigid ring and shaft. If the ring is held fixed and a torque T is applied to the shaft, determine the maximum shear stress in the rubber.
ro
ri
T h
T r
T F t = = = A 2prh 2p r2 h Shear stress is maximum when r is the smallest, i.e. r = ri. Hence, tmax =
T 2p ri 2 h
Ans.
300 N⭈m
5–27. The A-36 steel shaft is supported on smooth bearings that allow it to rotate freely. If the gears are subjected to the torques shown, determine the maximum shear stress developed in the segments AB and BC. The shaft has a diameter of 40 mm.
100 N⭈m A
The internal torque developed in segments AB and BC are shown in their respective FBDs, Figs. a and b. The polar moment of inertia of the shaft is J =
A tAB B max
200 N⭈m B
p (0.024) = 80(10-9)p m4. Thus, 2 C
300(0.02) TAB c = 23.87(106)Pa = 23.9 MPa = = J 80(10-9)p
A tBC B max =
200(0.02) TBC c = 15.92(106) Pa = 15.9 MPa = J 80(10-9)p
231
Ans.
Ans.
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300 N⭈m
*5–28. The A-36 steel shaft is supported on smooth bearings that allow it to rotate freely. If the gears are subjected to the torques shown, determine the required diameter of the shaft to the nearest mm if tallover = 60 MPa.
100 N⭈m
The internal torque developed in segments AB and BC are shown in their respective FBDs, Fig. a and b
A 200 N⭈m B
Here, segment AB is critical since its internal torque is the greatest. The polar p d 4 pd4 moment of inertia of the shaft is J = . Thus, a b = 2 2 32 C
tallow
TC = ; J
60(106) =
300(d>2) pd4>32
d = 0.02942 m = 30 mm
Ans.
•5–29.
When drilling a well at constant angular velocity, the bottom end of the drill pipe encounters a torsional resistance TA . Also, soil along the sides of the pipe creates a distributed frictional torque along its length, varying uniformly from zero at the surface B to tA at A. Determine the minimum torque TB that must be supplied by the drive unit to overcome the resisting torques, and compute the maximum shear stress in the pipe. The pipe has an outer radius ro and an inner radius ri . TA +
TB B
L
1 t L - TB = 0 2 A tA
2TA + tAL TB = 2
Ans.
Maximum shear stress: The maximum torque is within the region above the distributed torque. tmax =
tmax =
Tc J (2TA + tAL) ] (r0) 2 p 4 4 (r r i) 2 0
[
(2TA + tAL)r0 =
Ans.
p(r40 - r4i )
232
A
TA
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5–30. The shaft is subjected to a distributed torque along its length of t = 110x22 N # m>m, where x is in meters. If the maximum stress in the shaft is to remain constant at 80 MPa, determine the required variation of the radius c of the shaft for 0 … x … 3 m. x
x
T =
L
t dx =
Tc t = ; J
L0
10 x2dx = 6
80(10 ) =
3m c
10 3 x 3 t ⫽ (10x2) N⭈m/m
3 (10 3 )x c p 2
c4
c3 = 26.526(10-9) x3 c = (2.98 x) mm
Ans.
5–31. The solid steel shaft AC has a diameter of 25 mm and is supported by smooth bearings at D and E. It is coupled to a motor at C, which delivers 3 kW of power to the shaft while it is turning at 50 rev>s. If gears A and B remove 1 kW and 2 kW, respectively, determine the maximum shear stress developed in the shaft within regions AB and BC. The shaft is free to turn in its support bearings D and E.
TC =
3(103) P = = 9.549 N # m v 50(2p)
TA =
1 T = 3.183 N # m 3 C
3 kW
2 kW 25 mm
1 kW
A D
(tAB)max =
3.183 (0.0125) TC = 1.04 MPa = p 4 J 2 (0.0125 )
Ans.
(tBC)max =
9.549 (0.0125) TC = 3.11 MPa = p 4 J 2 (0.0125 )
Ans.
233
B
E
C
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*5–32. The pump operates using the motor that has a power of 85 W. If the impeller at B is turning at 150 rev>min, determine the maximum shear stress developed in the 20-mm-diameter transmission shaft at A.
150 rev/min A
Internal Torque: v = 150
rev 2p rad 1 min = 5.00p rad>s ¢ ≤ min rev 60 s
P = 85 W = 85 N # m>s T =
P 85 = = 5.411 N # m v 5.00p
Maximum Shear Stress: Applying torsion formula tmax =
Tc J 5.411 (0.01)
=
p 4 2 (0.01 )
= 3.44 MPa
Ans.
•5–33.
The gear motor can develop 2 hp when it turns at 450 rev>min. If the shaft has a diameter of 1 in., determine the maximum shear stress developed in the shaft. The angular velocity of the shaft is v = ¢ 450
rev 2p rad 1 min ≤ ¢ ≤ ¢ ≤ = 15p rad>s min 1 rev 60 s
and the power is P = 2 hp ¢
550 ft # lb>s ≤ = 1100 ft # lb>s 1 hp
Then T =
P 1100 12 in = = 23.34 lb # ft a b = 280.11 lb # in v 15p 1ft
The polar moment of inertia of the shaft is J =
tmax =
p (0.54) = 0.03125p in4. Thus, 2
280.11 (0.5) Tc = = 1426.60 psi = 1.43 ksi J 0.03125p
Ans.
234
B
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5–34. The gear motor can develop 3 hp when it turns at 150 rev>min. If the allowable shear stress for the shaft is tallow = 12 ksi, determine the smallest diameter of the shaft to the nearest 18 in. that can be used. The angular velocity of the shaft is v = a 150
rev 2p rad 1 min ba ba b = 5p rad>s min 1 rev 60 s
and the power is P = (3 hp) a
550 ft # lb>s b = 1650 ft # lb>s 1 hp
Then T =
P 1650 12 in = = (105.04 lb # ft)a b = 1260.51 lb # in v 5p 1 ft
The polar moment of inertia of the shaft is J =
tallow =
Tc ; J
12(103) =
p d 4 pd4 a b = . Thus, 2 2 32
1260.51 (d>2) pd4>32
d = 0.8118 in. =
7 in. 8
Ans.
5–35. The 25-mm-diameter shaft on the motor is made of a material having an allowable shear stress of tallow = 75 MPa . If the motor is operating at its maximum power of 5 kW, determine the minimum allowable rotation of the shaft. Allowable Shear Stress: The polar moment of inertia of the shaft is p J = A 0.01254 B = 38.3495(10-9) m4. 2 tallow =
Tc ; J
75(106) =
T(0.0125) 38.3495(10-9)
T = 230.10 N # m Internal Loading: T =
P ; v
230.10 =
5(103) v
v = 21.7 rad>s
Ans.
235
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*5–36. The drive shaft of the motor is made of a material having an allowable shear stress of tallow = 75 MPa. If the outer diameter of the tubular shaft is 20 mm and the wall thickness is 2.5 mm, determine the maximum allowable power that can be supplied to the motor when the shaft is operating at an angular velocity of 1500 rev>min. Internal Loading: The angular velocity of the shaft is v = a 1500
rev 2p rad 1 min ba ba b = 50p rad>s min 1 rev 60 s
We have T =
P P = v 50p
Allowable Shear Stress: The polar moment of inertia of the shaft is p J = A 0.014 - 0.00754 B = 10.7379(10-9) m4. 2
tallow =
Tc ; J
75(106) =
a
P b(0.01) 50p
10.7379(10-9)
P = 12 650.25 W = 12.7 kW
Ans.
•5–37.
A ship has a propeller drive shaft that is turning at 1500 rev>min while developing 1800 hp. If it is 8 ft long and has a diameter of 4 in., determine the maximum shear stress in the shaft caused by torsion.
Internal Torque: v = 1500
rev 2p rad 1 min a b = 50.0 p rad>s min 1 rev 60 s
P = 1800 hpa
T =
550 ft # lb>s b = 990 000 ft # lb>s 1 hp
990 000 P = = 6302.54 lb # ft v 50.0p
Maximum Shear Stress: Applying torsion formula tmax =
6302.54(12)(2) Tc = p 4 J 2 (2 ) = 6018 psi = 6.02 ksi
Ans.
236
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5–38. The motor A develops a power of 300 W and turns its connected pulley at 90 rev>min. Determine the required diameters of the steel shafts on the pulleys at A and B if the allowable shear stress is tallow = 85 MPa.
60 mm 90 rev/min
A B
150 mm
Internal Torque: For shafts A and B vA = 90
rev 2p rad 1 min a b = 3.00p rad>s min rev 60 s
P = 300 W = 300 N # m>s P 300 = = 31.83 N # m vA 3.00p
TA =
vB = vA a
rA 0.06 b = 3.00pa b = 1.20p rad>s rB 0.15
P = 300 W = 300 N # m>s
TB =
P 300 = = 79.58 N # m vB 1.20p
Allowable Shear Stress: For shaft A tmax = tallow = 85 A 106 B =
TA c J 31.83 A d2A B
A B
p dA 4 2 2
dA = 0.01240 m = 12.4 mm
Ans.
For shaft B tmax = tallow = 85 A 106 B =
TB c J 79.58 A d2B B
A B
p dB 4 2 2
dB = 0.01683 m = 16.8 mm
Ans.
237
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5–39. The solid steel shaft DF has a diameter of 25 mm and is supported by smooth bearings at D and E. It is coupled to a motor at F, which delivers 12 kW of power to the shaft while it is turning at 50 rev>s. If gears A, B, and C remove 3 kW, 4 kW, and 5 kW respectively, determine the maximum shear stress developed in the shaft within regions CF and BC. The shaft is free to turn in its support bearings D and E.
v = 50
3 kW 4 kW A
B
D
C
E
F
rev 2p rad c d = 100 p rad>s s rev
TF =
12(103) P = = 38.20 N # m v 100 p
TA =
3(103) P = = 9.549 N # m v 100 p
TB =
4(103) P = = 12.73 N # m v 100 p
(tmax)CF =
38.20(0.0125) TCF c = 12.5 MPa = p 4 J 2 (0.0125 )
Ans.
(tmax)BC =
22.282(0.0125) TBC c = 7.26 MPa = p 4 J 2 (0.0125 )
Ans.
*5–40. Determine the absolute maximum shear stress developed in the shaft in Prob. 5–39.
v = 50
12 kW
5 kW 25 mm
3 kW 4 kW
rev 2p rad c d = 100 p rad>s s rev
TF =
12(103) P = = 38.20 N # m v 100p
TA =
3(103) P = = 9.549 N # m v 100p
TB =
4(103) P = = 12.73 N # m v 100p
A D
Tmax = 38.2 N # m 38.2(0.0125) = 12.5 MPa tabs = Tc = max p 4 J 2 (0.0125 )
Ans.
238
B C
From the torque diagram,
12 kW
5 kW 25 mm
E
F
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•5–41.
The A-36 steel tubular shaft is 2 m long and has an outer diameter of 50 mm. When it is rotating at 40 rad> s, it transmits 25 kW of power from the motor M to the pump P. Determine the smallest thickness of the tube if the allowable shear stress is tallow = 80 MPa.
P
M
P
M
The internal torque in the shaft is T =
25(103) P = = 625 N # m v 40
The polar moment of inertia of the shaft is J =
tallow =
Tc ; J
80(106) =
p (0.0254 - Ci 4). Thus, 2
625(0.025) p 4 2 (0.025
- Ci 4)
Ci = 0.02272 m So that t = 0.025 - 0.02272 = 0.002284 m = 2.284 mm = 2.5 mm
Ans.
5–42. The A-36 solid tubular steel shaft is 2 m long and has an outer diameter of 60 mm. It is required to transmit 60 kW of power from the motor M to the pump P. Determine the smallest angular velocity the shaft can have if the allowable shear stress is tallow = 80 MPa. The polar moment of inertia of the shaft is J =
tallow =
Tc ; J
80(106) =
p (0.034) = 0.405(10-6)p m4. Thus, 2
T(0.03) 0.405(10-6)p
T = 3392.92 N # m P = Tv ;
60(103) = 3392.92 v v = 17.68 rad>s = 17.7 rad>s
Ans.
239
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5–43. A steel tube having an outer diameter of 2.5 in. is used to transmit 35 hp when turning at 2700 rev>min. Determine the inner diameter d of the tube to the nearest 1 8 in. if the allowable shear stress is tallow = 10 ksi.
v =
2700(2p) = 282.74 rad>s 60
d 2.5 in.
P = Tv 35(550) = T(282.74) T = 68.083 lb # ft tmax = tallow = 10(103) =
Tc J 68.083(12)(1.25) p 4 2 (12.5
- ci 4)
ci = 1.2416 in. d = 2.48 in. Use d = 212 in.
Ans.
*5–44. The drive shaft AB of an automobile is made of a steel having an allowable shear stress of tallow = 8 ksi. If the outer diameter of the shaft is 2.5 in. and the engine delivers 200 hp to the shaft when it is turning at 1140 rev>min, determine the minimum required thickness of the shaft’s wall.
v =
B
1140(2p) = 119.38 rad>s 60
P = Tv 200(550) = T(119.38) T = 921.42 lb # ft tallow = 8(103) =
Tc J 921.42(12)(1.25) p 4 2 (1.25
- r4i )
,
ri = 1.0762 in.
t = ro - ri = 1.25 - 1.0762 t = 0.174 in.
Ans.
240
A
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•5–45.
The drive shaft AB of an automobile is to be designed as a thin-walled tube. The engine delivers 150 hp when the shaft is turning at 1500 rev>min. Determine the minimum thickness of the shaft’s wall if the shaft’s outer diameter is 2.5 in. The material has an allowable shear stress of tallow = 7 ksi.
v =
B
A
1500(2p) = 157.08 rad>s 60 P = Tv 150(550) = T(157.08) T = 525.21 lb # ft
tallow = 7(103) =
Tc J 525.21(12)(1.25) p 4 2 (1.25
- r4i )
ri = 1.1460 in.
,
t = ro - ri = 1.25 - 1.1460 t = 0.104 in.
Ans.
5–46. The motor delivers 15 hp to the pulley at A while turning at a constant rate of 1800 rpm. Determine to the nearest 18 in. the smallest diameter of shaft BC if the allowable shear stress for steel is tallow = 12 ksi. The belt does not slip on the pulley.
B
C 3 in.
The angular velocity of shaft BC can be determined using the pulley ratio that is vBC
1.5 in.
rA 1.5 rev 2p rad 1 min = a b vA = a b a1800 ba ba b = 30p rad>s rC 3 min 1 rev 60 s
A
The power is P = (15 hp) a
550 ft # n>s b = 8250 ft # lb>s 1 hp
Thus, T =
P 8250 12 in. = = (87.54 lb # ft)a b = 1050.42 lb # in v 30p 1 ft
The polar moment of inertia of the shaft is J =
tallow =
Tc ; J
12(103) =
p d 4 pd4 a b = . Thus, 2 2 32
1050.42(d>2) pd4>32
d = 0.7639 in =
7 in. 8
Ans.
241
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5–47. The propellers of a ship are connected to a A-36 steel shaft that is 60 m long and has an outer diameter of 340 mm and inner diameter of 260 mm. If the power output is 4.5 MW when the shaft rotates at 20 rad>s, determine the maximum torsional stress in the shaft and its angle of twist.
T =
4.5(106) P = = 225(103) N # m v 20
tmax =
f =
225(103)(0.170) Tc = 44.3 MPa = p J [(0.170)4 - (0.130)4] 2
Ans.
225 A 103 B (60) TL = 0.2085 rad = 11.9° = p JG [(0.170)4 - (0.130)4)75(109) 2
Ans.
*5–48. A shaft is subjected to a torque T. Compare the effectiveness of using the tube shown in the figure with that of a solid section of radius c. To do this, compute the percent increase in torsional stress and angle of twist per unit length for the tube versus the solid section.
T c 2
T
c
Shear stress: For the tube, (tt)max =
c
Tc Jt
For the solid shaft, (ts)max =
Tc Js
% increase in shear stress =
=
(ts)max - (tt)max (100) = (tt)max Js - Jt (100) = Jt
p 2
Tc Jt
-
Tc Js
Tc Js
(100)
c4 - [p2 [c4 - (p2 )4]] p 2
[c4 - (p2 )4]
(100) Ans.
= 6.67 % Angle of twist: For the tube, ft =
TL Jt(G)
For the shaft, fs =
TL Js(G)
% increase in f =
ft - fs (100%) = fs
=
Js - Jt (100%) = Jt
TL Jt(G)
-
TL Js(G)
TL Js(G) p 2
(100%)
c4 - [p2 [c4 - (p2 )4]] p 2
[c4 - (p2 )4]
(100%)
= 6.67 %
Ans. 242
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•5–49.
The A-36 steel axle is made from tubes AB and CD and a solid section BC. It is supported on smooth bearings that allow it to rotate freely. If the gears, fixed to its ends, are subjected to 85-N # m torques, determine the angle of twist of gear A relative to gear D. The tubes have an outer diameter of 30 mm and an inner diameter of 20 mm. The solid section has a diameter of 40 mm.
400 mm 250 mm 400 mm B
fND = ©
TL JG
A
(85)(0.25)
2(85)(0.4) =
p 2
4
4
9
(0.015 - 0.01 )(75)(10 )
+
p 2
85 Nm
(0.024)(75)(109)
= 0.01534 rad = 0.879°
Ans.
5–50. The hydrofoil boat has an A-36 steel propeller shaft that is 100 ft long. It is connected to an in-line diesel engine that delivers a maximum power of 2500 hp and causes the shaft to rotate at 1700 rpm. If the outer diameter of the shaft is 8 in. and the wall thickness is 38 in., determine the maximum shear stress developed in the shaft. Also, what is the “wind up,” or angle of twist in the shaft at full power?
100 ft
Internal Torque: v = 1700
rev 2p rad 1 min a b = 56.67p rad>s min rev 60 s
P = 2500 hp a T =
550 ft # lb>s b = 1 375 000 ft # lb>s 1 hp
P 1 375 000 = = 7723.7 lb # ft v 56.67p
Maximum Shear Stress: Applying torsion Formula. tmax =
Tc J 7723.7(12)(4)
=
p 2
(44 - 3.6254)
Ans.
= 2.83 ksi
Angle of Twist: f =
TL = JG
7723.7(12)(100)(12) p 2
(44 - 3.6254)11.0(106)
= 0.07725 rad = 4.43°
Ans.
243
C
D 85 Nm
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5–51. The engine of the helicopter is delivering 600 hp to the rotor shaft AB when the blade is rotating at 1200 rev>min. Determine to the nearest 18 in. the diameter of the shaft AB if the allowable shear stress is tallow = 8 ksi and the vibrations limit the angle of twist of the shaft to 0.05 rad. The shaft is 2 ft long and made from L2 steel. A
B
1200(2)(p) = 125.66 rad>s v = 60 P = Tv 600(550) = T(125.66) T = 2626.06 lb # ft Shear - stress failure tallow = 8(103) =
Tc J
2626.06(12)c p 2
c4
c = 1.3586 in. Angle of twist limitation f =
0.05 =
TL JG 2626.06(12)(2)(12) p 2
c4(11.0)(106)
c = 0.967 in. Shear - stress failure controls the design. d = 2c = 2 (1.3586) = 2.72 in. Use d = 2.75 in.
Ans.
244
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*5–52. The engine of the helicopter is delivering 600 hp to the rotor shaft AB when the blade is rotating at 1200 rev>min. Determine to the nearest 18 in. the diameter of the shaft AB if the allowable shear stress is tallow = 10.5 ksi and the vibrations limit the angle of twist of the shaft to 0.05 rad. The shaft is 2 ft long and made from L2 steel. A
v =
B
1200(2)(p) = 125.66 rad>s 60 P = Tv
600(550) = T(125.66) T = 2626.06 lb # ft Shear - stress failure tallow = 10.5(10)3 =
2626.06(12)c p 2
c4
c = 1.2408 in. Angle of twist limitation f =
0.05 =
TL JG 2626.06(12)(2)(12) p 2
c4 (11.0)(106)
c = 0.967 in. Shear stress failure controls the design d = 2c = 2 (1.2408) = 2.48 in. Use d = 2.50 in.
Ans.
245
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•5–53. The 20-mm-diameter A-36 steel shaft is subjected to the torques shown. Determine the angle of twist of the end B.
A
Internal Torque: As shown on FBD.
D
Angle of Twist:
C B
TL fB = a JG
30 Nm 600 mm
200 mm
20 Nm 800 mm
1 [-80.0(0.8) + (-60.0)(0.6) + (-90.0)(0.2)] = p 4 9 (0.01 )(75.0)(10 ) 2 = -0.1002 rad = | 5.74° |
80 Nm
Ans.
5–54. The assembly is made of A-36 steel and consists of a solid rod 20 mm in diameter fixed to the inside of a tube using a rigid disk at B. Determine the angle of twist at D. The tube has an outer diameter of 40 mm and wall thickness of 5 mm.
A
B
The internal torques developed in segments AB and BD of the assembly are shown in Fig. a and b
0.4 m
C 0.1 m
p The polar moment of inertia of solid rod and tube are JAB = (0.024 - 0.0154) 2 p = 54.6875(10 - 9)p m4 and JBD = (0.014) = 5(10 - 9)p m4. Thus, 2 fD = ©
Ti Li TAB LAB TBD LBD = + Ji Gi JAB Gst JBD Gst -60 (0.4)
90(0.4) =
54.6875(10 - 9)p [75(109)]
+
5(10 - 9)p [75(109)]
= -0.01758 rad = 1.01°
Ans.
246
150 N⭈m
D 0.3 m 60 N⭈m
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5–55. The assembly is made of A-36 steel and consists of a solid rod 20 mm in diameter fixed to the inside of a tube using a rigid disk at B. Determine the angle of twist at C. The tube has an outer diameter of 40 mm and wall thickness of 5 mm.
A
B
The polar moment of inertia = 54.6875 (10 - 9)p m4. Thus,
fC = ©
of
the
tube
is
J =
150 N⭈m
0.4 m
The internal torques developed in segments AB and BC of the assembly are shown in Figs. a and b.
C 0.1 m
p (0.024 - 0.0154) 2
D 0.3 m 60 N⭈m
Ti Li TAB LAB TBC LBC = + Ji Gi JGst J Gst =
1 C 90(0.4) + 150(0.1) D 54.6875(10 )p [75(109)] -9
= 0.003958 rad = 0.227°
Ans.
*5–56. The splined ends and gears attached to the A-36 steel shaft are subjected to the torques shown. Determine the angle of twist of end B with respect to end A. The shaft has a diameter of 40 mm.
fB>A = ©
300 N⭈m
500 N⭈m
A
200 N⭈m
-300(0.3) 200(0.4) 400(0.5) TL = + + JG JG JG JG
C
400 N⭈m
190 = = JG
300 mm
190 p 4 (0.02 )(75)(109) 2
D B
400 mm
= 0.01008 rad = 0.578°
Ans. 500 mm
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•5–57.
The motor delivers 40 hp to the 304 stainless steel shaft while it rotates at 20 Hz. The shaft is supported on smooth bearings at A and B, which allow free rotation of the shaft. The gears C and D fixed to the shaft remove 25 hp and 15 hp, respectively. Determine the diameter of the shaft to the nearest 18 in. if the allowable shear stress is tallow = 8 ksi and the allowable angle of twist of C with respect to D is 0.20°.
External Applied Torque: Applying T =
TM =
40(550) = 175.07 lb # ft 2p(20)
TD =
15(550) = 65.65 lb # ft 2p(20)
A
D 10 in.
6 in.
25(550) = 109.42 lb # ft 2p(20)
Internal Torque: As shown on FBD. Allowable Shear Stress: Assume failure due to shear stress. By observation, section AC is the critical region. Tc J
tmax = tallow =
175.07(12) A d2 B
8(103) =
p 2
A d2 B
4
d = 1.102 in.
Angle of Twist: Assume failure due to angle of twist limitation. fC>D = 0.2(p) = 180
TCDLCD JG 65.65(12)(8) p 2
B 8 in.
P , we have 2pf TC =
C
A d2 B (11.0)(106) 4
d = 1.137 in. (controls !) 1 Use d = 1 in. 4
Ans.
248
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5–58. The motor delivers 40 hp to the 304 stainless steel solid shaft while it rotates at 20 Hz. The shaft has a diameter of 1.5 in. and is supported on smooth bearings at A and B, which allow free rotation of the shaft. The gears C and D fixed to the shaft remove 25 hp and 15 hp, respectively. Determine the absolute maximum stress in the shaft and the angle of twist of gear C with respect to gear D.
A
C D
10 in.
B 8 in. 6 in.
External Applied Torque: Applying T =
TM =
40(550) = 175.07 lb # ft 2p(20)
TD =
15(550) = 65.65 lb # ft 2p(20)
P , we have 2pf TC =
25(550) = 109.42 lb # ft 2p(20)
Internal Torque: As shown on FBD. Allowable Shear Stress: The maximum torque occurs within region AC of the shaft where Tmax = TAC = 175.07 lb # ft. abs = tmax
175.07(12)(0.75) Tmax c = 3.17 ksi = p 4 J 2 (0.75 )
Ans.
Angle of Twist: fC>D =
TCD LCD JG 65.65(12)(8)
=
p 2
(0.754)(11.0)(106)
= 0.001153 rad = 0.0661°
Ans.
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5–59. The shaft is made of A-36 steel. It has a diameter of 1 in. and is supported by bearings at A and D, which allow free rotation. Determine the angle of twist of B with respect to D.
A
B
60 lb⭈ft C
2 ft 60 lb⭈ft
2.5 ft
The internal torques developed in segments BC and CD are shown in Figs. a and b.
D 3 ft
p The polar moment of inertia of the shaft is J = (0.54) = 0.03125p in4. Thus, 2 TiLi TBC LBC TCD LCD = + FB/D = a JiGi J Gst J Gst -60(12)(2.5)(12) =
(0.03125p)[11.0(106)]
+ 0
= -0.02000 rad = 1.15°
Ans.
*5–60. The shaft is made of A-36 steel. It has a diameter of 1 in. and is supported by bearings at A and D, which allow free rotation. Determine the angle of twist of gear C with respect to B.
A
B
60 lb⭈ft C
2 ft 60 lb⭈ft
2.5 ft
The internal torque developed in segment BC is shown in Fig. a
D 3 ft
p The polar moment of inertia of the shaft is J = (0.54) = 0.03125p in4. Thus, 2 fC>B =
-60(12)(2.5)(12) TBC LBC = J Gst (0.03125p)[11.0(106)] = -0.02000 rad = 1.15°
Ans.
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•5–61.
The two shafts are made of A-36 steel. Each has a diameter of 1 in., and they are supported by bearings at A, B, and C, which allow free rotation. If the support at D is fixed, determine the angle of twist of end B when the torques are applied to the assembly as shown.
D
10 in.
C
80 lbft A
30 in.
40 lbft
8 in. 10 in. 12 in.
Internal Torque: As shown on FBD. Angle of Twist: TL fE = a JG 1 =
p 2
(0.54)(11.0)(105)
[-60.0(12)(30) + 20.0(12)(10)]
= -0.01778 rad = 0.01778 rad fF =
6 6 f = (0.01778) = 0.02667 rad 4 E 4
Since there is no torque applied between F and B then fB = fF = 0.02667 rad = 1.53°
Ans.
251
4 in.
6 in. B
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5–62. The two shafts are made of A-36 steel. Each has a diameter of 1 in., and they are supported by bearings at A, B, and C, which allow free rotation. If the support at D is fixed, determine the angle of twist of end A when the torques are applied to the assembly as shown.
D
10 in.
C
80 lbft A
30 in.
40 lbft
8 in. 10 in. 12 in.
Internal Torque: As shown on FBD. Angle of Twist: TL fE = a JG 1 =
p 2
(0.54)(11.0)(106)
[-60.0(12)(30) + 20.0(12)(10)]
= -0.01778 rad = 0.01778 rad
fF =
6 6 f = (0.01778) = 0.02667 rad 4 E 4
fA>F =
TGF LGF JG -40(12)(10)
=
p 2
(0.54)(11.0)(106)
= -0.004445 rad = 0.004445 rad fA = fF + fA>F = 0.02667 + 0.004445 = 0.03111 rad = 1.78°
Ans.
252
4 in.
6 in. B
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5–63. The device serves as a compact torsional spring. It is made of A-36 steel and consists of a solid inner shaft CB which is surrounded by and attached to a tube AB using a rigid ring at B. The ring at A can also be assumed rigid and is fixed from rotating. If a torque of T = 2 kip # in. is applied to the shaft, determine the angle of twist at the end C and the maximum shear stress in the tube and shaft.
12 in. 12 in.
B
T
1 in. A
0.5 in. C
Internal Torque: As shown on FBD. Maximum Shear Stress: (tBC)max =
2.00(0.5) TBC c = 10.2 ksi = p 4 J 2 (0.5 )
TBA c = J
(tBA)max =
2.00(1) p 2
(14 - 0.754)
Ans.
= 1.86 ksi
Ans.
Angle of Twist: fB =
TBA LBA JG (2.00)(12)
=
p 2
4
(1 - 0.754)11.0(103)
fC>B =
TBC LBC JG 2.00(24)
=
= 0.002032 rad
p 2
(0.54)11.0(103)
= 0.044448 rad
fC = fB + fC>B = 0.002032 + 0.044448 = 0.04648 rad = 2.66°
Ans.
253
0.75 in.
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*5–64. The device serves as a compact torsion spring. It is made of A-36 steel and consists of a solid inner shaft CB which is surrounded by and attached to a tube AB using a rigid ring at B. The ring at A can also be assumed rigid and is fixed from rotating. If the allowable shear stress for the material is tallow = 12 ksi and the angle of twist at C is limited to fallow = 3°, determine the maximum torque T that can be applied at the end C.
12 in. 12 in.
B
T
1 in. A
0.5 in. C
Internal Torque: As shown on FBD. Allowable Shear Stress: Assume failure due to shear stress. tmax = tallow =
12.0 =
TBC c J T (0.5) p 2
(0.54)
T = 2.356 kip # in tmax = tallow =
12.0 =
TBA c J T (1) p 2
(14 - 0.754)
T = 12.89 kip # in Angle of Twist: Assume failure due to angle of twist limitation. fB =
TBA LBA = JG
T(12) p 2
(14 - 0.754) 11.0(103)
= 0.001016T fC>B =
TBC LBC = JG
T(24) p 2
(0.54)11.0(103)
= 0.022224T (fC)allow = fB + fC>B 3(p) = 0.001016T + 0.022224T 180 T = 2.25 kip # in (controls !)
Ans.
254
0.75 in.
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•5–65.
The A-36 steel assembly consists of a tube having an outer radius of 1 in. and a wall thickness of 0.125 in. Using a rigid plate at B, it is connected to the solid 1-in-diameter shaft AB. Determine the rotation of the tube’s end C if a torque of 200 lb # in. is applied to the tube at this end. The end A of the shaft is fixed supported.
B C 200 lb⭈in.
4 in.
A 6 in.
fB =
TABL = JG
fC>B =
200(10) p 2
TCBL = JG
(0.5)4(11.0)(106)
= 0.001852 rad
-200(4) p 2
4
(1 - 0.8754)(11.0)(106)
= -0.0001119 rad
fC = fB + fC>B = 0.001852 + 0.0001119 = 0.001964 rad = 0.113°
Ans.
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5–66. The 60-mm diameter shaft ABC is supported by two journal bearings, while the 80-mm diameter shaft EH is fixed at E and supported by a journal bearing at H. If T1 = 2 kN # m and T2 = 4 kN # m, determine the angle of twist of gears A and C. The shafts are made of A-36 steel.
E A
600 mm D 100 mm H
T2 600 mm B 75 mm 900 mm
Equilibrium: Referring to the free - body diagram of shaft ABC shown in Fig. a ©Mx = 0; F(0.075) - 4(103) - 2(103) = 0
F = 80(103) N
Internal Loading: Referring to the free - body diagram of gear D in Fig. b, ©Mx = 0; 80(103)(0.1) - TDH = 0
TDH = 8(103)N # m
Also, from the free - body diagram of gear A, Fig. c, ©Mx = 0; TAB - 4(103) = 0
TAB = 4 A 103 B N # m
And from the free - body diagram of gear C, Fig. d, ©Mx = 0; -TBC - 2 A 103 B = 0
TBC = -2(103) N # m
Angle of Twist: The polar moment of inertia of segments AB, BC and DH p of the shaft are JAB = JBC = and A 0.034 B = 0.405(10 - 6)p m4 2 p 4 -6 4 JDH = A 0.04 B = 1.28(10 )p m . We have 2 fD =
8(103)(0.6) TDH LDH = 0.01592 rad = JDHGst 1.28(10 - 6)p(75)(109)
Then, using the gear ratio, fB = fD a
rD 100 b = 0.02122 rad b = 0.01592a rB 75
Also, fC>B =
-2(103)(0.9) TBC LBC = -0.01886 rad = 0.01886 rad = JBCGst 0.405(10 - 6)p(75)(109)
fA>B =
4(103)(0.6) TABLAB = 0.02515 rad = JAB Gst 0.405(10 - 6)p(75)(109)
Thus, fA = fB + fA>B fA = 0.02122 + 0.02515 = 0.04637 rad = 2.66°
Ans.
fC = fB + fC>B fC = 0.02122 + 0.01886 = 0.04008 rad = 2.30°
Ans.
256
T1
C
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5–66.
Continued
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5–67. The 60-mm diameter shaft ABC is supported by two journal bearings, while the 80-mm diameter shaft EH is fixed at E and supported by a journal bearing at H. If the angle of twist at gears A and C is required to be 0.04 rad, determine the magnitudes of the torques T1 and T2. The shafts are made of A-36 steel.
E A
600 mm D 100 mm H
T2 600 mm B 75 mm 900 mm
Equilibrium: Referring to the free - body diagram of shaft ABC shown in Fig. a ©Mx = 0; F(0.075) - T1 - T2 = 0
T1
F = 13.333 A T1 + T2 B
Internal Loading: Referring to the free - body diagram of gear D in Fig. b, ©Mx = 0; 13.333 A T1 + T2 B (0.1) - TDE = 0
TDE = 1.333 A T1 + T2 B
Also, from the free - body diagram of gear A, Fig. c, ©Mx = 0; TAB - T2 = 0
TAB = T2
and from the free - body diagram of gear C, Fig. d ©Mx = 0; TBC - T1 = 0
TBC = T1
Angle of Twist: The polar moments of inertia of segments AB, BC and DH p of the shaft are and JAB = JBC = A 0.034 B = 0.405(10 - 6)pm4 2 p JDH = A 0.044 B = 1.28(10 - 6)pm4. We have 2 fD =
1.333 A T1 + T2 B (0.6) TDE LDH = 2.6258(10 - 6) A T1 + T2 B = JDE Gst 1.28(10 - 6)p (75)(109)
Then, using the gear ratio, fB = fD ¢
rD 100 b = 3.5368(10 - 6) A T1 + T2 B ≤ = 2.6258(10 - 6) A T1 + T2 B a rB 75
Also, fC>B =
T1(0.9) TBC LBC = 9.4314(10 - 6)T1 = JBC Gst 0.405(10 - 6)p(75)(109)
fA>B =
T2(0.6) TAB LAB = 6.2876(10 - 6)T2 = JAB Gst 0.405(10 - 6)p(75)(109)
Here, it is required that fA = fC = 0.04 rad. Thus, fA = fB + fA>B 0.04 = 3.5368(10 - 6) A T1 + T2 B + 6.2876(10 - 6)T2 T1 + 2.7778T2 = 11309.73
(1)
fC = fB + fC>B 0.04 = 3.5368(10 - 6) A T1 + T2 B + 9.4314(10 - 6)T1 3.6667T1 + T2 = 11309.73
(2)
258
C
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5–67.
Continued
Solving Eqs. (1) and (2), T1 = 2188.98 N # m = 2.19 kN # m
Ans.
T2 = 3283.47 N # m = 3.28 kN # m
Ans.
*5–68. The 30-mm-diameter shafts are made of L2 tool steel and are supported on journal bearings that allow the shaft to rotate freely. If the motor at A develops a torque of T = 45 N # m on the shaft AB, while the turbine at E is fixed from turning, determine the amount of rotation of gears B and C.
A
45 Nm
B 1.5 m
50 mm
D C
Internal Torque: As shown on FBD. 0.5 m
Angle of Twist: fC =
TCE LCE JG 67.5(0.75)
=
p 2
(0.0154)75.0(103)
= 0.008488 rad = 0.486° fB =
75 50
Ans.
fC = 0.729°
Ans.
259
E 75 mm 0.75 m
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•5–69.
The shafts are made of A-36 steel and each has a diameter of 80 mm. Determine the angle of twist at end E.
0.6 m A
B 150 mm 10 kN⭈m
C
D
200 mm 0.6 m
E
Equilibrium: Referring to the free - body diagram of shaft CDE shown in Fig. a, ©Mx = 0; 10(103) - 2(103) - F(0.2) = 0
150 mm
F = 40(103) N
0.6 m 2 kN⭈m
Internal Loading: Referring to the free - body diagram of gear B, Fig. b, ©Mx = 0;
-TAB - 40(103)(0.15) = 0
TAB = -6(103) N # m
Referring to the free - body diagram of gear D, Fig. c, ©Mx = 0; 10(103) - 2(103) - TCD = 0
TCD = 8(103) N # m
Referring to the free - body diagram of shaft DE, Fig. d, ©Mx = 0;
-TDE - 2(103) = 0
Angle of Twist: The polar p J = A 0.044 B = 1.28(10 - 6)p m4. 2
TDE = -2(103) N # m moment
of
inertia
of
the
shafts
are
We have
fB =
-6(103)(0.6) TAB LAB = -0.01194 rad = 0.01194 rad = JGst 1.28(10 - 6)p(75)(109)
Using the gear ratio, fC = fB ¢
rB 150 b = 0.008952 rad ≤ = 0.01194 a rC 200
fE>C = ©
TiLi TCD LCD TDE LDE = + JiGi JGst JGst
Also,
0.6 =
-6
1.28(10 )p(75)(109)
b 8(103) + c -2(103) d r
= 0.01194 rad Thus, fE = fC + fE>C fE = 0.008952 + 0.01194 = 0.02089 rad = 1.20°
Ans.
260
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5–69.
Continued
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5–70. The shafts are made of A-36 steel and each has a diameter of 80 mm. Determine the angle of twist of gear D.
0.6 m
Equilibrium: Referring to the free-body diagram of shaft CDE shown in Fig. a, 3
3
A
©Mx = 0; 10(10 ) - 2(10 ) - F(0.2) = 0
F = 40(10 ) N
-TAB - 40(103)(0.15) = 0
150 mm 10 kN⭈m
C
Internal Loading: Referring to the free - body diagram of gear B, Fig. b, ©Mx = 0;
B
3
TAB = -6(103) N # m
D
200 mm 0.6 m
E
Referring to the free - body diagram of gear D, Fig. c, ©Mx = 0; 10(103) - 2(103) - TCD = 0
150 mm
TCD = 8(103) N # m
0.6 m 2 kN⭈m
Angle of Twist: The polar moment p J = A 0.044 B = 1.28(10 - 6)p m4. We have 2 fB =
of
inertia
of
the
shafts
are
-6(103)(0.6) TAB LAB = -0.01194 rad = 0.01194 rad = JGst 1.28(10 - 6)p(75)(109)
Using the gear ratio, fC = fB ¢
rB 150 b = 0.008952 rad ≤ = 0.01194a rC 200
Also, fD>C =
8(103)(0.6) TCD LCD = 0.01592 rad = JGst 1.28(10 - 6)p(75)(109)
Thus, fD = fC + fD>C fD = 0.008952 + 0.01592 = 0.02487 rad = 1.42°
Ans.
262
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*5–72. The 80-mm diameter shaft is made of 6061-T6 aluminum alloy and subjected to the torsional loading shown. Determine the angle of twist at end A.
0.6 m 0.6 m C
10 kN⭈m/m B A 2 kN⭈m
Equilibrium: Referring to the free - body diagram of segment AB shown in Fig. a, ©Mx = 0;
-TAB - 2(103) = 0
TAB = -2(103)N # m
And the free - body diagram of segment BC, Fig. b, ©Mx = 0;
Angle of Twist: The polar moment p J = A 0.042 B = 1.28(10 - 6)p m4. We have 2 fA = ©
1.28(10 - 6)p(26)(109)
0.6 m -
+
1 = -
of
inertia
of
the
shaft
is
LBC TiLi TABLAB TBC dx = + JiGi JGal JGal L0
-2(103)(0.6) =
TBC = - C 10(103)x + 2(103) D N # m
-TBC - 10(103)x - 2(103) = 0
1.28(10 - 6)p(26)(109)
L0
C 10(103)x + 2(103) D dx
1.28(10 - 6)p(26)(109)
b 1200 + C 5(103)x2 + 2(103)x D 2
0.6m 0
r
= -0.04017 rad = 2.30°
Ans.
263
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•5–73. The tapered shaft has a length L and a radius r at end A and 2r at end B. If it is fixed at end B and is subjected to a torque T, determine the angle of twist of end A. The shear modulus is G.
B 2r L
T
Geometry: r A
r rL + rx r(x) = r + x = L L p rL + rx 4 p r4 (L + x)4 a b = 2 L 2L4
J(x) = Angle of Twist:
L
T dx L0 J(x)G
f =
L
=
2TL4 dx p r4G L0 (L + x)4
=
L 1 2TL4 cd2 4 3 pr G 3(L + x) 0
=
7TL 12p r4G
Ans.
5–74. The rod ABC of radius c is embedded into a medium where the distributed torque reaction varies linearly from zero at C to t0 at B. If couple forces P are applied to the lever arm, determine the value of t0 for equilibrium. Also, find the angle of twist of end A. The rod is made from material having a shear modulus of G.
L 2 L 2
Equilibrium: Referring to the free-body diagram of the entire rod shown in Fig. a, 1 L ©Mx = 0; Pd - (t0)a b = 0 2 2 to =
Ans.
Internal Loading: The distributed torque expressed as a function of x, measured 4Pd>L to 8Pd from the left end, is t = ¢ ≤x = ¢ ≤ x = ¢ 2 ≤ x. Thus, the resultant L>2 L>2 L torque within region x of the shaft is
TR =
1 1 8Pd 4Pd 2 tx = B ¢ 2 ≤ x R x = x 2 2 L L2
Referring to the free - body diagram shown in Fig. b, ©Mx = 0; TBC -
4Pd 2 x = 0 L2
d 2
B P
4Pd L
TBC =
4Pd 2 x L2
Referring to the free - body diagram shown in Fig. c, ©Mx = 0; Pd - TAB = 0
TAB = Pd
264
d 2
t0
C
A
P
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5–74.
Continued
Angle of Twist: f = ©
LBC TAB LAB TBC dx TiLi = + JiGi JG JG L0
L>2
=
L0
4Pd 2 x dx L2
Pd(L>2) +
p 2
p 2
¢ c4 ≤ G
¢ c4 ≤ G L>2
8Pd x3 = £ ≥3 4 2 pc L G 3
+
PLd pc4G
0
=
4PLd 3pc4G
Ans.
265
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5–75. When drilling a well, the deep end of the drill pipe is assumed to encounter a torsional resistance TA . Furthermore, soil friction along the sides of the pipe creates a linear distribution of torque per unit length, varying from zero at the surface B to t0 at A. Determine the necessary torque TB that must be supplied by the drive unit to turn the pipe. Also, what is the relative angle of twist of one end of the pipe with respect to the other end at the instant the pipe is about to turn? The pipe has an outer radius ro and an inner radius ri . The shear modulus is G.
TB
B
L
t0 A
1 t L + TA - TB = 0 2 0 TB =
t0L + 2TA 2
T(x) +
t0 2 t0L + 2TA x = 0 2L 2
T(x) =
t0 2 t0 L + 2TA x 2 2L
f =
Ans.
T(x) dx L JG
=
L t0L + 2TA t0 2 1 ( x ) dx J G L0 2 2L
=
t0 3 L 1 t0 L + 2TA c x x dƒ JG 2 6L 0
=
t0 L2 + 3TAL 3JG
However, J =
f =
p (r 4 - ri 4) 2 o 2L(t0 L + 3TA)
Ans.
3p(ro 4 - ri 4)G
266
TA
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*5–76. A cylindrical spring consists of a rubber annulus bonded to a rigid ring and shaft. If the ring is held fixed and a torque T is applied to the rigid shaft, determine the angle of twist of the shaft. The shear modulus of the rubber is G. Hint: As shown in the figure, the deformation of the element at radius r can be determined from rdu = drg. Use this expression along with t = T>12pr2h2 from Prob. 5–26, to obtain the result.
ro r ri T h
gdr rdu
dr g du r
r du = g dr du =
gdr r
(1)
From Prob. 5-26, t =
T 2p r2h
g =
T 2p r2hG
and
g =
t G
From (1), du =
T dr 2p hG r3 r
u =
o dr T 1 ro T = cd| 3 2p hG Lri r 2p hG 2 r2 ri
=
1 1 T c- 2 + 2d 2p hG 2ro 2ri
=
1 1 T c - 2d 4p hG r2i ro
Ans.
267
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•5–77.
The A-36 steel shaft has a diameter of 50 mm and is fixed at its ends A and B. If it is subjected to the torque, determine the maximum shear stress in regions AC and CB of the shaft.
A
300 Nm
0.4 m C 0.8 m
Equilibrium: TA + TB - 300 = 0
[1]
Compatibility: fC>A = fC>B TA(0.4) TB(0.8) = JG JG TA = 2.00TB
[2]
Solving Eqs. [1] and [2] yields: TA = 200 N # m
TB = 100 N # m
Maximum Shear stress: (tAC)max =
200(0.025) TAc = 8.15 MPa = p 4 J 2 (0.025 )
Ans.
(tCB)max =
100(0.025) TBc = 4.07 MPa = p 4 J 2 (0.025 )
Ans.
268
B
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5–78. The A-36 steel shaft has a diameter of 60 mm and is fixed at its ends A and B.If it is subjected to the torques shown, determine the absolute maximum shear stress in the shaft.
200 N⭈m B 500 N⭈m
D 1.5 m
C A
Referring to the FBD of the shaft shown in Fig. a, TA + TB - 500 - 200 = 0
©Mx = 0;
(1)
Using the method of superposition, Fig. b fA = (fA)TA - (fA)T 0 =
500 (1.5) 700 (1) TA (3.5) - c + d JG JG JG TA = 414.29 N # m
Substitute this result into Eq (1), TB = 285.71 N # m Referring to the torque diagram shown in Fig. c, segment AC is subjected to maximum internal torque. Thus, the absolute maximum shear stress occurs here. tAbs =
414.29 (0.03) TAC c = = 9.77 MPa p J (0.03)4 2
Ans.
269
1m
1m
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5–79. The steel shaft is made from two segments: AC has a diameter of 0.5 in, and CB has a diameter of 1 in. If it is fixed at its ends A and B and subjected to a torque of determine the maximum shear stress in the shaft. Gst = 10.811032 ksi.
A
0.5 in. C D 500 lbft
5 in.
1 in.
8 in.
B 12 in.
Equilibrium: TA + TB - 500 = 0
(1)
Compatibility condition: fD>A = fD>B TA(5) p 2
4
(0.25 )G
TA(8) +
p 2
4
(0.5 )G
TB(12) =
p 2
(0.54)G
1408 TA = 192 TB
(2)
Solving Eqs. (1) and (2) yields TA = 60 lb # ft
TB = 440 lb # ft
tAC =
60(12)(0.25) TC = 29.3 ksi = p 4 J 2 (0.25 )
tDB =
440(12)(0.5) TC = 26.9 ksi = p 4 J 2 (0.5 )
Ans.
(max)
270
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*5–80. The shaft is made of A-36 steel, has a diameter of 80 mm, and is fixed at B while A is loose and can rotate 0.005 rad before becoming fixed. When the torques are applied to C and D, determine the maximum shear stress in regions AC and CD of the shaft.
2 kN⭈m B
4 kN⭈m D
600 mm C 600 mm A
Referring to the FBD of the shaft shown in Fig. a,
600 mm
TA + TB + 2 - 4 = 0
©Mx = 0;
(1)
Using the method of superposition, Fig. b, fA = (fA)T - (uA)TA 0.005 = B p 2
4(103)(0.6)
(0.04 ) C 75(10 ) D 4
9
2(103)(0.6)
+
p 2
(0.04 ) C 75(10 ) D 4
9
R -
TA (1.8)
p 2
(0.044) C 75(109) D
TA = 1162.24 N # m = 1.162 kN # m Substitute this result into Eq (1), TB = 0.838 kN # m Referring to the torque diagram shown in Fig. c, segment CD is subjected to a maximum internal torque. Thus, the absolute maximum shear stress occurs here. t$$$ =
2.838 (103)(0.04) TCD c = = 28.23 (106) Pa = 28.2 MPa p 4 J 2 (0.04)
271
Ans.
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•5–81. The shaft is made of A-36 steel and has a diameter of 80 mm. It is fixed at B and the support at A has a torsional stiffness of k = 0.5 MN # m>rad. If it is subjected to the gear torques shown, determine the absolute maximum shear stress in the shaft.
2 kN⭈m B
4 kN⭈m D
600 mm C 600 mm A 600 mm
Referring to the FBD of the shaft shown in Fig. a, TA + TB + 2 - 4 = 0
©Mx = 0;
(1)
Using the method of superposition, Fig. b, fA = (fA)T - (fA)TA TA 6
0.5(10 )
= D
4(103)(0.6)
p 2
(0.04 ) C 75(10 ) D 4
9
2(103)(0.6)
+
p 2
(0.04 ) C 75(10 ) D 4
9
T -
TA(1.8)
p 2
(0.044) C 75(109) D
TA = 1498.01 N # m = 1.498 kN # m Substituting this result into Eq (1), TB = 0.502 kN # m Referring to the torque diagram shown in Fig. c, segment CD subjected to maximum internal torque. Thus, the maximum shear stress occurs here. t$$$ =
2.502(103)(0.04) TCD C = = $$$ = 24.9 MPa p 4 J 2 (0.04)
272
Ans.
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5–82. The shaft is made from a solid steel section AB and a tubular portion made of steel and having a brass core. If it is fixed to a rigid support at A, and a torque of T = 50 lb # ft is applied to it at C, determine the angle of twist that occurs at C and compute the maximum shear stress and maximum shear strain in the brass and steel. Take Gst = 11.511032 ksi, Gbr = 5.611032 ksi.
3 ft
2 ft A 0.5 in. B 1 in.
Equilibrium: Tbr + Tst - 50 = 0
(1)
Both the steel tube and brass core undergo the same angle of twist fC>B fC>B =
TL = JG
Tst (2)(12)
Tst (2)(12) p 2
(0.54)(5.6)(104)
=
p 2
4
(1 - 0.54)(11.5)(106)
Tbr = 0.032464 Tst
(2)
Solving Eqs. (1) and (2) yields: Tst = 48.428 lb # ft; fC = ©
Tbr = 1.572 lb # ft
50(12)(3)(12) 1.572(12)(2)(12) TL + p 4 = p 4 6 6 JG 2 (0.5 )(5.6)(10 ) 2 (1 )(11.5)(10 ) = 0.002019 rad = 0.116°
Ans.
(tst)max AB =
50(12)(1) TABc = 382 psi = p 4 J 2 (1 )
(tst)max BC =
48.428(12)(1) Tst c = 394.63 psi = 395 psi (Max) = p 4 4 J 2 (1 - 0.5 )
Ans.
(gst)max =
(tst)max 394.63 = 343.(10 - 6) rad = G 11.5(106)
Ans.
(tbr)max =
1.572(12)(0.5) Tbr c = 96.07 psi = 96.1 psi (Max) = p 4 J 2 (0.5 )
Ans.
(gbr)max =
(tbr)max 96.07 = 17.2(10 - 6) rad = G 5.6(106)
Ans.
273
C
T 50 lbft
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5–83. The motor A develops a torque at gear B of 450 lb # ft, which is applied along the axis of the 2-in.-diameter steel shaft CD. This torque is to be transmitted to the pinion gears at E and F. If these gears are temporarily fixed, determine the maximum shear stress in segments CB and BD of the shaft. Also, what is the angle of twist of each of these segments? The bearings at C and D only exert force reactions on the shaft and do not resist torque. Gst = 1211032 ksi.
B E 4 ft
(1)
Compatibility condition: fB>C = fB>D TC(4) TD(3) = JG JG TC = 0.75 TD
(2)
Solving Eqs. (1) and (2), yields TD = 257.14 lb # ft TC = 192.86 lb # ft
(tBD)max =
f =
192.86(12)(1) p 2
(14)
257.14(12)(1) p 2
(14)
192.86(12)(4)(12) p 2
(14)(12)(106)
3 ft D A
TC + TD - 450 = 0
(tBC)max =
F
C
Equilibrium:
= 1.47 ksi
Ans.
= 1.96 ksi
Ans.
= 0.00589 rad = 0.338°
Ans.
274
450 lb⭈ft
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*5–84. A portion of the A-36 steel shaft is subjected to a linearly distributed torsional loading. If the shaft has the dimensions shown, determine the reactions at the fixed supports A and C. Segment AB has a diameter of 1.5 in. and segment BC has a diameter of 0.75 in.
300 lb⭈in./in. A
60 in.
B
C 48 in.
Equilibrium: TA + TC - 9000 = 0 TR = t x +
1 tx (300 - t)x = 150x + 2 2
300 t = ; 60 - x 60
But
(1)
TR = 150 x +
t = 5(60 - x)
1 [5(60 - x)]x 2
= (300x - 2.5x2) lb # in. Compatibility condition: fB>A = fB>C fB>A =
60 T(x) dx 1 = [TA - (300x - 2.5x2)] dx JG L0 L JG
=
60 1 [TAx - 150x2 + 0.8333x3] | JG 0
=
60TA - 360 000 JG TC(48)
60TA - 360 000 p 2
(0.754)G
=
p 2
(0.3754)G (2)
60TA - 768TC = 360 000 Solving Eqs. (1) and (2) yields: TC = 217.4 lb # in. = 18.1 lb # ft
Ans.
TA = 8782.6 lb # in. = 732 lb # ft
Ans.
275
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•5–85.
Determine the rotation of joint B and the absolute maximum shear stress in the shaft in Prob. 5–84.
300 lb⭈in./in. A
Equilibrium: TA + TC - 9000 = 0 TR = tx +
But
(1)
1 tx (300 - t)x = 150x + 2 2
300 t = ; 60 - x 60
C
t = 5(60 - x)
= (300x - 2.5x2) lb # in. Compatibility condition: fB>A = fB>C fB>A =
60 T(x) dx 1 = [TA - (300x - 2.5x2)] dx JG L0 L JG
=
60 1 [TAx - 150x2 + 0.8333x3] | JG 0
=
60TA - 360 000 JG TC(48)
60TA - 360 000 p 2
=
4
(0.75 )G
p 2
(0.3754)G
60TA - 768TC = 360 000
(2)
Solving Eqs. (1) and (2) yields: TC = 217.4 lb # in. = 18.1 lb # ft TA = 8782.6 lb # in. = 732 lb # ft For segment BC: fB = fB>C =
TCL = JG
217.4(48) p 2
(0.375)4(11.0)(106)
= 0.030540 rad
fB = 1.75° tmax =
Ans.
217.4(0.375) TC = p = 2.62 ksi 4 J 2 (0.375)
For segment AB, tmax =
B
48 in.
1 [5(60 - x)]x 2
TR = 150x +
60 in.
8782.6(0.75) TC = p = 13.3 ksi 4 J 2 (0.75)
abs = 13.3 ksi tmax
Ans.
276
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5–86. The two shafts are made of A-36 steel. Each has a diameter of 25 mm and they are connected using the gears fixed to their ends. Their other ends are attached to fixed supports at A and B. They are also supported by journal bearings at C and D, which allow free rotation of the shafts along their axes. If a torque of 500 N # m is applied to the gear at E as shown, determine the reactions at A and B.
B F
D
50 mm
0.75 m
100 mm 500 Nm E
C 1.5 m A
Equilibrium: TA + F(0.1) - 500 = 0
[1]
TB - F(0.05) = 0
[2]
TA + 2TB - 500 = 0
[3]
From Eqs. [1] and [2]
Compatibility: 0.1fE = 0.05fF fE = 0.5fF TA(1.5) TB(0.75) = 0.5 c d JG JG TA = 0.250TB
[4]
Solving Eqs. [3] and [4] yields: TB = 222 N # m
Ans.
TA = 55.6 N # m
Ans.
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5–87. Determine the rotation of the gear at E in Prob. 5–86.
B F
D
50 mm
0.75 m
100 mm 500 Nm E
C 1.5 m A
Equilibrium: TA + F(0.1) - 500 = 0
[1]
TB - F(0.05) = 0
[2]
TA + 2TB - 500 = 0
[3]
From Eqs. [1] and [2]
Compatibility: 0.1fE = 0.05fF fE = 0.5fF TA(1.5) TB(0.75) = 0.5 c d JG JG TA = 0.250TB
[4]
Solving Eqs. [3] and [4] yields: TB = 222.22 N # m
TA = 55.56 N # m
Angle of Twist: fE =
TAL = JG
55.56(1.5) p 2
(0.01254)(75.0)(109)
= 0.02897 rad = 1.66°
Ans.
278
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*5–88. The shafts are made of A-36 steel and have the same diameter of 4 in. If a torque of 15 kip # ft is applied to gear B, determine the absolute maximum shear stress developed in the shaft.
2.5 ft 2.5 ft
Equilibrium: Referring to the free - body diagrams of shafts ABC and DE shown in Figs. a and b, respectively, we have ©Mx = 0; TA + F(0.5) - 15 = 0
(1)
and
A B 6 in. 15 kip⭈ft
C
D
12 in.
©Mx = 0; F(1) - TE = 0
E
(2) 3 ft
Internal Loadings: The internal torques developed in segments AB and BC of shaft ABC and shaft DE are shown in Figs. c, d, and e, respectively. Compatibility Equation: fCrC = fDrD
¢
TABLAB TDE LDE TBCLBC + ≤ rC = ¢ ≤ rD JGst JGst JGst
C -TA(2.5) + F(0.5)(2.5) D (0.5) = -TE(3)(1) TA - 0.5F = 2.4TE
(3)
Solving Eqs. (1), (2), and (3), we have F = 4.412 kip
TE = 4.412 kip # ft
TA = 12.79 kip # ft
Maximum Shear Stress: By inspection, segment AB of shaft ABC is subjected to the greater torque.
A tmax B abs =
12.79(12)(2) TAB c = 12.2 ksi = Jst p 4 a2 b 2
Ans.
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•5–89.
The shafts are made of A-36 steel and have the same diameter of 4 in. If a torque of 15 kip # ft is applied to gear B, determine the angle of twist of gear B.
2.5 ft
Equilibrium: Referring to the free - body diagrams of shafts ABC and DE shown in Figs. a and b, respectively, ©Mx = 0; TA + F(0.5) - 15 = 0
2.5 ft A B
(1)
6 in. 15 kip⭈ft
and ©Mx = 0; F(1) - TE = 0
(2)
Internal Loadings: The internal torques developed in segments AB and BC of shaft ABC and shaft DE are shown in Figs. c, d, and e, respectively. Compatibility Equation: It is required that fCrC = fDrD
¢
TAB LAB TDE LDE TBC LBC + ≤ rC = ¢ ≤ rD JGst JGst JGst
C -TA(2.5) + F(0.5)(2.5) D (0.5) = -TE(3)(1) TA - 0.5F = 2.4TE
(3)
Solving Eqs. (1), (2), and (3), F = 4.412 kip
TE = 4.412 kip # ft
TA = 12.79 kip # ft
Angle of Twist: Here, TAB = -TA = -12.79 kip # ft fB =
-12.79(12)(2.5)(12) TAB LAB = JGst p 4 a 2 b(11.0)(103) 2
= -0.01666 rad = 0.955°
Ans.
280
C
D
12 in. E 3 ft
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5–90. The two 3-ft-long shafts are made of 2014-T6 aluminum. Each has a diameter of 1.5 in. and they are connected using the gears fixed to their ends. Their other ends are attached to fixed supports at A and B. They are also supported by bearings at C and D, which allow free rotation of the shafts along their axes. If a torque of 600 lb # ft is applied to the top gear as shown, determine the maximum shear stress in each shaft.
A B
E
2 in.
TA + F a
4 b - 600 = 0 12
(1)
TB - F a
2 b = 0 12
(2)
From Eqs. (1) and (2) TA + 2TB - 600 = 0
TAL TBL = 0.5 a b; JG JG
(3)
fE = 0.5fF TA = 0.5TB
(4)
Solving Eqs. (3) and (4) yields: TB = 240 lb # ft;
TA = 120 lb # ft
(tBD)max =
240(12)(0.75) TB c = 4.35 ksi = p 4 J 2 (0.75 )
Ans.
(tAC)max =
120(12)(0.75) TA c = 2.17 ksi = p 4 J 2 (0.75 )
Ans.
281
3 ft D
4 in.
4(fE) = 2(fF);
C
600 lbft
F
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5–91. The A-36 steel shaft is made from two segments: AC has a diameter of 0.5 in. and CB has a diameter of 1 in. If the shaft is fixed at its ends A and B and subjected to a uniform distributed torque of 60 lb # in.>in. along segment CB, determine the absolute maximum shear stress in the shaft.
A
0.5 in. C
5 in.
60 lbin./in. 1 in. 20 in.
Equilibrium: TA + TB - 60(20) = 0
(1)
Compatibility condition: fC>B = fC>A fC>B =
20 (TB - 60x) dx T(x) dx = p JG L L0 2 (0.54)(11.0)(106)
= 18.52(10-6)TB - 0.011112 18.52(10-6)TB - 0.011112 =
TA(5) p 4 6 2 (0.25 )(11.0)(10 )
18.52(10-6)TB - 74.08(10-6)TA = 0.011112 (2)
18.52TB - 74.08TA = 11112 Solving Eqs. (1) and (2) yields: TA = 120.0 lb # in. ;
TB = 1080 lb # in.
(tmax)BC =
1080(0.5) TB c = 5.50 ksi = p 4 J 2 (0.5 )
(tmax)AC =
120.0(0.25) TA c = 4.89 ksi = p 4 J 2 (0.25 )
abs = 5.50 ksi tmax
Ans.
282
B
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*5–92. If the shaft is subjected to a uniform distributed torque of t = 20 kN # m>m, determine the maximum shear stress developed in the shaft. The shaft is made of 2014-T6 aluminum alloy and is fixed at A and C.
400 mm
20 kN⭈m/m 600 mm a A 80 mm 60 mm
B a C
Equilibrium: Referring to the free - body diagram of the shaft shown in Fig. a, we have ©Mx = 0; TA + TC - 20(103)(0.4) = 0
(1)
Compatibility Equation: The resultant torque of the distributed torque within the region x of the shaft is TR = 20(103)x N # m. Thus, the internal torque developed in the shaft as a function of x when end C is free is T(x) = 20(103)x N # m, Fig. b. Using the method of superposition, Fig. c, fC = A fC B t - A fC B TC 0 =
0 =
0.4 m
T(x)dx TCL JG JG
0.4 m
20(103)xdx TC(1) JG JG
L0 L0
0 = 20(103) ¢
x2 2 0.4 m - TC ≤ 2 0
TC = 1600 N # m Substituting this result into Eq. (1), TA = 6400 N # m Maximum Shear Stress: By inspection, the maximum internal torque occurs at support A. Thus,
A tmax B abs =
6400(0.04) TA c = 93.1 MPa = J p 4 4 a0.04 - 0.03 b 2
Ans.
283
Section a–a
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•5–93.
The tapered shaft is confined by the fixed supports at A and B. If a torque T is applied at its mid-point, determine the reactions at the supports. 2c
T
A
B
Equilibrium:
c
TA + TB - T = 0
[1] L/2 L/ 2
Section Properties: r(x) = c +
J(x) =
c c x = (L + x) L L
4 p c pc4 c (L + x) d = (L + x)4 2 L 2L4
Angle of Twist: fT =
Tdx = Lp2 L J(x)G
L
Tdx pc4 2L4
(L + x)4 G L
=
dx 2TL4 pc4 G Lp2 (L + x)4
= -
=
fB =
L 1 2TL4 c d 2 4 3 p 3pc G (L + x) 2
37TL 324 pc4 G
Tdx = J(x)G L L0
L
TBdx pc4 2L4
(L + x)4G L
2TBL4 =
dx pc G L0 (L + x)4 4
2TBL4
= -
L 1 d 2 3 3pc G (L + x) 0 4
c
7TB L =
12pc4G
Compatibility: 0 = fT - fB 0 =
7TBL 37TL 4 324pc G 12pc4G
TB =
37 T 189
Ans.
Substituting the result into Eq. [1] yields: TA =
152 T 189
Ans.
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5–94. The shaft of radius c is subjected to a distributed torque t, measured as torque>length of shaft. Determine the reactions at the fixed supports A and B.
B t0
(
t t0 1
( Lx ) 2 )
x L
2t0
x
x2 x2 T(x) = t0 a1 + 2 b dx = t0 ax + b L 3L2 L0
(1)
By superposition: 0 = fB - fB L
0 =
L0 TB =
A
t0 a x +
x 3L2 b 3
2
dx -
JG
TB(L) 7t0L = - TB(L) JG 12
7t0 L 12
Ans.
From Eq. (1), TA = t0 a L + TA +
4t0 L L3 b = 2 3 3L
7t0 L 4t0 L = 0 12 3 TA =
3t0 L 4
Ans.
285
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5–95. Compare the values of the maximum elastic shear stress and the angle of twist developed in 304 stainless steel shafts having circular and square cross sections. Each shaft has the same cross-sectional area of 9 in2, length of 36 in., and is subjected to a torque of 4000 lb # in. r A
Maximum Shear Stress: For circular shaft 1
A = pc2 = 9;
(tc)max =
9 2 c = a b p
2(4000) Tc Tc 2T = = 525 psi = p 4 = 1 3 J pc p A 9x B 2 2 c
Ans.
For rectangular shaft A = a2 = 9 ; (tr)max =
a = 3 in.
4.81(4000) 4.81T = = 713 psi 3 a 33
Ans.
Angle of Twist: For circular shaft fc =
TL = JG
4000(36)
p 2
A B 11.0(106) 9 2 p
Ans.
= 0.001015 rad = 0.0582° For rectangular shaft fr =
7.10(4000)(36) 7.10 TL = 4 a4 G 3 (11.0)(106) Ans.
= 0.001147 rad = 0.0657° The rectangular shaft has a greater maximum shear stress and angle of twist.
286
a
A a
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*5–96. If a = 25 mm and b = 15 mm, determine the maximum shear stress in the circular and elliptical shafts when the applied torque is T = 80 N # m. By what percentage is the shaft of circular cross section more efficient at withstanding the torque than the shaft of elliptical cross section?
b
a
a
For the circular shaft: (tmax)c =
80(0.025) Tc = 3.26 MPa = p 4 J 2 (0.025 )
Ans.
For the elliptical shaft: (tmax)c =
2(80) 2T = 9.05 MPa = p a b2 p(0.025)(0.0152)
Ans.
(tmax)c - (tmax)c (100%) (tmax)c
% more efficient =
9.05 - 3.26 (100%) = 178 % 3.26
=
Ans.
•5–97.
It is intended to manufacture a circular bar to resist torque; however, the bar is made elliptical in the process of manufacturing, with one dimension smaller than the other by a factor k as shown. Determine the factor by which the maximum shear stress is increased.
kd
For the circular shaft: (tmax)c =
d
TA B Tc 16T = = 3 p J AdB4 p d d 2
2 2
For the elliptical shaft: (tmax)c =
d
2T 2T 16T = = 2 p a b2 p k2 d3 p A d2 B A kd B 2
Factor of increase in shear stress =
=
(tmax)c = (tmax)c
16T p k2 d3 16T p d3
1 k2
Ans.
287
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5–98. The shaft is made of red brass C83400 and has an elliptical cross section. If it is subjected to the torsional loading shown, determine the maximum shear stress within regions AC and BC, and the angle of twist f of end B relative to end A.
A 20 Nm
50 Nm
30 Nm 2m
Maximum Shear Stress:
C
(tBC)max =
2(30.0)
2TBC p a b2
=
B
p(0.05)(0.022) Ans.
= 0.955 MPa (tAC)max =
2(50.0)
2TAC 2
50 mm 20 mm
1.5 m
=
pab
p(0.05)(0.022) Ans.
= 1.59 MPa Angle of Twist: fB>A = a
(a2 + b2)T L p a3b3 G (0.052 + 0.022)
=
p(0.053)(0.023)(37.0)(109)
[(-30.0)(1.5) + (-50.0)(2)]
= -0.003618 rad = 0.207°
Ans.
5–99. Solve Prob. 5–98 for the maximum shear stress within regions AC and BC, and the angle of twist f of end B relative to C.
A 20 Nm
50 Nm
30 Nm 2m
Maximum Shear Stress:
C
(tBC)max =
2(30.0)
2TBC p a b2
=
B
p(0.05)(0.022)
= 0.955 MPa (tAC)max =
2
Ans.
2(50.0)
2TAC =
pab
p(0.05)(0.022)
= 1.59 MPa
Ans.
Angle of Twist: fB>C =
(a2 + b2) TBC L p a3 b3 G (0.052 + 0.022)(-30.0)(1.5)
=
50 mm 20 mm
1.5 m
p(0.053)(0.023)(37.0)(109)
= -0.001123 rad = | 0.0643°|
Ans.
288
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*5–100. Segments AB and BC of the shaft have circular and square cross sections, respectively. If end A is subjected to a torque of T = 2 kN # m, determine the absolute maximum shear stress developed in the shaft and the angle of twist of end A. The shaft is made from A-36 steel and is fixed at C.
600 mm C 600 mm
90 mm
B
30 mm
90 mm
Internal Loadings: The internal torques developed in segments AB and BC are shown in Figs. a, and b, respectively. Maximum Shear Stress: For segment AB,
A tmax B AB =
2(103)(0.03) TAB c = 47.2 MPa (max) J p a 0.034 b 2
For segment BC,
A tmax B BC =
4.81TBC 3
=
a
4.81 C 2(103) D (0.09)3
Ans.
= 13.20 MPa
Angle of Twist: fA =
7.10TBCLBC TABLAB + JG a4G 2(103)(0.6)
=
p a 0.034 b(75)(109) 2
7.10(2)(103)(0.6) +
(0.09)4(75)(109)
= 0.01431 rad = 0.820°
Ans.
289
A T
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•5–101. Segments AB and BC of the shaft have circular and square cross sections, respectively. The shaft is made from A-36 steel with an allowable shear stress of tallow = 75 MPa, and an angle of twist at end A which is not allowed to exceed 0.02 rad. Determine the maximum allowable torque T that can be applied at end A.The shaft is fixed at C.
600 mm C 600 mm
Internal Loadings: The internal torques developed in segments AB and BC are shown in Figs. a, and b, respectively.
90 mm
B
30 mm
90 mm
Allowable Shear Stress: For segment AB,
A T
tallow =
TAB c ; J
75(106) =
T(0.03) p a0.034 b 2
T = 3180.86 N # m For segment BC, tallow =
4.81TBC a3
75(106) =
;
4.81T (0.09)3 T = 11 366.94 N # m
Angle of Twist: fA =
TABLAB 7.10TBC LBC + JG a4G
0.02 =
T(0.6)
7.10T(0.6)
p a 0.034 b(75)(109) 2
+
(0.09)4 (75)(109)
T = 2795.90 N # m = 2.80 kN # m (controls)
Ans.
5–102. The aluminum strut is fixed between the two walls at A and B. If it has a 2 in. by 2 in. square cross section, and it is subjected to the torque of 80 lb # ft at C, determine the reactions at the fixed supports. Also, what is the angle of twist at C? Gal = 3.811032 ksi.
A C 2 ft 80 lb⭈ft
By superposition:
3 ft
0 = f - fB 0 =
7.10(TB)(5)
7.10(80)(2) 4
-
a G
a4 G
TB = 32 lb # ft
Ans.
TA + 32 - 80 = 0 TA = 48 lb # ft fC =
7.10(32)(12)(3)(12) (24)(3.8)(106)
Ans. = 0.00161 rad = 0.0925°
Ans.
290
B
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5–103. The square shaft is used at the end of a drive cable in order to register the rotation of the cable on a gauge. If it has the dimensions shown and is subjected to a torque of 8 N # m, determine the shear stress in the shaft at point A. Sketch the shear stress on a volume element located at this point.
5 mm A 5 mm
Maximum shear stress:
8 Nm
(tmax)A =
4.81(8) 4.81T = = 308 MPa a3 (0.005)3
Ans.
*5–104. The 6061-T6 aluminum bar has a square cross section of 25 mm by 25 mm. If it is 2 m long, determine the maximum shear stress in the bar and the rotation of one end relative to the other end.
C 1.5 m 20 N⭈m B
Maximum Shear Stress: tmax =
0.5 m A
4.81(80.0)
4.81Tmax a3
=
(0.0253)
Ans.
= 24.6 MPa
60 N·m 25 mm
Angle of Twist: 7.10(-20.0)(1.5) 7.10(-80.0)(0.5) 7.10TL fA>C = a 4 = + 4 9 aG (0.025 )(26.0)(10 ) (0.0254)(26.0)(109) = -0.04894 rad = | 2.80° |
Ans.
291
80 N⭈m 25 mm
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•5–105.
The steel shaft is 12 in. long and is screwed into the wall using a wrench. Determine the largest couple forces F that can be applied to the shaft without causing the steel to yield. tY = 8 ksi.
1 in. 12 in.
F(16) - T = 0 tmax = tY = 8(103) =
(1)
4.81T a3
F
8 in.
4.81T (1)3
1 in.
8 in.
T = 1663.2 lb # in.
F
From Eq. (1), F = 104 lb
Ans.
5–106. The steel shaft is 12 in. long and is screwed into the wall using a wrench. Determine the maximum shear stress in the shaft and the amount of displacement that each couple force undergoes if the couple forces have a magnitude of F = 30 lb, Gst = 10.811032 ksi.
1 in. 12 in.
T - 30(16) = 0 F
T = 480 lb # in. tmax =
4.81(480) 4.18T = a3 (1)3
1 in.
8 in.
Ans.
= 2.31 ksi f =
8 in.
7.10(480)(12) 7.10TL = = 0.00379 rad a4 G (1)4(10.8)(106)
dF = 8(0.00397) = 0.0303 in.
Ans.
292
F
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5–107. Determine the constant thickness of the rectangular tube if the average shear stress is not to exceed 12 ksi when a torque of T = 20 kip # in. is applied to the tube. Neglect stress concentrations at the corners. The mean dimensions of the tube are shown.
T 4 in.
Am = 2(4) = 8 in2 tavg =
12 =
T 2 t Am
2 in.
20 2 t (8)
t = 0.104 in.
Ans.
*5–108. Determine the torque T that can be applied to the rectangular tube if the average shear stress is not to exceed 12 ksi. Neglect stress concentrations at the corners. The mean dimensions of the tube are shown and the tube has a thickness of 0.125 in.
T 4 in.
Am = 2(4) = 8 in2 tavg =
T ; 2 t Am
12 =
2 in.
T 2(0.125)(8)
T = 24 kip # in. = 2 kip # ft
Ans.
293
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•5–109.
For a given maximum shear stress, determine the factor by which the torque carrying capacity is increased if the half-circular section is reversed from the dashed-line position to the section shown. The tube is 0.1 in. thick.
1.80 in. 0.6 in. 1.20 in. 0.5 in.
Am
p(0.552) = 1.4498 in2 = (1.10)(1.75) 2
Am ¿ = (1.10)(1.75) +
tmax =
p(0.552) = 2.4002 in2 2
T 2t Am
T = 2 t Am tmax Factor =
=
2t Am ¿ tmax 2t Am tmax Am ¿ 2.4002 = = 1.66 Am 1.4498
Ans.
5–110. For a given average shear stress, determine the factor by which the torque-carrying capacity is increased if the half-circular sections are reversed from the dashed-line positions to the section shown. The tube is 0.1 in. thick.
1.80 in. 0.6 in. 1.20 in. 0.5 in.
Section Properties: œ Am = (1.1)(1.8) - B
p (0.552) R (2) = 1.02967 in2 2
Am = (1.1)(1.8) + B
p (0.552) R (2) = 2.93033 in2 2
Average Shear Stress: tavg = Hence,
T ; 2 t Am
T = 2 t Am tavg œ tavg T¿ = 2 t Am
The factor of increase =
Am 2.93033 T = œ = T¿ Am 1.02967
= 2.85
Ans.
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5–111. A torque T is applied to two tubes having the cross sections shown. Compare the shear flow developed in each tube.
t t t
Circular tube:
a
T T 2T = = 2Am 2p (a>2)2 p a2
qct =
a
a
Square tube: qst =
T T = 2Am 2a2
qst T>(2a2) p = = qct 4 2T>(p a2) Thus; qst =
p q 4 ct
Ans.
*5–112. Due to a fabrication error the inner circle of the tube is eccentric with respect to the outer circle. By what percentage is the torsional strength reduced when the eccentricity e is one-fourth of the difference in the radii?
ab 2 a
Average Shear Stress:
e 2
For the aligned tube tavg =
T T = 2 t Am 2(a - b)(p) A a
T = tavg (2)(a - b)(p)a
B
+ b 2 2
a + b 2 b 2
For the eccentric tube tavg =
b
T¿ 2 t Am
t = a -
= a -
e e - a + bb = a - e - b 2 2 1 3 (a - b) - b = (a - b) 4 4
3 a + b 2 b T¿ = tavg (2)c (a - b) d(p)a 4 2 Factor =
tavg (2) C 34 (ab) D (p) A a T¿ = T tavg (2)(a - b)(p) A a
Percent reduction in strength = a1 -
B
+ b 2 2
B
+ b 2 2
=
3 4
3 b * 100 % = 25 % 4
295
Ans.
e 2
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•5–113.
The mean dimensions of the cross section of an airplane fuselage are shown. If the fuselage is made of 2014-T6 aluminum alloy having allowable shear stress of tallow = 18 ksi, and it is subjected to a torque of 6000 kip # ft, determine the required minimum thickness t of the cross section to the nearest 1>16 in. Also, find the corresponding angle of twist per foot length of the fuselage.
t 3 ft
4.5 ft
3 ft
Section Properties: Referring to the geometry shown in Fig. a, Am = p A 32 B + 4.5(6) = 55.2743 ft2 ¢ F
144 in2 ≤ = 7959.50 in2 1 ft2
ds = 2p(3) + 2(4.5) = 27.8496 fta
12 in. b = 334.19 in. 1 ft
Allowable Average Shear Stress:
A tavg B allow =
T ; 2tAm
18 =
6000(12) 2t(7959.50)
t = 0.2513 in. =
Angle of Twist: Using the result of t =
f = ©
Ans.
5 in, 16
ds TL 4Am 2G F t 6000(12)(1)(12)
=
5 in. 16
4(7959.502)(3.9)(103)
¢
334.19 ≤ 5>16
= 0.9349(10 - 3) rad = 0.0536°
Ans.
296
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5–114. The mean dimensions of the cross section of an airplane fuselage are shown. If the fuselage is made from 2014-T6 aluminum alloy having an allowable shear stress of tallow = 18 ksi and the angle of twist per foot length of fuselage is not allowed to exceed 0.001 rad>ft, determine the maximum allowable torque that can be sustained by the fuselage. The thickness of the wall is t = 0.25 in.
t 3 ft
4.5 ft
3 ft
Section Properties: Referring to the geometry shown in Fig. a, Am = p A 32 B + 4.5(6) = 55.2743 ft2 ¢ F
144 in2 ≤ = 7959.50 in2 1 ft2
ds = 2p(3) + 2(4.5) = 27.8496 fta
12 in. b = 334.19 in. 1 ft
Allowable Average Shear Stress:
A tavg B allow =
T ; 2tAm
18 =
T 2(0.25)(7959.50)
T = 71635.54 kip # ina
1ft b = 5970 kip # ft 12 in.
Angle of Twist: f =
ds TL 4Am 2G F t
0.001 =
T(1)(12) 4(7959.502)(3.9)(103)
T = 61610.65 kip # ina
a
334.19 b 0.25
1ft b = 5134 kip # ft (controls) 12 in.
Ans.
297
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5–115. The tube is subjected to a torque of 750 N # m. Determine the average shear stress in the tube at points A and B.
4 mm 6 mm
A
100 mm
Referring to the geometry shown in Fig. a,
6 mm 2
Am = 0.06 (0.1) = 0.006 m
B 750 N⭈m
Thus, (tavg)A
T 750 = = = 15.63(106)Pa = 15.6 MPa 2tA Am 2(0.004)(0.006)
Ans.
T 750 = = 10.42(106)Pa = 10.4 MPa 2tB Am 2(0.006)(0.006)
Ans.
(tavg)B =
4 mm 60 mm
*5–116. The tube is made of plastic, is 5 mm thick, and has the mean dimensions shown. Determine the average shear stress at points A and B if it is subjected to the torque of T = 5 N # m. Show the shear stress on volume elements located at these points. Am = (0.11)(0.08) +
tA = tB = tavg =
A
1 (0.08)(0.03) = 0.01 m2 2
B
50 mm 60 mm
T 5 = = 50 kPa 2tAm 2(0.005)(0.01)
T
Ans. 30 mm 40 mm
298
40 mm
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•5–117.
The mean dimensions of the cross section of the leading edge and torsion box of an airplane wing can be approximated as shown. If the wing is made of 2014-T6 aluminum alloy having an allowable shear stress of tallow = 125 MPa and the wall thickness is 10 mm, determine the maximum allowable torque and the corresponding angle of twist per meter length of the wing.
10 mm 0.5 m
10 mm
Section Properties: Referring to the geometry shown in Fig. a, Am =
F
p 1 a 0.52 b + A 1 + 0.5 B (2) = 1.8927 m2 2 2
ds = p(0.5) + 2 222 + 0.252 + 0.5 = 6.1019 m
Allowable Average Shear Stress:
A tavg B allow =
T ; 2tAm
125(106) =
T 2(0.01)(1.8927)
T = 4.7317(106)N # m = 4.73 MN # m
Ans.
Angle of Twist: f =
ds TL 2 4Am G F t 4.7317(106)(1)
=
4(1.89272)(27)(109)
¢
0.25 m 10 mm
6.1019 ≤ 0.01
= 7.463(10 - 3) rad = 0.428°>m
Ans.
299
2m
0.25 m
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5–118. The mean dimensions of the cross section of the leading edge and torsion box of an airplane wing can be approximated as shown. If the wing is subjected to a torque of 4.5 MN # m and the wall thickness is 10 mm, determine the average shear stress developed in the wing and the angle of twist per meter length of the wing. The wing is made of 2014-T6 aluminum alloy.
10 mm 0.5 m
10 mm
Section Properties: Referring to the geometry shown in Fig. a, Am =
F
p 1 a0.52 b + A 1 + 0.5 B (2) = 1.8927 m2 2 2
ds = p(0.5) + 2222 + 0.252 + 0.5 = 6.1019 m
Average Shear Stress: tavg =
4.5(106) T = = 119 MPa 2tAm 2(0.01)(1.8927)
Ans.
Angle of Twist: f =
ds TL 4Am 2G F t 4.5(106)(1)
=
4(1.89272)(27)(109)
¢
0.25 m 10 mm
6.1019 ≤ 0.01
= 7.0973(10 - 3) rad = 0.407°>m
Ans.
300
2m
0.25 m
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5–119. The symmetric tube is made from a high-strength steel, having the mean dimensions shown and a thickness of 5 mm. If it is subjected to a torque of T = 40 N # m, determine the average shear stress developed at points A and B. Indicate the shear stress on volume elements located at these points.
20 mm
30 mm
60 mm A B
40 Nm
Am = 4(0.04)(0.06) + (0.04)2 = 0.0112 m2 tavg =
T 2 t Am
(tavg)A = (tavg)B =
40 = 357 kPa 2(0.005)(0.0112)
Ans.
*5–120. The steel used for the shaft has an allowable shear stress of tallow = 8 MPa. If the members are connected with a fillet weld of radius r = 4 mm, determine the maximum torque T that can be applied.
50 mm
20 mm
T 2
Allowable Shear Stress: D 50 = = 2.5 d 20
and
4 r = = 0.20 d 20
From the text, K = 1.25 tmax = tallow = K
Tc J t 2 (0.01) 4 R 2 (0.01 )
8(10)4 = 1.25 B p T = 20.1 N # m
Ans.
301
T
20 mm
T 2
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•5–121.
The built-up shaft is to be designed to rotate at 720 rpm while transmitting 30 kW of power. Is this possible? The allowable shear stress is tallow = 12 MPa. v = 720
T =
75 mm
rev 2p rad 1 min a b = 24 p rad>s min 1 rev 60 s
60 mm
30(103) P = = 397.89 N # m v 24 p
tmax = K
Tc ; J
12(106) = Kc
397.89(0.03) p 4 2 (0.03 )
d;
K = 1.28
D 75 = = 1.25 d 60 From Fig. 5-32,
r = 0.133 d
r = 0.133 ; 60
r = 7.98 mm
Check: D - d 75 - 60 15 = = = 7.5 mm 6 7.98 mm 2 2 2 No, it is not possible.
Ans.
5–122. The built-up shaft is designed to rotate at 540 rpm. If the radius of the fillet weld connecting the shafts is r = 7.20 mm, and the allowable shear stress for the material is tallow = 55 MPa, determine the maximum power the shaft can transmit. D 75 = = 1.25; d 60
75 mm
60 mm
r 7.2 = = 0.12 d 60
From Fig. 5-32, K = 1.30 tmax = K
v = 540
Tc ; J
T(0.03) d; 55(106) = 1.30 c[ p 4 2 (0.03 )
T = 1794.33 N # m
rev 2p rad 1 min a b = 18 p rad>s min 1 rev 60 s
P = Tv = 1794.33(18p) = 101466 W = 101 kW
Ans.
302
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5–123. The steel shaft is made from two segments: AB and BC, which are connected using a fillet weld having a radius of 2.8 mm. Determine the maximum shear stress developed in the shaft.
C 50 mm
D
100 N⭈m
20 mm B
(tmax)CD =
100(0.025) TCDc = p 4 J 2 (0.025 )
40 N⭈m
A
= 4.07 MPa
60 N⭈m
For the fillet: D 50 = = 2.5; d 20
r 2.8 = = 0.14 d 20
From Fig. 5-32, K = 1.325 (tmax)f = K
60(0.01) TABc d = 1.325 c p 4 J 2 (0.01 ) = 50.6 MPa (max)
Ans.
*5–124. The steel used for the shaft has an allowable shear stress of tallow = 8 MPa. If the members are connected together with a fillet weld of radius r = 2.25 mm, determine the maximum torque T that can be applied.
30 mm
30 mm
15 mm
T T 2
Allowable Shear Stress: D 30 = = 2 d 15
r 2.25 = = 0.15 d 15
and
From the text, K = 1.30 tmax = tallow = K
Tc J
8(106) = 1.3 C
A 2r B (0.0075) p 4 2 (0.0075 )
S
T = 8.16 N # m
Ans.
303
T 2
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•5–125. The assembly is subjected to a torque of 710 lb # in. If the allowable shear stress for the material is tallow = 12 ksi, determine the radius of the smallest size fillet that can be used to transmit the torque.
tmax = tallow = K
0.75 in. A
Tc J
710 lb⭈in.
B 1.5 in.
3
12(10 ) =
K(710)(0.375) p 4 2 (0.375 )
C
K = 1.40 710 lb⭈ft
D 1.5 = = 2 d 0.75 From Fig. 5-32, r = 0.1; d
r = 0.1(0.75) = 0.075 in.
Ans.
Check: D - d 1.5 - 0.75 = = 0.375 7 0.075 in. 2 2
OK
5–126. A solid shaft is subjected to the torque T, which causes the material to yield. If the material is elastic plastic, show that the torque can be expressed in terms of the angle of twist f of the shaft as T = 43 TY11 - f3Y>4f32, where TY and fY are the torque and angle of twist when the material begins to yield. gY gL = L r rY
f =
rY =
gYL f
(1)
When rY = c, f = fY From Eq. (1), c =
gYL fY
(2)
Dividing Eq. (1) by Eq. (2) yields: rY fY = c f
(3)
Use Eq. 5-26 from the text. T =
r3Y p tY 2p tYc3 (4 c3 - r3Y) = a1 )b 6 3 4 c3
Use Eq. 5-24, TY =
T =
p t c3 from the text and Eq. (3) 2 Y
f3Y 4 TY a1 b 3 4 f3
QED
304
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5–127. A solid shaft having a diameter of 2 in. is made of elastic-plastic material having a yield stress of tY = 16 ksi and shear modulus of G = 1211032 ksi. Determine the torque required to develop an elastic core in the shaft having a diameter of 1 in. Also, what is the plastic torque?
Use Eq. 5-26 from the text: T =
p (16) p tY (4 c3 - rY 3) = [4(13) - 0.53] 6 6
= 32.46 kip # in. = 2.71 kip # ft
Ans.
Use Eq. 5-27 from the text: TP =
2p 2p t c3 = (16)(13) 3 Y 3
= 33.51 kip # in. = 2.79 kip # ft
Ans.
*5–128. Determine the torque needed to twist a short 3-mm-diameter steel wire through several revolutions if it is made from steel assumed to be elastic plastic and having a yield stress of tY = 80 MPa. Assume that the material becomes fully plastic.
When the material becomes fully plastic then, from Eq. 5-27 in the text, TP =
2 p (80)(106) 2 p tY 3 c = (0.00153) = 0.565 N # m 3 3
305
Ans.
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•5–129.
The solid shaft is made of an elastic-perfectly plastic material as shown. Determine the torque T needed to form an elastic core in the shaft having a radius of rY = 20 mm. If the shaft is 3 m long, through what angle does one end of the shaft twist with respect to the other end? When the torque is removed, determine the residual stress distribution in the shaft and the permanent angle of twist.
80 mm T
T
t (MPa) 160
Elastic-Plastic Torque: Applying Eq. 5-26 from the text T =
=
0.004
p tY A 4c3 - r3Y B 6 p(160)(106) C 4 A 0.043 B - 0.023 D 6
= 20776.40 N # m = 20.8 kN # m
Ans.
Angle of Twist: gY 0.004 L = a b(3) = 0.600 rad = 34.4° rY 0.02
f =
Ans.
When the reverse T = 20776.4 N # m is applied, G =
160(106) = 40 GPa 0.004
f¿ =
TL = JG
20776.4(3) p 4 9 2 (0.04 )(40)(10 )
= 0.3875 rad
The permanent angle of twist is, fr = f - f¿ = 0.600 - 0.3875 = 0.2125 rad = 12.2°
Ans.
Residual Shear Stress: (t¿)r = c =
20776.4(0.04) Tc = 206.67 MPa = p 4 J 2 (0.04 )
(t¿)r = 0.02 m =
20776.4(0.02) Tc = 103.33 MPa = p 4 J 2 (0.04 )
(tr)r = c = -160 + 206.67 = 46.7 MPa (tr)r = 0.02m = -160 + 103.33 = -56.7 MPa
306
g (rad)
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5–130. The shaft is subjected to a maximum shear strain of 0.0048 rad. Determine the torque applied to the shaft if the material has strain hardening as shown by the shear stress–strain diagram. 2 in.
From the shear - strain diagram, rY 2 = ; 0.0006 0.0048
T
rY = 0.25 in.
t (ksi) 12
From the shear stress–strain diagram, t1 =
6 r = 24r 0.25
6
t2 - 6 12 - 6 = ; r - 0.25 2 - 0.25
t2 = 3.4286 r + 5.1429
0.0006
0.0048
c
T = 2p
L0
t r2 dr 0.25
= 2p
2
24r3 dr + 2p
L0
= 2p[6r4] | + 2p c 0.25 0
L0.25
(3.4286r + 5.1429)r2 dr
3.4286r4 5.1429r3 2 + d | 4 3 0.25
= 172.30 kip # in. = 14.4 kip # ft
Ans.
5–131. An 80-mm diameter solid circular shaft is made of an elastic-perfectly plastic material having a yield shear stress of tY = 125 MPa. Determine (a) the maximum elastic torque TY; and (b) the plastic torque Tp. Maximum Elastic Torque. TY =
=
1 3 pc tY 2 1 pa 0.043 b A 125 B a 106 b 2
= 12 566.37 N # m = 12.6 kN # m
Ans.
Plastic Torque. TP =
=
2 3 pc tY 3 2 pa 0.043 b A 125 B a 106 b 3
= 16755.16 N # m = 16.8 kN # m
Ans. 307
g (rad)
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*5–132. The hollow shaft has the cross section shown and is made of an elastic-perfectly plastic material having a yield shear stress of tY. Determine the ratio of the plastic torque Tp to the maximum elastic torque TY.
c c 2
Maximum Elastic Torque. In this case, the torsion formula is still applicable. tY =
TY c J
TY =
J t c Y c 4 p 4 B c - a b R tY 2 2
=
=
c 15 3 pc tY 32
Plastic Torque. Using the general equation, with t = tY, c
TP = 2ptY
r2dr Lc>2 c
r3 = 2ptY ¢ ≤ ` 3 c>2 =
7 pc3tY 12
The ratio is 7 pc3tY TP 12 = = 1.24 TY 15 3 pc tY 32
Ans.
5–133. The shaft consists of two sections that are rigidly connected. If the material is elastic plastic as shown, determine the largest torque T that can be applied to the shaft. Also, draw the shear-stress distribution over a radial line for each section. Neglect the effect of stress concentration.
T 1 in.
0.75 in. T
0.75 in. diameter segment will be fully plastic. From Eq. 5-27 of the text: T = Tp =
2p tY 3 (c ) 3
t (ksi) 12
3
=
2p (12)(10 ) (0.3753) 3
= 1325.36 lb # in. = 110 lb # ft
Ans.
For 1 – in. diameter segment: tmax =
1325.36(0.5) Tc = p 4 J 2 (0.5)
= 6.75 ksi 6 tY
308
0.005
g (rad)
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5–134. The hollow shaft is made of an elastic-perfectly plastic material having a shear modulus of G and a yield shear stress of tY. Determine the applied torque Tp when the material of the inner surface is about to yield (plastic torque). Also, find the corresponding angle of twist and the maximum shear strain. The shaft has a length of L.
ci
Plastic Torque. Using the general equation with t = tY, co
TP = 2ptY
Lci
r2dr co
r3 = 2ptY ¢ ≤ ` 3 ci =
2 pt A c 3 - ci 3 B 3 Y o
Ans.
Angle of Twist. When the material is about to yield at the inner surface, g = gY at r = rY = ci. Also, Hooke’s Law is still valid at the inner surface. gY =
f =
tY G gY tY>G tYL L = L = rY ci ciG
Ans.
Shear Strain. Since the shear strain varies linearly along the radial line, Fig. a, gmax gY = co ci gmax = ¢
co co tY cotY ≤ gY = ¢ ≤ a b = ci ci G ciG
Ans.
309
c0
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5–135. The hollow shaft has inner and outer diameters of 60 mm and 80 mm, respectively. If it is made of an elasticperfectly plastic material, which has the t-g diagram shown, determine the reactions at the fixed supports A and C.
150 mm 450 mm B
C 15 kN⭈m
A
t (MPa)
120
Equation of Equilibrium. Refering to the free - body diagram of the shaft shown in Fig. a, ©Mx = 0; TA + TC - 15 A 103 B = 0
(1)
Elastic Analysis. It is required that fB>A = fB>C. Thus, the compatibility equation is fB>A = fB>C TCLBC TALAB = JG JG TA (0.45) = TC(0.15) TC = 3TA
(2)
Solving Eqs. (1) and (2), TA = 3750 N # m
TC = 11 250N # m
The maximum elastic torque and plastic torque in the shaft can be determined from p A 0.044 - 0.034 B J 2 T(120) A 106 B = 8246.68 N # m TY = tY = D c 0.04
co
TP = 2ptY
Lci
r2dr
= 2p(120) A 106 B ¢
g (rad) 0.0016
0.04 m
r3 = 9299.11 N # m ≤` 3 0.03 m
Since TC 7 TY, the results obtained using the elastic analysis are not valid. Plastic Analysis. Assuming that segment BC is fully plastic, TC = TP = 9299.11N # m = 9.3kN # m
Ans.
Substituting this result into Eq. (1), TA = 5700 N # m = 5.70 kN # m
Ans.
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5–135.
Continued
Since TA 6 TY, segment AB of the shaft is still linearly elastic. Here, 120 A 106 B = 75GPa. G = 0.0016 fB>C = fB>A =
fB>C =
gi L ; ci BC
5700.89(0.45) TALAB = 0.01244 rad = p JG A 0.044 - 0.034 B (75) A 109 B 2 0.01244 =
gi (0.15) 0.03
gi = 0.002489 rad Since gi 7 gY, segment BC of the shaft is indeed fully plastic.
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*5–136. The tubular shaft is made of a strain-hardening material having a t-g diagram as shown. Determine the torque T that must be applied to the shaft so that the maximum shear strain is 0.01 rad.
T 0.5 in. 0.75 in.
t (ksi) 15 10
0.005
From the shear–strain diagram, g 0.01 = ; 0.5 0.75
g = 0.006667 rad
From the shear stress–strain diagram, 15 - 10 t - 10 = ; t = 11.667 ksi 0.006667 - 0.005 0.01 - 0.005 15 - 11.667 t - 11.667 = ; r - 0.5 0.75 - 0.50
t = 13.333 r + 5
co
T = 2p
tr2 dr
Lci
0.75
= 2p
(13.333r + 5) r2 dr
L0.5 0.75
= 2p
L0.5
= 2p c
(13.333r3 + 5r2) dr
13.333r4 5r3 0.75 + d | 4 3 0.5
= 8.426 kip # in. = 702 lb # ft
Ans.
312
0.01
g (rad)
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•5–137.
The shear stress–strain diagram for a solid 50-mm-diameter shaft can be approximated as shown in the figure. Determine the torque T required to cause a maximum shear stress in the shaft of 125 MPa. If the shaft is 1.5 m long, what is the corresponding angle of twist?
T
1.5 m T t (MPa) 125 50
Strain Diagram: rg 0.0025
=
0.025 ; 0.01
0.0025
rg = 0.00625 m
Stress Diagram: t1 =
50(106) r = 8(109) r 0.00625
t2 - 50(106) 125(106) - 50(106) = r - 0.00625 0.025 - 0.00625 t2 = 4 A 109 B r + 25 A 106 B The Ultimate Torque: c
T = 2p
L0
t r2dr 0.00625 m
= 2p
L0
8 A 109 B r3 dr 0.025 m
+ 2p
L0.00625 m
9 6 C 4 A 10 B r + 25 A 10 B D r2dr
m = 2p C 2 A 109 B r4 D |0.00625 0
+ 2p B 1 A 109 B r4 +
25(106)r3 0.025 m R 2 3 0.00625 m
= 3269.30 N # m = 3.27 kN # m
Ans.
Angle of Twist: f =
gmax 0.01 L = a b (1.5) = 0.60 rad = 34.4° c 0.025
Ans.
313
0.010
g (rad)
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5–138. A tube is made of elastic-perfectly plastic material, which has the t-g diagram shown. If the radius of the elastic core is rY = 2.25 in., determine the applied torque T. Also, find the residual shear-stress distribution in the shaft and the permanent angle of twist of one end relative to the other when the torque is removed.
3 ft
3 in. T
T
6 in.
t (ksi)
Elastic - Plastic Torque. The shear stress distribution due to T is shown in Fig. a. The 10 r = 4.444r. Thus, linear portion of this distribution can be expressed as t = 2.25 tr = 1.5 in. = 4.444(1.5) = 6.667 ksi.
T = 2p
L
tr2dr
g (rad) 0.004
2.25 in.
= 2p
L1.5 in.
= 8.889p ¢
4.444r A r2dr B + 2p(10)
3 in.
L2.25 in.
r2dr
r4 2.25 in. r3 3 in. + 20p ¢ ≤ 2 ≤2 4 1.5 in. 3 2.25 in.
= 470.50 kip # in = 39.2 kip # ft
Ans.
Angle of Twist. f =
gY 0.004 L = (3)(12) = 0.064 rad rY 2.25
The process of removing torque T is equivalent to the application of T¿ , which is equal magnitude but opposite in sense to that of T. This process occurs in a linear 10 = 2.5 A 103 B ksi. manner and G = 0.004 f¿ =
10
T¿L = JG
470.50(3)(2)
p 2
A 34 - 1.54 B (2.5) A 103 B 470.50(3)
= 0.0568 rad
trœ = co =
T¿co = J
trœ = rY =
T¿rY 470.50(2.25) = = 8.875 ksi p 4 4 J 2 A 3 - 1.5 B
trœ = ci =
470.50(1.5) T¿ci = = 5.917 ksi p 4 4 J 2 A 3 - 1.5 B
p 2
A 34 - 1.54 B
= 11.83 ksi
Thus, the permanent angle of twist is fP = f - f¿ = 0.064 - 0.0568 = 0.0072 rad = 0.413°
Ans.
314
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5–138.
Continued
And the residual stresses are (tr)r = co = tr = c + trœ = c = -10 + 11.83 = 1.83 ksi (tr)r = rY = tr = rY + trœ = rY = -10 + 8.875 = -1.125 ksi (tr)r = ci = tr = ci + trœ = ci = -6.667 + 5.917 = -0.750 ksi The residual stress distribution is shown in Fig. a.
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5–139. The tube is made of elastic-perfectly plastic material, which has the t-g diagram shown. Determine the torque T that just causes the inner surface of the shaft to yield. Also, find the residual shear-stress distribution in the shaft when the torque is removed.
3 ft
3 in. T
T
6 in.
t (ksi)
Plastic Torque. When the inner surface of the shaft is about to yield, the shaft is about to become fully plastic. T = 2p
L
10
tr2dr 3 in.
= 2ptY
L1.5 in.
= 2p(10)a
g (rad)
r2dr
0.004
r3 3 in. b2 3 1.5 in.
= 494.80 kip # in. = 41.2 kip # ft
Ans.
Angle of Twist. f =
gY 0.004 (3)(12) = 0.096 rad L = rY 1.5
The process of removing torque T is equivalent to the application of T¿ , which is equal magnitude but opposite in sense to that of T. This process occurs in a linear 10 manner and G = = 2.5 A 103 B ksi. 0.004 f¿ =
494.80(3)(12) T¿L = = 0.05973 rad p 4 JG A 3 - 1.54 B (2.5) A 103 B 2
trœ = co =
trœ = ci =
494.80(3) T¿co = = 12.44 ksi p 4 J 4 3 1.5 A B 2 494.80(1.5) T¿ci = = 6.222 ksi p 4 J A 3 - 1.54 B 2
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5–139.
Continued
And the residual stresses are (tr)r = co = tr = c + trœ = c = -10 + 12.44 = 2.44 ksi
Ans.
(tr)r = ci = tr = ci + trœ = ci = -10 + 6.22 = -3.78 ksi
Ans.
The shear stress distribution due to T and T¿ and the residual stress distribution are shown in Fig. a.
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*5–140. The 2-m-long tube is made of an elastic-perfectly plastic material as shown. Determine the applied torque T that subjects the material at the tube’s outer edge to a shear strain of gmax = 0.006 rad. What would be the permanent angle of twist of the tube when this torque is removed? Sketch the residual stress distribution in the tube.
T 35 mm
30 mm t (MPa) 210
Plastic Torque: The tube is fully plastic if gi Ú gr = 0.003 rad. g 0.006 = ; 0.03 0.035
0.003
g = 0.005143 rad
Therefore the tube is fully plastic. co
TP = 2p
Lci
2p tg =
=
3
tg r2 dr
A c3o - c3i B
2p(210)(106) A 0.0353 - 0.033 B 3
= 6982.19 N # m = 6.98 kN # m
Ans.
Angle of Twist: fP =
gmax 0.006 L = a b(2) = 0.34286 rad co 0.035
When a reverse torque of TP = 6982.19 N # m is applied, G =
fPœ =
210(106) tY = = 70 GPa gY 0.003 TPL = JG
6982.19(2) p 4 2 (0.035
- 0.034)(70)(109)
= 0.18389 rad
Permanent angle of twist, fr = fP - fPœ = 0.34286 - 0.18389 = 0.1590 rad = 9.11°
Ans.
Residual Shear Stress: 6982.19(0.035)
tPœ o =
TP c = J
p 4 2 (0.035
tPœ i =
TP r = J
p 4 2 (0.035
- 0.034)
6982.19(0.03) - 0.034)
= 225.27 MPa
= 193.09 MPa
(tP)o = -tg + tPœ o = -210 + 225.27 = 15.3 MPa (tP)i = -tg + tPœ i = -210 + 193.09 = -16.9 MPa
318
g (rad)
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•5–141.
A steel alloy core is bonded firmly to the copper alloy tube to form the shaft shown. If the materials have the t-g diagrams shown, determine the torque resisted by the core and the tube.
450 mm A
100 mm 60 mm
B 15 kN⭈m
t (MPa)
Equation of Equilibrium. Refering to the free - body diagram of the cut part of the assembly shown in Fig. a, ©Mx = 0; Tc + Tt - 15 A 103 B = 0
180
(1)
Elastic Analysis. The shear modulus of steel and copper are Gst = 36 A 106 B and G q = = 18 GPa. Compatibility requires that 0.002
180 A 106 B 0.0024
g (rad) 0.0024
= 75 GPa
Steel Alloy t (MPa)
fC = ft 36
TcL TtL = JcGst JtG q
g (rad) 0.002
Tc
p 2
A 0.03 B (75) A 10 B 4
Tt
=
9
p 2
Copper Alloy
A 0.05 - 0.034 B (18) A 109 B 4
Tc = 0.6204Tt
(2)
Solving Eqs. (1) and (2), Tt = 9256.95 N # m
Tc = 5743.05 N # m
The maximum elastic torque and plastic torque of the core and the tube are (TY)c =
1 3 1 pc (tY)st = p A 0.033 B (180) A 106 B = 7634.07 N # m 2 2
(TP)c =
2 3 2 pc (tY)st = p A 0.033 B (180) A 106 B = 10 178.76 N # m 3 3
and p A 0.054 - 0.034 B J 2 T c(36) A 106 B d = 6152.49 N # m (TY)t = tY = D c 0.05
r2dr = 2p(36) A 106 B ¢
co
(TP)t = 2p(tY) q
Lci
r3 0.05 m = 7389.03 N # m ≤2 3 0.03 m
Since Tt 7 (TY)t, the results obtained using the elastic analysis are not valid.
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5–141.
Continued
Plastic Analysis. Assuming that the tube is fully plastic, Tt = (TP)t = 7389.03 N # m = 7.39 kN # m
Ans.
Substituting this result into Eq. (1), Tc = 7610.97 N # m = 7.61 kN # m
Ans.
Since Tc 6 (TY)c, the core is still linearly elastic. Thus, ft = ftc =
ft =
gi L; ci
TcL = JcGst
7610.97(0.45) p 4 9 2 (0.03 )(75)(10 )
0.3589 =
= 0.03589 rad
gi (0.45) 0.03
gi = 0.002393 rad Since gi 7 (gY) q = 0.002 rad, the tube is indeed fully plastic.
320
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5–142. A torque is applied to the shaft of radius r. If the material has a shear stress–strain relation of t = kg1>6, where k is a constant, determine the maximum shear stress in the shaft. r
r r g = gmax = gmax c r
T
1
gmax 6 1 b r6 t = kg = ka r 1 6
r
T = 2p
tr2 dr
L0
1
r
= 2p
L0
gmax = a
ka
1
1
gmax 6 13 gmax 6 6 12p kg6max r3 19 b r 6 dr = 2pk a b a b r6 = r r 19 19
6 19T b 3 12p kr
19T 12p r3
1
tmax = kg6max =
Ans.
5–143. Consider a thin-walled tube of mean radius r and thickness t. Show that the maximum shear stress in the tube due to an applied torque T approaches the average shear stress computed from Eq. 5–18 as r>t : q .
t
r
t 2r + t ; ro = r + = 2 2 J =
=
t 2r - t ri = r - = 2 2
2r - t 4 p 2r + t 4 ca b - a b d 2 2 2 p p [(2r + t)4 - (2r - t)4] = [64 r3 t + 16 r t3] 32 32
tmax =
Tc ; J
c = ro =
2r + t 2
T(2 r 2+ t) =
p 3 32 [64 r t 2r T(2r 2 +
=
+ 16 r t3]
2p r t[r2 + 14t2]
t 2 r2 )
2p r t c rr2 + 2
As
T(2 r 2+ t) =
1 t2 4 r2 d
t r : q , then : 0 r t
tmax =
=
T(1r + 0) 2p r t(1 + 0)
=
T 2p r2 t
T 2 t Am
QED
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*5–144. The 304 stainless steel shaft is 3 m long and has an outer diameter of 60 mm. When it is rotating at 60 rad>s, it transmits 30 kW of power from the engine E to the generator G. Determine the smallest thickness of the shaft if the allowable shear stress is tallow = 150 MPa and the shaft is restricted not to twist more than 0.08 rad.
E
Internal Torque: P = 30(103) W a T =
1 N # m>s b = 30(103) N # m>s W
30(103) P = = 500 N # m v 60
Allowable Shear Stress: Assume failure due to shear stress. tmax = tallow = 150(106) =
Tc J 500(0.03) p 4 2 (0.03
- r4i )
ri = 0.0293923 m = 29.3923 mm Angle of Twist: Assume failure due to angle of twist limitation. f =
0.08 =
TL JG 500(3)
p 2
A 0.03 - r4i B (75.0)(109) 4
ri = 0.0284033 m = 28.4033 mm Choose the smallest value of ri = 28.4033 mm t = ro - ri = 30 - 28.4033 = 1.60 mm
Ans.
322
G
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•5–145.
The A-36 steel circular tube is subjected to a torque of 10 kN # m. Determine the shear stress at the mean radius r = 60 mm and compute the angle of twist of the tube if it is 4 m long and fixed at its far end. Solve the problem using Eqs. 5–7 and 5–15 and by using Eqs. 5–18 and 5–20.
r 60 mm 4m
t 5 mm 10 kNm
Shear Stress: Applying Eq. 5-7, ro = 0.06 +
tr = 0.06 m =
0.005 = 0.0625 m 2
Tr = J
ri = 0.06 -
10(103)(0.06) p 4 2 (0.0625
- 0.05754)
0.005 = 0.0575 m 2
= 88.27 MPa
Ans.
Applying Eq. 5-18, tavg =
10(103) T = 88.42 MPa = 2 t Am 29(0.005)(p)(0.062)
Ans.
Angle of Twist: Applying Eq. 5-15, f =
TL JG 10(103)(4)
=
p 4 2 (0.0625
- 0.05754)(75.0)(109)
= 0.0785 rad = 4.495°
Ans.
Applying Eq. 5-20, f =
=
ds TL 4A2mG L t TL ds 4A2mG t L
Where
L
ds = 2pr
2pTLr =
4A2mG t 2p(10)(103)(4)(0.06)
=
4[(p)(0.062)]2 (75.0)(109)(0.005)
= 0.0786 rad = 4.503°
Ans.
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5–146. Rod AB is made of A-36 steel with an allowable shear stress of 1tallow2st = 75 MPa, and tube BC is made of AM1004-T61 magnesium alloy with an allowable shear stress of 1tallow2mg = 45 MPa. The angle of twist of end C is not allowed to exceed 0.05 rad. Determine the maximum allowable torque T that can be applied to the assembly.
0.3 m
0.4 m a A C 60 mm
Internal Loading: The internal torque developed in rod AB and tube BC are shown in Figs. a and b, respectively. Allowable Shear Stress: The polar moment of inertia of rod AB and tube p p a0.0154 b = 25.3125(10 - 9)p m4 and JBC = a0.034 - 0.0254 b BC are JAB = 2 2 = 0.2096875(10 - 6)p m4. We have
A tallow B st =
TAB cAB ; JAB
75(106) =
T(0.015) 25.3125(10 - 9)p
T = 397.61 N # m and
A tallow B mg =
TBC cBC ; JBC
45(106) =
T(0.03) 0.2096875(10 - 6)p
T = 988.13 N # m Angle of Twist: fB>A =
-T(0.7) TAB LAB = -0.11737(10 - 3)T = 0.11737(10 - 3)T = JAB Gst 25.3125(10 - 9)p(75)(109)
and fC>B =
T(0.4) TBC LBC = 0.03373(10 - 3)T = JBC Gmg 0.2096875(10 - 6)p(18)(109)
It is required that fC>A = 0.05 rad. Thus, fC>A = fB>A + fC>B 0.05 = 0.11737(10 - 3)T + 0.03373(10 - 3)T T = 331 N # m A controls B
Ans.
324
50 mm
30 mm Section a–a
T
a
B
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5–147. A shaft has the cross section shown and is made of 2014-T6 aluminum alloy having an allowable shear stress of tallow = 125 MPa. If the angle of twist per meter length is not allowed to exceed 0.03 rad, determine the required minimum wall thickness t to the nearest millimeter when the shaft is subjected to a torque of T = 15 kN # m.
30⬚ 30⬚
75 mm
Section Properties: Referring to the geometry shown in Fig. a, Am =
C
0.075 1 1 (0.15) ¢ ≤ + p A 0.0752 B = 0.01858 m2 2 tan 30° 2
ds = 2(0.15) + p(0.075) = 0.53562 m
Allowable Shear Stress:
A tavg B allow =
T ; 2tAm
125(106) =
15(103) 2t(0.01858)
t = 0.00323 m = 3.23 mm Angle of Twist: f =
ds TL 2 4Am G C t
0.03 =
15(103)(1) 4(0.018582)(27)(109)
a
0.53562 b t
t = 0.007184 m = 7.18 mm (controls) Use t = 8 mm
Ans.
325
t
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*5–148. The motor A develops a torque at gear B of 500 lb # ft, which is applied along the axis of the 2-in.diameter A-36 steel shaft CD. This torque is to be transmitted to the pinion gears at E and F. If these gears are temporarily fixed, determine the maximum shear stress in segments CB and BD of the shaft. Also, what is the angle of twist of each of these segments? The bearings at C and D only exert force reactions on the shaft.
B E
F 2 ft
1.5 ft
C
D A
Equilibrium: TC + TD - 500 = 0
[1]
Compatibility: fB>C = fB>D TC(2) TD(1.5) = JG JG TC = 0.75TD
[2]
Solving Eqs. [1] and [2] yields: TD = 285.71 lb # ft
TC = 214.29 lb # ft
Maximum Shear Stress: (tCB)max =
214.29(12)(1) TCc = 1.64 ksi = p 4 J 2 (1 )
Ans.
(tBD)max =
285.71(12)(1) TDc = 2.18 ksi = p 4 J 2 (1 )
Ans.
Angle of Twist: fCB = fBD =
TD LBD JG 285.71(12)(1.5)(12)
=
p 2
(14)(11.0)(106)
= 0.003572 rad = 0.205°
Ans.
326
500 lb·ft
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5–149. The coupling consists of two disks fixed to separate shafts, each 25 mm in diameter. The shafts are supported on journal bearings that allow free rotation. In order to limit the torque T that can be transmitted, a “shear pin” P is used to connect the disks together. If this pin can sustain an average shear force of 550 N before it fails, determine the maximum constant torque T that can be transmitted from one shaft to the other. Also, what is the maximum shear stress in each shaft when the “shear pin” is about to fail?
25 mm
P
130 mm 25 mm
T
Equilibrium: T - 550(0.13) = 0
©Mx = 0;
T = 71.5 N # m
Ans.
Maximum Shear Stress: tmax =
71.5(0.0125) Tc = 23.3 MPa = p 4 J 2 (0.0125 )
Ans.
5–150. The rotating flywheel and shaft is brought to a sudden stop at D when the bearing freezes. This causes the flywheel to oscillate clockwise–counterclockwise, so that a point A on the outer edge of the flywheel is displaced through a 10-mm arc in either direction. Determine the maximum shear stress developed in the tubular 304 stainless steel shaft due to this oscillation. The shaft has an inner diameter of 25 mm and an outer diameter of 35 mm. The journal bearings at B and C allow the shaft to rotate freely.
D 2m
B A
80 mm
Angle of Twist: f =
0.125 =
TL JG
Where
f =
10 = 0.125 rad 80
T(2) p 4 2 (0.0175
- 0.01254)(75.0)(109)
T = 510.82 N # m Maximum Shear Stress: tmax =
Tc = J
510.82(0.0175) p 4 2 (0.0175
- 0.01254)
Ans.
= 82.0 MPa
327
C
T
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5–151. If the solid shaft AB to which the valve handle is attached is made of C83400 red brass and has a diameter of 10 mm, determine the maximum couple forces F that can be applied to the handle just before the material starts to fail. Take tallow = 40 MPa. What is the angle of twist of the handle? The shaft is fixed at A.
B
A 150 mm 150 mm F 150 mm
tmax = tallow = 40(106) =
Tc J
F
0.3F(0.005) p 4 2 (0.005)
F = 26.18 N = 26.2 N
Ans.
T = 0.3F = 7.85 N # m f =
TL = JG
7.85(0.15) p 4 9 2 (0.005) (37)(10 )
= 0.03243 rad = 1.86°
Ans.
328
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6–1. Draw the shear and moment diagrams for the shaft. The bearings at A and B exert only vertical reactions on the shaft.
B
A
800 mm
250 mm
24 kN
6–2. Draw the shear and moment diagrams for the simply supported beam.
4 kN M 2 kNm A
B 2m
329
2m
2m
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6–3. The engine crane is used to support the engine, which has a weight of 1200 lb. Draw the shear and moment diagrams of the boom ABC when it is in the horizontal position shown. a + ©MA = 0;
4 F (3) - 1200(8) = 0; 5 A
+ c ©Fy = 0;
-Ay +
+ ©F = 0; ; x
Ax -
4 (4000) - 1200 = 0; 5
3 (4000) = 0; 5
A
3 ft
5 ft B
FA = 4000 lb
4 ft
Ay = 2000 lb
Ax = 2400 lb
*6–4. Draw the shear and moment diagrams for the cantilever beam.
2 kN/m
A
6 kNm 2m
The free-body diagram of the beam’s right segment sectioned through an arbitrary point shown in Fig. a will be used to write the shear and moment equations of the beam. + c ©Fy = 0;
C
V - 2(2 - x) = 0
V = {4 - 2x} kN‚
(1)
1 a + ©M = 0; -M - 2(2 - x)c (2 - x) d - 6 = 0 M = {-x2 + 4x - 10}kN # m‚(2) 2 The shear and moment diagrams shown in Figs. b and c are plotted using Eqs. (1) and (2), respectively. The value of the shear and moment at x = 0 is evaluated using Eqs. (1) and (2). Vx = 0 = 4 - 2(0) = 4 kN Mx = 0 = C -0 + 4(0) - 10 D = -10kN # m
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6–5. Draw the shear and moment diagrams for the beam.
10 kN
8 kN
15 kNm
2m
3m
6–6. Draw the shear and moment diagrams for the overhang beam.
8 kN/m
C
A B 2m
4m
6–7. Draw the shear and moment diagrams for the compound beam which is pin connected at B.
6 kip
8 kip
A C B 4 ft
331
6 ft
4 ft
4 ft
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*6–8. Draw the shear and moment diagrams for the simply supported beam.
150 lb/ft 300 lbft A
B 12 ft
The free-body diagram of the beam’s left segment sectioned through an arbitrary point shown in Fig. b will be used to write the shear and moment equations. The intensity of the triangular distributed load at the point of sectioning is w = 150 a
x b = 12.5x 12
Referring to Fig. b, + c ©Fy = 0; a + ©M = 0; M +
275 -
1 (12.5x)(x) - V = 0 2
V = {275 - 6.25x2}lb‚ (1)
x 1 (12.5x)(x)a b - 275x = 0 M = {275x - 2.083x3}lb # ft‚(2) 2 3
The shear and moment diagrams shown in Figs. c and d are plotted using Eqs. (1) and (2), respectively. The location where the shear is equal to zero can be obtained by setting V = 0 in Eq. (1). 0 = 275 - 6.25x2
x = 6.633 ft
The value of the moment at x = 6.633 ft (V = 0) is evaluated using Eq. (2). M x = 6.633 ft = 275(6.633) - 2.083(6.633)3 = 1216 lb # ft
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6–9. Draw the shear and moment diagrams for the beam. Hint: The 20-kip load must be replaced by equivalent loadings at point C on the axis of the beam.
15 kip 1 ft
A
C 4 ft
20 kip
B
4 ft
4 ft
6–10. Members ABC and BD of the counter chair are rigidly connected at B and the smooth collar at D is allowed to move freely along the vertical slot. Draw the shear and moment diagrams for member ABC. Equations of Equilibrium: Referring to the free-body diagram of the frame shown in Fig. a, + c ©Fy = 0;
P 150 lb
Ay - 150 = 0 C A
Ay = 150 lb a + ©MA = 0;
B
1.5 ft 1.5 ft
ND(1.5) - 150(3) = 0 D
ND = 300 lb Shear and Moment Diagram: The couple moment acting on B due to ND is MB = 300(1.5) = 450 lb # ft. The loading acting on member ABC is shown in Fig. b and the shear and moment diagrams are shown in Figs. c and d.
333
1.5 ft
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6–11. The overhanging beam has been fabricated with a projected arm BD on it. Draw the shear and moment diagrams for the beam ABC if it supports a load of 800 lb. Hint: The loading in the supporting strut DE must be replaced by equivalent loads at point B on the axis of the beam.
E
800 lb
B
Support Reactions: a + ©MC = 0;
5 ft
D 2 ft
C
A
800(10) -
3 4 FDE(4) - FDE(2) = 0 5 5
6 ft
4 ft
FDE = 2000 lb + c ©Fy = 0;
-800 +
+ ©F = 0; : x
-Cx +
3 (2000) - Cy = 0 5 4 (2000) = 0 5
Cy = 400 lb
Cx = 1600 lb
Shear and Moment Diagram:
*6–12. A reinforced concrete pier is used to support the stringers for a bridge deck. Draw the shear and moment diagrams for the pier when it is subjected to the stringer loads shown. Assume the columns at A and B exert only vertical reactions on the pier.
60 kN 60 kN 35 kN 35 kN 35 kN 1 m 1 m 1.5 m 1.5 m 1 m 1 m
A
334
B
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6–13. Draw the shear and moment diagrams for the compound beam. It is supported by a smooth plate at A which slides within the groove and so it cannot support a vertical force, although it can support a moment and axial load.
P
Support Reactions:
P
A
D
B C
From the FBD of segment BD a + ©MC = 0; + c ©Fy = 0; + ©F = 0; : x
By (a) - P(a) = 0 Cy - P - P = 0
By = P
a
a
a
a
Cy = 2P
Bx = 0
From the FBD of segment AB a + ©MA = 0; + c ©Fy = 0;
P(2a) - P(a) - MA = 0
MA = Pa
P - P = 0 (equilibrium is statisfied!)
6–14. The industrial robot is held in the stationary position shown. Draw the shear and moment diagrams of the arm ABC if it is pin connected at A and connected to a hydraulic cylinder (two-force member) BD. Assume the arm and grip have a uniform weight of 1.5 lbin. and support the load of 40 lb at C.
4 in. A
10 in. B
50 in.
120 D
335
C
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*6–16. Draw the shear and moment diagrams for the shaft and determine the shear and moment throughout the shaft as a function of x. The bearings at A and B exert only vertical reactions on the shaft.
500 lb 800 lb A
B x 3 ft
For 0 6 x 6 3 ft + c ©Fy = 0.
220 - V = 0
a + ©MNA = 0.
V = 220 lb‚
Ans.
M - 220x = 0 M = (220x) lb ft‚
Ans.
For 3 ft 6 x 6 5 ft + c ©Fy = 0;
220 - 800 - V = 0 V = -580 lb
a + ©MNA = 0;
Ans.
M + 800(x - 3) - 220x = 0 M = {-580x + 2400} lb ft‚
Ans.
For 5 ft 6 x … 6 ft + c ©Fy = 0; a + ©MNA = 0;
V - 500 = 0
V = 500 lb‚
Ans.
-M - 500(5.5 - x) - 250 = 0 M = (500x - 3000) lb ft
Ans.
336
2 ft
0.5 ft
0.5 ft
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•6–17.
Draw the shear and moment diagrams for the cantilevered beam.
300 lb
200 lb/ft
A 6 ft
The free-body diagram of the beam’s left segment sectioned through an arbitrary point shown in Fig. b will be used to write the shear and moment equations. The intensity of the triangular distributed load at the point of sectioning is x w = 200 a b = 33.33x 6 Referring to Fig. b, + c ©Fy = 0;
-300 -
a + ©M = 0; M +
1 (33.33x)(x) - V = 0 2
V = {-300 - 16.67x2} lb (1)
1 x (33.33x)(x)a b + 300x = 0 M = {-300x - 5.556x3} lb # ft (2) 2 3
The shear and moment diagrams shown in Figs. c and d are plotted using Eqs. (1) and (2), respectively.
337
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6–18. Draw the shear and moment diagrams for the beam, and determine the shear and moment throughout the beam as functions of x.
2 kip/ft
10 kip
8 kip 40 kip⭈ft
Support Reactions: As shown on FBD. Shear and Moment Function:
x 6 ft
For 0 … x 6 6 ft: + c ©Fy = 0;
4 ft
30.0 - 2x - V = 0 V = {30.0 - 2x} kip
Ans.
x a + ©MNA = 0; M + 216 + 2xa b - 30.0x = 0 2 M = {-x2 + 30.0x - 216} kip # ft
Ans.
For 6 ft 6 x … 10 ft: + c ©Fy = 0; a + ©MNA = 0;
V - 8 = 0
V = 8.00 kip
Ans.
-M - 8(10 - x) - 40 = 0 M = {8.00x - 120} kip # ft
Ans.
6–19. Draw the shear and moment diagrams for the beam.
2 kip/ ft 30 kip⭈ft
B A 5 ft
338
5 ft
5 ft
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*6–20. Draw the shear and moment diagrams for the simply supported beam.
10 kN 10 kN/m
A
B
3m
Since the area under the curved shear diagram can not be computed directly, the value of the moment at x = 3 m will be computed using the method of sections. By referring to the free-body diagram shown in Fig. b, a + ©M = 0; Mx= 3 m +
1 (10)(3)(1) - 20(3) = 0 2
Mx= 3m = 45 kN # m
339
Ans.
3m
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•6–21. The beam is subjected to the uniform distributed load shown. Draw the shear and moment diagrams for the beam.
2 kN/m
Equations of Equilibrium: Referring to the free-body diagram of the beam shown in Fig. a, a + ©MA = 0;
3 FBC a b (2) - 2(3)(1.5) = 0 5
B
A 1.5 m
FBC = 7.5 kN + c ©Fy = 0;
C
3 Ay + 7.5 a b - 2(3) = 0 5 Ay = 1.5 kN
3 Shear and Moment Diagram: The vertical component of FBC is A FBC B y = 7.5a b 5 = 4.5 kN. The shear and moment diagrams are shown in Figs. c and d.
340
2m
1m
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6–22. Draw the shear and moment diagrams for the overhang beam.
4 kN/m
A B 3m
Since the loading is discontinuous at support B, the shear and moment equations must be written for regions 0 … x 6 3 m and 3 m 6 x … 6 m of the beam. The free-body diagram of the beam’s segment sectioned through an arbitrary point within these two regions is shown in Figs. b and c. Region 0 … x 6 3 m, Fig. b + c ©Fy = 0;
-4 -
a + ©M = 0; M +
2 V = e - x2 - 4 f kN 3
1 4 a xb(x) - V = 0 2 3
1 4 x a xb(x)a b + 4x = 0 2 3 3
(1)
2 M = e - x3 - 4x f kN # m (2) 9
Region 3 m 6 x … 6 m, Fig. c + c ©Fy = 0;
V - 4(6 - x) = 0
1 a + ©M = 0; -M - 4(6 - x) c (6 - x) d = 0 2
V = {24 - 4x} kN
(3)
M = {-2(6 - x)2}kN # m
(4)
The shear diagram shown in Fig. d is plotted using Eqs. (1) and (3). The value of shear just to the left and just to the right of the support is evaluated using Eqs. (1) and (3), respectively. 2 Vx= 3 m - = - (32) - 4 = -10 kN 3 Vx=3 m + = 24 - 4(3) = 12 kN The moment diagram shown in Fig. e is plotted using Eqs. (2) and (4). The value of the moment at support B is evaluated using either Eq. (2) or Eq. (4). 2 Mx= 3 m = - (33) - 4(3) = -18 kN # m 9 or Mx= 3 m = -2(6 - 3)2 = -18 kN # m
341
3m
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6–23. Draw the shear and moment diagrams for the beam. It is supported by a smooth plate at A which slides within the groove and so it cannot support a vertical force, although it can support a moment and axial load.
w
B
A
L
*6–24. Determine the placement distance a of the roller support so that the largest absolute value of the moment is a minimum. Draw the shear and moment diagrams for this condition.
w
A
B a
wL2 wL - wx = 0 2a
+ c ©Fy = 0;
x = L -
L
L2 2a
x wL2 Mmax (+) + wxa b - a wL bx = 0 2 2a
a + ©M = 0;
Substitute x = L -
L2 ; 2a
Mmax (+) = a wL =
wL2 L2 w L2 2 b aL b aL b 2a 2a 2 2a
w L2 2 aL b 2 2a Mmax (-) - w(L - a)
©M = 0;
Mmax (-) =
(L - a) = 0 2
w(L - a)2 2
To get absolute minimum moment, Mmax (+) = Mmax (-) L2 2 w w (L ) = (L - a)2 2 2a 2 L a =
L2 = L - a 2a L 22
‚
Ans.
342
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6–25. The beam is subjected to the uniformly distributed moment m (moment>length). Draw the shear and moment diagrams for the beam.
m A L
Support Reactions: As shown on FBD. Shear and Moment Function: V = 0
+ c ©Fy = 0; a + ©MNA = 0;
M + mx - mL = 0
M = m(L - x)
Shear and Moment Diagram:
6–27. Draw the shear and moment diagrams for the beam.
+ c ©Fy = 0;
w0
w0L 1 w0x - a b(x) = 0 4 2 L B
x = 0.7071 L a + ©MNA = 0;
M +
w0L 1 w0x x L a b (x)a b ax - b = 0 2 L 3 4 3
Substitute x = 0.7071L, M = 0.0345 w0L2
343
L 3
A
2L 3
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*6–28. Draw the shear and moment diagrams for the beam.
w0
B
A L – 3
Support Reactions: As shown on FBD. Shear and Moment Diagram: Shear and moment at x = L>3 can be determined using the method of sections. + c ©Fy = 0;
w0 L w0 L - V = 0 3 6
a + ©MNA = 0;
M +
V =
w0 L 6
w0 L L w0 L L a b a b = 0 6 9 3 3 M =
5w0 L2 54
344
L – 3
L – 3
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•6–29.
Draw the shear and moment diagrams for the beam.
5 kN/m
5 kN/m
B
A 4.5 m
From FBD(a) + c ©Fy = 0; a + ©MNA = 0;
9.375 - 0.5556x2 = 0
x = 4.108 m
M + (0.5556) A 4.1082 B a
4.108 b - 9.375(4.108) = 0 3
M = 25.67 kN # m From FBD(b) a + ©MNA = 0;
M + 11.25(1.5) - 9.375(4.5) = 0 M = 25.31 kN # m
345
4.5 m
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6–30. Draw the shear and moment diagrams for the compound beam.
150 lb/ft
150 lb/ft
A 6 ft
Support Reactions: From the FBD of segment AB a + ©MB = 0;
450(4) - Ay (6) = 0
Ay = 300.0 lb
+ c ©Fy = 0;
By - 450 + 300.0 = 0
By = 150.0 lb
+ ©F = 0; : x
Bx = 0
From the FBD of segment BC a + ©MC = 0;
225(1) + 150.0(3) - MC = 0 MC = 675.0 lb # ft
+ c ©Fy = 0; + ©F = 0; : x
Cy - 150.0 - 225 = 0
Cy = 375.0 lb
Cx = 0
Shear and Moment Diagram: The maximum positive moment occurs when V = 0. + c ©Fy = 0; a + ©MNA = 0;
150.0 - 12.5x2 = 0
x = 3.464 ft
150(3.464) - 12.5 A 3.4642 B a
3.464 b - Mmax = 0 3
Mmax = 346.4 lb # ft
346
C
B 3 ft
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6–31. Draw the shear and moment diagrams for the beam and determine the shear and moment in the beam as functions of x.
w0
Support Reactions: As shown on FBD.
A
B x
Shear and Moment Functions:
L – 2
For 0 … x 6 L>2 + c ©Fy = 0;
3w0 L -w0x - V = 0 4 V =
a + ©MNA = 0;
w0 (3L - 4x) 4
Ans.
7w0 L2 3w0 L x x + w0 xa b + M = 0 24 4 2 M =
w0 A -12x2 + 18Lx - 7L2) 24
Ans.
For L>2 6 x … L + c ©Fy = 0;
V -
1 2w0 c (L - x) d(L - x) = 0 2 L V =
a + ©MNA = 0;
-M -
w0 (L - x)2 L
Ans.
1 2w0 L - x c (L - x) d(L - x)a b = 0 2 L 3
M = -
w0 (L - x)3 3L
Ans.
347
L – 2
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*6–32. The smooth pin is supported by two leaves A and B and subjected to a compressive load of 0.4 kNm caused by bar C. Determine the intensity of the distributed load w0 of the leaves on the pin and draw the shear and moment diagram for the pin.
0.4 kN/m C
A
+ c ©Fy = 0;
B
w0
1 2(w0)(20)a b - 60(0.4) = 0 2
20 mm 60 mm 20 mm
w0 = 1.2 kN>m
Ans.
•6–33.
The ski supports the 180-lb weight of the man. If the snow loading on its bottom surface is trapezoidal as shown, determine the intensity w, and then draw the shear and moment diagrams for the ski.
180 lb 3 ft
w 1.5 ft
Ski: + c ©Fy = 0;
1 1 w(1.5) + 3w + w(1.5) - 180 = 0 2 2 Ans.
w = 40.0 lb>ft Segment: + c ©Fy = 0;
30 - V = 0;
a + ©M = 0;
M - 30(0.5) = 0;
w0
V = 30.0 lb M = 15.0 lb # ft
348
w 3 ft
1.5 ft
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6–34. Draw the shear and moment diagrams for the compound beam.
5 kN 3 kN/m
A B 3m
6–35. Draw the shear and moment diagrams for the beam and determine the shear and moment as functions of x.
A x 3m
200 - V = 0
V = 200 N
Ans.
M - 200 x = 0 M = (200 x) N # m
Ans.
For 3 m 6 x … 6 m: 200 - 200(x - 3) V = e-
1 200 c (x - 3) d(x - 3) - V = 0 2 3
100 2 x + 500 f N 3
Ans.
Set V = 0, x = 3.873 m a + ©MNA = 0;
M +
1 200 x - 3 c (x - 3) d(x - 3)a b 2 3 3
+ 200(x - 3)a M = e-
1.5 m
B
For 0 … x 6 3 m:
+ c ©Fy = 0;
1.5 m
200 N/ m
Shear and Moment Functions:
a + ©MNA = 0;
3m
400 N/m
Support Reactions: As shown on FBD.
+ c ©Fy = 0;
D
C
x - 3 b - 200x = 0 2
100 3 x + 500x - 600 f N # m 9
Ans.
Substitute x = 3.87 m, M = 691 N # m
349
3m
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*6–36. Draw the shear and moment diagrams for the overhang beam.
18 kN 6 kN
A B 2m
6–37. Draw the shear and moment diagrams for the beam.
2m
M 10 kNm
2m
50 kN/m
50 kN/m
B A 4.5 m
350
4.5 m
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6–38. The dead-weight loading along the centerline of the airplane wing is shown. If the wing is fixed to the fuselage at A, determine the reactions at A, and then draw the shear and moment diagram for the wing.
3000 lb
400 lb/ft
250 lb/ft
A
8 ft
2 ft
Support Reactions:
3 ft
15 000 lb
-1.00 - 3 + 15 - 1.25 - 0.375 - Ay = 0
+ c ©Fy = 0; Ay = 9.375 kip
Ans.
a + ©MA = 0;
1.00(7.667) + 3(5) - 15(3) + 1.25(2.5) + 0.375(1.667) + MA = 0
MA = 18.583 kip # ft = 18.6 kip # ft
Ans.
+ ©F = 0; : x
Ans.
Ax = 0
Shear and Moment Diagram:
6–39. The compound beam consists of two segments that are pinned together at B. Draw the shear and moment diagrams if it supports the distributed loading shown. + c ©Fy = 0;
2wL 1w 2 x = 0 27 2L x =
a + ©M = 0;
w
B 2/3 L
4 L = 0.385 L A 27
M +
C
A
1w 1 2wL (0.385L)2 a b(0.385L) (0.385L) = 0 2L 3 27 M = 0.0190 wL2
351
1/3 L
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*6–40. Draw the shear and moment diagrams for the simply supported beam.
10 kN
10 kN
15 kNm A
B 2m
6–41. Draw the shear and moment diagrams for the compound beam. The three segments are connected by pins at B and E.
3 kN
2m
2m
3 kN
0.8 kN/m
B
E F
A C 2m
352
1m
1m
D 2m
1m
1m
2m
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6–42. Draw the shear and moment diagrams for the compound beam.
5 kN/m
Support Reactions:
A
From the FBD of segment AB a + ©MA = 0; + c ©Fy = 0;
B 2m
By (2) - 10.0(1) = 0
By = 5.00 kN
Ay - 10.0 + 5.00 = 0
Ay = 5.00 kN
C 1m
D
1m
From the FBD of segment BD a + ©MC = 0;
5.00(1) + 10.0(0) - Dy (1) = 0 Dy = 5.00 kN
+ c ©Fy = 0;
Cy - 5.00 - 5.00 - 10.0 = 0 Cy = 20.0 kN
+ ©F = 0; : x
Bx = 0
From the FBD of segment AB + ©F = 0; : x
Ax = 0
Shear and Moment Diagram:
6–43. Draw the shear and moment diagrams for the beam. The two segments are joined together at B.
8 kip
3 kip/ft
A
C B 3 ft
353
5 ft
8 ft
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*6–44. Draw the shear and moment diagrams for the beam.
w
8
FR =
x =
1 2 x dx = 21.33 kip 8 L0
1 8 3 8 10 x dx
21.33
8 kip/ft 1 w ⫽ x2 8
= 6.0 ft
x B
A 8 ft
•6–45.
Draw the shear and moment diagrams for the beam. L
FR =
dA =
LA
L0 w0
wdx =
w0
L
L L0 2
x2 dx =
w
w0 L 3
w
LA
2
x A
w0L w0x = 0 12 3L2
1 1>3 x = a b L = 0.630 L 4 w0L w0x3 1 a + ©M = 0; (x) a xb - M = 0 12 3L2 4
M =
B L
3
+ c ©Fy = 0;
w0
L
x3dx L L0 3L x = = = w0 L 4 dA 3 LA xdA
w0 2 x L2
w0Lx w0x4 12 12L2
Substitute x = 0.630L M = 0.0394 w0L2
354
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6–46. Draw the shear and moment diagrams for the beam. L
FR =
dA = w0
LA
L0
sina
w
2w0 L p xb dx = p L
w0
A L – 2
6–47. A member having the dimensions shown is used to resist an internal bending moment of M = 90 kN # m. Determine the maximum stress in the member if the moment is applied (a) about the z axis (as shown) (b) about the y axis. Sketch the stress distribution for each case.
200 mm y 150 mm
The moment of inertia of the cross-section about z and y axes are 1 (0.2)(0.153) = 56.25(10 - 6) m4 12
Iy =
1 (0.15)(0.23) = 0.1(10 - 3) m4 12
M z x
For the bending about z axis, c = 0.075 m. smax =
90(103) (0.075) Mc = 120(106)Pa = 120 MPa = Iz 56.25 (10 - 6)
Ans.
For the bending about y axis, C = 0.1 m. smax =
x
B L – 2
Iz =
p w w0 sin – x L
90(103) (0.1) Mc = 90 (106)Pa = 90 MPa = Iy 0.1 (10 - 3)
Ans.
The bending stress distribution for bending about z and y axes are shown in Fig. a and b respectively.
355
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*6–48. Determine the moment M that will produce a maximum stress of 10 ksi on the cross section.
0.5 in. A
3 in.
0.5 in.
0.5 in. B
C 3 in. M 10 in.
D 0.5 in.
Section Properties: y =
=
©yA ©A 0.25(4)(0.5) + 2[2(3)(0.5)] + 5.5(10)(0.5) = 3.40 in. 4(0.5) + 2[(3)(0.5)] + 10(0.5)
INA =
1 (4) A 0.53 B + 4(0.5)(3.40 - 0.25)2 12 + 2c
1 (0.5)(33) + 0.5(3)(3.40 - 2)2 d 12
+
1 (0.5) A 103 B + 0.5(10)(5.5 - 3.40)2 12
= 91.73 in4 Maximum Bending Stress: Applying the flexure formula smax = 10 =
Mc I M (10.5 - 3.4) 91.73
M = 129.2 kip # in = 10.8 kip # ft
Ans.
356
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•6–49.
Determine the maximum tensile and compressive bending stress in the beam if it is subjected to a moment of M = 4 kip # ft.
0.5 in. A 0.5 in.
3 in.
0.5 in. B
C 3 in. M 10 in.
D 0.5 in.
Section Properties: y =
=
©yA ©A 0.25(4)(0.5) + 2[2(3)(0.5)] + 5.5(10)(0.5) = 3.40 in. 4(0.5) + 2[(3)(0.5)] + 10(0.5)
INA =
1 (4) A 0.53 B + 4(0.5)(3.40 - 0.25)2 12 + 2c
1 (0.5)(33) + 0.5(3)(3.40 - 2)2 d 12
+
1 (0.5) A 103 B + 0.5(10)(5.5 - 3.40)2 12
= 91.73 in4 Maximum Bending Stress: Applying the flexure formula smax =
Mc I
(st)max =
4(103)(12)(10.5 - 3.40) = 3715.12 psi = 3.72 ksi 91.73
Ans.
(sc)max =
4(103)(12)(3.40) = 1779.07 psi = 1.78 ksi 91.73
Ans.
357
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6–50. The channel strut is used as a guide rail for a trolley. If the maximum moment in the strut is M = 30 N # m, determine the bending stress at points A, B, and C.
50 mm C 5 mm 5 mm
y =
B
2.5(50)(5) + 7.5(34)(5) + 2[20(5)(20)] + 2[(32.5)(12)(5)] 50(5) + 34(5) + 2[5(20)] + 2[(12)(5)]
30 mm
= 13.24 mm A
I = c
1 (50)(53) + 50(5)(13.24 - 2.5)2 d 12
+ c
5 mm
5 mm 5 mm 7 mm 10 mm 7 mm
1 (34)(53) + 34(5)(13.24 - 7.5)2 d 12
+ 2c
1 1 (5)(203) + 5(20)(20 - 13.24)2 d + 2c (12)(53) + 12(5)(32.5 - 13.24)2 d 12 12
= 0.095883(10 - 6) m4 30(35 - 13.24)(10 - 3)
sA =
0.095883(10 - 6) 30(13.24 - 10)(10 - 3)
sB =
0.095883(10 - 6)
= 6.81 MPa
Ans.
= 1.01 MPa
Ans.
6–51. The channel strut is used as a guide rail for a trolley. If the allowable bending stress for the material is sallow = 175 MPa, determine the maximum bending moment the strut will resist.
50 mm C 5 mm 5 mm B
-3
30(13.24)(10 )
sC =
-6
0.095883(10 )
= 4.14 MPa
©y2A 2.5(50)(5) + 7.5(34)(5) + 2[20(5)(20)] + 2[(32.5)(12)(5)] = = 13.24 mm y = ©A 50(5) + 34(5) + 2[5(20)] + 2[(12)(5)] I = c
1 1 (50)(53) + 50(5)(13.24 - 2.5)2 d + c (34)(53) + 34(5)(13.24 - 7.5)2 d 12 12
+ 2c
1 1 (5)(203) + 5(20)(20 - 13.24)2 d + 2c (12)(53) + 12(5)(32.5 - 13.24)2 d 12 12
= 0.095883(10 - 6) m4 s =
Mc ; I
175(106) =
30 mm
Ans.
M(35 - 13.24)(10 - 3) 0.095883(10 - 6)
M = 771 N # m
Ans.
358
A
5 mm
5 mm 5 mm 7 mm 10 mm 7 mm
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*6–52. The beam is subjected to a moment M. Determine the percentage of this moment that is resisted by the stresses acting on both the top and bottom boards, A and B, of the beam.
A 25 mm M
D
Section Property: I =
1 1 (0.2) A 0.23 B (0.15) A 0.153 B = 91.14583 A 10 - 6 B m4 12 12
150 mm 25 mm 25 mm
Bending Stress: Applying the flexure formula
B 150 mm
25 mm
My I
s =
sE =
sD =
M(0.1) 91.14583(10 - 6) M(0.075) 91.14583(10 - 6)
= 1097.143 M
= 822.857 M
Resultant Force and Moment: For board A or B F = 822.857M(0.025)(0.2) +
1 (1097.143M - 822.857M)(0.025)(0.2) 2
= 4.800 M M¿ = F(0.17619) = 4.80M(0.17619) = 0.8457 M sc a
M¿ b = 0.8457(100%) = 84.6 % M
Ans.
•6–53. Determine the moment M that should be applied to the beam in order to create a compressive stress at point D of sD = 30 MPa . Also sketch the stress distribution acting over the cross section and compute the maximum stress developed in the beam.
A 25 mm
Section Property:
150 mm
1 1 I = (0.2) A 0.23 B (0.15) A 0.153 B = 91.14583 A 10 - 6 B m4 12 12
25 mm 25 mm
Bending Stress: Applying the flexure formula s = 30 A 106 B =
My I M(0.075) 91.14583(10 - 6)
M = 36458 N # m = 36.5 kN # m smax =
M
D
Ans.
36458(0.1) Mc = 40.0 MPa = I 91.14583(10 - 6)
Ans.
359
B 150 mm
25 mm
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6–54. The beam is made from three boards nailed together as shown. If the moment acting on the cross section is M = 600 N # m, determine the maximum bending stress in the beam. Sketch a three-dimensional view of the stress distribution acting over the cross section.
25 mm
150 mm 20 mm
(0.0125)(0.24)(0.025) + 2 (0.1)(0.15)(0.2) = 0.05625 m y = 0.24 (0.025) + 2 (0.15)(0.02)
200 mm M 600 Nm
1 (0.24)(0.0253) + (0.24)(0.025)(0.043752) 12
I =
+ 2a
20 mm
1 b (0.02)(0.153) + 2(0.15)(0.02)(0.043752) 12
= 34.53125 (10 - 6) m4 smax = sB =
Mc I 600 (0.175 - 0.05625)
=
34.53125 (10 - 6)
= 2.06 MPa sC =
Ans.
My 600 (0.05625) = 0.977 MPa = I 34.53125 (10 - 6)
6–55. The beam is made from three boards nailed together as shown. If the moment acting on the cross section is M = 600 N # m, determine the resultant force the bending stress produces on the top board.
25 mm
150 mm
(0.0125)(0.24)(0.025) + 2 (0.15)(0.1)(0.02) = 0.05625 m 0.24 (0.025) + 2 (0.15)(0.02)
y =
20 mm 200 mm M 600 Nm
1 (0.24)(0.0253) + (0.24)(0.025)(0.043752) 12
I =
+ 2a
20 mm
1 b (0.02)(0.153) + 2(0.15)(0.02)(0.043752) 12
= 34.53125 (10 - 6) m4 s1 =
My 600(0.05625) = 0.9774 MPa = I 34.53125(10 - 6)
sb =
My 600(0.05625 - 0.025) = 0.5430 MPa = I 34.53125(10 - 6)
F =
1 (0.025)(0.9774 + 0.5430)(106)(0.240) = 4.56 kN 2
Ans.
360
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*6–56. The aluminum strut has a cross-sectional area in the form of a cross. If it is subjected to the moment M = 8 kN # m, determine the bending stress acting at points A and B, and show the results acting on volume elements located at these points.
A 100 mm 20 mm 100 mm
B
M ⫽ 8 kN⭈m
20 mm
50 mm 50 mm
Section Property: I =
1 1 (0.02) A 0.223 B + (0.1) A 0.023 B = 17.8133 A 10 - 6 B m4 12 12
Bending Stress: Applying the flexure formula s =
sA =
sB =
8(103)(0.11) 17.8133(10 - 6) 8(103)(0.01) 17.8133(10 - 6)
My I
= 49.4 MPa (C)
Ans.
= 4.49 MPa (T)
Ans.
•6–57. The aluminum strut has a cross-sectional area in the form of a cross. If it is subjected to the moment M = 8 kN # m, determine the maximum bending stress in the beam, and sketch a three-dimensional view of the stress distribution acting over the entire cross-sectional area.
A 100 mm 20 mm 100 mm
B 20 mm
M ⫽ 8 kN⭈m 50 mm
50 mm
Section Property: I =
1 1 (0.02) A 0.223 B + (0.1) A 0.023 B = 17.8133 A 10 - 6 B m4 12 12
Bending Stress: Applying the flexure formula smax =
smax =
8(103)(0.11) 17.8133(10 - 6)
sy = 0.01m =
My Mc and s = , I I
= 49.4 MPa
8(103)(0.01) 17.8133(10 - 6)
Ans.
= 4.49 MPa
361
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6–58. If the beam is subjected to an internal moment of M = 100 kip # ft, determine the maximum tensile and compressive bending stress in the beam.
3 in. 3 in. 6 in. M 2 in.
1.5 in.
Section Properties: The neutral axis passes through centroid C of the cross section as shown in Fig. a. The location of C is
©yA y = = ©A
4(8)(6) - 2 cp A 1.52 B d 8(6) - p A 1.52 B
= 4.3454 in.
Thus, the moment of inertia of the cross section about the neutral axis is I = ©I + Ad2
=
1 1 (6)a83 b + 6(8) A 4.3454 - 4 B 2 - B pa 1.54 b + pa 1.52 b A 4.3454 - 2 B 2 R 12 4
= 218.87 in4 Maximum Bending Stress: The maximum compressive and tensile bending stress occurs at the top and bottom edges of the cross section.
A smax B T =
100(12)(4.3454) Mc = = 23.8 ksi (T) I 218.87
Ans.
A smax B C =
My 100(12)(8 - 4.3454) = = 20.0 ksi (C) I 218.87
Ans.
362
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6–59. If the beam is made of material having an allowable tensile and compressive stress of (sallow)t = 24 ksi and (sallow)c = 22 ksi, respectively, determine the maximum allowable internal moment M that can be applied to the beam.
3 in. 3 in. 6 in. M 2 in.
1.5 in.
Section Properties: The neutral axis passes through centroid C of the cross section as shown in Fig. a. The location of C is
©yA y = = ©A
4(8)(6) - 2 cp A 1.52 B d 8(6) - p A 1.52 B
= 4.3454 in.
Thus, the moment of inertia of the cross section about the neutral axis is I = ©I + Ad2
=
1 1 (6) A 83 B + 6(8) A 4.3454 - 4 B 2 - B p A 1.54 B + p A 1.52 B A 4.3454 - 2 B 2 R 12 4
= 218.87 in4 Allowable Bending Stress: The maximum compressive and tensile bending stress occurs at the top and bottom edges of the cross section. For the top edge, (sallow)c =
My ; I
22 =
M(8 - 4.3454) 218.87 M = 1317.53 kip # ina
1 ft b = 109.79 kip # ft 12 in.
For the bottom edge,
A smax B t =
Mc ; I
24 =
M(4.3454) 218.87
M = 1208.82 kip # ina
1 ft b = 101 kip # ft (controls) 12 in.
363
Ans.
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*6–60. The beam is constructed from four boards as shown. If it is subjected to a moment of Mz = 16 kip # ft, determine the stress at points A and B. Sketch a three-dimensional view of the stress distribution.
y
A
C
1 in. 10 in. 1 in. 10 in.
2[5(10)(1)] + 10.5(16)(1) + 16(10)(1) y = 2(10)(1) + 16(1) + 10(1)
Mz 16 kipft z
= 9.3043 in.
14 in.
1 1 I = 2c (1)(103) + 1(10)(9.3043 - 5)2 d + (16)(13) + 16(1)(10.5 - 9.3043)2 12 12 +
B
1 in.
x
1 in.
1 (1)(103) + 1(10)(16 - 9.3043)2 = 1093.07 in4 12
sA =
16(12)(21 - 9.3043) Mc = = 2.05 ksi I 1093.07
Ans.
sB =
My 16(12)(9.3043) = = 1.63 ksi I 1093.07
Ans.
•6–61. The beam is constructed from four boards as shown. If it is subjected to a moment of Mz = 16 kip # ft, determine the resultant force the stress produces on the top board C.
y
A
C
1 in. 10 in. 1 in. 10 in.
y =
2[5(10)(1)] + 10.5(16)(1) + 16(10)(1) = 9.3043 in. 2(10)(1) + 16(1) + 10(1)
Mz 16 kipft z 14 in.
1 1 I = 2c (1)(103) + (10)(9.3043 - 5)2 d + (16)(13) + 16(1)(10.5 - 9.3043)2 12 12 +
1 (1)(103) + 1(10)(16 - 9.3043)2 = 1093.07 in4 12
sA =
16(12)(21 - 9.3043) Mc = = 2.0544 ksi I 1093.07
sD =
My 16(12)(11 - 9.3043) = = 0.2978 ksi I 1093.07
(FR)C =
1 (2.0544 + 0.2978)(10)(1) = 11.8 kip 2
Ans.
364
1 in.
B
1 in.
x
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6–62. A box beam is constructed from four pieces of wood, glued together as shown. If the moment acting on the cross section is 10 kN # m, determine the stress at points A and B and show the results acting on volume elements located at these points.
20 mm
160 mm
25 mm A 250 mm
25 mm
B
M 10 kNm
The moment of inertia of the cross-section about the neutral axis is I =
1 1 (0.2)(0.33) (0.16)(0.253) = 0.2417(10 - 3) m4. 12 12
For point A, yA = C = 0.15 m. sA =
10(103) (0.15) MyA = 6.207(106)Pa = 6.21 MPa (C) = I 0.2417(10 - 3)
Ans.
For point B, yB = 0.125 m. sB =
MyB 10(103)(0.125) = 5.172(106)Pa = 5.17 MPa (T) = I 0.2417(10 - 3)
Ans.
The state of stress at point A and B are represented by the volume element shown in Figs. a and b respectively.
365
20 mm
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6–63. Determine the dimension a of a beam having a square cross section in terms of the radius r of a beam with a circular cross section if both beams are subjected to the same internal moment which results in the same maximum bending stress.
a a
r
Section Properties: The moments of inertia of the square and circular cross sections about the neutral axis are 1 a4 a A a3 B = 12 12
IS =
IC =
1 4 pr 4
Maximum Bending Stress: For the square cross section, c = a>2.
A smax B S =
M(a>2) 6M Mc = 3 = 4 IS a >12 a
For the circular cross section, c = r.
A smax B c =
Mc Mr 4M = Ic 1 4 pr3 pr 4
It is required that
A smax B S = A smax B C 6M 4M = a3 pr3 a = 1.677r
Ans.
*6–64. The steel rod having a diameter of 1 in. is subjected to an internal moment of M = 300 lb # ft. Determine the stress created at points A and B. Also, sketch a three-dimensional view of the stress distribution acting over the cross section. I =
A B
p 4 p r = (0.54) = 0.0490874 in4 4 4
sA =
M ⫽ 300 lb⭈ft 45⬚
300(12)(0.5) Mc = = 36.7 ksi I 0.0490874
Ans. 0.5 in.
My 300(12)(0.5 sin 45°) sB = = = 25.9 ksi I 0.0490874
Ans.
366
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•6–65.
If the moment acting on the cross section of the beam is M = 4 kip # ft, determine the maximum bending stress in the beam. Sketch a three-dimensional view of the stress distribution acting over the cross section.
A
1.5 in. 12 in.
The moment of inertia of the cross-section about the neutral axis is
12 in.
1 1 (12)(153) (10.5)(123) = 1863 in4 I = 12 12
M 1.5 in.
1.5 in.
Along the top edge of the flange y = c = 7.5 in. Thus smax =
4(103)(12)(7.5) Mc = = 193 psi I 1863
Ans.
Along the bottom edge to the flange, y = 6 in. Thus s =
4(103)(12)(6) My = = 155 psi I 1863
6–66. If M = 4 kip # ft, determine the resultant force the bending stress produces on the top board A of the beam. A
1.5 in.
The moment of inertia of the cross-section about the neutral axis is
12 in.
1 1 (12)(153) (10.5)(123) = 1863 in4 12 12
I =
12 in. M
Along the top edge of the flange y = c = 7.5 in. Thus 1.5 in.
smax =
4(103)(12)(7.5) Mc = = 193.24 psi I 1863
Along the bottom edge of the flange, y = 6 in. Thus s =
4(103)(12)(6) My = = 154.59 psi I 1863
The resultant force acting on board A is equal to the volume of the trapezoidal stress block shown in Fig. a. FR =
1 (193.24 + 154.59)(1.5)(12) 2
= 3130.43 lb = 3.13 kip
Ans.
367
1.5 in.
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6–67. The rod is supported by smooth journal bearings at A and B that only exert vertical reactions on the shaft. If d = 90 mm, determine the absolute maximum bending stress in the beam, and sketch the stress distribution acting over the cross section.
12 kN/m d A
B 3m
Absolute Maximum Bending Stress: The maximum moment is Mmax = 11.34 kN # m as indicated on the moment diagram. Applying the flexure formula smax =
Mmax c I 11.34(103)(0.045)
=
p 4
(0.0454)
= 158 MPa
Ans.
368
1.5 m
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*6–68. The rod is supported by smooth journal bearings at A and B that only exert vertical reactions on the shaft. Determine its smallest diameter d if the allowable bending stress is sallow = 180 MPa.
12 kN/m d A
B 3m
1.5 m
Allowable Bending Stress: The maximum moment is Mmax = 11.34 kN # m as indicated on the moment diagram. Applying the flexure formula Mmax c I
smax = sallow =
11.34(103) A d2 B
180 A 106 B =
p 4
A d2 B 4
d = 0.08626 m = 86.3 mm
Ans.
•6–69.
Two designs for a beam are to be considered. Determine which one will support a moment of M = 150 kN # m with the least amount of bending stress. What is that stress?
200 mm
200 mm
30 mm
15 mm
300 mm 30 mm
Section Property:
300 mm 15 mm
For section (a) I =
1 1 (0.2) A 0.333 B (0.17)(0.3)3 = 0.21645(10 - 3) m4 12 12
15 mm (a)
For section (b) I =
1 1 (0.2) A 0.363 B (0.185) A 0.33 B = 0.36135(10 - 3) m4 12 12
Maximum Bending Stress: Applying the flexure formula smax =
Mc I
For section (a) smax =
150(103)(0.165) 0.21645(10 - 3)
= 114.3 MPa
For section (b) smax =
150(103)(0.18) 0.36135(10 - 3)
= 74.72 MPa = 74.7 MPa
Ans.
369
30 mm (b)
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6–70. The simply supported truss is subjected to the central distributed load. Neglect the effect of the diagonal lacing and determine the absolute maximum bending stress in the truss. The top member is a pipe having an outer diameter of 1 in. 3 and thickness of 16 in., and the bottom member is a solid rod having a diameter of 12 in.
y =
100 lb/ft
6 ft
5.75 in.
6 ft
6 ft
©yA 0 + (6.50)(0.4786) = = 4.6091 in. ©A 0.4786 + 0.19635
I = c
1 1 1 p(0.5)4 - p(0.3125)4 d + 0.4786(6.50 - 4.6091)2 + p(0.25)4 4 4 4
+ 0.19635(4.6091)2 = 5.9271 in4 Mmax = 300(9 - 1.5)(12) = 27 000 lb # in. smax =
27 000(4.6091 + 0.25) Mc = I 5.9271
= 22.1 ksi
Ans.
6–71. The axle of the freight car is subjected to wheel loadings of 20 kip. If it is supported by two journal bearings at C and D, determine the maximum bending stress developed at the center of the axle, where the diameter is 5.5 in. A
C
B
60 in. 10 in. 20 kip
smax =
200(2.75) Mc = 1 = 12.2 ksi 4 I 4 p(2.75)
Ans.
370
D
10 in. 20 kip
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*6–72. The steel beam has the cross-sectional area shown. Determine the largest intensity of distributed load w0 that it can support so that the maximum bending stress in the beam does not exceed smax = 22 ksi.
w0
12 ft
12 ft 8 in. 0.30 in. 10 in.
0.3 in.
Support Reactions: As shown on FBD.
0.30 in.
Internal Moment: The maximum moment occurs at mid span. The maximum moment is determined using the method of sections. Section Property: I =
1 1 (8) A 10.63 B (7.7) A 103 B = 152.344 in4 12 12
Absolute Maximum Bending Stress: The maximum moment is Mmax = 48.0w0 as indicated on the FBD. Applying the flexure formula smax = 22 =
Mmax c I 48.0w0 (12)(5.30) 152.344
w0 = 1.10 kip>ft
Ans.
•6–73.
The steel beam has the cross-sectional area shown. If w0 = 0.5 kip>ft, determine the maximum bending stress in the beam.
w0
12 ft
12 ft 8 in.
Support Reactions: As shown on FBD.
0.3 in.
0.30 in.
Internal Moment: The maximum moment occurs at mid span. The maximum moment is determined using the method of sections. Section Property: I =
1 1 (8) A 10.63 B (7.7) A 103 B = 152.344 in4 12 12
Absolute Maximum Bending Stress: The maximum moment is Mmax = 24.0 kip # ft as indicated on the FBD. Applying the flexure formula smax =
=
Mmax c I
24.0(12)(5.30) 152.344
= 10.0 ksi
Ans.
371
0.30 in. 10 in.
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6–74. The boat has a weight of 2300 lb and a center of gravity at G. If it rests on the trailer at the smooth contact A and can be considered pinned at B, determine the absolute maximum bending stress developed in the main strut of the trailer. Consider the strut to be a box-beam having the dimensions shown and pinned at C.
B 1 ft
G
C
A 3 ft
D
5 ft
4 ft
1.75 in.
1 ft
1.75 in.
3 in. 1.5 in.
Boat: + ©F = 0; : x a + ©MB = 0;
Bx = 0 -NA(9) + 2300(5) = 0 NA = 1277.78 lb
+ c ©Fy = 0;
1277.78 - 2300 + By = 0 By = 1022.22 lb
Assembly: a + ©MC = 0;
-ND(10) + 2300(9) = 0 ND = 2070 lb
+ c ©Fy = 0;
Cy + 2070 - 2300 = 0 Cy = 230 lb
I =
1 1 (1.75)(3)3 (1.5)(1.75)3 = 3.2676 in4 12 12
smax =
3833.3(12)(1.5) Mc = = 21.1 ksi I 3.2676
Ans.
372
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6–75. The shaft is supported by a smooth thrust bearing at A and smooth journal bearing at D. If the shaft has the cross section shown, determine the absolute maximum bending stress in the shaft.
40 mm A
B 0.75 m
D
C 1.5 m
25 mm
0.75 m
3 kN
3 kN
Shear and Moment Diagrams: As shown in Fig. a. Maximum Moment: Due to symmetry, the maximum moment occurs in region BC of the shaft. Referring to the free-body diagram of the segment shown in Fig. b. Section Properties: The moment of inertia of the cross section about the neutral axis is I =
p A 0.044 - 0.0254 B = 1.7038 A 10 - 6 B m4 4
Absolute Maximum Bending Stress:
sallow =
2.25 A 103 B (0.04) Mmaxc = = 52.8 MPa I 1.7038 A 10 - 6 B
Ans.
*6–76. Determine the moment M that must be applied to the beam in order to create a maximum stress of 80 MPa.Also sketch the stress distribution acting over the cross section.
300 mm
20 mm
The moment of inertia of the cross-section about the neutral axis is I =
M
1 1 (0.3)(0.33) (0.21)(0.263) = 0.36742(10 - 3) m4 12 12
260 mm
Thus, 20 mm 30 mm
smax
Mc = ; I
6
80(10 ) =
M(0.15) 0.36742(10 - 3)
M = 195.96 (103) N # m = 196 kN # m The bending stress distribution over the cross-section is shown in Fig. a.
373
Ans.
30 mm 30 mm
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•6–77.
The steel beam has the cross-sectional area shown. Determine the largest intensity of distributed load w that it can support so that the bending stress does not exceed smax = 22 ksi.
I = smax
w
1 1 (8)(10.6)3 (7.7)(103) = 152.344 in4 12 12
8 ft
w
8 ft
8 in.
Mc = I
22 =
8 ft
0.30 in. 10 in.
0.3 in.
0.30 in.
32w(12)(5.3) 152.344
w = 1.65 kip>ft
Ans.
6–78. The steel beam has the cross-sectional area shown. If w = 5 kip>ft, determine the absolute maximum bending stress in the beam.
w
8 ft
w
8 ft
8 ft 8 in. 0.3 in.
0.30 in. 10 in. 0.30 in.
From Prob. 6-78: M = 32w = 32(5)(12) = 1920 kip # in. I = 152.344 in4 smax =
1920(5.3) Mc = = 66.8 ksi I 152.344
Ans.
374
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6–79. If the beam ACB in Prob. 6–9 has a square cross section, 6 in. by 6 in., determine the absolute maximum bending stress in the beam.
15 kip 1 ft
A
20 kip
C 4 ft
B
4 ft
4 ft
Mmax = 46.7 kip # ft smax =
46.7(103)(12)(3) Mc = 15.6 ksi = 1 3 I 12 (6)(6 )
Ans.
*6–80. If the crane boom ABC in Prob. 6–3 has a rectangular cross section with a base of 2.5 in., determine its required height h to the nearest 14 in. if the allowable bending stress is sallow = 24 ksi.
A
a + ©MA = 0;
+ c ©Fy = 0;
-Ay +
+ ©F = 0; ; x
Ax -
smax =
4 (4000) - 1200 = 0; 5
3 (4000) = 0; 5
5 ft B
4 ft
4 F (3) - 1200(8) = 0; 5 B
3 ft
FB = 4000 lb
Ay = 2000 lb
Ax = 2400 lb
6000(12) A h2 B Mc = 24(10)3 = 1 3 I 12 (2.5)(h )
h = 2.68 in.
Ans.
Use h = 2.75 in.
Ans.
375
C
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•6–81.
If the reaction of the ballast on the railway tie can be assumed uniformly distributed over its length as shown, determine the maximum bending stress developed in the tie. The tie has the rectangular cross section with thickness t = 6 in.
15 kip 1.5 ft
15 kip 5 ft
1.5 ft
t
w
Support Reactions: Referring to the free - body diagram of the tie shown in Fig. a, we have + c ©Fy = 0;
w(8) - 2(15) = 0 w = 3.75 kip>ft
Maximum Moment: The shear and moment diagrams are shown in Figs. b and c. As indicated on the moment diagram, the maximum moment is Mmax = 7.5 kip # ft. Absolute Maximum Bending Stress: smax =
12 in.
7.5(12)(3) Mmaxc = 1.25 ksi = I 1 (12)(63) 12
Ans.
376
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6–82. The reaction of the ballast on the railway tie can be assumed uniformly distributed over its length as shown. If the wood has an allowable bending stress of sallow = 1.5 ksi, determine the required minimum thickness t of the rectangular cross sectional area of the tie to the nearest 18 in.
15 kip 1.5 ft
15 kip 5 ft
1.5 ft
t
w
Support Reactions: Referring to the free-body diagram of the tie shown in Fig. a, we have + c ©Fy = 0;
w(8) - 2(15) = 0 w = 3.75 kip>ft
Maximum Moment: The shear and moment diagrams are shown in Figs. b and c. As indicated on the moment diagram, the maximum moment is Mmax = 7.5 kip # ft. Absolute Maximum Bending Stress:
smax
t 7.5(12)a b 2 1.5 = 1 (12)t3 12
Mc = ; I
t = 5.48 in. Use
t = 5
12 in.
1 in. 2
Ans.
377
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6–83. Determine the absolute maximum bending stress in the tubular shaft if di = 160 mm and do = 200 mm.
15 kN/m 60 kN m d i do A
B 3m
Section Property: I =
p A 0.14 - 0.084 B = 46.370 A 10 - 6 B m4 4
Absolute Maximum Bending Stress: The maximum moment is Mmax = 60.0 kN # m as indicated on the moment diagram. Applying the flexure formula smax =
Mmaxc I
60.0(103)(0.1) =
46.370(10 - 6)
= 129 MPa
Ans.
378
1m
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*6–84. The tubular shaft is to have a cross section such that its inner diameter and outer diameter are related by di = 0.8do. Determine these required dimensions if the allowable bending stress is sallow = 155 MPa.
15 kN/m 60 kN m d i do A
B 3m
Section Property: I =
0.8do 4 do 4 dl 4 p do 4 p - a b R = 0.009225pd4o Ba b - a b R = B 4 2 2 4 16 2
Allowable Bending Stress: The maximum moment is Mmax = 60.0 kN # m as indicated on the moment diagram. Applying the flexure formula smax = sallow = 155 A 106 B =
Thus,
Mmax c I 60.0(103) A 2o B d
0.009225pd4o
do = 0.1883 m = 188 mm
Ans.
dl = 0.8do = 151 mm
Ans.
379
1m
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6–85. The wood beam has a rectangular cross section in the proportion shown. Determine its required dimension b if the allowable bending stress is sallow = 10 MPa.
500 N/m
1.5b A
B b 2m
Allowable Bending Stress: The maximum moment is Mmax = 562.5 N # m as indicated on the moment diagram. Applying the flexure formula smax = sallow = 10 A 106 B =
Mmax c I 562.5(0.75b) 1 12
(b)(1.5b)3
b = 0.05313 m = 53.1 mm
Ans.
380
2m
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6–86. Determine the absolute maximum bending stress in the 2-in.-diameter shaft which is subjected to the concentrated forces. The journal bearings at A and B only support vertical forces.
800 lb 600 lb
A
15 in.
B
15 in. 30 in.
The FBD of the shaft is shown in Fig. a. The shear and moment diagrams are shown in Fig. b and c, respectively. As indicated on the moment diagram, Mmax = 15000 lb # in. The moment of inertia of the cross-section about the neutral axis is I =
p 4 (1 ) = 0.25 p in4 4
Here, c = 1 in. Thus smax =
=
Mmax c I 15000(1) 0.25 p
= 19.10(103) psi Ans.
= 19.1 ksi
381
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6–87. Determine the smallest allowable diameter of the shaft which is subjected to the concentrated forces. The journal bearings at A and B only support vertical forces. The allowable bending stress is sallow = 22 ksi.
800 lb 600 lb
A
15 in.
B
15 in. 30 in.
The FBD of the shaft is shown in Fig. a The shear and moment diagrams are shown in Fig. b and c respectively. As indicated on the moment diagram, Mmax = 15,000 lb # in The moment of inertia of the cross-section about the neutral axis is I =
p 4 p d 4 a b = d 4 2 64
Here, c = d>2. Thus sallow =
Mmax c ; I
22(103) =
15000(d> 2) pd4>64
d = 1.908 in = 2 in.
Ans.
382
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*6–88. If the beam has a square cross section of 9 in. on each side, determine the absolute maximum bending stress in the beam.
1200 lb
800 lb/ft
B A 8 ft
Absolute Maximum Bending Stress: The maximum moment is Mmax = 44.8 kip # ft as indicated on moment diagram. Applying the flexure formula smax =
44.8(12)(4.5) Mmax c = = 4.42 ksi 1 3 I 12 (9)(9)
Ans.
•6–89.
If the compound beam in Prob. 6–42 has a square cross section, determine its dimension a if the allowable bending stress is sallow = 150 MPa.
Allowable Bending Stress: The maximum moments is Mmax = 7.50 kN # m as indicated on moment diagram. Applying the flexure formula smax = sallow = 150 A 106 B =
Mmax c I 7.50(103) A a2 B 1 12
a4
a = 0.06694 m = 66.9 mm
Ans.
383
8 ft
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6–90. If the beam in Prob. 6–28 has a rectangular cross section with a width b and a height h, determine the absolute maximum bending stress in the beam.
Absolute Maximum Bending Stress: The maximum moments is Mmax =
23w0 L2 216
as indicated on the moment diagram. Applying the flexure formula
smax
Mmax c = = I
A B
23w0 L2 h 2 216 1 3 12 bh
23w0 L2 =
Ans.
36bh2
384
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6–91. Determine the absolute maximum bending stress in the 80-mm-diameter shaft which is subjected to the concentrated forces. The journal bearings at A and B only support vertical forces.
A
0.5 m
B
0.4 m
0.6 m
12 kN 20 kN
The FBD of the shaft is shown in Fig. a The shear and moment diagrams are shown in Fig. b and c, respectively. As indicated on the moment diagram, Mmax = 6 kN # m. The moment of inertia of the cross-section about the neutral axis is I =
p (0.044) = 0.64(10 - 6)p m4 4
Here, c = 0.04 m. Thus smax =
6(103)(0.04) Mmax c = I 0.64(10 - 6)p = 119.37(106) Pa = 119 MPa
Ans.
385
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*6–92. Determine the smallest allowable diameter of the shaft which is subjected to the concentrated forces. The journal bearings at A and B only support vertical forces. The allowable bending stress is sallow = 150 MPa.
A
0.5 m
B
0.4 m
0.6 m
12 kN 20 kN
The FBD of the shaft is shown in Fig. a. The shear and moment diagrams are shown in Fig. b and c, respectively. As indicated on the moment diagram, Mmax = 6 kN # m. The moment of inertia of the cross-section about the neutral axis is I =
pd4 p d 4 a b = 4 2 64
Here, c = d>2. Thus sallow =
Mmax c ; I
150(106) =
6(103)(d> 2) pd4>64
d = 0.07413 m = 74.13 mm = 75 mm
386
Ans.
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•6–93. The man has a mass of 78 kg and stands motionless at the end of the diving board. If the board has the cross section shown, determine the maximum normal strain developed in the board. The modulus of elasticity for the material is E = 125 GPa. Assume A is a pin and B is a roller.
350 mm 30 mm A 1.5 m
Internal Moment: The maximum moment occurs at support B. The maximum moment is determined using the method of sections. Section Property: y =
=
I =
©yA ©A 0.01(0.35)(0.02) + 0.035(0.03)(0.03) = 0.012848 m 0.35(0.02) + 0.03(0.03) 1 (0.35) A 0.023 B + 0.35(0.02)(0.012848 - 0.01)2 12 +
1 (0.03) A 0.033 B + 0.03(0.03)(0.035 - 0.012848)2 12
= 0.79925 A 10 - 6 B m4 Absolute Maximum Bending Stress: The maximum moment is Mmax = 1912.95 N # m as indicated on the FBD. Applying the flexure formula smax =
Mmax c I 1912.95(0.05 - 0.012848)
=
0.79925(10 - 6)
= 88.92 MPa Absolute Maximum Normal Strain: Applying Hooke’s law, we have emax =
88.92(106) smax = 0.711 A 10 - 3 B mm>mm = E 125(109)
Ans.
387
B
2.5 m
C
20 mm 10 mm 10 mm 10 mm
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6–94. The two solid steel rods are bolted together along their length and support the loading shown. Assume the support at A is a pin and B is a roller. Determine the required diameter d of each of the rods if the allowable bending stress is sallow = 130 MPa.
20 kN/m
80 kN
A B
2m
Section Property: I = 2B
2m
p d 4 p d 2 5p 4 a b + d2 a b R = d 4 2 4 2 32
Allowable Bending Stress: The maximum moment is Mmax = 100 kN # m as indicated on moment diagram. Applying the flexure formula smax = sallow = 130 A 106 B =
Mmax c I 100(103)(d) 5p 32
d4
d = 0.1162 m = 116 mm
Ans.
6–95. Solve Prob. 6–94 if the rods are rotated 90° so that both rods rest on the supports at A (pin) and B (roller).
20 kN/m
Section Property: I = 2B
A
p d 4 p 4 a b R = d 4 2 32
smax = sallow =
2m
Mmax c I
100(103)(d) p 32
B
2m
Allowable Bending Stress: The maximum moment is Mmax = 100 kN # m as indicated on the moment diagram. Applying the flexure formula
130 A 106 B =
80 kN
d4
d = 0.1986 m = 199 mm
Ans.
388
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*6–96. The chair is supported by an arm that is hinged so it rotates about the vertical axis at A. If the load on the chair is 180 lb and the arm is a hollow tube section having the dimensions shown, determine the maximum bending stress at section a–a.
180 lb
1 in. a 3 in.
A a
0.5 in.
8 in.
c + ©M = 0;
M - 180(8) = 0 M = 1440 lb # in.
Ix =
1 1 (1)(33) (0.5)(2.53) = 1.59896 in4 12 12
smax =
1440 (1.5) Mc = = 1.35 ksi I 1.59896
Ans.
s (ksi)
•6–97.
A portion of the femur can be modeled as a tube having an inner diameter of 0.375 in. and an outer diameter of 1.25 in. Determine the maximum elastic static force P that can be applied to its center. Assume the bone to be roller supported at its ends. The s– P diagram for the bone mass is shown and is the same in tension as in compression.
P
2.30 1.25 4 in.
0.02
I =
1 p 4
0.375 4 4 4 C A 1.25 2 B - A 2 B D = 0.11887 in
Mmax =
P (4) = 2P 2
Require smax = 1.25 ksi smax =
Mc I
1.25 =
2P(1.25>2) 0.11887
P = 0.119 kip = 119 lb
Ans.
389
2.5 in.
0.05
P (in./ in.)
4 in.
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6–98. If the beam in Prob. 6–18 has a rectangular cross section with a width of 8 in. and a height of 16 in., determine the absolute maximum bending stress in the beam. 16 in.
Absolute Maximum Bending Stress: The maximum moment is Mmax = 216 kip # ft as indicated on moment diagram. Applying the flexure formula smax =
216(12)(8) Mmax c = 7.59 ksi = 1 3 I 12 (8)(16 )
8 in.
Ans.
6–99. If the beam has a square cross section of 6 in. on each side, determine the absolute maximum bending stress in the beam.
400 lb/ft
B A 6 ft
The maximum moment occurs at the fixed support A. Referring to the FBD shown in Fig. a, a + ©MA = 0;
Mmax - 400(6)(3) -
1 (400)(6)(8) = 0 2
Mmax = 16800 lb # ft The moment of inertia of the about the neutral axis is I =
smax =
1 (6)(63) = 108 in4. Thus, 12
16800(12)(3) Mc = I 108 = 5600 psi = 5.60 ksi
Ans.
390
6 ft
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*6–100. The steel beam has the cross-sectional area shown. Determine the largest intensity of the distributed load w0 that it can support so that the maximum bending stress in the beam does not exceed sallow = 22 ksi.
w0
9 ft
9 ft
9 in. 0.25 in.
0.25 in. 12 in. 0.25 in.
Support Reactions. The FBD of the beam is shown in Fig. a. The shear and moment diagrams are shown in Fig. a and b, respectively. As indicated on the moment diagram, Mmax = 27wo. The moment of inertia of the cross-section about the neutral axis is I =
1 1 (9)(12.53) (8.75)(123) 12 12
= 204.84375 in4 Here, ¢ = 6.25 in. Thus, sallow =
Mmax c ; I
22(103) =
(27wo)(12)(6.25) 204.84375
wo = 2 225.46 lb>ft = 2.23 kip>ft
Ans.
391
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•6–101.
The steel beam has the cross-sectional area shown. If w0 = 2 kip>ft, determine the maximum bending stress in the beam.
w0
9 ft
9 ft
9 in. 0.25 in.
0.25 in. 12 in. 0.25 in.
The FBD of the beam is shown in Fig. a The shear and moment diagrams are shown in Fig. b and c, respectively. As indicated on the moment diagram, Mmax = 54 kip # ft. The moment of inertia of the I cross-section about the bending axis is I =
1 1 (9) A 12.53 B (8.75) A 123 B 12 12
= 204.84375 in4 Here, c = 6.25 in. Thus smax =
=
Mmax c I 54 (12)(6.25) 204.84375
= 19.77 ksi = 19.8 ksi
Ans.
392
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6–102. The bolster or main supporting girder of a truck body is subjected to the uniform distributed load. Determine the bending stress at points A and B.
1.5 kip/ft
A 8 ft
B 12 ft F2
F1 0.75 in. 6 in.
12 in.
0.5 in. A B
0.75 in.
Support Reactions: As shown on FBD. Internal Moment: Using the method of sections. + ©MNA = 0;
M + 12.0(4) - 15.0(8) = 0 M = 72.0 kip # ft
Section Property: I =
1 1 (6) A 13.53 B (5.5) A 123 B = 438.1875 in4 12 12
Bending Stress: Applying the flexure formula s =
My I
sB =
72.0(12)(6.75) = 13.3 ksi 438.1875
Ans.
sA =
72.0(12)(6) = 11.8 ksi 438.1875
Ans.
393
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6–103. Determine the largest uniform distributed load w that can be supported so that the bending stress in the beam does not exceed s allow = 5 MPa .
w
The FBD of the beam is shown in Fig. a
0.5 m
The shear and moment diagrams are shown in Fig. b and c, respectively. As indicated on the moment diagram, |Mmax| = 0.125 w.
150 mm
The moment of inertia of the cross-section is, I =
1 (0.075) A 0.153 B = 21.09375 A 10 - 6 B m4 12
Here, c = 0.075 w. Thus, sallow = 5 A 106 B =
Mmax c ; I 0.125w(0.075)
21.09375 A 10 - 6 B
w = 11250 N>m = 11.25 kN>m
Ans.
394
1m 75 mm
0.5 m
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w
*6–104. If w = 10 kN>m, determine the maximum bending stress in the beam. Sketch the stress distribution acting over the cross section. Support Reactions. The FBD of the beam is shown in Fig. a
0.5 m
75 mm
The shear and moment diagrams are shown in Figs. b and c, respectively. As indicated on the moment diagram, |Mmax| = 1.25 kN # m.
150 mm
The moment of inertia of the cross-section is I =
1 (0.075) A 0.153 B = 21.09375 A 10 - 6 B m4 12
Here, c = 0.075 m. Thus smax =
=
Mmax c I 1.25 A 103 B (0.075) 21.09375 A 10 - 6 B
= 4.444 A 106 B Pa = 4.44 MPa
Ans.
The bending stress distribution over the cross section is shown in Fig. d
395
1m
0.5 m
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400 lb/ft
•6–105.
If the allowable bending stress for the wood beam is sallow = 150 psi, determine the required dimension b to the nearest 14 in. of its cross section. Assume the support at A is a pin and B is a roller.
B
A 3 ft
The FBD of the beam is shown in Fig. a The shear and moment diagrams are shown in Figs. b and c, respectively. As indicated on the moment diagram, Mmax = 3450 lb # ft.
2b b
The moment of inertia of the cross section is I =
2 1 (b)(2b)3 = b4 12 3
Here, c = 2b> 2 = b. Thus, sallow = 150 =
Mmax c ; I 3450(12)(b) > 3 b4
2
b = 7.453 in = 7
1 in. 2
Ans.
396
3 ft
3 ft
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400 lb/ft
6–106. The wood beam has a rectangular cross section in the proportion shown. If b 7.5 in., determine the absolute maximum bending stress in the beam.
B
A
The FBD of the beam is shown in Fig. a.
3 ft
The shear and moment diagrams are shown in Fig. b and c, respectively. As indicated on the moment diagram, Mmax = 3450 lb # ft.
2b b
The moment of inertia of the cross-section is I =
1 (7.5) A 153 B = 2109.375 in4 12
Here, c =
15 = 7.5 in. Thus 2
smax =
3450(12)(7.5) Mmax c = = 147 psi I 2109.375
Ans.
397
3 ft
3 ft
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6–107. A beam is made of a material that has a modulus of elasticity in compression different from that given for tension. Determine the location c of the neutral axis, and derive an expression for the maximum tensile stress in the beam having the dimensions shown if it is subjected to the bending moment M.
M h s
P
Ec(emax)t (h - c) c
Ec
Location of neutral axis: + ©F = 0; :
1 1 - (h - c)(smax)c (b) + (c)(smax)t (b) = 0 2 2
(h - c)(smax)c = c(smax)t (h - c)Ec (emax)t
[1]
(h - c) = cEt (emax)t ; c
Ec (h - c)2 = Etc2
Taking positive root: Ec c = h - c A Et Ec h A Et h2Ec c = = Ec 2Et + 2Ec 1 + A Et
[2] Ans.
©MNA = 0; 1 2 1 2 M = c (h - c)(smax)c (b) d a b (h - c) + c (c)(smax)t(b) d a b(c) 2 3 2 3 M =
1 1 (h - c)2 (b)(smax)c + c2b(smax)t 3 3
From Eq. [1]. (smax)c =
c (s ) h - c max t
M =
c 1 1 (h - c)2 (b)a b (smax)t + c2b(smax)t 3 h - c 3
M =
1 bc(smax)t (h - c + c) ; 3
(smax)t =
3M bhc
From Eq. [2]
(smax)t =
b
Et
(emax)t (h - c) (emax)c = c (smax)c = Ec(emax)c =
c
3M 2Et + 2Ec £ ≥ b h2 2Ec
Ans.
398
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*6–108. The beam has a rectangular cross section and is subjected to a bending moment M. If the material from which it is made has a different modulus of elasticity for tension and compression as shown, determine the location c of the neutral axis and the maximum compressive stress in the beam.
M h s
c b
Et
P Ec
See the solution to Prob. 6–107 c =
h2Ec
Ans.
2Et + 2Ec
Since (smax)c =
(smax)c =
c (s ) = h - c max t
2Ec 2Et
h2Ec ( 2Et + 2Ec)ch - a
h 1Ec 1Et + 1Ec
bd
(smax)t
(smax)t
(smax)c =
2Et + 2Ec 2Ec 3M ¢ 2≤¢ ≤ bh 2Et 2Ec
(smax)c =
3M 2Et + 2Ec ¢ ≤ bh2 2Et
Ans.
399
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•6–109.
The beam is subjected to a bending moment of M = 20 kip # ft directed as shown. Determine the maximum bending stress in the beam and the orientation of the neutral axis.
y 8 in. C
B
The y and z components of M are negative, Fig. a. Thus,
14 in. z
My = -20 sin 45° = -14.14 kip # ft
45 16 in.
Mz = -20 cos 45° = -14.14 kip # ft. The moments of inertia of the cross-section about the principal centroidal y and z axes are Iy =
1 1 (16) A 103 B (14) A 83 B = 736 in4 12 12
Iz =
1 1 (10) A 163 B (8) A 143 B = 1584 in4 12 12
My z
Mz y Iz
+
smax = sC = -
Iy -14.14(12)(8) -14.14(12)( - 5) + 1584 736
= 2.01 ksi smax = sA = -
(T)
Ans.
-14.14(12)(-8) -14.14(12)(5) + 1584 736
= -2.01 ksi = 2.01 ksi (C)
Ans.
Here, u = 180° + 45° = 225° tan a =
tan a =
Iz Iy
D 10 in. M
By inspection, the bending stress occurs at corners A and C are s = -
A
tan u
1584 tan 225° 736
a = 65.1°
Ans.
The orientation of neutral axis is shown in Fig. b.
400
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6–110. Determine the maximum magnitude of the bending moment M that can be applied to the beam so that the bending stress in the member does not exceed 12 ksi.
y 8 in. C
B
The y and z components of M are negative, Fig. a. Thus, 14 in.
My = -M sin 45° = -0.7071 M
z
45 16 in.
Mz = -M cos 45° = -0.7071 M The moments of inertia of the cross-section about principal centroidal y and z axes are Iy =
1 1 (16) A 103 B (14) A 83 B = 736 in4 12 12
Iz =
1 1 (10) A 163 B (8) A 143 B = 1584 in4 12 12
12 = -
Myzc
Mz yc Iz
+
D 10 in. M
By inspection, the maximum bending stress occurs at corners A and C. Here, we will consider corner C. sC = sallow = -
A
Iy
-0.7071 M(12)( -5) -0.7071 M (12)(8) + 1584 736
M = 119.40 kip # ft = 119 kip # ft
Ans.
401
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6–111. If the resultant internal moment acting on the cross section of the aluminum strut has a magnitude of M = 520 N # m and is directed as shown, determine the bending stress at points A and B. The location y of the centroid C of the strut’s cross-sectional area must be determined. Also, specify the orientation of the neutral axis.
y M 520 Nm 12
20 mm z
–y
5
13
B C
200 mm
20 mm
20 mm A 200 mm
Internal Moment Components: Mz = -
12 (520) = -480 N # m 13
My =
5 (520) = 200 N # m 13
Section Properties: 0.01(0.4)(0.02) + 2[(0.110)(0.18)(0.02)] ©yA = ©A 0.4(0.02) + 2(0.18)(0.02)
y =
= 0.057368 m = 57.4 mm Iz =
Ans.
1 (0.4) A 0.023 B + (0.4)(0.02)(0.057368 - 0.01)2 12 +
1 (0.04) A 0.183 B + 0.04(0.18)(0.110 - 0.057368)2 12
= 57.6014 A 10 - 6 B m4 Iy =
1 1 (0.2) A 0.43 B (0.18) A 0.363 B = 0.366827 A 10 - 3 B m4 12 12
Maximum Bending Stress: Applying the flexure formula for biaxial at points A and B s = -
Myz
Mzy +
Iz
Iy 200(-0.2)
-480(-0.142632) sA = -
+
-6
57.6014(10 )
0.366827(10 - 3)
= -1.298 MPa = 1.30 MPa (C) 200(0.2)
-480(0.057368) sB = -
Ans.
+
-6
57.6014(10 )
0.366827(10 - 3)
= 0.587 MPa (T)
Ans.
Orientation of Neutral Axis: tan a =
tan a =
Iz Iy
tan u
57.6014(10 - 6) 0.366827(10 - 3)
tan (-22.62°)
a = -3.74°
Ans.
402
200 mm
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*6–112. The resultant internal moment acting on the cross section of the aluminum strut has a magnitude of M = 520 N # m and is directed as shown. Determine maximum bending stress in the strut. The location y of the centroid C of the strut’s cross-sectional area must be determined. Also, specify the orientation of the neutral axis.
y M 520 Nm 12
20 mm z
–y
5
13
B C
200 mm
20 mm
20 mm A 200 mm
Internal Moment Components: Mz = -
12 (520) = -480 N # m 13
My =
5 (520) = 200 N # m 13
Section Properties: 0.01(0.4)(0.02) + 2[(0.110)(0.18)(0.02)] ©yA = ©A 0.4(0.02) + 2(0.18)(0.02)
y =
= 0.057368 m = 57.4 mm Iz =
Ans.
1 (0.4) A 0.023 B + (0.4)(0.02)(0.057368 - 0.01)2 12 1 (0.04) A 0.183 B + 0.04(0.18)(0.110 - 0.057368)2 12
+
= 57.6014 A 10 - 6 B m4 Iy =
1 1 (0.2) A 0.43 B (0.18) A 0.363 B = 0.366827 A 10 - 3 B m4 12 12
Maximum Bending Stress: By inspection, the maximum bending stress can occur at either point A or B. Applying the flexure formula for biaxial bending at points A and B s = -
My z
Mz y +
Iz
Iy 200(-0.2)
-480(-0.142632) sA = -
+
57.6014(10 - 6)
0.366827(10 - 3)
= -1.298 MPa = 1.30 MPa (C) (Max) 200(0.2)
-480(0.057368) sB = -
-6
57.6014(10 )
Ans.
+
0.366827(10 - 3)
= 0.587 MPa (T) Orientation of Neutral Axis: tan a =
tan a =
Iz Iy
tan u
57.6014(10 - 6) 0.366827(10 - 3)
tan (-22.62°)
a = -3.74°
Ans.
403
200 mm
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6–113. Consider the general case of a prismatic beam subjected to bending-moment components My and Mz, as shown, when the x, y, z axes pass through the centroid of the cross section. If the material is linear-elastic, the normal stress in the beam is a linear function of position such that s = a + by + cz. Using the equilibrium conditions 0 = 1A s dA, My = 1A zs dA, Mz = 1A - ys dA, determine the constants a, b, and c, and show that the normal stress can be determined from the equation s = [-1MzIy + MyIyz2y + 1MyIz + MzIyz2z]>1IyIz - Iyz22, where the moments and products of inertia are defined in Appendix A.
y z My dA sC y Mz z
Equilibrium Condition: sx = a + by + cz 0 =
LA
sx dA
0 =
LA
(a + by + cz) dA
0 = a
LA
dA + b
LA
y dA + c
My =
LA
z sx dA
=
LA
z(a + by + cz) dA
= a
Mz =
=
LA
= -a
LA
LA
z dA + b
LA
LA
z dA
yz dA + c
LA
[1]
z2 dA
[2]
-y sx dA
-y(a + by + cz) dA
LA
ydA - b
y2 dA - c
LA
LA
yz dA
[3]
Section Properties: The integrals are defined in Appendix A. Note that LA
y dA =
LA
z dA = 0.Thus,
From Eq. [1]
Aa = 0
From Eq. [2]
My = bIyz + cIy
From Eq. [3]
Mz = -bIz - cIyz
Solving for a, b, c: a = 0 (Since A Z 0) b = -¢
Thus,
MzIy + My Iyz
sx = - ¢
Iy Iz -
I2yz
≤
Mz Iy + My Iyz Iy Iz -
I2yz
c =
≤y + ¢
My Iz + Mz Iyz Iy Iz - I2yz My Iy + MzIyz Iy Iz - I2yz
≤z
(Q.E.D.)
404
x
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6–114. The cantilevered beam is made from the Z-section having the cross-section shown. If it supports the two loadings, determine the bending stress at the wall in the beam at point A. Use the result of Prob. 6–113.
50 lb 50 lb 3 ft
(My)max Iy =
= 50(3) + 50(5) = 400 lb # ft = 4.80(103)lb # in.
2 ft
0.25 in. 2 in.
1 1 (3.25)(0.25)3 + 2c (0.25)(2)3 + (0.25)(2)(1.125)2 d = 1.60319 in4 12 12
A
B 2.25 in.
1 1 Iz = (0.25)(3.25)3 + 2 c (2)(0.25)3 + (0.25)(2)(1.5)2 d = 2.970378 in4 12 12
0.25 in. 3 in.
0.25 in.
Iyz = 2[1.5(1.125)(2)(0.25)] = 1.6875 in4 Using the equation developed in Prob. 6-113. s = -a
sA =
Mz Iy + My Iyz Iy Iz -
I2yz
by + a
My Iz + Mz Iyz Iy Iz - I2yz
bz
{-[0 + (4.80)(103)(1.6875)](1.625) + [(4.80)(103)(2.970378) + 0](2.125)} [1.60319(2.970378) - (1.6875)2]
= 8.95 ksi
Ans.
6–115. The cantilevered beam is made from the Z-section having the cross-section shown. If it supports the two loadings, determine the bending stress at the wall in the beam at point B. Use the result of Prob. 6–113.
50 lb 50 lb 3 ft
3
(My)max = 50(3) + 50(5) = 400 lb # ft = 4.80(10 )lb # in. Iy =
1 1 (3.25)(0.25)3 + 2c (0.25)(2)3 + (0.25)(2)(1.125)2 d = 1.60319 in4 12 12
1 1 Iz = (0.25)(3.25)3 + 2 c (2)(0.25)3 + (0.25)(2)(1.5)2 d = 2.970378 in4 12 12
2.25 in.
sB =
Iy Iz -
I2yz
by + a
My Iz + Mz Iyz Iy Iz - I2yz
0.25 in. 3 in.
0.25 in.
Using the equation developed in Prob. 6-113. Mz Iy + My Iyz
A
B
Iyz = 2[1.5(1.125)(2)(0.25)] = 1.6875 in4
s = -a
2 ft
0.25 in. 2 in.
bz
-[0 + (4.80)(103)(1.6875)]( -1.625) + [(4.80)(103)(2.976378) + 0](0.125) [(1.60319)(2.970378) - (1.6875)2]
= 7.81 ksi
Ans.
405
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*6–116. The cantilevered wide-flange steel beam is subjected to the concentrated force P at its end. Determine the largest magnitude of this force so that the bending stress developed at A does not exceed sallow = 180 MPa.
200 mm 10 mm 150 mm 10 mm
Internal Moment Components: Using method of section
10 mm A
y
©Mz = 0;
Mz + P cos 30°(2) = 0
Mz = -1.732P
©My = 0;
My + P sin 30°(2) = 0
My = -1.00P
z
Section Properties:
x
2m
30
1 1 Iz = (0.2) A 0.173 B (0.19) A 0.153 B = 28.44583(10 - 6) m4 12 12 Iy = 2 c
P
1 1 (0.01) A 0.23 B d + (0.15) A 0.013 B = 13.34583(10 - 6) m4 12 12
Allowable Bending Stress: By inspection, maximum bending stress occurs at points A and B. Applying the flexure formula for biaxial bending at point A. sA = sallow = 180 A 106 B = -
Myz
Mzy Iz
+
Iy
(-1.732P)(0.085) 28.44583(10 - 6)
-1.00P(-0.1) +
13.34583(10 - 6)
P = 14208 N = 14.2 kN
Ans.
•6–117.
The cantilevered wide-flange steel beam is subjected to the concentrated force of P = 600 N at its end. Determine the maximum bending stress developed in the beam at section A.
200 mm 10 mm 150 mm 10 mm
Internal Moment Components: Using method of sections
A
y
©Mz = 0;
Mz + 600 cos 30°(2) = 0
Mz = -1039.23 N # m
©My = 0;
My + 600 sin 30°(2) = 0;
My = -600.0 N # m
z
Section Properties:
x
1 1 Iz = (0.2) A 0.173 B (0.19) A 0.153 B = 28.44583(10 - 6) m4 12 12 Iy = 2 c
Maximum Bending Stress: By inspection, maximum bending stress occurs at A and B. Applying the flexure formula for biaxial bending at point A s = -
Myz
Mzy Iz
+
Iy -600.0(-0.1)
-1039.32(0.085) sA = -
-6
28.44583(10 )
= 7.60 MPa (T)
+
13.34583(10 - 6)
(Max)
Ans.
406
2m
30 P
1 1 (0.01) A 0.23 B d + (0.15) A 0.013 B = 13.34583(10 - 6) m4 12 12
10 mm
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6–118. If the beam is subjected to the internal moment of M = 1200 kN # m, determine the maximum bending stress acting on the beam and the orientation of the neutral axis.
y 150 mm 150 mm
Internal Moment Components: The y component of M is positive since it is directed towards the positive sense of the y axis, whereas the z component of M, which is directed towards the negative sense of the z axis, is negative, Fig. a. Thus,
M 300 mm 30
My = 1200 sin 30° = 600 kN # m
150 mm
Mz = -1200 cos 30° = -1039.23 kN # m
z
x 150 mm
Section Properties: The location of the centroid of the cross-section is given by ©yA 0.3(0.6)(0.3) - 0.375(0.15)(0.15) = = 0.2893 m ©A 0.6(0.3) - 0.15(0.15)
y =
150 mm
The moments of inertia of the cross section about the principal centroidal y and z axes are Iy =
1 1 (0.6) A 0.33 B (0.15) A 0.153 B = 1.3078 A 10 - 3 B m4 12 12
Iz =
1 (0.3) A 0.63 B + 0.3(0.6)(0.3 - 0.2893)2 12 - c
1 (0.15) A 0.153 B + 0.15(0.15)(0.375 - 0.2893)2 d 12
= 5.2132 A 10 - 3 B m4 Bending Stress: By inspection, the maximum bending stress occurs at either corner A or B. s = -
Myz
Mzy
sA = -
+
Iz
Iy
c -1039.23 A 103 B d(0.2893) 5.2132 A 10 - 3 B
+
600 A 103 B (0.15) 1.3078 A 10 - 3 B
= 126 MPa (T)
sB = -
c -1039.23 A 103 B d(-0.3107) 5.2132 A 10 - 3 B
+
600 A 103 B ( -0.15) 1.3078 A 10 - 3 B
= -131 MPa = 131 MPa (C)(Max.)
Ans.
Orientation of Neutral Axis: Here, u = -30°. tan a =
tan a =
Iz Iy
tan u
5.2132 A 10 - 3 B
1.3078 A 10 - 3 B
tan(-30°)
a = -66.5°
Ans.
The orientation of the neutral axis is shown in Fig. b.
407
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6–119. If the beam is made from a material having an allowable tensile and compressive stress of (sallow)t = 125 MPa and (sallow)c = 150 MPa, respectively, determine the maximum allowable internal moment M that can be applied to the beam.
y 150 mm 150 mm M 300 mm
Internal Moment Components: The y component of M is positive since it is directed towards the positive sense of the y axis, whereas the z component of M, which is directed towards the negative sense of the z axis, is negative, Fig. a. Thus,
30 150 mm z
My = M sin 30° = 0.5M
x 150 mm
Mz = -M cos 30° = -0.8660M Section Properties: The location of the centroid of the cross section is y =
150 mm
0.3(0.6)(0.3) - 0.375(0.15)(0.15) ©yA = = 0.2893 m ©A 0.6(0.3) - 0.15(0.15)
The moments of inertia of the cross section about the principal centroidal y and z axes are Iy =
1 1 (0.6) A 0.33 B (0.15) A 0.153 B = 1.3078 A 10 - 3 B m4 12 12
Iz =
1 (0.3) A 0.63 B + 0.3(0.6)(0.3 - 0.2893)2 12 - c
1 (0.15) A 0.153 B + 0.15(0.15)(0.375 - 0.2893)2 d 12
= 5.2132 A 10 - 3 B m4 Bending Stress: By inspection, the maximum bending stress can occur at either corner A or B. For corner A which is in tension, sA = (sallow)t = 125 A 106 B = -
My zA
Mz yA Iz
+
Iy
(-0.8660M)(0.2893) 5.2132 A 10
-3
B
0.5M(0.15)
+
1.3078 A 10 - 3 B
M = 1185 906.82 N # m = 1186 kN # m (controls)
Ans.
For corner B which is in compression, sB = (sallow)c = -150 A 106 B = -
My zB
Mz yB Iz
+
Iy
(-0.8660M)(-0.3107) 5.2132 A 10 - 3 B
0.5M(-0.15)
+
1.3078 A 10 - 3 B
M = 1376 597.12 N # m = 1377 kN # m
408
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*6–120. The shaft is supported on two journal bearings at A and B which offer no resistance to axial loading. Determine the required diameter d of the shaft if the allowable bending stress for the material is sallow = 150 MPa .
z y
0.5 m
0.5 m
C
0.5 m 200 N
The FBD of the shaft is shown in Fig. a.
A
200 N 300 N
The shaft is subjected to two bending moment components Mz and My, Figs. b and c, respectively. Since all the axes through the centroid of the circular cross-section of the shaft are principal axes, then the resultant moment M = 2My 2 + Mz 2 can be used for design. The maximum moment occurs at D (x = 1m). Then, Mmax = 21502 + 1752 = 230.49 N # m Then, sallow =
Mmax C ; I
150(106) =
230.49(d>2) p 4
(d>2)4
d = 0.02501 m = 25 mm
Ans.
409
300 N
0.5 m D B E x 150 N 150 N
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•6–121.
The 30-mm-diameter shaft is subjected to the vertical and horizontal loadings of two pulleys as shown. It is supported on two journal bearings at A and B which offer no resistance to axial loading. Furthermore, the coupling to the motor at C can be assumed not to offer any support to the shaft. Determine the maximum bending stress developed in the shaft.
1m 1m 1m 1m A D
150 N 150 N
Support Reactions: As shown on FBD. Internal Moment Components: The shaft is subjected to two bending moment components My and Mz. The moment diagram for each component is drawn. Maximum Bending Stress: Since all the axes through the circle’s center for circular shaft are principal axis, then the resultant moment M = 2My 2 + Mz 2 can be used to determine the maximum bending stress. The maximum resultant moment occurs at E Mmax = 24002 + 1502 = 427.2 N # m. Applying the flexure formula Mmax c I 427.2(0.015) =
p 4
A 0.0154 B
= 161 MPa
Ans.
410
E
C
B 400 N
100 mm 400 N 60 mm
x
smax =
y
z
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6–122. Using the techniques outlined in Appendix A, Example A.5 or A.6, the Z section has principal moments of inertia of Iy = 0.060110-32 m4 and Iz = 0.471110-32 m4, computed about the principal axes of inertia y and z, respectively. If the section is subjected to an internal moment of M = 250 N # m directed horizontally as shown, determine the stress produced at point A. Solve the problem using Eq. 6–17.
50 mm y
A
200 mm
32.9
y¿ 250 Nm z
My = 250 cos 32.9° = 209.9 N # m
z¿ 300 mm
Mz = 250 sin 32.9° = 135.8 N # m
200 mm
50 mm B
50 mm
y = 0.15 cos 32.9° + 0.175 sin 32.9° = 0.2210 m z = -(0.175 cos 32.9° - 0.15 sin 32.9°) = -0.06546 m sA = -
Myz
Mzy +
Iz
Iy
209.9(-0.06546)
-135.8(0.2210) =
0.471(10 - 3)
+
60.0(10 - 6)
= -293 kPa = 293 kPa (C)
Ans.
6–123. Solve Prob. 6–122 using the equation developed in Prob. 6–113.
50 mm y
A
Internal Moment Components: My = 250 N # m
200 mm
Mz = 0
32.9
y¿
Section Properties: Iy =
250 Nm
1 1 (0.3) A 0.053 B + 2c (0.05) A 0.153 B + 0.05(0.15) A 0.12 B d 12 12
= 0.18125 A 10 Iz =
-3
z z¿ 300 mm
Bm
4
1 1 (0.05) A 0.33 B + 2c (0.15) A 0.053 B + 0.15(0.05) A 0.1252 B d 12 12
= 0.350(10 - 3) m4 Iyz = 0.15(0.05)(0.125)(-0.1) + 0.15(0.05)(-0.125)(0.1) = -0.1875 A 10 - 3 B m4 Bending Stress: Using formula developed in Prob. 6-113 s =
sA =
-(Mz Iy + My Iyz)y + (My Iz + MzIyz)z IyIz - I2yz -[0 + 250(-0.1875)(10 - 3)](0.15) + [250(0.350)(10 - 3) + 0](-0.175) 0.18125(10 - 3)(0.350)(10 - 3) - [0.1875(10 - 3)]2
= -293 kPa = 293 kPa (C)
Ans.
411
200 mm
50 mm B
50 mm
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*6–124. Using the techniques outlined in Appendix A, Example A.5 or A.6, the Z section has principal moments of inertia of Iy = 0.060110-32 m4 and Iz = 0.471110-32 m4, computed about the principal axes of inertia y and z, respectively. If the section is subjected to an internal moment of M = 250 N # m directed horizontally as shown, determine the stress produced at point B. Solve the problem using Eq. 6–17.
50 mm y
A
200 mm
32.9
y¿ 250 Nm z z¿ 300 mm
Internal Moment Components: My¿ = 250 cos 32.9° = 209.9 N # m Mz¿ = 250 sin 32.9° = 135.8 N # m Section Property: y¿ = 0.15 cos 32.9° + 0.175 sin 32.9° = 0.2210 m z¿ = 0.15 sin 32.9° - 0.175 cos 32.9° = -0.06546 m Bending Stress: Applying the flexure formula for biaxial bending s =
sB =
My¿z¿
Mz¿y¿ Iz¿
+
Iy¿ 209.9(-0.06546)
135.8(0.2210) 0.471(10 - 3)
-
0.060(10 - 3)
= 293 kPa = 293 kPa (T)
Ans.
412
200 mm
50 mm B
50 mm
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z
•6–125. Determine the bending stress at point A of the beam, and the orientation of the neutral axis. Using the method in Appendix A, the principal moments of inertia of the cross section are I¿z = 8.828 in4 and I¿y = 2.295 in4, where z¿ and y¿ are the principal axes. Solve the problem using Eq. 6–17.
1.183 in. 0.5 in.
z¿
A
4 in.
45 C y 1.183 in. 0.5 in.
M 3 kip ft
y′ 4 in.
Internal Moment Components: Referring to Fig. a, the y¿ and z¿ components of M are negative since they are directed towards the negative sense of their respective axes. Thus, Section Properties: Referring to the geometry shown in Fig. b, œ = 2.817 cos 45° - 1.183 sin 45° = 1.155 in. zA œ yA = -(2.817 sin 45° + 1.183 cos 45°) = -2.828 in.
Bending Stress: sA = -
= -
œ My¿zA
œ Mz¿yA
Iz¿
+
Iy¿
(-2.121)(12)(-2.828) (-2.121)(12)(1.155) + 8.828 2.295
= -20.97 ksi = 21.0 ksi (C)
Ans.
413
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z
6–126. Determine the bending stress at point A of the beam using the result obtained in Prob. 6–113. The moments of inertia of the cross sectional area about the z and y axes are Iz = Iy = 5.561 in4 and the product of inertia of the cross sectional area with respect to the z and y axes is Iyz 3.267 in4. (See Appendix A)
1.183 in. 0.5 in.
z¿
A
4 in.
45 C y 1.183 in. 0.5 in.
M 3 kip ft
y′ 4 in.
Internal Moment Components: Since M is directed towards the negative sense of the y axis, its y component is negative and it has no z component. Thus, My = -3 kip # ft
Mz = 0
Bending Stress:
sA =
=
- A MzIy + MyIyz B yA + A MyIz + MzIyz B zA IyIz - Iyz 2
- C 0(5.561) + (-3)(12)(-3.267) D (-1.183) + C -3(12)(5.561) + 0(-3.267) D (2.817) 5.561(5.561) - (-3.267)2
= -20.97 ksi = 21.0 ksi
Ans.
414
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6–127. The composite beam is made of 6061-T6 aluminum (A) and C83400 red brass (B). Determine the dimension h of the brass strip so that the neutral axis of the beam is located at the seam of the two metals. What maximum moment will this beam support if the allowable bending stress for the aluminum is 1sallow2al = 128 MPa and for the brass 1sallow2br = 35 MPa?
h B A 150 mm
Section Properties: n =
68.9(109) Eal = 0.68218 = Ebr 101(109)
bbr = nbal = 0.68218(0.15) = 0.10233 m y =
0.05 =
©yA ©A 0.025(0.10233)(0.05) + (0.05 + 0.5h)(0.15)h 0.10233(0.05) + (0.15)h
h = 0.04130 m = 41.3 mm INA =
Ans.
1 (0.10233) A 0.053 B + 0.10233(0.05)(0.05 - 0.025)2 12 +
1 (0.15) A 0.041303 B + 0.15(0.04130)(0.070649 - 0.05)2 12
= 7.7851 A 10 - 6 B m4 Allowable Bending Stress: Applying the flexure formula Assume failure of red brass (sallow)br = 35 A 106 B =
Mc INA M(0.04130) 7.7851(10 - 6)
M = 6598 N # m = 6.60 kN # m (controls!)
Ans.
Assume failure of aluminium (sallow)al = n
Mc INA
128 A 106 B = 0.68218c
M(0.05) 7.7851(10 - 6)
d
M = 29215 N # m = 29.2 kN # m
415
50 mm
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*6–128. The composite beam is made of 6061-T6 aluminum (A) and C83400 red brass (B). If the height h = 40 mm, determine the maximum moment that can be applied to the beam if the allowable bending stress for the aluminum is 1sallow2al = 128 MPa and for the brass 1sallow2br = 35 MPa.
h B A
Section Properties: For transformed section. 150 mm
68.9(109) Eal = 0.68218 = n = Ebr 101.0(109) bbr = nbal = 0.68218(0.15) = 0.10233 m y =
=
©yA ©A 0.025(0.10233)(0.05) + (0.07)(0.15)(0.04) 0.10233(0.05) + 0.15(0.04)
= 0.049289 m INA =
1 (0.10233) A 0.053 B + 0.10233(0.05)(0.049289 - 0.025)2 12 +
1 (0.15) A 0.043 B + 0.15(0.04)(0.07 - 0.049289)2 12
= 7.45799 A 10 - 6 B m4 Allowable Bending Stress: Applying the flexure formula Assume failure of red brass (sallow)br = 35 A 106 B =
Mc INA M(0.09 - 0.049289) 7.45799(10 - 6)
M = 6412 N # m = 6.41 kN # m (controls!)
Ans.
Assume failure of aluminium (sallow)al = n
Mc INA
128 A 106 B = 0.68218c
M(0.049289) 7.45799(10 - 6)
d
M = 28391 N # m = 28.4 kN # m
416
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•6–129. Segment A of the composite beam is made from 2014-T6 aluminum alloy and segment B is A-36 steel. If w = 0.9 kip>ft, determine the absolute maximum bending stress developed in the aluminum and steel. Sketch the stress distribution on the cross section.
w
15 ft A
3 in.
B
3 in. 3 in.
Maximum Moment: For the simply-supported beam subjected to the uniform 0.9 A 152 B wL2 = distributed load, the maximum moment in the beam is Mmax = 8 8 = 25.3125 kip # ft. Section Properties: The cross section will be transformed into that of steel as Eal 10.6 = = 0.3655. shown in Fig. a. Here, n = Est 29 Then bst = nbal = 0.3655(3) = 1.0965 in. The location of the centroid of the transformed section is y =
©yA 1.5(3)(3) + 4.5(3)(1.0965) = = 2.3030 in. ©A 3(3) + 3(1.0965)
The moment of inertia of the transformed section about the neutral axis is I = ©I + Ad2 =
1 (3) A 33 B + 3(3)(2.3030 - 1.5)2 12 +
1 (1.0965) A 33 B + 1.0965(3)(4.5 - 2.3030)2 12
= 30.8991 in4 Maximum Bending Stress: For the steel, (smax)st =
25.3125(12)(2.3030) Mmaxcst = = 22.6 ksi I 30.8991
Ans.
At the seam, ssty = 0.6970 in. =
Mmaxy 25.3125(12)(0.6970) = = 6.85 ksi I 30.8991
For the aluminium, (smax)al = n
25.3125(12)(6 - 2.3030) Mmaxcal = 0.3655c d = 13.3 ksi I 30.8991
Ans.
At the seam, saly = 0.6970 in. = n
Mmaxy 25.3125(12)(0.6970) = 0.3655c d = 2.50 ksi I 30.8991
The bending stress across the cross section of the composite beam is shown in Fig. b.
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6–130. Segment A of the composite beam is made from 2014-T6 aluminum alloy and segment B is A-36 steel. If the allowable bending stress for the aluminum and steel are (sallow)al = 15 ksi and (sallow)st = 22 ksi, determine the maximum allowable intensity w of the uniform distributed load.
w
15 ft A
3 in.
B
3 in. 3 in.
Maximum Moment: For the simply-supported beam subjected to the uniform distributed load, the maximum moment in the beam is w A 152 B wL2 = = 28.125w. Mmax = 8 8 Section Properties: The cross section will be transformed into that of steel as Eal 10.6 = = 0.3655. shown in Fig. a. Here, n = Est 29 Then bst = nbal = 0.3655(3) = 1.0965 in. The location of the centroid of the transformed section is y =
©yA 1.5(3)(3) + 4.5(3)(1.0965) = = 2.3030 in. ©A 3(3) + 3(1.0965)
The moment of inertia of the transformed section about the neutral axis is I = ©I + Ad2 =
1 1 (3) A 33 B + 3(3)(2.3030 - 1.5)2 + (1.0965) A 33 B 12 12 + 1.0965 A 33 B + 1.0965(3)(4.5 - 2.3030)2
= 30.8991 in4 Bending Stress: Assuming failure of steel, (sallow)st =
Mmax cst ; I
22 =
(28.125w)(12)(2.3030) 30.8991
w = 0.875 kip>ft (controls)
Ans.
Assuming failure of aluminium alloy, (sallow)al = n
Mmax cal ; I
15 = 0.3655c
(28.125w)(12)(6 - 2.3030) d 30.8991
w = 1.02 kip>ft
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6–131. The Douglas fir beam is reinforced with A-36 straps at its center and sides. Determine the maximum stress developed in the wood and steel if the beam is subjected to a bending moment of Mz = 7.50 kip # ft. Sketch the stress distribution acting over the cross section.
y
0.5 in.
0.5 in.
0.5 in.
z
6 in.
2 in.
Section Properties: For the transformed section. n =
1.90(103) Ew = 0.065517 = Est 29.0(103)
bst = nbw = 0.065517(4) = 0.26207 in. INA =
1 (1.5 + 0.26207) A 63 B = 31.7172 in4 12
Maximum Bending Stress: Applying the flexure formula (smax)st =
7.5(12)(3) Mc = = 8.51 ksi I 31.7172
(smax)w = n
Ans.
7.5(12)(3) Mc = 0.065517c d = 0.558 ksi I 31.7172
Ans.
419
2 in.
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*6–132. The top plate is made of 2014-T6 aluminum and is used to reinforce a Kevlar 49 plastic beam. Determine the maximum stress in the aluminum and in the Kevlar if the beam is subjected to a moment of M = 900 lb # ft.
6 in. 0.5 in. 0.5 in. 12 in. M
0.5 in. 0.5 in.
Section Properties: n =
10.6(103) Eal = 0.55789 = Ek 19.0(103)
bk = n bal = 0.55789(12) = 6.6947 in. y =
0.25(13)(0.5) + 2[(3.25)(5.5)(0.5)] + 5.75(6.6947)(0.5) ©yA = ©A 13(0.5) + 2(5.5)(0.5) + 6.6947(0.5) = 2.5247 in.
INA =
1 (13) A 0.53 B + 13(0.5)(2.5247 - 0.25)2 12 +
1 (1) A 5.53 B + 1(5.5)(3.25 - 2.5247)2 12 +
1 (6.6947) A 0.53 B + 6.6947(0.5)(5.75 - 2.5247)2 12
= 85.4170 in4 Maximum Bending Stress: Applying the flexure formula (smax)al = n
(smax)k =
900(12)(6 - 2.5247) Mc = 0.55789 c d = 245 psi I 85.4170
900(12)(6 - 2.5247) Mc = = 439 psi I 85.4168
Ans.
Ans.
420
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•6–133.
The top plate made of 2014-T6 aluminum is used to reinforce a Kevlar 49 plastic beam. If the allowable bending stress for the aluminum is (sallow)al = 40 ksi and for the Kevlar (sallow)k = 8 ksi, determine the maximum moment M that can be applied to the beam.
6 in. 0.5 in. 0.5 in.
Section Properties: n =
10.6(103) Eal = 0.55789 = Ek 19.0(103)
12 in.
bk = n bal = 0.55789(12) = 6.6947 in. y =
0.5 in.
© yA 0.25(13)(0.5) + 2[(3.25)(5.5)(0.5)] + 5.75(6.6947(0.5) = ©A 13(0.5) + 2(5.5)(0.5) + 6.6947(0.5) = 2.5247 in.
INA =
1 (13) A 0.53 B + 13(0.5)(2.5247 - 0.25)2 12 +
1 (1) A 5.53 B + 1(5.5)(3.25 - 2.5247)2 12 +
1 (6.6947) A 0.53 B + 6.6947(0.5)(5.75 - 2.5247)2 12
= 85.4170 in4 Maximum Bending Stress: Applying the flexure formula Assume failure of aluminium (sallow)al = n
Mc I
40 = 0.55789 c
M(6 - 2.5247) d 85.4170
M = 1762 kip # in = 146.9 kip # ft Assume failure of Kevlar 49 (sallow)k = 8 =
Mc I M(6 - 2.5247) 85.4170
M = 196.62 kip # in = 16.4 kip # ft
M
0.5 in.
(Controls!)
Ans.
421
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6–134. The member has a brass core bonded to a steel casing. If a couple moment of 8 kN # m is applied at its end, determine the maximum bending stress in the member. Ebr = 100 GPa, Est = 200 GPa.
8 kNm
3m 20 mm 100 mm 20 mm
n =
Ebr 100 = = 0.5 Est 200
I =
1 1 (0.14)(0.14)3 (0.05)(0.1)3 = 27.84667(10 - 6)m4 12 12
20 mm
100 mm
20 mm
Maximum stress in steel: (sst)max =
8(103)(0.07) Mc1 = 20.1 MPa = I 27.84667(10 - 6)
Ans.
(max)
Maximum stress in brass: (sbr)max =
0.5(8)(103)(0.05) nMc2 = 7.18 MPa = I 27.84667(10 - 6)
6–135. The steel channel is used to reinforce the wood beam. Determine the maximum stress in the steel and in the wood if the beam is subjected to a moment of M = 850 lb # ft. Est = 29(103) ksi, Ew = 1600 ksi.
y =
4 in.
0.5 in.
(0.5)(16)(0.25) + 2(3.5)(0.5)(2.25) + (0.8276)(3.5)(2.25) = 1.1386 in. 0.5(16) + 2(3.5)(0.5) + (0.8276)(3.5)
15 in. M 850 lbft
0.5 in.
1 1 I = (16)(0.53) + (16)(0.5)(0.88862) + 2 a b(0.5)(3.53) + 2(0.5)(3.5)(1.11142) 12 12 +
1 (0.8276)(3.53) + (0.8276)(3.5)(1.11142) = 20.914 in4 12
Maximum stress in steel: (sst) =
850(12)(4 - 1.1386) Mc = = 1395 psi = 1.40 ksi I 20.914
Ans.
Maximum stress in wood: (sw) = n(sst)max = 0.05517(1395) = 77.0 psi
Ans.
422
0.5 in.
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*6–136. A white spruce beam is reinforced with A-36 steel straps at its top and bottom as shown. Determine the bending moment M it can support if (sallow)st = 22 ksi and (sallow)w = 2.0 ksi.
y 0.5 in.
4 in.
M
0.5 in.
x z 3 in.
Section Properties: For the transformed section. n =
1.40(103) Ew = 0.048276 = Est 29.0(103)
bst = nbw = 0.048276(3) = 0.14483 in. INA =
1 1 (3) A 53 B (3 - 0.14483) A 43 B = 16.0224 in4 12 12
Allowable Bending Stress: Applying the flexure formula Assume failure of steel (sallow)st = 22 =
Mc I M(2.5) 16.0224
M = 141.0 kip # in = 11.7 kip # ft (Controls !)
Ans.
Assume failure of wood (sallow)w = n
My I
2.0 = 0.048276 c
M(2) d 16.0224
M = 331.9 kip # in = 27.7 kip # ft
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•6–137. If the beam is subjected to an internal moment of
M = 45 kN # m, determine the maximum bending stress developed in the A-36 steel section A and the 2014-T6 aluminum alloy section B.
A 50 mm
M 15 mm 150 mm
Section Properties: The cross section will be transformed into that of steel as shown in Fig. a. 73.1 A 109 B Eal = = 0.3655. Thus, bst = nbal = 0.3655(0.015) = 0.0054825 m. The Here, n = Est 200 A 109 B location of the transformed section is
©yA y = = ©A
0.075(0.15)(0.0054825) + 0.2cp A 0.052 B d 0.15(0.0054825) + p A 0.052 B
= 0.1882 m The moment of inertia of the transformed section about the neutral axis is I = ©I + Ad2 =
1 (0.0054825) A 0.153 B + 0.0054825(0.15)(0.1882 - 0.075)2 12 +
1 p A 0.054 B + p A 0.052 B (0.2 - 0.1882)2 4
= 18.08 A 10 - 6 B m4 Maximum Bending Stress: For the steel,
(smax)st =
45 A 103 B (0.06185) Mcst = = 154 MPa I 18.08 A 10 - 6 B
Ans.
For the aluminum alloy,
(smax)al = n
45 A 103 B (0.1882) Mcal = 0.3655 C S = 171 MPa I 18.08 A 10 - 6 B
424
Ans.
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6–138. The concrete beam is reinforced with three 20-mm diameter steel rods. Assume that the concrete cannot support tensile stress. If the allowable compressive stress for concrete is (sallow)con = 12.5 MPa and the allowable tensile stress for steel is (sallow)st = 220 MPa, determine the required dimension d so that both the concrete and steel achieve their allowable stress simultaneously. This condition is said to be ‘balanced’. Also, compute the corresponding maximum allowable internal moment M that can be applied to the beam. The moduli of elasticity for concrete and steel are Econ = 25 GPa and Est = 200 GPa, respectively.
200 mm M
Bending Stress: The cross section will be transformed into that of concrete as shown Est 200 = = 8. It is required that both concrete and steel in Fig. a. Here, n = Econ 25 achieve their allowable stress simultaneously. Thus, (sallow)con =
12.5 A 106 B =
Mccon ; I
Mccon I
M = 12.5 A 106 B ¢ (sallow)st = n
I ≤ ccon
220 A 106 B = 8 B
Mcst ; I
(1)
M(d - ccon) R I
M = 27.5 A 106 B ¢
I ≤ d - ccon
(2)
Equating Eqs. (1) and (2), 12.5 A 106 B ¢
I I ≤ = 27.5 A 106 B ¢ ≤ ccon d - ccon
ccon = 0.3125d (3) Section Properties: The area of the steel bars is Ast = 3c
(3) p A 0.022 B d = 0.3 A 10 - 3 B p m2. 4
Thus, the transformed area of concrete from steel is (Acon)t = nAs = 8 C 0.3 A 10 - 3 B p D
= 2.4 A 10 - 3 B p m2. Equating the first moment of the area of concrete above and below the neutral axis about the neutral axis, 0.2(ccon)(ccon>2) = 2.4 A 10 - 3 B p (d - ccon)
0.1ccon 2 = 2.4 A 10 - 3 B pd - 2.4 A 10 - 3 B pccon ccon 2 = 0.024pd - 0.024pccon
(4)
Solving Eqs. (3) and (4), d = 0.5308 m = 531 mm
Ans.
ccon = 0.1659 m Thus, the moment of inertia of the transformed section is I =
1 (0.2) A 0.16593 B + 2.4 A 10 - 3 B p(0.5308 - 0.1659)2 3
425
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6–138.
Continued
= 1.3084 A 10 - 3 B m4 Substituting this result into Eq. (1), M = 12.5 A 106 B C
1.3084 A 10 - 3 B 0.1659
S
= 98 594.98 N # m = 98.6 kN # m‚
Ans.
6–139. The beam is made from three types of plastic that are identified and have the moduli of elasticity shown in the figure. Determine the maximum bending stress in the PVC. (bbk)1 = n1 bEs =
160 (3) = 0.6 in. 800
(bbk)2 = n2 bpvc =
450 (3) = 1.6875 in. 800
500 lb
PVC EPVC 450 ksi Escon EE 160 ksi Bakelite EB 800 ksi 3 ft
y =
©yA (1)(3)(2) + 3(0.6)(2) + 4.5(1.6875)(1) = = 1.9346 in. ©A 3(2) + 0.6(2) + 1.6875(1)
I =
1 1 (3)(23) + 3(2)(0.93462) + (0.6)(23) + 0.6(2)(1.06542) 12 12 +
4 ft
1 in. 2 in. 2 in. 3 in.
1 (1.6875)(13) + 1.6875(1)(2.56542) = 20.2495 in4 12
(smax)pvc = n2
500 lb
450 1500(12)(3.0654) Mc = a b I 800 20.2495 = 1.53 ksi
Ans.
426
3 ft
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*6–140. The low strength concrete floor slab is integrated with a wide-flange A-36 steel beam using shear studs (not shown) to form the composite beam. If the allowable bending stress for the concrete is (sallow)con = 10 MPa, and allowable bending stress for steel is (sallow)st = 165 MPa, determine the maximum allowable internal moment M that can be applied to the beam.
1m
100 mm
15 mm 400 mm M 15 mm 15 mm
Section Properties: The beam cross section will be transformed into Econ 22.1 that of steel. Here, Thus, = = 0.1105. n = Est 200 bst = nbcon = 0.1105(1) = 0.1105 m. The location of the transformed section is y =
=
©yA ©A 0.0075(0.015)(0.2) + 0.2(0.37)(0.015) + 0.3925(0.015)(0.2) + 0.45(0.1)(0.1105) 0.015(0.2) + 0.37(0.015) + 0.015(0.2) + 0.1(0.1105)
= 0.3222 m The moment of inertia of the transformed section about the neutral axis is I = ©I + Ad2 =
1 (0.2) A 0.0153 B 12 + 0.2(0.015)(0.3222 - 0.0075)2 +
1 (0.015) A 0.373 B + 0.015(0.37)(0.3222 - 0.2)2 12
+
1 (0.2) A 0.0153 B + 0.2(0.015)(0.3925 - 0.3222)2 12
+
1 (0.1105) A 0.13 B + 0.1105(0.1)(0.45 - 0.3222)2 12 = 647.93 A 10 - 6 B m4
Bending Stress: Assuming failure of steel, (sallow)st =
M(0.3222) Mcst ; 165 A 106 B = I 647.93 A 10 - 6 B M = 331 770.52 N # m = 332 kN # m
Assuming failure of concrete,
(sallow)con = n
Mccon ; I
10 A 106 B = 0.1105C
M(0.5 - 0.3222) 647.93 A 10 - 6 B
S
M = 329 849.77 N # m = 330 kN # m (controls) Ans.
427
200 mm
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•6–141.
The reinforced concrete beam is used to support the loading shown. Determine the absolute maximum normal stress in each of the A-36 steel reinforcing rods and the absolute maximum compressive stress in the concrete. Assume the concrete has a high strength in compression and yet neglect its strength in supporting tension.
10 kip
8 in.
15 in. 4 ft
8 ft
Mmax = (10 kip)(4 ft) = 40 kip # ft Ast = 3(p)(0.5)2 = 2.3562 in2 Est = 29.0(103) ksi Econ = 4.20(103) ksi A¿ = nAst =
©yA = 0;
29.0(103) 4.20(103) 8(h¿)a
(2.3562) = 16.2690 in2
h¿ b - 16.2690(13 - h¿) = 0 2
h¿ 2 + 4.06724h - 52.8741 = 0 Solving for the positive root: h¿ = 5.517 in. I = c
1 (8)(5.517)3 + 8(5.517)(5.517>2)2 d + 16.2690(13 - 5.517)2 12
= 1358.781 in4 (scon)max =
My 40(12)(5.517) = = 1.95 ksi I 1358.781
(sst)max = na
10 kip
Ans.
My 29.0(103) 40(12)(13 - 5.517) ba b = a b = 18.3 ksi I 1358.781 4.20(103)
428
Ans.
4 ft
2 in. 1 in. diameter rods
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6–142. The reinforced concrete beam is made using two steel reinforcing rods. If the allowable tensile stress for the steel is (sst)allow = 40 ksi and the allowable compressive stress for the concrete is (sconc)allow = 3 ksi, determine the maximum moment M that can be applied to the section. Assume the concrete cannot support a tensile stress. Est = 29(103) ksi, Econc = 3.8(103) ksi.
8 in. 6 in. 4 in.
8 in.
M 18 in. 2 in. 1-in. diameter rods
Ast = 2(p)(0.5)2 = 1.5708 in2 A¿ = nAst = ©yA = 0;
29(103) 3.8(103)
(1.5708) = 11.9877 in2
22(4)(h¿ + 2) + h¿(6)(h¿>2) - 11.9877(16 - h¿) = 0 3h2 + 99.9877h¿ - 15.8032 = 0
Solving for the positive root: h¿ = 0.15731 in. I = c
1 1 (22)(4)3 + 22(4)(2.15731)2 d + c (6)(0.15731)3 + 6(0.15731)(0.15731>2)2 d 12 12 + 11.9877(16 - 0.15731)2 = 3535.69 in4
Assume concrete fails: (scon)allow =
My ; I
3 =
M(4.15731) 3535.69
M = 2551 kip # in. Assume steel fails: (sst)allow = na
My b; I
40 = ¢
29(103) 3
3.8(10 )
≤¢
M(16 - 0.15731) ≤ 3535.69
M = 1169.7 kip # in. = 97.5 kip # ft (controls) Ans.
429
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6–143. For the curved beam in Fig. 6–40a, show that when the radius of curvature approaches infinity, the curved-beam formula, Eq. 6–24, reduces to the flexure formula, Eq. 6–13. Normal Stress: Curved-beam formula M(R - r)
s =
where A¿ =
Ar(r - R)
dA LA r
and R =
A 1A
dA r
=
A A¿
M(A - rA¿)
s =
[1]
Ar(rA¿ - A)
r = r + y rA¿ = r
[2]
dA r = a - 1 + 1 b dA LA r + y LA r =
LA
a
= A -
r - r - y r + y y
+ 1b dA
dA
LA r + y
[3]
Denominator of Eq. [1] becomes, y
Ar(rA¿ - A) = Ar ¢ A -
LA r + y
dA - A ≤ = -Ar
y LA r + y
dA
Using Eq. [2], Ar(rA¿ - A) = -A
= A
=
¢
ry
LA r + y y2
LA r + y
+ y - y ≤ dA - Ay
LA r + y
dA - A 1A y dA - Ay
y LA r + y
as
y r
: 0
A I r
Then,
Ar(rA¿ - A) :
Eq. [1] becomes
s =
Mr (A - rA¿) AI
Using Eq. [2],
s =
Mr (A - rA¿ - yA¿) AI
Using Eq. [3],
s =
=
dA
dA
y2 y Ay A ¢ ¢ y ≤ dA - A 1A y dA r LA 1 + r r LA 1 + 1A y dA = 0,
But,
y
y Mr dA C A - ¢A dA ≤ - y S AI r + y r LA LA + y
y Mr dA C dA - y S AI LA r + y r LA + y
430
y≤ r
dA
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6–143. Continued y
=
y
As
r
Mr r C ¢ AI LA 1 +
y ≤ dA r
y -
r LA
¢
dA ≤S 1 + yr
=
: 0
¢
y r
LA 1 +
y≤ r
dA = 0
s =
Therefore,
and
y r LA
¢
y yA dA A dA = y≤ = 1 1 + r r r
yA My Mr b = aAI I r
(Q.E.D.)
*6–144. The member has an elliptical cross section. If it is subjected to a moment of M = 50 N # m, determine the stress at points A and B. Is the stress at point A¿ , which is located on the member near the wall, the same as that at A? Explain.
75 mm
150 mm A¿ 250 mm A
dA 2p b = (r - 2r2 - a2 ) a LA r
100 mm
2p(0.0375) = (0.175 - 20.1752 - 0.0752 ) = 0.053049301 m 0.075 A = p ab = p(0.075)(0.0375) = 2.8125(10 - 3)p R =
A 1A
dA r
=
B
2.8125(10 - 3)p = 0.166556941 0.053049301
r - R = 0.175 - 0.166556941 = 0.0084430586 sA =
sB =
M(R - rA)
50(0.166556941 - 0.1) =
2.8125(10 - 3)p (0.1)(0.0084430586)
=
2.8125(10 - 3)p (0.25)(0.0084430586)
ArA (r - R) M(R - rB) ArB (r - R)
50(0.166556941 - 0.25)
= 446k Pa (T)
= 224 kPa (C)
No, because of localized stress concentration at the wall.
Ans.
Ans. Ans.
431
M
Mr AI
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•6–145. The member has an elliptical cross section. If the allowable bending stress is sallow = 125 MPa determine the maximum moment M that can be applied to the member.
75 mm
150 mm A¿ 250 mm A 100 mm
B
b = 0.0375 m
a = 0.075 m;
A = p(0.075)(0.0375) = 0.0028125 p 2p(0.0375) dA 2pb (0.175 - 20.1752 - 0.0752) = (r - 2r2 - a2) = r a 0.075 LA = 0.053049301 m R =
A dA 1A r
=
0.0028125p = 0.166556941 m 0.053049301
r - R = 0.175 - 0.166556941 = 8.4430586(10 - 3) m s =
M(R - r) Ar(r - R)
Assume tension failure. 125(106) =
M(0.166556941 - 0.1) 0.0028125p(0.1)(8.4430586)(10 - 3)
M = 14.0 kN # m (controls)
Ans.
Assume compression failure: -125(106) =
M(0.166556941 - 0.25) 0.0028125p(0.25)(8.4430586)(10 - 3)
M = 27.9 kN # m
432
M
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6–146. Determine the greatest magnitude of the applied forces P if the allowable bending stress is (sallow)c = 50 MPa in compression and (sallow)t = 120 MPa in tension.
75 mm P
10 mm
10 mm 160 mm
10 mm
P 150 mm 250 mm
Internal Moment: M = 0.160P is positive since it tends to increase the beam’s radius of curvature. Section Properties: r =
=
©yA ©A 0.255(0.15)(0.01) + 0.335(0.15)(0.01) + 0.415(0.075)(0.01) 0.15(0.01) + 0.15(0.01) + 0.075(0.01)
= 0.3190 m A = 0.15(0.01) + 0.15(0.01) + 0.075(0.01) = 0.00375 m2 ©
dA 0.26 0.41 0.42 = 0.15 ln + 0.01 ln + 0.075 ln 0.25 0.26 0.41 LA r = 0.012245 m
R =
A © 1A dA r
=
0.00375 = 0.306243 m 0.012245
r - R = 0.319 - 0.306243 = 0.012757 m Allowable Normal Stress: Applying the curved-beam formula Assume tension failure (sallow)t = 120 A 106 B =
M(R - r) Ar(r - R) 0.16P(0.306243 - 0.25) 0.00375(0.25)(0.012757)
P = 159482 N = 159.5 kN Assume compression failure (sallow)t = -50 A 106 B =
M(R - r) Ar(r - R) 0.16P(0.306243 - 0.42) 0.00375(0.42)(0.012757)
P = 55195 N = 55.2 kN (Controls !)
Ans.
433
150 mm
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6–147. If P = 6 kN, determine the maximum tensile and compressive bending stresses in the beam.
75 mm P
10 mm
10 mm 160 mm
10 mm
P 150 mm 250 mm
Internal Moment: M = 0.160(6) = 0.960 kN # m is positive since it tends to increase the beam’s radius of curvature. Section Properties: r =
=
©yA ©A 0.255(0.15)(0.01) + 0.335(0.15)(0.01) + 0.415(0.075)(0.01) 0.15(0.01) + 0.15(0.01) + 0.075(0.01)
= 0.3190 m A = 0.15(0.01) + 0.15(0.01) + 0.075(0.01) = 0.00375 m2 ©
dA 0.41 0.42 0.26 = 0.15 ln + 0.01 ln + 0.075 ln 0.25 0.26 0.41 LA r = 0.012245 m
R =
A ©1A dA r
=
0.00375 = 0.306243 m 0.012245
r - R = 0.319 - 0.306243 = 0.012757 m Normal Stress: Applying the curved-beam formula (smax)t =
=
M(R - r) Ar(r - R) 0.960(103)(0.306243 - 0.25) 0.00375(0.25)(0.012757)
= 4.51 MPa (smax)c =
=
Ans.
M(R - r) Ar(r - R) 0.960(103)(0.306243 - 0.42) 0.00375(0.42)(0.012757)
= -5.44 MPa
Ans.
434
150 mm
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*6–148. The curved beam is subjected to a bending moment of M = 900 N # m as shown. Determine the stress at points A and B, and show the stress on a volume element located at each of these points.
A C B
100 mm C
A
30
20 mm
15 mm
150 mm
400 mm B M
Internal Moment: M = -900 N # m is negative since it tends to decrease the beam’s radius curvature. Section Properties: ©A = 0.15(0.015) + 0.1(0.02) = 0.00425 m2 ©rA = 0.475(0.15)(0.015) + 0.56(0.1)(0.02) = 2.18875(10 - 3) m3 r =
©
2.18875 (10 - 3) ©rA = = 0.5150 m ©A 0.00425
dA 0.57 0.55 = 0.015 ln + 0.1 ln = 8.348614(10 - 3) m 0.4 0.55 LA r
R =
A ©1A
dA r
=
0.00425 = 0.509067 m 8.348614(10 - 3)
r - R = 0.515 - 0.509067 = 5.933479(10 - 3) m Normal Stress: Applying the curved-beam formula sA =
M(R - rA)
-900(0.509067 - 0.57) =
ArA (r - R)
0.00425(0.57)(5.933479)(10 - 3) Ans.
= 3.82 MPa (T) sB =
M(R - rB)
-900(0.509067 - 0.4) =
ArB (r - R)
0.00425(0.4)(5.933479)(10 - 3)
= -9.73 MPa = 9.73 MPa (C)
Ans.
435
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•6–149.
The curved beam is subjected to a bending moment of M = 900 N # m. Determine the stress at point C. A C B
100 mm C
A
30
20 mm
15 mm
150 mm
400 mm B M
Internal Moment: M = -900 N # m is negative since it tends to decrease the beam’s radius of curvature. Section Properties: ©A = 0.15(0.015) + 0.1(0.02) = 0.00425 m2 ©rA = 0.475(0.15)(0.015) + 0.56(0.1)(0.02) = 2.18875(10 - 3) m r =
©
2.18875 (10 - 3) ©rA = = 0.5150 m ©A 0.00425
dA 0.57 0.55 = 0.015 ln + 0.1 ln = 8.348614(10 - 3) m 0.4 0.55 LA r
R =
A ©1A
dA r
=
0.00425 = 0.509067 m 8.348614(10 - 3)
r - R = 0.515 - 0.509067 = 5.933479(10 - 3) m Normal Stress: Applying the curved-beam formula sC =
M(R - rC)
-900(0.509067 - 0.55) =
ArC(r - R)
0.00425(0.55)(5.933479)(10 - 3)
= 2.66 MPa (T)
Ans.
436
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6–150. The elbow of the pipe has an outer radius of 0.75 in. and an inner radius of 0.63 in. If the assembly is subjected to the moments of M = 25 lb # in., determine the maximum stress developed at section a-a.
a 30
M 25 lbin.
1 in. a
dA = ©2p (r - 2r2 - c2) LA r = 2p(1.75 - 21.752 - 0.752) - 2p (1.75 - 21.752 - 0.632)
0.63 in. 0.75 in.
= 0.32375809 in. A = p(0.752) - p(0.632) = 0.1656 p R =
A dA 1A r
=
M = 25 lbin.
0.1656 p = 1.606902679 in. 0.32375809
r - R = 1.75 - 1.606902679 = 0.14309732 in. (smax)t =
M(R - rA) = ArA(r - R)
(smax)c = =
25(1.606902679 - 1) = 204 psi (T) 0.1656 p(1)(0.14309732)
M(R - rB) = ArB(r - R)
Ans.
25(1.606902679 - 2.5) = 120 psi (C) 0.1656p(2.5)(0.14309732)
Ans.
6–151. The curved member is symmetric and is subjected to a moment of M = 600 lb # ft. Determine the bending stress in the member at points A and B. Show the stress acting on volume elements located at these points.
0.5 in. B 2 in. A
1 A = 0.5(2) + (1)(2) = 2 in2 2 r =
1.5 in. 8 in.
9(0.5)(2) + 8.6667 A 12 B (1)(2) ©rA = = 8.83333 in. ©A 2
M
M
1(10) dA 10 10 = 0.5 ln + c cln d - 1 d = 0.22729 in. r 8 (10 - 8) 8 LA R =
A dA 1A r
=
2 = 8.7993 in. 0.22729
r - R = 8.83333 - 8.7993 = 0.03398 in. s =
M(R - r) Ar(r - R)
sA =
600(12)(8.7993 - 8) = 10.6 ksi (T) 2(8)(0.03398)
Ans.
sB =
600(12)(8.7993 - 10) = -12.7 ksi = 12.7 ksi (C) 2(10)(0.03398)
Ans.
437
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*6–152. The curved bar used on a machine has a rectangular cross section. If the bar is subjected to a couple as shown, determine the maximum tensile and compressive stress acting at section a-a. Sketch the stress distribution on the section in three dimensions.
a 75 mm a
50 mm
162.5 mm
250 N 60
150 mm
60 250 N 75 mm
a + ©MO = 0;
M - 250 cos 60° (0.075) - 250 sin 60° (0.15) = 0 M = 41.851 N # m
r2 dA 0.2375 = b ln = 0.05 ln = 0.018974481 m r r 0.1625 1 LA A = (0.075)(0.05) = 3.75(10 - 3) m2 R =
A 1A
dA r
=
3.75(10 - 3) = 0.197633863 m 0.018974481
r - R = 0.2 - 0.197633863 = 0.002366137 sA =
M(R - rA)
41.851(0.197633863 - 0.2375) =
ArA(r - R)
3.75(10 - 3)(0.2375)(0.002366137)
= -791.72 kPa Ans.
= 792 kPa (C) sB =
M(R - rB)
41.851 (0.197633863 - 0.1625) =
ArB(r - R)
3.75(10 - 3)(0.1625)(0.002366137)
= 1.02 MPa (T)
438
Ans.
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•6–153.
The ceiling-suspended C-arm is used to support the X-ray camera used in medical diagnoses. If the camera has a mass of 150 kg, with center of mass at G, determine the maximum bending stress at section A.
G
1.2 m A
200 mm 100 mm 20 mm 40 mm
Section Properties: r =
©
1.22(0.1)(0.04) + 1.25(0.2)(0.02) ©rA = = 1.235 m ©A 0.1(0.04) + 0.2(0.02)
dA 1.26 1.24 = 0.1 ln + 0.2 ln = 6.479051 A 10 - 3 B m r 1.20 1.24 LA
A = 0.1(0.04) + 0.2(0.02) = 0.008 m2 R =
A dA 1A r
=
0.008 = 1.234749 m 6.479051 (10 - 3)
r - R = 1.235 - 1.234749 = 0.251183 A 10 - 3 B m Internal Moment: The internal moment must be computed about the neutral axis as shown on FBD. M = -1816.93 N # m is negative since it tends to decrease the beam’s radius of curvature. Maximum Normal Stress: Applying the curved-beam formula sA =
M(R - rA) ArA (r - R) -1816.93(1.234749 - 1.26)
=
0.008(1.26)(0.251183)(10 - 3)
= 18.1 MPa (T) sB =
M(R - rB) ArB (r - R) -1816.93(1.234749 - 1.20)
=
0.008(1.20)(0.251183)(10 - 3)
= -26.2 MPa = 26.2 MPa (C)
Ans.
(Max)
439
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6–154. The circular spring clamp produces a compressive force of 3 N on the plates. Determine the maximum bending stress produced in the spring at A. The spring has a rectangular cross section as shown.
10 mm 20 mm
Internal Moment: As shown on FBD, M = 0.660 N # m is positive since it tends to increase the beam’s radius of curvature.
210 mm
200 mm A
Section Properties: 220 mm
0.200 + 0.210 r = = 0.205 m 2 r2 dA 0.21 = 0.02 ln = b ln = 0.97580328 A 10 - 3 B m r r 0.20 1 LA A = (0.01)(0.02) = 0.200 A 10 - 3 B m2 R =
0.200(10 - 3)
A 1A
dA r
=
0.97580328(10 - 3)
= 0.204959343 m
r - R = 0.205 - 0.204959343 = 0.040657 A 10 - 3 B m Maximum Normal Stress: Applying the curved-beam formula sC =
M(R - r2) Ar2(r - R) 0.660(0.204959343 - 0.21)
=
0.200(10 - 3)(0.21)(0.040657)(10 - 3)
= -1.95MPa = 1.95 MPa (C) st =
M(R - r1) Ar1 (r - R) 0.660(0.204959343 - 0.2)
=
0.200(10 - 3)(0.2)(0.040657)(10 - 3)
= 2.01 MPa (T)
(Max)
Ans.
440
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6–155. Determine the maximum compressive force the spring clamp can exert on the plates if the allowable bending stress for the clamp is sallow = 4 MPa.
10 mm 20 mm
210 mm
200 mm A
220 mm
Section Properties: r =
0.200 + 0.210 = 0.205 m 2
r2 dA 0.21 = b ln = 0.02 ln = 0.97580328 A 10 - 3 B m r1 0.20 LA r A = (0.01)(0.02) = 0.200 A 10 - 3 B m2 R =
0.200(10 - 3)
A 1A
dA r
=
0.97580328(10 - 3)
= 0.204959 m
r - R = 0.205 - 0.204959343 = 0.040657 A 10 - 3 B m Internal Moment: The internal moment must be computed about the neutral axis as shown on FBD. Mmax = 0.424959P is positive since it tends to increase the beam’s radius of curvature. Allowable Normal Stress: Applying the curved-beam formula Assume compression failure sc = sallow = -4 A 106 B =
M(R - r2) Ar2(r - R) 0.424959P(0.204959 - 0.21) 0.200(10 - 3)(0.21)(0.040657)(10 - 3)
P = 3.189 N Assume tension failure st = sallow = 4 A 106 B =
M(R - r1) Ar1 (r - R) 0.424959P(0.204959 - 0.2) 0.200(10 - 3)(0.2)(0.040657)(10 - 3)
P = 3.09 N (Controls !)
Ans.
441
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*6–156. While in flight, the curved rib on the jet plane is subjected to an anticipated moment of M = 16 N # m at the section. Determine the maximum bending stress in the rib at this section, and sketch a two-dimensional view of the stress distribution. 16 Nm 5 mm 20 mm 5 mm
0.6 m 5 mm
30 mm
LA
0.625 0.630 0.605 + (0.005)ln + (0.03)ln = 0.650625(10 - 3) in. 0.6 0.605 0.625
dA>r = (0.03)ln
A = 2(0.005)(0.03) + (0.02)(0.005) = 0.4(10 - 3) in2 R =
0.4(10 - 3)
A 1A dA>r
=
0.650625(10 - 3)
= 0.6147933
(sc)max =
M(R - rc) 16(0.6147933 - 0.630) = -4.67 MPa = ArA(r - R) 0.4(10 3)(0.630)(0.615 - 0.6147933)
(ss)max =
M(R - rs) 16(0.6147933 - 0.6) = 4.77 MPa = ArA(r - R) 0.4(10 - 3)(0.6)(0.615 - 0.6147933)
Ans.
If the radius of each notch on the plate is r = 0.5 in., determine the largest moment that can be applied. The allowable bending stress for the material is sallow = 18 ksi.
•6–157.
14.5 in.
M
b =
14.5 - 12.5 = 1.0 in. 2 r 0.5 = = 0.04 h 12.5
1 b = = 2.0 r 0.5 From Fig. 6-44: K = 2.60 smax = K
Mc I
18(103) = 2.60c
(M)(6.25) 1 3 12 (1)(12.5)
d
M = 180 288 lb # in. = 15.0 kip # ft
Ans.
442
1 in.
12.5 in.
M
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6–158. The symmetric notched plate is subjected to bending. If the radius of each notch is r = 0.5 in. and the applied moment is M = 10 kip # ft, determine the maximum bending stress in the plate.
14.5 in.
M
M
12.5 in.
r 0.5 = = 0.04 h 12.5
1 b = 2.0 = r 0.5
1 in.
From Fig. 6-44: K = 2.60 smax = K
(10)(12)(6.25) Mc = 2.60 c 1 d = 12.0 ksi 3 I 12 (1)(12.5)
Ans.
6–159. The bar is subjected to a moment of M = 40 N # m. Determine the smallest radius r of the fillets so that an allowable bending stress of sallow = 124 MPa is not exceeded.
80 mm 7 mm
20 mm r
M
M r
Allowable Bending Stress: sallow = K
Mc I
124 A 106 B = K B
40(0.01)
R 1 3 12 (0.007)(0.02 )
K = 1.45 Stress Concentration Factor: From the graph in the text w 80 r with = = 4 and K = 1.45, then = 0.25. h 20 h r = 0.25 20 r = 5.00 mm
Ans.
*6–160. The bar is subjected to a moment of M = 17.5 N # m. If r = 5 mm, determine the maximum bending stress in the material.
80 mm 7 mm
20 mm r
M
M
Stress Concentration Factor: From the graph in the text with r w 80 5 = = 4 and = = 0.25, then K = 1.45. h 20 h 20
r
Maximum Bending Stress: smax = K
Mc I
= 1.45 B
17.5(0.01)
R 1 3 12 (0.007)(0.02 )
= 54.4 MPa
Ans.
443
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•6–161. The simply supported notched bar is subjected to two forces P. Determine the largest magnitude of P that can be applied without causing the material to yield.The material is A-36 steel. Each notch has a radius of r = 0.125 in.
P
P 0.5 in. 1.75 in.
1.25 in.
20 in.
b =
20 in.
20 in.
20 in.
1.75 - 1.25 = 0.25 2
0.25 b = = 2; r 0.125
r 0.125 = = 0.1 h 1.25
From Fig. 6-44. K = 1.92 sY = K
Mc ; I
36 = 1.92 c
20P(0.625) 1 3 12 (0.5)(1.25)
d
P = 122 lb
Ans.
6–162. The simply supported notched bar is subjected to the two loads, each having a magnitude of P = 100 lb. Determine the maximum bending stress developed in the bar, and sketch the bending-stress distribution acting over the cross section at the center of the bar. Each notch has a radius of r = 0.125 in.
P
0.5 in.
1.75 - 1.25 = 0.25 2
b 0.25 = = 2; r 0.125
r 0.125 = = 0.1 h 1.25
From Fig. 6-44, K = 1.92 smax = K
1.75 in.
1.25 in.
20 in.
b =
P
2000(0.625) Mc = 1.92c 1 d = 29.5 ksi 3 I 12 (0.5)(1.25)
Ans.
444
20 in.
20 in.
20 in.
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6–163. Determine the length L of the center portion of the bar so that the maximum bending stress at A, B, and C is the same. The bar has a thickness of 10 mm.
7 mm
350 N 60 mm
A
r 7 = = 0.175 h 40
60 w = = 1.5 h 40
200 mm
40 mm 7 mm
C L 2
B L 2
200 mm
From Fig. 6-43, K = 1.5 (sA)max = K
(35)(0.02) MAc d = 19.6875 MPa = 1.5c 1 3 I 12 (0.01)(0.04 )
(sB)max = (sA)max = 19.6875(106) =
MB c I
175(0.2 + L2 )(0.03) 1 3 12 (0.01)(0.06 )
L = 0.95 m = 950 mm
Ans.
*6–164. The stepped bar has a thickness of 15 mm. Determine the maximum moment that can be applied to its ends if it is made of a material having an allowable bending stress of sallow = 200 MPa.
45 mm 30 mm 3 mm
M
M
Stress Concentration Factor: w 30 6 r = = 3 and = = 0.6, we have K = 1.2 h 10 h 10 obtained from the graph in the text. For the smaller section with
w 45 3 r = = 1.5 and = = 0.1, we have K = 1.75 h 30 h 30 obtained from the graph in the text. For the larger section with
Allowable Bending Stress: For the smaller section smax = sallow = K
Mc ; I
200 A 106 B = 1.2 B
M(0.005)
R 1 3 12 (0.015)(0.01 )
M = 41.7 N # m (Controls !)
Ans.
For the larger section smax = sallow = K
Mc ; I
200 A 106 B = 1.75 B
M(0.015)
R 1 3 12 (0.015)(0.03 )
M = 257 N # m
445
10 mm 6 mm
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•6–165.
The beam is made of an elastic plastic material for which sY = 250 MPa. Determine the residual stress in the beam at its top and bottom after the plastic moment Mp is applied and then released.
15 mm
1 1 (0.2)(0.23)3 (0.18)(0.2)3 = 82.78333(10 - 6)m4 12 12
20 mm 200 mm
Ix =
Mp
C1 = T1 = sY (0.2)(0.015) = 0.003sY
15 mm
C2 = T2 = sY (0.1)(0.02) = 0.002sY 200 mm
Mp = 0.003sY (0.215) + 0.002sY (0.1) = 0.000845 sY = 0.000845(250)(106) = 211.25 kN # m s =
Mp c
211.25(103)(0.115) =
I
82.78333(10 - 6)
y 0.115 = ; 250 293.5
= 293.5 MPa
y = 0.09796 m = 98.0 mm
stop = sbottom = 293.5 - 250 = 43.5 MPa
Ans.
6–166. The wide-flange member is made from an elasticplastic material. Determine the shape factor.
t
Plastic analysis: T1 = C1 = sY bt;
h
T2 = C2 = sY a
MP = sY bt(h - t) + sY a
h - 2t bt 2
t t
h - 2t h - 2t b(t) a b 2 2 b
t = sY c bt(h - t) + (h - 2t)2 d 4 Elastic analysis: I =
=
1 1 bh3 (b - t)(h - 2t)3 12 12 1 [bh3 - (b - t)(h - 2 t)3] 12
MY =
sy I c
=
=
1 sY A 12 B [bh3 - (b - t)(h - 2t)3] h 2
bh3 - (b - t)(h - 2t)3 sY 6h
Shape factor: k =
[bt(h - t) + 4t (h - 2t)2]sY MP = bh3 - (b - t)(h - 2t)3 MY s 6h
=
Y
3h 4bt(h - t) + t(h - 2t)2 c d 2 bh3 - (b - t)(h - 2t)3
Ans.
446
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6–167.
Determine the shape factor for the cross section.
Maximum Elastic Moment: The moment of inertia about neutral axis must be determined first.
a
1 1 (a)(3a)3 + (2a) A a3 B = 2.41667a4 12 12
INA =
a a
Applying the flexure formula with s = sY, we have sY =
MY c I
MY =
a
a
a
sY (2.41667a4) sYI = = 1.6111a3sY c 1.5a
Plastic Moment: MP = sY (a)(a)(2a) + sY (0.5a)(3a)(0.5a) = 2.75a3sY Shape Factor: k =
MP 2.75a3sY = = 1.71 MY 1.6111a3sY
Ans.
*6–168. The beam is made of elastic perfectly plastic material. Determine the maximum elastic moment and the plastic moment that can be applied to the cross section. Take a = 2 in. and sY = 36 ksi.
a a
Maximum Elastic Moment: The moment of inertia about neutral axis must be determined first. INA
a
1 1 (2) A 63 B + (4) A 23 B = 38.667 in4 = 12 12
Applying the flexure formula with s = sY, we have sY = = MY =
a
MY c I
36(38.667) sY I = c 3 = 464 kip # in = 38.7 kip # ft
Ans.
Plastic Moment: MP = 36(2)(2)(4) + 36(1)(6)(1) = 792 kip # in = 66.0 kip # ft
Ans.
447
a
a
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•6–169.
The box beam is made of an elastic perfectly plastic material for which sY = 250 MPa . Determine the residual stress in the top and bottom of the beam after the plastic moment Mp is applied and then released.
Plastic Moment: MP = 250 A 106 B (0.2)(0.025)(0.175) + 250 A 106 B (0.075)(0.05)(0.075)
25 mm
= 289062.5 N # m
150 mm
Modulus of Rupture: The modulus of rupture sr can be determined using the flexure formula with the application of reverse, plastic moment MP = 289062.5 N # m. I =
25 mm
25 mm 150 mm 25 mm
1 1 (0.2) A 0.23 B (0.15) A 0.153 B 12 12
= 91.14583 A 10 - 6 B m4 sr =
289062.5 (0.1) MP c = 317.41 MPa = I 91.14583 A 10 - 6 B
Residual Bending Stress: As shown on the diagram. œ œ = sbot = sr - sY stop
= 317.14 - 250 = 67.1 MPa
Ans.
6–170. Determine the shape factor for the wideflange beam.
15 mm
1 1 (0.2)(0.23)3 (0.18)(0.2)3 = 82.78333 A 10 - 6 B m4 12 12
Ix =
20 mm 200 mm
C1 = T1 = sY(0.2)(0.015) = 0.003sY
Mp
C2 = T2 = sY(0.1)(0.02) = 0.002sY Mp = 0.003sY(0.215) + 0.002sY(0.1) = 0.000845 sY
15 mm 200 mm
sY =
MY =
k =
MY c I sY A 82.78333)10 - 6 B 0.115
Mp MY
=
= 0.000719855 sY
0.000845sY = 1.17 0.000719855sY
Ans.
448
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6–171. Determine the shape factor of the beam’s cross section. 3 in.
Referring to Fig. a, the location of centroid of the cross-section is y =
7.5(3)(6) + 3(6)(3) ©yA = = 5.25 in. ©A 3(6) + 6(3)
6 in.
The moment of inertia of the cross-section about the neutral axis is I =
1 1 (3) A 63 B + 3(6)(5.25 - 3)2 + (6) A 33 B + 6(3)(7.5 - 5.25)2 12 12
1.5 in. 3 in. 1.5 in.
4
= 249.75 in
Here smax = sY and c = y = 5.25 in. Thus smax =
Mc ; I
sY =
MY (5.25) 249.75
MY = 47.571sY Referring to the stress block shown in Fig. b, sdA = 0; LA
T - C1 - C2 = 0
d(3)sY - (6 - d)(3)sY - 3(6)sY = 0 d = 6 in. Since d = 6 in., c1 = 0, Fig. c. Here T = C = 3(6) sY = 18 sY Thus, MP = T(4.5) = 18 sY (4.5) = 81 sY Thus, k =
MP 81 sY = = 1.70 MY 47.571 sY
Ans.
449
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*6–172. The beam is made of elastic-perfectly plastic material. Determine the maximum elastic moment and the plastic moment that can be applied to the cross section. Take sY = 36 ksi.
3 in.
Referring to Fig. a, the location of centroid of the cross-section is 6 in.
7.5(3)(6) + 3(6)(3) ©yA y = = = 5.25 in. ©A 3(6) + 6(3) The moment of inertia of the cross-section about the neutral axis is
1.5 in. 3 in. 1.5 in.
I =
1 1 (3)(63) + 3(6)(5.25 - 3)2 + (6)(33) + 6(3)(7.5 - 5.25)2 12 12
= 249.75 in4 Here, smax = sY = 36 ksi and ¢ = y = 5.25 in. Then smax =
Mc ; I
36 =
MY (5.25) 249.75
MY = 1712.57 kip # in = 143 kip # ft
Ans.
Referring to the stress block shown in Fig. b, sdA = 0; LA
T - C1 - C2 = 0
d(3) (36) - (6 - d)(3)(36) - 3(6) (36) = 0 d = 6 in. Since d = 6 in., c1 = 0, Here, T = C = 3(6)(36) = 648 kip Thus, MP = T(4.5) = 648(4.5) = 2916 kip # in = 243 kip # ft
Ans.
450
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•6–173.
Determine the shape factor for the cross section of the H-beam.
Ix =
1 1 (0.2)(0.023) + 2 a b(0.02)(0.23) = 26.8(10 - 6)m4 12 12
200 mm
C1 = T1 = sY(2)(0.09)(0.02) = 0.0036sy 20 mm
C2 = T2 = sY(0.01)(0.24) = 0.0024sy
Mp
20 mm
200 mm
Mp = 0.0036sY(0.11) + 0.0024sY(0.01) = 0.00042sY
20 mm
MYc sY = I MY =
k =
sY(26.8)(10 - 6) = 0.000268sY 0.1
Mp MY
=
0.00042sY = 1.57 0.000268sY
Ans.
6–174. The H-beam is made of an elastic-plastic material for which sY = 250 MPa. Determine the residual stress in the top and bottom of the beam after the plastic moment Mp is applied and then released. 200 mm
Ix =
1 1 (0.2)(0.023) + 2 a b (0.02)(0.23) = 26.8(10 - 6)m4 12 12 20 mm
C1 = T1 = sY(2)(0.09)(0.02) = 0.0036sy
200 mm
C2 = T2 = sY(0.01)(0.24) = 0.0024sy
20 mm
Mp = 0.0036sY(0.11) + 0.0024sY(0.01) = 0.00042sY Mp = 0.00042(250) A 106 B = 105 kN # m s¿ =
Mp c I
y 0.1 = ; 250 392
105(103)(0.1) =
26.8(10 - 6)
Mp
= 392 MPa
y = 0.0638 = 63.8 mm
sT = sB = 392 - 250 = 142 MPa
Ans.
451
20 mm
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6–175.
Determine the shape factor of the cross section. 3 in.
The moment of inertia of the cross-section about the neutral axis is I =
3 in.
1 1 (3)(93) + (6) (33) = 195.75 in4 12 12
3 in.
Here, smax = sY and c = 4.5 in. Then smax =
Mc ; I
sY =
MY(4.5) 195.75
3 in.
MY = 43.5 sY Referring to the stress block shown in Fig. a, T1 = C1 = 3(3)sY = 9 sY T2 = C2 = 1.5(9)sY = 13.5 sY Thus, MP = T1(6) + T2(1.5) = 9sY(6) + 13.5sY(1.5) = 74.25 sY k =
74.25 sY MP = = 1.71 MY 43.5 sY
Ans.
452
3 in.
3 in.
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*6–176. The beam is made of elastic-perfectly plastic material. Determine the maximum elastic moment and the plastic moment that can be applied to the cross section. Take sY = 36 ksi.
3 in. 3 in.
The moment of inertia of the cross-section about the neutral axis is I =
3 in.
1 1 (3)(93) + (6)(33) = 195.75 in4 12 12
Here, smax = sY = 36 ksi and c = 4.5 in. Then smax
Mc = ; I
3 in.
MY (4.5) 36 = 195.75 MY = 1566 kip # in = 130.5 kip # ft
Ans.
Referring to the stress block shown in Fig. a, T1 = C1 = 3(3)(36) = 324 kip T2 = C2 = 1.5(9)(36) = 486 kip Thus, MP = T1(6) + T2(1.5) = 324(6) + 486(1.5) = 2673 kip # in. = 222.75 kip # ft = 223 kip # ft
Ans.
453
3 in.
3 in.
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•6–177.
Determine the shape factor of the cross section for the tube.
The moment of inertia of the tube’s cross-section about the neutral axis is I =
5 in.
p 4 p A r - r4i B = A 64 - 54 B = 167.75 p in4 4 o 4
6 in.
Here, smax = sY and C = ro = 6 in, smax =
Mc ; I
sY =
MY (6) 167.75 p
MY = 87.83 sY The plastic Moment of the table’s cross-section can be determined by super posing the moment of the stress block of the solid circular cross-section with radius ro = 6 in and ri = 5 in. as shown in Figure a, Here, T1 = C1 =
1 p(62)sY = 18psY 2
T2 = C2 =
1 p(52)sY = 12.5p sY 2
Thus, MP = T1 b 2 c
4(6) 4(5) d r - T2 b 2 c dr 3p 3p
= (18psY)a
16 40 b - 12.5psY a b p 3p
= 121.33 sY k =
121.33 sY MP = = 1.38 MY 87.83 sY
Ans.
454
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6–178. The beam is made from elastic-perfectly plastic material. Determine the shape factor for the thick-walled tube. ro
Maximum Elastic Moment. The moment of inertia of the cross-section about the neutral axis is I =
p A r 4 - r4i B 4 o
With c = ro and smax = sY, smax =
Mc ; I
sY =
MY =
MY(ro) p A r 4 - ri 4 B 4 o p A r 4 - ri 4 B sY 4ro o
Plastic Moment. The plastic moment of the cross section can be determined by superimposing the moment of the stress block of the solid beam with radius r0 and ri as shown in Fig. a, Referring to the stress block shown in Fig. a, T1 = c1 =
p 2 r s 2 o Y
T2 = c2 =
p 2 r s 2 i Y
MP = T1 c2 a
4ro 4ri b d - T2 c2 a b d 3p 3p
=
8ro 8ri p 2 p r s a b - ri 2sY a b 2 o Y 3p 2 3p
=
4 A r 3 - ri 3 B sY 3 o
Shape Factor. 4 A r 3 - ri 3 B sY 16ro A ro 3 - ri 3 B MP 3 o k = = = p MY 3p A ro 4 - ri 4 B A ro 4 - ri 4 B sY 4ro
Ans.
455
ri
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6–179.
Determine the shape factor for the member.
Plastic analysis: T = C =
–h 2
h 1 bh (b)a bsY = s 2 2 4 Y –h 2
b h2 bh h MP = sY a b = s 4 3 12 Y Elastic analysis: I = 2c
1 h 3 b h3 (b)a b d = 12 2 48
b
sY A bh sYI 48 B b h2 = s = h c 24 Y 2 3
MY =
Shape factor: k =
Mp MY
=
bh2 12
sY
bh2 24
sY
= 2
Ans.
*6–180. The member is made from an elastic-plastic material. Determine the maximum elastic moment and the plastic moment that can be applied to the cross section. Take b = 4 in., h = 6 in., sY = 36 ksi.
–h 2
Elastic analysis: I = 2c
1 (4)(3)3 d = 18 in4 12
MY =
36(18) sYI = = 216 kip # in. = 18 kip # ft c 3
–h 2
Ans. b
Plastic analysis: T = C =
1 (4)(3)(36) = 216 kip 2
6 Mp = 2160 a b = 432 kip # in. = 36 kip # ft 3
Ans.
456
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•6–181.
The beam is made of a material that can be assumed perfectly plastic in tension and elastic perfectly plastic in compression. Determine the maximum bending moment M that can be supported by the beam so that the compressive material at the outer edge starts to yield.
h
sY
M
sdA = 0; LA
C - T = 0
sY
a
1 s (d)(a) - sY(h - d)a = 0 2 Y d =
M =
2 h 3
11 11a h2 2 1 sY a hb (a)a hb = sY 2 3 18 54
Ans.
6–182. The box beam is made from an elastic-plastic material for which sY = 25 ksi. Determine the intensity of the distributed load w0 that will cause the moment to be (a) the largest elastic moment and (b) the largest plastic moment.
w0
Elastic analysis: I =
9 ft
1 1 (8)(163) (6)(123) = 1866.67 in4 12 12
Mmax
sYI = ; c
9 ft
8 in.
25(1866.67) 27w0(12) = 8 Ans.
w0 = 18.0 kip>ft Plastic analysis:
16 in.
12 in.
6 in.
C1 = T1 = 25(8)(2) = 400 kip C2 = T2 = 25(6)(2) = 300 kip MP = 400(14) + 300(6) = 7400 kip # in. 27w0(12) = 7400 w0 = 22.8 kip>ft
Ans.
457
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6–183. The box beam is made from an elastic-plastic material for which sY = 36 ksi. Determine the magnitude of each concentrated force P that will cause the moment to be (a) the largest elastic moment and (b) the largest plastic moment.
P
From the moment diagram shown in Fig. a, Mmax = 6 P.
P
8 ft
6 ft
6 ft
The moment of inertia of the beam’s cross-section about the neutral axis is 6 in.
1 1 (6)(123) (5)(103) = 447.33 in4 I = 12 12
12 in.
10 in.
Here, smax = sY = 36 ksi and c = 6 in. smax =
Mc ; I
36 =
5 in.
MY (6) 447.33
MY = 2684 kip # in = 223.67 kip # ft It is required that Mmax = MY 6P = 223.67 P = 37.28 kip = 37.3 kip
Ans.
Referring to the stress block shown in Fig. b, T1 = C1 = 6(1)(36) = 216 kip T2 = C2 = 5(1)(36) = 180 kip Thus, MP = T1(11) + T2(5) = 216(11) + 180(5) = 3276 kip # in = 273 kip # ft It is required that Mmax = MP 6P = 273 P = 45.5 kip
Ans.
458
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*6–184. The beam is made of a polyester that has the stress–strain curve shown. If the curve can be represented by the equation s = [20 tan-1115P2] ksi, where tan-1115P2 is in radians, determine the magnitude of the force P that can be applied to the beam without causing the maximum strain in its fibers at the critical section to exceed Pmax = 0.003 in.>in.
P 2 in. 4 in.
8 ft s(ksi)
8 ft s 20 tan1(15 P)
P(in./in.)
Maximum Internal Moment: The maximum internal moment M = 4.00P occurs at the mid span as shown on FBD. Stress–Strain Relationship: Using the stress–strain relationship. the bending stress can be expressed in terms of y using e = 0.0015y. s = 20 tan - 1 (15e) = 20 tan - 1 [15(0.0015y)] = 20 tan - 1 (0.0225y) When emax = 0.003 in.>in., y = 2 in. and smax = 0.8994 ksi Resultant Internal Moment: The resultant internal moment M can be evaluated from the integal
M = 2
LA
ysdA.
ysdA
2in
= 2
LA
L0
y C 20 tan
-1
(0.0225y) D (2dy)
2in
= 80
L0
= 80 B
y tan - 1 (0.0225y) dy
1 + (0.0225)2y2 2(0.0225)2
tan - 1 (0.0225y) -
2in. y R2 2(0.0225) 0
= 4.798 kip # in Equating M = 4.00P(12) = 4.798 P = 0.100 kip = 100 lb
Ans.
459
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•6–185.
The plexiglass bar has a stress–strain curve that can be approximated by the straight-line segments shown. Determine the largest moment M that can be applied to the bar before it fails.
s (MPa)
20 mm M 20 mm
failure
60 40
tension
0.06 0.04
P (mm/mm) 0.02
compression 80 100
Ultimate Moment: LA
s dA = 0;
C - T2 - T1 = 0
1 1 d 1 d sc (0.02 - d)(0.02) d - 40 A 106 B c a b(0.02) d - (60 + 40) A 106 B c(0.02) d = 0 2 2 2 2 2 s - 50s d - 3500(106)d = 0 Assume.s = 74.833 MPa; d = 0.010334 m From the strain diagram, 0.04 e = 0.02 - 0.010334 0.010334
e = 0.037417 mm>mm
From the stress–strain diagram, 80 s = 0.037417 0.04
s = 74.833 MPa (OK! Close to assumed value)
Therefore, 1 C = 74.833 A 106 B c (0.02 - 0.010334)(0.02) d = 7233.59 N 2 T1 =
1 0.010334 (60 + 40) A 106 B c(0.02)a b d = 5166.85 N 2 2
1 0.010334 b d = 2066.74 N T2 = 40 A 106 B c (0.02)a 2 2
y1 =
2 (0.02 - 0.010334) = 0.0064442 m 3
y2 =
2 0.010334 a b = 0.0034445 m 3 2
y3 =
0.010334 1 2(40) + 60 0.010334 + c1 - a bda b = 0.0079225m 2 3 40 + 60 2
M = 7233.59(0.0064442) + 2066.74(0.0034445) + 5166.85(0.0079255) = 94.7 N # m
Ans.
460
0.04
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6–186. The stress–strain diagram for a titanium alloy can be approximated by the two straight lines. If a strut made of this material is subjected to bending, determine the moment resisted by the strut if the maximum stress reaches a value of (a) sA and (b) sB.
3 in. M
2 in. s (ksi) B
sB 180 sA 140
A
0.01
a) Maximum Elastic Moment : Since the stress is linearly related to strain up to point A, the flexure formula can be applied. sA =
Mc I M =
=
sA I c 1 140 C 12 (2)(33) D
1.5
= 420 kip # in = 35.0 kip # ft b)
Ans.
The Ultimate Moment : C1 = T1 =
1 (140 + 180)(1.125)(2) = 360 kip 2
C2 = T2 =
1 (140)(0.375)(2) = 52.5 kip 2
M = 360(1.921875) + 52.5(0.5) = 718.125 kip # in = 59.8 kip # ft
Ans.
Note: The centroid of a trapezodial area was used in calculation of moment.
461
0.04
P (in./in.)
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6–187. A beam is made from polypropylene plastic and has a stress–strain diagram that can be approximated by the curve shown. If the beam is subjected to a maximum tensile and compressive strain of P = 0.02 mm>mm, determine the maximum moment M.
M
s (Pa)
s 10(106)P1/ 4
emax = 0.02 smax = 10 A 106 B (0.02)1>4 = 3.761 MPa
M
100 mm
30 mm P (mm/mm)
e 0.02 = y 0.05 e = 0.4 y s = 10 A 106 B (0.4)1>4y1>4 y(7.9527) A 106 B y1>4(0.03)dy
0.05
M =
y s dA = 2
LA
M = 0.47716 A 106 B
L0
4 y5>4dy = 0.47716 A 106 B a b(0.05)9>4 5
0.05
L0
M = 251 N # m
Ans.
*6–188. The beam has a rectangular cross section and is made of an elastic-plastic material having a stress–strain diagram as shown. Determine the magnitude of the moment M that must be applied to the beam in order to create a maximum strain in its outer fibers of P max = 0.008.
400 mm M
200 mm
s(MPa)
200
0.004
C1 = T1 = 200 A 106 B (0.1)(0.2) = 4000 kN C2 = T2 =
1 (200) A 106 B (0.1)(0.2) = 2000 kN 2
M = 4000(0.3) + 2000(0.1333) = 1467 kN # m = 1.47 MN # m
Ans.
462
P (mm/mm)
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s(ksi) 90 80
•6–189.
The bar is made of an aluminum alloy having a stress–strain diagram that can be approximated by the straight line segments shown. Assuming that this diagram is the same for both tension and compression, determine the moment the bar will support if the maximum strain at the top and bottom fibers of the beam is P max = 0.03. 90 - 80 s - 80 = ; 0.03 - 0.025 0.05 - 0.025
60
4 in. M
s = 82 ksi
C1 = T1 =
1 (0.3333)(80 + 82)(3) = 81 kip 2
C2 = T2 =
1 (1.2666)(60 + 80)(3) = 266 kip 2
C3 = T3 =
1 (0.4)(60)(3) = 36 kip 2
0.006
0.025
0.05
P (in./ in.)
3 in.
M = 81(3.6680) + 266(2.1270) + 36(0.5333) = 882.09 kip # in. = 73.5 kip # ft
Ans.
Note: The centroid of a trapezodial area was used in calculation of moment areas.
6–190. The beam is made from three boards nailed together as shown. If the moment acting on the cross section is M = 650 N # m, determine the resultant force the bending stress produces on the top board. 15 mm
Section Properties: y =
0.0075(0.29)(0.015) + 2[0.0775(0.125)(0.02)] 0.29(0.015) + 2(0.125)(0.02)
M 650 Nm 20 mm
125 mm
= 0.044933 m INA
20 mm
1 = (0.29) A 0.0153 B + 0.29(0.015) (0.044933 - 0.0075)2 12 +
1 (0.04) A 0.1253 B + 0.04(0.125)(0.0775 - 0.044933)2 12
= 17.99037 A 10 - 6 B m4 Bending Stress: Applying the flexure formula s =
sB =
sA =
650(0.044933 - 0.015) 17.99037(10 - 6) 650(0.044933) 17.99037(10 - 6)
My I
= 1.0815 MPa
= 1.6234 MPa
Resultant Force: FR =
1 (1.0815 + 1.6234) A 106 B (0.015)(0.29) 2
= 5883 N = 5.88 kN
Ans.
463
250 mm
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6–191. The beam is made from three boards nailed together as shown. Determine the maximum tensile and compressive stresses in the beam. 15 mm M 650 Nm 20 mm
125 mm 20 mm
Section Properties: y =
0.0075(0.29)(0.015) + 2[0.0775(0.125)(0.02)] 0.29(0.015) + 2(0.125)(0.02)
= 0.044933 m INA =
1 (0.29) A 0.0153 B + 0.29(0.015)(0.044933 - 0.0075)2 12 +
1 (0.04) A 0.1253 B + 0.04(0.125)(0.0775 - 0.044933)2 12
= 17.99037 A 10 - 6 B m4 Maximum Bending Stress: Applying the flexure formula s =
(smax)t =
(smax)c =
650(0.14 - 0.044933) 17.99037(10 - 6) 650(0.044933) 17.99037(10 - 6)
My I Ans.
= 3.43 MPa (T)
= 1.62 MPa (C)
Ans.
464
250 mm
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*6–192. Determine the bending stress distribution in the beam at section a–a. Sketch the distribution in three dimensions acting over the cross section.
80 N
80 N
a
a 300 mm
400 mm
a + ©M = 0;
300 mm
400 mm
80 N
M - 80(0.4) = 0
80 N 15 mm
M = 32 N # m
100 mm
1 1 Iz = (0.075)(0.0153) + 2 a b (0.015)(0.13) = 2.52109(10 - 6)m4 12 12 smax =
32(0.05) Mc = 635 kPa = I 2.52109(10 - 6)
15 mm
•6–193. The composite beam consists of a wood core and two plates of steel. If the allowable bending stress for the wood is (sallow)w = 20 MPa, and for the steel (sallow)st = 130 MPa, determine the maximum moment that can be applied to the beam. Ew = 11 GPa, Est = 200 GPa.
n =
75 mm
Ans.
y
z 125 mm
200(109) Est = 18.182 = Ew 11(109) M
1 (0.80227)(0.1253) = 0.130578(10 - 3)m4 I = 12
x
75 mm
Failure of wood : (sw)max
20 mm
Mc = I
20(106) =
M(0.0625) 0.130578(10 - 3)
;
M = 41.8 kN # m
Failure of steel : (sst)max =
20 mm
nMc I 130(106) =
18.182(M)(0.0625) 0.130578(10 - 3)
M = 14.9 kN # m (controls)
Ans.
465
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6–194. Solve Prob. 6–193 if the moment is applied about the y axis instead of the z axis as shown.
y
z 125 mm
M
x
20 mm 75 mm 20 mm
n =
I =
11(109) 200(104)
= 0.055
1 1 (0.125)(0.1153) (0.118125)(0.0753) = 11.689616(10 - 6) 12 12
Failure of wood : (sw)max =
nMc2 I
20(106) =
0.055(M)(0.0375) 11.689616(10 - 6)
;
M = 113 kN # m
Failure of steel : (sst)max =
Mc1 I 130(106) =
M(0.0575) 11.689616(10 - 6)
M = 26.4 kN # m (controls)
Ans.
466
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6–195. A shaft is made of a polymer having a parabolic cross section. If it resists an internal moment of M = 125 N # m, determine the maximum bending stress developed in the material (a) using the flexure formula and (b) using integration. Sketch a three-dimensional view of the stress distribution acting over the cross-sectional area. Hint: The moment of inertia is determined using Eq. A–3 of Appendix A.
y
100 mm y 100 – z 2/ 25 M 125 N· m z
Maximum Bending Stress: The moment of inertia about y axis must be determined first in order to use Flexure Formula I =
LA
50 mm 50 mm
y2 dA 100mm
= 2
L0
y2 (2z) dy 100mm
= 20
L0
y2 2100 - y dy
100 mm 3 5 7 3 8 16 y (100 - y)2 (100 - y)2 R 2 = 20 B - y2 (100 - y)2 2 15 105 0
= 30.4762 A 10 - 6 B mm4 = 30.4762 A 10 - 6 B m4 Thus, smax =
125(0.1) Mc = 0.410 MPa = I 30.4762(10 - 6)
Ans.
Maximum Bending Stress: Using integration dM = 2[y(s dA)] = 2 b yc a
M =
smax by d(2z dy) r 100
smax 100mm 2 y 2100 - y dy 5 L0
125 A 103 B =
100 mm smax 3 5 7 3 8 16 y(100 - y)2 (100 - y)2 R 2 B - y2(100 - y)2 5 2 15 105 0
125 A 103 B =
smax (1.5238) A 106 B 5
smax = 0.410 N>mm2 = 0.410 MPa
Ans.
467
x
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*6–196. Determine the maximum bending stress in the handle of the cable cutter at section a–a. A force of 45 lb is applied to the handles. The cross-sectional area is shown in the figure.
20
45 lb
a
5 in. 4 in.
3 in. 0.75 in.
A a
0.50 in.
45 lb
a + ©M = 0;
M - 45(5 + 4 cos 20°) = 0 M = 394.14 lb # in.
394.14(0.375) Mc = 8.41 ksi = 1 3 I 12 (0.5)(0.75 )
smax =
Ans.
M 85 Nm
•6–197.
The curved beam is subjected to a bending moment of M = 85 N # m as shown. Determine the stress at points A and B and show the stress on a volume element located at these points.
100 mm
A
r2 0.57 0.59 dA 0.42 + 0.015 ln + 0.1 ln = b ln = 0.1 ln r1 0.40 0.42 0.57 LA r
400 mm
= 0.012908358 m
= LA
dA r
6.25(10 - 3) = 0.484182418 m 0.012908358
r - R = 0.495 - 0.484182418 = 0.010817581 m sA =
M(R - rA)
85(0.484182418 - 0.59) =
ArA(r - R)
6.25(10 - 3)(0.59)(0.010817581)
= -225.48 kPa
sA = 225 kPa (C) sB =
Ans.
M(R - rB)
85(0.484182418 - 0.40) =
ArB(r - R)
150 mm
6.25(10 - 3)(0.40)(0.010817581)
20 mm B
2
A = 2(0.1)(0.02) + (0.15)(0.015) = 6.25(10 ) m A
20 mm
30 -3
R =
A 15 mm
B
= 265 kPa (T)
468
Ans.
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6–198. Draw the shear and moment diagrams for the beam and determine the shear and moment in the beam as functions of x, where 0 … x 6 6 ft.
8 kip
2 kip/ ft 50 kipft
x 6 ft
+ c ©Fy = 0;
20 - 2x - V = 0 V = 20 - 2x
c + ©MNA = 0;
4 ft
Ans.
x 20x - 166 - 2xa b - M = 0 2 M = -x2 + 20x - 166
Ans.
6–199. Draw the shear and moment diagrams for the shaft if it is subjected to the vertical loadings of the belt, gear, and flywheel. The bearings at A and B exert only vertical reactions on the shaft.
300 N 450 N
A
B
200 mm
400 mm
300 mm
200 mm 150 N
469
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*6–200. A member has the triangular cross section shown. Determine the largest internal moment M that can be applied to the cross section without exceeding allowable tensile and compressive stresses of (sallow)t = 22 ksi and (sallow)c = 15 ksi, respectively.
4 in. 4 in.
M 2 in.
2 in.
y (From base) = I =
1 242 - 22 = 1.1547 in. 3
1 (4)(242 - 22)3 = 4.6188 in4 36
Assume failure due to tensile stress : smax =
My ; I
22 =
M(1.1547) 4.6188
M = 88.0 kip # in. = 7.33 kip # ft Assume failure due to compressive stress: smax =
Mc ; I
15 =
M(3.4641 - 1.1547) 4.6188
M = 30.0 kip # in. = 2.50 kip # ft
(controls)
Ans.
470
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•6–201.
The strut has a square cross section a by a and is subjected to the bending moment M applied at an angle u as shown. Determine the maximum bending stress in terms of a, M, and u. What angle u will give the largest bending stress in the strut? Specify the orientation of the neutral axis for this case.
y
a
z
x
a M
Internal Moment Components: Mz = -M cos u
My = -M sin u
Section Property: Iy = Iz =
1 4 a 12
Maximum Bending Stress: By Inspection, Maximum bending stress occurs at A and B. Applying the flexure formula for biaxial bending at point A s = -
My z
Mzy +
Iz
Iy
-M cos u (a2) = -
=
1 12
a4
-Msin u ( - a2) +
1 12
a4
6M (cos u + sin u) a3
Ans.
6M ds = 3 (-sin u + cos u) = 0 du a cos u - sin u = 0 u = 45°
Ans.
Orientation of Neutral Axis: tan a =
Iz Iy
tan u
tan a = (1) tan(45°) a = 45°
Ans.
471
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•7–1.
If the wide-flange beam is subjected to a shear of V = 20 kN, determine the shear stress on the web at A. Indicate the shear-stress components on a volume element located at this point.
200 mm
A
20 mm
20 mm B
V 300 mm 200 mm
The moment of inertia of the cross-section about the neutral axis is I =
1 1 (0.2)(0.343) (0.18)(0.33) = 0.2501(10 - 3) m4 12 12
From Fig. a, QA = y¿A¿ = 0.16 (0.02)(0.2) = 0.64(10 - 3) m3 Applying the shear formula, VQA 20(103)[0.64(10 - 3)] = tA = It 0.2501(10 - 3)(0.02) = 2.559(106) Pa = 2.56 MPa
Ans.
The shear stress component at A is represented by the volume element shown in Fig. b.
472
20 mm
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7–2. If the wide-flange beam is subjected to a shear of V = 20 kN, determine the maximum shear stress in the beam.
200 mm
A
20 mm
20 mm B
V 300 mm 200 mm
The moment of inertia of the cross-section about the neutral axis is I =
1 1 (0.2)(0.343) (0.18)(0.33) = 0.2501(10 - 3) m4 12 12
From Fig. a. Qmax = ©y¿A¿ = 0.16 (0.02)(0.2) + 0.075 (0.15)(0.02) = 0.865(10 - 3) m3 The maximum shear stress occurs at the points along neutral axis since Q is maximum and thicknest t is the smallest. tmax =
VQmax 20(103) [0.865(10 - 3)] = It 0.2501(10 - 3) (0.02) = 3.459(106) Pa = 3.46 MPa
Ans.
473
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7–3. If the wide-flange beam is subjected to a shear of V = 20 kN, determine the shear force resisted by the web of the beam.
200 mm
A
20 mm
20 mm B
V 300 mm 200 mm
The moment of inertia of the cross-section about the neutral axis is I =
1 1 (0.2)(0.343) (0.18)(0.33) = 0.2501(10 - 3) m4 12 12
For 0 … y 6 0.15 m, Fig. a, Q as a function of y is Q = ©y¿A¿ = 0.16 (0.02)(0.2) +
1 (y + 0.15)(0.15 - y)(0.02) 2
= 0.865(10 - 3) - 0.01y2 For 0 … y 6 0.15 m, t = 0.02 m. Thus. t =
20(103) C 0.865(10 - 3) - 0.01y2 D VQ = It 0.2501(10 - 3) (0.02) =
E 3.459(106) - 39.99(106) y2 F Pa.
The sheer force resisted by the web is, 0.15 m
Vw = 2
L0
0.15 m
tdA = 2
L0
C 3.459(106) - 39.99(106) y2 D (0.02 dy)
= 18.95 (103) N = 19.0 kN
Ans.
474
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*7–4. If the T-beam is subjected to a vertical shear of V = 12 kip, determine the maximum shear stress in the beam. Also, compute the shear-stress jump at the flangeweb junction AB. Sketch the variation of the shear-stress intensity over the entire cross section.
4 in. 4 in.
3 in.
4 in. B
6 in.
A V ⫽ 12 kip
Section Properties: y =
INA =
1.5(12)(3) + 6(4)(6) ©yA = = 3.30 in. ©A 12(3) + 4(6)
1 1 (12) A 33 B + 12(3)(3.30 - 1.5)2 + (4) A 63 B + 4(6)(6 - 3.30)2 12 12
= 390.60 in4 Qmax = y1œ A¿ = 2.85(5.7)(4) = 64.98 in3 QAB = y2œ A¿ = 1.8(3)(12) = 64.8 in3 Shear Stress: Applying the shear formula t =
tmax =
VQ It
VQmax 12(64.98) = = 0.499 ksi It 390.60(4)
Ans.
(tAB)f =
VQAB 12(64.8) = = 0.166 ksi Itf 390.60(12)
Ans.
(tAB)W =
VQAB 12(64.8) = = 0.498 ksi I tW 390.60(4)
Ans.
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•7–5.
If the T-beam is subjected to a vertical shear of V = 12 kip, determine the vertical shear force resisted by the flange.
4 in. 4 in.
3 in.
4 in. B
6 in.
A V ⫽ 12 kip
Section Properties: y =
©yA 1.5(12)(3) + 6(4)(6) = = 3.30 in. ©A 12(3) + 4(6)
INA =
1 1 (12) A 33 B + 12(3)(3.30 - 1.5)2 + (4) A 63 B + 6(4)(6 - 3.30)2 12 12
= 390.60 in4 Q = y¿A¿ = (1.65 + 0.5y)(3.3 - y)(12) = 65.34 - 6y2 Shear Stress: Applying the shear formula t =
VQ 12(65.34 - 6y2) = It 390.60(12) = 0.16728 - 0.01536y2
Resultant Shear Force: For the flange Vf =
tdA LA 3.3 in
=
L0.3 in
A 0.16728 - 0.01536y2 B (12dy)
= 3.82 kip
Ans.
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7–6. If the beam is subjected to a shear of V = 15 kN, determine the web’s shear stress at A and B. Indicate the shear-stress components on a volume element located at these points. Show that the neutral axis is located at y = 0.1747 m from the bottom and INA = 0.2182110-32 m4.
200 mm
A
30 mm
25 mm V
(0.015)(0.125)(0.03) + (0.155)(0.025)(0.25) + (0.295)(0.2)(0.03) y = = 0.1747 m 0.125(0.03) + (0.025)(0.25) + (0.2)(0.03) I =
1 (0.125)(0.033) + 0.125(0.03)(0.1747 - 0.015)2 12
+
1 (0.025)(0.253) + 0.25(0.025)(0.1747 - 0.155)2 12
+
1 (0.2)(0.033) + 0.2(0.03)(0.295 - 0.1747)2 = 0.218182 (10 - 3) m4 12
B
250 mm
30 mm
125 mm
œ QA = yAA = (0.310 - 0.015 - 0.1747)(0.2)(0.03) = 0.7219 (10 - 3) m3
QB = yABœ = (0.1747 - 0.015)(0.125)(0.03) = 0.59883 (10 - 3) m3 tA =
15(103)(0.7219)(10 - 3) VQA = 1.99 MPa = It 0.218182(10 - 3)(0.025)
Ans.
tB =
VQB 15(103)(0.59883)(10 - 3) = 1.65 MPa = It 0.218182(10 - 3)0.025)
Ans.
7–7. If the wide-flange beam is subjected to a shear of V = 30 kN, determine the maximum shear stress in the beam.
200 mm
A
30 mm
25 mm V B 250 mm 30 mm
Section Properties: I =
1 1 (0.2)(0.310)3 (0.175)(0.250)3 = 268.652(10) - 6 m4 12 12
Qmax = © y¿A = 0.0625(0.125)(0.025) + 0.140(0.2)(0.030) = 1.0353(10) - 3 m3 tmax =
VQ 30(10)3(1.0353)(10) - 3 = 4.62 MPa = It 268.652(10) - 6 (0.025)
Ans.
477
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*7–8. If the wide-flange beam is subjected to a shear of V = 30 kN, determine the shear force resisted by the web of the beam.
200 mm
A
30 mm
1 1 (0.2)(0.310)3 (0.175)(0.250)3 = 268.652(10) - 6 m4 12 12
I =
Q = a
25 mm V B
0.155 + y b (0.155 - y)(0.2) = 0.1(0.024025 - y2) 2
250 mm
30(10)3(0.1)(0.024025 - y2)
tf =
268.652(10)
-6
30 mm
200 mm
(0.2) 0.155
Vf =
L
tf dA = 55.8343(10)6
L0.125
= 11.1669(10)6[ 0.024025y -
(0.024025 - y2)(0.2 dy)
1 3 0.155 y ] 2 0.125
Vf = 1.457 kN Vw = 30 - 2(1.457) = 27.1 kN
Ans.
•7–9. Determine the largest shear force V that the member can sustain if the allowable shear stress is tallow = 8 ksi.
3 in. 1 in. V 3 in. 1 in.
1 in.
y =
(0.5)(1)(5) + 2 [(2)(1)(2)] = 1.1667 in. 1 (5) + 2 (1)(2)
I =
1 (5)(13) + 5 (1)(1.1667 - 0.5)2 12
+ 2a
1 b (1)(23) + 2 (1)(2)(2 - 1.1667)2 = 6.75 in4 12
Qmax = ©y¿A¿ = 2 (0.91665)(1.8333)(1) = 3.3611 in3 tmax = tallow = 8 (103) = -
VQmax It
V (3.3611) 6.75 (2)(1)
V = 32132 lb = 32.1 kip
Ans.
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7–10. If the applied shear force V = 18 kip, determine the maximum shear stress in the member. 3 in. 1 in. V 3 in. 1 in.
1 in.
y =
(0.5)(1)(5) + 2 [(2)(1)(2)] = 1.1667 in. 1 (5) + 2 (1)(2)
I =
1 (5)(13) + 5 (1)(1.1667 - 0.5)2 12
+ 2a
1 b (1)(23) + 2 (1)(2)(2 - 1.1667) = 6.75 in4 12
Qmax = ©y¿A¿ = 2 (0.91665)(1.8333)(1) = 3.3611 in3 tmax =
18(3.3611) VQmax = = 4.48 ksi It 6.75 (2)(1)
Ans.
7–11. The wood beam has an allowable shear stress of tallow = 7 MPa. Determine the maximum shear force V that can be applied to the cross section.
50 mm
50 mm 100 mm
50 mm
200 mm V 50 mm
I =
1 1 (0.2)(0.2)3 (0.1)(0.1)3 = 125(10 - 6) m4 12 12
tallow = 7(106) =
VQmax It V[(0.075)(0.1)(0.05) + 2(0.05)(0.1)(0.05)] 125(10 - 6)(0.1)
V = 100 kN
Ans.
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*7–12. The beam has a rectangular cross section and is made of wood having an allowable shear stress of tallow = 200 psi. Determine the maximum shear force V that can be developed in the cross section of the beam. Also, plot the shear-stress variation over the cross section.
V 12 in.
8 in.
Section Properties The moment of inertia of the cross-section about the neutral axis is I =
1 (8) (123) = 1152 in4 12
Q as the function of y, Fig. a, Q =
1 (y + 6)(6 - y)(8) = 4 (36 - y2) 2
Qmax occurs when y = 0. Thus, Qmax = 4(36 - 02) = 144 in3 The maximum shear stress occurs of points along the neutral axis since Q is maximum and the thickness t = 8 in. is constant. tallow =
VQmax ; It
200 =
V(144) 1152(8)
V = 12800 16 = 12.8 kip
Ans.
Thus, the shear stress distribution as a function of y is t =
12.8(103) C 4(36 - y2) D VQ = It 1152 (8) =
E 5.56 (36 - y2) F psi
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7–13. Determine the maximum shear stress in the strut if it is subjected to a shear force of V = 20 kN.
12 mm
Section Properties: INA
60 mm
1 1 = (0.12) A 0.0843 B (0.04) A 0.063 B 12 12
V
= 5.20704 A 10 - 6 B m4
12 mm 80 mm
Qmax = ©y¿A¿
20 mm
20 mm
= 0.015(0.08)(0.03) + 0.036(0.012)(0.12) = 87.84 A 10 - 6 B m3 Maximum Shear Stress: Maximum shear stress occurs at the point where the neutral axis passes through the section. Applying the shear formula tmax =
VQmax It 20(103)(87.84)(10 - 6)
=
5.20704(10 - 6)(0.08)
= 4 22 MPa
Ans.
7–14. Determine the maximum shear force V that the strut can support if the allowable shear stress for the material is tallow = 40 MPa.
12 mm
60 mm
Section Properties: INA =
V
1 1 (0.12) A 0.0843 B (0.04) A 0.063 B 12 12
12 mm
= 5.20704 A 10 - 6 B m4
80 mm
Qmax = ©y¿A¿
20 mm
= 0.015(0.08)(0.03) + 0.036(0.012)(0.12) = 87.84 A 10 - 6 B m3 Allowable shear stress: Maximum shear stress occurs at the point where the neutral axis passes through the section. Applying the shear formula tmax = tallow = 40 A 106 B =
VQmax It V(87.84)(10 - 6) 5.20704(10 - 6)(0.08)
V = 189 692 N = 190 kN
Ans.
481
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7–15. Plot the shear-stress distribution over the cross section of a rod that has a radius c. By what factor is the maximum shear stress greater than the average shear stress acting over the cross section? c y V
x = 2c2 - y2 ;
p 4 c 4
I =
t = 2 x = 2 2c2 - y2 dA = 2 x dy = 22c2 - y2 dy dQ = ydA = 2y 2c2 - y2 dy x
Q =
Ly
2y2c2 - y2 dy = -
3 x 2 2 2 2 (c - y2)2 | y = (c2 - y2)3 3 3
3
V[23 (c2 - y2)2] VQ 4V 2 t = = = [c - y2) p 4 2 2 It 3pc4 ( 4 c )(2 2c - y ) The maximum shear stress occur when y = 0 tmax =
4V 3 p c2
tavg =
V V = A p c2
The faector =
tmax = tavg
4V 3 pc2 V pc2
=
4 3
Ans.
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*7–16. A member has a cross section in the form of an equilateral triangle. If it is subjected to a shear force V, determine the maximum average shear stress in the member using the shear formula. Should the shear formula actually be used to predict this value? Explain.
I =
V
1 (a)(h)3 36
y h ; = x a>2 Q =
a
LA¿
Q = a
y =
y dA = 2c a
2h x a
1 2 2 b (x)(y) a h - yb d 2 3 3
4h2 2x b (x2)a 1 b a 3a
t = 2x t =
t =
V(4h2>3a)(x2)(1 - 2x VQ a) = It ((1>36)(a)(h3))(2x) 24V(x - a2 x2) a2h
24V 4 dt = 2 2 a 1 - xb = 0 a dx ah At x =
y =
a 4 h 2h a a b = a 4 2
tmax =
24V a 2 a a b a1 - a b b a 4 a2h 4
tmax =
3V ah
Ans.
No, because the shear stress is not perpendicular to the boundary. See Sec. 7-3.
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•7–17.
Determine the maximum shear stress in the strut if it is subjected to a shear force of V = 600 kN.
30 mm
150 mm
V 100 mm 100 mm 100 mm
The moment of inertia of the cross-section about the neutral axis is I =
1 1 (0.3)(0.213) (0.2)(0.153) = 0.175275(10 - 3) m4 12 12
From Fig. a, Qmax = ©y¿A¿ = 0.09(0.03)(0.3) + 0.0375(0.075)(0.1) = 1.09125(10 - 3) m3 The maximum shear stress occurs at the points along the neutral axis since Q is maximum and thickness t = 0.1 m is the smallest. tmax =
VQmax 600(103)[1.09125(10 - 3)] = It 0.175275(10 - 3) (0.1) = 37.36(106) Pa = 37.4 MPa
Ans.
484
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7–18. Determine the maximum shear force V that the strut can support if the allowable shear stress for the material is tallow = 45 MPa.
30 mm
150 mm
V 100 mm 100 mm 100 mm
The moment of inertia of the cross-section about the neutral axis is I =
1 1 (0.3)(0.213) (0.2)(0.153) = 0.175275 (10 - 3) m4 12 12
From Fig. a Qmax = ©y¿A¿ = 0.09(0.03)(0.3) + 0.0375 (0.075)(0.1) = 1.09125 (10 - 3) m3 The maximum shear stress occeurs at the points along the neutral axis since Q is maximum and thickness t = 0.1 m is the smallest. tallow =
VQmax ; It
45(106) =
V C 1.09125(10 - 3) D
0.175275(10 - 3)(0.1)
V = 722.78(103) N = 723 kN
Ans.
485
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7–19. Plot the intensity of the shear stress distributed over the cross section of the strut if it is subjected to a shear force of V = 600 kN.
30 mm
The moment of inertia of the cross-section about the neutral axis is I =
1 1 (0.3)(0.213) (0.2)(0.153) = 0.175275 (10 - 3) m4 12 12
For 0.075 m 6 y … 0.105 m, Fig. a, Q as a function of y is Q = y¿A¿ =
1 (0.105 + y) (0.105 - y)(0.3) = 1.65375(10 - 3) - 0.15y2 2
For 0 … y 6 0.075 m, Fig. b, Q as a function of y is Q = ©y¿A¿ = 0.09 (0.03)(0.3) +
1 (0.075 + y)(0.075 - y)(0.1) = 1.09125(10 - 3) - 0.05 y2 2
For 0.075 m 6 y … 0.105 m, t = 0.3 m. Thus, t =
600 (103) C 1.65375(10 - 3) - 0.15y2 D VQ = (18.8703 - 1711.60y2) MPa = It 0.175275(10 - 3) (0.3)
At y = 0.075 m and y = 0.105 m, t|y = 0.015 m = 9.24 MPa
ty = 0.105 m = 0
For 0 … y 6 0.075 m, t = 0.1 m. Thus, t =
VQ 600 (103) [1.09125(10 - 3) - 0.05 y2] = (37.3556 - 1711.60 y2) MPa = It 0.175275(10 - 3) (0.1)
At y = 0 and y = 0.075 m, t|y = 0 = 37.4 MPa
ty = 0.075 m = 27.7 MPa
The plot shear stress distribution over the cross-section is shown in Fig. c.
486
150 mm
V 100 mm 100 mm 100 mm
30 mm
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*7–20. The steel rod is subjected to a shear of 30 kip. Determine the maximum shear stress in the rod. The moment of inertia of the ciralor cross-section about the neutral axis (x axis) is p p I = r4 = (24) = 4 p in4 4 4
30 kip
dQ = ydA = y (2xdy) = 2xy dy 1
However, from the equation of the circle, x = (4 - y2)2 , Then 1
dQ = 2y(4 - y2)2 dy Thus, Q for the area above y is 2 in 1
2y (4 - y2)2 dy
Ly 3 2 in 2 = - (4 - y2)2 y 3 =
3 2 (4 - y2)2 3
1
Here, t = 2x = 2 (4 - y2)2 . Thus
30 C 23 (4 - y2)2 D VQ = t = 1 It 4p C 2(4 - y2)2 D 3
t =
5 (4 - y2) ksi 2p
By inspecting this equation, t = tmax at y = 0. Thus ¿= tmax
A 2 in.
Q for the differential area shown shaded in Fig. a is
Q =
1 in.
20 10 = 3.18 ksi = p 2p
Ans.
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•7–21.
The steel rod is subjected to a shear of 30 kip. Determine the shear stress at point A. Show the result on a volume element at this point. 1 in. A
The moment of inertia of the circular cross-section about the neutral axis (x axis) is I =
2 in.
p 4 p r = (24) = 4p in4 4 4
30 kip
Q for the differential area shown in Fig. a is dQ = ydA = y (2xdy) = 2xy dy 1
However, from the equation of the circle, x = (4 - y2)2 , Then 1
dQ = 2y (4 - y2)2 dy Thus, Q for the area above y is 2 in. 1
Q =
Ly
= -
2y (4 - y2)2 dy
2 in. 3 3 2 2 (4 - y2)2 ` = (4 - y2)2 3 3 y
1
Here t = 2x = 2 (4 - y2)2 . Thus,
30 C 23 (4 - y2)2 D VQ = t = 1 It 4p C 2(4 - y2)2 D 3
t =
5 (4 - y2) ksi 2p
For point A, y = 1 in. Thus tA =
5 (4 - 12) = 2.39 ksi 2p
Ans.
The state of shear stress at point A can be represented by the volume element shown in Fig. b.
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7–22. Determine the shear stress at point B on the web of the cantilevered strut at section a–a.
2 kN 250 mm
a
250 mm
4 kN 300 mm
a
20 mm 70 mm
(0.01)(0.05)(0.02) + (0.055)(0.07)(0.02) y = = 0.03625 m (0.05)(0.02) + (0.07)(0.02) I =
+
B
20 mm 50 mm
1 (0.05)(0.023) + (0.05)(0.02)(0.03625 - 0.01)2 12
1 (0.02)(0.073) + (0.02)(0.07)(0.055 - 0.03625)2 = 1.78625(10 - 6) m4 12
yBœ = 0.03625 - 0.01 = 0.02625 m QB = (0.02)(0.05)(0.02625) = 26.25(10 - 6) m3 tB =
6(103)(26.25)(10 - 6) VQB = It 1.78622(10 - 6)(0.02) = 4.41 MPa
Ans.
7–23. Determine the maximum shear stress acting at section a–a of the cantilevered strut.
2 kN 250 mm
a
250 mm
4 kN 300 mm
a
20 mm 70 mm
y =
(0.01)(0.05)(0.02) + (0.055)(0.07)(0.02) = 0.03625 m (0.05)(0.02) + (0.07)(0.02)
I =
1 (0.05)(0.023) + (0.05)(0.02)(0.03625 - 0.01)2 12
+
20 mm 50 mm
1 (0.02)(0.073) + (0.02)(0.07)(0.055 - 0.03625)2 = 1.78625(10 - 6) m4 12
Qmax = y¿A¿ = (0.026875)(0.05375)(0.02) = 28.8906(10 - 6) m3 tmax =
B
VQmax 6(103)(28.8906)(10 - 6) = It 1.78625(10 - 6)(0.02) = 4.85 MPa
Ans.
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*7–24. Determine the maximum shear stress in the T-beam at the critical section where the internal shear force is maximum.
10 kN/m
A 1.5 m
3m
The shear diagram is shown in Fig. b. As indicated, Vmax = 27.5 kN
150 mm
The neutral axis passes through centroid c of the cross-section, Fig. c. ' 0.075(0.15)(0.03) + 0.165(0.03)(0.15) © y A = y = ©A 0.15(0.03) + 0.03(0.15)
150 mm
1 (0.03)(0.153) + 0.03(0.15)(0.12 - 0.075)2 12 +
1 (0.15)(0.033) + 0.15(0.03)(0.165 - 0.12)2 12
= 27.0 (10 - 6) m4 From Fig. d, Qmax = y¿A¿ = 0.06(0.12)(0.03) = 0.216 (10 - 3) m3 The maximum shear stress occurs at points on the neutral axis since Q is maximum and thickness t = 0.03 m is the smallest. tmax =
27.5(103) C 0.216(10 - 3) D Vmax Qmax = It 27.0(10 - 6)(0.03) = 7.333(106) Pa = 7.33 MPa
Ans.
490
30 mm 30 mm
= 0.12 m I =
B
C
The FBD of the beam is shown in Fig. a,
1.5 m
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•7–25.
Determine the maximum shear stress in the T-beam at point C. Show the result on a volume element at this point.
10 kN/m
A
B
C
1.5 m
3m
150 mm
150 mm
30 mm
using the method of sections, + c ©Fy = 0;
VC + 17.5 -
1 (5)(1.5) = 0 2
VC = -13.75 kN The neutral axis passes through centroid C of the cross-section, 0.075 (0.15)(0.03) + 0.165(0.03)(0.15) ©yA = ©A 0.15(0.03) + 0.03(0.15)
y =
= 0.12 m I =
1 (0.03)(0.15) + 0.03(0.15)(0.12 - 0.075)2 12
+
1 (0.15)(0.033) + 0.15(0.03)(0.165 - 0.12)2 12
= 27.0 (10 - 6) m4 Qmax = y¿A¿ = 0.06 (0.12)(0.03) = 0.216 (10 - 3) m3 490 The maximum shear stress occurs at points on the neutral axis since Q is maximum and thickness t = 0.03 m is the smallest. tmax =
30 mm
13.75(103) C 0.216(10 - 3) D VC Qmax = It 27.0(10 - 6) (0.03)
= 3.667(106) Pa = 3.67 MPa
Ans.
491
1.5 m
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7–26. Determine the maximum shear stress acting in the fiberglass beam at the section where the internal shear force is maximum.
200 lb/ft
150 lb/ft
D
A 6 ft
6 ft
2 ft
4 in.
6 in.
0.5 in. 4 in.
Support Reactions: As shown on FBD. Internal Shear Force: As shown on shear diagram, Vmax = 878.57 lb. Section Properties: INA =
1 1 (4) A 7.53 B (3.5) A 63 B = 77.625 in4 12 12
Qmax = ©y¿A¿ = 3.375(4)(0.75) + 1.5(3)(0.5) = 12.375 in3 Maximum Shear Stress: Maximum shear stress occurs at the point where the neutral axis passes through the section. Applying the shear formula tmax =
=
VQmax It 878.57(12.375) = 280 psi 77.625(0.5)
Ans.
492
0.75 in.
0.75 in.
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7–27. Determine the shear stress at points C and D located on the web of the beam.
3 kip/ft
D
A
C
B 6 ft
6 ft
6 in.
0.75 in.
The FBD is shown in Fig. a. Using the method of sections, Fig. b, + c ©Fy = 0;
18 -
1 (3)(6) - V = 0 2
V = 9.00 kip. The moment of inertia of the beam’s cross section about the neutral axis is I =
1 1 (6)(103) (5.25)(83) = 276 in4 12 12
QC and QD can be computed by refering to Fig. c. QC = ©y¿A¿ = 4.5 (1)(6) + 2 (4)(0.75) = 33 in3 QD = y3œ A¿ = 4.5 (1)(6) = 27 in3 Shear Stress. since points C and D are on the web, t = 0.75 in. tC =
VQC 9.00 (33) = = 1.43 ksi It 276 (0.75)
Ans.
tD =
VQD 9.00 (27) = = 1.17 ksi It 276 (0.75)
Ans.
493
6 ft 1 in.
C D
4 in. 4 in.
6 in.
1 in.
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*7–28. Determine the maximum shear stress acting in the beam at the critical section where the internal shear force is maximum.
3 kip/ft
D
A
C
B 6 ft
6 ft
6 in.
The FBD is shown in Fig. a. The shear diagram is shown in Fig. b, Vmax = 18.0 kip.
0.75 in.
6 ft 1 in.
C D
4 in. 4 in.
6 in.
1 in.
The moment of inertia of the beam’s cross-section about the neutral axis is I =
1 1 (6)(103) (5.25)(83) 12 12
= 276 in4 From Fig. c Qmax = ©y¿A¿ = 4.5 (1)(6) + 2(4)(0.75) = 33 in3 The maximum shear stress occurs at points on the neutral axis since Q is the maximum and thickness t = 0.75 in is the smallest tmax =
Vmax Qmax 18.0 (33) = = 2.87 ksi It 276 (0.75)
Ans.
494
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7–30. The beam has a rectangular cross section and is subjected to a load P that is just large enough to develop a fully plastic moment Mp = PL at the fixed support. If the material is elastic-plastic, then at a distance x 6 L the moment M = Px creates a region of plastic yielding with an associated elastic core having a height 2y¿. This situation has been described by Eq. 6–30 and the moment M is distributed over the cross section as shown in Fig. 6–48e. Prove that the maximum shear stress developed in the beam is given by tmax = 321P>A¿2, where A¿ = 2y¿b, the crosssectional area of the elastic core.
P x Plastic region 2y¿
h
b Elastic region
Force Equilibrium: The shaded area indicares the plastic zone. Isolate an element in the plastic zone and write the equation of equilibrium. ; ©Fx = 0;
tlong A2 + sg A1 - sg A1 = 0 tlong = 0
This proves that the longitudinal shear stress. tlong, is equal to zero. Hence the corresponding transverse stress, tmax, is also equal to zero in the plastic zone. Therefore, the shear force V = P is carried by the malerial only in the elastic zone. Section Properties: INA =
1 2 (b)(2y¿)3 = b y¿ 3 12 3
Qmax = y¿ A¿ =
y¿ y¿ 2b (y¿)(b) = 2 2
Maximum Shear Stress: Applying the shear formula V A y¿2 b B 3
tmax However,
VQmax = = It
A¿ = 2by¿ tmax =
3P ‚ 2A¿
A by¿ B (b) 2 3
3
=
3P 4by¿
hence (Q.E.D.)
495
L
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7–31. The beam in Fig. 6–48f is subjected to a fully plastic moment Mp . Prove that the longitudinal and transverse shear stresses in the beam are zero. Hint: Consider an element of the beam as shown in Fig. 7–4c.
P x Plastic region 2y¿
h
b Elastic region
L
Force Equilibrium: If a fully plastic moment acts on the cross section, then an element of the material taken from the top or bottom of the cross section is subjected to the loading shown. For equilibrium ; ©Fx = 0;
sg A1 + tlong A2 - sg A1 = 0 tlong = 0
Thus no shear stress is developed on the longitudinal or transverse plane of the element. (Q. E. D.)
*7–32. The beam is constructed from two boards fastened together at the top and bottom with two rows of nails spaced every 6 in. If each nail can support a 500-lb shear force, determine the maximum shear force V that can be applied to the beam.
6 in. 6 in. 2 in. 2 in.
V
6 in.
Section Properties: I =
1 (6) A 43 B = 32.0 in4 12
Q = y¿A¿ = 1(6)(2) = 12.0 in4
Shear Flow: There are two rows of nails. Hence, the allowable shear flow 2(500) = 166.67 lb>in. q = 6 q =
166.67 =
VQ I V(12.0) 32.0
V = 444 lb
Ans.
496
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•7–33.
The beam is constructed from two boards fastened together at the top and bottom with two rows of nails spaced every 6 in. If an internal shear force of V = 600 lb is applied to the boards, determine the shear force resisted by each nail.
6 in. 6 in. 2 in. 2 in.
Section Properties: I =
1 (6) A 43 B = 32.0 in4 12
V
6 in.
Q = y¿A¿ = 1(6)(2) = 12.0 in4 Shear Flow: q =
VQ 600(12.0) = = 225 lb>in. I 32.0
There are two rows of nails. Hence, the shear force resisted by each nail is q 225 lb>in. F = a bs = a b(6 in.) = 675 lb 2 2
Ans.
7–34. The beam is constructed from two boards fastened together with three rows of nails spaced s = 2 in. apart. If each nail can support a 450-lb shear force, determine the maximum shear force V that can be applied to the beam. The allowable shear stress for the wood is tallow = 300 psi.
s s 1.5 in.
The moment of inertia of the cross-section about the neutral axis is I =
V
1 (6)(33) = 13.5 in4 12
6 in.
Refering to Fig. a, QA = Qmax = y¿A¿ = 0.75(1.5)(6) = 6.75 in3 The maximum shear stress occurs at the points on the neutral axis where Q is maximum and t = 6 in. tallow =
VQmax ; It
300 =
V(6.75) 13.5(6)
V = 3600 lb = 3.60 kips
Shear Flow: Since there are three rows of nails, F 450 b = 675 lb>in. qallow = 3 a b = 3 a s 2 VQA V(6.75) ; 675 = qallow = I 13.5 V = 1350 lb = 1.35 kip
497
Ans.
1.5 in.
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7–35. The beam is constructed from two boards fastened together with three rows of nails. If the allowable shear stress for the wood is tallow = 150 psi, determine the maximum shear force V that can be applied to the beam. Also, find the maximum spacing s of the nails if each nail can resist 650 lb in shear.
s s 1.5 in. V
6 in.
The moment of inertia of the cross-section about the neutral axis is I =
1 (6)(33) = 13.5 in4 12
Refering to Fig. a, QA = Qmax = y¿A¿ = 0.75(1.5)(6) = 6.75 in3 The maximum shear stress occurs at the points on the neutral axis where Q is maximum and t = 6 in. tallow =
VQmax ; It
150 =
V(6.75) 13.5(6)
V = 1800 lb = 1.80 kip Since there are three rows of nails, qallow = 3 a qallow =
VQA ; I
Ans.
F 650 1950 lb b = 3¢ b ≤ = a s s s in.
1800(6.75) 1950 = s 13.5
s = 2.167 in = 2
1 in 8
Ans.
498
1.5 in.
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*7–36. The beam is fabricated from two equivalent structural tees and two plates. Each plate has a height of 6 in. and a thickness of 0.5 in. If a shear of V = 50 kip is applied to the cross section, determine the maximum spacing of the bolts. Each bolt can resist a shear force of 15 kip.
0.5 in. s 3 in.
1 in. A
Section Properties: INA =
V
6 in.
1 1 (3) A 93 B (2.5) A 83 B 12 12
0.5 in.
N
1 1 (0.5) A 23 B + (1) A 63 B 12 12
3 in.
= 93.25 in4 Q = ©y¿A¿ = 2.5(3)(0.5) + 4.25(3)(0.5) = 10.125 in3 Shear Flow: Since there are two shear planes on the bolt, the allowable shear flow is 2(15) 30 . = q = s s VQ q = I 50(10.125) 30 = s 93.25 s = 5.53 in.
Ans.
•7–37.
The beam is fabricated from two equivalent structural tees and two plates. Each plate has a height of 6 in. and a thickness of 0.5 in. If the bolts are spaced at s = 8 in., determine the maximum shear force V that can be applied to the cross section. Each bolt can resist a shear force of 15 kip.
0.5 in. s 3 in.
1 in. A
Section Properties: INA
-
1 1 (0.5) A 23 B + (1) A 63 B 12 12
3 in.
Q = ©y¿A¿ = 2.5(3)(0.5) + 4.25(3)(0.5) = 10.125 in3 Shear Flow: Since there are two shear planes on the bolt, the allowable shear flow is 2(15) = 3.75 kip>in. q = 8
3.75 =
0.5 in.
N
= 93.25 in4
q =
V
6 in.
1 1 = (3) A 93 B (2.5) A 83 B 12 12
VQ I V(10.125) 93.25
y = 34.5 kip
Ans.
499
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7–38. The beam is subjected to a shear of V = 2 kN. Determine the average shear stress developed in each nail if the nails are spaced 75 mm apart on each side of the beam. Each nail has a diameter of 4 mm. The neutral axis passes through centroid C of the cross-section as shown in Fig. a. ' 0.175(0.05)(0.2) + 0.1(0.2)(0.05) © y A y = = = 0.1375 m ©A 0.05(0.2) + 0.2(0.05)
200 mm
25 mm
75 mm 50 mm 75 mm
V 200 mm
Thus, I =
1 (0.2)(0.053) + 0.2 (0.05)(0.175 - 0.1375)2 12 +
25 mm
1 (0.05)(0.23) + 0.05(0.2)(0.1375 - 0.1)2 12
= 63.5417(10 - 6) m4 Q for the shaded area shown in Fig. b is Q = y¿A¿ = 0.0375 (0.05)(0.2) = 0.375(10 - 3) m3 Since there are two rows of nails q = 2a
q =
VQ ; I
26.67 F =
F 2F b = = (26.67 F) N>m. s 0.075
2000 C 0.375 (10 - 3) D 63.5417 (10 - 6)
F = 442.62 N Thus, the shear stress developed in the nail is tn =
F 442.62 = = 35.22(106)Pa = 35.2 MPa p A 2 (0.004 ) 4
Ans.
500
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7–39. A beam is constructed from three boards bolted together as shown. Determine the shear force developed in each bolt if the bolts are spaced s = 250 mm apart and the applied shear is V = 35 kN.
25 mm 25 mm 100 mm 250 mm
2 (0.125)(0.25)(0.025) + 0.275 (0.35)(0.025) y = = 0.18676 m 2 (0.25)(0.025) + 0.35 (0.025) I = (2)a
+
1 b(0.025)(0.253) + 2 (0.025)(0.25)(0.18676 - 0.125)2 12
V
1 (0.025)(0.35)3 + (0.025)(0.35)(0.275 - 0.18676)2 12
350 mm
s = 250 mm
= 0.270236 (10 - 3) m4
25 mm -3
3
Q = y¿A¿ = 0.06176(0.025)(0.25) = 0.386(10 ) m q =
35 (0.386)(10 - 3) VQ = 49.997 kN>m = I 0.270236 (10 - 3)
F = q(s) = 49.997 (0.25) = 12.5 kN
Ans.
*7–40. The double-web girder is constructed from two plywood sheets that are secured to wood members at its top and bottom. If each fastener can support 600 lb in single shear, determine the required spacing s of the fasteners needed to support the loading P = 3000 lb. Assume A is pinned and B is a roller.
2 in. 2 in. s 10 in. A 4 ft 2 in. 2 in. 6 in. 0.5 in. 0.5 in.
Support Reactions: As shown on FBD. Internal Shear Force: As shown on shear diagram, Vmax = 1500 lb. Section Properties: INA =
P
1 1 (7) A 183 B (6) A 103 B = 2902 in4 12 12
Q = y¿A¿ = 7(4)(6) = 168 in3 Shear Flow: Since there are two shear planes on the bolt, the allowable shear flow is 2(600) 1200 = q = . s s VQ q = I 1500(168) 1200 = s 2902 s = 13.8 in.
Ans.
501
4 ft
B
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•7–41.
The double-web girder is constructed from two plywood sheets that are secured to wood members at its top and bottom. The allowable bending stress for the wood is sallow = 8 ksi and the allowable shear stress is tallow = 3 ksi. If the fasteners are spaced s = 6 in. and each fastener can support 600 lb in single shear, determine the maximum load P that can be applied to the beam.
2 in. 2 in. s 10 in. A 4 ft 2 in. 2 in. 6 in. 0.5 in. 0.5 in.
Support Reactions: As shown on FBD. Internal Shear Force and Moment: As shown on shear and moment diagram, Vmax = 0.500P and Mmax = 2.00P. Section Properties: INA =
P
1 1 (7) A 183 B (6) A 103 B = 2902 in4 12 12
Q = y2œ A¿ = 7(4)(6) = 168 in3 Qmax = ©y¿A¿ = 7(4)(6) + 4.5(9)(1) = 208.5 in3 Shear Flow: Assume bolt failure. Since there are two shear planes on the bolt, the 2(600) = 200 lb>in. allowable shear flow is q = 6 VQ q = I 0.500P(168) 200 = 2902 P = 6910 lb = 6.91 kip (Controls !)
Ans.
Shear Stress: Assume failure due to shear stress. VQmax It 0.500P(208.5) 3000 = 2902(1) tmax = tallow =
P = 22270 lb = 83.5 kip Bending Stress: Assume failure due to bending stress. Mc I 2.00P(12)(9) 8(103) = 2902
smax = sallow =
P = 107 ksi
502
4 ft
B
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7–42. The T-beam is nailed together as shown. If the nails can each support a shear force of 950 lb, determine the maximum shear force V that the beam can support and the corresponding maximum nail spacing s to the nearest 18 in. The allowable shear stress for the wood is tallow = 450 psi.
2 in.
s
The neutral axis passes through the centroid c of the cross-section as shown in Fig. a. ' 13(2)(12) + 6(12)(2) © y A y = = = 9.5 in. ©A 2(12) + 12(2) I =
1 (2)(123) + 2(12)(9.5 - 6)2 12 +
= 884 in4 Refering to Fig. a, Qmax and QA are Qmax = y1œ A1œ = 4.75(9.5)(2) = 90.25 in3 QA = y2œ A2œ = 3.5 (2)(12) = 84 in3 The maximum shear stress occurs at the points on the neutral axis where Q is maximum and t = 2 in. VQmax ; It
450 =
V (90.25) 884 (2)
V = 8815.51 lb = 8.82 kip Here, qallow =
F 950 = lb>in. Then s s VQA ; qallow = I
Ans.
8815.51(84) 950 = s 884 s = 1.134 in = 1
12 in. V
2 in.
1 (12)(23) + 12(2)(13 - 9.5)2 12
tallow =
s
12 in.
1 in 8
Ans.
503
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7–43. Determine the average shear stress developed in the nails within region AB of the beam. The nails are located on each side of the beam and are spaced 100 mm apart. Each nail has a diameter of 4 mm. Take P = 2 kN.
P 2 kN/m
A
B
C
1.5 m
The FBD is shown in Fig. a. As indicated in Fig. b, the internal shear force on the cross-section within region AB is constant that is VAB = 5 kN.
1.5 m
100 mm
The neutral axis passes through centroid C of the cross section as shown in Fig. c. ' 0.18(0.04)(0.2) + 0.1(0.2)(0.04) © y A = y = ©A 0.04(0.2) + 0.2(0.04)
40 mm
= 0.14 m
200 mm
1 I = (0.04)(0.23) + 0.04(0.2)(0.14 - 0.1)2 12 1 + (0.2)(0.043) + 0.2(0.04)(0.18 - 0.14)2 12
200 mm 20 mm 20 mm
= 53.333(10 - 6) m4 Q for the shaded area shown in Fig. d is Q = y¿A¿ = 0.04(0.04)(0.2) = 0.32(10 - 3) m3 Since there are two rows of nail, q = 2 a q =
VAB Q ; I
20F =
F F b = 2a b = 20F N>m. s 0.1
5(103) C 0.32(10 - 3) D 53.333(10 - 6)
F = 1500 N Thus, the average shear stress developed in each nail is
A tnail B avg =
F 1500 = = 119.37(106)Pa = 119 MPa p Anail 2 (0.004 ) 4
504
Ans.
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*7–44. The nails are on both sides of the beam and each can resist a shear of 2 kN. In addition to the distributed loading, determine the maximum load P that can be applied to the end of the beam. The nails are spaced 100 mm apart and the allowable shear stress for the wood is tallow = 3 MPa.
P 2 kN/m
A
B
C
1.5 m
1.5 m
100 mm
The FBD is shown in Fig. a. 40 mm
As indicated the shear diagram, Fig. b, the maximum shear occurs in region AB of Constant value, Vmax = (P + 3) kN. The neutral axis passes through Centroid C of the cross-section as shown in Fig. c. ' 0.18(0.04)(0.2) + 0.1(0.2)(0.04) © y A = y = ©A 0.04(0.2) + 0.2(0.04) = 0.14 m I =
1 (0.04)(0.23) + 0.04(0.2)(0.14 - 0.1)2 12 1 + (0.2)(0.043) + 0.2(0.04)(0.18 - 0.142) 12
Refering to Fig. d, Qmax = y1œ A1œ = 0.07(0.14)(0.04) = 0.392(10 - 3) m3 QA = y2œ A2œ = 0.04(0.04)(0.2) = 0.32(10 - 3) m3 The maximum shear stress occurs at the points on Neutral axis where Q is maximum and t = 0.04 m. Vmax Qmax ; It
3(106) =
(P + 3)(103) C 0.392(10 - 3) D 53.333(10 - 6)(0.04)
P = 13.33 kN Since there are two rows of nails qallow = 2 a qallow
Vmax QA = ; I
40 000 =
200 mm 20 mm 20 mm
= 53.333(10 - 6) m4
tallow =
200 mm
2(103) F d = 40 000 N>m. b = 2c s 0.1
(P + 3)(103) C 0.32(10 - 3) D 53.333(10 - 6)
P = 3.67 kN (Controls!)
Ans.
505
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7–44.
Continued
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•7–45.
The beam is constructed from four boards which are nailed together. If the nails are on both sides of the beam and each can resist a shear of 3 kN, determine the maximum load P that can be applied to the end of the beam.
3 kN
A
P
B
C
2m
2m
100 mm
30 mm
150 mm
30 mm
250 mm 30 mm 30 mm
Support Reactions: As shown on FBD. Internal Shear Force: As shown on shear diagram, VAB = (P + 3) kN. Section Properties: INA =
1 1 (0.31) A 0.153 B (0.25) A 0.093 B 12 12
= 72.0 A 10 - 6 B m4
Q = y¿A¿ = 0.06(0.25)(0.03) = 0.450 A 10 - 3 B m3 Shear Flow: There are two rows of nails. Hence the allowable shear flow is 3(2) = 60.0 kN>m. q = 0.1 VQ q = I (P + 3)(103)0.450(10 - 3) 60.0 A 103 B = 72.0(10 - 6) P = 6.60 kN
Ans.
507
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7–47. The beam is made from four boards nailed together as shown. If the nails can each support a shear force of 100 lb., determine their required spacing s and s if the beam is subjected to a shear of V = 700 lb.
D 1 in. 1 in. 2 in.
s¿ s¿ s
A
C s
10 in.
1 in.
10 in. V B 1.5 in.
Section Properties: y =
©yA 0.5(10)(1) + 1.5(2)(3) + 6(1.5)(10) = ©A 10(1) + 2(3) + 1.5(10) = 3.3548 in
INA =
1 (10) A 13 B + 10(1)(3.3548 - 0.5)2 12 1 + (2) A 33 B + 2(3)(3.3548 - 1.5)2 12
= 337.43 in4 QC = y1 ¿A¿ = 1.8548(3)(1) = 5.5645 in3 QD = y2 ¿A¿ = (3.3548 - 0.5)(10)(1) + 2 C (3.3548 - 1.5)(3)(1) D = 39.6774 in3 Shear Flow: The allowable shear flow at points C and D is qC = 100 , respectively. qB = s¿ VQC qC = I 700(5.5645) 100 = s 337.43 s = 8.66 in. VQD qD = I 700(39.6774) 100 = s¿ 337.43
100 and s
Ans.
s¿ = 1.21 in.
Ans.
508
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*7–48. The box beam is constructed from four boards that are fastened together using nails spaced along the beam every 2 in. If each nail can resist a shear of 50 lb, determine the greatest shear V that can be applied to the beam without causing failure of the nails.
1 in. 12 in. 5 in.
V 2 in.
1 in. 6 in.
1 in.
y =
©yA 0.5 (12)(1) + 2 (4)(6)(1) + (6.5)(6)(1) = = 3.1 in. ©A 12(1) + 2(6)(1) + (6)(1)
I =
1 (12)(13) + 12(1)(3.1 - 0.5)2 12 + 2a +
1 b (1)(63) + 2(1)(6)(4 - 3.1)2 12
1 (6)(13) + 6(1)(6.5 - 3.1)2 = 197.7 in4 12
QB = y1œ A¿ = 2.6(12)(1) = 31.2 in3
qB =
V(31.2) 1 VQB a b = = 0.0789 V 2 I 2(197.7)
qB s = 0.0789V(2) = 50 V = 317 lb (controls)
Ans.
QA = y2œ A¿ = 3.4(6)(1) = 20.4 in3
qA =
V(20.4) 1 VQA a b = = 0.0516 V 2 I 2(197.7)
qA s = 0.0516V(2) = 50 V = 485 lb
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7–50. A shear force of V = 300 kN is applied to the box girder. Determine the shear flow at points A and B. 90 mm
90 mm C
A
D
200 mm B
190 mm V
200 mm
10 mm 180 mm
10 mm
The moment of inertia of the cross-section about the neutral axis is I =
1 1 (0.2)(0.43) (0.18)(0.383) = 0.24359(10 - 3) m4 12 12
Refering to Fig. a Fig. b, QA = y1œ A1œ = 0.195 (0.01)(0.19) = 0.3705 (10 - 3) m3 QB = 2yzœ A2œ + y3œ A3œ = 2 [0.1(0.2)(0.01)] + 0.195(0.01)(0.18) = 0.751(10 - 3) m3 Due to symmety, the shear flow at points A and A¿ , Fig. a, and at points B and B¿ , Fig. b, are the same. Thus qA
3 -3 1 300(10 ) C 0.3705(10 ) D 1 VQA s b = c = a 2 I 2 0.24359(10 - 3)
= 228.15(103) N>m = 228 kN>m
qB =
Ans.
3 -3 1 VQB 1 300(10 ) C 0.751(10 ) D s a b = c 2 I 2 0.24359(10 - 3)
= 462.46(103) N>m = 462 kN>m
Ans.
510
100 mm
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7–51. A shear force of V = 450 kN is applied to the box girder. Determine the shear flow at points C and D. 90 mm
90 mm C
A
D
200 mm B
190 mm V
200 mm
10 mm 180 mm
10 mm
The moment of inertia of the cross-section about the neutral axis is I =
1 1 (0.2)(0.43) (0.18)(0.383) = 0.24359(10 - 3) m4 12 12
Refering to Fig. a, due to symmetry ACœ = 0. Thus QC = 0 Then refering to Fig. b, QD = y1œ A1œ + y2œ A2œ = 0.195 (0.01)(0.09) + 0.15(0.1)(0.01) = 0.3255(10 - 3) m3 Thus, qC =
qD =
VQC = 0 I
Ans.
450(103) C 0.3255(10 - 3) D VQD = I 0.24359(10 - 3)
= 601.33(103) N>m = 601 kN>m
Ans.
100 mm
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*7–52. A shear force of V = 18 kN is applied to the symmetric box girder. Determine the shear flow at A and B.
10 mm 30 mm 10 mm
A
100 mm C
B
100 mm
150 mm
10 mm 10 mm
V 150 mm 10 mm 125 mm 10 mm
Section Properties: INA =
1 1 (0.145) A 0.33 B (0.125) A 0.283 B 12 12 + 2c
1 (0.125) A 0.013 B + 0.125(0.01) A 0.1052 B d 12
= 125.17 A 10 - 6 B m4 QA = y2œ A¿ = 0.145(0.125)(0.01) = 0.18125 A 10 - 3 B m3 QB = y1œ A¿ = 0.105(0.125)(0.01) = 0.13125 A 10 - 3 B m3
Shear Flow: qA =
=
1 VQA c d 2 I 1 18(103)(0.18125)(10 - 3) d c 2 125.17(10 - 6)
= 13033 N>m = 13.0 kN>m qB =
=
Ans.
1 VQB c d 2 I 1 18(103)(0.13125)(10 - 3) d c 2 125.17(10 - 6)
= 9437 N>m = 9.44 kN>m
Ans.
512
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A shear force of V = 18 kN is applied to the box girder. Determine the shear flow at C.
•7–53.
10 mm 30 mm 10 mm
A
100 mm C
B
100 mm
150 mm
10 mm 10 mm
V 150 mm 10 mm 125 mm 10 mm
Section Properties: INA =
1 1 (0.145) A 0.33 B (0.125) A 0.283 B 12 12 +2 c
1 (0.125) A 0.013 B + 0.125(0.01) A 0.1052 B d 12
= 125.17 A 10 - 6 B m4 QC = ©y¿A¿ = 0.145(0.125)(0.01) + 0.105(0.125)(0.01) + 0.075(0.15)(0.02) = 0.5375 A 10 - 3 B m3 Shear Flow: qC =
=
1 VQC c d 2 I 1 18(103)(0.5375)(10 - 3) d c 2 125.17(10 - 4)
= 38648 N>m = 38.6 kN>m
Ans.
513
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7–54. The aluminum strut is 10 mm thick and has the cross section shown. If it is subjected to a shear of, V = 150 N, determine the shear flow at points A and B. 10 mm
2[0.005(0.03)(0.01)] + 2[0.03(0.06)(0.01)] + 0.055(0.04)(0.01) = 0.027727 m y = 2(0.03)(0.01) + 2(0.06)(0.01) + 0.04(0.01) I = 2c
1 (0.03)(0.01)3 + 0.03(0.01)(0.027727 - 0.005)2 d 12
+ 2c +
40 mm 10 mm 30 mm
1 (0.01)(0.06)3 + 0.01(0.06)(0.03 - 0.027727)2 d 12
B A V 40 mm 10 mm
30 mm 10 mm
1 (0.04)(0.01)3 + 0.04(0.01)(0.055 - 0.027727)2 = 0.98197(10 - 6) m4 12
yB ¿ = 0.055 - 0.027727 = 0.027272 m yA ¿ = 0.027727 - 0.005 = 0.022727 m QA = yA ¿A¿ = 0.022727(0.04)(0.01) = 9.0909(10 - 6) m3 QB = yB ¿A¿ = 0.027272(0.03)(0.01) = 8.1818(10 - 6) m3 qA =
VQA 150(9.0909)(10 - 6) = 1.39 kN>m = I 0.98197(10 - 6)
Ans.
qB =
VQB 150(8.1818)(10 - 6) = 1.25 kN>m = I 0.98197(10 - 6)
Ans.
7–55. The aluminum strut is 10 mm thick and has the cross section shown. If it is subjected to a shear of V = 150 N, determine the maximum shear flow in the strut.
y =
2[0.005(0.03)(0.01)] + 2[0.03(0.06)(0.01)] + 0.055(0.04)(0.01) 2(0.03)(0.01) + 2(0.06)(0.01) + 0.04(0.01)
10 mm 40 mm
B A
= 0.027727 m I = 2c
10 mm
1 (0.03)(0.01)3 + 0.03(0.01)(0.027727 - 0.005)2 d 12
30 mm
1 + 2 c (0.01)(0.06)3 + 0.01(0.06)(0.03 - 0.027727)2 d 12 +
10 mm
1 (0.04)(0.01)3 + 0.04(0.01)(0.055 - 0.027727)2 12
= 0.98197(10 - 6) m4
Qmax = (0.055 - 0.027727)(0.04)(0.01) + 2[(0.06 - 0.027727)(0.01)]a
0.06 - 0.0277 b 2
= 21.3(10 - 6) m3 qmax =
V 40 mm
1 150(21.3(10 - 6)) 1 VQmax b = 1.63 kN>m a b = a 2 I 2 0.98197(10 - 6)
Ans.
514
30 mm 10 mm
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*7–56. The beam is subjected to a shear force of V = 5 kip. Determine the shear flow at points A and B.
0.5 in. C
5 in. 5 in. 0.5 in.
0.5 in. 2 in.
0.25(11)(0.5) + 2[4.5(8)(0.5)] + 6.25(10)(0.5) ©yA y = = = 3.70946 in. ©A 11(0.5) + 2(8)(0.5) + 10(0.5)
A D
8 in.
1 1 (11)(0.53) + 11(0.5)(3.70946 - 0.25)2 + 2c (0.5)(83) + 0.5(8)(4.5 - 3.70946)2 d 12 12
I =
+
0.5 in.
V
B
1 (10)(0.53) + 10(0.5)(6.25 - 3.70946)2 12
= 145.98 in4 œ = 3.70946 - 0.25 = 3.45946 in. yA
yBœ = 6.25 - 3.70946 = 2.54054 in. œ QA = yA A¿ = 3.45946(11)(0.5) = 19.02703 in3
QB = yBœ A¿ = 2.54054(10)(0.5) = 12.7027 in3 qA =
1 VQA 1 5(103)(19.02703) a b = a b = 326 lb>in. 2 I 2 145.98
Ans.
qB =
1 VQB 1 5(103)(12.7027) a b = a b = 218 lb>in. 2 I 2 145.98
Ans.
•7–57.
The beam is constructed from four plates and is subjected to a shear force of V = 5 kip. Determine the maximum shear flow in the cross section.
0.5 in. C
5 in. 5 in. 0.5 in.
y =
©yA 0.25(11)(0.5) + 2[4.5(8)(0.5)] + 6.25(10)(0.5) = = 3.70946 in. ©A 11(0.5) + 2(8)(0.5) + 10(0.5)
I =
1 1 (11)(0.53) + 11(0.5)(3.45952) + 2 c (0.5)(83) + 0.5(8)(0.79052) d 12 12
0.5 in. 2 in.
A D
8 in. V
+
1 (10)(0.53) + 10(0.5)(2.54052) 12
= 145.98 in4 Qmax = 3.4594 (11)(0.5) + 2[(1.6047)(0.5)(3.7094 - 0.5)] = 24.177 in3 qmax =
1 VQmax 1 5(103)(24.177) a b = a b 2 I 2 145.98
= 414 lb>in.
Ans.
515
0.5 in.
B
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7–58. The channel is subjected to a shear of V = 75 kN. Determine the shear flow developed at point A.
30 mm 400 mm
A
200 mm
30 mm
V ⫽ 75 kN
30 mm
y =
©yA 0.015(0.4)(0.03) + 2[0.13(0.2)(0.03)] = = 0.0725 m ©A 0.4(0.03) + 2(0.2)(0.03)
I =
1 (0.4)(0.033) + 0.4(0.03)(0.0725 - 0.015)2 12 + 2c
1 (0.03)(0.23) + 0.03(0.2)(0.13 - 0.0725)2 d = 0.12025(10 - 3) m4 12
œ A¿ = 0.0575(0.2)(0.03) = 0.3450(10 - 3) m3 QA = yA
q =
qA =
VQ I 75(103)(0.3450)(10 - 3) 0.12025(10 - 3)
= 215 kN>m
Ans.
7–59. The channel is subjected to a shear of V = 75 kN. Determine the maximum shear flow in the channel.
30 mm 400 mm
A V ⫽ 75 kN
30 mm
y =
©yA 0.015(0.4)(0.03) + 2[0.13(0.2)(0.03)] = ©A 0.4(0.03) + 2(0.2)(0.03) = 0.0725 m
1 I = (0.4)(0.033) + 0.4(0.03)(0.0725 - 0.015)2 12 1 + 2c (0.03)(0.23) + 0.03(0.2)(0.13 - 0.0725)2 d 12 = 0.12025(10 - 3) m4 Qmax = y¿A¿ = 0.07875(0.1575)(0.03) = 0.37209(10 - 3) m3 qmax =
75(103)(0.37209)(10 - 3) 0.12025(10 - 3)
= 232 kN>m
Ans.
516
200 mm
30 mm
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*7–60. The angle is subjected to a shear of V = 2 kip. Sketch the distribution of shear flow along the leg AB. Indicate numerical values at all peaks.
A 5 in.
5 in. 45⬚ 45⬚
0.25 in.
Section Properties: b =
0.25 = 0.35355 in. sin 45°
h = 5 cos 45° = 3.53553 in. INA = 2c
1 (0.35355) A 3.535533 B d = 2.604167 in4 12
Q = y¿A¿ = [0.25(3.53553) + 0.5y]a2.5 -
y b(0.25) sin 45°
= 0.55243 - 0.17678y2 Shear Flow: VQ I 2(103)(0.55243 - 0.17678y2) = 2.604167
q =
= {424 - 136y2} lb>in. At y = 0,
Ans.
q = qmax = 424 lb>in.
Ans.
517
B V
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•7–61.
The assembly is subjected to a vertical shear of V = 7 kip. Determine the shear flow at points A and B and the maximum shear flow in the cross section.
A
0.5 in.
B V 2 in.
0.5 in.
0.5 in.
6 in.
6 in. 2 in. 0.5 in.
y =
©yA (0.25)(11)(0.5) + 2(3.25)(5.5)(0.5) + 6.25(7)(0.5) = = 2.8362 in. ©A 0.5(11) + 2(0.5)(5.5) + 7(0.5)
I =
1 1 (11)(0.53) + 11(0.5)(2.8362 - 0.25)2 + 2a b(0.5)(5.53) + 2(0.5)(5.5)(3.25 - 2.8362)2 12 12 +
1 (7)(0.53) + (0.5)(7)(6.25 - 2.8362)2 = 92.569 in4 12
QA = y1 ¿A1 ¿ = (2.5862)(2)(0.5) = 2.5862 in3 QB = y2 ¿A2 ¿ = (3.4138)(7)(0.5) = 11.9483 in3 Qmax = ©y¿A¿ = (3.4138)(7)(0.5) + 2(1.5819)(3.1638)(0.5) = 16.9531 in3 q =
VQ I
7(103)(2.5862) = 196 lb>in. 92.569 1 7(103)(11.9483) qB = a b = 452 lb>in. 2 92.569 1 7(103)(16.9531) b = 641 lb>in. qmax = a 2 92.569 qA =
Ans. Ans. Ans.
518
0.5 in.
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7–62. Determine the shear-stress variation over the cross section of the thin-walled tube as a function of elevation y and show that t max = 2V>A, where A = 2prt. Hint: Choose a differential area element dA = Rt du. Using dQ = y dA, formulate Q for a circular section from u to (p - u) and show that Q = 2R2t cos u, where cos u = 2R2 - y2>R.
ds du y
u t
dA = R t du dQ = y dA = yR t du Here y = R sin u Therefore dQ = R2 t sin u du p-u
Q =
p-u
R2 t sin u du = R2 t(-cos u) |
Lu
u
2
= R t [-cos (p - u) - (-cos u)] = 2R2 t cos u dI = y2 dA = y2 R t du = R3 t sin2 u du 2p
I =
L0
2p
R3 t sin2 u du = R3 t 2p
=
t =
sin 2u R3 t [u ] 2 2 0
R3 t [2p - 0] = pR3 t 2
VQ V(2R2t cos u) V cos u = = 3 It pR t pR t(2t)
Here cos u =
t =
=
L0
(1 - cos 2u) du 2
2R2 - y2 R
V 2R2 - y2 pR2t
Ans.
tmax occurs at y = 0; therefore tmax =
V pR t
A = 2pRt; therefore tmax =
2V A
QED
519
R
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7–63. Determine the location e of the shear center, point O, for the thin-walled member having the cross section shown where b2 7 b1. The member segments have the same thickness t.
t h
e
b2
Section Properties: I =
1 h 2 t h2 t h3 + 2c(b1 + b2)ta b d = C h + 6(b1 + b2) D 12 2 12 Q1 = y¿A¿ =
h ht (x )t = x 2 1 2 1
Q2 = y¿A¿ =
h ht (x )t = x 2 2 2 2
Shear Flow Resultant: VQ1 q1 = = I q2 =
VQ2 = I
P A ht2 x1 B P A ht2 x2 B
h C h + 6(b1 + b2) D h C h + 6(b1 + b2) D
6P
t h2 12
C h + 6(b1 + b2) D
=
t h2 12
C h + 6(b1 + b2) D
=
6P
b1
(Ff)1 =
L0
q1 dx1 =
6P
x1
x2
b1
h C h + 6(b1 + b2) D L0
x1 dx1
3Pb21
= b2
(Ff)2 =
L0
q2 dx2 =
h C h + 6(b1 + b2) D 6P
b2
h C h + 6(b1 + b2) D L0
x2 dx2
3Pb22
=
h C h + 6(b1 + b2) D
Shear Center: Summing moment about point A. Pe = A Ff B 2 h - A Ff B 1 h Pe =
e =
3Pb22
h C h + 6(b1 + b2) D
3(b22 - b21) h + 6(b1 + b2)
(h) -
3Pb21
h C h + 6(b1 + b2) D
(h)
Ans.
Note that if b2 = b1, e = 0 (I shape).
520
b1
O
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*7–64. Determine the location e of the shear center, point O, for the thin-walled member having the cross section shown. The member segments have the same thickness t.
b d 45⬚ O e
Section Properties: I =
=
t 1 a b(2d sin 45°)3 + 2 C bt(d sin 45°)2 D 12 sin 45° td2 (d + 3b) 3
Q = y¿A¿ = d sin 45° (xt) = (td sin 45°)x Shear Flow Resultant: qf =
P(td sin 45°)x VQ 3P sin 45° = = x td2 I d(d + 3b) (d + 3b) 3 b
Ff =
L0
b
qfdx =
2
3P sin 45° 3b sin 45° P xdx = d(d + 3b) L0 2d(d + 3b)
Shear Center: Summing moments about point A, Pe = Ff(2d sin 45°) Pe = c e =
3b2 sin 45° P d(2d sin 45°) 2d(d + 3b)
3b2 2(d + 3b)
Ans.
521
45⬚
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•7–65.
Determine the location e of the shear center, point O, for the thin-walled member having a slit along its side. Each element has a constant thickness t.
a e
a
t
a
Section Properties: I =
1 10 3 (2t)(2a)3 + 2 C at A a2 B D = a t 12 3
Q1 = y1œ A¿ =
y t (yt) = y2 2 2
Q2 = ©y¿A¿ =
a at (at) + a(xt) = (a + 2x) 2 2
Shear Flow Resultant: q1 =
P A 12 y2 B VQ1 3P 2 = 10 3 = y 3 I 20a a t 3
P C at2 (a + 2x) D VQ2 3P = = (a + 2x) q2 = 10 3 2 I 20a a t 3 a
(Fw)1 =
L0
a
q1 dy =
a
Ff =
L0
3P P y2 dy = 20 20a3 L0 a
q2 dx =
3P 3 (a + 2x)dx = P 2 10 20a L0
Shear Center: Summing moments about point A. Pe = 2(Fw)1 (a) + Ff(2a) Pe = 2 a e =
3 P b a + a Pb 2a 20 10
7 a 10
Ans.
522
O
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7–66. Determine the location e of the shear center, point O, for the thin-walled member having the cross section shown.
a 60⬚ O
a 60⬚ a e
Summing moments about A. Pe = F2 a I =
13 ab 2
t 1 1 1 (t)(a)3 + a b(a)3 = t a3 12 12 sin 30° 4
q1 =
V(a)(t)(a>4) 1 4
q2 = q1 +
F2 =
=
3
ta
V a
V(a>2)(t)(a>4) 1 4
ta
3
= q1 +
V 2a
V 4V 2 V (a) + a b (a) = a 3 2a 3 e =
223 a 3
Ans.
523
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7–67. Determine the location e of the shear center, point O, for the thin-walled member having the cross section shown. The member segments have the same thickness t.
b t h 2 O e h 2 b
Shear Flow Resultant: The shear force flows through as Indicated by F1, F2, and F3 on FBD (b). Hence, The horizontal force equilibrium is not satisfied (©Fx Z 0). In order to satisfy this equilibrium requirement. F1 and F2 must be equal to zero. Shear Center: Summing moments about point A. Pe = F2(0)
e = 0
Ans.
Also, The shear flows through the section as indicated by F1, F2, F3. + ©F Z 0 However, : x
To satisfy this equation, the section must tip so that the resultant of : : : : F1 + F2 + F3 = P Also, due to the geometry, for calculating F1 and F3, we require F1 = F3. Hence, e = 0
Ans.
524
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*7–68. Determine the location e of the shear center, point O, for the beam having the cross section shown. The thickness is t. 1 — r 2
e r
O
I = (2)c
1 r 2 (t)(r>2)3 + (r>2)(t)ar + b d + Isemi-circle 12 4
= 1.583333t r3 + Isemi-circle p>2
Isemi-circle =
p>2 2
L-p>2
(r sin u) t r du = t r3
L-p>2
sin2 u du
p Isemi-circle = t r3 a b 2 Thus, p I = 1.583333t r3 + t r3 a b = 3.15413t r3 2 r r Q = a b t a + rb + 2 4 Lu
p>2
r sin u (t r du)
Q = 0.625 t r2 + t r2 cos u q =
VQ P(0.625 + cos u)t r2 = I 3.15413 t r3
Summing moments about A: p>2
Pe =
L-p>2
(q r du)r p>2
Pe =
e =
Pr (0.625 + cos u)du 3.15413 L-p>2
r (1.9634 + 2) 3.15413
e = 1.26 r
Ans.
525
1 — r 2
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•7–69.
Determine the location e of the shear center, point O, for the thin-walled member having the cross section shown. The member segments have the same thickness t.
h1
h
O e h1 b
Summing moments about A. Pe = F(h) + 2V(b)
h 2 1 1 (t)(h3) + 2b(t)a b + (t)[h3 - (h - 2h1)3] 12 2 12
I =
=
(1)
t(h - 2h1)3 bth2 th3 + 6 2 12
Q1 = y¿A¿ =
t(hy - 2h1 y + y2) 1 (h - 2h1 + y)yt = 2 2
VQ Pt(hy - 2h1 y + y2) = I 2I
q1 =
V =
L
h1 Pt Pt hh1 2 2 (hy - 2h1 y + y2)dy = c - h31 d 2I L0 2I 2 3
q1 dy =
Q2 = ©y¿A¿ =
1 1 h (h - h1)h1 t + (x)(t) = t[h1 (h - h1) + hx] 2 2 2
VQ2 Pt = (h (h - h1) + hx) I 2I 1
q2 =
b
F =
L
q2 dx =
Pt Pt hb2 [h1 (h - h1) + hx]dx = ah1 hb - h21 b + b 2I L0 2I 2
From Eq, (1). Pe =
h2b2 4 Pt [h1 h2b - h21 hb + + hh21 b - h31 b] 2I 2 3
I =
t (2h3 + 6bh2 - (h - 2h1)3) 12
e =
b(6h1 h2 + 3h2b - 8h31) t (6h1 h2b + 3h2b2 - 8h1 3b) = 12I 2h3 + 6bh2 - (h - 2h1)3
526
Ans.
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7–70. Determine the location e of the shear center, point O, for the thin-walled member having the cross section shown.
t
r
a O
a e
Summing moments about A. Pe = r
dF L dA = t ds = t r du
(1)
y = r sin u dI = y2 dA = r2 sin2 u(t r du) = r3 t sin2 udu p+a
I = r3 t
L
sin2 u du = r3 t
Lp - a
1 - cos 2u du 2
=
sin 2u p + a r3 t (u ) 2 2 p - a
=
sin 2(p + a) sin 2(p - a) r3 t c ap + a b - ap - a bd 2 2 2
=
r3 t r3 t 2 (2a - 2 sin a cos a) = (2a - sin 2a) 2 2
dQ = y dA = r sin u(t r du) = r2 t sin u du u
Q = r2 t
q =
L
u
sin u du = r2 t (-cos u)|
Lp-a
= r2 t(-cos u - cos a) = -r2 t(cos u + cos a)
p-a
P(-r2t)(cos u + cos a) -2P(cos u + cos a) VQ = = r3t I r(2a - sin 2a) 2 (2a - sin 2a)
dF =
L
q ds =
L
q r du p+p
L =
dF =
2P r -2P (cos u + cos a) du = (2a cos a - 2 sin a) r(2a - sin 2a) Lp - a 2a - sin 2a
4P (sin a - a cos a) 2a - sin 2a
4P (sin a - a cos a) d 2a - sin 2a 4r (sin a - a cos a) e = 2a - sin 2a
From Eq. (1); P e = r c
Ans.
527
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7–71. Sketch the intensity of the shear-stress distribution acting over the beam’s cross-sectional area, and determine the resultant shear force acting on the segment AB. The shear acting at the section is V = 35 kip. Show that INA = 872.49 in4.
C
V 8 in.
B
A
6 in.
Section Properties: y =
4(8)(8) + 11(6)(2) ©yA = = 5.1053 in. ©A 8(8) + 6(2)
INA =
2 in.
1 (8) A 83 B + 8(8)(5.1053 - 4)2 12 +
1 (2) A 63 B + 2(6)(11 - 5.1053)2 12
= 872.49 in4 (Q.E.D) Q1 = y1œ A¿ = (2.55265 + 0.5y1)(5.1053 - y1)(8) = 104.25 - 4y21 Q2 = y2œ A¿ = (4.44735 + 0.5y2)(8.8947 - y2)(2) = 79.12 - y22 Shear Stress: Applying the shear formula t =
tCB =
VQ , It
VQ1 35(103)(104.25 - 4y21) = It 872.49(8) = {522.77 - 20.06y21} psi
At y1 = 0,
tCB = 523 psi
At y1 = -2.8947 in.
tCB = 355 psi
tAB =
VQ2 35(103)(79.12 - y22) = It 872.49(2)
= {1586.88 - 20.06y22} psi At y2 = 2.8947 in.
tAB = 1419 psi
Resultant Shear Force: For segment AB. VAB =
L
tAB dA 0.8947 in
=
L2.8947 in 0.8947 in
=
L2.8947 in
3 in. 3 in.
A 1586.88 - 20.06y22 B (2dy) A 3173.76 - 40.12y22 B dy
= 9957 lb = 9.96 kip
Ans.
528
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*7–72. The beam is fabricated from four boards nailed together as shown. Determine the shear force each nail along the sides C and the top D must resist if the nails are uniformly spaced at s = 3 in. The beam is subjected to a shear of V = 4.5 kip.
1 in. 1 in. 3 in.
10 in. A 1 in.
12 in. V
B
Section Properties: y =
0.5(10)(1) + 2(4)(2) + 7(12)(1) © yA = = 3.50 in. ©A 10(1) + 4(2) + 12(1)
INA =
1 (10) A 13 B + (10)(1)(3.50 - 0.5)2 12 +
1 (2) A 43 B + 2(4)(3.50 - 2)2 12 1 + (1) A 123 B + 1(12)(7 - 3.50)2 12
= 410.5 in4 QC = y1œ A¿ = 1.5(4)(1) = 6.00 in2 QD = y2œ A¿ = 3.50(12)(1) = 42.0 in2 Shear Flow: qC =
VQC 4.5(103)(6.00) = = 65.773 lb>in. I 410.5
qD =
VQD 4.5(103)(42.0) = = 460.41 lb>in. I 410.5
Hence, the shear force resisted by each nail is FC = qC s = (65.773 lb>in.)(3 in.) = 197 lb
Ans.
FD = qD s = (460.41 lb>in.)(3 in.) = 1.38 kip
Ans.
529
1 in.
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•7–73.
The member is subjected to a shear force of V = 2 kN. Determine the shear flow at points A, B, and C. The thickness of each thin-walled segment is 15 mm.
200 mm B 100 mm A C V ⫽ 2 kN
Section Properties: y =
=
© yA ©A 0.0075(0.2)(0.015) + 0.0575(0.115)(0.03) + 0.165(0.3)(0.015) 0.2(0.015) + 0.115(0.03) + 0.3(0.015)
= 0.08798 m 1 (0.2) A 0.0153 B + 0.2(0.015)(0.08798 - 0.0075)2 12 1 + (0.03) A 0.1153 B + 0.03(0.115)(0.08798 - 0.0575)2 12 1 + (0.015) A 0.33 B + 0.015(0.3)(0.165 - 0.08798)2 12
INA =
= 86.93913 A 10 - 6 B m4
QA = 0 ' QB = y 1œ A¿ = 0.03048(0.115)(0.015) = 52.57705 A 10 - 6 B m3
Ans.
QC = ©y¿A¿ = 0.03048(0.115)(0.015) + 0.08048(0.0925)(0.015) = 0.16424 A 10 - 3 B m3 Shear Flow: qA =
VQA = 0 I
Ans.
qB =
VQB 2(103)(52.57705)(10 - 6) = 1.21 kN>m = I 86.93913(10 - 6)
Ans.
qC =
VQC 2(103)(0.16424)(10 - 3) = 3.78 kN>m = I 86.93913(10 - 6)
Ans.
530
300 mm
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7–74. The beam is constructed from four boards glued together at their seams. If the glue can withstand 75 lb>in., what is the maximum vertical shear V that the beam can support? 3 in. 0.5 in.
Section Properties: INA =
1 1 (1) A 103 B + 2c (4) A 0.53 B + 4(0.5) A 1.752 B d 12 12
3 in. 0.5 in.
= 95.667 in4
V
Q = y¿A¿ = 1.75(4)(0.5) = 3.50 in3
4 in.
Shear Flow: There are two glue joints in this case, hence the allowable shear flow is 2(75) = 150 lb>in. q =
150 =
3 in.
0.5 in.
0.5 in.
VQ I V(3.50) 95.667
V = 4100 lb = 4.10 kip
Ans.
7–75. Solve Prob. 7–74 if the beam is rotated 90° from the position shown.
3 in. 0.5 in. 3 in. 0.5 in.
V
3 in.
4 in. 0.5 in.
Section Properties: INA =
1 1 (10) A 53 B (9) A 43 B = 56.167 in4 12 12
Q = y¿A¿ = 2.25(10)(0.5) = 11.25 in3 Shear Flow: There are two glue joints in this case, hence the allowable shear flow is 2(75) = 150 lb>in. q = 150 =
VQ I V(11.25) 56.167
V = 749 lb
Ans.
531
0.5 in.
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8–1. A spherical gas tank has an inner radius of r = 1.5 m. If it is subjected to an internal pressure of p = 300 kPa, determine its required thickness if the maximum normal stress is not to exceed 12 MPa. pr ; 2t
sallow =
12(106) =
300(103)(1.5) 2t
t = 0.0188 m = 18.8 mm
Ans.
8–2. A pressurized spherical tank is to be made of 0.5-in.-thick steel. If it is subjected to an internal pressure of p = 200 psi, determine its outer radius if the maximum normal stress is not to exceed 15 ksi.
sallow =
pr ; 2t
15(103) =
200 ri 2(0.5)
ri = 75 in. ro = 75 in. + 0.5 in. = 75.5 in.
Ans.
8–3. The thin-walled cylinder can be supported in one of two ways as shown. Determine the state of stress in the wall of the cylinder for both cases if the piston P causes the internal pressure to be 65 psi. The wall has a thickness of 0.25 in. and the inner diameter of the cylinder is 8 in.
P
Case (a): s1 =
pr ; t
s1 =
65(4) = 1.04 ksi 0.25
Ans.
s2 = 0
Ans.
Case (b): s1 =
pr ; t
s1 =
65(4) = 1.04 ksi 0.25
Ans.
s2 =
pr ; 2t
s2 =
65(4) = 520 psi 2(0.25)
Ans.
532
P
8 in.
8 in.
(a)
(b)
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*8–4. The tank of the air compressor is subjected to an internal pressure of 90 psi. If the internal diameter of the tank is 22 in., and the wall thickness is 0.25 in., determine the stress components acting at point A. Draw a volume element of the material at this point, and show the results on the element.
Hoop Stress for Cylindrical Vessels: Since
A
11 r = = 44 7 10, then thin wall t 0.25
analysis can be used. Applying Eq. 8–1
s1 =
pr 90(11) = = 3960 psi = 3.96 ksi t 0.25
Ans.
Longitudinal Stress for Cylindrical Vessels: Applying Eq. 8–2
s2 =
pr 90(11) = = 1980 psi = 1.98 ksi 2t 2(0.25)
Ans.
•8–5.
The spherical gas tank is fabricated by bolting together two hemispherical thin shells of thickness 30 mm. If the gas contained in the tank is under a gauge pressure of 2 MPa, determine the normal stress developed in the wall of the tank and in each of the bolts.The tank has an inner diameter of 8 m and is sealed with 900 bolts each 25 mm in diameter.
Normal Stress: Since
4 r = = 133.33 7 10, thin-wall analysis is valid. For the t 0.03
spherical tank’s wall,
s =
Referring
pr 2(4) = = 133 MPa 2t 2(0.03)
to
the free-body diagram p 2 6 P = pA = 2 A 10 B c A 8 B d = 32p A 10 B N. Thus, 4
Ans.
shown
in
Fig.
a,
6
+ c ©Fy = 0;
32p A 106 B - 450Pb - 450Pb = 0
Pb = 35.56 A 103 B p N
The normal stress developed in each bolt is then sb =
35.56 A 103 B p Pb = = 228 MPa p Ab A 0.0252 B 4
Ans.
533
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8–6. The spherical gas tank is fabricated by bolting together two hemispherical thin shells. If the 8-m inner diameter tank is to be designed to withstand a gauge pressure of 2 MPa, determine the minimum wall thickness of the tank and the minimum number of 25-mm diameter bolts that must be used to seal it. The tank and the bolts are made from material having an allowable normal stress of 150 MPa and 250 MPa, respectively.
Normal Stress: For the spherical tank’s wall, sallow =
pr 2t
150 A 106 B =
2 A 106 B (4) 2t
t = 0.02667 m = 26.7 mm Since
Ans.
r 4 = = 150 7 10, thin-wall analysis is valid. t 0.02667
Referring
the free-body diagram p P = pA = 2 A 106 B c A 82 B d = 32p A 106 B N. Thus, 4 + c ©Fy = 0;
to
32p A 106 B n =
shown
in
Fig.
a,
n n (P ) - (Pb)allow = 0 2 b allow 2
32p A 106 B
(1)
(Pb)allow
The allowable tensile force for each bolt is (Pb)allow = sallowAb = 250 A 106 B c
p A 0.0252 B d = 39.0625 A 103 B pN 4
Substituting this result into Eq. (1), n =
32p A 106 B
39.0625p A 103 B
= 819.2 = 820
Ans.
534
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8–7. A boiler is constructed of 8-mm thick steel plates that are fastened together at their ends using a butt joint consisting of two 8-mm cover plates and rivets having a diameter of 10 mm and spaced 50 mm apart as shown. If the steam pressure in the boiler is 1.35 MPa, determine (a) the circumferential stress in the boiler’s plate apart from the seam, (b) the circumferential stress in the outer cover plate along the rivet line a–a, and (c) the shear stress in the rivets.
a
8 mm
50 mm
a)
s1 =
pr 1.35(106)(0.75) = = 126.56(106) = 127 MPa t 0.008
Ans.
126.56 (106)(0.05)(0.008) = s1 ¿(2)(0.04)(0.008)
b)
s1 ¿ = 79.1 MPa
Ans.
c) From FBD(a) + c ©Fy = 0;
Fb - 79.1(106)[(0.008)(0.04)] = 0 Fb = 25.3 kN
(tavg)b =
Fb 25312.5 - p = 322 MPa 2 A 4 (0.01)
Ans.
535
0.75 m a
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*8–8. The gas storage tank is fabricated by bolting together two half cylindrical thin shells and two hemispherical shells as shown. If the tank is designed to withstand a pressure of 3 MPa, determine the required minimum thickness of the cylindrical and hemispherical shells and the minimum required number of longitudinal bolts per meter length at each side of the cylindrical shell. The tank and the 25 mm diameter bolts are made from material having an allowable normal stress of 150 MPa and 250 MPa, respectively. The tank has an inner diameter of 4 m. Normal Stress: For the cylindrical portion of the tank, the hoop stress is twice as large as the longitudinal stress. sallow =
pr ; t
150 A 106 B =
3 A 106 B (2) tc
tc = 0.04 m = 40 mm For the hemispherical cap, sallow
pr = ; t
150 A 10
6
B =
Ans. 3 A 106 B (2) 2ts
ts = 0.02 m = 20 mm Since
Ans.
r 6 10, thin-wall analysis is valid. t
Referring to the free-body diagram of the per meter length of the cylindrical portion, Fig. a, where P = pA = 3 A 106 B [4(1)] = 12 A 106 B N, we have + c ©Fy = 0;
12 A 106 B - nc(Pb)allow - nc(Pb)allow = 0 nc =
6 A 106 B
(1)
(Pb)allow
The allowable tensile force for each bolt is (Pb)allow = sallowAb = 250 A 106 B c
p A 0.0252 B d = 122.72 A 103 B N 4
Substituting this result into Eq. (1), nc = 48.89 = 49 bolts>meter
Ans.
536
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•8–9.
The gas storage tank is fabricated by bolting together two half cylindrical thin shells and two hemispherical shells as shown. If the tank is designed to withstand a pressure of 3 MPa, determine the required minimum thickness of the cylindrical and hemispherical shells and the minimum required number of bolts for each hemispherical cap. The tank and the 25 mm diameter bolts are made from material having an allowable normal stress of 150 MPa and 250 MPa, respectively. The tank has an inner diameter of 4 m.
Normal Stress: For the cylindrical portion of the tank, the hoop stress is twice as large as the longitudinal stress. sallow =
pr ; t
150 A 106 B =
3 A 106 B (2) tc
tc = 0.04 m = 40 mm For the hemispherical cap, sallow =
pr ; t
150 A 106 B =
Ans. 3 A 106 B (2) 2ts
ts = 0.02 m = 20 mm Since
Ans.
r 6 10, thin-wall analysis is valid. t
The allowable tensile force for each bolt is (Pb)allow = sallowAb = 250 A 106 B c
p A 0.0252 B d = 122.72 A 103 B N 4
Referring to the free-body diagram of the hemispherical cap, Fig. b, where p P = pA = 3 A 106 B c A 42 B d = 12p A 106 B N, 4 + ©F = 0; : x
12p A 106 B ns =
ns ns (Pb)allow (Pb)allow = 0 2 2
12p A 106 B
(1)
(Pb)allow
Substituting this result into Eq. (1), ns = 307.2 = 308 bolts
Ans.
537
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8–10. A wood pipe having an inner diameter of 3 ft is bound together using steel hoops each having a crosssectional area of 0.2 in2. If the allowable stress for the hoops is sallow = 12 ksi, determine their maximum spacing s along the section of pipe so that the pipe can resist an internal gauge pressure of 4 psi. Assume each hoop supports the pressure loading acting along the length s of the pipe.
s
4 psi
4 psi
s
s
Equilibrium for the steel Hoop: From the FBD + ©F = 0; : x
P = 72.0s
2P - 4(36s) = 0
Hoop Stress for the Steel Hoop: s1 = sallow = 12(103) =
P A 72.0s 0.2
s = 33.3 in.
Ans.
8–11. The staves or vertical members of the wooden tank are held together using semicircular hoops having a thickness of 0.5 in. and a width of 2 in. Determine the normal stress in hoop AB if the tank is subjected to an internal gauge pressure of 2 psi and this loading is transmitted directly to the hoops. Also, if 0.25-in.-diameter bolts are used to connect each hoop together, determine the tensile stress in each bolt at A and B. Assume hoop AB supports the pressure loading within a 12-in. length of the tank as shown.
18 in.
6 in. 6 in.
FR = 2(36)(12) = 864 lb ©F = 0; 864 - 2F = 0; F = 432 lb sh =
sb =
F 432 = = 432 psi Ah 0.5(2)
Ans.
F 432 = = 8801 psi = 8.80 ksi p Ab (0.25)2 4
Ans.
538
12 in. A
B 12 in.
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*8–12. Two hemispheres having an inner radius of 2 ft and wall thickness of 0.25 in. are fitted together, and the inside gauge pressure is reduced to -10 psi. If the coefficient of static friction is ms = 0.5 between the hemispheres, determine (a) the torque T needed to initiate the rotation of the top hemisphere relative to the bottom one, (b) the vertical force needed to pull the top hemisphere off the bottom one, and (c) the horizontal force needed to slide the top hemisphere off the bottom one.
0.25 in. 2 ft
Normal Pressure: Vertical force equilibrium for FBD(a). + c ©Fy = 0;
10 C p(242) D - N = 0
N = 5760p lb
The Friction Force: Applying friction formula Ff = ms N = 0.5(5760p) = 2880p lb a) The Required Torque: In order to initiate rotation of the two hemispheres relative to each other, the torque must overcome the moment produced by the friction force about the center of the sphere. T = Ffr = 2880p(2 + 0.125>12) = 18190 lb # ft = 18.2 kip # ft
Ans.
b) The Required Vertical Force: In order to just pull the two hemispheres apart, the vertical force P must overcome the normal force. P = N = 5760p = 18096 lb = 18.1 kip
Ans.
c) The Required Horizontal Force: In order to just cause the two hemispheres to slide relative to each other, the horizontal force F must overcome the friction force. F = Ff = 2880p = 9048 lb = 9.05 kip
Ans.
•8–13. The 304 stainless steel band initially fits snugly around the smooth rigid cylinder. If the band is then subjected to a nonlinear temperature drop of ¢T = 20 sin2 u °F, where u is in radians, determine the circumferential stress in the band.
1 64
10 in.
Compatibility: Since the band is fixed to a rigid cylinder (it does not deform under load), then dF - dT = 0 2p P(2pr) a¢Trdu = 0 AE L0
2pr P a b = 20ar E A L0 2p s = 10a E c L0
2p
sin2 udu
however,
P = sc A
2p
(1 - cos 2u)du
sc = 10aE = 10(9.60) A 10 - 6 B 28.0 A 103 B = 2.69 ksi
Ans.
539
u
in.
1 in.
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8–14. The ring, having the dimensions shown, is placed over a flexible membrane which is pumped up with a pressure p. Determine the change in the internal radius of the ring after this pressure is applied. The modulus of elasticity for the ring is E.
ro ri w p
Equilibrium for the Ring: Form the FBD + ©F = 0; : x
2P - 2pri w = 0
P = pri w
Hoop Stress and Strain for the Ring: s1 =
pri w pri P = = rs - ri A (rs - ri)w
Using Hooke’s Law e1 =
However,
e1 =
pri s1 = E E(rs - ri)
[1]
2p(ri)1 - 2pri (ri)1 - ri dri = = . ri ri 2pr
Then, from Eq. [1] pri dri = ri E(rs - ri) dri =
pr2i E(rs - ri)
Ans.
540
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8–15. The inner ring A has an inner radius r1 and outer radius r2. Before heating, the outer ring B has an inner radius r3 and an outer radius r4, and r2 7 r3. If the outer ring is heated and then fitted over the inner ring, determine the pressure between the two rings when ring B reaches the temperature of the inner ring. The material has a modulus of elasticity of E and a coefficient of thermal expansion of a.
r1 A
Equilibrium for the Ring: From the FBD + ©F = 0; : x
P = priw
2P - 2priw = 0
Hoop Stress and Strain for the Ring: s1 =
priw pri P = = ro - ri A (ro - ri)w
Using Hooke’s law e1 =
However,
e1 =
pri s1 = E E(ro - ri)
[1]
2p(ri)1 - 2pri (ri)1 - ri dri = = . ri ri 2pr
Then, from Eq. [1] pri dri = ri E(ro - ri) dri =
pr2i E(ro - ri)
Compatibility: The pressure between the rings requires dr2 + dr3 = r2 - r3
[2]
From the result obtained above dr2 =
pr22 E(r2 - r1)
dr3 =
pr23 E(r4 - r3)
Substitute into Eq. [2] pr22 pr23 + = r2 - r3 E(r2 - r1) E(r4 - r3) p =
r4
r2
E(r2 - r3)
Ans.
r22 r23 + r2 - r1 r4 - r3
541
r3 B
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*8–16. The cylindrical tank is fabricated by welding a strip of thin plate helically, making an angle u with the longitudinal axis of the tank. If the strip has a width w and thickness t, and the gas within the tank of diameter d is pressured to p, show that the normal stress developed along the strip is given by su = (pd>8t)(3 - cos 2u).
w u
Normal Stress: sh = s1 =
pr p(d>2) pd = = t t 2t
sl = s2 =
p(d>2) pd pr = = 2t 2t 4t
Equilibrium: We will consider the triangular element cut from the strip shown in Fig. a. Here, Ah = (w sin u)t and Thus, Al = (w cos u)t. pd pwd and (w sin u)t = sin u Fh = shAh = 2t 2 pwd pd (w cos u)t = cos u. 4t 4
Fl = slAl =
Writing the force equation of equilibrium along the x¿ axis, ©Fx¿ = 0;
c
pwd pwd sin u d sin u + c cos u d cos u - Nu = u 2 4 Nu =
pwd A 2 sin2 u + cos2 u B 4
However, sin2 u + cos2 u = 1. This equation becomes Nu = Also, sin2 u =
pwd A sin2 u + 1 B 4
1 (1 - cos 2u), so that 2 pwd Nu = (3 - cos 2u) 8
Since Au = wt, then Nu = su = Au su =
pwd (3 - cos 2u) 8 wt
pd (3 - cos 2u) 8t
(Q.E.D.)
542
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8–17. In order to increase the strength of the pressure vessel, filament winding of the same material is wrapped around the circumference of the vessel as shown. If the pretension in the filament is T and the vessel is subjected to an internal pressure p, determine the hoop stresses in the filament and in the wall of the vessel. Use the free-body diagram shown, and assume the filament winding has a thickness t and width w for a corresponding length of the vessel.
L w s1
t¿
T
p
t s1 T
Normal Stress in the Wall and Filament Before the Internal Pressure is Applied: The entire length w of wall is subjected to pretension filament force T. Hence, from equilibrium, the normal stress in the wall at this state is 2T - (sl ¿)w (2wt) = 0
(sl ¿)w =
T wt
and for the filament the normal stress is (sl ¿)fil =
T wt¿
Normal Stress in the Wall and Filament After the Internal Pressure is Applied: The stress in the filament becomes sfil = sl + (sl ¿)fil =
pr T + (t + t¿) wt¿
Ans.
sw = sl - (sl ¿)w =
pr T (t + t¿) wt
Ans.
And for the wall,
8–18. The vertical force P acts on the bottom of the plate having a negligible weight. Determine the shortest distance d to the edge of the plate at which it can be applied so that it produces no compressive stresses on the plate at section a–a. The plate has a thickness of 10 mm and P acts along the center line of this thickness.
300 mm a
a
200 mm
500 mm
sA = 0 = sa - sb 0 =
0 =
P Mc A I P (0.2)(0.01)
d
P(0.1 - d)(0.1) 1 12
P
(0.01)(0.23)
P(-1000 + 15000 d) = 0 d = 0.0667 m = 66.7 mm
Ans.
543
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8–19. Determine the maximum and minimum normal stress in the bracket at section a–a when the load is applied at x = 0.
100 kN 15 mm x 15 mm 200 mm 150 mm
a
Consider the equilibrium of the FBD of the top cut segment in Fig. a, + c ©Fy = 0; a + ©MC = 0;
N - 100 = 0
N = 100 kN
100(0.1) - M = 0
A = 0.2(0.03) = 0.006 m2
I =
M = 10 kN # m
1 (0.03)(0.23) = 20.0(10 - 6) m4 12
The normal stress developed is the combination of axial and bending stress. Thus, s =
My N ; A I
For the left edge fiber, y = C = 0.1 m. Then sL = -
100(103) 10(103)(0.1) 0.006 20.0(10 - 6)
= -66.67(106) Pa = 66.7 MPa (C) (Max)
Ans.
For the right edge fiber, y = 0.1 m. Then sR = -
100 (103) 10(103)(0.1) = 33.3 MPa (T) + 0.006 20.0(10 - 6)
Ans.
544
a
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*8–20. Determine the maximum and minimum normal stress in the bracket at section a–a when the load is applied at x = 300 mm.
100 kN 15 mm x 15 mm 200 mm 150 mm
a
Consider the equilibrium of the FBD of the top cut segment in Fig. a, + c ©Fy = 0; a + ©MC = 0;
N - 100 = 0
N = 100 kN
M - 100(0.2) = 0
A = 0.2 (0.03) = 0.006 m2
I =
M = 20 kN # m
1 (0.03)(0.23) = 20.0(10 - 6) m4 12
The normal stress developed is the combination of axial and bending stress. Thus, s =
My N ; A I
For the left edge fiber, y = C = 0.1 m. Then sC = -
100(103) 20.0(103)(0.1) + 0.006 20.0(10 - 6)
= 83.33(106) Pa = 83.3 MPa (T)(Min)
Ans.
For the right edge fiber, y = C = 0.1 m. Thus sR = -
100(103) 20.0(103)(0.1) 0.006 20.0(10 - 6)
= 117 MPa
Ans.
545
a
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•8–21.
The coping saw has an adjustable blade that is tightened with a tension of 40 N. Determine the state of stress in the frame at points A and B.
8 mm 75 mm
A
3 mm 8 mm 3 mm B
100 mm
50 mm
sA = -
sB =
P Mc 40 + = + A I (0.008)(0.003)
Mc = I
2(0.004) 1 12
(0.003)(0.008)3
4(0.004) 1 3 12 (0.003)(0.008)
= 123 MPa
Ans.
Ans.
= 62.5 MPa
8–22. The clamp is made from members AB and AC, which are pin connected at A. If it exerts a compressive force at C and B of 180 N, determine the maximum compressive stress in the clamp at section a–a. The screw EF is subjected only to a tensile force along its axis.
30 mm
40 mm
F C
180 N
15 mm 15 mm Section a – a
a
a
B
A E
There is no moment in this problem. Therefore, the compressive stress is produced by axial force only. smax =
P 240 = = 1.07 MPa A (0.015)(0.015)
Ans.
546
180 N
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8–23. The clamp is made from members AB and AC, which are pin connected at A. If it exerts a compressive force at C and B of 180 N, sketch the stress distribution acting over section a–a. The screw EF is subjected only to a tensile force along its axis.
30 mm
40 mm
F C
180 N
15 mm 15 mm Section a – a
a
a
180 N
B
A E
There is moment in this problem. Therefore, the compressive stress is produced by axial force only. smax =
240 P = = 1.07 MPa A (0.015)(0.015)
*8–24. The bearing pin supports the load of 700 lb. Determine the stress components in the support member at point A. The support is 0.5 in. thick.
0.75 in. A 2 in. 30⬚
A B
3 in.
©Fx = 0;
N - 700 cos 30° = 0;
N = 606.218 lb
©Fy = 0;
V - 700 sin 30° = 0;
V = 350 lb
a + ©M = 0;
M - 700(1.25 - 2 sin 30°) = 0; sA =
1.25 in. 700 lb
M = 175 lb # in.
(175)(0.375) N Mc 606.218 = - 1 3 A I (0.75)(0.5) 12 (0.5)(0.75)
sA = -2.12 ksi
Ans.
tA = 0
Ans.
(since QA = 0)
547
B
0.5 in.
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•8–25.
The bearing pin supports the load of 700 lb. Determine the stress components in the support member at point B. The support is 0.5 in. thick.
0.75 in. A 2 in. 30⬚
A B
B
0.5 in.
3 in.
©Fx = 0;
N - 700 cos 30° = 0;
N = 606.218 lb
©Fy = 0;
V - 700 sin 30° = 0;
V = 350 lb
a + ©M = 0;
M - 700(1.25 - 2 sin 30°) = 0; sB =
N Mc 606.218 + = + A I (0.75)(0.5)
1.25 in. 700 lb
M = 175 lb # in.
175(0.375) 1 12
(0.5)(0.75)3
sB = 5.35 ksi
Ans.
tB = 0
Ans.
(since QB = 0)
8–26. The offset link supports the loading of P = 30 kN. Determine its required width w if the allowable normal stress is sallow = 73 MPa. The link has a thickness of 40 mm.
P
s due to axial force: sa =
30(103) 750(103) P = = A (w)(0.04) w
sb =
w
50 mm
s due to bending: 30(103)(0.05 + w2)(w2) Mc = 1 3 I 12 (0.04)(w) 4500 (103)(0.05 + w2) =
w2 P
smax = sallow = sa + sb 73(106) =
4500(103)(0.05 + w2) 750(103) + w w2
73 w2 = 0.75 w + 0.225 + 2.25 w 73 w2 - 3 w - 0.225 = 0 w = 0.0797 m = 79.7 mm
Ans.
548
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8–27. The offset link has a width of w = 200 mm and a thickness of 40 mm. If the allowable normal stress is sallow = 75 MPa, determine the maximum load P that can be applied to the cables.
P
A = 0.2(0.04) = 0.008 m2 I =
s =
1 (0.04)(0.2)3 = 26.6667(10 - 6) m4 12
w
50 mm
P Mc + A I
75(106) =
0.150 P(0.1) P + 0.008 26.6667(10 - 6)
P = 109 kN
Ans.
P
*8–28. The joint is subjected to a force of P 80 lb and F 0. Sketch the normal-stress distribution acting over section a–a if the member has a rectangular cross-sectional area of width 2 in. and thickness 0.5 in. a B
s due to axial force: s =
0.5 in.
P 80 = = 80 psi A (0.5)(2)
2 in.
F
s due to bending: s =
A a
100(0.25) Mc = 1 = 1200 psi 3 I 12 (2)(0.5)
1.25 in. P
(smax)t = 80 + 1200 = 1280 psi = 1.28 ksi
Ans.
(smax)c = 1200 - 80 = 1120 psi = 1.12 ksi
Ans.
y (0.5 - y) = 1.28 1.12 y = 0.267 in.
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The joint is subjected to a force of P = 200 lb and F = 150 lb. Determine the state of stress at points A and B and sketch the results on differential elements located at these points. The member has a rectangular cross-sectional area of width 0.75 in. and thickness 0.5 in. •8–29.
a B
A a
0.5 in. 2 in.
F 1.25 in. P
A = 0.5(0.75) = 0.375 in2 œ QA = yA A¿ = 0.125(0.75)(0.25) = 0.0234375 in3 ;
I =
QB = 0
1 (0.75)(0.53) = 0.0078125 in4 12
Normal Stress: My N ; A I
s =
sA =
200 + 0 = 533 psi (T) 0.375
Ans.
sB =
50(0.25) 200 = -1067 psi = 1067 psi (C) 0.375 0.0078125
Ans.
Shear stress: t =
VQ It
tA =
150(0.0234375) = 600 psi (0.0078125)(0.75)
Ans.
tB = 0
Ans.
550
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8–30. If the 75-kg man stands in the position shown, determine the state of stress at point A on the cross section of the plank at section a–a. The center of gravity of the man is at G. Assume that the contact point at C is smooth.
C
G
600 mm A
a
50 mm
1.5 m B
12.5 mm
30 a 600 mm
300 mm
Support Reactions: Referring to the free-body diagram of the entire plank, Fig. a, a + ©MB = 0;
FC sin 30°(2.4) - 75(9.81) cos 30°(0.9) = 0 FC = 477.88 N
©Fx¿ = 0; Bx¿ - 75(9.81) sin 30° - 477.88 cos 30° = 0 Bx¿ = 781.73 N ©Fy¿ = 0; By¿ + 477.88 sin 30° - 75(9.81) cos 30° = 0 By¿ = 398.24 N Internal Loadings: Consider the equilibrium of the free-body diagram of the plank’s lower segment, Fig. b, ©Fx¿ = 0; 781.73 - N = 0
N = 781.73 N
©Fy¿ = 0; 398.24 - V = 0
V = 398.24 N
a + ©MO = 0;
M - 398.24(0.6) = 0
M = 238.94 N # m
Section Properties: The cross-sectional area and the moment of inertia about the centroidal axis of the plank’s cross section are A = 0.6(0.05) = 0.03 m2 I =
1 (0.6) A 0.053 B = 6.25 A 10 - 6 B m4 12
Referring to Fig. c, QA is QA = y¿A¿ = 0.01875(0.0125)(0.6) = 0.140625 A 10 - 3 B m3 Normal Stress: The normal stress is the combination of axial and bending stress. Thus, s =
My N ; A I
For point A, y = 0.0125 m. Then sA =
238.94(0.0125) -781.73 0.03 6.25 A 10 - 6 B
= -503.94 kPa = 504 kPa (C)
Ans.
551
Section a – a and b – b
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8–30.
Continued
Shear Stress: The shear stress is contributed by transverse shear stress. Thus,
tA
VQA = = It
398.24 c0.140625 A 10 - 3 B d 6.25 A 10 - 6 B (0.6)
Ans.
= 14.9 kPa
The state of stress at point A is represented on the element shown in Fig. d.
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8–31. Determine the smallest distance d to the edge of the plate at which the force P can be applied so that it produces no compressive stresses in the plate at section a–a. The plate has a thickness of 20 mm and P acts along the centerline of this thickness.
a
P
200 mm d 300 mm a
Consider the equilibrium of the FBD of the left cut segment in Fig. a, + : ©Fx = 0;
N - P = 0
a + ©MC = 0;
N = P
M - P(0.1 - d) = 0
A = 0.2 (0.02) = 0.004 m4
I =
M = P(0.1 - d)
1 (0.02)(0.23) = 13.3333(10 - 6) m4 12
The normal stress developed is the combination of axial and bending stress. Thus s =
My N ; A I
Since no compressive stress is desired, the normal stress at the top edge fiber must be equal to zero. Thus,
0 =
P(0.1 - d)(0.1) P ; 0.004 13.3333 (10 - 6)
0 = 250 P - 7500 P (0.1 - d) d = 0.06667 m = 66.7 mm
Ans.
553
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*8–32. The horizontal force of P = 80 kN acts at the end of the plate. The plate has a thickness of 10 mm and P acts along the centerline of this thickness such that d = 50 mm. Plot the distribution of normal stress acting along section a–a.
a
P
200 mm d 300 mm
Consider the equilibrium of the FBD of the left cut segment in Fig. a, + : ©Fx = 0; a + ©MC = 0;
N - 80 = 0
N = 80 kN
M - 80(0.05) = 0
A = 0.01(0.2) = 0.002 m2
I =
M = 4.00 kN # m
1 (0.01)(0.23) = 6.667(10 - 6) m4 12
The normal stress developed is the combination of axial and bending stress. Thus,
s =
My N ; A I
At point A, y = 0.1 m. Then sA =
80(103) 4.00(103)(0.1) 0.002 6.667(10 - 6)
= -20.0(106) Pa = 20.0 Mpa (C) At point B, y = 0.1 m. Then sB =
80(103) 4.00(103)(0.1) + 0.002 6.667(10 - 6)
= 100 (106) Pa = 100 MPa (T) The location of neutral axis can be determined using the similar triangles.
554
a
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•8–33.
The pliers are made from two steel parts pinned together at A. If a smooth bolt is held in the jaws and a gripping force of 10 lb is applied at the handles, determine the state of stress developed in the pliers at points B and C. Here the cross section is rectangular, having the dimensions shown in the figure.
0.18 in.
10 lb
D 0.2 in.
0.1 in. D
3 in.
30
E A B
0.2 in. 0.2 in.
B E
C C
0.2 in.
1.75 in. 2.5 in.
Q ©Fx = 0;
N - 10 sin 30° = 0;
N = 5.0 lb
a+ ©Fy = 0;
V - 10 cos 30° = 0;
V = 8.660 lb
+
a + ©MC = 0;
10 lb
M = 30 lb # in.
M - 10(3) = 0
A = 0.2(0.4) = 0.08 in2 I =
1 (0.2)(0.43) = 1.0667(10 - 3) in4 12
QB = 0 QC = y¿A¿ = 0.1(0.2)(0.2) = 4(10 - 3) in3 Point B: sB =
My 30(0.2) N -5.0 = 5.56 ksi(T) + = + A I 0.08 1.0667(10 - 3)
Ans.
VQ = 0 It
Ans.
My N -5.0 + = + 0 = -62.5 psi = 62.5 psi(C) A I 0.08
Ans.
VQ 8.660(4)(10 - 3) = 162 psi = It 1.0667(10 - 3)(0.2)
Ans.
tB = Point C: sC = Shear Stress : tC =
4 in.
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8–34.
Solve Prob. 8–33 for points D and E.
0.18 in.
10 lb
D 0.2 in.
0.1 in. D
3 in.
30
E A B
0.2 in. 0.2 in.
B E
C C
0.2 in.
1.75 in. 2.5 in.
a + ©MA = 0;
-F(2.5) + 4(10) = 0;
F = 16 lb
10 lb
Point D: sD = 0
tD =
Ans.
16(0.05)(0.1)(0.18) VQ = 667 psi = 1 It [12 (0.18)(0.2)3](0.18)
Ans.
Point E: sE =
My = I
28(0.1) 1 12
(0.18)(0.2)3
4 in.
Ans.
= 23.3 ksi (T)
tE = 0
Ans.
556
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8–35. The wide-flange beam is subjected to the loading shown. Determine the stress components at points A and B and show the results on a volume element at each of these points. Use the shear formula to compute the shear stress.
500 lb
3000 lb
2500 lb A B
2 ft
2 ft
2 ft
4 ft
6 ft
A 0.5 in.
1 1 I = (4)(73) (3.5)(63) = 51.33 in4 12 12
B
QB = ©y¿A¿ = 3.25(4)(0.5) + 2(2)(0.5) = 8.5 in3 QA = 0 -11500 (12)(3.5) -Mc = = -9.41 ksi I 51.33
Ans.
tA = 0 sB = tB =
4 in. 2 in. 4 in. 0.5 in.
A = 2(0.5)(4) + 6(0.5) = 7 in2
sA =
0.5 in.
Ans.
My 11500(12)(1) = = 2.69 ksi I 51.33
Ans.
VQB 2625(8.5) = = 0.869 ksi It 51.33(0.5)
Ans.
557
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*8–36. The drill is jammed in the wall and is subjected to the torque and force shown. Determine the state of stress at point A on the cross section of drill bit at section a–a.
y 400 mm a 20 N ·m x a 125 mm y A z
5 mm
B Section a – a
Internal Loadings: Consider the equilibrium of the free-body diagram of the drill’s right cut segment, Fig. a, 4 ©Fx = 0; N - 150 a b = 0 5
N = 120 N
3 ©Fy = 0; 150 a b - Vy = 0 5
Vy = 90 N T = 20 N # m
©Mx = 0; 20 - T = 0 4 3 ©Mz = 0; -150 a b (0.4) + 150 a b(0.125) + Mz = 0 5 5 Mz = 21 N # m
Section Properties: The cross-sectional area, the moment of inertia about the z axis, and the polar moment of inertia of the drill’s cross section are A = p A 0.0052 B = 25p A 10 - 6 B m2 Iz =
p A 0.0054 B = 0.15625p A 10 - 9 B m4 4
J =
p A 0.0054 B = 0.3125p A 10 - 9 B m4 2
Referring to Fig. b, QA is QA = 0 Normal Stress: The normal stress is a combination of axial and bending stress. Thus, s =
Mzy N A Iz
For point A, y = 0.005 m. Then sA =
-120
25p A 10
-6
B
21(0.005)
-
0.15625p A 10 - 9 B
= -215.43 MPa = 215 MPa (C)
558
Ans.
3
5 4
150 N
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8–36. Continued
Shear Stress: The transverse shear stress developed at point A is c A txy B V d
VyQA = A
Izt
Ans.
= 0
The torsional shear stress developed at point A is
C (txz)T D A =
20(0.005) Tc = 101.86 MPa = J 0.3125p A 10 - 9 B
Thus,
A txy B A = 0
Ans.
A txz B A = c A txz B T d = 102 MPa
Ans.
A
The state of stress at point A is represented on the element shown in Fig c.
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•8–37.
The drill is jammed in the wall and is subjected to the torque and force shown. Determine the state of stress at point B on the cross section of drill bit at section a–a.
y 400 mm a 20 N ·m x a 125 mm y
Internal Loadings: Consider the equilibrium of the free-body diagram of the drill’s right cut segment, Fig. a, 4 ©Fx = 0; N - 150 a b = 0 5
N = 120 N
3 ©Fy = 0; 150 a b - Vy = 0 5
Vy = 90 N
z
Section a – a
4 3 ©Mz = 0; -150 a b (0.4) + 150 a b (0.125) + Mz = 0 5 5 Mz = 21 N # m Section Properties: The cross-sectional area, the moment of inertia about the z axis, and the polar moment of inertia of the drill’s cross section are A = p A 0.0052 B = 25p A 10 - 6 B m2 Iz =
p A 0.0054 B = 0.15625p A 10 - 9 B m4 4
J =
p A 0.0054 B = 0.3125p A 10 - 9 B m4 2
Referring to Fig. b, QB is
QB = y¿A¿ =
4(0.005) p c A 0.0052 B d = 83.333 A 10 - 9 B m3 3p 2
Normal Stress: The normal stress is a combination of axial and bending stress. Thus, Mzy N A Iz
For point B, y = 0. Then
sB =
-120
25p A 10 - 6 B
- 0 = -1.528 MPa = 1.53 MPa(C)
560
5 mm
B
T = 20 N # m
©Mx = 0; 20 - T = 0
s =
A
Ans.
3
5 4
150 N
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8–37. Continued Shear Stress: The transverse shear stress developed at point B is c A txy B V d
90 c83.333 A 10 - 9 B d
VyQB = B
Izt
=
0.15625p A 10 - 9 B (0.01)
= 1.528 MPa
Ans.
The torsional shear stress developed at point B is c A txy B T d
= B
20(0.005) TC = 101.86 MPa = J 0.3125p A 10 - 9 B
Thus,
A tC B B = 0
Ans.
A txy B B = c A txy B T d - c A txy B V d B
B
Ans.
= 101.86 - 1.528 = 100.33 MPa = 100 MPa The state of stress at point B is represented on the element shown in Fig. d.
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8–38. Since concrete can support little or no tension, this problem can be avoided by using wires or rods to prestress the concrete once it is formed. Consider the simply supported beam shown, which has a rectangular cross section of 18 in. by 12 in. If concrete has a specific weight of 150 lb>ft3, determine the required tension in rod AB, which runs through the beam so that no tensile stress is developed in the concrete at its center section a–a. Neglect the size of the rod and any deflection of the beam.
a 16 in. B 2 in.
A a 4 ft
Support Reactions: As shown on FBD. Internal Force and Moment: + : ©Fx = 0; a + ©Mo = 0;
T - N = 0
N = T
M + T(7) - 900(24) = 0 M = 21600 - 7T
Section Properties: A = 18(12) = 216 in2 I =
1 (12) A 183 B = 5832 in4 12
Normal Stress: Requires sA = 0 sA = 0 = 0 =
N Mc + A I (21600 - 7T)(9) -T + 216 5832
T = 2160 lb = 2.16 kip
Ans.
562
4 ft
18 in. 6 in. 6 in.
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8–39. Solve Prob. 8–38 if the rod has a diameter of 0.5 in. Use the transformed area method discussed in Sec. 6.6. Est = 2911032 ksi, Ec = 3.6011032 ksi.
a 16 in. B 2 in.
A a 4 ft
Support Reactions: As shown on FBD. Section Properties: n =
29(103) Est = 8.0556 = Econ 3.6(103)
p Acon = (n - 1)Aat = (8.0556 - 1) a b A 0.52 B = 1.3854 in2 4 A = 18(12) + 1.3854 = 217.3854 in2 y =
©yA 9(18)(12) + 16(1.3854) = = 9.04461 in. ©A 217.3854
I =
1 (12) A 183 B + 12(18)(9.04461 - 9)2 + 1.3854(16 - 9.04461)2 12
= 5899.45 in4 Internal Force and Moment: + : ©Fx = 0;
T - N = 0
a + ©Mo = 0;
M + T(6.9554) - 900(24) = 0
N = T
M = 21600 - 6.9554T Normal Stress: Requires sA = 0 sA = 0 =
0 =
N Mc + A I (21600 - 6.9554T)(8.9554) -T + 217.3854 5899.45
T = 2163.08 lb = 2.16 kip
Ans.
563
4 ft
18 in. 6 in. 6 in.
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*8–40. Determine the state of stress at point A when the beam is subjected to the cable force of 4 kN. Indicate the result as a differential volume element.
4 kN
250 mm G 375 mm
D 2m
0.75 m
100 mm
B 1m
a + ©MD = 0;
A
150 mm
Cy = 0.6667 kN Cx - 4 = 0
Cx = 4.00 kN
Internal Forces and Moment: + : ©Fx = 0;
4.00 - N = 0
+ c ©Fy = 0;
V - 0.6667 = 0
a + ©Mo = 0;
N = 4.00 kN V = 0.6667 kN M = 0.6667 kN # m
M - 0.6667(1) = 0
Section Properties: A = 0.24(0.15) - 0.2(0.135) = 9.00 A 10 - 3 B m2 I =
1 1 (0.15) A 0.243 B (0.135) A 0.23 B = 82.8 A 10 - 6 B m4 12 12
QA = ©y¿A¿ = 0.11(0.15)(0.02) + 0.05(0.1)(0.015) = 0.405 A 10 - 3 B m3 Normal Stress: s = sA =
My N ; A I 4.00(103) -3
0.6667(103)(0) +
9.00(10 )
82.8(10 - 6)
= 0.444 MPa (T)
Ans.
Shear Stress: Applying shear formula.
tA =
=
200 mm
B
4(0.625) - Cy (3.75) = 0
+ : ©Fx = 0;
C
20 mm
15 mm
Support Reactions:
A
VQA It
0.6667(103) C 0.405(10 - 3) D 82.8(10 - 6)(0.015)
= 0.217 MPa
Ans.
564
20 mm
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•8–41.
Determine the state of stress at point B when the beam is subjected to the cable force of 4 kN. Indicate the result as a differential volume element.
4 kN
250 mm G 375 mm
D 2m
0.75 m
100 mm
B 1m
a + ©MD = 0;
A
150 mm
4(0.625) - Cy (3.75) = 0
Cx - 4 = 0
Cx = 4.00 kN
Internal Forces and Moment: + : ©Fx = 0;
4.00 - N = 0
+ c ©Fy = 0;
V - 0.6667 = 0
a + ©Mo = 0;
N = 4.00 kN V = 0.6667 kN M = 0.6667 kN # m
M - 0.6667(1) = 0
Section Properties: A = 0.24(0.15) - 0.2(0.135) = 9.00 A 10 - 3 B m2 I =
200 mm
B
Cy = 0.6667 kN + : ©Fx = 0;
C
20 mm
15 mm
Support Reactions:
A
1 1 (0.15) A 0.243 B (0.135) A 0.23 B = 82.8 A 10 - 6 B m 12 12
QB = 0 Normal Stress: s = sB =
My N ; A I 4.00(103) 9.00(10 - 3)
0.6667(103)(0.12) -
82.8(10 - 6)
= -0.522 MPa = 0.522 MPa (C)
Ans.
Shear Stress: Since QB = 0, then tB = 0
Ans.
565
20 mm
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8–42. The bar has a diameter of 80 mm. Determine the stress components that act at point A and show the results on a volume element located at this point. 200 mm 300 mm B A
©Fy = 0;
3 Vy - 5a b = 0 5
Vy = 3 kN
©Fz = 0;
4 Vz + 5 a b = 0 5
Vz = -4 kN
4
5 kN
©My = 0;
4 My + 5a b(0.3) = 0 5
My = -1.2 kN # m
©Mz = 0;
3 Mz + 5 a b(0.3) = 0 5
Mz = -0.9 kN # m
Iy = It =
p (0.044) = 0.64(10 - 6)p m4 4
Referring to Fig. b, (Qy)A = 0
(Qz)A = z¿A¿ =
4(0.04) p c (0.042) d = 42.67(10 - 6) m3 3p 2
The normal stress is contributed by bending stress only. Thus,
s = -
Myz
Mzy Iz
+
Iy
For point A, y = -0.04 m and z = 0. Then
s = -
-0.9(103)(-0.04) 0.64(10 - 6)p
+ 0 = -17.90(106)Pa = 17.9 MPa (C)
Ans.
The transverse shear stress developed at point A is
A txy B v =
Vy(Qy)A
A txz B v =
Vz(Qz)A
Iz t
Iy t
= 0
=
5
3
Consider the equilibrium of the FBD of bar’s left cut segment shown in Fig. a,
Ans.
4(103) C 42.67(10 - 6) D 0.64(10 - 6)p (0.08)
= 1.061(106) Pa = 1.06 MPa
Ans.
The state of stress for point A can be represented by the volume element shown in Fig. c, 566
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8–42. Continued
567
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8–43. The bar has a diameter of 80 mm. Determine the stress components that act at point B and show the results on a volume element located at this point. 200 mm 300 mm B A
Consider the equilibrium of the FBD of bar’s left cut segment shown in Fig. a,
5
3 4
©Fy = 0;
3 Vy - 5a b = 0 5
Vy = 3 kN
©Fz = 0;
4 Vz + 5 a b = 0 5
Vz = -4 kN
5 kN
©My = 0;
4 My + 5a b(0.3) = 0 5
My = -1.2 kN # m
©Mz = 0;
3 Mz + 5 a b (0.3) = 0 5
Mz = -0.9 kN # m
Iy = Iz =
p (0.044) = 0.64(10 - 6)p m4 4
Referring to Fig. b,
A Qy B B = y¿A¿ = c
4(0.04) p d c (0.042) d = 42.67(10 - 6) m3 3p 2
A Qz B B = 0 The normal stress is contributed by bending stress only. Thus, s = -
Myz
Mzy Iz
+
Iy
For point B, y = 0 and z = 0.04 m. Then
s = -0 +
-1.2(103)(0.04) 0.64(10 - 6)p
= -23.87(106) Pa = 23.9 MPa (C)
Ans.
The transverse shear stress developed at point B is
A txy B v =
Vy(Qy)B Iz t
=
3(103) C 42.67(10 - 6) D 0.64(10 - 6)p (0.08)
= 0.7958(106) MPa = 0.796 MPa
568
Ans.
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8–43. Continued
A txz B v =
Vz (Qz)B Iy t
= 0
Ans.
The state of stress for point B can be represented by the volume element shown in Fig. c
569
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*8–44. Determine the normal stress developed at points A and B. Neglect the weight of the block.
6 kip 3 in.
Referring to Fig. a,
12 kip
6 in.
©Fx = (FR)x;
-6 - 12 = F
©My = (MR)y;
6(1.5) - 12(1.5) = My
©Mz = (MR)z;
12(3) - 6(3) = Mz
a
F = -18.0 kip
A
My = -9.00 kip # in Mz = 18.0 kip # in
The cross-sectional area and moment of inertia about the y and z axes of the crosssection are A = 6(3) = 18 in2 Iy =
1 (6)(3)3 = 13.5 in4 12
Iz =
1 (3)(63) = 54.0 in4 12
The normal stress developed is the combination of axial and bending stress. Thus, s =
My z Mz y F + A Iz Iy
For point A, y = 3 in. and z = -1.5 in. sA =
18.0(3) -9.00(-1.5) -18.0 + 18.0 54.0 13.5 Ans.
= -1.00 ksi = 1.00 ksi (C) For point B, y = 3 in and z = 1.5 in.
sB =
18.0(3) -9.00(1.5) -18.0 + 18.0 54 13.5
= -3.00 ksi = 3.00 ksi (C)
Ans.
570
B a
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•8–45.
Sketch the normal stress distribution acting over the cross section at section a–a. Neglect the weight of the block.
6 kip 3 in.
12 kip
6 in. a A
B a
Referring to Fig. a, ©Fx = (FR)x;
-6 - 12 = F
F = -18.0 kip
©My = (MR)y;
6(1.5) - 12(1.5) = My
©Mz = (MR)z;
12(3) - 6(3) = Mz
My = -9.00 kip # in Mz = 18.0 kip # in
The cross-sectional area and the moment of inertia about the y and z axes of the cross-section are A = 3 (6) = 18.0 in2 Iy =
1 (6)(33) = 13.5 in4 12
Iz =
1 (3)(63) = 54.0 in4 12
The normal stress developed is the combination of axial and bending stress. Thus, s =
Myz Mzy F + A Iz Iy
For point A, y = 3 in. and z = -1.5 in. sA =
18.0(3) -9.00(-1.5) -18.0 + 18.0 54.0 13.5
= -1.00 ksi = 1.00 ksi (C) For point B, y = 3 in. and z = 1.5 in. sB =
18.0(3) -9.00(1.5) -18.0 + 18.0 54.0 13.5
= -3.00 ksi = 3.00 ksi (C)
571
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8–45. Continued
For point C, y = -3 in. and z = 1.5 in. sC =
18.0(-3) -9.00(1.5) -18.0 + 18.0 54.0 13.5
= -1.00 ksi = 1.00 ksi (C) For point D, y = -3 in. and z = -1.5 in. sD =
18.0(-3) -9.00(-1.5) -18.0 + 18.0 54.0 13.5
= 1.00 ksi (T) The normal stress distribution over the cross-section is shown in Fig. b
572
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8–46. The support is subjected to the compressive load P. Determine the absolute maximum and minimum normal stress acting in the material.
a — a 2 — 2 a a — 2 — 2
Section Properties: w = a +x A = a(a + x) I =
a 1 (a) (a + x)3 = (a + x)3 12 12
Internal Forces and Moment: As shown on FBD. Normal Stress: s =
=
N Mc ; A I
0.5Px C 12 (a + x) D -P ; a 3 a(a + x) 12 (a + x) =
P 3x -1 ; B R a a+x (a + x)2
P 1 3x + B R a a+x (a + x)2 P 4x + a = - B R a (a + x)2 P -1 3x sB = + B R a a+x (a + x)2
sA = -
=
[1]
P 2x - a B R a (a + x)2
In order to have maximum normal stress,
[2]
dsA = 0. dx
dsA P (a + x)2(4) - (4x + a)(2)(a + x)(1) = - B R = 0 a dx (a + x)4
-
Since
P (2a - 4x) = 0 a(a + x)3
P Z 0, then a(a + x)3 2a - 4x = 0
x = 0.500a
573
P
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8–46. Continued
Substituting the result into Eq. [1] yields smax = -
= -
P 4(0.500a) + a B R a (a + 0.5a)2 1.33P 1.33P = (C) 2 a a2
In order to have minimum normal stress,
Ans.
dsB = 0. dx
dsB P (a + x)2 (2) - (2x - a)(2)(a + x)(1) = B R = 0 a dx (a + x)4 P (4a - 2x) = 0 a(a + x)3 Since
P Z 0, then a(a + x)3 4a - 2x = 0
x = 2a
Substituting the result into Eq. [2] yields smin =
P 2(2a) - a P B R = 2 (T) a (a + 2a)2 3a
Ans.
574
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8–47. The support is subjected to the compressive load P. Determine the maximum and minimum normal stress acting in the material. All horizontal cross sections are circular.
P r
Section Properties: d¿ = 2r + x A = p(r + 0.5x)2 p (r + 0.5x)4 4
I =
Internal Force and Moment: As shown on FBD. Normal Stress: s =
Mc N ; A I
=
0.5Px(r + 0.5x) –P ; p 2 4 p(r + 0.5x) 4 (r + 0.5)
=
–1 P 2x ; B R p (r + 0.5x)2 (r + 0.5x)3
sA = -
= -
sB =
=
P 1 2x + B R p (r + 0.5x)2 (r + 0.5x)3 P r + 2.5x B R p (r + 0.5x)3
[1]
P –1 2x + B R p (r + 0.5x)2 (r + 0.5x)3 P 1.5x - r B R p (r + 0.5x)3
In order to have maximum normal stress,
[2] dsA = 0. dx
dsA P (r + 0.5x)3 (2.5) - (r + 2.5x)(3)(r + 0.5x)2 (0.5) = B R = 0 p dx (r + 0.5x)6 -
Since
P (r - 2.5x) = 0 p(r + 0.5x)4
P Z 0, then p(r + 0.5x)4 r - 2.5x = 0
x = 0.400r
Substituting the result into Eq. [1] yields smax = -
= -
P r + 2.5(0.400r) B R p [r + 0.5(0.400r)]3 0.368P 0.368P = (C) r2 r2
Ans.
575
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8–47. Continued
In order to have minimum normal stress,
dsB = 0. dx
dsB P (r + 0.5x)3 (1.5) - (1.5x - r)(3)(r + 0.5x)2 (0.5) = B R = 0 p dx (r + 0.5x)6 P (3r - 1.5x) = 0 p(r + 0.5x)4 Since
P Z 0, then p(r + 0.5x)4 x = 2.00r
3r - 1.5x = 0 Substituting the result into Eq. [2] yields smin =
P 1.5(2.00r) - r 0.0796P (T) B R = p [r + 0.5(2.00r)]3 r2
Ans.
*8–48. The post has a circular cross section of radius c. Determine the maximum radius e at which the load can be applied so that no part of the post experiences a tensile stress. Neglect the weight of the post.
P c e
Require sA = 0 sA = 0 =
P Mc + ; A I e =
0 =
(Pe)c -P + p 4 2 pc 4c
c 4
Ans.
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•8–49.
If the baby has a mass of 5 kg and his center of mass is at G, determine the normal stress at points A and B on the cross section of the rod at section a–a. There are two rods, one on each side of the cradle.
500 mm 15⬚ G a
75 mm a
6 mm A
B
Section a–a
Section Properties: The location of the neutral surface from the center of curvature of the rod, Fig. a, can be determined from A
R = ©
dA LA r
where A = p A 0.0062 B = 36p A 10 - 6 B m2 ©
dA = 2p ¢ r - 2r2 - c2 ≤ = 2p ¢ 0.081 - 20.0812 - 0.0062 ≤ = 1.398184 A 10 - 3 B m LA r
Thus, R =
36p A 10 - 6 B
1.398184 A 10 - 3 B
= 0.080889 m
Then e = r - R = 0.081 - 0.080889 = 0.111264 A 10 - 3 B m Internal Loadings: Consider the equilibrium of the free-body diagram of the cradle’s upper segment, Fig. b, + c ©Fy = 0;
-5(9.81) - 2N = 0
N = -24.525 N
a + ©MO = 0;
5(9.81)(0.5+ 0.080889) - 2M = 0
M = 14.2463 N # m
Normal Stress: The normal stress is the combination of axial and bending stress. Thus, s =
M(R - r) N + A Aer
Here, M = -14.1747 (negative) since it tends to increase the curvature of the rod. For point A, r = rA = 0.075 m. Then, sA =
-24.525
36p A 10
-6
B
-14.2463(0.080889 - 0.075)
+
36p A 10 - 6 B (0.111264) A 10 - 3 B (0.075)
= -89.1 MPa = 89.1 MPa (C)
Ans.
For point B, r = rB = 0.087 m. Then, sB =
-24.525
36p A 10
-6
B
-14.2463(0.080889 - 0.087)
+
36p A 10 - 6 B (0.111264) A 10 - 3 B (0.087) Ans.
= 79.3 kPa (T) dA 5 = 0.25 ln = 0.055786 4 LA r
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8–50. The C-clamp applies a compressive stress on the cylindrical block of 80 psi. Determine the maximum normal stress developed in the clamp.
1 in.
4 in.
4.5 in.
0.75 in.
R =
A dA 1 r
=
1(0.25) = 4.48142 0.055786
P = sbA = 80p (0.375)2 = 35.3429 lb M = 35.3429(8.98142) = 317.4205 lb # in. s =
M(R - r) P + Ar(r - R) A
(st)max =
317.4205(4.48142 - 4) 35.3429 + = 8.37 ksi (1)(0.25)(4)(4.5 - 4.48142) (1)(0.25)
(sc)max =
317.4205(4.48142 - 5) 35.3429 + = -6.95 ksi 1(0.25)(5)(4.5 - 4.48142) (1)(0.25)
578
Ans.
0.25 in.
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8–51. A post having the dimensions shown is subjected to the bearing load P. Specify the region to which this load can be applied without causing tensile stress to be developed at points A, B, C, and D.
x z a
a A
B
Equivalent Force System: As shown on FBD. Section Properties: 1 A = 2a(2a) + 2 B (2a)a R = 6a2 2 Iz =
1 1 1 a 2 (2a)(2a)3 + 2 B (2a) a3 + (2a) aa a + b R 12 36 2 3
= 5a4 Iy =
=
1 1 1 a 2 (2a)(2a)3 + 2 B (2a) a3 + (2a) aa b R 12 36 2 3 5 4 a 3
Normal Stress: s =
My z Mzy N + A Iz Iy
=
Peyy Pez z -P + 5 2 4 4 6a 5a 3a
=
P A -5a2 - 6eyy + 18ez z B 30a4
At point A where y = -a and z = a, we require sA 6 0. 0 7
P C -5a2 - 6(-a) ey + 18(a) ez D 30a4
0 7 -5a + 6ey + 18ez 6ey + 18ez 6 5a When
ez = 0,
When
ey = 0,
Ans.
5 a 6 5 ez 6 a 18
ey 6
Repeat the same procedures for point B, C and D. The region where P can be applied without creating tensile stress at points A, B, C and D is shown shaded in the diagram.
579
a P
a
D ez
ey C
a a
y
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*8–52. The hook is used to lift the force of 600 lb. Determine the maximum tensile and compressive stresses at section a–a. The cross section is circular and has a diameter of 1 in. Use the curved-beam formula to compute the bending stress.
300 lb
a
300 lb
2.5 in. a 1.5 in.
Section Properties: r = 1.5 + 0.5 = 2.00 in. dA = 2p A r - 2r2 - c2 B LA r
600 lb
= 2p A 2.00 - 22.002 - 0.52 B = 0.399035 in. A = p A 0.52 B = 0.25p in2 A
R =
dA 1A r
=
0.25p = 1.968246 in. 0.399035
r - R = 2 - 1.968246 = 0.031754 in. Internal Force and Moment: As shown on FBD. The internal moment must be computed about the neutral axis. M = 1180.95 lb # in. is positive since it tends to increase the beam’s radius of curvature. Normal Stress: Applying the curved-beam formula. For tensile stress (st)max =
=
M(R - r1) N + A Ar1 (r - R) 1180.95(1.968246 - 1.5) 600 + 0.25p 0.25p(1.5)(0.031754)
= 15546 psi = 15.5 ksi (T)
Ans.
For compressive stress, (sc)max =
=
M(R - r2) N + A Ar2 (r - R) 1180.95(1.968246 - 2.5) 600 + 0.25p 0.25p(2.5)(0.031754)
= -9308 psi = 9.31 ksi (C)
Ans.
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•8–53.
The masonry pier is subjected to the 800-kN load. Determine the equation of the line y = f1x2 along which the load can be placed without causing a tensile stress in the pier. Neglect the weight of the pier.
800 kN 1.5 m y 1.5 m
2.25 m y 2.25 m
x
x
C A B
A = 3(4.5) = 13.5 m2 Ix =
1 (3)(4.53) = 22.78125 m4 12
Iy =
1 (4.5)(33) = 10.125 m4 12
Normal Stress: Require sA = 0 sA =
0 =
Myx Mxy P + + A Ix Iy
800(103)y(2.25) 800(103)x(1.5) -800(103) + + 13.5 22.78125 10.125
0 = 0.148x + 0.0988y - 0.0741 y = 0.75 - 1.5 x
Ans.
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8–54. The masonry pier is subjected to the 800-kN load. If x = 0.25 m and y = 0.5 m, determine the normal stress at each corner A, B, C, D (not shown) and plot the stress distribution over the cross section. Neglect the weight of the pier.
800 kN 1.5 m y 1.5 m
2.25 m y 2.25 m
x
x
A = 3(4.5) = 13.5 m2 Ix =
1 (3)(4.53) = 22.78125 m4 12 C
1 Iy = (4.5)(33) = 10.125 m4 12 s =
sA =
A B
Myx Mxy P + + A Ix Iy
-800(103) 400(103)(2.25) 200(103)(1.5) + + 13.5 22.78125 10.125 Ans.
= 9.88 kPa (T) 3
sB =
3
3
400(10 )(2.25) 200(10 )(1.5) -800(10 ) + 13.5 22.78125 10.125 Ans.
= -49.4 kPa = 49.4 kPa (C) sC =
-800(103) 400(103)(2.25) 200(103)(1.5) + 13.5 22.78125 10.125 Ans.
= -128 kPa = 128 kPa (C) sD =
-800(103) 400(103)(2.25) 200(103)(1.5) + 13.5 22.78125 10.125
= -69.1 kPa = 69.1 kPa (C)
Ans.
582
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8–55. The bar has a diameter of 40 mm. If it is subjected to the two force components at its end as shown, determine the state of stress at point A and show the results on a differential volume element located at this point.
x 100 mm 150 mm z A
B
500 N
y 300 N
Internal Forces and Moment: ©Fx = 0;
Nx = 0
©Fy = 0;
Vy + 300 = 0
Vy = -300 N
©Fz = 0;
Vz - 500 = 0
Vz = 500 N
©Mx = 0;
Tx = 0
©My = 0;
My - 500(0.15) = 0
My = 75.0 N # m
©Mz = 0;
Mz - 300(0.15) = 0
Mz = 45.0 N # m
Section Properties: A = p A 0.022 B = 0.400 A 10 - 3 B p m2 p A 0.024 B = 40.0 A 10 - 9 B p m4 4
Ix = Iy =
J =
p A 0.024 B = 80.0 A 10 - 9 B p m4 2
(QA)z = 0 4(0.02) 1 c p A 0.022 B d = 5.333 A 10 - 6 B m3 3p 2
(QA)y = Normal Stress: s =
Myz Mzy N + A Iz Iy
sA = 0 -
75.0(0.02)
45.0(0) -9
+
40.0(10 )p
40.0(10 - 9)p
= 11.9 MPa (T)
Ans.
Shear Stress: The tranverse shear stress in the z and y directions can be obtained VQ using the shear formula, tV = . It (txy)A = -tVy = -
300 C 5.333(10 - 6) D
40.0(10 - 9)p (0.04) Ans.
= -0.318 MPa (txz)A = tVz = 0
Ans.
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*8–56.
Solve Prob. 8–55 for point B. x 100 mm 150 mm z A
B
500 N
y 300 N
Internal Forces and Moment: ©Fx = 0;
Nx = 0
©Fy = 0;
Vy + 300 = 0
Vy = -300 N
©Fz = 0;
Vz - 500 = 0
Vz = 500 N
©Mx = 0;
Tx = 0
©My = 0;
My - 500(0.15) = 0
My = 75.0 N # m
©Mz = 0;
Mz - 300(0.15) = 0
Mz = 45.0 N # m
Section Properties: A = p A 0.022 B = 0.400 A 10 - 3 B p m2 Ix = Iy =
J =
p A 0.024 B = 40.0 A 10 - 9 B p m4 4
p A 0.024 B = 80.0 A 10 - 9 B p m4 2
(QB)y = 0 (QB)z =
4(0.02) 1 c p A 0.022 B d = 5.333 A 10 - 6 B m3 3p 2
Normal Strees: s =
Myz Mzy N + A Iz Iy
sB = 0 -
75.0(0)
45.0(0.02) -9
40.0(10 ) p
+
40.0(10 - 9) p
= -7.16 MPa = 7.16 MPa (C)
Ans.
Shear Stress: The tranverse shear stress in the z and y directions can be obtained VQ using the shear formula, tV = . It (txz)B = tVz =
500 C 5.333(10 - 6) D
40.0(10 - 9) p (0.04) Ans.
= 0.531 MPa (txy)B = tVy = 0
Ans.
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•8–57. The 2-in.-diameter rod is subjected to the loads shown. Determine the state of stress at point A, and show the results on a differential element located at this point.
z B x
A
8 in.
y
600 lb 12 in.
Consider the equilibrium of the FBD of the right cut segment, Fig. a, ©Fy = 0 ;
Ny + 800 = 0
©Fz = 0 ;
Vz + 600 = 0
Vz = -600 lb
©Fx = 0 ;
Vx - 500 = 0
Vx = 500 lb
©My = 0 ;
Ty - 600(12) = 0
©Mz = 0 ;
Mz + 800(12) + 500(8) = 0
©Mx = 0 ;
Mx + 600(8) = 0
J =
Ny = -800 lb
p 4 p (1 ) = in4 4 4
Ix = Iz =
500 lb 800 lb
Ty = 7200 lb # in Mz = -13600 lb # in
Mx = -4800 lb # in A = p(12) = p in2
p 4 p (1 ) = in4 2 2
Referring to Fig. b, (Qz)A = y¿A¿ =
(Qx)A = 0
4(1) p 2 c (1 ) d = 0.6667 in3 3p 2
The normal stress is contributed by axial and bending stress. Thus, s =
Mzx Mxz N + A Ix Iz
For point A, z = 0 and x = 1 in. 4800(0) -13600(1) 800 p p>4 p>4
s =
= 17.57(103) psi = 17.6 ksi (T)
Ans.
The torsional shear stress developed at point A is (tyz)T =
TyC J
=
7200(1) = 4.584(103) psi = 4.584 ksi T p>2
The transverse shear stress developed at point A is. (tyz)g =
(txy)g =
Vz(Qz)A Ixt
=
600(0.6667) = 254.64 psi = 0.2546 ksi T p (2) 4
Vx(Qx)A 500(0) = = 0 p Izt (2) 4 585
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8–57. Continued Combining these two shear stress components, tyz = A tyz B T + A tyz B g = 4.584 + 0.2546 Ans.
= 4.838 ksi = 4.84 ksi txy = 0
Ans.
The state of stress of point A can be represented by the volume element shown in Fig. c.
586
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8–58. The 2-in.-diameter rod is subjected to the loads shown. Determine the state of stress at point B, and show the results on a differential element located at this point.
z B x
A
8 in.
y
600 lb 12 in.
Consider the equilibrium of the FBD of the right cut segment, Fig. a, ©Fy = 0;
Ny + 800 = 0
Ny = -800 lb
©Fz = 0;
Vz + 600 = 0
Vz = -600 lb
©Fx = 0;
Vx - 500 = 0
Vx = 500 lb
©My = 0;
Ty - 600(12) = 0
©Mz = 0;
Mz + 800(12) + 500(8) = 0 Mz = -13600 lb # in
©Mx = 0;
Mx + 600(8) = 0
800 lb
Ty = 7200 lb # in
Mx = -4800 lb # in.
The cross-sectional area the moment of inertia about x and Z axes and polar moment of inertia of the rod are ¿
A = p(12) = p in2
Ix = Iz =
p 4 p (1 ) = in4 4 4
J =
p 4 p (1 ) = in4 2 2
Referring to Fig. b, (Qz)B = 0 (Qx)B = z¿A¿ =
4(1) p 2 c (1 ) d = 0.6667 in4 3p 2
The normal stress is contributed by axial and bending stress. Thus, s =
Mzx Mxz N + A Ix Iz
For point B, x = 0 and z = 1 in. s =
4800 (1) 13600 (0) 800 + p p>4 p>4
= 5.86 ksi (C) The torsional shear stress developed at point B is (txy)T =
TyC J
=
7200(1) = 4.584(103) psi = 4.584 ksi : p>2
The transverse shear stress developed at point B is. (txy)v =
(tyz)v =
Vx(Qx)B 500 (0.6667) = = 212.21 psi = 0.2122 ksi : p Izt (2) 4 Vz(Qz)B Ixt
=
500 lb
600 (0) = 0 p (2) 4
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8–58.
Continued
Combining these two shear stress components, txy = (txy)T + (txy)v = 4.584 + 0.2122 = 4.796 ksi = 4.80 ksi
Ans.
tyz = 0
Ans.
The state of stress of point B can be represented by the volume element shown in Fig. c.
588
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8–59. If P = 60 kN, determine the maximum normal stress developed on the cross section of the column.
2P
150 mm 150 mm
Equivalent Force System: Referring to Fig. a, + c ©Fx = A FR B x;
F = 180 kN
-60 - 120 = -F
15 mm
©My = (MR)y;
-60(0.075) = -My
My = 4.5kN # m
©Mz = (MR)z;
-120(0.25) = -Mz
Mz = 30kN # m
100 mm 100 mm
Section Properties: The cross-sectional area and the moment of inertia about the y and z axes of the cross section are A = 0.2(0.3) - 0.185(0.27) = 0.01005 m2 Iz =
1 1 (0.2) A 0.33 B (0.185) A 0.273 B = 0.14655 A 10 - 3 B m4 12 12
Iy = 2c
1 1 (0.015) A 0.23 B d + (0.27) A 0.0153 B = 20.0759 A 10 - 6 B m4 12 12
Normal Stress: The normal stress is the combination of axial and bending stress. Here, F is negative since it is a compressive force. Also, My and Mz are negative since they are directed towards the negative sense of their respective axes. By inspection, point A is subjected to a maximum normal stress. Thus, s =
Myz Mzy N + A Iz Iy
smax = sA =
-180 A 103 B 0.01005
-
C -30 A 103 B D ( -0.15) 0.14655 A 10 - 3 B
15 mm 15 mm P
+
C -4.5 A 103 B D (0.1) 20.0759 A 10 - 6 B
= -71.0 MPa = 71.0 MPa(C)
Ans.
589
75 mm
100 mm
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*8–60. Determine the maximum allowable force P, if the column is made from material having an allowable normal stress of sallow = 100 MPa .
2P
150 mm 150 mm
Equivalent Force System: Referring to Fig. a, + c ©Fx = (FR)x;
15 mm
-P - 2P = -F
100 mm 100 mm
F = 3P ©My = (MR)y;
-P(0.075) = -My My = 0.075 P
©Mz = (MR)z;
-2P(0.25) = -Mz Mz = 0.5P
Section Properties: The cross-sectional area and the moment of inertia about the y and z axes of the cross section are A = 0.2(0.3) - 0.185(0.27) = 0.01005 m2 Iz =
1 1 (0.2) A 0.33 B (0.185) A 0.273 B = 0.14655 A 10 - 3 B m4 12 12
Iy = 2 c
1 1 (0.15) A 0.23 B d + (0.27) A 0.0153 B = 20.0759 A 10 - 6 B m4 12 12
Normal Stress: The normal stress is the combination of axial and bending stress. Here, F is negative since it is a compressive force. Also, My and Mz are negative since they are directed towards the negative sense of their respective axes. By inspection, point A is subjected to a maximum normal stress, which is in compression. Thus, s =
Myz Mzy N + A Iz Iy
-100 A 106 B = -
15 mm 15 mm P
(-0.5P)(-0.15) -0.075P(0.1) 3P + 0.01005 0.14655 A 10 - 3 B 20.0759 A 10 - 6 B
P = 84470.40 N = 84.5 kN
Ans.
590
75 mm
100 mm
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•8–61.
The beveled gear is subjected to the loads shown. Determine the stress components acting on the shaft at point A, and show the results on a volume element located at this point. The shaft has a diameter of 1 in. and is fixed to the wall at C.
z y A 200 lb
C B
x 8 in.
©Fx = 0;
Vx - 125 = 0;
©Fy = 0;
75 - Ny = 0;
Ny = 75 lb
©Fz = 0;
Vz - 200 = 0;
Vz = 200 lb
3 in.
Vx = 125 lb
75 lb
Mx = 1600 lb # in.
©Mx = 0;
200(8) - Mx = 0;
©My = 0;
200(3) - Ty = 0;
©Mz = 0;
Mz + 75(3) - 125(8) = 0;
Ty = 600 lb # in. Mz = 775 lb # in.
A = p(0.52) = 0.7854 in2 J =
p (0.54) = 0.098175 in4 2
I =
p (0.54) = 0.049087 in4 4
(QA)x = 0 (QA)z =
4(0.5) 1 a b (p)(0.52) = 0.08333 in3 3p 2
(sA)y = -
= -
Ny A
+
Mx c I
1600(0.5) 75 + 0.7854 0.049087
= 16202 psi = 16.2 ksi (T)
Ans.
(tA)yx = (tA)V - (tA)twist Tyc
Vx(QA)z =
=
It
-
J
600(0.5) 125(0.08333) 0.049087 (1) 0.098175
= -2843 psi = -2.84 ksi (tA)yz =
Vz(QA)x It
Ans.
= 0
Ans.
591
125 lb
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8–62. The beveled gear is subjected to the loads shown. Determine the stress components acting on the shaft at point B, and show the results on a volume element located at this point. The shaft has a diameter of 1 in. and is fixed to the wall at C.
z y A 200 lb
C B
x 8 in. 3 in. 75 lb
©Fx = 0;
Vx - 125 = 0;
Vx = 125 lb
©Fy = 0;
75 - Ny = 0;
Ny = 75 lb
©Fz = 0;
Vz - 200 = 0;
Vz = 200 lb Mx = 1600 lb # in.
©Mx = 0;
200(8) - Mx = 0;
©My = 0;
200(3) - Ty = 0;
©Mz = 0;
Mz + 75(3) - 125(8) = 0;
Ty = 600 lb # in. Mz = 775 lb # in.
A = p(0.52) = 0.7854 in2 J =
p (0.54) = 0.098175 in4 2
I =
p (0.54) = 0.049087 in4 4
(QB)z = 0 (QB)x =
4(0.5) 1 a b(p)(0.52) = 0.08333 in3 3p 2
(sB)y = -
= -
Ny A
Mz c +
I
775(0.5) 75 + 0.7854 0.049087
= 7.80 ksi (T)
Ans.
(tB)yz = (tB)V + (tB)twist Ty c
Vz(QB)x =
=
It
+
J
600(0.5) 200(0.08333) + 0.049087 (1) 0.098175 Ans.
= 3395 psi = 3.40 ksi (tB)yx =
Vx (QB)z It
Ans.
= 0
592
125 lb
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8–63. The uniform sign has a weight of 1500 lb and is supported by the pipe AB, which has an inner radius of 2.75 in. and an outer radius of 3.00 in. If the face of the sign is subjected to a uniform wind pressure of p = 150 lb>ft2, determine the state of stress at points C and D. Show the results on a differential volume element located at each of these points. Neglect the thickness of the sign, and assume that it is supported along the outside edge of the pipe.
12 ft B 150 lb/ft2 6 ft
F
E
3 ft
D C
A z
y x
Section Properties: A = p A 32 - 2.752 B = 1.4375p in2 Iy = Iz =
p 4 A 3 - 2.754 B = 18.6992 in4 4
(QC)z = (QD)y = 0 4(3) 1 4(2.75) 1 c (p) A 32 B d c (p) A 2.752 B d 3p 2 3p 2
(QC)y = (QD)z =
= 4.13542 in3 J =
p 4 A 3 - 2.754 B = 37.3984 in4 2
Normal Stress: s =
sC =
My z Mz y N + A Iz Iy (-64.8)(12)(0) 9.00(12)(2.75) -1.50 + 1.4375p 18.6992 18.6992 Ans.
= 15.6 ksi (T) sD =
(-64.8)(12)(3) 9.00(12)(0) -1.50 + 1.4375p 18.6992 18.6992 Ans.
= 124 ksi (T)
Shear Stress: The tranverse shear stress in the z and y directions and the torsional shear stress can be obtained using the shear formula and the torsion formula, VQ Tr and ttwist = , respectively. tV = It J (txz)D = ttwist =
64.8(12)(3) = 62.4 ksi 37.3984
Ans.
(txy)D = tVy = 0
Ans.
593
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8–63.
Continued
(txy)C = tVy - ttwist =
64.8(12)(2.75) 10.8(4.13542) 18.6992(2)(0.25) 37.3984 Ans.
= -52.4 ksi (txz)C = tVz = 0
Ans.
Internal Forces and Moments: As shown on FBD. ©Fx = 0;
1.50 + Nx = 0
Nx = -15.0 kip
©Fy = 0;
Vy - 10.8 = 0
Vy = 10.8 kip
©Fz = 0;
Vz = 0
©Mx = 0;
Tx - 10.8(6) = 0
Tx = 64.8 kip # ft
©My = 0;
My - 1.50(6) = 0
My = 9.00 kip # ft
©Mz = 0;
10.8(6) + Mz = 0
Mz = -64.8 kip # ft
594
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*8–64.
Solve Prob. 8–63 for points E and F. 12 ft B 150 lb/ft2 6 ft
F
E
3 ft
D C
A z
y x
Internal Forces and Moments: As shown on FBD. ©Fx = 0;
1.50 + Nx = 0
Nx = -1.50 kip
©Fy = 0;
Vy - 10.8 = 0
Vy = 10.8 kip
©Fz = 0;
Vz = 0
©Mx = 0;
Tx - 10.8(6) = 0
Tx = 64.8 kip # ft
©My = 0;
My - 1.50(6) = 0
My = 9.00 kip # ft
©Mz = 0;
10.8(6) + Mz = 0
Mz = -64.8 kip # ft
Section Properties: A = p A 32 - 2.752 B = 1.4375p in2 Iy = Iz =
p 4 A 3 - 2.754 B = 18.6992 in4 4
(QC)z = (QD)y = 0 (QC)y = (QD)z =
4(3) 1 4(2.75) 1 c (p) A 32 B d c (p) A 2.752 B d 3p 2 3p 2
= 4.13542 in3 J =
p 4 A 3 - 2.754 B = 37.3984 in4 2
Normal Stress: s =
sF =
My z Mzy N + A Iz Iy (-64.8)(12)(0) 9.00(12)(-3) -1.50 + 1.4375p 18.6992 18.6992 Ans.
= -17.7 ksi = 17.7 ksi (C) sE =
(-64.8)(12)(-3) 9.00(12)(0) -1.50 + 1.4375p 18.6992 18.6992
= -125 ksi = 125 ksi (C)
Ans.
595
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8–64.
Continued
Shear Stress: The tranverse shear stress in the z and y directions and the torsional shear stress can be obtained using the shear formula and the torsion formula, VQ Tr and ttwist = , respectively. tV = It J (txz)E = -ttwist = -
64.8(12)(3) = -62.4 ksi 37.3984
Ans.
(txy)E = tVy = 0
Ans.
(txy)F = tVy + ttwist =
64.8(12)(3) 10.8(4.13542) + 18.6992(2)(0.25) 37.3984
= 67.2 ksi
Ans.
(txy)F = tVy = 0
Ans.
596
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•8–65.
Determine the state of stress at point A on the cross section of the pipe at section a–a.
A 0.75 in. B y
50 lb
1 in. Section a–a
x a
60°
z
a
10 in.
Internal Loadings: Referring to the free - body diagram of the pipe’s right segment, Fig. a, ©Fy = 0; Vy - 50 sin 60° = 0
Vy = 43.30 lb
©Fz = 0; Vz - 50 cos 60° = 0
Vz = 25 lb T = -519.62 lb # in
©Mx = 0; T + 50 sin 60°(12) = 0 ©My = 0; My - 50 cos 60°(10) = 0
My = 250 lb # in
©Mz = 0; Mz + 50 sin 60° (10) = 0
Mz = -433.01 lb # in
Section Properties: The moment of inertia about the y and z axes and the polar moment of inertia of the pipe are Iy = Iz =
J =
p 4 A 1 - 0.754 B = 0.53689 in4 4
p 4 A 1 - 0.754 B = 1.07379 in4 2
Referring to Fig. b,
A Qy B A = 0 A Qz B A = y1œ A1œ - y2œ A2œ =
4(1) p 2 4(0.75) p c A1 B d c A 0.752 B d = 0.38542 in3 3p 2 3p 2
Normal Stress: The normal stress is contributed by bending stress only. Thus, s = -
Myz
Mzy Iz
+
Iy
For point A, y = 0.75 in and z = 0. Then sA =
-433.01(0.75) + 0 = 604.89 psi = 605 psi (T) 0.53689
Shear Stress: The torsional shear stress developed at point A is c A txz B T d
= A
TrA 519.62(0.75) = = 362.93 psi J 1.07379
597
Ans.
12 in.
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8–65.
Continued
The transverse shear stress developed at point A is c A txy B V d c A txz B V d
= 0 A
= A
Vz A Qz B A Iy t
=
25(0.38542) = 35.89 psi 0.53689(2 - 1.5)
Combining these two shear stress components,
A txy B A = 0
Ans.
A txz B A = c A txz B T d - c A txz B V d A
A
= 362.93 - 35.89 = 327 psi
Ans.
598
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8–66. Determine the state of stress at point B on the cross section of the pipe at section a–a.
A 0.75 in. B y
50 lb
1 in. Section a–a
x a
60°
z
a
10 in.
Internal Loadings: Referring to the free - body diagram of the pipe’s right segment, Fig. a, ©Fy = 0; Vy - 50 sin 60° = 0
Vy = 43.30 lb
©Fz = 0; Vz - 50 cos 60° = 0
Vz = 25 lb T = -519.62 lb # in
©Mx = 0; T + 50 sin 60°(12) = 0 ©My = 0; My - 50 cos 60°(10) = 0
My = 250 lb # in
©Mz = 0; Mz + 50 sin 60°(10) = 0
Mz = -433.01 lb # in
Section Properties: The moment of inertia about the y and z axes and the polar moment of inertia of the pipe are p 4 A 1 - 0.754 B = 0.53689 in4 4
Iy = Iz =
J =
p 4 A 1 - 0.754 B = 1.07379 in4 2
Referring to Fig. b,
A Qz B B = 0 A Qy B B = y1œ A1œ - y2œ A2œ =
4(1) p 2 4(0.75) p c A1 B d c A 0.752 B d = 0.38542 in3 3p 2 3p 2
Normal Stress: The normal stress is contributed by bending stress only. Thus, s = -
Myz
Mzy Iz
+
Iy
For point B, y = 0 and z = -1. Then sB = -0 +
250(1) = -465.64 psi = 466 psi (C) 0.53689
Ans.
Shear Stress: The torsional shear stress developed at point B is c A txy B T d
= B
519.62(1) TrC = = 483.91 psi J 1.07379
599
12 in.
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8–66.
Continued
The transverse shear stress developed at point B is c A txz B V d c A txy B V d
= 0 B
= B
Vy A Qy B B Izt
=
43.30(0.38542) = 62.17 psi 0.53689(2 - 1.5)
Combining these two shear stress components,
A txy B B = c A txy B T d - c A txy B V d B
B
Ans.
= 483.91 - 62.17 = 422 psi
A txz B B = 0
Ans.
600
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•8–67. The eccentric force P is applied at a distance ey from the centroid on the concrete support shown. Determine the range along the y axis where P can be applied on the cross section so that no tensile stress is developed in the material.
x
z
P
b 2
ey b 2
2h 3 h 3
Internal Loadings: As shown on the free - body diagram, Fig. a. Section Properties: The cross-sectional area and moment of inertia about the z axis of the triangular concrete support are A =
1 bh 2
Iz =
1 bh3 36
Normal Stress: The normal stress is the combination of axial and bending stress. Thus, s =
Mzy N A Iz
A Pey B y -P 1 1 bh bh3 2 36
s =
2P 2 A h + 18eyy B bh3
s = -
Here, it is required that sA … 0 and sB … 0. For point A, y =
(1) h , Then. Eq. (1) gives 3
2P 2 h ch + 18ey a b d 3 3 bh
0 Ú -
0 … h2 + 6hey ey Ú -
For Point B, y = -
h 6
2 h. Then. Eq. (1) gives 3 0 Ú -
2P 2 2 ch + 18ey a - hb d 3 3 bh
0 … h2 - 12hey ey …
h 12
Thus, in order that no tensile stress be developed in the concrete support, ey must be in the range of -
h h … ey … 6 12
Ans.
601
y
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*8–68. The bar has a diameter of 40 mm. If it is subjected to a force of 800 N as shown, determine the stress components that act at point A and show the results on a volume element located at this point.
150 mm 200 mm
1 1 I = p r4 = (p)(0.024) = 0.1256637 (10 - 6) m4 4 4 A
A = p r2 = p(0.022) = 1.256637 (10 - 3) m2 QA = y¿A¿ = a
z
B
4 (0.02) p (0.02)2 ba b = 5.3333 (10 - 6) m3 3p 2
y
x 30⬚
800 N
sA
P Mz = + A I 400 + 0 = 0.318 MPa 1.256637 (10 - 3)
Ans.
692.82 (5.3333) (10 - 6) VQA = 0.735 MPa = It 0.1256637 (10 - 6)(0.04)
Ans.
=
tA =
•8–69.
Solve Prob. 8–68 for point B.
150 mm 200 mm
A
z
B y
x 30⬚
800 N
1 1 I = p r4 = (p)(0.024) = 0.1256637 (10 - 6) m4 4 4 A = p r2 = p(0.022) = 1.256637 (10 - 3) m2 QB = 0 sB =
138.56 (0.02) P Mc 400 = -21.7 MPa = -3 A I 1.256637 (10 ) 0.1256637 (10 - 6)
tB = 0
Ans. Ans.
602
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8–70. The 43-in.-diameter shaft is subjected to the loading shown. Determine the stress components at point A. Sketch the results on a volume element located at this point. The journal bearing at C can exert only force components Cy and Cz on the shaft, and the thrust bearing at D can exert force components Dx , Dy , and Dz on the shaft.
D
z 125 lb 2 in.
8 in. 125 lb 2 in.
A
20 in.
8 in. B
C
10 in.
y
20 in.
x
A =
p (0.752) = 0.44179 in2 4
I =
p (0.3754) = 0.015531 in4 4
QA = 0 tA = 0 sA =
Ans.
My c I
=
-1250(0.375) = -30.2 ksi = 30.2 ksi (C) 0.015531
Ans.
8–71. Solve Prob. 8–70 for the stress components at point B.
D
z 125 lb 2 in.
8 in. 125 lb 2 in.
A
8 in. B
C
10 in. x
p A = (0.752) = 0.44179 in2 4 I =
p (0.3754) = 0.015531 in4 4
QB = y¿A¿ =
4(0.375) 1 a b(p)(0.3752) = 0.035156 in3 3p 2
sB = 0 tB =
Ans.
VzQB It
=
125(0.035156) = 0.377 ksi 0.015531(0.75)
Ans.
603
20 in.
20 in.
y
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*8–72. The hook is subjected to the force of 80 lb. Determine the state of stress at point A at section a–a. The cross section is circular and has a diameter of 0.5 in. Use the curved-beam formula to compute the bending stress.
80 lb 1.5 in. 45⬚
The location of the neutral surface from the center of curvature of the hook, Fig. a, can be determined from
where A = p(0.252) = 0.0625p in2 dA = 2p A r - 2r2 - c2 B = 2p A 1.75 - 21.752 - 0.252 B = 0.11278 in. LA r
Thus, R =
0.0625p = 1.74103 in. 0.11278
Then e = r - R = 1.75 - 1.74103 = 0.0089746 in. Referring to Fig. b, I and QA are I =
p (0.254) = 0.9765625(10 - 3)p in4 4
QA = 0 Consider the equilibrium of the FBD of the hook’s cut segment, Fig. c, + ©F = 0; ; x
N - 80 cos 45° = 0
N = 56.57 lb
+ c ©Fy = 0;
80 sin 45° - V = 0
V = 56.57 lb
a + ©Mo = 0;
M - 80 cos 45°(1.74103) = 0
M = 98.49 lb # in
The normal stress developed is the combination of axial and bending stress. Thus, s =
M(R - r) N + A Ae r
Here, M = 98.49 lb # in since it tends to reduce the curvature of the hook. For point A, r = 1.5 in. Then s =
(98.49)(1.74103 - 1.5) 56.57 + 0.0625p 0.0625p(0.0089746)(1.5)
= 9.269(103) psi = 9.27 ksi (T)
Ans.
The shear stress in contributed by the transverse shear stress only. Thus t =
VQA = 0 It
Ans.
The state of strees of point A can be represented by the element shown in Fig. d.
604
A B
B a
A R = dA © LA r
©
a A
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•8–73.
The hook is subjected to the force of 80 lb. Determine the state of stress at point B at section a–a. The cross section has a diameter of 0.5 in. Use the curved-beam formula to compute the bending stress.
80 lb 1.5 in. 45⬚
The location of the neutral surface from the center of curvature of the the hook, Fig. a, can be determined from R =
dA © LA r
dA = 2p A r - 2r2 - c2 B = 2p A 1.75 - 21.752 - 0.252 B = 0.11278 in. LA r
Thus, R =
0.0625p = 1.74103 in 0.11278
Then e = r - R = 1.75 - 1.74103 = 0.0089746 in Referring to Fig. b, I and QB are computed as p (0.254) = 0.9765625(10 - 3)p in4 4
I =
QB = y¿A¿ =
4(0.25) p c (0.252) d = 0.0104167 in3 3p 2
Consider the equilibrium of the FBD of the hook’s cut segment, Fig. c, + ©F = 0; ; x
N - 80 cos 45° = 0
+ c ©Fy = 0;
80 sin 45° - V = 0
a + ©Mo = 0;
N = 56.57 lb V = 56.57 lb
M - 80 cos 45° (1.74103) = 0
M = 98.49 lb # in
The normal stress developed is the combination of axial and bending stress. Thus, s =
M(R - r) N + A Ae r
Here, M = 98.49 lb # in since it tends to reduce. the curvature of the hook. For point B, r = 1.75 in. Then s =
(98.49)(1.74103 - 1.75) 56.57 + 0.0625p 0.0625 p (0.0089746)(1.75)
= 1.62 psi (T)
Ans.
The shear stress is contributed by the transverse shear stress only. Thus, t =
56.57 (0.0104167) VQB = 3.84 psi = It 0.9765625(10 - 3)p (0.5)
Ans.
The state of stress of point B can be represented by the element shown in Fig. d.
605
A B
B a
A
Where A = p(0.252) = 0.0625p in2 ©
a A
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8–74. The block is subjected to the three axial loads shown. Determine the normal stress developed at points A and B. Neglect the weight of the block.
100 lb 250 lb
50 lb
2 in.4 in.
5 in.
2 in.
3 in. 5 in.
A B
Mx = -250(1.5) - 100(1.5) + 50(6.5) = -200 lb # in. My = 250(4) + 50(2) - 100(4) = 700 lb # in. Ix =
1 1 (4)(133) + 2 a b (2)(33) = 741.33 in4 12 12
Iy =
1 1 (3)(83) + 2 a b (5)(43) = 181.33 in4 12 12
A = 4(13) + 2(2)(3) = 64 in2 s =
My x Mx y P + A Iy Ix
sA = -
700(4) -200 (-1.5) 400 + 64 181.33 741.33
= -21.3 psi sB = -
Ans.
700(2) -200 (-6.5) 400 + 64 181.33 741.33
= -12.2 psi
Ans.
606
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8–75. The 20-kg drum is suspended from the hook mounted on the wooden frame. Determine the state of stress at point E on the cross section of the frame at section a–a. Indicate the results on an element.
50 mm
25 mm
E 75 mm
Section a – a 0.5 m 0.5 m
1m
a B C
a 1m 30⬚
Support Reactions: Referring to the free-body diagram of member BC shown in Fig. a, a + ©MB = 0;
F sin 45°(1) - 20(9.81)(2) = 0
+ ©F = 0; : x
554.94 cos 45° - Bx = 0
Bx = 392.4 N
+ c ©Fy = 0;
554.94 sin 45° - 20(9.81) - By = 0
By = 196.2 N
1m b
F = 554.94 N
b
75 mm
1m D
F A 25 mm
Internal Loadings: Consider the equilibrium of the free - body diagram of the right segment shown in Fig. b.
Section b – b
+ ©F = 0; : x
N - 392.4 = 0
N = 392.4 N
+ c ©Fy = 0;
V - 196.2 = 0
V = 196.2 N
a + ©MC = 0;
196.2(0.5) - M = 0
M = 98.1 N # m
Section Properties: The cross -sectional area and the moment of inertia of the cross section are A = 0.05(0.075) = 3.75 A 10 - 3 B m2 I =
1 (0.05) A 0.0753 B = 1.7578 A 10 - 6 B m4 12
Referring to Fig. c, QE is
QE = y¿A¿ = 0.025(0.025)(0.05) = 3.125 A 10 - 6 B m3
Normal Stress: The normal stress is the combination of axial and bending stress. Thus, s =
My N ; A I
For point A, y = 0.0375 - 0.025 = 0.0125 m. Then sE =
392.4
3.75 A 10
-3
B
98.1(0.0125)
+
1.7578 A 10 - 6 B
= 802 kPa
Ans.
Shear Stress: The shear stress is contributed by transverse shear stress only. Thus, tE =
196.2 C 31.25 A 10 - 6 B D VQA = = 69.8 kPa It 1.7578 A 10 - 6 B (0.05)
Ans.
The state of stress at point E is represented on the element shown in Fig. d. 607
75 mm
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8–75.
Continued
608
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*8–76. The 20-kg drum is suspended from the hook mounted on the wooden frame. Determine the state of stress at point F on the cross section of the frame at section b–b. Indicate the results on an element.
50 mm
25 mm
E 75 mm
Section a – a 0.5 m 0.5 m
1m
a B C
a 1m 30⬚
1m b
FBD sin 30°(3) - 20(9.81)(2) = 0
+ c ©Fy = 0;
Ay - 261.6 cos 30° - 20(9.81) = 0
Ay = 422.75 N
+ ©F = 0; : x
Ax - 261.6 sin 30° = 0
Ax = 130.8 N
75 mm
1m
Support Reactions: Referring to the free-body diagram of the entire frame shown in Fig. a, a + ©MA = 0;
b
D
F A
FBD = 261.6 N 25 mm Section b – b
Internal Loadings: Consider the equilibrium of the free - body diagram of the lower cut segment, Fig. b, + ©F = 0; : x
130.8 - V = 0
V = 130.8 N
+ c ©Fy = 0;
422.75 - N = 0
N = 422.75 N
a + ©MC = 0;
130.8(1) - M = 0
M = 130.8 N # m
Section Properties: The cross -sectional area and the moment of inertia about the centroidal axis of the cross section are A = 0.075(0.075) = 5.625 A 10 - 3 B m2 I =
1 (0.075) A 0.0753 B = 2.6367 A 10 - 6 B m4 12
Referring to Fig. c, QE is
QF = y¿A¿ = 0.025(0.025)(0.075) = 46.875 A 10 - 6 B m3
Normal Stress: The normal stress is the combination of axial and bending stress. Thus, s =
My N ; A I
For point F, y = 0.0375 - 0.025 = 0.0125 m. Then sF =
-422.75
5.625 A 10
-3
B
130.8(0.0125) -
2.6367 A 10 - 6 B
= -695.24 kPa = 695 kPa (C)
Ans.
609
75 mm
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8–76.
Continued
Shear Stress: The shear stress is contributed by transverse shear stress only. Thus,
tA
130.8 c46.875 A 10 - 6 B d VQA = = = 31.0 kPa It 2.6367 A 10 - 6 B (0.075)
Ans.
The state of stress at point A is represented on the element shown in Fig. d.
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•8–77. The eye is subjected to the force of 50 lb. Determine the maximum tensile and compressive stresses at section a-a. The cross section is circular and has a diameter of 0.25 in. Use the curved-beam formula to compute the bending stress.
50 lb
0.25 in. 1.25 in. a
Section Properties: r = 1.25 +
0.25 = 1.375 in. 2
dA = 2p A r - 2r2 - c2 B LA r = 2p A 1.375 - 21.3752 - 0.1252 B = 0.035774 in. A = p A 0.1252 B = 0.049087 in2 R =
A dA 1A r
=
0.049087 = 1.372153 in. 0.035774
r - R = 1.375 - 1.372153 = 0.002847 in. Internal Force and Moment: As shown on FBD. The internal moment must be computed about the neutral axis. M = 68.608 lb # in is positive since it tends to increase the beam’s radius of curvature. Normal Stress: Applying the curved - beam formula, For tensile stress (st)max =
=
M(Rr1) N + A Ar1(r - R) 68.608(1.372153 - 1.25) 50.0 + 0.049087 0.049087(1.25)(0.002847)
= 48996 psi = 49.0 ksi (T)
Ans.
For compressive stress (sc)max =
=
M(R - r2) N + A Ar2(r - R) 68.608(1.372153 - 1.50) 50.0 + 0.049087 0.049087(1.50)(0.002847)
= -40826 psi = 40.8 ksi (C)
Ans.
611
a
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8–78. Solve Prob. 8–77 if the cross section is square, having dimensions of 0.25 in. by 0.25 in.
50 lb
0.25 in. 1.25 in. a
Section Properties: r = 1.25 +
0.25 = 1.375 in. 2
r2 dA 1.5 = bln = 0.25 ln = 0.45580 in. r1 1.25 LA r A = 0.25(0.25) = 0.0625 in2 R =
0.0625 A = = 1.371204 in. dA 0.045580 1A r
r - R = 1.375 - 1.371204 = 0.003796 in. Internal Force and Moment: As shown on FBD. The internal moment must be computed about the neutral axis. M = 68.560 lb # in. is positive since it tends to increase the beam’s radius of curvature. Normal Stress: Applying the curved -beam formula, For tensile stress (st)max =
=
M(R - r1) N + A Ar1(r - R) 68.560(1.371204 - 1.25) 50.0 + 0.0625 0.0625(1.25)(0.003796)
= 28818 psi = 28.8 ksi (T)
Ans.
For Compressive stress (sc)max =
=
M(R - r2) N + A Ar2 (r - R) 68.560(1.371204 - 1.5) 50.0 + 0.0625 0.0625(1.5)(0.003796)
= -24011 psi = 24.0 ksi (C)
Ans.
612
a
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8–79. If the cross section of the femur at section a–a can be approximated as a circular tube as shown, determine the maximum normal stress developed on the cross section at section a–a due to the load of 75 lb.
2 in.
75 lb
a
a
0.5 in. 1 in. Section a – a
M F
Internal Loadings: Considering the equilibrium for the free-body diagram of the femur’s upper segment, Fig. a, + c ©Fy = 0;
N - 75 = 0
N = 75 lb
a + ©MO = 0;
M - 75(2) = 0
M = 150 lb # in
Section Properties: The cross-sectional area, the moment of inertia about the centroidal axis of the femur’s cross section are A = p A 12 - 0.52 B = 0.75p in2 I =
p 4 A 1 - 0.54 B = 0.234375p in4 4
Normal Stress: The normal stress is a combination of axial and bending stress. Thus, s =
My N + A I
By inspection, the maximum normal stress is in compression. smax =
150(1) -75 = -236 psi = 236 psi (C) 0.75p 0.234375p
613
Ans.
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*8–80. The hydraulic cylinder is required to support a force of P = 100 kN. If the cylinder has an inner diameter of 100 mm and is made from a material having an allowable normal stress of sallow = 150 MPa, determine the required minimum thickness t of the wall of the cylinder.
P
t
100 mm
Equation of Equilibrium: The absolute pressure developed in the hydraulic cylinder can be determined by considering the equilibrium of the free-body diagram of the piston shown in Fig. a. The resultant force of the pressure on the p piston is F = pA = pc A 0.12 B d = 0.0025pp. Thus, 4 ©Fx¿ = 0; 0.0025pp - 100 A 103 B = 0
p = 12.732 A 106 B Pa
Normal Stress: For the cylinder, the hoop stress is twice as large as the longitudinal stress, sallow =
pr ; t
150 A 106 B =
12.732 A 106 B (50) t
t = 4.24 mm Since
Ans.
r 50 = = 11.78 7 10, thin -wall analysis is valid. t 4.24
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•8–81. The hydraulic cylinder has an inner diameter of 100 mm and wall thickness of t = 4 mm. If it is made from a material having an allowable normal stress of sallow = 150 MPa, determine the maximum allowable force P.
P
t
100 mm
Normal Stress: For the hydraulic cylinder, the hoop stress is twice as large as the longitudinal stress. Since
50 r = = 12.5 7 10, thin-wall analysis can be used. t 4
sallow =
pr ; t
150 A 106 B =
p(50) 4
p = 12 A 106 B MPa
Ans.
Equation of Equilibrium: The resultant force on the piston is F = pA = 12 A 106 B c
p A 0.12 B d = 30 A 103 B p. Referring to the free-body diagram of 4 the piston shown in Fig. a, ©Fx¿ = 0; 30 A 103 B p - P = 0
P = 94.247 A 103 B N = 94.2 kN
Ans.
615
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8–82. The screw of the clamp exerts a compressive force of 500 lb on the wood blocks. Determine the maximum normal stress developed along section a-a. The cross section there is rectangular, 0.75 in. by 0.50 in. 4 in.
Internal Force and Moment: As shown on FBD.
a
Section Properties: A = 0.5(0.75) = 0.375 in2 I =
a
1 (0.5) A 0.753 B = 0.017578 in4 12
0.75 in.
Maximum Normal Stress: Maximum normal stress occurs at point A. smax = sA =
=
Mc N + A I 2000(0.375) 500 + 0.375 0.017578
= 44000 psi = 44.0 ksi (T)
Ans.
8–83. Air pressure in the cylinder is increased by exerting forces P = 2 kN on the two pistons, each having a radius of 45 mm. If the cylinder has a wall thickness of 2 mm, determine the state of stress in the wall of the cylinder.
p =
s1 =
P
47 mm
2(103) P = 314 380.13 Pa = A p(0.0452) pr 314 380.13(0.045) = = 7.07 MPa t 0.002
Ans.
s2 = 0
P
Ans.
The pressure P is supported by the surface of the pistons in the longitudinal direction.
*8–84. Determine the maximum force P that can be exerted on each of the two pistons so that the circumferential stress component in the cylinder does not exceed 3 MPa. Each piston has a radius of 45 mm and the cylinder has a wall thickness of 2 mm.
s =
pr ; t
3(106) =
P
47 mm
p(0.045) 0.002
P = 133.3 kPa
Ans.
P = pA = 133.3 A 103 B (p)(0.045)2 = 848 N
Ans.
616
P
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•8–85.
The cap on the cylindrical tank is bolted to the tank along the flanges. The tank has an inner diameter of 1.5 m and a wall thickness of 18 mm. If the largest normal stress is not to exceed 150 MPa, determine the maximum pressure the tank can sustain. Also, compute the number of bolts required to attach the cap to the tank if each bolt has a diameter of 20 mm. The allowable stress for the bolts is 1sallow2b = 180 MPa.
Hoop Stress for Cylindrical Tank: Since
750 r = = 41.7 7 10, then thin wall t 18
analysis can be used. Applying Eq. 8–1
s1 = sallow = 150 A 106 B =
pr t p(750) 18
p = 3.60 MPa
Ans.
Force Equilibrium for the Cap: + c ©Fy = 0;
3.60 A 106 B C p A 0.752 B D - Fb = 0 Fb = 6.3617 A 106 B N
Allowable Normal Stress for Bolts: (sallow)b = 180 A 106 B =
P A 6.3617(106)
n C p4 (0.022) D
n = 112.5 Use n = 113 bolts
Ans.
617
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8–86. The cap on the cylindrical tank is bolted to the tank along the flanges. The tank has an inner diameter of 1.5 m and a wall thickness of 18 mm. If the pressure in the tank is p = 1.20 MPa, determine the force in each of the 16 bolts that are used to attach the cap to the tank. Also, specify the state of stress in the wall of the tank.
Hoop Stress for Cylindrical Tank: Since
750 r = = 41.7 7 10, then thin wall t 18
analysis can be used. Applying Eq. 8–1
s1 =
pr 1.20(106)(750) = = 50.0 MPa t 18
Ans.
Longitudinal Stress for Cylindrical Tank: s2 =
pr 1.20(106)(750) = = 25.0 MPa 2t 2(18)
Ans.
Force Equilibrium for the Cap: + c ©Fy = 0;
1.20 A 106 B C p A 0.752 B D - 16Fb = 0 Fb = 132536 N = 133 kN
Ans.
618
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9–1. Prove that the sum of the normal stresses sx + sy = sx¿ + sy¿ is constant. See Figs. 9–2a and 9–2b.
Stress Transformation Equations: Applying Eqs. 9-1 and 9-3 of the text. sx¿ + sy¿ =
sx + sy 2
sx - sy +
2
cos 2u + txy sin 2u
sx + sy +
2
sx - sy -
2
cos 2u - txy sin 2u
sx¿ + sy¿ = sx + sy
(Q.E.D.)
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9–2. The state of stress at a point in a member is shown on the element. Determine the stress components acting on the inclined plane AB. Solve the problem using the method of equilibrium described in Sec. 9.1.
A
8 ksi
2 ksi 5 ksi 60⬚ B
Referring to Fig a, if we assume that the areas of the inclined plane AB is ¢A, then the area of the horizontal and vertical of the triangular element are ¢A cos 60° and ¢A sin 60° respectively. The forces act acting on these two faces indicated on the FBD of the triangular element, Fig. b. ¢Fx¿ + 2¢A sin 60° cos 60° + 5¢ A sin 60° sin 60°
+Q©Fx¿ = 0;
+ 2¢A cos 60° sin 60° - 8¢A cos 60° cos 60° = 0 ¢Fx¿ = -3.482 ¢A ¢Fy¿ + 2¢A sin 60° sin 60° - 5¢ A sin 60° cos 60°
+a©Fy¿ = 0;
- 8¢A cos 60° sin 60° - 2¢A cos 60° cos 60° = 0 ¢Fy¿ = 4.629 ¢A From the definition, sx¿ = lim¢A:0
¢Fx¿ = -3.48 ksi ¢A
tx¿y¿ = lim¢A:0
¢Fy¿
Ans.
Ans.
= 4.63 ksi
¢A
The negative sign indicates that sx¿, is a compressive stress.
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9–3. The state of stress at a point in a member is shown on the element. Determine the stress components acting on the inclined plane AB. Solve the problem using the method of equilibrium described in Sec. 9.1.
500 psi B 60⬚
A
Referring to Fig. a, if we assume that the area of the inclined plane AB is ¢A, then the areas of the horizontal and vertical surfaces of the triangular element are ¢A sin 60° and ¢A cos 60° respectively. The force acting on these two faces are indicated on the FBD of the triangular element, Fig. b +R©Fx¿ = 0;
¢Fx¿ + 500 ¢A sin 60° sin 60° + 350¢A sin 60° cos 60° +350¢A cos 60° sin 60° = 0 ¢Fx¿ = -678.11 ¢A
+Q©Fy¿ = 0;
¢Fy¿ + 350¢A sin 60° sin 60° - 500¢A sin 60° cos 60° -350¢A cos 60° cos 60° = 0 ¢Fy¿ = 41.51 ¢A
From the definition sx¿ = lim¢A:0
tx¿y¿ = lim¢A:0
¢Fx¿ = -6.78 psi ¢A ¢Fy¿
Ans.
Ans.
= 41.5 psi
¢A
The negative sign indicates that sx¿, is a compressive stress.
621
350 psi
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*9–4. The state of stress at a point in a member is shown on the element. Determine the stress components acting on the inclined plane AB. Solve the problem using the method of equilibrium described in Sec. 9.1.
A
650 psi
¢Fx¿ - 400(¢Acos 60°)cos 60° + 650(¢ A sin 60°)cos 30° = 0
Q+ ©Fx¿ = 0
400 psi
60⬚
¢Fx¿ = -387.5¢A ¢Fy¿ - 650(¢Asin 60°)sin 30° - 400(¢ A cos 60°)sin 60° = 0
a+ ©Fy¿ = 0
B
¢Fy¿ = 455 ¢A sx¿ = lim¢A:0
sx¿y¿ = lim¢A:0
¢Fx¿ = -388 psi ¢A ¢Fy¿
Ans.
Ans.
= 455 psi
¢A
The negative sign indicates that the sense of sx¿, is opposite to that shown on FBD.
•9–5.
Solve Prob. 9–4 using the stress-transformation equations developed in Sec. 9.2. sy = 400 psi
sx = -650 psi sx¿ =
=
sx + sy
sx - sy +
2
2
txy = 0
A
400 psi
u = 30° 650 psi
cos 2u + txy sin 2u
60⬚
-650 + 400 -650 - 400 + cos 60° + 0 = -388 psi 2 2
Ans. B
The negative sign indicates sx¿, is a compressive stress. tx¿y¿ = = -a
sx - sy 2
sin 2u + txy cos 2u
-650 - 400 bsin 60° = 455 psi 2
Ans.
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9–6. The state of stress at a point in a member is shown on the element. Determine the stress components acting on the inclined plane AB. Solve the problem using the method of equilibrium described in Sec. 9.1.
90 MPa
A
35 MPa 60⬚ 30⬚
R+ ©Fy¿ = 0
B 50 MPa
¢Fy¿ - 50¢A sin 30° cos 30° - 35¢A sin 30° cos 60° + 90¢A cos 30° sin 30° + 35¢A cos 30° sin 60° = 0 ¢Fy¿ = -34.82¢A
b+ ©Fx¿ = 0
¢Fx¿ - 50¢A sin 30° sin 30° + 35¢A sin 30° sin 60° -90¢A cos 30° cos 30° + 35¢A cos 30° cos 60° = 0 ¢Fx¿ = 49.69 ¢A
sx¿ = lim¢A:0
¢Fx¿ = 49.7 MPa ¢A
tx¿y¿ = lim¢A:0
¢Fy¿
Ans.
Ans.
= -34.8 MPa
¢A
The negative signs indicate that the sense of sx¿, and tx¿y¿ are opposite to the shown on FBD.
9–7. Solve Prob. 9–6 using the stress-transformation equations developed in Sec. 9.2. Show the result on a sketch.
90 MPa
A
35 MPa 60⬚ 30⬚
sy = 50 MPa
sx = 90 MPa sx¿ =
=
sx + sy
sx - sy +
2
2
txy = -35 MPa
u = -150°
cos 2u + txy sin 2u
90 - 50 90 + 50 + cos(-300°) + (-35) sin ( -300°) 2 2
= 49.7 MPa tx¿y¿ = -
sx - sy
= -a
2
Ans. sin 2u + txy cos 2u
90 - 50 bsin(-300°) + ( -35) cos ( -300°) = -34.8 MPa 2
The negative sign indicates tx¿y¿ acts in -y¿ direction.
623
Ans.
B 50 MPa
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*9–8. Determine the normal stress and shear stress acting on the inclined plane AB. Solve the problem using the method of equilibrium described in Sec. 9.1.
45 MPa
B
80 MPa 45⬚ A
Force Equllibrium: Referring to Fig. a, if we assume that the area of the inclined plane AB is ¢A, then the area of the vertical and horizontal faces of the triangular sectioned element are ¢A sin 45° and ¢A cos 45°, respectively. The forces acting on the free-body diagram of the triangular sectioned element, Fig. b, are ©Fx¿ = 0;
¢Fx¿ + c45 A 106 B ¢A sin 45° dcos 45° + c45 A 106 B ¢A cos 45° dsin 45° - c80 A 106 B ¢A sin 45° dcos 45° = 0 ¢Fx¿ = -5 A 106 B ¢A
©Fy¿ = 0;
¢Fy¿ + c45 A 106 B ¢A cos 45° dcos 45° - c45 A 106 B ¢A sin 45° dsin 45° - c80 A 106 B ¢ A sin 45° dsin 45° = 0 ¢Fy¿ = 40 A 106 B ¢A
Normal and Shear Stress: From the definition of normal and shear stress, sx¿ = lim¢A:0
¢Fx¿ = -5 MPa ¢A
tx¿y¿ = lim¢A:0
¢Fy¿
Ans.
Ans.
= 40 MPa
¢A
The negative sign indicates that sx¿ is a compressive stress.
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•9–9.
Determine the normal stress and shear stress acting on the inclined plane AB. Solve the problem using the stress transformation equations. Show the result on the sectioned element.
45 MPa
80 MPa 45⬚
Stress Transformation Equations: u = +135° (Fig. a)
sx = 80 MPa
sy = 0
txy = 45 MPa
we obtain, sx¿ =
=
sx + sy
sx - sy +
2
2
cos u + txysin 2u
80 - 0 80 + 0 + cos 270 + 45 sin 270° 2 2 Ans.
= -5 MPa
tx¿y¿ = -
= -
sx - sy 2
B
sinu + txy cos 2u
80 - 0 sin 270° + 45 cos 270° 2
= 40 MPa
Ans.
The negative sign indicates that sx¿ is a compressive stress. These results are indicated on the triangular element shown in Fig. b.
625
A
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9–10. The state of stress at a point in a member is shown on the element. Determine the stress components acting on the inclined plane AB. Solve the problem using the method of equilibrium described in Sec. 9.1.
2 ksi
A 3 ksi 30⬚
4 ksi B
Force Equllibrium: For the sectioned element, ¢Fy¿ - 3(¢A sin 30°) sin 60° + 4(¢ A sin 30°)sin 30°
a+ ©Fy¿ = 0;
-2(¢A cos 30°) sin 30° - 4(¢A cos 30°) sin 60° = 0 ¢Fy¿ = 4.165 ¢A ¢Fx¿ + 3(¢A sin 30°) cos 60° + 4(¢ A sin 30°)cos 30°
Q+ ©Fx¿ = 0;
-2(¢A cos 30°) cos 30° + 4(¢A cos 30°) cos 60° = 0 ¢Fx¿ = -2.714 ¢A Normal and Shear Stress: For the inclined plane. sx = lim¢A:0
tx¿y¿ = lim¢A:0
¢Fx¿ = -2.71 ksi ¢A ¢Fy¿
Ans.
Ans.
= 4.17 ksi
¢A
Negative sign indicates that the sense of sx¿, is opposite to that shown on FBD.
9–11. Solve Prob. 9–10 using the stress-transformation equations developed in Sec. 9.2. Show the result on a sketch.
2 ksi
Normal and Shear Stress: In accordance with the established sign convention, u = +60°
sx = -3 ksi
sy = 2 ksi
A
txy = -4 ksi
3 ksi 30⬚
Stress Transformation Equations: Applying Eqs. 9-1 and 9-2. sx¿ =
=
sx + sy
sx - sy +
2
2
B
cos 2u + txy sin 2u
-3 - 2 -3 + 2 + cos 120° + (-4 sin 120°) 2 2 Ans.
= -2.71 ksi tx¿y¿ = -
= -
sx - sy 2
4 ksi
sin 2u + txy cos 2u
-3 - 2 sin 120° + (-4 cos 120°) 2
= 4.17 ksi
Ans.
Negative sign indicates sx¿, is a compressive stress
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*9–12. Determine the equivalent state of stress on an element if it is oriented 50° counterclockwise from the element shown. Use the stress-transformation equations.
10 ksi
16 ksi
sy = 0
sx = -10 ksi
txy = -16 ksi
u = +50° sx¿ =
=
sx + sy
= -a
=
2
cos 2u + txy sin 2u
-10 - 0 -10 + 0 + cos 100° + ( -16)sin 100° = -19.9 ksi 2 2
tx¿y¿ = - a
sy¿ =
sx - sy +
2
sx - sy 2
b sin 2u + txy cos 2u
-10 - 0 b sin 100° + (-16)cos 100° = 7.70 ksi 2
sx + sy 2
sx - sy -
Ans.
2
Ans.
cos 2u - txy sin 2u
-10 + 0 -10 - 0 - a bcos 100° - (-16)sin 100° = 9.89 ksi 2 2
627
Ans.
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•9–13.
Determine the equivalent state of stress on an element if the element is oriented 60° clockwise from the element shown. Show the result on a sketch.
350 psi
75 psi 200 psi
In accordance to the established sign covention, u = -60° (Fig. a)
sx = 200 psi
sy = -350 psi
txy = 75 psi
Applying Eqs 9-1, 9-2 and 9-3, sx¿ =
=
sx + sy
sx - sy +
2
2
cos 2u + txy sin 2u
200 - ( -350) 200 + (-350) + cos ( -120°) + 75 sin (-120°) 2 2
= -277.45 psi = -277 psi sy¿ =
=
sx + sy
sx - sy -
2
2
Ans.
cos 2u - txy sin 2u
200 - ( -350) 200 + (-350) cos ( -120°) - 75 sin ( -120°) 2 2
= 127.45 psi = 127 psi tx¿y¿ = -
= -
sx - sy 2
Ans.
sin 2u + txy cos 2u
200 - (-350) sin (-120°) + 75 cos (-120°) 2
= 200.66 psi = 201 psi
Ans.
Negative sign indicates that sx¿ is a compressive stress. These result, can be represented by the element shown in Fig. b.
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9–14. The state of stress at a point is shown on the element. Determine (a) the principal stress and (b) the maximum in-plane shear stress and average normal stress at the point. Specify the orientation of the element in each case. Show the results on each element.
30 ksi
12 ksi
sx = -30 ksi
sy = 0
txy = -12 ksi
a) sx + sy
s1, 2 =
;
2
C
a
sx - sy 2
2
b + txy 2 =
-30 + 0 -30 - 0 2 ; a b + (-12)2 2 C 2
s1 = 4.21 ksi
Ans.
s2 = -34.2 ksi
Ans.
Orientation of principal stress: txy
tan 2uP =
(sx - sy)>2
uP = 19.33° and
-12 = 0.8 (-30 -0)>2
=
-70.67°
Use Eq. 9-1 to determine the principal plane of s1 and s2. sx + sy
sx¿ =
sx - sy +
2
2
cos 2u + txy sin 2u
u = 19.33° sx¿ =
-30 + 0 -30 - 0 + cos 2(19.33°) + (-12)sin 2(19.33°) = -34.2 ksi 2 2
Therefore uP2 = 19.3°
Ans.
and uP1 = -70.7°
Ans.
b) tmaxin-plane = savg =
C
a
sx - sy 2
sx + sy 2
=
2
b + txy 2 =
-30 - 0 2 b + (-12)2 = 19.2 ksi C 2 a
-30 + 0 = -15 ksi 2
Ans.
Ans.
Orientation of max, in - plane shear stress: tan 2uP =
-(sx - sy)>2 =
txy
uP = -25.2°
and
-(-30 - 0)>2 = -1.25 -12 64.3°
Ans.
By observation, in order to preserve equllibrium along AB, tmax has to act in the direction shown in the figure.
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9–15. The state of stress at a point is shown on the element. Determine (a) the principal stress and (b) the maximum in-plane shear stress and average normal stress at the point. Specify the orientation of the element in each case. Show the results on each element.
80 MPa
50 MPa
60 MPa
In accordance to the established sign convention, sx = -60 MPa s1, 2 =
=
sy = -80 MPa
sx + sy ;
2
C
a
sx - sy 2
txy = 50 MPa
2
b + txy 2
-60 + (-80) -60 - (-80) 2 ; c d + 502 2 C 2
= -70 ; 22600 s2 = -121 MPa
s1 = -19.0 MPa txy
tan 2uP =
=
(sx - sy)>2
uP = 39.34°
Ans.
50 = 5 [-60 - (-80)]>2
and
-50.65°
Substitute u = 39.34° into Eq. 9-1, sx¿ =
=
sx + sy
sx - sy +
2
2
cos 2u + txy sin 2u
-60 + (-80) -60 - ( -80) + cos 78.69° + 50 sin 78.69° 2 2
= -19.0 MPa = s1 Thus, (uP)1 = 39.3°
Ans.
(uP)2 = -50.7°
The element that represents the state of principal stress is shown in Fig. a. t max
in-plane
=
C
a
sx - sy 2
tan 2uS =
2
b + txy 2 =
-(sx - sy)>2 =
txy
-60 - (-80) 2 d + 502 = 51.0 MPa C 2 c
-[-60 - (-80)]>2 = -0.2 50
uS = -5.65° and 84.3° By Inspection, t max
Ans.
Ans.
has to act in the sense shown in Fig. b to maintain
in-plane
equilibrium.
savg =
sx + sy 2
=
-60 + (-80) = -70 MPa 2
The element that represents the state of maximum in - plane shear stress is shown in Fig. c.
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9–15. Continued
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*9–16. The state of stress at a point is shown on the element. Determine (a) the principal stress and (b) the maximum in-plane shear stress and average normal stress at the point. Specify the orientation of the element in each case. Sketch the results on each element.
60 MPa
30 MPa
45 MPa
sx = 45 MPa
sy = -60 MPa
txy = 30 MPa
a) s1, 2 =
=
sx + sy ;
2
C
a
sx - sy 2
2
b + txy 2
45 - (-60) 2 45 - 60 a ; b + (30)2 2 C 2 s1 = 53.0 MPa
Ans.
s2 = -68.0 MPa
Ans.
Orientation of principal stress: tan 2uP =
txy (sx - sy)>2
uP = 14.87,
=
30 = 0.5714 (45 - (-60))>2
-75.13
Use Eq. 9-1 to determine the principal plane of s1 and s2: sx¿ =
=
sx + sy
sx - sy +
2
2
cos 2u + txy sin 2u,
where u = 14.87°
45 + (-60) 45 - (-60) + cos 29.74° + 30 sin 29.74° = 53.0 MPa 2 2
Therefore uP1 = 14.9°
Ans.
and uP2 = -75.1°
Ans.
b) tmaxin-plane = savg =
C
a
sx - sy 2
sx - sy 2
=
2
b + txy 2 =
45 - (-60) 2 b + 302 = 60.5 MPa C 2 a
45 + (-60) = -7.50 MPa 2
Ans.
Ans.
Orientation of maximum in - plane shear stress: tan 2uS =
-(sx - sy)>2 txy
=
-(45 - ( -60))>2 = -1.75 30
uS = -30.1°
Ans.
uS = 59.9°
Ans.
and
By observation, in order to preserve equilibrium along AB, tmax has to act in the direction shown.
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•9–17.
Determine the equivalent state of stress on an element at the same point which represents (a) the principal stress, and (b) the maximum in-plane shear stress and the associated average normal stress. Also, for each case, determine the corresponding orientation of the element with respect to the element shown. Sketch the results on each element.
75 MPa
125 MPa
50 MPa
Normal and Shear Stress: sx = 125 MPa
sy = -75 MPa
txy = -50 MPa
In - Plane Principal Stresses: s1,2 =
=
sx - sy ;
2
B
a
sx - sy 2
2
b + txy 2
125 + (-75) 125 - (-75) 2 a ; b + (-50)2 2 2 B
= 25; 212500 s2 = -86.8 MPa
s1 = 137 MPa
Ans.
Orientation of Principal Plane: tan 2uP =
txy
A sx - sy B >2
-50
=
A 125 -(-75) B >2
= -0.5
up = -13.28° and 76.72° Substitute u = -13.28° into sx¿ =
=
sx + sy
sx - sy +
2
2
cos 2u + txy sin 2u
125 + (-75) 125 - (-75) + cos(-26.57°)+(-50) sin(-26.57°) 2 2
= 137 MPa = s1 Thus,
A up B 1 = -13.3° and A up B 2 = 76.7°
Ans.
125 - (-75)>(-50) The element that represents the state of principal stress is shown in Fig. a. Maximum In - Plane Shear Stress: t max
in-plane
=
C
¢
sx - sy 2
2
≤ + txy 2 =
-100 - 0 2 b + 252 = 112 MPa 2 B a
Orientation of the Plane of Maximum In - Plane Shear Stress: tan 2us = -
A sx - sy B >2 txy
= -
A 125 - (-75) B >2
= 2
-50
us = 31.7° and 122°
633
Ans.
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9–17.
Continued
By inspection, t max
has to act in the same sense shown in Fig. b to maintain
in-plane
equilibrium. Average Normal Stress: savg =
sx + sy 2
=
125 + (-75) = 25 MPa 2
Ans.
The element that represents the state of maximum in - plane shear stress is shown in Fig. c.
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sy
9–18. A point on a thin plate is subjected to the two successive states of stress shown. Determine the resultant state of stress represented on the element oriented as shown on the right.
=
sx¿ + sy¿
sx¿ - sy¿ +
2
2
ⴙ
60⬚
Stress Transformation Equations: Applying Eqs. 9-1, 9-2, and 9-3 u = -30°, sx¿ = -200 MPa, to element (a) with sy¿ = -350 MPa and tx¿y¿ = 0. (sx)a =
350 MPa
cos 2u + tx¿y¿ sin 2u
-200 - (-350) -200 + (-350) + cos (-60°) + 0 2 2
= -237.5 MPa
A sy B a = =
sx¿ + sy¿
sx¿ - sy¿ -
2
2
cos 2u - tx¿y¿ sin 2u
-200 - (-350) -200 + (-350) cos (-60°) - 0 2 2
= -312.5 MPa
A txy B a = = -
sx¿ - sy¿ 2
sin 2u + tx¿y¿ cos 2u
-200 - (-350) sin (-60°) + 0 2
= 64.95 MPa For element (b), u = 25°, sx¿ = sy¿ = 0 and sx¿y¿ = 58 MPa. (sx)b =
sx¿ + sy¿
sx¿ - sy¿ +
2
2
cos 2u + tx¿y¿ sin 2u
= 0 + 0 + 58 sin 50° = 44.43 MPa
A sy B b =
sx¿ + sy¿
sx¿ - sy¿ -
2
2
cos 2u - tx¿y¿ sin 2u
= 0 - 0 - 58 sin 50° = -44.43 MPa
A txy B b = -
sx¿ - sy¿ 2
58 MPa
200 MPa
sin 2u + tx¿y¿ cos 2u
= -0 + 58 cos 50° = 37.28 MPa Combining the stress components of two elements yields ss = (sx)a + (sx)b = -237.5 + 44.43 = -193 MPa
Ans.
sy = A sy B a + A sy B b = -312.5 - 44.43 = -357 MPa
Ans.
txy = A txy B a + A txy B b = 64.95 + 37.28 = 102 MPa
Ans.
635
25⬚
ⴝ
txy sx
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9–19. The state of stress at a point is shown on the element. Determine (a) the principal stress and (b) the maximum in-plane shear stress and average normal stress at the point. Specify the orientation of the element in each case. Sketch the results on each element.
160 MPa
120 MPa
In accordance to the established sign Convention, sx = 0
sy = 160 MPa
s1, 2 =
=
sx + sy ;
2
B
a
txy = -120 MPa
sx - sy 2
2
b + t2xy
0 + 160 0 - 160 2 ; a b + (-120)2 2 2 B
= 80 ; 220800 s2 = -64.2 MPa
s1 = 224 MPa tan 2up =
txy (sx - sy)>2
up = 28.15°
=
Ans.
-120 = 1.5 (0 - 160)>2
and -61.85°
Substitute u = 28.15° into Eq. 9-1, sx¿ =
=
sx + sy
sx - sy +
2
2
cos 2u + txy sin 2u
0 + 160 0 - 160 + cos 56.31° + (-120) sin 56.31° 2 2
= -64.22 = s2 Thus, (up)1 = -61.8°
Ans.
(up)2 = 28.2°
The element that represents the state of principal stress is shown in Fig. a tmax
in-plane
=
B
a
sx - sy 2
tan 2us =
0 - 160 2 b + (-120)2 = 144 MPa 2 B a
-(sx - sy)>2
us = -16.8° By inspection, equilibrium.
2
b + t2xy =
tmax
in-plane
savg =
txy
Ans.
-(0 - 160)>2 = -0.6667 -120
=
Ans.
and 73.2°
has to act in the sense shown in Fig. b to maintain sx + sy 2
=
0 + 160 = 80 MPa 2
Ans.
The element that represents the state of Maximum in - plane shear stress is shown in Fig. (c)
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9–19. Continued
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*9–20. The stress acting on two planes at a point is indicated. Determine the normal stress sb and the principal stresses at the point.
a
4 ksi 60⬚ 45⬚
b
2 ksi sb a
Stress Transformation Equations: Applying Eqs. 9-2 and 9-1 with u = -135°, sy = 3.464 ksi, txy = 2.00 ksi, tx¿y¿ = -2 ksi, and sx¿ = sb¿., tx¿y¿ = -
-2 = -
sx - sy 2
sin 2u + txy cos 2u
sx - 3.464 sin (-270°) + 2cos ( -270°) 2
sx = 7.464 ksi sx¿ =
sy =
sx - sy
sx - sy +
2
2
cos 2u + txy sin 2u
7.464 - 3.464 7.464 + 3.464 + cos (-270°) + 2sin ( -270°) 2 2 Ans.
= 7.46 ksi In - Plane Principal Stress: Applying Eq. 9-5. s1, 2 =
=
sx + sy 2
;
B
a
sx - sy 2
2
b + t2xy
7.464 - 3.464 2 7.464 + 3.464 ; a b + 22 2 2 B
= 5.464 ; 2.828 s1 = 8.29 ksi
s2 = 2.64 ksi
Ans.
638
b
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•9–21. The stress acting on two planes at a point is indicated. Determine the shear stress on plane a–a and the principal stresses at the point.
b a
ta 45⬚
60 ksi 60⬚
txy = 60 cos 60° = 30 ksi sa =
80 =
sx + sy
sx - sy +
2
2
51.962 - sy
51.962 + sy +
2
cos 2u + txy sin 2u
2
cos (90°) + 30 sin (90°)
sy = 48.038 ksi ta = - a = -a
sx - sy 2
b sin 2u + txy cos u
51.962 - 48.038 bsin (90°) + 30 cos (90°) 2
ta = -1.96 ksi s1, 2 =
=
Ans.
sx + sy 2
;
C
a
sx - sy 2
2
b + t2xy
51.962 - 48.038 2 51.962 + 48.038 ; a b + (30)2 2 C 2
s1 = 80.1 ksi
Ans.
s2 = 19.9 ksi
Ans.
639
90⬚
a b
sx = 60 sin 60° = 51.962 ksi
80 ksi
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9–22. The T-beam is subjected to the distributed loading that is applied along its centerline. Determine the principal stress at point A and show the results on an element located at this point.
100 kN/m
A 1m
0.5 m 200 mm
75 mm
' ©yA 0.1(0.2)(0.02) + 0.21(0.02)(0.2) = = 0.155 m ©A 0.2(0.02) + 0.02(0.2) 1 (0.02)(0.2 3) + 0.02(0.2)(0.155 - 0.1)2 12
I =
+
1 (0.2)(0.023) + 0.2(0.02)(0.21 - 0.155)2 12
= 37.6667(10 - 6) m4 Referring to Fig. b, QA = y¿A¿ = 0.1175(0.075)(0.02) = 0.17625(10 - 3) m3 Using the method of sections and considering the FBD of the left cut segment of the beam, Fig. c, + c ©Fy = 0;
V - 100(1) = 0
a + ©MC = 0;
100(1)(0.5) - M = 0 M = 50 kN # m
V = 100 kN
The normal stress developed is contributed by bending stress only. For point A, y = 0.155 - 0.075 = 0.08 m. Thus s =
My 50(103) (0.08) = 106 MPa = I 37.6667(10 - 6)
The shear stress is contributed by the transverse shear stress only. Thus, t =
100(103)[0.17625(10 - 3)] VQA = 23.40(106)Pa = 23.40 MPa = It 37.6667(10 - 6) (0.02)
The state of stress of point A can be represented by the element shown in Fig. c. Here, sx = -106.19 MPa, sy = 0 and txy = 23.40 MPa. s1, 2 =
=
sx + sy 2
;
B
a
sx - sy 2
2
b + txy 2
-106.19 - 0 2 -106.19 + 0 ; b + 23.402 a 2 2 B
= -53.10 ; 58.02 s1 = 4.93 MPa
20 mm 200 mm
20 mm
The location of the centroid c of the T cross-section, Fig. a, is y =
A
s2 = -111 MPa
Ans.
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9–22.
Continued
tan 2up =
txy (sx - sy)>2
up = -11.89°
= ans
23.40 = -0.4406 ( -106.19 - 0)>2 78.11°
Substitute u = -11.89°, sx¿ =
=
sx + sy 2
sx - sy +
2
cos 2u + txy sin 2u
-106.19 + 0 -106.19 - 0 + cos (-23.78°) + 23.40 5m (-23.78°) 2 2
= -111.12 MPa = s2 Thus, (up)1 = 78.1°
Ans.
(up)2 = -11.9°
The state of principal stress can be represented by the element shown in Fig. e.
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•9–23.
The wood beam is subjected to a load of 12 kN. If a grain of wood in the beam at point A makes an angle of 25° with the horizontal as shown, determine the normal and shear stress that act perpendicular and parallel to the grain due to the loading.
I =
12 kN 1m
2m A 25⬚
300 mm 75 mm
1 (0.2)(0.3)3 = 0.45(10 - 3) m4 12
QA = yA¿ = 0.1125(0.2)(0.075) = 1.6875(10 - 3) m3 sA =
MyA 13.714(103)(0.075) = 2.2857 MPa (T) = I 0.45(10 - 3)
tA =
6.875(103)(1.6875)(10 - 3) VQA = 0.1286 MPa = It 0.45(10 - 3)(0.2)
sx = 2.2857 MPa sx¿ =
sx¿ =
sx + sy
sx - sy +
2
sy = 0
2
txy = -0.1286 MPa
u = 115°
cos 2u + txy sin 2u
2.2857 - 0 2.2857 + 0 + cos 230° + (-0.1286)sin 230° 2 2 Ans.
= 0.507 MPa tx¿y¿ = -
sx - sy
= -a
2
sin 2u + txy cos 2u
2.2857 - 0 b sin 230° + (-0.1286)cos 230° 2
= 0.958 MPa
Ans.
642
4m
200 mm
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*9–24. The wood beam is subjected to a load of 12 kN. Determine the principal stress at point A and specify the orientation of the element.
12 kN 1m
2m A 25⬚
I =
300 mm 75 mm
1 (0.2)(0.3)3 = 0.45(10 - 3) m4 12
QA = yA¿ = 0.1125(0.2)(0.075) = 1.6875(10 - 3) m3 sA =
13.714(103)(0.075) MyA = 2.2857 MPa (T) = I 0.45(10 - 3)
tA =
VQA 6.875(103)(1.6875)(10 - 3) = 0.1286 MPa = It 0.45(10 - 3)(0.2)
sx = 2.2857 MPa s1, 2 =
=
sy = 0
sx + sy ;
2
C
a
txy = -0.1286 MPa
sx - sy 2
2
b + t2xy
2.2857 - 0 2 2.2857 + 0 ; a b + (-0.1286)2 2 C 2
s1 = 2.29 MPa
Ans.
s2 = -7.20 kPa
Ans.
tan 2up =
txy (sx - sy)>2
=
-0.1286 (2.2857 - 0)>2
up = -3.21° Check direction of principal stress: sx¿ =
=
sx + sy 2
sx - sy +
2
cos 2u + txy sin 2u
2.2857 + 0 2.2857 - 0 + cos (-6.42°) - 0.1285 sin (-6.42) 2 2
= 2.29 MPa
643
4m
200 mm
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•9–25.
The bent rod has a diameter of 20 mm and is subjected to the force of 400 N. Determine the principal stress and the maximum in-plane shear stress that is developed at point A. Show the results on a properly oriented element located at this point.
100 mm
150 mm
400 N
400 N
250 mm
A
Using the method of sections and consider the FBD of the rod’s left cut segment, Fig. a. + ©F = 0; : x
N - 400 = 0 N = 400 N
a + ©MC = 0;
400(0.25) - M = 0 M = 100 N # m A = p(0.012) = 0.1(10 - 3) p m2 p (0.014) = 2.5(10 - 9)p m4 4
I =
The normal stress developed is the combination of axial and bending stress. Thus, My N ; A I
s = For point A, y = C = 0.01 m. s =
100(0.01) 400 -3 0.1(10 )p 2.5(10 - 9)p
= -126.05 (106)Pa = 126.05 MPa (C) Since no torque and transverse shear acting on the cross - section, t = 0 The state of stress at point A can be represented by the element shown in Fig. b Here, sx = -126.05 MPa, sy = 0 and txy = 0. Since no shear stress acting on the element s1 = sy = 0
s2 = sx = -126 MPa
Ans.
Thus, the state of principal stress can also be represented by the element shown in Fig. b. tmax
in-plane
=
B
a
sx - sy 2
tan 2us = -
2
b + t2xy =
(sx - sy)>2 = -
txy
us = 45° tx¿y¿ = -
-126.05 - 0 2 b + 02 = 63.0 MPa 2 B a
(-126.05 - 0)>2 = q 0
and -45°
sx - sy 2 = -
= 63.0 =
sin 2u + txy cos 2u
-126.05 - 0 sin 90° + 0 cos 90° 2 tmax
in-plane
644
Ans.
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9–25. Continued tmax This indicates that in-plane acts toward the positive sense of y¿ axis at the face of element defined by us = 45° savg =
sx + sy 2
=
-126.05 + 0 = -63.0 MPa 2
Ans.
The state of maximum In - plane shear stress can be represented by the element shown in Fig. c
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9–26. The bracket is subjected to the force of 3 kip. Determine the principal stress and maximum in-plane shear stress at point A on the cross section at section a–a. Specify the orientation of this state of stress and show the results on elements.
3 kip
3 kip
a 3 in.
a A
0.25 in.
2 in.
0.25 in.
Internal Loadings: Consider the equilibrium of the free - body diagram from the bracket’s left cut segment, Fig. a.
B 1 in.
+ ©F = 0; : x
N - 3 = 0
N = 3 kip Section a – a
M = 12 kip # in
©MO = 0; 3(4) - M = 0
Normal and Shear Stresses: The normal stress is the combination of axial and bending stress. Thus, s =
My N A I
The cross - sectional area and the moment of inertia about the z axis of the bracket’s cross section is A = 1(2) - 0.75(1.5) = 0.875 in2 I =
1 1 (1) A 23 B (0.75) A 1.53 B = 0.45573 in4 12 12
For point A, y = 1 in. Then sA =
(-12)(1) 3 = 29.76 ksi 0.875 0.45573
Since no shear force is acting on the section, tA = 0 The state of stress at point A can be represented on the element shown in Fig. b. In - Plane Principal Stress: sx = 29.76 ksi, sy = 0, and txy = 0. Since no shear stress acts on the element, s1 = sx = 29.8 ksi s2 = sy = 0
Ans.
The state of principal stresses can also be represented by the elements shown in Fig. b Maximum In - Plane Shear Stress: t max
in-plane
=
C
¢
sx - sy 2
2
≤ + txy 2 =
29.76 - 0 2 b + 02 = 14.9 ksi 2 B a
Ans.
Orientation of the Plane of Maximum In - Plane Shear Stress: tan 2us = -
A sx - sy B >2 txy
0.25 in.
= -
(29.76 - 0)>2 = -q 0
us = -45° and 45°
Ans.
646
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9–26.
Continued
Substituting u = -45° into tx¿y¿ = -
= -
sx - sy 2
sin 2u + txy cos 2u
29.76 - 0 sin(-90°) + 0 2
= 14.9 ksi = t max
in-plane
This indicates that t max
is directed in the positive sense of the y¿ axes on the ace
in-plane
of the element defined by us = -45°. Average Normal Stress: savg =
sx + sy 2
=
29.76 + 0 = 14.9 ksi 2
Ans.
The state of maximum in - plane shear stress is represented by the element shown in Fig. c.
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9–27. The bracket is subjected to the force of 3 kip. Determine the principal stress and maximum in-plane shear stress at point B on the cross section at section a–a. Specify the orientation of this state of stress and show the results on elements.
3 kip
3 kip
a 3 in.
a A
0.25 in.
2 in.
0.25 in.
Internal Loadings: Consider the equilibrium of the free - body diagram of the bracket’s left cut segment, Fig. a.
B 1 in.
+ ©F = 0; : x
N - 3 = 0
N = 3 kip Section a – a
M = 12 kip # in
©MO = 0; 3(4) - M = 0
Normal and Shear Stresses: The normal stress is the combination of axial and bending stress. Thus, s =
My N A I
The cross - sectional area and the moment of inertia about the z axis of the bracket’s cross section is A = 1(2) - 0.75(1.5) = 0.875 in2 I =
1 1 (1) A 23 B (0.75) A 1.53 B = 0.45573 in4 12 12
For point B, y = -1 in. Then sB =
(-12)(-1) 3 = -22.90 ksi 0.875 0.45573
Since no shear force is acting on the section, tB = 0 The state of stress at point A can be represented on the element shown in Fig. b. In - Plane Principal Stress: sx = -22.90 ksi, sy = 0, and txy = 0. Since no shear stress acts on the element, s1 = sy = 0
s2 = sx = -22.90 ksi
Ans.
The state of principal stresses can also be represented by the elements shown in Fig. b. Maximum In - Plane Shear Stress: t max
in-plane
=
C
¢
sx - sy 2
2
≤ + txy 2 =
-22.90 - 0 2 b + 02 = 11.5 ksi 2 B a
Ans.
Orientation of the Plane of Maximum In - Plane Shear Stress: tan 2us = -
A sx - sy B >2 txy
0.25 in.
= -
(-22.9 - 0)>2 = -q 0
us = 45° and 135°
Ans.
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9–27. Continued
Substituting u = 45° into tx¿y¿ = -
= -
sx - sy 2
sin 2u + txy cos 2u
-22.9 - 0 sin 90° + 0 2
= 11.5 ksi = t max
in-plane
is directed in the positive sense of the y¿ axes on the
This indicates that t max
in-plane
element defined by us = 45°. Average Normal Stress: savg =
sx + sy 2
=
-22.9 + 0 = -11.5 ksi 2
Ans.
The state of maximum in - plane shear stress is represented by the element shown in Fig. c.
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*9–28. The wide-flange beam is subjected to the loading shown. Determine the principal stress in the beam at point A and at point B. These points are located at the top and bottom of the web, respectively. Although it is not very accurate, use the shear formula to determine the shear stress.
8 kN/m A
B
1m
3m
B
Internal Forces and Moment: As shown on FBD(a).
200 mm
Section Properties: A = 0.2(0.22) - 0.19(0.2) = 6.00 A 10 - 3 B m2 1 1 (0.2) A 0.223 B (0.19) A 0.22 B = 50.8 A 10 - 6 B m4 12 12
I =
QA = QB = y¿A¿ = 0.105(0.01)(0.2) = 0.210 A 10 - 3 B m3 Normal Stress: s =
My N ; A I 21.65(103)
=
73.5(103)(0.1) ;
6.00(10 - 3)
50.8(10 - 6)
sA = 3.608 + 144.685 = 148.3 MPa sB = 3.608 - 144.685 = -141.1 MPa VQ . It
Shear Stress: Applying the shear formula t =
tA = tB =
36.5(103) C 0.210(10 - 3) D 50.8(10 - 6)(0.01)
= 15.09 MPa
In - Plane Principal Stress: sx = 148.3 MPa, sy = 0, and txy = -15.09 MPa for point A. Applying Eq. 9-5. s1, 2 =
=
sx + sy ;
2
C
sx - sy
a
2
2
b + t2xy
148.3 - 0 2 148.3 + 0 ; a b + (-15.09)2 2 C 2
= 74.147 ; 75.666 s1 = 150 MPa
s2 = -1.52 MPa
Ans.
sx = -141.1 MPa, sy = 0, and txy = -15.09 MPa for point B. Applying Eq. 9-5. s1, 2 =
=
sx + sy 2
;
C
a
sx - sy 2
2
b + t2xy
( -141.1) - 0 2 -141.1 + 0 ; a b + (-15.09)2 2 C 2
= -70.538 ; 72.134 s1 = 1.60 MPa
s2 = -143 MPa
Ans. 650
30⬚ 25 kN
A 10 mm
110 mm
10 mm 200 mm 10 mm
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•9–29.
The wide-flange beam is subjected to the loading shown. Determine the principal stress in the beam at point A, which is located at the top of the web. Although it is not very accurate, use the shear formula to determine the shear stress. Show the result on an element located at this point.
120 kN/m
30 kN
A 0.3 m
0.9 m
Using the method of sections and consider the FBD of the left cut segment of the bean, Fig. a V -
+ c ©Fy = 0;
1 2
a + ©MC = 0;
1 2
(90)(0.9)(0.3) + 30(0.9) - M = 0
M = 39.15 kN # m
150 mm
1 1 (0.15)(0.193) (0.13)(0.153) = 49.175(10 - 6) m4 12 12
Referring to Fig. b, QA = y¿A¿ = 0.085 (0.02)(0.15) = 0.255 (10 - 3) m3 The normal stress developed is contributed by bending stress only. For point A, y = 0.075 m. Thus, s =
My 39.15(103)(0.075) = 59.71(106)Pa = 59.71 MPa (T) = I 49.175(10 - 6)
The shear stress is contributed by the transverse shear stress only. Thus t =
70.5(103) C 0.255(10 - 3) D VQA = 18.28(106)Pa = 18.28 MPa = It 49.175(10 - 6) (0.02)
Here, sx = 59.71 MPa, sy = 0 and txy = 18.28 MPa. s1, 2 =
=
sx + sy ;
2
C
a
sx - sy 2
2
b + txy
59.71 - 0 2 59.71 + 0 a ; b + 18.282 2 C 2
= 29.86 ; 35.01 s2 = -5.15 MPa
s1 = 64.9 MPa tan 2uP =
txy (sx - sy)>2
uP = 15.74°
=
and
Ans.
18.28 = 0.6122 (59.71 - 0)>2 -74.26°
Substitute u = 15.74°, sx¿ =
=
sx + sy 2
sx - sy +
2
20 mm 150 mm 20 mm
V = 70.5 kN
(90)(0.9) - 30 = 0
The moment of inertia of the cross - section about the bending axis is I =
A 20 mm
cos 2u + txy sin 2u
59.71 - 0 59.71 + 0 + cos 31.48° + 18.28 sin 31.48° 2 2
= 64.9 MPa = s1
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9–29.
Continued
Thus,
A uP B 1 = 15.7°
A uP B 2 = -74.3°
Ans.
The state of principal stress can be represented by the element shown in Fig. d
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9–30. The cantilevered rectangular bar is subjected to the force of 5 kip. Determine the principal stress at points A and B. 1 I = (3)(63) = 54 in4 12
1.5 in. A 1.5 in.
A = (6)(3) = 18 in2
QA = 2.25(1.5)(3) = 10.125 in3
1.5 in. 1.5 in. B 1 in. 3 in.
QB = 2(2)(3) = 12 in3
1 in.
Point A:
15 in.
3 in.
3 5 4
5 kip
sA
45(1.5) Mxz 4 P = + = + = 1.472 ksi A I 18 54
tA =
Vz QA =
It
3(10.125) = 0.1875 ksi 54(3)
sx = 1.472 ksi s1, 2 =
=
sy = 0
sx + sy ;
2
C
a
sx - sy 2
txy = 0.1875 ksi 2
b + txy 2
1.472 - 0 2 1.472 + 0 ; a b + 0.18752 2 C 2
s1 = 1.50 ksi
Ans.
s2 = -0.0235 ksi
Ans.
Point B: sB =
tB =
45(1) Mxz 4 P = = -0.6111 ksi A I 18 54 Vz QB =
It
3(12) = 0.2222 ksi 54(3)
sx = -0.6111 ksi s1, 2 =
=
sy = 0
sx + sy 2
;
C
a
sx - sy 2
txy = 0.2222 ksi
2
b + txy 2
-0.6111 - 0 2 -0.611 + 0 ; a b + 0.2222 2 C 2
s1 = 0.0723 ksi
Ans.
s2 = -0.683 ksi
Ans.
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9–31. Determine the principal stress at point A on the cross section of the arm at section a–a. Specify the orientation of this state of stress and indicate the results on an element at the point.
7.5 mm A 50 mm
7.5 mm
Support Reactions: Referring to the free - body diagram of the entire arm shown in Fig. a, ©MB = 0; FCD sin 30°(0.3) - 500(0.65) = 0
FCD = 2166.67 N
+ ©F = 0; : x
Bx - 2166.67 cos 30° = 0
Bx = 1876.39 N
+ c ©Fy = 0;
2166.67 sin 30° - 500 - By = 0
By = 583.33 N
20 mm
Section a – a D
Internal Loadings: Consider the equilibrium of the free - body diagram of the arm’s left segment, Fig. b. + ©F = 0; : x
1876.39 - N = 0
N = 1876.39 N
+ c ©Fy = 0;
V - 583.33 = 0
V = 583.33 N
583.33(0.15) - M = 0
M = 87.5N # m
+ ©MO = 0;
0.15 m
Referring to Fig. b, QA = y¿A¿ = 0.02125(0.0075)(0.02) = 3.1875 A 10 - 6 B m3 Normal and Shear Stress: The normal stress is a combination of axial and bending stress. Thus,
-1876.39
0.5625 A 10
-3
B
MyA N + A I 87.5(0.0175)
+
0.16367 A 10 - 6 B
= 6.020 MPa
The shear stress is caused by transverse shear stress.
tA =
583.33 C 3.1875 A 10 - 6 B D VQA = = 1.515 MPa It 0.16367 A 10 - 6 B (0.0075)
The share of stress at point A can be represented on the element shown in Fig. d. In - Plane Principal Stress: sx = 6.020 MPa, sy = 0, and txy = 1.515 MPa. We have s1,2 =
=
sx + sy 2
;
C
¢
sx - sy 2
2
≤ + txy 2
6.020 - 0 2 6.020 + 0 ; a b + 1.5152 2 C 2
s1 = 6.38 MPa
s2 = -0.360 MPa
Ans.
654
C 0.15 m
0.35 m 500 N
1 1 (0.02) A 0.053 B (0.0125) A 0.0353 B = 0.16367 A 10 - 6 B m4 12 12
=
a
a
A = 0.02(0.05) - 0.0125(0.035) = 0.5625 A 10 - 3 B m2
sA =
60⬚
B
Section Properties: The cross - sectional area and the moment of inertia about the z axis of the arm’s cross section are
I =
7.5 mm
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9–31. Continued Orientation of the Principal Plane: tan 2uP =
txy
A sx - sy B >2
=
1.515 = 0.5032 (6.020 - 0)>2
up = 13.36° and 26.71° Substituting u = 13.36° into sx¿ =
=
sx + sy 2
sx - sy +
2
cos 2u + txy sin 2u
6.020 + 0 6.020 - 0 + cos 26.71° + 1.515 sin 26.71° 2 2
= 6.38 MPa = s1 Thus, A uP B 1 = 13.4 and A uP B 2 = 26.71°
Ans.
The state of principal stresses is represented by the element shown in Fig. e.
655
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*9–32. Determine the maximum in-plane shear stress developed at point A on the cross section of the arm at section a–a. Specify the orientation of this state of stress and indicate the results on an element at the point.
7.5 mm A 50 mm
7.5 mm
20 mm
7.5 mm
Section a – a
Support Reactions: Referring to the free - body diagram of the entire arm shown in Fig. a, ©MB = 0; FCD sin 30°(0.3) - 500(0.65) = 0
FCD = 2166.67 N
+ ©F = 0; : x
Bx - 2166.67 cos 30° = 0
Bx = 1876.39 N
+ c ©Fy = 0;
2166.67 sin 30° - 500 - By = 0
By = 583.33 N
D 60⬚
a
B
C
a 0.15 m
0.15 m
0.35 m 500 N
Internal Loadings: Considering the equilibrium of the free - body diagram of the arm’s left cut segment, Fig. b, + ©F = 0; : x
1876.39 - N = 0
N = 1876.39 N
+ c ©Fy = 0;
V - 583.33 = 0
V = 583.33 N
+ ©MO = 0;
583.33(0.15) - M = 0
M = 87.5 N # m
Section Properties: The cross - sectional area and the moment of inertia about the z axis of the arm’s cross section are A = 0.02(0.05) - 0.0125(0.035) = 0.5625 A 10 - 3 B m2 1 1 (0.02) A 0.053 B (0.0125) A 0.0353 B = 0.16367 A 10 - 6 B m4 12 12
I =
Referring to Fig. b, QA = y¿A¿ = 0.02125(0.0075)(0.02) = 3.1875 A 10 - 6 B m3 Normal and Shear Stress: The normal stress is a combination of axial and bending stress. Thus, sA =
MyA N + A I -1876.39
=
0.5625 A 10
-3
B
87.5(0.0175)
+
0.16367 A 10 - 6 B
= 6.020 MPa
The shear stress is contributed only by transverse shear stress. tA =
583.33 C 3.1875 A 10 - 6 B D VQA = = 1.515 MPa It 0.16367 A 10 - 6 B (0.0075)
Maximum In - Plane Shear Stress: sx = 6.020 MPa, sy = 0, and txy = 1.515 MPa. tmax
in-plane
=
C
¢
sx - sy 2
2
≤ + txy 2 =
6.020 - 0 2 b + 1.5152 = 3.37 MPa B 2 a
656
Ans.
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9–32.
Continued
Orientation of the Plane of Maximum In - Plane Shear Stress: tan 2us = -
A sx - sy B >2 txy
= -
(6.020 - 0)>2 = -1.9871 1.515
us = -31.6° and 58.4°
Ans.
Substituting u = -31.6° into tx¿y¿ = -
= -
sx - sy 2
sin 2u + txy cos 2u
6.020 - 0 sin(-63.29°) + 1.515 cos(-63.29°) 2
= 3.37 MPa = t max
in-plane
is directed in the positive sense of the y¿ axis on the face
This indicates that t max
in-plane
of the element defined by us = -31.6°. Average Normal Stress: savg =
sx + sy 2
=
6.020 + 0 = 3.01 MPa 2
Ans.
The state of maximum in - plane shear stress is represented on the element shown in Fig. e.
657
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•9–33. The clamp bears down on the smooth surface at E by tightening the bolt. If the tensile force in the bolt is 40 kN, determine the principal stress at points A and B and show the results on elements located at each of these points. The cross-sectional area at A and B is shown in the adjacent figure.
300 mm 50 mm 30 mm 100 mm
B A
Support Reactions: As shown on FBD(a). E
Internal Forces and Moment: As shown on FBD(b). Section Properties: I =
1 (0.03) A 0.053 B = 0.3125 A 10 - 6 B m4 12
QA = 0 QB = y¿A¿ = 0.0125(0.025)(0.03) = 9.375 A 10 - 6 B m3 Normal Stress: Applying the flexure formula s = -
sA = -
sB = -
2.40(103)(0.025)
= -192 MPa
0.3125(10 - 6) 2.40(103)(0) 0.3125(10 - 6)
= 0
Shear Stress: Applying the shear formula t =
tA =
tB =
My . I
24.0(103)(0) 0.3125(10 - 6)(0.03)
VQ It
= 0
24.0(103) C 9.375(10 - 6) D 0.3125(10 - 6)(0.03)
= 24.0 MPa
In - Plane Principal Stresses: sx = 0, sy = -192 MPa, and txy = 0 for point A. Since no shear stress acts on the element. s1 = sx = 0
Ans.
s2 = sy = -192 MPa
Ans.
sx = sy = 0 and txy = -24.0 MPa for point B. Applying Eq. 9-5 s1,2 =
sx + sy 2
;
C
a
sx - sy 2
2
b + t2xy
= 0 ; 20 + (-24.0)2 = 0 ; 24.0 s1 = 24.0
s2 = -24.0 MPa
Ans.
658
B
A
25 mm 100 mm 50 mm
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9–33. Continued Orientation of Principal Plane: Applying Eq. 9-4 for point B. tan 2up =
txy
A sx - sy B >2
up = -45.0°
=
and
-24.0 = -q 0 45.0°
Subsututing the results into Eq. 9-1 with u = -45.0° yields sx¿ =
sx + sy 2
sx - sy +
2
cos 2u + txy sin 2u
= 0 + 0 + [-24.0 sin (-90.0°)] = 24.0 MPa = s1 Hence, up1 = -45.0°
up2 = 45.0°
Ans.
659
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9–34. Determine the principal stress and the maximum inplane shear stress that are developed at point A in the 2-in.-diameter shaft. Show the results on an element located at this point. The bearings only support vertical reactions.
300 lb
Using the method of sections and consider the FBD of shaft’s left cut segment, Fig. a, + ©F = 0; : x
N - 3000 = 0
+ c ©Fy = 0;
75 - V = 0
a + ©MC = 0;
M - 75(24) = 0
N = 3000 lb V = 75 lb M = 1800 lb # in
A = p(12) = p in2
I =
p 4 p (1 ) = in4 4 4
Also, QA = 0 The normal stress developed is the combination of axial and bending stress. Thus My N ; A I
s = For point A, y = C = 1 in. Then s =
1800(1) 3000 p p>4
= -1.337 (103) psi = 1.337 ksi (c) The shear stress developed is due to transverse shear force. Thus, t =
VQA = 0 It
The state of stress at point A, can be represented by the element shown in Fig. b. Here, sx = -1.337 ksi, sy = 0 is txy = 0. Since no shear stress acting on the element, s1 = sy = 0
s2 = sx = -1.34 ksi
Ans.
Thus, the state of principal stress can also be represented by the element shown in Fig. b. t
max in-plane
=
C
a
sx - sy 2
tan 2us = us = 45°
2
b + t2xy =
-1.337 - 0 2 b + 02 = 0.668 ksi - 668 psi Ans. C 2 a
(sx - sy)>2 = -
txy and
(-1.337 - 0)>2 = q 0 Ans.
-45°
Substitute u = 45°, tx¿y¿ = -
= -
sx - sy 2
sin 2u + txy cos 2u
-1.337 - 0 sin 90° + 0 2
= 0.668 ksi = 668 psi =
A
3000 lb
tmax
in-plane
660
24 in.
3000 lb
12 in.
12 in.
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9–34. Continued tmax This indicates that in-plane acts toward the positive sense of y¿ axis at the face of the element defined by us = 45°. Average Normal Stress. The state of maximum in - plane shear stress can be represented by the element shown in Fig. c.
661
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9–35. The square steel plate has a thickness of 10 mm and is subjected to the edge loading shown. Determine the maximum in-plane shear stress and the average normal stress developed in the steel. sx = 5 kPa
sy = -5 kPa
tmax
sx - sy
in-plane
savg =
a
txy = 0
200 mm
b +
t2xy
C
=
5 + 5 2 b + 0 = 5 kPa C 2 a
sx + sy 3
50 N/m
2
=
2
50 N/m
Ans. 200 mm
5 - 5 = = 0 2
Ans.
Note: tan 2us =
tan 2us =
-(sx - sy)>2 txy -(5 + 5)>2 = q 0
us = 45°
*9–36. The square steel plate has a thickness of 0.5 in. and is subjected to the edge loading shown. Determine the principal stresses developed in the steel. sx = 0 s1,2 =
sy = 0
txy = 32 psi
sx + sy 2
;
16 lb/in.
C
a
sx - sy 2
2
b + t2xy
= 0 ; 20 + 322
4 in.
s1 = 32 psi
Ans.
s2 = -32 psi
Ans.
Note: tan 2up =
16 lb/in.
4 in.
txy (sx - sy)>2
=
32 = q 0
up = 45°
662
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•9–37.
The shaft has a diameter d and is subjected to the loadings shown. Determine the principal stress and the maximum in-plane shear stress that is developed at point A. The bearings only support vertical reactions.
P F
F A L 2
Support Reactions: As shown on FBD(a). Internal Forces and Moment: As shown on FBD(b). Section Properties: A =
p 2 d 4
p d 4 p 4 a b = d 4 2 64
I =
QA = 0
Normal Stress: N Mc ; A I
s =
-F ; d2
=
p 4
sA =
A B
pL d 4 2 p 4 d 64
4 2PL - Fb a 2 d pd
Shear Stress: Since QA = 0, tA = 0 In - Plane Principal Stress: sx =
4 2PL a - Fb. pd2 d
sy = 0 and txy = 0 for point A. Since no shear stress acts on the element, s1 = sx =
4 2PL a - Fb d pd2
Ans.
s2 = sy = 0
Ans.
Maximum In - Plane Shear Stress: Applying Eq. 9-7 for point A, t
max in-plane
=
=
=
Q
£
B
a
4 2 pd
sx - sy 2
2
b + t2xy
A 2PL d - FB - 0 2
2
≥ + 0
2PL 2 a - Fb d pd2
Ans.
663
L 2
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9–38. A paper tube is formed by rolling a paper strip in a spiral and then gluing the edges together as shown. Determine the shear stress acting along the seam, which is at 30° from the vertical, when the tube is subjected to an axial force of 10 N. The paper is 1 mm thick and the tube has an outer diameter of 30 mm. P = A
s =
p 4
= -
10 N
10 N 30 mm
10 = 109.76 kPa (0.032 - 0.0282)
sx = 109.76 kPa tx¿y¿ = -
30⬚
sx - sy 2
sy = 0
txy = 0
u = 30°
sin 2u + txy cos 2u
106.76 - 0 sin 60° + 0 = -47.5 kPa 2
Ans.
9–39. Solve Prob. 9–38 for the normal stress acting perpendicular to the seam.
30⬚
10 N
10 N 30 mm
s =
sn =
=
P = A
p 4
10 = 109.76 kPa (0.032 - 0.0282)
sx + sy 2
sx - sy +
2
cos 2u + txy sin 2u
109.76 - 0 109.76 + 0 + cos (60°) + 0 = 82.3 kPa 2 2
Ans.
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*9–40. Determine the principal stresses acting at point A of the supporting frame. Show the results on a properly oriented element located at this point.
800 mm B
A 300 mm
150 mm 12 mm
5
B 15 mm
130 mm A
y =
0.065(0.13)(0.015) + 0.136(0.15)(0.012) ©yA = = 0.0991 m ©A 0.13(0.015) + 0.15(0.012)
I =
1 (0.015)(0.133) + 0.015(0.13)(0.0991 - 0.065)2 12 1 (0.15)(0.012 3) + 0.15(0.012)(0.136 - 0.0991)2 = 7.4862(10 - 6) m4 12
+
QA = 0 A = 0.13(0.015) + 0.15(0.012) = 3.75(10 - 3) m2 Normal stress: s =
P Mc + A I
sA =
-3.6(103) -3
3.75(10 )
5.2767(103)(0.0991) -
7.4862(10 - 6)
= -70.80 MPa
Shear stress: tA = 0 Principal stress: s1 = 0
Ans.
s2 = -70.8 MPa
Ans.
665
4 3
6 kN
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•9–41. Determine the principal stress acting at point B, which is located just on the web, below the horizontal segment on the cross section. Show the results on a properly oriented element located at this point. Although it is not very accurate, use the shear formula to calculate the shear stress.
800 mm B
A 300 mm
150 mm 12 mm
y =
©yA 0.065(0.13)(0.015) + 0.136(0.15)(0.012) = = 0.0991 m ©A 0.13(0.015) + 0.15(0.012)
I =
1 (0.015)(0.133) + 0.015(0.13)(0.0991 - 0.065)2 12 +
130 mm A
1 (0.15)(0.0123) + 0.15(0.012)(0.136 - 0.0991)2 = 7.4862(10 - 6) m4 12
A = 0.13(0.015) + 0.15(0.012) = 3.75(10 - 3) m2 Normal stress: s =
Mc P + A I
sB = -
3.6(103) 3.75(10 - 3)
5.2767(103)(0.130 - 0.0991) +
7.4862(10 - 6)
= 20.834 MPa
Shear stress: tB =
VQ -4.8(103)(0.0369)(0.15)(0.012) = -2.84 MPa = It 7.4862(10 - 6)(0.015)
Principal stress: s1,2 = a
20.834 + 0 20.834 - 0 2 b ; a b + (-2.84)2 2 C 2
s1 = 21.2 MPa
Ans.
s2 = -0.380 MPa
Ans.
tan 2up =
A
-2.84 20.834 - 0 2
B
up = -7.63°
Ans.
666
5
B 15 mm
4 3
6 kN
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9–42. The drill pipe has an outer diameter of 3 in., a wall thickness of 0.25 in., and a weight of 50 lb>ft. If it is subjected to a torque and axial load as shown, determine (a) the principal stress and (b) the maximum in-plane shear stress at a point on its surface at section a.
1500 lb
800 lb⭈ft 20 ft
a 20 ft
Internal Forces and Torque: As shown on FBD(a). Section Properties: A =
p 2 A 3 - 2.52 B = 0.6875p in2 4
J =
p A 1.54 - 1.254 B = 4.1172 in4 2
s =
N -2500 = = -1157.5 psi A 0.6875p
Normal Stress:
Shear Stress: Applying the torsion formula. t =
800(12)(1.5) Tc = = 3497.5 psi J 4.1172
a) In - Plane Principal Stresses: sx = 0, sy = -1157.5 psi and txy = 3497.5 psi for any point on the shaft’s surface. Applying Eq. 9-5. s1,2 =
=
sx + sy 2
;
C
a
sx - sy 2
2
b + t2xy
0 - (-1157.5) 2 0 + (-1157.5) ; a b + (3497.5)2 2 C 2
= -578.75 ; 3545.08 s1 = 2966 psi = 2.97 ksi
Ans.
s2 = -4124 psi = -4.12 ksi
Ans.
b) Maximum In - Plane Shear Stress: Applying Eq. 9-7 t
max in-plane
a
sx - sy
2
b + t2xy
=
C
=
0 - (-1157.5) 2 ≤ + (3497.5)2 C 2
2
¢
= 3545 psi = 3.55 ksi
Ans.
667
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9–43. Determine the principal stress in the beam at point A.
60 kN 50 mm 150 kN
A
A 60 mm
0.5 m
Using the method of sections and consider the FBD of the beam’s left cut segment, Fig. a, + ©F = 0; : x
150 - N = 0
N = 150 kN
+ c ©Fy = 0;
V - 60 = 0
V = 60 kN
a + ©MC = 0;
M = 30 kN # m
60(0.5) - M = 0 A = 0.06(0.15) = 0.009 m2
1 (0.06)(0.153) = 16.875(10 - 6) m4 12
I = Referring to Fig. b,
QA = y¿A¿ = 0.05 (0.05)(0.06) = 0.15(10 - 3) m3 The normal stress developed is the combination of axial and bending stress. Thus My N ; A I
s =
For point A, y = 0.075 - 0.05 = 0.025 m. Then s =
30(103)(0.025) -150(103) 0.009 16.875(10 - 6)
= -61.11(106) Pa = 61.11 MPa (c) The shear stress developed is due to the transverse shear, Thus, t =
60(103) C 0.15(10 - 3) D VQA = 8.889 MPa = It 16.875(10 - 6) (0.06)
Here, sx = -61.11 MPa, sy = 0 and txy = 8.889 MPa, s1, 2 =
=
sx + sy ;
2
C
a
sx - sy 2
2
b + t2xy
-61.11 - 0 2 -61.11 + 0 ; a b + 8.8892 2 C 2
= -30.56 ; 31.82 s2 = -62.4 MPa
s1 = 1.27 MPa tan 2uP =
txy (sx - sy)>2
uP = -8.11°
=
and
Ans.
8.889 = -0.2909 (-61.11 - 0)>2 81.89°
668
0.25 m
150 mm
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9–43.
Continued
Substitute u = -8.11°, sx¿ =
=
sx + sy 2
sx - sy +
2
cos 2u + txy sin 2u
-61.11 + 0 -61.11 - 0 + cos (-16.22°) + 8.889 sin (-16.22°) 2 2
= -62.4 MPa = s2 Thus, (uP)1 = 81.9°
(uP)2 = -8.11°
The state of principal stresses can be represented by the elements shown in Fig. (c)
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*9–44. Determine the principal stress at point A which is located at the bottom of the web. Show the results on an element located at this point.
150 kN/m
Using the method of sections, consider the FBD of the bean’s left cut segment, Fig. a, V -
+ c ©Fy = 0;
I =
1 (100)(0.6) = 0 2
10 mm
M = 6 kN # m
A
1 1 (0.15)(0.223) (0.14)(0.23) = 39.7667(10 - 6) m4 12 12
150 mm
Referring to Fig. b QA = y¿A¿ = 0.105 (0.01)(0.15) = 0.1575(10 - 3) m3 The normal stress developed is due to bending only. For point A, y = 0.1 m. Then s =
My
6(103)(0.1) =
I
= 15.09(106)Pa = 15.09 MPa (c)
39.7667(10 - 6)
The shear stress developed is due to the transverse shear. Thus, t =
30(103) C 0.1575(10 - 3) D VQA = 11.88(106)Pa = 11.88 MPa = It 39.7667(10 - 6)(0.01)
Here, sx = -15.09 MPa, sy = 0 And txy = 11.88 MPa. s1, 2 =
=
sx + sy ;
2
C
a
sx - sy 2
2
b + t2xy
-15.09 - 0 2 -15.09 + 0 ; a b + 11.882 2 C 2
= -7.544 ; 14.074 s2 = -21.6 MPa
s1 = 6.53 MPa tan 2uP =
txy (sx - sy)>2
uP = -28.79°
=
Ans.
11.88 = -1.575 (-15.09 - 0)>2
and
61.21°
Substitute u = 61.21°, sx¿ =
=
sx + sy 2
sx - sy +
2
cos 2u + txy sin 2u
-15.09 + 0 -15.09 - 0 + cos 122.42° + 11.88 sin 122.42° 2 2
= 6.53 MPa = s1 Thus, (uP)1 = 61.2°
0.3 m
V = 30 kN
1 (100)(0.6)(0.2) - M = 0 2
a + ©MC = 0;
A 0.6 m
Ans.
(uP)2 = -28.8°
The state of principal stresses can be represented by the element shown in Fig. d.
670
10 mm 200 mm 10 mm
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9–44. Continued
671
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•9–45.
Determine the maximum in-plane shear stress in the box beam at point A. Show the results on an element located at this point.
10 kip 4 kip
A B
Using the method of section, consider the FBD, of bean’s left cut segment, Fig. a, 8 - 10 + V = 0
+ c ©Fy = 0; a + ©MC = 0;
M + 10(1.5) - 8(3.5) = 0
4 in. A
M = 13 kip # ft
4 in.
The moment of inertia of the cross - section about the neutral axis is
Referring to Fig. b, QA = 0 The normal stress developed is contributed by the bending stress only. For point A, y = C = 3 in. My =
I
13(12)(3) = 5.40 ksi (c) 86.6667
The shear stress is contributed by the transverse shear stress only. Thus t =
VQA = 0 It
The state of stress at point A can be represented by the element shown in Fig. c Here, sx = -5.40 ksi, sy = 0 and txy = 0. tmax
in-plane
=
C
a
sx - sy 2
tan 2us = -
2
b + txy 2 =
-5.40 - 0 2 b + 02 = 2.70 ksi C 2
(sx - sy)>2 = -
txy
us = 45°
and
a
Ans.
(-5.40 - 0)>2 = q 2
-45°
Substitute u = 45°, tx¿y¿ = -
sx - sy
= -
2
sin 2u + txy cos 2u
-5.40 - 0 sin 90° + 0 2
= 2.70 ksi =
tmax
in-plane
tmax This indicates that in-plane acts toward the positive sense of y¿ axis at the face of element defined by us = 45° savg =
sx + sy 2
=
-5.40 + 0 = -2.70 ksi 2
The state of maximum In - plane shear stress can be represented by the element shown in Fig. d. 672
B 6 in.
1 1 I = (6)(63) (4)(43) = 86.6667 in4 12 12
s =
1.5 ft
2 ft
V = 2 kip
2 ft 0.5 ft 3 in. 3 in.
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9–45. Continued
673
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9–46. Determine the principal stress in the box beam at point B. Show the results on an element located at this point.
10 kip 4 kip
A
Using the method of sections, consider the FBD of bean’s left cut segment, Fig. a, 8 - 10 + V = 0
+ c ©Fy = 0; a + ©MC = 0;
M = 13 kip # ft
M + 10(1.5) - 8(3.5) = 0 I =
B
V = 2 kip
4 in. A
1 1 (6)(63) (4)(43) = 86.6667 in4 12 12
4 in.
Referring to Fig. b, QB =
2y1œ A1œ
+
= 2 C 1(2)(1) D + 2.5(1)(6) = 19 in
The normal stress developed is contributed by the bending stress only. For point B, y = 0. My
s =
= 0
I
The shear stress is contributed by the transverse shear stress only. Thus 2(103)(19) VQB = = 219.23 psi It 86.6667(2)
t =
The state of stress at point B can be represented by the element shown in Fig. c Here, sx = sy = 0 and txy = 219.23 psi. s1, 2 =
sx + sy ;
2
C
a
sx - sy 2
2
b + txy 2
= 0 ; 20 + 219.232 s2 = -219 psi
s1 = 219 psi tan 2uP =
txy (sx - sy)>2
uP = 45°
=
and
Ans.
219.23 = q 0 -45°
Substitute u = 45°, sx¿ =
sx + sy 2
sx - sy +
2
cos 2u + txy sin 2u
= 0 + 0 + 219.23 sin 90° = 219 psi = s1 Thus, (uP)1 = 45°
Ans.
(uP)2 = -45°
The state of principal stress can be represented by the element shown in Fig. d.
674
B 6 in.
3
y2œ A2œ
1.5 ft
2 ft
2 ft 0.5 ft 3 in. 3 in.
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9–46. Continued
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9–47. The solid shaft is subjected to a torque, bending moment, and shear force as shown. Determine the principal stresses acting at point A. Ix = Iy =
J =
p (0.025)4 = 0.306796(10 - 6) m4 4
450 mm 300 N⭈m
p (0.025)4 = 0.613592(10 - 6) m4 2
25 mm
45 N⭈m
QA = 0 sA
800 N
60(0.025) Mx c = 4.889 MPa = = I 0.306796(10 - 6)
tA =
Ty c
45(0.025) =
J
0.613592(10 - 6)
sx = 4.889 MPa s1, 2 =
=
*9–48.
sy = 0
sx + sy ;
2
= 1.833 MPa
C
a
txy = -1.833 MPa
sx - sy 2
2
b + txy 2
4.889 - 0 2 4.889 + 0 ; b + (-1.833)2 a 2 C 2
s1 = 5.50 MPa
Ans.
s2 = -0.611 MPa
Ans.
Solve Prob. 9–47 for point B.
Ix = Iy =
p (0.025)4 = 0.306796(10 - 6) m4 4 450 mm
p J = (0.025)4 = 0.613592(10 - 6) m4 2 QB = yA¿ =
300 N⭈m
4(0.025) 1 a b p (0.0252) = 10.4167(10 - 6) m3 3p 2
800 N
Ty c
VzQB It
-
-6
800(10.4167)(10 ) =
J
0.306796(10 )(0.05)
sx = 0 s1, 2 =
sy = 0 sx + sy 2
;
C
a
sx - sy 2
45(0.025) -
-6
0.61359(10 - 6)
= -1.290 MPa
txy = -1.290 MPa 2
b + txy 2
= 0 ; 2(0)2 + (-1.290)2 s1 = 1.29 MPa
Ans.
s2 = -1.29 MPa
Ans.
676
A B 25 mm
45 N⭈m
sB = 0 tB =
A B
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•9–49.
The internal loadings at a section of the beam are shown. Determine the principal stress at point A. Also compute the maximum in-plane shear stress at this point.
50 mm A
200 mm 50 mm
50 mm y
200 mm
Section Properties:
z
40 kN⭈m 30 kN⭈m
A = 0.2(0.3) - 0.15(0.2) = 0.030 m4 800 kN
1 1 Iz = (0.2) A 0.33 B (0.15) A 0.23 B = 0.350 A 10 - 3 B m4 12 12 Iy =
1 1 (0.1) A 0.23 B + (0.2) A 0.053 B = 68.75 A 10 - 6 B m4 12 12
(QA)y = 0 Normal Stress: s =
sA =
Myz Mzy N + A Iz Iy -30(103)(0.1) -500(103) 40(103)(0.15) + 3 0.030 0.350(10 ) 68.75(10 - 6)
= -77.45 MPa tA = 0.
Shear Stress: Since (QA)y = 0,
In - Plane Principal Stresses: sx = -77.45 MPa. sy = 0. and txy = 0 for point A. Since no shear stress acts on the element. s1 = sy = 0
Ans.
s2 = sz = -77.4 MPa
Ans.
Maximum In-Plane Shear Stress: Applying Eq. 9–7. t
max in-plane
a
sx - sy
2
b + t2xy
=
C
=
-77.45 - 0 2 b + 0 C 2
2
a
= 38.7 MPa
Ans.
677
500 kN x
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9–50. The internal loadings at a section of the beam consist of an axial force of 500 N, a shear force of 800 N, and two moment components of 30 N # m and 40 N # m. Determine the principal stress at point A. Also calculate the maximum in-plane shear stress at this point. Ix =
A
1 (0.1)(0.2)3 = 66.67(10 - 6) in4 12
40 N⭈m
B C
50 mm
QA = 0 sA
200 mm
50 mm
100 mm
30(0.1) Mz P 500 = -20 kPa = = A Ix (0.1)(0.2) 66.67(10 - 6)
30 N⭈m
500 N 800 N
tA = 0 Here, the principal stresses are s1 = sy = 0
Ans.
s2 = sx = -20 kPa
Ans.
t
max in-plane=
=
C
a
sx - sy 2
2
b + txy 2
-20 - 0 2 b + 0 = 10 kPa C 2 a
Ans.
9–51. Solve Prob. 9–4 using Mohr’s circle.
A
400 psi
650 psi 60⬚
sx + sy 2
-650 + 400 = = -125 2
A(-650, 0)
B(400, 0)
C( -125, 0)
B
R = CA = = 650 - 125 = 525 sx¿ = -125 - 525 cos 60° = -388 psi
Ans.
tx¿y¿ = 525 sin 60° = 455 psi
Ans.
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*9–52.
Solve Prob. 9–6 using Mohr’s circle. 90 MPa
A
35 MPa 60⬚ 30⬚
sx = 90 MPa sx + sy 2
=
sy = 50 MPa
txy = -35 MPa
A(90, -35)
90 + 50 = 70 2
R = 2(90 - 70)2 + (35)2 = 40.311 Coordinates of point B: f = tan - 1 a
35 b = 60.255° 20
c = 300° - 180° - 60.255° = 59.745° sx¿ = 70 - 40.311 cos 59.745° = 49.7 MPa
Ans.
tx¿ = -40.311 sin 59.745° = -34.8 MPa
Ans.
679
B 50 MPa
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•9–53.
Solve Prob. 9–14 using Mohr’s circle. 30 ksi
12 ksi
sx + sy 2
=
-30 + 0 = -15 2
R = 2(30 - 15)2 + (12)2 = 19.21 ksi s1 = 19.21 - 15 = 4.21 ksi
Ans.
s2 = -19.21 - 15 = -34.2 ksi
Ans.
2uP2 = tan - 1 tmax
in-plane
12 ; (30 - 15)
uP2 = 19.3°
Ans.
= R = 19.2 ksi
Ans.
savg = -15 ksi 2uP2 = tan - 1
12 + 90°; (30 - 15)
Ans. us = 64.3°
Ans.
680
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9–54.
Solve Prob. 9–16 using Mohr’s circle.
350 psi
75 psi 200 psi
sx + sy 2
=
45 - 60 = -7.5 MPa 2
R = 2(45 + 7.5)2 + (30)2 = 60.467 MPa s1 = 60.467 - 7.5 = 53.0 MPa
Ans.
s2 = -60.467 - 7.5 = -68.0 MPa
Ans.
2uP1 = tan - 1 uP1 = 14.9° tmax
30 (45 + 7.5) Ans.
counterclockwise
= 60.5 MPa
Ans.
savg = -7.50 MPa
Ans.
in-plane
2uP1 = 90° - tan - 1 us1 = 30.1°
30 (45 + 7.5) Ans.
clockwise
681
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9–55.
Solve Prob. 9–12 using Mohr’s circle.
sx + sy 2
=
-10 + 0 = -5 ksi 2
10 ksi
16 ksi
R = 2(10 - 5)2 + (16)2 = 16.763 ksi f = tan - 1
16 = 72.646° (10 - 5)
a = 100 - 72.646 = 27.354° sx¿ = -5 - 16.763 cos 27.354° = -19.9 ksi
Ans.
tx¿y¿ = 16.763 sin 27.354° = 7.70 ksi
Ans.
sy¿ = 16.763 cos 27.354° - 5 = 9.89 ksi
*9–56. Solve Prob. 9–11 using Mohr’s circle.
2 ksi
Construction of the Circle: In accordance with the sign convention, sx = -3 ksi, sy = 2 ksi, and txy = -4 ksi. Hence, savg =
sx + sy 2
=
-3 + 2 = -0.500 ksi 2
3 ksi 30⬚
The coordinates for reference point A and C are A(-3, -4)
A
4 ksi B
C(-0.500, 0)
The radius of circle is R = 2(3 - 0.5)2 + 42 = 4.717 ksi Stress on the Inclined Plane: The normal and shear stress components A sx¿ and tx¿y¿ B are represented by the coordinates of point P on the circle. sx¿ = -0.500 - 4.717 cos 62.01° = -2.71 ksi
Ans.
tx¿y¿ = 4.717 sin 62.01° = 4.17 ksi
Ans.
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9–57. Mohr’s circle for the state of stress in Fig. 9–15a is shown in Fig. 9–15b. Show that finding the coordinates of point P1sx¿ , tx¿y¿2 on the circle gives the same value as the stress-transformation Eqs. 9–1 and 9–2.
A(sx, txy)
R =
sxœ =
C
Ca a
B(sy, -txy)
csx - a
sx + sy 2
sx + sy +
2
C
a
sx + sy 2
2
b d + t2xy =
sx - sy 2
C
a
b, 0b
sx - sy 2
2
b + t2xy
2
b + t2xy cos u¿
(1)
u¿ = 2uP - 2u (2)
cos (2uP - 2u) = cos 2uP cos 2u + sin 2up sin 2u From the circle: sx -
cos 2uP =
sin 2uP =
4A 4A
sx + sy 2
sx - sy 2
txy
sx - sy 2
B + 2
(3) t2xy (4)
B 2 + t2xy
Substitute Eq. (2), (3) and into Eq. (1) sx¿ =
tx¿y¿ =
sx + sy 2
C
a
sx - sy +
2
sx - sy 2
cos 2u + txy sin 2u
QED
2
b + t2xy sin u¿
(5)
sin u¿ = sin (2uP - 2u) (6)
= sin 2uP cos 2u - sin 2u cos 2uP Substitute Eq. (3), (4), (6) into Eq. (5), tx¿y¿ = -
sx - sy 2
sin 2u + txy cos 2u
QED
683
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9–58. Determine the equivalent state of stress if an element is oriented 25° counterclockwise from the element shown.
550 MPa
A(0, -550)
B(0, 550)
C(0, 0)
R = CA = CB = 550 sx¿ = -550 sin 50° = -421 MPa
Ans.
tx¿y¿ = -550 cos 50° = -354 MPa
Ans.
sy¿ = 550 sin 50° = 421 MPa
Ans.
9–59. Determine the equivalent state of stress if an element is oriented 20° clockwise from the element shown.
2 ksi
Construction of the Circle: In accordance with the sign convention, sx = 3 ksi, sy = -2 ksi, and tx¿y¿ = -4 ksi. Hence, savg =
sx + sy 2
=
3 + (-2) = 0.500 ksi 2
4 ksi
The coordinates for reference points A and C are A(3, -4)
3 ksi
C(0.500, 0)
The radius of the circle is R = 2(3 - 0.500)2 + 42 = 4.717 ksi Stress on the Rotated Element: The normal and shear stress components A sx¿ and tx¿y¿ B are represented by the coordinate of point P on the circle, sy¿, can be determined by calculating the coordinates of point Q on the circle. sx¿ = 0.500 + 4.717 cos 17.99° = 4.99 ksi
Ans.
tx¿y¿ = -4.717 sin 17.99° = -1.46 ksi
Ans.
sy¿ = 0.500 - 4.717 cos 17.99° = -3.99 ksi
Ans.
684
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*9–60. Determine the equivalent state of stress if an element is oriented 30° clockwise from the element shown. Show the result on the element.
9 ksi 4 ksi
In accordance to the established sign convention, sx = -6 ksi, sy = 9 ksi and txy = 4 ksi. Thus, savg =
sx + sy 2
=
-6 + 9 = 1.50 ksi 2
Then, the coordinates of reference point A and C are A(-6, 4)
C(1.5, 0)
The radius of the circle is R = CA = 2(-6 - 1.5)2 + 42 = 8.50 ksi Using these results, the circle shown in Fig. a can be constructed. Referring to the geometry of the circle, Fig. a, a = tan - 1 a
4 b = 28.07° 6 + 1.5
b = 60° - 28.07° = 31.93°
Then, sx¿ = 1.5 - 8.50 cos 31.93° = -5.71 ksi
Ans.
tx¿y¿ = -8.5 sin 31.95° = -4.50 ksi sy¿ = 8.71 ksi
Ans.
The results are shown in Fig. b.
685
6 ksi
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•9–61.
Determine the equivalent state of stress for an element oriented 60° counterclockwise from the element shown. Show the result on the element.
250 MPa 400 MPa
In accordance to the established sign convention, sx = -560 MPa, sy = 250 MPa and txy = -400 MPa. Thus, savg =
sx + sy 2
=
-560 + 250 = -155 MPa 2
Then, the coordinate of reference points A and C are A(-560, -400)
C(-155, 0)
The radius of the circle is R = CA = 3 C -560 - (-155) D 2 + (-400)2 = 569.23 MPa Using these results, the circle shown in Fig. a can be constructed. Referring to the geometry of the circle, Fig. a a = tan - 1 a
400 b = 44.64° 560 - 155
b = 120° - 44.64° = 75.36°
Then, sx¿ = -155 - 569.23 cos 75.36° = -299 MPa
Ans.
tx¿y¿ = 569.23 sin 75.36° = 551 MPa
Ans.
sy¿ = -155 + 569.23 cos 75.36° = -11.1 MPa
Ans.
The results are shown in Fig. b.
686
560 MPa
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9–62. Determine the equivalent state of stress for an element oriented 30° clockwise from the element shown. Show the result on the element.
5 ksi
In accordance to the established sign convention, sx = 2 ksi, sy = -5 ksi and txy = 0. Thus, savg =
sx + sy 2
=
2 + (-5) = -1.50 ksi 2
Then, the coordinate of reference points A and C are A(2, 0)
C(-1.5, 0)
The radius of the circle is R = CA = 3 C 2 - (-1.5) D 2 + 02 = 3.50 ksi Using these results, the circle shown in Fig. a can be constructed. Referring to the geometry of the circle, Fig. a, b = 60° Then, sx¿ = -1.50 + 3.50 cos 60° - 0.250 ksi
Ans.
tx¿y¿ = 3.50 sin 60° = 3.03 ksi
Ans.
sy¿ = -3.25 ksi
Ans.
The results are shown in Fig b.
687
2 ksi
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9–63. Determine the principal stress, the maximum in-plane shear stress, and average normal stress. Specify the orientation of the element in each case. 15 ksi 5 ksi
Construction of the Circle: In accordance with the sign convention, sx = 15 ksi, sy = 0 and txy = -5 ksi. Hence, sx + sy
savg =
=
2
15 + 0 = 7.50 ksi 2
Ans.
The coordinates for reference point A and C are A(15, -5)
C(7.50, 0)
The radius of the circle is R = 2(15 - 7.50)2 + 52 = 9.014 ksi a) In - Plane Principal Stress: The coordinates of points B and D represent s1 and s2, respectively. s1 = 7.50 + 9.014 = 16.5 ksi
Ans.
s2 = 7.50 - 9.014 = -1.51 ksi
Ans.
Orientation of Principal Plane: From the circle tan 2uP1 =
5 = 0.6667 15 - 7.50
uP1 = 16.8° (Clockwise)
Ans.
b) Maximum In - Plane Shear Stress: Represented by the coordinates of point E on the circle. tmax
in-plane
= -R = -9.01 ksi
Ans.
Orientation of the Plane for Maximum In - Plane Shear Stress: From the circle tan 2us =
15 - 7.50 = 1.500 5
us = 28.2° (Counterclockwise)
Ans.
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*9–64. Determine the principal stress, the maximum in-plane shear stress, and average normal stress. Specify the orientation of the element in each case.
20 MPa 80 MPa
30 MPa
In accordance to the established sign convention, sx = 30 MPa, sy = -20 MPa and txy = 80 MPa. Thus, savg =
sx + sy
30 + ( -20) = 5 MPa 2
=
2
Then, the coordinates of reference point A and the center C of the circle is A(30, 80)
C(5, 0)
Thus, the radius of circle is given by R = CA = 2(30 - 5)2 + (80 - 0)2 = 83.815 MPa Using these results, the circle shown in Fig. a, can be constructed. The coordinates of points B and D represent s1 and s2 respectively. Thus s1 = 5 + 83.815 = 88.8 MPa
Ans.
s2 = 5 - 83.815 = -78.8 MPa
Ans.
Referring to the geometry of the circle, Fig. a tan 2(uP)1 =
80 = 3.20 30 - 5
uP = 36.3° (Counterclockwise)
Ans.
The state of maximum in - plane shear stress is represented by the coordinate of point E. Thus tmax
in-plane
= R = 83.8 MPa
Ans.
From the geometry of the circle, Fig. a, tan 2us =
30 - 5 = 0.3125 80
us = 8.68° (Clockwise)
Ans.
The state of maximum in - plane shear stress is represented by the element in Fig. c
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9–64.
Continued
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•9–65. Determine the principal stress, the maximum inplane shear stress, and average normal stress. Specify the orientation of the element in each case.
120 psi
300 psi
A(300, 120)
B(0, -120)
C(150, 0)
R = 2(300 - 150)2 + 1202 = 192.094 s1 = 150 + 192.094 = 342 psi
Ans.
s2 = 150 - 192.094 = -42.1 psi
Ans.
tan 2uP =
120 = 0.8 300 - 150
uP1 = 19.3° Counterclockwise
Ans.
savg = 150 psi
Ans.
tmax
Ans.
in-plane
= 192 psi
tan 2us =
300 - 150 = 1.25 120
us = -25.7°
Ans.
9–66. Determine the principal stress, the maximum in-plane shear stress, and average normal stress. Specify the orientation of the element in each case. A(45, -50)
B(30, 50)
30 MPa
C(37.5, 0)
45 MPa
R = CA = CB = 27.52 + 502 = 50.56 50 MPa
a)
tan 2uP =
50 7.5
s1 = 37.5 + 50.56 = 88.1 MPa
Ans.
s2 = 37.5 - 50.56 = -13.1 MPa
Ans.
uP = -40.7°
2uP = 81.47°
b) t
max in-plane
= R = 50.6 MPa
Ans.
savg = 37.5 MPa
Ans.
2us = 90 - 2uP us = 4.27°
Ans.
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9–67. Determine the principal stress, the maximum in-plane shear stress, and average normal stress. Specify the orientation of the element in each case.
200 MPa 500 MPa
350 MPa
Construction of the Circle: In accordance with the sign convention, sx = 350 MPa, sy = -200 MPa, and txy = 500 MPa. Hence, savg =
sx + sy 2
=
350 + (-200) = 75.0 MPa 2
Ans.
The coordinates for reference point A and C are A(350, 500)
C(75.0, 0)
The radius of the circle is R = 2(350 - 75.0)2 + 5002 = 570.64 MPa a) In - Plane Principal Stresses: The coordinate of points B and D represent s1 and s2 respectively. s1 = 75.0 + 570.64 = 646 MPa
Ans.
s2 = 75.0 - 570.64 = -496 MPa
Ans.
Orientaion of Principal Plane: From the circle tan 2uP1 =
500 = 1.82 350 - 75.0
uP1 = 30.6° (Counterclockwise)
Ans.
b) Maximum In - Plane Shear Stress: Represented by the coordinates of point E on the circle. t
max in-plane
= R = 571 MPa
Ans.
Orientation of the Plane for Maximum In - Plane Shear Stress: From the circle tan 2us =
350 - 75.0 = 0.55 500
us = 14.4° (Clockwise)
Ans.
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*9–68. Draw Mohr’s circle that describes each of the following states of stress.
700 psi
4 ksi 40 MPa
600 psi
(a)
a) Here, sx = 600 psi, sy = 700 psi and txy = 0. Thus, savg =
sx + sy =
2
600 + 700 = 650 psi 2
Thus, the coordinate of reference point A and center of circle are A(600, 0)
C(650, 0)
Then the radius of the circle is R = CA = 650 - 600 = 50 psi The Mohr’s circle represents this state of stress is shown in Fig. a. b) Here, sx = 0, sy = 4 ksi and txy = 0. Thus, savg =
sx + sy =
2
0 + 4 = 2 ksi 2
Thus, the coordinate of reference point A and center of circle are A(0, 0)
C(2, 0)
Then the radius of the circle is R = CA = 2 - 0 = 2 psi c) Here, sx = sy = 0 and txy = -40 MPa. Thus, savg =
sx + sy 2
= 0
Thus, the coordinate of reference point A and the center of circle are A(0, -40)
C(0, 0)
Then, the radius of the circle is R = CA = 40 MPa The Mohr’s circle represents this state of stress shown in Fig. c
693
(b)
(c)
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9–68.
Continued
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9–69. The frame supports the distributed loading of 200 N兾m. Determine the normal and shear stresses at point D that act perpendicular and parallel, respectively, to the grain. The grain at this point makes an angle of 30° with the horizontal as shown.
200 N/ m B
30⬚ 1m
200 mm
75 mm D
1.5 m
C 100 mm
4m 60⬚
E
Support Reactions: As shown on FBD(a).
50 mm
30 mm
Internal Forces and Moment: As shown on FBD(b).
1.5 m 100 mm
Section Properties: I =
A
1 (0.1) A 0.23 B = 66.667 A 10 - 6 B m4 12
QD = y¿A¿ = 0.0625(0.075)(0.1) = 0.46875 A 10 - 3 B m3 Normal Stress: Applying the flexure formula. sD = -
My 150(-0.025) = 56.25 kPa = I 66.667(10 - 6)
Shear Stress: Applying the shear formula. tD =
50.0 C 0.46875(10 - 3) D VQD = 3.516 kPa = It 66.667(10 - 6)(0.1)
Construction of the Circle: In accordance to the established sign convention, sx = 56.25 kPa, sy = 0 and txy = -3.516 kPa. Hence. savg =
sx + sy 2
=
56.25 + 0 = 28.125 kPa 2
The coordinates for reference point A and C are A(56.25, -3.516)
C(28.125, 0)
The radius of the circle is R = 2(56.25 - 28.125)2 + 3.5162 = 28.3439 kPa Stresses on The Rotated Element: The normal and shear stress components A sx¿ and tx¿y¿ B are represented by the coordinates of point P on the circle. Here, u = 60°. sx¿ = 28.125 - 28.3439 cos 52.875° = 11.0 kPa
Ans.
tx¿y¿ = -28.3439 sin 52.875° = -22.6 kPa
Ans.
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9–70. The frame supports the distributed loading of 200 N兾m. Determine the normal and shear stresses at point E that act perpendicular and parallel, respectively, to the grain. The grain at this point makes an angle of 60° with the horizontal as shown.
200 N/ m B
30⬚ 1m
200 mm
75 mm D
1.5 m
C 100 mm
4m 60⬚
E
Support Reactions: As shown on FBD(a).
50 mm
30 mm
Internal Forces and Moment: As shown on FBD(b).
1.5 m 100 mm
Section Properties:
A
A = 0.1(0.05) = 5.00 A 10 - 3 B m2 Normal Stress: sE =
N -250 = -50.0 kPa = A 5.00(10 - 3)
Construction of the Circle: In accordance with the sign convention. sx = 0, sy = -50.0 kPa, and txy = 0. Hence. savg =
sx + sy 2
=
0 + (-50.0) = -25.0 kPa 2
The coordinates for reference points A and C are A(0, 0)
C(-25.0, 0)
The radius of circle is R = 25.0 - 0 = 25.0 kPa Stress on the Rotated Element: The normal and shear stress components A sx¿ and tx¿y¿ B are represented by coordinates of point P on the circle. Here, u = 150°. sx = -25.0 + 25.0 cos 60° = -12.5 kPa
Ans.
tx¿y¿ = 25.0 sin 60° = 21.7 kPa
Ans.
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9–71. The stair tread of the escalator is supported on two of its sides by the moving pin at A and the roller at B. If a man having a weight of 300 lb stands in the center of the tread, determine the principal stresses developed in the supporting truck on the cross section at point C. The stairs move at constant velocity.
1.25 ft 30⬚ A C 1.5 ft
30⬚
0.5 ft B 0.5 ft
2 in.
Support Reactions: As shown on FBD (a). Internal Forces and Moment: As shown on FBD (b). Section Properties: A = 2(0.5) = 1.00 in2 I =
1 (0.5) A 23 B = 0.3333 in4 12
QB = y¿A¿ = 0.5(1)(0.5) = 0.250 in3 Normal Stress: s =
sC =
My N ; A I 475.48(0) -137.26 + = -137.26 psi 1.00 0.3333
Shear Stress: Applying the shear formula t =
tC =
VQ . It
79.25(0.250) = 118.87 psi 0.3333(0.5)
Construction of the Circle: In accordance with the sign convention, sx = 0, sy = -137.26 psi, and txy = 118.87 psi. Hence, savg =
sx + sy 2
=
0 + (-137.26) = -68.63 psi 2
The coordinates for reference points A and C are A(0, 118.87)
1 in. C
C(-68.63, 0)
The radius of the circle is R = 2(68.63 - 0)2 + 118.872 = 137.26 psi In - Plane Principal Stress: The coordinates of point B and D represent s1 and s2, respectively. s1 = -68.63 + 137.26 = 68.6 psi
Ans.
s2 = -68.63 - 137.26 = -206 psi
Ans.
697
0.5 in.
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*9–72. The thin-walled pipe has an inner diameter of 0.5 in. and a thickness of 0.025 in. If it is subjected to an internal pressure of 500 psi and the axial tension and torsional loadings shown, determine the principal stress at a point on the surface of the pipe.
200 lb
200 lb 20 lb⭈ft
Section Properties: A = p A 0.2752 - 0.252 B = 0.013125p in2 J =
p A 0.2754 - 0.254 B = 2.84768 A 10 - 3 B in4 2
Normal Stress: Since
0.25 r = = 10, thin wall analysis is valid. t 0.025
slong =
pr 500(0.25) 200 N + = + = 7.350 ksi A 2t 0.013125p 2(0.025)
shoop =
pr 500(0.25) = = 5.00 ksi t 0.025
Shear Stress: Applying the torsion formula, t =
20(12)(0.275) Tc = 23.18 ksi = J 2.84768(10 - 3)
Construction of the Circle: In accordance with the sign convention sx = 7.350 ksi, sy = 5.00 ksi, and txy = -23.18 ksi. Hence, savg =
sx + sy 2
=
7.350 + 5.00 = 6.175 ksi 2
The coordinates for reference points A and C are A(7.350, -23.18)
C(6.175, 0)
The radius of the circle is R = 2(7.350 - 6.175)2 + 23.182 = 23.2065 ksi In - Plane Principal Stress: The coordinates of point B and D represent s1 and s2, respectively. s1 = 6.175 + 23.2065 = 29.4 ksi
Ans.
s2 = 6.175 - 23.2065 = -17.0 ksi
Ans.
698
20 lb⭈ft
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•9–73. The cantilevered rectangular bar is subjected to the force of 5 kip. Determine the principal stress at point A. 1.5 in. A 1.5 in.
1 in. B
1.5 in.
1.5 in.
3 in. 1 in.
15 in.
3 in.
3 5 4
5 kip
Internal Forces and Moment: As shown on FBD. Section Properties: A = 3(6) = 18.0 in2 I =
1 (3) A 63 B = 54.0 in4 12
QA = y¿A¿ = 2.25(1.5)(3) = 10.125 in3 Normal Stress: s =
sA =
My N ; A I 45.0(1.5) 4.00 + = 1.4722 ksi 18.0 54.0
Shear Stress: Applying the shear formula t =
tA =
VQ . It
3.00(10.125) = 0.1875 ksi 54.0(3)
Construction of the Circle: In accordance with the sign convention, sx = 1.4722 ksi, sy = 0, and txy = -0.1875 ksi. Hence, savg =
sx + sy 2
=
1.472 + 0 = 0.7361 ksi 2
The coordinates for reference points A and C are A(1.4722, -0.1875)
C(0.7361, 0)
The radius of the circle is R = 2(1.4722 - 0.7361)2 + 0.18752 = 0.7596 ksi In - Plane Principal Stress: The coordinates of point B and D represent s1 and s2, respectively. s1 = 0.7361 + 0.7596 = 1.50 ksi
Ans.
s2 = 0.7361 - 0.7596 = -0.0235 ksi
Ans.
699
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9–74. Solve Prob. 9–73 for the principal stress at point B. 1.5 in. A 1.5 in.
1 in. B
1.5 in.
1.5 in.
3 in. 1 in.
15 in.
3 in.
3 5 4
5 kip
Internal Forces and Moment: As shown on FBD. Section Properties: A = 3(6) = 18.0 in2 1 (3) A 63 B = 54.0 in4 12
I =
QB = y¿A¿ = 2(2)(3) = 12.0 in3 Normal Stress: My N ; A I
s =
45.0(1) 4.00 = -0.6111 ksi 18.0 54.0
sB =
Shear Stress: Applying the shear formula t =
tB =
VQ . It
3.00(12.0) = 0.2222 ksi 54.0(3)
Construction of the Circle: In accordance with the sign convention, sx = -0.6111 ksi, sy = 0, and txy = -0.2222 ksi. Hence. savg =
sx + sy 2
=
-0.6111 + 0 = -0.3055 ksi 2
The coordinates for reference points A and C are A(-0.6111, -0.2222)
C(-0.3055, 0)
The radius of the circle is R = 2(0.6111 - 0.3055)2 + 0.22222 = 0.3778 ksi In - Plane Principal Stress: The coordinates of point B and D represent s1 and s2, respectively. s1 = -0.3055 + 0.3778 = 0.0723 ksi
Ans.
s2 = -0.3055 - 0.3778 = -0.683 ksi
Ans.
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9–75. The 2-in.-diameter drive shaft AB on the helicopter is subjected to an axial tension of 10 000 lb and a torque of 300 lb # ft. Determine the principal stress and the maximum in-plane shear stress that act at a point on the surface of the shaft.
s =
10 000 P = = 3.183 ksi A p(1)2
t =
300(12)(1) Tc = = 2.292 ksi p 4 J 2 (1) s1, 2 =
=
t
sx + sy ;
2
A
(
sx - sy 2
B A
)2 + t2xy
3.183 - 0 2 3.183 + 0 ; ( ) + (2.292)2 2 A 2
s1 = 4.38 ksi
Ans.
s2 = -1.20 ksi
Ans.
max in-plane
=
A
(
=
A
(
sx - sy 2
)2 + t2xy
3.183 - 0 2 ) + (2.292)2 2
= 2.79 ksi
Ans.
701
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*9–76. The pedal crank for a bicycle has the cross section shown. If it is fixed to the gear at B and does not rotate while subjected to a force of 75 lb, determine the principal stress in the material on the cross section at point C.
75 lb B 3 in.
A 4 in.
C 0.4 in. 0.4 in.
0.2 in. 0.3 in.
Internal Forces and Moment: As shown on FBD Section Properties: I =
1 (0.3) A 0.83 B = 0.0128 in3 12
QC = y¿A¿ = 0.3(0.2)(0.3) = 0.0180 in3 Normal Stress: Applying the flexure formula. sC = -
My -300(0.2) = = 4687.5 psi = 4.6875 ksi I 0.0128
Shear Stress: Applying the shear formula. tC =
VQC 75.0(0.0180) = = 351.6 psi = 0.3516 ksi It 0.0128(0.3)
Construction of the Circle: In accordance with the sign convention, sx = 4.6875 ksi, sy = 0, and txy = 0.3516 ksi. Hence, savg =
sx + sy 2
=
4.6875 + 0 = 2.34375 ksi 2
The coordinates for reference points A and C are A(4.6875, 0.3516)
C(2.34375, 0)
The radius of the circle is R = 2(4.6875 - 2.34375)2 + 0.35162 = 2.3670 ksi In - Plane Principal Stress: The coordinates of point B and D represent s1 and s2, respectively. s1 = 2.34375 + 2.3670 = 4.71 ksi
Ans.
s2 = 2.34375 - 2.3670 = -0.0262 ksi
Ans.
702
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•9–77.
A spherical pressure vessel has an inner radius of 5 ft and a wall thickness of 0.5 in. Draw Mohr’s circle for the state of stress at a point on the vessel and explain the significance of the result. The vessel is subjected to an internal pressure of 80 psi.
Normal Stress: s1 = s2 =
pr 80(5)(12) = = 4.80 ksi 2t 2(0.5)
Mohr’s circle: A(4.80, 0)
B(4.80, 0)
C(4.80, 0)
Regardless of the orientation of the element, the shear stress is zero and the state of stress is represented by the same two normal stress components.
9–78. The cylindrical pressure vessel has an inner radius of 1.25 m and a wall thickness of 15 mm. It is made from steel plates that are welded along the 45° seam. Determine the normal and shear stress components along this seam if the vessel is subjected to an internal pressure of 8 MPa.
sx =
45⬚ 1.25 m
pr 8(1.25) = = 333.33 MPa 2t 2(0.015)
sy = 2sx = 666.67 MPa A(333.33, 0) sx¿ =
B(666.67, 0)
C(500, 0)
333.33 + 666.67 = 500 MPa 2
Ans.
tx¿y¿ = R = 666.67 - 500 = 167 MPa
Ans.
703
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•9–79.
Determine the normal and shear stresses at point D that act perpendicular and parallel, respectively, to the grains. The grains at this point make an angle of 30° with the horizontal as shown. Point D is located just to the left of the 10-kN force.
10 kN
A 100 mm
D
B 30⬚
1m
100 mm
D
100 mm
Using the method of section and consider the FBD of the left cut segment, Fig. a + c ©Fy = 0;
5 - V = 0
a + ©MC = 0;
V = 5 kN M = 5 kN # m
M - 5(1) = 0
The moment of inertia of the rectangular cross - section about the neutral axis is I =
1 (0.1)(0.33) = 0.225(10 - 3) m4 12
Referring to Fig. b, QD = y¿A¿ = 0.1(0.1)(0.1) = 0.001 m3 The normal stress developed is contributed by bending stress only. For point D, y = 0.05 m. Then s =
My 5(103)(0.05) = 1.111 MPa (T) = I 0.225(10 - 3)
The shear stress is contributed by the transverse shear stress only. Thus, t =
5(103)(0.001) VQD = 0.2222 MPa = It 0.225(10 - 3)(0.1)
The state of stress at point D can be represented by the element shown in Fig. c In accordance to the established sign convention, sx = 1.111 MPa, sy = 0 and txy = -0.2222 MPa, Thus. savg =
sx + sy 2
=
1.111 + 0 = 0.5556 MPa 2
Then, the coordinate of reference point A and the center C of the circle are A(1.111, -0.2222)
C(0.5556, 0)
Thus, the radius of the circle is given by R = 2(1.111 - 0.5556)2 + (-0.2222)2 = 0.5984 MPa Using these results, the circle shown in Fig. d can be constructed. Referring to the geometry of the circle, Fig. d, a = tan - 1 a
0.2222 b = 21.80° 1.111 - 0.5556
b = 180° - (120° - 21.80°) = 81.80°
704
1m 300 mm
2m C
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9–79. Continued Then sx¿ = 0.5556 - 0.5984 cos 81.80° = 0.4702 MPa = 470 kPa
Ans.
tx¿y¿ = 0.5984 sin 81.80° = 0.5922 MPa = 592 kPa
Ans.
705
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*9–80. Determine the principal stress at point D, which is located just to the left of the 10-kN force.
10 kN
A 100 mm
D
B 30⬚
1m
100 mm
D
100 mm
Using the method of section and consider the FBD of the left cut segment, Fig. a, + c ©Fy = 0;
5 - V = 0
a + ©MC = 0;
V = 5 kN M = 5 kN # m
M - 5(1) = 0 I =
1 (0.1)(0.33) = 0.225(10 - 3) m4 12
Referring to Fig. b, QD = y¿A¿ = 0.1(0.1)(0.1) = 0.001 m3 The normal stress developed is contributed by bending stress only. For point D, y = 0.05 m s =
My 5(103)(0.05) = 1.111 MPa (T) = I 0.225(10 - 3)
The shear stress is contributed by the transverse shear stress only. Thus, t =
5(103)(0.001) VQD = 0.2222 MPa = It 0.225(10 - 3)(0.1)
The state of stress at point D can be represented by the element shown in Fig. c. In accordance to the established sign convention, sx = 1.111 MPa, sy = 0, and txy = -0.2222 MPa. Thus, savg =
sx + sy 2
=
1.111 + 0 = 0.5556 MPa 2
Then, the coordinate of reference point A and center C of the circle are A(1.111, -0.2222)
C(0.5556, 0)
Thus, the radius of the circle is R = CA = 2(1.111 - 0.5556)2 + (-0.2222)2 = 0.5984 MPa Using these results, the circle shown in Fig. d. In-Plane Principal Stresses. The coordinates of points B and D represent s1 and s2, respectively. Thus, s1 = 0.5556 + 0.5984 = 1.15 MPa
Ans.
s2 = 0.5556 - 0.5984 = -0.0428 MPa
Ans.
706
1m 300 mm
2m C
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9–80. Continued Referring to the geometry of the circle, Fig. d, tan (2uP)1 =
0.2222 = 0.4 1.111 - 0.5556
(uP)1 = 10.9° (Clockwise)
Ans.
The state of principal stresses is represented by the element show in Fig. e.
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•9–81.
Determine the principal stress at point A on the cross section of the hanger at section a–a. Specify the orientation of this state of stress and indicate the result on an element at the point.
0.75 m
0.75 m a
250 mm a
900 N
Internal Loadings: Considering the equilibrium of the free - body diagram of the hanger’s left cut segment, Fig. a, + ©F = 0; : x
900 - N = 0
N = 900 N
+ c ©Fy = 0;
V - 900 = 0
V = 900 N
a + ©MO = 0;
900(1) - 900(0.25) - M = 0
M = 675 N # m
b
250 mm
Section Properties: The cross - sectional area and the moment of inertia about the centroidal axis of the hanger’s cross section are A = 0.05(0.1) - 0.04(0.09) = 1.4 A 10 - 3 B m2 1 1 (0.05) A 0.13 B (0.04) A 0.093 B = 1.7367 A 10 - 6 B m4 12 12
Referring to Fig. b, QA = 2y1œ A1œ + y2œ A2œ = 2[0.0375(0.025)(0.005)] + 0.0475(0.005)(0.04) = 18.875 A 10 - 6 B m3 Normal and Shear Stress: The normal stress is a combination of axial and bending stresses. Thus, sA =
675(0.025) MyA N 900 + = + = 9.074 MPa -3 A I 1.4 A 10 B 1.7367 A 10 - 6 B
The shear stress is caused by the transverse shear stress. tA =
900 C 18.875 A 10 - 6 B D VQA = = 0.9782 MPa It 1.7367 A 10 - 6 B (0.01)
The state of stress at point A is represented by the element shown in Fig. c. Construction of the Circle: sx = 9.074 MPa, sy = 0, and txy = 0.9782 MPa. Thus, savg =
sx + sy 2
=
9.074 + 0 = 4.537 MPa 2
The coordinates of reference points A and the center C of the circle are A(9.074, 0.9782)
C(4.537, 0)
Thus, the radius of the circle is R = CA = 2(9.074 - 4.537)2 + 0.97822 = 4.641 MPa Using these results, the circle is shown in Fig. d.
708
b
900 N 5 mm
25 mm A 100 mm
5 mm 50 mm
I =
0.5 m
5 mm
Sections a – a and b – b
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9–81. Continued In - Plane Principal Stress: The coordinates of point B and D represent s1 and s2, respectively. s1 = 4.537 + 4.641 = 9.18 MPa
Ans.
s2 = 4.537 - 4.641 = -0.104 MPa
Ans.
Orientaion of Principal Plane: Referring to the geometry of the circle, Fig. d, tan 2 A uP B 1 =
0.9782 = 0.2156 9.074 - 4.537
A uP B 1 = 6.08° (counterclockwise)
Ans.
The state of principal stresses is represented on the element shown in Fig. e.
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9–82. Determine the principal stress at point A on the cross section of the hanger at section b–b. Specify the orientation of the state of stress and indicate the results on an element at the point.
0.75 m
0.75 m a
250 mm a
900 N
Internal Loadings: Considering the equilibrium of the free - body diagram of the hanger’s left cut segment, Fig. a, V - 900 - 900 = 0
+ c ©Fy = 0; a + ©MO = 0;
900(2.25) + 900(0.25) - M = 0
Referring to Fig. b. QA = 2y1œ A1œ + y2œ A2œ = 2[0.0375(0.025)(0.005)] + 0.0475(0.005)(0.04) = 18.875 A 10 - 6 B m3 Normal and Shear Stress: The normal stress is contributed by the bending stress only. MyA 2250(0.025) = = 32.39 MPa I 1.7367 A 10 - 6 B
The shear stress is contributed by the transverse shear stress only. 1800 C 18.875 A 10 - 6 B D VQA = = 1.956 MPa It 1.7367 A 10 - 6 B (0.01)
The state stress at point A is represented by the element shown in Fig. c. Construction of the Circle: sx = 32.39 MPa, sy = 0, and txy = 1.956 MPa. Thus, savg =
sx + sy 2
=
32.39 + 0 = 16.19 MPa 2
The coordinates of reference point A and the center C of the circle are A(32.39, 1.956)
C(16.19, 0)
Thus, the radius of the circle is R = CA = 2(32.39 - 16.19)2 + 1.9562 = 16.313 MPa Using these results, the cricle is shown in Fig. d.
710
b
5 mm 25 mm A 100 mm
5 mm 50 mm
5 mm
Sections a – a and b – b
1 1 (0.05) A 0.13 B (0.04) A 0.093 B = 1.7367 A 10 - 6 B m4 12 12
tA =
250 mm
M = 2250 N # m
A = 0.05(0.1) - 0.04(0.09) = 1.4 A 10 - 3 B m2
sA =
b
900 N
V = 1800 N
Section Properties: The cross - sectional area and the moment of inertia about the centroidal axis of the hanger’s cross section are
I =
0.5 m
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9–82. Continued In - Plane Principal Stresses: The coordinates of reference point B and D represent s1 and s2, respectively. s1 = 16.19 + 16.313 = 32.5 MPa
Ans.
s2 = 16.19 - 16.313 = -0.118 MPa
Ans.
Orientaion of Principal Plane: Referring to the geometry of the circle, Fig. d, tan 2 A uP B 1 =
A uP B 1 = 3.44°
1.956 = 0.1208 32.39 - 16.19 Ans.
(counterclockwise)
The state of principal stresses is represented on the element shown in Fig. e.
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9–83. Determine the principal stresses and the maximum in-plane shear stress that are developed at point A. Show the results on an element located at this point. The rod has a diameter of 40 mm.
450 N
150 mm
Using the method of sections and consider the FBD of the member’s upper cut segment, Fig. a, + c ©Fy = 0;
450 - N = 0
a + ©MC = 0;
100 mm A
150 mm
N = 450 N
450(0.1) - M = 0
B
M = 45 N # m
A = p(0.022) = 0.4(10 - 3)p m2
I =
450 N
p (0.024) = 40(10 - 9)p m4 4
The normal stress is the combination of axial and bending stress. Thus, s =
My N + A I
For point A, y = C = 0.02 m. s =
45 (0.02) 450 + = 7.520 MPa 0.4(10 - 3)p 40(10 - 9)p
Since no transverse shear and torque is acting on the cross - section t = 0 The state of stress at point A can be represented by the element shown in Fig. b. In accordance to the established sign convention sx = 0, sy = 7.520 MPa and txy = 0. Thus savg =
sx + sy 2
=
0 + 7.520 = 3.760 MPa 2
Then, the coordinates of reference point A and the center C of the circle are A(0, 0)
C(3.760, 0)
Thus, the radius of the circle is R = CA = 3.760 MPa Using this results, the circle shown in Fig. c can be constructed. Since no shear stress acts on the element, s1 = sy = 7.52 MPa
s2 = sx = 0
Ans.
The state of principal stresses can also be represented by the element shown in Fig. b. The state of maximum in - plane shear stress is represented by point B on the circle, Fig. c. Thus. tmax
in-plane
= R = 3.76 MPa
Ans.
712
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9–83. Continued From the circle, 2us = 90° us = 45° (counter clockwise)
Ans.
The state of maximum In - Plane shear stress can be represented by the element shown in Fig. d.
713
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*9–84. Draw the three Mohr’s circles that describe each of the following states of stress.
5 ksi
(a) Here, smin = 0, sint = 3 ksi and smax = 5 ksi. The three Mohr’s circle of this state of stress are shown in Fig. a
3 ksi
(b) Here, smin = 0, sint = 140 MPa and smax = 180 MPa. The three Mohr’s circle of this state of stress are shown in Fig. b (a)
•9–85.
Draw the three Mohr’s circles that describe the following state of stress.
180 MPa
140 MPa (b)
300 psi
Here, smin = -300 psi, sint = 0 and smax = 400 psi. The three Mohr’s circle for this state of stress is shown in Fig. a. 400 psi
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z
9–86. The stress at a point is shown on the element. Determine the principal stress and the absolute maximum shear stress.
y
x 80 MPa
For y – z plane: A(0, -80)
B(90, 80)
C(45, 0)
R = 2452 + 802 = 91.79 s1 = 45 + 91.79 = 136.79 MPa s2 = 45 - 91.79 = -46.79 MPa Thus,
tabs
max
=
s1 = 0
Ans.
s2 = 137 MPa
Ans.
s3 = -46.8 MPa
Ans.
136.79 - (-46.79) smax - smin = = 91.8 MPa 2 2
Ans.
715
90 MPa
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z
9–87. The stress at a point is shown on the element. Determine the principal stress and the absolute maximum shear stress. Mohr’s circle for the element in y - 7 plane, Fig. a, will be drawn first. In accordance to the established sign convention, sy = 30 psi, sz = 120 psi and tyz = 70 psi. Thus savg =
sy + sz 2
=
x
y
120 psi 70 psi
30 + 120 = 75 psi 2 30 psi
Thus the coordinates of reference point A and the center C of the circle are A(30, 70)
C(75, 0)
Thus, the radius of the circle is R = CA = 2(75 - 30)2 + 702 = 83.217 psi Using these results, the circle shown in Fig. b. The coordinates of point B and D represent the principal stresses From the results, smax = 158 psi
smin = -8.22 psi
sint = 0 psi
Ans.
Using these results, the three Mohr’s circle are shown in Fig. c, From the geometry of the three circles, tabs
max
=
158.22 - ( -8.22) smax - smin = = 83.22 psi 2 2
716
Ans.
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z
*9–88. The stress at a point is shown on the element. Determine the principal stress and the absolute maximum shear stress. Mohr’s circle for the element in x - z plane, Fig. a, will be drawn first. In accordance to the established sign convention, sx = -2 ksi, sz = 0 and txz = 8 ksi. Thus savg =
sx + sz 2
=
-2 + 0 = -1 ksi 2
2 ksi 8 ksi
Thus, the coordinates of reference point A and the center C of the circle are A( -2, 8)
C(-1, 0)
Thus, the radius of the circle is R = CA = 2[-2 - (-1)]2 + 82 = 265 ksi Using these results, the circle in shown in Fig. b, The coordinates of points B and D represent s1 and s2, respectively. s = -1 + 265 = 7.062 ksi smax = 7.06 ksi sint = 0 smin = -9.06 ksi From the results obtained, sint = 0 ksi
smax = 7.06 ksi
smin = -9.06 ksi
Ans.
Using these results, the three Mohr’s circles are shown in Fig, c. From the geometry of the cricle, tabs
max
=
y
x
7.06 - (-9.06) smax - smin = = 8.06 ksi 2 2
Ans.
717
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•9–89.
The stress at a point is shown on the element. Determine the principal stress and the absolute maximum shear stress.
z
y
x
For x – y plane:
150 MPa
120 MPa
R = CA = 2(120 - 60)2 + 1502 = 161.55 s1 = 60 + 161.55 = 221.55 MPa s2 = 60 - 161.55 = -101.55 MPa s1 = 222 MPa tabs
max
=
s2 = 0 MPa
s3 = -102 MPa
Ans.
221.55 - (-101.55) smax - smin = = 162 MPa 2 2
Ans.
9–90. The state of stress at a point is shown on the element. Determine the principal stress and the absolute maximum shear stress.
z
x
For y - z plane: A(5, -4)
B(-2.5, 4)
4 ksi
s1 = 1.25 + 5.483 = 6.733 ksi
5 ksi
s2 = 1.25 - 5.483 = -4.233 ksi Thus,
tabs
max
=
2.5 ksi
C(1.25, 0)
R = 23.752 + 42 = 5.483
savg =
y
s1 = 6.73 ksi
Ans.
s2 = 0
Ans.
s3 = -4.23 ksi
Ans.
6.73 + (-4.23) = 1.25 ksi 2 6.73 - (-4.23) smax - smin = = 5.48 ksi 2 2
Ans.
718
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*9–92. The solid shaft is subjected to a torque, bending moment, and shear force as shown. Determine the principal stress acting at points A and B and the absolute maximum shear stress.
450 mm A B 300 N⭈m
25 mm
45 N⭈m 800 N
Internal Forces and Moment: As shown on FBD. Section Properties: Iz =
p A 0.0254 B = 0.306796 A 10 - 6 B m4 4
J =
p A 0.0254 B = 0.613592 A 10 - 6 B m4 2
(QA)x = 0 (QB)y = y¿A¿ =
4(0.025) 1 c (p) A 0.0252 B d = 10.417 A 10 - 6 B m3 3p 2
Normal stress: Applying the flexure formula. s = -
Mzy Iz -60.0(0.025)
sA = -
0.306796(10 - 6)
= 4.889 MPa
-60.0(0) sB = -
0.306796(10 - 6)
= 0
Shear Stress: Applying the torsion formula for point A, tA =
45.0(0.025) Tc = 1.833 MPa = J 0.613592(10 - 6)
The transverse shear stress in the y direction and the torsional shear stress can be VQ Tr obtained using shear formula and torsion formula. tv = and ttwist = , It J respectively. tB = (tv)y - ttwist =
800 C 10.417(10 - 6) D -6
0.306796(10 )(0.05)
45.0(0.025) -
0.613592(10 - 6)
= -1.290 MPa
Construction of the Circle: sx = 4.889 MPa, sz = 0, and txz = -1.833 MPa for point A. Hence, savg =
sx + sz 2
=
4.889 + 0 = 2.445 MPa 2
The coordinates for reference points A and C are A (4.889, –1.833) and C(2.445, 0).
719
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9–92.
Continued
The radius of the circle is R = 2(4.889 - 2.445)2 + 1.8332 = 3.056 MPa sx = sy = 0 and txy = -1.290 MPa for point B. Hence, savg =
sx + sz
= 0
2
The coordinates for reference points A and C are A(0. ‚–1.290) and C(0,0). The radius of the circle is R = 1.290 MPa In - Plane Principal Stresses: The coordinates of point B and D represent s1 and s2, respectively. For point A s1 = 2.445 + 3.056 = 5.50 MPa s2 = 2.445 - 3.506 = -0.611 MPa For point B s1 = 0 + 1.290 = 1.29 MPa s2 = 0 - 1.290 = -1.290 MPa Three Mohr’s Circles: From the results obtaired above, the principal stresses for point A are smax = 5.50 MPa
sint = 0
smin = -0.611 MPa
Ans.
sint = 0
smin = -1.29 MPa
Ans.
And for point B smax = 1.29 MPa
Absolute Maximum Shear Stress: For point A, tabs
max
=
5.50 - (-0.611) smax - smin = = 3.06 MPa 2 2
Ans.
1.29 - (-1.29) smax - smin = = 1.29 MPa 2 2
Ans.
For point B, tabs
max
=
720
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•9–93.
The propane gas tank has an inner diameter of 1500 mm and wall thickness of 15 mm. If the tank is pressurized to 2 MPa, determine the absolute maximum shear stress in the wall of the tank.
Normal Stress: Since
750 r = = 50 7 10, thin - wall analysis can be used. We have t 15 s1 =
2(750) pr = = 100 MPa t 15
s2 =
2(750) pr = = 50 MPa 2t 2(15)
The state of stress of any point on the wall of the tank can be represented on the element shown in Fig. a Construction of Three Mohr’s Circles: Referring to the element, smax = 100 MPa
sint = 50 MPa
smin = 0
Using these results, the three Mohr’s circles are shown in Fig. b. Absolute Maximum Shear Stress: From the geometry of three circles, tabs
max
=
smax - smin 100 - 0 = = 50 MPa 2 2
Ans.
721
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9–94. Determine the principal stress and absolute maximum shear stress developed at point A on the cross section of the bracket at section a–a.
12 in.
6 in.
5
3
a
4
a 0.5 in. B
0.25 in. A
0.25 in.
0.25 in. 1.5 in.1.5 in. Section a – a
Internal Loadings: Considering the equilibrium of the free - body diagram of the bracket’s upper cut segment, Fig. a, + c ©Fy = 0;
3 N - 500 a b = 0 5
N = 300 lb
+ ©F = 0; ; x
4 V - 500 a b = 0 5
V = 400 lb
3 4 ©MO = 0; M - 500 a b(12) - 500 a b(6) = 0 5 5
M = 6000 lb # in
Section Properties: The cross - sectional area and the moment of inertia of the bracket’s cross section are A = 0.5(3) - 0.25(2.5) = 0.875 in2 I =
1 1 (0.5) A 33 B (0.25) A 2.53 B = 0.79948 in4 12 12
Referring to Fig. b. QA = x1œ A1œ + x2œ A2œ = 0.625(1.25)(0.25) + 1.375(0.25)(0.5) = 0.3672 in3 Normal and Shear Stress: The normal stress is sA =
N 300 = = -342.86 psi A 0.875
The shear stress is contributed by the transverse shear stress. tA =
400(0.3672) VQA = = 734.85 psi It 0.79948(0.25)
The state of stress at point A is represented by the element shown in Fig. c. Construction of the Circle: sx = 0, sy = -342.86 psi, and txy = 734.85. Thus, savg =
sx + sy 2
=
0 + (-342.86) = -171.43 psi 2
The coordinates of reference point A and the center C of the circle are A(0, 734.85)
C(-171.43, 0)
Thus, the radius of the circle is R = CA = 2[0 - (-171.43)]2 + 734.852 = 754.58 psi 722
500 lb
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9–94. Continued Using these results, the cricle is shown in Fig. d. In - Plane Principal Stresses: The coordinates of reference point B and D represent s1 and s2, respectively. s1 = -171.43 + 754.58 = 583.2 psi s2 = -171.43 - 754.58 = -926.0 psi Three Mohr’s Circles: Using these results, smax = 583 psi
sint = 0 smin = -926 psi
Ans.
Absolute Maximum Shear Stress: tabs
max
=
583.2 - (-926.0) smax - smin = - 755 psi 2 2
Ans.
723
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9–95. Determine the principal stress and absolute maximum shear stress developed at point B on the cross section of the bracket at section a–a.
12 in.
Internal Loadings: Considering the equilibrium of the free - body diagram of the 6 in. bracket’s upper cut segment, Fig. a, a
+ c ©Fy = 0; + ©F = 0; ; x
3 N - 500 a b = 0 5
N = 300 lb
4 V - 500 a b = 0 5
V = 400 lb
1 1 (0.5) A 33 B (0.25) A 2.53 B = 0.79948 in4 12 12
Referring to Fig. b, QB = 0 Normal and Shear Stress: The normal stress is a combination of axial and bending stress. 6000(1.5) MxB N 300 + = + = 10.9 ksi A I 0.875 0.79948
Since QB = 0, tB = 0. The state of stress at point B is represented on the element shown in Fig. c. In - Plane Principal Stresses: Since no shear stress acts on the element, s2 = 0
Three Mohr’s Circles: Using these results, smax = 10.91 ksi
sint = smin = 0
Ans.
Absolute Maximum Shear Stress: tabs
max
=
0.25 in. 1.5 in.1.5 in. Section a – a
M = 6000 lb # in
smax - smin 10.91 - 0 = = 5.46 ksi 2 2
Ans.
724
500 lb
0.25 in. A
0.25 in.
A = 0.5(3) - 0.25(2.5) = 0.875 in2
s1 = 10.91 ksi
a 0.5 in.
Section Properties: The cross - sectional area and the moment of inertia about the centroidal axis of the bracket’s cross section are
sB =
4
B
4 3 ©MO = 0; M - 500 a b(12) - 500 a b(6) = 0 5 5
I =
5
3
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*9–96. The solid propeller shaft on a ship extends outward from the hull. During operation it turns at v = 15 rad>s when the engine develops 900 kW of power. This causes a thrust of F = 1.23 MN on the shaft. If the shaft has an outer diameter of 250 mm, determine the principal stresses at any point located on the surface of the shaft.
0.75 m A T
Power Transmission: Using the formula developed in Chapter 5, P = 900 kW = 0.900 A 106 B N # m>s 0.900(106) P = = 60.0 A 103 B N # m v 15
T0 =
Internal Torque and Force: As shown on FBD. Section Properties: A =
p A 0.252 B = 0.015625p m2 4
J =
p A 0.1254 B = 0.3835 A 10 - 3 B m4 2
Normal Stress: s =
-1.23(106) N = = -25.06 MPa A 0.015625p
Shear Stress: Applying the torsion formula, t =
60.0(103) (0.125) Tc = 19.56 MPa = J 0.3835(10 - 3)
In - Plane Principal Stresses: sx = -25.06 MPa, sy = 0 and txy = 19.56 MPa for any point on the shaft’s surface. Applying Eq. 9-5, s1,2 =
=
sx + sy 2
;
C
a
sx - sy 2
2
b + t2xy
-25.06 - 0 2 -25.06 + 0 ; a b + (19.56)2 2 C 2
= -12.53 ; 23.23 s1 = 10.7 MPa
s2 = -35.8 MPa
Ans.
725
F
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•9–97. The solid propeller shaft on a ship extends outward from the hull. During operation it turns at v = 15 rad>s when the engine develops 900 kW of power. This causes a thrust of F = 1.23 MN on the shaft. If the shaft has a diameter of 250 mm, determine the maximum in-plane shear stress at any point located on the surface of the shaft.
0.75 m A T
Power Transmission: Using the formula developed in Chapter 5, P = 900 kW = 0.900 A 106 B N # m>s T0 =
0.900(106) P = = 60.0 A 103 B N # m v 15
Internal Torque and Force: As shown on FBD. Section Properties: A =
p A 0.252 B = 0.015625p m2 4
J =
p A 0.1254 B = 0.3835 A 10 - 3 B m4 2
Normal Stress: s =
-1.23(106) N = = -25.06 MPa A 0.015625p
Shear Stress: Applying the torsion formula. t =
60.0(103) (0.125) Tc = 19.56 MPa = J 0.3835 (10 - 3)
Maximum In - Plane Principal Shear Stress: sx = -25.06 MPa, sy = 0, and txy = 19.56 MPa for any point on the shaft’s surface. Applying Eq. 9-7, t
max in-plane
a
sx - sy
2
b + t2xy
=
C
=
-25.06 - 0 2 b + (19.56)2 C 2
2
a
= 23.2 MPa
Ans.
726
F
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9–98. The steel pipe has an inner diameter of 2.75 in. and an outer diameter of 3 in. If it is fixed at C and subjected to the horizontal 20-lb force acting on the handle of the pipe wrench at its end, determine the principal stresses in the pipe at point A, which is located on the surface of the pipe.
20 lb 12 in.
10 in.
Internal Forces, Torque and Moment: As shown on FBD. A
Section Properties:
B
I =
p A 1.54 - 1.3754 B = 1.1687 in4 4
J =
p A 1.54 - 1.3754 B = 2.3374 in4 2
C y z
(QA)z = ©y¿A¿
x
4(1.5) 1 4(1.375) 1 = c p A 1.52 B d c p A 1.3752 B d 3p 2 3p 2 = 0.51693 in3 Normal Stress: Applying the flexure formula s =
sA =
My z Iy
,
200(0) = 0 1.1687
Shear Stress: The transverse shear stress in the z direction and the torsional shear VQ stress can be obtained using shear formula and torsion formula, tv = and It Tr ttwist = , respectively. J tA = (tv)z - ttwist =
20.0(0.51693) 240(1.5) 1.1687(2)(0.125) 2.3374
= -118.6 psi In - Plane Principal Stress: sx = 0, sz = 0 and txz = -118.6 psi for point A. Applying Eq. 9-5 s1,2 =
sx + sz 2
;
C
a
sx - sz 2
2
b + t2xz
= 0 ; 20 + (-118.6)2 s1 = 119 psi
s2 = -119 psi
Ans.
727
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9–99. Solve Prob. 9–98 for point B, which is located on the surface of the pipe.
20 lb 12 in.
10 in.
A B
Internal Forces, Torque and Moment: As shown on FBD. Section Properties:
C
I =
p A 1.54 - 1.3754 B = 1.1687 in4 4
y z x
p J = A 1.54 - 1.3754 B = 2.3374 in4 2 (QB)z = 0 Normal Stress: Applying the flexure formula s =
sB =
My z Iv
,
200(1.5) = 256.7 psi 1.1687
Shear Stress: Torsional shear stress can be obtained using torsion formula, Tr . ttwist = J tB = ttwist =
240(1.5) = 154.0 psi 2.3374
In - Plane Prinicipal Stress: sx = 256.7 psi, sy = 0, and txy = -154.0 psi for point B. Applying Eq. 9-5 s1,2 =
=
sx + sy 2
;
C
sx - sy
a
2
2
b + t2xy
256.7 - 0 2 256.7 + 0 ; a b + ( -154.0)2 2 C 2
= 128.35 ; 200.49 s1 = 329 psi
s2 = -72.1 psi
Ans.
728
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*9–100. The clamp exerts a force of 150 lb on the boards at G. Determine the axial force in each screw, AB and CD, and then compute the principal stresses at points E and F. Show the results on properly oriented elements located at these points. The section through EF is rectangular and is 1 in. wide.
A
150 lb
C
G 0.5 in. E
Support Reactions: FBD(a). a + ©MB = 0; + c ©Fy = 0;
F
FCD(3) - 150(7) = 0
FCD = 350 lb
Ans.
350 - 150 - FAB = 0
FAB = 200 lb
Ans.
B
1.5 in. 1.5 in.
Internal Forces and Moment: As shown on FBD(b). Section Properties: I =
1 (1) A 1.53 B = 0.28125 in4 12
QE = 0 QF = y¿A¿ = 0.5(0.5)(1) = 0.250 in3 Normal Stress: Applying the flexure formula s = -
My , I
sE = -
-300(0.75) = 800 psi 0.28125
sF = -
-300(0.25) = 266.67 psi 0.28125 VQ , It
Shear Stress: Applying the shear formula t =
tE =
200(0) = 0 0.28125(1)
tF =
200(0.250) = 177.78 psi 0.28125(1)
In - Plane Principal Stress: sx = 800 psi, sy = 0 and txy = 0 for point E. Since no shear stress acts upon the element. s1 = sx = 800 psi
Ans.
s2 = sy = 0
Ans.
sx = 266.67 psi, sy = 0, and txy = 177.78 psi for point F. Applying Eq. 9-5 s1,2 =
=
sx + sy 2
;
C
sx - sy
a
2
2
b + t2xy
266.67 - 0 2 266.67 + 0 ; a b + 177.782 2 C 2
= 133.33 ; 222.22 s1 = 356 psi
s2 = -88.9 psi
Ans.
729
150 lb
D 4 in.
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9–100.
Continued
Orientation of Principal Plane: Applying Eq. 9-4 for point F, tan 2up =
txy
A sx - sy B >2
up = 26.57°
=
and
177.78 = 1.3333 (266.67 - 0)>2 -63.43°
Substituting the results into Eq. 9-1 with u = 26.57° yields sx¿ =
=
sx + sy 2
sx - sy +
2
cos 2u + txy sin 2u
266.67 - 0 266.67 + 0 + cos 53.13° + 177.78 sin 53.13° 2 2
= 356 psi = s1 Hence, up1 = 26.6°
up2 = -63.4°
Ans.
730
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9–101. The shaft has a diameter d and is subjected to the loadings shown. Determine the principal stress and the maximum in-plane shear stress that is developed anywhere on the surface of the shaft.
F
T0
F T0
Internal Forces and Torque: As shown on FBD(b). Section Properties: A =
p 2 d 4
J =
p d 4 p 4 a b = d 2 2 32
Normal Stress: N -F 4F = p 2 = - 2 A pd 4 d
s =
Shear Stress: Applying the shear torsion formula, t =
T0 A d2 B 16T0 Tc = p 4 = J d pd3 32
16T0 4F , sy = 0, and txy = for any point on pd2 pd3 the shaft’s surface. Applying Eq. 9-5, In - Plane Principal Stress: sx = -
s1,2 =
sx + sy ;
2 - 4F2 pd
=
C
a
+ 0 ;
2
D
sx - sy 2
¢
- 4F2 pd
2
b + t2xy
- 0
2
2
≤ + a-
16T0 3
pd
b
2
=
64T20 2 -F ; F2 + ≤ 2 ¢ C pd d2
s1 =
64T20 2 -F + F2 + ≤ 2 ¢ C pd d2
Ans.
64T20 2 F + F2 + ≤ 2 ¢ C pd d2
Ans.
s2 = -
Maximum In - Plane Shear Stress: Applying Eq. 9-7, t
max in-plane
=
=
=
C
a
D
¢
sx - sy 2 - 4F2 pd 2
2
b + t2xy
- 0
2
≤ + a-
16T0 pd3
b
2
64T20 2 2 F + pd2 C d2
Ans.
731
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9–102. The state of stress at a point in a member is shown on the element. Determine the stress components acting on the plane AB.
A 50 MPa
30⬚
28 MPa
100 MPa B
Construction of the Circle: In accordance with the sign convention, sx = -50 MPa, sy = -100 MPa, and txy = -28 MPa. Hence, savg =
sx + sy 2
=
-50 + (-100) = -75.0 MPa 2
The coordinates for reference points A and C are A(–50, –28) and C(–75.0, 0). The radius of the circle is R = 2(75.0 - 50)2 + 282 = 37.54 MPa. Stress on the Rotated Element: The normal and shear stress components A sx¿ and tx¿y¿ B are represented by the coordinates of point P on the circle sx¿ = -75.0 + 37.54 cos 71.76° = -63.3 MPa
Ans.
tx¿y¿ = 37.54 sin 71.76° = 35.7 MPa
Ans.
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9–103. The propeller shaft of the tugboat is subjected to the compressive force and torque shown. If the shaft has an inner diameter of 100 mm and an outer diameter of 150 mm, determine the principal stress at a point A located on the outer surface. 10 kN
A 2 kN·m
Internal Loadings: Considering the equilibrium of the free - body diagram of the propeller shaft’s right segment, Fig. a, ©Fx = 0; 10 - N = 0
N = 10 kN
©Mx = 0; T - 2 = 0
T = 2 kN # m
Section Properties: The cross - sectional area and the polar moment of inertia of the propeller shaft’s cross section are A = p A 0.0752 - 0.052 B = 3.125p A 10 - 3 B m2 J =
p A 0.0754 - 0.054 B = 12.6953125p A 10 - 6 B m4 2
Normal and Shear Stress: The normal stress is a contributed by axial stress only. sA =
10 A 103 B N = = -1.019 MPa A 3.125p A 10 - 3 B
The shear stress is contributed by the torsional shear stress only. tA =
2 A 103 B (0.075) Tc = = 3.761 MPa J 12.6953125p A 10 - 6 B
The state of stress at point A is represented by the element shown in Fig. b. Construction of the Circle: sx = -1.019 MPa, sy = 0, and txy = -3.761 MPa. Thus, savg =
sx + sy 2
=
-1.019 + 0 = -0.5093 MPa 2
The coordinates of reference point A and the center C of the circle are A(-1.019, -3.761)
C(-0.5093, 0)
Thus, the radius of the circle is R = CA = 2[-1.019 - ( -0.5093)]2 + (-3.761)2 = 3.795 MPa Using these results, the circle is shown is Fig. c. In - Plane Principal Stress: The coordinates of reference points B and D represent s1 and s2, respectively. s1 = -0.5093 + 3.795 = 3.29 MPa
Ans.
s2 = -0.5093 - 3.795 = -4.30 MPa
Ans.
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9–103.
Continued
Orientation of the Principal Plane: Referring to the geometry of the circle, Fig. d, tan 2 A up B 2 =
3.761 = 7.3846 1.019 - 0.5093
A up B 2 = 41.1° (clockwise)
Ans.
The state of principal stresses is represented on the element shown in Fig. d.
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*9–104. The box beam is subjected to the loading shown. Determine the principal stress in the beam at points A and B.
6 in. A 6 in. B 8 in. 8 in.
Support Reactions: As shown on FBD(a). Internal Forces and Moment: As shown on FBD(b). Section Properties: I =
1 1 (8) A 83 B (6) A 63 B = 233.33 in4 12 12
QA = QB = 0 Normal Stress: Applying the flexure formula. s = -
My I
sA = -
-300(12)(4) = 61.71 psi 233.33
sB = -
-300(12)(-3) = -46.29 psi 233.33
1200 lb
800 lb
Shear Stress: Since QA = QB = 0, then tA = tB = 0. In - Plane Principal Stress: sx = 61.71 psi, sy = 0, and txy = 0 for point A. Since no shear stress acts on the element, s1 = sx = 61.7 psi
Ans.
s2 = sy = 0
Ans.
sx = -46.29 psi, sy = 0, and txy = 0 for point B. Since no shear stress acts on the element, s1 = sy = 0
Ans.
s2 = sx = -46.3 psi
Ans.
735
A B 3 ft
2.5 ft
2.5 ft
5 ft
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•9–105.
The wooden strut is subjected to the loading shown. Determine the principal stresses that act at point C and specify the orientation of the element at this point. The strut is supported by a bolt (pin) at B and smooth support at A.
50 N
50 N 60⬚ C
100 mm
40 N
40 N B
A 25 mm
50 mm 200 mm 200 mm 200 mm 200 mm 100 mm 100 mm
QC = y¿A¿ = 0.025(0.05)(0.025) = 31.25(10 - 6) m3 1 (0.025)(0.13) = 2.0833(10 - 6) m4 12
I =
Normal stress: sC = 0 Shear stress: VQC 44(31.25)(10 - 6) = 26.4 kPa = It 2.0833(10 - 6)(0.025)
t =
Principal stress: sx = sy = 0; s1,2 =
txy = -26.4 kPa
sx + sy ;
2
C
a
sx - sy 2
2
b + t2 xy
= 0 ; 20 + (26.4)2 s1 = 26.4 kPa
s2 = -26.4 kPa
;
Ans.
Orientation of principal stress: tan 2up =
txy (sx - sy)
= - q
2 up = +45° and -45° Use sx¿
Eq. 9-1 to determine the principal sx + sy sx - sy = + cos 2u + txy sin 2u 2 2
plane
of
s1
and
s2
u = up = -45° sx¿ = 0 + 0 + (-26.4) sin( -90°) = 26.4 kPa Therefore, up1 = -45°;
up2 = 45°
Ans.
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9–106. The wooden strut is subjected to the loading shown. If grains of wood in the strut at point C make an angle of 60° with the horizontal as shown, determine the normal and shear stresses that act perpendicular and parallel to the grains, respectively, due to the loading. The strut is supported by a bolt (pin) at B and smooth support at A.
50 N
50 N 60⬚ C
100 mm
40 N
40 N B
A 25 mm
50 mm 200 mm 200 mm 200 mm 200 mm 100 mm 100 mm
QC = y¿A¿ = 0.025(0.05)(0.025) = 31.25(10 - 6) m3 I =
1 (0.025)(0.13) = 2.0833(10 - 6) m4 12
Normal stress: sC = 0 Shear stress: t =
VQC 44(31.25)(10 - 6) = 26.4 kPa = It 2.0833(10 - 6)(0.025)
Stress transformation: sx = sy = 0; sx¿ =
sx + sy
sx - sy +
2
2
txy = -26.4 kPa;
u = 30°
cos 2u + txy sin 2u
= 0 + 0 + (-26.4) sin 60° = -22.9 kPa tx¿y¿ = -
sx - sy 2
Ans.
sin 2u + txy cos 2u
= -0 + (-26.4) cos 60° = -13.2 kPa
Ans.
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10–1. Prove that the sum of the normal strains in perpendicular directions is constant.
ex¿ =
ey¿ =
ex + ey 2
ex - ey +
ex + ey 2
2 ex - ey
-
2
cos 2u +
cos 2u -
gxy 2 gxy 2
sin 2u
(1)
sin 2u
(2)
Adding Eq. (1) and Eq. (2) yields: ex¿ + ey¿ = ex + ey = constant
QED
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10–2. The state of strain at the point has components of Px = 200 110-62, Py = -300 110-62, and gxy = 400(10-62. Use the strain-transformation equations to determine the equivalent in-plane strains on an element oriented at an angle of 30° counterclockwise from the original position. Sketch the deformed element due to these strains within the x–y plane.
y x
In accordance to the established sign convention, ex = 200(10 - 6), ex¿ =
ex + ey
ex - ey +
2
= c
ey = -300(10 - 6)
2
cos 2u +
gxy 2
gxy = 400(10 - 6)
u = 30°
sin 2u
200 - (-300) 200 + (-300) 400 + cos 60° + sin 60° d(10 - 6) 2 2 2
= 248 (10 - 6) gx¿y¿ 2
= -a
Ans.
ex - ey 2
b sin 2u +
gxy 2
cos 2u
gx¿y¿ = e - C 200 - ( -300) D sin 60° + 400 cos 60° f(10 - 6) = -233(10 - 6) ey¿ =
ex + ey
= c
2
Ans.
ex - ey -
2
cos 2u -
gxy 2
sin 2u
200 - ( -300) 200 + (-300) 400 cos 60° sin 60° d(10 - 6) 2 2 2
= -348(10 - 6)
Ans.
The deformed element of this equivalent state of strain is shown in Fig. a
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10–3. A strain gauge is mounted on the 1-in.-diameter A-36 steel shaft in the manner shown. When the shaft is rotating with an angular velocity of v = 1760 rev>min, the reading on the strain gauge is P = 800110-62. Determine the power output of the motor. Assume the shaft is only subjected to a torque.
v = (1760 rev>min)a
60⬚
2p rad 1 min ba b = 184.307 rad>s 60 sec 1 rev
ex = ey = 0 ex¿ =
ex + ey 2
ex - ey +
2
800(10 - 6) = 0 + 0 +
cos 2u +
gxy 2
gxy 2
sin 2u
sin 120°
gxy = 1.848(10 - 3) rad t = G gxy = 11(103)(1.848)(10 - 3) = 20.323 ksi
t =
Tc ; J
20.323 =
T(0.5) p 2
(0.5)4
;
T = 3.99 kip # in = 332.5 lb # ft P = Tv = 0.332.5 (184.307) = 61.3 kips # ft>s = 111 hp
Ans.
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*10–4. The state of strain at a point on a wrench has components Px = 120110-62, Py = -180110-62, gxy = 150110-62. Use the strain-transformation equations to determine (a) the in-plane principal strains and (b) the maximum in-plane shear strain and average normal strain. In each case specify the orientation of the element and show how the strains deform the element within the x–y plane. ex = 120(10 - 6) e1, 2 =
a)
ey = -180(10 - 6)
gxy = 150(10 - 6)
Ex - Ey 2 ex + ey gxy 2 ; a b + a b 2 A 2 2
120 + (-180) 120 - ( -180) 2 150 2 -6 ; a b + a b d 10 2 A 2 2 e1 = 138(10 - 6); e2 = -198(10 - 6) = c
Ans.
Orientation of e1 and e2 gxy 150 = = 0.5 tan 2up = ex - ey [120 - (-180)] up = 13.28° and -76.72° Use Eq. 10.5 to determine the direction of e1 and e2 ex¿ =
ex + ey
ex - ey +
2
2
cos 2u +
gxy 2
sin 2u
u = up = 13.28° ex¿ = c
120 + ( -180) 120 - ( -180) 150 + cos (26.56°) + sin 26.56° d 10 - 6 2 2 2
= 138 (10 - 6) = e1 Therefore up1 = 13.3° ; gmax
b)
=
2 in-plane
ex + ey 2
A
ex - ey
b + a 2
gxy
b
Ans. 2
2 2 150 2 120 - ( -180) 2 -6 -6 = 2c a b + a b d10 = 335 (10 ) 2 2 A
gmax
eavg =
a
in-plane
up2 = -76.7°
= c
120 + (-180) d 10 - 6 = -30.0(10 - 6) 2
Ans.
Ans.
Orientation of gmax tan 2us =
-(ex - ey) gxy
=
-[120 - ( -180)] = -2.0 150
us = -31.7° and 58.3°
Ans.
gmax Use Eq. 10–6 to determine the sign of in-plane gx¿y¿ ex - ey gxy = sin 2u + cos 2u 2 2 2 u = us = -31.7° gx¿y¿ = 2 c -
120 - (-180) 150 sin (-63.4°) + cos (-63.4°) d10 - 6 = 335(10 - 6) 2 2
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10–5. The state of strain at the point on the arm has components Px = 250110-62, Py = -450110-62, gxy = -825110-62. Use the strain-transformation equations to determine (a) the in-plane principal strains and (b) the maximum in-plane shear strain and average normal strain. In each case specify the orientation of the element and show how the strains deform the element within the x–y plane.
ex = 250(10 - 6)
ey = -450(10 - 6)
y
gxy = -825(10 - 6)
x
a) ex + ey
e1, 2 =
;
2
= c
A
ex - ey
a
2
2
b + a
gxy 2
b
2
250 - 450 250 - ( -450) 2 -825 2 -6 ; a b + a b d(10 ) 2 A 2 2
e1 = 441(10 - 6)
Ans.
e2 = -641(10 - 6)
Ans.
Orientation of e1 and e2 : gxy
tan 2up =
ex - ey
up = -24.84°
-825 250 - ( -450)
=
up = 65.16°
and
Use Eq. 10–5 to determine the direction of e1 and e2: ex¿ =
ex + ey
ex - ey +
2
2
cos 2u +
gxy 2
sin 2u
u = up = -24.84° ex¿ = c
250 - (-450) 250 - 450 -825 + cos (-49.69°) + sin (-49.69°) d(10 - 6) = 441(10 - 6) 2 2 2
Therefore, up1 = -24.8°
Ans.
up2 = 65.2°
Ans.
b) g
max in-plane
2 g
max in-plane
eavg =
=
A
= 2c
a
ex - ey 2
2
gxy 2
b
2
250 - (-450) 2 -825 2 -6 -3 b + a b d(10 ) = 1.08(10 ) A 2 2 a
ex + ey 2
b + a
= a
250 - 450 b (10 - 6) = -100(10 - 6) 2
Ans.
Ans.
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10–6. The state of strain at the point has components of Px = -100110-62, Py = 400110-62, and gxy = -300110-62. Use the strain-transformation equations to determine the equivalent in-plane strains on an element oriented at an angle of 60° counterclockwise from the original position. Sketch the deformed element due to these strains within the x–y plane.
y
x
In accordance to the established sign convention, ex = -100(10 - 6) ex¿ =
ex + ey
= c
ex - ey +
2
ey = 400(10 - 6)
2
gxy
cos 2u +
2
gxy = -300(10 - 6)
u = 60°
sin 2u
-100 - 400 -300 -100 + 400 + cos 120° + sin 120° d(10 - 6) 2 2 2
= 145(10 - 6) gx¿y¿ 2
= -a
Ans.
ex - ey 2
b sin 2u +
gxy 2
cos 2u
gx¿y¿ = c -(-100 - 400) sin 120° + (-300) cos 120° d(10 - 6) = 583(10 - 6) ey¿ =
ex + ey
= c
2
Ans.
ex - ey -
2
cos 2u -
gxy 2
sin 2u
-100 - 400 -300 -100 + 400 cos 120° sin 120° d(10 - 6) 2 2 2
= 155 (10 - 6)
Ans.
The deformed element of this equivalent state of strain is shown in Fig. a
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10–7. The state of strain at the point has components of Px = 100110-62, Py = 300110-62, and gxy = -150110-62. Use the strain-transformation equations to determine the equivalent in-plane strains on an element oriented u = 30° clockwise. Sketch the deformed element due to these strains within the x–y plane.
y
x
In accordance to the established sign convention, ex = 100(10 - 6) ex¿ =
ex + ey
ex - ey +
2
= c
ey = 300(10 - 6)
2
cos 2u +
gxy = -150(10 - 6) gxy 2
u = -30°
sin 2u
100 - 300 -150 100 + 300 + cos (-60°) + sin ( -60°) d (10 - 6) 2 2 2
= 215(10 - 6) gx¿y¿ 2
= -a
Ans.
ex - ey 2
b sin 2u +
gxy 2
cos 2u
gx¿y¿ = c -(100 - 300) sin ( -60°) + ( -150) cos ( -60°) d(10 - 6) = -248 (10 - 6) ey¿ =
ex + ey
= c
2
Ans.
ex - ey -
2
cos 2u -
gxy 2
sin 2u
100 - 300 -150 100 + 300 cos ( -60°) sin (-60°) d (10 - 6) 2 2 2
= 185(10 - 6)
Ans.
The deformed element of this equivalent state of strain is shown in Fig. a
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*10–8. The state of strain at the point on the bracket has components Px = -200110-62, Py = -650110-62, gxy ⫽ -175110-62. Use the strain-transformation equations to determine the equivalent in-plane strains on an element oriented at an angle of u = 20° counterclockwise from the original position. Sketch the deformed element due to these strains within the x–y plane.
ex = -200(10 - 6) ex¿ =
ex + ey
ex - ey +
2
= c
ey = -650(10 - 6)
2
cos 2u +
gxy 2
y
x
gxy = -175(10 - 6)
u = 20°
sin 2u
( -200) - (-650) (-175) -200 + (-650) + cos (40°) + sin (40°) d(10 - 6) 2 2 2
= -309(10 - 6) ey¿ =
ex + ey
ex - ey -
2
= c
Ans.
2
cos 2u -
gxy 2
sin 2u
-200 - ( -650) ( -175) -200 + (-650) cos (40°) sin (40°) d(10 - 6) 2 2 2
= -541(10 - 6) gx¿y¿ 2
ex - ey = -
2
Ans. sin 2u +
gxy 2
cos 2u
gx¿y¿ = [-(-200 - (-650)) sin (40°) + (-175) cos (40°)](10 - 6) = -423(10 - 6)
Ans.
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10–9. The state of strain at the point has components of Px = 180110-62, Py = -120110-62, and gxy = -100110-62. Use the strain-transformation equations to determine (a) the in-plane principal strains and (b) the maximum in-plane shear strain and average normal strain. In each case specify the orientation of the element and show how the strains deform the element within the x–y plane.
y
x
a)
In
accordance
to
the
established
sign
convention,
ex = 180(10 - 6),
ey = -120(10 - 6) and gxy = -100(10 - 6). ex + ey
e1, 2 =
;
2
= b
a
A
ex - ey 2
2
b + a
gxy 2
b
2
180 + (-120) 180 - ( -120) 2 -100 2 -6 ; c d + a b r (10 ) 2 A 2 2
= A 30 ; 158.11 B (10 - 6) e1 = 188(10 - 6) tan 2uP =
e2 = -128(10 - 6)
gxy
Ans.
-100(10 - 6)
ex - ey
C 180 - (-120) D (10 - 6)
=
uP = -9.217°
and
= -0.3333
80.78°
Substitute u = -9.217°, ex + ey
ex¿ =
2
= c
ex - ey +
2
cos 2u +
gxy 2
sin 2u
180 + ( -120) 180 - ( -120) -100 + cos (-18.43°) + sin (-18.43) d(10 - 6) 2 2 2
= 188(10 - 6) = e1 Thus, (uP)1 = -9.22°
(uP)2 = 80.8°
Ans.
The deformed element is shown in Fig (a). gmax ex - ey 2 gxy 2 in-plane = b) a b + a b 2 A 2 2 gmax
in-plane
tan 2us = - a
= b2
180 - (-120) 2 -100 2 -6 -6 d + a b r (10 ) = 316 A 10 B A 2 2
ex - ey gxy
c
b = -c
C 180 - (-120) D (10 - 6)
us = 35.78° = 35.8° and
-100(10 - 6)
Ans.
s = 3 Ans.
-54.22° = -54.2°
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10–9.
Continued
gmax The algebraic sign for in-plane when u = 35.78°. ex - ey gxy gx¿y¿ = -a b sin 2u + cos 2u 2 2 2 gx¿y¿ = e - C 180 - ( -120) D sin 71.56° + ( -100) cos 71.56° f(10 - 6)
eavg
= -316(10 - 6) ex + ey 180 + (-120) = = c d(10 - 6) = 30(10 - 6) 2 2
Ans.
The deformed element for the state of maximum In-plane shear strain is shown is shown in Fig. b
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10–10. The state of strain at the point on the bracket has components Px = 400110-62, Py = -250110-62, gxy ⫽ 310110-62. Use the strain-transformation equations to determine the equivalent in-plane strains on an element oriented at an angle of u = 30° clockwise from the original position. Sketch the deformed element due to these strains within the x–y plane.
ex = 400(10 - 6) ex¿ =
ex + ey
= c
ex - ey +
2
ey = -250(10 - 6)
2
cos 2u +
gxy 2
gxy = 310(10 - 6)
y
x
u = -30°
sin 2u
400 - ( -250) 400 + ( -250) 310 + cos (-60°) + a b sin (-60°) d(10 - 6) 2 2 2
= 103(10 - 6) ey¿ =
ex + ey
= c
ex - ey -
2
Ans.
2
cos 2u -
gxy 2
sin 2u
400 - (-250) 400 + (-250) 310 cos (60°) sin (-60°) d(10 - 6) 2 2 2
= 46.7(10 - 6) gx¿y¿ 2
ex - ey = -
2
Ans. sin 2u +
gxy 2
cos 2u
gx¿y¿ = [-(400 - (-250)) sin (-60°) + 310 cos ( -60°)](10 - 6) = 718(10 - 6)
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10–11. The state of strain at the point has components of Px = -100110-62, Py = -200110-62, and gxy = 100110-62. Use the strain-transformation equations to determine (a) the in-plane principal strains and (b) the maximum in-plane shear strain and average normal strain. In each case specify the orientation of the element and show how the strains deform the element within the x–y plane. In accordance to the established ey = -200(10 - 6) and gxy = 100(10 - 6). ex + ey
e1, 2 =
;
2
= b
A
a
ex - ey 2
b + a 2
gxy 2
b
sign
y
x
convention,
ex = -100(10 - 6),
2
-100 + (-200) 100 2 -100 - (-200) 2 -6 ; c d + a b r (10 ) 2 A 2 2
A -150 ; 70.71 B (10 - 6)
=
e1 = -79.3(10 - 6) tan 2uP =
e2 = -221(10 - 6)
gxy
100(10 - 6)
C -100 - (-200) D (10 - 6)
=
ex - ey
uP = 22.5°
and
Ans. = 1
-67.5°
Substitute u = 22.5, ex + ey
ex¿ =
ex - ey
cos 2u +
gxy
sin 2u 2 2 2 -100 + (-200) -100 - (-200) 100 + cos 45° + sin 45° d(10 - 6) = c 2 2 2 +
= -79.3(10 - 6) = e1 Thus, (uP)1 = 22.5°
(uP)2 = -67.5°
Ans.
The deformed element of the state of principal strain is shown in Fig. a gmax ex - ey 2 gxy 2 in-plane = a b + a b 2 A 2 2 gmax
in-plane
= b2
tan 2us = - a
c
-100 - (-200) 2 100 2 -6 -6 d + a b r (10 ) = 141(10 ) A 2 2
ex - ey gxy
b = -c
us = -22.5° The algebraic sign for gx¿y¿ 2
= -a
ex - ey 2
C -100 - ( -200) D (10 - 6) 100(10 - 6)
and gmax
in-plane
b sin 2u +
Ans.
s = -1 Ans.
67.5° when u = -22.5°. gxy 2
cos 2u
gx¿y¿ = - C -100 - (-200) D sin ( -45°) + 100 cos (-45°)
eavg
= 141(10 - 6) ex + ey -100 + ( -200) = = c d(10 - 6) = -150(10 - 6) 2 2
Ans.
The deformed element for the state of maximum In-plane shear strain is shown in Fig. b. 749
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10–11.
Continued
*10–12. The state of plane strain on an element is given by Px = 500110-62, Py = 300110-62, and gxy = -200110-62. Determine the equivalent state of strain on an element at the same point oriented 45° clockwise with respect to the original element.
y
Pydy dy gxy 2
Strain Transformation Equations: ex = 500 A 10 - 6 B
ey = 300 A 10 - 6 B
gxy = -200 A 10 - 6 B
u = -45°
We obtain ex¿ =
ex + ey +
2
= c
ex - ey 2
cos 2u +
gxy 2
sin 2u
500 - 300 -200 500 + 300 + cos (-90°) + a b sin (-90°) d A 10 - 6 B 2 2 2
= 500 A 10 - 6 B gx¿y¿ 2
= -a
Ans.
ex - ey 2
b sin 2u +
gxy 2
cos 2u
gx¿y¿ = [-(500 - 300) sin ( -90°) + (-200) cos ( -90°)] A 10 - 6 B = 200 A 10 - 6 B
ey¿ =
ex + ey
= c
2
Ans.
ex - ey -
2
cos 2u -
gxy 2
sin 2u
500 + 300 500 - 300 -200 cos ( -90°) - a b sin (-90°) d A 10 - 6 B 2 2 2
= 300 A 10 - 6 B
Ans.
The deformed element for this state of strain is shown in Fig. a.
750
gxy 2 dx
x Pxdx
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10–13. The state of plane strain on an element is Px = -300110-62, Py = 0, and gxy = 150110-62. Determine the equivalent state of strain which represents (a) the principal strains, and (b) the maximum in-plane shear strain and the associated average normal strain. Specify the orientation of the corresponding elements for these states of strain with respect to the original element.
y
gxy dy 2 x
In-Plane Principal Strains: ex = -300 A 10 - 6 B , ey = 0, and gxy = 150 A 10 - 6 B . We obtain ex + ey
e1, 2 =
;
2
= C
C
¢
ex - ey 2
2
≤ + ¢
gxy 2
≤
2
-300 + 0 -300 - 0 2 150 2 ; ¢ ≤ + ¢ ≤ S A 10 - 6 B 2 C 2 2
= ( -150 ; 167.71) A 10 - 6 B
e1 = 17.7 A 10 - 6 B
e2 = -318 A 10 - 6 B
Ans.
Orientation of Principal Strain: tan 2up =
gxy ex - ey
=
150 A 10 - 6 B
(-300 - 0) A 10 - 6 B
= -0.5
uP = -13.28° and 76.72° Substituting u = -13.28° into Eq. 9-1, ex¿ =
ex + ey
= c
ex - ey +
2
2
cos 2u +
gxy 2
sin 2u
-300 + 0 -300 - 0 150 + cos (-26.57°) + sin (-26.57°) d A 10 - 6 B 2 2 2
= -318 A 10 - 6 B = e2 Thus,
A uP B 1 = 76.7° and A uP B 2 = -13.3°
Ans.
The deformed element of this state of strain is shown in Fig. a. Maximum In-Plane Shear Strain: gmax ex - ey 2 gxy 2 in-plane = ¢ ≤ + ¢ ≤ 2 C 2 2 gmax
in-plane
-300 - 0 2 150 2 -6 -6 b + a b R A 10 B = 335 A 10 B A 2 2
= B2
a
Ans.
Orientation of the Maximum In-Plane Shear Strain: tan 2us = - ¢
ex - ey gxy
≤ = -C
(-300 - 0) A 10 - 6 B 150 A 10 - 6 B
S = 2
us = 31.7° and 122°
Ans.
751
gxy 2 dx
Pxdx
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10–13.
Continued
The algebraic sign for gx¿y¿ 2
= -¢
ex - ey 2
gmax
in-plane
≤ sin 2u +
when u = us = 31.7° can be obtained using gxy 2
cos 2u
gx¿y¿ = [-(-300 - 0) sin 63.43° + 150 cos 63.43°] A 10 - 6 B = 335 A 10 - 6 B
Average Normal Strain: eavg =
ex + ey 2
= a
-300 + 0 b A 10 - 6 B = -150 A 10 - 6 B 2
Ans.
The deformed element for this state of strain is shown in Fig. b.
752
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10–14. The state of strain at the point on a boom of an hydraulic engine crane has components of Px = 250110-62, Py = 300110-62, and gxy = -180110-62. Use the straintransformation equations to determine (a) the in-plane principal strains and (b) the maximum in-plane shear strain and average normal strain. In each case, specify the orientation of the element and show how the strains deform the element within the x–y plane.
y
a) In-Plane Principal Strain: Applying Eq. 10–9, ex + ey
e1, 2 =
;
2
= B
a
A
ex - ey 2
b + a 2
gxy 2
b
2
250 - 300 2 250 + 300 -180 2 -6 ; a b + a b R A 10 B 2 A 2 2
= 275 ; 93.41 e1 = 368 A 10 - 6 B
e2 = 182 A 10 - 6 B
Ans.
Orientation of Principal Strain: Applying Eq. 10–8, gxy
tan 2uP =
-180(10 - 6)
ex - ey
=
(250 - 300)(10 - 6)
uP = 37.24°
and
= 3.600
-52.76°
Use Eq. 10–5 to determine which principal strain deforms the element in the x¿ direction with u = 37.24°. ex¿ =
ex + ey
= c
2
ex - ey +
2
cos 2u +
gxy 2
sin 2u
250 + 300 250 - 300 -180 + cos 74.48° + sin 74.48° d A 10 - 6 B 2 2 2
= 182 A 10 - 6 B = e2 Hence, uP1 = -52.8°
and
uP2 = 37.2°
Ans.
b) Maximum In-Plane Shear Strain: Applying Eq. 10–11, g max ex - ey 2 gxy 2 in-plane = a b + a b 2 A 2 2 g
max in-plane
= 2B
-180 2 250 - 300 2 -6 b + a b R A 10 B A 2 2 a
= 187 A 10 - 6 B
Ans.
753
x
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10–14.
Continued
Orientation of the Maximum In-Plane Shear Strain: Applying Eq. 10–10, tan 2us = -
ex - ey
us = -7.76°
and
The proper sign of gx¿y¿ 2
ex - ey = -
= -
gxy
2
g
max in-plane
250 - 300 = -0.2778 -180
82.2°
Ans.
can be determined by substituting u = -7.76° into Eq. 10–6.
sin 2u +
gxy 2
cos 2u
gx¿y¿ = {-[250 - 300] sin (-15.52°) + (-180) cos (-15.52°)} A 10 - 6 B = -187 A 10 - 6 B
Normal Strain and Shear strain: In accordance with the sign convention, ex = 250 A 10 - 6 B
ey = 300 A 10 - 6 B
gxy = -180 A 10 - 6 B
Average Normal Strain: Applying Eq. 10–12, eavg =
ex + ey 2
= c
250 + 300 d A 10 - 6 B = 275 A 10 - 6 B 2
Ans.
754
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*10–16. The state of strain at a point on a support has components of Px = 350110-62, Py = 400110-62, gxy = -675110-62. Use the strain-transformation equations to determine (a) the in-plane principal strains and (b) the maximum in-plane shear strain and average normal strain. In each case specify the orientation of the element and show how the strains deform the element within the x–y plane.
a)
e1, 2 =
=
ex + ey ;
2
B
a
ex -ey 2
b + a 2
gxy 2
b
2
350 - 400 2 -675 2 350 + 400 ; a b + a b 2 A 2 2
e1 = 713(10 - 6)
Ans.
e2 = 36.6(10 - 6)
Ans.
tan 2uP =
gxy ex - ey
=
-675 (350 - 400)
uP = 42.9°
Ans.
b) (gx¿y¿)max =
2 (gx¿y¿)max
=
2
A
a
ex - ey 2
b + a 2
gxy 2
b
2
a
350 - 400 2 -675 2 b + a b A 2 2
(gx¿y¿)max = 677(10 - 6) eavg =
ex + ey
tan 2us =
2
=
Ans.
350 + 400 = 375(10 - 6) 2
-(ex - ey) gxy
=
Ans.
350 - 400 675
us = -2.12°
Ans.
755
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•10–17.
Solve part (a) of Prob. 10–4 using Mohr’s circle.
ex = 120(10 - 6)
ey = -180(10 - 6)
gxy = 150(10 - 6)
A (120, 75)(10 - 6) C (-30, 0)(10 - 6) R = C 2[120 - (-30)]2 + (75)2 D (10 - 6) = 167.71 (10 - 6) e1 = (-30 + 167.71)(10 - 6) = 138(10 - 6)
Ans.
e2 = (-30 - 167.71)(10 - 6) = -198(10 - 6)
Ans.
75 tan 2uP = a b , uP = 13.3° 30 + 120
Ans.
756
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10–18.
Solve part (b) of Prob. 10–4 using Mohr’s circle.
ex = 120(10 - 6)
ey = -180(10 - 6)
gxy = 150(10 - 6)
A (120, 75)(10 - 6) C (-30, 0)(10 - 6) R = C 2[120 - (-30)]2 + (75)2 D (10 - 6) = 167.71 (10 - 6) gxy max 2 in-plane
gxy
= R = 167.7(10 - 6)
max in-plane
= 335(10 - 6)
Ans.
eavg = -30 (10 - 6) tan 2us =
120 + 30 75
Ans. us = -31.7°
Ans.
757
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10–19.
Solve Prob. 10–8 using Mohr’s circle.
ex = -200(10 - 6)
ey = -650(10 - 6)
gxy = -175(10 - 6)
gxy 2
= -87.5(10 - 6)
u = 20°, 2u = 40° A(-200, -87.5)(10 - 6)
C(-425, 0)(10 - 6)
R = [2(-200 - (-425))2 + 87.52 ](10 - 6) = 241.41(10 - 6) tan a =
87.5 ; -200 - (-425)
a = 21.25°
f = 40 + 21.25 = 61.25° ex¿ = (-425 + 241.41 cos 61.25°)(10 - 6) = -309(10 - 6)
Ans.
ey¿ = (-425 - 241.41 cos 61.25°)(10 - 6) = -541(10 - 6)
Ans.
-gx¿y¿ 2
= 241.41(10 - 6) sin 61.25°
gx¿y¿ = -423(10 - 6)
Ans.
758
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*10–20.
Solve Prob. 10–10 using Mohr’s circle.
ex = 400(10 - 6) A(400, 155)(10 - 6)
ey = -250(10 - 6)
gxy = 310(10 - 6)
gxy 2
= 155(10 - 6)
C(75, 0)(10 - 6)
R = [2(400 - 75)2 + 1552 ](10 - 6) = 360.1(10 - 6) tan a =
155 ; 400 - 75
a = 25.50°
f = 60 + 25.50 = 85.5° ex¿ = (75 + 360.1 cos 85.5°)(10 - 6) = 103(10 - 6)
Ans.
ey¿ = (75 - 360.1 cos 85.5°)(10 - 6) = 46.7(10 - 6)
Ans.
gx¿y¿ 2
= (360.1 sin 85.5°)(10 - 6)
gx¿y¿ = 718(10 - 6)
Ans.
759
u = 30°
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•10–21.
Solve Prob. 10–14 using Mohr’s circle.
Construction of the Circle: In accordance with the sign convention, ex = 250 A 10 - 6 B , gxy ey = 300 A 10 - 6 B , and = -90 A 10 - 6 B . Hence, 2 eavg =
ex + ey 2
= a
250 + 300 b A 10 - 6 B = 275 A 10 - 6 B 2
Ans.
The coordinates for reference points A and C are A(250, -90) A 10 - 6 B
C(275, 0) A 10 - 6 B
The radius of the circle is R = a 2(275 - 250)2 + 902 b A 10 - 6 B = 93.408 In-Plane Principal Strain: The coordinates of points B and D represent e1 and e2, respectively. e1 = (275 + 93.408) A 10 - 6 B = 368 A 10 - 6 B
Ans.
e2 = (275 - 93.408) A 10 - 6 B = 182 A 10 - 6 B
Ans.
Orientation of Principal Strain: From the circle, tan 2uP2 =
90 = 3.600 275 - 250
2uP2 = 74.48°
2uP1 = 180° - 2uP2 uP1 =
180° - 74.78° = 52.8° (Clockwise) 2
Ans.
Maximum In-Plane Shear Strain: Represented by the coordinates of point E on the circle. g max in-plane
2 g
= -R = -93.408 A 10 - 6 B
max in-plane
= -187 A 10 - 6 B
Ans.
Orientation of the Maximum In-Plane Shear Strain: From the circle, tan 2us =
275 - 250 = 0.2778 90
us = 7.76° (Clockwise)
Ans.
760
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10–22. The strain at point A on the bracket has components Px = 300110-62, Py = 550110-62, gxy = -650110-62. Determine (a) the principal strains at A in the x– y plane, (b) the maximum shear strain in the x–y plane, and (c) the absolute maximum shear strain. ex = 300(10 - 6)
ey = 550(10 - 6)
A(300, -325)10 - 6
gxy = -650(10 - 6)
y
gxy 2
= -325(10 - 6)
A
C(425, 0)10 - 6
R = C 2(425 - 300)2 + (-325)2 D 10 - 6 = 348.2(10 - 6) a) e1 = (425 + 348.2)(10 - 6) = 773(10 - 6)
Ans.
e2 = (425 - 348.2)(10 - 6) = 76.8(10 - 6)
Ans.
b) g
max in-plane
= 2R = 2(348.2)(10 - 6) = 696(10 - 6)
Ans.
773(10 - 6) ; 2
Ans.
c) gabs max
=
2
gabs max
= 773(10 - 6)
10–23. The strain at point A on the leg of the angle has components Px = -140110-62, Py = 180110-62, gxy = -125110-62. Determine (a) the principal strains at A in the x–y plane, (b) the maximum shear strain in the x–y plane, and (c) the absolute maximum shear strain. ex = -140(10 - 6) A( -140, -62.5)10 - 6
ey = 180(10 - 6)
gxy = -125(10 - 6)
A
gxy 2
= -62.5(10 - 6)
C(20, 0)10 - 6
A 2(20 - ( -140))2 + (-62.5)2 B 10 - 6 = 171.77(10 - 6)
R = a)
e1 = (20 + 171.77)(10 - 6) = 192(10 - 6)
Ans.
e2 = (20 - 171.77)(10 - 6) = -152(10 - 6)
Ans.
(b, c) gabs max
=
g
max in-plane
= 2R = 2(171.77)(10 - 6) = 344(10 - 6)
Ans.
761
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*10–24. The strain at point A on the pressure-vessel wall has components Px = 480110-62, Py = 720110-62, gxy = 650110-62. Determine (a) the principal strains at A, in the x– y plane, (b) the maximum shear strain in the x–y plane, and (c) the absolute maximum shear strain.
ex = 480(10 - 6)
ey = 720(10 - 6)
A(480, 325)10 - 6
C(600, 0)10 - 6
gxy = 650(10 - 6)
y A
gxy 2
= 325(10 - 6)
R = (2(600 - 480)2 + 3252 )10 - 6 = 346.44(10 - 6) a) e1 = (600 + 346.44)10 - 6 = 946(10 - 6)
Ans.
e2 = (600 - 346.44)10 - 6 = 254(10 - 6)
Ans.
b) g
max in-plane
= 2R = 2(346.44)10 - 6 = 693(10 - 6)
Ans.
946(10 - 6) ; 2
Ans.
c) gabs max
2
=
gabs max
= 946(10 - 6)
762
x
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•10–25.
The 60° strain rosette is mounted on the bracket. The following readings are obtained for each gauge: Pa = -100110-62, Pb = 250110-62, and Pc = 150110-62. Determine (a) the principal strains and (b) the maximum inplane shear strain and associated average normal strain. In each case show the deformed element due to these strains.
b c 60⬚ 60⬚
This is a 60° strain rosette Thus, ex = ea = -100(10 - 6) 1 A 2eb + 2ec - ea B 3
ey =
1 C 2(250) + 2(150) - (-100) D (10 - 6) 3
=
= 300(10 - 6) gxy =
2 23
(eb - ec) =
2 23
(250 - 150)(10 - 6) = 115.47(10 - 6)
In accordance to the established sign convention, ex = -100(10 - 6), ey = 300(10 - 6) gxy and = 57.74(10 - 6). 2 Thus, eavg =
ex + ey 2
= a
-100 + 300 b(10 - 6) = 100(10 - 6) 2
Ans.
Then, the coordinates of reference point A and Center C of the circle are A( -100, 57.74)(10 - 6)
C(100, 0)(10 - 6)
Thus, the radius of the circle is R = CA = a 2(-100 - 100)2 + 208.16b(10 - 6) = 208.17(10 - 6) Using these result, the circle is shown in Fig. a. The coordinates of points B and D represent e1 and e2 respectively. e1 = (100 + 208.17)(10 - 6) = 308(10 - 6)
Ans.
e2 = (100 - 208.17)(10 - 6) = -108(10 - 6)
Ans.
Referring to the geometry of the circle, tan 2(uP)2 =
57.74(10 - 6) (100 + 100)(10 - 6)
= 0.2887
A uP B 2 = 8.05° (Clockwise)
Ans.
The deformed element for the state of principal strain is shown in Fig. b.
763
a
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10–25.
Continued gmax
The coordinates for point E represent eavg and
in-plane
2
. Thus,
gmax
in-plane
2
= R = 208.17(10 - 6)
gmax
in-plane
= 416(10 - 6)
Ans.
Referring to the geometry of the circle, tan 2us =
100 + 100 57.74
us = 36.9° (Counter Clockwise)
Ans.
The deformed element for the state of maximum In-plane shear strain is shown in Fig. c.
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10–26. The 60° strain rosette is mounted on a beam. The following readings are obtained for each gauge: Pa = 200110-62, Pb = - 450110-62, and Pc = 250110-62. Determine (a) the in-plane principal strains and (b) the maximum in-plane shear strain and average normal strain. In each case show the deformed element due to these strains.
b a 30⬚ 30⬚ c
With ua = 60°, ub = 120° and uc = 180°, ea = ex cos2 ua + ey sin2 ua + gxy sin ua cos ua 200(10 - 6) = ex cos2 60° + ey sin2 60° + gxy sin 60° cos 60° 0.25ex + 0.75ey + 0.4330 gxy = 200(10 - 6)
(1)
eb = ex cos2 ub + ey sin2 ub + gxy sin ub cos ub -450(10 - 6) = ex cos2 120° + ey sin2 120° + gxy sin 120° cos 120° 0.25ex + 0.75ey - 0.4330 gxy = -450(10 - 6)
(2)
ec = ex cos2 uc + ey sin2 uc + gxy sin uc cos uc 250(10 - 6) = ex cos2 180° + ey sin2 180° + gxy sin 180° cos 180° ex = 250(10 - 6) Substitute this result into Eqs. (1) and (2) and solve them, ey = -250 (10 - 6)
gxy = 750.56 (10 - 6)
In accordance to the established sign convention, ex = 250(10 - 6), ey = -250(10 - 6), gxy and = 375.28(10 - 6), Thus, 2 eavg =
ex + ey 2
= c
250 + ( -250) d(10 - 6) = 0 2
Ans.
Then, the coordinates of the reference point A and center C of the circle are A(250, 375.28)(10 - 6)
C(0, 0)
Thus, the radius of the circle is R = CA =
A 2(250 - 0)2 + 375.282 B (10 - 6) = 450.92(10 - 6)
Using these results, the circle is shown in Fig. a. The coordinates for points B and D represent e1 and e2, respectively. Thus, e1 = 451(10 - 6)
e2 = -451(10 - 6)
Ans.
Referring to the geometry of the circle, tan 2(uP)1 =
375.28 = 1.5011 250
(uP)1 = 28.2° (Counter Clockwise)
Ans.
The deformed element for the state of principal strains is shown in Fig. b.
765
60⬚
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10–26.
Continued gmax
in-plane
The coordinates of point E represent eavg and . Thus, 2 gmax in-plane gmax = 902(10 - 6) = R = 450.92(10 - 6) in-plane 2
Ans.
Referring to the geometry of the circle, tan 2us =
250 = 0.6662 375.28
us = 16.8° (Clockwise)
Ans.
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10–27. The 45° strain rosette is mounted on a steel shaft. The following readings are obtained from each gauge: Pa = 300110-62, Pb = -250110-62, and Pc = -450110-62. Determine (a) the in-plane principal strains and (b) the maximum in-plane shear strain and average normal strain. In each case show the deformed element due to these strains.
b
c 45⬚
With ua = 45°, ub = 90° and uc = 135°, ea = ex cos2 ua + ey sin2 ua + gxy sin ua cos ua 300(10 - 6) = ex cos2 45° + ey sin2 45° + gxy sin 45° cos 45° ex + ey + gxy = 600(10 - 6)
(1)
eb = ex cos2 ub + ey sin2 ub + gxy sin ub cos ub -250(10 - 6) = ex cos2 90° + ey sin2 90° + gxy sin 90° cos 90° ey = -250(10 - 6) ec = ex cos2 uc + ey sin2 uc + gxy sin uc cos uc -450(10 - 6) = ex cos2 135° + ey sin2 135° + gxy sin 135° cos 135° ex + ey - gxy = -900(10 - 6)
(2)
Substitute the result of ey into Eq. (1) and (2) and solve them ex = 100(10 - 6)
gxy = 750(10 - 6)
In accordance to the established sign convention, ex = 100(10 - 6), ey = -250(10 - 6) gxy and = 375(10 - 6). Thus, 2 eavg =
ex + ey 2
= c
100 + (-250) d(10 - 6) = -75(10 - 6) 2
Ans.
Then, the coordinates of the reference point A and the center C of the circle are A(100, 375)(10 - 6)
C(-75, 0)(10 - 6)
Thus, the radius of the circle is R = CA = a 2 C 100 - (-75) D 2 + 3752 b(10 - 6) = 413.82(10 - 6) Using these results, the circle is shown in Fig. a. The Coordinates of points B and D represent e1 and e2, respectively. Thus, e1 = e2 =
A -75 + 413.82 B (10 - 6) = 339(10 - 6)
Ans.
A -75 - 413.82 B (10 - 6) = -489(10 - 6)
Ans.
Referring to the geometry of the circle tan 2(uP)1 =
375 = 2.1429 100 + 75
(uP)1 = 32.5° (Counter Clockwise)
Ans.
767
45⬚
a
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10–27.
Continued
The deformed element for the state of principal strains is shown in Fig. b. gmax The coordinates of point E represent eavg and gmax
in-plane
2
= R = 413.82(106)
in-plane
gmax
2
in-plane
. Thus
= 828(10 - 6)
Ans.
Referring to the geometry of the circle tan 2us =
-100 + 75 = 0.4667 375
us = 12.5° (Clockwise)
Ans.
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*10–28. The 45° strain rosette is mounted on the link of the backhoe. The following readings are obtained from each gauge: Pa = 650110-62, Pb = -300110-62, Pc = 480110-62. Determine (a) the in-plane principal strains and (b) the maximum in-plane shear strain and associated average normal strain.
a 45⬚ b
ea = 650(10 - 6); ua = 180°;
eb = -300(10 - 6);
ub = 225°
c
uc = 270°
Applying Eq. 10–16, e = ex cos2 u + ey sin2 u + gxy sin u cos u 650(10 - 6) = ex cos2 (180°) + ey sin2 (180°) + gxy sin (180°) cos (180°) ex = 650 (10 - 6) 480 (10 - 6) = ex cos2 (270°) + ey sin2 (270°) + gxy sin (270°) cos (270°) ey = 480 (10 - 6) -300 (10 - 6) = 650 (10 - 6) cos2 (225°) + 480 (10 - 6) sin2 (225°) + gxy sin (225°) cos (225°) gxy = -1730 (10 - 6) Therefore, ex = 650 (10 - 6) gxy 2
ey = 480 (10 - 6)
gxy = -1730 (10 - 6)
= -865 (10 - 6)
Mohr’s circle: A(650, -865) 10 - 6
C(565, 0) 10 - 6
R = CA = C 2(650 - 565)2 + 8652 D 10 - 6 = 869.17 (10 - 6) (a)
(b)
e1 = [565 + 869.17]10 - 6 = 1434 (10 - 6)
Ans.
e2 = [565 - 869.17]10 - 6 = -304 (10 - 6)
Ans.
gmax
in-plane
45⬚
ec = 480(10 - 6)
= 2 R = 2(869.17) (10 - 6) = 1738 (10 - 6)
Ans.
eavg = 565(10 - 6)
Ans.
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10–30. For the case of plane stress, show that Hooke’s law can be written as sx =
E E 1Px + nPy2, sy = 1Py + nPx2 11 - n22 11 - n22
Generalized Hooke’s Law: For plane stress, sz = 0. Applying Eq. 10–18, ex =
1 A s - v sy B E x
vEex = A sx - v sy B v vEex = v sx - v2 sy ey =
[1]
1 (s - v sx) E y
E ey = -v sx + sy
[2]
Adding Eq [1] and Eq.[2] yields. vE ex - E ey = sy - v2 sy sy =
E A vex + ey B 1 - v2
(Q.E.D.)
Substituting sy into Eq. [2] E ey = -vsx +
sx =
E A v ex + ey B 1 - v2
E A v ex + ey B v (1 - v2)
Eey -
v
E v ex + E ey - E ey + Eey v2 =
=
v(1 - v2) E (ex + v ey) 1 - v2
(Q.E.D.)
770
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10–31. Use Hooke’s law, Eq. 10–18, to develop the straintransformation equations, Eqs. 10–5 and 10–6, from the stress-transformation equations, Eqs. 9–1 and 9–2.
Stress transformation equations: sx + sy
sx¿ =
2
tx¿y¿ =
sy¿ =
sx - sy +
sx - sy
2
(1)
sin 2u + txy cos 2u
2 sx + sy
sx - sy -
2
cos 2u + txy sin 2u
2
(2)
cos 2u - txy sin 2u
(3)
Hooke’s Law: ex =
v sy sx E E
(4)
ey =
sy -v sx + E E
(5)
txy = G gxy G =
(6)
E 2 (1 + v)
(7)
From Eqs. (4) and (5) ex + ey =
ex - ey =
(1 - v)(sx + sy)
(8)
E (1 + v)(sx - sy)
(9)
E
From Eqs. (6) and (7) txy =
E g 2 (1 + v) xy
(10)
From Eq. (4) ex¿ =
v sy¿ sx¿ E E
(11)
Substitute Eqs. (1) and (3) into Eq. (11) ex¿ =
(1 - v)(sx - sy)
(1 + v)(sx - sy) +
2E
2E
cos 2u +
(1 + v)txy sin 2u E
(12)
By using Eqs. (8), (9) and (10) and substitute into Eq. (12), ex¿ =
ex + ey 2
ex - ey +
2
cos 2u +
gxy 2
QED
sin 2u
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10–31.
Continued
From Eq. (6). gx¿y¿ = G gx¿y¿ =
E g 2 (1 + v) x¿y¿
(13)
Substitute Eqs. (13), (6) and (9) into Eq. (2), E (ex - ey) E E gx¿y¿ = sin 2u + g cos 2u 2 (1 + v) 2 (1 + v) 2 (1 + v) xy gx¿y¿
(ex - ey) = -
2
2
sin 2u +
gxy 2
QED
cos 2u
*10–32. A bar of copper alloy is loaded in a tension machine and it is determined that Px = 940110-62 and sx = 14 ksi, sy = 0, sz = 0. Determine the modulus of elasticity, Ecu, and the dilatation, ecu, of the copper. ncu = 0.35.
ex =
1 [s - v(sy + sz)] E x
940(10 - 6) =
ecu =
1 [14(103) - 0.35(0 + 0)] Ecu
Ecu = 14.9(103) ksi
Ans.
1 - 2(0.35) 1 - 2v (14 + 0 + 0) = 0.282(10 - 3) (sx + sy + sz) = E 14.9(103)
Ans.
•10–33.
The principal strains at a point on the aluminum fuselage of a jet aircraft are P1 = 780110-62 and P2 = 400110-62. Determine the associated principal stresses at the point in the same plane. Eal = 1011032 ksi, nal = 0.33. Hint: See Prob. 10–30.
Plane stress, s3 = 0 See Prob 10-30, s1 =
E (e1 + ve2) 1 - v2 10(103)
=
s2 =
1 - 0.332
Ans.
E (e2 + ve1) 1 - v2 10(103)
=
(780(10 - 6) + 0.33(400)(10 - 6)) = 10.2 ksi
1 - 0.332
(400(10 - 6) + 0.33(780)(10 - 6)) = 7.38 ksi
Ans.
772
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10–34. The rod is made of aluminum 2014-T6. If it is subjected to the tensile load of 700 N and has a diameter of 20 mm, determine the absolute maximum shear strain in the rod at a point on its surface.
700 N
Normal Stress: For uniaxial loading, sy = sz = 0. sx =
P = A
p 4
700 = 2.228 MPa (0.022)
Normal Strain: Applying the generalized Hooke’s Law. ex =
=
1 C s - v A sy + sz B D E x 1 C 2.228 A 106 B - 0 D 73.1(109)
= 30.48 A 10 - 6 B ey =
=
1 C s - v(sx + sz) D E y 1 C 0 - 0.35 A 2.228 A 106 B + 0 B D 73.1(109)
= -10.67 A 10 - 6 B ez =
=
1 C s - v A sx + sy B D E z 1 C 0 - 0.35 A 2.228 A 106 B + 0 B D 73.1(109)
= -10.67 A 10 - 6 B Therefore. emax = 30.48 A 10 - 6 B
emin = -10.67 A 10 - 6 B
Absolute Maximum Shear Strain: gabs
max
= emax - emin = [30.48 - (-10.67)] A 10 - 6 B = 41.1 A 10 - 6 B
Ans.
773
700 N
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10–35. The rod is made of aluminum 2014-T6. If it is subjected to the tensile load of 700 N and has a diameter of 20 mm, determine the principal strains at a point on the surface of the rod.
700 N
Normal Stress: For uniaxial loading, sy = sz = 0. sx =
P = A
p 4
700 = 2.228 MPa (0.022)
Normal Strains: Applying the generalized Hooke’s Law. ex =
=
1 C s - v A sy + sz B D E x 1 C 2.228 A 106 B - 0 D 73.1(109)
= 30.48 A 10 - 6 B ey =
=
1 C s - v(sx + sz) D E y 1 C 0 - 0.35 A 2.228 A 106 B + 0 B D 73.1(109)
= -10.67 A 10 - 6 B ez =
=
1 C s - v A sx + sy B D E z 1 C 0 - 0.35 A 2.228 A 106 B + 0 B D 73.1(109)
= -10.67 A 10 - 6 B Principal Strains: From the results obtained above, emax = 30.5 A 10 - 6 B
eint = emin = -10.7 A 10 - 6 B
Ans.
774
700 N
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*10–36. The steel shaft has a radius of 15 mm. Determine the torque T in the shaft if the two strain gauges, attached to the surface of the shaft, report strains of Px¿ = -80110-62 and Py¿ = 80110-62. Also, compute the strains acting in the x and y directions. Est = 200 GPa, nst = 0.3. ex¿ = -80(10 - 6)
y T
ex = ey = 0
Ans.
ex¿ = ex cos2 u + ey sin2 u + gxy sin u cos u u = 45° -80(10 - 6) = 0 + 0 + gxy sin 45° cos 45° gxy = -160(10 - 6)
Ans.
Also, u = 135° 80(10 - 6) = 0 + 0 + g sin 135° cos 135° gxy = -160(10 - 6) 200(109) E = = 76.923(109) 2(1 + V) 2(1 + 0.3)
t = Gg = 76.923(109)(160)(10 - 6) = 12.308(106) Pa 12.308(106) A
p B (0.015)4 2 = 65.2 N # m 0.015
Ans.
10–37. Determine the bulk modulus for each of the following materials: (a) rubber, Er = 0.4 ksi, nr = 0.48, and (b) glass, Eg = 811032 ksi, ng = 0.24. a) For rubber: Kr =
Er 0.4 = = 3.33 ksi 3 (1 - 2 vr) 3[1 - 2(0.48)]
Ans.
b) For glass: Kg =
Eg 3 (1 - 2 vg)
=
x T
Pure shear
tJ T = = c
x¿ 45⬚
ey¿ = 80(10 - 6)
G =
y¿
8(103) = 5.13 (103) ksi 3[1 - 2(0.24)]
Ans.
775
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10–38. The principal stresses at a point are shown in the figure. If the material is A-36 steel, determine the principal strains.
12 ksi
e1 =
1 1 e 12 - 0.32 C 8 + (-20) D f = 546 (10-6) C s - v(s2 + s3) D = E 1 29.0(103)
e2 =
1 1 e 8 - 0.32 C 12 + (-20) D f = 364 (10-6) C s - v(s1 + s3) D = E 2 29.0(103)
e3 =
20 ksi
8 ksi
1 1 C s3 - v(s1 + s2) D = C -20 - 0.32(12 + 8) D = -910 (10-6) E 29.0(103)
emax = 546 (10 - 6)
eint = 346 (10 - 6)
emin = -910 (10 - 6)
Ans.
10–39. The spherical pressure vessel has an inner diameter of 2 m and a thickness of 10 mm. A strain gauge having a length of 20 mm is attached to it, and it is observed to increase in length by 0.012 mm when the vessel is pressurized. Determine the pressure causing this deformation, and find the maximum in-plane shear stress, and the absolute maximum shear stress at a point on the outer surface of the vessel. The material is steel, for which Est = 200 GPa and nst = 0.3.
20 mm
1000 r = = 100 7 10, the thin wall analysis is valid to t 10 determine the normal stress in the wall of the spherical vessel. This is a plane stress Normal Stresses: Since
problem where smin = 0 since there is no load acting on the outer surface of the wall.
smax = slat =
pr p(1000) = = 50.0p 2t 2(10)
[1]
Normal Strains: Applying the generalized Hooke’s Law with emax = elat =
0.012 = 0.600 A 10 - 3 B mm>mm 20
emax =
1 C s - V (slat + smin) D E max
0.600 A 10 - 3 B =
1 [50.0p - 0.3 (50.0p + 0)] 200(104)
p = 3.4286 MPa = 3.43 MPa
Ans.
From Eq.[1] smax = slat = 50.0(3.4286) = 171.43 MPa Maximum In-Plane Shear (Sphere’s Surface): Mohr’s circle is simply a dot. As the result, the state of stress is the same consisting of two normal stresses with zero shear stress regardless of the orientation of the element. t
max in-plane
= 0
Ans.
smax - smin 171.43 - 0 = = 85.7MPa 2 2
Ans.
Absolute Maximum Shear Stress: tabs
max
=
776
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*10–40. The strain in the x direction at point A on the steel beam is measured and found to be Px = -100110-62. Determine the applied load P. What is the shear strain gxy at point A? Est = 2911032 ksi, nst = 0.3.
P y 3 in. A 3 ft
1 1 (5.5)(83) = 129.833 in4 (6)(9)3 12 12
Ix =
QA = (4.25)(0.5)(6) + (2.75)(0.5)(2.5) = 16.1875 in3 s = Eex = 29(103)(100)(10 - 6) = 2.90 ksi My , I
s =
2.90 =
1.5P(12)(1.5) 129.833
P = 13.945 = 13.9 kip tA =
0.5(13.945)(16.1875) VQ = = 1.739 ksi It 129.833(0.5)
G =
29(103) E = = 11.154(103) ksi 2(1 + v) 2(1 + 0.3)
gxy =
txy G
=
Ans.
1.739 = 0.156(10 - 3) rad 11.154(103)
Ans.
777
x
4 ft
7 ft
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•10–41.
The cross section of the rectangular beam is subjected to the bending moment M. Determine an expression for the increase in length of lines AB and CD. The material has a modulus of elasticity E and Poisson’s ratio is n.
C D B h
For line AB, sz = -
ey = -
A M
12My My My = 1 = 3 I b h3 12 b h v sz =
E
E b h3
h 2
¢LAB =
=
b
12 v My
L0
h
ey dy =
2 12 v M y dy 3 E b h L0
3vM 2Ebh
Ans.
For line CD, sz = -
ex = -
M h2 Mc 6M = - 1 = - 2 3 I bh b h 12 v sz E
=
6vM E b h2
¢LCD = ex LCD =
=
6vM (b) E b h2
6vM E h2
Ans.
10–42. The principal stresses at a point are shown in the figure. If the material is aluminum for which Eal = 1011032 ksi and nal = 0.33, determine the principal strains.
26 ksi
ex =
1 1 (s - v(sy + sz)) = (10 - 0.33(-15 - 26)) = 2.35(10 - 3) E x 10(103)
ey =
1 1 (s - v(sx + sz)) = (-15 - 0.33)(10 - 26)) = -0.972(10 - 3)Ans. E y 10(103)
ez =
Ans.
1 1 (s - v(sx + sy)) = (-26 - 0.33(10 - 15)) = -2.44(10 - 3) Ans. E z 10(103)
778
15 ksi 10 ksi
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10–43. A single strain gauge, placed on the outer surface and at an angle of 30° to the axis of the pipe, gives a reading at point A of Pa = -200(10-6). Determine the horizontal force P if the pipe has an outer diameter of 2 in. and an inner diameter of 1 in. The pipe is made of A-36 steel.
1.5 ft
Using the method of section and consider the equilibrium of the FBD of the pipe’s upper segment, Fig. a, Vz - p = 0
©Fz = 0;
Vz = p
©Mx = 0;
Tx - p(1.5) = 0 Tx = 1.5p
©My = 0;
My - p(2.5) = 0 My = 2.5p
30⬚ A
The normal strees is due to bending only. For point A, z = 0. Thus sx =
My z Iy
= 0
The shear stress is the combination of torsional shear stress and transverse shear stress. Here, J = p2 (14 - 0.54) = 0.46875 p in4. Thus, for point A tt =
1.5p(12)(1) 38.4 p Txc = = p J 0.46875p
Referring to Fig. b, (QA)z = y1œ A1œ - y2œ A2œ =
4 (1) p 2 4(0.5) p c (1 ) d c (0.52) d 3p 2 3p 2
= 0.5833 in3 Iy =
p 4
(14 - 0.54) = 0.234375 p in4
Combine these two shear stress components, t = tt + tv =
P 2.5 ft
38.4P 2.4889P 40.8889P + = p p p
Since no normal stress acting on point A, it is subjected to pure shear which can be represented by the element shown in Fig. c. For pure shear, ex = ez = 0, ea = ex cos3 ua + ez sin2 ua + gxz sin ua cos ua -200(10 - 6) = 0 + 0 + gxz sin 150° cos 150° gxz = 461.88(10 - 6) Applying the Hooke’s Law for shear, txz = G gxz 40.8889P = 11.0(103) C 461.88(10 - 6) D p P = 0.3904 kip = 390 lb
Ans.
779
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*10–44. A single strain gauge, placed in the vertical plane on the outer surface and at an angle of 30° to the axis of the pipe, gives a reading at point A of Pa = -200(10-6). Determine the principal strains in the pipe at point A. The pipe has an outer diameter of 2 in. and an inner diameter of 1 in. and is made of A-36 steel.
1.5 ft
P 2.5 ft
Using the method of sections and consider the equilibrium of the FBD of the pipe’s upper segment, Fig. a, Vz - P = 0
©Fz = 0;
Vz = P
©Mx = 0;
Tx - P(1.5) = 0 Tx = 1.5P
©My = 0;
My - P(2.5) = 0 My = 2.5P
By observation, no normal stress acting on point A. Thus, this is a case of pure shear. For the case of pure shear, ex = ez = ey = 0 ea = ex cos2 ua + ez sin2 ua + gxz sin ua cos ua -200(10 - 6) = 0 + 0 + gxz sin 150° cos 150° gxz = 461.88(10 - 6) e1, 2 =
ex + ez
= B
2
+
A
a
ex - ez 2
b + a 2
gxz 2
b
2
0 - 0 2 461.88 2 0 + 0 -6 ; a b + a b R (10 ) 2 A 2 2
e1 = 231(10 - 6)
e2 = -231(10 - 6)
Ans.
780
30⬚ A
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10–45. The cylindrical pressure vessel is fabricated using hemispherical end caps in order to reduce the bending stress that would occur if flat ends were used. The bending stresses at the seam where the caps are attached can be eliminated by proper choice of the thickness th and tc of the caps and cylinder, respectively. This requires the radial expansion to be the same for both the hemispheres and cylinder. Show that this ratio is tc>th = 12 - n2>11 - n2.Assume that the vessel is made of the same material and both the cylinder and hemispheres have the same inner radius. If the cylinder is to have a thickness of 0.5 in., what is the required thickness of the hemispheres? Take n = 0.3.
tc th r
For cylindrical vessel: s1 =
pr ; tc
e1 =
1 [s - v (s2 + s3)] E 1
=
s2 =
pr 2 tc s3 = 0
vpr pr 1 1 pr a b = a1 - v b E tc 2 tc E tc 2
d r = e1 r =
p r2 1 a1 - v b E tc 2
(1)
For hemispherical end caps: s1 = s2 =
e1 =
=
pr 2 th
1 [s - v (s2 + s3)] ; E 1
s3 = 0
vpr pr 1 pr a b = (1 - v) E 2 th 2 th 2 E th
d r = e1 r =
p r2 (1 - v) 2 E th
(2)
Equate Eqs. (1) and (2): p r2 p r2 1 a1 - vb = (1 - v) E tc 2 2 E th 2 (1 - 12 v) tc 2 - v = = th 1 - v 1 - v th =
QED
(1 - v) tc (1 - 0.3) (0.5) = = 0.206 in. 2 - v 2 - 0.3
Ans.
781
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10–46. The principal strains in a plane, measured experimentally at a point on the aluminum fuselage of a jet aircraft, are P1 = 630(10-6) and P2 = 350(10-6). If this is a case of plane stress, determine the associated principal stresses at the point in the same plane. Eal = 10(103) ksi and nal = 0.33. Normal Stresses: For plane stress, s3 = 0. Normal Strains: Applying the generalized Hooke’s Law. e1 =
1 C s - v (s2 + s3) D E 1
630 A 10 - 6 B =
1 [s1 - 0.33(s2 + 0)] 10(103)
6.30 = s1 - 0.33s2 e2 =
[1]
1 C s - v (s1 + s3) D E 2
350 A 10 - 6 B =
1 C s2 - 0.33(s1 + 0) D 10(103)
3.50 = s2 - 0.33s1
[2]
Solving Eqs.[1] and [2] yields: s1 = 8.37 ksi
s2 = 6.26 ksi
Ans.
10–47. The principal stresses at a point are shown in the figure. If the material is aluminum for which Eal = 1011032 ksi and nal = 0.33, determine the principal strains.
3 ksi
e1 =
1 1 e 8 - 0.33 C 3 + (-4) D f = 833 (10 - 6) C s - v(s2 + s3) D = E 1 10(103)
e2 =
1 1 e 3 - 0.33 C 8 + (-4) D f = 168 (10 - 6) C s - v(s1 + s3) D = E 2 10(103)
e3 =
1 1 C s3 - v(s1 + s2) D = C -4 - 0.33(8 + 3) D = -763 (10 - 6) E 10(103)
Using these results, e1 = 833(10 - 6)
e2 = 168(10 - 6)
e3 = -763(10 - 6)
782
8 ksi
4 ksi
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*10–48. The 6061-T6 aluminum alloy plate fits snugly into the rigid constraint. Determine the normal stresses sx and sy developed in the plate if the temperature is increased by ¢T = 50°C. To solve, add the thermal strain a¢T to the equations for Hooke’s Law.
y
400 mm
300 mm
x
Generalized Hooke’s Law: Since the sides of the aluminum plate are confined in the rigid constraint along the x and y directions, ex = ey = 0. However, the plate is allowed to have free expansion along the z direction. Thus, sz = 0. With the additional thermal strain term, we have ex =
0 =
1 cs - v A sy + sz B d + a¢T E x 1
68.9 A 109 B
csx - 0.35 A sy + 0 B d + 24a 10 - 6 b(50)
sx - 0.35sy = -82.68 A 106 B ey =
0 =
(1)
1 C s - v A sx + sz B D + a¢T E y 1 68.9a 10 b 9
C sy - 0.35(sx + 0) D + 24 A 10 - 6 B (50)
sy - 0.35sx = -82.68 A 106 B
(2)
Solving Eqs. (1) and (2), sx = sy = -127.2 MPa = 127.2 MPa (C)
Ans.
Since sx = sy and sy 6 sY, the above results are valid.
783
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•10–49.
Initially, gaps between the A-36 steel plate and the rigid constraint are as shown. Determine the normal stresses sx and sy developed in the plate if the temperature is increased by ¢T = 100°F. To solve, add the thermal strain a¢T to the equations for Hooke’s Law.
y 0.0015 in.
6 in.
8 in.
0.0025 in. x
Generalized Hooke’s Law: Since there are gaps between the sides of the plate and the rigid constraint, the plate is allowed to expand before it comes in contact with the constraint. dy dx 0.0025 0.0015 ey = = = 0.3125 A 10 - 3 B and = = 0.25 A 10 - 3 B . Thus, ex = Lx 8 Ly 6 However, the plate is allowed to have free expansion along the z direction. Thus, sz = 0. With the additional thermal strain term, we have ex =
1 csx - v A sy + sz B d + a¢T E
0.3125 a 10 - 3 b =
1 29.0 a103 b
C sx - 0.32 A sy + 0 B D + 6.60 A 10 - 6 B (100)
sx - 0.32sy = -10.0775 ey =
(1)
1 C s - v A sx + sz B D + a¢T E y
0.25 A 10 - 3 B =
1
29.0 A 103 B
C sy - 0.32(sx + 0) D + 6.60 A 10 - 6 B (100)
sy - 0.32sx = -11.89
(2)
Solving Eqs. (1) and (2), sx = -15.5 ksi = 15.5 ksi (C)
Ans.
sy = -16.8 ksi = 16.8 ksi (C)
Ans.
Since sx 6 sY and sy 6 sY, the above results are valid.
784
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10–50. Two strain gauges a and b are attached to a plate made from a material having a modulus of elasticity of E = 70 GPa and Poisson’s ratio n = 0.35. If the gauges give a reading of Pa = 450110-62 and Pb = 100110-62, determine the intensities of the uniform distributed load wx and wy acting on the plate. The thickness of the plate is 25 mm.
wy
b 45⬚
y
a
Normal Strain: Since no shear force acts on the plane along the x and y axes, gxy = 0. With ua = 0 and ub = 45°, we have 2
2
ea = ex cos ua + ey sin ua + gxy sin ua cos ua 450 A 10 - 6 B = ex cos2 0° + ey sin2 0°+0 ex = 450 A 10 - 6 B
eb = ex cos2 ub + ey sin2 ub + gxy sin ub cos ub 100 A 10 - 6 B = 450 A 10 - 6 B cos2 45° + ey sin2 45° + 0
ey = -250 A 10 - 6 B
Generalized Hooke’s Law: This is a case of plane stress. Thus, sz = 0. ex =
1 C s - v A sy + sz B D E x
450 A 10 - 6 B =
1
70 A 109 B
C sy - 0.35 A sy + 0 B D
sx - 0.35sy = 31.5 A 106 B ey =
(1)
1 C s - v A sx + sz B D E y
-250 A 10-6 B =
1
70 A 109 B
C sy - 0.35 A sy + 0 B D
sy - 0.35sx = -17.5 A 106 B
(2)
Solving Eqs. (1) and (2), sy = -7.379 A 106 B N>m2
sx = 28.917 A 106 B N>m2
Then, wy = syt = -7.379 A 106 B (0.025) = -184 N>m
Ans.
wx = sxt = 28.917 A 106 B (0.025) = 723 N>m
Ans.
785
z
x
wx
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10–51. Two strain gauges a and b are attached to the surface of the plate which is subjected to the uniform distributed load wx = 700 kN>m and wy = -175 kN>m. If the gauges give a reading of Pa = 450110-62 and Pb = 100110-62, determine the modulus of elasticity E, shear modulus G, and Poisson’s ratio n for the material.
wy
b 45⬚
y
Normal Stress and Strain: The normal stresses along the x, y, and z axes are sx =
700 A 103 B 0.025
sy = -
a
= 28 A 10 B N>m 6
175 A 103 B 0.025
2
= -7 A 106 B N>m2
z
sz = 0 (plane stress) Since no shear force acts on the plane along the x and y axes, gxy = 0. With ua = 0° and ub = 45°, we have ea = ex cos2 ua + ey sin2 ua + gxy sin ua cos ua 450 A 10 - 6 B = ex cos2 0° + ey sin2 0° + 0
ex = 450 A 10 - 6 B
eb = ex cos2 ub + ey sin2 ub + gxy sin ub cos ub 100 A 10 - 6 B = 450 A 10 - 6 B cos2 45°+ ey sin2 45° + 0 ey = -250 A 10 - 6 B Generalized Hooke’s Law: ex =
1 C s - v A sy + sz B D E x
450 A 10 - 6 B =
1 B 28 A 106 B - v C -7 A 106 B + 0 D R E
450 A 10 - 6 B E - 7 A 106 B v = 28 A 106 B ey =
(1)
1 [s - v(sx + sz)] E y
-250 A 10 - 6 B =
1 b -7 A 106 B - v C 28 A 106 B + 0 D r E
250 A 10 - 6 B E - 28 A 106 B v = 7 A 106 B
(2)
Solving Eqs. (1) and (2), E = 67.74 A 109 B N>m2 = 67.7 GPa
Ans.
v = 0.3548 = 0.355
Ans.
Using the above results, G =
67.74 A 109 B E = 2(1 + v) 2(1 + 0.3548)
= 25.0 A 109 B N>m2 = 25.0 GPa
Ans.
786
x
wx
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*10–52. The block is fitted between the fixed supports. If the glued joint can resist a maximum shear stress of tallow = 2 ksi, determine the temperature rise that will cause the joint to fail. Take E = 10 (103) ksi, n = 0.2, and Hint: Use Eq. 10–18 with an additional strain term of a¢T (Eq. 4–4).
40⬚
Normal Strain: Since the aluminum is confined along the y direction by the rigid frame, then ey = 0 and sx = sz = 0. Applying the generalized Hooke’s Law with the additional thermal strain, ey =
0 =
1 C s - v(sx + sz) D + a¢T E y
1 C sy - 0.2(0 + 0) D + 6.0 A 10 - 6 B (¢T) 10.0(103)
sy = -0.06¢T Construction of the Circle: In accordance with the sign convention. sx = 0, sy = -0.06¢T and txy = 0. Hence. savg =
sx + sy 2
=
0 + ( -0.06¢T) = -0.03¢T 2
The coordinates for reference points A and C are A (0, 0) and C( -0.03¢T, 0). The radius of the circle is R = 2(0 - 0.03¢T)2 + 0 = 0.03¢T Stress on The inclined plane: The shear stress components tx¿y¿, are represented by the coordinates of point P on the circle. tx¿y¿ = 0.03¢T sin 80° = 0.02954¢T Allowable Shear Stress: tallow = tx¿y¿ 2 = 0.02954¢T ¢T = 67.7 °F
Ans.
787
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z
•10–53.
The smooth rigid-body cavity is filled with liquid 6061-T6 aluminum. When cooled it is 0.012 in. from the top of the cavity. If the top of the cavity is covered and the temperature is increased by 200°F, determine the stress components sx , sy , and sz in the aluminum. Hint: Use Eqs. 10–18 with an additional strain term of a¢T (Eq. 4–4).
0.012 in.
4 in. 4 in.
6 in.
y
Normal Strains: Since the aluminum is confined at its sides by a rigid container and 0.012 allowed to expand in the z direction, ex = ey = 0; whereas ez = = 0.002. 6 Applying the generalized Hooke’s Law with the additional thermal strain,
ex =
0 =
1 C s - v(sy + sz) D + a¢T E x 1 C sx - 0.35 A sy + sz B D + 13.1 A 10 - 6 B (200) 10.0(103)
0 = sx - 0.35sy - 0.35sz + 26.2 ey =
0 =
[1]
1 C s - v(sx + sz) + a¢T E y 1 C sy - 0.35(sx + sz) D + 13.1 A 10 - 6 B (200) 10.0(103)
0 = sy - 0.35sx - 0.35sz + 26.2 ez =
0.002 =
[2]
1 C s - v A sx + sy B D + a¢T E z 1 C sz - 0.35 A sx + sy B D + 13.1 A 10 - 6 B (200) 10.0(103)
0 = sz - 0.35sx - 0.35sy + 6.20
[3]
Solving Eqs.[1], [2] and [3] yields: sx = sy = -70.0 ksi
sz = -55.2 ksi
Ans.
788
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z
10–54. The smooth rigid-body cavity is filled with liquid 6061-T6 aluminum. When cooled it is 0.012 in. from the top of the cavity. If the top of the cavity is not covered and the temperature is increased by 200°F, determine the strain components Px , Py , and Pz in the aluminum. Hint: Use Eqs. 10–18 with an additional strain term of a¢T (Eq. 4–4).
0.012 in.
4 in. 4 in.
6 in.
y
Normal Strains: Since the aluminum is confined at its sides by a rigid container, then ex = ey = 0
Ans.
and since it is not restrained in z direction, sz = 0. Applying the generalized Hooke’s Law with the additional thermal strain, ex =
0 =
1 C s - v A sy + sz B D + a¢T E x 1 C sx - 0.35 A sy + 0 B D + 13.1 A 10 - 6 B (200) 10.0(103)
0 = sx - 0.35sy + 26.2 ey =
0 =
[1]
1 C s - v(sx + sz) D + a¢T E y 1 C sy - 0.35(sx + 0) D + 13.1 A 10 - 6 B (200) 10.0(103)
0 = sy - 0.35sx + 26.2
[2]
Solving Eqs. [1] and [2] yields: sx = sy = -40.31 ksi ez =
=
1 C s - v A sx + sy B D + a¢T E z 1 {0 - 0.35[-40.31 + (-40.31)]} + 13.1 A 10 - 6 B (200) 10.0(103)
= 5.44 A 10 - 3 B
Ans.
789
x
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10–55. A thin-walled spherical pressure vessel having an inner radius r and thickness t is subjected to an internal pressure p. Show that the increase in the volume within the vessel is ¢V = 12ppr4>Et211 - n2. Use a small-strain analysis.
pr 2t
s1 = s2 = s3 = 0 e1 = e2 =
1 (s - vs2) E 1
e1 = e2 =
pr (1 - v) 2t E
e3 =
1 (-v(s1 + s2)) E e3 = -
V =
v pr tE
4pr3 3
V + ¢V =
4p 4pr3 ¢r 3 (r + ¢r)3 = (1 + ) r 3 3
where ¢V V V, ¢r V r V + ¢V -
eVol =
¢r 4p r3 a1 + 3 b r 3
¢V ¢r = 3a b V r
Since e1 = e2 =
eVol = 3e1 =
2p(r + ¢r) - 2p r ¢r = r 2p r
3pr (1 - v) 2t E
¢V = VeVol =
2pp r4 (1 - v) Et
QED
790
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*10–56. A thin-walled cylindrical pressure vessel has an inner radius r, thickness t, and length L. If it is subjected to an internal pressure p, show that the increase in its inner radius is dr = rP1 = pr211 - 12 n2>Et and the increase in its length is ¢L = pLr112 - n2>Et. Using these results, show that the change in internal volume becomes dV = pr211 + P12211 + P22L - pr2L. Since P1 and P2 are small quantities, show further that the change in volume per unit volume, called volumetric strain, can be written as dV>V = pr12.5 - 2n2>Et. Normal stress: pr ; t
s1 =
s2 =
pr 2t
Normal strain: Applying Hooke’s law e1 =
=
1 [s - v (s2 + s3)], E 1
vpr pr 1 1 pr a b = a1 - vb E t 2t Et 2
d r = et r =
e2 =
=
s3 = 0
p r2 1 a1 - v b Et 2
1 [s - v (s1 + s3)], E 2
QED
s3 = 0
vpr pr 1 1 pr a b = a - vb E 2t t Et 2
¢L = e2 L =
pLr 1 a - vb Et 2
V¿ = p(r + e1 r)2 (L + e2L) ;
QED V = p r2 L
dV = V¿ - V = pr2 (1 + e1)2 (1 + e2)L - pr2 L
QED
(1 + e1)2 = 1 + 2 e1 neglect e21 term (1 + e1)2 (1 + e2) = (1 + 2 e1)(1 + e2) = 1 + e2 + 2 e1 neglect e1 e2 term dV = 1 + e2 + 2 e1 - 1 = e2 + 2 e1 V =
2pr pr 1 1 a - vb + a1 - v b Et 2 Et 2
=
pr (2.5 - 2 v) Et
QED
791
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10–57. The rubber block is confined in the U-shape smooth rigid block. If the rubber has a modulus of elasticity E and Poisson’s ratio n, determine the effective modulus of elasticity of the rubber under the confined condition.
P
Generalized Hooke’s Law: Under this confined condition, ex = 0 and sy = 0. We have ex =
0 =
1 C s - v A sy + sz B D E x 1 (s - vsz) E x
sx = vsz
(1)
ez =
1 C s - v A sx + sy B D E z
ez =
1 [s - v(sx + 0)] E z
ez =
1 (s - vsx) E z
(2)
Substituting Eq. (1) into Eq. (2), ez =
sz E
A 1 - v2 B
The effective modulus of elasticity of the rubber block under the confined condition can be determined by considering the rubber block as unconfined but rather undergoing the same normal strain of ez when it is subjected to the same normal stress sz, Thus, sz = Eeff ez Eeff =
sz ez
sz =
sz E
A 1 - v2 B
=
E 1 - v2
Ans.
792
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z
10–58. A soft material is placed within the confines of a rigid cylinder which rests on a rigid support. Assuming that Px = 0 and Py = 0, determine the factor by which the modulus of elasticity will be increased when a load is applied if n = 0.3 for the material.
P
x
Normal Strain: Since the material is confined in a rigid cylinder. ex = ey = 0. Applying the generalized Hooke’s Law, ex =
1 C s - v(sy + sx) D E z
0 = sx - v(sy + sz) ey =
[1]
1 C s - v(sx + sz) D E y
0 = sy - v(sx + sz)
[2]
Solving Eqs.[1] and [2] yields: sx = sy =
v s 1 - v z
Thus, ez =
=
1 C s - v(sx + sy) D E z v v 1 csz - v a sz + s bd E 1 - v 1 - v z sz
=
E
c1 -
2v2 d 1 - v
=
sz 1 - v - 2v2 c d E 1 - v
=
sz (1 + v)(1 - 2v c d E 1 - v
Thus, when the material is not being confined and undergoes the same normal strain of ez, then the requtred modulus of elasticity is E¿ =
sz ez
=
The increased factor is k =
1 - v E (1 - 2v)(1 + v) E¿ 1 - v = E (1 - 2v)(1 + v) =
1 - 0.3 [1 - 2(0.3)](1 + 0.3)
= 1.35
Ans.
793
y
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10–59. A material is subjected to plane stress. Express the distortion-energy theory of failure in terms of sx , sy , and txy . Maximum distortion energy theory: (s21 - s1 s2 + s22) = s2Y s1,2 =
sx + sy ;
2
Let a =
sx + sy 2
s1 = a + b;
A
a
(1)
sx - sy 2
and b =
A
a
2
2 b + txy
sx - sy 2
2
2 b + txy
s2 = a - b
s21 = a2 + b2 + 2 a b;
s22 = a2 + b2 - 2 a b
s1 s2 = a2 - b2 From Eq. (1) (a2 + b2 + 2 a b - a2 + b2 + a2 + b2 - 2 a b) = s2y (a2 + 3 b2) = s2Y (sx + sy)2 4
+ 3
(sx - sy)2 4
+ 3 t2xy = s2Y
s2x + s2y - sxsy + 3 t2xy = s2Y
Ans.
*10–60. A material is subjected to plane stress. Express the maximum-shear-stress theory of failure in terms of sx , sy , and txy . Assume that the principal stresses are of different algebraic signs. Maximum shear stress theory: |s1 - s2| = sY s1,2 =
(1)
sx + sy ;
2
` s1 - s2 ` = 2
A
a
A
a
sx - sy 2
sx - sy 2
2
2 b + txy
2 b + txy 2
From Eq. (1) 4 ca
sx - sy 2
2
b + t2xy d = s2Y
2
(sx - sy) + 4 t2xy = s2Y
Ans.
794
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•10–61.
An aluminum alloy 6061-T6 is to be used for a solid drive shaft such that it transmits 40 hp at 2400 rev>min. Using a factor of safety of 2 with respect to yielding, determine the smallest-diameter shaft that can be selected based on the maximum-shear-stress theory. v = a2400 T =
2p rad 1 min rev ba ba b = 80 p rad>s min rev 60s
40 (550) (12) P 3300 # = = lb in. p v 80 p Tc J
Applying t =
t =
A 3300 p B c p 2
=
c4
6600 p3 c3
The principal stresses: s1 = t =
6600 ; p2 c3
s2 = -t =
6600 p2 c3
Maximum shear stress theory: Both principal stresses have opposite sign, hence,
` s1 - s2 ` =
2a
sY ; F.S.
37 (103) 6600 b = ` ` 2 3 2 pc
c = 0.4166 in. d = 0.833 in.
Ans.
10–62. Solve Prob. 10–61 using the maximum-distortionenergy theory. v = a2400 T =
2p rad 1 min rev ba ba b = 80 p rad>s min rev 60s
40 (550) (12) P 3300 = = lb.in. p v 80 p
Applying t =
t =
A 3300 p B c p 2
c4
=
Tc J 6600 p2 c3
The principal stresses: s1 = t =
6600 ; p2 c3
s2 = - t = -
6600 p2 c3
The maximum distortion-energy theory: s21 - s1 s2 + s22 = a 3B
sY 2 b F.S.
37(103) 2 6600 2 = a b R 2 p2 c3
c = 0.3971 in. d = 0.794 in.
Ans. 795
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10–63. An aluminum alloy is to be used for a drive shaft such that it transmits 25 hp at 1500 rev>min. Using a factor of safety of 2.5 with respect to yielding, determine the smallest-diameter shaft that can be selected based on the maximum-distortion-energy theory. sY = 3.5 ksi. 1500(2p) = 50p 60
T =
P v
T =
25(550)(12) 3300 = p 50p
t =
Tc , J
t =
3300 p c p 4 2c
s1 =
v =
J =
=
p 4 c 2
6600 p2c3
6600 p2c3
s2 =
s21 - s1 s2 + s22 = a 3a
-6600 p2c3
sY 2 b F.S.
3.5(103) 2 6600 2 b b = a 2.5 p2c3
c = 0.9388 in. d = 1.88 in.
Ans.
*10–64. A bar with a square cross-sectional area is made of a material having a yield stress of sY = 120 ksi. If the bar is subjected to a bending moment of 75 kip # in., determine the required size of the bar according to the maximumdistortion-energy theory. Use a factor of safety of 1.5 with respect to yielding. Normal and Shear Stress: Applying the flexure formula, s =
75 A a2 B Mc 450 = 1 4 = 3 I a 12 a
In-Plane Principal Stress: Since no shear stress acts on the element s1 = sx =
450 a3
s2 = sy = 0
Maximum Distortion Energy Theory: s21 - s1 s2 + s22 = s2allow a
120 2 450 2 b - 0 + 0 = a b 3 1.5 a a = 1.78 in.
Ans.
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•10–65.
Solve Prob. 10–64 using the maximum-shearstress theory.
Normal and Shear Stress: Applying the flexure formula, s =
75 A a2 B Mc 450 = 1 4 = 3 I a a 12
In-Plane Principal Stress: Since no shear stress acts on the element. s1 = sx =
450 a3
s2 = sx = 0
Maximum Shear Stress Theory: |s2| = 0 6 sallow =
120 = 80.0 ksi 1.5
(O.K!)
|s1| = sallow 120 450 = 1.5 a3 a = 1.78 in.
Ans.
10–66. Derive an expression for an equivalent torque Te that, if applied alone to a solid bar with a circular cross section, would cause the same energy of distortion as the combination of an applied bending moment M and torque T.
t =
Te c J
Principal stress: s1 = tx ¿ ud =
s2 = -t
1 + v 2 (s1 - s1 s2 + s22) 3E
(ud)1 =
1 + v 1 + v 3 T2x c2 ( 3 t2) = a b 3E 3E J2
Bending moment and torsion: s =
Mc ; I
t =
Tc J
Principal stress: s1, 2 =
s1 =
s + 0 s - 0 2 2 ; a b + t 2 A 2
s s2 + + t2 ; 2 A4
s2 =
s s2 + t2 2 A4
797
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10–66.
Continued
Let a =
s 2
b =
s2 + t2 A4
s21 = a2 + b2 + 2 a b s1 s2 = a2 - b2 s22 = a2 + b2 - 2 a b s21 - s1 s2 + s22 = 3 b2 + a2 1 + v 2 (s1 - s1 s2 + s22) 3E
ud =
(ud)2 =
=
1 + v 1 + v 3 s2 s2 (3 b2 + a2) = a + 3t2 + b 3E 3E 4 4 c2(1 + v) M2 1 + v 2 3 T2 b (s + 3 t2) = a 2 + 3E 3E I J2
(ud)1 = (ud)2 c3(1 + v) 3 Tx 2 c2(1 + v) M2 3 T2 = a + b 3E 3E J2 I2 J2 For circular shaft J = I
p 3 p 4
c4 c4
=2
Te =
J2 M2 + T2 A I2 3
Te =
4 2 M + T2 A3
Ans.
10–67. Derive an expression for an equivalent bending moment Me that, if applied alone to a solid bar with a circular cross section, would cause the same energy of distortion as the combination of an applied bending moment M and torque T. Principal stresses: s1 =
Me c ; I
ud =
1 + v 2 (s1 - s1 s2 + s22) 3E
(ud)1 =
s2 = 0
1 + v M2e c2 a 2 b 3E I
(1)
798
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10–67.
Continued
Principal stress: s + 0 s - 0 2 3 ; a b + t 2 A 2
s1, 2 =
s s2 + + t2; 2 A4
s1 =
s2 =
s s2 + t2 2 A4
Distortion Energy: s s2 ,b = + t2 2 A4
Let a =
s21 = a2 + b2 + 2 a b s1 s2 = a2 - b2 s22 = a2 + b2 - 2 a b s22 - s1 s2 + s22 = 3 b2 + a2 Apply s =
(ud)2 =
=
Mc ; I
t =
Tc J
3s2 1 + v 1 + v s2 (3 b2 + a2) = a + + 3 t2 b 3E 3E 4 4 1 + v 2 1 + v M2 c2 3 T2 c2 (s + 3 t2) = a 2 + b 3E 3E I J2
(2)
Equating Eq. (1) and (2) yields: (1 + v) Me c2 1 + v M2 c2 3T2 c2 a 2 b = a 2 + b 3E 3E I I J2 M2e 2
=
I
M1 3 T2 + 2 I J2
M2e = M1 + 3 T2 a
I 2 b J
For circular shaft I = J
p 4 p 2
c4 c4
=
1 2
1 2 Hence, M2e = M2 + 3 T2 a b 2 Me =
A
M2 +
3 2 T 4
Ans.
799
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*10–68. The short concrete cylinder having a diameter of 50 mm is subjected to a torque of 500 N # m and an axial compressive force of 2 kN. Determine if it fails according to the maximum-normal-stress theory. The ultimate stress of the concrete is sult = 28 MPa. A =
p (0.05)2 = 1.9635(10 - 3) m2 4
J =
p (0.025)4 = 0.61359(10 - 4) m4 2
2 kN
500 N⭈m
500 N⭈m
2 kN
3
s =
2(10 ) P = 1.019 MPa = A 1.9635(10 - 3)
t =
500(0.025) Tc = 20.372 MPa = J 0.61359(10 - 6)
sx = 0
sy = -1.019 MPa sx + sy
s1, 2 =
s1,2 =
2
;
A
a
sx - sy 2
txy = 20.372 MPa 2 b + txy 2
0 - 1.018 0 - (-1.019) 2 2 ; a b + 20.372 2 A 2
s1 = 19.87 MPa
s2 = -20.89 MPa
Failure criteria: |s1| 6 salt = 28 MPa
OK
|s2| 6 salt = 28 MPa
OK
No.
Ans.
•10–69.
Cast iron when tested in tension and compression has an ultimate strength of 1sult2t = 280 MPa and 1sult2c = 420 MPa, respectively. Also, when subjected to pure torsion it can sustain an ultimate shear stress of tult = 168 MPa. Plot the Mohr’s circles for each case and establish the failure envelope. If a part made of this material is subjected to the state of plane stress shown, determine if it fails according to Mohr’s failure criterion.
120 MPa
100 MPa
220 MPa
s1 = 50 + 197.23 = 247 MPa s2 = 50 - 197.23 = -147 MPa The principal stress coordinate is located at point A which is outside the shaded region. Therefore the material fails according to Mohr’s failure criterion. Yes.
Ans.
800
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10–69.
Continued
10–70. Derive an expression for an equivalent bending moment Me that, if applied alone to a solid bar with a circular cross section, would cause the same maximum shear stress as the combination of an applied moment M and torque T. Assume that the principal stresses are of opposite algebraic signs. Bending and Torsion: Mc Mc 4M = p 4 = ; I c p c3 4
s =
t =
Tc Tc 2T = p 4 = J c p c3 2
The principal stresses:
s1, 2 =
= tabs max
sx + sy 2
;
A
a
sx - sy 2
2
2 b + txy =
4M pc3
4M
+ 0 2
3
;
Q
¢pc
- 0 2
2
≤ + a
2T 3
pc
b
2
2 2M ; 2M2 + T2 p c3 p c3
= s1 - s2 = 2 c
2 2M2 + T2 d p c3
(1)
Pure bending: s1 = tabs max
Me c 4 Me Mc = p 4 = ; I c p c3 4
= s1 - s2 =
s2 = 0
4 Me
(2)
p c3
Equating Eq. (1) and (2) yields: 4 Me 4 2M2 + T2 = p c3 p c3 Me = 2M2 + T2
Ans.
801
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10–71. The components of plane stress at a critical point on an A-36 steel shell are shown. Determine if failure (yielding) has occurred on the basis of the maximum-shearstress theory.
60 MPa
40 MPa
In accordance to the established sign convention, sx = 70 MPa, sy = -60 MPa and txy = 40 MPa. s1, 2 =
=
sx + sy 2
;
A
a
sx - sy 2
70 MPa
2 b + txy 2
70 + (-60) 70 - (-60) 2 2 ; c d + 40 2 A 2
= 5 ; 25825 s1 = 81.32 MPa
s2 = -71.32 MPa
In this case, s1 and s2 have opposite sign. Thus, |s1 - s2| = |81.32 - (-71.32)| = 152.64 MPa 6 sy = 250 MPa Based on this result, the steel shell does not yield according to the maximum shear stress theory.
*10–72. The components of plane stress at a critical point on an A-36 steel shell are shown. Determine if failure (yielding) has occurred on the basis of the maximumdistortion-energy theory.
60 MPa
40 MPa
In accordance to the established sign convention, sx = 70 MPa, sy = -60 MPa and txy = 40 MPa. s1, 2 =
=
sx + sy 2
;
A
a
sx - sy 2
2 b + txy 2
70 + (-60) 70 - ( -60) 2 2 ; c d + 40 2 A 2
= 5 ; 25825 s1 = 81.32 MPa
s2 = -71.32 MPa
s1 2 - s1 s2 + s2 2 = 81.322 - 81.32(-71.32) + (-71.32)2 = 17,500 6 sy 2 = 62500 Based on this result, the steel shell does not yield according to the maximum distortion energy theory.
802
70 MPa
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•10–73.
If the 2-in. diameter shaft is made from brittle material having an ultimate strength of sult = 50 ksi for both tension and compression, determine if the shaft fails according to the maximum-normal-stress theory. Use a factor of safety of 1.5 against rupture.
30 kip 4 kip · ft
Normal Stress and Shear Stresses. The cross-sectional area and polar moment of inertia of the shaft’s cross-section are A = p A 12 B = pin2
J =
p 4 p A 1 B = in4 2 2
The normal stress is caused by axial stress. s =
N 30 = = -9.549 ksi p A
The shear stress is contributed by torsional shear stress. t =
4(12)(1) Tc = = 30.56 ksi p J 2
The state of stress at the points on the surface of the shaft is represented on the element shown in Fig. a. In-Plane Principal Stress. sx = -9.549 ksi, sy = 0 and txy = -30.56 ksi. We have s1, 2 =
=
sx + sy 2
;
A
a
sx - sy 2
2 b + txy 2
-9.549 - 0 2 -9.549 + 0 2 ; a b + ( -30.56) 2 A 2
= (-4.775 ; 30.929) ksi s1 = 26.15 ksi
s2 = -35.70 ksi
Maximum Normal-Stress Theory. sallow =
sult 50 = = 33.33 ksi F.S. 1.5
|s1| = 26.15 ksi 6 sallow = 33.33 ksi
(O.K.)
|s2| = 35.70 ksi 7 sallow = 33.33 ksi
(N.G.)
Based on these results, the material fails according to the maximum normal-stress theory.
803
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10–74. If the 2-in. diameter shaft is made from cast iron having tensile and compressive ultimate strengths of 1sult2t = 50 ksi and 1sult2c = 75 ksi, respectively, determine if the shaft fails in accordance with Mohr’s failure criterion. Normal Stress and Shear Stresses. The cross-sectional area and polar moment of inertia of the shaft’s cross-section are A = p A 12 B = p in2
J =
p 4 p A 1 B = in4 2 2
The normal stress is contributed by axial stress. s =
N 30 = = -9.549 ksi p A
The shear stress is contributed by torsional shear stress. t =
4(12)(1) Tc = = 30.56 ksi p J 2
The state of stress at the points on the surface of the shaft is represented on the element shown in Fig. a. In-Plane Principal Stress. sx = -9.549 ksi, sy = 0, and txy = -30.56 ksi. We have s1, 2 =
=
sx + sy 2
;
A
a
sx - sy 2
2 b + txy 2
-9.549 - 0 2 -9.549 + 0 2 ; a b + ( -30.56) 2 A 2
= ( -4.775 ; 30.929) ksi s1 = 26.15 ksi
s2 = -35.70 ksi
Mohr’s Failure Criteria. As shown in Fig. b, the coordinates of point A, which represent the principal stresses, are located inside the shaded region. Therefore, the material does not fail according to Mohr’s failure criteria.
804
30 kip 4 kip · ft
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10–75. If the A-36 steel pipe has outer and inner diameters of 30 mm and 20 mm, respectively, determine the factor of safety against yielding of the material at point A according to the maximum-shear-stress theory.
900 N
150 mm A
100 mm
200 mm
Internal Loadings. Considering the equilibrium of the free - body diagram of the post’s right cut segment Fig. a, ©Fy = 0; Vy + 900 - 900 = 0
Vy = 0 T = -360 N # m
©Mx = 0; T + 900(0.4) = 0
©Mz = 0; Mz + 900(0.15) - 900(0.25) = 0 Mz = 90 N # m Section Properties. The moment of inertia about the z axis and the polar moment of inertia of the pipe’s cross section are Iz =
p A 0.0154 - 0.014 B = 10.15625p A 10 - 9 B m4 4
J =
p A 0.0154 - 0.014 B = 20.3125p A 10 - 9 B m4 2
Normal Stress and Shear Stress. The normal stress is contributed by bending stress. Thus, sY = -
MyA 90(0.015) = = -42.31MPa Iz 10.15625p A 10 - 9 B
The shear stress is contributed by torsional shear stress. t =
360(0.015) Tc = = 84.62 MPa J 20.3125p A 10 - 9 B
The state of stress at point A is represented by the two - dimensional element shown in Fig. b. In - Plane Principal Stress. sx = -42.31 MPa, sz = 0 and txz = 84.62 MPa. We have s1, 2 =
=
sx + sz 2
;
A
a
sx - sz 2
2 b + txz 2
-42.31 - 0 2 -42.31 + 0 2 ; a b + 84.62 2 A 2
= (-21.16 ; 87.23) MPa s1 = 66.07 MPa
s2 = -108.38 MPa
805
200 mm
900 N
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10–75.
Continued
Maximum Shear Stress Theory. s1 and s2 have opposite signs. This requires |s1 - s2| = sallow 66.07 - (-108.38) = sallow sallow = 174.45 MPa The factor of safety is F.S. =
sY 250 = = 1.43 sallow 174.45
Ans.
806
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*10–76. If the A-36 steel pipe has an outer and inner diameter of 30 mm and 20 mm, respectively, determine the factor of safety against yielding of the material at point A according to the maximum-distortion-energy theory.
900 N
150 mm A
100 mm
200 mm
Internal Loadings: Considering the equilibrium of the free - body diagram of the pipe’s right cut segment Fig. a, ©Fy = 0; Vy + 900 - 900 = 0
Vy = 0 T = -360 N # m
©Mx = 0; T + 900(0.4) = 0
©Mz = 0; Mz + 900(0.15) - 900(0.25) = 0 Mz = 90 N # m Section Properties. The moment of inertia about the z axis and the polar moment of inertia of the pipe’s cross section are Iz =
p A 0.0154 - 0.014 B = 10.15625p A 10 - 9 B m4 4
J =
p A 0.0154 - 0.014 B = 20.3125p A 10 - 9 B m4 2
Normal Stress and Shear Stress. The normal stress is caused by bending stress. Thus, sY = -
MyA 90(0.015) = = -42.31MPa Iz 10.15625p A 10 - 9 B
The shear stress is caused by torsional stress. t =
360(0.015) Tc = = 84.62 MPa J 20.3125p A 10 - 9 B
The state of stress at point A is represented by the two -dimensional element shown in Fig. b. In - Plane Principal Stress. sx = -42.31 MPa, sz = 0 and txz = 84.62 MPa. We have s1, 2 =
=
sx + sz 2
;
A
a
sx - sz 2
2 b + txz 2
-42.31 - 0 2 -42.31 + 0 2 ; a b + 84.62 2 A 2
= (-21.16 ; 87.23) MPa s1 = 66.07 MPa
s2 = -108.38 MPa
807
200 mm
900 N
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10–76.
Continued
Maximum Distortion Energy Theory. s1 2 - s1s2 + s2 2 = sallow 2 66.072 - 66.07(-108.38) + (-108.38)2 = sallow 2 sallow = 152.55 MPa Thus, the factor of safety is F.S. =
sY 250 = = 1.64 sallow 152.55
Ans.
808
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•10–77.
The element is subjected to the stresses shown. If sY = 36 ksi, determine the factor of safety for the loading based on the maximum-shear-stress theory. sx = 4 ksi s1, 2 =
=
sy = -12 ksi
sx + sy ;
2
A
a
sx - sy 2
txy = -8 ksi 4 ksi
2 b + txy 2
8 ksi
4 - (-12) 2 4 - 12 2 ; a b + (-8) 2 A 2 s2 = -15.314 ksi
s1 = 7.314 ksi
tabsmax =
7.314 - (-15.314) s1 - s2 = = 11.314 ksi 2 2
tallow =
sY 36 = = 18 ksi 2 2
F.S. =
12 ksi
tallow 18 = = 1.59 abs tmax 11.314
Ans.
10–78. Solve Prob. 10–77 using the maximum-distortionenergy theory.
sx = 4 ksi s1, 2 =
=
sy = -12 ksi
sx + sy 2
;
A
a
sx - sy
txy = -8 ksi 4 ksi
2 b + txy 2
8 ksi
4 - (-12) 2 4 - 12 2 ; a b + (-8) 2 A 2
s1 = 7.314 ksi
s2 = -15.314 ksi
s1 2 - s1 s2 + s2 2 = a F.S. =
2
12 ksi
sY 2 b F.S.
362 = 1.80 A (7.314)2 - (7.314)(-15.314) + (-15.314)2
Ans.
809
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10–79. The yield stress for heat-treated beryllium copper is sY = 130 ksi. If this material is subjected to plane stress and elastic failure occurs when one principal stress is 145 ksi, what is the smallest magnitude of the other principal stress? Use the maximum-distortion-energy theory. Maximum Distortion Energy Theory : With s1 = 145 ksi, s21 - s1 s2 + s22 = s2Y 1452 - 145s2 + s22 = 1302 s22 - 145s2 + 4125 = 0 s2 =
-(-145) ; 2( -145)2 - 4(1)(4125) 2(1)
= 72.5 ; 33.634 Choose the smaller root, s2 = 38.9 ksi
Ans.
sy ⫽ 0.5sx
*10–80. The plate is made of hard copper, which yields at sY = 105 ksi. Using the maximum-shear-stress theory, determine the tensile stress sx that can be applied to the plate if a tensile stress sy = 0.5sx is also applied.
s1 = sx
sx
1 s2 = sx 2
|s1| = sY sx = 105 ksi
Ans.
sy ⫽ 0.5sx
•10–81.
Solve Prob. 10–80 using the maximum-distortionenergy theory. s1 = sx s2 =
sx 2
sx
s21 - s1 s2 + s22 = s2Y s2x -
s2x s2x + = (105)2 2 4
sx = 121 ksi
Ans.
810
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10–82. The state of stress acting at a critical point on the seat frame of an automobile during a crash is shown in the figure. Determine the smallest yield stress for a steel that can be selected for the member, based on the maximumshear-stress theory. Normal and Shear Stress: In accordance with the sign convention. sx = 80 ksi
sy = 0
25 ksi
txy = 25 ksi
80 ksi
In - Plane Principal Stress: Applying Eq. 9-5. s1,2 =
=
sx + sy ;
2
A
a
sx - sy 2
2 b + txy 2
80 - 0 2 80 + 0 2 ; a b + 25 2 A 2
= 40 ; 47.170 s1 = 87.170 ksi
s2 = -7.170 ksi
Maximum Shear Stress Theory: s1 and s2 have opposite signs so |s1 - s2| = sY |87.170 - (-7.170)| = sY sY = 94.3 ksi
Ans.
10–83. Solve Prob. 10–82 using the maximum-distortionenergy theory. Normal and Shear Stress: In accordance with the sign convention. sx = 80 ksi
sy = 0
txy = 25 ksi
In - Plane Principal Stress: Applying Eq. 9-5. s1,2 =
=
sx + sy ;
2
A
a
25 ksi
sx - s 2 b + txy 2 2
80 ksi
80 - 0 2 80 + 0 2 ; a b + 25 2 A 2
= 40 ; 47.170 s1 = 87.170 ksi
s2 = -7.170 ksi
Maximum Distortion Energy Theory: s21 - s1s2 + s22 = s2Y 87.1702 - 87.170(-7.170) + (-7.170)2 = s2Y sY = 91.0 ksi
Ans.
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*10–84. A bar with a circular cross-sectional area is made of SAE 1045 carbon steel having a yield stress of sY = 150 ksi. If the bar is subjected to a torque of 30 kip # in. and a bending moment of 56 kip # in., determine the required diameter of the bar according to the maximum-distortion-energy theory. Use a factor of safety of 2 with respect to yielding.
Normal and Shear Stresses: Applying the flexure and torsion formulas. 56 A d2 B Mc 1792 = = p d 4 I pd3 A B
s =
4
t =
Tc = J
2
30 A d2 B
A B
d 4 2
p 2
=
480 pd3
The critical state of stress is shown in Fig. (a) or (b), where sx =
1792 pd3
sy = 0
txy =
480 pd3
In - Plane Principal Stresses : Applying Eq. 9-5, s1,2 =
sx + sy 2 1792 3 pd
=
=
s1 =
;
A
+ 0 2
;
D
a
¢
sx - sy 2 1792 3 pd
- 0 2
2 b + txy 2
2
≤ + a
480 2 b pd3
896 1016.47 ; pd3 pd3
1912.47 pd3
s2 = -
120.47 pd3
Maximum Distortion Energy Theory : s21 - s1s2 + s22 = s2allow a
1912.47 2 1912.47 120.47 120.47 2 150 2 b - a bab + ab = a b 3 3 3 3 2 pd pd pd pd d = 2.30 in.
Ans.
812
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•10–85.
The state of stress acting at a critical point on a machine element is shown in the figure. Determine the smallest yield stress for a steel that might be selected for the part, based on the maximum-shear-stress theory.
10 ksi
The Principal stresses: s1,2 =
=
sx + sy ;
2
A
4 ksi
a
sx - sy 2
2 b + txy 2
8 ksi
8 - (-10) 2 8 - 10 2 ; a b + 4 2 A 2
s1 = 8.8489 ksi
s2 = -10.8489 ksi
Maximum shear stress theory: Both principal stresses have opposite sign. hence, |s1 - s2| = sY
8.8489 - (-10.8489) = sY
sY = 19.7 ksi
Ans.
10–86. The principal stresses acting at a point on a thinwalled cylindrical pressure vessel are s1 = pr>t, s2 = pr>2t, and s3 = 0. If the yield stress is sY, determine the maximum value of p based on (a) the maximum-shear-stress theory and (b) the maximum-distortion-energy theory.
a) Maximum Shear Stress Theory: s1 and s2 have the same signs, then |s2| = sg
2
pr 2 = sg 2t
p =
2t s r g
|s1| = sg
2
pr 2 = sg t
p =
t s (Controls!) r g
Ans.
b) Maximum Distortion Energy Theory : s21 - s1s2 + s22 = s2g a
pr 2 pr pr pr 2 b - a b a b + a b = s2g t t 2t 2t p =
2t 23r
sg
Ans.
813
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10–87. If a solid shaft having a diameter d is subjected to a torque T and moment M, show that by the maximumshear-stress theory the maximum allowable shear stress is tallow = 116>pd322M2 + T2. Assume the principal stresses to be of opposite algebraic signs.
T
T
M
M
Section properties : I =
p d 4 pd4 a b = ; 4 2 64
J =
p d 4 pd4 a b = 2 2 32
Thus, M(d2 ) Mc 32 M = = p d4 I pd3
s =
64
T (d2 ) Tc 16 T t = = = p d4 J pd3 32
The principal stresses : s1,2 =
=
sx + sy ;
2
A
a
sx - sy 2
2 b + txy 2
16 M 16 M 2 16 T 2 16 M 16 ; ; 2M2 + T2 a b + a b = 3 3 A pd pd p d3 pd3 p d3
Assume s1 and s2 have opposite sign, hence, tallow
2 C 163 2M2 + T2 D s1 - s2 16 pd = = = 2M2 + T2 2 2 pd3
QED
*10–88. If a solid shaft having a diameter d is subjected to a torque T and moment M, show that by the maximum-normalstress theory the maximum allowable principal stress is sallow = 116>pd321M + 2M2 + T22.
T M
M
Section properties : I =
p d4 ; 64
p d4 32
J =
Stress components : s =
M (d2 ) Mc 32 M = p 4 = ; I d p d3 64
t =
T(d2 ) Tc 16 T = p 4 = J d p d3 32
The principal stresses : s1,2 =
=
sx + sy 2
;
A
a
sx - sy 2
2 b + txy = 2
32 M 3 pd
32 M
+ 0
2
3
;
D
¢ pd
- 0
2
2
≤ + a
16 T 2 b p d3
16 M 16 ; 2M2 + T2 p d3 p d3
Maximum normal stress theory. Assume s1 7 s2 sallow = s1 =
=
16 M 16 + 2M2 + T2 p d3 p d3
16 [M + 2M2 + T2] p d3
QED
814
T
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•10–89.
The shaft consists of a solid segment AB and a hollow segment BC, which are rigidly joined by the coupling at B. If the shaft is made from A-36 steel, determine the maximum torque T that can be applied according to the maximum-shear-stress theory. Use a factor of safety of 1.5 against yielding.
A
B T C
Shear Stress: This is a case of pure shear, and the shear stress is contributed by p torsion. For the hollow segment, Jh = A 0.054 - 0.044 B = 1.845p A 10 - 6 B m4. Thus, 2 (tmax)h =
T(0.05) Tch = = 8626.28T Jh 1.845p A 10 - 6 B
For the solid segment, Js =
(tmax)s =
p A 0.044 B = 1.28p A 10 - 6 B m4. Thus, 2
T(0.04) Tcs = = 9947.18T Js 1.28p A 10 - 6 B
By comparision, the points on the surface of the solid segment are critical and their state of stress is represented on the element shown in Fig. a. In - Plane Principal Stress. sx = sy = 0 and txy = 9947.18T. We have s1,2 =
=
sx + sy 2
;
C
¢
sx - sy 2
2
≤ + t2xy
0 - 0 2 0 + 0 ; ¢ ≤ + (9947.18T)2 2 C 2
s1 = 9947.18T
s2 = -9947.18T
Maximum Shear Stress Theory. sallow =
80 mm
sY 250 = = 166.67 MPa F.S. 1.5
Since s1 and s2 have opposite sings, |s1 - s2| = sallow 9947.18T - (-9947.18T) = 166.67 A 106 B T = 8377.58 N # m = 8.38 kN # m
Ans.
815
80 mm 100 mm
T
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10–90. The shaft consists of a solid segment AB and a hollow segment BC, which are rigidly joined by the coupling at B. If the shaft is made from A-36 steel, determine the maximum torque T that can be applied according to the maximum-distortion-energy theory. Use a factor of safety of 1.5 against yielding.
A
B T C
Shear Stress. This is a case of pure shear, and the shear stress is contributed by p torsion. For the hollow segment, Jh = A 0.054 - 0.044 B = 1.845p A 10 - 6 B m4. Thus, 2 (tmax)h =
T(0.05) Tch = = 8626.28T Jh 1.845p A 10 - 6 B
For the solid segment, Js =
(tmax)s =
p A 0.044 B = 1.28p A 10 - 6 B m4. Thus, 2
T(0.04) Tcs = = 9947.18T Js 1.28p A 10 - 6 B
By comparision, the points on the surface of the solid segment are critical and their state of stress is represented on the element shown in Fig. a. In - Plane Principal Stress. sx = sy = 0 and txy = 9947.18T. We have s1,2 =
=
sx + sy 2
;
C
¢
sx - sy 2
2
≤ + t2xy
0 - 0 2 0 + 0 ; ¢ ≤ + (9947.18T)2 2 C 2
s1 = 9947.18T
s2 = -9947.18T
Maximum Distortion Energy Theory. sallow =
80 mm
sY 250 = = 166.67 MPa F.S. 1.5
Then, s1 2 - s1s2 + s2 2 = sallow 2 (9947.18T)2 - (9947.18T)( -9947.18T) + (-9947.18T)2 = C 166.67 A 106 B D 2 T = 9673.60 N # m = 9.67 kN # m
Ans.
816
80 mm 100 mm
T
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10–91. The internal loadings at a critical section along the steel drive shaft of a ship are calculated to be a torque of 2300 lb # ft, a bending moment of 1500 lb # ft, and an axial thrust of 2500 lb. If the yield points for tension and shear are sY = 100 ksi and tY = 50 ksi, respectively, determine the required diameter of the shaft using the maximum-shearstress theory. 2300 lb⭈ft 2500 lb
p I = c4 4
A = p c2
sA =
p J = c4 2
1500(12)(c) P Mc 2500 2500 72 000 + b = -a + b + = -a pc4 A I p c2 p c2 p c3 4
tA =
2300(12)(c) Tc 55 200 = = p c4 J p c3 2
s1,2 =
sx + sy
= -a
2
;
A
a
sx - sy 2
2 b + txy 2
2500 c + 72 000 2500c + 72 000 2 55200 2 b ; a b + a b 3 3 A 2p c 2p c p c3
(1)
Assume s1 and s2 have opposite signs: |s1 - s2| = sg 2500c + 72 000 2 55 200 2 3 b + a b = 100(10 ) 3 A 2p c p c3
2
a
(2500c + 72000)2 + 1104002 = 10 000(106)p2 c6 6.25c2 + 360c + 17372.16 - 10 000p2 c6 = 0 By trial and error: c = 0.750 57 in. Substitute c into Eq. (1): s1 = 22 193 psi
s2 = -77 807 psi
s1 and s2 are of opposite signs
OK
Therefore, d = 1.50 in.
Ans.
817
1500 lb⭈ft
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*10–92. The gas tank has an inner diameter of 1.50 m and a wall thickness of 25 mm. If it is made from A-36 steel and the tank is pressured to 5 MPa, determine the factor of safety against yielding using (a) the maximum-shear-stress theory, and (b) the maximum-distortion-energy theory.
(a) Normal Stress. Since
0.75 r = = 30 7 10, thin - wall analysis can be used.We have t 0.025
s1 = sh =
5(0.75) pr = = 150 MPa t 0.025
s2 = slong =
pr 5(0.75) = = 75 MPa 2t 2(0.025)
Maximum Shear Stress Theory. s1 and s2 have the sign. Thus, |s1| = sallow sallow = 150 MPa The factor of safety is F.S. =
sY 250 = = 1.67 sallow 150
Ans.
(b) Maximum Distortion Energy Theory. s1 2 - s1s2 + s2 2 = sallow 2 1502 - 150(75) + 752 = sallow 2 sallow = 129.90 MPa The factor of safety is F.S. =
sY 250 = = 1.92 sallow 129.90
Ans.
818
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•10–93.
The gas tank is made from A-36 steel and has an inner diameter of 1.50 m. If the tank is designed to withstand a pressure of 5 MPa, determine the required minimum wall thickness to the nearest millimeter using (a) the maximum-shear-stress theory, and (b) maximumdistortion-energy theory. Apply a factor of safety of 1.5 against yielding.
(a) Normal Stress. Assuming that thin - wall analysis is valid, we have s1 = sh =
5 A 106 B (0.75) 3.75 A 106 B pr = = t t t
s2 = slong =
5 A 106 B (0.75) 1.875 A 106 B pr = = 2t 2t t
Maximum Shear Stress Theory. sallow =
250 A 106 B sY = = 166.67 A 106 B Pa FS. 1.5
s1 and s2 have the same sign. Thus, |s1| = sallow 3.75 A 106 B
= 166.67 A 106 B
t
t = 0.0225 m = 22.5 mm
Since
Ans.
0.75 r = = 33.3 7 10, thin - wall analysis is valid. t 0.0225
(b) Maximum Distortion Energy Theory. sallow =
250 A 106 B sY = = 166.67 A 106 B Pa F.S. 1.5
Thus, s1 2 - s1s2 + s2 2 = sallow 2
C
3.75 A 106 B t
3.2476 A 106 B t
2
S - C
3.75 A 106 B t
SC
1.875 A 106 B t
S + C
1.875 A 106 B t
2
S = c166.67 A 106 B d
= 166.67 A 106 B
t = 0.01949 m = 19.5 mm
Since
2
Ans.
0.75 r = = 38.5 7 10, thin - wall analysis is valid. t 0.01949
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10–94. A thin-walled spherical pressure vessel has an inner radius r, thickness t, and is subjected to an internal pressure p. If the material constants are E and n, determine the strain in the circumferential direction in terms of the stated parameters. s1 = s2 =
pr 2t
e1 = e2 = e =
e =
1 (s - vs) E
pr 1 - v pr 1 - v s = a b = (1 - v) E E 2t 2Et
Ans.
10–95. The strain at point A on the shell has components Px = 250(10 - 6), Py = 400(10 - 6), gxy = 275(10 - 6), Pz = 0. Determine (a) the principal strains at A, (b) the maximum shear strain in the x–y plane, and (c) the absolute maximum shear strain. ex = 250(10 - 6) A(250, 137.5)10 - 6
ey = 400(10 - 6)
gxy = 275(10 - 6)
y A
gxy 2
= 137.5(10 - 6)
C(325, 0)10 - 6
R = a 2(325 - 250)2 + (137.5)2 b10 - 6 = 156.62(10 - 6) a) e1 = (325 + 156.62)10 - 6 = 482(10 - 6)
Ans.
e2 = (325 - 156.62)10 - 6 = 168(10 - 6)
Ans.
b) g
max in-plane
= 2R = 2(156.62)(10 - 6) = 313(10 - 6)
Ans.
c) gabs
max
2 gabs
max
=
482(10 - 6) 2
= 482(10 - 6)
Ans.
820
x
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*10–96. The principal plane stresses acting at a point are shown in the figure. If the material is machine steel having a yield stress of sY = 500 MPa, determine the factor of safety with respect to yielding if the maximum-shear-stress theory is considered.
100 MPa
150 MPa
Have, the in plane principal stresses are s1 = sy = 100 MPa
s2 = sx = -150 MPa
Since s1 and s2 have same sign, F.S =
sy =
|s1 - s2|
500 = 2 |100 - (-150)|
Ans.
•10–97.
The components of plane stress at a critical point on a thin steel shell are shown. Determine if failure (yielding) has occurred on the basis of the maximumdistortion-energy theory. The yield stress for the steel is sY = 650 MPa.
340 MPa
65 MPa
55 MPa
sx = -55 MPa s1, 2 =
=
sy = 340 MPa
sx + sy 2
;
A
a
sx - sy 2
txy = 65 MPa
2 b + txy 2
-55 - 340 2 -55 + 340 2 ; a b + 65 2 A 2
s1 = 350.42 MPa
s2 = -65.42 MPa
(s1 2 - s1s2 + s2 ) = [350.422 - 350.42(-65.42) + ( -65.42)2] = 150 000 6 s2Y = 422 500
OK
No.
Ans.
821
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10–98. The 60° strain rosette is mounted on a beam. The following readings are obtained for each gauge: Pa = 600110-62, Pb = -700110-62, and Pc = 350110-62. Determine (a) the in-plane principal strains and (b) the maximum in-plane shear strain and average normal strain. In each case show the deformed element due to these strains.
a 60⬚ 60⬚ b
Strain Rosettes (60º): Applying Eq. 10-15 with ex = 600 A 10 - 6 B , eb = -700 A 10
-6
c
B , ec = 350 A 10 B , ua = 150°, ub = -150° and uc = -90°, -6
350 A 10 - 6 B = ex cos2 (-90°) + ey sin2( -90°) + gxy sin (-90°) cos ( -90°) ey = 350 A 10 - 6 B
600 A 10 - 6 B = ex cos2 150° + 350 A 10 - 6 B sin2 150° + gxy sin 150° cos 150° 512.5 A 10 - 6 B = 0.75 ex - 0.4330 gxy
[1]
-787.5 A 10 - 6 B = 0.75ex + 0.4330 gxy
[2]
-700 A 10 - 6 B = ex cos2 ( -150°) + 350 A 10 - 6 B sin2(-150°) + gxy sin (-150°) cos (-150°)
Solving Eq. [1] and [2] yields ex = -183.33 A 10 - 6 B
gxy = -1501.11 A 10 - 6 B
Construction of she Circle: With ex = -183.33 A 10 - 6 B , ey = 350 A 10 - 6 B , and gxy = -750.56 A 10 - 6 B . 2 eavg =
ex + ey 2
= a
-183.33 + 350 b A 10 - 6 B = 83.3 A 10 - 6 B 2
Ans.
The coordinates for reference points A and C are A( -183.33, -750.56) A 10 - 6 B
C(83.33, 0) A 10 - 6 B
The radius of the circle is R = a 2(183.33 + 83.33)2 + 750.562 b A 10 - 6 B = 796.52 A 10 - 6 B a) In-plane Principal Strain: The coordinates of points B and D represent e1 and e2, respectively. e1 = (83.33 + 796.52) A 10 - 6 B = 880 A 10 - 6 B
Ans.
e2 = (83.33 - 796.52) A 10 - 6 B = -713 A 10 - 6 B
Ans.
Orientation of Principal Strain: From the circle, tan 2uP1 =
750.56 = 2.8145 183.33 + 83.33
2uP2 = 70.44°
2uP1 = 180° - 2uP2 uP =
180° - 70.44° = 54.8° (Clockwise) 2
Ans.
822
60⬚
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10–98.
Continued
b) Maximum In - Plane Shear Strain: Represented by the coordinates of point E on the circle. g max in-plane
2 g
max in-plane
= -R = -796.52 A 10 - 6 B = -1593 A 10 - 6 B
Ans.
Orientation of Maximum In-Plane Shear Strain: From the circle. tan 2uP =
183.33 + 83.33 = 0.3553 750.56
uP = 9.78° (Clockwise)
Ans.
10–99. A strain gauge forms an angle of 45° with the axis of the 50-mm diameter shaft. If it gives a reading of P = -200110-62 when the torque T is applied to the shaft, determine the magnitude of T. The shaft is made from A-36 steel.
T 45⬚
Shear Stress. This is a case of pure shear, and the shear stress developed is p contributed by torsional shear stress. Here, J = A 0.0254 B = 0.1953125p A 10 - 6 B m4. 2 Then 0.128 A 106 B T T(0.025) Tc = = t = p J 0.1953125p A 10 - 6 B
T
The state of stress at points on the surface of the shaft can be represented by the element shown in Fig. a. Shear Strain: For pure shear ex = ey = 0. We obtain, ea = ex cos2ua + ey sin2ua + gxysin ua cos ua -200 A 10 - 6 B = 0 + 0 + gxy sin 45° cos 45°
gxy = -400 A 10 - 6 B
Shear Stress and Strain Relation: Applying Hooke’s Law for shear, txy = Ggxy -
0.128 A 106 B T p
= 75 A 109 B C -400 A 10 - 6 B D
T = 736 N # m
Ans.
823
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*10–100. The A-36 steel post is subjected to the forces shown. If the strain gauges a and b at point A give readings of Pa = 300110-62 and Pb = 175110-62, determine the magnitudes of P1 and P2.
P1
P2 a
A
Internal Loadings: Considering the equilibrium of the free - body diagram of the 1 in. post’s segment, Fig. a, P2 - V = 0
V = P2
+ c ©Fy = 0;
N - P1 = 0
N = P1
a + ©MO = 0;
M + P2(2) = 0
M = 2P2
Section Properties: The cross - sectional area and the moment of inertia about the bending axis of the post’s cross - section are A = 4(2) = 8 in2 I =
1 (2) A 43 B = 10.667 in4 12
Referring to Fig. b,
A Qy B A = x¿A¿ = 1.5(1)(2) = 3 in3 Normal and Shear Stress: The normal stress is a combination of axial and bending stress. sA =
2P2(12)(1) MxA P1 N + = + = 2.25P2 - 0.125P1 A I 8 10.667
The shear stress is caused by transverse shear stress. tA =
VQA P2(3) = = 0.140625P2 It 10.667(2)
Thus, the state of stress at point A is represented on the element shown in Fig. c. Normal and Shear Strain: With ua = 90° and ub = 45°, we have ea = ex cos2ua + ey sin2ua + gxysin ua cos ua 300 A 10 - 6 B = ex cos2 90° + ey sin2 90° + gxysin 90° cos 90°
ey = 300 A 10 - 6 B eb = ex cos2ub + ey sin2 ub + gxysin ub cos ub
175 A 10 - 6 B = ex cos2 45° + 300 A 10 - 6 B sin2 45° + gxy sin 45°cos 45° ex + gxy = 50 A 10 - 6 B
(1)
824
A 1 in.
b 45⬚ c
+ ©F = 0; : x
2 in.
2 ft
A
4 in.
c
Section c– c
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10–100.
Continued
Since sy = sz = 0, ex = -vey = -0.32(300) A 10 - 6 B = -96 A 10 - 6 B Then Eq. (1) gives gxy = 146 A 10 - 6 B Stress and Strain Relation: Hooke’s Law for shear gives tx = Ggxy 0.140625P2 = 11.0 A 103 B C 146 A 10 - 6 B D P2 = 11.42 kip = 11.4 kip
Ans.
Since sy = sz = 0, Hooke’s Law gives sy = Eey 2.25(11.42) - 0.125P1 = 29.0 A 103 B C 300 A 10 - 6 B D P1 = 136 kip
Ans.
825
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10–101. A differential element is subjected to plane strain that has the following components: Px = 950110-62, Py = 420110-62, gxy = -325110-62. Use the strain-transformation equations and determine (a) the principal strains and (b) the maximum in-plane shear strain and the associated average strain. In each case specify the orientation of the element and show how the strains deform the element.
e1, 2 =
ex + ey ;
2
= c
A
a
ex - ey 2
2 b + gxy 2
950 - 420 2 -325 2 950 + 420 -6 ; a b + a b d(10 ) 2 A 2 2 e1 = 996(10 - 6)
Ans.
e2 = 374(10 - 6)
Ans.
Orientation of e1 and e2 : gxy
tan 2uP =
ex - ey
-325 950 - 420
=
uP = -15.76°, 74.24° Use Eq. 10.5 to determine the direction of e1 and e2. ex¿ =
ex + ey
ex - ey +
2
2
cos 2u +
gxy 2
sin 2u
u = uP = -15.76° ex¿ = b
( -325) 950 - 420 950 + 420 + cos (-31.52°) + sin (-31.52°) r (10 - 6) = 996(10 - 6) 2 2 2
uP1 = -15.8°
Ans.
uP2 = 74.2°
Ans.
b) gmax
in-plane
2 gmax
in-plane
eavg =
=
A
= 2c
a
ex - ey
A
2 a
ex + ey 2
b + a 2
gxy 2
b
2
950 - 420 2 -325 2 -6 -6 b + a b d(10 ) = 622(10 ) 2 2
Ans.
= a
Ans.
950 + 420 b (10 - 6) = 685(10 - 6) 2
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10–101.
Continued
Orientation of gmax : -(ex - ey)
tan 2uP =
-(950 - 420) -325
=
gxy
uP = 29.2° and uP = 119°
Ans.
Use Eq. 10.6 to determine the sign of ex - ey
gx¿y¿ = -
2
sin 2u +
2
gxy 2
g
max in-plane
:
cos 2u
u = uP = 29.2° gx¿y¿ = 2 c
-(950 - 420) -325 sin (58.4°) + cos (58.4°) d(10 - 6) 2 2
gxy = -622(10 - 6)
10–102. The state of plane strain on an element is Px = 400110-62, Py = 200110-62, and gxy = -300110-62. Determine the equivalent state of strain on an element at the same point oriented 30° clockwise with respect to the original element. Sketch the results on the element.
y
Pydy
dy
Stress Transformation Equations: ex = 400 A 10 - 6 B
ey = 200 A 10 - 6 B
gxy = -300 A 10 - 6 B
u = -30° gxy 2 dx
We obtain, ex¿ =
ex + ey +
2
= B
ex - ey 2
cos 2u +
gxy 2
sin 2u
400 + 200 400 - 200 -300 + cos (-60°) + a b sin (-60°) R A 10 - 6 B 2 2 2
= 480 A 10 - 6 B gx¿y¿ 2
= -¢
Ans.
ex - ey 2
≤ sin 2u +
gxy 2
cos 2u
gx¿y¿ = [-(400 - 200) sin ( -60°) + (-300) cos ( -60°)] A 10 - 6 B = 23.2 A 10 - 6 B
ey¿ =
ex + ey
= B
2
Ans.
ex - ey -
2
cos 2u -
gxy 2
sin 2u
400 + 200 400 - 200 -300 cos ( -60°) - a b sin (-60°) R A 10 - 6 B 2 2 2
= 120 A 10 - 6 B
Ans.
827
gxy 2 x Pxdx
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10–103. The state of plane strain on an element is Px = 400110-62, Py = 200110-62, and gxy = -300110-62. Determine the equivalent state of strain, which represents (a) the principal strains, and (b) the maximum in-plane shear strain and the associated average normal strain. Specify the orientation of the corresponding element at the point with respect to the original element. Sketch the results on the element.
y
Pydy
dy
Construction of the Circle: ex = 400 A 10 - 6 B , ey = 200 A 10 - 6 B , and
gxy 2
= -150 A 10 - 6 B .
Thus, eavg =
ex + ey 2
= a
400 + 200 b A 10 - 6 B = 300 A 10 - 6 B 2
Ans.
The coordinates for reference points A and the center C of the circle are A(400, -150) A 10 - 6 B
C(300, 0) A 10 - 6 B
The radius of the circle is R = CA = 2(400 - 300)2 + (-150)2 = 180.28 A 10 - 6 B Using these results, the circle is shown in Fig. a. In - Plane Principal Stresses: The coordinates of points B and D represent e1 and e2, respectively. Thus, e1 = (300 + 180.28) A 10 - 6 B = 480 A 10 - 6 B
Ans.
e2 = (300 - 180.28) A 10 - 6 B = 120 A 10 - 6 B
Ans.
Orientation of Principal Plane: Referring to the geometry of the circle, tan 2 A up B 1 =
150 = 1.5 400 - 300
A up B 1 = 28.2° (clockwise)
Ans.
The deformed element for the state of principal strains is shown in Fig. b. Maximum In - Plane Shear Stress: The coordinates of point E represent eavg and gmax . Thus in-plane gmax = -R = -180.28 A 10 - 6 B
in-plane
2 gmax
in-plane
= -361 A 10 - 6 B
Ans.
Orientation of the Plane of Maximum In - Plane Shear Strain: Referring to the geometry of the circle, tan 2us =
400 - 300 = 0.6667 150
uS = 16.8° (counterclockwise)
Ans.
The deformed element for the state of maximum in - plane shear strain is shown in Fig. c.
828
gxy 2
gxy 2 dx
x Pxdx
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10–103.
Continued
829
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11–1. The simply supported beam is made of timber that has an allowable bending stress of sallow = 6.5 MPa and an allowable shear stress of tallow = 500 kPa. Determine its dimensions if it is to be rectangular and have a height-towidth ratio of 1.25.
8 kN/m
2m
Ix =
1 (b)(1.25b)3 = 0.16276b4 12
Qmax = y¿A¿ = (0.3125b)(0.625b)(b) = 0.1953125b3 Assume bending moment controls: Mmax = 16 kN # m sallow =
Mmax c I
6.5(106) =
16(103)(0.625b) 0.16276b4
b = 0.21143 m = 211 mm
Ans.
h = 1.25b = 264 mm
Ans.
Check shear: Qmax = 1.846159(10 - 3) m3
I = 0.325248(10 - 3) m4 tmax =
VQmax 16(103)(1.846159)(10 - 3) = 429 kPa 6 500 kPa‚ OK = It 0.325248(10 - 3)(0.21143)
830
4m
2m
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11–2. The brick wall exerts a uniform distributed load of 1.20 kip>ft on the beam. If the allowable bending stress is sallow = 22 ksi and the allowable shear stress is tallow = 12 ksi, select the lightest wide-flange section with the shortest depth from Appendix B that will safely support the load.
1.20 kip/
4 ft
10 ft
ft
6 ft b
Bending Stress: From the moment diagram, Mmax = 44.55 kip # ft. Assuming bending controls the design and applying the flexure formula. Sreq d =
=
44.55 (12) = 24.3 in3 22
W12 * 22
A Sx = 25.4 in3, d = 12.31 in., tw = 0.260 in. B V for the W12 * 22 wide tw d = 6.60 kip.
Shear Stress: Provide a shear stress check using t = flange section. From the shear diagram, Vmax tmax =
=
Vmax tw d 6.60 0.260(12.31)
= 2.06 ksi 6 tallow = 12 ksi (O.K!) Hence,
Use
9 in. 0.5 in.
Mmax sallow
Two choices of wide flange section having the weight 22 lb>ft can be made. They are W12 * 22 and W14 * 22. However, W12 * 22 is the shortest. Select
0.5 in.
0.5 in.
Ans.
W12 * 22
831
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11–3. The brick wall exerts a uniform distributed load of 1.20 kip>ft on the beam. If the allowable bending stress is sallow = 22 ksi, determine the required width b of the flange to the nearest 14 in.
1.20 kip/
4 ft
10 ft
ft
6 ft b 0.5 in.
0.5 in. 9 in. 0.5 in.
Section Property: I =
1 1 (b) A 103 B (b - 0.5) A 93 B = 22.583b + 30.375 12 12
Bending Stress: From the moment diagram, Mmax = 44.55 kip # ft. sallow = 22 =
Mmax c I 44.55(12)(5) 22.583b + 30.375
b = 4.04 in. Use
b = 4.25 in.
Ans.
832
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*11–4. Draw the shear and moment diagrams for the shaft, and determine its required diameter to the nearest 1 4 in. if sallow = 7 ksi and tallow = 3 ksi. The bearings at A and D exert only vertical reactions on the shaft. The loading is applied to the pulleys at B, C, and E.
14 in.
20 in.
15 in.
12 in.
E A
C
B
D 35 lb
80 lb 110 lb
sallow = 7(103) =
Mmax c I 1196 c p 4 ; 4 c
c = 0.601 in.
d = 2c = 1.20 in. Use d = 1.25 in.
Ans.
Check shear: 2
tmax =
0.625 108(4(0.625) Vmax Q 3p )(p)( 2 ) = 117 psi 6 3 ksi OK = p 4 It 4 (0.625) (1.25)
•11–5.
Select the lightest-weight steel wide-flange beam from Appendix B that will safely support the machine loading shown. The allowable bending stress is sallow = 24 ksi and the allowable shear stress is tallow = 14 ksi.
2 ft
Bending Stress: From the moment diagram, Mmax = 30.0 kip # ft. Assume bending controls the design. Applying the flexure formula. Sreq¿d =
= Select
W12 * 16
Mmax sallow 30.0(12) = 15.0 in3 24
A Sx = 17.1 in3, d = 11.99 in., tw = 0.220 in. B V for the W12 * 16 wide tw d = 10.0 kip
Shear Stress: Provide a shear stress check using t = flange section. From the shear diagram, Vmax tmax =
=
Vmax tw d 10.0 0.220(11.99)
= 3.79 ksi 6 tallow = 14 ksi (O.K!) Hence, Use
5 kip
5 kip
Ans.
W12 * 16
833
2 ft
5 kip
2 ft
5 kip
2 ft
2 ft
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11–6. The compound beam is made from two sections, which are pinned together at B. Use Appendix B and select the lightest-weight wide-flange beam that would be safe for each section if the allowable bending stress is sallow = 24 ksi and the allowable shear stress is tallow = 14 ksi. The beam supports a pipe loading of 1200 lb and 1800 lb as shown.
C
A B 6 ft
Bending Stress: From the moment diagram, Mmax = 19.2 kip # ft for member AB. Assuming bending controls the design, applying the flexure formula. Sreq¿d =
= Select
Mmax sallow 19.2(12) = 9.60 in3 24
A Sx = 10.9 in3, d = 9.87 in., tw = 0.19 in. B
W10 * 12
For member BC, Mmax = 8.00 kip # ft. Sreq¿d =
= Select
Mmax sallow 8.00(12) = 4.00 in3 24
A Sx = 5.56 in3, d = 5.90 in., tw = 0.17 in. B
W6 * 9
V for the W10 * 12 widetw d flange section for member AB. From the shear diagram, Vmax = 2.20 kip. Shear Stress: Provide a shear stress check using t =
tmax =
=
Vmax tw d 2.20 0.19(9.87)
= 1.17 ksi 6 tallow = 14 ksi (O.K!) Use
Ans.
W10 * 12
For member BC (W6 * 9), Vmax = 1.00 kip. tmax =
=
Vmax tw d 1.00 0.17(5.90)
= 0.997 ksi 6 tallow = 14 ksi (O.K!) Hence, Use
1800 lb
1200 lb
W6 * 9
Ans.
834
6 ft
8 ft
10 ft
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11–7. If the bearing pads at A and B support only vertical forces, determine the greatest magnitude of the uniform distributed loading w that can be applied to the beam. sallow = 15 MPa, tallow = 1.5 MPa.
w
A
B 1m
1m 150 mm 25 mm 150 mm 25 mm
The location of c, Fig. b, is y =
0.1625(0.025)(0.15) + 0.075(0.15)(0.025) ©yA = ©A 0.025(0.15) + 0.15(0.025) = 0.11875 m
I =
+
1 (0.025)(0.153) + (0.025)(0.15)(0.04375)2 12 1 (0.15)(0.0253) + 0.15(0.025)(0.04375)2 12
= 21.58203125(10 - 6) m4 Referring to Fig. b, Qmax = y¿A¿ = 0.059375 (0.11875)(0.025) = 0.176295313(10 - 4) m3 Referring to the moment diagram, Mmax = 0.28125 w. Applying the Flexure formula with C = y = 0.11875 m, sallow =
Mmax c ; I
15(106) =
0.28125w(0.11875) 21.582(10 - 6)
W = 9.693(103) N>m Referring to shear diagram, Fig. a, Vmax = 0.75 w. tallow =
Vallow Qmax ; It
1.5(106) =
0.75w C 0.17627(10 - 3) D 21.582(10 - 6)(0.025)
W = 6.122(103) N>m = 6.12 kN>m (Control!)
Ans.
835
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*11–8. The simply supported beam is made of timber that has an allowable bending stress of sallow = 1.20 ksi and an allowable shear stress of tallow = 100 psi. Determine its smallest dimensions to the nearest 18 in. if it is rectangular and has a height-to-width ratio of 1.5.
12 kip/ft
B
A 3 ft
3 ft
1.5 b b
The moment of inertia of the beam’s cross-section about the neutral axis is 1 (b)(1.5b)3 = 0.28125b4. Referring to the moment diagram, I = 12 Mmax = 45.375 kip # ft. sallow =
Mmax c ; I
1.2 =
45.375(12)(0.75b) 0.28125b4
b = 10.66 in Referring to Fig. b, Qmax = y¿A¿ = 0.375b (0.75b)(b) = 0.28125b3. Referring to the shear diagram, Fig. a, Vmax = 33 kip. tmax =
Vmax Qmax ; It
100 =
33(103)(0.28125b3) 0.28125b4(b)
b = 18.17 in (Control!) Thus, use b = 18
1 in 4
Ans.
836
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•11–9.
Select the lightest-weight W12 steel wide-flange beam from Appendix B that will safely support the loading shown, where P = 6 kip. The allowable bending stress is sallow = 22 ksi and the allowable shear stress is tallow = 12 ksi.
P
P
9 ft
From the Moment Diagram, Fig. a, Mmax = 54 kip # ft. Mmax sallow
Sreq¿d =
54(12) 22
=
= 29.45 in3 Select W12 * 26
C Sx = 33.4 in3, d = 12.22 in and tw = 0.230 in. D
From the shear diagram, Fig. a, Vmax = 7.5 kip. Provide the shear-stress check for W 12 * 26, tmax =
=
Vmax tw d 7.5 0.230(12.22)
= 2.67 ksi 6 tallow = 12 ksi (O.K!) Hence Use
Ans.
W12 * 26
837
6 ft
6 ft
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11–10. Select the lightest-weight W14 steel wide-flange beam having the shortest height from Appendix B that will safely support the loading shown, where P = 12 kip. The allowable bending stress is sallow = 22 ksi and the allowable shear stress is tallow = 12 ksi.
P
P
9 ft
From the moment diagram, Fig. a, Mmax = 108 kip # ft. Mmax sallow
Sreq¿d =
108(12) 22
=
= 58.91 in3 Select W14 * 43
C Sx = 62.7 in3, d = 13.66 in and tw = 0.305 in. D
From the shear diagram, Fig. a, Vmax = 15 kip . Provide the shear-stress check for W14 * 43 , tmax =
=
Vmax tw d 15 0.305(13.66)
= 3.60 ksi 6 tallow = 12 ksi‚ (O.K!) Hence, Use
Ans.
W14 * 43
838
6 ft
6 ft
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11–11. The timber beam is to be loaded as shown. If the ends support only vertical forces, determine the greatest magnitude of P that can be applied. sallow = 25 MPa, tallow = 700 kPa.
150 mm 30 mm 120 mm
40 mm P 4m
A
y =
(0.015)(0.150)(0.03) + (0.09)(0.04)(0.120) = 0.05371 m (0.150)(0.03) + (0.04)(0.120)
I =
1 1 (0.150)(0.03)3 + (0.15)(0.03)(0.05371 - 0.015)2 + (0.04)(0.120)3 + 12 12
B
(0.04)(0.120)(0.09 - 0.05371)2 = 19.162(10 - 6) m4 Maximum moment at center of beam: Mmax =
P (4) = 2P 2
Mc ; I
s =
25(106) =
(2P)(0.15 - 0.05371) 19.162(10 - 6)
P = 2.49 kN Maximum shear at end of beam: Vmax =
P 2
VQ ; t = It
700(103) =
P 1 C (0.15 - 0.05371)(0.04)(0.15 - 0.05371) D 2 2 19.162(10 - 6)(0.04)
P = 5.79 kN Thus, P = 2.49 kN
Ans.
839
4m
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*11–12. Determine the minimum width of the beam to the nearest 14 in. that will safely support the loading of P = 8 kip. The allowable bending stress is sallow = 24 ksi and the allowable shear stress is tallow = 15 ksi.
P 6 ft
6 ft
6 in.
B A
Beam design: Assume moment controls. sallow =
Mc ; I
24 =
48.0(12)(3) 1 3 12 (b)(6 )
b = 4 in.
Ans.
Check shear: 8(1.5)(3)(4) VQ = 0.5 ksi 6 15 ksi OK = 1 3 It 12 (4)(6 )(4)
tmax =
•11–13.
Select the shortest and lightest-weight steel wideflange beam from Appendix B that will safely support the loading shown.The allowable bending stress is sallow = 22 ksi and the allowable shear stress is tallow = 12 ksi.
10 kip
6 kip
4 kip
A
B 4 ft
Beam design: Assume bending moment controls. Sreq¿d =
60.0(12) Mmax = = 32.73 in3 sallow 22
Select a W 12 * 26 Sx = 33.4 in3, d = 12.22 in., tw = 0.230 in. Check shear: tavg =
V 10.5 = = 3.74 ksi 6 12 ksi Aweb (12.22)(0.230)
Use W 12 * 26
Ans.
840
4 ft
4 ft
4 ft
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11–14. The beam is used in a railroad yard for loading and unloading cars. If the maximum anticipated hoist load is 12 kip, select the lightest-weight steel wide-flange section from Appendix B that will safely support the loading. The hoist travels along the bottom flange of the beam, 1 ft … x … 25 ft, and has negligible size. Assume the beam is pinned to the column at B and roller supported at A. The allowable bending stress is sallow = 24 ksi and the allowable shear stress is tallow = 12 ksi.
x
27 ft
A
B
12 kip
15 ft
C
Maximum moment occurs when load is in the center of beam. Mmax = (6 kip)(13.5 ft) = 81 lb # ft sallow =
M ; S
24 =
81(12) Sreq¿d Sreq¿d = 40.5 in3
Select a W 14 * 30, Sx = 42.0 in3, d = 13.84 in, tw = 0.270 in. At x = 1 ft, V = 11.56 kip t =
11.36 V = = 3.09 ksi 6 12 ksi Aweb (13.84)(0.270)
Use W14 * 30
Ans.
841
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11–15. The simply supported beam is made of timber that has an allowable bending stress of sallow = 960 psi and an allowable shear stress of tallow = 75 psi. Determine its dimensions if it is to be rectangular and have a heightto-width ratio of 1.25.
5 kip/ft
6 ft
1 I = (b)(1.25b)3 = 0.16276b4 12 Sreq¿d
b
Assume bending moment controls: Mmax = 60 kip # ft
960 =
Mmax Sreq¿d
60(103)(12) 0.26042 b3
b = 14.2 in. Check shear: tmax =
1.5(15)(103) 1.5V = = 88.9 psi 7 75 psi NO A (14.2)(1.25)(14.2)
Shear controls: tallow =
6 ft
1.25 b
I 0.16276b4 = = = 0.26042b3 c 0.625b
sallow =
B
A
1.5(15)(103) 1.5V = A (b)(1.25b)
b = 15.5 in.
Ans.
842
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*11–16. The simply supported beam is composed of two W12 * 22 sections built up as shown. Determine the maximum uniform loading w the beam will support if the allowable bending stress is sallow = 22 ksi and the allowable shear stress is tallow = 14 ksi.
w
Section properties: 24 ft
For W12 * 22 (d = 12.31 in. Ix = 156 in4 tw = 0.260 in. A = 6.48 in2) I = 2c 156 + 6.48a
S =
12.31 2 b d = 802.98 in4 2
I 802.98 = = 65.23 in3 c 12.31
Maximum Loading: Assume moment controls. M = sallowS(72 w)(12) = 22(65.23) w = 1.66 kip>ft Check Shear: tmax =
Ans. (Neglect area of flanges.)
12(1.66) Vmax = 3.11 ksi 6 tallow = 14 ksi OK = Aw 2(12.31)(0.26)
•11–17.
The simply supported beam is composed of two W12 * 22 sections built up as shown. Determine if the beam will safely support a loading of w = 2 kip>ft. The allowable bending stress is sallow = 22 ksi and the allowable shear stress is tallow = 14 ksi.
w
24 ft
Section properties: For W 12 * 22 (d = 12.31 in.
Ix = 156 in4
tw = 0.260 in.
A = 6.48 in2)
I = 2[156 + 6.48(6.1552)] = 802.98 in4 S =
802.98 I = = 65.23 in3 c 12.31
Bending stress: smax =
144 (12) Mallow = = 26.5 ksi 7 sallow = 22 ksi S 65.23
No, the beam falls due to bending stress criteria. Check shear: tmax =
Ans.
(Neglect area of flanges.)
Vmax 24 = = 3.75 ksi 6 tallow = 14 ksi OK Aw 2(12.31)(0.26)
843
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11–18. Determine the smallest diameter rod that will safely support the loading shown. The allowable bending stress is sallow = 167 MPa and the allowable shear stress is tallow = 97 MPa.
25 N/m 15 N/m
15 N/m
1.5 m
Bending Stress: From the moment diagram, Mmax = 24.375 N # m. Assume bending controls the design. Applying the flexure formula. sallow = 167 A 10
6
B =
Mmax c I 24.375 p 4
A d2 B
A d2 B 4
d = 0.01141 m = 11.4 mm
Ans.
Shear Stress: Provide a shear stress check using the shear formula with I =
p A 0.0057074 B = 0.8329 A 10 - 9 B m4 4
Qmax =
4(0.005707) 1 c (p) A 0.0057062 B d = 0.1239 A 10 - 6 B m3 3p 2
From the shear diagram, Vmax = 30.0 N. tmax =
=
Vmax Qmax It 30.0 C 0.1239(10 - 6) D
0.8329 (10 - 9)(0.01141)
= 0.391 MPa 6 tallow = 97 MPa (O.K!)
844
1.5 m
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11–19. The pipe has an outer diameter of 15 mm. Determine the smallest inner diameter so that it will safely support the loading shown. The allowable bending stress is sallow = 167 MPa and the allowable shear stress is tallow = 97 MPa.
25 N/m 15 N/m
15 N/m
1.5 m
Bending Stress: From the moment diagram, Mmax = 24.375 N # m. Q. Assume bending controls the design. Applying the flexure formula. sallow = 167 A 106 B =
Mmax c I 24.375(0.0075) p 4
C 0.00754 - A 2i B 4 D d
di = 0.01297 m = 13.0 mm
Ans.
Shear Stress: Provide a shear stress check using the shear formula with I =
p A 0.00754 - 0.0064864 B = 1.0947 A 10 - 9 B m4 4
Qmax =
4(0.0075) 1 4(0.006486) 1 c (p) A 0.00752 B d c (p) A 0.0064862 B d 3p 2 3p 2
= 99.306 A 10 - 9 B m3 From the shear diagram, Vmax = 30.0 N. Q tmax =
=
Vmax Qmax It 30.0 C 99.306(10 - 9) D
1.0947(10 - 9)(0.015 - 0.01297)
= 1.34 MPa 6 tallow = 97 MPa (O.K!)
845
1.5 m
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*11–20. Determine the maximum uniform loading w the W12 * 14 beam will support if the allowable bending stress is sallow = 22 ksi and the allowable shear stress is tallow = 12 ksi.
w 10 ft 10 ft
From the moment diagram, Fig. a, Mmax = 28.125 w. For W12 * 14, Sx = 14.9 in3, d = 11.91 in and tw = 0.200 in. sallow = 22 =
Mmax S 28.125 w (12) 14.9 Ans.
w = 0.9712 kip>ft = 971 lb>ft
From the shear diagram, Fig. a, Vmax = 7.5(0.9712) = 7.284 kip. Provide a shear stress check on W12 * 14, tmax =
=
Vmax tw d 7.284 0.200(11.91)
= 3.06 ksi 6 tallow = 12 ksi (O.K)
846
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•11–21.
Determine if the W14 * 22 beam will safely support a loading of w = 1.5 kip>ft. The allowable bending stress is sallow = 22 ksi and the allowable shear stress is tallow = 12 ksi.
w 10 ft 10 ft
For W14 * 22, Sx = 29.0 in3, d = 13.74 in and tw = 0.23 in. From the moment diagram, Fig. a, Mmax = 42.1875 kip # ft. smax =
=
Mmax S 42.1875(12) 29.0
= 17.46 ksi 6 sallow = 22 ksi (O.K!) From the shear diagram, Fig. a, Vmax = 11.25 kip. tmax =
=
Vmax tw d
11.25 0.23(13.74)
= 3.56 ksi 6 tallow = 12 ksi (O.K!) Based on the investigated results, we conclude that W14 * 22 can safely support the loading.
847
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11–22. Determine the minimum depth h of the beam to the nearest 18 in. that will safely support the loading shown. The allowable bending stress is sallow = 21 ksi and the allowable shear stress is tallow = 10 ksi. The beam has a uniform thickness of 3 in.
4 kip/ft
h
A B 12 ft
The section modulus of the rectangular cross-section is S =
I = C
1 12
(3)(h3) h>2
= 0.5 h2
From the moment diagram, Mmax = 72 kip # ft. Sreq¿d =
Mmax sallow
0.5h2 =
72(12) 21
h = 9.07 in Use
h = 9 18 in
Ans.
From the shear diagram, Fig. a, Vmax = 24 kip . Referring to Fig. b, 9.125 9.125 ba b (3) = 31.22 in3 and Qmax = y¿A¿ = a 4 2 1 I = (3) A 9.1253 B = 189.95 in4 . Provide the shear stress check by applying 12 shear formula, tmax =
=
Vmax Qmax It 24(31.22) 189.95(3)
= 1.315 ksi 6 tallow = 10 ksi (O.K!)
848
6 ft
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11–23. The box beam has an allowable bending stress of sallow = 10 MPa and an allowable shear stress of tallow = 775 kPa. Determine the maximum intensity w of the distributed loading that it can safely support. Also, determine the maximum safe nail spacing for each third of the length of the beam. Each nail can resist a shear force of 200 N.
w 30 mm 250 mm 30 mm 150 mm 30 mm
Section Properties: I =
1 1 (0.21) A 0.253 B (0.15) A 0.193 B = 0.1877 A 10 - 3 B m4 12 12
QA = y1 ¿A¿ = 0.11(0.03)(0.15) = 0.495 A 10 - 3 B m3 Qmax = ©y¿A¿ = 0.11(0.03)(0.15) + 0.0625(0.125)(0.06) = 0.96375 A 10 - 3 B m3 Bending Stress: From the moment diagram, Mmax = 4.50w. Assume bending controls the design. Applying the flexure formula. sallow = 10 A 106 B =
Mmax c I 4.50w (0.125) 0.1877 (10 - 3)
w = 3336.9 N>m Shear Stress: Provide a shear stress check using the shear formula. From the shear diagram, Vmax = 3.00w = 10.01 kN. tmax =
=
Vmax Qmax It 10.01(103) C 0.96375(10 - 3) D 0.1877(10 - 3)(0.06)
= 857 kPa 7 tallow = 775 kPa (No Good!) Hence, shear stress controls. tallow = 775 A 103 B =
Vmax Qmax It 3.00w C 0.96375(10 - 3) D 0.1877(10 - 3)(0.06)
w = 3018.8 N>m = 3.02 kN>m
Ans.
Shear Flow: Since there are two rows of nails, the allowable shear flow is 2(200) 400 = q = . s s
849
6m
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11–23.
Continued
For 0 … x 6 2 m and 4 m 6 x … 6 m, the design shear force is V = 3.00w = 9056.3 N. q =
VQA I
9056.3 C 0.495(10 - 3) D 400 = s 0.1877(10 - 3) s = 0.01675 m = 16.7 mm
Ans.
For 2 m 6 x 6 4 m, the design shear force is V = w = 3018.8 N. q =
VQA I
3018.8 C 0.495(10 - 3) D 400 = s 0.1877(10 - 3) s = 0.05024 m = 50.2 mm
Ans.
850
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*11–24. The simply supported joist is used in the construction of a floor for a building. In order to keep the floor low with respect to the sill beams C and D, the ends of the joists are notched as shown. If the allowable shear stress for the wood is tallow = 350 psi and the allowable bending stress is sallow = 1500 psi, determine the height h that will cause the beam to reach both allowable stresses at the same time. Also, what load P causes this to happen? Neglect the stress concentration at the notch.
P
2 in.
15 ft B
h
15 ft D
A 10 in. C
Bending Stress: From the moment diagram, Mmax = 7.50P. Applying the flexure formula. Mmax c I
salllow =
7.50P(12)(5)
1500 =
1 12
(2)(103)
P = 555.56 lb = 556 lb
Ans.
Shear Stress: From the shear diagram, Vmax = 0.500P = 277.78 lb. The notch is the critical section. Using the shear formula for a rectangular section. tallow = 350 =
3Vmax 2A 3(277.78) 2(2) h
h = 0.595 in.
Ans.
11–25. The simply supported joist is used in the construction of a floor for a building. In order to keep the floor low with respect to the sill beams C and D, the ends of the joists are notched as shown. If the allowable shear stress for the wood is tallow = 350 psi and the allowable bending stress is sallow = 1700 psi, determine the smallest height h so that the beam will support a load of P = 600 lb. Also, will the entire joist safely support the load? Neglect the stress concentration at the notch.
P
B
tallow =
1.5V ; A
350 =
D
A 10 in.
600 = 300 lb 2
1.5(300) (2)(h)
h = 0.643 in. smax =
Ans.
4500(12)(5) Mmax c = 1620 psi 6 1700 psi OK = 1 3 I 12 (2)(10)
Yes, the joist will safely support the load.
Ans.
851
h
15 ft
C
The reaction at the support is
2 in.
15 ft
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11–26. Select the lightest-weight steel wide-flange beam from Appendix B that will safely support the loading shown. The allowable bending stress is sallow = 22 ksi and the allowable shear stress is tallow = 12 ksi.
5 kip 18 kip ft B A 6 ft
From the moment diagram, Fig. a, Mmax = 48 kip # ft. Sreq¿d =
=
Mmax sallow 48(12) 22
= 26.18 in3 Select W 14 * 22 C Sx = 29.0 in3, d = 13.74 in. and tw = 0.230 in. D From the shear diagram, Fig. a, Vmax = 5 kip. Provide the shear stress check for W 14 * 22, tmax =
=
Vmax twd
5 0.230(13.74)
= 1.58 ksi 6 tallow = 12 ksi‚ (O.K!) Use
Ans.
W14 * 22
W12 * 22 would work also.
852
12 ft
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11–27. The T-beam is made from two plates welded together as shown. Determine the maximum uniform distributed load w that can be safely supported on the beam if the allowable bending stress is sallow = 150 MPa and the allowable shear stress is tallow = 70 MPa.
w
A 1.5 m
1.5 m 200 mm 20 mm 200 mm 20 mm
The neutral axis passes through centroid c of the beam’s cross-section. The location of c, Fig. b, is y =
0.21(0.02)(0.2) + 0.1(0.2)(0.02) ©yA = ©A 0.02(0.2) + 0.2(0.02) = 0.155 m
I =
1 (0.2)(0.023) + 0.2(0.02)(0.055)2 12
+
1 (0.02)(0.23) + 0.02(0.2)(0.055)2 12
= 37.667 (10 - 6) m4 Referring to Fig. b, Qmax = y¿A¿ = 0.0775(0.155)(0.02) = 0.24025(10 - 3) m3 Referring to the moment diagram, Mmax = -3.375 w. Applying the flexure formula with C = y = 0.155 m, sallow =
Mmax c ; I
150(106) =
3.375 w (0.155) 37.667(10 - 6)
w = 10.80(103) N>m = 10.8 kN>m (Control!)
Ans.
Referring to the shear diagram, Vmax = 1.5w. tallow =
Vmax Qmax ; It
70(106) =
1.5 w C 0.24025(10 - 3) D 37.667(10 - 6)(0.02)
w = 146.33(103) N>m = 146 kN>m
853
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*11–28. The beam is made of a ceramic material having an allowable bending stress of sallow = 735 psi and an allowable shear stress of tallow = 400 psi. Determine the width b of the beam if the height h = 2b.
15 lb 10 lb
6 lb/in.
2 in.
6 in.
2 in.
h b
Bending Stress: From the moment diagram, Mmax = 30.0 lb # in. Assume bending controls the design. Applying the flexure formula. sallow =
Mmax c I 30.0
735 =
1 12
A 2b2 B
(b) (2b)3
b = 0.3941 in. = 0.394 in.
Ans.
Shear Stress: Provide a shear stress check using the shear formula for a rectangular section. From the shear diagram, Vmax = 19.67 lb. tmax =
=
3Vmax 2A 3(19.67) 2(0.3941)(2)(0.3941)
= 94.95 psi 6 tallow = 400 psi (O.K!)
854
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•11–29.
The wood beam has a rectangular cross section. Determine its height h so that it simultaneously reaches its allowable bending stress of sallow = 1.50 ksi and an allowable shear stress of tallow = 150 psi. Also, what is the maximum load P that the beam can then support?
P
P
B
A 1.5 ft
3 ft
1.5 ft h
6 in.
The section modulus of the rectangular cross-section about the neutral axis is S =
I = C
1 12
(6) h3 h>2
= h2
From the moment diagram, Fig. a, Mmax = 1.5P. Mmax = sallow S 1.5P(12) = 1.50(103) h2 P = 83.33h2
(1)
h h 1 a b (6) = 0.75 h2 and I = (6) h3 = 0.5h3. 4 2 12 From the shear diagram, Fig. a, Vmax = P. Referring to Fig. b, Qmax = y¿A¿ =
tmax =
150 =
Vmax Qmax It P (0.75 h2) 0.5 h3 (6)
P = 600 h
(2)
Solving Eq (1) and (2) h = 7.20 in
P = 4320 lb = 4.32 kip
Ans.
855
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11–30. The beam is constructed from three boards as shown. If each nail can support a shear force of 300 lb, determine the maximum allowable spacing of the nails, s, s¿, s– , for regions AB, BC, and CD respectively. Also, if the allowable bending stress is sallow = 1.5 ksi and the allowable shear stress is tallow = 150 psi, determine if it can safely support the load.
1500 lb
500 lb s¿
s
A
s¿¿
C
B 6 ft
6 ft
6 ft
10 in. 4 in. 10 in.
The neutral axis passes through centroid c of the beam’s cross-section. The location of c, Fig. b, is y =
12(4)(10) + 2 C 5(10)(2) D ©yA = ©A 4(10) + 2(10)(2) = 8.50 in
The moment of inertia of the beam’s cross-section about the neutral axis is I = 2c +
1 (2)(103) + 2(10)(3.50)2 d 12 1 (10)(43) + 10(4)(3.50)2 12
= 1366.67 in4 Referring to Fig. b, Qmax = 2y2œ A2œ = 2 C 4.25(8.50)(2) D = 144.5 in3 QA = y1œ A1œ = 3.50(4)(10) = 140 in3 Referring to the moment diagram, Fig. a, Mmax = 9000 lb # ft. Applying flexure formula with C = y = 8.50 in, smax =
=
Mmax c I
9000(12)(8.50) 1366.67
= 671.70 psi 6 sallow = 1.50 ksi (O.K!) Referring to shear diagram, Fig. a, Vmax = 1500 lb. tmax =
=
Vmax Qmax It 1500 (144.5) = 39.65 psi 6 tallow = 150 psi (O.K!) 1366.67 (4)
1 S– = 11 in. Yes, it can support the load. 2
Ans.
856
2 in.
D
2 in.
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11–30.
Continued
Since there are two rows of nails, the allowable shear flow is 2(300) 600 2F qallow = = = . For region AB, V = 1500 lb. Thus S S S qallow =
VQA ; I
1500 (140) 600 = S 1366.67 Use
S = 3.904 in
S = 3 34 in
Ans.
For region BC, V = 1000 lb. Thus qallow =
VQA ; I
1000(140) 600 = S¿ 1366.67 Use
S¿ = 5.85 in
S¿ = 5 34 in
Ans.
For region CD, V = 500 lb. Thus qallow =
VQA ; I
500 (140) 600 = S– 1366.67 Use
S– = 11.71 in
S– = 1112 in
Ans.
857
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11–31. The tapered beam supports a concentrated force P at its center. If it is made from a plate that has a constant width b, determine the absolute maximum bending stress in the beam.
2h0
h0 L 2
L 2 P
Section Properties: h - h0 h0 = L x 2 I =
S =
h =
h0 (2x + L) L
h30 1 (b) a 3 b(2x + L)3 12 L 1 12
(b) A h3 2L
h30 3
L
B (2x + L)3
bh20 =
(2x + L)
6L2
(2x + L)2
Bending Stress: Applying the flexure formula. s =
M = S
Px 2 bh20 2
6L
= (2x + L)2
bh20
3PL2x (2x + L)2
In order to have the absolute maximum bending stress,
[1]
ds = 0. dx
3PL2 (2x + L)2(1) - x(2)(2x + L)(2) ds = c d = 0 dx bh20 (2x + L)4 x =
Substituting x =
L 2
L into Eq. [1] yields 2 smax =
3PL 8bh20
Ans.
858
h0
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*11–32. The beam is made from a plate that has a constant thickness b. If it is simply supported and carries a uniform load w, determine the variation of its depth as a function of x so that it maintains a constant maximum bending stress sallow throughout its length.
w
h0
y
x L –– 2
Moment Function: As shown on FBD(b). Section Properties: I =
1 3 by 12
S =
I = c
1 3 12 by y 2
=
1 2 by 6
Bending Stress: Applying the flexure formula. M = S
sallow =
w 2 2 8 (L - 4x ) 1 2 6 by
3w (L2 - 4x2)
sallow =
[1]
4by2
At x = 0, y = h0. From Eq. [1], sallow =
3wL2 4bh20
[2]
Equating Eq. [1] and [2] yields y2 =
h20 L2
y2 h20
+
A L2 - 4x2 B 4x2 = 1 L2
Ans.
The beam has a semi-elliptical shape.
859
L –– 2
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•11–33.
The beam is made from a plate having a constant thickness t and a width that varies as shown. If it supports a concentrated force P at its center, determine the absolute maximum bending stress in the beam and specify its location x, 0 6 x 6 L>2.
P P — 2
b0
L — 2
x
b L — 2
t P — 2
Section properties: b x = L; b0 2
2b0 x L
b =
I =
b0 t3 1 2b0 a xb t3 = x 12 L 6L
S =
I = c
b0 t
6L x t 2
=
b0 t2 x 3L
Bending stress: s =
M = S
P 2x b 0 t2
3L x
=
3PL 2b0t2
Ans.
The bending stress is independent of x. Therefore, the stress is constant throughout the span. Ans.
860
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11–34. The beam is made from a plate that has a constant thickness b. If it is simply supported and carries the distributed loading shown, determine the variation of its depth as a function of x so that it maintains a constant maximum bending stress sallow throughout its length.
w0
A
L –– 2
Moment Function: The distributed load as a function of x is w0 w = x L>2
w =
2w0 x L
The free-body diagram of the beam’s left cut segment is shown in Fig. b. Considering the moment equilibrium of this free-body diagram, d+ ©MO = 0;
M +
1 2w0 x 1 x R x ¢ ≤ - w0Lx = 0 B 2 L 3 4
M =
w0 A 3L2x - 4x3 B 12L
Section Properties: At position x, the height of the beam’s cross section is h. Thus 1 bh3 12
I = Then
1 bh3 I 12 1 S = = = bh2 c h>2 6 Bending Stress: The maximum bending stress smax as a function of x can be obtained by applying the flexure formula.
smax
At x =
w0 A 3L2x - 4x3 B w0 M 12L = = = A 3L2x - 4x3 B ‚ S 1 2 2bh2L bh 6
(1)
L , h = h0. From Eq. (1), 2 smax =
w0L2
(2)
2bh0 2
Equating Eqs. (1) and (2), w0 2
2bh L h =
A 3L2x - 4x3 B =
h0 L3>2
w0L2 2bh0 2
A 3L2x - 4x3 B 1>2
Ans.
861
h0 B
x
Support Reactions: As shown on the free-body diagram of the entire beam, Fig. a.
C
h
L –– 2
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11–35. The beam is made from a plate that has a constant thickness b. If it is simply supported and carries the distributed loading shown, determine the maximum bending stress in the beam.
w0
h0
h0 2h0 L – 2
Support Reactions: As shown on the free - body diagram of the entire beam, Fig. a. Moment Function: The distributed load as a function of x is w0 w = ; x L>2
w =
2w0 x L
The free - body diagram of the beam’s left cut segment is shown in Fig. b. Considering the moment equilibrium of this free - body diagram, d+ ©MO = 0;
M +
w0L x 1 2w0 a xbxa b x = 0 2 L 3 4
M =
w0 A 3L2x - 4x3 B 12L
Section Properties: Referring to the geometry shown in Fig. c, h - h0 h0 = ; x L>2
h =
h0 (2x + L) L
At position x, the height of the beam’s cross section is h. Thus I =
1 bh3 12
Then 1 bh3 bh0 2 12 I 1 = S = = bh2 = (2x + L)2 c h>2 6 6L2 Bending Stress: Applying the flexure formula,
smax
w0 A 3L2x - 4x3 B M 12L = = S bh0 2 (2x + L)2 6L2
smax =
w0L 2bh0 2
B
3L2x - 4x3 R (2x + L)2
In order to have absolute maximum bending stress,
(1) dsmax = 0. dx
2 2 2 2 3 dsmax w0L (2x + L) A 3L - 12x B - A 3L x - 4x B (2)(2x + L)(2) = C S = 0 dx 2bh0 2 (2x + L)4
w0L 2bh0
2
B
3L3 - 8x3 - 6L2x - 12Lx2 R = 0 (2x + L)3
862
L – 2
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11–35.
Since
Continued
w0L 2bh0 2
Z 0, then
3L3 - 8x3 - 6L2x - 12Lx2 = 0 Solving by trial and error, x = 0.2937L = 0.294L Substituting this result into Eq. (1), sabs = max
0.155w0L2
Ans.
bh0 2
863
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*11–36. Determine the variation of the radius r of the cantilevered beam that supports the uniform distributed load so that it has a constant maximum bending stress smax throughout its length.
w r0 r
Moment Function: As shown on FBD. Section Properties: I =
p 4 r 4
I = c
S =
p 4
r r
x
L
4
=
p 3 r 4
Bending Stress: Applying the flexure formula. smax =
wx2 2 p 3 4r
M = S
smax =
2wx2 pr3
[1]
At x = L, r = r0. From Eq. [1], smax =
2wL2 pr30
[2]
Equating Eq. [1] and [2] yields r3 =
r30 L2
x2
Ans.
•11–37.
Determine the variation in the depth d of a cantilevered beam that supports a concentrated force P at its end so that it has a constant maximum bending stress sallow throughout its length. The beam has a constant width b0 .
P d0
d
L
Section properties: I =
1 (b )(d3) 12 0
sallow =
S =
I = c
1 12
(b0)(d3) d>2
=
b0d2 6
M Px = S b0d2>6
(1)
PL b0d0 2>6
(2)
At x = L sallow =
Equate Eqs. (1) and (2): PL Px = b0d2>6 b0 d0 2>6 d2 = a
d0 2 bx ; L
x d = d0 AL
Ans.
864
x
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11–38. Determine the variation in the width b as a function of x for the cantilevered beam that supports a uniform distributed load along its centerline so that it has the same maximum bending stress sallow throughout its length. The beam has a constant depth t.
b —0 2 b —0 2 b — 2 w L
x t
Section properties: I =
1 b t3 12
S =
I = c
1 12
b t3 t 2
=
t2 b 6
Bending stress: sallow
M = = S
w x2 2 2
t 6b
=
3wx2 t2b
(1)
At x = L, b = b0 sallow =
3wL2 t2b0
(2)
Equating Eqs. (1) and (2) yields: 3wL2 3wx2 = 2 2 t b t b0 b =
b0 L2
x2
Ans.
865
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z
11–39. The shaft is supported on journal bearings that do not offer resistance to axial load. If the allowable normal stress for the shaft is sallow = 80 MPa, determine to the nearest millimeter the smallest diameter of the shaft that will support the loading. Use the maximum-distortionenergy theory of failure.
A
150 mm D
30 250 mm 50 N
x
30
C 30 150 N 100 mm 500 mm
100 N 30
250 N 250 mm
Torque and Moment Diagrams: As shown. In-Plane Principal Stresses: Applying Eq. 9–5 with sy = 0, sx = txy =
Mc 4M , and = I pc3
2T Tc . = J pc3 s1, 2 =
sx + sy ;
2
Aa
sx - sy 2
2
2 b + txy
=
2M 2M 2 2T 2 ; A a pc3 b + a pc3 b pc3
=
2M 2 ; 2M2 + T2 pc3 pc3
2 2M 2M2 + T2, then and b = pc3 pc3 s21 = a2 + b2 + 2ab, s1s2 = a2 - b2, s22 = a2 + b2 - 2ab, and s21 - s1 s2 + s22 = 3b2 + a2.
Maximum Distortion Energy Theory: Let a =
s21 - s1 s2 + s22 = s2allow 3a
2 2 2M 2 2 2 2M + T b + a b = s2allow pc3 pc3 6 4 A 4M2 + 3T2 B R p2s2allow 1
c = B
Shaft Design: By observation, the critical section is located just to the left of gear C, where M = 239.06252 + 46.012 = 60.354 N # m and T = 15.0 N # m. Using the maximum distortion energy theory, 6 4 A 4M2 + 3T2 B R p2s2allow 1
c = B
= b
4 p2 [80(106)]2
C 4(60.354)2 + 3(15.0)2 D r
1 6
= 0.009942 m d = 2c = 2(0.009942) = 0.01988 m = 19.88 mm Use
d = 20 mm
Ans.
866
B y
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z
*11–40. The shaft is supported on journal bearings that do not offer resistance to axial load. If the allowable shear stress for the shaft is tallow = 35 MPa, determine to the nearest millimeter the smallest diameter of the shaft that will support the loading. Use the maximum-shear-stress theory of failure.
A
x
150 mm D
30 250 mm 50 N
30
C 30 150 N 100 mm 500 mm
100 N 30
250 N 250 mm
Shaft Design: By observation, the critical section is located just to the left of gear C, where M = 239.06252 + 46.012 = 60.354 N # m and T = 15.0 N # m. Using the maximum shear stress theory. c = a = B
1
3 2 2M2 + T2 b ptallow
2 p(35)(106)
260.3542 + 15.02 R
1 3
= 0.01042 m d = 2c = 2(0.01042) = 0.02084 m = 20.84 mm Use
d = 21 mm
Ans.
867
B y
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z
•11–41.
The end gear connected to the shaft is subjected to the loading shown. If the bearings at A and B exert only y and z components of force on the shaft, determine the equilibrium torque T at gear C and then determine the smallest diameter of the shaft to the nearest millimeter that will support the loading. Use the maximum-shear-stress theory of failure with tallow = 60 MPa.
100 mm T 250 mm C 50 mm 150 mm
A
x
100 mm Fz 1.5 kN
From the free - body diagrams: T = 100 N # m
Ans.
Critical section is at support A. 1
1
3 3 2 2 c = c 22252 + 1502 d 2M2 + T2 d = c p tallow p(60)(106)
= 0.01421 m d = 2c = 0.0284 m = 28.4 mm Use d = 29 mm
Ans.
868
B
75 mm y
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z
11–42. The end gear connected to the shaft is subjected to the loading shown. If the bearings at A and B exert only y and z components of force on the shaft, determine the equilibrium torque T at gear C and then determine the smallest diameter of the shaft to the nearest millimeter that will support the loading. Use the maximum-distortionenergy theory of failure with sallow = 80 MPa.
100 mm T 250 mm C 50 mm 150 mm
A
x
T = 100 N # m
Ans.
Critical section is at support A. s1, 2 =
sx s2x 2 ; 2 A 4 + txy sx s2x 2 ,b = A 4 + txy 2
Let a =
s1 = a + b, s2 = a - b Require, s21 - s1 s2 + s22 = s2allowa2 + 2ab + b2 - [a2 - b2] + a2 - 2ab + b2 = s2allow a2 + 3b2 = s2allow s2x s2x + 3a + t2xy b = s2allow 4 4 s2x + 3t2xy = s2allow Mt 2 Tc 2 a p 4 b + 3a p 4 b = s2allow 4 c 2 c 4M 2 2T 2 1 ca b + 3a b d = s2allow 4 p p c c4 =
16 s2allow
c = a = c
2
p
M2 +
4 s2allow p2
12T2 p2
s2allow
(4M + 3T ) b 2
4 (80(106))2(p)2
2
100 mm Fz 1.5 kN
From the free-body diagrams:
1 2
(4(225) + 3(150) ) d 2
2
1 2
= 0.01605 m d = 2c = 0.0321 m = 32.1 mm Use d = 33 mm
Ans.
869
B
75 mm y
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11–43. The shaft is supported by bearings at A and B that exert force components only in the x and z directions on the shaft. If the allowable normal stress for the shaft is sallow = 15 ksi, determine to the nearest 18 in. the smallest diameter of the shaft that will support the loading. Use the maximum-distortion-energy theory of failure.
z C F¿x 100 lb
6 in.
A x
8 in. 12 in.
Critical moment is just to the right of D.
T = 1200 lb # in. Both states of stress will yield the same result.
Let
s s 2 2 ; a 2 A 2b + t
2 s = A and s + t2 = B 2 A4
s2a = (A + B)2, s2b = (A - B)2 sa sb = (A + B)(A - B) = A2 - B2 s2a - sa sb + s2b = A2 + B2 + 2AB - A2 + B2 + A2 + B2 - 2AB = A2 + 3B2 =
s2 s2 + 3a + t2 b = s2 + 3t2 4 4
s2a - sa sb + s2b = s2allow s2 + 3t2 = s2allow‚ s =
Mc Mc 4M = p 4 = I c pc3 4
t =
Tc Tc 2T = p 4 = J c p c3 2
(1)
From Eq. (1) 16M2 2
6
p c
c = a
12T2 +
p2 c6
= s2allow
16(2396)2 + 12(12002) 1>6 16M2 + 12T2 1>6 b = c d = 0.605 in. p2s2allow p2((15)(103))2
d = 2c = 1.210 in. Use d = 1
2 in. Fz 300 lb 10 in.
6 in.
M = 220572 + 12292 = 2396 lb # in.
sa, b =
F y 300 lb
D
1 in. 4
Ans.
870
4 in.
E
B y
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*11–44. The shaft is supported by bearings at A and B that exert force components only in the x and z directions on the shaft. If the allowable normal stress for the shaft is sallow = 15 ksi, determine to the nearest 18 in. the smallest diameter of the shaft that will support the loading. Use the maximum-shear-stress theory of failure. Take tallow = 6 ksi.
z C F¿x 100 lb
6 in.
A x
8 in. 12 in.
Critical moment is just to the right of D.
T = 1200 lb # in. Use Eq. 11-2, 1>3 2 2M2 + T2 b p tallow
c = a
1>3 2 2(2396)2 + (1200)2 b = 0.6576 in. 3 p(6)(10 )
2 in. Fz 300 lb 10 in.
6 in.
M = 2(2057)2 + (1229)2 = 2396 lb # in.
c = a
F y 300 lb
D
dreq¿d = 2c = 1.315 in. 3 Use d = 1 in. 8
Ans.
871
4 in.
E
B y
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z
•11–45. The bearings at A and D exert only y and z components of force on the shaft. If tallow = 60 MPa, determine to the nearest millimeter the smallest-diameter shaft that will support the loading. Use the maximum-shearstress theory of failure.
350 mm D 400 mm 200 mm B A
Critical moment is at point B: M = 2(473.7)2 + (147.4)2 = 496.1 N # m
x
T = 150 N # m
c = a
1>3 1>3 2 2 2 2 2496.1 2M2 + T2 b = a + 150 b = 0.0176 m p tallow p(60)(106)
c = 0.0176 m = 17.6 mm d = 2c = 35.3 mm Use d = 36 mm
Ans.
872
y
C 75 mm Fy ⫽ 3 kN 50 mm
Fz ⫽ 2 kN
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z
11–46. The bearings at A and D exert only y and z components of force on the shaft. If tallow = 60 MPa, determine to the nearest millimeter the smallest-diameter shaft that will support the loading. Use the maximumdistortion-energy theory of failure. sallow = 130 MPa.
350 mm D 400 mm 200 mm B A
The critical moment is at B. M = 2(473.7)2 + (147.4)2 = 496.1 N # m
x
T = 150 N # m Since, sa, b =
Let
s s 2 2 ; a 2 A 2b + t
s = A 2
and
s 2 2 = B a A 2b + t
s2a = (A + B)2
s2b = (A - B)2
sa sb = (A + B)(A - B) s2a - sa sb + s2b = A2 + B2 + 2AB - A2 + B2 + A2 + B2 - 2AB = A2 + 3B2 =
s2 s2 + 3a + t2 b 4 4
= s2 + 3t2 s2a - sasb + s2b = s2allow s2 + 3t2 = s2allow
(1)
s =
Mc Mc 4M = p 4 = I c pc3 4
t =
Tc Tc 2T = p 4 = J pc3 2 c
From Eq (1) 12T2 16M2 + 2 4 = s2allow 2 4 pc pc c = a = a
16M2 + 12T2 1>6 b p2s2allow 16(496.1)2 + 12(150)2 2
4
2
p ((130)(10 ))
b
1>4
= 0.01712 m
d = 2c = 34.3 mm
Ans.
873
y
C 75 mm Fy ⫽ 3 kN 50 mm
Fz ⫽ 2 kN
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11–47. Draw the shear and moment diagrams for the shaft, and then determine its required diameter to the nearest millimeter if sallow = 140 MPa and tallow = 80 MPa. The bearings at A and B exert only vertical reactions on the shaft.
1500 N 800 N A
B
600 mm 125 mm
Bending Stress: From the moment diagram, Mmax = 111 N # m. Assume bending controls the design. Applying the flexure formula. sallow = 140 A 106 B =
Mmax c I 111 A d2 B p 4
A d2 B 4
d = 0.02008 m = 20.1 mm d = 21 mm
Use
Ans.
Shear Stress: Provide a shear stress check using the shear formula with I =
p A 0.01054 B = 9.5466 A 10 - 9 B m4 4
Qmax =
4(0.0105) 1 c (p)(0.0105)2 d = 0.77175 A 10 - 6 B m3 3p 2
From the shear diagram, Vmax = 1484 N. tmax =
=
Vmax Qmax It
1484 C 0.77175(10 - 6) D 9.5466(10 - 9)(0.021)
= 5.71 MPa 6 tallow = 80 MPa (O.K!)
874
75 mm
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*11–48. The overhang beam is constructed using two 2-in. by 4-in. pieces of wood braced as shown. If the allowable bending stress is sallow = 600 psi, determine the largest load P that can be applied. Also, determine the associated maximum spacing of nails, s, along the beam section AC if each nail can resist a shear force of 800 lb. Assume the beam is pin-connected at A, B, and D. Neglect the axial force developed in the beam along DA.
D
2 ft 3 ft A
2 ft
Section properties: I =
1 (4)(4)3 = 21.33 in4 12
S =
21.33 I = = 10.67 in3 c 2
Mmax = sallow S 3P(12) = 600(10.67) P = 177.78 = 178 lb
Ans.
Nail Spacing: V = P = 177.78 lb Q = (4)(2)(1) = 8 in3 q =
177.78(8) VQ = = 66.67 lb>in. I 21.33
S =
800 lb = 12.0 in. 66.67 lb>in.
Ans.
875
2 in. 2 in.
s B
MA = Mmax = 3P
P
C
4 in.
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z
•11–49. The bearings at A and B exert only x and z components of force on the steel shaft. Determine the shaft’s diameter to the nearest millimeter so that it can resist the loadings of the gears without exceeding an allowable shear stress of tallow = 80 MPa. Use the maximum-shear-stress theory of failure.
Fx 5 kN A 75 mm
x
50 mm
150 mm 350 mm
B Fz 7.5 kN 250 mm
Maximum resultant moment M = 212502 + 2502 = 1274.75 N # m 1
1
3 3 2 2 21274.752 + 3752 d = 0.0219 m 2M2 + T2 d = c c = c 6 p tallow p(80)(10 )
d = 2c = 0.0439 m = 43.9 mm Use d = 44 mm
Ans.
876
y
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z
11–50. The bearings at A and B exert only x and z components of force on the steel shaft. Determine the shaft’s diameter to the nearest millimeter so that it can resist the loadings of the gears without exceeding an allowable shear stress of tallow = 80 MPa. Use the maximumdistortion-energy theory of failure with sallow = 200 MPa.
Fx 5 kN A 75 mm
x
50 mm
150 mm 350 mm
Maximum resultant moment M = 212502 + 2502 = 1274.75 N # m s1, 2 =
sx s2x 2 ; 2 A 4 + txy sx s2x 2 ,b = 2 A 4 + txy
Let a =
s1 = a + b,
s2 = a - b
Require, s21 - s1 s2 + s22 = s2allow a2 + 2ab + b2 - [a2 - b2] + a2 - 2ab + b2 = s2allow a2 + 3b2 = s2allow s2x s2x + 3a + t2xy b = s2allow 4 4 s2x + 3t2xy = s2allow Mc 2 Tc 2 a p 4 b + 3a p 4 b = s2allow 4 c 2 c 1 6
c
Ba
c6 =
4M 2 2T 2 b + 3a b R = s2allow p p 16
s2allow p2
c = B
= B
M2 +
4 s2allow p2
12T2 s2allow p2
(4M2 + 3 T2) R
4
1 4
(4(1274.75)2 + 3(375)2) R (200(106))2(p)2
1 4
= 0.0203 m = 20.3 mm d = 40.6 mm
Ans.
Use d = 41 mm
Ans.
877
B Fz 7.5 kN 250 mm
y
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11–51. Draw the shear and moment diagrams for the beam. Then select the lightest-weight steel wide-flange beam from Appendix B that will safely support the loading. Take sallow = 22 ksi, and tallow = 12 ksi.
3 kip/ft 1.5 kip ft A B 12 ft
Bending Stress: From the moment diagram, Mmax = 18.0 kip # ft. Assume bending controls the design. Applying the flexure formula. Sreq¿d =
= Select
Mmax sallow 18.0(12) = 9.82 in3 22
A Sx = 10.9 in3, d = 9.87 in., tw = 0.19 in. B
W10 * 12
V for the W10 * 12 wide twd = 9.00 kip
Shear Stress: Provide a shear stress check using t = flange section. From the shear diagram, Vmax tmax =
=
Vmax tw d 9.00 0.19(9.87)
= 4.80 ksi 6 tallow = 12 ksi (O.K!) Hence,
Use
W10 * 12
Ans.
878
6 ft
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*11–52. The beam is made of cypress having an allowable bending stress of sallow = 850 psi and an allowable shear stress of tallow = 80 psi. Determine the width b of the beam if the height h = 1.5b.
300 lb
75 lb/ft
B
A 5 ft
5 ft
h 1.5b b
Ix =
1 (b)(1.5b)3 = 0.28125 b4 12
Qmax = y¿A¿ = (0.375b) (0.75b)(b) = 0.28125 b3 Assume bending controls. Mmax = 527.34 lb # ft sallow =
Mmax c ; I
850 =
527.34(12)(0.75 b) 0.28125 b4
b = 2.71 in.
Ans.
Check shear: I = 15.12 in4 tmax =
Qmax = 5.584 in3
VQmax 281.25(5.584) = It 15.12(2.71)
= 38.36 psi 6 80 psi
OK
879
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•11–53.
The tapered beam supports a uniform distributed load w. If it is made from a plate and has a constant width b, determine the absolute maximum bending stress in the beam.
w h0 L –– 2
Support Reactions: As shown on FBD(a). Moment Function: As shown on FBD(b). Section Properties: h - h0 h0 = L x 2 I =
S =
h =
h0 (2x + L) L
h30 1 (b) a 3 b(2x + L)3 12 L 1 12
(b) A h0 2L
B (2x + L)3
h30 3
L
bh20 =
(2x + L)
6L2
(2x + L)2
Bending Stress: Applying the flexure formula. s =
M = S
w 2
(Lx - x2)
bh20 2
6L
3wL2 (Lx - x2) =
(2x + L)2
[1]
bh20 (2x + L)2
In order to have the absolute maximum bending stress,
ds = 0. dx
3wL2 (2x + L)2(L - 2x) - (Lx - x2)(2)(2x + L)(2) ds = c d = 0 dx bh20 (2x + L)4 x =
Substituting x =
L 4
L into Eq. [1] yields 4 smax =
wL2 4bh20
Ans.
880
h0
2 h0 L –– 2
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11–54. The tubular shaft has an inner diameter of 15 mm. Determine to the nearest millimeter its outer diameter if it is subjected to the gear loading. The bearings at A and B exert force components only in the y and z directions on the shaft. Use an allowable shear stress of tallow = 70 MPa, and base the design on the maximum-shear-stress theory of failure.
z
100 mm B 500 N 150 mm
A 200 mm 150 mm
x
I =
p 4 p (c - 0.00754) and J = (c4 - 0.00754) 4 2
tallow =
Aa
sx - sy
tallow =
Aa
Mc 2 Tc 2 b + a b 2I J
t2allow =
M2 c2 T2 c2 + 2 4I J2
¢
2
100 mm
2
b + t2xy
c4 - 0.00754 2 4M2 4T2 ≤ = 2 + 2 c p p
c4 - 0.00754 2 = 2M2 + T2 c p tallow c4 - 0.00754 2 2752 + 502 = c p(70)(106) c4 - 0.00754 = 0.8198(10 - 6)c Solving, c = 0.0103976 m d = 2c = 0.0207952 m = 20.8 mm Use d = 21 mm
Ans.
881
500 N
y
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11–55. Determine to the nearest millimeter the diameter of the solid shaft if it is subjected to the gear loading. The bearings at A and B exert force components only in the y and z directions on the shaft. Base the design on the maximum-distortion-energy theory of failure with sallow = 150 MPa.
z
100 mm B 500 N 150 mm
A 200 mm 150 mm
x
s1, 2 =
100 mm
sx 2 ; 2 A 4 + txy s2x
sx s2x 2 ,b = A 4 + txy 2
Let a =
s1 = a + b, s2 = a - b Require, s21 - s1 s2 + s21 = s2allow a2 + 2ab + b2 - [a2 - b2] + a2 - 2ab + b2 = sallow a2 + 3b2 = s2allow s2x s2x + 3a + t2xy b = s2allow 4 4 s2x + 3t2xy = s2allow Mc 2 Tc 2 a p 4 b + 3a p 4 b = s2allow 4 c 2 c 1 6
c
ca
c6 =
4M 2 2T 2 b + 3a b d = s2allow p p 16 s2allow p2
c = a = c
M2 +
4 s2allow p2
12T2 s2allow p2
(4M2 + 3T2) b
4 (150(106))2(p)2
1 4
1 4
(4(75) + 3(50) ) d = 0.009025 m 2
2
d = 2c = 0.0181 m Use d = 19 mm
Ans.
882
500 N
y
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•12–1. An A-36 steel strap having a thickness of 10 mm and a width of 20 mm is bent into a circular arc of radius r = 10 m. Determine the maximum bending stress in the strap.
Moment-Curvature Relationship: M 1 = r EI
however,
M =
I s c
1 1 c s = r EI
s =
0.005 c E = a b C 200 A 109 B D = 100 MPa r 10
12–2. A picture is taken of a man performing a pole vault, and the minimum radius of curvature of the pole is estimated by measurement to be 4.5 m. If the pole is 40 mm in diameter and it is made of a glass-reinforced plastic for which Eg = 131 GPa, determine the maximum bending stress in the pole.
r ⫽ 4.5 m
Moment-Curvature Relationship: M 1 = r EI
however,
M =
I s c
I 1 c s = r EI
s =
0.02 c E = a b C 131 A 109 B D = 582 MPa r 4.5
Ans.
883
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12–3. When the diver stands at end C of the diving board, it deflects downward 3.5 in. Determine the weight of the diver. The board is made of material having a modulus of elasticity of E = 1.5(103) ksi.
B
A
3.5 in.
2 in.
C 9 ft
3 ft
Support Reactions and Elastic Curve. As shown in Fig. a. Moment Functions. Referring to the free-body diagrams of the diving board’s cut segments, Fig. b, M A x1 B is a + ©MO = 0; and M A x2 B is a + ©MO = 0;
M A x1 B + 3Wx1 = 0
M A x1 B = -3Wx1
-M A x2 B - Wx2 = 0
M A x2 B = -Wx2
Equations of Slope and Elastic Curve. EI
d2v = M(x) dx2
For coordinate x1, EI
d2v1 dx1 2
= -3Wx1
d2v1 3 = - Wx1 2 + C1 dx1 2
(1)
1 EIv1 = - Wx1 3 + C1x1 + C2 2
(2)
EI
For coordinate x2 EI
EI
d2v2 dx2 2
= -Wx2
dv2 1 = - Wx2 2 + C3 dx2 2
EIv2 = -
(3)
1 Wx2 3 + C3x2 + C4 6
(4)
Boundary Conditions. At x1 = 0, v1 = 0. Then, Eq. (2) gives 1 EI(0) = - W A 03 B + C1(0) + C2 2
C2 = 0
At x1 = 3 ft, v1 = 0. Then, Eq. (2) gives 1 EI(0) = - W A 33 B + C1(3) + 0 2
C1 = 4.5W
884
18 in.
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12-3.
Continued
At x2 = 9 ft, v2 = 0. Then, Eq. (4) gives 1 EI(0) = - W A 93 B + C3(9) + C4 6 9C3 + C4 = 121.5W Continuity Conditions. At x1 = 3 ft and x2 = 9 ft,
(5) dv2 dv1 . Thus, Eqs. (1) and = dx1 dx2
(3) give 1 3 - W A 32 B + 4.5W = - c - W A 92 B + C3 d 2 2
C3 = 49.5W
Substituting the value of C3 into Eq. (5), C4 = -324W Substituting the values of C3 and C4 into Eq. (4), v2 =
1 1 a - Wx2 3 + 49.5Wx2 - 324Wb EI 6
At x2 = 0, v2 = -3.5 in. Then, -324W(1728) -3.5 =
1.5 A 106 B c
1 (18) A 2 3 B d 12
W = 112.53 lb = 113 lb
Ans.
885
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*12–4. Determine the equations of the elastic curve using the x1 and x2 coordinates. EI is constant.
P A
EI
d2v1 dx1 2
= M1 (x)
M1(x) = 0;
EI
EI
d v1 dx1
2
L
= 0
x3
dv1 = C1 dx1
(1)
EI v1 = C1x1 + C2
(2)
M2(x) = Px2 - P(L - a) EI
EI
d2 v2 dx2 2
= Px2 - P(L - a)
dv2 P 2 = x - P(L - a)x2 + C3 dx2 2 2
EI v2 =
(3)
P(L - a)x22 P 3 x2 + C3x2 + C4 6 2
(4)
Boundary conditions: At x2 = 0,
dv2 = 0 dx2
From Eq. (3), 0 = C3 At x2 = 0, v2 = 0 0 = C4 Continuity condition: At x1 = a, x2 = L - a;
dv1 dv2 = dx1 dx2
From Eqs. (1) and (3), C1 = - c
P(L - a)2 - P(L - a)2 d ; 2
C1 =
P(L - a)2 2
At x1 = a, x2 = L - a, v1 = v2 From Eqs. (2) and (4), a
P(L - a)3 P(L - a)3 P(L - a)2 b a + C2 = 2 6 2
C2 = -
Pa(L - a)2 P(L - a)3 2 3
From Eq. (2), v1 =
P [3(L - a)2x1 - 3a(L - a)2 - 2(L - a)3] 6EI
Ans.
For Eq. (4), v2 =
B
x1 2
P [x2 - 3(L - a)x33] 6EI 2
Ans.
886
L 2
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•12–5.
Determine the equations of the elastic curve for the beam using the x1 and x2 coordinates. EI is constant.
P A
B
x1 L
Moment Functions. Referring to the FBDs of the beam’s cut segments shown in Fig. b and c, 1 P(x1) = 0 2
a + ©MO = 0;
M(x1) +
a + ©MO = 0;
-Px2 - M(x2) = 0
M(x1) = -
P x 2 1
And
EI
M(x2) = -Px2
d2v = M(x) dx2
For coordinate x1, EI
EI
d2v1 dx1 2
= -
P x 2 1
dv1 P = - x1 2 + C1 dx1 4
EI v1 = -
(1)
P 3 x + C1x + C2 12 1
(2)
For coordinate x2, EI
EI
d2v2 dx2 2
= -Px2
dv2 P = - x2 2 + C3 dx2 2
EI v2 = -
(3)
P 3 x + C3x2 + C4 6 2
(4)
At x1 = 0, v1 = 0. Then, Eq (2) gives EI(0) = -
P (0) + C1(0) + C2 12
C2 = 0
At x1 = L, v1 = 0. Then, Eq (2) gives EI(0) = At x2 =
P (L3) + C1L + 0 12
C1 =
PL2 12
L , v2 = 0. Then Eq (4) gives 2 EI(0) = -
P L 3 L a b + C3 a b + C4 6 2 2
C3L + 2C4 =
PL3 24
(5)
887
x2 L 2
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•12–5.
Continued
At x1 = L and x2 =
-
dv2 L dv1 , = . Thus, Eqs. (1) and (3) gives 2 dx1 dx2
P 2 P L 2 PL2 = - c - a b + C3 d AL B + 4 12 2 2 C3 =
7PL2 24
Substitute the result of C3 into Eq. (5) C4 = -
PL3 8
Substitute the values of C1 and C2 into Eq. (2) and C3 and C4 into Eq (4), v1 =
P A -x1 3 + L2x1 B 12EI
Ans.
v2 =
P A -4x2 3 + 7L2x2 - 3L3 B 24EI
Ans.
888
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12–6. Determine the equations of the elastic curve for the beam using the x1 and x3 coordinates. Specify the beam’s maximum deflection. EI is constant.
P A
Support Reactions and Elastic Curve: As shown on FBD(a). L
Slope and Elastic Curve:
For M(x1) = -
x3
d2v = M(x) dx2
EI
P x. 2 1 EI
d2y1 dx21
EI y1 = For M(x3) = Px3 -
= -
P x 2 1
dy1 P = - x21 + C1 dx1 4
EI
[1]
P 3 x + C1x1 + C2 12 1
[2]
3PL . 2 EI
d2y3 dx23
= Px3 -
3PL 2
dy3 P 2 3PL = x3 x3 + C3 dx3 2 2
EI
EI y3 =
[3]
P 3 3PL 3 x x3 + C3x3 + C4 6 3 4
[4]
Boundary Conditions: y1 = 0 at x1 = 0. From Eq. [2], C2 = 0 y1 = 0 at x1 = L. From Eq. [2]. 0 = -
PL3 + C1L 12
C1 =
PL2 12
y3 = 0 at x3 = L. From Eq. [4]. 0 =
PL3 3PL3 + C3L + C4 6 4
0 = -
7PL3 + C3L + C4 12
[5]
Continuity Condition: At x1 = x3 = L,
-
dy1 dy3 . From Eqs. [1] and [3], = dx1 dx3
PL2 PL2 PL2 3PL2 + = + C3 4 12 2 2
From Eq. [5], C4 = -
B
x1
Moment Function: As shown on FBD(b) and (c).
C3 =
5PL2 6
PL3 4
889
L 2
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12–6.
Continued
The Slope: Substitute the value of C1 into Eq. [1], dy1 P = A L2 - 3x21 B dx1 12EI dy1 P = 0 = A L2 - 3x21 B dx1 12EI
x1 =
L 23
The Elastic Curve: Substitute the values of C1, C2, C3, and C4 into Eqs. [2] and [4], respectively. y1 =
Px1 A -x21 + L2 B 12EI
yO = y1 |x1 =
y3 =
L 23
=
PA
L 23
B
12EI
Ans. a-
0.0321PL3 L3 + L2 b = 3 EI
P A 2x33 - 9Lx23 + 10L2x3 - 3L3 B 12EI
Ans.
yC = y3 |x3 = 32 L =
2 P 3 3 3 3 c2 a L b - 9La Lb + 10L2 a L b - 3L3 d 12EI 2 2 2
= -
PL3 8EI
Hence, ymax =
PL3 8EI
Ans.
890
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12–7. The beam is made of two rods and is subjected to the concentrated load P. Determine the maximum deflection of the beam if the moments of inertia of the rods are IAB and IBC , and the modulus of elasticity is E. EI
P B A
C l
d2y = M(x) dx2
L
M1(x) = - Px1 EIBC
EIBC
d2y1 dx1 2
= - Px1
dy1 Px21 = + C1 dx1 2
EIBC y1 = -
(1)
Px31 + C1x1 + C2 6
(2)
M2(x) = - Px2 EIAB
EIAB
d2y2 dx2 2
= - Px2
dy2 P = - x2 2 + C3 dx2 2
EIAB y2 = -
(3)
P 3 x + C3x2 + C4 2 2
(4)
Boundary conditions: At x2 = L,
0 = -
dy2 = 0 dx2
PL2 + C3; 2
C3 =
PL2 2
At x2 = L, y = 0 0 = -
PL3 PL3 + + C4; 6 2
C4 = -
PL3 3
Continuity Conditions: At x1 = x2 = l,
dy1 dy2 = dx1 dx2
891
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12–7.
Continued
From Eqs. (1) and (3), 1 PI 2 1 PI 2 PL2 cc+ C1 d = + d EIBC 2 EIAB 2 2 C1 =
IBC PL2 Pl2 Pl2 c+ d + IAB 2 2 2
At x1 = x2 = l, y1 = y2 From Eqs. (2) and (4), IBC PL2 Pl2 1 Pl3 Pl2 e+ c a+ b + dl + C2 f EIBC 6 IAB 2 2 2 =
1 PL2l PL3 Pl3 c+ d EIAB 6 2 3
C2 =
IBC PL3 IBC Pl3 Pl3 IAB 3 IAB 3 3
Therefore, y1 =
Px1 3 IBC 1 Pl2 PL2 Pl2 e+ c a+ b + dx1 EIBC 6 IAB 2 2 2 +
IBC PL3 IBC Pl3 Pl3 f IAB 3 IAB 3 3
At x1 = 0, y1 |x = 0 = ymax ymax =
=
IBC Pl3 IBC PL3 IAB 3 I Pl3 P e f = e l3 - L3 - a bl f EIBC IAB 3 IAB 3 3 3EIAB IBC IAB 3 P e a1 b l - L3 f 3EIAB IBC
Ans.
892
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*12–8. Determine the equations of the elastic curve for the beam using the x1 and x2 coordinates. EI is constant.
P
Referring to the FBDs of the beam’s cut segments shown in Fig. b and c, a + ©MO = 0;
M(x1) +
PL - Px1 = 0 2
M(x1) = Px1 -
PL 2
x1 x2
And a + ©MO = 0; EI
L 2
M(x2) = 0
d2v = M(x) dx2
For coordinate x1, EI
EI
d2v1 dx21
= Px1 -
PL 2
dv1 P 2 PL = x x + C1 dx1 2 1 2 1
EI v1 =
(1)
P 3 PL 2 x x + C1x1 + C2 6 1 4 1
(2)
For coordinate x2, EI
EI
d2v2 dx22
= 0
dv2 = C3 dx2
(3)
EI v2 = C3x2 = C4 At x1 = 0,
(4)
dv1 = 0. Then, Eq.(1) gives dx1
EI(0) =
PL P 2 (0 ) (0) + C1 2 2
C1 = 0
At x1 = 0, v1 = 0. Then, Eq(2) gives EI(0) = At x1 = x2 =
PL 2 P 3 (0 ) (0 ) + 0 + C2 6 4
C2 = 0
dv2 L dv1 = , . Thus, Eqs.(1) and (3) gives 2 dx1 dx2
P L 2 PL L a b a b = C3 2 2 2 2 Also, at x1 = x2 =
C3 = -
PL2 8
L , v = v2. Thus, Eqs, (2) and (4) gives 2 1
PL L 2 PL2 L P L 3 a b a b = ab a b + C4 6 2 4 2 8 2
C4 =
PL3 48
Substitute the values of C1 and C2 into Eq. (2) and C3 and C4 into Eq (4), v1 =
P A 2x31 - 3Lx21 B 12EI
Ans.
v2 =
PL2 (-6x2 + L) 48EI
Ans.
893
L 2
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•12–9.
Determine the equations of the elastic curve using the x1 and x2 coordinates. EI is constant. EI
d2y = M(x) dx2
M1 =
EI
EI
P
A
x1
Pb x L 1
d2y1 dx21
=
a
EI y3 =
L
(1)
Pb 3 x + C3x1 + C2 6L 1
(2)
Pb x - P(x2 - a) L 2
But b = L - a. Thus M2 = Paa1 -
EI
EI
d2y2 dx2 2
x2 b L
= Pa a1 -
x2 b L
dy2 x22 = Paax2 b + C3 dx2 2L
EI y2 = Pa a
(3)
x22 x22 b + C3x2 + C4 2 6L
(4)
Applying the boundary conditions: y1 = 0 at x1 = 0 Therefore,C2 = 0, y2 = 0 at x2 = L 0 =
b
x2
Pb x L 1
dy1 Pb 2 = x + C1 dx1 2L 1
M2 =
B
Pa L2 + C3L + C4 3
(5)
Applying the continuity conditions: y1 |x1 = a = y2 |x2 = a Pb 3 a2 a3 a + C1a = Pa a b + C3a + C4 6L 2 6L
(6)
dy1 dy2 2 2 = dx1 x1 = a dx2 x2 = a a2 Pb 2 a + C1 = Pa a a b + C3 2L 2L
(7)
894
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•12–9.
Continued
Solving Eqs. (5), (6) and (7) simultaneously yields, C1 = C4 =
Pb 2 A L - b2 B ; 6L
C3 = -
Pa A 2L2 + a2 B 6L
Pa3 6
Thus, EIy1 =
Pb 3 Pb 2 x A L - b 2 B x1 6L 1 6L
or v1 =
Pb A x3 - A L2 - b2 B x1 B 6EIL 1
Ans.
and EIy2 = Pa a y2 =
x22 x32 Pa Pa3 b A 2L2 + a2 B x2 + 2 6L 6L 6
Pa C 3x22 L - x32 - A 2L2 + a2 B x2 + a2L D 6EIL
Ans.
895
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12–10. Determine the maximum slope and maximum deflection of the simply supported beam which is subjected to the couple moment M0 . EI is constant.
M0
A
B L
Support Reactions and Elastic Curve: As shown on FBD(a). Moment Function: As shown on FBD(b). Slope and Elastic Curve: EI
d2y = M(x) dx2
EI
M0 d2y = x 2 L dx
EI
M0 2 dy = x + C1 dx 2L
EI y =
[1]
M0 3 x + C1x + C2 6L
[2]
Boundary Conditions: y = 0 at x = 0. From Eq. [2]. 0 = 0 + 0 + C2
C2 = 0
y = 0 at x = L. From Eq. [2]. 0 =
M0 3 A L B + C1 (L) 6L
C1 = -
M0L 6
The Slope: Substitute the value of C1 into Eq. [1], M0 dy = A 3x2 - L2 B dx 6LEI M0 dy = 0 = A 3x2 - L2 B dx 6LEI uB =
x =
23 L 3
M0L dy 2 = dx x = 0 6EI
umax = uA =
M0L dy 2 = dx x = L 3EI
Ans.
The Elastic Curve: Substituting the values of C1 and C2 into Eq. [2], y =
ymax occurs at x =
M0 A x3 - L2x B 6LEI
23 L, 3 ymax = -
23M0L2 Ans 27EI
The negative sign indicates downward displacement.
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12–11. Determine the equations of the elastic curve for the beam using the x1 and x2 coordinates. Specify the beam’s maximum deflection. EI is constant.
P
Referring to the FBDs of the beam’s cut segments shown in Fig. b and c, 2 M(x1) - Px1 = 0 3
a + ©M0 = 0;
A B
x1
2P x M(x1) = 3 1
a
2a x2
And 1 P P(3a - x2) - M(x2) = 0 M(x2) = Pa x 3 3 2
a + ©M0 = 0; EI
d2y = M(x) dx2
For coordinate x1, EI
EI
d2y1 dx21
=
2P x 3 1
dy1 P 2 = x + C1 dx1 3 1
EI y1 =
(1)
P 3 x = C1x1 + C2 9 1
(2)
For coordinate x2, EI
EI
d2y2 dx2
2
= Pa -
P x 3 2
dy2 P 2 = Pax2 x + C3 dx2 6 2
EI y2 =
(3)
Pa 2 P 3 x x + C3x2 + C4 2 2 18 2
(4)
At x1 = 0, y1 = 0. Then, Eq (2) gives EI(0) =
P 3 A 0 B + C1(0) + C2 9
C2 = 0
At x2 = 3a, y2 = 0. Then Eq (4) gives EI(0) =
Pa P (3a)2 (3a)3 + C3(3a) + C4 2 18 C3(3a) + C4 = -3Pa3
At x1 = x2 = a,
(5)
dy1 dy2 = . Thus, Eq. (1) and (3) gives dx1 dx2
P 2 P 2 a + C1 = Pa(a) a + C3 3 6 C1 - C3 =
Pa2 2
(6)
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12–11.
Continued
Also, At x1 = x2 = a, y1 = y2. Thus, Eqs, (2) and (4) gives. P 3 Pa 2 P 3 a + C1a = (a ) a + C3a + C4 9 2 18 C1a - C3a - C4 =
Pa3 3
(7)
Solving Eqs. (5), (6) and (7), C4 =
Pa3 6
C3 = -
19 Pa2 18
C1 = -
5Pa2 9
Substitute the values of C1 into Eq. (1) and C3 into Eq. (3), dy1 P = A 3x1 2 - 5a2 B dx1 9EI dy1 P = 0 = A 3x1 2 - 5a2 B dx1 9EI
x1 =
5 a 7 a (Not Valid) A3
And dy2 P = A 18ax2 - 3x2 2 - 19a2 B dx2 18EI dy2 P = 0 = A 18ax2 - 3x2 2 - 19a2) dx2 18EI x2 = 4.633a 7 3a (Not Valid)
x2 = 1.367a
Substitute the values of C1 and C2 into Eq. (2) and C3 and C4 into Eq.(4), y1 =
P A x 3 - 5a2x1 B 9EI 1
Ans.
y2 =
P A -x2 3 + 9ax2 2 - 19a2x2 + 3a3 B 18EI
Ans.
Vmax occurs at x2 = 1.367a. Thus. ymax = -
0.484 Pa3 0.484 Pa3 = T EI EI
Ans.
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*12–12. Determine the equations of the elastic curve for the beam using the x1 and x2 coordinates. Specify the slope at A and the maximum displacement of the shaft. EI is constant.
P
P
a
a
A
B
Referring to the FBDs of the beam’s cut segments shown in Fig. b and c, a + ©M0 = 0;
M(x1) - Px1 = 0
x1
M(x1) = Px1
x2
And
L
a + ©M0 = 0;
M(x2) - Pa = 0 EI
M(x2) = Pa
d2y = M(x) dx2
For coordinate x1, EI
EI
d2y1
= Px1
dx21
dy1 P 2 = x + C1 dx1 2 1
(1)
P 3 x + C1x1 + C2 6 1
EI y1 =
(2)
For coordinate x2, EI
EI
d2y2 dx2 2
= Pa
dy2 = Pax2 + C3 dx2
(3)
Pa 2 x + C3x2 + C4 2 2
EI y2 =
(4)
At x1 = 0, y1 = 0. Then, Eq. (2) gives EI (0) = Due to symmetry, at x2 =
P 3 (0 ) + C1(0) + C2 6
L dv2 = 0. Then, Eq. (3) gives , 2 dx2
EI (0) = Pa a At x1 = x2 = a,
C2 = 0
L b + C3 2
C3 = -
PaL 2
dy1 dy2 = . Thus, Eqs(1) and (3) give dx1 dx2 P 2 PaL a + C1 = Pa (a) + a b 2 2 C1 =
Pa2 PaL 2 2
899
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*12–12.
Continued
Also, at x1 = x2 = a, y1 = y2. Thus, Eq. (2) and (4) give P 3 Pa2 PaL Pa 2 PaL a + a ba = (a ) + a ba + C4 6 2 2 2 2 C4 =
Pa3 6
Substituting the value of C1 and C2 into Eq. (2) and C3 and C4 into Eq.(4), y1 =
P C x 3 + a(3a - 3L)x1 D 6EI 1
Ans.
y2 =
Pa A 3x2 2 - 3Lx2 + a2 B 6EI
Ans.
Due to symmetry, ymax occurs at x2 = ymax =
L . Thus 2
Pa Pa A 4a2 - 3L2 B = A 3L2 - 4a2 B T 24EI 24EI
Ans.
Substitute the value C1 into Eq (1), dy1 P = A x 2 + a2 - aL B dx1 2EI 1 At point A, x1 = 0. Then uA =
dy1 Pa Pa 2 = (a - L) = (L - a) T dx1 x1 = 0 2EI 2EI
900
Ans.
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12–13. The bar is supported by a roller constraint at B, which allows vertical displacement but resists axial load and moment. If the bar is subjected to the loading shown, determine the slope at A and the deflection at C. EI is constant d2y1
EI
EI
dx21
P C A
L 2
= M1 = Px1
dy1 Px21 = + C1 dx1 2
EI y1 =
Px31 + C1x1 + C2 6
EI
d2y2 PL = M2 = dx2 2
EI
dy2 PL = x + C3 dx2 2 2
EI y2 =
PL 2 x + C3x2 + C4 4 2
Boundary conditions: At x1 = 0, y1 = 0 0 = 0 + 0 + C2 ; At x2 = 0,
At x1 = P(L2 )3
C3 = 0
P A L2 B 2 2
dy1 dy2 L L = , x = , y1 = y2, 2 2 2 dx1 dx2
+ C1 a
6
PL(L2 )2 L b = + C4 2 4
+ C1 = -
C4 = -
C2 = 0
dy2 = 0 dx2
0 + C3 = 0 ;
PL A L2 B 2
;
3 C1 = - PL2 8
11 PL3 48
At x1 = 0 dy1 3 PL2 = uA = dx1 8 EI At x1 =
yC =
yC =
Ans.
L 2
P A L2 B 3 6EI
- a
B
3 PL2 L ba b + 0 8 EI 2
-PL2 6EI
Ans.
901
L 2
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12–14. The simply supported shaft has a moment of inertia of 2I for region BC and a moment of inertia I for regions AB and CD. Determine the maximum deflection of the beam due to the load P. M1 (x) =
P x 2 1
M2(x) =
P x 2 2
P B
A
L – 4
Elastic curve and slope: EI
EI
EI
d2v = M(x) dx2 d2v1 dx1
=
2
P x 2 1
dv1 Px21 = + C1 dx1 4
EIv1 =
2EI
2EI
(1)
Px31 + C1x1 + C2 12
d2v2
=
dx2 2
(2)
P x 2 2
dv2 Px22 = + C3 dx1 4
2EIv2 =
(3)
Px32 + C3x2 + C4 12
(4)
Boundary Conditions: v1 = 0 at x1 = 0 From Eq. (2), C2 = 0 dv2 L = 0 at x2 = dx2 2 From Eq. (3), 0 =
PL2 + C3 16
C3 =
PL2 16
Continuity conditions: dv1 dv2 L = at x1 = x2 = dx1 dx2 4
902
C
L – 4
L – 4
D
L – 4
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12–14.
Continued
From Eqs. (1) and (3), PL2 1 PL2 PL2 + C1 = - a b 64 128 2 16 C1 =
-5PL2 128
v1 = v2 at x1 = x2 =
L 4
From Eqs. (2) and (4) PL3 5PL2 L PL3 1 PL2 L 1 a b = - a b a b + C4 768 128 4 1536 2 16 4 2 C4 = v2 =
-PL3 384 P A 32x32 - 24L2 x2 - L3 B 768EI
vmax = v2 2
= x2 = L2
-3PL3 3PL3 = T 256EI 256EI
Ans.
903
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12–15. Determine the equations of the elastic curve for the shaft using the x1 and x3 coordinates. Specify the slope at A and the deflection at the center of the shaft. EI is constant.
P
P
A
Support Reactions and Elastic Curve: As shown on FBD(a).
x1
Moment Function: As shown on FBD(b) and (c).
x3 a
Slope and Elastic Curve: EI
d2 y = M(x) dx2
For M(x1) = -Px1, EI
EI
d2y1 dx21
= -Px1
dy1 P = - x21 + C1 dx1 2
EI y1 = -
[1]
P 3 x + C1x1 + C2 6 1
[2]
For M(x3) = -Pa, EI
EI
d2y3 dx23
= -Pa
dy3 = -Pax3 + C3 dx3
EI y3 = -
[3]
Pa 2 x + C3x3 + C4 2 3
[4]
Boundary Conditions: y1 = 0 at x1 = a. From Eq. [2], 0 = Due to symmetry,
Pa3 + C1a + C2 6
[5]
dy3 b = 0 at x3 = . From Eq. [3] dx3 2
b 0 = -Paa b + C3 2
C3 =
Pab 2
y3 = 0 at x3 = 0 From Eq.[4]. C4 = 0 Continuity Condition: At x1 = a and x3 = 0,
-
From Eq. [5]
dy1 dy3 = . From Eqs. [1] and [3], dx1 dx3
Pa2 Pab + C1 = 2 2 C2 = -
C1 =
B
Pa (a + b) 2
Pa2 (2a + 3b) 6
904
b
a
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12–15. Continued The Slope: Either Eq. [1] or [3] can be used. Substitute the value of C1 into Eq. [1], dy1 P = C -x21 + a(a + b) D dx1 2EI uA =
dy1 P Pab 2 = C -a2 + a(a + b) D = dx1 x1 = a 2EI 2EI
Ans.
The Elastic Curve: Substitute the values of C1, C2, C3, and C4 into Eqs. [2] and [4], respectively, y1 =
P C -x31 + 3a(a + b)x1 - a2(2b + 3b) D 6EI
Ans.
y3 =
Pax3 ( -x3 + b) 2EI
Ans.
yC = y3 |x3 = b2
=
=
Pa A b2 B 2EI
a-
b + bb 2
Pab2 8EI
Ans.
905
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*12–16. The fence board weaves between the three smooth fixed posts. If the posts remain along the same line, determine the maximum bending stress in the board. The board has a width of 6 in. and a thickness of 0.5 in. E = 1.60(103) ksi. Assume the displacement of each end of the board relative to its center is 3 in.
4 ft
3 in. A
Support Reactions and Elastic Curve: As shown on FBD(a). Moment Function: As shown on FBD(b). Slope and Elastic Curve: EI
d2y = M(x) dx2
EI EI
d2y P = x 2 dx2
dy P 2 = x + C1 dx 4
EI y =
[1]
P 3 x + C1x + C2 12
Boundary Conditions: Due to symmetry,
[2]
L dy = 0 at x = . dx 2
Also, y = 0 at x = 0. From Eq. [1] 0 =
P L 2 a b + C1 4 2
From Eq. [2] 0 = 0 + 0 + C2
C1 = -
PL2 16
C2 = 0
The Elastic Curve: Substitute the values of C1 and C2 into Eq. [2], y =
Px A 4x2 - 3L2 B 48EI
[1]
Require at x = 48 in., y = -3 in. From Eq.[1], P(48)
-3 =
1 48(1.60) A 106 B A 12 B (6) A 0.53 B
C 4 A 482 B - 3 A 962 B D
P = 16.28 lb Maximum Bending Stress: From the moment diagram, the maximum moment is Mmax = 390.625 lb # in. Applying the flexure formula, smax =
4 ft
390.625(0.25) Mc = = 1562.5 psi = 1.56 ksi 1 3 I 12 (6) A 0.5 B
Ans.
906
B
C
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•12–17.
Determine the equations of the elastic curve for the shaft using the x1 and x2 coordinates. Specify the slope at A and the deflection at C. EI is constant.
a + ©MO = 0;
a + ©MO = 0;
MO - M(x2) = 0
M(x2) = MO
d2y = M(x) dx2
For coordinate x1, EI
EI
d2y1 dx1 2
=
MO x L 1
dy1 MO 2 = x + C1 dx1 2L 1
EI y1 =
(1)
MO 3 x + C1x1 + C2 6L 1
(2)
For coordinate x2, EI
EI
d2y2 dx2 2
= MO
dy2 = MOx2 + C3 dx2
EI y2 =
(3)
MO 2 x + C3x2 + C4 2 2
(4)
At x1 = 0, y1 = 0. Then, Eq. (2) gives EI (0) =
MO 3 (0 ) + C1(0) + C2 6L
C2 = 0
At x1 = L, y1 = 0. Then, Eq. (2) gives C1 = Also, at x2 =
-ML 6
L , y = 0. Then Eq. (4) gives. 2 2 EI(0) =
MO L 2 L a b + C3 a b + C4 2 2 2
C3L + 2C4 = -
MOL2 4
(5)
907
C
x2 L
MO M(x1) = x L 1
And
EI
B
x1
Referring to the FBDs of the shaft’s cut segments shown in Fig. b and c, MO M(x1) x = 0 L 1
M0
A
L 2
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•12–17.
Continued
At x1 = L and x2 =
dy2 L dy1 , = . Then, Eq. (1) and (3) give 2 dx1 dx2 MO 2 MOL L = - cMO a b + C3 d AL B 2L 6 2
C3 = -
5MOL 6
Substitute the result of C3 into Eq. (5), C4 =
7M0L2 24
Substitute the value of C1 into Eq. (1), dy1 MO = A 3x1 2 - L2 B dx1 6LEI At A, x1 = 0. Thus uA =
MOL dy1 MO 2 = = dx1 x1 = 0 6EI 6EI
Ans.
Substitute the values of C1 and C2 into Eq (2) and C3 and C4 into Eq. (4), y1 =
MO A x 3 - L2x1 B 6EIL 1
Ans.
y2 =
MO A 12x2 2 - 20 Lx2 + 7L2 B 24EI
Ans.
At C, x2 = 0. Thus yC = y2 2
= x2 = 0
7MOL2 24EI
Ans.
c
908
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12–18. Determine the equation of the elastic curve for the beam using the x coordinate. Specify the slope at A and the maximum deflection. EI is constant.
M0
A
M0
x
B L
Referring to the FBD of the beam’s cut segment shown in Fig. b, a + ©MO = 0;
M(x) +
2MO x - MO = 0 L
EI
d2y = M(x) dx2
EI
2MO d2y = MO x L dx2
EI
MO 2 dy = MOx x + C1 dx L
M(x) = MO -
2MO x L
(1)
MO 3 MO 2 x x + C1x + C2 2 3L
EI y =
(2)
At x = 0, y = 0. Then Eq (2) gives EI(0) =
MO 3 MO 2 A0 B A 0 B + C1(0) + C2 2 3L
C2 = 0
Also, at x = L, y = 0. Then Eq (2) gives EI(0) =
MO 2 MO 3 AL B A L B + C1L + 0 2 3L
C1 = -
MOL 6
Substitute the value of C1 into Eq (1), MO dy = A 6Lx - 6x2 - L2 B dx 6EIL MO dy = 0 = A 6Lx - 6x2 - L2 B dx 6EIL x = 0.2113 L
and
0.7887 L
At A, x = 0. Thus uA = -
MOL 6EI
Ans.
Substitute the values of C1 and C2 into Eq (2) y =
MO A 3Lx2 - 2x3 - L2x B 6EIL
Ans.
909
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12–18.
Continued
vmax occurs at x = 0.21132 L ymax =
or
0.7887 L. Thus,
MO c3L(0.2113L)2 - 2(0.2113L)3 - L2(0.2113L) d 6EIL
= -
0.0160 MOL2 0.0160 MOL2 = EI EI
Ans.
T
and ymax =
=
MO c3L(0.7887L)2 - 2(0.7887L)3 - L2(0.7887L) d 6EIL 0.0160 MOL2 EI
Ans.
c
910
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12–19. Determine the deflection at the center of the beam and the slope at B. EI is constant.
M0
A
M0
x
B L
Referring to FBD of the beam’s cut segment shown in Fig. b, a + ¢Mo = 0;
M(x) +
2Mo x - Mo = 0 L
EI
d2v = M(x) dx2
EI
2Mo d2v = Mo x L dx2
EI
Mo 2 dv = Mox x + C1 dx L
EI v =
M(x) = Mo -
2Mo x L
(1)
Mo 3 Mo 2 x x + C1x + C2 2 3L
(2)
At x = 0, v = 0. Then Eq. (2) gives EI(0) =
Mo 3 Mo 2 (0 ) A 0 B + C1(0) + C2 2 3L
C2 = 0
Also, at x = L, v = 0. Then Eq. (2) gives EI (0) =
Mo 2 Mo 3 AL B A L B + C1L + 0 2 3L
C1 = -
MoL 6
Substitute the value of C1 into Eq. (1), Mo dv = A 6Lx - 6x2 - L2 B dx 6EIL At B, x = L. Thus uB =
MoL MoL dv 2 = = dx x = L 6EI 6EI
Ans.
Substitute the values of C1 and C2 into Eq. (2), v =
Mo A 3Lx2 - 2x3 - L2x B 6EIL
At the center of the beam, x =
L . Thus 2
v冷x = L2 = 0
Ans.
911
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*12–20. Determine the equations of the elastic curve using the x1 and x2 coordinates, and specify the slope at A and the deflection at C. EI is constant.
8 kip
A
C
B
x1
x2 20 ft
Referring to the FBDs of the beam’s cut segments shown in Fig. b, and c, M(x1) = (-5x1) kip # ft
a + ©Mo = 0;
M(x1) + 5x1 = 0
a + ©Mo = 0;
-M(x2) - 8x2 - 20 = 0 M(x2) = ( -8x2 - 20) kip # ft
And
EI
d2v = M(x) dx2
For coordinate x1, EI
EI
d2v1 dx21
= (-5x1) kip # ft
dv1 5 = a - x21 + C1 b kip # ft2 dx1 2
(1)
5 EI v1 = a - x1 3 + C1x1 + C2 b kip # ft3 6
(2)
For coordinate x2, EI
EI
d2v2 dx2 2
= (-8x2 - 20) kip # ft
dv2 = dx2
A -4x2 2 - 20x2 + C3 B kip # ft2
(3)
4 EI v2 = a - x2 3 - 10x2 2 + C3x2 + C4 b kip # ft3 3
(4)
At x1 = 0, v1 = 0. Then, Eq (2) gives EI(0) = -
5 3 A 0 B + C1(0) + C2 6
C2 = 0
Also, at x1 = 20 ft, v1 = 0. Then, Eq (2) gives EI(0) = -
5 A 203 B + C1 (20) + 0 6
C1 = 333.33 kip # ft2
Also, at x2 = 10 ft, v2 = 0. Then, Eq. (4) gives EI(0) = -
4 A 103 B - 10 A 102 B + C3(10) + C4 3
10C3 + C4 = 2333.33
(5)
912
10 ft
20 kip⭈ft
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*12–20.
Continued
At x1 = 20 ft and x2 = 10 ft,
-
dv1 dv2 = . Then Eq. (1) and (3) gives dx1 dx2
5 A 202 B + 333.33 = - C -4 A 102 B - 20(10) + C3 D 2 C3 = 1266.67 kip # ft2
Substitute the value of C3 into Eq (5), C4 = -10333.33 kip # ft3 Substitute the value of C1 into Eq. (1), dv1 5 1 a - x1 2 + 333.33b kip # ft2 = dx1 EI 2 At A, x1 = 0. Thus, uA =
333 kip # ft2 dv1 2 = dx1 x1 = 0 EI
uA
Ans.
Substitute the values of C1 and C2 into Eq. (2) and C3 and C4 into Eq (4), v1 =
1 5 a - x1 3 + 333 x1 b kip # ft3 EI 6
Ans.
v2 =
4 1 a - x2 3 - 10x2 2 + 1267x2 - 10333b kip # ft3 EI 3
Ans.
At C, x2 = 0. Thus vC = v2 冷x2 = 0 = -
10 333 kip # ft3 10 333 kip # ft3 = T EI EI
913
Ans.
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•12–21.
Determine the elastic curve in terms of the and coordinates and the deflection of end C of the overhang beam. EI is constant.
w
A
C
Support Reactions and Elastic Curve. As shown in Fig. a.
B
Moment Functions. Referring to the free-body diagrams of the beam’s cut segments, Fig. b, M(x1) is a + ©MO = 0;
M(x1) +
wL x = 0 8 1
M(x1) = -
-M(x2) - wx2 a
x2 b = 0 2
M(x2) = -
w 2 x 2 2
Equations of Slope and Elastic Curve. EI
d2v = M(x) dx2
For coordinate x1, EI
EI
d2v1 dx1 2
= -
wL x 8 1
dv1 wL 2 = x + C1 dx1 16 1
EIv1 = -
(1)
wL 3 x + C1x1 + C2 48 1
(2)
For coordinate x2, EI
EI
d2v2 dx2 2
= -
w 2 x 2 2
dv2 w = - x2 3 + C3 dx2 6
EIv2 = -
(3)
w 4 x + C3x2 + C4 24 2
(4)
Boundary Conditions. At x1 = 0, v1 = 0. Then, Eq. (2) gives EI(0) = -
wL 3 A 0 B + C1(0) + C2 48
C2 = 0
At x1 = L, v1 = 0. Then, Eq. (2) gives EI(0) = At x2 =
wL 3 A L B + C1L + 0 48
C1 =
wL3 48
L , v = 0. Then, Eq. (4) gives 2 2
EI(0) = -
w L 4 L a b + C3 a b + C4 24 2 2
L wL4 C3 + C4 = 2 384
(5)
914
x2 L
wL x 8 1
and M(x2) is a + ©MO = 0;
x1
L 2
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•12–21.
Continued
Continuity Conditions. At x1 = L Land x2 =
dv2 L dv1 , = . Thus, Eqs. (1) and 2 dx1 dx2
(3) give
-
w L 3 wL3 wL 2 = - C - a b + C3 S AL B + 16 48 6 2
C3 =
wL3 16
Substituting the value of C3 into Eq. (5), C4 = -
11wL4 384
Substituting the values of C3 and C4 into Eq. (4), v2 =
w A -16x2 4 + 24L3x2 - 11L4 B 384EI
At C, x2 = 0. Thus, vC = v2冷x2 = 0 = -
11wL4 11wL4 = T 384EI 384EI
Ans.
915
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12–22. Determine the elastic curve for the cantilevered beam using the x coordinate. Specify the maximum slope and maximum deflection. E = 29(103) ksi.
3 kip/ft
A
Referring to the FBD of the beam’s cut segment shown in Fig. b, a + ©Mo = 0; M(x) + 81 +
1 2
x 3
9 ft
M(x) = A 13.5x - 0.05556x3 - 81 B kip # ft. EI
d2v = M(x) dx2
EI
d2v = A 13.5x - 0.05556x3 - 81 B kip # ft dx2
EI
dv = A 6.75x2 - 0.01389x4 - 81x + C1 B kip # ft2 dx
(1)
EI v = A 2.25x3 - 0.002778x5 - 40.5x2 + C1x + C2 B kip # ft3 (2) At x = 0,
dv = 0. Then, Eq (1) gives dx
EI(0) = 6.75 A 02 B - 0.01388 A 04 B - 81(0) + C1
C1 = 0
Also, at x = 0, v = 0. Then Eq. (2) gives EI(0) = 2.25 A 03 B - 0.002778 A 05 B - 40.5 A 02 B + 0 + C2
C2 = 0
Substitute the value of C1 into Eq (1) gives. 1 dv = A 6.75x2 - 0.01389x4 - 81x B kip # ft2 dx EI The Maximum Slope occurs at x = 9 ft. Thus, umax =
273.375 kip # ft2 dv 2 = dx x = 9ft EI =
273.375 kip # ft2 EI
umax
For W14 * 30, I = 291 in4. Thus u = 273.375 A 12 2 B = 0.00466 rad
Ans.
Substitute the values of C1 and C2 into Eq (2), v =
1 A 2.25x3 - 0.002778x5 - 40.5x2 B kip # ft3 EI
The maximum deflection occurs at x = 9 ft, Thus, vmax = v 冷x = 9 ft = -
=
=
B
x
A x B (x) A B - 13.5x = 0 1 3
1804.275 kip # ft3 EI
1804.275 kip # ft3 T EI 1804.275 A 12 3 B
29.0 A 103 B (291)
= 0.369 in T
Ans.
916
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12–23. The beam is subjected to the linearly varying distributed load. Determine the maximum slope of the beam. EI is constant.
w0
A
B
x L
EI
d2y = M(x) dx2
EI
w0 d2y = A L2x - x3 B 2 6L dx
EI
w0 L2x2 x4 dy = a b + C1 dx 6L 2 4
EI y =
(1)
w0 L2x3 x5 a b + C1x + C2 6L 6 20
(2)
Boundary conditions: At x = 0, y = 0. From Eq. (2), C2 = 0 At x = L, y = 0 From Eq. (2), 0 =
w0 L5 L5 a b + C1L ; 6L 6 20
C1 = -
7w0L3 360
The slope: From Eq.(1), w0 dy L2x2 x4 7L4 = a b dx 6EIL 2 4 60 umax =
w0L3 dy 2 = dx x = L 45EI
Ans.
917
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*12–24. The beam is subjected to the linearly varying distributed load. Determine the maximum deflection of the beam. EI is constant.
w0
A
B
x L
EI
d2y = M(x) dx2
EI
w0 d2y = A L2x - x3 B 2 6L dx
EI
w0 L2x2 x4 dy = a b + C1 dx 6L 2 4
EI y =
(1)
w0 L2x3 x5 a b + C1x + C2 6L 6 20
(2)
Boundary conditions: y = 0 at x = 0. From Eq. (2), C2 = 0 y = 0 at x = L. From Eq. (2), 0 =
w0 L2 L5 a b + C1L; 6L 6 20
C1 = -
7w0L3 360
w0 dy L2x2 x4 7L4 = a b dx 6EIL 2 4 60 dy L2x2 x4 7L4 = 0 = a b dx 2 4 60 15x4 - 30L2x2 + 7L4 = 0; y =
x = 0.5193L
w0x A 10L2x2 - 3x4 - 7L4 B 360EIL
Substitute x = 0.5193L into y, ymax = -
0.00652w0L4 EI
Ans.
918
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•12–25.
Determine the equation of the elastic curve for the simply supported beam using the x coordinate. Determine the slope at A and the maximum deflection. EI is constant.
12 kN/m
A
B
x 6m
Referring to the FBD of the beam’s cut segment shown in Fig. b, a + ©Mo = 0; M(x) +
1 1 (2x)(x) A x3 B - 36x = 0 M(x) = a36x - x3 bkN # m 2 3
EI
d2v = M(x) dx2
EI
d2v 1 = a36x - x3 b kN # m 2 3 dx dv 1 4 = a18x2 x + C1 b kN # m2 dx 12
EI
EI v = a6x3 Due to the Symmetry,
(1)
1 5 x + C1x + C2 b kN # m3 60
(2)
dv = 0 at x = 6 m. Then, Eq (1) gives dx
EI(0) = 18 A 62 B -
1 A 64 B + C1 12
C1 = -540 kN # m2
Also, at x = 0, v = 0. Then, Eq (2) gives EI(0) = 6 A 03 B -
1 A 05 B + C1(0) + C2 60
C2 = 0
Substitute the value of C1 into Eq. (1), dv 1 1 4 = a18x2 x - 540b kN # m2 dx EI 12 At A, x = 0. Then uA =
dv 540 kN # m2 540kN # m2 2 = = dx x = 0 EI EI
Ans.
Substitute the values of C1 and C2 into Eq (2) v =
1 1 5 a6x3 x - 540xb kN # m3 EI 60
Ans.
Due to Symmetry, vmax occurs at mind span x = 6 m. Thus, vmax =
1 1 c6 A 63 B A 65 B - 540(6) d EI 60
= -
2074 kN # m3 2073.6 kN # m3 = EI EI
Ans.
T
919
6m
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12–26. Determine the equations of the elastic curve using the coordinates x1 and x2 , and specify the slope and deflection at B. EI is constant.
w
C
A
B
x1 a x2 L
EI
d2y = M(x) dx2
For M1(x) = -
EI
EI
d2y1 dx21
= -
w 2 wa2 x1 + wax1 2 2
w 2 wa2 x1 + wax1 2 2
dy1 w wa 2 wa2 = - x21 + x1 x + C1 dx1 6 2 2 1
EI y1 = -
w 4 wa 3 wa2 2 x1 + x1 x1 + C1x1 + C2 24 6 4
For M2(x) = 0 ;
EI
(1)
EI
d2y2 dx2 2
(2)
= 0
dy2 = C3 dx2
(3)
EI y2 = C3x2 + C4
(4)
Boundary conditions: At x1 = 0.
dy1 = 0 dx1
From Eq. (1), C1 = 0 At x1 = 0. y1 = 0 From Eq. (2): C2 = 0 Continuity conditions: At x1 = a,
dy1 dy2 = dx1 dx2
x2 = a ;
From Eqs. (1) and (3), -
wa3 wa3 wa3 + = C3; 6 2 2
C3 = -
wa3 6
From Eqs. (2) and (4), At x1 = a, x2 = a -
y1 = y2
wa4 wa4 wa4 wa4 + = + C4 ; 24 6 4 6
C4 =
wa4 24
920
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12–26.
Continued
The slope, from Eq. (3). uB =
dy2 wa3 = dx2 6EI
Ans.
The elastic curve: y1 =
w a -x41 + 4ax31 - 6a2 x21 b 24EI
Ans.
y2 =
wa3 a -4x2 + ab 24EI
Ans.
yB = y2 2
= x3 = L
wa3 a -4L + ab 24EI
Ans.
921
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12–27. Wooden posts used for a retaining wall have a diameter of 3 in. If the soil pressure along a post varies uniformly from zero at the top A to a maximum of 300 lb>ft at the bottom B, determine the slope and displacement at the top of the post. Ew = 1.6(103) ksi.
A
6 ft
Moment Function: As shown on FBD. Slope and Elastic Curve: B
d2y EI 2 = M(y) dy EI
EI
d2y = -8.333y3 dy2
dy = -2.0833y4 + C1 dy
[1]
EI y = -0.4167y5 + C1y + C2 Boundary Conditions:
[2]
dy = 0 at y = 6 ft and y = 0 at y = 6 ft dy
From Eq. [1], 0 = -2.0833 A 64 B + C1
C1 = 2700
From Eq. [2], 0 = -0.4167 A 65 B + 2700(6) + C2
C2 = -12960
The Slope: Substituting the value of C1 into Eq. [1], 1 dy = b A -2.0833y4 + 2700 B r lb # ft2 dy EI uA =
dy 2 2700 lb # ft2 = dy y = 0 EI 2700(144)
=
1.6 A 106 B A p4 B A 1.54 B
= 0.0611 rad
Ans.
The Elastic Curve: Substituting the values of C1 C2 into Eq. [2], y =
1 EI
E A -0.4167y5 + 2700y - 12960 B F lb # ft3
yA = y|y = 0 = -
12960 lb # ft3 EI 12960(1728)
= -
1.6 A 106 B A p4 B A 1.54 B
= -3.52 in.
Ans.
The negative sign indicates leftward displacement.
922
300 lb/ft
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*12–28. Determine the slope at end B and the maximum deflection of the cantilevered triangular plate of constant thickness t. The plate is made of material having a modulus of elasticity E.
b 2 b 2 L
Section Properties. Referring to the geometry shown in Fig. a, b(x) b = ; x L
b(x) =
A
b x L t
Thus, the moment of the plate as a function of x is I(x) =
1 bt3 x C b(x) D t3 = 12 12L
x
Moment Functions. Referring to the free-body diagram of the plate’s cut segments, Fig. b, + ©MO = 0;
x -M(x) - w(x) a b = 0 2
M(x) = -
w 2 x 2
Equations of Slope and Elastic Curve. E
M(x) d 2v = I(x) dx2
d 2v E 2 = dx
E
w 2 x 6wL 2 = - 3 x 3 bt bt x 12L
-
dv 3wL = - 3 x2 + C1 dx bt
Ev = -
(1)
wL 3 x + C1x + C2 bt3
Boundary Conditions. At x = L, E(0) = -
(2)
dv = 0. Then Eq. (1) gives dx
3wL 2 A L B + C1 bt3
C1 =
3wL3 bt3
At x = L, v = 0. Then Eq. (2) gives E(0) = -
wL 3 A L B + C1(L) + C2 bt3
C2 = -
2wL4 bt3
Substituting the value of C1 into Eq. (1), dv 3wL = A -x2 + L2 B dx Ebt3 At B, x = 0. Thus, uB =
dv 2 3wL3 = dx x = 0 Ebt3
Substituting the values of C1 and C2 into Eq. (2), v =
wL A -x3 + 3L2x - 2L3 B Ebt3
vmax occurs at x = 0. Thus, vmax = v冷x = 0 = -
2wL4 2wL4 = 3 Ebt Ebt3
w
Ans.
T
923
B
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•12–29.
The beam is made of a material having a specific weight g. Determine the displacement and slope at its end A due to its weight. The modulus of elasticity for the material is E.
L
Section Properties: h
bh 2 1 h a xb(x)(b) = x 2 L 2L
h(x) =
h x L
I(x) =
3 h bh3 3 1 (b)a xb = x 12 L 12L3
V(x) =
Moment Function: As shown on FBD. Slope and Elastic Curve: E
E
M(x) d2y = I(x) dx2
d2y = dx2
E
bhg 6L
x3
2gL2 = -
3
bh 3 3x 12L
h2
2gL2 dy = - 2 x + C1 dx h
Ey = -
gL2
x2 + C1 x + C2
h2
From Eq. [2], 0 = -
[2]
dy = 0 at x = L and y = 0 at x = L. dx
Boundary Conditions: From Eq. [1], 0 = -
[1]
2gL2 2
h
gL2 2
h
(L) + C1
A L2 B +
C1 =
2gL3 2
h
2gL3
(L) + C2
h3 C2 = -
gL4 h2
The Slope: Substituting the value of C1 into Eq. [1], 2gL2 dy = 2 ( -x + L) dx hE uA =
2gL3 dy 2 = 2 dx x = 0 hE
Ans.
The Elastic Curve: Substituting the values of C1 and C2 into Eq. [2], y =
gL2 h2E
A -x2 + 2Lx - L2 B
yA |x = 0 = -
gL4
Ans.
h2E
The negative sign indicates downward displacement.
924
A
b
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12–30. The beam is made of a material having a specific weight of g. Determine the displacement and slope at its end A due to its weight. The modulus of elasticity for the material is E.
r
Section Properties: r(x) =
r x L 4
I(x) =
A
2 p r pr2 3 a xb x = x 3 L 3L2
V(x) =
L
4
pr 4 p r a xb = x 4 L 4L4
Moment Function: As shown on FBD. Slope and Elastic Curve: M(x) d2y = I(x) dx2
E
2
pr g2 4 gL2 d 2y 12L x E 2 = - 4 = - 2 pr 4 4 dx 3r x 4L
E
gL2 dy = - 2 x + C1 dx 3r
Ey = Boundary Conditions: From Eq. [1], 0 = -
From Eq. [2], 0 = -
6r2
x2 + C1x + C2
[2]
dy = 0 at x = L and y = 0 at x = L. dx
gL2 3r
gL2
[1]
2
(L) + C1
L2 B + ¢ 2 A
C1 =
gL2
gL3
6r
3r
gL3 3r2
L + C2 2 ≤
C2 = -
gL4 6r2
The Slope: Substituting the value of C1 into Eq. [1], gL2 dy = 2 (-x + L) dx 3r E uA =
gL3 dy 2 = 2 dx x = 0 3r E
Ans.
The Elastic Curve: Substituting the values of C1 and C2 into Eq. [2], y =
gL2 6r2E
A -x2 + 2Lx - L2 B
yA |x = 0 = -
gL4
Ans.
6r2E
The negative sign indicates downward displacement.
925
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12–31. The tapered beam has a rectangular cross section. Determine the deflection of its free end in terms of the load P, length L, modulus of elasticity E, and the moment of inertia I0 of its fixed end.
b
A
P
Moment function: M(x) = -Px
L
Moment of inertia: w =
b x; L
I =
l0 1 b 1 x a xb t3 = b t3 a b = x 12 L 12 L L
Slope and elastic curve: EI(x)
d2y = M(x) dx2
Ea
l0 d2y bx 2 = -Px ; L dx
El0
dy = -PLx + C1 dx
El0 y =
El0
d2y = -PL dx2 (1)
-PL 2 x + C1x + C2 2
(2)
Boundary conditions: dy = 0, x = L dx From Eq. (1), 0 = -PL2 + C1 ;
C1 = PL2
y = 0, x = L From Eq. (2), 0 = y =
PL3 + PL3 + C2 ; 2
C2 = -
PL3 2
PL ( -x2 + Lx - L2) 2El0
x = 0,
ymax = y 2
= u=0
PL3 2El0
Ans.
The negative sign indicates downward displacement.
926
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*12–32. The beam is made from a plate that has a constant thickness t and a width that varies linearly. The plate is cut into strips to form a series of leaves that are stacked to make a leaf spring consisting of n leaves. Determine the deflection at its end when loaded. Neglect friction between the leaves.
P
b
L
Use the triangular plate for the calculation. M = Px I =
1 b a xb(t)3 12 L
d2v M Px = = 1 EI dx2 E A 12 B A Lb B x(t)3 d2v 12PL = 2 dx Ebt3 dv 12PL = x + C1 dx Ebt3 v =
6PL 2 x + C1x + C2 Ebt3
dv = 0 at x = L dx C1 =
-12PL2 Ebt3
v = 0 at x = L C2 =
6PL3 Ebt3
When x = 0 vmax =
6PL3 Ebt3
Ans.
927
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•12–33.
The tapered beam has a rectangular cross section. Determine the deflection of its center in terms of the load P, length L, modulus of elasticity E, and the moment of inertia Ic of its center.
P
b
L — 2
Moment of inertia: L — 2
2b w = x L I =
2IC 1 2b 1 2x a xb (t3) = (b) A t3 B a b = a bx 12 L 12 L L
Elastic curve and slope: EI(x)
Ea
d2v = M(x) dx2
2IC d2v P b(x) 3 = x L 2 dx
EIC
dv PL = x + C1 dx 4
(1)
PL 2 x + C1x + C2 8
EICv1 =
(2)
Boundary condition: Due to symmetry: dv = 0 dx
x =
at
L 2
From Eq. (1), 0 =
PL2 + C1 8
C1 = -
PL2 8
v = 0 at x = 0 C2 = 0 v =
PLx (x - L) 8EIC
vC = v `
= x = L2
PL3 32EIC
Ans.
The negative sign indicates downward displacement.
928
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12–34. The leaf spring assembly is designed so that it is subjected to the same maximum stress throughout its length. If the plates of each leaf have a thickness t and can slide freely between each other, show that the spring must be in the form of a circular arc in order that the entire spring becomes flat when a large enough load P is applied. What is the maximum normal stress in the spring? Consider the spring to be made by cutting the n strips from the diamond-shaped plate of thickness t and width b.The modulus of elasticity for the material is E. Hint: Show that the radius of curvature of the spring is constant.
nb
b x P
L 2
Section Properties: Since the plates can slide freely relative to each other, the plates resist the moment individually. At an arbitrary distance x from the support, the 2nx nx numbers, of plates is L = . Hence, L 2 I(x) =
1 2nx nbt3 a b (b) A t3 B = x 12 L 6L
Moment Function: As shown on FBD. Bending Stress: Applying the flexure formula,
smax
M(x) c = = I(x)
Px 2
A 2t B
nbt3 6L x
=
3PL 2nbt2
Ans.
Moment - Curvature Relationship: Px
M(x) 1 3PL 2 = Constant (Q.E.D.) = = = 3 nbt3 r EI(x) nbt E E A 6L x B
929
x L 2
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12–35. The shaft is made of steel and has a diameter of 15 mm. Determine its maximum deflection. The bearings at A and B exert only vertical reactions on the shaft. Est = 200 GPa.
15 mm A
B
200 mm
300 mm 250 N
M = -(-201.43) 6 x - 0 7 -250 6 x - 0.2 7 - 80 6 x - 0.5 7 M = 201.43x - 250 6 x - 0.2 7 - 80 6 x - 0.5 7 Elastic curve and slope: EI
d2v = M = 201.43x - 250 6 x - 0.2 7 - 80 6 x - 0.5 7 dx3
EI
dv = 100.71x2 - 125 6 x - 0.2 7 2 - 40 6 x - 0.5 7 2 + C1 dx
EIv = 33.57x3 - 41.67 6 x - 0.2 7 3 - 13.33 6 x - 0.5 7 3 + C1x + C2 (1) Boundary conditions: v = 0
at
x = 0
From Eq. (1) C2 = 0 v = 0
at
x = 0.7 m
0 = 11.515 - 5.2083 - 0.1067 + 0.7C1 C1 = -8.857 dv 1 = C 100.71x2 - 125 6 x - 0.2 7 2 - 40 6 x - 0.5 7 2 - 8.857 D dx EI Assume vmax occurs at 0.2 m 6 x 6 0.5 m dv 1 = 0 = C 100.71x2 - 125(x - 0.2)2 - 8.857 D dx EI 24.28x2 - 50x + 13.857 = 0 x = 0.3300 m v =
O.K.
1 C 33.57x3 - 41.67 6 x - 0.2 7 3 - 13.33 6 x - 0.5 7 3 - 8.857x D EI
Substitute x = 0.3300 m into the elastic curve: vmax = -
1.808N # m3 1.808 = = -0.00364 = -3.64 mm Ans. 9 p EI 200 A 10 B 4 (0.0075)4
The negative sign indicates downward displacement.
930
200 mm 80 N
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*12–36. The beam is subjected to the loads shown. Determine the equation of the elastic curve. EI is constant.
4 kip
2 kip
4 kip⭈ft
A B x
8 ft
M = -(-2.5) 6 x - 0 7 - 2 6 x - 8 7 - 4 6 x - 16 7 M = 2.5x - 2 6 x - 8 7 - 4 6 x - 16 7 Elastic curve and slope: EI
d2v = M = 2.5x - 2 6 x - 8 7 - 4 6 x - 16 7 dx2
EI
dv = 1.25x2 - 6 x - 8 7 2 - 2 6 x - 16 7 2 + C1 dx
EIv = 0.417x3 - 0.333 6 x - 8 7 3 - 0.667 6 x - 16 7 3 + C1x + C2
(1)
Boundary conditions: v = 0
at
x = 0
From Eq. (1), C2 = 0 v = 0
at
x = 24 ft
0 = 5760 - 1365.33 - 341.33 + 24C1 C1 = -169 v =
1 C 0.417x3 - 0.333 6 x - 8 7 3 - 0.667 6 x - 16 7 3 - 169x D kip # ft3 Ans. EI
931
8 ft
8 ft
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•12–37.
Determine the deflection at each of the pulleys C, D, and E. The shaft is made of steel and has a diameter of 30 mm. The bearings at A and B exert only vertical reactions on the shaft. Est = 200 GPa.
C
E
D
A
B
250 mm
250 mm
250 mm
250 mm
M = -(-180) 6 x - 0 7 - 150 6 x - 0.25 7 150 N
- 60 6 x - 0.5 7 -150 6 x - 0.75 7 M = 180x - 150 6 x - 0.25 7 - 60 6 x - 0.5 7 -150 6 x - 0.75 7 Elastic curve and slope: EI
d2v = M = 180x - 150 6 x - 0.25 7 - 60 6 x - 0.5 7 dx2 -150 6 x - 0.75 7
EI
dv = 90x2 - 75 6 x - 0.25 7 2 - 30 6 x - 0.50 7 2 dx - 75 6 x - 0.75 7 2 + C1
(1)
EIv = 30x3 - 25 6 x - 0.25 7 3 - 10 6 x - 0.50 7 3 - 25 6 x - 0.75 7 3 + C1x + x2
(2)
Boundary conditions: v = 0
at
x = 0
From Eq. (2) C2 = 0 v = 0
x = 1.0 m
at
0 = 30 - 10.55 - 1.25 - 0.39 + C1 C1 = -17.8125 dv 1 = C 90x2 - 75 6 x - 0.25 7 2 - 30 6 x - 0.5 7 2 dx EI - 75 6 x - 0.75 7 2 - 17.8125 D v =
(3)
1 C 30x3 - 25 6 x - 0.25 7 3 - 10 6 x - 0.5 7 3 EI - 25 6 x - 0.75 7 3 - 17.8125x D
vC = v `
= x = 0.25m
-3.984 -3.984 = = -0.000501 m EI 200 A 109 B p4 (0.015)4 Ans.
= -0.501 mm vD = v `
x = 0.5m
vE = v `
x = 0.75 m
-5.547
=
=
200 A 109 B p4 (0.015)4
= -0.000698 m = -0.698 mm
-3.984 = -0.501 mm EI
Ans.
Ans. (symmetry check !)
The negative signs indicate downward displacement.
932
60 N
150 N
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12–38. The shaft supports the two pulley loads shown. Determine the equation of the elastic curve.The bearings at A and B exert only vertical reactions on the shaft. EI is constant.
A
B
x 20 in.
20 in. 40 lb
M = -10 6 x - 0 7 -40 6 x - 20 7 -(-110) 6 x - 40 7 M = -10x - 40 6 x - 20 7 + 110 6 x - 40 7 Elastic curve and slope: EI
d2v = M dx2
EI
d2v = -10x - 40 6 x - 20 7 + 110 6 x - 40 7 dx2
EI
dv = -5x2 - 20 6 x - 20 7 2 + 55 6 x - 40 7 2 + C1 dx
EIv = -1.667x3 - 6.667 6 x - 20 7 3 + 18.33 6 x - 40 7 3 + C1x + C2 (1) Boundary conditions: v = 0 at x = 0 From Eq. (1): C2 = 0 v = 0 at x = 40 in. 0 = -106,666.67 - 53,333.33 + 0 + 40C1. C1 = 4000 v =
1 C -1.67x3 - 6.67 6 x - 20 7 3 + 18.3 6 x - 40 7 3 + 4000x D lb # in3 EI
933
Ans.
20 in. 60 lb
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12–39. Determine the maximum deflection of the simply supported beam. E = 200 GPa and I = 65.0(106) mm4.
30 kN 15 kN
Support Reactions and Elastic Curve. As shown in Fig. a. Moment Function. From Fig. a, we obtain
A
M = -(-25)(x - 0) - 30(x - 2) - 15(x - 4)
2m
= 25x - 30(x - 2) - 15(x - 4) Equations of Slope and Elastic Curve. EI
d2v = M dx2
EI
d2v = 25x - 30(x - 2) - 15(x - 4) dx2
EI
dv = 12.5x2 - 15(x - 2)2 - 7.5(x - 4)2 + C1 dx
(1)
EIv = 4.1667x3 - 5(x - 2)3 - 2.5(x - 4)3 + C1x + C2
(2)
Boundary Conditions. At x = 0, v = 0. Then, Eq. (2) gives 0 = 0 - 0 - 0 + C1(0) + C2
C2 = 0
At x = 6 m, v = 0. Then Eq. (2) gives 0 = 4.1667 A 63 B - 5(6 - 2)3 - 2.5(6 - 4)3 + C1(6) + C2 C1 = -93.333 kN # m3 Substituting the value of C1 into Eq. (1), dv 1 = c12.5x2 - 15(x - 2)2 - 7.5(x - 4)2 - 93.333 d dx EI Assuming that
B
dv = 0 occurs in the region 2 m 6 x 6 4 m. Then dx
dv 1 = 0 = c12.5x2 - 15(x - 2)2 - 93.333 d dx EI 12.5x2 - 15(x - 2)2 - 93.333 = 0 2.5x2 - 60x + 153.333 = 0 Solving for the root 2 m 6 x 6 4 m, x = 2.9079 ft O.K.
934
2m
2m
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12–39. Continued Substituting the values of C1 and C2 into Eq. (2), v =
1 c4.1667x3 - 5(x - 2)3 - 2.5(x - 4)3 - 93.333x d EI
vmax occurs at x = 2.9079 m, where
Ans.
dv = 0. Thus, dx
vmax = v|x = 2.9079 ft =
1 c4.1667 A 2.90793 B - 5(2.9079 - 2)3 - 0 - 93.333(2.9079) d EI
= -
172.69 A 103 B 172.69kN # m3 = EI 200 A 109 B C 65.0 A 10 - 6 B D
= -0.01328 m = 13.3 mm T
Ans.
935
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*12–40. Determine the eqution of the elastic curve, the slope at A, and the deflection at B of the simply supported beam. EI is constant.
M0
M0
A B L 3
Support Reactions and Elastic Curve. As shown in Fig. a. Moment Function. M = -(-MO)a x = MO a x -
L 0 2 0 b - MO ax - L b 3 3
L 0 2 0 b - MO a x - L b 3 3
Equations of Slope and Elastic Curve. EI
d2v = M dx2
EI
0 d2v L 0 2 = M a x b M a x Lb O O 3 3 dx2
EI
dv L 2 = MO ax - b - MO a x - Lb + C1 dx 3 3
EIv
(1)
2 MO MO L 2 2 ax - b a x - Lb + C1x + C2 2 3 2 3
Boundary Conditions. Due to symmetry, EI(0) = MO a
L L - b - 0 + C1 2 3
(2)
dv L = 0 at x = . Then Eq. (1) gives dx 2 C1 = -
MOL 6
At x = 0, v = 0. Then, Eq. (2) gives EI(0) = 0 - 0 + C1(0) + C2
C2 = 0
Substituting the value of C1 into Eq. (1), 2 MO L 2 dv = B 6 ax - b - 6 ax - Lb - L R dx 6EI 3 3
At A, x = 0. Thus, uA =
MO MO L MOL dv = = C 6(0) - 6(0) - L D = ` dx x = 0 6EI 6EI 6EI
Ans.
Substituting the values of C1 and C2 into Eq. (2), v =
MO L 2 2 2 B 3 a x - b - 3 ax - L b - Lx R 6EI 3 3
At B, x =
Ans.
L . Thus, 3
vB = v|x = L3 =
= -
MO L B 3(0) - 3(0) - L a b R 6EI 3 MOL2 MOL2 = 18EI 18EI
Ans.
T
936
D
C L 3
L 3
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•12–41.
Determine the equation of the elastic curve and the maximum deflection of the simply supported beam. EI is constant.
M0
M0
A B
Support Reactions and Elastic Curve. As shown in Fig. a.
L 3
Moment Function. M = -(-MO)a x -
L 0 2 b - MO ax - L b 3 3
L 0 2 b - MO a x - L b 3 3
= MO a x -
Equations of Slope and Elastic Curve. EI
d2v = M dx2
EI
d2v L 0 2 = M a x b - MO a x - Lb O 3 3 dx2
EI
dv L 2 = MO ax - b - MO a x - Lb + C1 dx 3 3
EIv
(1)
2 MO MO L 2 2 ax - b a x - Lb + C1x + C2 2 3 2 3
Boundary Conditions. Due to symmetry, EI(0) = MO a
(2)
dv L = 0 at x = . Then Eq. (1) gives dx 2
L L - b - 0 + C1 2 3
C1 = -
MOL 6
At x = 0, v = 0. Then, Eq. (2) gives EI(0) = 0 - 0 + C1(0) + C2
C2 = 0
Substituting the values of C1 and C2 into Eq. (2), v =
2 MO L 2 2 B 3 a x - b - 3 ax - Lb - Lx R 6EI 3 3
vmax occurs at x =
L dv = 0. Then, , where 2 dx
vmax = v|x = L2 =
= -
Ans.
MO L L 2 L B3a - b - 0 - La b R 6EI 2 3 2
5MOL2 5MOL2 = T 72EI 72EI
Ans.
937
D
C L 3
L 3
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12–42. Determine the equation of the elastic curve, the slope at A, and the maximum deflection of the simply supported beam. EI is constant.
P
Moment Function.
= Px - Pax -
L 3
L 2 b - Pa x - Lb 3 3
L 2 b - Pax - Lb 3 3
Equations of Slope and Elastic Curve. EI
d2v = M dx2
EI
2 d2v L = Px - Pax - b - Pax - Lb 3 3 dx2
EI
2 P dv P L 2 P 2 = x2 ax - b a x - Lb + C1 dx 2 2 3 2 3
EIv =
(1)
3 P 3 P L 3 P 2 x ax - b a x - Lb + C1x + C2 6 6 3 6 3
Boundary Conditions. Due to symmetry,
EI(0) =
(2)
dv L = 0 at x = . Then Eq. (1) gives dx 2
P L 2 P L L 2 a b a - b - 0 + C1 2 2 2 2 3
C1 = -
PL2 9
At x = 0, v = 0. Then, Eq. (2) gives EI(0) = 0 - 0 - 0 + C1(0) + C2
C2 = 0
Substituting the value of C1 into Eq. (1), 2 P L 2 2 dv = B 9x2 - 9a x - b - 9ax - Lb - 2L2 R dx 18EI 3 3
At A, x = 0. Thus, uA =
dv P PL2 PL2 = = C 0 - 0 - 0 - 2L2 D = ` dx x = 0 18EI 9EI 9EI
Ans.
SubStituting the values of C1 and C2 into Eq. (2), v =
3 P L 3 2 B 3x3 - 3ax - b - 3ax - Lb - 2L2x R 18EI 3 3
vmax occurs at x =
= -
Ans.
L dv , where = 0. Then, 2 dx
vmax = v|x = L2 =
B
A
Support Reactions and Elastic Curve. As shown in Fig. a.
M = -(-P)(x - 0) - Pax -
P
P L 3 L L 3 L B 3 a b - 3a - b - 0 - 2L2 a b R 18EI 2 2 3 2
23PL3 23PL3 = T 648EI 648EI
Ans.
938
L 3
L 3
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12–43. Determine the maximum deflection of the cantilevered beam. The beam is made of material having an E = 200 GPa and I = 65.0(106) mm6.
15kN
30 kN/m
A
Support Reactions and Elastic Curve. As shown in Fig. a. Moment Function. From Fig. b, we obtain M = -(-37.5)(x - 0) - 67.5(x - 0)0 - a-
1.5 m
20 (x - 0)3 6
20 30 b (x - 1.5)3 - a - b(x - 1.5)2 6 2
= 37.5x - 67.5 -
10 3 10 x + (x - 1.5)3 + 15(x - 1.5)2 3 3
Equations of Slope and Elastic Curve. EI
d2v = M dx2
EI
d2v 10 3 10 = 37.5x - 67.5 x + (x - 1.5)3 + 15(x - 1.5)2 3 3 dx2
EI
dv 5 5 = 18.75x2 - 67.5x - x4 + (x - 1.5)4 + 5(x - 1.5)3 + C1 dx 6 6
EIv = 6.25x3 - 33.75x2 -
(1)
1 5 1 5 x + (x - 1.5)5 + (x - 1.5)4 + C1x + C2 (2) 6 6 4
Boundary Conditions. At x = 0,
dv = 0 Then Eq. (1) gives dx
0 = 0 - 0 - 0 + 0 + 0 + C1
C1 = 0
At x = 0, v = 0. Then Eq. (2) gives 0 = 0 - 0 - 0 + 0 + 0 + 0 + C2
C2 = 0
Substituting the values of C1 and C2 into Eq. (2), v =
1 1 1 5 c6.25x3 - 33.75x2 - x5 + (x - 1.5)5 + (x - 1.5)4 d EI 6 6 4
Ans.
vmax occurs at x = 3 m Thus vmax = v|x = 3 m =
1 1 1 5 c6.25 A 33 B - 33.75 A 32 B - A 35 B + (3 - 1.5)5 + (3 - 1.5)4 d EI 6 6 4
= -
167.91kN # m3 = EI
167.91 A 103 B
200 A 109 B c65.0 A 10 - 6 B d
= -0.01292 m = 12.9 mm T
Ans.
939
1.5 m
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*12–44. The beam is subjected to the load shown. Determine the equation of the elastic curve. EI is constant.
50 kN 3 kN/m
Support Reactions and Elastic Curve: As shown on FBD. Moment Function: Using discontinuity function,
B
A x
M = 24.6 6 x - 0 7 - 1.5 6 x - 0 7 2 - (-1.5) 6 x - 4 7 2
4m
- 50 6 x - 7 7 = 24.6x - 1.5x2 + 1.5 6 x - 47 2 - 50 6 x - 7 7 Slope and Elastic Curve: EI
EI EI
d2 y = M dx2
d2 y = 24.6x - 1.5x2 + 1.5 6 x - 47 2 - 50 6 x - 7 7 dx2
dy = 12.3x2 - 0.5x3 + 0.5 6 x - 4 7 3 - 25 6 x - 7 7 2 + C1 dx
[1]
EI y = 4.10x3 - 0.125x4 + 0.125 6 x - 4 7 4 - 8.333 6 x - 7 7 3 + C1x + C2
[2]
Boundary Conditions: y = 0 at x = 0. From Eq.[2], C2 = 0 y = 0 at x = 10 m. From Eq.[2], 0 = 4.10 A 103 B - 0.125 A 104 B + 0.125(10 - 4)4 - 8.333(10 - 7)3 + C1 (10) C1 = -278.7 The Elastic Curve: Substituting the values of C1 and C2 into Eq.[2], y =
1 {4.10x3 - 0.125x4 + 0.125 6 x - 4 7 4 EI - 8.33 6 x - 7 7 3 - 279x} kN # m3
Ans.
940
3m
3m
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•12–45.
The beam is subjected to the load shown. Determine the displacement at x = 7 m and the slope at A. EI is constant.
50 kN 3 kN/m
Support Reactions and Elastic Curve: As shown on FBD.
B
A
Moment Function: Using the discontinuity function.
x
2
M = 24.6 6 x - 0 7 - 1.5 6 x - 0 7 - (-1.5) 6 x - 4 7
4m
2
- 50 6 x - 7 7 = 24.6x - 1.5x2 + 1.5 6 x - 47 2 - 50 6 x - 7 7 Slope and Elastic Curve: EI
EI EI
d2 y = M dx2
d2 y = 24.6x - 1.5x2 + 1.5 6 x - 47 2 - 50 6 x - 7 7 dx2
dy = 12.3x2 - 0.5x3 + 0.5 6 x - 4 7 3 - 25 6 x - 7 7 2 + C1 dx
[1]
EI y = 4.10x3 - 0.125x4 + 0.125 6 x - 4 7 4 - 8.333 6 x - 7 7 3 + C1x + C2
[2]
Boundary Conditions: y = 0 at x = 0. From Eq.[2], C2 = 0 y = 0 at x = 10 m. From Eq.[2], 0 = 4.10 A 103 B - 0.125 A 104 B + 0.125(10 - 4)4 - 8.333(10 - 7)3 + C1 (10) C1 = -278.7 The Slope: Substituting the value of C1 into Eq.[1], 1 dy E 12.3x2 - 0.5x3 + 0.5 6 x - 4 7 3 - 25 6 x - 7 7 2 - 278.7 F kN # m2 = dx EI uA =
279kN # m2 dy 1 {0 - 0 + 0 - 0 - 278.7} = = ` dx x = 0 EI EI
Ans.
The Elastic Curve: Substituting the values of C1 and C2 into Eq.[2], y =
1 E 4.10x3 - 0.125x4 + 0.125 6 x - 4 7 4 - 8.33 6 x - 7 7 3 EI - 278.7x F kN # m3
y |x = 7 m =
1 E 4.10 A 73 B - 0.125 A 74 B + 0.125(7 - 4)4 - 0 - 278.7(7) F kN # m3 EI
= -
835 kN # m3 EI
Ans.
941
3m
3m
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12–46. Determine the maximum deflection of the simply supported beam. E = 200 GPa and I = 65.0(106) mm4.
20 kN
15 kN/m
Support Reactions and Elastic Curve. As shown in Fig. a. A
Moment Function. From Fig. b, we obtain M = -(-22.5)(x - 0) - 20(x - 1.5) -
15 5 (x - 3)2 - a - b(x - 3)3 2 6
= 22.5x - 20(x - 1.5) - 7.5(x - 3)2 +
1.5 m
5 (x - 3)3 6
Equations of Slope and Elastic Curve. EI
d2v = M dx2
EI
d2v 5 = 22.5x - 20(x - 1.5) - 7.5(x - 3)2 + (x - 3)3 2 6 dx
EI
dv 5 (x - 3)4 + C1 = 11.25x2 - 10(x - 1.5)2 - 2.5(x - 3)3 + dx 24
EIv = 3.75x3 -
(1)
10 1 (x - 1.5)3 - 0.625(x - 3)4 + (x - 3)5 + C1x + C2 (2) 3 24
Boundary Conditions. At x = 0, v = 0. Then, Eq. (2) gives 0 = 0 - 0 - 0 + 0 + C1(0) + C2
C2 = 0
At x = 6 m, v = 0. Then Eq. (2) gives 0 = 3.75 A 63 B -
10 1 (6 - 1.5)3 - 0.625(6 - 3)4 + (6 - 3)5 + C1(6) + C2 3 24
C1 = -77.625 kN # m2 Substituting the value of C1 into Eq. (1), dv 1 5 = c11.25x2 - 10(x - 1.5)2 - 2.5(x - 3)3 + (x - 3)4 - 77.625 d dx EI 24 Assuming that
dv = 0 occurs in the region 1.5 m 6 x 6 3 m, then dx
dv 1 = 0 = c11.25x2 - 10(x - 1.5)2 - 0 + 0 - 77.625 d dx EI Solving for the root 1.5 m 6 x 6 3 m, x = 2.970 m O.K. Substituting the values of C1 and C2 into Eq. (2), v =
1 10 1 c3.75x3 (x - 1.5)3 - 0.625(x - 3)4 + (x - 3)5 - 77.625x d Ans. EI 3 24
vmax occurs at x = 2.970 m, where
dv = 0. Thus, dx
942
1.5 m
3m
B
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12–46.
Continued
vmax = v ƒ x = 2.970 m =
1 10 c3.75 A 2.9703 B (2.970 - 1.5)3 - 0 + 0 - 77.625(2.970) d EI 3
= -
142.89kN # m3 = EI
142.89 A 103 B
200 A 109 B c65.0 A 10 - 6 B d
= -0.01099 m = 11.0 mm T
Ans.
943
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12–47. The wooden beam is subjected to the load shown. Determine the equation of the elastic curve. If Ew = 12 GPa, determine the deflection and the slope at end B. M = -63 6 x - 0 7 0 - (-16) 6 x - 0 7 -
6 kN
4 kN
2 kN/m A
B x
2 6 x - 0 72 2
3m
1.5 m
1.5 m
2 - a - b 6 x - 3 7 2 - 4 6 x - 4.5 7 2
400 mm
M = -63 + 16x - x2 + 6 x - 3 7 2 - 4 6 x - 4.5 7
200 mm
Elastic curve and Slope: EI
d2v = M = -63 + 16x - x2 + 6 x - 3 7 2 - 4 6 x - 4.5 7 dx2
EI
1 dv x3 = -63x + 8x2 + 6 x - 3 7 3 - 2 6 x - 4.5 7 2 + C1 dx 3 3
EIv = -31.5x2 +
(1)
1 8 3 x4 2 x + 6 x - 3 7 4 - 6 x - 4.5 7 3 3 12 12 3
+ C1x + C2
(2)
Boundary condition: dv = 0 dx
at
x = 0
From Eq. (1), C1 = 0 v = 0
x = 0
at
From Eq. (2), C2 = 0 1 x3 1 dv = c -63x + 8x2 + 6 x - 3 7 3 - 2 6 x - 4.5 7 2 d dx EI 3 3 v =
(3)
1 8 x4 1 c -31.5x2 + x3 + 6 x - 3 74 EI 3 12 12 -
2 6 x - 4.5 7 3 d kN # m3 (4) 3
Ans.
1 (0.20)(0.40)3 = 1.067 A 10 - 3 B m4 12
I =
At point B, x = 6m
uB =
-157.5 A 103 B dv -157.5 = = = -0.0123 rad = -0.705° Ans. ` dx x = 6m EI 12 A 103 B (1.067) A 10 - 3 B
The negative sign indicates clockwise rotation.
vB =
-661.5 A 103 B -661.5 = = -0.0517m = -51.7 mm EI 12 A 103 B (1.067) A 10 - 3 B
944
Ans.
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*12–48. The beam is subjected to the load shown. Determine the slopes at A and B and the displacement at C. EI is constant.
30 kN
A
The negative sign indicates downward displacement.
C
Moment Function: Using the discontinuity function,
3m
M = 66.75 6 x - 0 7 -6 6 x - 0 7 2 - 30 6 x - 3 7 = 66.75x - 6x2 - 30 6 x - 3 7 Slope and Elastic Curve:
EI EI
d2y = M dx2
d2y = 66.75x - 6x2 - 30 6 x - 3 7 dx2
dy = 33.375x2 - 2x3 - 15 6 x - 3 7 2 + C1 dx
[1]
EI y = 11.125x3 - 0.5x4 - 5 6 x - 3 7 3 + C1x + C2
[2]
Boundary Conditions: y = 0 at x = 0. From Eq.[2], C2 = 0 y = 0 at x = 8 m. From Eq.[2], 0 = 11.125 A 83 B - 0.5 A 84 B - 5(8 - 3)3 + C1 (8) C1 = -377.875 The Slope: Substituting the value of C1 into Eq.[1], 1 dy = E 33.375x2 - 2x3 - 15 6 x - 3 7 2 - 377.875 F kN # m2 dx EI uA =
dy 1 378 kN # m2 = {0 - 0 - 0 - 377.875} = ` dx x = 0 EI EI
uB =
dy ` dx x = 8 m
=
1 E 33.375 A 82 B - 2 A 83 B - 15(8 - 3)2 - 377.875 F EI
=
359 kN # m2 EI
Ans.
Ans.
The Elastic Curve: Substituting the values of C1 and C2 into Eq.[2], y =
1 E 11.125x3 - 0.5x4 - 5 6 x - 3 7 3 - 377.875 x F kN # m3 EI
yC = y|x = 3 m =
1 E 11.125 A 33 B - 0.5 A 34 B - 0 - 377.875(3) F EI
= -
B
x
Support Reactions and Elastic Curve: As shown on FBD.
EI
12 kN/m
874 kN # m3 EI
Ans.
945
5m
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•12–49.
Determine the equation of the elastic curve of the simply supported beam and then find the maximum deflection. The beam is made of wood having a modulus of elasticity E = 1.5(103) ksi.
600 lb 500 lb/ft 3 in.
A
B 6 ft
Support Reactions and Elastic Curve. As shown in Fig. a. Moment Function. From Fig. b, we obtain M = -(-2400)(x - 0) - 600(x - 9) -
500 500 (x - 0)2 - ¢ ≤ (x - 6) 2 2
= 2400x - 600(x - 9) - 250x2 + 250(x - 6)2 Equations of Slope and Elastic Curve. EI
d2v = M dx2
EI
d2v = 2400x - 600(x - 9) - 250x2 + 250(x - 6)2 dx2
EI
dv 250 3 250 = 1200x2 - 300(x - 9)2 x + (x - 6)3 + C1 dx 3 3
EIv = 400x3 - 100(x - 9)3 -
125 4 125 x + (x - 6)4 + C1x + C2 6 6
Boundary Conditions. At x = 0, v = 0. Then Eq.(2) gives C2 = 0 At x = 12ft, v = 0. Then Eq.(2) gives 0 = 400 A 12 3 B - 100(12 - 9)3 -
125 125 (12)4 + (12 - 6)4 + C1(12) 6 6
C1 = -23625 lb # ft2 Substituting the value of C1 into Eq.(1), dv 1 250 3 250 = x + (x - 6)3 - 23625 R B 1200x2 - 300(x - 9)2 dx EI 3 3 Assuming that
dv = 0 occurs in the region 0 6 x 6 6 ft. Then dx
dv 1 250 3 = 0 = c1200x2 x - 23625 d dx EI 3 1200x2 -
250 3 x - 23625 = 0 3
Solving x = 5.7126 ft O.K.
946
(1) (2)
3 ft
3 ft
6 in.
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•12–49.
Continued
Substituting the values of C1 and C2 into Eq.(2), v =
1 125 4 125 c400x3 - 100(x - 9)3 x + (x - 6)4 - 23625x d EI 6 6
vmax occurs at x = 5.7126 ft, where
Ans.
dv = 0. Thus, dx
vmax = v|x = 5.7126 ft =
1 125 c400 A 5.71263 B - 0 A 5.71264 B + 0 - 23625(5.7126) d EI 6
= -
82.577.41lb # ft3 = EI
82577.41 A 12 3 B
1.5 A 106 B c
1 (3) A 63 B d 12
= -1.76 in = 1.76 in T
Ans.
947
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12–50. The beam is subjected to the load shown. Determine the equations of the slope and elastic curve. EI is constant.
2 kN/m
8 kN⭈m
A B
Support Reactions and Elastic Curve: As shown on FBD. x
Moment Function: Using the discontinuity function,
5m
1 1 M = 0.200 6 x - 0 7 - (2) 6 x - 0 7 2 - ( -2) 6 x - 5 7 2 2 2 - (-17.8) 6 x - 5 7 = 0.200x - x2 + 6 x - 5 7 2 + 17.8 6 x - 5 7 Slope and Elastic Curve: EI
EI EI
d2y = M dx2
d2y = 0.200x - x2 + 6 x - 5 7 2 + 17.8 6 x - 5 7 dx2
dy = 0.100x2 - 0.3333x3 + 0.3333 6 x - 5 7 3 + 8.90 6 x - 5 7 2 + C1 [1] dx
EI y = 0.03333x3 - 0.08333x4 + 0.08333 6 x - 5 7 4 + 2.9667 6 x - 5 7 3 + C1x + C2
[2]
Boundary Conditions: y = 0 at x = 0. From Eq.[2], C2 = 0 y = 0 at x = 5 m. From Eq.[2], 0 = 0.03333 A 53 B - 0.08333 A 54 B + 0 + 0 + C1 (5) C1 = 9.5833 The Slope: Substituting the value of C1 into Eq.[1], dy 1 = E 0.100x2 - 0.333x3 + 0.333 6 x - 5 7 3 + 8.90 6 x - 5 7 2 dx EI + 9.58 F kN # m2
Ans.
948
3m
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12–51. The beam is subjected to the load shown. Determine the equation of the elastic curve. EI is constant. The Elastic Curve: Substituting the values of C1 and C2 into Eq.[2], y =
+ 9.58x F kN # m3
1.5 m
Ans.
6 6 6 x - 0 7 2 -(-1.25) 6 x - 1.5 7 -(- ) 6 x - 1.5 7 2 2 2
- (-27.75) 6 x - 4.5 7 M = -3x2 + 1.25 6 x - 1.5 7 +3 6 x - 1.5 7 2 + 27.75 6 x - 4.5 7 Elastic curve and slope: EI
d2v = M dx2 = -3x2 + 1.25 6 x - 1.5 7 + 3 6 x - 1.5 7 2 + 27.75 6 x - 4.5 7
EI
dv = -x3 + 0.625 6 x - 1.5 7 2 + 6 x - 1.5 7 3 dx + 13.875 6 x - 4.5 7 2 + C1
EIv = -0.25x4 + 0.208 6 x - 1.5 7 3 + 0.25 6 x - 1.5 7 4 + 4.625 6 x - 4.5 7 3 + C1x + C2
(1)
Boundary conditions: v = 0
at
x = 1.5 m
From Eq.(1) 0 = -1.266 + 1.5C1 + C2 1.5C1 + C2 = 1.266 v = 0
at
(2)
x = 4.5 m
From Eq.(1) 0 = -102.516 + 5.625 + 20.25 + 4.5C1 + C2 4.5C1 + C2 = 76.641
(3)
Solving Eqs. (2) and (3) yields: C1 = 25.12 C2 = -36.42 v =
Ans.
1 C -0.25x4 + 0.208 6 x - 1.5 7 3 + 0.25 6 x - 1.5 7 4 EI + 4.625 6 x - 4.5 7 3 + 25.1x - 36.4 D kN # m3
949
B
A
1 E 0.0333x3 - 0.0833x4 + 0.0833 6 x - 5 7 4 + 2.97 6 x - 5 7 3 EI
M = -
20 kN
6 kN/m
3m
1.5 m
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*12–52. The wooden beam is subjected to the load shown. Determine the equation of the elastic curve. Specify the deflection at the end C. Ew = 1.6(103) ksi.
0.8 kip/ft
1.5 kip
A
12 in.
C
B x 9 ft
M = -0.3 6 x - 0 7 -a-
1 1.6 a b 6 x - 0 7 3 -(-5.4) 6 x - 9 7 6 18
0.8 0.8 1 b 6 x - 9 72 - a b 6 x - 9 73 2 6 9
M = -0.3x - 0.0148x3 + 5.4 6 x - 9 7 +0.4 6 x - 9 7 2 + 0.0148 6 x - 9 7 3 Elastic curve and slope: EI
d2v = M = -0.3x - 0.0148x3 + 5.4 6 x - 9 7 +0.4 6 x - 9 7 2 dx2 +0.0148 6 x - 9 7 3
EI
dv = -0.15x2 - 0.003704x4 + 2.7 6 x - 9 7 2 + 0.1333 6 x - 9 7 3 dx + 0.003704 6 x - 9 7 4 + C1
EIv = -0.05x3 + 0.0007407x5 + 0.9 6 x - 9 7 3 + 0.03333 6 x - 9 7 4 + 0.0007407 6 x - 9 7 5 + C1x + C2
(1)
Boundary conditions: v = 0
at
x = 0
at
x = 9 ft
From Eq.(1) C2 = 0 v = 0 From Eq.(1) 0 = -36.45 - 43.74 + 0 + 0 + 0 + 9C1 C1 = 8.91 v =
1 C -0.05x3 - 0.000741x5 + 0.9 6 x - 9 7 3 + 0.0333 6 x - 9 7 4 EI + 0.000741 6 x - 9 7 5 + 8.91x D kip # ft3
Ans.
At point C, x = 18 ft
vC =
-612.29 A 12 3 B -612.29kip # ft3 = = -0.765 in. 1 EI 1.6 A 103 B A 12 B (6) A 12 3 B
Ans.
The negative sign indicates downward displacement.
950
9 ft
6 in.
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12–53. Determine the displacement at C and the slope at A of the beam.
8 kip/ft
Support Reactions and Elastic Curve: As shown on FBD.
C
Moment Function: Using the discontinuity function, M = -
x
1 1 8 (8) 6 x - 0 7 2 - a - b 6 x - 6 7 3 - (-88) 6 x - 6 7 2 6 9
= -4x2 +
6 ft
4 6 x - 6 7 3 + 88 6 x - 6 7 27
Slope and Elastic Curve: EI
EI EI
d2y = M dx2
d 2y 4 = -4x2 + 6 x - 6 7 3 + 88 6 x - 6 7 27 dx2
dy 4 1 = - x3 + 6 x - 6 7 4 + 44 6 x - 6 7 2 + C1 dx 3 27
1 1 44 EI y = - x4 + 6 x - 6 75 + 6 x - 6 7 3 + C1x + C2 3 135 3
[1] [2]
Boundary Conditions: y = 0 at x = 6 ft. From Eq.[2], 0 = -
1 4 A 6 B + 0 + 0 + C1 (6) + C2 3 432 = 6C1 + C2
[3]
y = 0 at x = 15 ft. From Eq.[2], 0 = -
1 1 44 (15 - 6)3 + (15 - 6)3 + C1 (15) + C2 A 154 B + 3 135 3 5745.6 = 15C1 + C2
[4]
Solving Eqs. [3] and [4] yields, C1 = 590.4
C2 = -3110.4
The Slope: Substitute the value of C1 into Eq.[1], dy 1 1 4 = 6 x - 6 7 4 + 44 6 x - 6 7 2 + 590.4 r kip # ft2 b - x3 + dx EI 3 27
uA =
302 kip # ft2 dy 1 4 = b - A 63 B + 0 + 0 + 590.4 r = ` dx x = 6 ft EI 3 EI
Ans.
The Elastic Curve: Substitute the values of C1 and C2 into Eq. [2], y =
1 1 1 44 6 x - 6 75 + 6 x - 6 73 b - x4 + EI 3 135 3 + 590.4x - 3110.4 r kip # ft3
yC = y |x = 0 =
B A
3110kip # ft3 1 {-0 + 0 + 0 + 0 - 3110.4} kip # ft3 = EI EI
951
Ans.
9 ft
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12–54. The beam is subjected to the load shown. Determine the equation of the elastic curve. EI is constant.
6 kip/ft B A x 9 ft
M = -
1 16 1 10 a b 6 x - 0 7 3 -(-77.4) 6 x - 9 7 - a - b a b 6 x - 9 7 3 6 24 6 15
M = -0.1111x3 + 77.4 6 x - 9 7 +0.1111 6 x - 9 7 3 Elastic curve and slope: EI
d2v = M = -0.1111x3 + 77.4 6 x - 9 7 +0.1111 6 x - 9 7 3 dx2
EI
dv = -0.02778x4 + 38.7 6 x - 9 7 2 + 0.02778 6 x - 9 7 4 + C1 dx
EIv = -0.005556x5 + 12.9 6 x - 9 7 3 + 0.005556 6 x - 9 7 5 + C1x + C2
(1)
Boundary conditions: v = 0
at
x = 9 ft
From Eq.(1) 0 = -328.05 + 0 + 0 + 9C1 + C2 9C1 + C2 = 328.05 v = 0
at
(2)
x = 24 ft
0 = -44236.8 + 43537.5 + 4218.75 + 24C1 + C2 24C1 + C2 = -3519.45
(3)
Solving Eqs. (2) and (3) yields, C1 = -256.5 C2 = 2637 v =
1 C -0.00556x5 + 12.9 6 x - 9 7 3 + 0.00556 6 x - 9 7 5 EI - 265.5x + 2637 D kip # ft3
Ans.
952
15 ft
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12–55. Determine the slope and deflection at C. EI is constant.
15 kip
A C B 30 ft
|tB>A|
uA =
30 -33 750 1 -225 a b(30)(10) = 2 EI EI
tB>A =
1125 EI
uA = uC>A =
1 -225 -5062.5 5062.5 1 -225 a b(30) + a b(15) = = 2 EI 2 EI EI EI
uC = uC>A + uA uC =
5062.5 1125 3937.5 = EI EI EI
¢ C = |tC>A| tC>A =
¢C =
Ans.
45 |t | 30 B>A
1 -225 1 225 101 250 ab(30)(25) + a b(15)(10) = 2 EI 2 EI EI 45 33 750 50 625 101.250 b = a EI 30 EI EI
Ans.
953
15 ft
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*12–56. Determine the slope and deflection at C. EI is constant.
10 kN
A C B 6m
Referring to Fig. b, |uC>A| =
1 30 135 kN # m2 a b A9B = 2 EI EI
|tB>A| =
6 1 30 180 kN # m3 c a b A6B d = 3 2 EI EI
|tC>A| = a =
6 1 30 2 1 30 + 3b c a b A 6 B d + c (3) d c a b A3B d 3 2 EI 3 2 EI
540 kN # m3 EI
From the geometry shown in Fig. b, uA =
|tB>A| =
6
180>EI 30 kN # m2 = 6 EI
Here, + buC = uA + uC>A uC = uC =
135 30 + EI EI
105 kN # m2 EI
uC
Ans.
9 yC = 2 tC>A 2 - 2 tB>A 2 a b 6 =
540 180 9 a b EI EI 6
=
270 kN # m3 T EI
Ans.
954
3m
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•12–57. Determine the deflection of end B of the cantilever beam. E is constant.
P
P
B
A L 2
Support Reactions and
M Diagram. As shown in Fig. a. EI
Moment Area Theorem. Since A is a fixed support, uA = 0. Referring to the geometry of the elastic curve, Fig. b, uB = |uB>A| =
=
1 3PL PL L 1 PL L + B R¢ ≤ + B R¢ ≤ 2 2EI 2EI 2 2 2EI 2
5PL2 8 EI
Ans.
¢ B = |tB>A| - ¢
=
7PL3 16EI
3L PL L L 5L 1 PL L L 1 PL ≤¢ ≤¢ ≤ + B ¢ ≤¢ ≤R + B ¢ ≤¢ ≤R 4 2EI 2 6 2 EI 2 3 2 2EI 2 Ans.
T
955
L 2
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12–58. Determine the slope at A and the maximum deflection. EI is constant.
20 kip⭈ft
20 kip⭈ft
A 6 ft
Point D is located at the mid span of the beam. Due to symmetry, the slope at D is zero. Referring to Fig. b, |uD>A| = a
120 kip # ft2 20 b(6) = EI EI
|tD>A| = 3 a
360 kip # ft3 20 b A6B = EI EI
|tC>D| = 6a
1440 kip # ft3 20 b A 12 B = EI EI
From the geomtry shown in Fig. b uA = |uD>A| =
120 kip # ft2 EI
uA
Ans.
y D = uA(6) - |tD>A| =
360 120 (6) EI EI
=
360 kip # ft3 EI
c
yC = |tC>D| - 4D =
1440 360 EI EI
=
1080 kip # ft EI
T (max)
Ans.
956
C
B 12 ft
6 ft
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12–59. Determine the slope and deflection at C. EI is constant.
20 kip⭈ft
20 kip⭈ft
A
Referring to Fig. b, |uC>A| = a
360 kip # ft2 20 b A 18 B = EI EI
|uB>A| = 6a |tC>A| = 9 a
6 ft
b
1440 kip # ft3 20 b A 12 B = = EI EI
3240 kip # ft3 20 b A 18 B = EI EI
From the geometry shown in Fig. b uA =
|tB>A| 12
=
1440>EI 12
=
120 kip # ft2 EI
uA
Here, + b uC = uA + uC>A uC = -
uC =
360 120 + EI EI
240 kip # ft2 EI
yC = |tC>A| - |tB>A| a
uC
Ans.
18 b 12
=
1440 18 3240 a b EI EI 12
=
1080 kip # ft3 T EI
Ans.
957
C
B 12 ft
6 ft
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*12–60. If the bearings at A and B exert only vertical reactions on the shaft, determine the slope at A and the maximum deflection of the shaft. EI is constant.
50 lb⭈ft
B C 2 ft
Point E is located at the mid span of the shaft. Due to symmetry, the slope at E is zero. Referring to Fig. b, |uE>A| =
100 lb # ft2 50 (2) = EI EI
|tE>A| = (1)a
100 lb # ft3 50 b (2) = EI EI
Here, uA = |uE>A| =
100 lb # ft2 EI
uA
Ans.
ymax = uA (4) - |tE>A| =
100 100 (4) EI EI
=
300 lb # ft3 EI
50 lb⭈ft
A
Ans.
c
958
D 4 ft
2 ft
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•12–61.
Determine the maximum slope and the maximum deflection of the beam. EI is constant.
M0
B
A
M0L M0 L = a b = EI 2 2EI
uC>A
M0
L
uC = uC>A + uA 0 =
M0 L + uA 2EI
umax = uA =
M0 L -M0 L = 2EI 2EI
¢ max = |tB>C| =
Ans.
M0 L2 M0 L L a ba b = EI 2 4 8EI
Ans.
12–62. Determine the deflection and slope at C. EI is constant.
A
B
C M0
tB>A
M0L2 1 -M0 1 = a b(L)a b(L) = 2 EI 3 6EI
L
¢ C = |tC>A| - 2|tB>A| -M0 7M0L2 L L 1 -M0 a b(L)aL + b + a b (L) a b = 2 EI 3 EI 2 6EI
tC>A =
¢C =
uA =
7M0 L2 M0L2 5M0L2 - (2) a b = 6EI 6EI 6EI |tB>A| L
uC>A =
=
Ans.
M0L 6EI
M0 M0 3M0L 3M0L 1 ab(L) + a b(L) = = 2 EI EI 2EI 2EI
uC = uC>A + uA uC =
3M0L M0L 4M0L = 2EI 6EI 3EI
Ans.
959
L
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12–63. Determine the slope at A of the overhang beam. E = 200 GPa and I = 45.5(106) mm4.
30 kN 30 kN⭈m
M Support Reactions and Diagram. As shown in Fig. a. EI
A
Moment Area Theorem. Referring to Fig. b,
4m
1 30 30 1 1 |tB>A| = c (4) d B ¢ (4) R ≤ (4) R + c (4) d B 3 2 EI 2 EI
=
320 kN # m3 EI
From the geometry of the elastic curve, Fig. b, uA =
=
|tB>A| LAB
=
320>EI 80 kN # m2 = 4 EI
80 A 103 B
200 A 109 B C 45.5 A 10 - 6 B D
C B
Ans.
= 0.00879 rad
960
2m
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*12–64. Determine the deflection at C of the overhang beam. E = 200 GPa and I = 45.5(106) mm4.
30 kN 30 kN⭈m
1 1 30 1 30 |tB>A| = c (4) d B ¢ (4) R ≤ (4) R + c (4) d B 3 2 EI 2 EI
=
A
320 kN # m3 EI
4m
1 30 30 1 1 |tC>A| = c (4) + 2 d B ¢ (4) R ≤ (4) R + c (4) + 2 d B 3 2 EI 2 EI 1 60 2 + c (2) d B ¢ ≤ (2) R 3 2 EI
=
760 kN # m3 EI
¢ C = |tC>A| - |tB>A ¢
C B
L ≤ LAB
=
760 320 6 ¢ ≤ EI EI 4
=
280 A 103 B 280 kN # m3 = EI 200 A 109 B C 45.5 A 10 - 6 B D
= 0.03077 m = 30.8 mm T
Ans.
961
2m
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•12–65. Determine the position a of roller support B in terms of L so that the deflection at end C is the same as the maximum deflection of region AB of the overhang beam. EI is constant.
P L C
A B
M Support Reactions and Diagram. As shown in Fig. a. EI Moment Area Theorem. Referring to Fig. b, |tB>A| =
Pa2(L - a) a 1 P(L - a) B ¢ ≤ (a) R = 3 2 EI 6EI
|tC>A| = aL -
=
a
2(L - a) 1 P(L - a) 1 P(L - a) 2 ab B ¢ ≤ (a) R + B ¢ ≤ (L - a) R 3 2 EI 3 2 EI
P(L - a) A 2L2 - aL B 6EI
From the geometry shown in Fig. b, ¢ C = |tC>A| -
=
=
uA =
|tB>A| a
L
PL(L - a) A 2L - a B 6EI
-
Pa2 (L - a) L ¢ ≤ a 6EI
PL(L - a)2 3EI |tB>A| a
Pa2(L - a) Pa(L - a) 6EI = = a 6EI
The maximum deflection in region AB occurs at point D, where the slope of the elastic curve is zero (uD = 0). Thus, |uD>A| = uA Pa(L - a) 1 P(L - a) x R (x) = B 2 EIa 6EI x =
23 a 3
Also, ¢ D = |t4>D| = a
23Pa2(L - a) 2 23 1 P(L - a) 23 23 ab B c a ab d R a ab = 9 2 EIa 3 3 27EI
It is required that ¢C = ¢D PL(L - a)2 23Pa2(L - a) = 3EI 27EI 23 2 a + La - L2 = 0 9 Solving for the positive root, a = 0.858L
Ans.
962
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12–66. Determine the slope at A of the simply supported beam. EI is constant. Support Reactions and
P
M Diagram. As shown in Fig. a. EI
A
Moment Area Theorem.
2L 3
1 2PL 2 2 1 2PL L 5 tB>A = a Lb c a b a L b d + Lc a ba bd 9 2 9EI 3 9 2 9EI 3 =
4PL3 81EI
Referring to the geometry of the elastic curve, Fig. b,
uA
B
4PL3 |tB>A| 81EI 4PL2 = = = L L 81EI
Ans.
963
L 3
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12–67. The beam is subjected to the load P as shown. Determine the magnitude of force F that must be applied at the end of the overhang C so that the deflection at C is zero. EI is constant.
F P B
A
C 3
3
tB>A =
1 Fa 2 2Fa 1 Pa Pa a b(2a)(a) + a b(2a)a ab = 2 2EI 2 EI 3 2EI 3EI
tC>A =
1 -Fa 2a 1 -Fa 2a 1 Pa a b(2a)(2a) + a b(2a)a a + b + a b(a) a b 2 2EI 2 EI 3 2 EI 3
=
a
a
a
2Fa3 Pa3 EI EI
¢ C = tC>A -
3 t = 0 2 B>A
Pa3 2Fa3 3 Pa3 2Fa3 - a b = 0 EI EI 2 2EI 3EI F =
P 4
Ans.
M0 ⫽ Pa
*12–68. If the bearings at A and B exert only vertical reactions on the shaft, determine the slope at A and the maximum deflection.
tB>A
uA =
A C a
a Pa 17Pa3 1 Pa b(a)a3a + b + a b (2a)(a + a) = = a 2 EI 3 EI 3EI |tB>A| 4a
=
B
Ans.
Assume ¢ max is at point E located at 0 6 x 6 2a uE>A =
1 Pa Pa Pa2 Pax a b(a) + a b(x) = + 2 EI EI 2EI EI
uE = 0 = uE>A + uA 0 =
Pax -17Pa2 Pa2 + + a b 2EI EI 12EI
x =
11 a 12
¢ max = |tB>E| = a
(2a Pa 11 b a 2a ab c EI 12 2
11 12 a)
+ ad =
2a P
17Pa2 12EI
481Pa3 288EI
964
Ans.
D a
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•12–69.
The beam is subjected to the loading shown. Determine the slope at A and the displacement at C. Assume the support at A is a pin and B is a roller. EI is constant.
P
Support Reactions and Elastic Curve: As shown.
P
A
C a
M/EI Diagram: As shown.
P
a
B a
a
Moment - Area Theorems: Due to symmetry, the slope at midspan (point C) is zero. Hence the slope at A is uA = uA>C =
=
1 3Pa 3Pa 1 Pa a b(a) + a b(a) + a b(a) 2 2EI 2EI 2 2EI 5Pa2 2EI
Ans.
The displacement at C is ¢ C = tA>C =
=
2a 3Pa a 1 Pa 2a 1 3Pa a b (a)a b + a b aa + b + a b(a)a a + b 2 2EI 3 2EI 2 2 2EI 3 19Pa3 T 6EI
Ans.
12–70. The shaft supports the gear at its end C. Determine the deflection at C and the slopes at the bearings A and B. EI is constant.
A
L –– 2
1 -PL L L -PL3 = a ba ba b = 2 2EI 2 6 48EI
tB>A
L -PL3 1 -PL a b(L)a b = 2 2EI 2 8EI
tC>A =
L ¢ C = |tC>A| - a L b|tB>A| 2
=
uA =
PL3 PL3 PL3 - 2a b = 8EI 48EI 12EI |tB>A| L 2
uB>A =
=
PL3 48 EI L 2
=
Ans.
PL2 24EI
Ans.
1 -PL L -PL2 PL2 a ba b = = 2 2EI 2 8EI 8EI
uB = uB>A + uA uB =
B
PL2 PL2 PL2 = 8EI 24EI 12EI
Ans.
965
C L –– 2
P
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12–71. The shaft supports the gear at its end C. Determine its maximum deflection within region AB. EI is constant. The bearings exert only vertical reactions on the shaft.
A
B
L –– 2
uD>A =
C L –– 2
P
tB>A
A L2 B
1 Px a bx = 2 EI
PL A B A 2EI B A 13 B A L2 B
1 L 2 2
A L2 B
;
x = 0.288675 L
¢ max =
2 1 P(0.288675 L) a b (0.288675 L)a b(0.288675 L) 2 EI 3
¢ max =
0.00802PL3 EI
Ans.
*12–72. Determine the value of a so that the displacement at C is equal to zero. EI is constant.
P
P A
C
B
Moment-Area Theorems: (¢ C)1 = (tA>C)1 =
(tB>A)2 =
a
1 Pa 2 PaL2 ab(L)a Lb = 2 EI 3 3EI
(tC>A)2 = a -
(¢ C)2 =
1 PL L L PL3 a ba ba b = 2 4EI 2 3 48EI
Pa Pa L L 1 L L 5PaL2 b¢ ≤¢ ≤ + ab¢ ≤¢ ≤ = 2EI 2 4 2 2EI 2 3 48EI
PaL2 1 1 PaL2 5PaL2 |(tB>A)2| - |(tC>A)2| = ¢ = ≤ 2 2 3EI 48EI 16EI
Require, ¢ C = 0 = (¢ C)1 - (¢ C)2 0 =
PaL2 PL3 48EI 16EI
a =
L 3
Ans.
966
L 2
L 2
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•12–73.
The shaft is subjected to the loading shown. If the bearings at A and B only exert vertical reactions on the shaft, determine the slope at A and the displacement at C. EI is constant.
A
B C a
M/EI Diagram: As shown. Moment-Area Theorems: M0 M0 1 a 1 a ab(a)a b + a b(a)aa + b 2 EI 3 2 EI 3
tB>A =
= -
5M0 a2 6EI
M0 M0 a2 a 1 ab(a)a b = 2 EI 3 6EI
tC>A =
The slope at A is
uA =
5M0a 2
|tB>A| L
=
6EI
2a
=
5M0 a 12EI
Ans.
The displacement at C is, ¢C = `
1 t ` - |tC>A| 2 B>A
=
M0 a2 1 5M0 a2 ¢ ≤ 2 6EI 6EI
=
M0 a2 4EI
M0
M0
Ans.
c
967
a
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12–74. Determine the slope at A and the maximum deflection in the beam. EI is constant.
12 kip 24 kip⭈ft
A
B
6 ft
Here, tB>A = 20 c
=
1 12 36 1 24 1 60 a b(6) d + 12 c (12) d + 10 c a b(12) d + 4c a b(6) d 2 EI EI 2 EI 2 EI
8064 kip # ft3 EI
From the geometry of the elastic curve diagram, Fig. b, uA =
tB>A =
L
8064>EI 336 kip # ft2 = uA 24 EI
Ans.
Assuming that the zero slope of the elastic curve occurs in the region 6ft 6 x = 18ft such as point C where the maximum deflection occurs, then uC>A = uA 1 12 36 1 2x 336 a b (6) + a bx + a b(x) = 2 EI EI 2 EI EI x2 + 36x - 300 = 0 Solving for the root 0 6 x 6 12 ft, x = 6.980 ft O.K. Thus, ymax = tA>C = 4 c
=
1 12 36 1 a b(6) d + 9.490 c (6.980) d + 10.653 c (13.960)(6.980) d 2 EI EI 2
3048 kip # ft3 T EI
Ans.
968
12 ft
6 ft
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12–75. The beam is made of a ceramic material. In order to obtain its modulus of elasticity, it is subjected to the elastic loading shown. If the moment of inertia is I and the beam has a measured maximum deflection ¢ , determine E. The supports at A and D exert only vertical reactions on the beam.
P
P B
C
A
D a
a L
Moment-Area Theorems: Due to symmetry, the slope at midspan (point E) is zero. Hence the maximum displacement is, ¢ max = tA>E = a =
L - 2a 1 Pa 2 Pa L - 2a ba b aa + b + a b(a)a ab EI 2 4 2 EI 3
Pa A 3L2 - 4a2 B 24EI
Require, ¢ max = ¢ , then, ¢ =
Pa A 3L2 - 4a2 B 24EI
E =
Pa A 3L2 - 4a2 B 24¢I
Ans.
*12–76. The bar is supported by a roller constraint at B, which allows vertical displacement but resists axial load and moment. If the bar is subjected to the loading shown, determine the slope at A and the deflection at C. EI is constant.
P C A
B
L — 2
uA>B =
1 PL L PL L 3PL2 a ba b + a b = 2 2EI 2 2EI 2 8EI
uA = uA>B uA =
3PL2 8EI
Ans.
tA>B =
1 PL L L PL L L L 11PL3 a ba ba b + a ba + b = 2 2EI 2 3 2EI 2 2 4 48EI
tC>B =
PL L L PL3 a ba b = 2EI 2 4 16EI
¢ C = tA>B - tC>B =
11PL3 PL3 PL3 = 48EI 16EI 6EI
Ans.
969
L — 2
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•12–77. The bar is supported by the roller constraint at C, which allows vertical displacement but resists axial load and moment. If the bar is subjected to the loading shown, determine the slope and displacement at A. EI is constant.
P
B C
A
Support Reactions and Elastic Curve: As shown. a
M/EI Diagram: As shown.
2a
Moment-Area Theorems: uA>C = a -
Pa 1 Pa 5Pa2 b (2a) + a b(a) = EI 2 EI 2EI
tB>C = a -
Pa 2Pa3 b(2a)(a) = EI EI
tA>C = a -
1 2 Pa Pa 13Pa3 b(2a)(2a) + a b (a)a ab = EI 2 EI 3 3EI
Due to the moment constraint, the slope at support C is zero. Hence, the slope at A is uA = |uA>C| =
5Pa2 2EI
Ans.
and the displacement at A is ¢ A = |tA>C| - |tB>C| 2Pa3 7Pa3 13Pa3 = 3EI EI 3EI
=
Ans.
T
12–78. The rod is constructed from two shafts for which the moment of inertia of AB is I and of BC is 2I. Determine the maximum slope and deflection of the rod due to the loading. The modulus of elasticity is E.
P
L 2
uA>C =
1 -PL L -PL L -5PL2 5PL2 1 -PL L a ba b + a ba b + a ba b = = 2 2EI 2 2 4EI 2 4EI 2 16EI 16EI
uA = uA>C + uC umax = uA =
5PL2 5PL2 + 0 = 16EI 16EI
Ans.
¢ max = ¢ A = |tA>C| = `
1 -PL L L 1 -PL L L L a ba ba b + a ba ba + b 2 2 EI 2 3 2 4EI 2 2 3 + a
=
-PL L L L ba ba + b ` 4EI 2 2 4
3PL3 16EI
Ans.
970
C
B
A
L 2
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12–79. Determine the slope at point D and the deflection at point C of the simply supported beam. The beam is made of material having a modulus of elasticity E. The moment of inertia of segments AB and CD of the beam is I, while the moment of inertia of segment BC of the beam is 2I.
A L 4
Moment Area Theorem. Referring to Fig. b,
=
tC>D =
=
L L PL L 5L 1 PL L L 1 PL c a ba bd + c a bd + c a ba bd 6 2 4EI 4 2 8EI 2 6 2 4EI 4 PL3 16EI L 1 PL L c a ba bd 12 2 4EI 4 PL3 384EI
From the geometry of Fig. b,
uD =
|tA>D| L
PL3 PL2 18EI = = L 16EI
¢ C + tC>D =
Ans.
tA>D 4
PL3 PL 16EI = = 384EI 4 3
¢C
¢C =
D C
B
M Support Reactions and Diagram. As shown in Fig. a. EI
tA>D =
P
P
5PL3 384EI
Ans.
971
L 2
L 4
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*12–80. Determine the slope at point A and the maximum deflection of the simply supported beam. The beam is made of material having a modulus of elasticity E. The moment of inertia of segments AB and CD of the beam is I, while the moment of inertia of segment BC is 2I.
A L 4
Moment Area Theorem. Due to symmetry, the slope at the midspan of the beam, i.e., point E, is zero (uE = 0). Thus the maximum deflection occurs here. Referring to the geometry of the elastic curve, Fig. b,
=
1 PL L PL L a ba b + a b 2 4EI 4 8EI 4
PL2 16EI
¢ max = ¢ E = |tA>E| =
=
D C
B
M Support Reactions and Diagram. As shown in Fig. a. EI
uA = |uA>E| =
P
P
Ans. L 1 PL L 3 PL L Lc a bd + c a ba bd 8 8EI 4 6 2 4EI 4
13PL3 T 768EI
Ans.
972
L 2
L 4
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•12–81. Determine the position a of roller support B in terms of L so that deflection at end C is the same as the maximum deflection of region AB of the simply supported overhang beam. EI is constant.
Support Reactions and
A B a L
M Diagram. As shown in Fig. a. EI
Moment Area Theorem. Referring to Fig. b, |tB>A| =
MOa2 a 1 MO c a b (a) d = 3 2 EI 6EI
|tC>A| = aL =
2 1 MO L - a MO ab c a b (a) d + a bc (La) d 3 2 EI 2 EI
MO 2 A a + 3L2 - 3La B 6EI
From the geometry shown in Fig. b, ¢ C = |tC>A| -
uA
|tB>A| a
L
=
MO 2 MOa2 L a b A a + 3L2 - 3La B 6EI 6EI a
=
MO 2 A a + 3L2 - 4La B 6EI
MOa2 |tB>A| MO a 6EI = = = a a 6EI
The maximum deflection in region AB occurs at point D, where the slope of the elastic curve is zero (uD = 0). Thus, |uD>A| = uA MOa 1 MO a b (x)2 = 2 EIa 6EI x =
23 a 3
Also, ¢ D = |tA>D| =
23MOa2 2 23 1 MO 23 23 a≤ B a b¢ a≤ R ¢ a≤ = ¢ 3 3 2 EIa 3 3 27EI
It is required that ¢C = ¢D MO 2 23MO a2 A a + 3L2 - 4La B = 6EI 27EI 0.6151a2 - 4La + 3L2 = 0 Solving for the root 6 L, a = 0.865L
Ans.
973
C
M0
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12–82. The W10 * 15 cantilevered beam is made of A-36 steel and is subjected to the loading shown. Determine the slope and displacement at its end B.
3 kip/ft
B A
Here, uB = 冷uB>A冷 =
1 54 a b (6) 3 EI
=
108 kip # ft2 EI
6 ft
uC
For W 10 * 15 I = 68.9 in4, and for A36 steel E = 29.0 A 103 B ksi. Thus uB =
108 A 12 2 B
29 A 103 B (68.9)
= 0.00778 rad uB
Ans.
1 54 3 b (6) d yB = 冷 tB>A冷 = c (6) + 6 d c a 4 3 EI =
=
1134 kip # ft3 EI 1134 A 12 3 B
29 A 103 B (68.9)
= 0.981 in.
Ans.
T
974
6 ft
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12–83. The cantilevered beam is subjected to the loading shown. Determine the slope and displacement at C. Assume the support at A is fixed. EI is constant.
P w
Support Reactions and Elastic Curve: As shown.
A
C
B a
M/EI Diagrams: The M/EI diagrams for the uniform distributed load and concentrated load are drawn separately as shown.
a
Moment-Area Theorems: The slope at support A is zero. The slope at C is 2Pa 1 wa 2 1 ab (2a) + a b(a) 2 EI 3 2EI
uC = 冷uC>A冷 =
a2 (12P + wa) 6EI
=
Ans.
The displacement at C is ¢ C = 冷tC>A 冷 =
=
2Pa 4 1 wa 2 3 1 ab(2a) a ab + a b(a)a a + a b 2 EI 3 3 2EI 4 a3 (64P + 7wa) 24EI
Ans.
T
*12–84. Determine the slope at C and deflection at B. EI is constant.
w C A
B
Support Reactions and Elastic Curve: As shown.
a
M/EI Diagram: As shown. Moment-Area Theorems: The slope at support A is zero. The slope at C is uC = 冷uC>A冷 =
=
1 wa2 wa2 ab(a) + a b (a) 2 EI 2EI wa3 EI
Ans.
The displacement at B is ¢ B = 冷tB>A冷 =
1 wa2 2 wa2 a 1 wa2 3 ab(a)aa + ab + a b(a)a a + b + a b(a)a ab 2 EI 3 2EI 2 3 2EI 4
=
41wa4 24EI
Ans.
T
975
a
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•12–85.
Determine the slope at B and the displacement at C. The member is an A-36 steel structural tee for which I = 76.8 in4.
5 kip 1.5 kip/ft
B
A
C 3 ft
3 ft
Support Reactions and Elastic Curve: As shown. M/EI Diagrams: The M/EI diagrams for the uniform distributed load and concentrated load are drawn separately as shown. Moment-Area Theorems: Due to symmetry, the slope at midspan C is zero. Hence the slope at B is uB = 冷uB>C冷 =
=
2 6.75 1 7.50 a b (3) + a b (3) 2 EI 3 EI 24.75 kip # ft2 EI 24.75(144)
=
29.0 A 103 B (76.8)
= 0.00160 rad
Ans.
The dispacement at C is ¢ C = 冷tA>C冷 =
=
2 2 6.75 5 1 7.50 a b (3)a b (3) + a b (3) a b(3) 2 EI 3 3 EI 8 47.8125 kip # ft3 EI 47.8125(1728)
=
29.0 A 103 B (76.8)
= 0.0371 in.
Ans.
T
12–86. The A-36 steel shaft is used to support a rotor that exerts a uniform load of 5 kN兾m within the region CD of the shaft. Determine the slope of the shaft at the bearings A and B. The bearings exert only vertical reactions on the shaft.
uE>A =
5 kN/m A
1 75 4.6875 2 3.5156 4.805 a b(0.1) + a b(0.15) + a b(0.15) = 2 EI EI 3 EI EI
uA = uE>A =
4.805 = EI
4.805 = 0.00306rad = 0.175° 200 (109)(0.01)4
976
B
C 20 mm 100 mm
Ans.
40 mm 300 mm
D 20 mm 100 mm
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12–87. The W12 * 45 simply supported beam is made of A-36 steel and is subjected to the loading shown. Determine the deflection at its center C.
12 kip 50 kip⭈ft B
A C 12 ft
A ¢C B 1 = ¢ 2 (x) =
12 A 24 3 B 3456 PL3 = = T 48EI 48EI EI Mx A L2 - x2 B 6LEI
At point C, x =
A ¢C B 2 = =
L 2
MN A L2 B 6LEI
A L2 - A L2 B 2 B
50 A 24 2 B 1800 ML2 = = T 16EI 16EI EI
¢C = A ¢C B 1 + A ¢C B 2 = 5256(1728)
=
29 A 103 B (350)
3456 1800 5256 + = EI EI EI
= 0.895 in. T
Ans.
977
12 ft
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*12–88. The W10 * 15 cantilevered beam is made of A-36 steel and is subjected to the loading shown. Determine the displacement at B and the slope at B.
6 kip
4 kip
A
B 6 ft
Using the table in appendix, the required slopes and deflections for each load case are computed as follow: (¢ B)1 =
(uB)1 =
(¢ B)2 =
(uB)2 =
5(4) A 12 3 B 720 kip # in.3 5PL3 = = T 48EI 48EI EI 4 A 12 2 B 72 kip # in.2 PL2 = = 8EI 8EI EI
(uB)1
6 A 12 3 B 3456 kip # in.3 PL3 = = T 3EI 3EI EI 6 A 12 2 B 432 kip # in.2 PL2 = = 2EI 2EI EI
(uB)2
Then the slope and deflection at B are uB = (uB)1 + (uB)2 =
432 72 + EI EI
=
504 kip # ft2 EI
¢ B = (¢ B)1 + (¢ B)2 =
3456 720 + EI EI
=
4176 kip # in.3 EI
For A36 steel W10 * 15, I = 68.9 in4 And E = 29.0 A 103 B ksi uB =
504
29.0 A 103 B (68.9)
= 0.252(10-3) rad
Ans.
4176
¢B =
29.0 A 103 B (68.9)
= 0.00209 in
Ans.
978
6 ft
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•12–89.
Determine the slope and deflection at end C of the overhang beam. EI is constant.
w
A
C D a
Elastic Curves. The uniform distributed load on the beam is equivalent to the sum of the seperate loadings shown in Fig.a. The elastic curve for each seperate loading is shown Fig. a. Method of Superposition. Using the table in the appendix, the required slopes and deflections are (uC)1 = (uB)1 =
w(2a)3 wa3 wL3 = = 24EI 24EI 3EI
(¢ C)1 = (uB)1(a) =
wa3 wa4 (a) = 3EI 3EI
(uC)2 =
wa3 wL3 = 6EI 6EI
(¢ C)2 =
wL4 wa4 = 8EI 8EI
T
MOL (uC)3 = (uB)3 = = 3EI (¢ C)3 = (uB)3 (a) =
c
¢
wa2 2
≤ (2a)
3EI
wa3 wa4 (a) = 3EI 3EI
=
wa3 3EI
T
Then the slope and deflection of C are uC = (uC)1 + (uC)2 + (uC)3 = -
=
wa3 wa 3 wa3 + + 3EI 6EI 3EI
wa3 6EI
Ans.
¢ C = (¢ C)1 + (¢ C)2 + (¢ C)3 = -
=
wa4 wa 4 wa4 + + 3EI 8EI 3EI
wa4 T 8EI
Ans.
979
B a
a
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12–90. Determine the slope at A and the deflection at point D of the overhang beam. EI is constant.
w
A
C D a
Elastic Curves. The uniform distributed load on the deformation of span AB is equivalent to the sum of the seperate loadings shown in Fig. a. The elastic curve for each seperate loading is shown in Fig. a. Method of Superposition. Using the table in the appendix, the required slope and deflections are (uA)1 =
w(2a)3 wL3 wa3 = = 24EI 24EI 3EI
(¢ D)1 =
5w(2a)4 5wa4 5wL4 = = 384EI 384EI 24EI
T
wa2 (2a) MOL 2 wa3 (uA)2 = = = 6EI 6EI 6EI
¢
(¢ D)2 =
=
wa2 ≤ (a) 2
MOx A L2 - x2 B = C (2a)2 - a2 D 6EIL 6EI(2a) wa4 8EI
c
Then the slope and deflection of point D are uA = (uA)1 + (uA)2 =
wa3 wa3 wa3 = 3EI 6EI 6EI
Ans.
¢ D = (¢ D)1 + (¢ D)2 =
5wa4 wa4 wa4 = T 24EI 8EI 12EI
Ans.
980
B a
a
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12–91. Determine the slope at B and the deflection at point C of the simply supported beam. E = 200 GPa and I = 45.5(106) mm4.
9 kN/m
10 kN
A
B
C 3m
Elastic Curves. The loading system on the beam is equivalent to the sum of the seperate loadings shown in Fig. a. The elastic curves for each loading are shown in Fig. a. Method of Superposition. Using the table in the appendix, the required slope and deflections are
(uB)1 = (¢ C)1 =
9 A 63 B wOL3 43.2kN # m2 = = 45EI 45EI EI wOx A 3x4 - 10L2x2 + 7L4 B 360EIL
=
9(3) C 3 A 34 B - 10 A 62 B A 32 B + 7 A 64 B D 360EI(6)
=
75.9375kN # m3 EI
(uB)2 =
(¢ C)2 =
T
10 A 62 B PL2 22.5kN # m2 = = 16EI 16EI EI 10 A 63 B PL3 45kN # m3 = = 48EI 48EI EI
T
Then the slope at B and deflection at C are uB = (uB)1 + (uB)2
=
65.7 A 103 B 43.2 22.5 65.7kN # m2 + = = = 0.00722 rad EI EI EI 200 A 109 B C 45.5 A 10 - 6 B D
Ans.
¢ C = (¢ C)1 + (¢ C)2 120.9375 A 10 B 75.9375 45 120.9375 kN # m3 + = = EI EI EI 200 A 109 B C 45.5 A 10 - 6 B D 3
=
= 0.01329 m = 13.3 mm T
Ans.
981
3m
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*12–92. Determine the slope at A and the deflection at point C of the simply supported beam. The modulus of elasticity of the wood is E = 10 GPa .
3 kN
100 m C
A
1.5 m
Method of Superposition. Using the table in the appendix, the required slopes and deflections are (uA)1 =
Pab(L + b) 3(1.5)(4.5)(6 + 4.5) 5.90625kN # m2 = = 6EIL 6EI(6) EI
(¢ C)1 =
3(4.5)(1.5) 2 Pbx A L2 - b2 - x2 B = A 6 - 4.52 - 1.52 B 6EIL 6EI(6)
(uA)2 = (¢ C)2 =
7.594kN # m3 EI
T
3 A 62 B PL2 6.75 kN # m2 = = 16EL 16EI EI 3(1.5) Px 9.281 a3(6)2 - 4(1.5)2 b = A 3L2 - 4x2 B = 48EI 48EI EI
Then the slope and deflection at C are uA = (uA)1 + (uA)2 =
6.75 5.90625 + EI EI
12.65625kN # m2 = = EI
12.6525 A 103 B
10 A 109 B c
1 (0.1) A 0.2 3 B d 12
Ans.
= 0.0190 rad
and ¢ C = (¢ C)1 + (¢ C)2
=
9.281 7.594 + = EI EI
16.88 A 103 B
10 A 109 B c
1 (0.1) A 0.2 3 B d 12
= 0.0253 m = 25.3 mm
982
B
1.5 m
Elastic Curves. The two concentrated forces P are applied seperately on the beam and the resulting elastic curves are shown in Fig. a.
=
3 kN
Ans.
3m
200 m
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•12–93.
The W8 * 24 simply supported beam is made of A-36 steel and is subjected to the loading shown. Determine the deflection at its center C.
6 kip/ft 5 kip⭈ft A
B
I = 82.8 in4
C 8 ft
5(6) A 164 B 2560 5wL4 (¢ C)1 = = = T 768EI 768EI EI ¢ 2 (x) =
Mx A L2 - x2 B 6LEI
At point C, x =
(¢ C)2 =
=
M A L2 B
6LEI
L 2
A L2 - A L2 B 2 B
5 A 162 B 80 ML2 = = 16EI 16EI EI
¢ C = (¢ C)1 + (¢ C)2 = 2640(1728)
=
8 ft
29 A 103 B (82.8)
T
80 2640 2560 + = EI EI EI Ans.
= 1.90 in.
12–94. Determine the vertical deflection and slope at the end A of the bracket. Assume that the bracket is fixed supported at its base, and neglect the axial deformation of segment AB. EI is constant.
3 in. B
6 in.
A 3
¢A =
uA =
3
8(3) 72 PL = = 3EI 3EI EI
Ans.
8 A 32 B PL2 36 = = 2EI 2EI EI
Ans.
983
8 kip
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12–95. The simply supported beam is made of A-36 steel and is subjected to the loading shown. Determine the deflection at its center C. I = 0.1457(10-3) m4.
20 kN 4 kN/m
A 5m
Using the table in appendix, the required deflections for each load case are computed as follow: (yC)1 =
5(4) A 104 B 5wL4 = 768EI 768 EI =
(yC)2 =
260.42 kN # m3 EI
T
20N A 103 B PL3 416.67 kN # m3 = = T 48EI 48EI EI
Then the deflection of point C is yC = (yC)1 + (yC)2 =
260.42 416.67 + EI EI
=
677.08 kN # m3 EI
T
= 0.1457 A 10 - 3 B m4 and E = 200GPa
¢C =
677.08 A 103 B
200 A 109 B C 0.1457 A 10 - 3 B D
Ans.
= 0.0232 m = 23.2 m T
984
B
C 5m
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*12–96. Determine the deflection at end E of beam CDE. The beams are made of wood having a modulus of elasticity of E = 10 GPa .
2m
1.5 m A
C
1m
3 kN
Method of Superposition. Referring to the table in the appendix, the deflection of point D is 4.5 A 33 B 2.53125 kN # m3 PL3 = = 48EI 48EI EI
T
Subsequently, 3 2.53125 3 3.796875 kN # m3 (¢ E)1 = ¢ D a b = a b = T 2 EI 2 EI Also,
(¢ E)2 = (uD)3 =
3 A 13 B 1 kN # m3 PL3 = = T 3EI 3EI EI 3(2) MOL 2 kN # m2 = = 3EI 3EI EI
(¢ E)3 = (uD)3L =
2 2 kN # m3 (1) = T EI EI
Thus, the deflection of end E is ¢ E = (¢ E)1 + (¢ E)2 + (¢ E)3
=
3.796875 1 2 6.796875kN # m3 + + = = EI EI EI EI
6.796875 A 103 B
10 A 109 B c
1 (0.075) A 0.153 B d 12
= 0.03222 m = 32.2 mm T
Ans.
985
a a
E
75 mm
D
a
¢D =
1.5 m
a
150 mm Section a – a B
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•12–97.
The pipe assembly consists of three equal-sized pipes with flexibility stiffness EI and torsional stiffness GJ. Determine the vertical deflection at point A.
L – 2
C
¢B =
P A L2 B 3 3EI
(¢ A)1 =
=
P A L2 B 3 3EI
PL3 24EI
=
L – 2
P
PL3 24EI
L – 2
A
B
(PL>2) A L2 B PL2 TL u = = = JG JG 4JG (¢ A)2 = u a
L PL3 b = 2 8JG
¢ A = ¢ B + (¢ A)1 + (¢ A)2 =
PL3 PL3 PL3 + + 24EI 24EI 8JG
= PL3 a
1 1 + b 12EI 8JG
Ans.
12–98. Determine the vertical deflection at the end A of the bracket. Assume that the bracket is fixed supported at its base B and neglect axial deflection. EI is constant. u =
a
A
ML Pab = EI EI
b
Pa2b (¢ A)1 = u(a) = EI (¢ A)2 =
B
PL3 Pa3 = 3EI 3EI
¢ A = (¢ A)1 + (¢ A)2 =
P
Pa2 (3b + a) Pa3 Pa2b + = EI 3EI 3EI
Ans.
986
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12–99. Determine the vertical deflection and slope at the end A of the bracket. Assume that the bracket is fixed supported at its base, and neglect the axial deformation of segment AB. EI is constant.
20 lb/in. 80 lb
B 4 in. 3 in.
C
Elastic Curve: The elastic curves for the concentrated load, uniform distibuted load, and couple moment are drawn separately as shown. Method of Superposition: Using the table in Appendix C, the required slopes and displacements are
(uA)1 =
20 A 4 3 B wL3AB 213.33 lb # in2 = = 6EI 6EI EI
(uA)2 = (uB)2 =
(uA)3 = (uB)3 =
(¢ A)v1 =
160(3) M0 LBC 480 lb # in2 = = EI EI EI 80 A 32 B PL2BC 360 lb # in2 = = 2EI 2EI EI
20 A 4 4 B wL4AB 640 lb # in3 = = 8EI 8EI EI
T
(¢ A)v2 = (uB)2 (LAB) =
480 1920lb # in3 (4) = EI EI
T
(¢ A)v3 = (uB)3 (LAB) =
1440lb # in3 360 (4) = EI EI
T
The slope at A is uA = (uA)1 + (uA)2 + (uA)3 =
213.33 480 360 + + EI EI EI
=
1053 lb # in2 EI
Ans.
The vertical displacement at A is (¢ A)v = (¢ A)v1 + (¢ A)v2 (¢ A)v3 =
640 1920 1440 + + EI EI EI
=
4000 lb # in3 EI
Ans.
T
987
A
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*12–100. The framework consists of two A-36 steel cantilevered beams CD and BA and a simply supported beam CB. If each beam is made of steel and has a moment of inertia about its principal axis of Ix = 118 in4, determine the deflection at the center G of beam CB.
A
15 kip B
D C
G 8 ft
16 ft
¢C =
8 ft
7.5 A 163 B PL3 10.240 = = T 3EI 3EI EI 15 A 163 B PL3 1.280 = = T 48EI 48EI EI
¢ ¿G =
¢G = ¢C + ¢ ¿G =
1,280 11,520 10,240 + = EI EI EI 11,520(1728)
=
29 A 103 B (118)
Ans.
= 5.82 in. T
•12–101.
The wide-flange beam acts as a cantilever. Due to an error it is installed at an angle u with the vertical. Determine the ratio of its deflection in the x direction to its deflection in the y direction at A when a load P is applied at this point. The moments of inertia are Ix and Iy. For the solution, resolve P into components and use the method of superposition. Note: The result indicates that large lateral deflections (x direction) can occur in narrow beams, Iy V Ix, when they are improperly installed in this manner. To show this numerically, compute the deflections in the x and y directions for an A-36 steel W10 * 15, with P = 1.5 kip, u = 10°, and L = 12 ft.
ymax =
xmax = ymax
P cos L3 ; 3EIx P sin u L3 3 EIy Pcosu L3 3 EIx
W 10 * 15 ymax =
xmax =
=
xmax =
u Vertical P y u
L A x
P sinu L3 3EIy
Ix tan u Iy
Ans.
Ix = 68.9 in4
1.5( cos 10°)(144)3 3(29) A 103 B (68.9)
1.5( sin 10°)(144)3 3(29) A 103 B (2.89)
Iy = 2.89 in4
= 0.736 in.
Ans.
= 3.09 in.
Ans.
988
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12–102. The simply supported beam carries a uniform load of 2 kip>ft. Code restrictions, due to a plaster ceiling, require the maximum deflection not to exceed 1>360 of the span length. Select the lightest-weight A-36 steel wide-flange beam from Appendix B that will satisfy this requirement and safely support the load. The allowable bending stress is sallow = 24 ksi and the allowable shear stress is tallow = 14 ksi. Assume A is a pin and B a roller support.
8 kip
A
Strength criterion:
24 =
M Sreq’d
96(12) Sreq’d
Sreq’d = 48 in3 Choose W14 * 34, S = 48.6 in3, tw = 0.285 in., d = 13.98 in., I = 340 in4. tallow = 14 Ú
V A web
24 = 6.02 ksi O.K. (13.98)(0.285)
Deflection criterion: Maximum is at center. vmax =
P(4)(8) 5wL4 + (2) C (16)2 - (4)2 - (8)2) D (12)3 384EI 6EI(16)
= c
117.33(8) 5(2)(16)4 + d(12)3 384EI EI 4.571(106)
=
29(106)(340)
= 0.000464 in. 6
B 4 ft
Mmax = 96 kip # ft
sallow =
8 kip 2 kip/ft
1 (16)(12) = 0.533 in. O.K. 360
Use W14 * 34
Ans.
989
8 ft
4 ft
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12–103. Determine the reactions at the supports A and B, then draw the moment diagram. EI is constant.
M0 A B L
Support Reactions: FBD(a). + ©F = 0; : x
Ax = 0 A y - By = 0
[1]
M0 - A y L + MB = 0
[2]
+ c ©Fy = 0; a + ©MB = 0;
Ans.
Moment Function: FBD(b) a + ©MNA = 0;
M(x) + M0 - A y x = 0 M(x) = A y x - M0
Slope and Elastic Curve: EI
EI
EI
d2y = M(x) dx2
d 2y = A y x - M0 dx2
Ay dy = x2 - M0x + C1 dx 2
EI y =
Ay 6
x3 -
[3]
M0 2 x + C1x + C2 2
[4]
Boundary Conditions: At x = 0, y = 0. From Eq.[4], C2 = 0 At x = L,
0 =
dy = 0. From Eq. [3], dx
A y L2
- M0 L + C1
2
[5]
At x = L, y = 0. From Eq. [4], 0 =
A y L3 -
6
M0 L2 + C1 L 2
[6]
Solving Eqs. [5] and [6] yields, Ay =
3M0 2L
C1 =
Ans.
M0 L 4
Substituting Ay, into Eqs. [1] and [2] yields: By =
3M0 2L
MB =
M0 2
Ans.
990
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*12–104. Determine the value of a for which the maximum positive moment has the same magnitude as the maximum negative moment. EI is constant.
P
a L
A y + By - P = 0
+ c ©Fy = 0; a + ©MA = 0;
[1]
MA + By L - Pa = 0
[2]
Moment Functions: FBD(b) and (c). M(x1) = Byx1 M(x2) = Byx2 - Px2 + PL - Pa Slope and Elastic Curve: EI
d2y = M(x) dx2
For M(x1) = Byx1, EI
EI
d2y1 dx21
= Byx1
By dy1 = x2 + C1 dx1 2 1
EI y1 =
By 6
[3]
x31 + C1x1 + C2
[4]
For M(x2) = Byx2 - Px2 + PL - Pa, EI
EI
d2y2 dx22
= Byx2 - Px2 + PL - Pa
By dy2 P = x2 - x22 + PLx2 - Pax2 + C3 dx2 2 2 2
EI y2 =
By 6
x32 -
[5]
P 3 PL 2 Pa 2 x + x x + C3x2 + C4 6 2 2 2 2 2
[6]
Boundary Conditions: y1 = 0 at x1 = 0. From Eq.[4], C2 = 0 dy2 dx2
= 0 at x2 = L. From Eq.[5]
0 =
ByL2 2
-
C3 = -
PL2 + PL2 - PaL + C3 2 By L2 2
-
PL2 + PaL 2
991
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*12–104.
Continued
y2 = 0 at x2 = L. From Eq.[6], 0 =
By L3 6
By L2 PL3 PL3 PaL2 PL2 + + a+ PaL bL + C4 6 2 2 2 2 C4 =
By L3 3
+
PL3 PaL2 6 2
Continuity Conditions: At x1 = x2 = L - a, By 2
dy1 dy2 . From Eqs.[3] and [5], = dx1 dx2
(L - a)2 + C1 =
By 2
(L - a)2 -
P (L - a)2 + PL(L - a) 2
- Pa(L - a) + a C1 =
By L2 2
-
PL2 + PaL b 2
By L2 Pa2 2 2
At x1 = x2 = L - a, y1 = y2. From Eqs.[4] and [6], By L2 Pa2 (L - a) + a b(L - a) 6 2 2
By
3
By =
6
(L - a)3 -
+ a-
By L2 2
-
P PL Pa (L - a)3 + (L - a)2 (L - a)2 6 2 2 By L3 PL2 PL3 PaL2 + PaL b(L - a) + + 2 3 6 2
By L3 Pa3 Pa2L + = 0 6 2 3 By =
3Pa2 Pa3 Pa2 = (3L - a) 2 3 2L 2L 2L3
Substituting By into Eqs.[1] and [2], we have Ay =
P A 2L3 - 3a2L + a3 B 2L3
MA =
Pa A -3aL + a2 + 2L2 B 2L2
Require |Mmax( + )| = |Mmax( - )|. From the moment diagram, Pa2 Pa (3L - a)(L - a) = A -3aL + a2 + 2L2 B 2L3 2L2 a2 - 4aL + 2L2 = 0 a = A 2 - 22 B L
Ans.
992
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•12–105. Determine the reactions at the supports A, B, and C; then draw the shear and moment diagrams. EI is constant.
P
A L 2
Support Reactions: FBD(a). + ©Fx = 0; :
Ax = 0
+ c ©Fy = 0;
A y + By + Cy - 2P = 0
a + ©MA = 0;
By L + Cy (2L) - Pa
Ans. [1]
3L L b - Pa b = 0 2 2
[2]
Moment Function: FBD(b) and (c). M(x1) = Cy x1 M(x2) = Cy x2 - Px2 +
PL 2
Slope and Elastic Curve: d2y = M(x) dx2
EI
For M(x1) = Cy x1, EI
d2y1
= Cyx1
dx21
Cy dy1 = x2 + C1 dx1 2 1
EI
EI y1 =
Cy 6
x31 + C1x1 + C2
For M(x2) = Cyx2 - Px2 +
EI
EI
[3]
d2y2
PL , 2
= Cyx2 - Px2 +
dx22
[4]
PL 2
Cy dy2 P PL = x22 - x22 + x + C3 dx2 2 2 2 2
EI y2 =
Cy 6
x32 -
[5]
P 3 PL 2 x + x + C3x2 + C4 6 2 4 2
[6]
Boundary Conditions: y1 = 0 at x1 = 0. From Eq.[4], C2 = 0 Due to symmetry,
0 =
Cy L2 2
-
dy2 = 0 at x2 = L. From Eq.[5], dx2 PL2 PL2 + + C3 2 2
C3 = -
Cy L2 2
993
P
B L 2
C L 2
L 2
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•12–105.
Continued
y2 = 0 at x2 = L. From Eq. [6], 0 =
Cy L3
2
-
6
Cy L PL3 PL3 + + abL + C4 6 4 2 Cy L3
C4 =
-
3
PL3 12
Continuity Conditions: At x1 = x2 =
dy2 L dy1 , . From Eqs.[3] and [5], = 2 dx1 dx2 Cy 2
a
Cy L 2 Cy L2 L 2 P L 2 PL L b + C1 = a b - a b + a b 2 2 2 2 2 2 2 2 C1 =
At x1 = x2 = Cy
a
6
Cy =
6
Cy L2 PL2 8 2
L , y = y2. From Eqs.[4] and [6], 2 1
Cy L2 L PL2 L 3 b + a ba b 2 8 2 2 a
Cy L2 L Cy L3 L 3 P L 3 PL L 2 PL3 b a b + a b + aba b + 2 6 2 4 2 2 2 3 12 Cy =
5 P 16
Ans.
Substituting Cy into Eqs.[1] and [2], By =
11 P 8
Ay =
5 P 16
Ans.
994
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12–106. Determine the reactions at the supports, then draw the shear and moment diagram. EI is constant.
P
A
B L
Support Reactions: FBD(a). + ©F = 0; : x
Ax = 0
Ans.
By - A y - P = 0
+ c ©Fy = 0; a + ©MB = 0;
[1]
A y L - MA - PL = 0
[2]
Moment Functions: FBD(b) and (c). M(x1) = -Px1 M(x2) = MA - A yx2 Slope and Elastic Curve: EI
d2y = M(x) dx2
For M(x1) = -Px1, EI
EI
d2y1 dx21
= -Px1
dy1 P = - x21 + C1 dx1 2
[3]
P EI y1 = - x31 + C1x1 + C2 6
[4]
For M(x2) = MA - A yx2, EI
EI
EI y2 =
d2y2 dx22
= MA - A yx2
Ay dy2 = MAx2 x2 + C3 dx2 2 2
[5]
Ay MA 2 x2 x3 + C3x2 + C4 2 6 2
[6]
995
L
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12–106.
Continued
Boundary Conditions: y2 = 0 at x2 = 0. From Eq.[6], C4 = 0 dy2 = 0 at x2 = 0. From Eq.[5], C3 = 0 dx2 y2 = 0 at x2 = L. From Eq. [6], 0 =
A y L3 MA L2 2 6
[7]
Solving Eqs.[2] and [7] yields, MA =
PL 2
Ay =
3P 2
Ans.
Substituting the value of Ay into Eq.[1], By =
5P 2
Ans.
Note: The other boundary and continuity conditions can be used to determine the constants C1 and C2 which are not needed here.
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12–107. Determine the moment reactions at the supports A and B. EI is constant.
P
P
A
B a
Support Reactions: FBD(a).
a L
a + ©MB = 0;
Pa + P(L - a) + MA - A y L - MB = 0 PL + MA - A y L - MB = 0
[1]
Moment Functions: FBD(b) and (c). M(x1) = A y x1 - MA M(x2) = A y x2 - Px2 + Pa - MA Slope and Elastic Curve: d2y = M(x) dx2
EI
For M(x1) = A y x1 - MA, EI
EI
d2y1 dx21
= A y x1 - MA
Ay dy1 = x2 - MA x1 + C1 dx1 2 1
EI y1 =
Ay 6
x31 -
[2]
MA 2 x + C1x1 + C2 2 1
[3]
For M(x2) = A y x2 - Px2 + Pa - Ma, EI
EI
EI y2 =
d2y = A y x2 - Px2 + Pa - MA dx22
Ay dy2 P = x2 - x22 + Pax2 - MA x2 + C3 dx2 2 2 2 Ay 6
x32 -
[4]
MA 2 P 3 Pa 2 x + x x + C3 x2 + C4 6 2 2 2 2 2
[5]
Boundary Conditions: dy1 = 0 at x1 = 0. From Eq.[2], C1 = 0 dx1 y1 = 0 at x1 = 0. From Eq.[3], C2 = 0 Due to symmetry,
0 =
Ay 2
a
dy2 dx2
= 0 at x2 =
L . From Eq.[4], 2
L 2 P L 2 L L b a b + Paa b - MA a b + C3 2 2 2 2 2
C3 = -
A y L2 8
+
MA L PL2 PaL + 8 2 2
997
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12–107.
Continued
Due to symmetry, A y a2
dy1 dy2 = at x1 = a and x2 = L - a. From Eqs.[2] and [4], dx1 dx2
- MA a = -
2
Ay 2
(L - a)2 +
+ MA (L - a) +
-A y a2 -
3A y L2 8
+ A y aL +
P (L - a)2 - Pa(L - a) 2
A y L2 8
-
MA L PL2 PaL + 8 2 2
MAL 3PaL 3Pa2 3PL2 + + = 0 8 2 2 2
[6]
Continuity Conditions: At x1 = x2 = a, A y a2 2
dy1 dy2 . From Eqs.[2] and [4], = dx1 dx2
- MA a
A y a2 =
2
A y L2 MA L Pa2 PL2 PaL + Pa2 - MA a + + 2 8 8 2 2
-
A y L2 MA L Pa2 PL2 PaL + + = 0 2 8 8 2 2
[7]
Solving Eqs.[6] and [7] yields, MA =
Pa (L - a) L
Ans.
Ay = P Substitute the value of MA and Ay obtained into Eqs.[1], MB =
Pa (L - a) L
Ans.
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*12–108. Determine the reactions at roller support A and fixed support B.
w
Equations of Equilibrium. Referring to the free-body diagram of the entire beam, Fig. a, A
+ ©F = 0; : x
Bx = 0
+ c ©Fy = 0;
A y + By - wL = 0
a+ ©MB = 0;
wL a
(1)
L 2 b - A y a Lb - MB = 0 2 3
MB =
wL2 2 - A yL 2 3
(2)
Moment Functions. Referring to the free-body diagram of the beam’s segment, Fig. b, x L L M(x) + wxa b + wa b ¢ x + ≤ - A yx = 0 2 3 6
a+ ©MO = 0;
M(x) = A yx -
w 2 wL wL2 x x 2 3 18
Equations of Slope and Elastic Curves. EI
d2v = M(x) dx2
EI
wL2 d2v w 2 wL x x = A x y 2 3 18 dx2
EI
Ay dv w 3 wL 2 wL2 = x2 x x x + C1 dx 2 6 6 18
EIv =
Ay 6
x3 -
(3)
w 4 wL 3 wL2 2 x x x + C1x + C2 24 18 36
(4)
Boundary Conditions. At x = 0, v = 0. Then Eq. (4) gives 0 = 0 - 0 - 0 - 0 + 0 + C2 0 =
At x =
0 =
C2 = 0At x =
2 dv L, = 0. Then Eq. (3) gives 3 dx
Ay
2 2 w 2 3 wL 2 2 wL2 2 a Lb a Lb a Lb a L b + C1 2 3 6 3 6 3 18 3
C1 =
2A yL2 13wL3 81 9
(5)
2 L, v = 0. Then Eq. (4) gives 3
Ay
C1 =
B
Ans.
2 3 w 2 4 wL 2 3 wL2 2 2 2 a Lb a Lb a Lb a Lb + C1 a Lb 6 3 24 3 18 3 36 3 3 2A yL2 wL3 18 27
(6)
Solving Eqs. (5) and (6), Ay =
17wL wL Ans.C1 = Substituting the result of Ay into Eqs. (1) and (2), 24 324
By =
7wL 24
MB =
wL2 36
Ans.
The shear and moment diagrams are shown in Figs. c and d, respectively. 999
L 3
2L 3
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•12–109.
Use discontinuity functions and determine the reactions at the supports, then draw the shear and moment diagrams. EI is constant.
3 kip/ft C A
B 8 ft
+ ©F = 0 : x
Cx = 0
+ c ©Fy = 0
A y + By + Cy - 24 = 0
a + ©MA = 0
Ans. (1)
18 Cy + 8By - 24(4) = 0
(2)
Bending Moment M(x): M(x) = -(-Cy) 6 x - 0 7 -( -By) 6 x - 10 7 = Cyx + By 6 x - 10 7 -
3 6 x - 10 7 2 2
3 6 x - 10 7 2 2
Elastic curve and slope: EI
d2v 3 = M(x) = Cyx + By 6 x - 10 7 - 6 x - 10 7 2 2 dx2
EI
Cyx2 By dv 1 = + 6 x - 10 7 2 - 6 x - 10 7 3 + C1 dx 2 2 2
EIv =
Cyx3 6
By +
6
6 x - 10 7 3 -
1 6 x - 10 7 4 + C1x + C2 8
(3)
(4)
Boundary conditions: v = 0
at
x = 0
From Eq. (4) C2 = 0 v = 0
at
x = 10 ft
From Eq. (4) 0 = 166.67 Cy + 10C1 v = 0
at
(5)
x = 18 ft
0 = 972Cy + 85.33By - 512 + 18C1
(6)
Solving Eqs. (2),(5) and (6) yields: By = 14.4 kip
Ans.
Cy = -1.07 kip = 1.07 kip T
Ans.
C1 = 17.78 From Eq. (1): A y = 10.7 kip
Ans.
1000
10 ft
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12–110. Determine the reactions at the supports, then draw the shear and moment diagrams. EI is constant.
w0
A
C
B L
Support Reaction: FBD(b). + ©F = 0; : x
Ax = 0
Ans.
A y + By + Cy - w0L = 0
+ c ©Fy = 0; a + ©MA = 0;
[1]
By L + Cy (2L) - w0 L(L) = 0
[2]
Moment Function: FBD(b). a + ©MNA = 0;
-M(x) -
x 1 w0 a xbx a b + Cyx = 0 2 L 3
M(x) = Cyx -
w0 3 x 6L
Slope and Elastic Curve: EI
EI
EI
d 2y = M(x) dx2
w0 3 d 2y = Cyx x 2 6L dx
Cy w0 4 dy = x2 x + C1 dx 2 24L
EI y =
Cy 6
x3 -
[3]
w0 5 x + C1x + C2 120L
[4]
Boundary Conditions: At x = 0, y = 0. From Eq.[4], C2 = 0 Due to symmetry,
0 =
Cy L2 -
2
C1 = -
dy = 0 at x = L. From Eq. [3], dx w0L3 + C1 24
CyL2 2
+
w0L3 24
At x = L, y = 0. From Eq. [4], 0 =
Cy L3 6
2
-
Cy L w0L4 w0L3 + a+ bL 120 2 24 Cy =
w0L 10
Ans.
1001
L
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12–110.
Continued
Substituting Cy into Eqs. [1] and [2] yields: By =
4w0L 5
Ay =
w0L 10
Ans.
Shear and Moment diagrams: The maximum span (positive) moment occurs when the shear force V = 0. From FBD(c), + c ©Fy = 0;
w0L 1 w0 - a xbx = 0 10 2 L x =
+ ©MNA = 0;
M +
25 L 5
w0L 1 w0 x a xb (x) a b (x) = 0 2 L 3 10 M =
At x =
25 L, 5
At x = L,
M =
w0L w0 3 x x 10 6L 25w0L2 75
M = -
w0L2 15
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12–111. Determine the reactions at pin support A and roller supports B and C. EI is constant.
w
Equations of Equilibrium. Referring to the free-body diagram of the entire beam, Fig. a, + ©F = 0; : x
Ax = 0
+ c ©Fy = 0;
A y + By + Cy - wL = 0
a + ©MB = 0;
Ans.
Cy (L) + wLa A y - Cy =
(1)
L b - A y(L) = 0 2
wL 2
(2)
Moment Functions. Referring to the free-body diagram of the beam’s segment, Fig. b, M(x1) is M(x1) + wx1 ¢
a + ©MO = 0;
x1 ≤ - A yx1 = 0 2
M(x1) = A yx1 -
w 2 x 2 1
and M(x2) is given by a + ©MO = 0;
Cyx2 - M(x2) = 0 M(x2) = Cyx2
Equations of Slope and Elastic Curves. EI
d2v = M(x) dx2
For coordinate x1, EI
d2v w 2 = A yx1 x 2 1 dx21
EI
Ay dv w 3 x 2 x + C1 = dx1 2 1 6 1
EIv =
Ay 6
x1 3 -
(3)
w 4 x + C1x1 + C2 24 1
(4)
For coordinate x2, EI
d2v = Cyx2 dx22
EI
Cy dv = x 2 + C3 dx2 2 2
EIv =
Cy 6
(5)
x2 3 + C3x2 + C4
(6)
1003
A
C
B L
L
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12–111.
Continued
Boundary Conditions. At x1 = 0, v1 = 0. Then Eq.(4) gives 0 = 0 - 0 + 0 + C2
C2 = 0
At x1 = L, v1 = 0. Then Eq. (4) gives 0 =
Ay 6
A L3 B -
w A L4 B + C1L 24
C1 =
A yL2 wL3 24 6
At x2 = 0, v2 = 0. Then Eq. (6) gives 0 = 0 + 0 + C4
C4 = 0
At x2 = L, v2 = 0. Then Eq. (6) gives 0 =
Cy 6
A L B + C3L 3
Continuity Conditions. At x1 = x2 = L, Ay 2
A L2 B -
A y + Cy =
C3 = -
CyL2 6
dv1 dv2 . Then Eqs.(3) and (5) give = dx1 dx2
Cy A yL2 CyL2 wL3 w 3 ≤ = - B A L2 B R AL B + ¢ 6 24 6 2 6 3wL 8
(7)
Solving Eqs. (2) and (7), Ay =
7wL 16
Cy = -
wL 16
Ans.
The negative sign indicates that Cy acts in the opposite sense to that shown on freebody diagram. Substituting these results into Eq. (1), By =
5wL 8
Ans.
The shear and moment diagrams are shown in Figs. c and d, respectively.
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*12–112. Determine the moment reactions at fixed supports A and B. EI is constant.
w0
Equations of Equilibrium. Due to symmetry, A y = By = R and MA = MB = M. Referring to the free-body diagram of the entire beam, Fig. a, + c ©Fy = 0;
2R R =
A
1 wL = 0 2 0
L 2
w0L 4
Moment Function. Referring to the free-body diagram of the beam’s segment, Fig. b, M(x) + c
w0L x 1 2w0 a xb (x) d a b + M x = 0 2 L 3 4 M(x) =
w0L w0 3 x x - M 4 3L
Equations of Slope and Elastic Curves. EI
EI
EI
d2v = M(x) dx2 d2v dx 2
=
w0L w0 3 x x - M 4 3L
w0L 2 w0 4 dv = x x - Mx + C1 dx 8 12L
EIv =
Due to symmetry,
(2)
dv = 0. Then Eq. (1) gives dx
0 = 0 - 0 - 0 + C1
0 =
(1)
w0L 3 w0 5 M 2 x x x + C1x + C2 24 60L 2
Boundary Conditions. At x = 0,
C1 = 0
dv L = 0 at x = . Then Eq. (1) gives dx 2
w0 L 4 w0L L 2 L a b a b - Ma b 8 2 12L 2 2
MA = MB = M =
B
5w0L2 96
Note. The boundary condition v = 0 at x = 0 can be used to determine C2 using Eq.(2).
1005
L 2
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The beam has a constant E1I1 and is supported by the fixed wall at B and the rod AC. If the rod has a cross-sectional area A2 and the material has a modulus of elasticity E2 , determine the force in the rod.
•12–113.
C w
L2
B
A L1
TAC + By - wL1 = 0
+ c ©Fy = 0 c + ©MB = 0
TAC(L1) + MB MB =
wL1 2 = 0 2
(1)
wL1 2 - TACL1 2
(2)
Bending Moment M(x): wx2 2
M(x) = TACx -
Elastic curve and slope: EI
d2v wx2 = M(x) = TACx 2 2 dx
EI
TACx2 dv wx3 = + C1 dx 2 6
EIv =
(3)
TACx3 wx4 + C1x + C2 6 24
(4)
Boundary conditions: v =
TACL2 A 2E2
x = 0
From Eq. (4) -E2I1 a
TACL2 b = 0 - 0 + 0 + C2 A 2E2
C2 = a v = 0
-E1I1L2 b TAC A 2E2
at
x = L1
From Eq. (4) 0 =
TACL1 3 wL1 4 E1I1L2 + C1L1 T 6 24 A 2E2 AC dv = 0 dx
at
(5)
x = L1
From Eq. (3) 0 =
TACL1 2 wL1 3 + C1 2 6
(6)
Solving Eqs. (5) and (6) yields: TAC =
3A 2E2wL1 4
8 A A 2E2L1 3 + 3E1I1L2 B
Ans.
1006
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12–114. The beam is supported by a pin at A, a roller at B, and a post having a diameter of 50 mm at C. Determine the support reactions at A, B, and C. The post and the beam are made of the same material having a modulus of elasticity E = 200 GPa, and the beam has a constant moment of inertia I = 255(106) mm4.
15 kN/m
A 1m 6m
Equations of Equilibrium. Referring to the free-body diagram of the entire beam, Fig. a, + ©F = 0; : x
Ax = 0
+ c ©Fy = 0;
A y + By + FC - 15(12) = 0
a+ ©MB = 0;
15(12)(6) - FC(6) - A y(12) = 0
Ans. (1)
2A y + FC = 180
(2)
Moment Functions. Referring to the free-body diagram of the beam’s segment, Fig. b, x M(x) + 15xa b - A yx = 0 2
a+ ©MO = 0;
M(x) = A yx - 7.5x2 Equations of Slope and Elastic Curves. EI
d2v = M(x) dx2
EI
d2v = A yx - 7.5x2 dx2
EI
Ay dv = x2 - 2.5x3 + C1 dx 2
EIv =
Ay 6
(3)
x3 - 0.625x4 + C1x + C2
(4)
Boundary Conditions. At x = 0, v = 0. Then Eq. (4) gives C2 = 0
0 = 0 - 0 + 0 + C2 At x = 6 m, v = - ¢ C = -
E C 255 A 10 - 6 B D a -
FC(1) FCLC 1600FC . Then Eq. (4) gives = = p A CE pE 2 0.05 E A B 4
Ay 1600FC b = A 63 B - 0.625 A 64 B + C1(6) pE 6
C1 = 135 - 6A y - 0.02165FC Due to symmetry,
0 =
Ay 2
dv = 0 at x = 6 m. Then Eq. (3) gives dx
A 62 B - 2.5 A 63 B + 135 - 6A y - 0.02165FC (5)
12A y - 0.02165FC = 405 Solving Eqs. (2) and (5), FC = 112.096 kN = 112 kN
A y = 33.95 kN = 34.0 kN
Ans.
Substituting these results into Eq. (1), By = 33.95 kN = 34.0 kN
Ans.
1007
B
C 6m
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12–115. Determine the moment reactions at the supports A and B, then draw the shear and moment diagrams. EI is constant.
A B L
Support Reaction: FBD(a). + ©F = 0; : x + c ©Fy = 0; a + ©MA = 0;
Ax = 0
Ans.
By - A y = 0
[1]
ByL - MA - M0 = 0
[2]
Elastic Curve: As shown. M/EI Diagrams: M/EI diagrams for By and M0 acting on a cantilever beam are shown. Moment-Area Theorems: From the elastic curve, tB>A = 0. tB>A = 0 =
M0 2 L 1 By L a b(L)a Lb + a b (L)a b 2 EI 3 EI 2 By =
3M0 2L
Ans.
Substituting the value of By into Eqs.[1] and [2] yields, Ay =
3M0 2L
MA =
M0 2
Ans.
1008
M0
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*12–116. The rod is fixed at A, and the connection at B consists of a roller constraint which allows vertical displacement but resists axial load and moment. Determine the moment reactions at these supports. EI is constant.
w
B
A L
Support Reaction: FBD(a). a + ©MA = 0;
MB + MA - wLa
L b = 0 2
[1]
Elastic Curve: As shown. M/EI Diagrams: M/EI diagrams for MB and the uniform distributed load acting on a cantilever beam are shown. Moment-Area Theorems: Since both tangents at A and B are horizontal (parallel), uB>A = 0. uB>A = 0 = a
MB 1 wL2 b(L) + a b(L) EI 3 2EI
MB =
wL2 6
Ans.
Substituting MB into Eq.[1], MA =
wL2 3
Ans.
1009
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•12–117. Determine the value of a for which the maximum positive moment has the same magnitude as the maximum negative moment. EI is constant.
P a
L
(tA>B)1 =
2(L - a) -P(L - a)2(2L + a) 1 -P(L - a) a b (L - a)a a + b = 2 EI 3 6EI
(tA>B)2 =
A yL3 2L 1 A yL a b(L)a b = 2 EI 3 3EI
tA>B = 0 = (tA>B)1 + (tA>B)2 0 =
A yL3 -P(L - a)2(2L + a) + 6EI 3EI Ay =
P(L - a)2(2L + a) 2L3
Require: |M1| = |M2| Pa(L - a)2(2L + a) 3
2L
Pa(L - a)(L + a) =
2L2
a2 + 2La - L2 = 0 a = 0.414L
Ans.
1010
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12–118. Determine the reactions at the supports, then draw the shear and moment diagrams. EI is constant.
M0
M0
A L
Require: tA>B = 0 = a 0 =
M0 L 1 -A y L 2L b(L)a b + a b(L)a b EI 2 2 EI 3
A y L3 M0L2 ; 2EI 3EI
Ay =
3M0 2L
Ans.
Equilibrium: a+ ©MB = 0;
3M0 (L) - Cy (L) = 0 2L Cy =
+ c ©Fy = 0;
By -
Ans.
3M0 3M0 = 0 2L 2L
By = + ©F = 0; : x
3M0 2L
3M0 L
Ans.
Cx = 0
Ans.
1011
C
B L
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12–119. Determine the reactions at the supports, then draw the shear and moment diagrams. EI is constant. Support B is a thrust bearing.
P A
B
L
Support Reactions: FBD(a). + ©F = 0; : x + c ©Fy = 0; a+ ©MA = 0;
Bx = 0
Ans.
-A y + By + Cy - P = 0 By (L) + Cy (2L) - Pa
[1]
3L b = 0 2
[2]
Elastic Curve: As shown. M/EI Diagrams: M/EI diagrams for P and By acting on a simply supported beam are drawn separately. Moment-Area Theorems: (tA>C)1 =
=
1 3PL L 3L L 1 3PL 3L 2 3L a ba ba ba b + a ba ba + b 2 8EI 2 3 2 2 8EI 2 2 6 7PL3 16EI
(tA>C)2 =
By L By L3 1 ab (2L)(L) = 2 2EI 2EI
(tB>C)1 =
1 PL L 2 L PL L L a ba ba ba b + a ba ba b 2 8EI 2 3 2 4EI 2 4 +
=
(tB>C)2 =
1 3PL L L L a ba ba + b 2 8EI 2 2 6
5PL3 48EI By L By L3 1 L ab(L)a b = 2 2EI 3 12EI
tA>C = (tA>C)1 + (tA>C)2 =
By L3 7PL3 16EI 2EI
tB>C = (tB>C)1 + (tB>C)2 =
By L3 5PL3 48EI 12EI
From the elastic curve, tA>C = 2tB>C By L3 By L3 5PL3 7PL3 = 2a b 16EI 2EI 48EI 12EI By =
11P 16
Ans.
Substituting By into Eqs. [1] and [2] yields, Cy =
13P 32
Ay =
3P 32
Ans.
1012
C
L 2
L 2
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*12–120. Determine the moment reactions at the supports A and B. EI is constant.
w
B
A L – 2
-MA 1 -wL2 L 1 Ay L a b(L) + a b (L) + a ba b 2 EI EI 3 8EI 2
uB>A = 0 =
Ay L
0 =
tB>A = 0 =
0 =
2
- MA -
wL2 48
(1)
-MA L L 1 -wL2 L L 1 Ay L a b(L)a b + a b(L)a b + a ba ba b 2 EI 3 EI 2 3 8EI 2 8 Ay L 6
-
MA wL2 2 384
(2)
Solving Eqs. (1) and (2) yields: Ay =
3wL 32
MA =
5wL2 192
c + ©MB = 0;
Ans.
MB +
3wL 5wL2 wL L (L) a b = 0 32 192 2 4
MB =
11wL2 192
Ans.
1013
L – 2
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•12–121.
Determine the reactions at the bearing supports A, B, and C of the shaft, then draw the shear and moment diagrams. EI is constant. Each bearing exerts only vertical reactions on the shaft.
A
1m
1m 400 N
Support Reactions: FBD(a). + c ©Fy = 0; a + ©MA = 0;
A y + By + Cy - 800 = 0
[1]
By (2) + Cy (4) - 400(1) - 400(3) = 0
[2]
Method of superposition: Using the table in Appendix C, the required displacements are yB œ =
yB
fl
Pbx A L2 - b2 - x2 B 6EIL
=
400(1)(2) 2 A 4 - 12 - 2 2 B 6EI(4)
=
366.67 N # m3 EI
T
By A 4 3 B 1.3333By m3 PL3 = = = 48EI 48EI EI
c
The compatibility condition requires (+ T)
0 = 2yB ¿ + yB – 0 = 2a
1.3333By 366.67 b + ab EI EI
By = 550 N
Ans.
Substituting By into Eqs. [1] and [2] yields, A y = 125 N
Cy = 125 N
Ans.
1014
C
B
1m
1m 400 N
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12–122. Determine the reactions at the supports A and B. EI is constant.
P
A
B L
Referring to the FBD of the beam, Fig. a + ©F = 0; : x
Ax = 0
Ans.
By - P - A y = 0
+ c ©Fy = 0;
A y = By - P
(1)
3 a+ ©MA = 0; -MA + By L - Pa Lb = 0 2 MA = By L -
3 PL 2
(2)
Referring to Fig. b and the table in appendix, the necessary deflections are computed as follow: yP =
Px2 (3LAC - x) 6EI
=
P(L2) 3 c 3a L b - L d 6EI 2
=
7PL3 12EI
yBy =
T
By L3 PL3AB c = 3EI 3EI
The compatibility condition at support B requires that (+ T)
0 = vP + vBy 0 =
-By L3 7PL3 + a b 12EI 3EI
By =
7P 4
Ans.
Substitute this result into Eq (1) and (2) Ay =
3P 4
MA =
PL 4
Ans.
1015
L 2
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12–123. Determine the reactions at the supports A, B, and C, then draw the shear and moment diagrams. EI is constant.
12 kip
A
Support Reaction: FBD(b).
+ c ©Fy = 0; a + ©MA = 0;
Cx = 0
Ans.
A y + By + Cy - 12 - 36.0 = 0
[1]
By (12) + Cy (24) - 12(6) - 36.0(18) = 0
[2]
Method of superposition: Using the table in Appendix C, the required displacements are
yB ¿ = yB – =
=
yB –¿ =
5(3) A 24 4 B 6480 kip # ft3 5wL4 = = 768EI 768EI EI
T
Pbx A L2 - b2 - x2 B 6EIL 2376 kip # ft3 12(6)(12) A 24 2 - 62 - 12 2 B = 6EI(24) EI By A 24 3 B 288By ft3 PL3 = = 48EI 48EI EI
T
c
The compatibility condition requires (+ T)
0 = yB ¿ + yB – + yB –¿ 0 =
288By 2376 6480 + + ab EI EI EI
By = 30.75 kip
Ans.
Substituting By into Eqs.[1] and [2] yields, A y = 2.625 kip
Cy = 14.625 kip
Ans.
1016
C
B 6 ft
+ ©F = 0; : x
3 kip/ft
6 ft
12 ft
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*12–124. The assembly consists of a steel and an aluminum bar, each of which is 1 in. thick, fixed at its ends A and B, and pin connected to the rigid short link CD. If a horizontal force of 80 lb is applied to the link as shown, determine the moments created at A and B. Est = 2911032 ksi, Eal = 1011032 ksi.
C
80 lb
D
1 in.
Steel
30 in. Aluminum
0.5 in.
A
; ©Fx = 0
Pal + Pst - 80 = 0
(1)
Compatibility condition: ¢ st = ¢ al PstL3 Pal L3 = 3EstIst 3EalIal Pst = a
1 (29) A 103 B A 12 B (1) A 0.53 B EstIst b (Pal) = Pal 1 EalIal (10) A 103 B A 12 B (1) A 13 B (N)
Pst = 0.3625 Pal
(2)
Solving Eqs. (1) and (2) yields: Pal = 58.72 lb
Pst = 21.28 lb
MA = Pst (30) = 639 lb # in. = 0.639 kip # in.
Ans.
MB = Pal (30) = 1761 lb # in. = 1.76 kip # in.
Ans.
1017
B
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•12–125. Determine the reactions at the supports A, B, and C, then draw the shear and moment diagrams. EI is constant. A 3m
Cx = 0
Ans.
a+ ©MC = 0; A y(12) + By(16) - 10(3) - 10(9) = 0 2A y + By = 20 + c ©Fy = 0;
(1)
A y + By + Cy - 10 - 10 = 0 A y + By + Cy = 20
(2)
Referring to Fig. b and table in appendix, the necessary deflections are: (vP)1 = (vP)2 =
Pbx A L2 - b2 - x 2 B 6EILAC AC
=
10(3)(6) A 12 2 - 32 - 62 B 6EI(12)
=
247.5 kN # m3 EI
T
By(12 3) 36 By PL3AC c (vB)y = = = 48EI 48EI EI The compatibility condition at support B requires that (+ T) 0 = (vP)1 + (vP)2 + (vB)y 0 =
36 By 247.5 247.5 + + ab EI EI EI
By = 13.75 kN
Ans.
Substitute this result into Eq. (1) and (2) and solve, A y = Cy = 3.125 kN
Ans.
The shear And moment diagrams are shown in Fig. b and c respectively.
1018
C
B
Referring to the FBD of the beam, Fig. a, + ©F = 0; ; x
10 kN
10 kN
3m
3m
3m
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12–126. Determine the reactions at the supports A and B. EI is constant.
M0 A B L
Referring to the FBD of the beam, Fig. a, + ©F = 0; : x
Ax = 0
+ c ©Fy = 0;
By - A y = 0
Ans. (1)
By(L) - Mo - MA = 0
a+ ©MA = 0;
MA = ByL - Mo
(2)
Referring to Fig. b and the table in the appendix, the necessary deflections are: vMo =
vBy
MoL2 2EI
T
ByL3 PL3 = = 3EI 3EI
c
Compatibility condition at roller support B requires (+ T)
0 = vMo + (vB)y 0 =
By =
ByL3 MoL2 + ab 2EI 3EI 3Mo 2L
Ans.
Substitute this result into Eq. (1) and (2) Ay =
3Mo 2L
MA =
Mo 2
Ans.
1019
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12–127. Determine the reactions at support C. EI is constant for both beams.
P D
B
A
C L 2
Support Reactions: FBD (a). + ©F = 0; : x
Cx = 0
a + ©MA = 0;
Cy(L) - By a
Ans. L b = 0 2
[1]
Method of superposition: Using the table in Appendix C, the required displacements are yB =
yB ¿ =
yB – =
By L3 PL3 = 48EI 48EI
T
P A L2 B 3 PL3BD PL3 = = 3EI 3EI 24EI By L3 PL3BD = 3EI 24EI
T
c
The compatibility condition requires yB = yB ¿ + yB –
(+ T)
By L3 48EI
=
By =
By L3 PL3 + ab 24EI 24EI 2P 3
Substituting By into Eq. [1] yields, Cy =
P 3
Ans.
1020
L 2
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*12–128. The compound beam segments meet in the center using a smooth contact (roller). Determine the reactions at the fixed supports A and B when the load P is applied. EI is constant.
P
A C
B L
L
(P - R)L3 RL3 = 3EI 3EI
¢C =
R =
P 2
Member AC: ©Fy = 0;
Ay =
P 2
Ans.
©Fx = 0;
Ax = 0
Ans.
© MA = 0;
MA =
PL 2
Ans.
Member BC: ©Fy = 0;
By =
P 2
Ans.
©Fx = 0;
Bx = 0
Ans.
©MB = 0;
MB =
PL 2
Ans.
The beam has a constant E1 I1 and is supported by the fixed wall at B and the rod AC. If the rod has a crosssectional area A2 and the material has a modulus of elasticity E2, determine the force in the rod.
•12–129.
(¢ A)¿ =
dA =
wL41 ; 8E1I1
¢A =
C L2
TACL2 A 2E2
L1
By superposition: ¢ A = (¢ A)¿ - dA
TACL2 wL41 TACL31 = A 2E2 8E1I1 3E1I1 TAC a
L2 L31 wL41 + b = A 2E2 3E1I1 8E1I1
TAC =
B
A
TAC L31 3E1I1
(+ T)
w
3wA 2E2L41
8 C 3E1I1L2 + A 2E2L31 D
Ans.
1021
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12–130. Determine the reactions at A and B. Assume the support at A only exerts a moment on the beam. EI is constant.
P
A 2
(uA)1 =
PL ; 8EI
(uA)2 =
B
MAL EI
L – 2
By superposition:
L – 2
0 = (uA)1 - (uA)2 0 =
MAL PL2 8EI EI
MA =
PL 8
Ans.
Equilibrium: a + ©MB = 0; MB =
-
PL PL + - MB = 0 8 2
3PL 8
Ans.
+ ©F = 0 ; : x
Bx = 0
Ans.
+ c ©Fy = 0 ;
By = P
Ans.
12–131. The beam is supported by the bolted supports at its ends. When loaded these supports do not provide an actual fixed connection, but instead allow a slight rotation a before becoming fixed. Determine the moment at the connections and the maximum deflection of the beam.
P
u - u¿ = a L — 2
ML ML PL2 = a 16EI 3EI 6EI ML = a M = a
PL2 - a b(2EI) 16EI
2EI PL ab 8 L
¢ max = ¢ - ¢ ¿ =
Ans.
M(L2 ) PL3 - 2c C L2 - (L>2)2 D d 48EI 6EIL
¢ max =
PL3 L2 PL 2EIa a b 48EI 8EI 8 L
¢ max =
aL PL3 + 192EI 4
Ans.
1022
L — 2
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*12–132. The beam is supported by a pin at A, a spring having a stiffness k at B, and a roller at C. Determine the force the spring exerts on the beam. EI is constant.
w
A B
Method of Superposition: Using the table in appendix C, the required displacements are 5w(2L)4 5wL4AC 5wL4 = = 384EI 384EI 24EI
T
Fsp (2L)3 Fsp L3 PL3AC = = yB – = 48EI 48EI 6EI
c
yB ¿ =
Using the spring formula, ysp =
Fsp k
L
C k L
.
The compatibility condition requires ysp = yB ¿ + yB –
(+ T)
3
Fsp k
=
Fsp =
Fsp L 5wL4 + ab 24EI 6EI 5wkL4
4 A 6EI + kL3 B
Ans.
•12–133.
The beam is made from a soft linear elastic material having a constant EI. If it is originally a distance ¢ from the surface of its end support, determine the distance a at which it rests on this support when it is subjected to the uniform load w0 , which is great enough to cause this to happen.
w0 ⌬
a L
The curvature of the beam in region BC is zero, therefore there is no bending moment in the region BC, The reaction F is at B where it touches the support. The slope is zero at this point and the deflection is ¢ where ¢ =
R(L - a)3 w0(L - a)4 8EI 3EI
u1
=
w0(L - a)3 R(L - a)2 6EI 2EI
Thus, 1
8¢EI 4 R = a b 9w30
Ans. 1
72¢EI 4 L - a = a b w0 1
72¢EI 4 a = L - a b w0
Ans.
1023
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12–134. Before the uniform distributed load is applied on the beam, there is a small gap of 0.2 mm between the beam and the post at B. Determine the support reactions at A, B, and C. The post at B has a diameter of 40 mm, and the moment of inertia of the beam is I = 875(106) mm4. The post and the beam are made of material having a modulus of elasticity of E = 200 GPa .
30 kN/m
A
6m
Equations of Equilibrium. Referring to the free-body diagram of the beam, Fig. a, + ©F = 0; : x
Ax = 0
+ c ©Fy = 0;
A y + FB + Cy - 30(12) = 0
a + ©MA = 0;
Ans. (1)
FB(6) + Cy(12) - 30(12)(6) = 0
(2)
Method of superposition: Referring to Fig. b and the table in the Appendix, the necessary deflections are
(vB)1 =
(vB)2 =
5(30) A 12 4 B 8100kN # m3 5wL4 = = T 384EI 384EI EI FB A 12 3 B 36FB PL3 = = 48EI 48EI EI
c
The deflection of point B is vB = 0.2 A 10 - 3 B +
FB(a) FBLB = 0.2 A 10 - 3 B + AE AE
T
The compatibility condition at support B requires
A+TB
vB = (vB)1 + (vB)2 0.2 A 10 - 3 B +
FB (1) 36FB 8100 = + ab AE EI EI
0.2 A 10 - 3 B E + FB
p A 0.04 2 B 4
+
FB 36FB 8100 = A I I 36FB
875 A 10 - 6 B
8100
=
875 A 10 - 6 B
-
C 1m
0.2 A 10 - 3 B C 200 A 109 B D 1000
FB = 219.78 kN = 220 kN
Ans.
Substituting the result of FB into Eqs. (1) and (2), A y = Cy = 70.11 kN = 70.1 kN
Ans.
1024
B
0.2 mm 6m
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12–135. The 1-in.-diameter A-36 steel shaft is supported by unyielding bearings at A and C. The bearing at B rests on a simply supported steel wide-flange beam having a moment of inertia of I = 500 in4. If the belt loads on the pulley are 400 lb each, determine the vertical reactions at A, B, and C.
3 ft 5 ft
A
2 ft 5 ft
B
For the shaft: (¢ b)1 =
(¢ b)2 =
400 lb
800(3)(5) 13200 A -52 - 32 + 102 B = 6EIs(10) EIs By A 103 B 48EIs
C
5 ft
20.833By =
EIs
For the beam:
¢b =
By A 103 B 48EIb
20.833By =
EIb
Compatibility condition: + T ¢ b = (¢ b)1 - (¢ b)2 20.833By EIb Is =
=
20.833By 13200 EIs EIs
p (0.5)4 = 0.04909 in4 4
20.833By (0.04909) 500
400 lb
= 13200 - 20.833By
By = 634 lb
Ans.
Form the free-body digram, A y = 243 lb
Ans.
Cy = 76.8 lb
Ans.
1025
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*12–136. If the temperature of the 75-mm-diameter post CD is increased by 60°C, determine the force developed in the post. The post and the beam are made of A-36 steel, and the moment of inertia of the beam is I = 255(106) mm4.
3m
3m
A B
C 3m
D
Method of Superposition. Referring to Fig. a and the table in the Appendix, the necessary deflections are (vC)1 =
FCD A 33 B 9FCD PLBC 3 c = = 3EI 3EI EI
(vC)2 = (uB)2LBC =
3FCD (3) 9FCD MOLAB c (LBC) = (3) = 3EI 3EI EI
The compatibility condition at end C requires
A+cB
vC = (vC)1 + (vC)2 =
9FCD 9FCD 18FCD c + = EI EI EI
Referring to Fig. b, the compatibility condition of post CD requires that dFCD + vC = dT dFCD =
(1)
FCD (3) FCD LCD = AE AE
dT = a¢TL = 12 A 10 - 6 B (60)(3) = 2.16 A 10 - 3 B m Thus, Eq. (1) becomes 3FCD 18FCD + = 2.16 A 10 - 3 B AE EI 3FCD
p A 0.0752 B 4
+
18FCD
255 A 10 - 6 B
= 2.16 A 10 - 3 B C 200 A 109 B D
FCD = 6061.69N = 6.06 kN
Ans.
1026
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•12–137.
The shaft supports the two pulley loads shown. Using discontinuity functions, determine the equation of the elastic curve. The bearings at A and B exert only vertical reactions on the shaft. EI is constant.
x A
12 in.
B
12 in. 70 lb
180 lb
M = -180 6 x - 0 7 - (-277.5) 6 x - 12 7 - 70 6 x - 24 7 M = -180x + 277.5 6 x - 12 7 - 70 6 x - 24 7 Elastic curve and slope: EI
d2v = M = -180x + 277.5 6 x - 12 7 - 70 6 x - 24 7 dx2
EI
dv = -90x2 + 138.75 6 x - 12 7 dx
EIv = -30x3 + 46.25 6 x - 12 7
3
2
- 35(x - 24 7
2
- 11.67 6 x - 24 7
+ C1 3
+ C1x + C2 (1)
Boundary conditions: v = 0
at
x = 12 in,
From Eq. (1) 0 = -51,840 + 12C1 + C2 12C1 + C2 = 51 840 v = 0
at
(2)
x = 60 in.
From Eq.(1) 0 = -6 480 000 + 5 114 880 - 544 320 + 60C1 + C2 60C1 + C2 = 1909440
(3)
Solving Eqs. (2) and (3) yields: C1 = 38 700 v =
C2 = -412 560
1 [-30x3 + 46.25 6 x - 12 7 EI
3
- 11.7 6 x - 24 7
3
+ 38 700x - 412 560]
Ans.
1027
36 in.
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12–138. The shaft is supported by a journal bearing at A, which exerts only vertical reactions on the shaft, and by a thrust bearing at B, which exerts both horizontal and vertical reactions on the shaft. Draw the bending-moment diagram for the shaft and then, from this diagram, sketch the deflection or elastic curve for the shaft’s centerline. Determine the equations of the elastic curve using the coordinates x1 and x2 . EI is constant.
80 lb A
x1
EI
d2v1 dx21
= 26.67x1
dv1 = 13.33x21 + C1 dx1
(1)
EIv1 = 4.44x31 + C1x1 + C2
(2)
EI
For M2 (x) = -26.67x2 EI
d2v2 dx22
= -26.67x2
dv2 = -13.33x22 + C3 dx2
(3)
EIv2 = -4.44x32 + C3x2 + C4
(4)
EI
Boundary conditions: v1 = 0
at
x1 = 0
at
x2 = 0
From Eq.(2) C2 = 0 v2 = 0 C4 = 0 Continuity conditions: dv1 dv2 = dx1 dx2
at
x1 = x2 = 12
From Eqs. (1) and (3) 1920 + C1 = -( -1920 + C3) C1 = -C3 v1 = v2
(5)
x1 = x2 = 12
at
7680 + 12C1 = -7680 + 12C3 C3 - C1 = 1280
(6)
Solving Eqs. (5) and (6) yields: C3 = 640
80 lb 12 in.
For M1 (x) = 26.67 x1
C1 = -640
v1 =
1 A 4.44x31 - 640x1 B lb # in3 EI
Ans
v2 =
1 A -4.44x32 + 640x2 B lb # in3 EI
Ans.
1028
B
4 in. 4 in. x2 12 in.
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12–139. The W8 * 24 simply supported beam is subjected to the loading shown. Using the method of superposition, determine the deflection at its center C. The beam is made of A-36 steel.
6 kip/ft 5 kip⭈ft A
B C 8 ft
Elastic Curves: The elastic curves for the uniform distributed load and couple moment are drawn separately as shown. Method of superposition: Using the table in Appendix C, the required displacements are
(¢ C)1 =
-5(6) A 164 B 2560 kip # ft3 -5wL4 = = T 768EI 768EI EI
(¢ C)2 = -
= -
=
M0x A L2 - x2 B 6EIL 5(8) C (16)2 - (8)2 D 6EI(16)
80 kip # ft3 EI
T
The displacement at C is ¢ C = (¢ C)1 + (¢ C)2 =
80 2560 + EI EI
=
2640 kip # ft3 EI 2640(1728)
=
29 A 103 B (82.8)
= 1.90 in.
Ans.
T
1029
8 ft
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*12–140. Using the moment-area method, determine the slope and deflection at end C of the shaft. The 75-mmdiameter shaft is made of material having E = 200 GPa. Support Reactions and
B
A
M Diagram. As shown in Fig. a. EI
1m
1m
15 kN
= (1) B
=
冷tC>A 冷
1 7.5 1 3 1 a b (2) R + c (2) d B a b(2) R 2 EI 3 2 EI
5.5 kN # m3 EI
= (1 + 1) B
1 7.5 1 1 3 a b(2) R + c (2) + 1 d B a b(2) R 2 EI 3 2 EI 2 3 1 + c (1) d B a b(1) R 3 2 EI
冷uC>A 冷
=
9 kN # m3 EI
=
1 3 1 7.5 a b(2) + a b(3) 2 EI 2 EI
=
3 kN # m3 EI
Referring to the geometry of the elastic curve, Fig. b,
uA =
冷tB>A 冷 LAB
5.5 EI 2.75kN # m2 = = 2 EI
uC = uC>A - uA =
=
3 2.75 EI EI
0.25 kN # m2 = EI
0.25 A 103 B
200 A 109 B c
p A 0.03754 B d 4
= 0.805 A 10 - 3 B rad
Ans.
and ¢ C = 冷 tC>A冷 - 冷 tB>A ¢
=
9 5.5 3 a b EI EI 2
=
0.75 kN # m3 = EI
LAC ≤ LAB
0.75 A 103 B
p 200 A 10 B c A 0.03754 B d 4
1m 3 kN
Moment Area Theorem. Referring to Fig. b,
冷tB>A 冷
C
= 0.002414 m = 2.41 mm c Ans.
9
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•12–141. Determine the reactions at the supports. EI is constant. Use the method of superposition.
w A
wL C L3 - 2(3L)L2 + (3L)3 D 24EI
¢B = ¢C =
D B L
4
11wL 12EI
=
Due to symmetry, By = Cy By (L)(2L) ¢ BB = ¢ CC =
6EI(3L)
C (3L)2 - (2L)2 - L2 D
4By L3 =
9EI By (L)(L)
¢ BC = ¢ CB =
6EI(3L)
C -L2 - L2 + (3L)2 D
7By L3 =
18EI
By superposition: +T
0 =
0 = ¢ B - ¢ BB - ¢ BC 4By L3 7By L3 11wL4 12EI 9EI 18EI
By = Cy =
11wL 10
Ans.
Equilibrium: a+ ©MD = 0; Ay = c + ©Fy = 0; Dy = + ©F = 0; ; x
3wLa
3L 11wL 11wL b (L) (2L) - A y (3L) = 0 2 10 10
2wL 5
Ans.
2wL 11wL 11wL + + + Dy - 3wL = 0 5 10 10 2wL 5
Ans.
Dx = 0
Ans.
1031
C L
L
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12–142. Determine the moment reactions at the supports A and B. Use the method of integration. EI is constant.
w0
A
B L
Support Reactions: FBD(a). A y + By -
+ c ©Fy = 0; a + ©MA = 0;
w0L = 0 2
[1]
ByL + MA - MB -
w0L L a b = 0 2 3
[2]
Moment Function: FBD(b). a+ ©MNA = 0;
-M(x) -
x 1 w0 a xb x a b - MB + Byx = 0 2 L 3
M(x) = Byx -
w0 3 x - MB 6L
Slope and Elastic Curve: EI
EI
EI
EI y =
d2y = M(x) dx2
w0 3 d2y = Byx x - MB 6L dx2
By w0 4 dy = x2 x - MBx + C1 dx 2 24L By 6
x3 -
[3]
w0 5 MB 2 x x + C1x + C2 120L 2
[4]
Boundary Conditions: At x = 0,
dy = 0 dx
From Eq.[3],
At x = 0, y = 0. At x = L,
0 =
From Eq.[4],
dy = 0. dx
By L2 2
-
C1 = 0 C2 = 0
From Eq. [3].
w0L3 - MBL 24
0 = 12By L - w0 L2 - 24MB At x = L, y = 0. 0 =
By L3 6
-
[5]
From Eq. [4], w0 L4 MB L2 120 2
0 = 20By L - w0 L2 - 60MB
[6]
1032
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12–142.
Continued
Solving Eqs. [5] and [6] yields, MB = By =
w0 L2 30
Ans.
3w0L 20
Substituting By and MB into Eqs. [1] and [2] yields, MA =
w0L2 20
Ay =
7w0 L 20
Ans.
1033
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12–143. If the cantilever beam has a constant thickness t, determine the deflection at end A. The beam is made of material having a modulus of elasticity E.
L w0 x
Section Properties: Referring to the geometry shown in Fig. a, A
h(x) h0 = ; x L
h0 x h(x) = L
h0 B
Thus, the moment of inertia of the tapered beam as a function of x is I(x) =
3 h0 th0 3 3 1 1 t C h(x) D 3 = t¢ x≤ = x 12 12 L 12L3
Moment Function. Referring to the free-body diagram of the beam’s segment, Fig. b, M(x) + B
a + ©MO = 0;
1 w0 x a xbx R a b = 0 2 L 3
M(x) = -
w0 3 x 6L
Equations of slope and Elastic Curve. E
M(x) d2v = 2 I(x) dx
w0 3 x 2w0L2 dv 6L E 2 = = dx th0 3 3 th0 3 x 3 12L -
2
E
2w0L2 dv = x + C1 dx th0 3
Ev = -
w0L2 th0 3
(1)
x2 + C1x + C2
Boundary conditions. At x = L,
0 = -
2w0L2 th0 3
(2)
dv = 0. Then Eq. (1) gives dx
(L) + C1
C1 =
2w0L3 th0 3
At x = L, v = 0. Then Eq. (2) gives 0 = -
w0L2 th0
3
A L2 B +
2w0L3 th0
3
(L) + C2
C2 = -
w0L4 th0 3
Substituting the results of C1 and C2 into Eq. (2), v =
w0L2 Eth0 3
A -x2 + 2Lx - L2 B
At A, x = 0. Then vA = v冷x = 0 = -
w0L4 Eth0 3
=
w0L4 Eth0 3
Ans.
T
1034
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*12–144. Beam ABC is supported by beam DBE and fixed at C. Determine the reactions at B and C. The beams are made of the same material having a modulus of elasticity E = 200 GPa, and the moment of inertia of both beams is I = 25.0(106) mm4.
100 lb/ft
a
A
B
C
D
E a 4 ft
4 ft 6 ft
6 in. Section a – a
+ ©F = 0; : x
Cx = 0
+ c ©Fy = 0;
By + Cy - 9(6) = 0
a+ ©MC = 0;
9(6)(3) - By(4) - MC = 0
Ans. (1)
MC = 162 - 4By
(2)
Method of superposition: Referring to Fig. b and the table in the appendix, the deflections are
vB =
By A 63 B 4.5By PLDE 3 = = T 48EI 48EI EI
(vB)1 =
=
(vB)2 =
9 A 42 B wx2 A x2 - 4Lx + 6L2 B = C 4 2 - 4(6)(4) + 6 A 62 B D 24EI 24EI 816 kN # m3 T EI By A 4 3 B 21.3333By PLBC 3 c = = 3EI 3EI EI
The compatibility condition at support B requires that
A+TB
vB = (vB)1 + (vB)2 4.5By EI
=
21.3333By 816 + ab EI EI
By = 31.59 kN = 31.6 kN
Ans.
Substituting the result of By into Eqs. (1) and (2), MC = 35.65 kN # m = 35.7 kN # m
Ans.
Cy = 22.41 kN = 22.4 kN
Ans.
1035
a
6 ft
3 in.
Equation of Equilibrium. Referring to the free-body diagram of the beam, Fig. a,
a
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•12–145.
Using the method of superposition, determine the deflection at C of beam AB. The beams are made of wood having a modulus of elasticity of E = 1.5(103) ksi.
100 lb/ft
a
A
B
C
D
E a 4 ft
4 ft 6 ft
6 in. Section a – a
Method of superposition. Referring to Fig. b and the table in the appendix, the deflection of point B is
¢B =
600 A 83 B PLDE 3 6400 lb # ft3 = = T 48EI 48EI EI
Subsequently, referring to Fig. c, (¢ C)1 = ¢ B a
(¢ C)2 =
6 6400 6 3200 lb # ft3 b = a b = T 12 EI 12 EI
5(100) A 12 4 B 5wL4 27000 lb # ft3 = = T 384EI 384EI EI
Thus, the deflection of point C is
A+TB
¢ C = (¢ C)1 + (¢ C)2 =
3200 27000 + EI EI
30200 lb # ft3 = = EI
30200 A 12 3 B
1.5 A 106 B c
1 (3) A 63 B d 12
= 0.644 in T
Ans.
1036
a
6 ft
3 in.
Support Reactions: The reaction at B is shown on the free-body diagram of beam AB, Fig. a.
a
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12–146. The rim on the flywheel has a thickness t, width b, and specific weight g. If the flywheel is rotating at a constant rate of v, determine the maximum moment developed in the rim. Assume that the spokes do not deform. Hint: Due to symmetry of the loading, the slope of the rim at each spoke is zero. Consider the radius to be sufficiently large so that the segment AB can be considered as a straight beam fixed at both ends and loaded with a uniform centrifugal force per unit length. Show that this force is w ⫽ btgv2r>g.
A
B
v r
Centrifugal Force: The centrifugal force action on a unit length of the rim rotating at a constant rate of v is g btgv2r w = mv2 r = bta bv2r = g g
(Q.E.D.)
Elastic Curve: Member AB of the rim is modeled as a straight beam with both of its ends fixed and subjected to a uniform centrifigal force w. Method of Superposition: Using the table in Appendix C, the required displacements are uB ¿ =
wL3 6EI
yB ¿ =
wL4 c 8EI
uB – =
yB – =
MBL EI
uB ¿– =
MBL2 c 2EI
yB –¿ =
ByL2 2EI
ByL3 3EI
T
Computibility requires, 0 = uB ¿ + uB – + uB ¿– 2
0 =
By L MBL wL3 + + ab 6EI EI 2EI
0 = wL2 + 6MB - 3By L
[1]
0 = yB ¿ + yB – + yB –¿
(+ c )
3
0 =
By L MB L2 wL4 + + ab 8EI 2EI 3EI
0 = 3wL2 + 12MB - 8By L
[2]
Solving Eqs. [1] and [2] yields, By =
wL 2
MB =
Due to symmetry, A y =
wL 2
wL2 12 MA =
wL2 12
Maximum Moment: From the moment diagram, the maximum moment occurs at btgv2r pr the two fixed end supports. With w = and L = ru = . g 3
Mmax
wL2 = = 12
A B
btgv2r pr 2 g 3
12
=
t
p2btgv2r3 108g
Ans.
1037
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•13–1.
Determine the critical buckling load for the column. The material can be assumed rigid.
P
L 2 k
Equilibrium: The disturbing force F can be determined by summing moments about point A. a + ©MA = 0;
P(Lu) - F a
L b = 0 2
A
F = 2Pu Spring Formula: The restoring spring force F1 can be determine using spring formula Fs = kx. Fs = ka
L kLu ub = 2 2
Critical Buckling Load: For the mechanism to be on the verge of buckling, the disturbing force F must be equal to the restoring spring force F1. 2Pcr u =
Pcr =
L 2
kLu 2 kL 4
Ans.
1038
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13–2. Determine the critical load Pcr for the rigid bar and spring system. Each spring has a stiffness k.
P
Equilibrium: The disturbing forces F1 and F2 can be related to P by writing the moment equation of equlibrium about point A. Using small angle ananlysis, where cos u ⬵ 1 and sin u = u, + ©MA = 0;
F2 a
L 3 k
L 2 b + F1 a Lb - PLu = 01 3 3
L 3
F2 + 2F1 = 3Pu
k
(1)
Spring Force. The restoring spring force A Fsp B 1 and A Fsp B 2 can be determined using the spring formula, 2 1 Lu and x2 = Lu, Fig. b. Thus, 3 3 2 2 = kx1 = ka Lu b = kLu 3 3
L 3 A
Fsp = kx, where x1 =
A Fsp B 1
A Fsp B 2 = kx2 = ka Lu b = 1 3
Critical Buckling Load. When the mechanism is on the verge of buckling the disturbing force F must be equal to the restoring force of the spring Fsp. Thus, F1 = A Fsp B 1 =
2 kLu 3
F2 = A Fsp B 2 =
1 kLu 3
Substituting this result into Eq. (1), 2 1 kLu + 2 a kLu b = 3Pcr u 3 3 Pcr =
5 kL 9
Ans.
1039
1 kLu 3
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13–3. The leg in (a) acts as a column and can be modeled (b) by the two pin-connected members that are attached to a torsional spring having a stiffness k (torque兾rad). Determine the critical buckling load. Assume the bone material is rigid.
a + ©MA = 0;
-P(u)a
P
L b + 2ku = 0 2
L — 2
Require:
k
Pcr =
4k L
Ans.
L — 2
(a)
*13–4. Rigid bars AB and BC are pin connected at B. If the spring at D has a stiffness k, determine the critical load Pcr for the system.
(b)
P A
Equilibrium. The disturbing force F can be related P by considering the equilibrium of joint A and then the equilibrium of member BC,
a B
Joint A (Fig. b) + c ©Fy = 0;
FAB cos f - P = 0
FAB =
a
P cos f
k D
Member BC (Fig. c)
a
©MC = 0; F(a cos u) -
P P cos f (2a sin u) sin f(2a cos u) = 0 cos f cos f
C
F = 2P(tan u + tan f) Since u and f are small, tan u ⬵ u and tan f ⬵ f. Thus, F = 2P(u + f)
(1)
Also, from the geometry shown in Fig. a, 2au = af
f = 2u
Thus Eq. (1) becomes F = 2P(u + 2u) = 6Pu Spring Force. The restoring spring force Fsp can be determined using the spring formula, Fsp = kx, where x = au, Fig. a. Thus, Fsp = kx = kau
1040
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13–4.
Continued
Critical Buckling Load. When the mechanism is on the verge of buckling the disturbing force F must be equal to the restoring spring force Fsp. F = Fsp
6Pcru = kau
Pcr =
ka 6
Ans.
1041
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•13–5.
An A-36 steel column has a length of 4 m and is pinned at both ends. If the cross sectional area has the dimensions shown, determine the critical load. 25 mm
Section Properties: A = 0.01(0.06) + 0..05(0.01) = 1.10 A 10 - 3 B m2 Ix = Iy =
10 mm
1 1 (0.01) A 0.063 B + (0.05) A 0.013 B = 0.184167 A 10 - 6 B m4 12 12
25 mm
Critical Buckling Load: K = 1 for pin supported ends column. Applying Euler’s formula, Pcr =
25 mm
25 mm 10 mm
p2EI (KL)2 p2 (200)(109)(0.184167)(10 - 6)
=
[1(4)]2 Ans.
= 22720.65 N = 22.7 kN Critical Stress: Euler’s formula is only valid if scr 6 sg. scr =
Pcr 22720.65 = 20.66 MPa 6 sg = 250 MPa = A 1.10(10 - 3)
O.K.
13–6. Solve Prob. 13–5 if the column is fixed at its bottom and pinned at its top. 25 mm
Section Properties: A = 0.01(0.06) + 0.05(0.01) = 1.10 A 10 - 3 B m2
10 mm
1 1 Ix = Iy = (0.01) A 0.063 B + (0.05) A 0.013 B = 0.184167 A 10 - 6 B m4 12 12
25 mm
Critical Buckling Load: K = 0.7 for one end fixed and the other end pinned column. Applying Euler’s formula, Pcr =
p EI (EL)2 p2 (200)(109)(0.184167)(10 - 6)
=
[0.7(4)]2 Ans.
= 46368.68 N = 46.4 kN Critical Stress: Euler’s formula is only valid if scr 6 sg. scr =
25 mm
25 mm 10 mm
2
Pcr 46368.68 = 42.15 MPa 6 sg = 250 MPa = A 1.10(10 - 3)
1042
O.K.
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13–7. A column is made of A-36 steel, has a length of 20 ft, and is pinned at both ends. If the cross-sectional area has the dimensions shown, determine the critical load.
6 in. 0.25 in.
The cross sectional area and moment of inertia of the square tube is
5.5 in.
A = 6(6) - 5.5(5.5) = 5.75 in2 I =
0.25 in.
1 1 (6)(63) (5.5)(5.53) = 31.74 in4 12 12
0.25 in.
0.25 in.
The column is pinned at both of its end, k = 1. For A36 steel, E = 29.0(103) ksi and sg = 36 ksi (table in appendix). Applying Euler’s formula, Pcr =
p2 C 29.0(103) D (31.74) p2EI = (KL)2 C 1(20)(12) D 2
Ans.
= 157.74 kip = 158 Critical Stress. Euler’s formula is valid only if scr 6 sg. scr =
Pcr 157.74 = = 27.4 ksi 6 sg = 36 ksi A 5.75
O.K.
*13–8. A column is made of 2014-T6 aluminum, has a length of 30 ft, and is fixed at its bottom and pinned at its top. If the cross-sectional area has the dimensions shown, determine the critical load.
6 in. 0.25 in.
5.5 in.
The cross-sectional area and moment of inertia of the square tube is 0.25 in.
A = 6(6) - 5.5(5.5) = 5.75 in2
0.25 in.
1 1 I = (6)(63) (5.5)(5.53) = 31.74 in4 12 12 The column is fixed at one end, K = 0.7. For 2014–76 aluminium, E = 10.6(103) ksi and sg = 60 ksi (table in appendix). Applying Euler’s formula, Pcr =
p2 C 10.6(103) D (31.74) p2EI = (KL)2 C 0.7(30)(12) D 2
Ans.
= 52.29 kip = 52.3 kip Critical Stress. Euler’s formula is valid only if scr 6 sg. scr =
Pcr 52.3 = = 9.10 ksi 6 sg = 60 ksi A 5.75
O.K.
1043
0.25 in.
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•13–9.
The W14 * 38 column is made of A-36 steel and is fixed supported at its base. If it is subjected to an axial load of P = 15 kip, determine the factor of safety with respect to buckling.
P
From the table in appendix, the cross-sectional area and moment of inertia about weak axis (y-axis) for W14 * 38 are 20 ft
A = 11.2 in2
Iy = 26.7 in4
The column is fixed at its base and free at top, k = 2. Here, the column will buckle about the weak axis (y axis). For A36 steel, E = 29.0(103) ksi and sy = 36 ksi. Applying Euler’s formula, p2 C 29.0(103) D (26.7)
p2EIy
Pcr =
C 2 (20)(12) D 2
=
(KL)2
= 33.17 kip
Thus, the factor of safety with respect to buckling is F.S =
Pcr 33.17 = = 2.21 P 15
Ans.
The Euler’s formula is valid only if scr 6 sg. scr =
Pcr 33.17 = = 2.96 ksi 6 sg = 36 ksi A 11.2
O.K.
13–10. The W14 * 38 column is made of A-36 steel. Determine the critical load if its bottom end is fixed supported and its top is free to move about the strong axis and is pinned about the weak axis.
P
From the table in appendix, the cross-sectional area and moment of inertia about weak axis (y-axis) for W14 * 38 are A = 11.2 in2
Ix = 385 in4
Iy = 26.7 in4
The column is fixed at its base and free at top about strong axis. Thus, kx = 2. For A36 steel, E = 29.0(103) ksi and sg = 36 ksi. Pcr =
p2EIx (KxLx)
2
=
p2 C 29.0(103) D (385)
C 2 (20)(12) D 2
= 478.28 kip
The column is fixed at its base and pinned at top about weak axis. Thus, ky = 0.7. Pcr =
p2EIy 2
(KyLy)
=
p2 C 29.0(103) D (26.7)
C 0.7(20)(12) D 2
Ans.
= 270.76 kip = 271 kip (Control) The Euler’s formula is valid only if scr 6 sg. scr =
Pcr 270.76 = = 24.17 ksi 6 sg = 36 ksi A 11.2
O.K.
1044
20 ft
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13–11. The A-36 steel angle has a cross-sectional area of A = 2.48 in2 and a radius of gyration about the x axis of rx = 1.26 in. and about the y axis of ry = 0.879 in. The smallest radius of gyration occurs about the z axis and is rz = 0.644 in. If the angle is to be used as a pin-connected 10-ft-long column, determine the largest axial load that can be applied through its centroid C without causing it to buckle.
y z C
x
x z
y
The least radius of gyration: r2 = 0.644 in. scr =
p2E
2 A KL r B
controls. K = 1.0
;
p2 (29)(103) (120) 2 C 1.00.644 D
=
= 8.243 ksi 6 sg
O.K.
Pcr = scr A = 8.243 (2.48) = 20.4 kip
Ans.
*13–12. An A-36 steel column has a length of 15 ft and is pinned at both ends. If the cross-sectional area has the dimensions shown, determine the critical load.
8 in. 0.5 in.
0.5 in. 6 in. 0.5 in.
Ix =
1 1 (8)(73) (7.5)(63) = 93.67 in4 12 12
Iy = 2 a Pcr =
1 1 b(0.5)(83) + (6)(0.53) = 42.729 in4 (controls) 12 12
p2(29)(103)(42.729) p2EI = 2 (EL) [(1.0)(15)(12)]2 = 377 kip
Ans.
Check: A = (2)(8)(0.5) + 6(0.5) = 11 in2 scr =
Pcr 377 = = 34.3 ksi 6 sg A 11
Therefore, Euler’s formula is valid
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•13–13.
An A-36 steel column has a length of 5 m and is fixed at both ends. If the cross-sectional area has the dimensions shown, determine the critical load. I =
10 mm 10 mm
50 mm
1 1 (0.1)(0.053) (0.08)(0.033) = 0.86167 (10 - 6) m4 12 12
Pcr =
100 mm
p2(200)(109)(0.86167)(10 - 6) p2EI = 2 (KL) [(0.5)(5)]2 = 272 138 N = 272 kN
scr =
=
Pcr ; A
Ans.
A = (0.1)(0.05) - (0.08)(0.03) = 2.6(10 - 3) m2
272 138 = 105 MPa 6 sg 2.6 (10 - 3)
Therefore, Euler’s formula is valid.
13–14. The two steel channels are to be laced together to form a 30-ft-long bridge column assumed to be pin connected at its ends. Each channel has a cross-sectional area of A = 3.10 in2 and moments of inertia Ix = 55.4 in4, Iy = 0.382 in4. The centroid C of its area is located in the figure. Determine the proper distance d between the centroids of the channels so that buckling occurs about the x–x and y¿ – y¿ axes due to the same load. What is the value of this critical load? Neglect the effect of the lacing. Est = 2911032 ksi, sY = 50 ksi.
y 0.269 in.
C d
y
In order for the column to buckle about x - x and y - y at the same time, Iy must be equal to Ix Iy = Ix 0.764 + 1.55 d2 = 110.8 d = 8.43 in.
Ans.
Check: d 7 2(1.231) = 2.462 in.
O.K. 3
p (29)(10 )(110.8) p2 EI = 2 (KL) [1.0(360)]2
= 245 kip
Ans.
Check stress: scr =
x C
d 2 Iy = 2(0.382) + 2 (3.10)a b = 0.764 + 1.55 d2 2
Pcr =
1.231 in.
x
Ix = 2(55.4) = 110.8 in.4
2
y¿
Pcr 245 = = 39.5 ksi 6 sg A 2(3.10)
Therefore, Euler’s formula is valid. 1046
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13–15. An A-36-steel W8 * 24 column is fixed at one end and free at its other end. If it is subjected to an axial load of 20 kip, determine the maximum allowable length of the column if F.S. = 2 against buckling is desired. Section Properties. From the table listed in the appendix, the cross-sectional area and moment of inertia about the y axis for a W8 * 24 are A = 7.08 in2
Iy = 18.3 in4
Critical Buckling Load. The critical buckling load is Pcr = Pallow (F.S) = 20(2) = 40 kip Applying Euler’s formula, p2 EIy
Pcr =
40 =
(KL)2
p2 C 29 A 103 B D (18.3) (2L)2
L = 180.93 in = 15.08 ft = 15.1 ft
Ans.
Critical Stress. Euler’s formula is valid only if scr 6 sY. scr =
Pcr 40 = = 5.65 ksi 6 sY = 36 ksi A 7.08
O.K.
*13–16. An A-36-steel W8 * 24 column is fixed at one end and pinned at the other end. If it is subjected to an axial load of 60 kip, determine the maximum allowable length of the column if F.S. = 2 against buckling is desired. Section Properties. From the table listed in the appendix, the cross-sectional area and moment of inertia about the y axis for a W8 * 24 are A = 7.08 in2
Iy = 18.3 in4
Critical Buckling Load. The critical buckling load is Pcr = Pallow (F.S.) = 60(2) = 120 kip Applying Euler’s formula, Pcr =
120 =
p2EIy (KL)2
p2 C 24 A 103 B D (18.3) (0.7L)2
L = 298.46 in = 24.87 ft = 24.9 ft
Ans.
Critical Stress. Euler’s formula is valid only if scr 6 sY. scr =
Pcr 120 = = 16.95 ksi 6 sY = 36 ksi A 7.08
O.K.
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•13–17.
The 10-ft wooden rectangular column has the dimensions shown. Determine the critical load if the ends are assumed to be pin connected. Ew = 1.611032 ksi, sY = 5 ksi.
Section Properties: 10 ft
A = 4(2) = 8.00 in2
4 in.
Ix =
1 (2) A 43 B = 10.667 in4 12
Iy =
1 (4) A 23 B = 2.6667 in4 (Controls !) 12
2 in.
Critical Buckling Load: K = 1 for pin supported ends column. Applying Euler’s formula,. Pcr =
p2EI (KL)2 p2(1.6)(103)(2.6667)
=
[1(10)(12)]2 Ans.
= 2.924 kip = 2.92 kip Critical Stress: Euler’s formula is only valid if scr 6 sg. scr =
Pcr 2.924 = = 0.3655 ksi 6 sg = 5 ksi A 8.00
O.K.
13–18. The 10-ft column has the dimensions shown. Determine the critical load if the bottom is fixed and the top is pinned. Ew = 1.611032 ksi, sY = 5 ksi. Section Properties: A = 4(2) = 8.00 in2 10 ft
1 (2) A 43 B = 10.667 in4 Ix = 12
4 in. 2 in.
1 Iy = (4) A 23 B = 2.6667 in4 (Controls!) 12 Critical Buckling Load: K = 0.7 for column with one end fixed and the other end pinned. Applying Euler’s formula. Pcr =
p2EI (KL)2 p2 (1.6)(103)(2.6667)
=
[0.7(10)(12)]2 Ans.
= 5.968 kip = 5.97 kip Critical Stress: Euler’s formula is only valid if scr 6 sg. scr =
Pcr 5.968 = = 0.7460 ksi 6 sg = 5 ksi A 8.00
O.K.
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13–19. Determine the maximum force P that can be applied to the handle so that the A-36 steel control rod BC does not buckle. The rod has a diameter of 25 mm.
P 350 mm A 250 mm
45⬚
Support Reactions: a + ©MA = 0;
P(0.35) - FBC sin 45°(0.25) = 0 FBC = 1.9799P
Section Properties: A =
p A 0.0252 B = 0.15625 A 10 - 3 B p m2 4
I =
p A 0.01254 B = 19.17476 A 10 - 9 B m4 4
Critical Buckling Load: K = 1 for a column with both ends pinned. Appyling Euler’s formula, Pcr = FBC =
1.9799P =
p2EI (KLBC)2
p2(200)(109) C 19.17476(10 - 9) D [1(0.8)]2
P = 29 870 N = 29.9 kN
Ans.
Critical Stress: Euler’s formula is only valid if scr 6 sg. scr =
1.9799(29 870) Pcr = 120.5 MPa 6 sg = 250 MPa = A 0.15625(10 - 3)p
1049
O.K.
C
B
800 mm
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*13–20. The W10 * 45 is made of A-36 steel and is used as a column that has a length of 15 ft. If its ends are assumed pin supported, and it is subjected to an axial load of 100 kip, determine the factor of safety with respect to buckling.
P
Critical Buckling Load: Iy = 53.4 in4 for a W10 * 45 wide flange section and K = 1 for pin supported ends column. Applying Euler’s formula, Pcr =
15 ft
p2EI (KL)2 p2 (29)(103)(53.4)
=
[1(15)(12)]2 P
= 471.73 kip Critical Stress: Euler’s formula is only valid if scr 6 sg. A = 13.3 in2 for the W10 * 45 wide-flange section. scr =
Pcr 471.73 = = 35.47 ksi 6 sg = 36 ksi A 13.3
O.K.
Pcr 471.73 = = 4.72 P 100
Ans.
Factor of Safety: F.S =
The W10 * 45 is made of A-36 steel and is used as a column that has a length of 15 ft. If the ends of the column are fixed supported, can the column support the critical load without yielding?
•13–21.
P
Critical Buckling Load: Iy = 53.4 in4 for W10 * 45 wide flange section and K = 0.5 for fixed ends support column. Applying Euler’s formula, Pcr =
15 ft
p2EI (KL)2 p2 (29)(103)(53.4)
=
[0.5(15)(12)]2 P
= 1886.92 kip Critical Stress: Euler’s formula is only valid if scr 6 sg. A = 13.3 in2 for W10 * 45 wide flange section. scr =
Pcr 1886.92 = = 141.87 ksi 7 sg = 36 ksi (No!) A 13.3
Ans.
The column will yield before the axial force achieves the critical load Pcr and so Euler’s formula is not valid.
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13–22. The W12 * 87 structural A-36 steel column has a length of 12 ft. If its bottom end is fixed supported while its top is free, and it is subjected to an axial load of P = 380 kip, determine the factor of safety with respect to buckling.
W 12 * 87
A = 25.6 in2
Ix = 740 in4
P
Iy = 241 in4 (controls) 12 ft
K = 2.0 Pcr =
p2(29)(103)(241) p2EI = = 831.63 kip 2 (KL) [(2.0)(12)(12)]2 Pcr 831.63 = = 2.19 P 380
F.S. =
Ans.
Check: scr =
=
Pcr A 831.63 = 32.5 ksi 6 sg 25.6
O.K.
13–23. The W12 * 87 structural A-36 steel column has a length of 12 ft. If its bottom end is fixed supported while its top is free, determine the largest axial load it can support. Use a factor of safety with respect to buckling of 1.75.
W 12 * 87
A = 25.6 in2
Ix = 740 in4
P
Iy = 241 in4
(controls)
K = 2.0 12 ft
Pcr
p2(29)(103)(241) p2EI = = = 831.63 kip 2 (KL) (2.0(12)(12))2
P =
Pcr 831.63 = = 475 ksi F.S 1.75
Ans.
Check: scr =
P 831.63 = = 32.5 ksi 6 sg A 25.6
O.K.
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*13–24. An L-2 tool steel link in a forging machine is pin connected to the forks at its ends as shown. Determine the maximum load P it can carry without buckling. Use a factor of safety with respect to buckling of F.S. = 1.75. Note from the figure on the left that the ends are pinned for buckling, whereas from the figure on the right the ends are fixed.
P
P
1.5 in.
0.5 in.
24 in.
Section Properties: A = 1.5(0.5) = 0.750 in2 Ix =
1 (0.5) A 1.53 B = 0.140625 in4 12
Iy =
1 (1.5) A 0.53 B = 0.015625 in4 12
P
Critical Buckling Load: With respect to the x - x axis, K = 1 (column with both ends pinned). Applying Euler’s formula, Pcr =
p2EI (KL)2 p2(29.0)(103)(0.140625)
=
[1(24)]2
= 69.88 kip With respect to the y - y axis, K = 0.5 (column with both ends fixed). Pcr =
p2EI (KL)2 p2(29.0)(103)(0.015625)
=
[0.5(24)]2
= 31.06 kip
(Controls!)
Critical Stress: Euler’s formula is only valid if scr 6 sg. scr =
Pcr 31.06 = = 41.41 ksi 6 sg = 102 ksi A 0.75
O.K.
Factor of Safety: F.S =
1.75 =
Pcr P 31.06 P
P = 17.7 kip
Ans.
1052
P
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The W14 * 30 is used as a structural A-36 steel column that can be assumed pinned at both of its ends. Determine the largest axial force P that can be applied without causing it to buckle. •13–25.
P
From the table in appendix, the cross-sectional area and the moment of inertia about weak axis (y-axis) for W14 * 30 are A = 8.85 in2
Iy = 19.6 in4
25 ft
Critical Buckling Load: Since the column is pinned at its base and top, K = 1. For A36 steel, E = 29.0(103) ksi and sg = 36 ksi. Here, the buckling occurs about the weak axis (y-axis). P = Pcr =
p2EIy (KL)2
=
p2 C 29.0(103) D (19.6)
C 1(25)(12) D 2
Ans.
= 62.33 kip = 62.3 kip Euler’s formula is valid only if scr 6 sg. scr =
Pcr 62.33 = = 7.04 ksi 6 sg = 36 ksi A 8.85
O.K.
13–26. The A-36 steel bar AB has a square cross section. If it is pin connected at its ends, determine the maximum allowable load P that can be applied to the frame. Use a factor of safety with respect to buckling of 2. a + ©MA = 0;
C
FBC sin 30°(10) - P(10) = 0 FBC = 2 P
+ : ©Fx = 0;
A
1.5 in.
30⬚ B
1.5 in.
FA - 2P cos 30° = 0 1.5 in.
10 ft
FA = 1.732 P
P
Buckling load: Pcr = FA(F.S.) = 1.732 P(2) = 3.464 P L = 10(12) = 120 in. I =
1 (1.5)(1.5)3 = 0.421875 in4 12
Pcr =
p2 EI (KL)2
3.464 P =
p2 (29)(103)(0.421875) [(1.0)(120)]2
P = 2.42 kip
Ans.
Pcr = FA(F.S.) = 1.732(2.42)(2) = 8.38 kip Check: scr =
Pcr 8.38 = = 3.72 ksi 6 sg A 1.5 (1.5)
O.K.
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13–27. Determine the maximum allowable intensity w of the distributed load that can be applied to member BC without causing member AB to buckle. Assume that AB is made of steel and is pinned at its ends for x–x axis buckling and fixed at its ends for y–y axis buckling. Use a factor of safety with respect to buckling of 3. Est = 200 GPa, sY = 360 MPa.
w
C
1.5 m
B 0.5 m
2m 30 mm x
Ix =
1 (0.02)(0.033) = 45.0(10 - 9)m4 12
Iy =
1 (0.03)(0.023) = 20(10 - 9) m4 12
x
x-x axis: Pcr = FAB (F.S.) = 1.333w(3) = 4.0 w K = 1.0, Pcr =
L = 2m
p2EI (KL)2
4.0w =
p2(200)(109)(45.0)(10 - 9) [(1.0)(2)]2
w = 5552 N>m = 5.55 kN>m
Ans.
(controls)
y-y axis K = 0.5, 4.0w =
L = 2m
p2 (200)(109)(20)(10 - 9) [(0.5)(2)]2
w = 9870 N>m = 9.87 kN>m Check: scr =
20 mm y
y
Moment of inertia:
4(5552) Pcr = = 37.0 MPa 6 sg A (0.02)(0.03)
O.K.
1054
30 mm
A
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*13–28. Determine if the frame can support a load of w = 6 kN>m if the factor of safety with respect to buckling of member AB is 3. Assume that AB is made of steel and is pinned at its ends for x–x axis buckling and fixed at its ends for y–y axis buckling. Est = 200 GPa, sY = 360 MPa.
w
C
B
1.5 m
0.5 m
Check x-x axis buckling: Ix =
1 (0.02)(0.03)3 = 45.0(10 - 9) m4 12
K = 1.0 Pcr
2m 30 mm x 20 mm y
y
L = 2m
p2(200)(109)(45.0)(10 - 9) p2EI = = 2 (KL) ((1.0)(2))2
x
A
30 mm
Pcr = 22.2 kN a + ©MC = 0;
FAB(1.5) - 6(2)(1) = 0 FAB = 8 kN
Preq’d = 8(3) = 24 kN 7 22.2 kN No, AB will fail.
Ans.
The beam supports the load of P = 6 kip. As a result, the A-36 steel member BC is subjected to a compressive load. Due to the forked ends on the member, consider the supports at B and C to act as pins for x–x axis buckling and as fixed supports for y–y axis buckling. Determine the factor of safety with respect to buckling about each of these axes. •13–29.
a + ©MA = 0;
P 4 ft
A 3 ft
1 )(1)(3)3 p2(29)(103)(12 p2EI = = 178.9 kip 2 (KL) (1.0(5)(12))2
178.9 = 8.94 20
Ans.
y-y axis buckling: Pcr =
F.S. =
3 in.
y
x-x axis buckling:
F.S. =
B
C x
3 FBC a b(4) - 6000(8) = 0 5 FBC = 20 kip
Pcr =
4 ft
1 )(3)(1)3 p2 (29)(103)(12 p2EI = = 79.51 2 (KL) (0.5(5)(12))2
79.51 = 3.98 20
Ans.
1055
1 in.
x
y
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13–30. Determine the greatest load P the frame will support without causing the A-36 steel member BC to buckle. Due to the forked ends on the member, consider the supports at B and C to act as pins for x–x axis buckling and as fixed supports for y–y axis buckling.
P 4 ft
A 3 ft
3 FBC a b(4) - P(8) = 0 5
a + ©MA = 0;
4 ft
B
y
3 in. C x
1 in.
FBC = 3.33 P y
x-x axis buckling: Pcr =
x
1 )(1)(3)3 p2(29)(103)(12 p2EI = = 178.9 kip (KL)2 (1.0(5)(12))2
y -y axis buckling: Pcr =
1 )(3)(1)3 p2(29)(103)(12 p2EI = = 79.51 kip (KL)2 (0.5(5)(12))2
Thus, 3.33 P = 79.51 P = 23.9 kip
Ans.
13–31. Determine the maximum distributed load that can be applied to the bar so that the A-36 steel strut AB does not buckle. The strut has a diameter of 2 in. It is pin connected at its ends.
w
C
A
2 ft
The compressive force developed in member AB can be determined by writing the moment equation of equilibrium about C. a + ©MC = 0;
FAB(2) - w(2)(3) = 0
A = p(12) = p in2
I =
FAB = 3w
4 ft
p 4 p (1 ) = in4 4 4
Since member AB is pinned at both ends, K = 1. For A36 steel, E = 29.0(103) ksi and sg = 36 ksi. Pcr =
p EI ; (KL)2
p C 29.0(10 ) D (p>4) 2
2
3w =
3
C 1(4)(12) D 2
Ans.
w = 32.52 kip>ft = 32.5 kip>ft The Euler’s formula is valid only if scr 6 sg. scr =
3(32.52) Pcr = = 31.06 ksi 6 sg = 36 ksi p A
O.K.
1056
B
2 ft
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*13–32. The members of the truss are assumed to be pin connected. If member AC is an A-36 steel rod of 2 in. diameter, determine the maximum load P that can be supported by the truss without causing the member to buckle.
P
C B
4 ft
D
A 3 ft
Section the truss through a-a, the FBD of the top cut segment is shown in Fig. a. The compressive force developed in member AC can be determined directly by writing the force equation of equilibrium along x axis. + : ©Fx = 0;
3 FAC a b - P = 0 5
A = p(12) = p in2
I =
FAC =
5 P (C) 3
p 4 p (1 ) = in4 4 4
Since both ends of member AC are pinned, K = 1. For A-36 steel, E = 29.0(103) ksi and sg = 36 ksi. The length of member AC is LAC = 232 + 42 = 5 ft. Pcr =
p2EI ; (KL)2
p2 C 29.0(103) D (p>4) 5 P = 3 C 1(5)(12) D 2
P = 37.47 kip = 37.5 kip
Ans.
Euler’s formula is valid only if scr 6 sg.
scr
5 (37.47) Pcr 3 = = = 19.88 ksi 6 sg = 36 ksi p A
O.K.
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•13–33.
The steel bar AB of the frame is assumed to be pin connected at its ends for y–y axis buckling. If w = 3 kN>m, determine the factor of safety with respect to buckling about the y–y axis due to the applied loading. Est = 200 GPa, sY = 360 MPa.
6m w B C 40 mm 40 mm
3m
40 mm
y
x A
4m
The force with reference to the FBD shown in Fig. a. a + ©MC = 0;
3 3(6)(3) - FAB a b(6) = 0 5
A = 0.04(0.08) = 3.2(10 - 3) m2
Iy =
FAB = 15 kN
1 (0.08)(0.043) = 0.4267(10 - 6)m4 12
The length of member AB is L = 232 + 42 = 5m. Here, buckling will occur about the weak axis, (y-axis). Since both ends of the member are pinned, Ky = 1. Pcr =
p2EIy (KyLy)2
=
p2 C 200(109) D C 0.4267(10 - 6) D
C 1.0(5) D 2
= 33.69 kN
Euler’s formula is valid only if scr 6 sg. scr =
33.69(103) Pcr = 10.53(106)Pa = 10.53 MPa 6 sg = 360 MPa = A 3.2(10 - 3)
O.K.
Thus, the factor of safety against buckling is F.S =
Pcr 33.69 = = 2.25 FAB 15
Ans.
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13–34. The members of the truss are assumed to be pin connected. If member AB is an A-36 steel rod of 40 mm diameter, determine the maximum force P that can be supported by the truss without causing the member to buckle.
2m
C
E
D 1.5 m B A 2m P
By inspecting the equilibrium of joint E, FAB = 0. Then, the compressive force developed in member AB can be determined by analysing the equilibrium of joint A, Fig. a. + c ©Fy = 0;
3 FAC a b - P = 0 5
+ : ©Fx = 0;
5 4 P a b - FAB = 0 3 5
A = p(0.022) = 0.4(10 - 3)p m2
I =
FAC =
5 P (T) 3
FAB =
4 P(c) 3
p (0.024) = 40(10 - 9) p m4 4
Since both ends of member AB are pinned, K = 1. For A36 steel, E = 200 GPa and sg = 250 MPa. Pcr =
p2EI ; (KL)2
p2 C 200(109) D C 40(10 - 9)p D 4 P = 3 C 1(2) D 2 P = 46.51(103) N = 46.5 kN
Ans.
The Euler’s formula is valid only if scr 6 sg.
scr
4 (46.51)(103) Pcr 3 = 49.35(106) Pa = 49.35 MPa 6 sg = 250 MPa O.K. = = A 0.4(10 - 3)p
1059
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13–35. The members of the truss are assumed to be pin connected. If member CB is an A-36 steel rod of 40 mm diameter, determine the maximum load P that can be supported by the truss without causing the member to buckle.
2m
C
E
D 1.5 m B A 2m P
Section the truss through a–a, the FBD of the left cut segment is shown in Fig. a. The compressive force developed in member CB can be obtained directly by writing the force equation of equilibrium along y axis. + c ©Fy = 0;
FCB - P = 0
A = p(0.022) = 0.4(10 - 3)p m2
FCB = P (C) I =
p (0.024) = 40(10 - 9)p m4 4
Since both ends of member CB are pinned, K = 1. For A36 steel, E = 200 GPa and sg = 250 MPa. Pcr =
p2EI ; (KL)2
P =
p2 C 200(109) D C 40(10 - 9)p D
C 1(1.5) D 2
= 110.24(103) N = 110 kN
Ans.
The Euler’s formula is valid only if scr 6 sg. scr =
110.24(103) Pcr = 87.73(106) Pa = 87.73 MPa 6 sg = 250 MPa = A 0.4(10 - 3)p
1060
O.K.
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*13–36. If load C has a mass of 500 kg, determine the required minimum diameter of the solid L2-steel rod AB to the nearest mm so that it will not buckle. Use F.S. = 2 against buckling.
A
45°
4m
D
Equilibriun. The compressive force developed in rod AB can be determined by analyzing the equilibrium of joint A, Fig. a. ©Fy¿ = 0; FAB sin 15° - 500(9.81) cos 45° = 0
FAB = 13 400.71 N
Section Properties. The cross-sectional area and moment of inertia of the solid rod are A =
p 2 d 4
I =
p d 4 p 4 a b = d 4 2 64
Critical Buckling Load. Since the rod is pinned at both of its ends, K = 1. Here, Pcr = FAB (F.S.) = 13400.71(2) = 26801.42 N. Applying Euler’s formula,
Pcr =
p2EIy (KL)2
26801.42 =
p2 C 200 A 109 B D c
p 4 d d 64
[1(4)]2
d = 0.04587 m = 45.87 mm Use d = 46 mm
Ans.
Critical Stress. Euler’s formula is valid only if scr 6 sY. scr =
Pcr 26801.42 = = 16.13 MPa 6 sY = 703 MPa p A 2 A 0.046 B 4
1061
O.K.
60° B
C
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•13–37. If the diameter of the solid L2-steel rod AB is 50 mm, determine the maximum mass C that the rod can support without buckling. Use F.S. = 2 against buckling.
A
45°
4m
D
Equilibrium. The compressive force developed in rod AB can be determined by analyzing the equilibrium of joint A, Fig. a. ©Fy¿ = 0; FAB sin 15° - m(9.81) cos 45° = 0
FAB = 26.8014m
B
Section Properties. The cross-sectional area and moment of inertia of the rod are A =
I =
p A 0.052 B = 0.625 A 10 - 3 B pm2 4 p A 0.0254 B = 97.65625 A 10 - 9 B pm4 4
Critical Buckling Load. Since the rod is pinned at both of its ends, K = 1. Here, Pcr = FAB (F.S.) = 26.8014m(2) = 53.6028m. Applying Euler’s formula, Pcr =
p2EIy (KL)2
53.6028m =
p2 c200 A 109 B d c97.65625 A 10 - 9 B p d [1(4)]2
m = 706.11 kg = 7.06 kg
Ans.
Critical Stress. Euler’s formula is valid only if scr 6 sY. scr =
53.6028(706.11) Pcr = = 19.28 MPa 6 sY = 703 MPa A p 0.625 A 10 - 3 B
1062
60°
O.K.
C
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13–38. The members of the truss are assumed to be pin connected. If member GF is an A-36 steel rod having a diameter of 2 in., determine the greatest magnitude of load P that can be supported by the truss without causing this member to buckle.
H
16 ft P
Support Reactions: As shown on FBD(a). Member Forces: Use the method of sections [FBD(b)]. FGF = 1.3333P (C)
Section Properties: A =
I =
p 2 A 2 B = p in2 4 p 4 A 1 B = 0.250p in4 4
Critical Buckling Load: K = 1 for a column with both ends pinned. Applying Euler’s formula, Pcr = FGF =
1.3333P =
p2EI (KLGF)2 p2 (29)(103)(0.250p) [1(16)(12)]2
P = 4.573 kip = 4.57 kip
Ans.
Critical Stress: Euler’s formula is only valid if scr 6 sg. scr =
1.3333(4.573) Pcr = = 1.94 ksi 6 sg = 36 ksi p A
1063
O.K.
D
C
B 16 ft
FGF (12) - P(16) = 0
E
12 ft
A
+ ©MB = 0;
F
G
16 ft P
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13–39. The members of the truss are assumed to be pin connected. If member AG is an A-36 steel rod having a diameter of 2 in., determine the greatest magnitude of load P that can be supported by the truss without causing this member to buckle.
H
16 ft P
Support Reactions: As shown on FBD(a). Member Forces: Use the method of joints [FBD(b)]. 3 = 0 F 5 AG
FAG = 1.6667P (C)
Section Properties: LAG = 2162 + 122 = 20.0 ft
A =
I =
p 2 A 2 B = p in2 4 p 4 A 1 B = 0.250p in4 4
Critical Buckling Load: K = 1 for a column with both ends pinned. Applying Euler’s formula, Pcr = FGF =
1.6667P =
p2EI (KLGF)2 p2 (29)(103)(0.250p) [1(20)(12)]2
P = 2.342 kip = 2.34 kip
Ans.
Critical Stress: Euler’s formula is only valid if scr = sg. scr =
1.6667(2.342) Pcr = = 1.24 ksi 6 sg = 36 ksi p A
1064
O.K.
D
C
B 16 ft
P -
E
12 ft
A
+ c ©Fy = 0;
F
G
16 ft P
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*13–40. The column is supported at B by a support that does not permit rotation but allows vertical deflection. Determine the critical load Pcr . EI is constant.
L
B
Pcr A
Elastic curve: EI
d2y = M = -P y dx2
P d2y + y = 0 EI dx2 y = C1 sin c
P P x d + C2 cos c xd A EI A EI
Boundry conditions: At x = 0; 0 = 0 + C2; At x = L;
y = 0 C2 = 0 dv = 0 dx
P P dv = C1 cos c L] d = 0; dx A EI A EI cos c
P L d = 0; A EI
For n = 1 ;
Pcr =
C1
P p L = na b A EI 2
P Z 0 A EI n = 1, 3, 5
p2 P = EI 4L2
p2EI 4L2
Ans.
1065
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w
The ideal column has a weight w (force兾length) and rests in the horizontal position when it is subjected to the axial load P. Determine the maximum moment in the column at midspan. EI is constant. Hint: Establish the differential equation for deflection, Eq. 13–1, with the origin at the mid span. The general solution is v = C1 sin kx + C2 cos kx + 1w>12P22x2 - 1wL>12P22x - 1wEI>P22 where k2 = P>EI. •13–41.
P
L
Moment Functions: FBD(b). a + ©Mo = 0; M(x) =
wL x wx a b - M(x) - a bx - Pv = 0 2 2
w 2 A x - Lx B - Pv 2
[1]
Differential Equation of The Elastic Curve: EI
d2y = M(x) dx2
EI
d2y w 2 = A x - Lx B - Py 2 dx2
w d2y P y = + A x2 - Lx B EI 2EI dx2 The solution of the above differential equation is of the form v = C1 sin a
P P w 2 wL wEI xb + C2 cos ¢ xb + x x A EI A EI 2P 2P P2
[2]
dv P P P P w wL = C1 cos ¢ x ≤ - C2 sin ¢ x≤ + xdx A EI A EI A EI A EI P 2P
[3]
and
The integration constants can be determined from the boundary conditions. Boundary Condition: At x = 0, y = 0. From Eq. [2], 0 = C2 -
wEI P2
C2 =
wEI P2
At x =
L dy = 0. From Eq.[3], , 2 dx
0 = C1
P wEI P w L P L P L wL cos ¢ sin ¢ ≤ ≤ + a b A EI A EI 2 A EI 2 P 2 2P P2 A EI C1 =
wEI P L tan ¢ ≤ A EI 2 P2
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13–41.
Continued
Elastic Curve: y =
w EI P L P EI P x2 L EI x≤ + x≤ + tan ¢ cos ¢ - x B ≤ sin ¢ R P P A EI 2 A EI P A EI 2 2 P
However, y = ymax at x =
ymax =
=
L . Then, 2
P L P L P L EI w EI EI L2 tan ¢ cos ¢ B ≤ sin ¢ ≤ + ≤ R P P A EI 2 A EI 2 P A EI 2 8 P wEI P L PL2 sec - 1R B ¢ ≤ A EI 2 8EI P2
Maximum Moment: The maximum moment occurs at x =
Mmax =
L . From, Eq.[1], 2
w L2 L - L a b R - Pymax B 2 4 2
= -
wL2 wEI P L PL2 - P b 2 B sec ¢ - 1R r ≤ 8 A EI 2 8EI P
= -
PL wEI B sec ¢ ≤ - 1R P A EI 2
Ans.
13–42. The ideal column is subjected to the force F at its midpoint and the axial load P. Determine the maximum moment in the column at midspan. EI is constant. Hint: Establish the differential equation for deflection, Eq. 13–1. The general solution is v = C1 sin kx + C2 cos kx - c2x>k2, where c2 = F>2EI, k2 = P>EI.
F P
L 2
Moment Functions: FBD(b). a + ©Mo = 0;
M(x) +
F x + P(v) = 0 2 M(x) = -
F x - Pv 2
[1]
Differential Equation of The Elastic Curve: EI
d2y = M(x) dx2
EI
F d2y = - x - Py 2 2 dx
d2y F P y = x + EI 2EI dx2 The solution of the above differential equation is of the form, v = C1 sin a
P P F xb + C2 cos ¢ xb x A EI A EI 2P
1067
[2]
L 2
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13–42.
Continued
and dv P P P P F = C1 cos ¢ x ≤ - C2 sin ¢ x≤ dx A EI A EI A EI A EI 2P
[3]
The integration constants can be determined from the boundary conditions. Boundary Conditions: At x = 0, y = 0. From Eq.[2], C2 = 0 At x =
L dy = 0. From Eq.[3], , 2 dx 0 = C1
C1 =
P L P F cos ¢ ≤ A EI A EI 2 2P F EI P L sec ¢ ≤ 2P A P A EI 2
Elastic Curve: y =
F EI P L P F sec ¢ x≤ x ≤ sin ¢ 2P A P A EI 2 A EI 2P
=
F EI P L P sec ¢ x≤ - xR B ≤ sin ¢ 2P A P A EI 2 A EI
However, y = ymax at x =
ymax =
=
L . Then, 2
F EI P L P L L sec ¢ B ≤ sin ¢ ≤ - R 2P A P A EI 2 A EI 2 2 F EI P L L tan ¢ B ≤ - R 2P A P A EI 2 2
Maximum Moment: The maximum moment occurs at x =
Mmax = -
L . From Eq.[1], 2
F L a b - Pymax 2 2
= -
FL F EI P L L - Pb tan ¢ B ≤ - Rr 4 2P A P A EI 2 2
= -
P L F EI tan ¢ ≤ 2 AP A EI 2
Ans.
1068
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13–43. The column with constant EI has the end constraints shown. Determine the critical load for the column.
P
L
Moment Function. Referring to the free-body diagram of the upper part of the deflected column, Fig. a, a + ©MO = 0;
M + Pv = 0
M = -Pv
Differential Equation of the Elastic Curve. EI
d2v = M dx2
EI
d2v = -Pv dx2
d2v P + v = 0 2 EI dx The solution is in the form of v = C1 sin a
P P xb + C2 cos ¢ xb A EI A EI
(1)
dv P P P P = C1 cos ¢ x ≤ - C2 sin ¢ x≤ dx A EI A EI A EI A EI
(2)
Boundary Conditions. At x = 0, v = 0. Then Eq. (1) gives 0 = 0 + C2 At x = L,
C2 = 0
dv = 0. Then Eq. (2) gives dx 0 = C1
P P cos ¢ L≤ A EI A EI
C1 = 0 is the trivial solution, where v = 0. This means that the column will remain straight and buckling will not occur regardless of the load P. Another possible solution is cos ¢
P L≤ = 0 A EI
P np L = A EI 2
n = 1, 3, 5
The smallest critical load occurs when n = 1, then p Pcr L = A EI 2 Pcr =
p2EI 4L2
Ans.
1069
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*13–44. Consider an ideal column as in Fig. 13–10c, having both ends fixed. Show that the critical load on the column is given by Pcr = 4p2EI>L2. Hint: Due to the vertical deflection of the top of the column, a constant moment M¿ will be developed at the supports. Show that d2v>dx2 + 1P>EI2v = M¿>EI. The solution is of the form v = C1 sin11P>EIx2 + C2 cos11P>EIx2 + M¿>P. Moment Functions: M(x) = M¿ - Py Differential Equation of The Elastic Curve: EI
d2y = M(x) dx2
d2y = M¿ - Py dx2
EI
M¿ d2y P + y = 2 EI EI dx
(Q.E.D.)
The solution of the above differential equation is of the form v = C1 sin a
P P M¿ xb + C2 cos ¢ xb + A EI P A EI
[1]
and dv P P P P = C1 cos ¢ x ≤ - C2 sin ¢ x≤ dx A EI A EI A EI A EI
[2]
The integration constants can be determined from the boundary conditions. Boundary Conditions: At x = 0, y = 0. From Eq.[1], C2 = At x = 0,
M¿ P
dy = 0. From Eq.[2], C1 = 0 dx
Elastic Curve: y =
M¿ P x≤ R B 1 - cos ¢ P A EI
and M¿ P P dy = sin ¢ x≤ dx P A EI A EI However, due to symmetry sin B
L dy = 0 at x = . Then, dx 2
P L a bR = 0 A EI 2
or
P L a b = np A EI 2
where n = 1, 2, 3,...
The smallest critical load occurs when n = 1. Pce =
4p2EI L2
(Q.E.D.)
1070
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•13–45. Consider an ideal column as in Fig. 13–10d, having one end fixed and the other pinned. Show that the critical load on the column is given by Pcr = 20.19EI>L2. Hint: Due to the vertical deflection at the top of the column,a constant moment M¿ will be developed at the fixed support and horizontal reactive forces R¿ will be developed at both supports. Show that d2v>dx2 + 1P>EI2v = 1R¿>EI21L - x2. The solution is of the form v = C1 sin 11P>EIx2 + C2 cos 11P>EIx2 + 1R¿>P21L - x2. After application of the boundary conditions show that tan 11P>EIL2 = 1P>EI L. Solve by trial and error for the smallest nonzero root.
Equilibrium. FBD(a). Moment Functions: FBD(b). M(x) = R¿(L - x) - Py Differential Equation of The Elastic Curve: EI
d2y = M(x) dx2
EI
d2y = R¿(L - x) - Py dx2
d2y P R¿ + y = (L - x) 2 EI EI dx
(Q.E.D.)
The solution of the above differential equation is of the form v = C1 sin a
P P R¿ (L - x) xb + C2 cos ¢ xb + A EI P A EI
[1]
and dv P P P P R¿ = C1 cos ¢ x ≤ - C2 sin ¢ x≤ dx A EI A EI A EI A EI P The integration constants can be determined from the boundary conditions. Boundary Conditions: At x = 0, y = 0. From Eq.[1], C2 = -
At x = 0,
R¿L P
dy R¿ EI = 0. From Eq.[2], C1 = dx P AP
Elastic Curve: y =
=
R¿ EI P R¿L P R¿ sin ¢ x≤ cos ¢ x≤ + (L - x) P AP A EI P A EI P EI P P R¿ sin ¢ x ≤ - L cos ¢ x ≤ + (Lx) R B P AP A EI A EI
1071
[2]
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13–45.
Continued
However, y = 0 at x = L. Then, 0 =
P P EI sin ¢ L ≤ - L cos ¢ L≤ A EI A EI AP tan ¢
P P L≤ = L A EI A EI
(Q.E.D.)
By trial and error and choosing the smallest root, we have P L = 4.49341 A EI Then, Pcr =
20.19EI L2
(Q.E.D.)
13–46. Determine the load P required to cause the A-36 steel W8 * 15 column to fail either by buckling or by yielding. The column is fixed at its base and free at its top.
1 in.
P
Section properties for W8 * 15: A = 4.44 in2
Ix = 48.0 in4
rx = 3.29 in.
d = 8.11 in.
Iy = 3.41 in4
8 ft
Buckling about y-y axis: K = 2.0 P = Pcr =
L = 8(12) = 96 in. p2EIy (KL)2
Check: scr =
p2(29)(103)(3.41) =
[(2.0)(96)]2
= 26.5 kip
(controls)
Pcr 26.5 = = 5.96 ksi 6 sg A 4.44
Ans.
O.K.
Check yielding about x-x axis: smax =
P ec KL P c1 + 2 sec a bd A 2r A EA r
26.5 P = = 5.963 ksi A 4.44
(1) A 8.11 ec 2 B = = 0.37463 2 r (3.29)2
2.0(96) P 26.5 KL = = 0.4184 2r A EA 2(3.29) A 29(103)(4.44) smax = 5.963[1 + 0.37463 sec (0.4184)] = 8.41 ksi 6 sg = 36 ksi
1072
O.K.
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13–47. The hollow red brass C83400 copper alloy shaft is fixed at one end but free at the other end. Determine the maximum eccentric force P the shaft can support without causing it to buckle or yield. Also, find the corresponding maximum deflection of the shaft.
2m a
a
P 150 mm
30 mm
Section Properties. A = p A 0.032 - 0.022 B = 0.5 A 10 - 3 B p m2 I =
20 mm Section a – a
p A 0.034 - 0.024 B = 0.1625 A 10 - 6 B p m4 4
0.1625 A 10 B p I = = 0.01803 m C 0.5 A 10 - 3 B p AA -6
r =
e = 0.15 m
c = 0.03 m
For a column that is fixed at one end and free at the other, K = 2. Thus, KL = 2(2) = 4 m Yielding. In this case, yielding will occur before buckling. Applying the secant formula, smax =
P ec KL P B 1 + 2 sec ¢ ≤R A 2rx A EA rx
70.0 A 106 B = 70.0 A 106 B =
P
0.5 A 10
-3
Bp
P
0.5 A 10 - 3 B p
D1 +
0.15(0.03) 0.018032
secC
P 4 ST 2(0.01803)A 101 A 109 B C 0.5 A 10 - 3 B p D
a1 + 13.846 sec 8.8078 A 10 - 3 B 2Pb
Solving by trial and error, P = 5.8697 kN = 5.87 kN
Ans.
Maximum Deflection. vmax = e B sec ¢
P KL ≤ - 1R A EI 2
= 0.15 D sec C
5.8697 A 103 B
4 a b S - 1T
C 101 A 109 B C 0.1625 A 10 - 6 B p D 2
= 0.04210 m = 42.1 mm
Ans.
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*13–48. The hollow red brass C83400 copper alloy shaft is fixed at one end but free at the other end. If the eccentric force P = 5 kN is applied to the shaft as shown, determine the maximum normal stress and the maximum deflection.
2m a
a
P 150 mm
30 mm
Section Properties. A = p A 0.032 - 0.022 B = 0.5 A 10 - 3 B p m2 I =
20 mm Section a – a
p A 0.034 - 0.024 B = 0.1625 A 10 - 6 B p m4 4
0.1625 A 10 - 6 B p I = 0.01803 m = r = C 0.5 A 10 - 3 B p AA e = 0.15 m
c = 0.03 m
For a column that is fixed at one end and free at the other, K = 2. Thus, KL = 2(2) = 4 m Yielding. Applying the secant formula, smax =
P ec KL P B 1 + 2 sec ¢ ≤R A 2r A EA r 5 A 103 B
5 A 103 B 4 ST secC D1 + = 2(0.01803) C 101 A 109 B C 0.5 A 10 - 3 B p D 0.018032 0.5 A 10 - 3 B p 0.15(0.03)
Ans.
= 57.44 MPa = 57.4 MPa Since smax 6 sY = 70 MPa, the shaft does not yield. Maximum Deflection. vmax = e B sec ¢
P KL ≤ - 1R A EI 2
= 0.15 Dsec C
5 A 103 B
4 a b S - 1T
C 101 A 109 B C 0.1625 A 10 - 6 B p D 2
= 0.03467 m = 34.7 mm
Ans.
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•13–49.
The tube is made of copper and has an outer diameter of 35 mm and a wall thickness of 7 mm. Using a factor of safety with respect to buckling and yielding of F.S. = 2.5, determine the allowable eccentric load P. The tube is pin supported at its ends. Ecu = 120 GPa, sY = 750 MPa.
2m
P 14 mm
Section Properties: A =
p (0.0352 - 0.0212) = 0.61575(10 - 3) m2 4
I =
p (0.01754 - 0.01054) = 64.1152(10 - 9) m4 4
r =
I 64.1152(10 - 9) = = 0.010204 m AA A 0.61575(10 - 3)
For a column pinned at both ends, K = 1. Then KL = 1(2) = 2 m. Buckling: Applying Euler’s formula, Pmax = Pcr =
p2 (120)(109) C 64.1152(10 - 9) D p2EI = = 18983.7 N = 18.98 kN (KL)2 22
Critical Stress: Euler’s formula is only valid if scr 6 sg. scr =
Pcr 18983.7 = 30.83 MPa 6 sg = 750 MPa = A 0.61575(10 - 3)
O.K.
Yielding: Applying the secant formula, smax =
(KL) Pmax Pmax ec B 1 + 2 sec ¢ ≤R A 2r A EA r
750 A 106 B =
0.61575(10 - 3)
750 A 106 B =
0.61575(10 - 3)
Pmax
Pmax
B1 +
0.014(0.0175) 0.0102042
sec ¢
Pmax 2 ≤R 2(0.010204)A 120(109)[0.61575(10 - 3)]
A 1 + 2.35294 sec 0.01140062Pmax B
Solving by trial and error, Pmax = 16 885 N = 16.885 kN (Controls!) Factor of Safety: P =
Pmax 16.885 = = 6.75 kN F.S. 2.5
Ans.
1075
P
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13–50. The tube is made of copper and has an outer diameter of 35 mm and a wall thickness of 7 mm. Using a factor of safety with respect to buckling and yielding of F.S. = 2.5, determine the allowable eccentric load P that it can support without failure. The tube is fixed supported at its ends. Ecu = 120 GPa, sY = 750 MPa.
2m
P 14 mm
Section Properties: A =
p A 0.0352 - 0.0212 B = 0.61575 A 10 - 3 B m2 4
I =
p A 0.01754 - 0.01054 B = 64.1152 A 10 - 9 B m4 4
r =
I 64.1152(10 - 9) = = 0.010204 ms AA A 0.61575(10 - 3)
For a column fixed at both ends, K = 0.5. Then KL = 0.5(2) = 1 m. Buckling: Applying Euler’s formula, Pmax = Pcr =
p2(120)(109) C 64.1152(10 - 9) D p2EI = = 75 935.0 N = 75.93 kN (KL)2 12
Critical Stress: Euler’s formula is only valid if scr 6 sg. scr =
Pcr 75 935.0 = 123.3 MPa 6 sg = 750 MPa = A 0.61575(10 - 3)
O. K.
Yielding: Applying the secant formula, smax =
(KL) Pmax Pmax ec B 1 + 2 sec ¢ ≤R A 2r A EA r
750 A 106 B =
0.61575(10 - 3)
750 A 106 B =
0.61575(10 - 3)
Pmax
Pmax
B1 +
0.014(0.0175) 0.0102042
sec ¢
2 Pmax ≤R 2(0.010204)A 120(109)[0.61575(10 - 3)]
A 1 + 2.35294 sec 5.70032 A 10 - 3 B 2P B
Solving by trial and error, Pmax = 50 325 N = 50.325 kN (Controls!) Factor of Safety: P =
Pmax 50.325 = = 20.1 kN F.S. 2.5
Ans.
1076
P
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13–51. The wood column is fixed at its base and can be assumed pin connected at its top. Determine the maximum eccentric load P that can be applied without causing the column to buckle or yield. Ew = 1.811032 ksi, sY = 8 ksi.
P
y 4 in. x
x P y 10 in.
10 ft
Section Properties: A = 10(4) = 40 in2
ry =
Iy =
1 (4)(103) = 333.33 in4 12
Ix =
1 (10)(43) = 53.33 in4 12
Ix 333.33 = = 2.8868 in. AA A 40
Buckling about x-x axis: P = Pcr =
p2(1.8)(103)(53.33) p2EI = = 134 kip (KL)2 [(0.7)(10)(12)]2
Check: scr =
Pcr 134 = = 3.36 ksi 6 sg A 40
O.K.
Yielding about y -y axis: smax =
P ec KL P a1 + 2 seca b b A 2r A EA r
5(5) ec = = 3.0 r2 2.88682 a
0.7(10)(12) P P KL b = = 0.0542212P 2r A EA 2(2.8868) A 1.8(103)(40)
8(40) = P[1 + 3.0 sec (0.0542212P)] By trial and error: P = 73.5 kip
Ans.
(controls)
1077
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*13–52. The wood column is fixed at its base and can be assumed fixed connected at its top. Determine the maximum eccentric load P that can be applied without causing the column to buckle or yield. Ew = 1.811032 ksi, sY = 8 ksi.
P
y 4 in. x
x P y 10 in.
10 ft
Section Properties: A = 10(4) = 40 in2
ry =
Iy =
1 (4)(103) = 333.33 in4 12
Ix =
1 (10)(43) = 53.33 in4 12
Iy 333.33 = = 2.8868 in. AA A 40
Buckling about x-x axis: P = Pcr =
p2(1.8)(103)(53.33) p2EI = = 263 kip 2 (KL) [(0.5)(10)(12)]2
Check: scr =
Pcr 263 = = 6.58 ksi 6 sg A 40
O.K.
Yielding about y-y axis: smax =
P ec KL P a1 + 2 seca bb A 2r A EA r
5(5) ec = = 3.0 r2 2.88682 a
0.5(10)(12) P P KL b = = 0.0387292P 2r A EA 2(2.8868) A 1.8(103)(40)
8(40) = P[1 + 3.0 sec (0.0387292P)] By trial and error: P = 76.5 kip
Ans.
(controls)
1078
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The W200 * 22 A-36-steel column is fixed at its base. Its top is constrained to rotate about the y–y axis and free to move along the y–y axis. Also, the column is braced along the x–x axis at its mid-height. Determine the allowable eccentric force P that can be applied without causing the column either to buckle or yield. Use F.S. = 2 against buckling and F.S. = 1.5 against yielding.
•13–53.
100 mm
y
A = 2860 mm2 = 2.86 A 10 - 3 B m2
ry = 22.3 mm = 0.0223 m
Ix = 20.0 A 106 B mm4 = 20.0 A 10 - 6 B m4
c =
bf 2
=
102 = 51 mm = 0.051 m 2
e = 0.1m Buckling About the Strong Axis. Since the column is fixed at the base and free at the top, Kx = 2. Applying Euler’s formula,
Pcr =
p2EIx (KL)x 2
=
p2 c200 A 109 B d c20.0 A 10 - 6 B d [2(10)]2
= 98.70kN
Euler’s formula is valid if scr 6 sY. scr =
98.70 A 103 B Pcr = = 34.51 MPa 6 sY = 250MPa A 2.86 A 10 - 3 B
O.K.
Then, Pallow =
Pcr 98.70 = = 49.35 kN F.S. 2
Yielding About Weak Axis. Since the support provided by the bracing can be considered a pin connection, the upper portion of the column is pinned at both of its ends. Then Ky = 1 and L = 5 m. Applying the secant formula, smax =
A KL B y Pmax Pmax ec C 1 + 2 sec B RS A 2ry A EA ry
250 A 106 B =
250 A 106 B =
Pmax
2.86 A 10
-3
Pmax
B
2.86 A 10 - 3 B
D1 +
0.1(0.051) 0.02232
secC
1(5) Pmax ST 2(0.0223)A 200 A 109 B C 2.86 A 10 - 3 B D
c1 + 10.2556 sec 4.6875 A 10 - 3 B 2Pmax d
Solving by trial and error, Pmax = 39.376 kN Then, Pallow =
Pmax 39.376 = = 26.3 kN (controls) 1.5 1.5
Ans.
1079
5m
y
x
5m
Section Properties. From the table listed in the appendix, the necessary section properties for a W200 * 22 are
P
x
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13–54. The W200 * 22 A-36-steel column is fixed at its base. Its top is constrained to rotate about the y–y axis and free to move along the y–y axis. Also, the column is braced along the x–x axis at its mid-height. If P = 25 kN, determine the maximum normal stress developed in the column.
100 mm
y
Section Properties. From the table listed in the appendix, necessary section properties for a W200 * 22 are A = 2860 mm2 = 2.86 A 10 - 3 B m2
ry = 22.3 mm = 0.0223 m
Ix = 20.0 A 106 B mm4 = 20.0 A 10 - 6 B m4
c =
bf 2
=
102 = 51 mm = 0.051 m 2
e = 0.1m Buckling About the Strong Axis. Since the column is fixed at the base and free at the top, Kx = 2. Applying Euler’s formula,
Pcr =
p2EIx (KL)x 2
=
p2 c200 A 109 B d c20.0 A 10 - 6 B d [2(10)]2
= 98.70kN
Euler’s formula is valid only if scr 6 sY. scr =
98.70 A 103 B Pcr = = 34.51 MPa 6 sY = 250 MPa A 2.86 A 10 - 3 B
O.K.
Since P = 25 kN 6 Pcr, the column does not buckle. Yielding About Weak Axis. Since the support provided by the bracing can be considered a pin connection, the upper portion of the column is pinned at both of its ends. Then Ky = 1 and L = 5 m. Applying the secant formula, smax =
=
(KL) P P ec C 1 + 2 sec B RS A 2ry A EA ry 2.5 A 103 B
2.86 A 10 - 3 B
D1 +
0.1(0.051) 0.02232
secC
25 A 103 B 1(5) ST 2(0.0223) C 200 A 109 B C 2.86 A 10 - 3 B D
= 130.26 MPa = 130 MPa
Ans.
Since smax 6 sY = 250 MPa, the column does not yield.
1080
y
x
5m
5m
P
x
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13–55. The wood column is fixed at its base, and its top can be considered pinned. If the eccentric force P = 10 kN is applied to the column, investigate whether the column is adequate to support this loading without buckling or yielding. Take E = 10 GPa and sY = 15 MPa.
x
P 150 mm
25 mm yx 25 mm 75 mm
5m
Section Properties. A = 0.05(0.15) = 7.5 A 10 - 3 B m2 Ix =
rx =
1 (0.05) A 0.153 B = 14.0625 A 10 - 6 B m4 12
14.0625 A 10 - 6 B Ix = 0.04330 m = AA C 7.5 A 10 - 3 B
1 (0.15) A 0.053 B = 1.5625 A 10 - 6 B m4 12 e = 0.15 m c = 0.075 m
Iy =
For a column that is fixed at one end and pinned at the other K = 0.7. Then, (KL)x = (KL)y = 0.7(5) = 3.5 m Buckling About the Weak Axis. Applying Euler’s formula, Pcr =
p2EIy (KL)y 2
=
p2 C 10 A 109 B D C 1.5625 A 10 - 6 B D 3.52
= 12.59 kN
Euler’s formula is valid if scr 6 sY. scr =
12.59 A 103 B Pcr = = 1.68 MPa 6 sY = 15 MPa A 7.5 A 10 - 3 B
O.K.
Since Pcr 7 P = 10 kN, the column will not buckle. Yielding About Strong Axis. Applying the secant formula. smax =
=
(KL)x P ec P C 1 + 2 sec B RS A 2rx A EA rx 10 A 103 B
7.5 A 10 - 3 B
D1 +
0.15(0.075) 2
0.04330
secC
10 A 103 B 3.5 ST 2(0.04330) C 10 A 109 B C 7.5 A 10 - 3 B D
= 10.29 MPa Since smax 6 sY = 15 MPa , the column will not yield.
Ans.
1081
75 mm
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*13–56. The wood column is fixed at its base, and its top can be considered pinned. Determine the maximum eccentric force P the column can support without causing it to either buckle or yield. Take E = 10 GPa and sY = 15 MPa .
x
P 150 mm
25 mm yx 25 mm 75 mm
5m
Section Properties. A = 0.05(0.15) = 7.5 A 10 - 3 B m2 Ix =
1 (0.05) A 0.153 B = 14.0625 A 10 - 6 B m4 12
14.0625 A 10 Ix = C 7.5 A 10 - 3 B AA
-6
rx =
B
= 0.04330 m
1 (0.15) A 0.053 B = 1.5625 A 10 - 6 B m4 12 e = 0.15 m c = 0.075 m
Iy =
For a column that is fixed at one end and pinned at the other K = 0.7. Then, (KL)x = (KL)y = 0.7(5) = 3.5 m Buckling About the Weak Axis. Applying Euler’s formula, Pcr =
p2EIy (KL)y
2
=
p2 C 10 A 109 B D C 1.5625 A 10 - 6 B D 3.52
= 12.59 kN = 12.6 kN
Ans.
Euler’s formula is valid if scr 6 sY. scr
12.59 A 103 B Pcr = = = 1.68 MPa 6 sY = 15 MPa A 7.5 A 10 - 3 B
O.K.
Yielding About Strong Axis. Applying the secant formula with P = Pcr = 12.59 kN, smax =
=
(KL)x P P ec C B 1 + 2 sec B RS A 2rx A EA rx 12.59 A 103 B 7.5 A 10 - 3 B
D1 +
0.15(0.075) 0.043302
secC
12.59 A 103 B B 3.5 ST 2(0.04330) C 10 A 109 B C 7.5 A 10 - 3 B D
= 13.31 MPa 6 sY = 15 MPa
O.K.
1082
75 mm
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The W250 * 28 A-36-steel column is fixed at its base. Its top is constrained to rotate about the y–y axis and free to move along the y–y axis. If e = 350 mm, determine the allowable eccentric force P that can be applied without causing the column either to buckle or yield. Use F.S. = 2 against buckling and F.S. = 1.5 against yielding.
•13–57.
P e x
x y
6m
Section Properties. From the table listed in the appendix, necessary section properties for a W250 * 28 are A = 3620 mm2 = 3.62 A 10 - 3 B m2
rx = 105 mm = 0.105 m
Iy = 1.78 A 106 B mm4 = 1.78 A 10 - 6 B m4
c =
260 d = = 130 mm = 0.13 m 2 2
e = 0.35 m Buckling About the Strong Axis. Since the column is fixed at the base and pinned at the top, Kx = 0.7. Applying Euler’s formula, Pcr =
p2EIy (KL)y 2
=
p2 C 200 A 109 B D C 1.78 A 10 - 6 B D [0.7(6)]2
= 199.18 kN
Euler’s formula is valid only if scr 6 sY. scr =
199.18 A 103 B Pcr = = 55.02 MPa 6 sY = 250 MPa A 3.62 A 10 - 3 B
O.K.
Thus, Pallow =
Pcr 199.18 = = 99.59 kN F.S. 2
Yielding About Strong Axis. Since the column is fixed at its base and free at its top, Kx = 2. Applying the secant formula, smax =
(KL)x Pmax Pmax ec C 1 + 2 sec B RS A 2rx A EA rx
250 A 106 B =
250 A 106 B =
Pmax
3.62 A 10
-3
Pmax
B
3.62 A 10 - 3 B
D1 +
0.35(0.13) 0.1052
secC
2(6) Pmax ST 2(0.105)A 200 A 109 B C 3.62 A 10 - 3 B D
A 1 + 4.1270 sec (0.0021237)2Pmax B
Solving by trial and error, Pmax = 133.45 kN Then, Pallow =
y
Pmax 133.45 = = 88.97 kN = 89.0 kN (controls) 1.5 1.5
Ans.
1083
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13–58. The W250 * 28 A-36-steel column is fixed at its base. Its top is constrained to rotate about the y–y axis and free to move along the y–y axis. Determine the force P and its eccentricity e so that the column will yield and buckle simultaneously.
P e x
x y
6m
Section Properties. From the table listed in the appendix, the necessary section properties for a W250 * 28 are A = 3620 mm2 = 3.62 A 10 - 3 B m2
rx = 105 mm = 0.105 m
Iy = 1.78 A 106 B mm4 = 1.78 A 10 - 6 B m4
c =
d 260 = = 130 mm = 0.13 m 2 2
Buckling About the Weak Axis. Since the column is fixed at the base and pinned at its top, Kx = 0.7. Applying Euler’s formula, Pcr =
p2EIy (KL)y 2
=
p2 C 200 A 109 B D C 1.78 A 10 - 6 B D [0.7(6)]2
= 199.18 kN = 199 kN
Ans.
Euler’s formula is valid only if scr 6 sY. scr =
199.18 A 103 B Pcr = = 55.02 MPa 6 sY = 250 MPa A 3.62 A 10 - 3 B
O.K.
Yielding About Strong Axis. Since the column is fixed at its base and free at its top, Kx = 2. Applying the secant formula with P = Pcr = 199.18 kN, smax =
(KL)x P P ec C 1 + 2 sec B RS A 2rx A EA rx
250 A 106 B =
199.18 A 103 B 3.62 A 10 - 3 B
D1 +
e(0.13) 0.1052
secC
y
199.18 A 103 B 2(6) ST 2(0.105) C 200 A 109 B C 3.62 A 10 - 3 B D
e = 0.1753 m = 175 mm
Ans.
1084
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13–59. The steel column supports the two eccentric loadings. If it is assumed to be pinned at its top, fixed at the bottom, and fully braced against buckling about the y–y axis, determine the maximum deflection of the column and the maximum stress in the column. Est = 200 GPa, sY = 360 MPa.
130 kN 50 kN 80 mm 120 mm
100 mm 10 mm 6m
A = 0.12(0.1) - (0.1)(0.09) = 3.00 A 10 - 3 B m2 Ix =
1 1 (0.1) A 0.123 B (0.09) A 0.13 B = 6.90 A 10 - 6 B m4 12 12
rx =
Ix 6.90(10 - 6) = = 0.047958 m AA A 3.00(10 - 3)
For a column fixed at one end and pinned at the other end, K = 0.7. (KL)x = 0.7(6) = 4.2 m The eccentricity of the two applied loads is, e =
130(0.12) - 50(0.08) = 0.06444 m 180
Yielding About x–x Axis: Applying the secant formula, (KL)x P P ec B 1 + 2 sec ¢ ≤R A 2rx A EA rx 180(103) =
-3
3.00(10 )
y
B1 +
0.06444(0.06) 0.047958
2
sec ¢
180(103) 4.2 ≤R 2(0.047958)A 200(109)(3.00)(10 - 3)
= 199 MPa
Ans.
Since smax 6 sg = 360 MPa, the column does not yield. Maximum Displacement: ymax = e B sec ¢
P KL ≤ - 1R A EI 2
= 0.06444 B sec ¢
4.2 180(103) a b ≤ - 1R A 200(109)[6.90(10 - 6)] 2
= 0.02433 m = 24.3 mm
Ans.
1085
10 mm 100 mm
x
Section Properties:
smax =
x
10 mm
y
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*13–60. The steel column supports the two eccentric loadings. If it is assumed to be fixed at its top and bottom, and braced against buckling about the y–y axis, determine the maximum deflection of the column and the maximum stress in the column. Est = 200 GPa, sY = 360 MPa.
130 kN 50 kN 80 mm 120 mm
100 mm 10 mm 6m
A = 0.12(0.1) - (0.1)(0.09) = 3.00 A 10 - 3 B m2 Ix =
1 1 (0.1) A 0.123 B (0.09) A 0.013 B = 6.90 A 10 - 6 B m4 12 12
rx =
6.90(10 - 6) Ix = = 0.047958 m AA A 3.00(10 - 3)
For a column fixed at both ends, K = 0.5. (KL)x = 0.5(6) = 3.00 m The eccentricity of the two applied loads is, e =
130(0.12) - 50(0.08) = 0.06444 m 180
Yielding About x–x Axis: Applying the secant formula, (KL)x P P ec B 1 + 2 sec ¢ ≤R A 2rx A EA rx 180(103) =
-3
3.00(10 )
y
B1 +
0.06444(0.06) 0.047958
2
sec ¢
180(103) 3.00 ≤R 2(0.047958)A 200(109)(3.00)(10 - 3)
= 178 MPa
Ans.
Since smax 6 sg = 360 MPa, the column does not yield. Maximum Displacement: ymax = e B sec ¢
P KL ≤ - 1R A EI 2
= 0.06444 B sec ¢
3 180(103) a b ≤ - 1R 9 6 A 200(10 )[6.90(10 )] 2
= 0.01077 m = 10.8 mm
Ans.
1086
10 mm 100 mm
x
Section Properties:
smax =
x
10 mm
y
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13–61. The W250 * 45 A-36-steel column is pinned at its top and fixed at its base. Also, the column is braced along its weak axis at mid-height. If P = 250 kN, investigate whether the column is adequate to support this loading. Use F.S. = 2 against buckling and F.S. = 1.5 against yielding.
P 4 250 mm
P 250 mm
4m
Section Properties. From the table listed in the appendix, the necessary section properties for a W250 * 45 are A = 5700 mm = 5.70 A 10 2
-3
Bm
2
rx = 112 mm = 0.112 m
Iy = 7.03 A 106 B mm4 = 7.03 A 10 - 6 B m4
c =
266 d = = 133 mm = 0.133 m 2 2
The eccentricity of the equivalent force P¿ = 250 +
250 = 312.5 kN is 4
250 (0.25) 4 = 0.15 m 250 250 + 4
250(0.25) e =
Buckling About the Weak Axis. The column is braced along the weak axis at midheight and the support provided by the bracing can be considered as a pin. The top portion of the column is critical is since the top is pinned so Ky = 1 and L = 4 m Applying Euler’s formula, Pcr =
p2EIy (KL)y 2
=
p2 C 200 A 109 B D C 7.03 A 10 - 6 B D [1(4)]2
= 867.29 kN
Euler’s equation is valid only if scr 6 sY. scr =
867.29 A 103 B Pcr = = 152.16 MPa 6 sY = 250 MPa A 5.70 A 10 - 3 B
O.K.
Then, œ Pallow =
Pcr 867.29 = = 433.65 kN F.S. 2
œ Since Pallow 7 P¿ , the column does not buckle.
Yielding About Strong Axis. Since the column is fixed at its base and pinned at its top, Kx = 0.7 and L = 8 m. Applying the secant formula with œ = P¿(F.S.) = 312.5(1.5) = 468.75 kN Pmax smax =
=
œ œ (KL)x Pmax Pmax ec C 1 + 2 sec B RS A 2rx A EA rx
468.75 A 103 B 5.70 A 10 - 3 B
C1 +
0.15(0.133) 0.112
2
sec B
468.75 A 103 B 0.7(8) RS 2(0.112) C 200 A 109 B C 5.70 A 10 - 3 B D
= 231.84 MPa Since smax 6 sY = 250 MPa, the column does not yield.
1087
4m
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The W250 * 45 A-36-steel column is pinned at its top and fixed at its base. Also, the column is braced along its weak axis at mid-height. Determine the allowable force P that the column can support without causing it either to buckle or yield. Use F.S. = 2 against buckling and F.S. = 1.5 against yielding.
•13–62.
P 4 250 mm
P 250 mm
4m
Section Properties. From the table listed in the appendix, the necessary section properties for a W250 * 45 are A = 5700 mm2 = 5.70 A 10 - 3 B m2
rx = 112 mm = 0.112 m
Iy = 7.03 A 106 B mm4 = 7.03 A 10 - 6 B m4
c =
The eccentricity of the equivalent force P¿ = P +
d 266 = = 133 mm = 0.133 m 2 2 P = 1.25P is 4
P (0.25) 4 = 0.15 m P P + 4
P(0.25) e =
Buckling About the Weak Axis. The column is braced along the weak axis at midheight and the support provided by the bracing can be considered as a pin. The top portion of the column is critical is since the top is pinned so Ky = 1 and L = 4 m. Applying Euler’s formula, Pcr =
p2EIy (KL)y
2
=
p2 C 200 A 109 B D C 7.03 A 10 - 6 B D [1(4)]2
= 867.29 kN
Euler’s equation is valid only if scr 6 sY. scr
867.29 A 103 B Pcr = = = 152.16 MPa 6 sY = 250 MPa A 5.70 A 10 - 3 B
O.K.
Then, œ = Pallow
Pcr F.S.
867.29 2 = 346.92 kN
1.25Pallow = Pallow
Yielding About Strong Axis. Since the column is fixed at its base and pinned at its top, Kx = 0.7 and L = 8 m. Applying the secant formula, smax =
œ œ (KL)x Pmax Pmax ec C 1 + 2 sec B RS A 2rx A EA rx
250 A 106 B = 250 A 106 B =
1.25Pmax
5.70 A 10
-3
1.25Pmax
B
5.70 A 10 - 3 B
C1 +
0.15(0.133) 0.1122
sec B
0.7(8) 1.25Pmax RS 2(0.112) A 200 A 109 B C 5.70 A 10 - 3 B D
A 1 + 1.5904 sec (0.00082783)2Pmax B
Solving by trial and error, Pmax = 401.75 kN Then, Pallow =
401.75 = 267.83 kN = 268 kN (controls) 1.5
Ans.
1088
4m
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13–63. The W14 * 26 structural A-36 steel member is used as a 20-ft-long column that is assumed to be fixed at its top and fixed at its bottom. If the 15-kip load is applied at an eccentric distance of 10 in., determine the maximum stress in the column.
15 kip
Section Properties for W 14 * 26 A = 7.69 in2
d = 13.91 in.
10 in.
20 ft
rx = 5.65 in.
Yielding about x-x axis: smax =
P ec KL P b d; c1 + 2 sec a A 2 r AE A r
15 P = = 1.9506 ksi; A 7.69
K = 0.5
10 A 13.91 ec 2 B = = 2.178714 r2 (5.65)2
0.5 (20)(12) 15 KL P = = 0.087094 2 r A EA 2(5.65) A 29 (103)(7.69) smax = 1.9506[1 + 2.178714 sec (0.087094)] = 6.22 ksi 6 sg = 36 ksi
O.K.
Ans.
*13–64. The W14 * 26 structural A-36 steel member is used as a column that is assumed to be fixed at its top and pinned at its bottom. If the 15-kip load is applied at an eccentric distance of 10 in., determine the maximum stress in the column.
15 kip
Section Properties for W 14 * 26 A = 7.69 in2
d = 13.91 in.
20 ft
rx = 5.65 in.
Yielding about x-x axis: smax =
P ec KL P c1 + 2 sec a b d; A 2 r AE A r
15 P = = 1.9506 ksi ; A 7.69
K = 0.7
10 A 13.91 ec 2 B = = 2.178714 2 r (5.65)2
0.7 (20)(12) 15 KL P = = 0.121931 2 r A EA 2(5.65) A 29 (103)(7.69) smax = 1.9506[1 + 2.178714 sec (0.121931)] = 6.24 ksi 6 sg = 36 ksi
10 in.
O.K.
Ans.
1089
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•13–65. Determine the maximum eccentric load P the 2014-T6-aluminum-alloy strut can support without causing it either to buckle or yield. The ends of the strut are pin-connected.
P 150 mm
100 mm
a a 3m 50 mm
100 mm Section a – a
Section Properties. The necessary section properties are A = 0.05(0.1) = 5 A 10 - 3 B m2 Iy =
1 (0.1) A 0.053 B = 1.04167 A 10 - 6 B m4 12
4.1667 A 10 Ix = C 5 A 10 - 3 B AA
-6
rx =
B
= 0.02887 m
For a column that is pinned at both of its ends K = 1. Thus, (KL)x = (KL)y = 1(3) = 3 m Buckling About the Weak Axis. Applying Euler’s formula, Pcr =
p2EIy (KL)y 2
=
p2 C 73.1 A 109 B D C 1.04167 A 10 - 6 B D 32
= 83.50 kN = 83.5 kN
Ans.
Critical Stress: Euler’s formula is valid only if scr 6 sY. scr =
83.50 A 103 B Pcr = = 16.70 MPa 6 sY = 414 MPa A 5 A 10 - 3 B
O.K.
Yielding About Strong Axis. Applying the secant formula, smax =
=
(KL)x P P ec C 1 + 2 sec B RS A 2rx A EA rx
83.50 A 103 B 5 A 10 - 3 B
D1 +
0.15(0.05) 0.028872
83.50 A 10 B 3 ST 2(0.02887) C 73.1 A 109 B C 5 A 10 - 3 B D 3
secC
= 229.27 MPa 6 sY = 414 MPa
O.K.
1090
P 150 mm
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13–66. The W8 * 48 structural A-36 steel column is fixed at its bottom and free at its top. If it is subjected to the eccentric load of 75 kip, determine the factor of safety with respect to either the initiation of buckling or yielding.
75 kip
y
12 ft
Section Properties: For a wide flange section W8 * 48, A = 14.1 in2
rx = 3.61 in.
Iy = 60.9 in4
d = 8.50 in.
For a column fixed at one end and free at the other and, K = 2. (KL)y = (KL)x = 2(12)(12) = 288 in. Buckling About y–y Axis: Applying Euler’s formula, P = Pcr =
p2EIy (KL)2y p2 (29.0)(103)(60.9)
=
2882
= 210.15 kip Critical Stress: Euler’s formula is only valid if scr 6 sg. scr =
Pcr 210.15 = = 14.90 ksi 6 sg = 36 ksi A 14.1
O. K.
Yielding About x–x Axis: Applying the secant formula, smax =
36 =
(KL)x Pmax Pmax ec B 1 + 2 sec ¢ ≤R A 2rx A EA rx
8 A 1.50 Pmax Pmax 288 2 B sec ¢ B1 + ≤R 14.1 2(3.61)A 29.0(103)(14.1) 3.612
36(14.1) = Pmax A 1 + 2.608943 sec 0.06238022Pmax B Solving by trial and error, Pmax = 117.0 kip (Controls!) Factor of Safety: F.S. =
Pmax 117.0 = = 1.56 P 75
Ans.
1091
8 in. y x
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13–67. The W8 * 48 structural A-36 steel column is fixed at its bottom and pinned at its top. If it is subjected to the eccentric load of 75 kip, determine if the column fails by yielding. The column is braced so that it does not buckle about the y–y axis.
75 kip
y
12 ft
Section Properties: For a wide flange section W8 * 48, A = 14.1 in2
rx = 3.61 in.
d = 8.50 in.
For a column fixed at one end and pinned at the other end, K = 0.7. (KL)x = 0.7(12)(12) = 100.8 in. Yielding About x–x Axis: Applying the secant formula, smax =
=
(KL)x P P ec B 1 + 2 sec ¢ ≤R A 2rx A EA rx
8 A 1.50 75 100.8 75 2 B sec ¢ B1 + ≤R 2 14.1 2(3.61)A 3.61 29.0(103)(14.1)
= 19.45 ksi 6 sg = 36 ksi
O.K.
Hence, the column does not fail by yielding.
Ans.
1092
8 in. y x
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*13–68. Determine the load P required to cause the steel W12 * 50 structural A-36 steel column to fail either by buckling or by yielding. The column is fixed at its bottom and the cables at its top act as a pin to hold it.
2 in.
P
25 ft
Section Properties: For a wide flange section W12 * 50, A = 14.7 in2
rx = 5.18 in.
Iy = 56.3 in4
d = 12.19 in.
For a column fixed at one end and pinned at the other end, K = 0.7. (KL)y = (KL)x = 0.7(25)(12) = 210 in. Buckling About y–y Axis: Applying Euler’s formula, P = Pcr =
p2EIy (KL)2y p2 (29.0)(103)(56.3)
=
2102
= 365.40 kip Critical Stress: Euler’s formula is only valid if scr 6 sg. scr =
Pcr 365.4 = = 24.86 ksi 6 sg = 36 ksi A 14.7
O. K.
Yielding About x–x Axis: Applying the secant formula, smax =
36 =
(KL)x P P ec B 1 + 2 sec ¢ ≤R A 2rx A EA rx
2 A 12.19 P 210 P 2 B sec ¢ B1 + ≤R 14.7 2(5.18)A 29.0(103)(14.7) 5.182
36(14.7) = P A 1 + 0.454302 sec 0.03104572P B Solving by trial and error, Pmax = 343.3 kip = 343 kip (Controls!)
Ans.
1093
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•13–69.
Solve Prob. 13–68 if the column is an A-36 steel W12 * 16 section.
2 in.
P
25 ft
Section Properties: For a wide flange section W12 * 16, A = 4.71 in2
rx = 4.67 in.
Iy = 2.82 in4
d = 11.99 in.
For a column fixed at one end and pinned at the other end, K = 0.7. (KL)y = (KL)x = 0.7(25)(12) = 210 in. Buckling About y–y Axis: Applying Euler’s formula, P = Pcr =
p 2EIy (KL)2y p2 (29.0)(103)(2.82)
=
2102
= 18.30 kip = 18.3 kip‚
Ans.
Critical Stress: Euler’s formula is only valid if scr 6 sg. scr =
Pcr 18.30 = = 3.89 ksi 6 sg = 36 ksi A 4.71
Yielding About x–x Axis: Applying the secant formula, smax =
=
(KL)x P P ec B 1 + 2 sec ¢ ≤R A 2rx A EA rx
2 A 11.99 18.30 210 18.30 2 B seca bR B1 + 4.71 2(4.67)A 29.0(103)(4.71) 4.672
= 6.10 ksi 6 sg = 36 ksi
O.K.
1094
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13–70. A column of intermediate length buckles when the compressive stress is 40 ksi. If the slenderness ratio is 60, determine the tangent modulus. p2 Et
A KrL B 2
scr =
40 =
;
a
KL b = 60 r
p2 Et (60)2
Et = 14590 ksi = 14.6 (103) ksi ‚
Ans
s(ksi)
13–71. The 6-ft-long column has the cross section shown and is made of material which has a stress-strain diagram that can be approximated as shown. If the column is pinned at both ends, determine the critical load Pcr for the column.
0.5 in. 55 0.5 in.
5 in.
25
0.5 in. 3 in.
Section Properties: The neccessary section properties are A = 2[0.5(3)] + 5(0.5) = 5.5 in2
P (in./in.) 0.001
I = 2B ry =
1 1 (0.5) A 33 B R + (5) A 0.53 B = 2.3021 in4 12 12
2.3021 Iy = = 0.6470 in. AA A 5.5
For the column pinned at both of its ends, K = 1. Thus, 1(6)(12) KL = 111.29 = ry 0.6470 Critical Stress. Applying Engesser’s equation,
scr =
p2Et a
KL
r
b
=
p2Et 111.292
= 0.7969 A 10 - 3 B Et
(1)
From the stress - strain diagram, the tangent moduli are (Et)1 =
25 ksi = 25 A 103 B ksi 0.001
(Et)2 =
(55 - 25) ksi = 10 A 103 B (ksi) 0.004 - 0.001
0 … s 6 25 ksi
25ksi 6 s … 40 ksi
Substituting (Et)1 = 25 A 103 B into Eq. (1),
scr = 0.7969 A 10 - 3 B c25 A 103 B d = 19.92 ksi
Since scr 6 sY = 25 ksi, elastic buckling occurs. Thus, Pcr = scr A = 19.92(5.5) = 109.57 kip = 110 kip‚
Ans. 1095
0.004
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s(ksi)
*13–72. The 6-ft-long column has the cross section shown and is made of material which has a stress-strain diagram that can be approximated as shown. If the column is fixed at both ends, determine the critical load Pcr for the column.
0.5 in. 55 0.5 in.
5 in.
25
0.5 in. 3 in.
Section Properties. The neccessary section properties are A = 2[0.5(3)] + 5(0.5) = 5.5 in2
P (in./in.) 0.001
I = 2B ry =
1 1 (0.5) A 33 B R + (5) A 0.53 B = 2.3021 in4 12 12
Iy 2.3021 = = 0.6470 in. AA A 5.5
For the column fixed at its ends, K = 0.5. Thus, 0.5(6)(12) KL = 55.64 = ry 0.6470 Critical Stress. Applying Engesser’s equation, From the stress - strain diagram, the tangent moduli are (Et)1 =
25 ksi = 25 A 103 B ksi 0.001
(Et)2 =
(55 - 25)ksi = 10 A 103 B ksi 0.004 - 0.001
0 … s 6 25 ksi
25 ksi 6 s … 40 ksi
Substituting (Et)1 = 25 A 103 B ksi into Eq. (1),
scr = 3.1875 A 10 - 3 B c25 A 103 B d = 79.69 ksi
Since scr 7 sY = 25 ksi, the inelastic buckling occurs. Substituting (Et)2 into Eq. (1), scr = 3.1875 A 10 - 3 B c10 A 103 B d = 31.88 ksi Since 25 ksi 6 scr 6 55 ksi, this result can be used to calculate the critical load. Pcr = scr A = 31.88(5.5) = 175.31 kip = 175 kip
Ans.
1096
0.004
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s (MPa)
•13–73.
The stress-strain diagram of the material of a column can be approximated as shown. Plot P兾A vs. KL兾r for the column.
350
200
Tangent Moduli. From the stress - strain diagram, (Et)1 =
(Et)2 =
200 A 106 B 0.001
0 … s 6 200MPa
= 200 GPa
(350 - 200) A 106 B 0.004 - 0.001
= 50 GPa
0
P (in./in.) 0.001
0.004
200MPa 6 s … 350 MPa
Critical Stress. Applying Engesser’s equation, scr
P = A
p2Et
(1)
KL 2 a b r
If Et = (Et)1 = 200 GPa, Eq. (1) becomes p2 C 200 A 109 B D 1.974 A 106 B P = = MPa A KL 2 KL 2 a b a b r r
when scr =
P = sY = 200 MPa, this equation becomes A
200 A 106 B =
p2 C 200 A 109 B D a
KL 2 b r
KL = 99.346 = 99.3 r
If Et = (Et)2 = 50 GPa, Eq. (1) becomes P = A
p2 c50 A 109 B d
0.4935 A 106 B
MPa KL 2 KL 2 ¢ ≤ ¢ ≤ r r P when scr = = sY = 200 MPa, this A equation gives 200 A 106 B =
=
p2 C 50 A 109 B D a
KL 2 b r
KL = 49.67 = 49.7 r Using these results, the graphs of
P KL vs. is shown in Fig. a can be plotted. r A 1097
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s (MPa)
13–74. Construct the buckling curve, P兾A versus L兾r, for a column that has a bilinear stress–strain curve in compression as shown. The column is pinned at its ends.
260
140
0.001
Tangent modulus: From the stress–strain diagram, (Et)1 =
140(106) = 140 GPa 0.001
(Et)2 =
(260 - 140)(106) = 40 GPa 0.004 - 0.001
Critical Stress: Applying Engesser’s equation, scr
p2Et P = A L 2 a b r
[1]
Substituting (Et)1 = 140 GPa into Eq. [1], we have p2 C 140(109) D P = A ALB2 r
P = A When
1.38(106)
A Lr B 2
MPa
L P = 140 MPa, = 99.3 r A
Substitute (Et)2 = 40 GPa into Eq. [1], we have p2 C 40(109) D P = A ALB2 r
P = A When
0.395(106)
A Lr B 2
MPa
L P = 140 MPa, = 53.1 r A
1098
0.004
P (mm/mm)
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13–75. The stress-strain diagram for a material can be approximated by the two line segments shown. If a bar having a diameter of 80 mm and a length of 1.5 m is made from this material, determine the critical load provided the ends are pinned. Assume that the load acts through the axis of the bar. Use Engesser’s equation.
s (MPa) 1100
200 0.001
E1 =
200 (106) = 200 GPa 0.001
E2 =
1100 (106) - 200 (106) = 150 GPa 0.007 - 0.001
Section properties: I =
p 4 c; 4
r =
p 4 I 0.04 c 4 c = = = = 0.02 m AA C p c2 2 2
A = pc2
Engesser’s equation: 1.0(1.5) KL = = 75 r 0.02 scr =
p2 Et
A
B
KL 2 r
=
p2 Et (75)2
= 1.7546(10 - 3) Et
Assume Et = E1 = 200 GPa scr = 1.7546 (10 - 3)(200)(109) = 351 MPa 7 200 MPa Therefore, inelastic buckling occurs: Assume Et = E2 = 150 GPa scr = 1.7546 (10 - 3)(150)(109) = 263.2 MPa 200 MPa 6 scr 6 1100 MPa
O.K.
Critical load: Pcr = scr A = 263.2 (106)(p)(0.042) = 1323 kN‚
Ans
1099
0.007
P (mm/mm)
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*13–76. The stress-strain diagram for a material can be approximated by the two line segments shown. If a bar having a diameter of 80 mm and a length of 1.5 m is made from this material, determine the critical load provided the ends are fixed. Assume that the load acts through the axis of the bar. Use Engesser’s equation.
s (MPa) 1100
200 0.001
E1 =
200 (106) = 200 GPa 0.001
E2 =
1100 (106) - 200 (106) = 150 GPa 0.007 - 0.001
Section properties: I =
p 4 c; 4
r =
p 4 I 0.04 c 4 c = = = = 0.02 m AA C p c2 2 2
A = pc2
Engesser’s equation: 0.5 (1.5) KL = = 37.5 r 0.02 scr =
p2 Et
A
B
KL 2 r
=
p2 Et (37.5)2
= 7.018385(10 - 3) Et
Assume Et = E1 = 200 GPa scr = 7.018385 (10 - 3)(200)(109) = 1403.7 MPa 7 200 MPa
NG
Assume Et = E2 = 150 GPa scr = 7.018385 (10 - 3)(150)(109) = 1052.8 MPa 200 MPa 6 scr 6 1100 MPa
O.K.
Critical load: Pcr = scr A = 1052.8 (106)(p)(0.042) = 5292 kN
Ans.
1100
0.007
P (mm/mm)
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•13–77.
The stress-strain diagram for a material can be approximated by the two line segments shown. If a bar having a diameter of 80 mm and length of 1.5 m is made from this material, determine the critical load provided one end is pinned and the other is fixed. Assume that the load acts through the axis of the bar. Use Engesser’s equation.
s (MPa) 1100
200 0.001
E1 =
200 (106) = 200 GPa 0.001
E2 =
1100 (106) - 200 (106) = 150 GPa 0.007 - 0.001
Section properties: I =
p 4 c ; 4
r =
p 4 I 0.04 c 4 c = = = = 0.02 m AA C pc2 2 2
A = pc2
Engesser’s equation: 0.7 (1.5) KL = = 52.5 r 0.02 scr =
p2 Et
A
B
KL 2 r
=
p2Et (52.5)2
= 3.58081 (10 - 3) Et
Assume Et = E1 = 200 GPa scr = 3.58081 (10 - 3)(200)(109) = 716.2 MPa 7 200 MPa
NG
Assume Et = E2 = 150 GPa scr = 3.58081 (10 - 3)(150)(109) = 537.1 MPa 200 MPa 6 scr 6 1100 MPa
O.K.
Critical load: Pcr = scr A = 537.1 (106)(p)(0.042) = 2700 kN
Ans.
1101
0.007
P (mm/mm)
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13–78. Determine the largest length of a structural A-36 steel rod if it is fixed supported and subjected to an axial load of 100 kN. The rod has a diameter of 50 mm. Use the AISC equations.
Section Properties: A = p A 0.0252 B = 0.625 A 10 - 3 B p m2 I =
p A 0.0254 B = 97.65625 A 10 - 9 B p m4 4
r =
97.65625(10 - 9)p I = = 0.0125 m AA A 0.625(10 - 3)p
Slenderness Ratio: For a column fixed at both ends, K = 0.5. Thus, 0.5L KL = = 40.0L r 0.0125 AISC Column Formula: Assume a long column. sallow = 100(103) 0.625(10 - 3)p
=
12p2E
2 23 A KL r B
12p2 C 200(109) D 23(40.0L)3
L = 3.555 m KL KL 2p2E = 40.0(3.555) = 142.2 and for A–36 steel, a b = r r e A sg KL KL 2p2[200(109)] = = 125.7. Since a b … … 200, the assumption is correct. r e r A 250(106)
Here,
Thus, L = 3.56 m
Ans.
1102
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13–79. Determine the largest length of a W10 * 45 structural steel column if it is pin supported and subjected to an axial load of 290 kip. Est = 29(103) ksi, sY = 50 ksi. Use the AISC equations.
Section Properties: For a W10 * 45 wide flange section, A = 13.3 in2
ry = 2.01 in
Slenderness Ratio: For a column pinned at both ends, K = 1. Thus, a
1(L) KL b = = 0.49751L r y 2.01
AISC Column Formula: Assume a long column, sallow =
12p2E
2 23 A KL r B
12p2 C 29(103) D 290 = 13.3 23(0.49751L)2 L = 166.3 in. KL KL 2p2E = 0.49751 (166.3) = 82.76 and for grade 50 steel, a b = r r c A sg KL KL 2p2[29(103)] 6 a b , the assumption is not correct. = = 107.0. Since r r c 50 A Thus, the column is an intermediate column. Here,
Applying Eq. 13–23,
B1 sallow =
(KL>r)2 2(KL>r)2c
R sg
3(KL>r) (KL>r)3 5 + 3 8(KL>r)c 8(KL>r)3c
B1 -
(0.49751L)2
R (50) 2(107.02) 290 = 13.3 3(0.49751L) (0.49751L)3 5 + 3 8(107.0) 8(107.03)
0 = 12.565658 A 10 - 9 B L3 - 24.788132 A 10 - 6 B L2 - 1.743638 A 10 - 3 B L + 0.626437
Solving by trial and error, L = 131.12 in. = 10.9 ft
Ans.
1103
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*13–80. Determine the largest length of a W10 * 12 structural A-36 steel section if it is pin supported and is subjected to an axial load of 28 kip. Use the AISC equations. For a W 10 * 12,
A = 3.54 in2
ry = 0.785 in. s =
28 P = = 7.91 ksi A 3.54
Assume a long column: sallow = a a
12p2E 23(KL>r)2
KL 12p2(29)(103) 12p2E = = 137.4 b = r A 23(7.91) A 23sallow
KL 2p2(29)(103) 2p2E = = 126.1, b = r c A sg A 36
KL KL 7 a b r r c
Long column. KL = 137.4 r L = 137.4 ¢
r 0.785 ≤ = 137.4 ¢ ≤ = 107.86 in. K 1
= 8.99 ft
Ans.
•13–81.
Using the AISC equations, select from Appendix B the lightest-weight structural A-36 steel column that is 14 ft long and supports an axial load of 40 kip.The ends are pinned. Take sY = 50 ksi.
Try, W6 * 15 (A = 4.43 in2
ry = 1.46 in.)
a
KL 2p2(29)(103) 2p2E b = = = 107 r c A sY 50 A
a
(1.0)(14)(12) KL b = = 115.1, ry 1.46
a
KL KL b 7 a b ry r c
Long column sallow =
12p2(29)(103) 12 p2E = = 11.28 ksi 2 23(KL>r) 23(115.1)2
Pallow = sallowA = 11.28(4.43) = 50.0 kip 7 40 kip
O.K.
Use W6 * 15
Ans.
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13–82. Using the AISC equations, select from Appendix B the lightest-weight structural A-36 steel column that is 12 ft long and supports an axial load of 40 kip. The ends are fixed. Take sY = 50 ksi.
A = 2.68 in2
Try W6 * 9 a
ry = 0.905 in.
KL 2p2(29)(103) 2p2E b = = = 107 r c A sY A 50
0.5(12)(12) KL = = 79.56 ry 0.905 KL KL 6 a b ry r c Intermediate column sallow =
KL>r 2 C 1 - 12 A (KL>r)c B D sg
KL>r KL>r 3 C 53 + 38 A (KL>r)c B - 18 A (KL>r)c B D
=
2 C 1 - 12 A 79.56 126.1 B D 36 ksi
1 79.56 3 C 53 + 38 A 79.56 126.1 B - 8 A 12.61 B D
= 15.40 ksi
Pallow = sallowA = 15.40(2.68) = 41.3 kip 7 40 kip
O.K.
Use W6 * 9
Ans.
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13–83. Using the AISC equations, select from Appendix B the lightest-weight structural A-36 steel column that is 24 ft long and supports an axial load of 100 kip.The ends are fixed.
Section Properties: Try a W8 * 24 wide flange section, A = 7.08 in2
ry = 1.61 in
Slenderness Ratio: For a column fixed at both ends, K = 0.5. Thus, a AISC
Column
0.5(24)(12) KL b = = 89.44 r y 1.61 Formula:
For
A–36
steel,
a
KL 2p2E b = r c A sg
KL KL 2p2[29(103)] 6 a b , the column is an intermediate = 126.1. Since r r c A 36 column. Applying Eq. 13–23, =
(KL>r)2
B1 sallow =
R sg
3(KL>r) (KL>r)3 5 + 3 8(KL>r)c 8(KL>r)3c
B1 =
2(KL>r)2c
(89.442) 2(126.12)
R (36)
3(89.44) (89.443) 5 + 3 8(126.1) 8(126.13)
= 14.271 ksi The allowable load is Pallow = sallowA = 14.271(7.08) = 101 kip 7 P = 100 kip Thus, Use
O.K. Ans.
W8 * 24
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*13–84. Using the AISC equations, select from Appendix B the lightest-weight structural A-36 steel column that is 30 ft long and supports an axial load of 200 kip.The ends are fixed.
a
A = 14.1 in2
ry = 2.08 in.
Try W8 * 48
KL 2 p2 (29)(103) 2 p2E b = = = 126.1 r c A sg A 36
0.5 (30)(12) KL = = 86.54 ry 2.08 a
KL KL b 6 a b intermediate column. ry r c
b 1 - 12 B sallow =
b 53 + 38 B
KL r
A KL r Bc
e1 =
e 53
+
3 8
C
1 2
KL r
A KL r Bc
2
R r sg
R - 18 B
KL r
A KL r Bc
2 C 86.54 126.1 D f36
86.54 126.1
D - C
D
1 86.54 3 f 8 126.1
3
R r
= 14.611 ksi
Pallow = sallow A = 14.611 (14.1) = 206 kip 7 P = 200 kip Use
O.K.
W 8 * 48
Ans.
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•13–85.
A W8 * 24 A-36-steel column of 30-ft length is pinned at both ends and braced against its weak axis at midheight. Determine the allowable axial force P that can be safely supported by the column. Use the AISC column design formulas.
Section Properties. From the table listed in the appendix, the necessary section properties for a W8 * 24 are A = 7.08 in2
rx = 3.42 in.
ry = 1.61 in.
Slenderness Ratio. For a column pinned at both of its ends, K = 1. Here, Lx = 30(12) = 360 in. and Ly = 15(12) = 180 in. Thus,
¢
1(360) KL = 105.26 ≤ = r x 3.42
¢
1(180) KL = 111.80 (controls) ≤ = r y 1.61
AISC
=
C
Column 2p2 C 29 A 103 B D 36
Formulas.
For
= 126.10. Since
¢
A-36
steel
¢
KL KL ≤ 6 ¢ ≤ , the r y r c
KL 2p2E ≤ = r c A sY column
is
an
intermediate column. (KL>r)2
B1 sallow =
R sY
3(KL>r) (KL>r)3 5 + 3 8(KL>r)c 8(KL>r)c 3 C1 -
=
2(KL>r)c 2
111.802
2 A 126.102 B
S(36)
3(111.80) 5 111.803 + 3 8(126.10) 8 A 126.103 B
= 11.428 ksi Thus, the allowable force is Pallow = sallowA = 11.428(7.08) = 80.91 kip = 80.9 kip
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13–86. Check if a W10 * 39 column can safely support an axial force of P = 250 kip. The column is 20 ft long and is pinned at both ends and braced against its weak axis at mid-height. It is made of steel having E = 29(103) ksi and sY = 50 ksi. Use the AISC column design formulas.
Section Properties. From the table listed in the appendix, the necessary section properties for a W10 * 39 are A = 11.5 in2
rx = 4.27 in.
ry = 1.98 in.
Slenderness Ratio. For a column pinned at both of its ends, K = 1. Here, Lx = 20(12) = 240 in. and Ly = 10(12) = 120 in. Thus,
¢
1(240) KL = 56.21 ≤ = r x 4.27
¢
1(120) KL = 60.606 (controls) ≤ = r y 1.98
AISC
=
Column 2p2 c29 A 103 B d
S
50
Formulas.
For
A-36
= 107.00 . Since ¢
steel
¢
KL 2p2E ≤ = r c A sY
KL KL ≤ 6 ¢ ≤ , the column is an r y r c
intermediate column.
B1 sallow =
2(KL>r)c 2
R sY
3(KL>r) (KL>r)3 5 + 3 8(KL>r)c 8(KL>r)c 3 C1 -
=
(KL>r)2
60.6062
2 A 107.002 B
S(50)
3(60.606) 5 60.6063 + 3 8(107.00) 8 A 107.003 B = 22.614 ksi
Thus, the allowable force is Pallow = sallowA = 22.614(11.5) = 260.06 kip 7 P = 250 kip Thus, a W10 * 39 column is adequate.
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13–87. A 5-ft-long rod is used in a machine to transmit an axial compressive load of 3 kip. Determine its smallest diameter if it is pin connected at its ends and is made of a 2014-T6 aluminum alloy.
Section properties: A =
p 2 d ; 4
I =
p d 4 pd4 a b = 4 2 64
pd4
r =
I d 64 = = AA C p4 d2 4
sallow =
P = A
p 4
3 3.820 = d2 d2
Assume long column: 1.0 (5)(12) 240 KL = = d r d 4
sallow =
54 000
A
B
KL 2 r
;
3.820 54000 = d2 C 240 D 2 d
d = 1.42 in.
Ans.
KL 240 = = 169 7 55 r 1.42
O.K.
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*13–88. Check if a W10 * 45 column can safely support an axial force of P = 200 kip. The column is 15 ft long and is pinned at both of its ends. It is made of steel having E = 29(103) ksi and sY = 50 ksi. Use the AISC column design formulas.
Section Properties. Try W10 * 45. From the table listed in the appendix, the necessary section properties are A = 13.3 in2
ry = 2.01 in.
Slenderness Ratio. For a column pinned at both of its ends, K = 1. Thus, a
1(15)(12) KL b = = 89.552 r y 2.01
KL 2p2E AISC Column Formulas. Here, a b = = r c A sY S KL KL Since a b 6 a b , the r y r c
2p2 c29 A 103 B d 50
= 107.00.
column is an intermediate column. (KL>r)2
B1 sallow =
R sY
(KL>r)3 3(KL>r) 5 + 3 8(KL>r)c 8(KL>r)c 3 C1 -
=
2(KL>r)c 2
89.5522
2 A 107.002 B
S(50)
3(89.552) 5 89.5523 + 3 8(107.00) 8 A 107.003 B
= 17.034 ksi Thus, the allowable force is Pallow = sallowA = 17.034(13.3) = 226.55 kip 7 P = 200 kip
O.K.
Thus, A W10 * 45 can be used
Ans.
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•13–89.
Using the AISC equations, check if a column having the cross section shown can support an axial force of 1500 kN. The column has a length of 4 m, is made from A-36 steel, and its ends are pinned.
20 mm
350 mm
300 mm 10 mm
Section Properties: A = 0.3(0.35) - 0.29(0.31) = 0.0151 m2 Iy =
1 1 (0.04) A 0.33 B + (0.31) A 0.013 B = 90.025833 A 10 - 6 B m4 12 12
ry =
Iy 90.02583(10 - 6) = = 0.077214 m AA A 0.0151
Slenderness Ratio: For a column pinned at both ends, K = 1. Thus, a
AISC =
Column 2p2[200(109)]
1(4) KL b = = 51.80 r y 0.077214
Formula: = 125.7. Since
A 250(10 ) column. Applying Eq. 13–23, 6
For
a
KL 2p2E b = r c A sg
(KL>r)2 2(KL>r)2c
R sg
3(KL>r) (KL>r)3 5 + 3 8(KL>r)c 8(KL>r)3c
B1 =
steel,
KL KL 6 a b , the column is an intermediate r r c
B1 sallow =
A–36
(51.802) 2(125.72)
20 mm
R (250)(106)
3(51.80) (51.803) 5 + 3 8(125.7) 8(125.73)
= 126.2 MPa The allowable load is Pallow = sallowA = 126.2 A 106 B (0.0151) = 1906 kN 7 P = 1500 kN
O.K.
Thus, the column is adequate.
Ans.
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13–90. The A-36-steel tube is pinned at both ends. If it is subjected to an axial force of 150 kN, determine the maximum length that the tube can safely support using the AISC column design formulas.
100 mm
80 mm
Section Properties. A = p A 0.052 - 0.042 B = 0.9 A 10 - 3 B p m2 I =
r =
p A 0.054 - 0.044 B = 0.9225 A 10 - 6 B p m4 4
0.9225 A 10 - 6 B p I = = 0.03202 m AA C 0.9 A 10 - 3 B p
Slenderness Ratio. For a column pinned at both of its ends, K = 1. Thus, 1(L) KL = = 31.23L r 0.03202 AISC Column Formulas. sallow =
12p2E 23(KL>r)2
150 A 103 B
.9 A 10 - 3 B p
=
12p2 C 200 A 109 B D 23(31.23L)2
L = 4.4607 m = 4.46 m
Here,
KL = 31.23(4.4607) = 139.33. r 2p2 C 200 A 109 B D
= 125.66. Since a
250 A 10 B long column is correct. =
C
6
Ans.
For
A-36
steel
a
KL 2p2E b = r c A sY
KL KL b 6 6 200, the assumption of a r c r
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13–91. The bar is made of a 2014-T6 aluminum alloy. Determine its smallest thickness b if its width is 5b. Assume that it is pin connected at its ends.
600 lb b 5b
8 ft
Section Properties: A = b(5b) = 5b2
Iy =
1 5 4 (5b) A b3 B = b 12 12
ry =
5 4 Iy 23 12 b = = b AA C 5b2 6
600 lb
Slenderness Ratio: For a column pinned at both ends, K = 1. Thus, a
1(8)(12) 332.55 KL = b = 23 b r y 6 b
Aluminum (2014 - T6 alloy) Column Formulas: Assume a long column and apply Eq. 13–26. sallow =
54 000 (KL>r)2
0.600 54 000 = 2 5b A 332.55 B 2 b
b = 0.7041 in. Here,
KL KL 332.55 = 7 55, the assumption is correct. Thus, = 472.3. Since r r 0.7041 b = 0.704 in.
Ans.
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*13–92. The bar is made of a 2014-T6 aluminum alloy. Determine its smallest thickness b if its width is 5b. Assume that it is fixed connected at its ends.
600 lb b 5b
8 ft
Section Properties: A = b(5b) = 5b2 Iy =
1 5 4 (5b) A b3 B = b 12 12
ry =
Iy 23 12 b = = b AA C 5b2 6
5
600 lb
4
Slenderness Ratio: For a column fixed at both ends, K = 0.5. Thus, a
0.5(8)(12) 166.28 KL = b = r y b 23 6 b
Aluminum (2014 - T6 alloy) Column Formulas: Assume a long column and apply Eq. 13–26. sallow =
54 000 (KL>r)2
0.600 54 000 = 5b2 A 166.28 B 2 b
b = 0.4979 in. Here,
KL KL 166.28 = 334.0. Since = 7 55, the assumption is correct. r r 0.4979
Thus, b = 0.498 in.
Ans.
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•13–93.
The 2014-T6 aluminum column of 3-m length has the cross section shown. If the column is pinned at both ends and braced against the weak axis at its mid-height, determine the allowable axial force P that can be safely supported by the column.
15 mm
170 mm
15 mm
15 mm 100 mm
Section Properties. A = 0.1(0.2) - 0.085(0.17) = 5.55 A 10 - 3 B m2 Ix =
1 1 (0.1) A 0.23 B (0.085) A 0.173 B = 31.86625 A 10 - 6 B m4 12 12
Iy = 2 c rx =
ry =
1 1 (0.015) A 0.13 B d + (0.17) A 0.0153 B = 2.5478 A 10 - 6 B m4 12 12
31.86625 A 10 - 6 B Ix = = 0.07577 AA C 5.55 A 10 - 3 B
2.5478 A 10 - 6 B Iy = = 0.02143 m AA C 5.55 A 10 - 3 B
Slenderness Ratio. For a column pinned at both of its ends, K = 1. Here, Lx = 3 m and Ly = 1.5 m. Thus, a
(1)(3) KL b = = 39.592 r x 0.07577
a
(1)(1.5) KL b = = 70.009 (controls) r y 0.02143
2014-T6 Alumimum Alloy Column Formulas. Since a
KL b 7 55, the column can r y
be classified a long column, sallow = D
373 A 103 B
T Mpa
= C
373 A 103 B
S MPa
a
KL 2 b r
70.0092
= 76.103 MPa Thus, the allowed force is Pallow = sallowA = 76.103 A 106 B C 5.55 A 10 - 3 B D = 422.37 kN = 422 kN
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13–94. The 2014-T6 aluminum column has the cross section shown. If the column is pinned at both ends and subjected to an axial force P = 100 kN, determine the maximum length the column can have to safely support the loading.
15 mm
170 mm
15 mm
15 mm 100 mm
Section Properties. A = 0.1(0.2) - 0.085(0.17) = 5.55 A 10 - 3 B m2 Iy = 2c
ry =
1 1 (0.015) A 0.13 B d + (0.17) A 0.0153 B = 2.5478 A 10 - 6 B m4 12 12
2.5478 A 10 - 6 B Iy = = 0.02143 m AA C 5.55 A 10 - 3 B
Slenderness Ratio. For a column pinned at both of its ends, K = 1. Then, a
1(L) KL b = = 46.6727L r y 0.02143
2014-T6 Alumimum Alloy Column Formulas. Assuming a long column, sallow = D 100 A 103 B
373 A 103 B
5.55 A 10 - 3 B
a
KL 2 b r
= C
T MPa
373 A 103 B
(46.672L)2
S A 106 B Pa
L = 3.083 m = 3.08 m Since a
Ans.
KL b = 46.6727(3.083) = 143.88 7 55, the assumption is correct. r y
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13–95. The 2014-T6 aluminum hollow section has the cross section shown. If the column is 10 ft long and is fixed at both ends, determine the allowable axial force P that can be safely supported by the column.
4 in.
3 in.
Section Properties. A = p A 22 - 1.52 B = 1.75p in2 r =
I =
p 4 A 2 - 1.54 B = 2.734375p in4 4
I 2.734375p = = 1.25 in. AA A 1.75p
Slenderness Ratio. For a column fixed at both of its ends, K = 0.5. Thus, 0.5(10)(12) KL = = 48 r 1.25 2014-T6 Aluminum Alloy Column Formulas. Since 12 6
KL 6 55, the column can r
be classified as an intermediate column. sallow = c30.7 - 0.23 a
KL b d ksi r
= [30.7 - 0.23(48)] ksi = 19.66 ksi Thus, the allowable load is Pallow = sallowA = 19.66 A 106 B (1.75p) = 108.09 kip = 108 kip
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*13–96. The 2014-T6 aluminum hollow section has the cross section shown. If the column is fixed at its base and pinned at its top, and is subjected to the axial force P = 100 kip, determine the maximum length of the column for it to safely support the load.
4 in.
3 in.
Section Properties. A = p A 22 - 1.52 B = 1.75p in2 r =
I =
p 4 A 2 - 1.54 B = 2.734375p in4 4
I 2.734375p = = 1.25 in. AA A 1.75p
Slenderness Ratio. For a column fixed at its base and pinned at its top, K = 0.7. Thus, 0.7(L) KL = = 0.56L r 1.25 2014-T6 Aluminum Alloy Column Formulas. Assuming an intermediate column, sallow = c30.7 - 0.23a
KL b d ksi r
100 = 30.7 - 0.23(0.56L) 1.75p L = 97.13 in. = 8.09 ft
Ans.
KL = 0.56(97.13) = 54.39 6 55, the assumption of an intermediate column r is correct. Since
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•13–97.
The tube is 0.25 in. thick, is made of a 2014-T6 aluminum alloy, and is fixed at its bottom and pinned at its top. Determine the largest axial load that it can support.
P x
y 6 in. x
6 in. y
10 ft
Section Properties: P
A = 6(6) - 5.5(5.5) = 5.75 in2 I =
1 1 (6) A 63 B (5.5) A 5.53 B = 31.7448 in4 12 12
r =
I 31.7448 = = 2.3496 in. A 5.75 AA
Slenderness Ratio: For a column fixed at one end and pinned at the other end, K = 0.7. Thus, 0.7(10)(12) KL = = 35.75 r 2.3496 Aluminium (2014 –∑ T6 alloy) Column Formulas: Since 12 6
KL 6 55, the r
column is classified as an intermediate column. Applying Eq. 13–25, sallow = c30.7 - 0.23 a
KL b d ksi r
= [30.7 - 0.23(33.75)] = 24.48 ksi The allowable load is Pallow = sallowA = 22.48(5.75) = 129 kip
Ans.
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13–98. The tube is 0.25 in. thick, is made of a 2014-T6 aluminum alloy, and is fixed connected at its ends. Determine the largest axial load that it can support.
P x
y 6 in. x
6 in. y
10 ft
Section Properties: P
A = 6(6) - 5.5(5.5) = 5.75 in2 I =
1 1 (6) A 63 B (5.5) A 5.53 B = 31.7448 in4 12 12
r =
I 31.7448 = = 2.3496 in. AA A 5.75
Slenderness Ratio: For column fixed at both ends, K = 0.5. Thus, 0.5(10)(12) KL = = 25.54 r 2.3496 Aluminium (2014 – T6 alloy) Column Formulas: Since 12 6
KL 6 55, the r
column is classified as an intermediate column. Applying Eq. 13–25, sallow = c30.7 - 0.23 a
KL b d ksi r
= [30.7 - 0.23(25.54)] = 24.83 ksi The allowable load is Pallow = sallowA = 24.83(5.75) = 143 kip
Ans.
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13–99. The tube is 0.25 in. thick, is made of 2014-T6 aluminum alloy and is pin connected at its ends. Determine the largest axial load it can support.
P x
y 6 in. x
6 in. y
Section Properties: A = 6(6) - 5.5(5.5) = 5.75 in2 I =
1 1 (6) A 63 B (5.5) A 5.53 B = 31.7448 in4 12 12
r =
I 31.7448 = = 2.3496 in. AA A 5.75
10 ft
P
Slenderness Ratio: For a column pinned as both ends, K = 1. Thus, 1(10)(12) KL = = 51.07 r 2.3496 Aluminum (2014 – T6 alloy) Column Formulas: Since 12 6
KL 6 55, the r
column is classified as an intermediate column. Applying Eq. 13–25, sallow = c30.7 - 0.23 a
KL b d ksi r
= [30.7 - 0.23(51.07)] = 18.95 ksi The allowable load is Pallow = sallowA = 18.95(5.75) = 109 kip
Ans.
*13–100. A rectangular wooden column has the cross section shown. If the column is 6 ft long and subjected to an axial force of P = 15 kip, determine the required minimum 1 dimension a of its cross-sectional area to the nearest 16 in. so that the column can safely support the loading. The column is pinned at both ends.
a
2a
Slenderness Ratio. For a column pinned at both of its ends, K = 1. Then, (1)(6)(12) KL 72 = = a a d NFPA Timber Column Formula. Assuming an intermediate column, sallow = 1.20 c1 -
1 KL>d 2 a b d ksi 3 26.0
15 1 72>a 2 = 1.20 c1 - a b d 2a(a) 3 26.0 a = 2.968 in. Use a = 3 in.
Ans.
KL 72 KL = = 24. Since 11 6 6 26, the assumption is correct. d 3 d
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•13–101.
A rectangular wooden column has the cross section shown. If a = 3 in. and the column is 12 ft long, determine the allowable axial force P that can be safely supported by the column if it is pinned at its top and fixed at its base.
a
2a
Slenderness Ratio. For a column fixed at its base and pinned at its top K = 0.7. Then, 0.7(12)(12) KL = = 33.6 d 3 NFPA Timer Column Formula. Since 26 6
KL 6 50, the column can be classified d
as a long column.
sallow =
540 ksi 540 = = 0.4783 ksi 2 (KL>d) 33.62
The allowable force is Pallow = sallowA = 0.4783(3)(6) = 8.61 kip
Ans.
13–102. A rectangular wooden column has the cross section shown. If a = 3 in. and the column is subjected to an axial force of P = 15 kip, determine the maximum length the column can have to safely support the load. The column is pinned at its top and fixed at its base.
a
2a
Slenderness Ratio. For a column fixed at its base and pinned at its top, K = 0.7. Then, KL 0.7L = = 0.2333L d 3 NFPA Timber Column Formula. Assuming an intermediate column, sallow = 1.20 c1 -
1 KL>d 2 a b d ksi 3 26.0
15 1 0.2333L 2 = 1.20c1 - a b d 3(6) 3 26.0 L = 106.68 in. = 8.89 ft
Ans.
KL KL = 0.2333(106.68) = 24.89. Since 11 6 6 26, the assumption is d d correct. Here,
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13–103. The timber column has a square cross section and is assumed to be pin connected at its top and bottom. If it supports an axial load of 50 kip, determine its smallest side dimension a to the nearest 12 in. Use the NFPA formulas.
14 ft a
Section properties: A = a2
sallow = s =
50 P = 2 A a
Assume long column: sallow =
50 = a2
C
540
2 A KL d B
540 (1.0)(14)(12) a
D
2
a = 7.15 in. (1.0)(14)(12) KL KL = = 23.5, 6 26 d 7.15 d
Assumption NG
Assume intermediate column: sallow = 1.20 B 1 -
1 KL>d 2 a b R 3 26.0
2 50 1 a b R = 1.20 B 1 - a 2 a 26.0 3 1.0(14)(12)
a = 7.46 in. 1.0(14)(12) KL KL = = 22.53, 11 6 6 26 d 7.46 d
Assumption O.K.
1 Use a = 7 in. 2
Ans.
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*13–104. The wooden column shown is formed by gluing together the 6 in. * 0.5 in. boards. If the column is pinned at both ends and is subjected to an axial load P = 20 kip, determine the required number of boards needed to form the column in order to safely support the loading.
P 6 in.
0.5 in.
9 ft
Slenderness Ratio. For a column pinned at both of its ends, K = 1. If the number of the boards required is n and assuming that n(0.5) 6 6 in. Then, d = n(0.5). Thus, (1)(9)(12) KL 216 = = n d n(0.5)
P
NFPA Timber Column Formula. Assuming an intermediate column, sallow = 1.20 B 1 -
1 KL>d 2 a b R ksi 3 26.0
1 216>n 2 20 = 1.20 B 1 - a b R [n(0.5)](6) 3 26.0 n2 - 5.5556n - 23.01 = 0 Solving for the positive root, n = 8.32 Use n = 9
Ans.
KL KL 216 = = 24. Since n(0.5) = 9(0.5) = 4.5 in. 6 6 in. and 11 6 6 26, d 9 d the assumptions made are correct. Here,
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•13–105.
The column is made of wood. It is fixed at its bottom and free at its top. Use the NFPA formulas to determine its greatest allowable length if it supports an axial load of P = 2 kip.
P 2 in. y
y
x
x 4 in.
Slenderness Ratio: For a column fixed at one end and free at the other end, K = 2. Thus, L
2(L) KL = = 1.00L d 2 NFPA Timber Column Formulas: Assume a long column. Apply Eq. 13–29, sallow =
540 ksi (KL>d)2
2 540 = 2(4) (1.00L)2 L = 46.48 in Here,
KL KL = 1.00(46.48) = 46.48. Since 26 6 6 50, the assumption is correct. d d
Thus, L = 46.48 in. = 3.87 ft
Ans.
13–106. The column is made of wood. It is fixed at its bottom and free at its top. Use the NFPA formulas to determine the largest allowable axial load P that it can support if it has a length L = 4 ft.
P 2 in. y
x
y x 4 in.
Slenderness Ratio: For a column fixed at one end and free at the other end, K = 2. Thus, L
2(4)(12) KL = = 48.0 d 2 NFPA Timber Column Formulas: Since 26 6
KL 6 50, it is a long column. Apply d
Eq. 13–29,
sallow =
=
540 ksi (KL>d)2 540 48.02
= 0.234375 ksi The allowable axial force is Pallow = sallowA = 0.234375[2(4)] = 1.875 kip
Ans.
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13–107. The W14 * 53 structural A-36 steel column supports an axial load of 80 kip in addition to an eccentric load P. Determine the maximum allowable value of P based on the AISC equations of Sec. 13.6 and Eq. 13–30. Assume the column is fixed at its base, and at its top it is free to sway in the x–z plane while it is pinned in the y–z plane.
z 80 kip x y
12 ft
Section Properties: For a W14 * 53 wide flange section. A = 15.6 in2
Ix = 541 in4
d = 13.92 in.
rx = 5.89 in.
ry = 1.92 in. Slenderness Ratio: By observation, the largest slenderness ratio is about y-y axis. For a column fixed at one end and free at the other end, K = 2. Thus, a
2(12)(12) KL b = = 150 r y 1.92
Allowable Stress: The allowable stress can be determined using AISC Column 2p2[29(103)] 2p2E KL b = = = 126.1. Since Formulas. For A–36 steel, a r c B sY B 36 KL KL b … … 200, the column is a long column. Applying Eq. 13–21, a r c r sallow =
12p2E 23(KL>r)2 12p2(29.0)(103)
=
23(1502)
= 6.637 ksi Maximum Stress: Bending is about x-x axis. Applying we have smax = sallow =
6.637 =
Mc P + A I
P(10) A 13.92 P + 80 2 B + 15.6 541
P = 7.83 kip
Ans.
1127
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y x 10 in.
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*13–108. The W12 * 45 structural A-36 steel column supports an axial load of 80 kip in addition to an eccentric load of P = 60 kip. Determine if the column fails based on the AISC equations of Sec. 13.6 and Eq. 13–30. Assume that the column is fixed at its base, and at its top it is free to sway in the x–z plane while it is pinned in the y–z plane.
z 80 kip x y
12 ft
Section Properties: For a W12 * 45 wide flange section, A = 13.2 in2
d2 = 12.06 in.
Ix = 350 in4
rx = 5.15 in.
ry = 1.94 in. Slenderness Ratio: By observation, the largest slenderness ratio is about y -y axis. For a column fixed at one end and free at the other end, K = 2. Thus, a
2(12)(12) KL b = = 148.45 r y 1.94
Allowable Stress: The allowable stress can be determined using AISC Column 2p2[29(103)] 2p2E KL b = = = 126.1. Since Formulas. For A–36 steel, a r c B sY B 36 KL KL b … … 200, the column is a long column. Applying Eq. 13–21, a r c r sallow =
12p2E 23(KL>r)2 12p2(29.0)(103)
=
23(148.452)
= 6.776 ksi Maximum Stress: Bending is about x-x axis. Applying Eq. 1 we have smax =
=
Mc P + A I
60(10) A 12.06 140 2 B + 13.2 350
= 20.94 ksi Since smax 7 sallow, the column is not adequate.
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•13–109.
The W14 * 22 structural A-36 steel column is fixed at its top and bottom. If a horizontal load (not shown) causes it to support end moments of M = 10 kip # ft, determine the maximum allowable axial force P that can be applied. Bending is about the x–x axis. Use the AISC equations of Sec. 13.6 and Eq. 13–30.
P x
M y
y
x
12 ft
Section properties for W14 * 22: A = 6.49 in2
d = 13.74 in2
Ix = 199 in4
ry = 1.04 in. M
Allowable stress method:
P
0.5(12)(12) KL = = 69.231 ry 1.04 a
KL KL 2p2E KL 2p2(29)(103) b = = = 126.1, 6 a b r c r r c B sY B 36 y
Hence,
B1 (sa)allow =
B 53
+
3 8
smax = (sa)allow =
16.510 =
1 2
¢ KL r
2 A KL r B 2 A KL r Bc
A KL r Bc
-
c1 -
≤ R sY
1 8
3 A KL r B 3 A KL r Bc
=
R
c 53 +
3 8
1 2
2 A 69.231 126.1 B d36
1 69.231 3 A 69.231 126.1 B - 8 A 126.1 B d
= 16.510 ksi
My c P + A Ix
10(12)(13.74 P 2 ) + 6.49 199
P = 80.3 kip
Ans.
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13–110. The W14 * 22 column is fixed at its top and bottom. If a horizontal load (not shown) causes it to support end moments of M = 15 kip # ft, determine the maximum allowable axial force P that can be applied. Bending is about the x–x axis. Use the interaction formula with 1sb2allow = 24 ksi.
P x
M y
y
x
12 ft
Section Properties for W 14 * 22: A = 6.49 in2
d = 13.74 in2
Ix = 199 in4
ry = 1.04 in. M P
Interaction method: 0.5(12)(12) KL = = 69.231 ry 1.04 a
KL 2p2E KL KL 2p2(29)(103) b = = = 126.1, 6 a b r c ry r c B sY B 36
Hence,
B1 (sa)allow =
B 53 sa =
+
3 8
1 2
¢ KL r
2 A KL r B 2 A KL r Bc
A KL r Bc
-
c1 -
≤ R sY
1 8
P P = = 0.15408 P A 6.49
3 A KL r B 3 A KL r Bc
=
R
c 53 +
3 8
A
1 2
2 A 69.231 126.1 B d36
69.231 126.1
B
-
1 8
A
B
69.231 3 d 126.1
= 16.510 ksi
15(12) A 13.74 Mxc 2 B = = 6.214 ksi sb = Ix 199 sb sa + = 1.0 (sa)allow (sb)allow 0.15408 P 6.2141 + = 1.0 16.510 24 P = 79.4 kip
Ans.
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13–111. The W14 * 43 structural A-36 steel column is fixed at its bottom and free at its top. Determine the greatest eccentric load P that can be applied using Eq. 13–30 and the AISC equations of Sec. 13.6.
40 kip
16 in.
10 ft
Section properties for W14 * 43: A = 12.6 in2
d = 13.66 in.
Iy = 45.2 in4
ry = 1.89 in.
b = 7.995 Allowable stress method: 2(10)(12) KL = = 126.98 ry 1.89 a
2p2E KL KL KL 2p2 (29)(103) b = = = 126.1, 200 7 7 a b r c ry r c B sY B 36
(sa)allow =
12p2(29)(103) 12p2E = = 9.26 ksi 2 23(KL>r) 23(126.98)2
smax = (sa)allow =
P
My c P + A Iy
P(16) A 7.995 P + 40 A B 9.26 = + 12.6 45.2 P = 4.07 kip
Ans.
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*13–112. The W10 * 45 structural A-36 steel column is fixed at its bottom and free at its top. If it is subjected to a load of P = 2 kip, determine if it is safe based on the AISC equations of Sec. 13.6 and Eq. 13–30.
40 kip
16 in.
10 ft
Section Properties for W10 * 45: A = 13.3 in2
d = 10.10 in.
Iy = 53.4 in4
ry = 2.01 in.
b = 8.020 in. Allowable stress method: 2.0(10)(12) KL = = 119.4 ry 2.01 a
KL 2p2E 2p2(29)(103) b = = = 126.1 r c B sY B 36
KL KL 6 a b r r c
(sa)allow Ú
(sa)allow =
10.37 Ú
c1 5 3
+
A
1 (KL>r) 2 (KL>r)3 dsY
3 KL>r 8 KL>rc
My c P + A Iy
2 C 1 - 12 A 119.4 126.1 B D 36
2
B -
c
3
1 (KL>r) 8 (KL>rc)3
= 5 3
+
3 8
1 119.4 3 A 119.4 136.1 B - 8 A 126.1 B
P
= 10.37 ksi
2(16) A 8.020 42 2 B + 13.3 53.4 O.K.
10.37 Ú 5.56 Column is safe. Yes.
Ans.
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•13–113.
The A-36-steel W10 * 45 column is fixed at its base. Its top is constrained to move along the x–x axis but free to rotate about and move along the y–y axis. Determine the maximum eccentric force P that can be safely supported by the column using the allowable stress method.
12 in.
y
Section Properties. From the table listed in the appendix, the section properties for a W10 * 45 are bf = 8.02 in.
rx = 4.32 in.
Iy = 53.4 in4
ry = 2.01 in.
Slenderness Ratio. Here, Lx = 24(12) = 288 in. and for a column fixed at its base and free at its top, Kx = 2. Thus, a
2(288) KL b = = 133.33 (controls) r x 4.32
Since the column is fixed at its base and pinned at its top, Ky = 0.7 and Ly = 24(12) = 288 in. Then, a
0.7(288) KL b = = 100.30 r y 2.01
Allowable Stress. The allowable stress will be determined using the AISC column formulas. For A-36 steel, 2p2 C 29 A 103 B D KL 2p2E KL KL b = = = 126.10. Since a b 6 a b 6 200, r c r c r x B sY C 36 the column is classified as a long column. a
sallow =
=
12p2E 23(KL>r)2
12p2 C 29 A 103 B D 23(133.332)
= 8.400 ksi
Maximum Stress. Bending is about the weak axis. Since M = P(12) and bf 8.02 = = 4.01 in, c = 2 2 sallow =
Mc P + A I
8.400 =
[P(12)](4.01) P + 13.3 53.4
y
x
24 ft
A = 13.3 in2
P
P = 8.604 kip = 8.60 kip
Ans.
1133
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13–114. The A-36-steel W10 * 45 column is fixed at its base. Its top is constrained to move along the x–x axis but free to rotate about and move along the y–y axis. Determine the maximum eccentric force P that can be safely supported by the column using an interaction formula. The allowable bending stress is (sb)allow = 15 ksi.
12 in.
bf = 8.02 in.
rx = 4.32 in.
y 24 ft
Iy = 53.4 in4
ry = 2.01 in.
Slenderness Ratio. Here, Lx = 24(12) = 288 in and for a column fixed at its base and free at its top, Kx = 2. Thus, a
2(288) KL b = = 133.33 (controls) r x 4.32
Since the column is fixed at its base and pinned at its top, Ky = 0.7 and Ly = 24(12) = 288 in. Then, a
0.7(288) KL b = = 100.30 r y 2.01
Allowable Stress. The allowable stress will be determined using the AISC column 2p2 C 29 A 103 B D 2p2E KL = 126.10. Since formulas. For A-36 steel, a b = = C r c B sY 36 a
KL KL b 6 a b 6 200, the column is classified as a long column. r c r x sallow =
=
12p2E 23(KL>r)2
12p2 C 29 A 103 B D 23 A 133.332 B
= 8.400 ksi
Interaction Formula. Bending is about the weak axis. Here, M = P(12) and bf 8.02 = = 4.01 in. c = 2 2 P>A Mc>Ar2 + = 1 (sa)allow (sb)allow P>13.3 + 8.400
P(12)(4.01) n C 13.3 A 2.012 B D 15
y
x
Section Properties. From the table listed in the appendix, the section properties for a W10 * 45 are A = 13.3 in2
P
= 1
P = 14.57 kip = 14.6 kip
Ans.
14.57>13.3 sa = = 0.1304 6 0.15 (sa)allow 8.400
O.K.
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13–115. The A-36-steel W12 * 50 column is fixed at its base. Its top is constrained to move along the x–x axis but free to rotate about and move along the y–y axis. If the eccentric force P = 15 kip is applied to the column, investigate if the column is adequate to support the loading. Use the allowable stress method.
12 in.
bf = 8.08 in.
rx = 5.18 in.
y 24 ft
Iy = 56.3 in4
ry = 1.96 in. Slenderness Ratio. Here, Lx = 24(12) = 288 in. and for a column fixed at its base and pinned at its top, Kx = 2. Thus, a
2(288) KL b = = 111.20 (controls) r x 5.18
Since the column is fixed at its base and pinned at its top, Ky = 0.7 and Ly = 24(12) = 288 in. Thus, a
0.7(288) KL b = = 102.86 r y 1.96
Allowable Stress. The allowable stress will be determined using the AISC column formulas. For A-36 steel, a a
2p2 C 29 A 103 B D KL 2p2E = 126.10. Since b = = C r c A sY 36
KL KL b 6 a b , the column can be classified as an intermediate column. r x r c
B1 sallow =
2(KL>r)C 2
R sY
(KL>r)3 3(KL>r) 5 + 3 8(KL>r)C 8(KL>r)C 3 C1 -
=
(KL>r)2
111.202
2 A 126.102 B
S(36)
3(111.20) 5 111.203 + 3 8(126.10) 8 A 126.103 B
= 11.51 ksi Maximum Stress. Bending is about the weak axis. Since, M = 15(12) = 180 kip # in. bf 8.08 and c = = = 4.04 in., 2 2 smax =
y
x
Section Properties. From the table listed in the appendix, the section properties for a W12 * 50 are A = 14.7 in2
P
180(4.04) P Mc 15 + = + = 13.94 ksi A I 14.7 56.3
Since smax 7 sallow, the W12 * 50 column is inadequate according to the allowable stress method.
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*13–116. The A-36-steel W12 * 50 column is fixed at its base. Its top is constrained to move along the x–x axis but free to rotate about and move along the y–y axis. If the eccentric force P = 15 kip is applied to the column, investigate if the column is adequate to support the loading. Use the interaction formula. The allowable bending stress is (sb)allow = 15 ksi.
12 in.
bf = 8.08 in.
rx = 5.18 in.
y 24 ft
ry = 1.96 in.
Slenderness Ratio. Here, Lx = 24(12) = 288 in and for a column fixed at its base and pinned at its top, Kx = 2. Thus, a
2(288) KL b = = 111.20 (controls) r x 5.18
Since the column is fixed at its base and pinned at its top, Ky = 0.7 and Ly = 24(12) = 288 in. Then, a Allowable =
C
Axial
2p2 C 29 A 103 B D 36
0.7(288) KL b = = 102.86 r y 1.96 Stress.
For
= 126.10. Since a
A-36
steel,
a
KL 2p2E b = r c A sY
KL KL b 6 a b , the column can be r x r c
classified as an intermediate column. C1 sallow =
2(KL>r)c 2
SsY
3(KL>r) (KL>r)3 5 + 3 8(KL>r)c 8(KL>r)c 3 C1 -
=
(KL>r)2
111.202
2 A 126.102 B
S(36)
3(111.20) 5 111.203 + 3 8(126.10) 8 A 126.103 B
= 11.51 ksi Interaction Formula. Bending is about the weak axis. Here, M = 15(12) bf 8.08 = 180 kip # in. and c = = = 4.04 in. 2 2
P>A Mc>Ar2 15>14.7 + + = (sa)allow (sb)allow 11.51
180(4.04) n C 14.7 A 1.962 B D 15
y
x
Section Properties. From the table listed in the appendix, the section properties for a W12 * 50 are A = 14.7 in2
P
= 0.9471 6 1
P = 14.57 kip = 14.6 kip
Ans.
15>14.7 sa = = 0.089 6 0.15 (sa)allow 11.51
O.K.
Thus, a W12 * 50 column is adequate according to the interaction formula.
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•13–117.
A 16-ft-long column is made of aluminum alloy 2014-T6. If it is fixed at its top and bottom, and a compressive load P is applied at point A, determine the maximum allowable magnitude of P using the equations of Sec. 13.6 and Eq. 13–30.
P A 4.25 in. x 0.5 in. y 8 in.
Section properties: A = 2(0.5)(8) + 8(0.5) = 12 in2 Ix =
1 1 (8)(93) (7.5)(83) = 166 in4 12 12
Iy = 2 a ry =
1 1 b(0.5)(83) + (8)(0.53) = 42.75 in4 12 12
Iy 42.75 = = 1.8875 in. AA A 12
Allowable stress method: 0.5(16)(12) KL KL = 50.86, 12 6 = 6 55 ry ry 1.8875 sallow = c30.7 - 0.23a
KL bd r
= [30.7 - 0.23(50.86)] = 19.00 ksi smax = sallow =
19.00 =
Mx c P + A Ix
P(4.25)(4.5) P + 12 166
P = 95.7 kip
Ans.
1137
y x 8 in. 0.5 in.
0.5 in.
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13–118. A 16-ft-long column is made of aluminum alloy 2014-T6. If it is fixed at its top and bottom, and a compressive load P is applied at point A, determine the maximum allowable magnitude of P using the equations of Sec. 13.6 and the interaction formula with 1sb2allow = 20 ksi.
P A 4.25 in. x 0.5 in. y 8 in.
Section Properties: A = 2(0.5)(8) + 8(0.5) = 12 in2 Ix =
1 1 (8)(93) (7.5)(83) = 166 in4 12 12
Iy = 2 a ry =
1 1 b(0.5)(83) + (8)(0.53) = 42.75 in4 12 12
Iy 42.75 = = 1.8875 in. AA A 12
Interaction method: 0.5(16)(12) KL KL = 50.86, 12 6 = 6 55 ry ry 1.8875 sallow = c30.7 - 0.23a
KL bd r
= [30.7 - 0.23(50.86)] = 19.00 ksi sa =
P P = = 0.08333P A 12
sb =
P(4.25)(4.50) Mc = = 0.1152P Ix 166 sb sa + = 1.0 (sa)allow (sb)allow 0.08333P 0.1152P + = 1 19.00 20
P = 98.6 kip
Ans.
1138
y x 8 in. 0.5 in.
0.5 in.
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13–119. The 2014-T6 hollow column is fixed at its base and free at its top. Determine the maximum eccentric force P that can be safely supported by the column. Use the allowable stress method. The thickness of the wall for the section is t = 0.5 in.
6 in.
P
3 in. 6 in.
8 ft
Section Properties. A = 6(3) - 5(2) = 8 in2 Ix =
1 1 (3) A 63 B (2) A 53 B = 33.1667 in4 12 12
rx =
33.1667 Ix = = 2.036 in. AA A 8
Iy =
1 1 (6) A 33 B (5) A 23 B = 10.1667 in4 12 12
ry =
Iy 10.1667 = = 1.127 in. AA A 8
Slenderness Ratio. For a column fixed at its base and free at its top, K = 2. Thus, a
2(8)(12) KL b = = 170.32 r y 1.127
Allowable Stress. Since a
KL b 7 55, the column can be classified as a long r y
column.
sallow =
54 000 ksi 54 000 ksi = = 1.862 ksi (KL>r)2 170.312
Maximum Stress. Bending occurs about the strong axis so that M = P(6) and 6 c = = 3 in. 2 sallow =
1.862 =
Mc P + A I
C P(6) D (3) P + 8 33.1667
P = 2.788 kip = 2.79 kip
Ans.
1139
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*13–120. The 2014-T6 hollow column is fixed at its base and free at its top. Determine the maximum eccentric force P that can be safely supported by the column. Use the interaction formula. The allowable bending stress is (sb)allow = 30 ksi. The thickness of the wall for the section is t = 0.5 in.
6 in.
P
3 in. 6 in.
8 ft
Section Properties. A = 6(3) - 5(2) = 8 in2 Ix =
1 1 (3) A 63 B (2) A 53 B = 33.1667 in4 12 12
rx =
33.1667 Ix = = 2.036 in. AA A 8
Iy =
1 1 (6) A 33 B (5) A 23 B = 10.1667 in4 12 12
ry =
Iy 10.1667 = = 1.127 in. AA A 8
Slenderness Ratio. For a column fixed at its base and pinned at its top, K = 2. Thus, a
2(8)(12) KL b = = 170.32 r y 1.127
KL b 7 55, the column can be classified as the column is r y classified as a long column. Allowable Stress. Since a
sallow =
54000 ksi 54000 ksi = = 1.862 ksi (KL>r)2 170.312
Interaction Formula. Bending is about the strong axis. Since M = P(6) and 6 c = = 3 in, 2 P>A Mc>Ar2 + = 1 (sa)allow (sb)allow P>8 + 1.862
[P(6)](3) n C 8 A 2.0362 B D 30
= 1
P = 11.73 kip = 11.7 kip
Ans.
1140
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•13–121. The 10-ft-long bar is made of aluminum alloy 2014-T6. If it is fixed at its bottom and pinned at the top, determine the maximum allowable eccentric load P that can be applied using the formulas in Sec. 13.6 and Eq. 13–30.
P x 1.5 in. 1.5 in. x
Section Properties: A = 6(4) = 24.0 in2 Ix =
1 (4) A 63 B = 72.0 in4 12
Iy =
1 (6) A 43 B = 32.0 in4 12
ry =
Iy 32.0 = = 1.155 in. AA A 24
Slenderness Ratio: The largest slenderness ratio is about y-y axis. For a column pinned at one end fixed at the other end, K = 0.7. Thus, a
0.7(10)(12) KL b = = 72.75 r y 1.155
Allowable Stress: The allowable stress can be determined using aluminum KL (2014 –T6 alloy) column formulas. Since 7 55, the column is classified as a long r column. Applying Eq. 13–26, sallow = c =
54 000 d ksi (KL>r)2
54 000 72.752
= 10.204 ksi Maximum Stress: Bending is about x-x axis. Applying Eq. 13–30, we have smax = sallow =
10.204 =
P Mc + A I P(1.5)(3) P + 24.0 72.0
P = 98.0 kip
Ans.
1141
y 3 in.
2 in. y 2 in.
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13–122. The 10-ft-long bar is made of aluminum alloy 2014-T6. If it is fixed at its bottom and pinned at the top, determine the maximum allowable eccentric load P that can be applied using the equations of Sec. 13.6 and the interaction formula with 1sb2allow = 18 ksi.
P x 1.5 in. 1.5 in. x
Section Properties: A = 6(4) = 24.0 in2 Ix =
1 (4) A 63 B = 72.0 in4 12
Iy =
1 (6) A 43 B = 32.0 in4 12
rx =
Ix 72.0 = = 1.732 in. AA A 24.0
ry =
Iy 32.0 = = 1.155 in. AA A 24.0
Slenderness Ratio: The largest slenderness radio is about y-y axis. For a column pinned at one end and fixed at the other end, K = 0.7. Thus a
0.7(10)(12) KL b = = 72.75 r y 1.155
Allowable Stress: The allowable stress can be determined using aluminum KL (2014 –T6 alloy) column formulas. Since 7 55, the column is classified as a long r column. Applying Eq. 13–26, (sa)allow = c =
54 000 d ksi (KL>r)2
54 000 72.752
= 10.204 ksi Interaction Formula: Bending is about x-x axis. Applying Eq. 13–31, we have Mc>Ar2 P>A + = 1 (sa)allow (sb)allow P(1.5)(3)>24.0(1.7322) P>24.0 + = 1 10.204 18 P = 132 kip
Ans.
1142
y 3 in.
2 in. y 2 in.
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13–123. The rectangular wooden column can be considered fixed at its base and pinned at its top. Also, the column is braced at its mid-height against the weak axis. Determine the maximum eccentric force P that can be safely supported by the column using the allowable stress method.
6 in.
6 in.
Section Properties. Ix =
dx = 6 in.
5 ft
1 (3) A 63 B = 54 in4 12
dy = 3 in.
Slenderness Ratio. Here, Lx = 10(12) = 120 in. and for a column fixed at its base and pinned at its top, K = 0.7. Thus, a
0.7(120) KL b = = 14 d x 6
Since the bracing provides support equivalent to a pin, Ky = 1 and Ly = 5(12) = 60 in. Then a
1(60) KL b = = 20 (controls) d y 3
KL 6 26, the column can be classified as the column d is classified as an intermediate column. Allowable Stress. Since 11 6
sallow = 1.20c 1 = 1.20 c 1 -
1 KL>d 2 a b d ksi 3 26.0 1 20 2 a b d ksi = 0.9633 ksi 3 26.0
Maximum Stress. Bending occurs about the strong axis. Here, M = P(6) and 6 c = = 3 in. 2 sallow =
0.9633 =
6 in. 3 in.
5 ft
A = 6(3) = 18 in2
P
P Mc + A I [P(6)](3) P + 18 54
P = 2.477 kip = 2.48 kip
Ans.
1143
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*13–124. The rectangular wooden column can be considered fixed at its base and pinned at its top. Also, the column is braced at its mid-height against the weak axis. Determine the maximum eccentric force P that can be safely supported by the column using the interaction formula. The allowable bending stress is (sb)allow = 1.5 ksi.
6 in.
6 in.
5 ft
Section Properties.
rx =
Ix =
1 (3) A 63 B = 54 in4 12
Ix 54 = = 1.732 in. AA A 18
dx = 6 in.
dy = 3 in.
Slenderness Ratio. Here, Lx = 10(12) = 120 in. and for a column fixed at its base pinned at its top, K = 0.7. Thus, a
0.7(120) KL b = = 14 d x 6
Since the bracing provides support equivalent to a pin, Ky = 1 and Ly = 5(12) = 60 in. Then a
1(60) KL b = = 20 (controls) d y 3
KL 6 26, the column can be classified as the d column is classified as an intermediate column. Allowable Axial Stress. Since 11 6
sallow = 1.20c 1 = 1.20 c 1 -
1 KL>d 2 a b d ksi 3 26.0 1 20 2 a b d ksi = 0.9633 ksi 3 26.0
Interaction Formula. Bending occurs about the strong axis. Since M = P(6) and 6 c = = 3 in. 2 P>A Mc>Ar2 + = 1 (sa)allow (sb)allow P>18 + 0.9633
[P(6)](3) n C 18 A 1.7322 B D 1.5
6 in. 3 in.
5 ft
A = 6(3) = 18 in2
P
= 1
P = 3.573 kip = 3.57 kip
Ans.
1144
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•13–125.
The 10-in.-diameter utility pole supports the transformer that has a weight of 600 lb and center of gravity at G. If the pole is fixed to the ground and free at its top, determine if it is adequate according to the NFPA equations of Sec. 13.6 and Eq. 13–30.
G
15 in.
18 ft
2(18)(12) KL = = 43.2 in. d 10 26 6 43.2 … 50 Use Eq. 13–29, sallow =
540 540 = = 0.2894 ksi (KL>d) (43.2)2
smax =
Mc P + A I
smax =
(600)(15)(5) 600 + 2 p (5) A p4 B (5)4
smax = 99.31 psi 6 0.289 ksi
O.K.
Yes.
Ans.
1145
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13–126. Using the NFPA equations of Sec. 13.6 and Eq. 13–30, determine the maximum allowable eccentric load P that can be applied to the wood column. Assume that the column is pinned at both its top and bottom.
P 0.75 in. 6 in. 3 in.
12 ft
Section Properties: A = 6(3) = 18.0 in2
Iy =
1 (6) A 33 B = 13.5 in4 12
Slenderness Ratio: For a column pinned at both ends, K = 1.0. Thus, a
1.0(12)(12) KL b = = 48.0 d y 3
Allowable Stress: The allowable stress can be determined using NFPA timber KL column formulas. Since 26 6 6 50, it is a long column. Applying Eq. 13–29, d sallow =
=
540 ksi (KL>d)2 540 = 0.234375 ksi 48.02
Maximum Stress: Bending is about y -y axis. Applying Eq. 13–30, we have smax = sallow =
0.234375 =
P Mc + A I P(0.75)(1.5) P + 18.0 13.5
P = 1.69 kip
Ans.
1146
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13–127. Using the NFPA equations of Sec. 13.6 and Eq. 13–30, determine the maximum allowable eccentric load P that can be applied to the wood column. Assume that the column is pinned at the top and fixed at the bottom.
P 0.75 in. 6 in. 3 in.
12 ft
Section Properties: A = 6(3) = 18.0 in2 Iy =
1 (6) A 33 B = 13.5 in4 12
Slenderness Ratio: For a column pinned at one end and fixed at the other end, K = 0.7. Thus, a
0.7(12)(12) KL b = = 33.6 d y 3
Allowable Stress: The allowable stress can be determined using NFPA timber KL column formulas. Since 26 6 6 50, it is a long column. Applying Eq. 13–29, d sallow =
=
540 ksi (KL>d)2 540 = 0.4783 ksi 33.62
Maximum Stress: Bending is about y -y axis. Applying Eq. 13–30, we have smax = sallow =
0.4783 =
Mc P + A I P(0.75)(1.5) P + 18.0 13.5
P = 3.44 kip
Ans.
1147
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*13–128. The wood column is 4 m long and is required to support the axial load of 25 kN. If the cross section is square, determine the dimension a of each of its sides using a factor of safety against buckling of F.S. = 2.5. The column is assumed to be pinned at its top and bottom. Use the Euler equation. Ew = 11 GPa, and sY = 10 MPa.
25 kN
a
4m a
1 a4 (a) A a3 B = , P = (2.5)25 = 62.5 kN and K = 1 12 12 cr for pin supported ends column. Applying Euler’s formula, Critical Buckling Load: I =
Pcr = 62.5 A 10
3
B =
p2EI (KL)2
a p2(11)(109) A 12 B 4
[1(4)]2
a = 0.1025 m = 103 mm
Ans.
Critical Stress: Euler’s formula is only valid if scr 6 sY. scr =
62.5(103) Pcr = = 5.94 MPa 6 s Y = 10 MPa A 0.1025(0.1025)
1148
O.K.
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•13–129. If the torsional springs attached to ends A and C of the rigid members AB and BC have a stiffness k, determine the critical load Pcr.
P
k
A
L 2 B
Equilibrium. When the system is given a slight lateral disturbance, the configuration shown in Fig. a is formed. The couple moment M can be related to P by considering the equilibrium of members AB and BC. Member AB + c ©Fy = 0; a + ©MA = 0;
By - P = 0
(1)
By a
(2)
L L sin u b + Bx a cos ub - M = 0 2 2
Member BC a + ©MC = 0; -By a
L L sin u b + Bx a cos ub + M = 0 2 2
(3)
Solving Eqs. (1), (2), and (3), we obtain Bx = 0
By =
2M L sin u
M =
PL sin u 2
Since u is very small, the small angle analysis gives sin u ⬵ u. Thus, M =
PL u 2
(4)
Torslonal Spring Moment. The restoring couple moment Msp can be determined using the torsional spring formula, M = ku. Thus, Msp = ku Critical Buckling Load. When the mechanism is on the verge of bucklling M must equal Msp. M = Msp Pcr L u = ku 2 Pcr =
2k L
Ans.
1149
L 2 k
C
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13–129.
Continued
1150
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13–130. Determine the maximum intensity w of the uniform distributed load that can be applied on the beam without causing the compressive members of the supporting truss to buckle. The members of the truss are made from A-36-steel rods having a 60-mm diameter. Use F.S. = 2 against buckling.
w
B A
C 2m
Equilibrium. The force developed in member BC can be determined by considering the equilibrium of the free-body diagram of the beam AB, Fig. a. 3 w(5.6)(2.8) - FBC a b(5.6) = 0 FBC = 4.6667w 5
a + ©MA = 0;
The Force developed in member CD can be obtained by analyzing the equilibrium of joint C, Fig. b, + c ©Fy = 0;
FAC a
+ : ©Fx = 0;
4 12 4.6667wa b + 7.28a b w-FCD = 0 5 13
5 3 b - 4.6667wa b = 0 13 5
FAC = 7.28w (T)
FCD = 10.4533w (C)
Section Properties. The cross-sectional area and moment of inertia of the solid circular rod CD are A = p A 0.032 B = 0.9 A 10 - 3 B p m2
I =
p A 0.034 B = 0.2025 A 10 - 6 B p m4 4
Critical Buckling Load. Since both ends of member CD are pinned, K = 1. The critical buckling load is Pcr = FCD (F.S.) = 10.4533w(2) = 20.9067w Applying Euler’s formula, Pcr =
p2EI (KL)2
20.9067w =
p2 C 200 A 109 B D C 0.2025 A 10 - 6 B p D [1(3.6)]2
Ans.
w = 4634.63 N>m = 4.63 kN>m Critical Stress: Euler’s formula is valid only if scr 6 sY. scr =
20.907(4634.63) Pcr = = 34.27 MPa 6 sY = 250 MPa A 0.9 A 10 - 3 B p
1151
O.K.
3.6 m
D
1.5 m
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13–131. The W10 * 45 steel column supports an axial load of 60 kip in addition to an eccentric load P. Determine the maximum allowable value of P based on the AISC equations of Sec. 13.6 and Eq. 13–30. Assume that in the x–z plane Kx = 1.0 and in the y–z plane Ky = 2.0. Est = 2911032 ksi, sY = 50 ksi.
z P 60 kip
x y
10 ft
Section properties for W 10 * 45: A = 13.3 in2
d = 10.10 in.
rx = 4.32 in.
ry = 2.01 in.
Ix = 248 in4
Allowable stress method: a
1.0(10)(12) KL b = = 27.8 r x 4.32
a
2.0(10)(12) KL b = = 119.4 r y 2.01
a
2p2(29)(103) KL 2p2E b = = = 107 r c B sg B 50
(controls)
KL KL 7 a b r r c (sa)allow =
12p2(29)(103) 12p2E = = 10.47 ksi 2 23(KL>r) 23(119.4)4
smax = (sa)allow =
Mc P + A I
P(8) A 10.10 P + 60 2 B + 10.47 = 13.3 248 P = 25.0 kip
Ans.
1152
y x 8 in.
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*13–132. The A-36-steel column can be considered pinned at its top and fixed at its base. Also, it is braced at its mid-height along the weak axis. Investigate whether a W250 * 45 section can safely support the loading shown. Use the allowable stress method.
600 mm 10 kN
4.5 m
Section Properties. From the table listed in the appendix, the necessary section properties for a W250 * 45 are A = 5700 mm2 = 5.70 A 10 - 3 B m2
d = 266 mm = 0.266 m
Ix = 71.1 A 106 B mm4 = 71.1 A 10 - 6 B m4
rx = 112 mm = 0.112 m
ry = 35.1 mm = 0.0351 mm
Slenderness Ratio. Here, Lx = 9 m and for a column fixed at its base and pinned at its top, Kx = 0.7. Thus,
¢
0.7(9) KL = 56.25 ≤ = r x 0.112
Since the bracing provides support equivalent to a pin, Ky = 1 and Ly = 4.5 m. Then,
¢
1(4.5) KL = 128.21 (controls) ≤ = r y 0.0351
Allowable Stress. For A-36 steel, a Since a
2p2 C 200 A 109 B D KL 2p2E b = = = 125.66. r c C 250 A 106 B B sY
KL KL b 6 a b 6 200, the column can be classified as a long column. r c r y sallow =
12p2 C 200 A 109 B D 12p2E = = 62.657 MPa 2 23(KL>r) 23(128.21)2
Maximum Stress. Bending occurs about the strong axis. Here, P = 10 + 40 0.266 d = 50 kN, M = 40(0.6) = 24 kN # m and c = = = 0.133 m, 2 2 smax =
50 A 103 B 24 A 103 B (0.133) P Mc + = + = 53.67 MPa A I 5.70 A 10 - 3 B 71.1 A 10 - 6 B
Since smax 6 sallow, the column is adequate according to the allowable stress method.
1153
4.5 m
40 kN
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•13–133.
The A-36-steel column can be considered pinned at its top and fixed at its base. Also, it is braced at its mid-height along the weak axis. Investigate whether a W250 * 45 section can safely support the loading shown. Use the interaction formula. The allowable bending stress is (sb)allow = 100 MPa.
600 mm 10 kN
4.5 m
Section Properties. From the table listed in the appendix, the necessary section properties for a W250 * 45 are A = 5700 mm2 = 5.70 A 10 - 3 B m2 Ix = 71.1 A 10 B mm = 71.1 A 10 6
4
-6
d = 266 mm = 0.266 m 4.5 m
Bm
4
rx = 112 mm = 0.112 m
ry = 35.1 mm = 0.0351 mm
Slenderness Ratio. Here, Lx = 9 m and for a column fixed at its base and pinned at its top, Kx = 0.7. Thus, a
0.7(9) KL b = = 56.25 r x 0.112
Since the bracing provides support equivalent to a pin, Ky = 1 and Ly = 4..5 m. Then, a
Allowable
=
C
1(4.5) KL b = = 128.21 (controls) r y 0.0351
Axial
2p2 C 200 A 109 B D 250 A 10
6
B
Stress.
For
= 125.66. Since a
A–36
steel,
a
KL 2p2E b = r c B sY
KL KL b 6 a b 6 200, the column can be r c r y
classified as a long column.
sallow =
12p2 C 200 A 109 B D 12p2E = = 62.657 MPa 23(KL>r)2 23(128.21)2
Interaction Formula. Bending is about the strong axis. Here, P = 10 + 40 = 50 kN, d 0.266 M = 40(0.6) = 24 kN # m and c = = = 0.133 m, 2 2 Mc>Ar2 P>A + = (sa)allow (sb)allow
50 A 103 B n 5.70 A 10 - 3 B 62.657 A 106 B
+
24 A 103 B (0.133) n C 5.70 A 10 - 3 B A 0.1122 B D 100 A 106 B
= 0.5864 6 1 sa = (sa)allow
50 A 103 B n 5.7 A 10 - 3 B 62.657 A 106 B
O.K.
= 0.140 6 0.15
O.K.
Thus, a W250 * 45 column is adequate according to the interaction formula.
1154
40 kN
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13–134. The member has a symmetric cross section. If it is pin connected at its ends, determine the largest force it can support. It is made of 2014-T6 aluminum alloy.
0.5 in. 2 in.
P
5 ft
Section properties: A = 4.5(0.5) + 4(0.5) = 4.25 in2 P
1 1 (0.5)(4.53) + (4)(0.5)3 = 3.839 in4 I = 12 12 r =
I 3.839 = = 0.9504 in. AA A 4.25
Allowable stress: 1.0(5)(12) KL = = 63.13 r 0.9504 KL 7 55 r Long column sallow =
54000 54000 = = 13.55 ksi (KL>r)2 63.132
Pallow = sallowA = 13.55(4.25) = 57.6 kip
Ans.
1155
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13–135. The W200 * 46 A-36-steel column can be considered pinned at its top and fixed at its base. Also, the column is braced at its mid-height against the weak axis. Determine the maximum axial load the column can support without causing it to buckle. 6m
Section Properties. From the table listed in the appendix, the section properties for a W200 * 46 are A = 5890 mm2 = 5.89 A 10 - 3 B m2
Iy = 15.3 A 106 B mm4 = 15.3 A 10 - 6 B m4
Ix = 45.5 A 106 B mm4 = 45.5 A 10 - 6 B m4
Critical Buckling Load. For buckling about the strong axis, Kx = 0.7 and Lx = 12 m. Since the column is fixed at its base and pinned at its top,
Pcr =
p2EIx (KL)x 2
=
p2 c200 A 109 B d c45.5 A 10 - 6 B d [0.7(12)]2
= 1.273 A 106 B N = 1.27 MN
For buckling about the weak axis, Ky = 1 and Ly = 6 m since the bracing provides a support equivalent to a pin. Applying Euler’s formula,
Pcr =
p2EIy (KL)y 2
=
p2 c200 A 109 B d c15.3 A 10 - 6 B d [1(6)]2
= 838.92 kN = 839 kN (controls)Ans.
Critical Stress. Euler’s formula is valid only if scr 6 sY. scr =
838.92 A 103 B Pcr = = 142.43 MPa 6 sY = 250 MPa A 5.89 A 10 - 3 B
1156
O.K.
6m
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*13–136. The structural A-36 steel column has the cross section shown. If it is fixed at the bottom and free at the top, determine the maximum force P that can be applied at A without causing it to buckle or yield. Use a factor of safety of 3 with respect to buckling and yielding.
P 20 mm A 10 mm
Section properties:
100 mm -3
2
4m
©A = 0.2(0.01) + 0.15 (0.01) + 0.1(0.01) = 4.5(10 ) m
150 mm A
100 mm 10 mm
0.005 (0.2)(0.01) + 0.085 (0.15)(0.01) + 0.165 (0.1)(0.01) ©xA x = = ©A 4.5(10 - 3) = 0.06722 m 1 (0.2)(0.013) + 0.2 (0.01)(0.06722 - 0.005)2 12
Iy =
+
1 (0.01)(0.153) + 0.01 (0.15)(0.085 - 0.06722)2 12
+
1 (0.1)(0.013) + 0.1 (0.01)(0.165 - 0.06722)2 12
= 20.615278 (10 - 6) m4 1 1 1 (0.01)(0.23) + (0.15)(0.013) + (0.01)(0.13) 12 12 12
Ix =
= 7.5125 (10 - 6) m4 ry =
Iy 20.615278(10 - 6) = = 0.0676844 AA A 4.5 (10 - 3)
Buckling about x-x axis: Pcr =
p2(200)(109)(7.5125)(10 - 6) p2 EI = 2 (KL) [2.0(4)]2
= 231.70 kN scr =
(controls)
231.7 (103) Pcr = 51.5 MPa 6 sg = 250 MPa = A 4.5 (10 - 3)
Yielding about y-y axis: smax =
P ec KL P c1 + 2 sec a b d; A 2r A EA r
e = 0.06722 - 0.02 = 0.04722 m
0.04722 (0.06722) ec = = 0.692919 0.0676844 r2 2.0 (4) P P KL = = 1.96992 P (10 - 3) 2P 2r A EA 2(0.0676844) A 200 (109)(4.5)(10 - 3) 250(106)(4.5)(10 - 3) = P[1 + 0.692919 sec (1.96992P (10 - 3)2P)] By trial and error: P = 378.45 kN Hence, Pallow =
231.70 = 77.2 kN 3
Ans.
1157
10 mm 100 mm
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•13–137.
The structural A-36 steel column has the cross section shown. If it is fixed at the bottom and free at the top, determine if the column will buckle or yield when the load P = 10 kN. Use a factor of safety of 3 with respect to buckling and yielding.
P 20 mm A 10 mm 100 mm
Section properties: 4m
©A = 0.2 (0.01) + 0.15 (0.01) + 0.1 (0.01) = 4.5 (10 - 3) m2 0.005 (0.2)(0.01) + 0.085 (0.15)(0.01) + 0.165 (0.1)(0.01) ©xA = 0.06722 m = ©A 4.5 (10 - 3)
Iy =
1 (0.2)(0.013) + 0.2 (0.01)(0.06722 - 0.005)2 12
ry =
+
1 (0.01)(0.153) + 0.01 (0.15)(0.085 - 0.06722)2 12
+
1 (0.1)(0.013) + 0.1 (0.01)(0.165 - 0.06722)2 = 20.615278 (10 - 6) m4 12
1 1 1 (0.01)(0.23) + (0.15)(0.013) + (0.01)(0.13) = 7.5125 (10 - 6) m4 12 12 12 Iy BA
20.615278 (10 - 6) =
B
4.5 (10 - 3)
= 0.067843648 m
Buckling about x-x axis: Pcr =
p2(200)(109)(7.5125)(10 - 6) p2 EI = = 231.70 kN 2 (KL) [2.0(4)]2
scr =
231.7 (103) Pcr = 51.5 MPa 6 sg = 250 MPa = A 4.5 (10 - 3)
Pallow =
O.K.
Pcr 231.7 = = 77.2 kN 7 P = 10 kN FS 3
Hence the column does not buckle. Yielding about y-y axis: smax =
P =
P KL ec P bd c1 + 2 sec a A 2r A EA r
A
100 mm 10 mm
x =
Ix =
150 mm
e = 0.06722 - 0.02 = 0.04722 m
10 = 3.333 kN 3
3.333 (103) P = 0.7407 MPa = A 4.5 (10 - 3) 0.04722 (0.06722) ec = 0.689815 = (0.067844) r2 2.0 (4) P KL 3.333 (103) = = 0.1134788 2 r AE A 2(0.06783648) A 200 (109)(4.5)(10 - 3) smax = 0.7407 [1 + 0.692919 sec (0.1134788)] = 1.25 MPa 6 sg = 250 MPa Hence the column does not yield! No.
Ans. 1158
10 mm 100 mm
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sy
14–1. A material is subjected to a general state of plane stress. Express the strain energy density in terms of the elastic constants E, G, and n and the stress components sx , sy , and txy .
txy
sx
‘Strain Energy Due to Normal Stresses: We will consider the application of normal stresses on the element in two successive stages. For the first stage, we apply only sx on the element. Since sx is a constant, from Eq. 14-8, we have s2x s2x V dV = 2E Lv 2E
(Ui)1 =
When sy is applied in the second stage, the normal strain ex will be strained by ex ¿ = -vey = -
vsy E
. Therefore, the strain energy for the second stage is
(Ui)2 =
=
s2y
¢
Lv 2E
B
s2y
Lv 2E
+ sx ex ¿ ≤ dV + sx a -
vsy E
b R dV
Since sx and sy are constants, (Ui)2 =
V (s2 - 2vsx sy) 2E y
Strain Energy Due to Shear Stresses: The application of txy does not strain the element in normal direction. Thus, from Eq. 14–11, we have (Ui)3 =
t2xy Lv 2G
dV =
t2xy V 2G
The total strain energy is Ui = (Ui)1 + (Ui)2 + (Ui)3 =
t2xy V s2x V V + (s2y - 2vsx sy) + 2E 2E 2G
=
t2xy V V (s2x + s2y - 2vsx sy) + 2E 2G
and the strain energy density is t2xy Ui 1 = (s2x + s2y - 2vsx sy) + V 2E 2G
Ans.
1159
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14–2. The strain-energy density must be the same whether the state of stress is represented by sx , sy , and txy , or by the principal stresses s1 and s2 . This being the case, equate the strain–energy expressions for each of these two cases and show that G = E>[211 + n2].
U =
1 v 1 2 (s2x + s2y) - sxsy + t R dV E 2 G xy Lv 2 E
U =
1 v (s21 + s22) s s R dV B E 1 2 Lv 2 E
B
Equating the above two equations yields. v 1 2 1 v 1 (s2x + s2y) sxsy + txy = (s21 + s22) s s 2E E 2G 2E E 1 2 However, s1, 2 =
sx + sy 2
;
A
a
sx - sy 2
(1)
2 b + txy 2
Thus, A s21 + s22 B = s2x + s2y + 2 t2xy s1 s2 = sxsy - t2xy Substitute into Eq. (1) v 1 2 1 v v 2 1 t = (s2 + s2y + 2t2xy) ss + t A s2 + s2y B - sxsy + 2E x E 2 G xy 2E x E x y E xy t2xy v 2 1 2 txy = + t 2G E E xy 1 v 1 = + 2G E E 1 1 = (1 + v) 2G E G =
E 2(1 + v)
QED
1160
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14–3. Determine the strain energy in the stepped rod assembly. Portion AB is steel and BC is brass. Ebr = 101 GPa, Est = 200 GPa, (sY)br = 410 MPa, (sY)st = 250 MPa.
100 mm A
B
30 kN
30 kN 1.5 m
Referring to the FBDs of cut segments in Fig. a and b, + ©F = 0; : x
NBC - 20 = 0
+ ©F = 0; : x
NAB - 30 - 30 - 20 = 0
NBC = 20 kN NAB = 80 kN
p The cross-sectional area of segments AB and BC are AAB = (0.12) = 2.5(10 - 3)p m2 and 4 p ABC = (0.0752) = 1.40625(10 - 3)p m2. 4 (Ui)a = ©
NAB 2LAB NBC 2LBC N2L = + 2AE 2AAB Est 2ABC Ebr =
C 80(103) D 2 (1.5)
2 C 2.5(10 - 3)p D C 200(109) D
+
C 20(103) D 2(0.5)
2 C 1.40625(10 - 3) p D C 101(109) D
= 3.28 J
Ans.
This result is valid only if s 6 sy. sAB =
80(103) NAB = 10.19(106)Pa = 10.19 MPa 6 (sy)st = 250 MPa = AAB 2.5(10 - 3)p
O.K.
sBC =
20 (103) NBC = 4.527(106)Pa = 4.527 MPa 6 (sy)br = 410 MPa = ABC 1.40625(10 - 3) p
O.K.
1161
0.5 m
75 mm C 20 kN
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*14–4. Determine the torsional strain energy in the A-36 steel shaft. The shaft has a diameter of 40 mm.
900 N⭈m 200 N⭈m
0.5 m
Referring to the FBDs of the cut segments shown in Fig. a, b and c,
300 N⭈m
0.5 m
TAB = 300 N # m
©Mx = 0;
TAB - 300 = 0
©Mx = 0;
TBC - 200 - 300 = 0
©Mx = 0;
TCD - 200 - 300 + 900 = 0 TCD = -400 N # m
0.5 m
TBC = 500 N # m
The shaft has a constant circular cross-section and its polar moment of inertia is p J = (0.024) = 80(10 - 9)p m4. 2 (Ui)t = ©
TAB 2 LAB TBC 2LBC TCD LCD T2L = + + 2GJ 2GJ 2GJ 2GJ 1
=
2 C 75(10 ) 80 (10 - 9)p D 9
c3002 (0.5) + 5002 (0.5) + (-400)2 (0.5) d
= 6.63 J
Ans.
•14–5.
Determine the strain energy in the rod assembly. Portion AB is steel, BC is brass, and CD is aluminum. Est = 200 GPa, Ebr = 101 GPa, and Eal = 73.1 GPa.
15 mm A
20 mm 2 kN B
25 mm D 5 kN C
2 kN
5 kN
3 kN
300 mm
N2 L Ui = © 2AE [3 (103) ]2 (0.3) =
2 (p4 )(0.0152)(200)(109)
[7 (103) ]2 (0.4) +
2(p4 )(0.022)(101)(109) [-3 (103) ]2 (0.2) + 2
(p4 )(0.0252)(73.1)(109)
= 0.372 N # m = 0.372 J
Ans.
1162
400 mm
200 mm
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14–6. If P = 60 kN, determine the total strain energy stored in the truss. Each member has a cross-sectional area of 2.511032 mm2 and is made of A-36 steel.
2m B C
Normal Forces. The normal force developed in each member of the truss can be determined using the method of joints.
1.5 m
Joint A (Fig. a) D
+ ©F = 0; : x
FAD = 0
+ c ©Fy = 0;
FAB - 60 = 0
FAB = 60 kN (T)
P
Joint B (Fig. b) + c ©Fy = 0;
3 FBD a b - 60 = 0 5
FBD = 100 kN (C)
+ ©F = 0; : x
4 100 a b - FBC = 0 5
FBC = 80 kN (T)
Axial Strain Energy. LBD = 222 + 1.52 = 2.5 m (Ui)a = ©
=
A = 2.5 A 103 B mm2 = 2.5 A 10 - 3 B m2
and
N2L 2AE
2 C 2.5 A 10
1
-3
B D C 200 A 109 B D
c C 60 A 103 B D 2 (1.5) + C 100 A 103 B D 2 (2.5) + C 80 A 103 B D 2 (2) d
= 43.2 J
Ans.
This result is only valid if s 6 sY. We only need to check member BD since it is subjected to the greatest normal force sBD =
A
100 A 103 B FBD = = 40 MPa 6 sY = 250 MPa A 2.5 A 10 - 3 B
O.K.
1163
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14–7. Determine the maximum force P and the corresponding maximum total strain energy stored in the truss without causing any of the members to have permanent deformation. Each member has the crosssectional area of 2.511032 mm2 and is made of A-36 steel.
2m B C
1.5 m
D
Normal Forces. The normal force developed in each member of the truss can be determined using the method of joints. Joint A (Fig. a) + ©F = 0; : x
FAD = 0
+ c ©Fy = 0;
FAB - P = 0
FAB = P (T)
Joint B (Fig. b) + c ©Fy = 0;
3 FBD a b - P = 0 5
+ ©F = 0; : x
4 1.6667Pa b - FBC = 0 5
FBD = 1.6667P (C)
FBC = 1.3333P(T)
Axial Strain Energy. A = 2.5 A 103 B mm2 = 2.5 A 10 - 3 B m2. Member BD is critical since it is subjected to the greatest force. Thus, sY =
FBD A
250 A 106 B =
1.6667P
2.5 A 10 - 3 B
P = 375 kN
Ans.
Using the result of P FAB = 375 kN
FBD = 625 kN
FBC = 500 kN
Here, LBD = 21.52 + 22 = 2.5 m. (Ui)a = ©
N2L = 2AE 1
=
2 C 2.5 A 10 - 3 B D C 200 A 109 B D
c C 375 A 103 B D 2 (1.5) + C 625 A 103 B D 2 (2.5) + C 500 A 103 B D 2 (2) d
= 1687.5 J = 1.6875 kJ
Ans.
1164
A P
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*14–8. Determine the torsional strain energy in the A-36 steel shaft. The shaft has a radius of 30 mm.
4 kN⭈m 3 kN⭈m
0.5 m
T2L 1 = [02(0.5) + ((3)(103))2(0.5) + ((1)(103))2(0.5)] Ui = © 2JG 2JG =
0.5 m
2.5(106) JG 2.5(106)
=
0.5 m
75(109)(p2 )(0.03)4
= 26.2 N # m = 26.2 J
Ans.
•14–9.
Determine the torsional strain energy in the A-36 steel shaft. The shaft has a radius of 40 mm.
12 kN⭈m 6 kN⭈m
0.5 m
Internal Torsional Moment: As shown on FBD. 8 kN⭈m
Torsional Strain Energy: With polar moment p J = A 0.044 B = 1.28 A 10 - 6 B p m4. Applying Eq. 14–22 gives 2
of
inertia
T2L Ui = a 2GJ =
1 C 80002 (0.6) + 20002 (0.4) + 2GJ
=
45.0(106) N2 # m3 GJ
A -100002 B (0.5) D
45.0(106) =
9
75(10 )[1.28(10 - 6) p]
= 149 J
Ans.
1165
0.4 m 0.6 m
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14–10. Determine the torsional strain energy stored in the tapered rod when it is subjected to the torque T. The rod is made of material having a modulus of rigidity of G. L
2r0
Internal Torque. The internal torque in the shaft is constant throughout its length as shown in the free-body diagram of its cut segment, Fig. a, Torsional Strain Energy. Referring to the geometry shown in Fig. b, r = r0 +
T
r0 r0 (x) = (L + x) L L
The polar moment of inertia of the bar in terms of x is J(x) =
4 pr0 4 p 4 p r0 (L + x)4 r = c (L + x) d = 2 2 L 2L4
We obtain, L
(Ui)t =
T2dx dx = L0 2GJ L0
L
T2 dx 2G B
pr0 4 2L4
(L + x)4 R
L
=
T2L4 dx pr0 4G L0 (L + x)4
=
L T2L4 1 B R ` pr0 4G 3(L + x)3 0
=
7 T2L 24pr0 4 G
Ans.
1166
r0
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14–11. The shaft assembly is fixed at C. The hollow segment BC has an inner radius of 20 mm and outer radius of 40 mm, while the solid segment AB has a radius of 20 mm. Determine the torsional strain energy stored in the shaft. The shaft is made of 2014-T6 aluminum alloy. The coupling at B is rigid.
600 mm 20 mm
600 mm C 40 mm B 60 N⭈m
Internal Torque. Referring to the free-body diagram of segment AB, Fig. a, TAB = -30 N # m
©Mx = 0; TAB + 30 = 0
Referring to the free-body diagram of segment BC, Fig. b, ©Mx = 0; TBC + 30 + 60 = 0
TAB = -90 N # m
p Torsional Strain Energy. Here, JAB = A 0.024 B = 80 A 10 - 9 B p m4 2 p JBC = A 0.044 - 0.024 B = 1200 A 10 - 9 B p m4, 2 (Ui)t = ©
TAB 2LAB TBC 2LBC T2L = + 2GJ 2GJAB 2GJBC (-30)2(0.6)
=
and
2 C 27 A 109 B D C 80 A 10 - 9 B p D
(-90)2(0.6)
+
2 C 27 A 109 B D C 1200 A 10 - 9 B p D
= 0.06379 J
Ans.
1167
A 20 mm 30 N⭈m
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*14–12. Consider the thin-walled tube of Fig. 5–28. Use the formula for shear stress, tavg = T>2tAm, Eq. 5–18, and the general equation of shear strain energy, Eq. 14–11, to show that the twist of the tube is given by Eq. 5–20, Hint: Equate the work done by the torque T to the strain energy in the tube, determined from integrating the strain energy for a differential element, Fig. 14–4, over the volume of material.
Ui =
t2 dV Lv 2 G
but t =
T 2 t Am
Thus, Ui =
T2 dV 2 2 Lv 8 t AmG L
=
2
T2 dV T2 dA TL dA = dx = 2 2 2 2 2 2 8 A m G Lv t 8 A m G LA t L0 8 A mG LA t
However, dA = t ds. Thus, Ui =
ds T2L 2 8 AmG L t
Ue =
1 Tf 2
Ue = Ui ds T2L 1 Tf = 2 8 A2mG L t f =
ds TL 4 A2mG L t
QED
1168
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•14–13.
Determine the ratio of shearing strain energy to bending strain energy for the rectangular cantilever beam when it is subjected to the loading shown. The beam is made of material having a modulus of elasticity of E and Poisson’s ratio of n.
P L a
a
b h Section a – a
Internal Moment. Referring to the free-body diagram of the left beam’s cut segment, Fig. a, + c ©Fy = 0;
-V - P = 0
V = -P
+ ©MO = 0;
M + Px = 0
M = -Px
Shearing Strain Energy. For the rectangular cross section, the form factor is fs =
6 5
6 2 2 L (-P) dx L fsV dx 5 3P2 3P2L (Ui)v = = = dx = 2GA 2G(bh) 5bhG L0 5bhG L0 L0 L
However, G =
E , then 2(1 + v) (Ui)v =
6(1 + v)P2L 5bhE
Bending Strain Energy. L
(Ui)b =
M2dx = L0 2EI L0
L
(-Px)2dx2 2Ea
1 bh3 b 12
6P2 6P2 x3 L 2P2L3 2 x dx = = ¢ ≤ ` bh3E L0 bh3E 3 0 bh3E L
=
Then, the ratio is 6(1 + v)P2L 3(1 + v) h 2 (Ui)v 5bhE a b = = 2 3 (Ui)b 5 L 2P L bh3E
Ans.
From this result, we can conclude that the proportion of the shearing strain energy stored in the beam increases if the depth h of the beam’s cross section increases but (Ui)v 1 = 0.009. the decreases if L increases. Suppose that v = and L = 10h, then 2 (Ui)b shearing strain energy is only 0.09% of the bending strain energy. Therefore, the effect of the shearing strain energy is usually neglected if L 7 10h.
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14–14. Determine the bending strain-energy in the beam due to the loading shown. EI is constant.
M0 A
L
Ui =
L0
C B
2
M dx 2EI
=
1 c 2EI L0
=
M20L 24EI
L>2
L — 2
a
L — 2
L>2 2 2 -M0 M0 x1 b dx1 + a x2 b dx2] L L L0
Ans.
Note: Strain energy is always positive regardless of the sign of the moment function.
14–15. Determine the bending strain energy in the beam. EI is constant.
P
P
Referring to the FBD of the entire beam, Fig. a, a + ©MB = 0;
Pa
L 3L b + Pa b - Ay (L) = 0 4 4
L 4
Ay = P
Using the coordinates, x1 and x2, the FBDs of the beam’s cut segments in Figs. b and c are drawn. For coordinate x1, a + ©Mc = 0;
M(x1) - Px1 = 0
M(x1) = Px1
For coordinate x2 coordinate, a + ©Mc = 0;
M(x2) - Pa
L b = 0 4
L
(Ui)b = ©
1 M2dx = c2 2EI 2EI L0 L0
M(x2) =
PL 4
L>4
L>2
(Px1)2dx1 +
L0
a
=
P2 3 P2L2 2 1 c2 a x1 b ` + x ` d 2EI 3 16 2 0 0
=
P2L3 48EI
L 4
PL 2 b dx2 d 4
L
Ans.
1170
L 2
L 4
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*14–16. Determine the bending strain energy in the A-36 structural steel W10 * 12 beam. Obtain the answer using the coordinates 1a2 x1 and x4, and 1b2 x2 and x3.
6 kip
x1
x4 x2 12 ft
Support Reactions: As shown on FBD(a). Internal Moment Function: As shown on FBD(b), (c), (d) and (e). Bending Strain Energy: Using coordinates x1 and x4 and applying Eq. 14–17 gives L
Ui =
M2dx L0 2EI 1 c 2EI L0
12ft
=
6ft
1 c 2EI L0
12ft
=
=
3888 kip2 # ft3 EI
(-3.00x1)2 dx1 +
L0
(-6.00x4)2 dx4 d
6ft
9.00x21dx1 +
L0
36.0x24 dx4 d
For W10 * 12 wide flange section, I = 53.8 in4. Ui =
3888(123) 29.0(103)(53.8)
= 4.306 in # kip = 359 ft # lb
Ans.
b) Using coordinates x2 and x3 and applying Eq. 14–17 gives L
Ui =
M2dx L0 2EI 1 c 2EI L0
12ft
=
6ft
1 c 2EI L0
12ft
=
=
3888 kip2 # ft3 EI
(3.00x2 - 36.0)2dx2 +
L0
(6.00x3 - 36.0)2 dx3 d
A 9.00x22 - 216x + 1296 B dx2 +
6ft
L0
A 36.0x23 - 432x + 1296 B dx3 d
For W 10 * 12 wide flange section, I = 53.8 in4. Ui =
3888(123) 29.0(103)(53.8)
= 4.306 in # kip = 359 ft # lb
Ans.
1171
x3 6 ft
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•14–17.
Determine the bending strain energy in the A-36 steel beam. I = 99.2 (106) mm4.
9 kN/m
6m
Referring to the FBD of the entire beam, Fig. a, 1 (9)(6)(2) - Ay (6) = 0 2
a + ©MB = 0;
Ay = 9 kN
Referring to the FBD of the beam’s left cut segment, Fig. b,
a + ©M0 = 0;
M(x) +
1 3 a xb (x) (x>3) - 9x = 0 2 2
M(x) = a 9x L
(Ui)b =
For
A
36
1 3 x b 4
kN # m
M2 dx 1 = 2EI L0 L0 2EI
6m
a 9x -
1 2EI L0
6m
=
a 81x2 +
=
1 7 9 5 2 6m 1 c a27x3 + x x b d 2EI 112 10 0
=
666.51 kN2 # m3 EI
steel,
E = 200 GPa.
Here,
= 99.2(10 - 6) m4. Then
(Ui)b =
666.51 (10002)
200(109) C 99.2(10 - 6) D
1 3 2 x b dx 4 1 6 9 x - x4 bdx 16 2
I = C 99.2 (106) mm4 D a
4 1m b 1000 mm
Ans.
= 33.6 J
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14–18. Determine the bending strain energy in the A-36 steel beam due to the distributed load. I = 122 (106) mm4.
15 kN/m
A
B 3m
Referring to the FBD of the entire beam, Fig. a, 1 (15)(3) = 0 2
Ay -
+ c ©Fy = 0; a + ©MA = 0;
MA -
Ay = 22.5 kN
1 (15)(3)(2) = 0 2
MA = 45 kN # m
Referring to the FBD of the beam’s left cut segment, Fig. b, a + ©M0 = 0;
M(x) +
1 (5x)(x)(x>3) - 22.5x + 45 = 0 2
M(x) = (22.5x - 0.8333x3 - 45) kN # m L
(Ui)b =
=
1 M2dx = c 2EI 2EI L0 L0 1 c 2EI L0
3m
(22.5x - 0.8333x3 - 45)2 dx
3m
0.6944x6 - 37.5x4 + 75x3 + 506.25x2 - 2025x + 2025)dx d
=
1 a0.09921x7 - 7.5x5 + 18.75x4 + 168.75x3 2EI - 1012.5x2 + 2025xb 2
3m 0
715.98 kN2 # m2 = EI For
A
36
steel,
E = 200 GPa.
Here,
I = c122(106) mm4 d a
4 1m b 1000 mm
= 122(10 - 6) m4. Thus,
(Ui)b =
715.98 (10002)
200(109) C 122 (10 - 6) D
Ans.
= 29.3 J
1173
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14–19. Determine the strain energy in the horizontal curved bar due to torsion. There is a vertical force P acting at its end. JG is constant.
r 90⬚
P
T = Pr(1 - cos u) Strain energy: L
Ui =
T2 ds L0 2JG
However, ds = rdu
s = ru; u
Ui =
T2rdu r = 2JG L0 L0 2JG P2r3 2JG L0
p>2
=
P2r3 2JG L0
p>2
=
P2r3 2JG L0
p>2
=
=
P2r3 3p a - 1b JG 8
p>2
[Pr(1 - cos u)]2du
(1 - cos u)2 du (1 + cos2 u - 2 cos u)du
(1 +
cos 2u + 1 - 2 cos u) du 2 Ans.
1174
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*14–20. Determine the bending strain energy in the beam and the axial strain energy in each of the two rods. The beam is made of 2014-T6 aluminum and has a square cross section 50 mm by 50 mm. The rods are made of A-36 steel and have a circular cross section with a 20-mm diameter.
2m 8 kN
Support Reactions: As shown on FBD(a).
8 kN
Internal Moment Function: As shown on FBD(b) and (c). Axial Strain Energy: Applying Eq. 14–16 gives
1m
N2L (Ui)a = 2AE
C 8.00(103) D 2 (2)
=
2AE
64.0(106) N2 # m AE
=
64.0(106) =
p 4
(0.022) [200(109)] Ans.
= 1.02 J Bending Strain Energy: Applying Eq. 14–17 gives L
(Ui) b =
M2dx L0 2EI
=
1 B2 2EI L0
=
85.333 kN2 # m3 EI
1m
(8.00x1)2 dx 1 +
2m
L0
8.002 dx2 R
85.333(106)
=
1 73.1(109) C 12 (0.05) (0.053) D
= 2241.3 N # m = 2.24 kJ
Ans.
1175
2m
1m
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•14–21.
The pipe lies in the horizontal plane. If it is subjected to a vertical force P at its end, determine the strain energy due to bending and torsion. Express the results in terms of the cross-sectional properties I and J, and the material properties E and G.
z
L
C x
B L — 2
A
P
Ui =
M2 dx T2 dx + L 2E I L 2JG L 2
=
L L PL 2 ( 2 ) dx (P x)2 dx (P x)2 dx + + 2 E I 2 EI L0 L0 2 J G L0
=
L 31 P2 L3 P2L2 P2 a b + + (L) 2EI 2 3 2EI 3 8JG
=
P2 L3 3 P2 L3 + 16 E I 8JG
= P2 L3 c
3 1 + d 16 E I 8JG
Ans.
1176
y
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14–22. The beam shown is tapered along its width. If a force P is applied to its end, determine the strain energy in the beam and compare this result with that of a beam that has a constant rectangular cross section of width b and height h.
b
h
P L
Moment of Inertia: For the beam with the uniform section, I =
bh3 = I0 12
For the beam with the tapered section, I =
I0 bh3 1 b a xb A h3 B = x = x 12 L 12L L
Internal Moment Function: As shown on FBD. Bending Strain Energy: For the beam with the tapered section, applying Eq. 14–17 gives L
UI =
M2 dx L0 2EI
=
L (-Px)2 1 dx I0 2E L0 L x
=
P2L xdx 2EI0 L0
=
P2L3 3P2 L3 = 4EI0 bh3 E
L
Ans.
For the beam with the uniform section, L
Ui =
M2dx L0 2EI L
=
1 (-Px)2 dx 2EI0 L0
=
P3 L3 6EI0
The strain energy in the capered beam is 1.5 times as great as that in the beam having a uniform cross section. Ans.
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14–23. Determine the bending strain energy in the cantilevered beam due to a uniform load w. Solve the problem two ways. (a) Apply Eq. 14–17. (b) The load w dx acting on a segment dx of the beam is displaced a distance y, where y = w1-x4 + 4L3x - 3L42>124EI2, the equation of the elastic curve. Hence the internal strain energy in the differential segment dx of the beam is equal to the external work, i.e., dUi = 121w dx21-y2. Integrate this equation to obtain the total strain energy in the beam. EI is constant.
w dx w
dx
Internal Moment Function: As shown on FBD. Bending Strain Energy: a) Applying Eq. 14–17 gives L
Ui =
M2dx L0 2EI L
=
1 w 2 c - x2 d dx R B 2EI L0 2
=
w2 x4 dx R B 8EI L0
=
w2 L5 40EI
L
b) Integrating dUi =
1 w (wdx) B A -x4 + 4L3x - 3L4 B R 2 24EI
dUi =
w2 A x4 -4L3x + 3L4 B dx 48EI
Ui =
=
Ans.
1 (wdx)( -y) 2
dUi =
w2 48EI L0
L
x L
A x4 - 4L3x + 3L4 B dx
w2L5 40EI
Ans.
1178
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*14–24. Determine the bending strain energy in the simply supported beam due to a uniform load w. Solve the problem two ways. (a) Apply Eq. 14–17. (b) The load w dx acting on the segment dx of the beam is displaced a distance y, where y = w1-x4 + 2Lx3 - L3x2>124EI2, the equation of the elastic curve. Hence the internal strain energy in the differential segment dx of the beam is equal to the external work, i.e., dUi = 121w dx21-y2. Integrate this equation to obtain the total strain energy in the beam. EI is constant.
w dx w
x
Support Reactions: As shown on FBD(a). Internal Moment Function: As shown on FBD(b). Bending Strain Energy: a) Applying Eq. 14–17 gives L
Ui =
M2dx L0 2EI L
=
2 1 w 2 c (Lx - x ) d dx R B 2EI L0 2
=
w2 (L2x2 + x4 - 2Lx3) dx R B 8EI L0
=
w2L5 240EI
L
b) Integrating dUi =
Ans.
1 (wdx) (-y) 2
dUi =
1 w (wdx) B (-x4 + 2Lx3 - L3x) R 2 24EI
dUi =
w2 (x4 - 2Lx3 + L3x) dx 48EI L
Ui =
=
w2 (x4 - 2Lx3 + L3x) dx 48EI L0 w2L5 240EI
Ans.
1179
dx L
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•14–25. Determine the horizontal displacement of joint A. Each bar is made of A-36 steel and has a cross-sectional area of 1.5 in2.
2 kip
A
3 ft D 3 ft C
B 4 ft
Member Forces: Applying the method of joints to joint at A, we have + ©F = 0; : x
4 - 2 = 0 F 5 AD
+ c ©Fy = 0;
FAB -
FAD = 2.50 kip (T)
3 (2.50) = 0 5
FAB = 1.50 kip (C)
At joint D + ©F = 0; : x
4 4 F - (2.50) = 0 5 DB 5
+ c ©Fy = 0;
3 3 ( 2.50) + (2.50) - FDC = 0 5 5
FDB = 2.50 kip (C)
FDC = 3.00 kip (T) Axial Strain Energy: Applying Eq. 14–16, we have N2L Ui = a 2AE =
1 [2.502 (5) + (-1.50)2 (6) + (-2.50)2 (5) + 3.002(3)] 2AE
=
51.5 kip2 # ft AE 51.5(12)
=
1.5[29.0(103)]
= 0.014207 in # kip
External Work: The external work done by 2 kip force is Ue =
1 (2) (¢ A)h = (¢ A)h 2
Conservation of Energy: Ue = Ui (¢ A)h = 0.014207 = 0.0142 in.
Ans.
1180
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14–26. Determine the horizontal displacement of joint C. AE is constant.
C P
L
L
A
B L
Member Forces: Applying the method of joints to C, we have + c ©Fy = 0;
FBC cos 30° - FAC cos 30° = 0
+ ©F = 0; : x
P - 2F sin 30° = 0
Hence,
FBC = P (C)
FBC = FAC = F
F = P
FAC = P (T)
Axial Strain Energy: Applying Eq. 14–16, we have N2L Ui = a 2AE =
1 C P2L + (-P)2 L D 2AE
=
P2L AE
External Work: The external work done by force P is Ui =
1 P(¢ c)k 2
Conservation of Energy: Ue = Ui P2L 1 P(¢ C)k = 2 AE (¢ C)k =
2PL AE
Ans.
1181
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14–27. Determine the vertical displacement of joint C. AE is constant.
A
L
L
C
L
B
Joint C: + ©F = 0 : x
FCB cos 30° - FCA cos 30° = 0 FCB = FCA
+ c © Fy = 0
FCA sin 30° + FCB sin 30° - P = 0 FCB = FCA = P
Conservation of energy: Ue = Ui N2L 1 P¢ C = © 2 2EA L 1 2 P¢ C = [F 2 + FCA ] 2 2EA CB P¢ C =
¢C =
L (P2 + P2) EA 2PL AE
Ans.
1182
P
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*14–28. Determine the horizontal displacement of joint D. AE is constant.
D
C
P
Joint B:
L
+ c ©Fy = 0;
FBC = 0.75P
+ ©F = 0; ; x
FBA = P
0.6 L
B
A
Joint D:
0.8 L
+ T ©Fy = 0;
FDA = 0
+ ©Fx = 0; :
FDC = P
Joint A: + T ©Fy = 0;
3 F - 0.75P = 0 5 AC FAC = 1.25P
Conservation of energy: Ue = Ui N2L 1 P¢ D = © 2 2AE 1 1 P¢ D = [(0.75P)2(0.6L) + (P)2(0.8L) + (02)(0.6L) 2 2AE + (P2)(0.8L) + (1.25P)2(L)] ¢D =
3.50PL AE
Ans.
•14–29.
The cantilevered beam is subjected to a couple moment M0 applied at its end. Determine the slope of the beam at B. EI is constant.
M0
A B
Ui =
2 L M20L M dx 1 = M20 dx = 2EI L0 2EI L0 2EI
Ue =
1 (M0 uB) 2
L
L
Conservation of energy: Ue = Ui M0 2L 1 M0 uB = 2 2EI uB =
M0L EI
Ans.
1183
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14–30. Determine the vertical displacement of point C of the simply supported 6061-T6 aluminum beam. Consider both shearing and bending strain energy.
100 kip a B
A C
1.5 ft
Support Reactions. Referring to the free-body diagram of the entire beam, Fig. a, a + ©MB = 0;
100(1.5) - Ay(3) = 0
Ay = 50 kip
50 - V = 0
+ c ©Fy = 0; a + ©MO = 0;
V = 50 kip
M - 50x = 0
M = 50x
Shearing Strain Energy. For the rectangular beam, the form factor is fs =
6 5
6 A 502 B dx 18 in. fsV2dx 5 (Ul)v = = 2 = 0.3041 in # kip L0 2GA L0 2 C 3.7 A 103 B D [4(12)] L
Bending Strain Energy. I =
1 (4) A 123 B = 576 in4. We obtain 12
L
(Ui)b =
L (50x)2 dx M2dx = 2 L0 2EI L0 2 C 10.0 A 103 B D (576)
= 0.4340 A 10-3 B
18 in.
L0
= 0.4340 A 10-3 B ¢
x2 dx
x3 18in. ≤` 3 0
= 0.84375 in # kip Thus, the total strain energy stored in the beam is Ui = (Ui)v + (Ui)b = 0.3041 + 0.84375 = 1.1478 in # kip
Ans.
External Work. The external work done by the external force (100 kip) is Ue =
1 1 P¢ = (100)¢ C = 50¢ C 2 2
Conservation of Energy. Ue = Ui 50¢ C = 1.1478 Ans.
¢ C = 0.02296 in. = 0.0230 in. 1184
1.5 ft 4 in.
12 in.
Internal Loading. Referring to the free-body diagram of the beam’s left cut segment, Fig. b,
a
Section a – a
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14–31. Determine the slope at the end B of the A-36 steel beam. I = 8011062 mm4.
6 kN⭈m A
B
M = -750x 8m L
2
M dx 1 M uB = 2 L0 2EI B ( -750x)2dx 1 (6(103)) uB = 2 2EI L0
uB =
16000 200 (109)(80)(10-6)
= 1 (10-3) rad
Ans.
*14–32. Determine the deflection of the beam at its center caused by shear. The shear modulus is G.
P b
Support Reactions: As shown on FBD(a).
h
Shear Functions: As shown on FBD(b). Shear Strain Energy: Applying 14–19 with fe =
L 2
6 for a rectangular section, we have 5
L
Ui =
feV2dx L0 2GA L
2 6 P 2 1 a b a b dx R = B2 2bhG 2 L0 5
=
3P2L 20bhG
External Work: The external work done by force P is Ue =
1 (P) ¢ 2
Conservation of Energy: Ue = Ui 3P2L 1 (P)¢ = 2 20bhG ¢ =
3PL 10bhG
Ans.
1185
L 2
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•14–33. The A-36 steel bars are pin connected at B and C. If they each have a diameter of 30 mm, determine the slope at E. 3
2
C D
A E 2m
3m L
300 N⭈m
B
2m
3m
2
Ui =
M dx 1 1 65625 (75x1)2dx1 + (2) (-75x2)2dx2 = = (2) 2EI 2EI 2EI EI L0 L0 L0
Ue =
1 1 (M¿)u = (300) uE = 150 uE 2 2
Conservation of energy: Ue = Ui
150 uE =
uE =
65625 EI
473.5 473.5 = 0.0550 rad = 3.15° = EI (200)(109)(p4 )(0.0154)
14–34. The A-36 steel bars are pin connected at B. If each has a square cross section, determine the vertical displacement at B.
Ans.
800 lb 2 in. A
B
C
D 2 in.
Support Reactions: As shown on FBD(a). 8 ft
Moment Functions: As shown on FBD(b) and (c). Bending Strain Energy: Applying 14–17, we have L
Ui =
M2dx L0 2EI 4ft
10ft
=
1 B 2EI L0
=
23.8933(106) lb2 # ft3 EI
( -800x1)2 dx1 +
23.8933(106)(123)
=
1 29.0(106) C 12 (2)(23) D
L0
( -320x2)2 dx2 R
= 1067.78 in # lb
External Work: The external work done by 800 lb force is Ue =
1 (800)(¢ B) = 400¢ B 2
Conservation of Energy: Ue = Ui 400¢ B = 1067.78 ¢ B = 2.67 in.
Ans.
1186
4 ft
10 ft
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14–35. Determine the displacement of point B on the A-36 steel beam. I = 8011062 mm4.
20 kN
A
C B 3m
L
3
5m
5
1 M2dx = [(12.5)(103)(x1)]2dx1 + [(7.5)(103)(x2)]2 dx2 R Ui = B 2EI L0 L0 2EI L0 = Ue =
1.875(109) EI
1 1 P¢ = (20)(103)¢ B = 10(103)¢ B 2 2
Conservation of energy: Ue = Ui 10(103)¢ B =
¢B =
1.875(109) EI
187500 187500 = 0.0117 m = 11.7 mm = EI 200(109)(80)(10 - 6)
Ans.
*14–36. The rod has a circular cross section with a moment of inertia I. If a vertical force P is applied at A, determine the vertical displacement at this point. Only consider the strain energy due to bending. The modulus of elasticity is E.
r A
Moment function: P
a + ©MB = 0;
P[r(1 - cos u)] - M = 0;
M = P r (1 - cos u)
Bending strain energy: s
Ui =
M2 ds L0 2 E I
ds = r du
u
p
=
r M2 r du = [P r (1 - cos u) ]2 du 2 E I L0 L0 2 E I
=
P2 r3 (1 + cos2 u - 2 cos u)du 2 E I L0
=
1 cos 2u P2 r3 a1 + + - 2 cos ub du 2 E I L0 2 2
=
P2 r3 3 cos 2u P2 r3 3 3 p P2 r3 a + - 2 cos u b du = a pb = 2 E I L0 2 2 2 EI 2 4 EI
p
p
p
Conservation of energy: Ue = Ui ;
¢A =
3 p P2 r3 1 P ¢A = 2 4 EI
3 p P r3 2 EI
Ans.
1187
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•14–37.
The load P causes the open coils of the spring to make an angle u with the horizontal when the spring is stretched. Show that for this position this causes a torque T = PR cos u and a bending moment M = PR sin u at the cross section. Use these results to determine the maximum normal stress in the material.
P
R
d
u
T = P R cos u;
M = P R sin u
Bending: smax =
P R sin u d Mc = d4 I 2 (p4 )(16 )
tmax =
P R cos u d2 Tc = p d4 J ( )
P
2 16
smax =
sx + sy 2
;
C
a
sx - sy 2
2
b + t2xy
=
16 P R sin u 2 16 P R cos u 2 16 P R sin u ; a b + a b C p d3 pd3 p d3
=
16 P R [sin u + 1] p d3
Ans.
14–38. The coiled spring has n coils and is made from a material having a shear modulus G. Determine the stretch of the spring when it is subjected to the load P. Assume that the coils are close to each other so that u L 0° and the deflection is caused entirely by the torsional stress in the coil.
P
R u
Bending Strain Energy: Applying 14–22, we have Ui =
P2R2L T2L 16P2R2L = = p 2GJ pd4G 2G C 32 (d4) D P
However, L = n(2pR) = 2npR. Then Ui =
32nP2R3 d4G
External Work: The external work done by force P is Ue =
1 P¢ 2
Conservation of Energy: Ue = Ui 1 32nP2R3 P¢ = 2 d4G ¢ =
64nPR3 d4G
Ans.
1188
d
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14–39. The pipe assembly is fixed at A. Determine the vertical displacement of end C of the assembly. The pipe has an inner diameter of 40 mm and outer diameter of 60 mm and is made of A-36 steel. Neglect the shearing strain energy.
A
800 mm
600 N
B
400 mm
Internal Loading: Referring to the free-body diagram of the cut segment BC, Fig. a, ©My = 0; My + 600x = 0
My = -600x
Referring to the free-body diagram of the cut segment AB, Fig. b, ©Mx = 0; Mx - 600y = 0
Mx = 600y
©My = 0; 600(0.4) - Ty = 0
Ty = 240 N # m
p A 0.034 - 0.024 B = 0.325 A 10 - 6 B pm4. We obtain 2
Torsional Strain Energy. J = L
(Ui)t =
T2dx = L0 2GJ L0
Bending Strain Energy. I =
0.8 m
2 C 75 A 109 B D C 0.325 A 10 - 6 B p D
1 M2dx = B 2EI 2EI L0 L0
0.4 m
0.8 m
(-600x)2 dx +
=
0.4 m 0.8 m 1 + 120 A 103 B y3 2 B 120 A 103 B x3 2 R 2EI 0 0
=
34 560 N2 # m3 EI 34 560
=
200 A 10 B c0.1625 A 10 9
= 0.3009 J
p A 0.034 - 0.024 B = 0.1625 A 10 - 6 B pm4. We obtain 4
L
(Ui)b =
2402 dx
-6
Bp d
L0
A 600y)2 dy R
= 0.3385 J
Thus, the strain energy stored in the pipe is Ui = (Ui)t + (Ui)b = 0.3009 + 0.3385 = 0.6394 J External Work. The work done by the external force P = 600 N is Ue =
1 1 P¢ = (600)¢ C = 300¢ C 2 2
Conservation of Energy. Ue = Ut 300¢ C = 0.6394 ¢ C = 2.1312 A 10 - 3 B = 2.13 mm
Ans.
1189
C
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*14–40. The rod has a circular cross section with a polar moment of inertia J and moment of inertia I. If a vertical force P is applied at A, determine the vertical displacement at this point. Consider the strain energy due to bending and torsion. The material constants are E and G.
z
y r P
T = Pr(1 - cos u);
M = Pr sin u
Torsion strain energy:
x s
Ui =
u
T2 ds T2 rdu = L0 2GJ L0 2GJ p
=
r [Pr(1 - cos u)]2 du 2GJ L0
=
P2 r3 (1 + cos2 u - 2 cos u)du 2GJ L0
=
cos 2u + 1 P2 r3 a1 + - 2 cos u bdu 2GJ L0 2
=
3P2r3 p 4GJ
p
p
Bending strain energy: s
Ui =
M2ds L0 2EI u
p
=
M2r du r = [Pr sin u]2 du 2EI 2EI L0 L0
=
P2 r3 1 - cos 2u P2 r3 p a bdu = 2EI L0 2 4EI
p
Conservation of energy: Ue = Ui 3P2 r3 p P2 r3 p 1 P¢ = + 2 4GJ 4EI ¢ =
1 Pr3 p 3 a + b 2 GJ EI
Ans.
1190
A
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•14–41. Determine the vertical displacement of end B of the frame. Consider only bending strain energy. The frame is made using two A-36 steel W460 * 68 wide-flange sections.
3m B
Internal Loading. Using the coordinates x1 and x2, the free-body diagrams of the frame’s segments in Figs. a and b are drawn. For coordinate x1, -M1 - 20 A 103 B x1 = 0
+ ©MO = 0; For coordinate x2, + ©MO = 0;
M1 = -20 A 103 B x1
M2 - 20 A 103 B (3) = 0
M2 = 60 A 103 B N # m
20 kN
A
Bending Strain Energy. L
(Ub)i =
M2dx 1 = B 2EI L0 L0 2EI
400 A 10 1 D£ = 2EI 3
6
B
3m
c -20 A 103 B x1 d dx1 +
3m
x1 ≥ 3 3
4m
2
+ 3.6 A 109 B x 2
4m
L0
c60 A 103 B d dx2 R 2
T
0
0
=
9 A 109 B N2 # m2 EI
For a W460 * 68, I = 297 A 106 B mm4 = 297 A 10 - 6 B m4. Then (Ub)i =
9 A 109 B
200 A 109 B (297) A 10 - 6 B
= 151.52 J
External Work. The work done by the external force P = 20 kN is Ue =
4m
1 1 P¢ = c20 A 103 B d ¢ B = 10 A 103 B ¢ B 2 2
Conservation of Energy. Ue = Ui 10 A 103 B ¢ B = 151.52 ¢ B = 0.01515 m = 15.2 mm
Ans.
1191
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14–42. A bar is 4 m long and has a diameter of 30 mm. If it is to be used to absorb energy in tension from an impact loading, determine the total amount of elastic energy that it can absorb if (a) it is made of steel for which Est = 200 GPa, sY = 800 MPa, and (b) it is made from an aluminum alloy for which Eal = 70 GPa, sY = 405 MPa.
a) eg =
800(106) sY = 4(10 - 3) m>m = E 200(109)
ur =
1 1 (s )(e ) = (800)(106)(N>m2)(4)(10 - 3)m>m = 1.6 MJ>m3 2 Y g 2
V =
p (0.03)2(4) = 0.9(10 - 3)p m2 4
ui = 1.6(106)(0.9)(10 - 3)p = 4.52 kJ
Ans.
b)
eg =
405(106) sY = 5.786(10 - 3) m>m = E 70(109)
ur =
1 1 (s )(e ) = (405)(106)(N>m2)(5.786)(10 - 3)m>m = 1.172 MJ>m3 2 Y g 2
V =
p (0.03)2 (4) = 0.9(10 - 3)p m3 4
ui = 1.172(106)(0.9)(10 - 3)p = 3.31 kJ
Ans.
14–43. Determine the diameter of a red brass C83400 bar that is 8 ft long if it is to be used to absorb 800 ft # lb of energy in tension from an impact loading. No yielding occurs. Elastic Strain Energy: The yielding axial force is PY = sgA. Applying Eq. 14–16, we have Ui =
(sgA)2L s2gAL N2L = = 2AE 2AE 2E
Substituting, we have Ui =
0.8(12) =
s2gAL 2E
11.42 C p4 (d2) D (8)(12) 2[14.6(103)]
d = 5.35 in.
Ans.
1192
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*14–44. A steel cable having a diameter of 0.4 in. wraps over a drum and is used to lower an elevator having a weight of 800 lb. The elevator is 150 ft below the drum and is descending at the constant rate of 2 ft兾s when the drum suddenly stops. Determine the maximum stress developed in the cable when this occurs. Est = 2911032 ksi, sY = 50 ksi. k =
AE = L
p 4
(0.42)(29)(103) 150 (12)
150 ft
= 2.0246 kip>in.
Ue = Ui 1 1 mv2 + W ¢ max = k¢ 2max 2 2 800 1 1 c d[(12) (2)]2 + 800 ¢ max = (2.0246)(103)¢ 2max 2 32.2 (12) 2 596.27 + 800 ¢ max = 1012.29 ¢ 2max ¢ max = 1.2584 in. Pmax = k¢ max = 2.0246 (1.2584) = 2.5477 kip smax =
Pmax 2.5477 = p = 20.3 ksi 6 sg 2 A 4 (0.4)
O.K.
Ans.
•14–45.
The composite aluminum bar is made from two segments having diameters of 5 mm and 10 mm. Determine the maximum axial stress developed in the bar if the 5-kg collar is dropped from a height of h = 100 mm. Eal = 70 GPa, sY = 410 MPa.
5 mm
200 mm
300 mm
¢ st = ©
WL = AE
Pmax = W B 1 +
(0.0052)(70)(109)
C
1 + 2a
= 5(9.81) B 1 + smax =
Pmax = A
5(9.81)(0.3)
5(9.81)(0.2) p 4
p 4
C
+
p 4
(0.012)(70)(109)
h
= 9.8139(10 - 4) m
10 mm
h bR ¢ st
1 + 2a
0.1 9.8139(10 - 6)
7051 = 359 MPa 6 sy (0.0052)
b R = 7051 N O.K.
1193
Ans.
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14–46. The composite aluminum bar is made from two segments having diameters of 5 mm and 10 mm. Determine the maximum height h from which the 5-kg collar should be dropped so that it produces a maximum axial stress in the bar of s max = 300 MPa, Eal = 70 GPa, sY = 410 MPa.
5 mm
200 mm
300 mm h 10 mm
¢ st = ©
WL = AE
Pmax = W B 1 +
5(9.81)(0.3)
5(9.81)(0.2) p 2 9 4 (0.005 )(70)(10 )
C
1 + 2a
+
p 4
(0.012)(70)(109)
= 9.8139(10 - 6) m
h bR ¢ st
p h bR 300(106)a b (0.0052) = 5(9.81) B 1 + 1 + 2a 4 C 9.8139(10 - 6) 120.1 = 1 + 21 + 203791.6 h h = 0.0696 m = 69.6 mm
Ans.
1194
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14–47. The 5-kg block is traveling with the speed of v = 4 m>s just before it strikes the 6061-T6 aluminum stepped cylinder. Determine the maximum normal stress developed in the cylinder.
C
B 40 mm
Equilibrium. The equivalent spring constant for segments AB and BC are
kAB
kBC
p A 0.022 B c68.9 A 109 B d AAB E 4 = = = 72.152 A 106 B N>m LAB 0.3 p A 0.042 B c68.9 A 109 B d ABC E 4 = = = 288.608 A 106 B N>m LBC 0.3
Equilibrium requires FAB = FBC kAB ¢ AB = kBC ¢ BC 72.152 A 106 B ¢ AB = 288.608 A 106 B ¢ BC ¢ BC =
1 ¢ 4 AB
(1)
Conservation of Energy. Ue = Ui 1 1 1 mv2 = kAB ¢ AB 2 + kBC ¢ BC 2 2 2 2
(2)
Substituting Eq. (1) into Eq. (2), 2 1 1 1 1 mv2 = kAB ¢ AB 2 + kBC a ¢ AB b 2 2 2 4
1 1 1 mv2 = kAB ¢ AB 2 + k ¢ 2 2 2 32 BC AB 1 1 1 (5) A 42 B = c 72.152 A 106 B d ¢ AB 2 + c288.608 A 106 B d ¢ AB 2 2 2 32 ¢ AB = 0.9418 A 10 - 3 B m Maximum Stress. The force developed in segment AB is FAB = kAB ¢ AB = 72.152 A 106 B c0.9418 A 10 - 3 B d = 67.954 A 103 B N.
Thus, smax = sAB =
300 mm
300 mm
67.954 A 103 B FAB = = 216.30 MPa = 216 MPa p AAB 2 0.02 A B 4
Since smax 6 sY = 255 MPa, this result is valid.
1195
Ans.
v A
20 mm
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*14–48. Determine the maximum speed v of the 5-kg block without causing the 6061-T6 aluminum stepped cylinder to yield after it is struck by the block.
C
B 40 mm
Equilibrium. The equivalent spring constant for segments AB and BC are
kAB
kBC
p A 0.022 B C 68.9 A 109 B D AAB E 4 = = = 72.152 A 106 B N>m LAB 0.3 p A 0.042 B C 68.9 A 109 B D ABC E 4 = = = 288.608 A 106 B N>m LBC 0.3
Equilibrium requires FAB = FBC kAB ¢ AB = kBC ¢ BC 72.152 A 106 B ¢ AB = 288.608 A 106 B ¢ BC ¢ BC =
1 ¢ 4 AB
(1)
Conservation of Energy. Ue = Ui 1 1 1 mv2 = kAB ¢ AB 2 + kBC ¢ BC 2 2 2 2
(2)
Substituting Eq. (1) into Eq. (2), 2 1 1 1 1 mv2 = kAB ¢ AB 2 + kBC a ¢ AB b 2 2 2 4
1 1 1 mv2 = kAB ¢ AB 2 + k ¢ 2 2 2 32 BC AB 1 1 1 (5)v2 = c72.152 A 106 B d ¢ AB 2 + c288.608 A 106 B d ¢ AB 2 2 2 32 ¢ AB = 0.23545 A 10 - 3 B v Maximum Stress. The force developed in segment AB is FAB = kAB ¢ AB = 72.152 A 106 B C 0.23545 A 10 - 3 B v D = 16988.46v. Thus, smax = sAB = 255 A 106 B =
300 mm
300 mm
FAB AAB
16988.46v p A 0.022 B 4
v = 4.716 m>s = 4.72 m>s
Ans.
1196
v A
20 mm
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•14–49. The steel beam AB acts to stop the oncoming railroad car, which has a mass of 10 Mg and is coasting towards it at v = 0.5 m>s . Determine the maximum stress developed in the beam if it is struck at its center by the car. The beam is simply supported and only horizontal forces occur at A and B. Assume that the railroad car and the supporting framework for the beam remains rigid. Also, compute the maximum deflection of the beam. Est = 200 GPa,sY = 250 MPa.
v ⫽ 0.5 m/s
k =
1m B
10(103)(9.81)(23) PL3 = 0.613125(10 - 3) m = 1 48EI 48(200)(104)(12 )(0.2)(0.23)
10(103)(9.81) W = 160(106) N>m = ¢ st 0.613125(10 - 3)
¢ max =
0.613125(10 - 3)(0.52) ¢ st v2 = = 3.953(10 - 3) m = 3.95 mm C g C 9.81
Ans.
W¿ = k¢ max = 160(106)(3.953)(10 - 3) = 632455.53 N M¿ =
632455.53(2) w¿L = = 316228 N # m 4 4
smax =
316228(0.1) M¿c = 237 MPa 6 sg = 1 3 I 12 (0.2)(0.2 )
A 1m
From Appendix C: ¢ st =
200 mm 200 mm
O.K.
Ans.
1197
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14–50. The aluminum bar assembly is made from two segments having diameters of 40 mm and 20 mm. Determine the maximum axial stress developed in the bar if the 10-kg collar is dropped from a height of h = 150 mm. Take Eal = 70 GPa, sY = 410 MPa.
C
1.2 m 40 mm B 0.6 m
kAB =
kBC =
AAB E = LAB
p (0.012) C 70(109) D 0.6
= 11.667(106) p N>m
A
p (0.022) C 70(109) D ABC E = = 23.333 (106) p N>m LBC 1.2
Equilibrium requires FAB = FBC kAB ¢ AB = kBC ¢ BC 11.667(106) p ¢ AB = 23.333(106) p ¢ BC ¢ BC = 0.5 ¢ AB
(1)
Ue = Ui mg (h + ¢ AB + ¢ BC) =
1 1 k ¢ 2 + kBC ¢ 2BC 2 AB AB 2
(2)
Substitute Eq. (1) into (2), mg (h + ¢ AB + 0.5 ¢ AB) =
mg (h + 1.5¢ AB) =
1 1 k ¢ 2 + kBC (0.5¢ AB)2 2 AB AB 2 1 k ¢ 2 + 0.125 kBC ¢ 2AB 2 AB AB
10(9.81)(0.15 + 1.5¢ AB) =
1 C 11.667(106)p D ¢ 2AB + 0.125 C 23.333(106)p D ¢ 2AB 2
27.4889 (106)¢ 2AB - 147.15 ¢ AB - 14.715 = 0 ¢ AB = 0.7343(10 - 3) m The force developed in segment AB C 11.667(106)p D C 0.7343(10 - 3) D = 26.915(103) N. Thus smax = sAB =
h
20 mm
is
FAB = kAB ¢ AB =
26.915(103) FAB = 85.67(106)Pa = 85.7 MPa = AAB p (0.012)
Since smax 6 sy = 410 MPa, this result is valid
1198
Ans.
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14–51. The aluminum bar assembly is made from two segments having diameters of 40 mm and 20 mm. Determine the maximum height h from which the 60-kg collar can be dropped so that it will not cause the bar to yield. Take Eal = 70 GPa, sY = 410 MPa.
C
1.2 m 40 mm B 0.6 m
kAB =
kBC =
p(0.012) C 70(109) D AAB E = = 11.667(106) p N>m LAB 0.6
A
p(0.022) C 70(109) D ABC E = = 23.333(106) p N>m LBC 1.2
Here, FAB = kAB ¢ AB = C 11.667(106)p D ¢ AB. It is required that smax = sAB = sy. sy =
FAB ; AAB
410(106) =
C 11.667(106)p D ¢ AB p(0.012)
¢ AB = 0.003514 m Equilibrium requires that FAB = FBC kAB ¢ AB = kBC ¢ BC 11.6667(106)p ¢ AB = 23.333(106)p ¢ BC ¢ BC = 0.5 ¢ B = 0.5(0.003514) = 0.001757 m Ue = Ui mg(h + ¢ AB + ¢ BC) =
1 1 k ¢ 2 + kBC ¢ 2BC 2 AB AB 2
60(9.81)(h + 0.003514 + 0.001757) =
1 C 11.667(106)p D (0.0035142) 2 +
h
20 mm
The equivalent spring constants for segment AB and BC are
1 C 23.333(106)p D (0.0017572) 2
h = 0.571 m
Ans.
1199
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*14–52. The 50-lb weight is falling at 3 ft>s at the instant it is 2 ft above the spring and post assembly. Determine the maximum stress in the post if the spring has a stiffness of k = 200 kip>in. The post has a diameter of 3 in. and a modulus of elasticity of E = 6.8011032 ksi. Assume the material will not yield.
3 ft/s
2 ft
k
2 ft
Equilibrium: This requires Fsp = FP. Hence ksp ¢ sp = kP ¢ P
and
¢ sp = =
P
ksp
[1]
¢P
Conservation of Energy: The equivalent spring constant for the post is kp =
AE = L
p 4
(32) C 6.80(103) D 2(12)
= 2.003 A 106 B lb>in. Ue = Ui
1 1 1 my2 + W(h + ¢ max) = kP ¢ 2P + ksp ¢ 2sp 2 2 2
[2]
However, ¢ max = ¢ P + ¢ sp. Then, Eq. [2] becomes 1 1 1 my2 + W A h + ¢ P + ¢ sp B = kP ¢ 2P + ksp ¢ 2sp 2 2 2
[3]
Substituting Eq. [1] into [3] yields kp 1 1 1 k2P 2 my2 + W ¢ h + ¢ P + ¢ P ≤ = kP¢ 2P + ¢ ¢ ≤ 2 ksp 2 2 ksp P 2.003(106) 1 50 ¢P R ¢ ≤ A 32 B (12) + 50 B 24 + ¢ p + 2 32.2 200(103) =
1 [2.003(106)]2 1 ≤ ¢ 2P C 2.003 A 106 B D ¢ 2P + ¢ 2 2 200(103)
11.029 A 106 B ¢ 2P - 550.69¢ P - 1283.85 = 0 Solving for positive root, we have ¢ P = 0.010814 in. Maximum Stress: The maximum axial force for the post is Pmax = kp ¢ p = 2.003 A 106 B (0.010814) = 21.658 kip. smax =
Pmax 21.658 = p 2 = 3.06 ksi A 4 (3 )
Ans.
1200
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The 50-kg block is dropped from h = 600 mm onto the bronze C86100 tube. Determine the minimum length L the tube can have without causing the tube to yield.
•14–53.
A 30 mm 20 mm
Maximum Stress. A = p A 0.03 - 0.02 2
h ⫽ 600 mm
B = 0.5 A 10 B p
2
-3
¢ st =
50(9.81)L WL = 3.0317 A 10 - 6 B L = AE C 0.5 A 10 - 3 B p D C 103 A 109 B D
sst =
50(9.81) W = = 0.3123 MPa A 0.5 A 10 - 3 B p
Section a – a
a
a L
B
Using these results, n = 1 +
C
1 + 2a
h 0.6 395 821.46 b = 1 + 1 + 2B R = 1 + 1 + -6 ¢ st C C L 3.0317 A 10 B L
Then, smax = sY = nsst 345 = ¢ 1 +
A
1 +
395 821.46 ≤ (0.3123) L
L = 0.3248 m = 325 mm
Ans.
14–54. The 50-kg block is dropped from h = 600 mm onto the bronze C86100 tube. If L = 900 mm, determine the maximum normal stress developed in the tube.
A 30 mm 20 mm
Maximum Stress. A = p A 0.03 - 0.02 2
2
B = 0.5 A 10 B p
¢ st =
50(9.81)(0.9) WL = = 2.7285 A 10 - 6 B AE C 0.5 A 10 - 3 B p D C 103 A 109 B D
sst =
50(9.81) W = = 0.3123 MPa A 0.5 A 10 - 3 B p
Section a – a
C
1 + 2a
B
h 0.6 b = 1 + 1 + 2B R = 664.18 ¢ st C 2.7285 A 10 - 6 B
Thus, smax = nsst = 664.18(0.3123) = 207.40 MPa = 207 MPa Since smax 6 sY = 345 MPa, this result is valid.
1201
a
a L
Using these results, n = 1 +
h ⫽ 600 mm
-3
Ans.
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14–55. The steel chisel has a diameter of 0.5 in. and a length of 10 in. It is struck by a hammer that weighs 3 lb, and at the instant of impact it is moving at 12 ft兾s. Determine the maximum compressive stress in the chisel, assuming that 80% of the impacting energy goes into the chisel. Est = 2911032 ksi, sY = 100 ksi. k =
AE = L
p 4
(0.52)(29)(103) 10
= 569.41 kip>in.
10 in.
0.8 Ue = Ui 0.8 c
3 1 1 a b((12)(12))2 + 3¢ max d = (569.41)(103)¢ 2max 2 (32.2)(12) 2
¢ max = 0.015044 in. P = k¢ max = 569.41(0.015044) = 8.566 kip smax =
Pmax 8.566 = 43.6 ksi 6 sg = p 2 A 4 (0.5)
O.K.
Ans.
*14–56. The sack of cement has a weight of 90 lb. If it is dropped from rest at a height of h = 4 ft onto the center of the W10 * 39 structural steel A-36 beam, determine the maximum bending stress developed in the beam due to the impact. Also, what is the impact factor?
h
12 ft
Impact Factor: From the table listed in Appendix C, ¢ st =
90[24(12)]3 PL3 = 7.3898 A 10 - 3 B in. = 48EI 48[29.0(106)](209) n = 1 +
= 1 +
C
1 + 2a
C
1 + 2a
h b ¢ st 4(12) 7.3898(10 - 3)
b
= 114.98 = 115
Ans.
Maximum Bending Stress: The maximum moment occurs at mid-span where 90(24)(12) PL = = 6480 lb # in. Mmax = 4 4 sst =
6480(9.92>2) Mmax c = = 153.78 psi I 209
Thus, smax = nsst = 114.98(153.78) = 17.7 ksi
Ans.
Since smax 6 sg = 36 ksi, the above analysis is valid.
1202
12 ft
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•14–57.
The sack of cement has a weight of 90 lb. Determine the maximum height h from which it can be dropped from rest onto the center of the W10 * 39 structural steel A-36 beam so that the maximum bending stress due to impact does not exceed 30 ksi.
h
12 ft
Maximum Bending Stress: The maximum moment occurs at mid-span where 90(24)(12) PL = = 6480 lb # in. Mmax = 4 4 sst =
6480(9.92>2) Mmax c = = 153.78 psi I 209
However, smax = nsst 30 A 103 B = n(153.78) n = 195.08 Impact Factor: From the table listed in Appendix C, ¢ st =
90[24(12)]3 PL3 = 7.3898 A 10 - 3 B in. = 48EI 48[29.0(106)](209) n = 1 +
C
1 + 2a
h b ¢ st
195.08 = 1 +
C
1 + 2a
h b 7.3898(10 - 3)
h = 139.17 in. = 11.6 ft
Ans.
1203
12 ft
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14–58. The tugboat has a weight of 120 000 lb and is traveling forward at 2 ft兾s when it strikes the 12-in.-diameter fender post AB used to protect a bridge pier. If the post is made from treated white spruce and is assumed fixed at the river bed, determine the maximum horizontal distance the top of the post will move due to the impact. Assume the tugboat is rigid and neglect the effect of the water.
3 ft
A C
12 ft
B
From Appendix C: Pmax =
3EI(¢ C)max (LBC)3
Conservation of energy: 1 1 mv2 = Pmax (¢ C)max 2 2 1 1 3EI(¢ C)2max mv2 = a b 2 2 (LBC)3 (¢ C)max =
(¢ C)max =
Pmax =
uC =
mv2L3BC C 3EI (120 000>32.2)(2)2(12)3
C (3)(1.40)(106)(144)(p4 )(0.5)4
3[1.40(106)](p4 )(6)4(11.177) (144)3
= 0.9315 ft = 11.177 in.
= 16.00 kip
16.00(103)(144)2 PmaxL2BC = = 0.11644 rad 2EI 2(1.40)(106)(p4 )(6)4
(¢ A)max = (¢ C)max + uC(LCA) (¢ A)max = 11.177 + 0.11644(36) = 15.4 in.
Ans.
1204
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14–59. The wide-flange beam has a length of 2L, a depth 2c, and a constant EI. Determine the maximum height h at which a weight W can be dropped on its end without exceeding a maximum elastic stress s max in the beam.
W
h
A
2c B L
L
1 1 (-Px)2 dx P¢ C = 2a b 2 2EI L0 ¢C =
2PL3 3EI
¢ st =
2WL3 3EI
n = 1 +
C
1 + 2a
h b ¢ st
smax = n(sst)max smax = B 1 + a
C
1 + 2a
(sst)max =
WLc I
h WLc b R ¢ st I
2 smax I 2h - 1b = 1 + WLc ¢ st
h =
=
2 ¢ st smax I - 1b - 1R Ba 2 WLc
smax I 2 2smaxI smax L2 smax I WL3 b - 2R Ba R = B 3EI WLc WLc 3Ec WLc
Ans.
1205
L
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*14–60. The 50-kg block C is dropped from h = 1.5 m onto the simply supported beam. If the beam is an A-36 steel W250 * 45 wide-flange section, determine the maximum bending stress developed in the beam.
C h
Equilibrium. Referring to the free-body diagram of the beam under static A condition, Fig. a a + ©MA = 0;
By(6) - P(4) = 0
By =
2 P 3
4m
Then, the maximum moment in the beam occurs at the position where P is applied. Thus, Mmax = By(2) =
2 4 P(2) = P 3 3
Impact Factor. From the table listed in the appendix, the deflection of the beam at Pba the point of application of P is ¢ = A L2 - b2 - a2 B , where P = 50(9.81) 6EIL = 490.5 N, L = 6 m, a = 4 m, and b = 2 m. From the table listed in the appendix, the necessary section properties for a W250 * 45 are d = 266 mm = 0.266 m and Ix = 71.1 A 106 B mm4 = 71.1 A 10 - 6 B m4. Then 490.5(2)(4) ¢ st =
6c 200 A 10 B d c 71.1 A 10 9
-6
B d(6)
A 62 - 22 - 42 B = 0.1226 A 10 - 3 B m
We have,
n = 1 +
Maximum
C
1 + 2¢
Stress.
h 1.5 1 + 2C S = 157.40 ≤ = 1 + ¢ st Q 0.1226 A 10 - 3 B
The
maximum
= 157.40(490.5) = 77.21 A 10 B N. 3
= 102.94 A 10
smax =
3
force Then,
on
B N # m. Applying the flexure formula,
beam is Pmax = nP 4 4 = Pmax = C 77.21 A 103 B D 3 3
the
Mmax
102.94 A 103 B (0.266>2) Mmaxc = = 192.56 MPa = 193 MPa I 71.1 A 10 - 6 B
Since smax 6 sY = 250 MPa, this result is valid.
1206
B
Ans.
2m
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•14–61.
Determine the maximum height h from which the 50-kg block C can be dropped without causing yielding in the A-36 steel W310 * 39 wide flange section when the block strikes the beam.
C h A
B 4m
Equilibrium. Referring to the free-body diagram of the beam under static condition, Fig. a
a + ¢MA = 0;
By(6) - P(4) = 0
By =
2 P 3
Then, the maximum moment in the beam occurs at the position where P is applied. Thus, Mmax = By(2) =
4 2 P(2) = P 3 3
Maximum Stress. Since P = 50(9.81) = 490.5 N. Then the maximum force on the 4 4 beam is Pmax = nP = 490.5n and Mmax = P = (490.5n) = 654n. From the 3 3 table listed in the appendix, the necessary section properties for a W310 * 39 are d = 310 mm = 0.31 m and Ix = 84.8 A 106 B mm4 = 84.8 A 10 - 6 B m4. Applying the
flexure formula,
smax =
Mmax c I
250 A 106 B =
654n(0.31>2) 84.8 A 10 - 6 B
n = 209.13 Impact Factor. From the table listed in the appendix, the deflection of the beam at Pba the point of where P is applied is ¢ = A L2 - b2 - a2 B , where L = 6 m, 6EIL a = 4 m, and b = 2 m. Then 490.5(2)(4)
¢ st =
6 C 200 A 109 B D C 84.8 A 10 - 6 B D (6)
A 62 - 22 - 42 B = 0.1028 A 10 - 3 B m
We have, n = 1 +
C
1 + 2¢
209.13 = 1 +
S
h ≤ ¢ st
1 + 2C
h
0.1028 A 10 - 3 B
S
h = 2.227 m = 2.23 m
Ans.
Since smax 6 sY = 250 MPa, this result is valid.
1207
2m
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14–62. The diver weighs 150 lb and, while holding himself rigid, strikes the end of a wooden diving board 1h = 02 with a downward velocity of 4 ft>s. Determine the maximum bending stress developed in the board. The board has a thickness of 1.5 in. and width of 1.5 ft. Ew = 1.811032 ksi, sY = 8 ksi.
v
h 4 ft
Static Displacement: The static displacement at the end of the diving board can be determined using the conservation of energy. L
M2 dx 1 P¢ = 2 L0 2EI 1 1 (150)¢ st = c 2 2EI L0 ¢ st =
4 ft
10 ft
(-375x1)2 dx1 +
L0
(-150x2) dx2 d
70.0(103) lb # ft3 EI 70.0(103)(123)
=
1 1.8(106) C 12 (18)(1.53) D
= 13.274 in. Conservation of Energy: The equivalent spring constant for the board is W 150 k = = = 11.30 lb>in., ¢ st 13.274 Ue = Ui 1 1 my2 + W¢ max = k¢ 2max 2 2 c
1 150 1 a b A 42 B d (12) + 150¢ max = (11.30)¢ 2max 2 32.2 2
Solving for the positive root, we have ¢ max = 29.2538 in. Maximum Stress: The maximum force on to the beam is Pmax = k¢ max = 11.30(29.2538) = 330.57 lb. The maximum moment occurs at the middle support Mmax = 330.57(10)(12) = 39668.90 lb # in. smax =
39668.90(0.75) Mmax c = 5877 psi = 5.88 ksi = 1 3 I 12 (18)(1.5 )
Ans.
Note: The result will be somewhat inaccurate since the static displacement is so large.
1208
10 ft
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14–63. The diver weighs 150 lb and, while holding himself rigid, strikes the end of the wooden diving board. Determine the maximum height h from which he can jump onto the board so that the maximum bending stress in the wood does not exceed 6 ksi. The board has a thickness of 1.5 in. and width of 1.5 ft. Ew = 1.811032 ksi.
v
Static Displacement: The static displacement at the end of the diving board can be determined using the conservation of energy.
h
L
M2 dx 1 P¢ = 2 L0 2EI 1 1 (150)¢ st = c 2 2EI L0 ¢ st =
4 ft
4 ft
10 ft
(-375x1)2 dx1 +
L0
(-150x2) dx2 d
70.0(103) lb # ft3 EI 70.0(103)(123)
=
1 1.8(106) C 12 (18)(1.53) D
= 13.274 in. Maximum Stress: The maximum force on the beam is Pmax. The maximum moment occurs at the middle support Mmax = Pmax (10)(12) = 120Pmax. smax = 6 A 103 B =
Mmax c I 120Pmax (0.75) 1 12
(18)(1.53)
Pmax = 337.5 lb Conservation of Energy: The equivalent spring constant for the board is 150 W = 11.30 lb>in.. The maximum displacement at the end of the = k = ¢ st 13.274 Pmax 337.5 board is ¢ max = = = 29.687 in. k 11.30 Ue = Ui W(h + ¢ max) =
150(h + 29.867) =
1 k¢ 2max 2
1 (11.30) A 29.8672 B 2
h = 3.73 in.
Ans.
Note: The result will be somewhat at inaccurate since the static displacement is so large.
1209
10 ft
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*14–64. The weight of 175 lb is dropped from a height of 4 ft from the top of the A-36 steel beam. Determine the maximum deflection and maximum stress in the beam if the supporting springs at A and B each have a stiffness of k = 500 lb>in. The beam is 3 in. thick and 4 in. wide.
4 ft A
k 8 ft
8 ft
From Appendix C: ¢ beam =
PL3 48EI
kbeam =
1 )(4)(33) 48(29)(103)(12 48EI = = 1.7700 kip>in. 3 L (16(12))3
From equilibrium (equivalent system): 2Fsp = Fbeam 2ksp ¢ sp = kbeam ¢ beam ¢ sp =
1.7700(103) ¢ beam 2(500)
¢ sp = 1.7700¢ beam
(1)
Conservation of energy: Ue = Ui W(h + ¢ sp + ¢ beam) =
1 1 k ¢2 + 2a bksp ¢ 2sp 2 beam beam 2
From Eq. (1): 175[(4)(12) + 1.770¢ beam + ¢ beam] =
1 (1.7700)(103)¢ 2beam + 500(1.7700¢ beam)2 2
2451.5¢ 2beam - 484.75¢ beam - 8400 = 0 ¢ beam = 1.9526 in. From Eq. (1): ¢ sp = 3.4561 in. ¢ max = ¢ sp + ¢ beam = 3.4561 + 1.9526 = 5.41 in.
Ans.
Fbeam = kbeam ¢ beam = 1.7700(1.9526) = 3.4561 kip Mmax =
smax =
3.4561(16)(12) Fbeam L = = 165.893 kip # in. 4 4 165.893(1.5) Mmax c = 27.6 ksi 6 sg = 1 3 I 12 (4)(3 )
O.K.
Ans.
1210
3 in.
B k
4 in.
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•14–65.
The weight of 175 lb, is dropped from a height of 4 ft from the top of the A-36 steel beam. Determine the load factor n if the supporting springs at A and B each have a stiffness of k = 300 lb/in. The beam is 3 in. thick and 4 in. wide.
4 ft A
k 8 ft
8 ft
From Appendix C: ¢ beam =
PL3 48EI
kbeam =
1 )(4)(33) 48(29)(103)(12 48EI = = 1.7700 kip>in. L3 (16(12))3
From equilibrium (equivalent system): 2Fsp = Fbeam 2ksp ¢ sp = kbeam ¢ beam ¢ sp =
1.7700(103) ¢ beam 2(300) (1)
¢ sp = 2.95¢ beam Conservation of energy: Ue = Ui W(h + ¢ beam + ¢ sp) =
1 1 k ¢2 + 2a bksp ¢ 2sp 2 beam beam 2
From Eq. (1): 175[(4)(12) + ¢ beam + 2.95¢ beam] =
1 (1.7700)(103)¢ 2beam + 300(2.95¢ beam)2 2
3495.75¢ 2beam - 691.25¢ beam - 8400 = 0 ¢ beam = 1.6521 in. Fbeam = kbeam ¢ beam = 1.7700(1.6521) = 2.924 kip n =
2.924(103) = 16.7 175
smax = n(sst)max = na
M =
Ans. Mc b I
175(16)(12) = 8.40 kip # in. 4
smax = 16.7 a
8.40(1.5) 1 12
(4)(33)
b = 23.4 ksi 6 sg
O.K.
1211
3 in.
B k
4 in.
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14–66. Block C of mass 50 kg is dropped from height h = 0.9 m onto the spring of stiffness k = 150 kN>m mounted on the end B of the 6061-T6 aluminum cantilever beam. Determine the maximum bending stress developed in the beam.
C h a
k
100 mm B
Conservation of Energy. From the table listed in the appendix, the
A
a
200 mm
3
PL . Thus, the = 3EI
displacement of end B under static conditions is ¢ st
equivalent spring constant for the beam is kb = I =
1 (0.1) A 0.23 B = 66.6667 A 10 - 6 B m4, 12
L = 3 m,
3m
3EI , where L3
and
E = Eal
= 68.9 GPa. Thus,
3EI kb = = L3
3 c68.9 A 109 B d c66.6667 A 10 - 6 B d 33
= 510.37 A 103 B N>m
Equilibrium requires, Fsp = P ksp ¢ sp = kb ¢ b 150 A 103 B ¢ sp = 510.37 A 103 B ¢ b ¢ sp = 3.4025¢ b
(1)
We have, Ue = Ui mg A h + ¢ sp + ¢ b B =
1 1 k ¢ 2 + ksp ¢ sp 2 2 b b 2
Substituting Eq. (1) into this equation, 50(9.81)(0.9 + 3.4025¢ b + ¢ b) =
1 1 c510.37 A 103 B d ¢ b 2 + c150 A 103 B d(3.4025¢ b)2 2 2
1123444.90¢ b 2 - 2159.41¢ b - 441.45 = 0 Solving for the positive root, ¢ b = 0.020807 m Maximum
Stress.
The
maximum
force
Pmax = kb ¢ b = 510.37 A 10 B (0.020807) = 10.619 A 10 3
occurs
at
= 31.858 A 103 B smax =
3
on
the
beam
is
B N. The maximum moment
Mmax = Pmax L = 10.619 A 103 B (3) 0.2 N # m. Applying the flexure formula with c = = 0.1 m, 2
fixed
support
A,
where
31.858 A 103 B (0.1) Mmax c = = 47.79 MPa = 47.8 MPa I 66.6667 A 10 - 6 B
Since smax 6 sY = 255 MPa, this result is valid. 1212
Ans.
Section a – a
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14–67. Determine the maximum height h from which 200-kg block C can be dropped without causing the 6061-T6 aluminum cantilever beam to yield. The spring mounted on the end B of the beam has a stiffness of k = 150 kN>m.
C h a
k
100 mm B
A
a 3m
Maximum Stress. From the table listed in the appendix, the displacement of end B PL3 under static conditions is ¢ st = . Thus, the equivalent spring constant for the 3EI 3EI 1 beam is kb = , where I = (0.1) A 0.23 B = 66.6667 A 10 - 6 B m4, L = 3 m, and 12 L3 E = Eal = 68.9 GPa. Thus, 3EI kb = = L3
3 c68.9 A 109 B d c66.6667 A 10 - 6 B d 33
= 510.37 A 103 B N>m
The maximum force on the beam is Pmax = kb ¢ b = 510.37 A 103 B ¢ b. The maximum moment occurs at the fixed support A, where Mmax = Pmax L = 510.37 A 103 B ¢ b(3)
= 1.5311 A 106 B ¢ b. Applying the flexure formula with smax = sY = 255 MPa and 0.2 = 0.1 m, c = 2 Mmax c I
smax = sY = 255 A 106 B =
1.5311 A 106 B ¢ b(0.1) 66.6667 A 10 - 6 B
¢ b = 0.11103 m Equilibrium requires, Fsp = P ksp ¢ sp = kb ¢ b 150 A 103 B ¢ sp = 510.37 A 103 B (0.11103) ¢ sp = 0.37778 m Conservation of Energy. Ue = Ui mg A h + ¢ sp + ¢ b B =
1 1 kb ¢ b 2 + ksp ¢ sp 2 2 2
200(9.81)(h + 0.37778 + 0.11103) =
1 1 c510.37 A 103 B d(0.11103)2 + c150 A 103 B d(0.37778)2 2 2
h = 6.57 m
Ans.
1213
200 mm Section a – a
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*14–68. The 2014-T6 aluminum bar AB can slide freely along the guides mounted on the rigid crash barrier. If the railcar of mass 10 Mg is traveling with a speed of v = 1.5 m>s, determine the maximum bending stress developed in the bar. The springs at A and B have a stiffness of k = 15 MN>m.
300 mm k
A
v
2m
2m a
a
k
B
Equilibrium. Referring to the free-body diagram of the bar for static conditions, Fig. a, + ©F = 0; : x
2Fsp - P = 0
Fsp =
P 2
(1)
300 mm 400 mm
Referring to the table listed in the appendix, the displacement of the bar at the PL3 position where P is applied under static conditions is ¢ st = . Thus, the 48EI 48EI 1 equivalent spring constant for the bar is kb = , where I = (0.4) A 0.33 B 12 L3 and Thus, = 0.9 A 10 - 3 B m4, L = 4 m, E = Eal = 73.1 GPa. kb =
48 c73.1 A 109 B d c0.9 A 10 - 3 B d 43
= 49.3425 A 106 B N>m
Using Eq. (1) Fsp =
P 2
ksp ¢ sp =
1 k ¢ 2 b b
1 kb 1 49.3425 A 10 ¢ ≤ ¢b = C 2 ksp 2 15 A 106 B
6
¢ sp =
B
S ¢ b = 1.64475¢ b
Conservation of Energy. 1 1 1 mv2 = kb ¢ b 2 + 2c ksp ¢ sp 2 d 2 2 2 Substituting Eq. (2) into this equation, 1 1 mv2 = kb ¢ b 2 + ksp (1.64475¢ b)2 2 2 1 1 mv2 = kb ¢ b 2 + 2.7052ksp ¢ b 2 2 2 1 1 c10 A 103 B d A 1.52 B = c49.3425 A 106 B d ¢ b 2 + 2.7052c15 A 106 B d ¢ b 2 2 2 ¢ b = 0.01313 m
1214
(2)
Section a – a
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14–68.
Continued
Maximum Stress. The maximum force on the bar is (Pb)max = kb ¢ b = 49.3425 A 106 B (0.01313) = 647.90 A 103 B N. The maximum moment occurs at the midspan of the bar, where Mmax = = 647.90 A 103 B N # m. Applying the flexure formula, smax =
647.90 A 103 B (4) (Pb)max L = 4 4
647.90 A 103 B (0.15) Mmax c = = 107.98 MPa = 108 MPa I 0.9 A 10 - 3 B
Since smax 6 sY = 414 MPa, this result is valid.
1215
Ans.
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•14–69.
The 2014-T6 aluminum bar AB can slide freely along the guides mounted on the rigid crash barrier. Determine the maximum speed v the 10-Mg railcar without causing the bar to yield when it is struck by the railcar. The springs at A and B have a stiffness of k = 15 MN>m.
300 mm k
A
v
2m
2m a
a
k
B
Equilibrium. Referring to the free-body diagram of the bar for static conditions, Fig. a, + ©F = 0; : x
2Fsp - P = 0
Fsp
P = 2
(1)
300 mm 400 mm
Referring to the table listed in the appendix, the displacement of the bar at the PL3 position where P is applied under static conditions is ¢ st = . Thus, the 48EI 48EI 1 equivalent spring constant for the bar is kb = , where I = (0.4) A 0.33 B 12 L3 and Thus, = 0.9 A 10 - 3 B m4, L = 4 m, E = Eal = 73.1 GPa.
kb =
48c73.1 A 109 B d c 0.9 A 10 - 3 B d 43
= 49.3425 A 106 B N>m
Using Eq. (1) Fsp =
P 2
ksp ¢ sp =
1 k ¢ 2 b b
1 kb 1 49.3425 A 10 ¢ ≤ ¢b = C 2 ksp 2 15 A 106 B
6
B
S ¢ b = 1.64475¢ b
(2)
Maximum Stress. The maximum force on the bar is
(Pb)max = kb ¢ b
¢ sp =
= 49.3425 A 106 B ¢ b. The maximum moment occurs at the midspan of the bar, where 49.3425 A 106 B ¢ b(4) (Pb)max L = = 49.3425 A 106 B ¢ b. Applying the flexure Mmax = 4 4 formula with smax = sY = 414 MPa,
smax =
Mmax c I
414 A 106 B =
49.3425 A 106 B ¢ b (0.15) 0.9 A 10 - 3 B
¢ b = 0.050342 m
1216
Section a – a
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14–69. Continued
Substituting this result into Eq. (2), ¢ sp = 0.0828 m Conservation of Energy. 1 1 1 mv2 = kb ¢ b 2 + 2 B ksp ¢ sp 2 R 2 2 2 1 1 1 c 10 A 103 B dv2 = c49.3425 A 106 B d A 0.0503422 B + 2 B c15 A 106 B d A 0.08282 B R 2 2 2 v = 5.75 m>s
Ans.
1217
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14–70. The simply supported W10 * 15 structural A-36 steel beam lies in the horizontal plane and acts as a shock absorber for the 500-lb block which is traveling toward it at 5 ft兾s. Determine the maximum deflection of the beam and the maximum stress in the beam during the impact. The spring has a stiffness of k = 1000 lb>in. For W 10 * 15: I = 68.9 in4
12 ft v ⫽ 5 ft/s
d = 9.99 in. k
From Appendix C: ¢ beam =
PL3 48EI
kbeam =
48(29)(103)(68.9) 48EI = = 4.015 kip>in. L3 (24(12))3
12 ft
Equilibrium (equivalent system): Fsp = Fbeam ksp ¢ sp = kbeam ¢ beam ¢ sp =
4.015(103) ¢ beam 1000
¢ sp = 4.015¢ beam
(1)
Conservation of energy: Ue = Ui 1 1 1 mv2 = kbeam ¢ 2beam + ksp ¢ 2sp 2 2 2 From Eq. (1): 500 1 1 1 a b (5(12))2 = (4.015)(103)¢ 2beam + (1000)(4.015¢ beam)2 2 32.2(12) 2 2 10067.6¢ 2beam = 2329.2 ¢ beam = 0.481 in.
Ans.
Fbeam = kbeam ¢ beam = 4.015(0.481) = 1.931 kip Mmax = a smax =
1.931 b (12) (12) = 139.05 kip # in. 2
139.05(9.99 Mmax c 2 ) = = 10.1 ksi 6 sg I 68.9
O.K.
Ans.
1218
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14–71. The car bumper is made of polycarbonatepolybutylene terephthalate. If E = 2.0 GPa, determine the maximum deflection and maximum stress in the bumper if it strikes the rigid post when the car is coasting at v = 0.75 m/s. The car has a mass of 1.80 Mg, and the bumper can be considered simply supported on two spring supports connected to the rigid frame of the car. For the bumper take I = 30011062 mm4, c = 75 mm, sY = 30 MPa and k = 1.5 MN>m.
0.9 m
k
0.9 m
k
v ⫽ 0.75 m/s
Equilibrium: This requires Fsp =
ksp ¢ sp =
k¢ beam 2
Pbeam . Then 2 or
¢ sp =
k ¢ 2ksp beam
[1]
Conservation of Energy: The equivalent spring constant for the beam can be determined using the deflection table listed in the Appendix C. k =
48 C 2(109) D C 300(10 - 6) D 48EI = = 4 938 271.6 N>m L3 1.83
Thus, Ue = Ui 1 1 1 my2 = k¢ 2beam + 2a ksp ¢ 2sp b 2 2 2
[2]
Substitute Eq. [1] into [2] yields 1 1 k2 mv2 = k¢ 2beam + ¢2 2 2 4ksp beam (4 93 8271.6)2 2 1 1 ¢ beam (1800) A 0.752 B = (493 8271.6) ¢ 2beam + 2 2 4[1.5(106)] ¢ beam = 8.8025 A 10 - 3 B m Maximum Displacement: From Eq. [1] ¢ sp =
4 938 271.6 2[1.5(106)]
C 8.8025 A 10 - 3 B D =
0.014490 m. ¢ max = ¢ sp + ¢ beam = 0.014490 + 8.8025 A 10 - 3 B = 0.02329 m = 23.3 mm
Ans.
Maximum Stress: The maximum force on the beam is Pbeam = k¢ beam = 4 938 271.6 C 8.8025 A 10 - 3 B D = 43 469.3 N. The maximum moment
occurs at mid-span. Mmax =
smax =
43 469.3(1.8) Pbeam L = = 19 561.2 N # m. 4 4
19 561.2(0.075) Mmax c = 4.89 MPa = I 300(10 - 6)
Ans.
Since smax 6 sg = 30 MPa, the above analysis is valid.
1219
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*14–72. Determine the horizontal displacement of joint B on the two-member frame. Each A-36 steel member has a cross-sectional area of 2 in2.
800 lb B
30⬚
60⬚
A
Member
n
N
L
nNL
AB
1.1547
800
120
11085.25
BC
–0.5774
0
60
0
5 ft
C
© = 110851.25 1 # ¢ Bh = ©
¢ Bh =
nNL AE
110851.25 110851.25 = 0.00191 in. = AE 29(106)(2)
Ans.
•14–73. Determine the horizontal displacement of point B. Each A-36 steel member has a cross-sectional area of 2 in2.
B
200 lb
Member Real Forces N: As shown on figure(a). 8 ft
Member Virtual Forces n: As shown on figure(b).
6 ft
nNL 1#¢ = a AE 1 lb # (¢ B)h =
1 [0.8333(166.67)(10)(12) AE +( -0.8333)(-166.67)(10)(12) +0.500(100)(12)(12)]
1 lb # (¢ B)h =
(¢ B)h =
40533.33 lb2 # in AE 40533.33 2[29.0(106)]
A
C
Virtual-Work Equation:
= 0.699 A 10 - 3 B in. :
Ans.
1220
6 ft
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14–74. Determine the vertical displacement of point B. Each A-36 steel member has a cross-sectional area of 2 in2.
B
200 lb
Member Real Forces N: As shown on figure(a). 8 ft
Member Virtual Forces n: As shown on figure(b). A
C
Virtual-Work Equation: nNL 1#¢ = a AE 1 lb # (¢ B)v =
6 ft
6 ft
1 [(-0.625)(166.67)(10)(12) AE +(-0.625)(-166.67)(10)(12) +0.375(100)(12)(12)]
1 lb # (¢ B)v =
(¢ B)v =
5400 lb2 # in AE 5400 2[29.0(106)]
= 0.0931 A 10 - 3 B in. T
Ans.
14–75. Determine the vertical displacement of joint C on the truss. Each A-36 steel member has a cross-sectional area of A = 300 mm2. Member
n
N
L
20 kN 3m
C A
1.50
45.0
3
202.5
AD
0
18.03
213
0
BC
1.50
45.0
3
202.5
BD
0
–20.0
2
0
CD
–1.803
–54.08
213
351.56
DE
–1.803
–72.11
213
468.77
4m D
E
© = 1225.33 nNL AE 1225.33(103) ¢ Cv =
300(10 - 6)(200)(109)
3m B
nNL
AB
1 # ¢ Cv = ©
30 kN
Ans.
= 0.0204 m = 20.4 mm
1221
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*14–76. Determine the vertical displacement of joint D on the truss. Each A-36 steel member has a cross-sectional area of A = 300 mm2.
20 kN
30 kN
3m
3m B C
A
Member
n
N
L
nNL
4m
AB
0
45.0
3
0
AD
0.9014
18.03
213
58.60
BC
0
45.0
3
0
BD
0
–20.0
2
0
CD
0
–54.08
213
0
DE
–0.9014
–72.11
213
234.36
D
E
© = 292.96 1 # ¢ Dv = ©
nNL AE 292.96(103)
¢ Dv =
300(10 - 6)(200)(109)
= 4.88(10 - 3) m = 4.88 mm
Ans.
•14–77. Determine the vertical displacement of point B. Each A-36 steel member has a cross-sectional area of 4.5 in2.
F
E
D
Virtual-Work Equation: Applying Eq. 14–39, we have Member
n
N
AB
0.6667
BC
0.6667
6 ft
L
nNL
3.333
96
213.33
3.333
96
213.33
A
C
B
CD
0
0
72
0
DE
0
0
96
0
EF
0
0
96
0
AF
0
0
72
0
AE
–0.8333
–4.167
120
416.67
CE
–0.8333
–4.167
120
416.67
BE
1.00
5.00
72
360.00
8 ft
©1620 kip2 # in. nNL 1#¢ = a AE 1 kip # (¢ B)v =
(¢ B)v =
8 ft 5 kip
1620 kip2 # in. AE 1620 = 0.0124 in. T 4.5[29.0(103)]
Ans.
1222
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14–78. Determine the vertical displacement of point E. Each A-36 steel member has a cross-sectional area of 4.5 in2.
F
E
D
6 ft
Virtual-Work Equation: Applying Eq. 14–39, we have Member
n
N
L
nNL
AB
0.6667
3.333
96
213.33
BC
0.6667
3.333
96
213.33
CD
0
0
72
0
DE
0
0
96
0
EF
0
0
96
0
AF
0
0
72
0
AE
–0.8333
–4.167
120
416.67
CE
–0.8333
–4.167
120
416.67
BE
0
5.00
72
0
A
C
B 8 ft
8 ft 5 kip
©1260 kip2 # in. nNL 1#¢ = a AE 1 kip # (¢ E)v =
(¢ E)v =
1260 kip2 # in. AE 1260 = 0.00966 in. T 4.5[29.0(103)]
Ans.
14–79. Determine the horizontal displacement of joint B of the truss. Each A-36 steel member has a cross-sectional area of 400 mm2.
5 kN 4 kN
2m C
B
Member
n
N
L
nNL
AB
0
0
1.5
0
AC
–1.25
–5.00
2.5
15.625
AD
1.00
4.00
2.0
8.000
BC
1.00
4.00
2.0
8.000
CD
0.75
–2.00
1.5
–2.25
1.5 m
D A
© = 29.375 1 # ¢ Bh = ©
nNL AE 29.375(103)
¢ Bh =
400(10 - 6)(200)(109)
= 0.3672(10 - 3)m = 0.367 mm
Ans.
1223
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*14–80. Determine the vertical displacement of joint C of the truss. Each A-36 steel member has a cross-sectional area of 400 mm2.
5 kN 2m
4 kN
C
B
Member
n
N
L
nNL
AB
0
0
1.5
0
AC
0
–5.00
2.5
0
AD
0
4.00
2.0
0
BC
0
4.00
2.0
0
CD
–1.00
–2.00
1.5
3.00
1.5 m
D A
© = 3.00 1 # ¢ Cv = ©
nNL AE 3.00 (103)
¢ Cv =
400(10 - 6)(200)(109)
= 37.5(10 - 6)m = 0.0375 mm
Ans.
•14–81. Determine the vertical displacement of point A. Each A-36 steel member has a cross-sectional area of 400 mm2.
E
D
Virtual-Work Equation: Member
n
AB
–0.750
BC
–0.750
AE
1.25
CE
–1.25
BE
0
DE
2m
N
1.50
–22.5 A 103 B –22.5 A 103 B 37.5 A 10
3
–62.5 A 10
3
60.0 A 10
3
B B
22.0 A 103 B
B
L 1.5 1.5 2.5 2.5 2 1.5
nNL
25.3125 A 103 B
C A
25.3125 A 103 B
117.1875 A 10
3
195.3125 A 10
3
135.00 A 10
3
B 1.5 m
B B
30 kN
0
B
© 498.125 A 103 B N2 # m
nNL 1#¢ = a AE 1 N # (¢ A)v =
(¢ A)v =
498.125(103) N2 # m AE 498.125(103) 0.400(10 - 3)[200(109)]
= 6.227 A 10 - 3 B m = 6.23 mm T
Ans.
1224
1.5 m 20 kN
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14–82. Determine the vertical displacement of point B. Each A-36 steel member has a cross-sectional area of 400 mm2.
E
D
2m C A B 1.5 m
Virtual-Work Equation: Member AB
n
N
0
BC
0
AE
0
CE
–1.25
BE
1.00
DE
0.750
–22.5 A 103 B –22.5 A 10
3
–62.5 A 10
3
B
37.5 A 103 B
B
22.0 A 103 B 60.0 A 103 B
L
30 kN
nNL
1.5
0
1.5
0
2.5 2.5 2 1.5
195.3125 A 10
0 3
B
40.0 A 103 B 67.5 A 103 B
© 302.8125 A 103 B N2 # m
nNL 1#¢ = a AE 1 N # (¢ B)v =
(¢ B)v =
302.8125(103) N2 # m AE 302.8125(103) 0.400(10 - 3)[200(109)]
= 3.785 A 10 - 3 B m = 3.79 mm T
Ans.
1225
1.5 m 20 kN
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14–83. Determine the vertical displacement of joint C. Each A-36 steel member has a cross-sectional area of 4.5 in2. 1 # ¢ Cv = ©
¢ Cv =
J
A
nNL AE
B 12 ft
Ans.
*14–84. Determine the vertical displacement of joint H. Each A-36 steel member has a cross-sectional area of 4.5 in2.
¢ Hv =
H
G
F 9 ft
21 232 = 0.163 in. 4.5 (29(103))
1 # ¢ Nv = ©
I
C 12 ft
12 ft
12 ft
6 kip
8 kip
6 kip
I
H
G
J
E
D
F 9 ft
A
nNL AE
12 ft
20 368 = 0.156 in. 4.5 (29(103))
Ans.
1226
C
B 12 ft 6 kip
12 ft 8 kip
E
D 12 ft 6 kip
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•14–85.
Determine the vertical displacement of joint C. The truss is made from A-36 steel bars having a cross- sectional area of 150 mm2.
G 2m H
Member Real Forces N. As indicated in Fig. a.
F 2m
Member Virtual Forces n. As indicated in Fig. b. 18 A 103 B
0.15 A 10 - 3 B
Virtual Work Equation. Since smax = Member AB
n(N)
N(N)
9 A 10
0.375
DE
0.375
BC
0.375
CD
0.375
AH
–0.625
EF
–0.625
BH
0
DF
0
CH
0
CF
0
GH
–0.625
FG
–0.625
CG
1
9 A 10
3 3
B B
9 A 103 B 9 A 103 B
–15 A 103 B –15 A 103 B 6 A 103 B 6 A 10
3
–3.75 A 10
3
B
–3.75 A 103 B
B
–11.25 A 103 B –11.25 A 103 B 18 A 103 B
nNL(N
L(m)
2
# m)
5.0625 A 10
1.5
5.0625 A 10
1.5
3 3
B B
5.0625 A 103 B
1.5
24.4375 A 103 B
2.5 2.5
24.4375 A 103 B
2
0
2
0
2.5
0
2.5
0
17.578125 A 103 B
2.5
17.578125 A 103 B
2.5
72 A 103 B
©174.28125 A 103 B
Then 1#¢ = ©
nNL AE
1N # (¢ C)v =
174.28125 A 103 B
0.15 A 10 - 3 B C 200 A 109 B D
(¢ C)v = 5.809 A 10 - 3 B m = 5.81 mm T
Ans.
1227
B 1.5 m
1.5 m 6 kN
5.0625 A 103 B
1.5
4
E
A
= 120 MPa 6 sY = 250 MPa,
C 1.5 m 12 kN
D 1.5 m 6 kN
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14–86. Determine the vertical displacement of joint G. The truss is made from A-36 steel bars having a cross-sectional area of 150 mm2.
G 2m H
Member Real Forces N. As indicated in Fig. a.
F 2m
Member Virtual Forces n. As indicated in Fig. b. E
A
Virtual
Work
Equation.
Since
18 A 10 B FCG = = 120 MPa 6 sY = 250 MPa, = A 0.15 A 10 - 3 B 3
smax
Member
n(N)
AB
0.375
DE
0.375
BC
0.375
CD
0.375
AH
–0.625
EF
–0.625
BH
0
DF
0
CH
0
CF
0
GH FG CG
–0.625 –0.625
N(N)
9 A 103 B 9 A 103 B 9 A 103 B 9 A 103 B
–15 A 103 B –15 A 103 B 6 A 103 B 6 A 103 B
–3.75 A 10
3
–11.25 A 10
3
B
–3.75 A 103 B
–11.25 A 10 18 A 10
0
3 3
B B B
5.0625 A 103 B
1.5
5.0625 A 103 B
1.5
5.0625 A 103 B
1.5
5.0625 A 103 B
1.5
24.4375 A 103 B
2.5 2.5
24.4375 A 103 B
2
0
2
0
2.5
0
2.5
17.578125 A 10
2.5
17.578125 A 10
2.5 4
0 3 3
B B
0
©102.28125 A 103 B
Then 1#¢ = ©
nNL AE
1N # (¢ G)v =
1.5 m
1.5 m 6 kN
nNL(N 2 # m)
L(m)
B
102.28125 A 103 B
0.15 A 10 - 3 B C 200 A 109 B D
(¢ G)v = 3.409 A 10 - 3 B m = 3.41 mm T
Ans.
1228
C 1.5 m 12 kN
D 1.5 m 6 kN
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14–87. Determine the displacement at point C. EI is constant.
P
P
A
B
a
a– 2
C a– 2
a
L
1 # ¢C =
mM dx L0 EI
¢C = 2 a =
a
1 1 bc a x1 b(Px1)dx1 + EI L0 2 L0
a>2
1 (a + x2)(Pa)dx2 d 2
23Pa3 24EI
Ans.
*14–88. The beam is made of southern pine for which Ep = 13 GPa. Determine the displacement at A.
15 kN 4 kN/m
A
L
1 # ¢A =
mM L0 EI
B 1.5 m 1.5
¢A =
=
C 3m
3
1 c (x1)(15x1)dx1 + (0.5x2)(2x22 + 1.5x2)dx2 d EI L0 L0
43.875(103) 43.875 kN # m3 = 0.0579 m = 57.9 mm = 1 9 EI 13(10 )(12 )(0.12)(0.18)3
1229
180 mm
Ans.
120 mm
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•14–89. Determine the displacement at C of the A-36 steel beam. I = 7011062 mm4.
2 kN/m
Real Moment Function M(x): As shown on figure(a).
C B
A
Virtual Moment Functions m(x): As shown on figure(b).
10 m
5m
Virtual Work Equation: For the displacement at point C. L
1#¢ = 1 kN # ¢ C =
¢C =
mM dx L0 EI 1 EI L0
10 m
0.500x1 (2.50x1)dx1 +
1 EI L0
5m
x2 A x22 B dx2
572.92 kN # m3 EI 572.92(1000)
=
200(109)[70(10 - 6)]
= 0.04092 m = 40.9 mm T
Ans.
14–90. Determine the slope at A of the A-36 steel beam. I = 7011062 mm4.
2 kN/m
C
Real Moment Function M(x): As shown on figure(a). B
A
Virtual Moment Functions mu (x): As shown on figure(b).
10 m
Virtual Work Equation: For the slope at point A. L
1#u = 1 kN # m # uA =
uA =
muM dx L0 EI 1 EI L0
10 m
(1 - 0.100x1)(2.50x1) dx1 +
5m
0 A 1.00x22 B dx2
41.667 kN # m2 EI 41.667(1000)
=
1 EI L0
200(109)[70(10 - 6)]
= 0.00298 rad
Ans.
1230
5m
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14–91. Determine the slope at B of the A-36 steel beam. I = 7011062 mm4.
2 kN/m
C
Real Moment Function M(x): As shown on figure(a). B
A
Virtual Moment Functions mU (x): As shown on figure(b).
10 m
5m
Virtual Work Equation: For the slope at point B. L
muM dx L0 EI
1#u = 1 kN # m # uB =
uB =
1 EI L0
10 m
0.100x1(2.50x1) dx1 +
0 A 1.00x22 B dx2
5m
83.333 kN # m2 EI 83.333(1000)
=
1 EI L0
200(109)[70(10 - 6)]
Ans.
= 0.00595 rad
*14–92. Determine the displacement at B of the 1.5-indiameter A-36 steel shaft.
2 ft 2 ft
A
3 ft
D B
L
1 # ¢B =
mM dx L0 EI
140 lb
2
¢B
1.5 ft
140 lb C
2
1 = c (0.5294x1)(327.06x1)dx1 + 0.5294(2 + x2)(654.12 + 47.06x2)dx2 EI L0 L0 1.5
+
=
L0
3
(0.4706x3)(592.94x3)dx3 +
L0
0.4706(x4 + 1.5)(889.41 - 47.06x4)dx4 d
6437.67(123) 6437.67 lb # ft3 = = 1.54 in. EI 29(106) p4 (0.75)4
320 lb 320 lb
Ans.
1231
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•14–93.
Determine the slope of the 1.5-in-diameter A-36 steel shaft at the bearing support A.
2 ft
L
1 # uA =
muM dx L0 EI 2
uA =
1 (1 - 0.1176x1)(327.06x1)dx1 + B EI L0 L0 L0
3 ft
D
1.5
(0.1176x3)(592.94x3)dx3 140 lb
0.1176(x4 + 1.5)(889.41 - 47.06x4)dx4 R
2387.53(122) 2387.53 lb # ft2 = = 0.0477 rad = 2.73° EI 29(106) A p4 B (0.754)
1.5 ft
B
5
+
=
2 ft
A
140 lb C
Ans. 320 lb 320 lb
14–94. The beam is made of Douglas fir. Determine the slope at C.
8 kN
Virtual Work Equation: For the slope at point C. A
L
1#u =
muM dx L0 EI
1 kN # m # uC = 0 +
1 EI L0 +
uC =
B 1.5 m
1.5 m
1.5 m
1.5 m
1 EI L0
(0.3333x2)(4.00x2) dx2 180 mm 1.5 m
(1 - 0.3333x3)(4.00x3)dx3
120 mm
4.50 kN # m3 EI 4.50(1000)
= -
C
1 13.1(10 ) C 12 (0.12)(0.183) D 9
= 5.89 A 10 - 3 B rad
Ans.
1232
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14–95. The beam is made of oak, for which Eo = 11 GPa. Determine the slope and displacement at A.
200 mm 400 mm
4 kN/m
B
A 3m
Virtual Work Equation: For the displacement at point A, L
1#¢ = 1 kN # ¢ A =
+
¢A =
L0
mM dx EI
1 EI L0
1 EI L0
3m
3m
2 x1 a x31 bdx1 9
(x2 + 3) A 2.00x22 + 6.00x2 + 6.00 B dx2
321.3 kN # m3 EI 321.3(103)
=
1 11(109) C 12 (0.2)(0.43) D
Ans.
= 0.02738 m = 27.4 mm T For the slope at A. L
muM dx L0 EI
1#u = 1 kN # m # uA =
1 EI L0
3m
3m
+
uA = =
L0
2 1.00a x31 bdx1 9
1.00 A 2.00x22 + 6.00x2 + 6.00 B dx2
67.5 kN # m2 EI 67.5(1000)
1 11(109) C 12 (0.2)(0.43) D
= 5.75 A 10 - 3 B rad
Ans.
1233
3m
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*14–96. Determine the displacement at point C. EI is constant.
P
A
C B a
a
L
1 # ¢C =
mM dx L0 EI a
¢C =
=
•14–97.
a
1 c (x1)(Px1)dx1 + (x2)(Px2)dx2 d EI L0 L0 2Pa3 3EI
Ans.
Determine the slope at point C. EI is constant.
P
A
C B a
L
1 # uC =
L0 a
uC =
=
L0
muMdx EI
A a1 B Px1 dx1 x
EI
a
+
(1)Px2dx2 EI L0
Pa2 5Pa2 Pa2 + = 3EI 2EI 6EI
Ans.
1234
a
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14–98.
Determine the slope at point A. EI is constant.
P
A
L
1 # uA =
uA
muM dx L0 EI
C B a
a a x1 1 Pa2 = a1 (0)(Px2)dx2 R = b (Px1)dx1 + B a EI L0 6EI L0
14–99. Determine the slope at point A of the simply supported Douglas fir beam.
a
Ans.
3 kN 0.6 kN⭈m
a
A
B
C
Real Moment Function M. As indicated in Fig. a. 1.5 m
a 0.5 m
Virtual Moment Functions m. As indicated in Fig. b. 75 mm
Virtual Work Equation. 150 mm
L
1#u =
mu M dx EI L0
1kN # m # uA =
Section a – a
1 B EI L0
2m
(1 - 0.3333x1)(0.8x1 + 0.6)dx1 1m
+
uA =
=
=
1 B EI L0
2m
L0
(0.3333x2)(2.2x2)dx2 R
A -0.2667x1 2 + 0.6x1 + 0.6 B dx1 +
1m
L0
0.7333x2 2dx2 R
1.9333 kN # m2 EI
1.9333 A 103 B
1 13.1 A 109 B c (0.075) A 0.153 B d 2
= 0.00700 rad
Ans.
1235
1m
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14–99.
Continued
1236
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*14–100. Determine the displacement at C of the simply supported Douglas fir beam.
3 kN 0.6 kN⭈m
a
A
Real Moment Function M. As indicated in Fig. a.
B
C
Virtual Moment Functions m. As indicated in Fig. b.
1.5 m
Virtual Work Equation. L
1#¢ =
L0
1 kN # ¢ C =
150 mm
1 B EI L0
1.5 m
1m
(0.5x1)(0.8x1 + 0.6)dx1 +
L0
Section a – a
(0.5x2)(2.2x2)dx2
0.5 m
¢C =
1 B EI L0
1.5 m
(0.5x3 + 0.5)(2.2 - 0.8x3)dx3 R
L0
A 0.4x1 2 + 0.3x1 B dx1 + 0.5 m
+
=
=
L0
1m
L0
1.1x2 2 dx2
A -0.4x3 2 + 0.7x3 + 1.1 B dx3 R
1.775kN # m3 EI
1.775 A 103 B
13.1 A 109 B c
0.5 m 75 mm
mM dx EI
+
a
1 (0.075) A 0.153 B d 12
= 6.424 A 10 - 3 B m = 6.42 mm T
Ans.
1237
1m
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•14–101.
Determine the slope of end C of the overhang beam. EI is constant.
w
C
A
Real Moment Function M. As indicated in Fig. a.
B
D L 2
Virtual Moment Function mu. As indicated in Fig. b. Virtual Work Equation. L
1#u = 1 # uC =
uC =
mu M dx L0 EI L L>2 x1 w w 1 (1) ¢ x 3 ≤ dx2 R B ¢ - ≤ c A 11Lx1 - 12x1 2 B ddx1 + EI L0 L 24 3L 2 L0
1 w B EI 24L L0
uC = -
L
A 12x1 3 - 11Lx1 2 B dx1 +
w 3L L0
L>2
x2 3 dx2 R
13wL3 13wL3 = 576EI 576EI
Ans.
1238
L 2
L 2
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14–102. Determine the displacement of point D of the overhang beam. EI is constant.
w
L 2
Virtual Work Equation. L
1 # ¢D =
L0
mM dx EI
1 B EI L0
L>2
¢
x1 w ≤ c A 11Lx1 - 12x1 2 B ddx1 2 24
L>2
+
L0
¢D =
w B 48EI L0
¢D =
wL4 T 96EI
L>2
¢
x2 w ≤ c A 13Lx2 - 12x2 2 - L2 B ddx2 R 2 24
A 11Lx1 2 - 12x1 3 B dx1 +
L>2
L0
B
D
Virtual Moment Function m. As indicated in Fig. b.
1#¢ =
C
A
Real Moment Function M. As indicated in Fig. a.
A 13Lx2 2 - 12x2 3 - L2x2 B dx2 R Ans.
1239
L 2
L 2
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14–103. Determine the displacement of end C of the overhang Douglas fir beam.
400 lb a
Real Moment Functions M. As indicated in Fig. a.
400 lb⭈ft
A B
a
Virtual Moment Functions m. As indicated in Fig. b. 8 ft
Virtual Work Equation.
4 ft 3 in.
L
mM 1#¢ = dx L0 EI 1 lb # ¢ C =
¢C =
=
=
1 B EI L0
1 B EI L0
6 in.
8 ft
8 ft
4 ft
125x1 2dx1 +
Section a – a
4 ft
(0.5x1)(250x1)dx1 +
L0
L0
x2(400x2 + 400)dx2 R
A 400x2 2 + 400x2 B dx2 R
33066.67 lb # ft3 EI 33066.67 A 12 3 B
1.90 A 106 B c
1 (3) A 63 B d 12
= 0.5569 in. = 0.557 in. T
Ans.
1240
C
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*14–104. Determine the slope at A of the overhang white spruce beam.
400 lb a 400 lb⭈ft
A
8 ft
Virtual Moment Functions m. indicated in Fig. b.
6 in.
L
muM 1#u = dx L0 EI
uA =
=
=
1 B EI L0
1 B EI L0
8 ft
Section a – a 8 ft
4 ft
(1 - 0.125x1)(250x1)dx1 +
L0
0(400x2 + 400)dx2 R
A 250x1 - 31.25x1 2 B dx1 + 0 R
2666.67 lb # ft2 EI 2666.67 A 12 2 B
1.940 A 106 B c
1 (3) A 63 B d 12
= 0.00508 rad = 0.00508 rad
Ans.
1241
C 4 ft
3 in.
Virtual Work Equation.
1 lb # ft # uA =
B
a
Real Moment Functions M. As indicated in Fig. a.
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•14–105. Determine the displacement at point B. The moment of inertia of the center portion DG of the shaft is 2I, whereas the end segments AD and GC have a moment of inertia I. The modulus of elasticity for the material is E.
w A
C D a
Real Moment Function M(x): As shown on Fig. a. Virtual Moment Functions m(x): As shown on Fig. b. Virtual Work Equation: For the slope at point B, apply Eq. 14–42. L
1#¢ =
mM dx L0 EI
1 # ¢B = 2 B
a x1 1 a b(w ax1)dx1 R EI L0 2
a
+ 2B
¢B =
1 1 w 2 (x + a) B wa(a + x2) x R dx2 R 2EI L0 2 2 2 2
65wa4 48EI
Ans.
T
1242
B a
G a
a
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14–106. Determine the displacement of the shaft at C. EI is constant.
w0
A
B C L – 2
L – 2
L
1 # ¢C =
mM dx L0 E I L
¢C
2 w0 L w0 3 1 1 = 2a b a x1 b a x1 x bdx1 E I L0 2 4 3L 1
=
w0 L4 120 E I
Ans.
14–107. Determine the slope of the shaft at the bearing support A. EI is constant.
w0
A
B C L – 2
L
1 # uA =
mu M dx L0 E I L
uA
2 w0 L w0 3 1 1 = C a1 x ba x1 x bdx1 S E I L0 L 1 4 3L 1
L 2
+
=
L0
a
w0L w0 3 1 x ba x x bdx2 L 2 4 2 3L 2
5 w0L3 192 E I
Ans.
1243
L – 2
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*14–108. Determine the slope and displacement of end C of the cantilevered beam. The beam is made of a material having a modulus of elasticity of E. The moments of inertia for segments AB and BC of the beam are 2I and I, respectively.
P
L 2
Real Moment Function M. As indicated in Fig. a. Virtual Moment Functions mu and M. As indicated in Figs. b and c. Virtual Work Equation. For the slope at C, L
1#u =
L0
1 # uC =
uC =
mu M dx EI
1 EI L0
L>2
1(Px1)dx1 +
1 2 EI L0
L>2
1 B Pa x2 +
L b R dx2 2
5PL2 16 EI
Ans.
For the displacement at C, L
1#¢ = 1 # ¢C =
¢C =
mM dx L0 EI 1 EI L0
L>2
x1(Px1)dx1 +
1 2EI L0
L>2
¢ x2 +
L L ≤ B P ¢ x2 + ≤ R dx2 2 2
3PL T 16EI
Ans.
1244
C
B
A
L 2
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•14–109.
Determine the slope at A of the A-36 steel W200 * 46 simply supported beam.
12 kN/m 6 kN/m
Real Moment Function M. As indicated in Fig. a. A 3m
Virtual Work Equation. L
1#u =
mu M dx EI L0
1kN # m # uA =
1 B EI L0
3m
(1 - 0.1667x1) A 31.5x1 - 6x1 2 B dx1
3m
+
uA =
=
=
1 B EI L0
3m
L0
(0.1667x2) A 22.5x2 - 3x2 2 B dx2 R
A x1 3 - 11.25x1 2 + 31.5x1 B dx1 +
3m
L0
B
C
Virtual Moment Functions m. As indicated in Fig. b.
A 3.75x2 2 - 0.5x2 3 B dx2 R
84.375 kN # m2 EI
84.375 A 103 B
200 A 109 B c45.5 A 10 - 6 B d
= 0.009272 rad = 0.00927 rad
Ans.
1245
3m
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14–110. Determine the displacement at point C of the A-36 steel W200 * 46 simply supported beam.
12 kN/m 6 kN/m
Real Moment Functions M. As indicated in Fig. a. A 3m
Virtual Work Equation. L
1#¢ =
mM dx L0 EI
1kN # ¢ C =
1 B EI L0
(0.5x1) A 31.5x1 - 6x1 2 B dx1
3m
3m
+
¢C =
=
=
1 B EI L0
3m
L0
(0.5x2) A 22.5x2 - 3x2 2 B dx2 R
A 15.75x1 2 - 3x1 3 B dx1 +
3m
L0
B
C
Virtual Moment Functions m. As indicated in Figs. b.
A 11.25x2 2 - 1.5x2 3 B dx2 R
151.875 kN # m3 EI
151.875 A 103 B
200 A 109 B c45.5 A 10 - 6 B d
= 0.01669 m = 16.7 mm T
Ans.
1246
3m
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14–111. The simply supported beam having a square cross section is subjected to a uniform load w. Determine the maximum deflection of the beam caused only by bending, and caused by bending and shear. Take E = 3G.
w
a L
For bending and shear, L
1#¢ =
L fsvV mM dx + dx EI L0 GA L0 L>2
¢ = 2
L0
x A 12 x B A wL 2 x - w 2 B dx 2
EI
L>2
+ 2
L0
A 65 B A 12 B A wL 2 - wx B dx GA
A B wL wx2 2 L>2 wx4 2 L>2 1 wL 3 a x b + a x b EI 6 8 GA 2 2 0 0 6 5
=
=
5wL4 3wL2 + 384EI 20 GA 5wL4
¢ =
=
1 384(3G) A 12 B a4
+
3wL2 20(G)a2
20wL4 3wL2 + 4 384Ga 20Ga2
= a
L 2 L 2 w 5 3 ba b Ba ba b + R a a G 96 20
Ans.
For bending only, ¢ =
5w L 4 a b 96G a
Ans.
1247
a
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*14–112. The frame is made from two segments, each of length L and flexural stiffness EI. If it is subjected to the uniform distributed load determine the vertical displacement of point C. Consider only the effect of bending.
w B C L
L
A
Real Moment Function M(x): As shown on Fig. a. Virtual Moment Functions m(x): As shown on Fig. b. Virtual Work Equation: For the vertical displacement at point C, L
1#¢ =
L0
mM dx EI L
1 # (¢ C)v =
(¢ C)v =
L
1 w 2 1 wL2 (1.00x1)a x1 b dx1 + (1.00L)a b dx2 EI L0 2 EI L0 2 5wL4 8EI
Ans.
T
1248
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•14–113.
The frame is made from two segments, each of length L and flexural stiffness EI. If it is subjected to the uniform distributed load, determine the horizontal displacement of point B. Consider only the effect of bending.
w B C L
Real Moment Function M(x): As shown on Fig. a. L
Virtual Moment Functions m(x): As shown on Fig. b. Virtual Work Equation: For the horizontal displacement at point B,
A
L
1#¢
mM = dx L0 EI L
1 # (¢ B)h =
(¢ B)h =
L
1 w 2 1 wL2 (0)a x1 b dx1 + (1.00L - 1.00x2)a b dx2 EI L0 2 EI L0 2 wL4 : 4EI
Ans.
14–114. Determine the vertical displacement of point A on the angle bracket due to the concentrated force P. The bracket is fixed connected to its support. EI is constant. Consider only the effect of bending.
P L A
L
1 # ¢ Av =
mM dx L0 EI L L
¢ Av =
=
L
1 C (x1)(Px1)dx1 + (1L)(PL)dx2 S EI L0 L0 4PL3 3EI
Ans.
1249
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14–115. Beam AB has a square cross section of 100 mm by 100 mm. Bar CD has a diameter of 10 mm. If both members are made of A-36 steel, determine the vertical displacement of point B due to the loading of 10 kN.
C 10 kN
2m D
A 3m
Real Moment Function M(x): As shown on Fig. a. Virtual Moment Functions m(x): As shown on Fig. b. Virtual Work Equation: For the displacement at point B, L
1#¢ =
1 kN # ¢ B =
L0
nNL mM dx + EI AE
1 EI L0
+
3m
(0.6667x1)(6.667x1)dx1
1 EI L0
+
¢B =
2m
(1.00x2)(10.0x2)dx2
1.667(16.667)(2) AE
66.667 kN # m3 55.556 kN # m + EI AE 66.667(1000)
=
200(10 ) C 9
1 12
(0.1) A 0.1
3
BD
= 0.04354 m = 43.5 mm
55.556(1000)
+
C A 0.012 B D C 200 A 109 B D p 4
Ans.
T
1250
B 2m
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*14–116. Beam AB has a square cross section of 100 mm by 100 mm. Bar CD has a diameter of 10 mm. If both members are made of A-36 steel, determine the slope at A due to the loading of 10 kN.
C 10 kN
2m
Real Moment Function M(x): As shown on Fig. a.
D
A
Virtual Moment Functions mu(x): As shown on Fig. b.
B
3m
2m
Virtual Work Equation: For the slope at point A, L
1#u =
1 kN # m # uA =
muM nNL dx + EI AE L0 3m
1 EI L0
(1 - 0.3333x1)(6.667x1)dx1
+
uA =
1 EI L0
2m
0(10.0x2)dx2 +
10.0 kN # m2 11.111 kN EI AE 11.111(1000)
10.0(1000)
=
(-0.3333)(16.667)(2) AE
200 A 10
9
B C (0.1) A 0.1 B D 1 12
3
-
C A 0.012 B D C 200 A 109 B D p 4
= 0.00529 rad
Ans.
14–117. Bar ABC has a rectangular cross section of 300 mm by 100 mm. Attached rod DB has a diameter of 20 mm. If both members are made of A-36 steel, determine the vertical displacement of point C due to the loading. Consider only the effect of bending in ABC and axial force in DB.
D
4m 20 kN
300 mm 100 mm A
Real Moment Function M(x): As shown on Fig. a.
3m
Virtual Moment Functions m(x): As shown on Fig. b. Virtual Work Equation: For the displacement at point C, L
1#¢ =
L0
mM nNL dx + EI AE
1 kN # ¢ C = 2c
¢C =
1 EI L0
3m
(1.00x)(20.0x) dx d +
2.50(50.0) (5) AE
360 kN # m3 625 kN # m + EI AE 625(1000)
360(1000)
=
200 A 10
9
B C (0.1) A 0.3 B D 1 12
+
3
B
C A 0.02 2 B D C 200 A 109 B D p 4
= 0.017947 m = 17.9 mm T
Ans.
1251
3m
C
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14–118. Bar ABC has a rectangular cross section of 300 mm by 100 mm. Attached rod DB has a diameter of 20 mm. If both members are made of A-36 steel, determine the slope at A due to the loading. Consider only the effect of bending in ABC and axial force in DB.
D
4m 20 kN
300 mm 100 mm A
B 3m
Real Moment Function M(x): As shown on Fig. a. Virtual Moment Functions mu (x): As shown on Fig. b. Virtual Work Equation: For the slope at point A, L
1#u =
muM nNL dx + AE L0 EI
1 kN # m # uA =
uA =
1 EI L0
3m
(1 - 0.3333x)(20.0x)dx +
30.0 kN # m2 104.167 kN EI AE 104.167(1000)
30.0(1000)
=
( -0.41667)(50.0)(5) AE
200 A 10
9
B C (0.1) A 0.3 B D 1 12
3
-
C A 0.02 2 B D C 200 A 109 B D p 4
= -0.991 A 10 - 3 B rad = 0.991 A 10 - 3 B rad
Ans.
1252
3m
C
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14–119. Determine the vertical displacement of point C. The frame is made using A-36 steel W250 * 45 members. Consider only the effects of bending.
15 kN/m 15 kN
D B C 2.5 m
5m
A
Real Moment Functions M. As indicated in Fig. a. Virtual Moment Functions m. As indicated in Fig. b. Virtual Work Equation. L
1#¢ =
L0
mM dx EI
1 kN # (¢ C)v =
1 B EI L0
(0.5x1) A 52.5x1 - 7.5x1 2 B dx1
2.5 m
5m
+
0(15x2)dx2
L0
2.5 m
+
(¢ C)v =
1 B EI L0
2.5 m
L0
A 26.25x1 2 - 3.75x1 3 B dx1 + 0
2.5 m
+
=
=
(0.5x3) A 75 + 22.5x3 - 7.5x3 2 B dx2 R
L0
A 37.5x3 + 11.25x3 2 - 3.75x3 3 B dx3 R
239.26 kN # m3 EI
239.26 A 103 B
200 A 109 B c71.1 A 10 - 6 B d
= 0.01683 m = 16.8 mm T
Ans.
1253
2.5 m
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*14–120. Determine the horizontal displacement of end B. The frame is made using A-36 steel W250 * 45 members. Consider only the effects of bending.
15 kN/m 15 kN
D B C 2.5 m
5m
Real Moment Functions M. As indicated in Fig. a. A
Virtual Moment Functions m. As indicated in Fig. b. Virtual Work Equation. L
1#¢ =
L0
mM dx EI
1 kN # (¢ B)h =
(¢ B)h =
=
=
1 B EI L0
1 B EI L0
5m
x1 A 52.5x1 - 7.5x1 2 B dx1 +
5m
A 52.5x1 2 - 7.5x1 3 B dx1 +
5m
L0
x2(15x2)dx2 R
5m
L0
15x2 2dx2 R
1640.625 kN # m3 EI
1640.625 A 103 B
200 A 109 B c71.1 A 10 - 6 B d
= 0.1154 m = 115 mm :
Ans.
1254
2.5 m
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•14–121. Determine the displacement at point C. EI is constant.
A
B
a
C
M0
C
M0
a
L
mM dx L0 E I
1 # ¢C =
a
¢C =
L0
(1x) A
EI
5 M0 a 6EI
=
14–122.
M0 a
xB
a
dx +
(1x) M0 dx EI L0
2
Ans.
Determine the slope at B. EI is constant.
A
B
a L
mu M dx L0 E I
1 # uB =
a
uB =
=
L0
A xa B A Ma0 x B EI
dx
M0 a 3EI
14–123.
Ans.
Solve Prob. 14–72 using Castigliano’s theorem.
Member
N
0N>0P
N(P = 0)
L
N(0N>0P)L
AB
1.1547P + 800
1.1547
800
120
110851.25
BC
–0.5774P
–0.5774
0
60
0 © = 110851.25
¢ Bb = ©N a
0N L 110851.25 110851.25 = 0.00191 in. b = = 0P AE AE (2)(29)(106)
1255
Ans.
a
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*14–124.
Solve Prob. 14–73 using Castigliano’s theorem.
Member Force N: Member forces due to external force P and external applied forces are shown on the figure. Castigliano’s Second Theorem: N(P = 200 lb)
L
–0.8333P
0N 0P –0.8333
–166.67
10.0
0N bL 0P 1388.89
BC
0.8333P
0.8333
166.67
10.0
1388.89
AC
0.500P
0.500
100.00
12
600.00
Member
N
AB
Na
© 3377.78 lb # ft 0N L b ¢ = a Na 0P AE (¢ B)h =
3377.78 lb # ft AE 3377.78(12)
=
•14–125.
2 C 29.0 A 106 B D
= 0.699 A 10 - 3 B in. :
Ans.
Solve Prob. 14–75 using Castigliano’s theorem.
Member
N
N(P = 30)
0N>0P
L
N(0N>0P)L
AB
1.50P
1.50
45.00
3.0
202.50
AD
5 213
0
5213
213
0
BD
–20
0
–20
2.0
0
BC
1.5P
1.5
45.00
3.0
202.50
CD
–0.5 213P
–0.5 213
-15 213
213
351.54
–0.5 213
-20 213
213
468.72
DE
– A 0.5 213P + 5 213 B
1225.26 A 10 B 0N L b = 0P AE 300 A 10 - 6 B (200) A 109 B
© = 1225.26
3
¢ Cv = ©Na
= 0.02.04 m = 20.4 mm
Ans.
1256
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14–126. Solve Prob. 14–76 using Castigliano’s theorem.
Member
N
N(P = 0)
0N>0P
N(0N>0P)L
L
AB
45
0
45.00
3
0
AD
0.25 213P + 5 213
0.25213
5213
213
58.59
BC
45
0
45
3
0
BD
–20
0
–20
2
0
CD
–15 213
0
-15 213
213
0
–0.25213
-20 213
213
234.36
– A 0.25 213P + 20 213 B
DE
¢ Dv = ©Na
© = 292.95
292.95 A 103 B 292.95 0N L b = = 0P AE AE 300 A 10 - 6 B (200) A 109 B
= 4.88 A 10 - 3 B m = 4.88 mm
14–127.
Ans.
Solve Prob. 14–77 using Castigliano’s theorem.
Member Forces N: Member forces due to external force P and external applied forces are shown on the figure. Castigliano’s Second Theorem:
3.333
96
0N bL 0P 213.33
0.6667
3.333
96
213.33
0
0
0
72
0
DE
0
0
0
96
0
EF
0
0
0
96
0
AF
0
0
0
72
0
AE
– 0.8333P
– 0.8333
–4.167
120
416.67
CE
– 0.8333P
– 0.8333
–4.167
120
416.67
BE
1.00P
1.00
5.00
72
360.00
Member
N
AB
0.6667P
0N 0P 0.6667
BC
0.6667P
CD
N(P = 5 kip)
L
Na
©1620 kip # in 0N L b ¢ = a Na 0P AE (¢ B)v =
1620 kip # in. AE 1620
=
4.5 C 29.0 A 103 B D
= 0.0124 in. T
Ans.
1257
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*14–128.
Solve Prob. 14–78 using Castigliano’s theorem.
Member Forces N: Member forces due to external force P and external applied forces are shown on the figure. Castigliano’s Second Theorem: N(P = 0)
L
3.333
96
0N bL 0P 213.33
AB
0.6667P+3.333
0N 0P 0.6667
BC
0.6667P+3.333
0.6667
3.333
96
213.33
CD
0
0
0
72
0
DE
0
0
0
96
0
EF
0
0
0
96
0
AF
0
0
0
72
0
AE
–(0.8333P + 4.167)
– 0.8333
–4.167
120
416.67
CE
–(0.8333P + 4.167)
– 0.8333
–4.167
120
416.67
BE
5.0
0
5.00
72
0
Member
N
Na
©1260 kip # in 0N L b ¢ = a Na 0P AE (¢ E)v =
1260 kip # in. AE 1260
=
•14–129.
4.5 C 29.0 A 103 B D
= 0.00966 in. T
Ans.
Solve Prob. 14–79 using Castigliano’s theorem.
Member
N
0N>0P
N(P = 4)
L
N(0N>0P)L
AB
0
0
0
1.5
0
AC
–1.25P
–1.25
–5
2.5
15.625
AD
P
1
4
2.0
8.00
BC
P
1
4
2.0
8.00
CD
–(5 -0.75P)
0.75
–2
1.5
–2.25 © = 29.375
¢ Bh = ©Na
29.375 A 103 B 0N L ba b = = 0.367 A 10 - 3 B m 0P AE 400 A 10 - 6 B (200) A 109 B
= 0.367 mm
Ans.
1258
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14–130.
Solve Prob. 14–80 using Castigliano’s theorem.
Member
N
0N>0P
N(P = 5)
L
N(0N>0P)L
AB
0
0
0
1.5
0
AC
–5
0
–5
2.5
0
AD
4
0
4
2.0
0
BC
4
0
4
2.0
0
CD
–(P - 3)
–1
–2
1.5
3 © = 3
3 A 10 B 3 0N L b = = 0P AE AE 400 A 10 6 B (200) A 109 B 3
¢ Cv = ©Na
= 37.5 A 10 - 6 B m = 0.0375 mm
14–131.
Ans.
Solve Prob. 14–81 using Castigliano’s theorem.
Member Forces N: Member forces due to external force P and external applied forces are shown on the figure. Castigliano’s Second Theorem: N(P = 30 kN)
L
–22.5
1.5
0N bL 0P 25.3125
AB
– 0.750P
0N 0P – 0.750
BC
– 0.750P
– 0.750
–22.5
1.5
25.3125
AE
1.25P
1.25
37.5
2.5
117.1875
CE
–(1.25P + 25.0)
– 1.25
–62.5
2.5
195.3125
BE
20.0
0
20.0
2
0
DE
1.50P+15.0
1.50
60.0
1.5
135.00
Member
N
Na
# a 498.125 kN m 0N L b ¢ = a Na 0P AE (¢ A)v =
=
498.125 kN # m AE
498.125 A 103 B
0.400 A 10 - 3 B C 200 A 109 B D
= 6.227 A 10 - 3 B m = 6.23 mm T
Ans.
1259
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*14–132.
Solve Prob. 14–82 using Castigliano’s theorem.
Member Forces N: Member forces due to external force P and external applied forces are shown on the figure. Castigliano’s Second Theorem: 0N bL 0P 0
AB
– 22.5
0N 0P 0
BC
– 22.5
0
–22.5
1.5
0
AE
37.5
0
37.5
2.5
0
CE
–(1.25P + 37.5)
– 1.25
–62.5
2.5
195.3125
BE
1.00P
1.00
20.0
2
40.0
DE
0.750P + 45
0.750
60.0
1.5
67.50
Member
N
N(P = 20 kN)
L
–22.5
1.5
Na
# a 302.8125 kN m 0N L b ¢ = a Na 0P AE (¢ B)v =
=
302.8125 kN # m AE
302.8125 A 103 B
0.400 A 10 - 3 B C 200 A 109 B D
= 3.785 A 10 - 3 B m = 3.79 mm T
•14–133.
Ans.
Solve Prob. 14–83 using Castigliano’s theorem.
¢ Cv = ©N a
21232 21232 0N L = 0.163 in. b = = 0P AE AE 4.5 (29)(103)
Ans.
1260
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14–134.
Solve Prob. 14–84 using Castigliano’s theorem.
¢ Hv = ©N a
20368 20368 0N L b = = 0P AE AE 4.5 (29) A 103 B
= 0.156 in.
14–135.
Ans.
Solve Prob. 14–87 using Castigliano’s theorem.
0M1 x1 = 0P¿ 2
0M2 x2 a = + 0P¿ 2 2
Set P¿ = 0 M1 = Px1
M2 = Pa
a
¢C =
L0
Ma
0M dx b 0P EI a
= (2)
=
1 1 (Px1) a x1 b dx + B EI L0 2 L0
a>2
(Pa) a
1 a + x2 bdx2 R 2 2
23Pa3 24 EI
Ans.
1261
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*14–136.
Solve Prob. 14–88 using Castigliano’s theorem.
0M1 = x1 0P
0M2 = -0.5 x2 0P
Set P = 15 kN M2 = -1.5x2 - 2x22
M1 = 15x1 L
¢A =
Ma
L0
0M dx b 0P EI
1.5
3
A -1.5x2 - 2x22 B (-0.5x2)dx2 R
=
1 B EI L0
=
43.875(103) 43.875 kN # m3 = = 0.0579 m 1 9 EI 13(10 ) 12 (0.12)(0.18)3
(15x1)(x1)dx +
L0
= 57.9 mm
•14–137.
Ans.
Solve Prob. 14–90 using Castigliano’s theorem.
Internal Moment Function M(x): The internal moment function in terms of the couple moment M¿ and the applied load are shown on the figure. Castigliano’s Second Theorem: The slope at A can be determined with 0M(x1) 0M(x2) = 1 - 0.100x1, = 0 and setting M¿ = 0. 0M¿ 0M¿ L
u =
uA =
=
=
L0
Ma
1 EI L0
0M dx b 0M¿ EI
10 m
(2.50x1)(1 - 0.100x1)dx1 +
1 EI L0
5m
A 1.00x22 B (0)dx2
41.667 kN # m2 EI
41.667 A 103 B
200 A 109 B C 70 A 10 - 6 B D
Ans.
= 0.00298 rad
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14–138.
Solve Prob. 14–92 using Castigliano’s theorem.
0M1 = 0.5294x1 0P
0M2 = 0.5294x2 + 1.0588 0P
0M3 = 0.4706x3 0P
0M4 = 0.4706x4 + 0.7059 0P
Set P = 0 M1 = 327.06x1
M2 = 47.06x2 + 654.12
M3 = 592.94x3
M4 = 889.41 - 47.06x4
L
¢B =
L0
Ma
0M dx b 0P EI 2
1 (327.06x1)(0.5294x1)dx1 c EI L0
=
2
+
(47.06x2 + 654.12)(0.5294x2 + 1.0588)dx2
L0
1.5
+
(592.94x3)(0.4706x3)dx3
L0 3
+
=
14–139.
L0
(889.41 - 47.06x4)(0.4706x4 + 0.7059)dx4 d
6437.69 A 12 3 B 6437.69 lb # ft3 = = 1.54 in. EI 29 A 106 B A p4 B A 0.754 B
Ans.
Solve Prob. 14–93 using Castigliano’s theorem.
0M1 = 1 - 0.1176 x1 0M¿
0M2 = 0.1176 x2 0M¿
0M3 = 0.1176x3 + 0.1764 0M¿
Set M¿ = 0 M1 = 327.06x1
uA =
L
Ma
M2 = 592.94x2
M3 = 889.41 - 47.06x3
2
0M dx 1 b = c (327.06x1)(1 - 0.1176x1)dx1 0M¿ EI EI L0
1.5
+
(592.94x2)(0.1176x2)dx2 +
L0 5
+
=
L0
(889.41 - 47.06x3)(0.1176x3 + 0.1764)dx3 d
2387.54(12 2) 2387.54 lb # ft2 = 0.0477 rad = 2.73° = EI 29(106)(p4 )(0.754)
1263
Ans.
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*14–140.
Solve Prob. 14–96 using Castigliano’s theorem.
0M1 = x1 0P¿
0M2 = x2 0P¿
Set P = P¿ M1 = Px1
M2 = Px2
L
¢C =
=
a
Ma
L0
a
1 0M bdx = (Px1)(x1)dx1 + (Px2)(x2)dx2 R B 0P¿ EI L0 L0
2Pa3 3EI
14–141.
Ans.
Solve Prob. 14–89 using Castigliano’s theorem.
Set M¿ = 0 L
uC =
Ma
L0
0M dx b 0M¿ EI
1 (Px1)a x1 b dx1 a (Px2)(1)dx2 a + = EI EI L0 L0 a
=
Pa2 5Pa2 Pa2 + = 3EI 2EI 6EI
14–142.
Ans.
Solve Prob. 14–98 using Castigliano’s theorem.
0M1 x1 = 1 a 0M¿
0M2 = 0 0M¿
Set M¿ = 0 M1 = -Px1 L
uA =
L0
Ma
M2 = Px2 0M dx b 0M¿ EI
=
a a x1 -Pa2 1 bdx1 + ( -Px1)a 1 (Px2)(0)dx2 R = B a EI L0 6EI L0
=
Pa2 6EI
Ans.
1264
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14–143.
Solve Prob. 14–112 using Castigliano’s theorem.
Internal Moment Function M(x): The internal moment function in terms of the load P and external applied load are shown on the figure. Castigliano’s Second Theorem: The vertical displacement at C can be determined 0M(x1) 0M(x2) with = 1.00x1, = 1.00L and setting P = 0. 0P 0P L
¢ =
Ma
L0
0M dx b 0P EI
L
(¢ C)v =
=
*14–144.
L
2
1 w 2 1 wL a x1 b(1.00x1) dx1 + a b(1.00L) dx2 EI L0 2 EI L0 2 5wL4 8EI
Ans.
T
Solve Prob. 14–114 using Castigliano’s theorem.
Castigliano’s Second Theorem: The horizontal displacement at A can be determined 0M(x1) 0M(x2) using = 1.00x1, = 1.00L and setting P¿ = P. 0P¿ 0P¿ L
¢ =
L0
Ma L
(¢ A)h =
=
0M dx b 0P EI L
1 1 (Px1)(1.00x1) dx1 + (PL)(1.00L) dx2 EI L0 EI L0 4PL3 3EI
Ans.
1265
14 Solutions 46060_Part2
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•14–145.
Solve Prob. 14–121 using Castigliano’s theorem.
L
0M dx b = = Ma 0P E I L0 L0
¢C
=
A
a M0 a
a M0 (1x) x B (1x) dx + dx EI EI L0
5 M0 a2 6EI
Ans.
14–146. Determine the bending strain energy in the beam due to the loading shown. EI is constant.
P
a
L
Ui =
=
a
a
1 M 2 dx = c2 (Px1)2 dx1 + (Pa)2 dx2 d 2EI 2EI L0 L0 L0 5P2a3 6EI
Ans.
1266
P
a
a
14 Solutions 46060_Part2
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Page 1267
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14–147. The 200-kg block D is dropped from a height h = 1 m onto end C of the A-36 steel W200 * 36 overhang beam. If the spring at B has a stiffness k = 200 kN>m, determine the maximum bending stress developed in the beam.
D h A B 4m
Equilibrium. The support reactions and the moment functions for regions AB and BC of the beam under static conditions are indicated on the free-body diagram of the beam, Fig. a, Ue = Ui L
1 M 2dx P¢ st = © 2 L0 2EI 1 1 P¢ st = B 2 2EI L0
¢ st =
4m
a
2 P x2 b dx + 2 L0
2m
(Px1)2 dx R
8P EI
Here, I = 34.4 A 106 B mm4 = 34.4 A 10 - 6 B m4 (see the appendix) and E = Est = 200 GPa. Then, the equivalent spring constant can be determined from P = kb ¢ st P = kb a
8P b EI
EI = kb = 8
200 A 109 B c34.4 A 10 - 6 B d 8
= 860 A 103 B N>m
From the free-body diagram, Fsp =
3 P 2
ksp ¢ sp =
¢ sp =
3 (k ¢ ) 2 b b
3 3 kb 3 860 A 10 B ≥ ¢ b = 6.45 ¢ b ¢ ≤ ¢b = £ 2 ksp 2 200 A 103 B
(1)
Conservation of Energy. mga h + ¢ b +
3 1 1 ¢ b = ksp ¢ sp 2 + kb ¢ b 2 2 sp 2 2
1267
C
k 2m
14 Solutions 46060_Part2
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8:19 AM
Page 1268
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14–147.
Continued
Substiuting Eq. (1) into this equation. 200(9.81)c1 + ¢ b +
3 1 1 (6.45¢ b) d = c200 A 103 B d(6.45¢ b)2 + c860 A 103 B d ¢ b 2 2 2 2
4590.25 A 103 B ¢ b 2 - 20944.35¢ b - 1962 = 0 Solving for the positive root ¢ b = 0.02308 m Maximum
Stress.
The
maximum
force
on
the
beam
is
Pmax = kb ¢ b
= 860 A 103 B (0.02308) = 19.85 A 103 B N. The maximum moment occurs at the
Mmax = Pmax L = 19.85 A 103 B (2) = 39.70 A 103 B N # m. 0.201 d Applying the flexure formula with c = = = 0.1005 m. 2 2 supporting
smax =
spring,
where
39.70 A 103 B (0.1005) Mmax c = = 115.98 MPa = 116 MPa I 34.4 A 10 - 6 B
Since smax 6 sY = 250 MPa, this result is valid.
1268
Ans.
14 Solutions 46060_Part2
6/11/10
8:19 AM
Page 1269
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*14–148. Determine the maximum height h from which the 200-kg block D can be dropped without causing the A-36 steel W200 * 36 overhang beam to yield. The spring at B has a stiffness k = 200 kN>m.
D h A B 4m
Equilibrium. The support reactions and the moment functions for regions AB and BC of the beam under static conditions are indicated on the free-body diagram of the beam, Fig. a, Ue = Ui L
1 M 2dx P¢ st = © 2 L0 2EI 1 1 P¢ st = B 2 2EI L0 ¢ st =
4m
a
2 P x2 b dx + 2 L0
2m
(Px1)2 dx R
8P EI
Here, I = 34.4 A 106 B mm4 = 34.4 A 10 - 6 B m4 (see the appendix) and E = Est = 200 GPa. Then, the equivalent spring constant can be determined from P = kb ¢ st P = kb a
8P b EI
EI kb = = 8
200 A 109 B c34.4 A 10 - 6 B d 8
= 860 A 103 B N>m
From the free-body diagram, Fsp =
3 P 2
ksp ¢ sp =
¢ sp =
3 (k ¢ ) 2 b b
3 3 kb 3 860 A 10 B ≥ ¢ b = 6.45¢ b ¢ ≤ ¢b = £ 2 ksp 2 200 A 103 B
(1)
1269
C
k 2m
14 Solutions 46060_Part2
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Page 1270
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*14–148.
Continued
Maximum Stress. The maximum force on the beam is Pmax = kb ¢ b = 860 A 103 B ¢ b. The maximum moment occurs at the supporting spring, where Mmax = Pmax L = 860 A 103 B ¢ b(2) = 1720 A 103 B ¢ b. 0.201 d = = 0.1005 m, c = 2 2 smax =
Applying
the
flexure
formula
with
Mmaxc I
250 A 106 B =
1720 A 103 B ¢ b(0.1005) 34.4 A 10 - 6 B
¢ b = 0.04975 m Substituting this result into Eq. (1), ¢ sp = 0.3209 m Conservation of Energy. mga h + ¢ b +
3 1 1 ¢ b = ksp ¢ sp 2 + kb ¢ b 2 2 sp 2 2
200(9.81)ch + 0.04975 +
3 1 (0.3209) d = c200 A 103 B d(0.3209)2 2 2 +
1 c860 A 103 B d(0.04975)2 2
h = 5.26 m
Ans.
1270
14 Solutions 46060_Part2
6/11/10
8:19 AM
Page 1271
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•14–149.
The L2 steel bolt has a diameter of 0.25 in., and the link AB has a rectangular cross section that is 0.5 in. wide by 0.2 in. thick. Determine the strain energy in the link AB due to bending, and in the bolt due to axial force. The bolt is tightened so that it has a tension of 350 lb. Neglect the hole in the link.
6 in.
4 in.
0.2 in. A
B
8 in.
Bending strain energy: L
(Ub)i =
=
6
4
1 M 2 dx = c (140x1)2dx1 + (210x2)2dx2 d 2EI 2EI L0 L0 L0 1.176(106) 1.176(106) = 122 in # lb = 10.1 ft # lb = 1 EI 29(106)(12 )(0.5)(0.2 3)
Ans.
Axial force strain energy: L
(Ua)i =
(350)2(8) N 2 dx N 2L = 0.344 in # lb = = 2AE 2(29)(106)(p4 )(0.252) L0 2EA
Ans.
14–150. Determine the vertical displacement of joint A. Each bar is made of A-36 steel and has a cross-sectional area of 600 mm2. Use the conservation of energy.
B
C
Joint A:
2m
+ c ©Fy = 0;
4 F - 5 = 0 5 AB
FAB = 6.25 kN
D A
+ ©F = 0; ; x
FAD -
3 (6.25) = 0 5
1.5 m
FAD = 3.75 kN 5 kN
Joint B: + c ©Fy = 0;
4 4 F - (6.25) = 0 5 BD 5
+ ©F = 0; : x
3 FBC - 2a b (6.25) = 0 5
FBD = 6.25 kN
FBC = 7.5 kN
Conservation of energy: Ue = Ui N 2L 1 P¢ = © 2 2AE 1 1 (5) A 103 B ¢ A c A 6.25 A 103 B B 2(2.5) + A 3.75 A 103 B B 2(3) 2 2AE + A 6.25 A 103 B B 2(2.5) + A 7.5 A 103 B B 2(1.5) d ¢A =
64 375 64 375 = = 0.5364 A 10 - 3 B m = 0.536 mm -6 AE 600 A 10 B (200) A 109 B
1271
Ans.
1.5 m
14 Solutions 46060_Part2
6/11/10
8:19 AM
Page 1272
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14–151. Determine the total strain energy in the A-36 steel assembly. Consider the axial strain energy in the two 0.5-in.-diameter rods and the bending strain energy in the beam for which I = 43.4 in4. 3 ft
500 lb
Support Reactions: As shown FBD(a). Internal Moment Function: As shown on FBD(b). Total Strain Energy: L
(Ui)T =
4 ft
2
NL M dx + 2EI 2AE L0
= 2B
1 2EI L0
4 ft
(250x)2 dx R + 2 B
=
1.3333 A 106 B lb2 # ft3
=
1.3333 A 106 B A 12 3 B
+
EI
29.0 A 106 B (43.4)
+
p 4
2502(3) R 2AE
0.1875 A 106 B lb2 # ft AE
0.1875 A 106 B (12)
A 0.52 B C 29.0 A 106 B D
= 2.23 in # lb
Ans.
*14–152. Determine the vertical displacement of joint E. For each member A = 400 mm2, E = 200 GPa. Use the method of virtual work. Member
n
N
L
F
0
1.5
0
AE
– 0.8333
– 37.5
2.5
78.125
AB
0.6667
30.0
2.0
40.00
EF
0
0
2.0
0
EB
– 0.50
22.5
1.5
–16.875
ED
– 0.6667
– 30.0
2.0
40.00
BC
0
0
2.0
0
BD
0.8333
37.5
2.5
78.125
CD
– 0.5
– 22.5
1.5
16.875
A
2m
nNL AE
400 A 10 - 6 B (200) A 109 B
C
B 45 kN
© = 236.25
236.25 A 103 B
D
1.5 m
0
¢ Bv =
E
nNL
AF
1 # ¢ Bv = ©
4 ft
2
= 2.95 A 10 - 3 B = 2.95 mm
Ans.
1272
2m
14 Solutions 46060_Part2
6/11/10
8:19 AM
Page 1273
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•14–153.
Solve Prob. 14–152 using Castigliano’s theorem.
F
E
D
1.5 m
A
C
B 45 kN 2m
Member
N
0N>0P
N(P = 45)
L
2m
N(0N>0P)L
AF
0
0
0
1.5
0
AE
–(0.8333P + 37.5)
– 0.8333
–37.5
2.5
78.125
AB
0.6667P + 30
0.6667
30.0
2.0
40.00
BE
22.5–0.5P
– 0.5
22.5
1.5
–16.875
BD
0.8333P + 37.5
0.8333
37.5
2.5
78.125
BC
0
0
0
2.0
0
CD
–(0.5P + 22.5)
– 0.5
–22.5
1.5
16.875
DE
–(0.6667P + 30)
– 0.6667
–30.0
2.0
40.00
EF
0
0
0
2.0
0 © = 236.25
¢ Bv = ©N
=
236.25 0N L = 0P AE AE
236.25 A 103 B
400 A 10 - 6 B (200) A 109 B
= 2.95 A 10 - 3 B m = 2.95 mm
Ans.
14–154. The cantilevered beam is subjected to a couple moment M0 applied at its end. Determine the slope of the beam at B. EI is constant. Use the method of virtual work.
M0
A B L
L
uB =
=
L (1) M0 muM dx = dx EI L0 EI L0
M0L EI
Ans.
1273
14 Solutions 46060_Part2
6/11/10
8:19 AM
Page 1274
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14–155. Solve Prob. 14–154 using Castigliano’s theorem. M0
A L
uB =
=
L0
ma
B
L
M0(1) dm dy b = dx dm¿ EI EI L0
L
M0L EI
Ans.
*14–156. Determine the displacement of point B on the 3 aluminum beam. Eal = 10.6110 2 ksi. Use the conser
3 in.
3 kip
C B 12 ft
L
12(12)
M 2 dx 1 = (2) 2EI 2EI L0 L0
Ue =
1 1 P¢ = (3)¢ B = 1.5¢ B 2 2
(1.5x)2 dx =
2239488 EI
Conservation of energy: Ue = Ui
1.5¢ B =
2 239 488 EI
¢B =
1 492 992 EI
y =
0.5(7)(1) + (4)(6)(1) = 2.1154 in. 7(1) + 6(1)
I =
1 1 (7) A 13 B + (7)(1)(2.1154 - 0.5)2 + (1) A 63 B + (1)(6)(4 - 2.1154)2 12 12
= 58.16 in4 ¢B =
3 in. 1 in. 6 in.
A
Ui =
1 in.
1 492 992 = 2.42 in. (10.6)(103)(58.16)
Ans.
1274
12 ft
14 Solutions 46060_Part2
6/11/10
8:19 AM
Page 1275
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14–157. A 20-lb weight is dropped from a height of 4 ft onto the end of a cantilevered A-36 steel beam. If the beam is a W12 * 50, determine the maximum stress developed in the beam. 4 ft
12 ft
From Appendix C: ¢ st =
20(12(12))3 PL3 = = 1.742216 A 10 - 3 B in. 3EI 3(29) A 106 B (394)
n = 1 +
A
1 + 2a
h 4(12) b = 235.74 b = 1 + 1 + 2a ¢ st A 1.742216(10 - 3)
smax = nsst = 235.74 £
20(12)(12) A 12.19 2 B 394
≥ = 10503 psi = 10.5 ksi 6 sg O.K. Ans.
1275
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