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01 Solutions 46060
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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 3kip 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 800lb 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 freebody 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 600kg 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 freebody 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 freebody 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 600kg 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 freebody 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 freebody 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 freebody 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 freebody 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 freebody 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 freebody 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 300N forces act in the z direction and the 500N 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 freebody 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 freebody 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 crosssectional 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 (103) 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 crosssectional 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 builtup 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 crosssectional 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 crosssectional 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° fishmouth 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 crosssectional 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 crosssection 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 175lb woman stands on a vinyl floor wearing stiletto highheel 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 flatheeled 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.
Flatheeled 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 pinconnected members having the crosssectional 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 twotine grapple that supports the log having a mass of 3 Mg. The crosssectional 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 twotine 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 freebody diagram shown in Fig. a. + c ©Fy = 0;
P P +  N = 0 2 2
4 in.
N = P
Referring to the freebody 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 crosssectional 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 6mm 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 freebody 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 6mm 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 freebody 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 crosssectional 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 crosssectional 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 crosssectional 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 crosssectional 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 crosssectional 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 Cclamp 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 freebody 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 Cclamp 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 freebody 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
crosssectional
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 crosssectional 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 crosssectional 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 crosssectional 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 freebody 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 freebody 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 softride 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 freebody diagram of the softride 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 softride 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 freebody diagram of the softride 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 0.40in.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 200mmdiameter 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 30mmthick 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 20mmdiameter 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 yokeandrod 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 crosssectional 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 airfilled 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 = tan1 ¢
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 ex dx L 2
L0
x ex dx
L 1 1 1 2 2 =  c ex d 冷 =  c eL  d 2 2 2 0
=
x
2
x x
¢L =
F
1 2 [1  eL ] 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 secondorder 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. = 0.211032 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 linearelastic 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 crosssectional 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 A36 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 crosssectional 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 crosssectional 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 linearelastic 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 A36 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 A36 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 crosssectional 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 2014T6 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 2014T6 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 = 0.4511062 s ⫹ 0.36110122 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(106)s + 0.36(1012)s3, dP = A 0.45(106) + 1.08(1012) 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 = 0.4511062 s ⫹ 0.36110122 s3, where s is in kPa. Determine the modulus of toughness and the amount of elongation of a 200mmlong 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(106)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(106)s  0.36(1012)s3)ds 6873.52
= 0.12 s  0.225(106)s2  0.09(1012)s40 = 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 RambergOsgood 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(106)
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 2014T6 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 1.5in. side changed its length to 1.500132 in., determine Poisson’s ratio and the new length of the 2in. 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 Ti6A14V 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 crosssectional 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 6061T6 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 2014T6 aluminum cylinders having the unloaded lengths shown. If each cylinder has a diameter of 30 mm, determine the placement x of the applied 80kN 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 A36 steel and have a crosssectional 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 8mmdiameter 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 A36 steel wire AB has a crosssectional 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 A36 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 crosssectional 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 A36 steel rod is subjected to the loading shown. If the crosssectional 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
crosssectional 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 A36 steel rod is subjected to the loading shown. If the crosssectional 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 crosssectional 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 crosssectional 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 crosssectional 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 crosssectional 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 (Ti6A14V) rods and a rigid bar AC. The crosssectional 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 (Ti6A14V) rods and a rigid bar AC. The crosssectional 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 500lb load if the members were originally horizontal when the load was applied. Each wire has a crosssectional 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 500lb load is applied. The members were originally horizontal, and each wire has a crosssectional 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.
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•4–13. The bar has a length L and crosssectional 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 pinconnected A36 steel members, each having a crosssectional 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 pinconnected A36 steel members, each having a crosssectional 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 A36 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 crosssectional 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 A36 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 crosssectional 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 pinconnected rod CB that has a crosssectional area of 500 mm2 and is made of A36 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 springsupported 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 springsupported 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 A36 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. 424 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 crosssectional 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 500lb 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 + y21
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 300mm 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 freebody 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 crosssectional 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 highstrength concrete and six A36 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 highstrength concrete and six A36 steel reinforcing rods. If it is subjected to an axial force of 30 kip, determine the required diameter of each rod so that onefourth of the load is carried by the concrete and threefourths 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 20mmdiameter A36 steel segment AB and 50mmdiameter 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 20mmdiameter A36 steel segment AB and 50mmdiameter 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 A36 steel column, having a crosssectional area of 18 in2, is encased in highstrength 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 A36 steel column is encased in highstrength 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 A36 steel tie rods. Each rod has an unstretched length of 0.75 m and a crosssectional 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 A36 steel reinforcing rods. If it is subjected to an axial force of 900 kN, determine the required diameter of each rod so that onefifth of the load is carried by the steel and fourfifths 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 crosssectional 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 A36 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 freebody 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 A36 steel wires are used to support the 650lb 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 freebody 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 2014T6 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 crosssectional 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 crosssectional 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. 451.
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. 451, 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 A36 steel 12in.diameter rods. A 6061T6solidaluminum 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 onehalf turn, determine the average normal stress in the rods and in the cylinder. The singlethreaded 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 A36 steel 12in.diameter rods. A 6061T6solidaluminum 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 singlethreaded 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 A36 steel and have equal crosssectional 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 800lb load. Determine the normal stress in each A36 steel cable if each cable has a crosssectional 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 A36 steel cables each having a crosssectional area of 0.04 in2. Determine the rotation of the bar when the 800lb 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 A36 steel and having a crosssectional area of 1000 mm2, and a 2014T6 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 crosssectional 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 crosssectional 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 crosssectional 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 crosssectional 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 A36 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 crosssectional 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 A36 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 crosssectional 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 crosssectional 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 = 9.6011062>°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 crosssectional 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 A36 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 6ftlong steam pipe is made of A36 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 6ftlong steam pipe is made of A36 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 A36 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 40ftlong A36 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 crosssectional 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 A36 steel and 2014T6 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 crosssectional 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 A36 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 A36 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 A36 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 crosssectional 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 A36 steel and form a pinconnected truss. If the truss is constructed when T1 = 50°F, determine the force in each bar when T2 = 110°F. Each bar has a crosssectional 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 A36 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 crosssectional 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 150lb 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(106)>°F, acu = 9.60(106)>°F. Each wire has a crosssectional 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 AM1004T61 magnesium alloy tube AB is capped with a rigid plate E. The gap between E and end C of the 6061T6 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 aa
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 AM1004T61 magnesium alloy tube AB is capped with a rigid plate. The gap between E and end C of the 6061T6 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 aa
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(106)>°C, aal = 23(106)>°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. 1024. 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. 425. 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. 424. 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. 425. 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 1in. 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. 425, 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
187
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4–90. The A36 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. 424, 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. 425, 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. 424, 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. 425, 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.
189
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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 stressconcentration 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 stressconcentration 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 stressconcentration 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 300kip weight is slowly set on the top of a post made of 2014T6 aluminum with an A36 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.
191
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4–98. The bar has a crosssectional 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 A36 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 A36 steel as an elasticperfectly plastic material.
P
450 mm 150 mm 150 mm 30⬚ A
C
300 mm B
Equation of Equilibrium. Refering to the freebody 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.)
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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.
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4–102. The rigid lever arm is supported by two A36 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 A36 steel as an elasticperfectly plastic material.
P
450 mm 150 mm 150 mm 30⬚ A
C
300 mm B
Equation of Equilibrium. Refering to the freebody 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),
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4–103. The three bars are pinned together and subjected to the load P. If each bar has a crosssectional 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 25mm diameter A36 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 perfectlyplastic material.
D
F
E
600 mm P
Equation of Equilibrium. Referring to the freebody 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 25mm diameter A36 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 perfectlyplastic material.
D 600 mm
P A
400 mm
Equation of Equilibrium. Referring to the freebody 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.
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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 crosssectional 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 stressstrain 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 crosssectional 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 stressstrain 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, crosssectional 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 2014T6 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 2014T6 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 25mm diameter and 600mm length. If they are made of A36 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 freebody 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 A36 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 forcefitted 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 crosssectional 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 A36 steel wires, each having an unstretched length of 12 in. and crosssectional 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 1in.diameter hole is bored through the shaft? Sketch the shearstress 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 onehalf of the applied torque 1T>22. Solve the problem two ways: (a) by using the torsion formula, (b) by finding the resultant of the shearstress 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 Crosssections 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 shearstress 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
217
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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 30mmdiameter 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 50mmdiameter 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 shearstress 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 crosssection 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 freebody 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 60mmdiameter 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 crosssection 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 SaintVenant’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 A36 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(109)p m4. Thus, 2 C
300(0.02) TAB c = 23.87(106)Pa = 23.9 MPa = = J 80(109)p
A tBC B max =
200(0.02) TBC c = 15.92(106) Pa = 15.9 MPa = J 80(109)p
231
Ans.
Ans.
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300 N⭈m
*5–28. The A36 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(109) 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 20mmdiameter 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 25mmdiameter 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(109) m4. 2 tallow =
Tc ; J
75(106) =
T(0.0125) 38.3495(109)
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(109) m4. 2
tallow =
Tc ; J
75(106) =
a
P b(0.01) 50p
10.7379(109)
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 A36 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 A36 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(106)p m4. Thus, 2
T(0.03) 0.405(106)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 thinwalled 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 A36 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 A36 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 85N # 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 A36 steel propeller shaft that is 100 ft long. It is connected to an inline 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.
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•5–53. The 20mmdiameter A36 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 A36 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 A36 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 A36 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 A36 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 A36 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 A36 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 A36 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 A36 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 A36 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 A36 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 1indiameter 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 60mm diameter shaft ABC is supported by two journal bearings, while the 80mm 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 A36 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 60mm diameter shaft ABC is supported by two journal bearings, while the 80mm 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 A36 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 30mmdiameter 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 A36 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 A36 steel and each has a diameter of 80 mm. Determine the angle of twist of gear D.
0.6 m
Equilibrium: Referring to the freebody 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 80mm diameter shaft is made of 6061T6 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 freebody 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. 526, 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 A36 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 A36 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 A36 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$$$ =
2.838 (103)(0.04) TCD c = = 28.23 (106) Pa = 28.2 MPa p 4 J 2 (0.04)
271
Ans.
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•5–81. The shaft is made of A36 steel and has a diameter of 80 mm. It is fixed at B and the support at A has a torsional stiffness of k = 0.5 MN # m>rad. If it is subjected to the gear torques shown, determine the absolute maximum shear stress in the shaft.
2 kN⭈m B
4 kN⭈m D
600 mm C 600 mm A 600 mm
Referring to the FBD of the shaft shown in Fig. a, TA + TB + 2  4 = 0
©Mx = 0;
(1)
Using the method of superposition, Fig. b, fA = (fA)T  (fA)TA TA 6
0.5(10 )
= D
4(103)(0.6)
p 2
(0.04 ) C 75(10 ) D 4
9
2(103)(0.6)
+
p 2
(0.04 ) C 75(10 ) D 4
9
T 
TA(1.8)
p 2
(0.044) C 75(109) D
TA = 1498.01 N # m = 1.498 kN # m Substituting this result into Eq (1), TB = 0.502 kN # m Referring to the torque diagram shown in Fig. c, segment CD subjected to maximum internal torque. Thus, the maximum shear stress occurs here. t$$$ =
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 2in.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 A36 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 A36 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.
277
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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 A36 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 A36 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 3ftlong shafts are made of 2014T6 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 A36 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(106)TB  0.011112 18.52(106)TB  0.011112 =
TA(5) p 4 6 2 (0.25 )(11.0)(10 )
18.52(106)TB  74.08(106)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 2014T6 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 midpoint, 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 crosssectional 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 A36 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 A36 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 6061T6 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 halfcircular section is reversed from the dashedline 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 torquecarrying capacity is increased if the halfcircular sections are reversed from the dashedline 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 onefourth 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 2014T6 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 2014T6 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 2014T6 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 2014T6 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 highstrength 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 builtup 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. 532,
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 builtup 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. 532, 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. 532, 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. 532, 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. 526 from the text. T =
r3Y p tY 2p tYc3 (4 c3  r3Y) = a1 )b 6 3 4 c3
Use Eq. 524, 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 elasticplastic 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. 526 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. 527 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 3mmdiameter 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. 527 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 elasticperfectly 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
ElasticPlastic Torque: Applying Eq. 526 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 80mm diameter solid circular shaft is made of an elasticperfectly 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 elasticperfectly 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 shearstress 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. 527 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 elasticperfectly 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 tg 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 strainhardening material having a tg 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 50mmdiameter 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 elasticperfectly plastic material, which has the tg diagram shown. If the radius of the elastic core is rY = 2.25 in., determine the applied torque T. Also, find the residual shearstress 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 elasticperfectly plastic material, which has the tg diagram shown. Determine the torque T that just causes the inner surface of the shaft to yield. Also, find the residual shearstress 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.
317
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*5–140. The 2mlong tube is made of an elasticperfectly 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 tg 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 thinwalled 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 A36 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. 57, 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. 518, tavg =
10(103) T = 88.42 MPa = 2 t Am 29(0.005)(p)(0.062)
Ans.
Angle of Twist: Applying Eq. 515, f =
TL JG 10(103)(4)
=
p 4 2 (0.0625
 0.05754)(75.0)(109)
= 0.0785 rad = 4.495°
Ans.
Applying Eq. 520, 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 A36 steel with an allowable shear stress of 1tallow2st = 75 MPa, and tube BC is made of AM1004T61 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 2014T6 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 2in.diameter A36 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 10mm 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 freebody 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 freebody 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 20kip 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 freebody 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 (twoforce 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 freebody 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 freebody 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 freebody 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 freebody 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 180lb 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 deadweight 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 crosssection 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.
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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 threedimensional 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 crosssectional 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 crosssectional 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 threedimensional view of the stress distribution acting over the entire crosssectional 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 threedimensional 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 crosssection 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 threedimensional 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 threedimensional view of the stress distribution acting over the cross section.
A
1.5 in. 12 in.
The moment of inertia of the crosssection 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 crosssection 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 crosssectional 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 crosssectional 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 boxbeam 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 freebody 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 crosssection 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 crosssection is shown in Fig. a.
373
Ans.
30 mm 30 mm
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•6–77.
The steel beam has the crosssectional 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 crosssectional 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. 678: 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 freebody 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 2in.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 crosssection 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 crosssection 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 80mmdiameter 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 crosssection 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 crosssection 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 crosssectional 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 crosssection 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 crosssectional 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 crosssection 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 crosssection 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 crosssection 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 crosssection 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 crosssection 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 crosssection 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 crosssectional 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 crosssectional 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 bendingmoment components My and Mz, as shown, when the x, y, z axes pass through the centroid of the cross section. If the material is linearelastic, 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 Zsection having the crosssection 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. 6113. 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 Zsection having the crosssection 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. 6113. 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 wideflange 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 wideflange 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 crosssection 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.
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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 crosssection 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 30mmdiameter 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 = 0.06011032 m4 and Iz = 0.47111032 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. 6113 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 = 0.06011032 m4 and Iz = 0.47111032 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 6061T6 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 6061T6 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
50 mm
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•6–129. Segment A of the composite beam is made from 2014T6 aluminum alloy and segment B is A36 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 simplysupported 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 2014T6 aluminum alloy and segment B is A36 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 simplysupported 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 A36 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 2014T6 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 2014T6 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 A36 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 A36 steel section A and the 2014T6 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 20mm 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 wideflange A36 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 A36 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. 1in. 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 curvedbeam formula, Eq. 6–24, reduces to the flexure formula, Eq. 6–13. Normal Stress: Curvedbeam 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 curvedbeam 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 curvedbeam 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 curvedbeam 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 curvedbeam 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 aa.
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 aa. 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 ceilingsuspended Carm is used to support the Xray 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 curvedbeam 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 curvedbeam 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 curvedbeam 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 twodimensional 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. 644: 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. 644: 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 A36 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. 644. 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 bendingstress 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. 644, 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. 643, 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 wideflange 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 crosssection 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 crosssection 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 elasticperfectly 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 crosssection 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 crosssection 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 Hbeam.
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 Hbeam is made of an elasticplastic 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 crosssection 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 elasticperfectly 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 crosssection 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 crosssection 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 crosssection can be determined by super posing the moment of the stress block of the solid circular crosssection 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 elasticperfectly plastic material. Determine the shape factor for the thickwalled tube. ro
Maximum Elastic Moment. The moment of inertia of the crosssection 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 elasticplastic 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 elasticplastic 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 elasticplastic 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 crosssection 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 tan1115P2] ksi, where tan1115P2 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 straightline 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 elasticplastic 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 threedimensional view of the stress distribution acting over the crosssectional 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 crosssectional 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 wideflange beam is subjected to a shear of V = 20 kN, determine the shear stress on the web at A. Indicate the shearstress 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 crosssection 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 wideflange 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 crosssection 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.
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7–3. If the wideflange 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 crosssection 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.
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*7–4. If the Tbeam is subjected to a vertical shear of V = 12 kip, determine the maximum shear stress in the beam. Also, compute the shearstress jump at the flangeweb junction AB. Sketch the variation of the shearstress 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 Tbeam 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 shearstress 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 = 0.218211032 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 wideflange 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 wideflange 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 shearstress variation over the cross section.
V 12 in.
8 in.
Section Properties The moment of inertia of the crosssection 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 shearstress 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. 73.
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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 crosssection 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.
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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 crosssection 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 crosssection 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, ty = 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, ty = 0 = 37.4 MPa
ty = 0.075 m = 27.7 MPa
The plot shear stress distribution over the crosssection is shown in Fig. c.
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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 crosssection 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 crosssection 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 Tbeam 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 crosssection, 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 Tbeam 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 crosssection, 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 crosssection 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 elasticplastic, 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 500lb 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 450lb 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 crosssection 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 crosssection 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 crosssection 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 doubleweb 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 doubleweb 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 Tbeam 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 crosssection 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 crosssection 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 crosssection 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 crosssection 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 crosssection 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 shearstress variation over the cross section of the thinwalled 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 pu
Q =
pu
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 thinwalled 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 thinwalled 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 thinwalled 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 thinwalled 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 thinwalled 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 + Isemicircle 12 4
= 1.583333t r3 + Isemicircle p>2
Isemicircle =
p>2 2
Lp>2
(r sin u) t r du = t r3
Lp>2
sin2 u du
p Isemicircle = 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 =
Lp>2
(q r du)r p>2
Pe =
e =
Pr (0.625 + cos u)du 3.15413 Lp>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 thinwalled 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 thinwalled 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)
Lpa
= r2 t(cos u  cos a) = r2 t(cos u + cos a)
pa
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 shearstress distribution acting over the beam’s crosssectional 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 thinwalled 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 0.5in.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 thinwalled 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, thinwall 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 freebody 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 8m 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 25mm 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, thinwall analysis is valid. t 0.02667
Referring
the freebody 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 8mm thick steel plates that are fastened together at their ends using a butt joint consisting of two 8mm 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, thinwall analysis is valid. t
Referring to the freebody 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, thinwall 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 freebody 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 0.25in.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 12in. 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 freebody 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 normalstress distribution acting over section a–a if the member has a rectangular crosssectional 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 crosssectional 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 75kg 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 freebody 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 freebody 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 crosssectional 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 wideflange 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 freebody 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 crosssectional 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 freebody 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 crosssectional 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.
561
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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 crosssectional 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 crosssectional area and the moment of inertia about the y and z axes of the crosssection 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 crosssection 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 freebody 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 Cclamp 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 curvedbeam 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 curvedbeam 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 800kN 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 800kN 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.
584
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•8–57. The 2in.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 2in.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 crosssectional 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 crosssectional 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 crosssectional 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 crosssectional 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 43in.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 curvedbeam 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 curvedbeam 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 20kg 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 freebody 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 20kg 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 freebody 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 aa. The cross section is circular and has a diameter of 0.25 in. Use the curvedbeam 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 freebody 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 crosssectional 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 freebody 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, thinwall 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 freebody 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 aa. 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 fla