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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
(b) + c ©Fy = 0;
8 in.
8 in.
3 kip
3 kip
Ans.
FA - 4.5 - 4.5 - 5.89 - 6 - 6 - 8 = 0
FA = 34.9 kN
3m
200 mm
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
3 ft
C
500 lbft
1 ft
TC = 500 lb # ft
Ans.
2 ft
2 ft
1
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*1–4. A force of 80 N is supported by the bracket as
shown. Determine the resultant internal loadings acting on
the section through point A.
0.3 m
A
30
0.1 m
80 N
Equations of Equilibrium:
+
Q©Fx¿ = 0;
NA - 80 cos 15° = 0
NA = 77.3 N
a+ ©Fy¿ = 0;
Ans.
VA - 80 sin 15° = 0
VA = 20.7 N
a+
©MA = 0;
Ans.
MA + 80 cos 45°(0.3 cos 30°)
- 80 sin 45°(0.1 + 0.3 sin 30°) = 0
MA = - 0.555 N # m
Ans.
or
a+
©MA = 0;
MA + 80 sin 15°(0.3 + 0.1 sin 30°)
-80 cos 15°(0.1 cos 30°) = 0
MA = - 0.555 N # m
Ans.
Negative sign indicates that MA acts in the opposite direction to that shown on FBD.
2
45
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•1–5.
Determine the resultant internal loadings in the
beam at cross sections through points D and E. Point E is
just to the right of the 3-kip load.
3 kip
1.5 kip/ ft
A
D
6 ft
Support Reactions: For member AB
a + ©MB = 0;
+
: ©Fx = 0;
+ c ©Fy = 0;
9.00(4) - A y(12) = 0
A y = 3.00 kip
Bx = 0
By + 3.00 - 9.00 = 0
By = 6.00 kip
Equations of Equilibrium: For point D
+
: ©Fx = 0;
+ c ©Fy = 0;
ND = 0
Ans.
3.00 - 2.25 - VD = 0
VD = 0.750 kip
a + ©MD = 0;
Ans.
MD + 2.25(2) - 3.00(6) = 0
MD = 13.5 kip # ft
Ans.
Equations of Equilibrium: For point E
+
: ©Fx = 0;
+ c ©Fy = 0;
NE = 0
Ans.
- 6.00 - 3 - VE = 0
VE = - 9.00 kip
a + ©ME = 0;
Ans.
ME + 6.00(4) = 0
ME = - 24.0 kip # ft
Ans.
Negative signs indicate that ME and VE act in the opposite direction to that shown
on FBD.
3
E
B
6 ft
4 ft
4 ft
C
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1–6. Determine the normal force, shear force, and moment
at a section through point C. Take P = 8 kN.
B
0.1 m
Support Reactions:
a + ©MA = 0;
8(2.25) - T(0.6) = 0
+
: ©Fx = 0;
30.0 - A x = 0
A x = 30.0 kN
+ c ©Fy = 0;
Ay - 8 = 0
A y = 8.00 kN
0.5 m
C
T = 30.0 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
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.
+ c ©Fy = 0;
7.50 - 6 - VC = 0
VC = 1.50 kN
a + ©MC = 0;
MC + 6(0.5) - 7.5(1) = 0
Ans.
MC = 4.50 kN # m
Ans.
•1–9.
Determine the resultant internal loadings on the
cross section through point D. Assume the reactions at
the supports A and B are vertical.
6 kN
3 kN/m
Referring to the FBD of the entire beam, Fig. a,
B
A
a + ©MA = 0;
By(4) - 6(0.5) -
1
(3)(3)(2) = 0
2
By = 3.00 kN
0.5 m 0.5 m
Referring to the FBD of this segment, Fig. b,
+
: ©Fx = 0;
+ c ©Fy = 0;
ND = 0
VD -
1
(1.5)(1.5) + 3.00 = 0
2
a + ©MD = 0; 3.00(1.5) -
C
Ans.
VD = - 1.875 kN
Ans.
1
(1.5)(1.5)(0.5) - MD = 0 MD = 3.9375 kN # m
2
= 3.94 kN # m
5
Ans.
D
1.5 m
1.5 m
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1–10. The boom DF of the jib crane and the column DE
have a uniform weight of 50 lb/ft. If the hoist and load weigh
300 lb, determine the resultant internal loadings in the crane
on cross sections through points A, B, and C.
D
2 ft
F
A
B
8 ft
3 ft
5 ft
C
300 lb
7 ft
E
Equations of Equilibrium: For point A
+
; © Fx = 0;
+ c © Fy = 0;
NA = 0
Ans.
VA - 150 - 300 = 0
VA = 450 lb
a + ©MA = 0;
Ans.
- MA - 150(1.5) - 300(3) = 0
MA = - 1125 lb # ft = - 1.125 kip # ft
Ans.
Negative sign indicates that MA acts in the opposite direction to that shown on FBD.
Equations of Equilibrium: For point B
+
; © Fx = 0;
NB = 0
+ c © Fy = 0;
VB - 550 - 300 = 0
Ans.
VB = 850 lb
a + © MB = 0;
Ans.
- MB - 550(5.5) - 300(11) = 0
MB = - 6325 lb # ft = - 6.325 kip # ft
Ans.
Negative sign indicates that MB acts in the opposite direction to that shown on FBD.
Equations of Equilibrium: For point C
+
; © Fx = 0;
+ c © Fy = 0;
VC = 0
Ans.
- NC - 250 - 650 - 300 = 0
NC = - 1200 lb = - 1.20 kip
a + ©MC = 0;
Ans.
- MC - 650(6.5) - 300(13) = 0
MC = - 8125 lb # ft = - 8.125 kip # ft
Ans.
Negative signs indicate that NC and MC act in the opposite direction to that shown
on FBD.
6
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1–11. The force F = 80 lb acts on the gear tooth.
Determine the resultant internal loadings on the root of the
tooth, i.e., at the centroid point A of section a–a.
a
F 80 lb
30
Equations of Equilibrium: For section a–a
+
Q©Fx¿ = 0;
VA - 80 cos 15° = 0
0.23 in.
VA = 77.3 lb
a+ ©Fy¿ = 0;
Ans.
A
NA - 80 sin 15° = 0
0.16 in.
NA = 20.7 lb
a + ©MA = 0;
Ans.
- MA - 80 sin 15°(0.16) + 80 cos 15°(0.23) = 0
MA = 14.5 lb # in.
45
Ans.
*1–12. The sky hook is used to support the cable of a
scaffold over the side of a building. If it consists of a smooth
rod that contacts the parapet of a wall at points A, B, and C,
determine the normal force, shear force, and moment on
the cross section at points D and E.
a
0.2 m
0.2 m
B
0.2 m
0.2 m
D
E
Support Reactions:
+ c ©Fy = 0;
NB - 18 = 0
d+ ©MC = 0;
18(0.7) - 18.0(0.2) - NA(0.1) = 0
0.2 m
0.3 m
NB = 18.0 kN
A
NA = 90.0 kN
+
: ©Fx = 0;
NC - 90.0 = 0
0.3 m
NC = 90.0 kN
Equations of Equilibrium: For point D
+
: © Fx = 0;
18 kN
VD - 90.0 = 0
VD = 90.0 kN
+ c © Fy = 0;
Ans.
ND - 18 = 0
ND = 18.0 kN
d+ © MD = 0;
Ans.
MD + 18(0.3) - 90.0(0.3) = 0
MD = 21.6 kN # m
Ans.
Equations of Equilibrium: For point E
+
: © Fx = 0;
90.0 - VE = 0
VE = 90.0 kN
+ c © Fy = 0;
d + © ME = 0;
Ans.
NE = 0
Ans.
90.0(0.2) - ME = 0
ME = 18.0 kN # m
Ans.
7
C
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•1–13.
The 800-lb load is being hoisted at a constant speed
using the motor M, which has a weight of 90 lb. Determine
the resultant internal loadings acting on the cross section
through point B in the beam. The beam has a weight of
40 lb>ft and is fixed to the wall at A.
M
1.5 ft
A
D
4 ft
+
: ©Fx = 0;
3 ft
3 ft
4 ft
0.25 ft
Ans.
VB - 0.8 - 0.16 = 0
VB = 0.960 kip
a + ©MB = 0;
B
- NB - 0.4 = 0
NB = - 0.4 kip
+ c ©Fy = 0;
4 ft
C
Ans.
- MB - 0.16(2) - 0.8(4.25) + 0.4(1.5) = 0
MB = - 3.12 kip # ft
Ans.
1–14. Determine the resultant internal loadings acting on
the cross section through points C and D of the beam in
Prob. 1–13.
M
1.5 ft
A
D
4 ft
4 ft
C
B
3 ft
3 ft
4 ft
For point C:
+
; ©Fx = 0;
+ c ©Fy = 0;
a + ©MC = 0;
NC + 0.4 = 0;
NC = - 0.4kip
VC - 0.8 - 0.04 (7) = 0;
Ans.
VC = 1.08 kip
Ans.
- MC - 0.8(7.25) - 0.04(7)(3.5) + 0.4(1.5) = 0
MC = - 6.18 kip # ft
Ans.
ND = 0
Ans.
For point D:
+
; ©Fx = 0;
+ c ©Fy = 0;
VD - 0.09 - 0.04(14) - 0.8 = 0;
VD = 1.45 kip
a + ©MD = 0;
- MD - 0.09(4) - 0.04(14)(7) - 0.8(14.25) = 0
MD = - 15.7 kip # ft
Ans.
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;
- VC + 60 = 0;
VC = 60 N
+
: ©Fx = 0;
NC = 0
+d ©MC = 0;
- MC + 60(0.015) = 0;
Ans.
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;
Vb - b - 5 sin 45° = 0
Vb - b = 3.536 kN = 3.54 kN
a + ©MC = 0;
5.303 sin 45° (3) - 5(1.5) - Mb - b = 0
Mb - b = 3.75 kN # m
10
Ans.
Ans.
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
a + ©MC = 0;
MC + 80(6) = 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
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
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
1
(6)(6) - VD = 0
2
+ c ©Fy = 0;
18.0 -
a + ©MA = 0;
MD - 18.0 (2) = 0
6 kip/ft
3 ft
B
D
3 ft
6 ft
Ans.
VD = 0
Ans.
MD = 36.0 kip # ft
Ans.
The forged steel clamp exerts a force of F = 900 N
on the wooden block. Determine the resultant internal
loadings acting on section a–a passing through point A.
•1–21.
200 mm
F 900 N
Internal Loadings: Referring to the free-body diagram of the section of the clamp
shown in Fig. a,
©Fy¿ = 0;
900 cos 30° - Na - a = 0
Na - a = 779 N
Ans.
©Fx¿ = 0;
Va - a - 900 sin 30° = 0
Va - a = 450 N
Ans.
a + ©MA = 0;
900(0.2) - Ma - a = 0
Ma - a = 180 N # m
Ans.
12
a
30
a
A
F 900 N
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1–22. The floor crane is used to lift a 600-kg concrete pipe.
Determine the resultant internal loadings acting on the
cross section at G.
0.2 m
0.2 m
0.4 m
0.6 m
G
E
B
F
0.3 m
H
C
0.5 m
75
A
Support Reactions: We will only need to compute FEF by writing the moment
equation of equilibrium about D with reference to the free-body diagram of the
hook, Fig. a.
a + ©MD = 0;
FEF(0.3) - 600(9.81)(0.5) = 0
FEF = 9810 N
Internal Loadings: Using the result for FEF, section FG of member EF will be
considered. Referring to the free-body diagram, Fig. b,
+
: ©Fx = 0;
+ c ©Fy = 0;
a + ©MG = 0;
9810 - NG = 0
NG = 9810 N = 9.81 kN
VG = 0
Ans.
Ans.
MG = 0
Ans.
13
D
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1–23. The floor crane is used to lift a 600-kg concrete pipe.
Determine the resultant internal loadings acting on the
cross section at H.
0.2 m
0.2 m
0.4 m
0.6 m
G
E
B
F
0.3 m
H
C
0.5 m
75
A
Support Reactions: Referring to the free-body diagram of the hook, Fig. a.
a + ©MF = 0;
Dx(0.3) - 600(9.81)(0.5) = 0
Dx = 9810 N
+ c ©Fy = 0;
Dy - 600(9.81) = 0
Dy = 5886 N
Subsequently, referring to the free-body diagram of member BCD, Fig. b,
a + ©MB = 0;
FAC sin 75°(0.4) - 5886(1.8) = 0
+
: ©Fx = 0;
Bx + 27 421.36 cos 75° - 9810 = 0 Bx = 2712.83 N
+ c ©Fy = 0;
27 421.36 sin 75° - 5886 - By = 0
FAC = 27 421.36 N
By = 20 601 N
Internal Loadings: Using the results of Bx and By, section BH of member BCD will
be considered. Referring to the free-body diagram of this part shown in Fig. c,
+
: ©Fx = 0;
NH + 2712.83 = 0
NH = - 2712.83 N = - 2.71 kN
Ans.
+ c ©Fy = 0;
- VH - 2060 = 0
VH = - 20601 N = - 20.6 kN
Ans.
a + ©MD = 0;
MH + 20601(0.2) = 0
MH = - 4120.2 N # m
= - 4.12 kN # m Ans.
The negative signs indicates that NH, VH, and MH act in the opposite sense to that
shown on the free-body diagram.
14
D
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*1–24. The machine is moving with a constant velocity. It
has a total mass of 20 Mg, and its center of mass is located at
G, excluding the front roller. If the front roller has a mass of
5 Mg, determine the resultant internal loadings acting on
point C of each of the two side members that support the
roller. Neglect the mass of the side members. The front
roller is free to roll.
2m
G
C
B
A
1.5 m
Support Reactions: We will only need to compute NA by writing the moment
equation of equilibrium about B with reference to the free-body diagram of the
steamroller, Fig. a.
a + ©MB = 0; NA (5.5) - 20(103)(9.81)(1.5) = 0
NA = 53.51(103) N
Internal Loadings: Using the result for NA, the free-body diagram of the front roller
shown in Fig. b will be considered.
+
; ©Fx = 0; 2NC = 0
NC = 0
+ c ©Fy = 0; 2VC + 53.51(103) - 5(103)(9.81) = 0
VC = - 2229.55 N
= - 2.23 kN
Ans.
Ans.
a + ©MC = 0; 53.51(103)(2) - 5(103)(9.81)(2) - 2MC = 0 MC = 4459.10 N # m
= 4.46 kN # m Ans.
15
4m
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z
•1–25.
Determine the resultant internal loadings acting on
the cross section through point B of the signpost. The post is
fixed to the ground and a uniform pressure of 7 lb>ft2 acts
perpendicular to the face of the sign.
3 ft
2 ft
©Fx = 0;
(VB)x - 105 = 0;
(VB)x = 105 lb
©Fy = 0;
(VB)y = 0
Ans.
©Fz = 0;
(NB)z = 0
Ans.
©Mx = 0;
(MB)x = 0
Ans.
©My = 0;
(MB)y - 105(7.5) = 0;
(MB)y = 788 lb # ft
Ans.
©Mz = 0;
(TB)z - 105(0.5) = 0;
(TB)z = 52.5 lb # ft
Ans.
Ans.
3 ft
7 lb/ft2
6 ft
B
A
4 ft
y
x
1–26. The shaft is supported at its ends by two bearings A
and B and is subjected to the forces applied to the pulleys
fixed to the shaft. Determine the resultant internal
loadings acting on the cross section located at point C. The
300-N forces act in the z direction and the 500-N forces
act in the x direction. The journal bearings at A and B
exert only x and z components of force on the shaft.
z
A
400 mm
150 mm
200 mm
C
250 mm
x
300 N
300 N
B
500 N
500 N
y
©Fx = 0;
(VC)x + 1000 - 750 = 0;
(VC)x = - 250 N
©Fy = 0;
(NC)y = 0
©Fz = 0;
(VC)z + 240 = 0;
©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.
Ans.
(VC)z = - 240 N
(MC)x = - 108 N # m
Ans.
Ans.
(MC)z = - 138 N # m Ans.
16
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1–27. The pipe has a mass of 12 kg>m. If it is fixed to the
wall at A, determine the resultant internal loadings acting on
the cross section at B. Neglect the weight of the wrench CD.
z
A
300 mm
200 mm
B
60 N
x
D
400 mm
60 N
150 mm
C
©Fx = 0;
(NB)x = 0
Ans.
©Fy = 0;
(VB)y = 0
Ans.
©Fz = 0;
(VB)z - 60 + 60 - (0.2)(12)(9.81) - (0.4)(12)(9.81) = 0
Ans.
(VB)z = 70.6 N
©Mx = 0;
(TB)x + 60(0.4) - 60(0.4) - (0.4)(12)(9.81)(0.2) = 0
(TB)x = 9.42 N # m
©My = 0;
Ans.
(MB)y + (0.2)(12)(9.81)(0.1) + (0.4)(12)(9.81)(0.2) - 60(0.3) = 0
(MB)y = 6.23 N # m
©Mz = 0;
Ans.
Ans.
(MB)z = 0
17
150 mm
y
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z
*1–28. The brace and drill bit is used to drill a hole at O. If
the drill bit jams when the brace is subjected to the forces
shown, determine the resultant internal loadings acting on
the cross section of the drill bit at A.
O
x
Internal Loading: Referring to the free-body diagram of the section of the drill and
brace shown in Fig. a,
©Fx = 0;
©Fy = 0;
©Fz = 0;
©Mx = 0;
©My = 0;
©Mz = 0;
A VA B x - 30 = 0
A NA B y - 50 = 0
A VA B z - 10 = 0
A MA B x - 10(2.25) = 0
A TA B y - 30(0.75) = 0
A MA B z + 30(1.25) = 0
A VA B x = 30 lb
Ans.
A NA B y = 50 lb
Ans.
A VA B z = 10 lb
Ans.
A MA B x = 22.5 lb # ft
A TA B y = 22.5 lb # ft
A MA B z = - 37.5 lb # ft
Ans.
Ans.
Ans.
The negative sign indicates that (MA)Z acts in the opposite sense to that shown on
the free-body diagram.
18
3 in. 9 in.
Fx 30 lb
A
Fz 10 lb
9 in.
6 in.
6 in.
6 in.
Fy 50 lb
y
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•1–29. The curved rod has a radius r and is fixed to the
wall at B. Determine the resultant internal loadings acting
on the cross section through A which is located at an angle u
from the horizontal.
B
A
r
U
Equations of Equilibrium: For point A
R+ ©Fx = 0;
P
P cos u - NA = 0
NA = P cos u
Q+ ©Fy = 0;
Ans.
VA - P sin u = 0
VA = P sin u
d+ ©MA = 0;
Ans.
MA - P[r(1 - cos u)] = 0
MA = Pr(1 - cos u)
Ans.
19
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1–30. A differential element taken from a curved bar is
shown in the figure. Show that dN>du = V, dV>du = - N,
dM>du = - T, and dT>du = M.
M dM
V dV
©Fx = 0;
(1)
N
©Fy = 0;
N sin
du
du
du
du
- V cos
+ (N + dN) sin
+ (V + dV) cos
= 0
2
2
2
2
(2)
©Mx = 0;
T cos
du
du
du
du
+ M sin
- (T + dT) cos
+ (M + dM) sin
= 0
2
2
2
2
(3)
©My = 0;
du
du
du
du
- M cos
+ (T + dT) sin
+ (M + dM) cos
= 0
2
2
2
2
du
du
du
du
Since
is can add, then sin
, cos
=
= 1
2
2
2
2
T sin
Eq. (1) becomes Vdu - dN +
dVdu
= 0
2
Neglecting the second order term, Vdu - dN = 0
dN
= V
du
Eq. (2) becomes Ndu + dV +
QED
dNdu
= 0
2
Neglecting the second order term, Ndu + dV = 0
dV
= -N
du
Eq. (3) becomes Mdu - dT +
QED
dMdu
= 0
2
Neglecting the second order term, Mdu - dT = 0
dT
= M
du
Eq. (4) becomes Tdu + dM +
QED
dTdu
= 0
2
Neglecting the second order term, Tdu + dM = 0
dM
= -T
du
N dN
M V
du
du
du
du
+ V sin
- (N + dN) cos
+ (V + dV) sin
= 0
N cos
2
2
2
2
QED
20
(4)
T
T dT
du
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1–31. The column is subjected to an axial force of 8 kN,
which is applied through the centroid of the cross-sectional
area. Determine the average normal stress acting at section
a–a. Show this distribution of stress acting over the area’s
cross section.
8 kN
75 mm
75 mm
10 mm
10 mm
70 mm
10 mm
70 mm
a
a
A = (2)(150)(10) + (140)(10)
= 4400 mm2 = 4.4 (10-3) m2
s =
8 (103)
P
= 1.82 MPa
=
A
4.4 (10 - 3)
Ans.
*1–32. The lever is held to the fixed shaft using a tapered
pin AB, which has a mean diameter of 6 mm. If a couple is
applied to the lever, determine the average shear stress in
the pin between the pin and lever.
B 12 mm
A
250 mm
20 N
a + ©MO = 0;
tavg =
- F(12) + 20(500) = 0;
F = 833.33 N
V
833.33
= p 6
= 29.5 MPa
A
( )2
Ans.
4 1000
21
250 mm
20 N
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•1–33.
The bar has a cross-sectional area A and is
subjected to the axial load P. Determine the average
normal and average shear stresses acting over the shaded
section, which is oriented at u from the horizontal. Plot the
variation of these stresses as a function of u 10 … u … 90°2.
P
P
u
A
Equations of Equilibrium:
R+ ©Fx = 0;
V - P cos u = 0
V = P cos u
Q+ ©Fy = 0;
N - P sin u = 0
N = P sin u
Average Normal Stress and Shear Stress: Area at u plane, A¿ =
s =
A
.
sin u
P sin u
N
P
=
=
sin2 u
A
A¿
A
sin u
tavg =
Ans.
V
P cos u
=
A
A¿
sin u
=
P
P
sin u cos u =
sin 2u
A
2A
Ans.
1–34. The built-up shaft consists of a pipe AB and solid
rod BC. The pipe has an inner diameter of 20 mm and outer
diameter of 28 mm. The rod has a diameter of 12 mm.
Determine the average normal stress at points D and E and
represent the stress on a volume element located at each of
these points.
4 kN
D
At D:
sD =
4(103)
P
= 13.3 MPa (C)
= p
2
2
A
4 (0.028 - 0.02 )
Ans.
8(103)
P
= 70.7 MPa (T)
= p
2
A
4 (0.012 )
Ans.
At E:
sE =
B
A
22
6 kN
6 kN E
C
8 kN
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1–35. The bars of the truss each have a cross-sectional
area of 1.25 in2. Determine the average normal stress in
each member due to the loading P = 8 kip. State whether
the stress is tensile or compressive.
B
C
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.
Joint E:
sED =
FED
10.67
=
= 8.53 ksi
A ED
1.25
(C)
Ans.
sEB =
FEB
6.0
=
= 4.80 ksi
A EB
1.25
(T)
Ans.
Joint B:
sBC =
FBC
29.33
= 23.5 ksi
=
A BC
1.25
(T)
Ans.
sBD =
FBD
23.33
=
= 18.7 ksi
A BD
1.25
(C)
Ans.
23
E
0.75 P
4 ft
D
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*1–36. The bars of the truss each have a cross-sectional
area of 1.25 in2. If the maximum average normal stress in
any bar is not to exceed 20 ksi, determine the maximum
magnitude P of the loads that can be applied to the truss.
B
3 ft
A
4 ft
P
Joint A:
+ c ©Fy = 0;
3
- P + a b FAB = 0
5
FAB = (1.667)P
+
: ©Fx = 0;
4
- FAE + (1.667)Pa b = 0
5
FAE = (1.333)P
Joint E:
+ c ©Fy = 0;
FEB - (0.75)P = 0
FEB = (0.75)P
+
: ©Fx = 0;
(1.333)P - FED = 0
FED = (1.333)P
Joint B:
+ c ©Fy = 0;
3
3
a b FBD - (0.75)P - (1.667)Pa b = 0
5
5
FBD = (2.9167)P
+
: ©Fx = 0;
4
4
FBC - (2.9167)Pa b - (1.667)P a b = 0
5
5
FBC = (3.67)P
The highest stressed member is BC:
sBC =
C
(3.67)P
= 20
1.25
P = 6.82 kip
Ans.
24
E
0.75 P
4 ft
D
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•1–37.
The plate has a width of 0.5 m. If the stress distribution at the support varies as shown, determine the force P
applied to the plate and the distance d to where it is applied.
4m
P
d
x
s (15x1/2) MPa
The resultant force dF of the bearing pressure acting on the plate of area dA = b dx
= 0.5 dx, Fig. a,
1
1
dF = sb dA = (15x2)(106)(0.5dx) = 7.5(106)x2 dx
+ c ©Fy = 0;
L
dF - P = 0
4m
1
L0
7.5(106)x2 dx - P = 0
P = 40(106) N = 40 MN
Ans.
Equilibrium requires
a + ©MO = 0;
L
xdF - Pd = 0
4m
1
L0
x[7.5(106)x2 dx] - 40(106) d = 0
d = 2.40 m
Ans.
25
30 MPa
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1–38. The two members used in the construction of an
aircraft fuselage are joined together using a 30° fish-mouth
weld. Determine the average normal and average shear
stress on the plane of each weld. Assume each inclined
plane supports a horizontal force of 400 lb.
N - 400 sin 30° = 0;
N = 200 lb
400 cos 30° - V = 0;
V = 346.41 lb
1.5 in.
1 in.
1 in.
800 lb
800 lb
30
A¿ =
1.5(1)
= 3 in2
sin 30°
s =
N
200
=
= 66.7 psi
A¿
3
Ans.
t =
V
346.41
=
= 115 psi
A¿
3
Ans.
1–39. If the block is subjected to the centrally applied
force of 600 kN, determine the average normal stress in the
material. Show the stress acting on a differential volume
element of the material.
150 mm
600 kN
150 mm
150 mm
150 mm
The cross-sectional area of the block is A = 0.6(0.3) - 0.3(0.2) = 0.12 m2.
savg =
30
600(103)
P
=
= 5(106) Pa = 5 MPa
A
0.12
Ans.
The average normal stress distribution over the cross-section of the block and the
state of stress of a point in the block represented by a differential volume element
are shown in Fig. a
26
50 mm
100 mm
100 mm
50 mm
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*1–40. The pins on the frame at B and C each have a
diameter of 0.25 in. If these pins are subjected to double
shear, determine the average shear stress in each pin.
3 ft
500 lb
3 ft
A
3 ft
Support Reactions: FBD(a)
a + ©Mg = 0;
500(6) + 300(3) - Dy (6) = 0
Dy = 650 lb
+
; ©Fx = 0;
500 - Ex = 0
Ex = 500 lb
+ c ©Fy = 0;
650 - 300 - Ey = 0
Ey = 350 lb
a + ©MB = 0;
Cy (3) - 300(1.5) = 0
Cy = 150 lb
+ c ©Fy = 0;
By + 150 - 300 = 0
By = 150 lb
1.5 ft
From FBD (b)
150(1.5) + Bx(3) - 650(3) = 0
Bx = 575 lb
From FBD (c),
+
: ©Fx = 0;
Cx - 575 = 0
Cx = 575 lb
Hence, FB = FC = 2 5752 + 1502 = 594.24 lb
Average shear stress: Pins B and C are subjected to double shear as shown on FBD (d)
(tB)avg = (tC)avg =
1.5 ft
300 lb
D
From FBD (c),
a + ©MA = 0;
C
B
V
297.12
= p
2
A
4 (0.25 )
= 6053 psi = 6.05 ksi
Ans.
27
3 ft
E
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•1–41.
Solve Prob. 1–40 assuming that pins B and C are
subjected to single shear.
3 ft
500 lb
3 ft
A
3 ft
C
B
Support Reactions: FBD(a)
a + ©Mg = 0;
1.5 ft
500(6) + 300(3) - Dy (6) = 0
300 lb
D
Dy = 650 lb
+
; ©Fx = 0;
500 - Ex = 0
Ex = 500 lb
+ c ©Fy = 0;
650 - 300 - Ey = 0
Ey = 350 lb
From FBD (c),
a + ©MB = 0;
Cy (3) - 300(1.5) = 0
+ c ©Fy = 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).
F
For Pin D, FD = Dy = 650 lb then VD = zD = 325 lb
(pD)avg =
VD
325
= p
2
AD
4 (0.25)
Ans.
= 6621 psi = 6.62 ksi
Fg
For Pin E, FE = 2 5002 + 3502 = 610.32 lb then VE = z = 305.16 lb
(tE)avg =
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
C
B
Dy = 650 lb
+
; ©Fx = 0;
500 - Ex = 0
Ex = 500 lb
+ c ©Fy = 0;
650 - 300 - Ey = 0
Ey = 350 lb
1.5 ft
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
650
= p
2
AD
(0.25
)
4
Ans.
= 13242 psi = 13.2 ksi
For Pin E, VE = FE = 2 5002 + 3502 = 610.32 lb
(tE)avg =
VE
610.32
= p
2
AE
4 (0.25 )
= 12433 psi = 12.4 ksi
Ans.
*1–44. A 175-lb woman stands on a vinyl floor wearing
stiletto high-heel shoes. If the heel has the dimensions
shown, determine the average normal stress she exerts on
the floor and compare it with the average normal stress
developed when a man having the same weight is wearing
flat-heeled shoes. Assume the load is applied slowly, so that
dynamic effects can be ignored. Also, assume the entire
weight is supported only by the heel of one shoe.
Stiletto shoes:
A =
1.2 in.
1
(p)(0.3)2 + (0.6)(0.1) = 0.2014 in2
2
0.3 in.
0.1 in.
0.5 in.
P
175 lb
=
s =
= 869 psi
A
0.2014 in2
Ans.
Flat-heeled shoes:
A =
1
(p)(1.2)2 + 2.4(0.5) = 3.462 in2
2
s =
P
175 lb
=
= 50.5 psi
A
3.462 in2
Ans.
30
3 ft
E
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•1–45.
The truss is made from three pin-connected
members having the cross-sectional areas shown in the
figure. Determine the average normal stress developed in
each member when the truss is subjected to the load shown.
State whether the stress is tensile or compressive.
500 lb
C
sBC =
FBC
375
=
= 469 psi
A BC
0.8
(T)
Ans.
(T)
Ans.
4 ft
Joint A:
FAC
500
=
= 833 psi
A AC
0.6
in 2
.
A
1–46. Determine the average normal stress developed in
links AB and CD of the smooth two-tine grapple that
supports the log having a mass of 3 Mg. The cross-sectional
area of each link is 400 mm2.
+ c ©Fy = 0;
1.5
Ans.
(C)
AB
FAB
625
=
= 417 psi
A AB
1.5
A
sAB =
AAC 0.6 in.2
ABC 0.8 in.2
Joint B:
œ
sAC
=
3 ft
A
C
20
B
E
D
2(F sin 30°) - 29.43 = 0
0.2 m
F = 29.43 kN
a + ©ME = 0; P cos 20°(0.2) - (29.43 cos 30°)(1.2) + (29.43 sin 30°)(0.4 cos 30°)
1.2 m
= 0
P = 135.61 kN
30
30
0.4 m
s =
135.61(103)
P
= 339 MPa
=
A
400(10 - 6)
Ans.
31
B
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1–47. Determine the average shear stress developed in pins
A and B of the smooth two-tine grapple that supports the log
having a mass of 3 Mg. Each pin has a diameter of 25 mm and
is subjected to double shear.
+ c ©Fy = 0;
A
C
20
E
B
D
2(F sin 30°) - 29.43 = 0
0.2 m
F = 29.43 kN
1.2 m
a + ©ME = 0; P cos 20°(0.2) - (29.43 cos 30°)(1.2) + (29.43 sin 30°)(0.4 cos 30°)
= 0
P = 135.61 kN
tA = tB =
V
=
A
135.61(103)
2
p
2
(0.025)
4
30
30
0.4 m
= 138 MPa
Ans.
*1–48. The beam is supported by a pin at A and a short
link BC. If P = 15 kN, determine the average shear stress
developed in the pins at A, B, and C. All pins are in double
shear as shown, and each has a diameter of 18 mm.
P
4P
4P
2P
0.5m
0.5 m
1m
1.5 m
1.5 m
C
30
B
For pins B and C:
82.5 (103)
V
tB = tC =
= p 18 2 = 324 MPa
A
4 (1000 )
A
Ans.
For pin A:
FA = 2 (82.5)2 + (142.9)2 = 165 kN
tA =
82.5 (103)
V
= p 18 2 = 324 MPa
A
4 (1000 )
Ans.
32
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•1–49. The beam is supported by a pin at A and a short
link BC. Determine the maximum magnitude P of the loads
the beam will support if the average shear stress in each pin
is not to exceed 80 MPa. All pins are in double shear as
shown, and each has a diameter of 18 mm.
P
4P
4P
2P
0.5m
0.5 m
1m
1.5 m
1.5 m
C
30
B
a + ©MA = 0;
A
2P(0.5) + 4P(2) + 4P(3.5) + P(4.5) - (TCB sin 30°)(5) = 0
TCB = 11P
+
: ©Fx = 0;
A x - 11P cos 30° = 0
A x = 9.5263P
+ c ©Fy = 0;
A y - 11P + 11P sin 30° = 0
A y = 5.5P
FA = 2 (9.5263P)2 + (5.5P)2 = 11P
Require;
t =
V
;
A
80(106) = p
11P>2
2
4 (0.018)
P = 3.70 kN
Ans.
33
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1–50. The block is subjected to a compressive force of
2 kN. Determine the average normal and average shear
stress developed in the wood fibers that are oriented along
section a–a at 30° with the axis of the block.
50 mm
a
150 mm
2 kN
2 kN
30
a
Force equilibrium equations written perpendicular and parallel to section a–a gives
+Q©Fx¿ = 0;
Va - a - 2 cos 30° = 0
Va - a = 1.732 kN
+a©Fy¿ = 0;
2 sin 30° - Na - a = 0
Na - a = 1.00 kN
The cross sectional area of section a–a is A = a
0.15
b (0.05) = 0.015 m2. Thus
sin 30°
(sa - a)avg =
1.00(103)
Na - a
=
= 66.67(103)Pa = 66.7 kPa
A
0.015
Ans.
(ta - a)avg =
1.732(103)
Va - a
=
= 115.47(103)Pa = 115 kPa
A
0.015
Ans.
34
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1–51. During the tension test, the wooden specimen is
subjected to an average normal stress of 2 ksi. Determine
the axial force P applied to the specimen. Also, find the
average shear stress developed along section a–a of
the specimen.
P
a
4 in.
a
2 in.
1 in.
Internal Loading: The normal force developed on the cross section of the middle
portion of the specimen can be obtained by considering the free-body diagram
shown in Fig. a.
+ c ©Fy = 0;
P
P
+
- N = 0
2
2
4 in.
N = P
Referring to the free-body diagram shown in fig. b, the shear force developed in the
shear plane a–a is
+ c ©Fy = 0;
P
- Va - a = 0
2
Va - a =
P
2
Average Normal Stress and Shear Stress: The cross-sectional area of the specimen is
A = 1(2) = 2 in2. We have
savg =
N
;
A
2(103) =
P
2
P = 4(103)lb = 4 kip
Ans.
4(103)
P
=
= 2(103) lb. The area of the shear plane is
2
2
A a - a = 2(4) = 8 in2. We obtain
Using the result of P, Va - a =
A ta - a B avg =
2(103)
Va - a
=
= 250 psi
Aa - a
8
Ans.
35
P
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*1–52. If the joint is subjected to an axial force of
P = 9 kN, determine the average shear stress developed in
each of the 6-mm diameter bolts between the plates and the
members and along each of the four shaded shear planes.
P
P
100 mm
Internal Loadings: The shear force developed on each shear plane of the bolt and
the member can be determined by writing the force equation of equilibrium along
the member’s axis with reference to the free-body diagrams shown in Figs. a. and b,
respectively.
©Fy = 0; 4Vb - 9 = 0
Vb = 2.25 kN
©Fy = 0; 4Vp - 9 = 0
Vp = 2.25 kN
Average Shear Stress: The areas of each shear plane of the bolt and the member
p
are A b = (0.0062) = 28.274(10 - 6)m2 and A p = 0.1(0.1) = 0.01 m2, respectively.
4
We obtain
A tavg B b =
2.25(103)
Vb
= 79.6 MPa
=
Ab
28.274(10 - 6)
A tavg B p =
Vp
Ap
=
Ans.
2.25(103)
= 225 kPa
0.01
Ans.
36
100 mm
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•1–53.
The average shear stress in each of the 6-mm diameter
bolts and along each of the four shaded shear planes is not
allowed to exceed 80 MPa and 500 kPa, respectively.
Determine the maximum axial force P that can be applied
to the joint.
P
P
100 mm
100 mm
Internal Loadings: The shear force developed on each shear plane of the bolt and
the member can be determined by writing the force equation of equilibrium along
the member’s axis with reference to the free-body diagrams shown in Figs. a. and b,
respectively.
©Fy = 0;
4Vb - P = 0
Vb = P>4
©Fy = 0;
4Vp - P = 0
Vp = P>4
Average Shear Stress: The areas of each shear plane of the bolts and the members
p
are A b = (0.0062) = 28.274(10 - 6)m2 and A p = 0.1(0.1) = 0.01m2, respectively.
4
We obtain
A tallow B b =
Vb
;
Ab
80(106) =
P>4
28.274(10 - 6)
P = 9047 N = 9.05 kN (controls)
A tallow B p =
Vp
Ap
;
500(103) =
Ans.
P>4
0.01
P = 20 000 N = 20 kN
37
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1–54. The shaft is subjected to the axial force of 40 kN.
Determine the average bearing stress acting on the collar C
and the normal stress in the shaft.
40 kN
30 mm
C
40 mm
Referring to the FBDs in Fig. a,
+ c ©Fy = 0;
Ns - 40 = 0
Ns = 40 kN
+ c ©Fy = 0;
Nb - 40 = 0
Nb = 40 kN
Here, the cross-sectional area of the shaft and the bearing area of the collar are
p
p
A s = (0.032) = 0.225(10 - 3)p m2 and A b = (0.04 2) = 0.4(10 - 3)p m2. Thus,
4
4
A savg B s =
40(103)
Ns
= 56.59(106) Pa = 56.6 MPa
=
As
0.225(10 - 3)p
Ans.
A savg B b =
40(103)
Nb
= 31.83(106)Pa = 31.8 MPa
=
Ab
0.4(10 - 3)p
Ans.
38
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1–55. Rods AB and BC each have a diameter of 5 mm. If
the load of P = 2 kN is applied to the ring, determine the
average normal stress in each rod if u = 60°.
A
u
P
B
Consider the equilibrium of joint B, Fig. a,
+
: ©Fx = 0;
2 - FAB sin 60° = 0
+ c ©Fy = 0;
2.309 cos 60° - FBC = 0
FAB = 2.309 kN
C
FBC = 1.155 kN
The cross-sectional area of wires AB and BC are A AB = A BC =
p
(0.0052)
4
= 6.25(10 - 6)p m2. Thus,
A savg B AB =
2.309(103)
FAB
= 117.62(106) Pa = 118 MPa
=
A AB
6.25(10 - 6)p
Ans.
A savg B BC =
1.155(103)
FBC
= 58.81(106) Pa = 58.8 MPa
=
A BC
6.25(10 - 6)p
Ans.
39
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*1–56. Rods AB and BC each have a diameter of 5 mm.
Determine the angle u of rod BC so that the average
normal stress in rod AB is 1.5 times that in rod BC. What is
the load P that will cause this to happen if the average
normal stress in each rod is not allowed to exceed 100 MPa?
A
u
P
B
Consider the equilibrium of joint B, Fig. a,
FAB cos u - FBC = 0
+ c ©Fy = 0;
+
: ©Fx = 0;
(1)
P - FAB sin u = 0
(2)
p
(0.0052)
4
= 6.25(10 - 6)p m2. Since the average normal stress in rod AB is required to be
The cross-sectional area of rods AB and BC are A AB = A BC =
1.5 times to that of rod BC, then
A savg B AB = 1.5 A savg B BC
FAB
FBC
= 1.5 a
b
A AB
A BC
FAB
6.25(10 - 6)p
= 1.5 c
FBC
6.25(10 - 6)p
d
FAB = 1.5 FBC
(3)
Solving Eqs (1) and (3),
u = 48.19° = 48.2°
Ans.
Since wire AB will achieve the average normal stress of 100 MPa first when P
increases, then
FAB = sallow A AB = C 100(106) D C 6.25(10 - 6)p D = 1963.50 N
Substitute the result of FAB and u into Eq (2),
P = 1.46 kN
Ans.
40
C
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•1–57.
The specimen failed in a tension test at an angle of
52° when the axial load was 19.80 kip. If the diameter of the
specimen is 0.5 in., determine the average normal and
average shear stress acting on the area of the inclined
failure plane. Also, what is the average normal stress acting
on the cross section when failure occurs?
+
b © Fx = 0;
52
0.5 in.
V - 19.80 cos 52° = 0
V = 12.19 kip
+a © Fy = 0;
N - 19.80 sin 52° = 0
N = 15.603 kip
Inclined plane:
s¿ =
P
;
A
15.603
s¿ = p(0.25)2 = 62.6 ksi
V
;
A
12.19
œ
tavg
= p(0.25)2 = 48.9 ksi
œ
tavg
=
Ans.
sin 52°
Ans.
sin 52°
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 B
2
3 MPa
3 MPa
B
2
= 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;
N - 50 cos 30° = 0
N = 43.30 kip
Q© Fx = 0;
- V + 50 sin 30° = 0
V = 25.0 kip
+
B
A
6 in.
50 kip
Average Normal and Shear Stress:
A¿ = a
2
b (6) = 13.86 in2
sin 60°
N
43.30
=
= 3.125 ksi
A¿
13.86
V
25.0
tavg =
=
= 1.80 ksi
A¿
13.86
s =
Ans.
Ans.
*1–60. If P = 20 kN, determine the average shear stress
developed in the pins at A and C. The pins are subjected to
double shear as shown, and each has a diameter of 18 mm.
C
Referring to the FBD of member AB, Fig. a
a + ©MA = 0;
30
FBC sin 30° (6) - 20(2) - 20(4) = 0
FBC = 40 kN
A
+
: ©Fx = 0;
A x - 40 cos 30° = 0
+ c ©Fy = 0;
A y - 20 - 20 + 40 sin 30°
A x = 34.64 kN
A y = 20 kN
2m
P
Thus, the force acting on pin A is
FA = 2 A x 2 + A y 2 = 2 34.64 2 + 202 = 40 kN
Pins A and C are subjected to double shear. Referring to their FBDs in Figs. b and c,
FA
FBC
40
40
VA =
VC =
=
= 20 kN
=
= 20 kN
2
2
2
2
p
The cross-sectional area of Pins A and C are A A = A C = (0.0182)
4
= 81(10 - 6)p m2. Thus
tA =
20(103)
VA
= 78.59(106) Pa = 78.6 MPa
=
AA
81(10 - 6)p
Ans.
tC =
20(103)
VC
= 78.59(106) Pa = 78.6 MPa
=
AC
81(10 - 6)p
Ans.
42
B
2m
2m
P
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•1–61. Determine the maximum magnitude P of the load
the beam will support if the average shear stress in each pin
is not to allowed to exceed 60 MPa. All pins are subjected to
double shear as shown, and each has a diameter of 18 mm.
C
30
Referring to the FBD of member AB, Fig. a,
a + ©MA = 0;
FBC sin 30°(6) - P(2) - P(4) = 0
+
: ©Fx = 0;
A x - 2P cos 30° = 0
A x = 1.732P
A y - P - P + 2P sin 30° = 0
+ c ©Fy = 0;
A
FBC = 2P
FA = 2 A x 2 + A y 2 = 2 (1.732P)2 + P2 = 2P
All pins are subjected to same force and double shear. Referring to the FBD of the
pin, Fig. b,
2P
F
=
= P
2
2
The cross-sectional area of the pin is A =
tallow =
V
;
A
60(106) =
2m
P
Ay = P
Thus, the force acting on pin A is
V =
B
2m
p
(0.0182) = 81.0(10 - 6)p m2. Thus,
4
P
81.0(10 - 6)p
P = 15 268 N = 15.3 kN
Ans.
43
2m
P
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1–62. The crimping tool is used to crimp the end of the
wire E. If a force of 20 lb is applied to the handles,
determine the average shear stress in the pin at A. The pin is
subjected to double shear and has a diameter of 0.2 in. Only
a vertical force is exerted on the wire.
20 lb
C
E
A
Support Reactions:
5 in.
From FBD(a)
a + ©MD = 0;
B D
1.5 in. 2 in. 1 in.
20(5) - By (1) = 0
+
: ©Fx = 0;
20 lb
By = 100 lb
Bx = 0
From FBD(b)
+
: ©Fx = 0;
Ax = 0
a + ©ME = 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:
C
E
From FBD(a)
A
a + ©MD = 0;
20(5) - By (1) = 0
+
: ©Fx = 0;
Bx = 0
By = 100 lb
5 in.
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
50.0
= p
2
AB
4 (0.2 )
= 1592 psi = 1.59 ksi
Ans.
44
20 lb
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*1–64. The triangular blocks are glued along each side of
the joint. A C-clamp placed between two of the blocks is
used to draw the joint tight. If the glue can withstand a
maximum average shear stress of 800 kPa, determine the
maximum allowable clamping force F.
50 mm
F
glue
45
25 mm
Internal Loadings: The shear force developed on the glued shear plane can be
obtained by writing the force equation of equilibrium along the x axis with
reference to the free-body diagram of the triangular block, Fig. a.
+
: ©Fx = 0;
F cos 45° - V = 0
V =
F
2 2
F
2
Average Normal and Shear Stress: The area of the glued shear plane is
A = 0.05(0.025) = 1.25(10 - 3)m2. We obtain
tavg =
V
;
A
2 2
F
2
800(103) =
1.25(10 - 3)
F = 1414 N = 1.41 kN
Ans.
•1–65.
The triangular blocks are glued along each side of
the joint. A C-clamp placed between two of the blocks is
used to draw the joint tight. If the clamping force is
F = 900 N, determine the average shear stress developed
in the glued shear plane.
50 mm
45
F
glue
25 mm
Internal Loadings: The shear force developed on the glued shear plane can be
obtained by writing the force equation of equilibrium along the x axis with
reference to the free-body diagram of the triangular block, Fig. a.
+
: ©Fx = 0;
900 cos 45° - V = 0
V = 636.40 N
Average Normal and Shear Stress: The area of the glued shear plane is
A = 0.05(0.025) = 1.25(10 - 3)m2. We obtain
tavg =
V
636.40
= 509 kPa
=
A
1.25(10 - 3)
Ans.
45
F
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1–66. Determine the largest load P that can be a applied
to the frame without causing either the average normal
stress or the average shear stress at section a–a to exceed
s = 150 MPa and t = 60 MPa, respectively. Member CB
has a square cross section of 25 mm on each side.
B
Analyse the equilibrium of joint C using the FBD Shown in Fig. a,
+ c ©Fy = 0;
4
FBC a b - P = 0
5
FBC = 1.25P
2m
a
Referring to the FBD of the cut segment of member BC Fig. b.
+
: ©Fx = 0;
+ c ©Fy = 0;
The
3
Na - a - 1.25Pa b = 0
5
Na - a = 0.75P
4
1.25Pa b - Va - a = 0
5
Va - a = P
cross-sectional
area
of
section
a
a–a
is
A
A a - a = (0.025) a
0.025
b
3>5
= 1.0417(10 - 3)m2. For Normal stress,
sallow =
Na - a
;
Aa - a
150(106) =
0.75P
1.0417(10 - 3)
P = 208.33(103) N = 208.33 kN
For Shear Stress
Va - a
;
tallow =
Aa - a
60(106) =
P
1.0417(10 - 3)
P = 62.5(103) N = 62.5 kN (Controls!)
Ans.
46
C
1.5 m
P
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1–67. The prismatic bar has a cross-sectional area A. If it
is subjected to a distributed axial loading that increases
linearly from w = 0 at x = 0 to w = w0 at x = a, and then
decreases linearly to w = 0 at x = 2a, determine the average
normal stress in the bar as a function of x for 0 … x 6 a.
w0
x
a
a
Equation of Equilibrium:
+
: ©Fx = 0;
-N +
1 w0
1
a x + w0 b(a - x) + w0a = 0
2 a
2
N =
w0
A 2a2 - x2 B
2a
Average Normal Stress:
w
0
w0
N
2a (2a - x )
s =
=
=
A 2a2 - x2 B
A
A
2aA
2
2
Ans.
*1–68. The prismatic bar has a cross-sectional area A. If it is
subjected to a distributed axial loading that increases linearly
from w = 0 at x = 0 to w = w0 at x = a, and then decreases
linearly to w = 0 at x = 2a, determine the average normal
stress in the bar as a function of x for a 6 x … 2a.
w0
x
a
a
Equation of Equilibrium:
+
: ©Fx = 0;
-N +
1 w0
c (2a - x) d(2a - x) = 0
2 a
N =
w0
(2a - x)2
2a
Average Normal Stress:
w
s =
2
0
w0
N
2a (2a - x)
=
=
(2a - x)2
A
A
2aA
Ans.
The tapered rod has a radius of r = (2 - x>6) in.
and is subjected to the distributed loading of
w = (60 + 40x) lb>in. Determine the average normal stress
at the center of the rod, B.
•1–69.
A = pa 2 -
r
w (60 40x) lb/ in.
x
r = (2 — ) in.
6
3 2
b = 7.069 in2
6
x
B
6
© Fx = 0;
N -
L3
(60 + 40x) dx = 0;
3 in.
N = 720 lb
720
s =
= 102 psi
7.069
Ans.
47
3 in.
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1–70. The pedestal supports a load P at its center. If the
material has a mass density r, determine the radial
dimension r as a function of z so that the average normal
stress in the pedestal remains constant. The cross section
is circular.
P
r1
z
r
Require:
P + W1
P + W1 + dW
s =
=
A
A + dA
P dA + W1 dA = A dW
P + W1
dW
=
= s
dA
A
(1)
dA = p(r + dr)2 - pr2 = 2p r dr
dW = pr2(rg) dt
From Eq. (1)
pr2(rg) dz
= s
2p r dr
r rg dz
= s
2 dr
z
r
rg
dr
dz =
2s L0
Lr1 r
rg z
r
= ln ;
r1
2s
p
r = r1 e(2a)z
However,
s =
P
p r21
r = r1 e
(p r12rg )z
Ans.
2P
48
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1–71. Determine the average normal stress at section a–a
and the average shear stress at section b–b in member AB.
The cross section is square, 0.5 in. on each side.
150 lb/ft
Consider the FBD of member BC, Fig. a,
a + ©MC = 0;
B
4 ft
C
60
a
FAB sin 60°(4) - 150(4)(2) = 0
FAB = 346.41 lb
a
Referring to the FBD in Fig. b,
+
b©Fx¿ = 0;
Na - a + 346.41 = 0
b
Na - a = - 346.41 lb
Referring to the FBD in Fig. c.
+ c ©Fy = 0;
Vb - b - 346.41 sin 60° = 0
b
Vb - b = 300 lb
The cross-sectional areas of section a–a and b–b are A a - a = 0.5(0.5) = 0.25 in2 and
0.5
b = 0.5 in2. Thus
A b - b = 0.5 a
cos 60°
Na - a
346.41
Ans.
=
= 1385.64 psi = 1.39 ksi
sa - a =
Aa - a
0.25
tb - b =
Vb - b
300
=
= 600 psi
Ab - b
0.5
Ans.
49
A
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•1–73. Member B is subjected to a compressive force of
800 lb. If A and B are both made of wood and are 38 in. thick,
determine to the nearest 14 in. the smallest dimension h of
the horizontal segment so that it does not fail in shear. The
average shear stress for the segment is tallow = 300 psi.
tallow = 300 =
B
307.7
13
(32) h
12
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) =
80 kN
20(10 )
40 kN
p 2
4 d
d = 0.0135 m = 13.5 mm
Ans.
50
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*1–76. The lapbelt assembly is to be subjected to a force
of 800 N. Determine (a) the required thickness t of
the belt if the allowable tensile stress for the material
is (st)allow = 10 MPa, (b) the required lap length dl
if the glue can sustain an allowable shear stress of
(tallow)g = 0.75 MPa, and (c) the required diameter dr of
the pin if the allowable shear stress for the pin is
(tallow)p = 30 MPa.
800 N
45 mm
t
dl
dr
800 N
Allowable Normal Stress: Design of belt thickness.
10 A 106 B =
P
;
A
(st)allow =
800
(0.045)t
t = 0.001778 m = 1.78 mm
Ans.
Allowable Shear Stress: Design of lap length.
VA
400
;
0.750 A 106 B =
(tallow)g =
A
(0.045) dt
dt = 0.01185 m = 11.9 mm
Ans.
Allowable Shear Stress: Design of pin size.
VB
400
;
30 A 106 B = p 2
(tallow)P =
A
4 dr
dr = 0.004120 m = 4.12 mm
Ans.
•1–77.
The wood specimen is subjected to the pull of
10 kN in a tension testing machine. If the allowable normal
stress for the wood is (st)allow = 12 MPa and the allowable
shear stress is tallow = 1.2 MPa, determine the required
dimensions b and t so that the specimen reaches these
stresses simultaneously. The specimen has a width of 25 mm.
10 kN
Allowable Shear Stress: Shear limitation
tallow =
V
;
A
1.2 A 106 B =
t
A
b
3
5.00(10 )
(0.025) t
t = 0.1667 m = 167 mm
Ans.
Allowable Normal Stress: Tension limitation
sallow =
P
;
A
12.0 A 106 B =
10 kN
10(103)
(0.025) b
b = 0.03333 m = 33.3 mm
Ans.
51
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1–78. Member B is subjected to a compressive force of
600 lb. If A and B are both made of wood and are 1.5 in.
thick, determine to the nearest 1>8 in. the smallest dimension
a of the support so that the average shear stress along the
blue line does not exceed tallow = 50 psi. Neglect friction.
600 lb
3
5
4
B
A
Consider the equilibrium of the FBD of member B, Fig. a,
+
: ©Fx = 0;
4
600 a b - Fh = 0
5
Fh = 480 lb
Referring to the FBD of the wood segment sectioned through glue line, Fig. b
+
: ©Fx = 0;
480 - V = 0
V = 480 lb
The area of shear plane is A = 1.5(a). Thus,
tallow =
V
;
A
50 =
480
1.5a
a = 6.40 in
Use a = 612 in.
Ans.
52
a
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1–79. The joint is used to transmit a torque of
T = 3 kN # m. Determine the required minimum diameter
of the shear pin A if it is made from a material having a
shear failure stress of tfail = 150 MPa. Apply a factor of
safety of 3 against failure.
100 mm
T
A
T
Internal Loadings: The shear force developed on the shear plane of pin A can be
determined by writing the moment equation of equilibrium along the y axis with
reference to the free-body diagram of the shaft, Fig. a.
©My = 0; V(0.1) - 3(103) = 0
V = 30(103)N
Allowable Shear Stress:
tfail
150
tallow =
=
= 50 MPa
F.S.
3
Using this result,
tallow =
V
;
A
50(106) =
30(103)
p
d 2
4 A
dA = 0.02764 m = 27.6 mm
Ans.
*1–80. Determine the maximum allowable torque T that
can be transmitted by the joint. The shear pin A has a
diameter of 25 mm, and it is made from a material having a
failure shear stress of tfail = 150 MPa. Apply a factor of
safety of 3 against failure.
100 mm
T
Internal Loadings: The shear force developed on the shear plane of pin A can be
determined by writing the moment equation of equilibrium along the y axis with
reference to the free-body diagram of the shaft, Fig. a.
©My = 0; V(0.1) - T = 0
T
V = 10T
Allowable Shear Stress:
tfail
150
tallow =
=
= 50 MPa
F.S.
3
The area of the shear plane for pin A is A A =
p
(0.0252) = 0.4909(10 - 3)m2. Using
4
these results,
tallow =
V
;
AA
50(106) =
10T
0.4909(10 - 3)
T = 2454.37 N # m = 2.45 kN # m
Ans.
53
A
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•1–81.
The tension member is fastened together using two
bolts, one on each side of the member as shown. Each bolt
has a diameter of 0.3 in. Determine the maximum load P
that can be applied to the member if the allowable shear
stress for the bolts is tallow = 12 ksi and the allowable
average normal stress is sallow = 20 ksi.
60
P
P
N - P sin 60° = 0
a+ ©Fy = 0;
P = 1.1547 N
(1)
V - P cos 60° = 0
b+ ©Fx = 0;
P = 2V
(2)
Assume failure due to shear:
tallow = 12 =
V
(2) p4 (0.3)2
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
FBC = 4.392 kN
For wire BD,
FBD
;
sallow =
A BD
6(103)
165(106) = 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
2
4 dAB
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
2
4 dBC
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
P
2
4 (0.007 )
P = 6349.94 N = 6.350 kN
For wire AB,
FAB
;
sallow =
A AB
0.8966 P
165(106) = p
2
4 (0.006 )
P = 5203.42 N = 5.203 kN
For wire BC,
FBC
sallow =
;
A BC
0.7321 P
165(106) = p
2
4 (0.005 )
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 2
4 d1
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
T
;
2
(0.25)
4
24(103) = p
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
2
4 dC
dC = 0.01128 m = 11.28 mm
Use dC = 12 mm
For pin D,
VD
;
tallow =
AD
40(106) =
Ans.
6.093(103)
p
2
4 dD
dD = 0.01393 m = 13.93 mm
Use
dD = 14 mm
Ans.
59
A
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•1–89. The eye bolt is used to support the load of 5 kip.
Determine its diameter d to the nearest 18 in. and the required
thickness h to the nearest 18 in. of the support so that the
washer will not penetrate or shear through it. The allowable
normal stress for the bolt is sallow = 21 ksi and the allowable
shear stress for the supporting material is tallow = 5 ksi.
1 in.
h
d
5 kip
Allowable Normal Stress: Design of bolt size
5(103)
P
sallow =
;
21.0(103) = p 2
Ab
4 d
d = 0.5506 in.
Use d =
5
in.
8
Ans.
Allowable Shear Stress: Design of support thickness
5(103)
V
;
5(103) =
tallow =
A
p(1)(h)
Use h =
3
in.
8
Ans.
60
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1–90. The soft-ride suspension system of the mountain
bike is pinned at C and supported by the shock absorber
BD. If it is designed to support a load P = 1500 N,
determine the required minimum diameter of pins B and C.
Use a factor of safety of 2 against failure. The pins are made
of material having a failure shear stress of tfail = 150 MPa,
and each pin is subjected to double shear.
P
A
Internal Loadings: The forces acting on pins B and C can be determined by
considering the equilibrium of the free-body diagram of the soft-ride suspension
system shown in Fig. a.
a + ©MC = 0;
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,
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) =
100 mm
300 mm
Ans.
2333.49
p 2
d
4 C
dC = 0.006294 m = 6.29 mm
Ans.
61
30 mm
B
C
60
D
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1–91. The soft-ride suspension system of the mountain
bike is pinned at C and supported by the shock absorber
BD. If it is designed to support a load of P = 1500 N,
determine the factor of safety of pins B and C against
failure if they are made of a material having a shear failure
stress of tfail = 150 MPa. Pin B has a diameter of 7.5 mm,
and pin C has a diameter of 6.5 mm. Both pins are subjected
to double shear.
P
A
Internal Loadings: The forces acting on pins B and C can be determined by
considerning the equilibrium of the free-body diagram of the soft-ride suspension
system shown in Fig. a.
+ ©MC = 0;
1500(0.4) - FBD sin 60°(0.1) - FBD cos 60°(0.03) = 0
FBD = 5905.36 N
+
: ©Fx = 0;
Cx - 5905.36 cos 60° = 0
+ c ©Fy = 0;
5905.36 sin 60° - 1500 - Cy = 0 Cy = 3614.20 N
Cx = 2952.68 N
Thus,
FB = FBD = 5905.36 N
FC = 2 Cx 2 + Cy 2 = 2 2952.682 + 3614.202
= 4666.98 N
Since both pins are in double shear,
VB =
FB
5905.36
=
= 2952.68N
2
2
VC =
FC
4666.98
=
= 2333.49 N
2
2
Allowable Shear Stress: The areas of the shear plane for pins B and C are
p
p
A B = (0.00752) = 44.179(10 - 6)m2 and A C = (0.00652) = 33.183(10 - 6)m2.
4
4
We obtain
A tavg B B =
VB
2952.68
= 66.84 MPa
=
AB
44.179(10 - 6)
A tavg B C =
VC
2333.49
= 70.32 MPa
=
AC
33.183(10 - 6)
100 mm
300 mm
Using these results,
tfail
150
=
= 2.24
(F.S.)B =
66.84
A tavg B B
tfail
150
(F.S.)C =
=
= 2.13
70.32
A tavg B C
Ans.
Ans.
62
30 mm
B
C
60
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) = p
4.40(103)
2
2
4 (d w - 0.00611 )
dw = 0.0154 m = 15.4 mm
Ans.
•1–93.
The assembly is used to support the distributed
loading of w = 500 lb>ft. Determine the factor of safety with
respect to yielding for the steel rod BC and the pins at B and
C if the yield stress for the steel in tension is sy = 36 ksi
and in shear ty = 18 ksi. The rod has a diameter of 0.40 in.,
and the pins each have a diameter of 0.30 in.
C
4 ft
For rod BC:
s =
A
1.667
P
= 13.26 ksi
= p
2
A
4 (0.4 )
F. S. =
sy
s
=
36
= 2.71
13.26
Ans.
3 ft
For pins B and C:
w
0.8333
V
= 11.79 ksi
= p
t =
2
A
4 (0.3 )
F. S. =
ty
t
=
B
1 ft
18
= 1.53
11.79
Ans.
63
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1–94. If the allowable shear stress for each of the 0.30in.-diameter steel pins at A, B, and C is tallow = 12.5 ksi,
and the allowable normal stress for the 0.40-in.-diameter
rod is sallow = 22 ksi, determine the largest intensity w of
the uniform distributed load that can be suspended from
the beam.
C
4 ft
Assume failure of pins B and C:
1.667w
tallow = 12.5 = p
2
4 (0.3 )
w = 0.530 kip>ft
A
Ans.
(controls)
B
Assume failure of pins A:
3 ft
FA = 2 (2w)2 + (1.333w)2 = 2.404 w
w
1.202w
tallow = 12.5 = p
2
4 (0.3 )
1 ft
w = 0.735 kip>ft
Assume failure of rod BC:
3.333w
sallow = 22 = p
2
4 (0.4 )
w = 0.829 kip>ft
1–95. If the allowable bearing stress for the material
under the supports at A and B is 1sb2allow = 1.5 MPa,
determine the size of square bearing plates A¿ and B¿
required to support the load. Dimension the plates to the
nearest mm. The reactions at the supports are vertical. Take
P = 100 kN.
40 kN/m
Referring to the FBD of the bean, Fig. a
1.5 m
A
a + ©MA = 0;
NB(3) + 40(1.5)(0.75) - 100(4.5) = 0
NB = 135 kN
a + ©MB = 0;
40(1.5)(3.75) - 100(1.5) - NA(3) = 0
NA = 25.0 kN
For plate A¿ ,
NA
(sb)allow =
;
A A¿
1.5(106) =
25.0(103)
a2A¿
aA¿ = 0.1291 m = 130 mm
Ans.
For plate B¿ ,
sallow =
NB
;
A B¿
1.5(106) =
135(103)
a2B¿
aB¿ = 0.300 m = 300 mm
Ans.
64
P
A¿
B¿
3m
B
1.5 m
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40 kN/m
*1–96. If the allowable bearing stress for the material under
the supports at A and B is 1sb2allow = 1.5 MPa, determine
the maximum load P that can be applied to the beam. The
bearing plates A¿ and B¿ have square cross sections of
150 mm * 150 mm and 250 mm * 250 mm, respectively.
A
P
A¿
B¿
3m
1.5 m
B
1.5 m
Referring to the FBD of the beam, Fig. a,
a + ©MA = 0;
NB(3) + 40(1.5)(0.75) - P(4.5) = 0
NB = 1.5P - 15
a + ©MB = 0;
40(1.5)(3.75) - P(1.5) - NA(3) = 0
NA = 75 - 0.5P
For plate A¿ ,
NA
(sb)allow =
;
A A¿
1.5(106) =
(75 - 0.5P)(103)
0.15(0.15)
P = 82.5 kN
For plate B¿ ,
NB
;
(sb)allow =
A B¿
1.5(106) =
(1.5P - 15)(103)
0.25(0.25)
P = 72.5 kN (Controls!)
Ans.
•1–97.
The rods AB and CD are made of steel having a
failure tensile stress of sfail = 510 MPa. Using a factor of
safety of F.S. = 1.75 for tension, determine their smallest
diameter so that they can support the load shown. The
beam is assumed to be pin connected at A and C.
B
D
6 kN
5 kN
4 kN
Support Reactions:
a + ©MA = 0;
FCD(10) - 5(7) - 6(4) - 4(2) = 0
A
FCD = 6.70 kN
a + ©MC = 0;
4(8) + 6(6) + 5(3) - FAB(10) = 0
FAB = 8.30 kN
Allowable Normal Stress: Design of rod sizes
For rod AB
sallow =
sfail
FAB
=
;
F.S
A AB
510(106)
8.30(103)
= p 2
1.75
4 d AB
dAB = 0.006022 m = 6.02 mm
Ans.
For rod CD
sallow =
FCD
sfail
=
;
F.S
A CD
C
2m
510(106)
6.70(103)
= p 2
1.75
4 d CD
dCD = 0.005410 m = 5.41 mm
Ans.
65
2m
3m
3m
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1–98. The aluminum bracket A is used to support the
centrally applied load of 8 kip. If it has a constant thickness
of 0.5 in., determine the smallest height h in order to
prevent a shear failure. The failure shear stress is
tfail = 23 ksi. Use a factor of safety for shear of F.S. = 2.5.
A
Equation of Equilibrium:
+ c ©Fy = 0;
V - 8 = 0
V = 8.00 kip
Allowable Shear Stress: Design of the support size
tallow =
tfail
V
= ;
F.S
A
23(103)
8.00(103)
=
2.5
h(0.5)
8 kip
h = 1.74 in.
Ans.
1–99. The hanger is supported using the rectangular pin.
Determine the magnitude of the allowable suspended load
P if the allowable bearing stress is (sb)allow = 220 MPa, the
allowable tensile stress is (st)allow = 150 MPa, and the
allowable shear stress is tallow = 130 MPa. Take t = 6 mm,
a = 5 mm, and b = 25 mm.
20 mm
75 mm
10 mm
a
a
b
Allowable Normal Stress: For the hanger
(st)allow =
P
;
A
150 A 106 B =
P
(0.075)(0.006)
P
P = 67.5 kN
37.5 mm
Allowable Shear Stress: The pin is subjected to double shear. Therefore, V =
tallow =
130 A 106 B =
V
;
A
P
2
P>2
(0.01)(0.025)
P = 65.0 kN
Allowable Bearing Stress: For the bearing area
(sb)allow =
P
;
A
220 A 106 B =
37.5 mm
t
P>2
(0.005)(0.025)
P = 55.0 kN (Controls!)
Ans.
66
h
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*1–100. The hanger is supported using the rectangular
pin. Determine the required thickness t of the hanger, and
dimensions a and b if the suspended load is P = 60 kN.
The allowable tensile stress is (st)allow = 150 MPa, the
allowable bearing stress is (sb)allow = 290 MPa, and the
allowable shear stress is tallow = 125 MPa.
20 mm
75 mm
10 mm
a
a
37.5 mm
t
P
37.5 mm
Allowable Normal Stress: For the hanger
(st)allow =
P
;
A
150 A 106 B =
60(103)
(0.075)t
t = 0.005333 m = 5.33 mm
Ans.
Allowable Shear Stress: For the pin
tallow =
125 A 106 B =
V
;
A
30(103)
(0.01)b
b = 0.0240 m = 24.0 mm
Ans.
Allowable Bearing Stress: For the bearing area
(sb)allow =
P
;
A
290 A 106 B =
30(103)
(0.0240) a
a = 0.00431 m = 4.31 mm
Ans.
67
b
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•1–101. The 200-mm-diameter aluminum cylinder supports
a compressive load of 300 kN. Determine the average normal
and shear stress acting on section a–a. Show the results on a
differential element located on the section.
300 kN
a
Referring to the FBD of the upper segment of the cylinder sectional through a–a
shown in Fig. a,
+Q©Fx¿ = 0;
Na - a - 300 cos 30° = 0
Na - a = 259.81 kN
+a©Fy¿ = 0;
Va - a - 300 sin 30° = 0
Va - a = 150 kN
30
a
0.1
Section a–a of the cylinder is an ellipse with a = 0.1 m and b =
m. Thus,
cos 30°
0.1
b = 0.03628 m2.
A a - a = pab = p(0.1)a
cos 30°
A sa - a B avg =
259.81(103)
Na - a
=
= 7.162(106) Pa = 7.16 MPa
Aa - a
0.03628
Ans.
A ta - a B avg =
150(103)
Va - a
=
= 4.135(106) Pa = 4.13 MPa
Aa - a
0.03628
Ans.
d
The differential element representing the state of stress of a point on section a–a is
shown in Fig. b
1–102. The long bolt passes through the 30-mm-thick
plate. If the force in the bolt shank is 8 kN, determine the
average normal stress in the shank, the average shear stress
along the cylindrical area of the plate defined by the section
lines a–a, and the average shear stress in the bolt head along
the cylindrical area defined by the section lines b–b.
8 mm
a
7 mm
b
8 kN
18 mm
b
a
8 (103)
P
= p
= 208 MPa
ss =
2
A
4 (0.007)
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.
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
A
4 ft
D
B
E
Ans.
2.5 ft
MD + 0.225(0.75) + 0.4(1.5) = 0
C
MD = - 0.769 kip # ft
Ans.
3 ft
5 ft
Segment CE:
Q+ ©Fx = 0;
NE + 2.0 = 0;
NE = - 2.00 kip
R+ ©Fy = 0;
VE = 0
Ans.
a + ©ME = 0;
ME = 0
Ans.
Ans.
•1–105.
The pulley is held fixed to the 20-mm-diameter
shaft using a key that fits within a groove cut into the pulley
and shaft. If the suspended load has a mass of 50 kg,
determine the average shear stress in the key along section
a–a. The key is 5 mm by 5 mm square and 12 mm long.
a + ©MO = 0;
a
75 mm
F (10) - 490.5 (75) = 0
F = 3678.75 N
tavg =
a
3678.75
V
=
= 61.3 MPa
A
(0.005)(0.012)
Ans.
70
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1–106. The bearing pad consists of a 150 mm by 150 mm
block of aluminum that supports a compressive load of
6 kN. Determine the average normal and shear stress acting
on the plane through section a–a. Show the results on a
differential volume element located on the plane.
6 kN
a
Equation of Equilibrium:
+Q©Fx = 0;
Va - a - 6 cos 60° = 0
Va - a = 3.00 kN
a+ ©Fy = 0;
Na - a - 6 sin 60° = 0
Na - a = 5.196 kN
30
a
150 mm
Averge Normal Stress And Shear Stress: The cross sectional Area at section a–a is
0.15
b (0.15) = 0.02598 m2.
A = a
sin 60°
sa - a =
5.196(103)
Na - a
=
= 200 kPa
A
0.02598
Ans.
ta - a =
3.00(103)
Va - a
=
= 115 kPa
A
0.02598
Ans.
1–107. The yoke-and-rod connection is subjected to a
tensile force of 5 kN. Determine the average normal stress
in each rod and the average shear stress in the pin A
between the members.
5 kN
40 mm
For the 40 – mm – dia rod:
30 mm
5 (103)
P
s40 =
= p
= 3.98 MPa
2
A
4 (0.04)
Ans.
For the 30 – mm – dia rod:
5 kN
3
s30 =
5 (10 )
V
= p
= 7.07 MPa
2
A
4 (0.03)
Ans.
Average shear stress for pin A:
tavg =
A
25 mm
2.5 (103)
P
= p
= 5.09 MPa
2
A
4 (0.025)
Ans.
71
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*1–108. The cable has a specific weight g (weight>volume)
and cross-sectional area A. If the sag s is small, so that its
length is approximately L and its weight can be distributed
uniformly along the horizontal axis, determine the average
normal stress in the cable at its lowest point C.
A
s
C
L/2
Equation of Equilibrium:
a + ©MA = 0;
Ts -
gAL L
a b = 0
2
4
T =
B
gAL2
8s
Average Normal Stress:
gAL2
gL2
T
8s
s =
=
=
A
A
8s
Ans.
72
L/2
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2–1. An air-filled rubber ball has a diameter of 6 in. If
the air pressure within it is increased until the ball’s
diameter becomes 7 in., determine the average normal
strain in the rubber.
d0 = 6 in.
d = 7 in.
e =
pd - pd0
7 - 6
=
= 0.167 in./in.
pd0
6
Ans.
2–2. A thin strip of rubber has an unstretched length of
15 in. If it is stretched around a pipe having an outer diameter
of 5 in., determine the average normal strain in the strip.
L0 = 15 in.
L = p(5 in.)
e =
L - L0
5p - 15
=
= 0.0472 in.>in.
L0
15
Ans.
2–3. The rigid beam is supported by a pin at A and wires
BD and CE. If the load P on the beam causes the end C to
be displaced 10 mm downward, determine the normal strain
developed in wires CE and BD.
D
E
4m
P
¢LBD
¢LCE
=
3
7
A
3 (10)
= 4.286 mm
7
¢LCE
10
=
= 0.00250 mm>mm
eCE =
L
4000
3m
¢LBD =
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
œ
- LAC
LAC
301.734 - 300
eAC = eAB =
=
= 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
eAD =
= 0.0281(10 - 3) mm>mm
400
eDB =
Ans.
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 =
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
=
eBD =
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
2°
xA = (1) ¢
≤ p = 0.03491 m
180°
A
2°
xB = (4) ¢
≤ p = 0.13963 m
180°
1m
x = 4 + xB - xA = 4.10472 m
A¿B¿ = 232 + 4.104722 = 5.08416 m
AB = 232 + 42 = 5.00 m
Average Normal Strain:
eAB =
=
A¿B¿ - AB
AB
5.08416 - 5
= 16.8 A 10 - 3 B m>m
5
Ans.
14
4m
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2–26. The material distorts into the dashed position
shown. Determine (a) the average normal strains along
sides AC and CD and the shear strain gxy at F, and (b) the
average normal strain along line BE.
y
15 mm
C
25 mm
D
10 mm
B
E
75 mm
90 mm
A
Referring to Fig. a,
LBE = 2(90 - 75)2 + 802 = 26625 mm
LAC¿ = 21002 + 152 = 210225 mm
LC¿D¿ = 80 - 15 + 25 = 90 mm
f = tan-1 ¢
25
p rad
≤ = 14.04° ¢
≤ = 0.2450 rad.
100
180°
When the plate deforms, the vertical position of point B and E do not change.
LBB¿
15
=
;
LBB¿ = 13.5 mm
90
100
LEE¿
25
=
;
75
100
LEE¿ = 18.75 mm
LB¿E¿ = 2(90 - 75)2 + (80 - 13.5 + 18.75)2 = 27492.5625 mm
Thus,
A eavg B AC =
LAC¿ - LAC
210225 - 100
=
= 0.0112 mm>mm
LAC
100
Ans.
A eavg B CD =
LC¿D¿ - LCD
90 -80
=
= 0.125 mm>mm
LCD
80
Ans.
A eavg B BE =
LB¿E¿ - LBE
27492.5625 - 26625
=
= 0.0635 mm>mm
LBE
26625
Ans.
Referring to Fig. a, the angle at corner F becomes larger than 90° after the plate
deforms. Thus, the shear strain is negative.
0.245 rad
Ans.
15
80 mm
F
x
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2–27. The material distorts into the dashed position
shown. Determine the average normal strain that occurs
along the diagonals AD and CF.
y
15 mm
25 mm
D
C
10 mm
The undeformed length of diagonals AD and CF are
B
E
LAD = LCF = 280 + 100 = 216400 mm
2
2
The deformed length of diagonals AD and CF are
75 mm
90 mm
LAD¿ = 2(80 + 25) + 100 = 221025 mm
2
2
LC¿F = 2(80 - 15)2 + 1002 = 214225 mm
A
Thus,
A eavg B AD =
LAD¿ - LAD
221025 - 216400
=
= 0.132 mm>mm
LAD
216400
Ans.
A eavg B CF =
LC¿F - LCF
214225 - 216400
=
= -0.0687 mm>mm
LCF
216400
Ans.
*2–28. The wire is subjected to a normal strain that is
2
defined by P = xe - x , where x is in millimeters. If the wire
has an initial length L, determine the increase in its length.
80 mm
P ⫽ xe⫺x
L
2
dL = e dx = x e-x dx
L
2
L0
x e-x dx
L
1
1
1
2
2
= - c e-x d 冷 = - c e-L - d
2
2
2
0
=
x
2
x
x
¢L =
F
1
2
[1 - e-L ]
2
Ans.
16
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•2–29. The curved pipe has an original radius of 2 ft. If it is
heated nonuniformly, so that the normal strain along its length
is P = 0.05 cos u, determine the increase in length of the pipe.
e = 0.05 cos u
¢L =
L
2 ft
e dL
=
A
u
90°
(0.05 cos u)(2 du)
L0
90°
= 0.1
90°
cos u du = [0.1[sin u] 0冷 ] = 0.100 ft
L0
Ans.
Solve Prob. 2–29 if P = 0.08 sin u.
2–30.
dL = 2 due = 0.08 sin u
¢L =
e dL
L
90°
=
2 ft
(0.08 sin u)(2 du)
L0
= 0.16
L0
90°
sin u du = 0.16[-cos u] 0冷 = 0.16 ft
Ans.
2–31. The rubber band AB has an unstretched length of
1 ft. If it is fixed at B and attached to the surface at point A¿,
determine the average normal strain in the band. The surface
is defined by the function y = (x2) ft, where x is in feet.
y
y ⫽ x2
A¿
Geometry:
1 ft
L =
L0
A
1 + a
dy 2
b dx
dx
However y = x2 then
1 ft
dy
= 2x
dx
B
1 ft
L =
=
L0
A
u
90°
21 + 4 x2 dx
1
1 ft
C 2x21 + 4 x2 + ln A 2x + 21 + 4x2 B D 冷0
4
= 1.47894 ft
Average Normal Strain:
L - L0
1.47894 - 1
eavg =
=
= 0.479 ft>ft
L0
1
Ans.
17
A
1 ft
x
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*2–32. The bar is originally 300 mm long when it is flat. If it
is subjected to a shear strain defined by gxy = 0.02x, where
x is in meters, determine the displacement ¢y at the end of
its bottom edge. It is distorted into the shape shown, where
no elongation of the bar occurs in the x direction.
y
⌬y
x
300 mm
Shear Strain:
dy
= tan gxy ;
dx
dy
= tan (0.02 x)
dx
300 mm
¢y
dy =
L0
L0
tan (0.02 x)dx
¢y = -50[ln cos (0.02x)]|0300 mm
= 2.03 mm
Ans.
The fiber AB has a length L and orientation u. If its
ends A and B undergo very small displacements uA and vB ,
respectively, determine the normal strain in the fiber when
it is in position A¿B¿.
•2–33.
y
B¿
vB
B
L
Geometry:
LA¿B¿ = 2(L cos u - uA) + (L sin u + yB)
2
2
A
= 2L3 + u2A + y2B + 2L(yB sin u - uA cos u)
Average Normal Strain:
LA¿B¿ - L
eAB =
L
=
A
1 +
2(yB sin u - uA cos u)
u2A + y2B
+
- 1
L
L2
Neglecting higher terms u2A and y2B
1
2(yB sin u - uA cos u) 2
eAB = B 1 +
R - 1
L
Using the binomial theorem:
eAB = 1 +
=
2uA cos u
1 2yB sin u
¢
≤ + ... - 1
2
L
L
yB sin u
uA cos u
L
L
Ans.
18
uA A¿
u
x
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2–34. If the normal strain is defined in reference to the
final length, that is,
Pnœ = lim a
p : p¿
¢s¿ - ¢s
b
¢s¿
instead of in reference to the original length, Eq. 2–2, show
that the difference in these strains is represented as a
second-order term, namely, Pn - Pnœ = PnPnœ .
eB =
¢S¿ - ¢S
¢S
œ
=
eB - eA
¢S¿ - ¢S
¢S¿ - ¢S
¢S
¢S¿
¢S¿ 2 - ¢S¢S¿ - ¢S¿¢S + ¢S2
¢S¢S¿
2
2
¢S¿ + ¢S - 2¢S¿¢S
=
¢S¢S¿
=
=
(¢S¿ - ¢S)2
¢S¿ - ¢S
¢S¿ - ¢S
= ¢
≤¢
≤
¢S¢S¿
¢S
¢S¿
= eA eBœ (Q.E.D)
19
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•3–1.
A concrete cylinder having a diameter of 6.00 in. and
gauge length of 12 in. is tested in compression. The results of
the test are reported in the table as load versus contraction.
Draw the stress–strain diagram using scales of 1 in. = 0.5 ksi
and 1 in. = 0.2110-32 in.>in. From the diagram, determine
approximately the modulus of elasticity.
Stress and Strain:
s =
P
(ksi)
A
e =
dL
(in./in.)
L
0
0
0.177
0.00005
0.336
0.00010
0.584
0.000167
0.725
0.000217
0.902
0.000283
1.061
0.000333
1.220
0.000375
1.362
0.000417
1.645
0.000517
1.768
0.000583
1.874
0.000625
Modulus of Elasticity: From the stress–strain diagram
Eapprox =
1.31 - 0
= 3.275 A 103 B ksi
0.0004 - 0
Ans.
1
Load (kip)
Contraction (in.)
0
5.0
9.5
16.5
20.5
25.5
30.0
34.5
38.5
46.5
50.0
53.0
0
0.0006
0.0012
0.0020
0.0026
0.0034
0.0040
0.0045
0.0050
0.0062
0.0070
0.0075
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3–2. Data taken from a stress–strain test for a ceramic are
given in the table. The curve is linear between the origin and
the first point. Plot the diagram, and determine the modulus
of elasticity and the modulus of resilience.
Modulus of Elasticity: From the stress–strain diagram
E =
33.2 - 0
= 55.3 A 103 B ksi
0.0006 - 0
S (ksi)
P (in./in.)
0
33.2
45.5
49.4
51.5
53.4
0
0.0006
0.0010
0.0014
0.0018
0.0022
S (ksi)
P (in./in.)
0
33.2
45.5
49.4
51.5
53.4
0
0.0006
0.0010
0.0014
0.0018
0.0022
Ans.
Modulus of Resilience: The modulus of resilience is equal to the area under the
linear portion of the stress–strain diagram (shown shaded).
ut =
1
lb
in.
in # lb
(33.2) A 103 B ¢ 2 ≤ ¢ 0.0006
≤ = 9.96
2
in.
in
in3
Ans.
3–3. Data taken from a stress–strain test for a ceramic are
given in the table. The curve is linear between the origin
and the first point. Plot the diagram, and determine
approximately the modulus of toughness. The rupture stress
is sr = 53.4 ksi.
Modulus of Toughness: The modulus of toughness is equal to the area under the
stress–strain diagram (shown shaded).
(ut)approx =
lb
in.
1
(33.2) A 103 B ¢ 2 ≤ (0.0004 + 0.0010) ¢ ≤
2
in.
in
+ 45.5 A 103 B ¢
+
1
lb
in.
(7.90) A 103 B ¢ 2 ≤ (0.0012) ¢ ≤
2
in.
in
+
= 85.0
lb
in.
≤ (0.0012) ¢ ≤
in.
in2
1
lb
in.
(12.3) A 103 B ¢ 2 ≤ (0.0004) ¢ ≤
2
in.
in
in # lb
in3
Ans.
2
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*3–4. A tension test was performed on a specimen having
an original diameter of 12.5 mm and a gauge length of
50 mm. The data are listed in the table. Plot the stress–strain
diagram, and determine approximately the modulus of
elasticity, the ultimate stress, and the fracture stress. Use a
scale of 20 mm = 50 MPa and 20 mm = 0.05 mm>mm.
Redraw the linear-elastic region, using the same stress scale
but a strain scale of 20 mm = 0.001 mm>mm.
Stress and Strain:
s =
dL
P
(MPa) e =
(mm/mm)
A
L
0
0
90.45
0.00035
259.9
0.00120
308.0
0.00204
333.3
0.00330
355.3
0.00498
435.1
0.02032
507.7
0.06096
525.6
0.12700
507.7
0.17780
479.1
0.23876
Modulus of Elasticity: From the stress–strain diagram
(E)approx =
228.75(106) - 0
= 229 GPa
0.001 - 0
Ans.
Ultimate and Fracture Stress: From the stress–strain diagram
(sm)approx = 528 MPa
Ans.
(sf)approx = 479 MPa
Ans.
3
Load (kN)
Elongation (mm)
0
11.1
31.9
37.8
40.9
43.6
53.4
62.3
64.5
62.3
58.8
0
0.0175
0.0600
0.1020
0.1650
0.2490
1.0160
3.0480
6.3500
8.8900
11.9380
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3–5. A tension test was performed on a steel specimen
having an original diameter of 12.5 mm and gauge length
of 50 mm. Using the data listed in the table, plot the
stress–strain diagram, and determine approximately the
modulus of toughness. Use a scale of 20 mm = 50 MPa and
20 mm = 0.05 mm>mm.
Stress and Strain:
s =
P
dL
(MPa) e =
(mm/mm)
A
L
0
0
90.45
0.00035
259.9
0.00120
308.0
0.00204
333.3
0.00330
355.3
0.00498
435.1
0.02032
507.7
0.06096
525.6
0.12700
507.7
0.17780
479.1
0.23876
Modulus of Toughness: The modulus of toughness is equal to the
total area under the stress–strain diagram and can be
approximated by counting the number of squares. The total
number of squares is 187.
(ut)approx = 187(25) A 106 B ¢
N
m
≤ a 0.025 b = 117 MJ>m3
m
m2
Ans.
4
Load (kN)
Elongation (mm)
0
11.1
31.9
37.8
40.9
43.6
53.4
62.3
64.5
62.3
58.8
0
0.0175
0.0600
0.1020
0.1650
0.2490
1.0160
3.0480
6.3500
8.8900
11.9380
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3–6. A specimen is originally 1 ft long, has a diameter of
0.5 in., and is subjected to a force of 500 lb. When the force
is increased from 500 lb to 1800 lb, the specimen elongates
0.009 in. Determine the modulus of elasticity for the
material if it remains linear elastic.
Normal Stress and Strain: Applying s =
dL
P
and e =
.
A
L
0.500
= 2.546 ksi
s1 = p
2
4 (0.5 )
1.80
= 9.167 ksi
2
4 (0.5 )
s2 = p
¢e =
0.009
= 0.000750 in.>in.
12
Modulus of Elasticity:
E =
¢s
9.167 - 2.546
=
= 8.83 A 103 B ksi
¢e
0.000750
Ans.
3–7. A structural member in a nuclear reactor is made of a
zirconium alloy. If an axial load of 4 kip is to be supported
by the member, determine its required cross-sectional area.
Use a factor of safety of 3 relative to yielding. What is the
load on the member if it is 3 ft long and its elongation is
0.02 in.? Ezr = 14(103) ksi, sY = 57.5 ksi. The material has
elastic behavior.
Allowable Normal Stress:
F.S. =
3 =
sy
sallow
57.5
sallow
sallow = 19.17 ksi
sallow =
P
A
19.17 =
4
A
A = 0.2087 in2 = 0.209 in2
Ans.
Stress–Strain Relationship: Applying Hooke’s law with
e =
0.02
d
=
= 0.000555 in.>in.
L
3 (12)
s = Ee = 14 A 103 B (0.000555) = 7.778 ksi
Normal Force: Applying equation s =
P
.
A
P = sA = 7.778 (0.2087) = 1.62 kip
Ans.
5
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*3–8. The strut is supported by a pin at C and an A-36
steel guy wire AB. If the wire has a diameter of 0.2 in.,
determine how much it stretches when the distributed load
acts on the strut.
A
60⬚
200 lb/ft
a + ©MC = 0;
1
FAB cos 60°(9) - (200)(9)(3) = 0
2
9 ft
FAB = 600 lb
The normal stress the wire is
sAB =
FAB
600
= 19.10(103) psi = 19.10 ksi
= p
2
AAB
(0.2
)
4
Since sAB 6 sy = 36 ksi, Hooke’s Law can be applied to determine the strain
in wire.
sAB = EPAB;
19.10 = 29.0(103)PAB
PAB = 0.6586(10 - 3) in>in
9(12)
The unstretched length of the wire is LAB =
= 124.71 in. Thus, the wire
sin 60°
stretches
dAB = PAB LAB = 0.6586(10 - 3)(124.71)
= 0.0821 in.
Ans.
6
B
C
Here, we are only interested in determining the force in wire AB.
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The s –P diagram for a collagen fiber bundle from
which a human tendon is composed is shown. If a segment
of the Achilles tendon at A has a length of 6.5 in. and an
approximate cross-sectional area of 0.229 in2, determine its
elongation if the foot supports a load of 125 lb, which causes
a tension in the tendon of 343.75 lb.
•3–9.
s =
s (ksi)
4.50
A
3.75
3.00
2.25
1.50
P
343.75
=
= 1.50 ksi
A
0.229
125 lb
0.75
0.05
From the graph e = 0.035 in.>in.
d = eL = 0.035(6.5) = 0.228 in.
0.10
P (in./in.)
Ans.
s (ksi)
3–10. The stress–strain diagram for a metal alloy having an
original diameter of 0.5 in. and a gauge length of 2 in. is given
in the figure. Determine approximately the modulus of
elasticity for the material, the load on the specimen that causes
yielding, and the ultimate load the specimen will support.
105
90
75
60
From the stress–strain diagram, Fig. a,
45
60 ksi - 0
E
=
;
1
0.002 - 0
sy = 60 ksi
E = 30.0(103) ksi
Ans.
30
15
su>t = 100 ksi
0
Thus,
PY = sYA = 60 C p4 (0.52) D = 11.78 kip = 11.8 kip
Ans.
Pu>t = su>t A = 100 C p4 (0.52) D = 19.63 kip = 19.6 kip
Ans.
7
0
0
0.05 0.10 0.15 0.20 0.25 0.30 0.35
0.001 0.002 0.003 0.004 0.005 0.006 0.007
P (in./in.)
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s (ksi)
3–11. The stress–strain diagram for a steel alloy having an
original diameter of 0.5 in. and a gauge length of 2 in. is
given in the figure. If the specimen is loaded until it is
stressed to 90 ksi, determine the approximate amount of
elastic recovery and the increase in the gauge length after it
is unloaded.
105
90
75
60
45
30
15
0
From the stress–strain diagram Fig. a, the modulus of elasticity for the steel alloy is
E
60 ksi - 0
=
;
1
0.002 - 0
E = 30.0(103) ksi
when the specimen is unloaded, its normal strain recovered along line AB, Fig. a,
which has a gradient of E. Thus
Elastic Recovery =
90
90 ksi
= 0.003 in>in
=
E
30.0(103) ksi
Ans.
Thus, the permanent set is
PP = 0.05 - 0.003 = 0.047 in>in
Then, the increase in gauge length is
¢L = PPL = 0.047(2) = 0.094 in
Ans.
8
0
0
0.05 0.10 0.15 0.20 0.25 0.30 0.35
0.001 0.002 0.003 0.004 0.005 0.006 0.007
P (in./in.)
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*3–12. The stress–strain diagram for a steel alloy having an
original diameter of 0.5 in. and a gauge length of 2 in.
is given in the figure. Determine approximately the modulus
of resilience and the modulus of toughness for the material.
The Modulus of resilience is equal to the area under the stress–strain diagram up to
the proportional limit.
sPL = 60 ksi
PPL = 0.002 in>in.
Thus,
(ui)r =
1
1
in # lb
sPLPPL = C 60(103) D (0.002) = 60.0
2
2
in3
Ans.
The modulus of toughness is equal to the area under the entire stress–strain
diagram. This area can be approximated by counting the number of squares. The
total number is 38. Thus,
C (ui)t D approx = 38 c 15(103)
lb
in
in # lb
d a0.05 b = 28.5(103)
2
in
in
in3
s (ksi)
105
90
75
60
45
30
15
0
0
0
0.05 0.10 0.15 0.20 0.25 0.30 0.35
0.001 0.002 0.003 0.004 0.005 0.006 0.007
P (in./in.)
9
Ans.
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•3–13.
A bar having a length of 5 in. and cross-sectional
area of 0.7 in2 is subjected to an axial force of 8000 lb. If the
bar stretches 0.002 in., determine the modulus of elasticity
of the material. The material has linear-elastic behavior.
8000 lb
8000 lb
5 in.
Normal Stress and Strain:
s =
8.00
P
=
= 11.43 ksi
A
0.7
e =
dL
0.002
=
= 0.000400 in.>in.
L
5
Modulus of Elasticity:
E =
s
11.43
=
= 28.6(103) ksi
e
0.000400
Ans.
3–14. The rigid pipe is supported by a pin at A and an
A-36 steel guy wire BD. If the wire has a diameter of
0.25 in., determine how much it stretches when a load of
P = 600 lb acts on the pipe.
B
Here, we are only interested in determining the force in wire BD. Referring 4 ft
to the FBD in Fig. a
a + ©MA = 0;
FBD A 45 B (3) - 600(6) = 0
FBD = 1500 lb
A
D
C
The normal stress developed in the wire is
3 ft
FBD
1500
sBD =
= 30.56(103) psi = 30.56 ksi
= p
2
ABD
(0.25
)
4
Since sBD 6 sy = 36 ksi, Hooke’s Law can be applied to determine the strain in
the wire.
sBD = EPBD;
P
30.56 = 29.0(103)PBD
PBD = 1.054(10 - 3) in.>in.
The unstretched length of the wire is LBD = 232 + 42 = 5ft = 60 in. Thus, the
wire stretches
dBD = PBD LBD = 1.054(10 - 3)(60)
= 0.0632 in
Ans.
10
3 ft
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3–15. The rigid pipe is supported by a pin at A and an
A-36 guy wire BD. If the wire has a diameter of 0.25 in.,
determine the load P if the end C is displaced 0.075 in.
downward.
B
4 ft
P
A
D
C
3 ft
Here, we are only interested in determining the force in wire BD. Referring to the
FBD in Fig. a
FBD A 45 B (3) - P(6) = 0
a + ©MA = 0;
FBD = 2.50 P
The unstretched length for wire BD is LBD = 232 + 42 = 5 ft = 60 in. From the
geometry shown in Fig. b, the stretched length of wire BD is
LBD¿ = 2602 + 0.0752 - 2(60)(0.075) cos 143.13° = 60.060017
Thus, the normal strain is
PBD =
LBD¿ - LBD
60.060017 - 60
=
= 1.0003(10 - 3) in.>in.
LBD
60
Then, the normal stress can be obtain by applying Hooke’s Law.
sBD = EPBD = 29(103) C 1.0003(10 - 3) D = 29.01 ksi
Since sBD 6 sy = 36 ksi, the result is valid.
sBD =
FBD
;
ABD
2.50 P
29.01(103) = p
2
4 (0.25 )
P = 569.57 lb = 570 lb
Ans.
11
3 ft
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s (MPa)
*3–16. Determine the elongation of the square hollow bar
when it is subjected to the axial force P = 100 kN. If this
axial force is increased to P = 360 kN and released, find
the permanent elongation of the bar. The bar is made of a
metal alloy having a stress–strain diagram which can be
approximated as shown.
500
600 mm
P
250
50 mm
5 mm
0.00125
Normal Stress and Strain: The cross-sectional area of the hollow bar is
A = 0.052 - 0.042 = 0.9(10 - 3)m2. When P = 100 kN,
s1 =
100(103)
P
= 111.11 MPa
=
A
0.9(10 - 3)
From the stress–strain diagram shown in Fig. a, the slope of the straight line OA
which represents the modulus of elasticity of the metal alloy is
E =
250(106) - 0
= 200 GPa
0.00125 - 0
Since s1 6 250 MPa, Hooke’s Law can be applied. Thus
s1 = Ee1; 111.11(106) = 200(109)e1
e1 = 0.5556(10 - 3) mm>mm
Thus, the elongation of the bar is
d1 = e1L = 0.5556(10 - 3)(600) = 0.333 mm
Ans.
When P = 360 kN,
s2 =
360(103)
P
= 400 MPa
=
A
0.9(10 - 3)
From the geometry of the stress–strain diagram, Fig. a,
e2 - 0.00125
0.05 - 0.00125
=
400 - 250
500 - 250
e2 = 0.0305 mm>mm
When P = 360 kN is removed, the strain recovers linearly along line BC, Fig. a,
parallel to OA. Thus, the elastic recovery of strain is given by
s2 = Eer;
400(106) = 200(109)er
er = 0.002 mm>mm
The permanent set is
eP = e2 - er = 0.0305 - 0.002 = 0.0285 mm>mm
Thus, the permanent elongation of the bar is
dP = ePL = 0.0285(600) = 17.1 mm
Ans.
12
0.05
P (mm/mm)
50 mm
P
5 mm
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3–16. Continued
13
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s (ksi)
3–17. A tension test was performed on an aluminum
2014-T6 alloy specimen. The resulting stress–strain diagram
is shown in the figure. Estimate (a) the proportional limit,
(b) the modulus of elasticity, and (c) the yield strength
based on a 0.2% strain offset method.
70
60
50
40
30
20
10
0
0.02
0.002
Proportional Limit and Yield Strength: From the stress–strain diagram, Fig. a,
spl = 44 ksi
Ans.
sY = 60 ksi
Ans.
Modulus of Elasticity: From the stress–strain diagram, the corresponding strain for
sPL = 44 ksi is epl = 0.004 in.>in. Thus,
E =
44 - 0
= 11.0(103) ksi
0.004 - 0
Ans.
Modulus of Resilience: The modulus of resilience is equal to the area under the
14
0.04
0.004
0.06
0.006
0.08
0.008
0.10
0.010
P (in./in.)
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s (ksi)
3–18. A tension test was performed on an aluminum
2014-T6 alloy specimen. The resulting stress–strain
diagram is shown in the figure. Estimate (a) the modulus of
resilience; and (b) modulus of toughness.
70
60
50
40
30
20
10
0
0.02
0.002
0.04
0.004
0.06
0.006
0.08
0.008
0.10
0.010
P (in./in.)
stress–strain diagram up to the proportional limit. From the stress–strain diagram,
spl = 44 ksi
epl = 0.004 in.>in.
Thus,
A Ui B r = splepl = (44)(103)(0.004) = 88
1
2
1
2
in # lb
in3
Ans.
Modulus of Toughness: The modulus of toughness is equal to the area under the
entire stress–strain diagram. This area can be approximated by counting the number
of squares. The total number of squares is 65. Thus,
C A Ui B t D approx = 65 B 10(103)
lb
in.
in # lb
c0.01 d = 6.50(103)
2R
in.
in
in3
Ans.
The stress–strain diagram for a bone is shown, and can be described by the equation
3–19. The stress–strain diagram for a bone is shown, and
can be described by the equation P = 0.45110-62 s ⫹
0.36110-122 s3, where s is in kPa. Determine the yield
strength assuming a 0.3% offset.
P
s
P ⫽ 0.45(10⫺6)s + 0.36(10⫺12)s3
P
e = 0.45(10-6)s + 0.36(10-12)s3,
dP = A 0.45(10-6) + 1.08(10-12) s2 B ds
E =
ds
1
2 =
= 2.22 MPa
dP
0.45(10 - 6)
Ans.
s=0
15
P
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*3–20. The stress–strain diagram for a bone is shown and
can be described by the equation P = 0.45110-62 s ⫹
0.36110-122 s3, where s is in kPa. Determine the modulus
of toughness and the amount of elongation of a 200-mmlong region just before it fractures if failure occurs at
P = 0.12 mm>mm.
P
s
P ⫽ 0.45(10⫺6)s + 0.36(10⫺12)s3
P
When e = 0.12
120(103) = 0.45 s + 0.36(10-6)s3
Solving for the real root:
s = 6873.52 kPa
6873.52
ut =
LA
dA =
L0
(0.12 - e)ds
6873.52
ut =
L0
(0.12 - 0.45(10-6)s - 0.36(10-12)s3)ds
6873.52
= 0.12 s - 0.225(10-6)s2 - 0.09(10-12)s4|0
= 613 kJ>m3
Ans.
d = eL = 0.12(200) = 24 mm
Ans.
16
P
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•3–21.
The stress–strain diagram for a polyester resin
is given in the figure. If the rigid beam is supported by a
strut AB and post CD, both made from this material, and
subjected to a load of P = 80 kN, determine the angle
of tilt of the beam when the load is applied. The diameter of
the strut is 40 mm and the diameter of the post is 80 mm.
B
2m
P
A
C
0.75 m 0.75 m
D
0.5 m
From the stress–strain diagram,
E =
32.2(10)6
= 3.22(109) Pa
0.01
s (MPa)
100
95
Thus,
40(10 )
FAB
= p
= 31.83 MPa
2
AAB
4 (0.04)
70
60
31.83(106)
sAB
= 0.009885 mm>mm
=
eAB =
E
3.22(109)
40
32.2
sAB =
50
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
20
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
P>2
2
4 (0.012)
A
;
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
P>2
100
95
2
4 (0.04)
80
6
95(10 ) = p
compression
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
20
s =
= p 2 = 6.366 ksi
A
(2
)
4
0
0
Normal Strain:
e =
0.25
= 0.0500 in.>in.
5
From the stress–strain diagram, the copolymer will satisfy both stress and strain
requirements.
Ans.
18
0.10
0.20
0.30
P (in./in.)
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*3–24. The stress–strain diagram for many metal alloys
can be described analytically using the Ramberg-Osgood
three parameter equation P = s>E + ksn, where E, k, and
n are determined from measurements taken from the
diagram. Using the stress–strain diagram shown in the
figure, take E = 3011032 ksi and determine the other two
parameters k and n and thereby obtain an analytical
expression for the curve.
s (ksi)
80
60
40
20
0.1
0.2
0.3
0.4
0.5
P (10–6)
Choose,
s = 40 ksi,
e = 0.1
s = 60 ksi,
e = 0.3
0.1 =
40
+ k(40)n
30(103)
0.3 =
60
+ k(60)n
30(103)
0.098667 = k(40)n
0.29800 = k(60)n
0.3310962 = (0.6667)n
ln (0.3310962) = n ln (0.6667)
n = 2.73
Ans.
k = 4.23(10-6)
Ans.
•3–25.
The acrylic plastic rod is 200 mm long and 15 mm in
diameter. If an axial load of 300 N is applied to it, determine
the change in its length and the change in its diameter.
Ep = 2.70 GPa, np = 0.4.
s =
200 mm
P
300
= 1.697 MPa
= p
2
A
4 (0.015)
elong =
300 N
300 N
1.697(106)
s
= 0.0006288
=
E
2.70(109)
d = elong L = 0.0006288 (200) = 0.126 mm
Ans.
elat = -Velong = -0.4(0.0006288) = -0.0002515
¢d = elatd = -0.0002515 (15) = -0.00377 mm
Ans.
19
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3–26. The short cylindrical block of 2014-T6 aluminum,
having an original diameter of 0.5 in. and a length of 1.5 in.,
is placed in the smooth jaws of a vise and squeezed until the
axial load applied is 800 lb. Determine (a) the decrease in its
length and (b) its new diameter.
800 lb
800 lb
a)
s =
P
800
= 4074.37 psi
= p
2
A
4 (0.5)
elong =
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 =
50(103)
P
= 376.70 Mpa
= p
2
A
4 (0.013 )
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 =
20(103)
P
= 150.68Mpa
= p
2
A
4 (0.013 )
0.002
P(mm/mm)
Normal Strain: From the Stress–Strain diagram, the modulus of elasticity
400(106)
E =
= 200 GPa. Applying Hooke’s Law
0.002
elong =
150.68(106)
s
= 0.7534 A 10 - 3 B mm>mm
=
E
200(109)
Thus,
dL = elong L0 = 0.7534 A 10 - 3 B (50) = 0.03767 mm
L = L0 + dL = 50 + 0.03767 = 50.0377 mm
Ans.
Poisson’s Ratio: The lateral and longitudinal can be related using poisson’s ratio.
elat = -velong = -0.4(0.7534) A 10 - 3 B
= -0.3014 A 10 - 3 B mm>mm
dd = elat d = -0.3014 A 10 - 3 B (13) = -0.003918 mm
d = d0 + dd = 13 + ( -0.003918) = 12.99608 mm
Ans.
•3–29.
The aluminum block has a rectangular cross
section and is subjected to an axial compressive force of
8 kip. If the 1.5-in. side changed its length to 1.500132 in.,
determine Poisson’s ratio and the new length of the 2-in.
side. Eal ⫽ 10(103) ksi.
s =
elat =
2 in.
8 kip
8 kip
3 in.
P
8
=
= 2.667 ksi
A
(2)(1.5)
elong =
v =
1.5 in.
s
-2.667
= -0.0002667
=
E
10(103)
1.500132 - 1.5
= 0.0000880
1.5
-0.0000880
= 0.330
-0.0002667
Ans.
h¿ = 2 + 0.0000880(2) = 2.000176 in.
Ans.
21
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3–30. The block is made of titanium Ti-6A1-4V and is
subjected to a compression of 0.06 in. along the y axis, and its
shape is given a tilt of u = 89.7°. Determine Px, Py, and gxy.
y
Normal Strain:
ey =
4 in. u
dLy
Ly
=
-0.06
= -0.0150 in.>in.
4
Ans.
Poisson’s Ratio: The lateral and longitudinal strain can be related using Poisson’s ratio.
x
5 in.
ex = -vey = -0.36(-0.0150)
= 0.00540 in. >in.
Ans.
Shear Strain:
b = 180° - 89.7° = 90.3° = 1.576032 rad
gxy =
p
p
- b =
- 1.576032 = -0.00524 rad
2
2
Ans.
3–31. The shear stress–strain diagram for a steel alloy is
shown in the figure. If a bolt having a diameter of 0.75 in.
is made of this material and used in the double lap joint,
determine the modulus of elasticity E and the force P
required to cause the material to yield. Take n = 0.3.
P/2
P/2
P
t(ksi)
60
The shear force developed on the shear planes of the bolt can be determined by
considering the equilibrium of the FBD shown in Fig. a
+ ©F = 0;
:
x
V + V - P = 0
V = =
g(rad)
P
0.00545
2
From the shear stress–strain diagram, the yield stress is ty = 60 ksi. Thus,
ty =
Vy
A
;
60 =
P>2
p
4
A 0.752 B
P = 53.01 kip = 53.0 kip
Ans.
From the shear stress–strain diagram, the shear modulus is
G =
60 ksi
= 11.01(103) ksi
0.00545
Thus, the modulus of elasticity is
G =
E
;
2(1 + y)
11.01(103) =
E
2(1 + 0.3)
E = 28.6(103) ksi
Ans.
22
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*3–32. A shear spring is made by bonding the rubber
annulus to a rigid fixed ring and a plug. When an axial load
P is placed on the plug, show that the slope at point y in
the rubber is dy>dr = -tan g = -tan1P>12phGr22. For small
angles we can write dy>dr = -P>12phGr2. Integrate this
expression and evaluate the constant of integration using
the condition that y = 0 at r = ro. From the result compute
the deflection y = d of the plug.
P
h
ro
y
d
ri
r
y
Shear Stress–Strain Relationship: Applying Hooke’s law with tA =
g =
P
.
2p r h
tA
P
=
G
2p h G r
dy
P
= -tan g = -tan a
b
dr
2p h G r
(Q.E.D)
If g is small, then tan g = g. Therefore,
dy
P
= dr
2p h G r
At r = ro,
y = -
dr
P
2p h G L r
y = -
P
ln r + C
2p h G
0 = -
P
ln ro + C
2p h G
y = 0
C =
Then, y =
ro
P
ln
r
2p h G
At r = ri,
y = d
d =
P
ln ro
2p h G
ro
P
ln
ri
2p h G
Ans.
23
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•3–33.
The support consists of three rigid plates, which
are connected together using two symmetrically placed
rubber pads. If a vertical force of 5 N is applied to plate
A, determine the approximate vertical displacement of
this plate due to shear strains in the rubber. Each pad
has cross-sectional dimensions of 30 mm and 20 mm.
Gr = 0.20 MPa.
C
B
40 mm
40 mm
A
tavg =
g =
V
2.5
=
= 4166.7 Pa
A
(0.03)(0.02)
5N
t
4166.7
= 0.02083 rad
=
G
0.2(106)
d = 40(0.02083) = 0.833 mm
Ans.
3–34. A shear spring is made from two blocks of rubber,
each having a height h, width b, and thickness a. The
blocks are bonded to three plates as shown. If the plates
are rigid and the shear modulus of the rubber is G,
determine the displacement of plate A if a vertical load P is
applied to this plate. Assume that the displacement is small
so that d = a tan g L ag.
P
d
A
h
Average Shear Stress: The rubber block is subjected to a shear force of V =
P
.
2
P
t =
V
P
2
=
=
A
bh
2bh
Shear Strain: Applying Hooke’s law for shear
P
g =
t
P
2bh
=
=
G
G
2bhG
Thus,
d = ag = =
Pa
2bhG
Ans.
24
a
a
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s (ksi)
3–35. The elastic portion of the tension stress–strain
diagram for an aluminum alloy is shown in the figure. The
specimen used for the test has a gauge length of 2 in. and a
diameter of 0.5 in. When the applied load is 9 kip, the new
diameter of the specimen is 0.49935 in. Compute the shear
modulus Gal for the aluminum.
70
0.00614
From the stress–strain diagram,
P (in./in.)
70
s
=
= 11400.65 ksi
e
0.00614
Eal =
When specimen is loaded with a 9 - kip load,
s =
9
P
= 45.84 ksi
= p
2
A
(0.5)
4
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.
70
0.00614
P
10
s =
= p
= 50.9296 ksi
2
A
(0.5)
4
From the stress–strain diagram
E =
70
= 11400.65 ksi
0.00614
elong =
G =
s
50.9296
=
= 0.0044673 in.>in.
E
11400.65
E
;
2(1 + v)
3.8(103) =
11400.65
;
2(1 + v)
v = 0.500
elat = - velong = - 0.500(0.0044673) = - 0.002234 in.>in.
¢d = elat d = - 0.002234(0.5) = - 0.001117 in.
d¿ = d + ¢d = 0.5 - 0.001117 = 0.4989 in.
Ans.
25
P (in./in.)
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s(psi)
3–37. The s – P diagram for elastic fibers that make up
human skin and muscle is shown. Determine the modulus
of elasticity of the fibers and estimate their modulus of
toughness and modulus of resilience.
55
11
1
E =
11
= 5.5 psi
2
Ans.
ut =
1
1
(2)(11) + (55 + 11)(2.25 - 2) = 19.25 psi
2
2
Ans.
ut =
1
(2)(11) = 11 psi
2
Ans.
3–38. A short cylindrical block of 6061-T6 aluminum,
having an original diameter of 20 mm and a length of
75 mm, is placed in a compression machine and squeezed
until the axial load applied is 5 kN. Determine (a) the
decrease in its length and (b) its new diameter.
a)
s =
-5(103)
P
= p
= - 15.915 MPa
2
A
4 (0.02)
s = E elong ;
- 15.915(106) = 68.9(109) elong
elong = - 0.0002310 mm>mm
d = elong L = - 0.0002310(75) = - 0.0173 mm
b)
v = -
elat
;
elong
0.35 = -
Ans.
elat
-0.0002310
elat = 0.00008085 mm>mm
¢d = elat d = 0.00008085(20) = 0.0016 mm
d¿ = d + ¢d = 20 + 0.0016 = 20.0016 mm
Ans.
26
2 2.25
P(in./in.)
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3–39. The rigid beam rests in the horizontal position on
two 2014-T6 aluminum cylinders having the unloaded lengths
shown. If each cylinder has a diameter of 30 mm, determine
the placement x of the applied 80-kN load so that the beam
remains horizontal. What is the new diameter of cylinder A
after the load is applied? nal = 0.35.
a + ©MA = 0;
FB(3) - 80(x) = 0;
a + ©MB = 0;
-FA(3) + 80(3 - x) = 0;
FB =
80 kN
x
A
80x
3
FA =
B
210 mm
220 mm
(1)
80(3 - x)
3
3m
(2)
Since the beam is held horizontally, dA = dB
s =
P
;
A
d = eL = a
P
e =
P
A
E
dA = dB ;
s
A
=
E
E
bL =
PL
AE
80(3 - x)
(220)
3
80x
3 (210)
=
AE
AE
80(3 - x)(220) = 80x(210)
x = 1.53 m
Ans.
From Eq. (2),
FA = 39.07 kN
sA =
39.07(103)
FA
= 55.27 MPa
= p
2
A
4 (0.03 )
elong =
55.27(106)
sA
= -0.000756
= E
73.1(109)
elat = -velong = -0.35(-0.000756) = 0.0002646
œ
dA
= dA + d elat = 30 + 30(0.0002646) = 30.008 mm
Ans.
*3–40. The head H is connected to the cylinder of a
compressor using six steel bolts. If the clamping force in
each bolt is 800 lb, determine the normal strain in the
3
bolts. Each bolt has a diameter of 16
in. If sY = 40 ksi and
3
Est = 29110 2 ksi, what is the strain in each bolt when the
nut is unscrewed so that the clamping force is released?
C
L
H
Normal Stress:
s =
P
800
= 28.97 ksi 6 sg = 40 ksi
=
p
A
A 3 B2
4 16
Normal Strain: Since s 6 sg, Hooke’s law is still valid.
e =
28.97
s
= 0.000999 in.>in.
=
E
29(103)
Ans.
If the nut is unscrewed, the load is zero. Therefore, the strain e = 0
27
Ans.
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•3–41. The stone has a mass of 800 kg and center of gravity
at G. It rests on a pad at A and a roller at B. The pad is fixed
to the ground and has a compressed height of 30 mm, a
width of 140 mm, and a length of 150 mm. If the coefficient
of static friction between the pad and the stone is ms = 0.8,
determine the approximate horizontal displacement of the
stone, caused by the shear strains in the pad, before the
stone begins to slip. Assume the normal force at A acts
1.5 m from G as shown. The pad is made from a material
having E = 4 MPa and n = 0.35.
0.4 m
B
Equations of Equilibrium:
a + ©MB = 0;
+ ©F = 0;
:
x
FA(2.75) - 7848(1.25) - P(0.3) = 0
[1]
P - F = 0
[2]
Note: The normal force at A does not act exactly at A. It has to shift due to friction.
Friction Equation:
F = ms FA = 0.8 FA
[3]
Solving Eqs. [1], [2] and [3] yields:
FA = 3908.37 N
F = P = 3126.69 N
Average Shear Stress: The pad is subjected to a shear force of V = F = 3126.69 N.
t =
V
3126.69
=
= 148.89 kPa
A
(0.14)(0.15)
Modulus of Rigidity:
G =
E
4
=
= 1.481 MPa
2(1 + v)
2(1 + 0.35)
Shear Strain: Applying Hooke’s law for shear
g =
148.89(103)
t
= 0.1005 rad
=
G
1.481(106)
Thus,
dh = hg = 30(0.1005) = 3.02 mm
Ans.
28
P
G
1.25 m
0.3 m
1.5 m
A
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3–42. The bar DA is rigid and is originally held in the
horizontal position when the weight W is supported from C.
If the weight causes B to be displaced downward 0.025 in.,
determine the strain in wires DE and BC. Also, if the wires
are made of A-36 steel and have a cross-sectional area of
0.002 in2, determine the weight W.
E
3 ft
2 ft
D
3 ft
B
5
3
=
0.025
d
A
4 ft
d = 0.0417 in
eDE =
C
0.0417
d
=
= 0.00116 in.>in.
L
3(12)
Ans.
W
3
sDE = EeDE = 29(10 )(0.00116) = 33.56 ksi
FDE = sDEADE = 33.56 (0.002) = 0.0672 kip
a + ©MA = 0;
-(0.0672) (5) + 3(W) = 0
W = 0.112 kip = 112 lb
sBC =
W
0.112
=
= 55.94 ksi
ABC
0.002
eBC =
sBC
55.94
= 0.00193 in.>in.
=
E
29 (103)
Ans.
Ans.
3–43. The 8-mm-diameter bolt is made of an aluminum
alloy. It fits through a magnesium sleeve that has an inner
diameter of 12 mm and an outer diameter of 20 mm. If the
original lengths of the bolt and sleeve are 80 mm and
50 mm, respectively, determine the strains in the sleeve and
the bolt if the nut on the bolt is tightened so that the tension
in the bolt is 8 kN. Assume the material at A is rigid.
Eal = 70 GPa, Emg = 45 GPa.
50 mm
A
30 mm
Normal Stress:
sb =
8(103)
P
= 159.15 MPa
= p
2
Ab
4 (0.008 )
ss =
8(103)
P
= 39.79 MPa
= p
2
2
As
4 (0.02 - 0.012 )
Normal Strain: Applying Hooke’s Law
eb =
159.15(106)
sb
= 0.00227 mm>mm
=
Eal
70(109)
Ans.
es =
39.79(106)
ss
= 0.000884 mm>mm
=
Emg
45(109)
Ans.
29
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*3–44. The A-36 steel wire AB has a cross-sectional area
of 10 mm2 and is unstretched when u = 45.0°. Determine
the applied load P needed to cause u = 44.9°.
A
400 mm
u
400 m
m
B
P
¿
LAB
400
=
sin 90.2°
sin 44.9°
¿
= 566.67 mm
LAB
LAB =
e =
400
= 565.69
sin 45°
¿
- LAB
LAB
566.67 - 565.69
=
= 0.001744
LAB
565.69
s = Ee = 200(109) (0.001744) = 348.76 MPa
a + ©MA = 0
P(400 cos 0.2°) - FAB sin 44.9° (400) = 0
(1)
However,
FAB = sA = 348.76(106)(10)(10 - 6) = 3.488 kN
From Eq. (1),
P = 2.46 kN
Ans.
30
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•4–1. The ship is pushed through the water using an A-36
steel propeller shaft that is 8 m long, measured from the
propeller to the thrust bearing D at the engine. If it has an
outer diameter of 400 mm and a wall thickness of 50 mm,
determine the amount of axial contraction of the shaft
when the propeller exerts a force on the shaft of 5 kN. The
bearings at B and C are journal bearings.
A
Internal Force: As shown on FBD.
B
C
D
5 kN
Displacement:
8m
dA =
-5.00 (103)(8)
PL
= p
2
2
9
AE
4 (0.4 - 0.3 ) 200(10 )
= -3.638(10 - 6) m
= -3.64 A 10 - 3 B mm
Ans.
Negative sign indicates that end A moves towards end D.
4–2. The copper shaft is subjected to the axial loads
shown. Determine the displacement of end A with respect
to end D. The diameters of each segment are dAB = 3 in.,
dBC = 2 in., and dCD = 1 in. Take Ecu = 1811032 ksi.
50 in.
A
p
The cross-sectional area of segment AB, BC and CD are AAB = (32) = 2.25p in2,
4
p
p
ABC = (22) = p in2 and ACD = (12) = 0.25p in2.
4
4
Thus,
PCD LCD
PiLi
PAB LAB
PBC LBC
=
+
+
AiEi
AAB ECu
ABC ECu
ACD ECu
2.00 (75)
6.00 (50)
=
(2.25p) C 18(10 ) D
3
+
p C 18(10 ) D
3
-1.00 (60)
+
(0.25p) C 18(103) D
= 0.766(10 - 3) in.
Ans.
The positive sign indicates that end A moves away from D.
122
60 in.
2 kip
6 kip
The normal forces developed in segment AB, BC and CD are shown in the
FBDS of each segment in Fig. a, b and c respectively.
dA>D = ©
75 in.
B 2 kip
1 kip
C
3 kip
D
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4–3. The A-36 steel rod is subjected to the loading shown.
If the cross-sectional area of the rod is 50 mm2, determine
the displacement of its end D. Neglect the size of the
couplings at B, C, and D.
1m
A 9 kN B
The normal forces developed in segments AB, BC and CD are shown in the FBDS
of each segment in Fig. a, b and c, respectively.
The
cross-sectional
areas
of
all
2
1
m
b = 50.0(10 - 6) m2.
A = A 50 mm2 B a
1000 mm
dD = ©
the
segments
are
PiLi
1
=
a PAB LAB + PBC LBC + PCD LCD b
AiEi
A ESC
1
=
50.0(10 ) C 200(109) D
-6
c -3.00(103)(1) + 6.00(103)(1.5) + 2.00(103)(1.25) d
= 0.850(10 - 3) m = 0.850 mm
Ans.
The positive sign indicates that end D moves away from the fixed support.
123
1.5 m
1.25 m
C
4 kN
D
2 kN
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*4–4. The A-36 steel rod is subjected to the loading
shown. If the cross-sectional area of the rod is 50 mm2,
determine the displacement of C. Neglect the size of the
couplings at B, C, and D.
1m
1.5 m
1.25 m
C
A 9 kN B
4 kN
D
2 kN
The normal forces developed in segments AB and BC are shown the FBDS of each
segment in Fig. a and b, respectively. The cross-sectional area of these two segments
2
1m
are A = A 50 mm2 B a
b = 50.0 (10 - 6) m2. Thus,
10.00 mm
dC = ©
PiLi
1
=
A P L + PBC LBC B
AiEi
A ESC AB AB
1
=
50.0(10 - 6) C 200(109) D
c -3.00(103)(1) + 6.00(103)(1.5) d
= 0.600 (10 - 3) m = 0.600 mm
Ans.
The positive sign indicates that coupling C moves away from the fixed support.
4–5. The assembly consists of a steel rod CB and an
aluminum rod BA, each having a diameter of 12 mm. If the rod
is subjected to the axial loadings at A and at the coupling B,
determine the displacement of the coupling B and the end
A. The unstretched length of each segment is shown in the
figure. Neglect the size of the connections at B and C, and
assume that they are rigid. Est = 200 GPa, Eal = 70 GPa.
dB =
C
18 kN
6 kN
3m
12(103)(3)
PL
= 0.00159 m = 1.59 mm
= p
2
9
AE
4 (0.012) (200)(10 )
dA = ©
A
B
2m
Ans.
18(103)(2)
12(103)(3)
PL
+ p
= p
2
9
2
9
AE
4 (0.012) (200)(10 )
4 (0.012) (70)(10 )
= 0.00614 m = 6.14 mm
Ans.
4–6. The bar has a cross-sectional area of 3 in2, and
E = 3511032 ksi. Determine the displacement of its end A
when it is subjected to the distributed loading.
x
w ⫽ 500x1/3 lb/in.
A
4 ft
P(x) =
L0
x
x
w dx = 500
x3 dx =
L0
1
1500 43
x
4
L
dA =
4(12)
P(x) dx
1
3
1
1500 4
1500
b a b(48)3
=
x3 dx = a
AE
4
(3)(35)(108)(4) 7
(3)(35)(106) L0
L0
dA = 0.0128 in.
Ans.
124
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4–7. The load of 800 lb is supported by the four 304 stainless
steel wires that are connected to the rigid members AB and
DC. Determine the vertical displacement of the load if the
members were horizontal before the load was applied. Each
wire has a cross-sectional area of 0.05 in2.
E
F
4 ft
H
D
C
2 ft
Referring to the FBD of member AB, Fig. a
5 ft
4.5 ft
a + ©MA = 0;
FBC (5) - 800(1) = 0
FBC = 160 lb
a + ©MB = 0;
800(4) - FAH (5) = 0
FAH = 640 lb
800 lb
A
B
1 ft
Using the results of FBC and FAH, and referring to the FBD of member DC, Fig. b
a + ©MD = 0;
FCF (7) - 160(7) - 640(2) = 0
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)(2)
FDE LDE
=
= 0.01567 in T
A Est
0.05 C 28.0 (106) D
dC =
342.86 (4)(12)
FCF LCF
=
= 0.01176 in T
A Est
0.05 C 28.0 (106) D
From the geometry shown in Fig. c,
dH = 0.01176 +
5
(0.01567 - 0.01176) = 0.01455 in T
7
Subsequently,
dA>H =
640(4.5)(12)
FAH LAH
=
= 0.02469 in T
A Est
0.05 C 28.0(106) D
dB>C =
160(4.5)(12)
FBC LBC
=
= 0.006171 in T
A Est
0.05 C 28.0(106) D
Thus,
A + T B dA = dH + dA>H = 0.01455 + 0.02469 = 0.03924 in T
A + T B dB = dC + dB>C = 0.01176 + 0.006171 = 0.01793 in T
From the geometry shown in Fig. d,
dP = 0.01793 +
4
(0.03924 - 0.01793) = 0.0350 in T
5
125
Ans.
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*4–8. The load of 800 lb is supported by the four
304 stainless steel wires that are connected to the rigid
members AB and DC. Determine the angle of tilt of each
member after the load is applied.The members were originally
horizontal, and each wire has a cross-sectional area of 0.05 in2.
E
F
4 ft
H
D
C
2 ft
Referring to the FBD of member AB, Fig. a,
5 ft
4.5 ft
a + ©MA = 0;
FBC (5) - 800(1) = 0
FBC = 160 lb
a + ©MB = 0;
800(4) - FAH (5) = 0
FAH = 640 lb
800 lb
A
B
1 ft
Using the results of FBC and FAH and referring to the FBD of member DC, Fig. b,
a + ©MD = 0;
FCF (7) - 160(7) - 640(2) = 0
FCF = 342.86 lb
a + ©MC = 0;
640(5) - FDE (7) = 0
FDE = 457.14 lb
Since E and F are fixed,
dD =
457.14 (4)(12)
FDE LDE
=
= 0.01567 in T
A Est
0.05 C 28.0(106) D
dC =
342.86 (4)(12)
FCF LCF
=
= 0.01176 in T
A Est
0.05 C 28.0(106) D
From the geometry shown in Fig. c
dH = 0.01176 +
u =
5
(0.01567 - 0.01176) = 0.01455 in T
7
0.01567 - 0.01176
= 46.6(10 - 6) rad
7(12)
Ans.
Subsequently,
dA>H =
640 (4.5)(12)
FAH LAH
=
= 0.02469 in T
A Est
0.05 C 28.0(106) D
dB>C =
160 (4.5)(12)
FBC LBC
=
= 0.006171 in T
A Est
0.05 C 28.0(106) D
Thus,
A + T B dA = dH + dA>H = 0.01455 + 0.02469 = 0.03924 in T
A + T B dB = dC + dB>C = 0.01176 + 0.006171 = 0.01793 in T
From the geometry shown in Fig. d
f =
0.03924 - 0.01793
= 0.355(10 - 3) rad
5(12)
Ans.
126
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4–8. Continued
127
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•4–9.
The assembly consists of three titanium (Ti-6A1-4V)
rods and a rigid bar AC. The cross-sectional area of each rod
is given in the figure. If a force of 6 kip is applied to the ring
F, determine the horizontal displacement of point F.
D
4 ft
C
ACD ⫽ 1 in2
2 ft
E
Internal Force in the Rods:
a + ©MA = 0;
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
6 ft
B
1 ft F
6 kip
2
1 ft AEF ⫽ 2 in
A
Displacement:
dC =
2.00(4)(12)
FCD LCD
= 0.0055172 in.
=
ACD E
(1)(17.4)(103)
dA =
4.00(6)(12)
FAB LAB
= 0.0110344 in.
=
AAB E
(1.5)(17.4)(103)
dF>E =
6.00(1)(12)
FEF LEF
= 0.0020690 in.
=
AEF E
(2)(17.4)(103)
œ
dE
0.0055172
=
;
2
3
œ
dE
= 0.0036782 in.
œ
dE = dC + dE
= 0.0055172 + 0.0036782 = 0.0091954 in.
dF = dE + dF>E
= 0.0091954 + 0.0020690 = 0.0113 in.
Ans.
4–10. The assembly consists of three titanium (Ti-6A1-4V)
rods and a rigid bar AC. The cross-sectional area of each rod
is given in the figure. If a force of 6 kip is applied to the ring
F, determine the angle of tilt of bar AC.
D
C
4 ft
ACD ⫽ 1 in
2
2 ft
E
Internal Force in the Rods:
a + ©MA = 0;
FCD(3) - 6(1) = 0
FCD = 2.00 kip
+ ©F = 0;
:
x
6 - 2.00 - FAB = 0
FAB = 4.00 kip
AAB ⫽ 1.5 in2
B
Displacement:
dC =
2.00(4)(12)
FCD LCD
= 0.0055172 in.
=
ACD E
(1)(17.4)(103)
dA =
4.00(6)(12)
FAB LAB
= 0.0110344 in.
=
AAB E
(1.5)(17.4)(103)
u = tan - 1
dA - dC
0.0110344 - 0.0055172
= tan - 1
3(12)
3(12)
= 0.00878°
Ans.
128
6 ft
A
1 ft F
6 kip
2
1 ft AEF ⫽ 2 in
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4–11. The load is supported by the four 304 stainless steel
wires that are connected to the rigid members AB and DC.
Determine the vertical displacement of the 500-lb load if
the members were originally horizontal when the load was
applied. Each wire has a cross-sectional area of 0.025 in2.
E
F
G
3 ft
5 ft
H
D
C
1 ft
2 ft
1.8 ft
I
Internal Forces in the wires:
A
FBD (b)
FBC(4) - 500(3) = 0
+ c ©Fy = 0;
FAH + 375.0 - 500 = 0
FAH = 125.0 lb
a + ©MD = 0;
FCF(3) - 125.0(1) = 0
FCF = 41.67 lb
+ c ©Fy = 0;
FDE + 41.67 - 125.0 = 0
FDE = 83.33 lb
FBC = 375.0 lb
FBD (a)
Displacement:
dD =
83.33(3)(12)
FDELDE
= 0.0042857 in.
=
ADEE
0.025(28.0)(106)
dC =
41.67(3)(12)
FCFLCF
= 0.0021429 in.
=
ACFE
0.025(28.0)(106)
œ
dH
= 0.0014286 in.
dH = 0.0014286 + 0.0021429 = 0.0035714 in.
dA>H =
125.0(1.8)(12)
FAHLAH
= 0.0038571 in.
=
AAHE
0.025(28.0)(106)
dA = dH + dA>H = 0.0035714 + 0.0038571 = 0.0074286 in.
dB =
375.0(5)(12)
FBGLBG
= 0.0321428 in.
=
ABGE
0.025(28.0)(106)
dlœ
0.0247143
=
;
3
4
1 ft
500 lb
a + ©MA = 0;
œ
dH
0.0021429
=
;
2
3
B
3 ft
dlœ = 0.0185357 in.
dl = 0.0074286 + 0.0185357 = 0.0260 in.
Ans.
129
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*4–12. The load is supported by the four 304 stainless
steel wires that are connected to the rigid members AB and
DC. Determine the angle of tilt of each member after the
500-lb load is applied. The members were originally
horizontal, and each wire has a cross-sectional area of
0.025 in2.
E
F
G
3 ft
5 ft
H
D
C
1 ft
2 ft
1.8 ft
I
Internal Forces in the wires:
A
FBD (b)
FBG(4) - 500(3) = 0
+ c ©Fy = 0;
FAH + 375.0 - 500 = 0
FAH = 125.0 lb
a + ©MD = 0;
FCF(3) - 125.0(1) = 0
FCF = 41.67 lb
+ c ©Fy = 0;
FDE + 41.67 - 125.0 = 0
FDE = 83.33 lb
FBG = 375.0 lb
FBD (a)
Displacement:
dD =
83.33(3)(12)
FDELDE
= 0.0042857 in.
=
ADEE
0.025(28.0)(106)
dC =
41.67(3)(12)
FCFLCF
= 0.0021429 in.
=
ACFE
0.025(28.0)(106)
œ
dH
0.0021429
=
;
2
3
œ
dH
= 0.0014286 in.
œ
+ dC = 0.0014286 + 0.0021429 = 0.0035714 in.
dH = dH
tan a =
0.0021429
;
36
dA>H =
125.0(1.8)(12)
FAHLAH
= 0.0038571 in.
=
AAHE
0.025(28.0)(106)
a = 0.00341°
Ans.
dA = dH + dA>H = 0.0035714 + 0.0038571 = 0.0074286 in.
375.0(5)(12)
FBGLBG
= 0.0321428 in.
=
ABGE
0.025(28.0)(106)
tan b =
1 ft
500 lb
a + ©MA = 0;
dB =
B
3 ft
0.0247143
;
48
b = 0.0295°
Ans.
130
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•4–13. The bar has a length L and cross-sectional area A.
Determine its elongation due to the force P and its own
weight.The material has a specific weight g (weight>volume)
and a modulus of elasticity E.
d =
=
L
P(x) dx
1
=
(gAx + P) dx
AE L0
L A(x) E
L
gAL2
gL2
1
PL
a
+ PLb =
+
AE
2
2E
AE
Ans.
P
4–14. The post is made of Douglas fir and has a diameter
of 60 mm. If it is subjected to the load of 20 kN and the soil
provides a frictional resistance that is uniformly distributed
along its sides of w = 4 kN>m, determine the force F at its
bottom needed for equilibrium.Also, what is the displacement
of the top of the post A with respect to its bottom B?
Neglect the weight of the post.
20 kN
A
y
w
2m
Equation of Equilibrium: For entire post [FBD (a)]
+ c ©Fy = 0;
F + 8.00 - 20 = 0
B
F = 12.0 kN
Ans.
Internal Force: FBD (b)
+ c ©Fy = 0;
-F(y) + 4y - 20 = 0
F(y) = {4y - 20} kN
Displacement:
L
dA>B =
2m
F(y)dy
1
=
(4y - 20)dy
AE L0
L0 A(y)E
=
2m
1
A 2y2 - 20y B 冷0
AE
= -
32.0 kN # m
AE
32.0(103)
= -p
2
9
4 (0.06 ) 13.1 (10 )
= - 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
9
4 (0.06 ) 13.1 (10 )
= -1.026 A 10 - 3 B m
Ans.
= -1.03 mm
Negative sign indicates that end A moves toward end B.
132
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*4–16. The linkage is made of two pin-connected A-36
steel members, each having a cross-sectional area of 1.5 in2.
If a vertical force of P = 50 kip is applied to point A,
determine its vertical displacement at A.
P
A
2 ft
B
C
1.5 ft
Analysing the equilibrium of Joint A by referring to its FBD, Fig. a,
+ ©F = 0 ;
:
x
3
3
FAC a b - FAB a b = 0
5
5
+ c ©Fy = 0
4
-2Fa b - 50 = 0
5
The
initial
FAC = FAB = F
F = -31.25 kip
length
of
members
AB
and
AC
is
12
in
b = 30 in. The axial deformation of members
L = 21.52 + 22 = (2.50 ft)a
1 ft
AB and AC is
d =
(-31.25)(30)
FL
=
= -0.02155 in.
AE
(1.5) C 29.0(103) D
The negative sign indicates that end A moves toward B and C. From the geometry
1.5
shown in Fig. b, u = tan - 1 a
b = 36.87°. Thus,
2
A dA B g =
d
0.02155
=
= 0.0269 in. T
cos u
cos 36.87°
Ans.
133
1.5 ft
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•4–17.
The linkage is made of two pin-connected A-36
steel members, each having a cross-sectional area of 1.5 in2.
Determine the magnitude of the force P needed to displace
point A 0.025 in. downward.
P
A
2 ft
Analysing the equilibrium of joint A by referring to its FBD, Fig. a
+ ©F = 0;
:
x
3
3
FAC a b - FAB a b = 0
5
5
+ c ©Fy = 0;
4
-2Fa b - P = 0
5
The
initial
B
1.5 ft
F = -0.625 P
of
members
AB
and
AC
are
12 in
b = 30 in. The axial deformation of members
L = 21.5 + 2 = (2.50 ft)a
1 ft
AB and AC is
2
length
2
d =
C
FAC = FAB = F
-0.625P(30)
FL
=
= -0.4310(10 - 3) P
AE
(1.5) C 29.0(103) D
The negative sign indicates that end A moves toward B and C. From the geometry
1.5
shown in Fig. b, we obtain u = tan - 1 a
b = 36.87°. Thus
2
(dA)g =
d
cos u
0.025 =
0.4310(10 - 3) P
cos 36.87°
P = 46.4 kips
Ans.
134
1.5 ft
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4–18. The assembly consists of two A-36 steel rods and a
rigid bar BD. Each rod has a diameter of 0.75 in. If a force
of 10 kip is applied to the bar as shown, determine the
vertical displacement of the load.
C
A
3 ft
2 ft
B
Here, FEF = 10 kip. Referring to the FBD shown in Fig. a,
1.25 ft
a + ©MB = 0;
FCD (2) - 10(1.25) = 0
FCD = 6.25 kip
a + ©MD = 0;
10(0.75) - FAB(2) = 0
FAB = 3.75 kip
The cross-sectional area of the rods is A =
A and C are fixed,
3.75 (2)(12)
FAB LAB
=
= 0.007025 in. T
A Est
0.140625p C 29.0(103) D
dD =
6.25(3)(12)
FCD LCD
=
= 0.01756 in T
A Est
0.140625p C 29.0(103) D
From the geometry shown in Fig. b
dE = 0.007025 +
1.25
(0.01756 - 0.00725) = 0.01361 in. T
2
Here,
dF>E =
10 (1) (12)
FEF LEF
=
= 0.009366 in T
A Est
0.140625p C 29.0(103) D
Thus,
A + T B dF = dE + dF>E = 0.01361 + 0.009366 = 0.0230 in T
Ans.
135
D
0.75 ft
F
p
(0.752) = 0.140625p in2. Since points
4
dB =
E
10 kip
1 ft
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4–19. The assembly consists of two A-36 steel rods and a
rigid bar BD. Each rod has a diameter of 0.75 in. If a force
of 10 kip is applied to the bar, determine the angle of tilt of
the bar.
C
A
Here, FEF = 10 kip. Referring to the FBD shown in Fig. a,
3 ft
a + ©MB = 0;
FCD (2) - 10(1.25) = 0
FCD = 6.25 kip
a + ©MD = 0;
10(0.75) - FAB (2) = 0
FAB = 3.75 kip
2 ft
B
p
The cross-sectional area of the rods is A = (0.752) = 0.140625p in2. Since points
4
A and C are fixed then,
dB =
3.75 (2)(12)
FAB LAB
=
= 0.007025 in T
A Est
0.140625p C 29.0(103) D
dD =
6.25 (3)(12)
FCD LCD
=
= 0.01756 in T
A Est
0.140625p C 29.0(103) D
0.01756 - 0.007025
= 0.439(10 - 3) rad
2(12)
Ans.
136
D
0.75 ft
F
10 kip
From the geometry shown in Fig. b,
u =
1.25 ft
E
1 ft
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*4–20. The rigid bar is supported by the pin-connected
rod CB that has a cross-sectional area of 500 mm2 and is
made of A-36 steel. Determine the vertical displacement of
the bar at B when the load is applied.
C
3m
45 kN/m
Force In The Rod. Referring to the FBD of member AB, Fig. a
a + ©MA = 0;
3
1
1
FBC a b (4) - (45)(4) c (4) d = 0
5
2
3
4m
Displacement. The initial length of rod BC is LBC = 232 + 42 = 5 m. The axial
deformation of this rod is
dBC =
50.0(103)(5)
FBC LBC
=
= 2.50 (10 - 3) m
ABC Est
0.5(10 - 3) C 200(109) D
3
From the geometry shown in Fig. b, u = tan - 1 a b = 36.87°. Thus,
4
(dB)g =
2.50(10 - 3)
dBC
=
= 4.167 (10 - 3) m = 4.17 mm
sin u
sin 36.87°
137
B
A
FBC = 50.0 kN
Ans.
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•4–21.
A spring-supported pipe hanger consists of two
springs which are originally unstretched and have a stiffness
of k = 60 kN>m, three 304 stainless steel rods, AB and CD,
which have a diameter of 5 mm, and EF, which has a
diameter of 12 mm, and a rigid beam GH. If the pipe and
the fluid it carries have a total weight of 4 kN, determine the
displacement of the pipe when it is attached to the support.
F
B
D
k
G
0.75 m
0.75 m
k
H
E
A
C
Internal Force in the Rods:
0.25 m 0.25 m
FBD (a)
a + ©MA = 0;
FCD (0.5) - 4(0.25) = 0
+ c ©Fy = 0;
FAB + 2.00 - 4 = 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 =
4.00(103)(750)
FEFLEF
= 0.1374 mm
= p
2
9
AEFE
4 (0.012) (193)(10 )
dA>B = dC>D =
2(103)(750)
PCDLCD
= 0.3958 mm
= p
2
9
ACDE
4 (0.005) (193)(10 )
dC = dD + dC>D = 0.1374 + 0.3958 = 0.5332 mm
Displacement of the spring
dsp =
Fsp
k
=
2.00
= 0.0333333 m = 33.3333 mm
60
dlat = dC + dsp
= 0.5332 + 33.3333 = 33.9 mm
Ans.
138
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4–22. A spring-supported pipe hanger consists of two
springs, which are originally unstretched and have a
stiffness of k = 60 kN>m, three 304 stainless steel rods, AB
and CD, which have a diameter of 5 mm, and EF, which has
a diameter of 12 mm, and a rigid beam GH. If the pipe is
displaced 82 mm when it is filled with fluid, determine the
weight of the fluid.
F
B
D
k
G
0.75 m
0.75 m
k
H
E
Internal Force in the Rods:
A
C
FBD (a)
a + ©MA = 0;
FCD(0.5) - W(0.25) = 0
FCD =
+ c ©Fy = 0;
FAB +
W
- W = 0
2
W
2
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 =
W(750)
FEFLEF
= p
2
9
AEFE
4 (0.012) (193)(10 )
= 34.35988(10 - 6) W
dA>B = dC>D =
W
FCDLCD
2 (750)
= p
2
9
ACDE
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
h
P(x) dx
P
dx
=
d =
[d1h + ( d 2 - d1 )x ] t
A(x)E
E
L0
L
h
h
=
Ph
dx
E t L0 d1 h + (d2 - d1)x
d1
P
h
dx
Ph
=
E t d1 h L0 1 + d2 - d1
d1 h
h
d1 h
d2 - d1
Ph
=
a
b cln a1 +
xb d ƒ
E t d1 h d2 - d1
d1 h
0
x
d2 - d1
d1 + d2 - d1
Ph
Ph
cln a1 +
bd =
cln a
bd
=
E t(d2 - d1)
d1
E t(d2 - d1)
d1
=
d2
Ph
cln d
E t(d2 - d1)
d1
Ans.
140
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4–25. Determine the elongation of the A-36 steel member
when it is subjected to an axial force of 30 kN. The member
is 10 mm thick. Use the result of Prob. 4–24.
20 mm
30 kN
30 kN
75 mm
0.5 m
Using the result of prob. 4-24 by substituting d1 = 0.02 m, d2 = 0.075 m t = 0.01 m
and L = 0.5 m.
d = 2c
= 2c
d2
PL
ln d
Est t(d2 - d1) d1
30(103) (0.5)
9
200(10 )(0.01)(0.075 - 0.02)
ln a
0.075
bd
0.02
= 0.360(10 - 3) m = 0.360 mm
Ans.
4–26. The casting is made of a material that has a specific
weight g and modulus of elasticity E. If it is formed into a
pyramid having the dimensions shown, determine how far
its end is displaced due to gravity when it is suspended in
the vertical position.
b0
b0
L
Internal Forces:
+ c ©Fz = 0;
P(z) -
1
gAz = 0
3
P(z) =
1
gAz
3
Displacement:
L
d =
P(z) dz
L0 A(z) E
L1
3 gAz
=
L0 A E
dz
=
L
g
z dz
3E L0
=
gL2
6E
Ans.
141
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4–27. The circular bar has a variable radius of r = r0eax
and is made of a material having a modulus of elasticity
of E. Determine the displacement of end A when it is
subjected to the axial force P.
L
Displacements: The cross-sectional area of the bar as a function of x is
A(x) = pr2 = pr0 2e2ax. We have
x
B
L
d =
L
P(x)dx
P
dx
=
2
A(x)E
pr0 E L0 e2ax
L0
r0
r ⫽ r0 eax
L
1
P
2
c
d
=
pr0 2E 2ae2ax 0
= -
A
P
a1 - e - 2aL b
2apr0 2E
P
Ans.
*4–28. The pedestal is made in a shape that has a radius
defined by the function r = 2>12 + y1>22 ft, where y is
in feet. If the modulus of elasticity for the material is
E = 1411032 psi, determine the displacement of its top
when it supports the 500-lb load.
y
500 lb
0.5 ft
r⫽
2
(2 ⫹ y 1/ 2)
4 ft
d =
=
P(y) dy
L A(y) E
y
4
dy
500
3
14(10 )(144) L0 p(2 + y2-1
2
)
2
1 ft
4
-3
= 0.01974(10 )
L0
r
1
2
(4 + 4y + y) dy
4
2 3
1
= 0.01974(10 - 3)c4y + 4 a y2 b + y2 d
3
2
0
= 0.01974(10 - 3)(45.33)
= 0.8947(10 - 3) ft = 0.0107 in.
Ans.
142
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•4–29.
The support is made by cutting off the two
opposite sides of a sphere that has a radius r0 . If the original
height of the support is r0>2, determine how far it shortens
when it supports a load P. The modulus of elasticity is E.
P
r0
Geometry:
A = p r2 = p(r0 cos u)2 = p r20 cos2 u
y = r0 sin u;
dy = r0 cos u du
Displacement:
L
P(y) dy
L0 A(y) E
d =
= 2B
When y =
u
u
r0 cos u du
P
P
du
=
2
R
B
R
E L0 p r20 cos2 u
p r0 E L0 cos u
=
u
2P
[ln (sec u + tan u)] 2
p r0 E
0
=
2P
[ln (sec u + tan u)]
p r0 E
r0
;
4
u = 14.48°
d =
=
2P
[ln (sec 14.48° + tan 14.48°)]
p r0 E
0.511P
p r0 E
Ans.
Also,
Geometry:
A (y) = p x2 = p (r20 - y2)
Displacement:
L
d =
P(y) dy
L0 A(y) E
0
0
r0 + y p
2P 1
2P p dy
=
ln
=
B
R 2
2
2
pE L0 r0 - y
p E 2r0 r0 - y 0
=
P
[ln 1.667 - ln 1]
p r0 E
=
0.511 P
p r0 E
Ans.
143
r0
2
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4–30. The weight of the kentledge exerts an axial force of
P ⫽ 1500 kN on the 300-mm diameter high strength
concrete bore pile. If the distribution of the resisting skin
friction developed from the interaction between the soil
and the surface of the pile is approximated as shown, and
the resisting bearing force F is required to be zero,
determine the maximum intensity p0 kN>m for equilibrium.
Also, find the corresponding elastic shortening of the pile.
Neglect the weight of the pile.
P
p0
12 m
Internal Loading: By considering the equilibrium of the pile with reference to its
entire free-body diagram shown in Fig. a. We have
1
p (12) - 1500 = 0
2 0
+ c ©Fy = 0;
p0 = 250 kN>m
Ans.
Thus,
p(y) =
250
y = 20.83y kN>m
12
The normal force developed in the pile as a function of y can be determined by
considering the equilibrium of a section of the pile shown in Fig. b.
1
(20.83y)y - P(y) = 0
2
+ c ©Fy = 0;
P(y) = 10.42y2 kN
Displacement: The cross-sectional area of the pile is A =
p
(0.32) = 0.0225p m2.
4
We have
L
d =
12 m
P(y)dy
10.42(103)y2dy
=
0.0225p(29.0)(109)
L0
L0 A(y)E
12 m
=
L0
5.0816(10 - 6)y2dy
= 1.6939(10 - 6)y3 冷 0
12 m
= 2.9270(10 - 3)m = 2.93 mm
Ans.
144
F
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4–31. The column is constructed from high-strength
concrete and six A-36 steel reinforcing rods. If it is subjected
to an axial force of 30 kip, determine the average normal
stress in the concrete and in each rod. Each rod has a
diameter of 0.75 in.
4 in.
30 kip
Equations of Equilibrium:
6Pst + Pcon - 30 = 0
+ c ©Fy = 0;
3 ft
[1]
Compatibility:
dst = dcon
Pcon(3)(12)
Pst(3)(12)
p
2
(0.75
)(29.0)(103)
4
=
[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:
Pst
1.388
= 3.14 ksi
= p
2
Ast
(0.75
)
4
Ans.
Pcon
21.670
=
= 0.455 ksi
p 2
Acon
(8 ) - 6 A p B (0.752)
Ans.
sst =
scon =
4
4
*4–32. The column is constructed from high-strength
concrete and six A-36 steel reinforcing rods. If it is subjected
to an axial force of 30 kip, determine the required diameter
of each rod so that one-fourth of the load is carried by the
concrete and three-fourths by the steel.
4 in.
30 kip
Equilibrium: The force of 30 kip is required to distribute in such a manner that 3/4
of the force is carried by steel and 1/4 of the force is carried by concrete. Hence
Pst =
3
(30) = 22.5 kip
4
Pcon =
1
(30) = 7.50 kip
4
3 ft
Compatibility:
dst = dcon
PstL
Pcon L
=
AstEst
Acon Econ
Ast =
22.5Acon Econ
7.50 Est
3 C 4 (8 ) - 6 A 4 B d D (4.20)(10 )
p
6 a bd2 =
4
29.0(103)
p
2
p
2
3
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
Pst L
p
2
2
(0.08
0.07
) (200) (109)
4
= p
2
9
4 (0.07 ) (24) (10 )
Pst = 2.5510 Pcon
(2)
Solving Eqs. (1) and (2) yields
Pst = 57.47 kN
sst =
Pcon = 22.53 kN
57.47 (103)
Pst
= 48.8 MPa
= p
2
2
Ast
4 (0.08 - 0.07 )
Ans.
22.53 (103)
Pcon
= 5.85 MPa
= p
2
Acon
4 (0.07 )
Ans.
scon =
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 =
Pbr
2.9457
= 0.341 ksi
= p 2
Abr
(6
- 52)
4
Ans.
sst =
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 =
AbrEbr
Est
p
2
2
3
p
4 (6 - 5 )(14.6)(10 )
a b d2 =
4
28.0(103)
d = 2.39 in.
Ans.
*4–36. The composite bar consists of a 20-mm-diameter
A-36 steel segment AB and 50-mm-diameter red brass
C83400 end segments DA and CB. Determine the average
normal stress in each segment due to the applied load.
+ ©F = 0;
;
x
(1)
0 = ¢ D - dD
50(0.25)
150(0.5)
0 = p
2
9
4 (0.02) (200)(10 )
- p
2
9
4 (0.05 )(101)(10 )
FD(0.5)
FD(0.5)
- p
2
9
4 (0.05 )(101)(10 )
500 mm
50 mm
20 mm
75 kN 100 kN
D
FC - FD + 75 + 75 - 100 - 100 = 0
FC - FD - 50 = 0
+
;
250 mm
- 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
A
75 kN
100 kN B
250 mm
C
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•4–37.
The composite bar consists of a 20-mm-diameter
A-36 steel segment AB and 50-mm-diameter red brass
C83400 end segments DA and CB. Determine
the displacement of A with respect to B due to the
applied load.
250 mm
500 mm
50 mm
20 mm
75 kN 100 kN
D
+
;
A
75 kN
250 mm
100 kN B
C
0 = ¢ D - dD
150(103)(500)
0 = p
50(103)(250)
2
9
4 (0.02 )(200)(10 )
- p
2
9
4 (0.05 )(101)(10 )
FD(500)
FD(500)
- p
2
9
4 (0.05 )(101)(10 )
- p
2
9
4 (0.02) (200)(10 )
FD = 107.89 kN
Displacement:
dA>B =
42.11(103)(500)
PABLAB
= p
2
9
AABEst
4 (0.02 )200(10 )
= 0.335 mm
Ans.
4–38. The A-36 steel column, having a cross-sectional area
of 18 in2, is encased in high-strength concrete as shown. If
an axial force of 60 kip is applied to the column, determine
the average compressive stress in the concrete and in the
steel. How far does the column shorten? It has an original
length of 8 ft.
16 in.
Pst + Pcon - 60 = 0
+ c ©Fy = 0;
dst = dcon ;
60 kip
Pst(8)(12)
18(29)(103)
(1)
Pcon(8)(12)
=
[(9)(16) - 18](4.20)(103)
Pst = 0.98639 Pcon
(2)
Solving Eqs. (1) and (2) yields
Pst = 29.795 kip;
sst =
Pcon = 30.205 kip
Pst
29.795
=
= 1.66 ksi
Ast
18
Ans.
Pcon
30.205
=
= 0.240 ksi
Acon
9(16) - 18
Ans.
scon =
Either the concrete or steel can be used for the deflection calculation.
d =
29.795(8)(12)
PstL
= 0.0055 in.
=
AstE
18(29)(103)
Ans.
148
9 in.
8 ft
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4–39. The A-36 steel column is encased in high-strength
concrete as shown. If an axial force of 60 kip is applied to
the column, determine the required area of the steel so that
the force is shared equally between the steel and concrete.
How far does the column shorten? It has an original length
of 8 ft.
60 kip
16 in.
9 in.
8 ft
The force of 60 kip is shared equally by the concrete and steel. Hence
Pst = Pcon = P = 30 kip
dcon = dst;
Ast =
PL
PL
=
Acon Econ
Ast Est
[9(16) - Ast] 4.20(103)
AconEcon
=
Est
29(103)
= 18.2 in2
d =
Ans.
30(8)(12)
PstL
= 0.00545 in.
=
AstEst
18.2(29)(103)
Ans.
*4–40. The rigid member is held in the position shown by
three A-36 steel tie rods. Each rod has an unstretched length
of 0.75 m and a cross-sectional area of 125 mm2. Determine
the forces in the rods if a turnbuckle on rod EF undergoes
one full turn. The lead of the screw is 1.5 mm. Neglect the
size of the turnbuckle and assume that it is rigid. Note: The
lead would cause the rod, when unloaded, to shorten 1.5 mm
when the turnbuckle is rotated one revolution.
B
D
0.75 m
E
A
0.5 m
0.5 m
C
0.75 m
a + ©ME = 0;
-TAB(0.5) + TCD(0.5) = 0
F
TAB = TCD = T
+ T ©Fy = 0;
(1)
TEF - 2T = 0
TEF = 2T
(2)
Rod EF shortens 1.5mm causing AB (and DC) to elongate. Thus:
0.0015 = dA>B + dE>F
0.0015 =
T(0.75)
-6
2T(0.75)
9
(125)(10 )(200)(10 )
+
(125)(10 - 6)(200)(109)
2.25T = 37500
T = 16666.67 N
TAB = TCD = 16.7 kN
Ans.
TEF = 33.3 kN
Ans.
149
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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
+ c ©Fy = 0;
Pcon + 6Pst - 900 = 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
Pcon L
0.25(0.375) - 6(p4 )(0.022)
D C 25(10 ) D
Pst L
=
9
C
(p4 )(0.022) 200(109)
D
Pcon = 36.552 Pst
(2)
Solving Eqs (1) and (2) yields
Pst = 21.15 kN
Pcon = 773.10 kN
Thus,
scon =
773.10(103)
Pcon
= 8.42 MPa
=
Acon
0.15(0.375) - 6(p4 )(0.022)
Ans.
sst =
21.15(103)
Pst
= 67.3 MPa
= p
2
Ast
4 (0.02 )
Ans.
150
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4–42. The post is constructed from concrete and six A-36
steel reinforcing rods. If it is subjected to an axial force of
900 kN, determine the required diameter of each rod so that
one-fifth of the load is carried by the steel and four-fifths by
the concrete. Est = 200 GPa, Ec = 25 GPa.
900 kN
250 mm
375 mm
The normal force in each steel rod is
Pst =
1
5 (900)
6
= 30 kN
The normal force in concrete is
Pcon =
4
(900) = 720 kN
5
Since the steel rods and the concrete are firmly bonded, their deformation must be
the same. Thus
dcon = dst
Pcon L
Pst L
=
Acon Econ
Ast Est
720(103) L
30(103)L
C 0.25(0.375) - 6(p4 d2) D C 25(109) D
=
p 2
4 d
49.5p d2 = 0.09375
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,
2F + FEF - 2 C 40(103) D = 0
+ ©F = 0;
:
x
(1)
Compatibility Equation: Using the method of superposition, Fig. b,
+ B 0 = -d + d
A:
P
EF
0 = -p
40(103)(300)
2
9
4 (0.03 )(101)(10 )
+ cp
FEF (450)
2
9
4 (0.04 )(193)(10 )
+ p
A FEF>2 B (300)
2
9
4 (0.03 )(101)(10 )
FEF = 42 483.23 N
Substituting this result into Eq. (1),
F = 18 758.38 N
151
d
30 mm
40 kN
D
G
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4–43.
Continued
Normal Stress: We have,
sAB = sCD =
sEF =
F
18 758.38
= 26.5 MPa
= p
2
ACD
4 (0.03 )
Ans.
FEF
42 483.23
= 33.8 MPa
= p
2
AEF
4 (0.04 )
Ans.
152
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*4–44. The two pipes are made of the same material and
are connected as shown. If the cross-sectional area of BC is
A and that of CD is 2A, determine the reactions at B and D
when a force P is applied at the junction C.
B
L
–
2
Equations of Equilibrium:
+ ©F = 0;
;
x
FB + FD - P = 0
[1]
Compatibility:
+ B
A:
0 = dP - dB
0 =
0 =
P A L2 B
2AE
- C
FB A L2 B
AE
+
FB A L2 B
2AE
S
3FBL
PL
4AE
4AE
FB =
P
3
Ans.
From Eq. [1]
FD =
C
2
P
3
Ans.
153
D
P
L
–
2
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•4–45.
The bolt has a diameter of 20 mm and passes
through a tube that has an inner diameter of 50 mm and an
outer diameter of 60 mm. If the bolt and tube are made of
A-36 steel, determine the normal stress in the tube and bolt
when a force of 40 kN is applied to the bolt. Assume the end
caps are rigid.
160 mm
40 kN
Referring to the FBD of left portion of the cut assembly, Fig. a
+ ©F = 0;
:
x
40(103) - Fb - Ft = 0
(1)
Here, it is required that the bolt and the tube have the same deformation. Thus
dt = db
Ft(150)
C
p
2
2
9
4 (0.06 - 0.05 ) 200(10 )
D
Fb(160)
=
C
p
2
9
4 (0.02 ) 200(10 )
D
Ft = 2.9333 Fb
(2)
Solving Eqs (1) and (2) yields
Fb = 10.17 (103) N
40 kN
150 mm
Ft = 29.83 (103) N
Thus,
sb =
10.17(103)
Fb
= 32.4 MPa
= p
2
Ab
4 (0.02 )
Ans.
st =
29.83 (103)
Ft
= 34.5 MPa
= p
2
2
At
4 (0.06 - 0.05 )
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.
50 mm
Equation of Equilibrium: Referring to the free-body diagram of the assembly
shown in Fig. a,
200(103) - FD - FA = 0
(1)
Compatibility Equation: Using the method of superposition, Fig. b,
+ B
A:
d = dP - dFD
0.15 = p
200(103)(600)
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)
+ p
2
9
4 (0.025 )(200)(10 )
FD = 20 365.05 N = 20.4 kN
Ans.
Substituting this result into Eq. (1),
FA = 179 634.95 N = 180 kN
Ans.
155
S
D
B
25 mm
C
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4–47. Two A-36 steel wires are used to support the 650-lb
engine. Originally, AB is 32 in. long and A¿B¿ is 32.008 in.
long. Determine the force supported by each wire when the
engine is suspended from them. Each wire has a crosssectional area of 0.01 in2.
B¿ B
A¿ A
TA¿B¿ + TAB - 650 = 0
+ c ©Fy = 0;
(1)
dAB = dA¿B¿ + 0.008
TA¿B¿ (32.008)
TAB (32)
(0.01)(29)(106)
=
(0.01)(29)(106)
+ 0.008
32TAB - 32.008TA¿B¿ = 2320
TAB = 361 lb
Ans.
TA¿B¿ = 289 lb
Ans.
*4–48. Rod AB has a diameter d and fits snugly between
the rigid supports at A and B when it is unloaded. The
modulus of elasticity is E. Determine the support reactions
at A and B if the rod is subjected to the linearly distributed
axial load.
p⫽
A
1
p L - FA - FB = 0
2 0
+ ©F = 0;
:
x
(1)
Compatibility Equation: Using the method of superposition, Fig. b,
+ B
A:
0 = dP - dFA
L
0 =
L0
FA (L)
P(x)dx
AE
AE
L
0 =
L0
B
x
Equation of Equilibrium: Referring to the free-body diagram of rod AB shown in
Fig. a,
P(x)dx - FAL
156
p0
p0
x
L
L
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4–48.
Continued
Here, P(x) =
p0 2
1 p0
a xb x =
x . Thus,
2 L
2L
0 =
L
p0
x2 dx - FAL
2L L0
0 =
p0 x3 L
¢ ≤ ` - FAL
2L 3 0
FA =
p0L
6
Ans.
Substituting this result into Eq. (1),
FB =
p0L
3
Ans.
157
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•4–49.
The tapered member is fixed connected at its ends
A and B and is subjected to a load P = 7 kip at x = 30 in.
Determine the reactions at the supports. The material is
2 in. thick and is made from 2014-T6 aluminum.
A
B
3 in.
P
6 in.
x
60 in.
y
1.5
=
120 - x
60
y = 3 - 0.025 x
+ ©F = 0;
:
x
FA + FB - 7 = 0
(1)
dA>B = 0
30
-
L0
60
FA dx
FBdx
+
= 0
2(3 - 0.025 x)(2)(E)
L30 2(3 - 0.025 x)(2)(E)
30
-FA
L0
60
dx
dx
+ FB
= 0
(3 - 0.025 x)
L30 (3 - 0.025x)
60
40 FA ln(3 - 0.025 x)|30
0 - 40 FB ln(3 - 0.025x)|30 = 0
-FA(0.2876) + 0.40547 FB = 0
FA = 1.40942 FB
Thus, from Eq. (1).
FA = 4.09 kip
Ans.
FB = 2.91 kip
Ans.
158
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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.
159
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4–51. The rigid bar supports the uniform distributed load
of 6 kip>ft. Determine the force in each cable if each cable
has a cross-sectional area of 0.05 in2, and E = 3111032 ksi.
C
6 ft
6 kip/ft
A
D
B
3 ft
a + ©MA = 0;
TCB a
u = tan - 1
6
= 45°
6
2
25
b(3) - 54(4.5) + TCD a
2
25
b9 = 0
(1)
L2B¿C¿ = (3)2 + (8.4853)2 - 2(3)(8.4853) cos u¿
Also,
L2D¿C¿ = (9)2 + (8.4853)2 - 2(9)(8.4853) cos u¿
(2)
Thus, eliminating cos u¿ .
-L2B¿C¿(0.019642) + 1.5910 = -L2D¿C¿(0.0065473) + 1.001735
L2B¿C¿(0.019642) = 0.0065473 L2D¿C¿ + 0.589256
L2B¿C¿ = 0.333 L2D¿C¿ + 30
But,
LB¿C = 245 + dBC¿ ,
LD¿C = 245 + dDC¿
Neglect squares or d¿ B since small strain occurs.
L2D¿C = (245 + dBC)2 = 45 + 2 245 dBC
L2D¿C = (245 + dDC)2 = 45 + 2 245 dDC
45 + 2245 dBC = 0.333(45 + 2245 dDC) + 30
2 245 dBC = 0.333(2245 dDC)
dDC = 3dBC
Thus,
TCD 245
TCB 245
= 3
AE
AE
TCD = 3 TCB
From Eq. (1).
TCD = 27.1682 kip = 27.2 kip
Ans.
TCB = 9.06 kip
Ans.
160
3 ft
3 ft
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*4–52. The rigid bar is originally horizontal and is
supported by two cables each having a cross-sectional area
of 0.05 in2, and E = 3111032 ksi. Determine the slight
rotation of the bar when the uniform load is applied.
C
See solution of Prob. 4-51.
6 ft
TCD = 27.1682 kip
dDC =
TCD 245
0.05(31)(103)
27.1682245
= 0.1175806 ft
0.05(31)(103)
=
6 kip/ft
A
D
B
Using Eq. (2) of Prob. 4-51,
3 ft
3 ft
3 ft
(245 + 0.1175806)2 = (9)2 + (8.4852)2 - 2(9)(8.4852) cos u¿
u¿ = 45.838°
Thus,
¢u = 45.838° - 45° = 0.838°
Ans.
•4–53.
The press consists of two rigid heads that are held
together by the two A-36 steel 12-in.-diameter rods. A 6061T6-solid-aluminum cylinder is placed in the press and the
screw is adjusted so that it just presses up against the
cylinder. If it is then tightened one-half turn, determine
the average normal stress in the rods and in the cylinder.
The single-threaded screw on the bolt has a lead of 0.01 in.
Note: The lead represents the distance the screw advances
along its axis for one complete turn of the screw.
12 in.
2 in.
10 in.
+ ©F = 0;
:
x
2Fst - Fal = 0
dst = 0.005 - dal
Fst(12)
p
( 4 )(0.5)2(29)(103)
= 0.005 -
Fal(10)
p(1)2(10)(103)
Solving,
Fst = 1.822 kip
Fal = 3.644 kip
srod =
Fst
1.822
= p
= 9.28 ksi
Ast
( 4 )(0.5)2
Ans.
scyl =
Fal
3.644
=
= 1.16 ksi
Aal
p(1)2
Ans.
161
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4–54. The press consists of two rigid heads that are held
together by the two A-36 steel 12-in.-diameter rods. A 6061T6-solid-aluminum cylinder is placed in the press and the
screw is adjusted so that it just presses up against the
cylinder. Determine the angle through which the screw can
be turned before the rods or the specimen begin to yield.
The single-threaded screw on the bolt has a lead of 0.01 in.
Note: The lead represents the distance the screw advances
along its axis for one complete turn of the screw.
12 in.
2 in.
10 in.
+ ©F = 0;
:
x
2Fst - Fal = 0
dst = d - dal
Fst(12)
(p4 )(0.5)2(29)(103)
= d-
Fal(10)
(1)
p(1)2(10)(103)
Assume steel yields first,
sY = 36 =
Fst
(p4 )(0.5)2
;
Fst = 7.068 kip
Then Fal = 14.137 kip;
sal =
14.137
= 4.50 ksi
p(1)2
4.50 ksi 6 37 ksi steel yields first as assumed. From Eq. (1),
d = 0.01940 in.
Thus,
0.01940
u
=
360°
0.01
u = 698°
Ans.
162
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4–55. The three suspender bars are made of A-36 steel
and have equal cross-sectional areas of 450 mm2.
Determine the average normal stress in each bar if the rigid
beam is subjected to the loading shown.
A
2m
1m
+ c ©Fy = 0;
FAD + FBE + FCF - 50(103) - 80(103) = 0
(1)
a + ©MD = 0;
FBE(2) + FCF(4) - 50(103)(1) - 80(103)(3) = 0
(2)
Referring to the geometry shown in Fig. b,
dBE = dAD + a
dBE =
dCF - dAD
b(2)
4
1
A d + dCF B
2 AD
FBE L
FCF L
1 FADL
= a
+
b
AE
2
AE
AE
FAD + FCF = 2 FBE
(3)
Solving Eqs. (1), (2) and (3) yields
FBE = 43.33(103) N
FAD = 35.83(103) N
FCF = 50.83(103) N
Thus,
sBE =
43.33(103)
FBE
= 96.3 MPa
=
A
0.45(10 - 3)
Ans.
sAD =
35.83(103)
FAD
= 79.6 MPa
=
A
0.45(10 - 3)
Ans.
sCF = 113 MPa
Ans.
163
80 kN
50 kN
E
D
Referring to the FBD of the rigid beam, Fig. a,
C
B
1m
1m
F
1m
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*4–56. The rigid bar supports the 800-lb load. Determine
the normal stress in each A-36 steel cable if each cable has a
cross-sectional area of 0.04 in2.
C
12 ft
800 lb
B
A
5 ft
Referring to the FBD of the rigid bar, Fig. a,
FBC a
a + ©MA = 0;
12
3
b(5) + FCD a b (16) - 800(10) = 0
13
5
(1)
The unstretched length of wires BC and CD are LBC = 2122 + 52 = 13 ft and
LCD = 2122 + 162 = 20 ft. The stretches of wires BC and CD are
dBC =
FBC (13)
FBC LBC
=
AE
AE
dCD =
FCD(20)
FCD LCD
=
AE
AE
Referring to the geometry shown in Fig. b, the vertical displacement of the points on
d
12
3
the rigid bar is dg =
. For points B and D, cos uB =
and cos uD = . Thus,
cos u
13
5
the vertical displacement of points B and D are
A dB B g =
FBC (13)>AE
dBC
169 FBC
=
=
cos uB
12>13
12AE
A dD B g =
FCD (20)>AE
dCD
100 FCD
=
=
cos uD
3>5
3 AE
The similar triangles shown in Fig. c give
A dB B g
5
=
A dD B g
16
1 169 FBC
1 100 FCD
b =
b
a
a
5 12 AE
16
3AE
FBC =
125
F
169 CD
(2)
Solving Eqs. (1) and (2), yields
FCD = 614.73 lb
FBC = 454.69 lb
Thus,
sCD =
FCD
614.73
= 15.37(103) psi = 15.4 ksi
=
ACD
0.04
Ans.
sBC =
FBC
454.69
=
= 11.37(103) psi = 11.4 ksi
ABC
0.04
Ans.
164
D
5 ft
6 ft
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4–56.
Continued
165
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•4–57.
The rigid bar is originally horizontal and is
supported by two A-36 steel cables each having a crosssectional area of 0.04 in2. Determine the rotation of the bar
when the 800-lb load is applied.
C
12 ft
800 lb
B
A
Referring to the FBD of the rigid bar Fig. a,
a + ©MA = 0;
FBC a
12
3
b(5) + FCD a b (16) - 800(10) = 0
13
5
5 ft
(1)
The unstretched length of wires BC and CD are LBC = 2122 + 52 = 13 ft and
LCD = 2122 + 162 = 20 ft. The stretch of wires BC and CD are
dBC =
FBC (13)
FBC LBC
=
AE
AE
dCD =
FCD(20)
FCD LCD
=
AE
AE
Referring to the geometry shown in Fig. b, the vertical displacement of the points on
d
12
3
the rigid bar is dg =
. For points B and D, cos uB =
and cos uD = . Thus,
cos u
13
5
the vertical displacement of points B and D are
A dB B g =
FBC (13)>AE
dBC
169 FBC
=
=
cos uB
12>13
12AE
A dD B g =
FCD (20)>AE
dCD
100 FCD
=
=
cos uD
3>5
3 AE
The similar triangles shown in Fig. c gives
A dB B g
5
=
A dD B g
16
1 169 FBC
1 100 FCD
a
b =
a
b
5 12 AE
16
3 AE
FBC =
125
F
169 CD
(2)
Solving Eqs (1) and (2), yields
FCD = 614.73 lb
FBC = 454.69 lb
Thus,
A dD B g =
100(614.73)
3(0.04) C 29.0 (106) D
= 0.01766 ft
Then
u = a
0.01766 ft 180°
ba
b = 0.0633°
p
16 ft
Ans.
166
D
5 ft
6 ft
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4–58. The horizontal beam is assumed to be rigid and
supports the distributed load shown. Determine the vertical
reactions at the supports. Each support consists of a wooden
post having a diameter of 120 mm and an unloaded
(original) length of 1.40 m. Take Ew = 12 GPa.
18 kN/m
A
B
C
1.40 m
2m
a + ©MB = 0;
FC(1) - FA(2) = 0
(1)
+ c ©Fy = 0;
FA + FB + FC - 27 = 0
(2)
dB - dA
dC - dA
=
;
2
3
1m
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.
18 kN/m
A
a + ©MB = 0;
FC(1) - FA(2) = 0
(1)
c + ©Fy = 0;
FA + FB + FC - 27 = 0
(2)
dB - dA
dC - dA
=
;
2
3
3dB - dA = 2dC
3FBL
FAL
2FCL
=
;
AE
AE
AE
3FB - FA = 2FC
(3)
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 =
0.1194 - 0.0597
(10 - 3)
3
u = 0.0199(10 - 3) rad = 1.14(10 - 3)°
Ans.
167
C
1.40 m
2m
Solving Eqs. (1)–(3) yields :
FA = 5.7857 kN;
B
1m
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*4–60. The assembly consists of two posts AD and CF
made of A-36 steel and having a cross-sectional area of
1000 mm2, and a 2014-T6 aluminum post BE having a crosssectional area of 1500 mm2. If a central load of 400 kN is
applied to the rigid cap, determine the normal stress in each
post. There is a small gap of 0.1 mm between the post BE
and the rigid member ABC.
400 kN
0.5 m
A
Equation of Equilibrium. Due to symmetry, FAD = FCF = F. Referring to the
FBD of the rigid cap, Fig. a,
FBE + 2F - 400(103) = 0
(1)
Compatibility Equation. Referring to the initial and final position of rods AD (CF)
and BE, Fig. b,
d = 0.1 + dBE
F(400)
1(10 ) C 200(10 ) D
-3
9
= 0.1 +
FBE (399.9)
1.5(10 - 3) C 73.1(109) D
F = 1.8235 FBE + 50(103)
(2)
Solving Eqs (1) and (2) yield
FBE = 64.56(103) N
F = 167.72(103) N
Normal Stress.
sAD = sCF =
sBE =
B
C
0.4 m
D
+ c ©Fy = 0;
0.5 m
167.72(103)
F
= 168 MPa
=
Ast
1(10 - 3)
Ans.
64.56(103)
FBE
= 43.0 MPa
=
Aal
1.5(10 - 3)
Ans.
168
E
F
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•4–61.
The distributed loading is supported by the three
suspender bars. AB and EF are made of aluminum and CD
is made of steel. If each bar has a cross-sectional area of
450 mm2, determine the maximum intensity w of the
distributed loading so that an allowable stress of 1sallow2st =
180 MPa in the steel and 1sallow2al = 94 MPa in the
aluminum is not exceeded. Est = 200 GPa, Eal = 70 GPa.
Assume ACE is rigid.
1.5 m
1.5 m
B
al
D
st
A
F
al
C
2m
E
w
a + ©MC = 0;
FEF(1.5) - FAB(1.5) = 0
FEF = FAB = F
+ c ©Fy = 0;
2F + FCD - 3w = 0
(1)
Compatibility condition :
dA = dC
FCDL
FL
=
;
9
A(70)(10 )
A(200)(109)
F = 0.35 FCD
(2)
Assume failure of AB and EF:
F = (sallow)al A
= 94(106)(450)(10 - 6)
= 42300 N
From Eq. (2) FCD = 120857.14 N
From Eq. (1) w = 68.5 kN>m
Assume failure of CD:
FCD = (sallow)st A
= 180(106)(450)(10 - 6)
= 81000 N
From Eq. (2) F = 28350 N
From Eq. (1) w = 45.9 kN>m
(controls)
Ans.
169
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4–62. The rigid link is supported by a pin at A, a steel wire
BC having an unstretched length of 200 mm and crosssectional area of 22.5 mm2, and a short aluminum block
having an unloaded length of 50 mm and cross-sectional
area of 40 mm2. If the link is subjected to the vertical load
shown, determine the average normal stress in the wire and
the block. Est = 200 GPa, Eal = 70 GPa.
C
200 mm
B
100 mm
D
450(250) - FBC(150) - FD(150) = 0
50 mm
750 - FBC - FD = 0
[1]
Compatibility:
dBC = dD
FD(50)
FBC(200)
22.5(10 - 6)200(109)
=
40(10 - 6)70(109)
FBC = 0.40179 FD
[2]
Solving Eqs. [1] and [2] yields:
FD = 535.03 N
150 mm
450 N
Equations of Equilibrium:
a + ©MA = 0;
A
150 mm
FBC = 214.97 N
Average Normal Stress:
sD =
FD
535.03
= 13.4 MPa
=
AD
40(10 - 6)
Ans.
sBC =
FBC
214.97
= 9.55 MPa
=
ABC
22.5(10 - 6)
Ans.
170
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4–63. The rigid link is supported by a pin at A, a steel wire
BC having an unstretched length of 200 mm and crosssectional area of 22.5 mm2, and a short aluminum block
having an unloaded length of 50 mm and cross-sectional area
of 40 mm2. If the link is subjected to the vertical load shown,
determine the rotation of the link about the pin A. Report
the answer in radians. Est = 200 GPa, Eal = 70 GPa.
C
200 mm
B
100 mm
D
450(250) - FBC(150) - FD(150) = 0
50 mm
750 - FBC - FD = 0
[1]
Compatibility:
dBC = dD
FD(50)
FBC(200)
-6
9
22.5(10 )200(10 )
=
40(10 - 6)70(109)
FBC = 0.40179 FD
[2]
Solving Eqs. [1] and [2] yields :
FD = 535.03 N
FBC = 214.97 N
Displacement:
dD =
535.03(50)
FDLD
= 0.009554 mm
=
ADEal
40(10 - 6)(70)(109)
tan u =
150 mm
450 N
Equations of Equilibrium:
a + ©MA = 0;
A
150 mm
dD
0.009554
=
150
150
u = 63.7(10 - 6) rad = 0.00365°
Ans.
171
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*4–64. The center post B of the assembly has an original
length of 124.7 mm, whereas posts A and C have a length of
125 mm. If the caps on the top and bottom can be
considered rigid, determine the average normal stress in
each post. The posts are made of aluminum and have a
cross-sectional area of 400 mm2. Eal = 70 GPa.
800 kN/m
A
a + ©MB = 0;
100 mm
B
C
-FA(100) + FC(100) = 0
(1)
2F + FB - 160 = 0
(2)
dA = dB + 0.0003
F (0.125)
FB (0.1247)
400 (10 - 6)(70)(106)
=
400 (10 - 6)(70)(106)
+ 0.0003
0.125 F - 0.1247FB = 8.4
(3)
Solving Eqs. (2) and (3)
F = 75.762 kN
FB = 8.547 kN
sA = sC =
sB =
75.726 (103)
400(10 - 6)
8.547 (103)
400 (10 - 6)
= 189 MPa
Ans.
= 21.4 MPa
Ans.
•4–65.
The assembly consists of an A-36 steel bolt and a
C83400 red brass tube. If the nut is drawn up snug against
the tube so that L = 75 mm, then turned an additional
amount so that it advances 0.02 mm on the bolt, determine
the force in the bolt and the tube. The bolt has a diameter of
7 mm and the tube has a cross-sectional area of 100 mm2.
L
Equilibrium: Since no external load is applied, the force acting on the tube and the
bolt is the same.
Compatibility:
0.02 = dt + db
0.02 =
P(75)
P(75)
-6
9
100(10 )(101)(10 )
125 mm
800 kN/m
FA = FC = F
+ c ©Fy = 0;
100 mm
+ p
2
9
4 (0.007 )(200)(10 )
P = 1164.83 N = 1.16 kN
Ans.
172
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4–66. The assembly consists of an A-36 steel bolt and a
C83400 red brass tube. The nut is drawn up snug against
the tube so that L = 75 mm. Determine the maximum
additional amount of advance of the nut on the bolt so that
none of the material will yield. The bolt has a diameter of
7 mm and the tube has a cross-sectional area of 100 mm2.
L
Allowable Normal Stress:
(sg)st = 250 A 106 B = p
Pst
2
4 (0.007)
Pst = 9.621 kN
(sg)br = 70.0 A 106 B =
Pbr
100(10 - 6)
Pbr = 7.00 kN
Since Pst 7 Pbr, by comparison he brass will yield first.
Compatibility:
a = dt + db
7.00(103)(75)
=
100(10 - 6)(101)(109)
7.00(103)(75)
+ p
2
9
4 (0.007) (200)(10 )
= 0.120 mm
Ans.
173
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4–67. The three suspender bars are made of the same
material and have equal cross-sectional areas A. Determine
the average normal stress in each bar if the rigid beam ACE
is subjected to the force P.
a + ©MA = 0;
D
d
FCD(d) + FEF(2d) - Pa b = 0
2
FCD + 2FEF =
+ c ©Fy = 0;
B
F
L
P
P
2
(1)
FAB + FCD + FEF - P = 0
A
C
d
2
(2)
d
2
E
d
dC - dE
dA - dE
=
d
2d
2dC = dA + dE
2FCDL
FABL
FEFL
=
+
AE
AE
AE
2FCD - FAB - FEF = 0
(3)
Solving Eqs. (1), (2) and (3) yields
P
3
P
12
FAB =
7P
12
sAB =
7P
12A
Ans.
sCD =
P
3A
Ans.
sEF =
P
12A
Ans.
FCD =
FEF =
*4–68. A steel surveyor’s tape is to be used to measure the
length of a line. The tape has a rectangular cross section of
0.05 in. by 0.2 in. and a length of 100 ft when T1 = 60°F and
the tension or pull on the tape is 20 lb. Determine the
true length of the line if the tape shows the reading to
be 463.25 ft when used with a pull of 35 lb at T2 = 90°F. The
ground on which it is placed is flat. ast = 9.60110-62>°F,
Est = 2911032 ksi.
P
P
0.2 in.
0.05 in.
dT = a¢TL = 9.6(10 - 6)(90 - 60)(463.25) = 0.133416 ft
d =
(35 - 20)(463.25)
PL
= 0.023961 ft
=
AE
(0.2)(0.05)(29)(106)
L = 463.25 + 0.133416 + 0.023961 = 463.41 ft
Ans.
174
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•4–69.
Three bars each made of different materials are
connected together and placed between two walls when the
temperature is T1 = 12°C. Determine the force exerted
on the (rigid) supports when the temperature becomes
T2 = 18°C. The material properties and cross-sectional
area of each bar are given in the figure.
+ )
(;
Copper
Steel
Brass
Est ⫽ 200 GPa
Ebr ⫽ 100 GPa
Ecu ⫽ 120 GPa
ast ⫽ 12(10⫺6)/⬚C abr ⫽ 21(10⫺6)/°C acu ⫽ 17(10⫺6)/⬚C
Acu ⫽ 515 mm2
Ast ⫽ 200 mm2
Abr ⫽ 450 mm2
300 mm
200 mm
100 mm
0 = ¢T - d
0 = 12(10 - 6)(6)(0.3) + 21 (10 - 6)(6)(0.2) + 17 (10 - 6)(6)(0.1)
F(0.3)
-
-6
F(0.2)
9
200(10 )(200)(10 )
-
-6
F(0.1)
9
-
450(10 )(100)(10 )
515(10 - 6)(120)(109)
F = 4203 N = 4.20 kN
Ans.
k ⫽ 1000 lb/in.
4–70. The rod is made of A-36 steel and has a diameter of
0.25 in. If the rod is 4 ft long when the springs are compressed
0.5 in. and the temperature of the rod is T = 40°F, determine
the force in the rod when its temperature is T = 160°F.
k ⫽ 1000 lb/ in.
4 ft
Compatibility:
+ B
A:
x = dT - dF
x = 6.60(10 - 6)(160 - 40)(2)(12) - p
1.00(0.5)(2)(12)
2
3
4 (0.25 )(29.0)(10 )
x = 0.01869 in.
F = 1.00(0.01869 + 0.5) = 0.519 kip
Ans.
4–71. A 6-ft-long steam pipe is made of A-36 steel with
sY = 40 ksi. It is connected directly to two turbines A
and B as shown. The pipe has an outer diameter of 4 in.
and a wall thickness of 0.25 in. The connection was made
at T1 = 70°F. If the turbines’ points of attachment are
assumed rigid, determine the force the pipe exerts on
the turbines when the steam and thus the pipe reach a
temperature of T2 = 275°F.
6 ft
A
Compatibility:
+ B
A:
0 = dT - dF
0 = 6.60(10 - 6)(275 - 70)(6)(12) - p
F(6)(12)
2
4 (4
- 3.52)(29.0)(103)
F = 116 kip
Ans.
175
B
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*4–72. A 6-ft-long steam pipe is made of A-36 steel with
sY = 40 ksi. It is connected directly to two turbines A and
B as shown. The pipe has an outer diameter of 4 in. and a
wall thickness of 0.25 in. The connection was made at
T1 = 70°F. If the turbines’ points of attachment are
assumed to have a stiffness of k = 8011032 kip>in., determine
the force the pipe exerts on the turbines when the steam
and thus the pipe reach a temperature of T2 = 275°F.
6 ft
A
B
Compatibility:
x = dT - dF
80(103)(x)(3)(12)
x = 6.60(10 - 6)(275 - 70)(3)(12) - p
2
2
3
4 (4 - 3.5 )(29.0)(10 )
x = 0.001403 in.
F = k x = 80(103)(0.001403) = 112 kip
Ans.
•4–73.
The pipe is made of A-36 steel and is connected to
the collars at A and B. When the temperature is 60° F, there
is no axial load in the pipe. If hot gas traveling through the
pipe causes its temperature to rise by ¢T = 140 + 15x2°F,
where x is in feet, determine the average normal stress in the
pipe. The inner diameter is 2 in., the wall thickness is 0.15 in.
A
Compatibility:
L
0 = dT - dF
Where
0 = 6.60 A 10 - 6 B
8ft
L0
dT =
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
Where
0 = 9.60 A 10 - 6 B
2ft
0 = 9.60 A 10 - 6 B
L0
dT = 1 a¢Tdx
2
2
3
4 (1.4 - 1 )15.0(10 )
2ft
L0
F(8)
[(200 - 17.5x) - 60] dx - p
(140 - 17.5x) dx - p
F(8)
2
2
3
4 (1.4 - 1 ) 15.0(10 )
F = 7.60 kip
Ans.
4–75. The 40-ft-long A-36 steel rails on a train track are laid
with a small gap between them to allow for thermal
expansion. Determine the required gap d so that the rails just
touch one another when the temperature is increased from
T1 = -20°F to T2 = 90°F. Using this gap, what would be the
axial force in the rails if the temperature were to rise to
T3 = 110°F? The cross-sectional area of each rail is 5.10 in2.
d
40 ft
Thermal Expansion: Note that since adjacent rails expand, each rail will be
d
required to expand on each end, or d for the entine rail.
2
d = a¢TL = 6.60(10 - 6)[90 - (-20)](40)(12)
Ans.
= 0.34848 in. = 0.348 in.
Compatibility:
+ B
A:
0.34848 = dT - dF
0.34848 = 6.60(10 - 6)[110 - (-20)](40)(12) -
d
F(40)(12)
5.10(29.0)(103)
F = 19.5 kip
Ans.
177
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*4–76. The device is used to measure a change in
temperature. Bars AB and CD are made of A-36 steel and
2014-T6 aluminum alloy respectively. When the temperature
is at 75°F, ACE is in the horizontal position. Determine the
vertical displacement of the pointer at E when the
temperature rises to 150°F.
0.25 in.
A
3 in.
C
E
1.5 in.
Thermal Expansion:
A dT B CD = aal ¢TLCD = 12.8(10 - 6)(150 - 75)(1.5) = 1.44(10 - 3) in.
B
D
A dT B AB = ast ¢TLAB = 6.60(10 - 6)(150 - 75)(1.5) = 0.7425(10 - 3) in.
From the geometry of the deflected bar AE shown Fig. b,
dE = A dT B AB + C
= 0.7425(10 - 3) + B
A dT B CD - A dT B AB
0.25
S(3.25)
1.44(10 - 3) - 0.7425(10 - 3)
R (3.25)
0.25
= 0.00981 in.
Ans.
•4–77. The bar has a cross-sectional area A, length L,
modulus of elasticity E, and coefficient of thermal
expansion a. The temperature of the bar changes uniformly
along its length from TA at A to TB at B so that at any point
x along the bar T = TA + x1TB - TA2>L. Determine the
force the bar exerts on the rigid walls. Initially no axial force
is in the bar and the bar has a temperature of TA.
+
:
x
A
TA
0 = ¢ T - dF
(1)
However,
d¢ T = a¢ T dx = a(TA +
TB - TA
x - TA)dx
L
L
¢T = a
= ac
L
TB - TA
TB - TA 2
x dx = ac
x d冷
L
2L
L0
0
TB - TA
aL
Ld =
(TB - TA)
2
2
From Eq.(1).
0 =
FL
aL
(TB - TA) 2
AE
F =
a AE
(TB - TA)
2
B
Ans.
178
TB
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4–78. The A-36 steel rod has a diameter of 50 mm and is
lightly attached to the rigid supports at A and B when
T1 = 80°C. If the temperature becomes T2 = 20°C and an
axial force of P = 200 kN is applied to its center, determine
the reactions at A and B.
0.5 m
FB - FA + 200(103) = 0
(1)
When the rod is unconstrained at B, it has a free contraction of
dT = ast ¢ TL = 12(10 - 6)(80 - 20)(1000) = 0.72 mm. Also, under force P and FB
with unconstrained at B, the deformation of the rod are
dP =
200(103)(500)
PLAC
= 0.2546 mm
=
p
AE
(0.052) C 200(109) D
4
dFB =
FB (1000)
FB LAB
= 2.5465(10 - 6) FB
=
p
AE
(0.052) C 200(109) D
4
Using the method of super position, Fig. b,
+ B
A:
B
P
Referring to the FBD of the rod, Fig. a
+ ©F = 0;
:
x
C
A
0 = -dT + dP + dFB
0 = -0.72 + 0.2546 + 2.5465(10 - 6) FB
FB = 182.74(103) N = 183 kN
Ans.
Substitute the result of FB into Eq (1),
FA = 382.74(103) N = 383 kN
Ans.
179
0.5 m
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4–79. The A-36 steel rod has a diameter of 50 mm and is
lightly attached to the rigid supports at A and B when
T1 = 50°C. Determine the force P that must be applied to
the collar at its midpoint so that, when T2 = 30°C, the
reaction at B is zero.
C
A
B
P
0.5 m
0.5 m
When the rod is unconstrained at B, it has a free contraction of
dT = ast ¢TL = 12(10 - 6)(50 - 30)(1000) = 0.24 mm. Also, under force P with
unconstrained at B, the deformation of the rod is
dP =
P(500)
PLAC
= 1.2732(10 - 6) P
=
p
AE
(0.052) C 200(109) D
4
Since FB is required to be zero, the method of superposition, Fig. b, gives
+ B
A:
0 = -dT + dP
0 = -0.24 + 1.2732(10 - 6)P
P = 188.50(103) N = 188 kN
Ans.
*4–80. The rigid block has a weight of 80 kip and is to be
supported by posts A and B, which are made of A-36 steel,
and the post C, which is made of C83400 red brass. If all the
posts have the same original length before they are loaded,
determine the average normal stress developed in each post
when post C is heated so that its temperature is increased
by 20°F. Each post has a cross-sectional area of 8 in2.
A
Equations of Equilibrium:
a + ©MC = 0;
FB(3) - FA(3) = 0
+ c ©Fy = 0;
2F + FC - 80 = 0
FA = FB = F
[1]
Compatibility:
(dC)F - (dC)T = dF
(+ T)
FCL
8(14.6)(103)
- 9.80 A 10 - 5 B (20)L =
FL
8(29.0)(103)
8.5616 FC - 4.3103 F = 196
[2]
Solving Eqs. [1] and [2] yields:
F = 22.81 kip
FC = 34.38 kip
average Normal Sress:
sA = sB =
sC =
F
22.81
=
= 2.85 ksi
A
8
Ans.
FC
34.38
=
= 4.30 ksi
A
8
Ans.
180
C
B
3 ft
3 ft
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•4–81.
The three bars are made of A-36 steel and form
a pin-connected truss. If the truss is constructed when
T1 = 50°F, determine the force in each bar when
T2 = 110°F. Each bar has a cross-sectional area of 2 in2.
t
5f
5f
t
A
4 ft
B
D
3 ft
(dT ¿)AB - (dF ¿)AB = (dT)AD + (dF)AD
(1)
œ
cos u;
However, dAB = dAB
œ
dAB
=
dAB
5
= dAB
cos u
4
Substitute into Eq. (1)
5
5
(dT)AB - (dF)AB = (dT)AD + (dF)AD
4
4
FAB(5)(12)
5
d
c6.60(10 - 6)(110° - 50°)(5)(12) 4
2(29)(103)
= 6.60(10 - 6)(110° - 50°)(4)(12) +
FAD(4)(12)
2(29)(103)
620.136 = 75FAB + 48FAD
+ ©F = 0;
:
x
3
3
F
- FAB = 0;
5 AC
5
+ c ©Fy = 0;
4
FAD - 2a FAB b = 0
5
(2)
FAC = FAB
(3)
Solving Eqs. (2) and (3) yields :
FAD = 6.54 kip
Ans.
FAC = FAB = 4.09 kip
Ans.
181
C
3 ft
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4–82. The three bars are made of A-36 steel and form a pinconnected truss. If the truss is constructed when T1 = 50°F,
determine the vertical displacement of joint A when
T2 = 150°F. Each bar has a cross-sectional area of 2 in2.
t
5f
5f
t
A
4 ft
(dT ¿)AB - (dF ¿)AB = (dT)AD + (dF)AD
(1)
œ
However, dAB = dAB
cos u;
œ
dAB
=
B
dAB
5
= dAB
cos u
4
3 ft
Substitute into Eq. (1)
5
5
(d )
- (dT)AB = (dT)AD + (dF)AD
4 T AB
4
FAB(5)(12)
5
d
c6.60(10 - 6)(150° - 50°)(5)(12) 4
2(29)(103)
= 6.60(10 - 6)(150° - 50°)(4)(12) +
FAD(4)(12)
2(29)(103)
239.25 - 6.25FAB = 153.12 + 4 FAD
4 FAD + 6.25FAB = 86.13
+ © F = 0;
:
x
3
3
F
- FAB = 0;
5 AC
5
+ c © Fy = 0;
4
FAD - 2 a FAB b = 0;
5
(2)
FAC = FAB
FAD = 1.6FAB
(3)
Solving Eqs. (2) and (3) yields:
FAB = 6.8086 kip:
FAD = 10.8939 kip
(dA)r = (dT)AD + (dT)AD
= 6.60(10 - 6)(150° - 50°)(4)(12) +
D
10.8939(4)(12)
2(29)(103)
= 0.0407 in. c
Ans.
182
C
3 ft
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4–83. The wires AB and AC are made of steel, and wire
AD is made of copper. Before the 150-lb force is applied,
AB and AC are each 60 in. long and AD is 40 in. long. If
the temperature is increased by 80°F, determine the
force in each wire needed to support the load. Take
Est = 29(103) ksi, Ecu = 17(103) ksi, ast = 8(10-6)>°F,
acu = 9.60(10-6)>°F. Each wire has a cross-sectional area of
0.0123 in2.
40 in.
60 in.
45⬚
45⬚
A
150 lb
Equations of Equilibrium:
+ ©F = 0;
:
x
FAC cos 45° - FAB cos 45° = 0
FAC = FAB = F
2F sin 45° + FAD - 150 = 0
+ c ©Fy = 0;
[1]
Compatibility:
(dAC)T = 8.0 A 10 - 6 B (80)(60) = 0.03840 in.
(dAC)Tr =
(dAC)T
0.03840
=
= 0.05431 in.
cos 45°
cos 45°
(dAD)T = 9.60 A 10 - 6 B (80)(40) = 0.03072 in.
d0 = (dAC)Tr - (dAD)T = 0.05431 - 0.03072 = 0.02359 in.
(dAD)F = (dAC)Fr + d0
F(60)
FAD (40)
6
0.0123(17.0)(10 )
=
0.0123(29.0)(106) cos 45°
C
D
B
+ 0.02359
0.1913FAD - 0.2379F = 23.5858
[2]
Solving Eq. [1] and [2] yields:
FAC = FAB = F = 10.0 lb
Ans.
FAD = 136 lb
Ans.
183
60 in.
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*4–84. The AM1004-T61 magnesium alloy tube AB is
capped with a rigid plate E. The gap between E and end C of
the 6061-T6 aluminum alloy solid circular rod CD is 0.2 mm
when the temperature is at 30° C. Determine the normal
stress developed in the tube and the rod if the temperature
rises to 80° C. Neglect the thickness of the rigid cap.
25 mm
a
Section a-a
E
B
A
20 mm
C
25 mm
a
0.2 mm
300 mm
Compatibility Equation: If tube AB and rod CD are unconstrained, they will have a
free expansion of A dT B AB = amg ¢TLAB = 26(10 - 6)(80 - 30)(300) = 0.39 mm and
A dT)CD = aal ¢TLCD = 24(10 - 6)(80 - 30)(450) = 0.54 mm. Referring
deformation diagram of the tube and the rod shown in Fig. a,
to
the
d = 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 -
A
F(450)
B
p
2
9
4 0.025 (68.9)(10 )
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
ACD
A 0.0252 B
Ans.
Ans.
4
F = 107 442.47 N
184
450 mm
D
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•4–85. The AM1004-T61 magnesium alloy tube AB is
capped with a rigid plate. The gap between E and end C of
the 6061-T6 aluminum alloy solid circular rod CD is 0.2 mm
when the temperature is at 30° C. Determine the highest
temperature to which it can be raised without causing
yielding either in the tube or the rod. Neglect the thickness
of the rigid cap.
25 mm
a
Section a-a
E
B
A
20 mm
C
25 mm
a
0.2 mm
300 mm
Then
sCD =
F
107 442.47
=
= 218.88MPa 6 (sY)al
p
ACD
A 0.0252 B
(O.K.!)
4
Compatibility Equation: If tube AB and rod CD are unconstrained, they will have
a free expansion of A dT B AB = amg ¢TLAB = 26(10 - 6)(T - 30)(300) = 7.8(10 - 6)
(T - 30) and
A dT B CD = aal ¢TLCD = 24(10 - 6)(T - 30)(450) = 0.0108(T - 30).
Referring to the deformation diagram of the tube and the rod shown in Fig. a,
d = C A dT B AB - A dF B AB D + C A dT B CD - A dF B CD D
0.2 = C 7.8(10 - 3)(T - 30) -
+ C 0.0108(T - 30) -
107 442.47(300)
p A 0.0252 - 0.022 B (44.7)(109)
107 442.47(450)
p
4
A 0.0252 B (68.9)(109)
S
S
T = 172° C
Ans.
185
450 mm
D
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4–86. The steel bolt has a diameter of 7 mm and fits
through an aluminum sleeve as shown. The sleeve has an
inner diameter of 8 mm and an outer diameter of 10 mm.
The nut at A is adjusted so that it just presses up against
the sleeve. If the assembly is originally at a temperature
of T1 = 20°C and then is heated to a temperature of
T2 = 100°C, determine the average normal stress in the
bolt and the sleeve. Est = 200 GPa, Eal = 70 GPa, ast =
14(10-6)>°C, aal = 23(10-6)>°C.
A
Compatibility:
(ds)T - (db)T = (ds)F + (db)F
23(10 - 6)(100 - 20)L - 14(10 - 6)(100 - 20)L
FL
FL
+ p
2
2
9
2
(0.01
0.008
)70(10
)
(0.007
)200(109)
4
4
= p
F = 1133.54 N
Average Normal Stress:
ss =
F
1133.54
= 40.1 MPa
= p
2
As
(0.01
- 0.0082)
4
Ans.
sb =
F
1133.54
= 29.5 MPa
= p
2
Ab
4 (0.007 )
Ans.
4–87. Determine the maximum normal stress developed
in the bar when it is subjected to a tension of P = 8 kN.
5 mm
40 mm
20 mm
P
P
For the fillet:
r ⫽ 10 mm
20 mm
r
10
=
= 0.5
h
20
w
40
=
= 2
h
20
From Fig. 10-24. K = 1.4
smax = Ksavg
= 1.4 a
8 (103)
b
0.02 (0.005)
= 112 MPa
For the hole:
r
10
=
= 0.25
w
40
From Fig. 4-25. K = 2.375
smax = Ksavg
= 2.375 a
8 (103)
b
(0.04 - 0.02)(0.005)
= 190 MPa
Ans.
186
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*4–88. If the allowable normal stress for the bar is
sallow = 120 MPa, determine the maximum axial force P
that can be applied to the bar.
5 mm
40 mm
20 mm
P
P
Assume failure of the fillet.
r ⫽ 10 mm
20 mm
r
10
=
= 0.5
h
20
w
40
=
= 2;
h
20
From Fig. 4-24. K = 1.4
sallow = smax = Ksavg
120 (106) = 1.4 a
P
b
0.02 (0.005)
P = 8.57 kN
Assume failure of the hole.
r
10
=
= 0.25
w
20
From Fig. 4-25. K = 2.375
sallow = smax = Ksavg
120 (104) = 2.375 a
P
b
(0.04 - 0.02) (0.005)
P = 5.05 kN (controls)
Ans.
•4–89.
The member is to be made from a steel plate that is
0.25 in. thick. If a 1-in. hole is drilled through its center,
determine the approximate width w of the plate so that it
can support an axial force of 3350 lb. The allowable stress is
sallow = 22 ksi.
0.25 in.
w
3350 lb
sallow = smax = Ksavg
1 in.
3.35
d
22 = K c
(w - 1)(0.25)
w =
3.35K + 5.5
5.5
By trial and error, from Fig. 4-25, choose
w =
r
= 0.2;
w
K = 2.45
3.35(2.45) + 5.5
= 2.49 in.
5.5
Since
r
0.5
=
= 0.2
w
2.49
3350 lb
Ans.
OK
187
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4–90. The A-36 steel plate has a thickness of 12 mm. If
there are shoulder fillets at B and C, and sallow = 150 MPa,
determine the maximum axial load P that it can support.
Calculate its elongation, neglecting the effect of the fillets.
r = 30 mm
120 mm
r = 30 mm
60 mm
P
A
w
120
=
= 2
h
60
and
60 mm
P
D
B
Maximum Normal Stress at fillet:
r
30
=
= 0.5
h
60
C
800 mm
200 mm
200 mm
From the text, K = 1.4
smax = sallow = Ksavg
150(106) = 1.4 B
P
R
0.06(0.012)
P = 77142.86 N = 77.1 kN
Ans.
Displacement:
d = ©
PL
AE
77142.86(800)
77142.86(400)
=
9
+
(0.06)(0.012)(200)(10 )
(0.12)(0.012)(200)(109)
= 0.429 mm
Ans.
4–91. Determine the maximum axial force P that can be
applied to the bar. The bar is made from steel and has an
allowable stress of sallow = 21 ksi.
0.125 in.
1.25 in.
1.875 in.
P
Assume failure of the fillet.
r
0.25
=
= 0.2
h
1.25
P
w
1.875
=
= 1.5
h
1.25
0.75 in.
From Fig. 4-24, K = 1.73
sallow = smax = Ksavg
21 = 1.73 a
P
b
1.25 (0.125)
P = 1.897 kip
Assume failure of the hole.
r
0.375
=
= 0.20
w
1.875
From Fig. 4-25, K = 2.45
sallow = smax = Ksavg
21 = 2.45 a
P
b
(1.875 - 0.75)(0.125)
P = 1.21 kip (controls)
Ans.
188
r ⫽ 0.25 in.
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*4–92. Determine the maximum normal stress developed
in the bar when it is subjected to a tension of P = 2 kip.
0.125 in.
1.25 in.
1.875 in.
At fillet:
P
r
0.25
=
= 0.2
h
1.25
P
w
1.875
=
= 1.5
h
1.25
0.75 in.
From Fig. 4-24, K = 1.73
smax = Ka
r ⫽ 0.25 in.
P
2
d = 22.1 ksi
b = 1.73 c
A
1.25(0.125)
At hole:
r
0.375
=
= 0.20
w
1.875
From Fig. 4-25, K = 2.45
smax = 2.45 c
2
d = 34.8 ksi
(1.875 - 0.75)(0.125)
(Controls)
Ans.
•4–93.
Determine the maximum normal stress developed
in the bar when it is subjected to a tension of P = 8 kN.
5 mm
60 mm
P
Maximum Normal Stress at fillet:
r
15
=
= 0.5
h
30
P
ht
= 1.4 B
8(103)
R = 74.7 MPa
(0.03)(0.005)
Maximum Normal Stress at the hole:
r
6
=
= 0.1
w
60
From the text, K = 2.65
smax = K savg = K
P
(w - 2r) t
= 2.65 B
8(103)
R
(0.06 - 0.012)(0.005)
= 88.3 MPa
P
r = 15 mm
12 mm
w
60
=
= 2
h
30
and
From the text, K = 1.4
smax = Ksavg = K
30 mm
(Controls)
Ans.
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 stress-concentration factor for this geometry?
0.5 in.
A
P
4 in.
1 in.
B
12 ksi
P =
L
3 ksi
sdA = Volume under curve
Number of squares = 10
P = 10(3)(1)(0.5) = 15 kip
savg =
K =
Ans.
15 kip
P
=
= 7.5 ksi
A
(4 in.)(0.5 in.)
smax
12 ksi
=
= 1.60
savg
7.5 ksi
Ans.
4–95. The resulting stress distribution along section AB
for the bar is shown. From this distribution, determine the
approximate resultant axial force P applied to the bar. Also,
what is the stress-concentration factor for this geometry?
0.5 in.
A
0.6 in.
0.8 in.
Number of squares = 28
P = 28(6)(0.2)(0.5) = 16.8 kip
0.6 in.
Ans.
B
P
16.8
=
= 28 ksi
savg =
A
2(0.6)(0.5)
K =
6 ksi
36 ksi
smax
36
=
= 1.29
savg
28
Ans.
*4–96. The resulting stress distribution along section AB
for the bar is shown. From this distribution, determine the
approximate resultant axial force P applied to the bar. Also,
what is the stress-concentration factor for this geometry?
10 mm
A
20 mm
80 mm
B
5 MPa
Number of squares = 19
30 MPa
6
P = 19(5)(10 )(0.02)(0.01) = 19 kN
savg =
K =
P
0.2 in.
Ans.
19(103)
P
=
= 23.75 MPa
A
0.08(0.01)
smax
30 MPa
=
= 1.26
savg
23.75 MPa
Ans.
190
P
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•4–97.
The 300-kip weight is slowly set on the top of a post
made of 2014-T6 aluminum with an A-36 steel core. If both
materials can be considered elastic perfectly plastic,
determine the stress in each material.
Aluminum
1 in.
2 in.
Steel
Equations of Equilibrium:
+ c ©Fy = 0;
Pst + Pal - 300 = 0
[1]
Elastic Analysis: Assume both materials still behave elastically under the load.
dst = dal
Pst L
p
2
(2)
(29)(103)
4
Pal L
= p
2
2
3
4 (4 - 2 )(10.6)(10 )
Pst = 0.9119 Pal
Solving Eqs. [1] and [2] yields:
Pal = 156.91 kip
Pst = 143.09 kip
Average Normal Stress:
sal =
Pal
156.91
= p 2
Aal
(4
- 22)
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
186.90
= 19.83 ksi 6 (sg)al = 60.0 ksi
= p 2
2
Aal
4 (4 - 2 )
Then sal = 19.8 ksi
Ans.
191
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4–98. The bar has a cross-sectional area of 0.5 in2 and is
made of a material that has a stress–strain diagram that
can be approximated by the two line segments shown.
Determine the elongation of the bar due to the applied
loading.
A
B
5 ft
8 kip
C 5 kip
2 ft
s(ksi)
40
20
Average Normal Stress and Strain: For segment BC
sBC =
0.001
PBC
5
=
= 10.0 ksi
ABC
0.5
10.0
20
=
;
eBC
0.001
eBC =
0.001
(10.0) = 0.00050 in.>in.
20
Average Normal Stress and Strain: For segment AB
sAB =
PAB
13
=
= 26.0 ksi
AAB
0.5
40 - 20
26.0 - 20
=
eAB - 0.001
0.021 - 0.001
eAB = 0.0070 in.>in.
Elongation:
dBC = eBCLBC = 0.00050(2)(12) = 0.0120 in.
dAB = eAB LAB = 0.0070(5)(12) = 0.420 in.
dTot = dBC + dAB = 0.0120 + 0.420 = 0.432 in.
Ans.
192
0.021
P (in./in.)
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4–99. The rigid bar is supported by a pin at A and two
steel wires, each having a diameter of 4 mm. If the yield
stress for the wires is sY = 530 MPa, and Est = 200 GPa,
determine the intensity of the distributed load w that can
be placed on the beam and will just cause wire EB to
yield. What is the displacement of point G for this case?
For the calculation, assume that the steel is elastic
perfectly plastic.
E
D
800 mm
A
B
C
G
w
400 mm
Equations of Equilibrium:
a + ©MA = 0;
FBE(0.4) + FCD(0.65) - 0.8w (0.4) = 0
0.4 FBE + 0.65FCD = 0.32w
[1]
Plastic Analysis: Wire CD will yield first followed by wire BE. When both wires yield
FBE = FCD = (sg)A
p
= 530 A 106 B a b A 0.0042 B = 6.660 kN
4
Substituting the results into Eq. [1] yields:
w = 21.9 kN>m
Ans.
Displacement: When wire BE achieves yield stress, the corresponding yield strain is
eg =
sg
E
530(106)
=
200(109)
= 0.002650 mm>mm
dBE = eg LBE = 0.002650(800) = 2.120 mm
From the geometry
dBE
dG
=
0.8
0.4
dG = 2dBE = 2(2.120) = 4.24 mm
Ans.
193
250 mm
150 mm
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*4–100. The rigid bar is supported by a pin at A and two
steel wires, each having a diameter of 4 mm. If the yield
stress for the wires is sY = 530 MPa, and Est = 200 GPa,
determine (a) the intensity of the distributed load w that
can be placed on the beam that will cause only one of the
wires to start to yield and (b) the smallest intensity of
the distributed load that will cause both wires to yield. For
the calculation, assume that the steel is elastic perfectly
plastic.
E
D
800 mm
A
B
C
G
w
400 mm
Equations of Equilibrium:
a + ©MA = 0;
FBE(0.4) + FCD(0.65) - 0.8w (0.4) = 0
0.4 FBE + 0.65 FCD = 0.32w
[1]
(a) By observation, wire CD will yield first.
p
Then FCD = sg A = 530 A 106 B a b A 0.0042 B = 6.660 kN.
4
From the geometry
dCD
dBE
=
;
0.4
0.65
dCD = 1.625dBE
FBEL
FCDL
= 1.625
AE
AE
FCD = 1.625 FBE
[2]
Using FCD = 6.660 kN and solving Eqs. [1] and [2] yields:
FBE = 4.099 kN
w = 18.7 kN>m
Ans.
(b) When both wires yield
FBE = FCD = (sg)A
p
= 530 A 106 B a b A 0.0042 B = 6.660 kN
4
Substituting the results into Eq. [1] yields:
w = 21.9 kN>m
Ans.
194
250 mm
150 mm
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•4–101.
The rigid lever arm is supported by two A-36 steel
wires having the same diameter of 4 mm. If a force of
P = 3 kN is applied to the handle, determine the force
developed in both wires and their corresponding elongations.
Consider A-36 steel as an elastic-perfectly plastic material.
P
450 mm
150 mm
150 mm
30⬚
A
C
300 mm
B
Equation of Equilibrium. Refering to the free-body diagram of the lever shown
in Fig. a,
FAB (300) + FCD (150) - 3 A 103 B (450) = 0
a + ©ME = 0;
2FAB + FCD = 9 A 103 B
(1)
Elastic Analysis. Assuming that both wires AB and CD behave as linearly elastic,
the compatibility equation can be written by referring to the geometry of Fig. b.
dAB = a
300
bd
150 CD
dAB = 2dCD
(2)
FAB L
FCD L
= 2a
b
AE
AE
FAB = 2FCD
(3)
Solving Eqs. (1) and (3),
FCD = 1800 N
FAB = 3600 N
Normal Stress.
sCD =
FCD
1800
=
= 143.24 MPa 6 (sY)st
p
ACD
A 0.0042 B
(O.K.)
FAB
3600
=
= 286.48 MPa 7 (sY)st
p
2
AAB
0.004
A
B
(N.G.)
4
sAB =
4
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
ACD
A 0.0042 B
(O.K.)
4
195
D
E
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4–101.
Continued
Since wire CD is linearly elastic, its elongation can be determined by
dCD =
2716.81(300)
FCD LCD
=
p
ACD Est
A 0.0042 B (200) A 109 B
4
Ans.
= 0.3243 mm = 0.324 mm
From Eq. (2),
dAB = 2dCD = 2(0.3243) = 0.649 mm
Ans.
196
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4–102. The rigid lever arm is supported by two A-36 steel
wires having the same diameter of 4 mm. Determine the
smallest force P that will cause (a) only one of the wires to
yield; (b) both wires to yield. Consider A-36 steel as an
elastic-perfectly plastic material.
P
450 mm
150 mm
150 mm
30⬚
A
C
300 mm
B
Equation of Equilibrium. Refering to the free-body diagram of the lever arm shown
in Fig. a,
a + ©ME = 0;
FAB (300) + FCD (150) - P(450) = 0
2FAB + FCD = 3P
(1)
Elastic Analysis. The compatibility equation can be written by referring to the
geometry of Fig. b.
dAB = a
300
bd
150 CD
dAB = 2dCD
FAB L
FCD L
= 2a
b
AE
AE
FCD =
1
F
2 AB
(2)
Assuming that wire AB is about to yield first,
FAB = (sY)st AAB = 250 A 106 B c
p
A 0.0042 B d = 3141.59 N
4
From Eq. (2),
FCD =
1
(3141.59) = 1570.80 N
2
Substituting the result of FAB and FCD into Eq. (1),
P = 2618.00 N = 2.62 kN
Ans.
Plastic Analysis. Since both wires AB and CD are required to yield,
FAB = FCD = (sY)st A = 250 A 106 B c
p
A 0.0042 B d = 3141.59 N
4
Substituting this result into Eq. (1),
197
D
E
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4–103. The three bars are pinned together and subjected
to the load P. If each bar has a cross-sectional area A, length
L, and is made from an elastic perfectly plastic material, for
which the yield stress is sY, determine the largest load
(ultimate load) that can be supported by the bars, i.e., the
load P that causes all the bars to yield. Also, what is the
horizontal displacement of point A when the load reaches
its ultimate value? The modulus of elasticity is E.
B
L
u
C
L
P
u
L
D
P = 3141.59 N = 3.14 kN
A
Ans.
When all bars yield, the force in each bar is, FY = sYA
+ ©F = 0;
:
x
P - 2sYA cos u - sYA = 0
P = sYA(2 cos u + 1)
Ans.
Bar AC will yield first followed by bars AB and AD.
dAB = dAD =
dA =
FY(L)
sYAL
sYL
=
=
AE
AE
E
dAB
sYL
=
cos u
E cos u
Ans.
*4–104. The rigid beam is supported by three 25-mm
diameter A-36 steel rods. If the beam supports the force of
P = 230 kN, determine the force developed in each rod.
Consider the steel to be an elastic perfectly-plastic material.
D
F
E
600 mm
P
Equation of Equilibrium. Referring to the free-body diagram of the beam shown in
Fig. a,
+ c ©Fy = 0;
FAD + FBE + FCF - 230 A 103 B = 0
(1)
FBE + 3FCF = 460 A 103 B
(2)
FBE(400) + FCF(1200) - 230 A 103 B (800) = 0
a + ©MA = 0;
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
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
ACF
A 0.0252 B
(N.G.)
FBE
65714.29
=
= 133.87 MPa 6 (sY)st
p
ABE
A 0.0252 B
(O.K.)
FAD
32857.14
=
= 66.94 MPa 6 (sY)st
p
AAD
A 0.0252 B
(O.K.)
4
sBE =
4
sAD =
4
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
sBE =
Ans.
FBE
91844.61
=
= 187.10 MPa 6 (sY)st
p
ABE
A 0.0252 B
(O.K.)
FAD
15436.93
=
= 31.45 MPa 6 (sY)st
p
AAD
A 0.0252 B
(O.K.)
4
sAD =
4
199
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•4–105.
The rigid beam is supported by three 25-mm
diameter A-36 steel rods. If the force of P = 230 kN is
applied on the beam and removed, determine the residual
stresses in each rod. Consider the steel to be an elastic
perfectly-plastic material.
D
600 mm
P
Equation of Equilibrium. Referring to the free-body diagram of the beam shown in
Fig. a,
+ c ©Fy = 0;
FAD + FBE + FCF - 230 A 103 B = 0
(1)
FBE + 3FCF = 460 A 103 B
(2)
FBE(400) + FCF(1200) - 230 A 103 B (800) = 0
a + ©MA = 0;
Elastic Analysis. Referring to the deflection diagram of the beam shown in Fig. b,
the compatibility equation can be written as
dBE = dAD + a
dBE =
dCF - dAD
b(400)
1200
2
1
d
+ dCF
3 AD
3
(3)
FBE L
2 FCD L
1 FCF L
= a
b + a
b
AE
3 AE
3 AE
FBE =
2
1
F
+ FCF
3 AD
3
(4)
Solving Eqs. (1), (2), and (4)
FCF = 131428.57 N
FBE = 65714.29 N
FAD = 32857.14 N
Normal Stress.
sCF =
FCF
131428.57
=
= 267.74 MPa (T) 7 (sY)st
p
ACF
A 0.0252 B
(N.G.)
FBE
65714.29
=
= 133.87 MPa (T) 6 (sY)st
p
ABE
A 0.0252 B
(O.K.)
FAD
32857.14
=
= 66.94 MPa (T) 6 (sY)st
p
AAD
A 0.0252 B
(O.K.)
4
sBE =
4
sAD =
4
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
A
B
400 mm
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
ABE
A 0.0252 B
(O.K.)
FAD
15436.93
=
= 31.45 MPa (T) 6 (sY)st
p
AAD
A 0.0252 B
(O.K.)
4
sAD =
4
Residual Stresses. The process of removing P can be represented by applying the
force P¿ , which has a magnitude equal to that of P but is opposite in sense, Fig. c.
Since the process occurs in a linear manner, the corresponding normal stress must
have the same magnitude but opposite sense to that obtained from the elastic
analysis. Thus,
œ
sCF
= 267.74 MPa (C)
œ
sBE
= 133.87 MPa (C)
œ
sAD
= 66.94 MPa (C)
Considering the tensile stress as positive and the compressive stress as negative,
œ
= 250 + (-267.74) = -17.7 MPa = 17.7 MPa (C)
(sCF)r = sCF + sCF
Ans.
œ
= 187.10 + (-133.87) = 53.2 MPa (T)
(sBE)r = sBE + sBE
Ans.
œ
(sAD)r = sAD + sAD
= 31.45 + (-66.94) = -35.5 MPa = 35.5 MPa (C)
Ans.
201
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4–106. The distributed loading is applied to the rigid
beam, which is supported by the three bars. Each bar has a
cross-sectional area of 1.25 in2 and is made from a
s(ksi)
material having a stress–strain diagram that can be
approximated by the two line segments shown. If a load of
60
w = 25 kip>ft is applied to the beam, determine the stress
in each bar and the vertical displacement of the beam.
4 ft
A
B
5 ft
36
A
0.0012
a + ©MB = 0;
0.2
FC(4) - FA(4) = 0;
FA = FC = F
+ c ©Fy = 0;
2F + FB - 200 = 0
(1)
Since the loading and geometry are symmetrical, the bar will remain horizontal.
Therefore, the displacement of the bars is the same and hence, the force in each bar
is the same. From Eq. (1).
F = FB = 66.67 kip
Thus,
sA = sB = sC =
66.67
= 53.33 ksi
1.25
Ans.
From the stress-strain diagram:
60 - 36
53.33 - 36
=
:
e - 0.0012
0.2 - 0.0012
e = 0.14477 in.>in.
d = eL = 0.14477(5)(12) = 8.69 in.
Ans.
202
4 ft
∋ (in./in.)
B
C
w
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4–107. The distributed loading is applied to the
rigid beam, which is supported by the three bars.
Each bar has a cross-sectional area of 0.75 in2 and is
s(ksi)
made from a material having a stress–strain diagram
that can be approximated by the two line segments
60
shown. Determine the intensity of the distributed
loading w needed to cause the beam to be displaced
36
downward 1.5 in.
0.0012
a + ©MB = 0;
FC(4) - FA(4) = 0;
+ c ©Fy = 0;
2F + FB - 8 w = 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),
(2)
From the stress-strain diagram:
e =
1.5
= 0.025 in.>in.
5 (12)
60 - 36
s - 36
=
;
0.025 - 0.0012
0.2 - 0.0012
s = 38.87 ksi
Hence F = sA = 38.87 (0.75) = 29.15 kip
From Eq. (2), w = 10.9 kip>ft
Ans.
203
B
5 ft
FA = FC = F
FB = F = 2.6667 w
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
720
= 58.67 ksi
= p
2
ABC
4 (0.125 )
P (in./in.)
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
960
= p
= 78.23 ksi
2
ABC
4 (0.125)
From Stress–Strain diagram
78.23 - 70
80 - 70
=
eBC - 0.007
0.12 - 0.007
eBC = 0.09997 in.>in.
Displacement:
dBC = eBCLBC = 0.09997(40) = 3.9990 in.
dD
dBC
=
;
80
50
dD =
8
(3.9990) = 6.40 in.
5
Ans.
206
0.12
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4–111. The bar having a diameter of 2 in. is fixed
connected at its ends and supports the axial load P. If the
material is elastic perfectly plastic as shown by the
stress–strain diagram, determine the smallest load P needed
to cause segment CB to yield. If this load is released,
determine the permanent displacement of point C.
P
A
C
2 ft
B
3 ft
s (ksi)
20
0.001
When P is increased, region AC will become plastic first, then CB will become
plastic. Thus,
FA = FB = sA = 20(p)(1)2 = 62.832 kip
+ ©F = 0;
:
x
FA + FB - P = 0
(1)
P = 2(62.832) = 125.66 kip
P = 126 kip
Ans.
The deflection of point C is,
dC = eL = (0.001)(3)(12) = 0.036 in. ;
Consider the reverse of P on the bar.
FB ¿(3)
FA ¿(2)
=
AE
AE
FA ¿ = 1.5 FB ¿
So that from Eq. (1)
FB ¿ = 0.4P
FA ¿ = 0.6P
dC ¿ =
0.4(P)(3)(12)
0.4(125.66)(3)(12)
FB ¿L
= 0.02880 in. :
=
=
AE
AE
p(1)2(20>0.001)
¢d = 0.036 - 0.0288 = 0.00720 in. ;
Ans.
207
P (in./in.)
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*4–112. Determine the elongation of the bar in
Prob. 4–111 when both the load P and the supports are
removed.
P
A
C
2 ft
B
3 ft
s (ksi)
20
0.001
When P is increased, region AC will become plastic first, then CB will become
plastic. Thus,
FA = FB = sA = 20(p)(1)2 = 62.832 kip
+ ©F = 0;
:
x
FA + FB - P = 0
(1)
P = 2(62.832) = 125.66 kip
P = 126 kip
Ans.
The deflection of point C is,
dC = eL = (0.001)(3)(12) = 0.036 in. ;
Consider the reverse of P on the bar.
FB ¿(3)
FA ¿(2)
=
AE
AE
FA ¿ = 1.5 FB ¿
So that from Eq. (1)
FB ¿ = 0.4P
FA ¿ = 0.6P
The resultant reactions are
FA ¿¿ = FB ¿¿ = -62.832 + 0.6(125.66) = 62.832 - 0.4(125.66) = 12.568 kip
When the supports are removed the elongation will be,
d =
12.568(5)(12)
PL
= 0.0120 in.
=
AE
p(1)2(20>0.001)
Ans.
208
P (in./in.)
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s
•4–113.
A material has a stress–strain diagram that can be
described by the curve s = cP1>2. Determine the deflection
d of the end of a rod made from this material if it has a
length L, cross-sectional area A, and a specific weight g.
L
1
2
A
s2 = c2 e
s = ce ;
s2(x) = c2e(x)
P
(1) d
P(x)
;
A
However s(x) =
e(x) =
dd
dx
From Eq. (1),
P2(x)
= c2
A2
P2(x)
dd
=
dx
A2c2
dd
;
dx
L
d =
1
1
P2(x) dx = 2 2
(gAx)2 dx
2 2
Ac L
A c L0
g2
=
d =
L
c2 L0
x2 dx =
g2 x3 L
冷
c2 3 0
g3L3
Ans.
3c2
4–114. The 2014-T6 aluminum rod has a diameter of 0.5 in.
and is lightly attached to the rigid supports at A and B when
T1 = 70°F. If the temperature becomes T2 = -10°F, and
an axial force of P = 16 lb is applied to the rigid collar as
shown, determine the reactions at A and B.
A
B
P/2
P/2
5 in.
8 in.
+ 0 = ¢ - ¢ + d
:
B
T
B
0 = p
0.016(5)
2
3
4 (0.5 )(10.6)(10 )
- 12.8(10 - 6)[70° - (-10°)](13) + p
FB(13)
2
3
4 (0.5 )(10.6)(10 )
FB = 2.1251 kip = 2.13 kip
+ ©F = 0;
:
x
Ans.
2(0.008) + 2.1251 - FA = 0
FA = 2.14 kip
Ans.
4–115. The 2014-T6 aluminum rod has a diameter of
0.5 in. and is lightly attached to the rigid supports at A
and B when T1 = 70°F. Determine the force P that must
be applied to the collar so that, when T = 0°F, the
reaction at B is zero.
+
:
A
P/2
5 in.
0 = ¢ B - ¢ T + dB
0 = p
P(5)
2
3
4 (0.5 )(10.6)(10 )
B
P/2
- 12.8(10 - 6)[(70)(13)] + 0
P = 4.85 kip
Ans.
209
8 in.
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*4–116. The rods each have the same 25-mm diameter
and 600-mm length. If they are made of A-36 steel,
determine the forces developed in each rod when the
temperature increases to 50° C.
C
600 mm
60⬚
B
A
60⬚
600 mm
Equation of Equilibrium: Referring to the free-body diagram of joint A shown in
Fig. a,
+ c ©Fx = 0;
FAD sin 60° - FAC sin 60° = 0
+ ©F = 0;
:
x
FAB - 2F cos 60° = 0
FAC = FAD = F
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
A
FAB (600)
B
p
2
9
4 0.025 (200)(10 )
- 0.36 = 2(0.36) - 2 C
A
F(600)
B
p
2
9
4 0.025 (200)(10 )
FAB + 2F = 176 714.59
S
(2)
Solving Eqs. (1) and (2),
FAB = FAC = FAD = 58 904.86 N = 58.9 kN
Ans.
210
D
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•4–117.
Two A-36 steel pipes, each having a crosssectional area of 0.32 in2, are screwed together using a
union at B as shown. Originally the assembly is adjusted so
that no load is on the pipe. If the union is then tightened
so that its screw, having a lead of 0.15 in., undergoes two full
turns, determine the average normal stress developed in the
pipe. Assume that the union at B and couplings at A and C
are rigid. Neglect the size of the union. Note: The lead would
cause the pipe, when unloaded, to shorten 0.15 in. when the
union is rotated one revolution.
B
A
3 ft
C
2 ft
The loads acting on both segments AB and BC are the same since no external load
acts on the system.
0.3 = dB>A + dB>C
0.3 =
P(2)(12)
P(3)(12)
3
0.32(29)(10 )
+
0.32(29)(103)
P = 46.4 kip
P
46.4
=
= 145 ksi
A
0.32
sAB = sBC =
Ans.
4–118. The brass plug is force-fitted into the rigid casting.
The uniform normal bearing pressure on the plug is
estimated to be 15 MPa. If the coefficient of static friction
between the plug and casting is ms = 0.3, determine the
axial force P needed to pull the plug out. Also, calculate the
displacement of end B relative to end A just before the plug
starts to slip out. Ebr = 98 GPa.
100 mm
B
15 MPa
P - 4.50(106)(2)(p)(0.02)(0.1) = 0
P = 56.549 kN = 56.5 kN
Ans.
Displacement:
dB>A = a
PL
AE
0.1 m
56.549(103)(0.15)
=
2
9
p(0.02 )(98)(10 )
+
L0
P
A
Equations of Equilibrium:
+ ©F = 0;
:
x
150 mm
0.56549(106) x dx
p(0.022)(98)(109)
= 0.00009184 m = 0.0918 mm
Ans.
211
20 mm
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4–119. The assembly consists of two bars AB and CD of
the same material having a modulus of elasticity E1 and
coefficient of thermal expansion a1, and a bar EF having a
modulus of elasticity E2 and coefficient of thermal
expansion a2. All the bars have the same length L and
cross-sectional area A. If the rigid beam is originally
horizontal at temperature T1, determine the angle it makes
with the horizontal when the temperature is increased
to T2.
D
B
L
A
C
d
Equations of Equilibrium:
a + ©MC = 0;
FAB = FEF = F
+ c ©Fy = 0;
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 +
FCD (L)
R
AE1
= a2 (T2 - T1) L -
F(L)
F(L)
+ a1 (T2 - T1) L AE2
AE1
[2]
Substitute Eq. [1] into [2].
2a1 (T2 - T1) L +
4FL
FL
FL
= a2 (T2 - T1)L + a1 (T2 - T1)L AE1
AE2
AE1
F
5F
+
= a2 (T2 - T1) - a1 (T2 - T1)
AE1
AE2
F¢
5E2 + E1
b = (T2 - T1)(a2 - a1) ;
AE1E2
F =
AE1E2 (T2 - T1)(a2 - a1)
5E2 + E1
(dEF)T = a2 (T2 - T1) L
(dEF)F =
AE1E2 (T2 - T1)(a2 - a1)(L)
E1 (T2 - T1)(a2 - a1)(L)
=
AE2 (5E2 + E1)
5E2 + E1
dEF = (dEF)T - (dEF)F =
a2 L(T2 - T1)(5E2 - E1) - E1L(T2 - T1)(a2 - a1)
5E2 + E1
(dAB)T = a1 (T2 - T1) L
(dAB)F =
AE1E2 (T2 - T1)(a2 - a1)(L)
E2 (T2 - T1)(a2 - a1)(L)
=
AE1 (5E2 + E1)
5E2 + E1
dAB = (dAB)T - (dAB)F =
F
a1 L(5E2 + E1)(T2 - T1) - E2 L(T2 - T1)(a2 - a1)
5E2 + E1
212
E
d
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4–119.
Continued
dEF - dAB =
L(T2 - T1)
[a2 (5E2 + E1) - E1 (a2 - a1) - a1 (5E2 + E1)
5E2 + E1
+ E2 (a2 - a1)]
=
L(T2 - T1)
C (5E2 + E1)(a2 - a1) + (a2 - a1)(E2 - E1) D
5E2 + E1
=
L(T2 - T1)(a2 - a1)
(5E2 + E1 + E2 - E1)
5E2 + E1
=
L(T2 - T1)(a2 - a1)(6E2)
5E2 + E1
u =
3E2L(T2 - T1)(a2 - a1)
dEF - dAB
=
2d
d(5E2 + E1)
Ans.
*4–120. The rigid link is supported by a pin at A and two
A-36 steel wires, each having an unstretched length of 12 in.
and cross-sectional area of 0.0125 in2. Determine the force
developed in the wires when the link supports the vertical
load of 350 lb.
12 in.
C
5 in.
B
Equations of Equilibrium:
a + ©MA = 0;
4 in.
A
-FC(9) - FB (4) + 350(6) = 0
[1]
Compatibility:
6 in.
dC
dB
=
4
9
350 lb
FC(L)
FB (L)
=
4AE
9AE
9FB - 4FC = 0‚
[2]
FB = 86.6 lb
Ans.
FC = 195 lb
Ans.
Solving Eqs. [1] and [2] yields:
213
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•5–1.
A shaft is made of a steel alloy having an allowable
shear stress of tallow = 12 ksi. If the diameter of the shaft is
1.5 in., determine the maximum torque T that can be
transmitted. What would be the maximum torque T¿ if a
1-in.-diameter hole is bored through the shaft? Sketch the
shear-stress distribution along a radial line in each case.
T
T¿
Allowable Shear Stress: Applying the torsion formula
tmax = tallow =
Tc
J
T (0.75)
12 = p
4
2 (0.75 )
T = 7.95 kip # in.
Ans.
Allowable Shear Stress: Applying the torsion formula
tmax = tallow =
T¿c
J
12 = p
T¿ (0.75)
4
4
2 (0.75 - 0.5 )
T¿ = 6.381 kip # in. = 6.38 kip # in.
tr = 0.5 in =
Ans.
T¿r
6.381(0.5)
= 8.00 ksi
= p
4
4
J
2 (0.75 - 0.5 )
214
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5–2. The solid shaft of radius r is subjected to a torque T.
Determine the radius r¿ of the inner core of the shaft that
resists one-half of the applied torque 1T>22. Solve the
problem two ways: (a) by using the torsion formula, (b) by
finding the resultant of the shear-stress distribution.
r¿
r
T
a)
tmax =
Tc
Tr
2T
= p 4 =
J
p r3
2 r
(T2 )r¿
T
t = p
=
4
3
(r¿)
p(r¿)
2
Since t =
r¿ =
T
r¿ 2T
=
a
b
r pr3
p(r¿)3
r¿
t
;
r max
r
1
= 0.841 r
Ans.
24
r
2
b)
r¿
dT = 2p
L0
r
2
r¿
dT = 2p
L0
r
2
L0
tr2 dr
L0
r
tmax r2 dr
L0 r
r¿
dT = 2p
r 2T 2
a 3 br dr
L0 r pr
r¿
4T
T
= 4
r3 dr
2
r L0
r¿ =
r
1
Ans.
= 0.841r
24
215
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5–3. The solid shaft is fixed to the support at C and
subjected to the torsional loadings shown. Determine the
shear stress at points A and B and sketch the shear stress on
volume elements located at these points.
10 kN⭈m
C
A
50 mm
The internal torques developed at Cross-sections pass through point B and A are
shown in Fig. a and b, respectively.
The polar moment of inertia of the shaft is J =
p
(0.0754) = 49.70(10 - 6) m4. For
2
point B, rB = C = 0.075 Thus,
tB =
4(103)(0.075)
TB c
= 6.036(106) Pa = 6.04 MPa
=
J
49.70(10 - 6)
Ans.
From point A, rA = 0.05 m.
tA =
TArA
6(103)(0.05)
= 6.036(106) Pa = 6.04 MPa.
=
J
49.70 (10 - 6)
216
Ans.
B
75 mm
4 kN⭈m
75 mm
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*5–4. The tube is subjected to a torque of 750 N # m.
Determine the amount of this torque that is resisted by the
gray shaded section. Solve the problem two ways: (a) by
using the torsion formula, (b) by finding the resultant of the
shear-stress distribution.
75 mm
100 mm
750 Nm
25 mm
a) Applying Torsion Formula:
tmax =
750(0.1)
Tc
= 0.4793 MPa
= p
4
4
J
2 (0.1 - 0.025 )
tmax = 0.4793 A 106 B = p
T¿(0.1)
4
4
2 (0.1 - 0.075 )
T¿ = 515 N # m
Ans.
b) Integration Method:
r
t = a b tmax
c
and
dA = 2pr dr
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 =
A
30 N⭈m
90(0.02)
Tmax c
= p
4
J
(0.02
- 0.01854)
2
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 30-mm-diameter shaft is used to transmit
the torques applied to the gears. Determine the absolute
maximum shear stress on the shaft.
500 Nm
A
200 Nm
Internal Torque: As shown on torque diagram.
C
Maximum Shear Stress: From the torque diagram Tmax = 400 N # m. Then, applying
torsion Formula.
400 Nm
300 mm
abs =
tmax
Tmax c
J
Ans.
•5–9.
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
400(0.015)
= p
= 75.5 MPa
4
2 (0.015 )
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 - nFR = 0;
T
F =
nR
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
tavg = tmax ;
n =
4T
2T
=
nRpd2
p r3
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 =
8 in.
210(0.375)
Tc
= 7.82 ksi
= p
4
4
J
2 (0.375 - 0.34 )
Ans.
15 lb
210(0.5)
Tc
= 2.36 ksi
tBC =
= p
4
J
(0.5
- 0.434)
2
Ans.
220
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*5–12. The motor delivers a torque of 50 N # m to the shaft
AB. This torque is transmitted to shaft CD using the gears
at E and F. Determine the equilibrium torque Tⴕ on shaft
CD and the maximum shear stress in each shaft. The
bearings B, C, and D allow free rotation of the shafts.
A
50 mm
30 mm
Equilibrium:
B
a + ©ME = 0;
a + ©MF = 0;
50 - F(0.05) = 0
F = 1000 N
35 mm
T¿
T¿ - 1000(0.125) = 0
T¿ = 125 N # m
C
E
125 mm
D
F
Ans.
Internal Torque: As shown on FBD.
Maximum Shear Stress: Applying torsion Formula.
(tAB)max =
50.0(0.015)
TAB c
= 9.43 MPa
= p
4
J
2 (0.015 )
Ans.
(tCD)max =
125(0.0175)
TCDc
= 14.8 MPa
= p
4
J
2 (0.0175 )
Ans.
•5–13. If the applied torque on shaft CD is T¿ = 75 N # m,
determine the absolute maximum shear stress in each shaft.
The bearings B, C, and D allow free rotation of the shafts,
and the motor holds the shafts fixed from rotating.
A
50 mm
Equilibrium:
30 mm
a + ©MF = 0;
75 - F(0.125) = 0;
a + ©ME = 0;
600(0.05) - TA = 0
B
F = 600 N
35 mm
T¿
TA = 30.0 N # m
Internal Torque: As shown on FBD.
Maximum Shear Stress: Applying the torsion formula
(tEA)max =
30.0(0.015)
TEA c
= 5.66 MPa
= p
4
J
2 (0.015 )
Ans.
(tCD)max =
75.0(0.0175)
TCDc
= 8.91 MPa
= p
4
J
2 (0.0175 )
Ans.
221
C
E
125 mm
D
F
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250 N⭈m
5–14. The solid 50-mm-diameter shaft is used to transmit
the torques applied to the gears. Determine the absolute
maximum shear stress in the shaft.
75 N⭈m
A
325 N⭈m
150 N⭈ m
B
500 mm
The internal torque developed in segments AB , BC and CD of the shaft are shown
in Figs. a, b and c.
C
D
400 mm
500 mm
The maximum torque occurs in segment AB. Thus, the absolute maximum shear
stress occurs in this segment. The polar moment of inertia of the shaft is
p
J = (0.0254) = 0.1953p(10 - 6)m4. Thus,
2
A tmax B abs =
250(0.025)
TAB c
= 10.19(106)Pa = 10.2 MPa
=
J
0.1953p(10 - 6)
222
Ans.
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5–15. The solid shaft is made of material that has an
allowable shear stress of tallow = 10 MPa. Determine the
required diameter of the shaft to the nearest mm.
15 N⭈m
25 N⭈m
A
30 N⭈m
B
60 N⭈m
C
70 N⭈m
D
E
The internal torques developed in each segment of the shaft are shown in the torque
diagram, Fig. a.
Segment DE is critical since it is subjected to the greatest internal torque. The polar
p d 4
p 4
moment of inertia of the shaft is J =
a b =
d . Thus,
2 2
32
TDE c
tallow =
;
J
d
70a b
2
10(106) =
p 4
d
32
d = 0.03291 m = 32.91 mm = 33 mm
223
Ans.
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*5–16. The solid shaft has a diameter of 40 mm.
Determine the absolute maximum shear stress in the shaft
and sketch the shear-stress distribution along a radial line
of the shaft where the shear stress is maximum.
15 N⭈m
25 N⭈m
A
30 N⭈m
B
The internal torque developed in each segment of the shaft are shown in the torque
diagram, Fig. a.
60 N⭈m
C
70 N⭈m
D
E
Since segment DE subjected to the greatest torque, the absolute maximum shear
p
stress occurs here. The polar moment of inertia of the shaft is J = (0.024)
2
= 80(10 - 9)p m4. Thus,
tmax =
70(0.02)
TDE c
= 5.57(106) Pa = 5.57 MPa
=
J
80(10 - 9)p
Ans.
The shear stress distribution along the radial line is shown in Fig. b.
•5–17.
The rod has a diameter of 1 in. and a weight of
10 lb/ft. Determine the maximum torsional stress in the rod
at a section located at A due to the rod’s weight.
4.5 ft
B
Here, we are only interested in the internal torque. Thus, other components of the
internal loading are not indicated in the FBD of the cut segment of the rod, Fig. a.
©Mx = 0;
TA - 10(4)(2) = 0
TA = 80 lb # ft a
The polar moment of inertia of the cross section at A is J =
12in
b = 960 lb # in.
1ft
p
(0.54) = 0.03125p in4.
2
Thus
tmax =
960 (0.5)
TA c
=
= 4889.24 psi = 4.89 ksi
J
0.03125p
Ans.
224
4 ft
A
1.5 ft
1.5 ft
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5–18. The rod has a diameter of 1 in. and a weight of
15 lb/ft. Determine the maximum torsional stress in the rod
at a section located at B due to the rod’s weight.
4.5 ft
B
4 ft
Here, we are only interested in the internal torque. Thus, other components of the
internal loading are not indicated in the FBD of the cut segment of the rod, Fig. a.
©Mx = 0;
TB - 15(4)(2) = 0
TB = 120 lb # ft a
12 in
b = 1440 lb # in.
1ft
p
The polar moment of inertia of the cross-section at B is J = (0.54)
2
= 0.03125p in4. Thus,
tmax =
1440(0.5)
TB c
=
= 7333.86 psi = 7.33 ksi
J
0.03125p
Ans.
225
A
1.5 ft
1.5 ft
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5–19. Two wrenches are used to tighten the pipe. If P =
300 N is applied to each wrench, determine the maximum
torsional shear stress developed within regions AB and BC.
The pipe has an outer diameter of 25 mm and inner
diameter of 20 mm. Sketch the shear stress distribution for
both cases.
P
B
Internal Loadings: The internal torque developed in segments AB and BC of the
pipe can be determined by writing the moment equation of equilibrium about the x
axis by referring to their respective free - body diagrams shown in Figs. a and b.
©Mx = 0; TAB - 300(0.25) = 0
TAB = 75 N # m
TBC = 150 N # m
Allowable Shear Stress: The polar moment of inertia of the pipe is
p
J =
A 0.01254 - 0.014 B = 22.642(10 - 9)m4.
2
A tmax B AB =
75(0.0125)
TAB c
= 41.4 MPa
=
J
22.642(10 - 9)
A tAB B r = 0.01 m =
A tmax B BC =
Ans.
TAB r
75(0.01)
= 33.1 MPa
=
J
22.642(10 - 9)
150(0.0125)
TBC c
= 82.8 MPa
=
J
22.642(10 - 9)
A tBC B r = 0.01 m =
Ans.
TBC r
150(0.01)
= 66.2 MPa
=
J
22.642(10 - 9)
The shear stress distribution along the radial line of segments AB and BC of the
pipe is shown in Figs. c and d, respectively.
226
A
250 mm
P
And
©Mx = 0; TBC - 300(0.25) - 300(0.25) = 0
250 mm
C
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*5–20. Two wrenches are used to tighten the pipe. If the
pipe is made from a material having an allowable shear stress
of tallow = 85 MPa, determine the allowable maximum force
P that can be applied to each wrench. The pipe has an outer
diameter of 25 mm and inner diameter of 20 mm.
P
250 mm
C
B
A
250 mm
Internal Loading: By observation, segment BC of the pipe is critical since it is
subjected to a greater internal torque than segment AB. Writing the moment
equation of equilibrium about the x axis by referring to the free-body diagram
shown in Fig. a, we have
©Mx = 0; TBC - P(0.25) - P(0.25) = 0
TBC = 0.5P
Allowable Shear Stress: The polar moment of inertia of the pipe is
p
J =
A 0.01254 - 0.014 B = 22.642(10 - 9)m4
2
tallow =
TBC c
;
J
85(106) =
0.5P(0.0125)
22.642(10 - 9)
P = 307.93N = 308 N
Ans.
227
P
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•5–21.
The 60-mm-diameter solid shaft is subjected to the
distributed and concentrated torsional loadings shown.
Determine the absolute maximum and minimum shear
stresses on the outer surface of the shaft and specify their
locations, measured from the fixed end A.
A
2 kN⭈m/m
1.5 m
1200 N⭈m
C
The internal torque for segment BC is Constant TBC = 1200 N # m, Fig. a. However,
the internal for segment AB varies with x, Fig. b.
TAB - 2000x + 1200 = 0
TAB = (2000x - 1200) N # m
The minimum shear stress occurs when the internal torque is zero in segment AB.
By setting TAB = 0,
0 = 2000x - 1200
x = 0.6 m
Ans.
And
d = 1.5 m - 0.6 m = 0.9 m
Ans.
tmin = 0
Ans.
The maximum shear stress occurs when the internal torque is the greatest. This
occurs at fixed support A where
d = 0
Ans.
At this location,
(TAB)max = 2000(1.5) - 1200 = 1800 N # m
The polar moment of inertia of the rod is J =
tmax =
p
(0.034) = 0.405(10 - 6)p. Thus,
2
(TAB)max c
1800(0.03)
= 42.44(106)Pa = 42.4 MPa
=
J
0.405(10 - 6)p
228
Ans.
B
0.8 m
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5–22. The solid shaft is subjected to the distributed and
concentrated torsional loadings shown. Determine the
required diameter d of the shaft to the nearest mm if the
allowable shear stress for the material is tallow = 50 MPa.
A
2 kN⭈m/m
1.5 m
1200 N⭈m
C
The internal torque for segment BC is constant TBC = 1200 N # m, Fig. a. However,
the internal torque for segment AB varies with x, Fig. b.
TAB - 2000x + 1200 = 0 TAB = (2000x - 1200) N # m
For segment AB, the maximum internal torque occurs at fixed support A where
x = 1.5 m. Thus,
A TAB B max = 2000(1.5) - 1200 = 1800 N # m
Since A TAB B max 7 TBC, the critical cross-section is at A. The polar moment of inertia
p d 4
pd4
of the rod is J =
. Thus,
a b =
2 2
32
tallow =
Tc
;
J
50(106) =
1800(d>2)
pd4>32
d = 0.05681 m = 56.81 mm = 57 mm
229
Ans.
B
0.8 m
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*5–24. The copper pipe has an outer diameter of 2.50 in.
and an inner diameter of 2.30 in. If it is tightly secured to the
wall at C and a uniformly distributed torque is applied to it
as shown, determine the shear stress developed at points A
and B. These points lie on the pipe’s outer surface. Sketch
the shear stress on volume elements located at A and B.
B
A
C
125 lbft/ft
4 in.
9 in.
12 in.
Internal Torque: As shown on FBD.
Maximum Shear Stress: Applying the torsion formula
tA =
TA c
J
125.0(12)(1.25)
= 1.72 ksi
= p
4
4
2 (1.25 - 1.15 )
tB =
Ans.
TB c
J
218.75(12)(1.25)
= p
= 3.02 ksi
4
4
2 (1.25 - 1.15 )
Ans.
•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
= 3.59 ksi
4
4
2 (1.25 - 1.15 )
Ans.
According to Saint-Venant’s principle, application of the torsion formula should be
as points sufficiently removed from the supports or points of concentrated loading.
230
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5–26. A cylindrical spring consists of a rubber annulus
bonded to a rigid ring and shaft. If the ring is held fixed and
a torque T is applied to the shaft, determine the maximum
shear stress in the rubber.
ro
ri
T
h
T
r
T
F
t =
=
=
A
2prh
2p r2 h
Shear stress is maximum when r is the smallest, i.e. r = ri. Hence,
tmax =
T
2p ri 2 h
Ans.
300 N⭈m
5–27. The A-36 steel shaft is supported on smooth
bearings that allow it to rotate freely. If the gears are
subjected to the torques shown, determine the maximum
shear stress developed in the segments AB and BC. The
shaft has a diameter of 40 mm.
100 N⭈m
A
The internal torque developed in segments AB and BC are shown in their
respective FBDs, Figs. a and b.
The polar moment of inertia of the shaft is J =
200 N⭈m
B
p
(0.024) = 80(10-9)p m4. Thus,
2
C
300(0.02)
TAB c
= 23.87(106)Pa = 23.9 MPa
=
A tAB B max =
J
80(10-9)p
Ans.
A tBC B max =
Ans.
200(0.02)
TBC c
= 15.92(106) Pa = 15.9 MPa
=
J
80(10-9)p
231
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300 N⭈m
*5–28. The A-36 steel shaft is supported on smooth
bearings that allow it to rotate freely. If the gears are
subjected to the torques shown, determine the required
diameter of the shaft to the nearest mm if tallover = 60 MPa.
100 N⭈m
The internal torque developed in segments AB and BC are shown in their
respective FBDs, Fig. a and b
A
200 N⭈m
B
Here, segment AB is critical since its internal torque is the greatest. The polar
p d 4
pd4
moment of inertia of the shaft is J =
. Thus,
a b =
2 2
32
C
TC
tallow =
;
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 4
2 c
c3 = 26.526(10-9) x3
c = (2.98 x) mm
Ans.
5–31. The solid steel shaft AC has a diameter of 25 mm and
is supported by smooth bearings at D and E. It is coupled to
a motor at C, which delivers 3 kW of power to the shaft
while it is turning at 50 rev>s. If gears A and B remove 1 kW
and 2 kW, respectively, determine the maximum shear stress
developed in the shaft within regions AB and BC. The shaft
is free to turn in its support bearings D and E.
TC =
3(103)
P
=
= 9.549 N # m
v
50(2p)
TA =
1
T = 3.183 N # m
3 C
3 kW
2 kW
25 mm
1 kW
A
D
(tAB)max =
3.183 (0.0125)
TC
= 1.04 MPa
= p
4
J
2 (0.0125 )
Ans.
(tBC)max =
9.549 (0.0125)
TC
= 3.11 MPa
= p
4
J
2 (0.0125 )
Ans.
233
B
E
C
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*5–32. The pump operates using the motor that has a
power of 85 W. If the impeller at B is turning at 150 rev>min,
determine the maximum shear stress developed in the
20-mm-diameter transmission shaft at A.
150 rev/min
A
Internal Torque:
v = 150
rev 2p rad 1 min
= 5.00p rad>s
¢
≤
min
rev
60 s
P = 85 W = 85 N # m>s
T =
P
85
=
= 5.411 N # m
v
5.00p
Maximum Shear Stress: Applying torsion formula
tmax =
Tc
J
5.411 (0.01)
=
p
4
2 (0.01 )
= 3.44 MPa
Ans.
•5–33.
The gear motor can develop 2 hp when it turns at
450 rev>min. If the shaft has a diameter of 1 in., determine
the maximum shear stress developed in the shaft.
The angular velocity of the shaft is
v = ¢ 450
rev
2p rad
1 min
≤ ¢
≤ ¢
≤ = 15p rad>s
min
1 rev
60 s
and the power is
P = 2 hp ¢
550 ft # lb>s
≤ = 1100 ft # lb>s
1 hp
Then
T =
P
1100
12 in
=
= 23.34 lb # ft a
b = 280.11 lb # in
v
15p
1ft
The polar moment of inertia of the shaft is J =
tmax =
p
(0.54) = 0.03125p in4. Thus,
2
280.11 (0.5)
Tc
=
= 1426.60 psi = 1.43 ksi
J
0.03125p
Ans.
234
B
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5–34. The gear motor can develop 3 hp when it turns at
150 rev>min. If the allowable shear stress for the shaft is
tallow = 12 ksi, determine the smallest diameter of the shaft
to the nearest 18 in. that can be used.
The angular velocity of the shaft is
v = a 150
rev
2p rad 1 min
ba
ba
b = 5p rad>s
min
1 rev
60 s
and the power is
P = (3 hp) a
550 ft # lb>s
b = 1650 ft # lb>s
1 hp
Then
T =
P
1650
12 in
=
= (105.04 lb # ft)a
b = 1260.51 lb # in
v
5p
1 ft
The polar moment of inertia of the shaft is J =
tallow =
Tc
;
J
12(103) =
p d 4
pd4
a b =
. Thus,
2 2
32
1260.51 (d>2)
pd4>32
d = 0.8118 in. =
7
in.
8
Ans.
5–35. The 25-mm-diameter shaft on the motor is made
of a material having an allowable shear stress of
tallow = 75 MPa . If the motor is operating at its maximum
power of 5 kW, determine the minimum allowable rotation
of the shaft.
Allowable Shear Stress: The polar moment of inertia of the shaft is
p
J =
A 0.01254 B = 38.3495(10-9) m4.
2
tallow =
Tc
;
J
75(106) =
T(0.0125)
38.3495(10-9)
T = 230.10 N # m
Internal Loading:
T =
P
;
v
230.10 =
5(103)
v
v = 21.7 rad>s
Ans.
235
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*5–36. The drive shaft of the motor is made of a material
having an allowable shear stress of tallow = 75 MPa. If the
outer diameter of the tubular shaft is 20 mm and the wall
thickness is 2.5 mm, determine the maximum allowable
power that can be supplied to the motor when the shaft is
operating at an angular velocity of 1500 rev>min.
Internal Loading: The angular velocity of the shaft is
v = a 1500
rev
2p rad 1 min
ba
ba
b = 50p rad>s
min
1 rev
60 s
We have
T =
P
P
=
v
50p
Allowable Shear Stress: The polar moment of inertia of the shaft is
p
J = A 0.014 - 0.00754 B = 10.7379(10-9) m4.
2
tallow =
Tc
;
J
75(106) =
a
P
b(0.01)
50p
10.7379(10-9)
P = 12 650.25 W = 12.7 kW
Ans.
•5–37.
A ship has a propeller drive shaft that is turning at
1500 rev>min while developing 1800 hp. If it is 8 ft long and
has a diameter of 4 in., determine the maximum shear stress
in the shaft caused by torsion.
Internal Torque:
v = 1500
rev 2p rad 1 min
a
b
= 50.0 p rad>s
min 1 rev
60 s
P = 1800 hpa
T =
550 ft # lb>s
b = 990 000 ft # lb>s
1 hp
990 000
P
=
= 6302.54 lb # ft
v
50.0p
Maximum Shear Stress: Applying torsion formula
tmax =
6302.54(12)(2)
Tc
=
p 4
J
2 (2 )
= 6018 psi = 6.02 ksi
Ans.
236
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5–38. The motor A develops a power of 300 W and turns
its connected pulley at 90 rev>min. Determine the required
diameters of the steel shafts on the pulleys at A and B if the
allowable shear stress is tallow = 85 MPa.
60 mm
90 rev/min
A
B
150 mm
Internal Torque: For shafts A and B
vA = 90
rev 2p rad 1 min
a
b
= 3.00p rad>s
min
rev
60 s
P = 300 W = 300 N # m>s
P
300
=
= 31.83 N # m
vA
3.00p
TA =
vB = vA a
rA
0.06
b = 3.00pa
b = 1.20p rad>s
rB
0.15
P = 300 W = 300 N # m>s
TB =
P
300
=
= 79.58 N # m
vB
1.20p
Allowable Shear Stress: For shaft A
tmax = tallow =
85 A 106 B =
TA c
J
31.83 A d2A B
A B
p dA 4
2 2
dA = 0.01240 m = 12.4 mm
Ans.
For shaft B
tmax = tallow =
85 A 106 B =
TB c
J
79.58 A d2B B
A B
p dB 4
2 2
dB = 0.01683 m = 16.8 mm
Ans.
237
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5–39. The solid steel shaft DF has a diameter of 25 mm
and is supported by smooth bearings at D and E. It is
coupled to a motor at F, which delivers 12 kW of power to
the shaft while it is turning at 50 rev>s. If gears A, B, and C
remove 3 kW, 4 kW, and 5 kW respectively, determine the
maximum shear stress developed in the shaft within regions
CF and BC. The shaft is free to turn in its support bearings
D and E.
v = 50
3 kW 4 kW
A
B
D
C
E
F
rev 2p rad
c
d = 100 p rad>s
s
rev
TF =
12(103)
P
=
= 38.20 N # m
v
100 p
TA =
3(103)
P
=
= 9.549 N # m
v
100 p
TB =
4(103)
P
=
= 12.73 N # m
v
100 p
(tmax)CF =
38.20(0.0125)
TCF c
= 12.5 MPa
= p
4
J
2 (0.0125 )
Ans.
(tmax)BC =
22.282(0.0125)
TBC c
= 7.26 MPa
= p
4
J
2 (0.0125 )
Ans.
*5–40. Determine the absolute maximum shear stress
developed in the shaft in Prob. 5–39.
v = 50
12 kW
5 kW
25 mm
3 kW 4 kW
rev 2p rad
c
d = 100 p rad>s
s
rev
TF =
12(103)
P
=
= 38.20 N # m
v
100p
TA =
3(103)
P
=
= 9.549 N # m
v
100p
TB =
4(103)
P
=
= 12.73 N # m
v
100p
A
D
Tmax = 38.2 N # m
tabs = Tc =
max
J
38.2(0.0125) = 12.5 MPa
Ans.
p
4
2 (0.0125 )
238
B
C
From the torque diagram,
12 kW
5 kW
25 mm
E
F
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•5–41.
The A-36 steel tubular shaft is 2 m long and has an
outer diameter of 50 mm. When it is rotating at 40 rad> s, it
transmits 25 kW of power from the motor M to the pump
P. Determine the smallest thickness of the tube if the
allowable shear stress is tallow = 80 MPa.
P
M
P
M
The internal torque in the shaft is
T =
25(103)
P
=
= 625 N # m
v
40
The polar moment of inertia of the shaft is J =
tallow =
Tc
;
J
80(106) = p
p
(0.0254 - Ci 4). Thus,
2
625(0.025)
4
4
2 (0.025 - Ci )
Ci = 0.02272 m
So that
t = 0.025 - 0.02272
= 0.002284 m = 2.284 mm = 2.5 mm
Ans.
5–42. The A-36 solid tubular steel shaft is 2 m long and has
an outer diameter of 60 mm. It is required to transmit
60 kW of power from the motor M to the pump P.
Determine the smallest angular velocity the shaft can have
if the allowable shear stress is tallow = 80 MPa.
The polar moment of inertia of the shaft is J =
tallow =
Tc
;
J
80(106) =
p
(0.034) = 0.405(10-6)p m4. Thus,
2
T(0.03)
0.405(10-6)p
T = 3392.92 N # m
P = Tv ;
60(103) = 3392.92 v
v = 17.68 rad>s = 17.7 rad>s
Ans.
239
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5–43. A steel tube having an outer diameter of 2.5 in. is
used to transmit 35 hp when turning at 2700 rev>min.
Determine the inner diameter d of the tube to the nearest
1
8 in. if the allowable shear stress is tallow = 10 ksi.
v =
2700(2p)
= 282.74 rad>s
60
d
2.5 in.
P = Tv
35(550) = T(282.74)
T = 68.083 lb # ft
tmax = tallow =
10(103) =
Tc
J
68.083(12)(1.25)
p
4
4
2 (12.5 - ci )
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
4
2 (1.25 - ri )
,
ri = 1.0762 in.
t = ro - ri = 1.25 - 1.0762
t = 0.174 in.
Ans.
240
A
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•5–45.
The drive shaft AB of an automobile is to be
designed as a thin-walled tube. The engine delivers 150 hp
when the shaft is turning at 1500 rev>min. Determine the
minimum thickness of the shaft’s wall if the shaft’s outer
diameter is 2.5 in. The material has an allowable shear stress
of tallow = 7 ksi.
v =
B
A
1500(2p)
= 157.08 rad>s
60
P = Tv
150(550) = T(157.08)
T = 525.21 lb # ft
tallow =
7(103) =
Tc
J
525.21(12)(1.25)
p
4
4
2 (1.25 - ri )
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
1.5 in.
rA
1.5
rev
2p rad 1 min
vBC = a b vA = a
b a1800
ba
ba
b = 30p rad>s
rC
3
min
1 rev
60 s
A
The power is
P = (15 hp) a
550 ft # n>s
b = 8250 ft # lb>s
1 hp
Thus,
T =
P
8250
12 in.
=
= (87.54 lb # ft)a
b = 1050.42 lb # in
v
30p
1 ft
The polar moment of inertia of the shaft is J =
tallow =
Tc
;
J
12(103) =
p d 4
pd4
a b =
. Thus,
2 2
32
1050.42(d>2)
pd4>32
d = 0.7639 in =
7
in.
8
Ans.
241
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5–47. The propellers of a ship are connected to a A-36
steel shaft that is 60 m long and has an outer diameter of
340 mm and inner diameter of 260 mm. If the power output is
4.5 MW when the shaft rotates at 20 rad>s, determine the
maximum torsional stress in the shaft and its angle of twist.
T =
4.5(106)
P
=
= 225(103) N # m
v
20
tmax =
f =
225(103)(0.170)
Tc
= 44.3 MPa
=
p
J
[(0.170)4 - (0.130)4]
2
Ans.
225 A 103 B (60)
TL
= 0.2085 rad = 11.9°
=
p
JG
[(0.170)4 - (0.130)4)75(109)
2
Ans.
*5–48. A shaft is subjected to a torque T. Compare the
effectiveness of using the tube shown in the figure with that
of a solid section of radius c. To do this, compute the percent
increase in torsional stress and angle of twist per unit length
for the tube versus the solid section.
T
c
2
T
c
Shear stress:
For the tube,
(tt)max =
c
Tc
Jt
For the solid shaft,
(ts)max =
Tc
Js
% increase in shear stress =
Tc
Tc
(ts)max - (tt)max
Jt - Js
(100) =
(100)
Tc
(tt)max
J
s
p 4
p
p 4
4
Js - Jt
2 c - [ 2 [c - ( 2 ) ]]
(100)
(100) =
=
p
p 4
4
Jt
2 [c - ( 2 ) ]
Ans.
= 6.67 %
Angle of twist:
For the tube,
ft =
TL
Jt(G)
For the shaft,
fs =
TL
Js(G)
TL
% increase in f =
TL
Jt(G) - Js(G)
ft - fs
(100%) =
(100%)
TL
fs
J (G)
s
=
Js - Jt
(100%) =
Jt
p 4
p
p 4
4
2 c - [ 2 [c - ( 2 ) ]]
(100%)
p
p
4
4
2 [c - ( 2 ) ]
= 6.67 %
Ans.
242
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•5–49.
The A-36 steel axle is made from tubes AB and CD
and a solid section BC. It is supported on smooth bearings
that allow it to rotate freely. If the gears, fixed to its ends, are
subjected to 85-N # m torques, determine the angle of twist
of gear A relative to gear D. The tubes have an outer
diameter of 30 mm and an inner diameter of 20 mm. The
solid section has a diameter of 40 mm.
400 mm
250 mm
400 mm
B
fND = ©
TL
JG
A
(85)(0.25)
2(85)(0.4)
= p
4
2 (0.015
4
9
- 0.01 )(75)(10 )
85 Nm
+ p
4
9
2 (0.02 )(75)(10 )
= 0.01534 rad = 0.879°
Ans.
5–50. The hydrofoil boat has an A-36 steel propeller
shaft that is 100 ft long. It is connected to an in-line diesel
engine that delivers a maximum power of 2500 hp and
causes the shaft to rotate at 1700 rpm. If the outer
diameter of the shaft is 8 in. and the wall thickness is 38 in.,
determine the maximum shear stress developed in the
shaft. Also, what is the “wind up,” or angle of twist in the
shaft at full power?
100 ft
Internal Torque:
v = 1700
rev 2p rad 1 min
a
b
= 56.67p rad>s
min
rev
60 s
P = 2500 hp a
T =
550 ft # lb>s
b = 1 375 000 ft # lb>s
1 hp
P
1 375 000
=
= 7723.7 lb # ft
v
56.67p
Maximum Shear Stress: Applying torsion Formula.
tmax =
Tc
J
7723.7(12)(4)
= p 4
= 2.83 ksi
4
2 (4 - 3.625 )
Ans.
Angle of Twist:
f =
7723.7(12)(100)(12)
TL
= p 4
JG
(4
- 3.6254)11.0(106)
2
= 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 4
2 c
c = 1.3586 in.
Angle of twist limitation
f =
0.05 =
TL
JG
2626.06(12)(2)(12)
p 4
6
2 c (11.0)(10 )
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 4
2 c
c = 1.2408 in.
Angle of twist limitation
f =
0.05 =
TL
JG
2626.06(12)(2)(12)
p 4
6
2 c (11.0)(10 )
c = 0.967 in.
Shear stress failure controls the design
d = 2c = 2 (1.2408) = 2.48 in.
Use d = 2.50 in.
Ans.
245
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•5–53. The 20-mm-diameter A-36 steel shaft is subjected to
the torques shown. Determine the angle of twist of the end B.
A
Internal Torque: As shown on FBD.
D
Angle of Twist:
C
B
TL
fB = a
JG
30 Nm
600 mm
200 mm
20 Nm
800 mm
1
[-80.0(0.8) + (-60.0)(0.6) + (-90.0)(0.2)]
= p
4
9
(0.01
)(75.0)(10
)
2
= -0.1002 rad = | 5.74° |
80 Nm
Ans.
5–54. The assembly is made of A-36 steel and consists of a
solid rod 20 mm in diameter fixed to the inside of a tube using
a rigid disk at B. Determine the angle of twist at D. The tube
has an outer diameter of 40 mm and wall thickness of 5 mm.
A
B
The internal torques developed in segments AB and BD of the assembly are shown
in Fig. a and b
0.4 m
C
0.1 m
p
The polar moment of inertia of solid rod and tube are JAB = (0.024 - 0.0154)
2
p
= 54.6875(10 - 9)p m4 and JBD = (0.014) = 5(10 - 9)p m4. Thus,
2
fD = ©
Ti Li
TAB LAB
TBD LBD
=
+
Ji Gi
JAB Gst
JBD Gst
-60 (0.4)
90(0.4)
=
54.6875(10 - 9)p [75(109)]
+
5(10 - 9)p [75(109)]
= -0.01758 rad = 1.01°
Ans.
246
150 N⭈m
D
0.3 m
60 N⭈m
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5–55. The assembly is made of A-36 steel and consists of a
solid rod 20 mm in diameter fixed to the inside of a tube using
a rigid disk at B. Determine the angle of twist at C. The tube
has an outer diameter of 40 mm and wall thickness of 5 mm.
A
B
The polar moment of inertia
= 54.6875 (10 - 9)p m4. Thus,
fC = ©
of
the
tube
is
J =
150 N⭈m
0.4 m
The internal torques developed in segments AB and BC of the assembly are shown
in Figs. a and b.
C
0.1 m
p
(0.024 - 0.0154)
2
D
0.3 m
60 N⭈m
Ti Li
TAB LAB
TBC LBC
=
+
Ji Gi
JGst
J Gst
=
1
C 90(0.4) + 150(0.1) D
54.6875(10 )p [75(109)]
-9
= 0.003958 rad = 0.227°
Ans.
*5–56. The splined ends and gears attached to the A-36
steel shaft are subjected to the torques shown. Determine
the angle of twist of end B with respect to end A. The shaft
has a diameter of 40 mm.
fB>A = ©
300 N⭈m
500 N⭈m
A
200 N⭈m
-300(0.3)
200(0.4)
400(0.5)
TL
=
+
+
JG
JG
JG
JG
C
400 N⭈m
300 mm
190
190
= p
=
4
JG
(0.02
)(75)(109)
2
D
B
400 mm
= 0.01008 rad = 0.578°
Ans.
500 mm
247
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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.
tmax = tallow =
8(103) =
Tc
J
175.07(12) A d2 B
p
2
A d2 B
4
d = 1.102 in.
Angle of Twist: Assume failure due to angle of twist limitation.
fC>D =
B
8 in.
P
, we have
2pf
TC =
C
TCDLCD
JG
65.65(12)(8)
0.2(p)
=
180
p d 4
6
2 A 2 B (11.0)(10 )
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
4
6
2 (0.75 )(11.0)(10 )
= 0.001153 rad = 0.0661°
Ans.
249
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5–59. The shaft is made of A-36 steel. It has a diameter of
1 in. and is supported by bearings at A and D, which allow
free rotation. Determine the angle of twist of B with
respect to D.
A
B
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
FB/D = a
60 lb⭈ft
TiLi TBC LBC TCD LCD
=
+
JiGi
J Gst
J Gst
-60(12)(2.5)(12)
=
(0.03125p)[11.0(106)]
+ 0
= -0.02000 rad = 1.15°
Ans.
*5–60. The shaft is made of A-36 steel. It has a diameter of
1 in. and is supported by bearings at A and D, which allow
free rotation. Determine the angle of twist of gear C with
respect to B.
A
B
60 lb⭈ft
C
2 ft
60 lb⭈ft
2.5 ft
The internal torque developed in segment BC is shown in Fig. a
D
3 ft
p
The polar moment of inertia of the shaft is J = (0.54) = 0.03125p in4. Thus,
2
fC>B =
-60(12)(2.5)(12)
TBC LBC
=
J Gst
(0.03125p)[11.0(106)]
= -0.02000 rad
= 1.15°
Ans.
250
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•5–61.
The two shafts are made of A-36 steel. Each has a
diameter of 1 in., and they are supported by bearings at A,
B, and C, which allow free rotation. If the support at D is
fixed, determine the angle of twist of end B when the
torques are applied to the assembly as shown.
D
10 in.
C
80 lbft
A
30 in.
40 lbft
8 in.
10 in.
12 in.
Internal Torque: As shown on FBD.
Angle of Twist:
fE = a
TL
JG
1
= p
4
5
2 (0.5 )(11.0)(10 )
[-60.0(12)(30) + 20.0(12)(10)]
= -0.01778 rad = 0.01778 rad
fF =
6
6
f = (0.01778) = 0.02667 rad
4 E
4
Since there is no torque applied between F and B then
fB = fF = 0.02667 rad = 1.53°
Ans.
251
4 in.
6 in.
B
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5–62. The two shafts are made of A-36 steel. Each has a
diameter of 1 in., and they are supported by bearings at A,
B, and C, which allow free rotation. If the support at D is
fixed, determine the angle of twist of end A when the
torques are applied to the assembly as shown.
D
10 in.
C
80 lbft
A
30 in.
40 lbft
8 in.
10 in.
12 in.
Internal Torque: As shown on FBD.
Angle of Twist:
fE = a
TL
JG
1
= p
4
6
2 (0.5 )(11.0)(10 )
[-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
4
6
2 (0.5 )(11.0)(10 )
= -0.004445 rad = 0.004445 rad
fA = fF + fA>F
= 0.02667 + 0.004445
= 0.03111 rad = 1.78°
Ans.
252
4 in.
6 in.
B
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5–63. The device serves as a compact torsional spring. It is
made of A-36 steel and consists of a solid inner shaft CB
which is surrounded by and attached to a tube AB using a
rigid ring at B. The ring at A can also be assumed rigid
and is fixed from rotating. If a torque of T = 2 kip # in. is
applied to the shaft, determine the angle of twist at the end C
and the maximum shear stress in the tube and shaft.
12 in.
12 in.
B
T
1 in.
A
0.5 in. C
Internal Torque: As shown on FBD.
Maximum Shear Stress:
2.00(0.5)
TBC c
= 10.2 ksi
= p
4
J
2 (0.5 )
Ans.
2.00(1)
TBA c
= 1.86 ksi
= p 4
4
J
2 (1 - 0.75 )
Ans.
(tBC)max =
(tBA)max =
Angle of Twist:
fB =
TBA LBA
JG
(2.00)(12)
= p
4
2 (1
fC>B =
- 0.754)11.0(103)
= 0.002032 rad
TBC LBC
JG
2.00(24)
= p
4
3
2 (0.5 )11.0(10 )
= 0.044448 rad
fC = fB + fC>B
= 0.002032 + 0.044448
= 0.04648 rad = 2.66°
Ans.
253
0.75 in.
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*5–64. The device serves as a compact torsion spring. It is
made of A-36 steel and consists of a solid inner shaft CB
which is surrounded by and attached to a tube AB using a
rigid ring at B. The ring at A can also be assumed rigid and
is fixed from rotating. If the allowable shear stress for the
material is tallow = 12 ksi and the angle of twist at C is
limited to fallow = 3°, determine the maximum torque T
that can be applied at the end C.
12 in.
12 in.
B
T
1 in.
A
0.5 in. C
Internal Torque: As shown on FBD.
Allowable Shear Stress: Assume failure due to shear stress.
tmax = tallow =
TBC c
J
T (0.5)
12.0 = p
4
2 (0.5 )
T = 2.356 kip # in
tmax = tallow =
TBA c
J
12.0 = p
T (1)
4
4
2 (1 - 0.75 )
T = 12.89 kip # in
Angle of Twist: Assume failure due to angle of twist limitation.
fB =
T(12)
TBA LBA
= p 4
4
JG
(1
0.75
) 11.0(103)
2
= 0.001016T
fC>B =
T(24)
TBC LBC
= p
4
JG
(0.5
)11.0(103)
2
= 0.022224T
(fC)allow = fB + fC>B
3(p)
= 0.001016T + 0.022224T
180
T = 2.25 kip # in (controls !)
Ans.
254
0.75 in.
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•5–65.
The A-36 steel assembly consists of a tube having
an outer radius of 1 in. and a wall thickness of 0.125 in. Using
a rigid plate at B, it is connected to the solid 1-in-diameter
shaft AB. Determine the rotation of the tube’s end C if a
torque of 200 lb # in. is applied to the tube at this end. The
end A of the shaft is fixed supported.
B
C
200 lb⭈in.
4 in.
A
6 in.
fB =
200(10)
TABL
= 0.001852 rad
= p
4
6
JG
2 (0.5) (11.0)(10 )
fC>B =
-200(4)
TCBL
= -0.0001119 rad
= p 4
4
6
JG
2 (1 - 0.875 )(11.0)(10 )
fC = fB + fC>B
= 0.001852 + 0.0001119
= 0.001964 rad = 0.113°
Ans.
255
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5–66. The 60-mm diameter shaft ABC is supported by two
journal bearings, while the 80-mm diameter shaft EH is
fixed at E and supported by a journal bearing at H. If
T1 = 2 kN # m and T2 = 4 kN # m, determine the angle of
twist of gears A and C. The shafts are made of A-36 steel.
E
A
600 mm
D
100 mm
H
T2
600 mm
B
75 mm
900 mm
Equilibrium: Referring to the free - body diagram of shaft ABC shown in Fig. a
©Mx = 0; F(0.075) - 4(103) - 2(103) = 0
F = 80(103) N
Internal Loading: Referring to the free - body diagram of gear D in Fig. b,
©Mx = 0; 80(103)(0.1) - TDH = 0
TDH = 8(103)N # m
Also, from the free - body diagram of gear A, Fig. c,
©Mx = 0; TAB - 4(103) = 0
TAB = 4 A 103 B N # m
And from the free - body diagram of gear C, Fig. d,
©Mx = 0; -TBC - 2 A 103 B = 0
TBC = -2(103) N # m
Angle of Twist: The polar moment of inertia of segments AB, BC and DH
p
of
the
shaft
are
JAB = JBC =
and
A 0.034 B = 0.405(10 - 6)p m4
2
p
4
-6
4
JDH =
A 0.04 B = 1.28(10 )p m . We have
2
fD =
8(103)(0.6)
TDH LDH
= 0.01592 rad
=
JDHGst
1.28(10 - 6)p(75)(109)
Then, using the gear ratio,
fB = fD a
rD
100
b = 0.02122 rad
b = 0.01592a
rB
75
Also,
fC>B =
-2(103)(0.9)
TBC LBC
= -0.01886 rad = 0.01886 rad
=
JBCGst
0.405(10 - 6)p(75)(109)
fA>B =
4(103)(0.6)
TABLAB
= 0.02515 rad
=
JAB Gst
0.405(10 - 6)p(75)(109)
Thus,
fA = fB + fA>B
fA = 0.02122 + 0.02515
= 0.04637 rad = 2.66°
Ans.
fC = fB + fC>B
fC = 0.02122 + 0.01886
= 0.04008 rad = 2.30°
Ans.
256
T1
C
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5–66.
Continued
257
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5–67. The 60-mm diameter shaft ABC is supported by two
journal bearings, while the 80-mm diameter shaft EH is fixed
at E and supported by a journal bearing at H. If the angle
of twist at gears A and C is required to be 0.04 rad,
determine the magnitudes of the torques T1 and T2. The
shafts are made of A-36 steel.
E
A
600 mm
D
100 mm
H
T2
600 mm
B
75 mm
900 mm
Equilibrium: Referring to the free - body diagram of shaft ABC shown in Fig. a
©Mx = 0; F(0.075) - T1 - T2 = 0
T1
F = 13.333 A T1 + T2 B
Internal Loading: Referring to the free - body diagram of gear D in Fig. b,
©Mx = 0; 13.333 A T1 + T2 B (0.1) - TDE = 0
TDE = 1.333 A T1 + T2 B
Also, from the free - body diagram of gear A, Fig. c,
©Mx = 0; TAB - T2 = 0
TAB = T2
and from the free - body diagram of gear C, Fig. d
©Mx = 0; TBC - T1 = 0
TBC = T1
Angle of Twist: The polar moments of inertia of segments AB, BC and DH
p
of
the
shaft
are
and
JAB = JBC =
A 0.034 B = 0.405(10 - 6)pm4
2
p
JDH =
A 0.044 B = 1.28(10 - 6)pm4. We have
2
fD =
1.333 A T1 + T2 B (0.6)
TDE LDH
= 2.6258(10 - 6) A T1 + T2 B
=
JDE Gst
1.28(10 - 6)p (75)(109)
Then, using the gear ratio,
fB = fD ¢
rD
100
b = 3.5368(10 - 6) A T1 + T2 B
≤ = 2.6258(10 - 6) A T1 + T2 B a
rB
75
Also,
fC>B =
T1(0.9)
TBC LBC
= 9.4314(10 - 6)T1
=
JBC Gst
0.405(10 - 6)p(75)(109)
fA>B =
T2(0.6)
TAB LAB
= 6.2876(10 - 6)T2
=
JAB Gst
0.405(10 - 6)p(75)(109)
Here, it is required that fA = fC = 0.04 rad. Thus,
fA = fB + fA>B
0.04 = 3.5368(10 - 6) A T1 + T2 B + 6.2876(10 - 6)T2
T1 + 2.7778T2 = 11309.73
(1)
fC = fB + fC>B
0.04 = 3.5368(10 - 6) A T1 + T2 B + 9.4314(10 - 6)T1
3.6667T1 + T2 = 11309.73
(2)
258
C
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5–67.
Continued
Solving Eqs. (1) and (2),
T1 = 2188.98 N # m = 2.19 kN # m
Ans.
T2 = 3283.47 N # m = 3.28 kN # m
Ans.
*5–68. The 30-mm-diameter shafts are made of L2 tool
steel and are supported on journal bearings that allow the
shaft to rotate freely. If the motor at A develops a torque of
T = 45 N # m on the shaft AB, while the turbine at E is fixed
from turning, determine the amount of rotation of gears B
and C.
A
45 Nm
B
1.5 m
50 mm
D
C
Internal Torque: As shown on FBD.
0.5 m
Angle of Twist:
fC =
TCE LCE
JG
67.5(0.75)
= p
4
3
2 (0.015 )75.0(10 )
= 0.008488 rad = 0.486°
fB =
75
50
Ans.
fC = 0.729°
Ans.
259
E
75 mm
0.75 m
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•5–69.
The shafts are made of A-36 steel and each has a
diameter of 80 mm. Determine the angle of twist at end E.
0.6 m
A
B
150 mm 10 kN⭈m
C
D
200 mm
0.6 m
E
Equilibrium: Referring to the free - body diagram of shaft CDE shown in Fig. a,
©Mx = 0; 10(103) - 2(103) - F(0.2) = 0
150 mm
F = 40(103) N
0.6 m
2 kN⭈m
Internal Loading: Referring to the free - body diagram of gear B, Fig. b,
©Mx = 0;
-TAB - 40(103)(0.15) = 0
TAB = -6(103) N # m
Referring to the free - body diagram of gear D, Fig. c,
©Mx = 0; 10(103) - 2(103) - TCD = 0
TCD = 8(103) N # m
Referring to the free - body diagram of shaft DE, Fig. d,
©Mx = 0;
-TDE - 2(103) = 0
Angle of Twist: The polar
p
J =
A 0.044 B = 1.28(10 - 6)p m4.
2
TDE = -2(103) N # m
moment
of
inertia
of
the
shafts
are
We have
fB =
-6(103)(0.6)
TAB LAB
= -0.01194 rad = 0.01194 rad
=
JGst
1.28(10 - 6)p(75)(109)
Using the gear ratio,
fC = fB ¢
rB
150
b = 0.008952 rad
≤ = 0.01194 a
rC
200
fE>C = ©
TiLi
TCD LCD
TDE LDE
=
+
JiGi
JGst
JGst
Also,
0.6
=
-6
1.28(10 )p(75)(109)
b 8(103) + c -2(103) d r
= 0.01194 rad
Thus,
fE = fC + fE>C
fE = 0.008952 + 0.01194
= 0.02089 rad = 1.20°
Ans.
260
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5–69.
Continued
261
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5–70. The shafts are made of A-36 steel and each has a
diameter of 80 mm. Determine the angle of twist of gear D.
0.6 m
Equilibrium: Referring to the free-body diagram of shaft CDE shown in Fig. a,
3
3
A
©Mx = 0; 10(10 ) - 2(10 ) - F(0.2) = 0
F = 40(10 ) N
-TAB - 40(103)(0.15) = 0
150 mm 10 kN⭈m
C
Internal Loading: Referring to the free - body diagram of gear B, Fig. b,
©Mx = 0;
B
3
TAB = -6(103) N # m
D
200 mm
0.6 m
E
Referring to the free - body diagram of gear D, Fig. c,
©Mx = 0; 10(103) - 2(103) - TCD = 0
150 mm
TCD = 8(103) N # m
0.6 m
2 kN⭈m
Angle of Twist: The polar moment
p
J =
A 0.044 B = 1.28(10 - 6)p m4. We have
2
fB =
of
inertia
of
the
shafts
are
-6(103)(0.6)
TAB LAB
= -0.01194 rad = 0.01194 rad
=
JGst
1.28(10 - 6)p(75)(109)
Using the gear ratio,
fC = fB ¢
rB
150
b = 0.008952 rad
≤ = 0.01194a
rC
200
Also,
fD>C =
8(103)(0.6)
TCD LCD
= 0.01592 rad
=
JGst
1.28(10 - 6)p(75)(109)
Thus,
fD = fC + fD>C
fD = 0.008952 + 0.01592
= 0.02487 rad = 1.42°
Ans.
262
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*5–72. The 80-mm diameter shaft is made of 6061-T6
aluminum alloy and subjected to the torsional loading
shown. Determine the angle of twist at end A.
0.6 m
0.6 m
C
10 kN⭈m/m
B
A
2 kN⭈m
Equilibrium: Referring to the free - body diagram of segment AB shown in Fig. a,
©Mx = 0;
-TAB - 2(103) = 0
TAB = -2(103)N # m
And the free - body diagram of segment BC, Fig. b,
©Mx = 0;
Angle of Twist: The polar moment
p
J =
A 0.042 B = 1.28(10 - 6)p m4. We have
2
fA = ©
1.28(10 - 6)p(26)(109)
0.6 m -
+
1
= -
of
inertia
of
the
shaft
is
LBC
TiLi
TABLAB
TBC dx
=
+
JiGi
JGal
JGal
L0
-2(103)(0.6)
=
TBC = - C 10(103)x + 2(103) D N # m
-TBC - 10(103)x - 2(103) = 0
1.28(10 - 6)p(26)(109)
L0
C 10(103)x + 2(103) D dx
1.28(10 - 6)p(26)(109)
b 1200 + C 5(103)x2 + 2(103)x D 2
0.6m
0
r
= -0.04017 rad = 2.30°
Ans.
263
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•5–73. The tapered shaft has a length L and a radius r at
end A and 2r at end B. If it is fixed at end B and is subjected
to a torque T, determine the angle of twist of end A. The
shear modulus is G.
B
2r
L
T
Geometry:
r
A
r
rL + rx
r(x) = r +
x =
L
L
p rL + rx 4
p r4
(L + x)4
a
b =
2
L
2L4
J(x) =
Angle of Twist:
L
T dx
L0 J(x)G
f =
L
=
2TL4
dx
p r4G L0 (L + x)4
=
L
1
2TL4
cd2
4
3
pr G
3(L + x) 0
=
7TL
12p r4G
Ans.
5–74. The rod ABC of radius c is embedded into a medium
where the distributed torque reaction varies linearly from
zero at C to t0 at B. If couple forces P are applied to the lever
arm, determine the value of t0 for equilibrium. Also, find the
angle of twist of end A. The rod is made from material
having a shear modulus of G.
L
2
L
2
Equilibrium: Referring to the free-body diagram of the entire rod shown in Fig. a,
1
L
©Mx = 0; Pd - (t0)a b = 0
2
2
to =
Ans.
Internal Loading: The distributed torque expressed as a function of x, measured
4Pd>L
to
8Pd
from the left end, is t = ¢
≤x = ¢
≤ x = ¢ 2 ≤ x. Thus, the resultant
L>2
L>2
L
torque within region x of the shaft is
TR =
1
1
8Pd
4Pd 2
tx = B ¢ 2 ≤ x R x =
x
2
2
L
L2
Referring to the free - body diagram shown in Fig. b,
©Mx = 0; TBC -
4Pd 2
x = 0
L2
d
2
B
P
4Pd
L
TBC =
4Pd 2
x
L2
Referring to the free - body diagram shown in Fig. c,
©Mx = 0; Pd - TAB = 0
TAB = Pd
264
d
2
t0
C
A
P
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5–74.
Continued
Angle of Twist:
f = ©
LBC
TAB LAB
TBC dx
TiLi
=
+
JiGi
JG
JG
L0
L>2
=
L0
4Pd 2
x dx
L2
Pd(L>2)
+
p
2
p
2
¢ c4 ≤ G
¢ c4 ≤ G
L>2
8Pd
x3
=
£ ≥3
4 2
pc L G 3
+
PLd
pc4G
0
=
4PLd
3pc4G
Ans.
265
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5–75. When drilling a well, the deep end of the drill pipe
is assumed to encounter a torsional resistance TA .
Furthermore, soil friction along the sides of the pipe creates
a linear distribution of torque per unit length, varying from
zero at the surface B to t0 at A. Determine the necessary
torque TB that must be supplied by the drive unit to turn
the pipe. Also, what is the relative angle of twist of one end
of the pipe with respect to the other end at the instant the
pipe is about to turn? The pipe has an outer radius ro and an
inner radius ri . The shear modulus is G.
TB
B
L
t0
A
1
t L + TA - TB = 0
2 0
TB =
t0L + 2TA
2
T(x) +
t0 2
t0L + 2TA
x = 0
2L
2
T(x) =
t0 2
t0 L + 2TA
x
2
2L
f =
Ans.
T(x) dx
L JG
=
L
t0L + 2TA
t0 2
1
(
x ) dx
J G L0
2
2L
=
t0 3 L
1 t0 L + 2TA
c
x x dƒ
JG
2
6L
0
=
t0 L2 + 3TAL
3JG
However, J =
f =
p
(r 4 - ri 4)
2 o
2L(t0 L + 3TA)
Ans.
3p(ro 4 - ri 4)G
266
TA
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*5–76. A cylindrical spring consists of a rubber annulus
bonded to a rigid ring and shaft. If the ring is held fixed and
a torque T is applied to the rigid shaft, determine the angle
of twist of the shaft. The shear modulus of the rubber is G.
Hint: As shown in the figure, the deformation of the
element at radius r can be determined from rdu = drg. Use
this expression along with t = T>12pr2h2 from Prob. 5–26,
to obtain the result.
ro
r
ri
T
h
gdr rdu
dr
g
du
r
r du = g dr
du =
gdr
r
(1)
From Prob. 5-26,
t =
T
2p r2h
g =
T
2p r2hG
and
g =
t
G
From (1),
du =
T dr
2p hG r3
r
u =
o
dr
T
1 ro
T
=
cd|
3
2p hG Lri r
2p hG
2 r2 r
i
=
1
1
T
c- 2 + 2d
2p hG
2ro
2ri
=
1
1
T
c - 2d
4p hG r2i
ro
Ans.
267
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•5–77.
The A-36 steel shaft has a diameter of 50 mm and is
fixed at its ends A and B. If it is subjected to the torque,
determine the maximum shear stress in regions AC and CB
of the shaft.
A
300 Nm
0.4 m
C
0.8 m
Equilibrium:
TA + TB - 300 = 0
[1]
Compatibility:
fC>A = fC>B
TA(0.4)
TB(0.8)
=
JG
JG
TA = 2.00TB
[2]
Solving Eqs. [1] and [2] yields:
TA = 200 N # m
TB = 100 N # m
Maximum Shear stress:
(tAC)max =
200(0.025)
TAc
= 8.15 MPa
= p
4
J
2 (0.025 )
Ans.
(tCB)max =
100(0.025)
TBc
= 4.07 MPa
= p
4
J
2 (0.025 )
Ans.
268
B
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5–78. The A-36 steel shaft has a diameter of 60 mm and is
fixed at its ends A and B.If it is subjected to the torques shown,
determine the absolute maximum shear stress in the shaft.
200 N⭈m
B
500 N⭈m
D
1.5 m
C
A
Referring to the FBD of the shaft shown in Fig. a,
TA + TB - 500 - 200 = 0
©Mx = 0;
(1)
Using the method of superposition, Fig. b
fA = (fA)TA - (fA)T
0 =
500 (1.5)
700 (1)
TA (3.5)
- c
+
d
JG
JG
JG
TA = 414.29 N # m
Substitute this result into Eq (1),
TB = 285.71 N # m
Referring to the torque diagram shown in Fig. c, segment AC is subjected to
maximum internal torque. Thus, the absolute maximum shear stress occurs here.
tAbs =
414.29 (0.03)
TAC c
=
= 9.77 MPa
p
J
(0.03)4
2
Ans.
269
1m
1m
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5–79. The steel shaft is made from two segments: AC has a
diameter of 0.5 in, and CB has a diameter of 1 in. If it is
fixed at its ends A and B and subjected to a torque of
determine the maximum shear stress in the shaft.
Gst = 10.811032 ksi.
A
0.5 in.
C
D 500 lbft
5 in.
1 in.
8 in.
B
12 in.
Equilibrium:
TA + TB - 500 = 0
(1)
Compatibility condition:
fD>A = fD>B
TA(5)
p
4
2 (0.25 )G
TA(8)
TB(12)
= p
4
2 (0.5 )G
2 (0.5 )G
+ p
4
1408 TA = 192 TB
(2)
Solving Eqs. (1) and (2) yields
TA = 60 lb # ft
TB = 440 lb # ft
tAC =
60(12)(0.25)
TC
= 29.3 ksi
= p
4
J
2 (0.25 )
tDB =
440(12)(0.5)
TC
= 26.9 ksi
=
p
4
J
2 (0.5 )
Ans.
(max)
270
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*5–80. The shaft is made of A-36 steel, has a diameter of
80 mm, and is fixed at B while A is loose and can rotate
0.005 rad before becoming fixed. When the torques are
applied to C and D, determine the maximum shear stress in
regions AC and CD of the shaft.
2 kN⭈m
B
4 kN⭈m
D
600 mm
C
600 mm
A
Referring to the FBD of the shaft shown in Fig. a,
600 mm
TA + TB + 2 - 4 = 0
©Mx = 0;
(1)
Using the method of superposition, Fig. b,
fA = (fA)T - (uA)TA
0.005 = B
4(103)(0.6)
C
p
4
9
2 (0.04 ) 75(10 )
D
2(103)(0.6)
+
C
p
4
9
2 (0.04 ) 75(10 )
D
R - p
TA (1.8)
C
4
9
2 (0.04 ) 75(10 )
D
TA = 1162.24 N # m = 1.162 kN # m
Substitute this result into Eq (1),
TB = 0.838 kN # m
Referring to the torque diagram shown in Fig. c, segment CD is subjected to a
maximum internal torque. Thus, the absolute maximum shear stress occurs here.
t$$$ =
2.838 (103)(0.04)
TCD c
=
= 28.23 (106) Pa = 28.2 MPa
p
4
J
2 (0.04)
271
Ans.
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•5–81. The shaft is made of A-36 steel and has a diameter
of 80 mm. It is fixed at B and the support at A has a torsional
stiffness of k = 0.5 MN # m>rad. If it is subjected to the gear
torques shown, determine the absolute maximum shear stress
in the shaft.
2 kN⭈m
B
4 kN⭈m
D
600 mm
C
600 mm
A
600 mm
Referring to the FBD of the shaft shown in Fig. a,
TA + TB + 2 - 4 = 0
©Mx = 0;
(1)
Using the method of superposition, Fig. b,
fA = (fA)T - (fA)TA
TA
6
0.5(10 )
= D
4(103)(0.6)
C
p
4
9
2 (0.04 ) 75(10 )
D
2(103)(0.6)
+
C
p
4
9
2 (0.04 ) 75(10 )
D
T -
TA(1.8)
C
p
4
9
2 (0.04 ) 75(10 )
D
TA = 1498.01 N # m = 1.498 kN # m
Substituting this result into Eq (1),
TB = 0.502 kN # m
Referring to the torque diagram shown in Fig. c, segment CD subjected to maximum
internal torque. Thus, the maximum shear stress occurs here.
t$$$ =
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 =
Tst (2)(12)
Tst (2)(12)
TL
= p 4
= p
4
4
4
6
JG
(0.5
)(5.6)(10
)
2
2 (1 - 0.5 )(11.5)(10 )
Tbr = 0.032464 Tst
(2)
Solving Eqs. (1) and (2) yields:
Tst = 48.428 lb # ft;
fC = ©
Tbr = 1.572 lb # ft
50(12)(3)(12)
1.572(12)(2)(12)
TL
+ p 4
= p
4
6
6
JG
2 (0.5 )(5.6)(10 )
2 (1 )(11.5)(10 )
= 0.002019 rad = 0.116°
Ans.
(tst)max AB =
50(12)(1)
TABc
= 382 psi
= p 4
J
2 (1 )
(tst)max BC =
48.428(12)(1)
Tst c
= 394.63 psi = 395 psi (Max)
= p 4
4
J
2 (1 - 0.5 )
Ans.
(gst)max =
(tst)max
394.63
= 343.(10 - 6) rad
=
G
11.5(106)
Ans.
(tbr)max =
1.572(12)(0.5)
Tbr c
= 96.07 psi = 96.1 psi (Max)
=
p
4
J
2 (0.5 )
Ans.
(gbr)max =
(tbr)max
96.07
= 17.2(10 - 6) rad
=
G
5.6(106)
Ans.
273
C
T 50 lbft
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5–83. The motor A develops a torque at gear B of 450 lb # ft,
which is applied along the axis of the 2-in.-diameter steel shaft
CD. This torque is to be transmitted to the pinion gears at E
and F. If these gears are temporarily fixed, determine the
maximum shear stress in segments CB and BD of the shaft.
Also, what is the angle of twist of each of these segments? The
bearings at C and D only exert force reactions on the shaft
and do not resist torque. Gst = 1211032 ksi.
B
E
4 ft
(1)
Compatibility condition:
fB>C = fB>D
TC(4)
TD(3)
=
JG
JG
TC = 0.75 TD
(2)
Solving Eqs. (1) and (2), yields
TD = 257.14 lb # ft
TC = 192.86 lb # ft
(tBD)max =
f =
192.86(12)(1)
p
4
2 (1 )
257.14(12)(1)
p
4
2 (1 )
192.86(12)(4)(12)
p
4
6
2 (1 )(12)(10 )
3 ft
D
A
TC + TD - 450 = 0
(tBC)max =
F
C
Equilibrium:
= 1.47 ksi
Ans.
= 1.96 ksi
Ans.
= 0.00589 rad = 0.338°
Ans.
274
450 lb⭈ft
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*5–84. A portion of the A-36 steel shaft is subjected to a
linearly distributed torsional loading. If the shaft has the
dimensions shown, determine the reactions at the fixed
supports A and C. Segment AB has a diameter of 1.5 in. and
segment BC has a diameter of 0.75 in.
300 lb⭈in./in.
A
60 in.
B
C
48 in.
Equilibrium:
TA + TC - 9000 = 0
TR = t x +
But
(1)
1
tx
(300 - t)x = 150x +
2
2
300
t
=
;
60 - x
60
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
4
2 (0.75 )G
= p
4
2 (0.375 )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
p
4
2 (0.75 )G
4
2 (0.375 )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 =
217.4(48)
TCL
= 0.030540 rad
= p
4
6
JG
2 (0.375) (11.0)(10 )
fB = 1.75°
tmax =
Ans.
217.4(0.375)
TC
= p
= 2.62 ksi
4
J
2 (0.375)
For segment AB,
tmax =
B
48 in.
1
[5(60 - x)]x
2
TR = 150x +
60 in.
8782.6(0.75)
TC
= p
= 13.3 ksi
4
J
2 (0.75)
abs = 13.3 ksi
tmax
Ans.
276
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5–86. The two shafts are made of A-36 steel. Each has a
diameter of 25 mm and they are connected using the gears
fixed to their ends. Their other ends are attached to fixed
supports at A and B. They are also supported by journal
bearings at C and D, which allow free rotation of the shafts
along their axes. If a torque of 500 N # m is applied to the
gear at E as shown, determine the reactions at A and B.
B
F
D
50 mm
0.75 m
100 mm
500 Nm
E
C
1.5 m
A
Equilibrium:
TA + F(0.1) - 500 = 0
[1]
TB - F(0.05) = 0
[2]
TA + 2TB - 500 = 0
[3]
From Eqs. [1] and [2]
Compatibility:
0.1fE = 0.05fF
fE = 0.5fF
TA(1.5)
TB(0.75)
= 0.5 c
d
JG
JG
TA = 0.250TB
[4]
Solving Eqs. [3] and [4] yields:
TB = 222 N # m
Ans.
TA = 55.6 N # m
Ans.
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 =
55.56(1.5)
TAL
= p
4
JG
(0.0125
)(75.0)(109)
2
= 0.02897 rad = 1.66°
Ans.
278
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*5–88. The shafts are made of A-36 steel and have the
same diameter of 4 in. If a torque of 15 kip # ft is applied to
gear B, determine the absolute maximum shear stress
developed in the shaft.
2.5 ft
2.5 ft
Equilibrium: Referring to the free - body diagrams of shafts ABC and DE shown in
Figs. a and b, respectively, we have
©Mx = 0; TA + F(0.5) - 15 = 0
(1)
and
A
B
6 in.
15 kip⭈ft
C
D
12 in.
©Mx = 0; F(1) - TE = 0
E
(2)
3 ft
Internal Loadings: The internal torques developed in segments AB and BC of shaft
ABC and shaft DE are shown in Figs. c, d, and e, respectively.
Compatibility Equation:
fCrC = fDrD
¢
TABLAB
TDE LDE
TBCLBC
+
≤ rC = ¢
≤ rD
JGst
JGst
JGst
C -TA(2.5) + F(0.5)(2.5) D (0.5) = -TE(3)(1)
TA - 0.5F = 2.4TE
(3)
Solving Eqs. (1), (2), and (3), we have
F = 4.412 kip
TE = 4.412 kip # ft
TA = 12.79 kip # ft
Maximum Shear Stress: By inspection, segment AB of shaft ABC is subjected to the
greater torque.
A tmax B abs =
12.79(12)(2)
TAB c
= 12.2 ksi
=
Jst
p 4
a2 b
2
Ans.
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•5–89.
The shafts are made of A-36 steel and have the
same diameter of 4 in. If a torque of 15 kip # ft is applied to
gear B, determine the angle of twist of gear B.
2.5 ft
Equilibrium: Referring to the free - body diagrams of shafts ABC and DE shown in
Figs. a and b, respectively,
©Mx = 0; TA + F(0.5) - 15 = 0
2.5 ft
A
B
(1)
6 in.
15 kip⭈ft
and
©Mx = 0; F(1) - TE = 0
(2)
Internal Loadings: The internal torques developed in segments AB and BC of shaft
ABC and shaft DE are shown in Figs. c, d, and e, respectively.
Compatibility Equation: It is required that
fCrC = fDrD
¢
TAB LAB
TDE LDE
TBC LBC
+
≤ rC = ¢
≤ rD
JGst
JGst
JGst
C -TA(2.5) + F(0.5)(2.5) D (0.5) = -TE(3)(1)
TA - 0.5F = 2.4TE
(3)
Solving Eqs. (1), (2), and (3),
F = 4.412 kip
TE = 4.412 kip # ft
TA = 12.79 kip # ft
Angle of Twist: Here, TAB = -TA = -12.79 kip # ft
fB =
-12.79(12)(2.5)(12)
TAB LAB
=
JGst
p 4
a 2 b(11.0)(103)
2
= -0.01666 rad = 0.955°
Ans.
280
C
D
12 in.
E
3 ft
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5–90. The two 3-ft-long shafts are made of 2014-T6
aluminum. Each has a diameter of 1.5 in. and they are
connected using the gears fixed to their ends. Their other
ends are attached to fixed supports at A and B. They are
also supported by bearings at C and D, which allow free
rotation of the shafts along their axes. If a torque of 600 lb # ft
is applied to the top gear as shown, determine the maximum
shear stress in each shaft.
A
B
E
2 in.
TA + F a
4
b - 600 = 0
12
(1)
TB - F a
2
b = 0
12
(2)
From Eqs. (1) and (2)
TA + 2TB - 600 = 0
TAL
TBL
= 0.5 a
b;
JG
JG
(3)
fE = 0.5fF
TA = 0.5TB
(4)
Solving Eqs. (3) and (4) yields:
TB = 240 lb # ft;
TA = 120 lb # ft
(tBD)max =
240(12)(0.75)
TB c
= 4.35 ksi
=
p
4
J
2 (0.75 )
Ans.
(tAC)max =
120(12)(0.75)
TA c
= 2.17 ksi
=
p
4
J
2 (0.75 )
Ans.
281
3 ft
D
4 in.
4(fE) = 2(fF);
C
600 lbft
F
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5–91. The A-36 steel shaft is made from two segments: AC
has a diameter of 0.5 in. and CB has a diameter of 1 in. If
the shaft is fixed at its ends A and B and subjected to a uniform
distributed torque of 60 lb # in.>in. along segment CB,
determine the absolute maximum shear stress in the shaft.
A
0.5 in.
C
5 in.
60 lbin./in.
1 in.
20 in.
Equilibrium:
TA + TB - 60(20) = 0
(1)
Compatibility condition:
fC>B = fC>A
fC>B =
20
(TB - 60x) dx
T(x) dx
=
p
JG
L
L0 2 (0.54)(11.0)(106)
= 18.52(10-6)TB - 0.011112
18.52(10-6)TB - 0.011112 = p
TA(5)
4
6
2 (0.25 )(11.0)(10 )
18.52(10-6)TB - 74.08(10-6)TA = 0.011112
(2)
18.52TB - 74.08TA = 11112
Solving Eqs. (1) and (2) yields:
TA = 120.0 lb # in. ;
TB = 1080 lb # in.
(tmax)BC =
1080(0.5)
TB c
= 5.50 ksi
= p
4
J
2 (0.5 )
(tmax)AC =
120.0(0.25)
TA c
= 4.89 ksi
= p
4
J
2 (0.25 )
abs = 5.50 ksi
tmax
Ans.
282
B
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*5–92. If the shaft is subjected to a uniform distributed
torque of t = 20 kN # m>m, determine the maximum shear
stress developed in the shaft. The shaft is made of 2014-T6
aluminum alloy and is fixed at A and C.
400 mm
20 kN⭈m/m
600 mm
a
A
80 mm
60 mm
B
a
C
Equilibrium: Referring to the free - body diagram of the shaft shown in Fig. a, we
have
©Mx = 0; TA + TC - 20(103)(0.4) = 0
(1)
Compatibility Equation: The resultant torque of the distributed torque within the
region x of the shaft is TR = 20(103)x N # m. Thus, the internal torque developed in
the shaft as a function of x when end C is free is T(x) = 20(103)x N # m, Fig. b. Using
the method of superposition, Fig. c,
fC = A fC B t - A fC B TC
0 =
0 =
0.4 m
T(x)dx
TCL
JG
JG
0.4 m
20(103)xdx
TC(1)
JG
JG
L0
L0
0 = 20(103) ¢
x2 2 0.4 m
- TC
≤
2 0
TC = 1600 N # m
Substituting this result into Eq. (1),
TA = 6400 N # m
Maximum Shear Stress: By inspection, the maximum internal torque occurs at
support A. Thus,
A tmax B abs =
6400(0.04)
TA c
= 93.1 MPa
=
J
p
4
4
a0.04 - 0.03 b
2
Ans.
283
Section a–a
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•5–93.
The tapered shaft is confined by the fixed supports
at A and B. If a torque T is applied at its mid-point,
determine the reactions at the supports.
2c
T
A
B
Equilibrium:
c
TA + TB - T = 0
[1]
L/2
L/ 2
Section Properties:
r(x) = c +
J(x) =
c
c
x =
(L + x)
L
L
4
p c
pc4
c (L + x) d =
(L + x)4
2 L
2L4
Angle of Twist:
fT =
Tdx
=
Lp2
L J(x)G
L
Tdx
pc4
4
2L4 (L + x) 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
4
2L4 (L + x) G
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
x
t0 a x + 3L
2b
3
L
0 =
L0
TB =
A
2
dx
-
JG
TB(L)
7t0L
=
- TB(L)
JG
12
7t0 L
12
Ans.
From Eq. (1),
TA = t0 a L +
TA +
4t0 L
L3
b =
2
3
3L
7t0 L
4t0 L
= 0
12
3
TA =
3t0 L
4
Ans.
285
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5–95. Compare the values of the maximum elastic shear
stress and the angle of twist developed in 304 stainless steel
shafts having circular and square cross sections. Each shaft
has the same cross-sectional area of 9 in2, length of 36 in.,
and is subjected to a torque of 4000 lb # in.
r
A
Maximum Shear Stress:
For circular shaft
1
A = pc2 = 9;
(tc)max =
9 2
c = a b
p
2(4000)
Tc
Tc
2T
=
= 525 psi
= p 4 =
1
3
J
pc
pA9B2
2 c
Ans.
x
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 =
4000(36)
TL
=
p 9 2
6
JG
2 A p B 11.0(10 )
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)
% more efficient =
=
Ans.
(tmax)c - (tmax)c
(100%)
(tmax)c
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
A B
Tc
16T
=
=
3
p
J
AdB4 p d
T d2
2 2
For the elliptical shaft:
(tmax)c =
2T
2T
16T
=
=
p a b2
p k2 d3
p A d B A kd B 2
2
d
2
Factor of increase in shear stress =
16T
(tmax)c
p k2 d3
= 16T
(tmax)c
3
pd
=
1
k2
Ans.
287
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5–98. The shaft is made of red brass C83400 and has an
elliptical cross section. If it is subjected to the torsional
loading shown, determine the maximum shear stress within
regions AC and BC, and the angle of twist f of end B
relative to end A.
A
20 Nm
50 Nm
30 Nm
2m
Maximum Shear Stress:
C
(tBC)max =
2(30.0)
2TBC
p a b2
=
B
p(0.05)(0.022)
Ans.
= 0.955 MPa
(tAC)max =
2(50.0)
2TAC
2
50 mm
20 mm
1.5 m
=
pab
p(0.05)(0.022)
Ans.
= 1.59 MPa
Angle of Twist:
fB>A = a
(a2 + b2)T L
p a3b3 G
(0.052 + 0.022)
=
p(0.053)(0.023)(37.0)(109)
[(-30.0)(1.5) + (-50.0)(2)]
= -0.003618 rad = 0.207°
Ans.
5–99. Solve Prob. 5–98 for the maximum shear stress
within regions AC and BC, and the angle of twist f of end B
relative to C.
A
20 Nm
50 Nm
30 Nm
2m
Maximum Shear Stress:
C
(tBC)max =
2(30.0)
2TBC
p a b2
=
B
p(0.05)(0.022)
= 0.955 MPa
(tAC)max =
2
Ans.
2(50.0)
2TAC
=
pab
p(0.05)(0.022)
= 1.59 MPa
Ans.
Angle of Twist:
fB>C =
(a2 + b2) TBC L
p a3 b3 G
(0.052 + 0.022)(-30.0)(1.5)
=
50 mm
20 mm
1.5 m
p(0.053)(0.023)(37.0)(109)
= -0.001123 rad = | 0.0643°|
Ans.
288
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*5–100. Segments AB and BC of the shaft have circular and
square cross sections, respectively. If end A is subjected to a
torque of T = 2 kN # m, determine the absolute maximum
shear stress developed in the shaft and the angle of twist of
end A. The shaft is made from A-36 steel and is fixed at C.
600 mm
C
600 mm
90 mm
B
30 mm
90 mm
Internal Loadings: The internal torques developed in segments AB and BC are
shown in Figs. a, and b, respectively.
Maximum Shear Stress: For segment AB,
A tmax B AB =
2(103)(0.03)
TAB c
= 47.2 MPa (max)
J
p
a 0.034 b
2
For segment BC,
A tmax B BC =
4.81TBC
3
=
a
4.81 C 2(103) D
(0.09)3
Ans.
= 13.20 MPa
Angle of Twist:
fA =
7.10TBCLBC
TABLAB
+
JG
a4G
2(103)(0.6)
=
p
a 0.034 b(75)(109)
2
7.10(2)(103)(0.6)
+
(0.09)4(75)(109)
= 0.01431 rad = 0.820°
Ans.
289
A
T
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•5–101. Segments AB and BC of the shaft have circular and
square cross sections, respectively. The shaft is made from
A-36 steel with an allowable shear stress of tallow = 75 MPa,
and an angle of twist at end A which is not allowed to exceed
0.02 rad. Determine the maximum allowable torque T that
can be applied at end A.The shaft is fixed at C.
600 mm
C
600 mm
Internal Loadings: The internal torques developed in segments AB and BC are
shown in Figs. a, and b, respectively.
90 mm
B
30 mm
90 mm
Allowable Shear Stress: For segment AB,
A
T
tallow =
TAB c
;
J
75(106) =
T(0.03)
p
a0.034 b
2
T = 3180.86 N # m
For segment BC,
tallow =
4.81TBC
a3
75(106) =
;
4.81T
(0.09)3
T = 11 366.94 N # m
Angle of Twist:
fA =
TABLAB
7.10TBC LBC
+
JG
a4G
0.02 =
T(0.6)
7.10T(0.6)
p
a 0.034 b(75)(109)
2
+
(0.09)4 (75)(109)
T = 2795.90 N # m = 2.80 kN # m (controls)
Ans.
5–102. The aluminum strut is fixed between the two walls
at A and B. If it has a 2 in. by 2 in. square cross section, and
it is subjected to the torque of 80 lb # ft at C, determine the
reactions at the fixed supports. Also, what is the angle of
twist at C? Gal = 3.811032 ksi.
A
C
2 ft
80 lb⭈ft
By superposition:
3 ft
0 = f - fB
0 =
7.10(TB)(5)
7.10(80)(2)
4
-
a G
a4 G
TB = 32 lb # ft
Ans.
TA + 32 - 80 = 0
TA = 48 lb # ft
fC =
7.10(32)(12)(3)(12)
(24)(3.8)(106)
Ans.
= 0.00161 rad = 0.0925°
Ans.
290
B
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5–103. The square shaft is used at the end of a drive cable in
order to register the rotation of the cable on a gauge. If it has
the dimensions shown and is subjected to a torque of 8 N # m,
determine the shear stress in the shaft at point A. Sketch the
shear stress on a volume element located at this point.
5 mm
A
5 mm
Maximum shear stress:
8 Nm
(tmax)A =
4.81(8)
4.81T
=
= 308 MPa
a3
(0.005)3
Ans.
*5–104. The 6061-T6 aluminum bar has a square cross
section of 25 mm by 25 mm. If it is 2 m long, determine the
maximum shear stress in the bar and the rotation of one
end relative to the other end.
C
1.5 m
20 N⭈m
B
Maximum Shear Stress:
tmax =
4.81(80.0)
4.81Tmax
a3
=
(0.0253)
Ans.
= 24.6 MPa
60 N·m
25 mm
Angle of Twist:
fA>C = a
0.5 m
A
7.10(-20.0)(1.5)
7.10(-80.0)(0.5)
7.10TL
=
+
4
4
9
aG
(0.025 )(26.0)(10 )
(0.0254)(26.0)(109)
= -0.04894 rad = | 2.80° |
Ans.
291
80 N⭈m
25 mm
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•5–105.
The steel shaft is 12 in. long and is screwed into
the wall using a wrench. Determine the largest couple
forces F that can be applied to the shaft without causing the
steel to yield. tY = 8 ksi.
1 in.
12 in.
F(16) - T = 0
tmax = tY =
8(103) =
(1)
4.81T
a3
F
8 in.
4.81T
(1)3
1 in.
8 in.
T = 1663.2 lb # in.
F
From Eq. (1),
F = 104 lb
Ans.
5–106. The steel shaft is 12 in. long and is screwed into the
wall using a wrench. Determine the maximum shear stress
in the shaft and the amount of displacement that each
couple force undergoes if the couple forces have a
magnitude of F = 30 lb, Gst = 10.811032 ksi.
1 in.
12 in.
T - 30(16) = 0
F
T = 480 lb # in.
tmax =
4.81(480)
4.18T
=
a3
(1)3
1 in.
8 in.
Ans.
= 2.31 ksi
f =
8 in.
7.10(480)(12)
7.10TL
=
= 0.00379 rad
a4 G
(1)4(10.8)(106)
dF = 8(0.00397) = 0.0303 in.
Ans.
292
F
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5–107. Determine the constant thickness of the rectangular
tube if the average shear stress is not to exceed 12 ksi when a
torque of T = 20 kip # in. is applied to the tube. Neglect stress
concentrations at the corners. The mean dimensions of the
tube are shown.
T
4 in.
Am = 2(4) = 8 in2
tavg =
12 =
T
2 t Am
2 in.
20
2 t (8)
t = 0.104 in.
Ans.
*5–108. Determine the torque T that can be applied to the
rectangular tube if the average shear stress is not to exceed
12 ksi. Neglect stress concentrations at the corners. The
mean dimensions of the tube are shown and the tube has a
thickness of 0.125 in.
T
4 in.
Am = 2(4) = 8 in2
tavg =
T
;
2 t Am
12 =
2 in.
T
2(0.125)(8)
T = 24 kip # in. = 2 kip # ft
Ans.
293
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•5–109.
For a given maximum shear stress, determine the
factor by which the torque carrying capacity is increased if
the half-circular section is reversed from the dashed-line
position to the section shown. The tube is 0.1 in. thick.
1.80 in.
0.6 in.
1.20 in.
0.5 in.
p(0.552)
= 1.4498 in2
Am = (1.10)(1.75) 2
Am ¿ = (1.10)(1.75) +
tmax =
p(0.552)
= 2.4002 in2
2
T
2t Am
T = 2 t Am tmax
Factor =
=
2t Am ¿ tmax
2t Am tmax
Am ¿
2.4002
=
= 1.66
Am
1.4498
Ans.
5–110. For a given average shear stress, determine the
factor by which the torque-carrying capacity is increased if
the half-circular sections are reversed from the dashed-line
positions to the section shown. The tube is 0.1 in. thick.
1.80 in.
0.6 in.
1.20 in.
0.5 in.
Section Properties:
œ
Am
= (1.1)(1.8) - B
p (0.552)
R (2) = 1.02967 in2
2
Am = (1.1)(1.8) + B
p (0.552)
R (2) = 2.93033 in2
2
Average Shear Stress:
tavg =
Hence,
T
;
2 t Am
T = 2 t Am tavg
œ
tavg
T¿ = 2 t Am
The factor of increase =
Am
2.93033
T
= œ =
T¿
Am
1.02967
= 2.85
Ans.
294
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5–111. A torque T is applied to two tubes having the
cross sections shown. Compare the shear flow developed in
each tube.
t
t
t
Circular tube:
a
T
T
2T
=
=
2Am
2p (a>2)2
p a2
qct =
a
a
Square tube:
qst =
T
T
=
2Am
2a2
qst
T>(2a2)
p
=
=
qct
4
2T>(p a2)
Thus;
qst =
p
q
4 ct
Ans.
*5–112. Due to a fabrication error the inner circle of the
tube is eccentric with respect to the outer circle. By what
percentage is the torsional strength reduced when the
eccentricity e is one-fourth of the difference in the radii?
ab
2
a
Average Shear Stress:
e
2
For the aligned tube
tavg =
T
T
=
2 t Am
2(a - b)(p) A a + b B 2
2
T = tavg (2)(a - b)(p)a
a + b 2
b
2
For the eccentric tube
tavg =
b
T¿
2 t Am
t = a -
e
e
- a + bb = a - e - b
2
2
= a -
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 +2 b B 2
3
T¿
=
=
a + b 2
T
4
tavg (2)(a - b)(p) A
B
2
3
Percent reduction in strength = a1 - b * 100 % = 25 %
4
295
Ans.
e
2
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•5–113.
The mean dimensions of the cross section of an
airplane fuselage are shown. If the fuselage is made of
2014-T6 aluminum alloy having allowable shear stress of
tallow = 18 ksi, and it is subjected to a torque of 6000 kip # ft,
determine the required minimum thickness t of the cross
section to the nearest 1>16 in. Also, find the corresponding
angle of twist per foot length of the fuselage.
t
3 ft
4.5 ft
3 ft
Section Properties: Referring to the geometry shown in Fig. a,
Am = p A 32 B + 4.5(6) = 55.2743 ft2 ¢
F
144 in2
≤ = 7959.50 in2
1 ft2
ds = 2p(3) + 2(4.5) = 27.8496 fta
12 in.
b = 334.19 in.
1 ft
Allowable Average Shear Stress:
A tavg B allow =
T
;
2tAm
18 =
6000(12)
2t(7959.50)
t = 0.2513 in. =
Angle of Twist: Using the result of t =
f = ©
Ans.
5
in,
16
ds
TL
4Am 2G F t
6000(12)(1)(12)
=
5
in.
16
4(7959.502)(3.9)(103)
¢
334.19
≤
5>16
= 0.9349(10 - 3) rad = 0.0536°
Ans.
296
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5–114. The mean dimensions of the cross section of an
airplane fuselage are shown. If the fuselage is made from
2014-T6 aluminum alloy having an allowable shear stress of
tallow = 18 ksi and the angle of twist per foot length of
fuselage is not allowed to exceed 0.001 rad>ft, determine
the maximum allowable torque that can be sustained by the
fuselage. The thickness of the wall is t = 0.25 in.
t
3 ft
4.5 ft
3 ft
Section Properties: Referring to the geometry shown in Fig. a,
Am = p A 32 B + 4.5(6) = 55.2743 ft2 ¢
F
144 in2
≤ = 7959.50 in2
1 ft2
ds = 2p(3) + 2(4.5) = 27.8496 fta
12 in.
b = 334.19 in.
1 ft
Allowable Average Shear Stress:
A tavg B allow =
T
;
2tAm
18 =
T
2(0.25)(7959.50)
T = 71635.54 kip # ina
1ft
b = 5970 kip # ft
12 in.
Angle of Twist:
f =
ds
TL
4Am 2G F t
0.001 =
T(1)(12)
4(7959.502)(3.9)(103)
T = 61610.65 kip # ina
a
334.19
b
0.25
1ft
b = 5134 kip # ft (controls)
12 in.
Ans.
297
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5–115. The tube is subjected to a torque of 750 N # m.
Determine the average shear stress in the tube at points A
and B.
4 mm
6 mm
A
100 mm
Referring to the geometry shown in Fig. a,
6 mm
2
Am = 0.06 (0.1) = 0.006 m
B
750 N⭈m
Thus,
T
750
=
= 15.63(106)Pa = 15.6 MPa
(tavg)A =
2tA Am
2(0.004)(0.006)
Ans.
T
750
=
= 10.42(106)Pa = 10.4 MPa
2tB Am
2(0.006)(0.006)
Ans.
(tavg)B =
4 mm
60 mm
*5–116. The tube is made of plastic, is 5 mm thick, and has
the mean dimensions shown. Determine the average shear
stress at points A and B if it is subjected to the torque of
T = 5 N # m. Show the shear stress on volume elements
located at these points.
Am = (0.11)(0.08) +
tA = tB = tavg =
A
1
(0.08)(0.03) = 0.01 m2
2
B
50 mm
60 mm
T
5
=
= 50 kPa
2tAm
2(0.005)(0.01)
T
Ans.
30 mm
40 mm
298
40 mm
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•5–117.
The mean dimensions of the cross section of the
leading edge and torsion box of an airplane wing can be
approximated as shown. If the wing is made of 2014-T6
aluminum alloy having an allowable shear stress of
tallow = 125 MPa and the wall thickness is 10 mm,
determine the maximum allowable torque and the
corresponding angle of twist per meter length of the wing.
10 mm
0.5 m
10 mm
Section Properties: Referring to the geometry shown in Fig. a,
Am =
F
p
1
a 0.52 b + A 1 + 0.5 B (2) = 1.8927 m2
2
2
ds = p(0.5) + 2 222 + 0.252 + 0.5 = 6.1019 m
Allowable Average Shear Stress:
A tavg B allow =
T
;
2tAm
125(106) =
T
2(0.01)(1.8927)
T = 4.7317(106)N # m = 4.73 MN # m
Ans.
Angle of Twist:
f =
ds
TL
2
4Am G F t
4.7317(106)(1)
=
4(1.89272)(27)(109)
¢
0.25 m
10 mm
6.1019
≤
0.01
= 7.463(10 - 3) rad = 0.428°>m
Ans.
299
2m
0.25 m
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5–118. The mean dimensions of the cross section of the
leading edge and torsion box of an airplane wing can
be approximated as shown. If the wing is subjected to a
torque of 4.5 MN # m and the wall thickness is 10 mm,
determine the average shear stress developed in the wing
and the angle of twist per meter length of the wing. The
wing is made of 2014-T6 aluminum alloy.
10 mm
0.5 m
10 mm
Section Properties: Referring to the geometry shown in Fig. a,
Am =
F
p
1
a0.52 b + A 1 + 0.5 B (2) = 1.8927 m2
2
2
ds = p(0.5) + 2222 + 0.252 + 0.5 = 6.1019 m
Average Shear Stress:
tavg =
4.5(106)
T
=
= 119 MPa
2tAm
2(0.01)(1.8927)
Ans.
Angle of Twist:
f =
ds
TL
4Am 2G F t
4.5(106)(1)
=
4(1.89272)(27)(109)
¢
0.25 m
10 mm
6.1019
≤
0.01
= 7.0973(10 - 3) rad = 0.407°>m
Ans.
300
2m
0.25 m
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5–119. The symmetric tube is made from a high-strength
steel, having the mean dimensions shown and a thickness of
5 mm. If it is subjected to a torque of T = 40 N # m,
determine the average shear stress developed at points A
and B. Indicate the shear stress on volume elements located
at these points.
20 mm
30 mm
60 mm
A
B
40 Nm
Am = 4(0.04)(0.06) + (0.04)2 = 0.0112 m2
tavg =
T
2 t Am
(tavg)A = (tavg)B =
40
= 357 kPa
2(0.005)(0.0112)
Ans.
*5–120. The steel used for the shaft has an allowable shear
stress of tallow = 8 MPa. If the members are connected with
a fillet weld of radius r = 4 mm, determine the maximum
torque T that can be applied.
50 mm
20 mm
T
2
Allowable Shear Stress:
D
50
=
= 2.5
d
20
and
4
r
=
= 0.20
d
20
From the text, K = 1.25
tmax = tallow = K
Tc
J
t
2 (0.01)
4 R
2 (0.01 )
8(10)4 = 1.25 B p
T = 20.1 N # m
Ans.
301
T
20 mm
T
2
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•5–121.
The built-up shaft is to be designed to rotate
at 720 rpm while transmitting 30 kW of power. Is this
possible? The allowable shear stress is tallow = 12 MPa.
v = 720
T =
75 mm
rev 2p rad 1 min
a
b
= 24 p rad>s
min
1 rev
60 s
60 mm
30(103)
P
=
= 397.89 N # m
v
24 p
tmax = K
Tc
;
J
12(106) = Kc
397.89(0.03)
p
4
2 (0.03 )
d;
K = 1.28
D
75
=
= 1.25
d
60
From Fig. 5-32,
r
= 0.133
d
r
= 0.133 ;
60
r = 7.98 mm
Check:
D - d
75 - 60
15
=
=
= 7.5 mm 6 7.98 mm
2
2
2
No, it is not possible.
Ans.
5–122. The built-up shaft is designed to rotate at 540 rpm.
If the radius of the fillet weld connecting the shafts
is r = 7.20 mm, and the allowable shear stress for the
material is tallow = 55 MPa, determine the maximum power
the shaft can transmit.
D
75
=
= 1.25;
d
60
75 mm
60 mm
r
7.2
=
= 0.12
d
60
From Fig. 5-32, K = 1.30
tmax = K
v = 540
Tc
;
J
T(0.03)
d;
55(106) = 1.30 c[ p
4
2 (0.03 )
T = 1794.33 N # m
rev 2p rad 1 min
a
b
= 18 p rad>s
min
1 rev
60 s
P = Tv = 1794.33(18p) = 101466 W = 101 kW
Ans.
302
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5–123. The steel shaft is made from two segments: AB and
BC, which are connected using a fillet weld having a radius
of 2.8 mm. Determine the maximum shear stress developed
in the shaft.
C
50 mm
D
100 N⭈m
20 mm B
(tmax)CD =
100(0.025)
TCDc
= p
4
J
2 (0.025 )
40 N⭈m
A
= 4.07 MPa
60 N⭈m
For the fillet:
D
50
=
= 2.5;
d
20
r
2.8
=
= 0.14
d
20
From Fig. 5-32, K = 1.325
(tmax)f = K
60(0.01)
TABc
d
= 1.325 c p
4
J
2 (0.01 )
= 50.6 MPa (max)
Ans.
*5–124. The steel used for the shaft has an allowable shear
stress of tallow = 8 MPa. If the members are connected
together with a fillet weld of radius r = 2.25 mm, determine
the maximum torque T that can be applied.
30 mm
30 mm
15 mm
T
T
2
Allowable Shear Stress:
D
30
=
= 2
d
15
r
2.25
=
= 0.15
d
15
and
From the text, K = 1.30
tmax = tallow = K
Tc
J
A 2r B (0.0075)
8(106) = 1.3 C p
4
2 (0.0075 )
S
T = 8.16 N # m
Ans.
303
T
2
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•5–125. The assembly is subjected to a torque of 710 lb # in.
If the allowable shear stress for the material is tallow = 12 ksi,
determine the radius of the smallest size fillet that can be used
to transmit the torque.
tmax = tallow = K
0.75 in.
A
Tc
J
710 lb⭈in.
B
1.5 in.
3
12(10 ) =
K(710)(0.375)
p
4
2 (0.375 )
C
K = 1.40
710 lb⭈ft
D
1.5
=
= 2
d
0.75
From Fig. 5-32,
r
= 0.1;
d
r = 0.1(0.75) = 0.075 in.
Ans.
Check:
D - d
1.5 - 0.75
=
= 0.375 7 0.075 in.
2
2
OK
5–126. A solid shaft is subjected to the torque T, which
causes the material to yield. If the material is elastic plastic,
show that the torque can be expressed in terms of the angle
of twist f of the shaft as T = 43 TY11 - f3Y>4f32, where TY
and fY are the torque and angle of twist when the material
begins to yield.
f =
gY
gL
=
L
r
rY
rY =
gYL
f
(1)
When rY = c, f = fY
From Eq. (1),
c =
gYL
fY
(2)
Dividing Eq. (1) by Eq. (2) yields:
rY
fY
=
c
f
(3)
Use Eq. 5-26 from the text.
T =
r3Y
p tY
2p tYc3
(4 c3 - r3Y) =
a1 )b
6
3
4 c3
Use Eq. 5-24, TY =
T =
p
t c3 from the text and Eq. (3)
2 Y
f3Y
4
TY a1 b
3
4 f3
QED
304
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5–127. A solid shaft having a diameter of 2 in. is made
of elastic-plastic material having a yield stress of
tY = 16 ksi and shear modulus of G = 1211032 ksi.
Determine the torque required to develop an elastic core
in the shaft having a diameter of 1 in. Also, what is the
plastic torque?
Use Eq. 5-26 from the text:
T =
p (16)
p tY
(4 c3 - rY 3) =
[4(13) - 0.53]
6
6
= 32.46 kip # in. = 2.71 kip # ft
Ans.
Use Eq. 5-27 from the text:
TP =
2p
2p
t c3 =
(16)(13)
3 Y
3
= 33.51 kip # in. = 2.79 kip # ft
Ans.
*5–128. Determine the torque needed to twist a short
3-mm-diameter steel wire through several revolutions if it is
made from steel assumed to be elastic plastic and having a
yield stress of tY = 80 MPa. Assume that the material
becomes fully plastic.
When the material becomes fully plastic then, from Eq. 5-27 in the text,
TP =
2 p (80)(106)
2 p tY 3
c =
(0.00153) = 0.565 N # m
3
3
305
Ans.
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•5–129.
The solid shaft is made of an elastic-perfectly
plastic material as shown. Determine the torque T needed
to form an elastic core in the shaft having a radius of
rY = 20 mm. If the shaft is 3 m long, through what angle
does one end of the shaft twist with respect to the other end?
When the torque is removed, determine the residual stress
distribution in the shaft and the permanent angle of twist.
80 mm
T
T
t (MPa)
160
Elastic-Plastic Torque: Applying Eq. 5-26 from the text
T =
=
0.004
p tY
A 4c3 - r3Y B
6
p(160)(106)
C 4 A 0.043 B - 0.023 D
6
= 20776.40 N # m = 20.8 kN # m
Ans.
Angle of Twist:
gY
0.004
L = a
b(3) = 0.600 rad = 34.4°
rY
0.02
f =
Ans.
When the reverse T = 20776.4 N # m is applied,
G =
160(106)
= 40 GPa
0.004
f¿ =
20776.4(3)
TL
= 0.3875 rad
= p
4
9
JG
2 (0.04 )(40)(10 )
The permanent angle of twist is,
fr = f - f¿
= 0.600 - 0.3875 = 0.2125 rad = 12.2°
Ans.
Residual Shear Stress:
(t¿)r = c =
20776.4(0.04)
Tc
= 206.67 MPa
=
p
4
J
2 (0.04 )
(t¿)r = 0.02 m =
20776.4(0.02)
Tc
= 103.33 MPa
=
p
4
J
2 (0.04 )
(tr)r = c = -160 + 206.67 = 46.7 MPa
(tr)r = 0.02m = -160 + 103.33 = -56.7 MPa
306
g (rad)
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5–130. The shaft is subjected to a maximum shear strain
of 0.0048 rad. Determine the torque applied to the shaft
if the material has strain hardening as shown by the shear
stress–strain diagram.
2 in.
From the shear - strain diagram,
rY
2
=
;
0.0006
0.0048
T
rY = 0.25 in.
t (ksi)
12
From the shear stress–strain diagram,
t1 =
6
r = 24r
0.25
6
t2 - 6
12 - 6
=
;
r - 0.25
2 - 0.25
t2 = 3.4286 r + 5.1429
0.0006
0.0048
c
T = 2p
L0
t r2 dr
0.25
= 2p
2
24r3 dr + 2p
L0
= 2p[6r4] | + 2p c
0.25
0
L0.25
(3.4286r + 5.1429)r2 dr
3.4286r4
5.1429r3 2
+
d |
4
3
0.25
= 172.30 kip # in. = 14.4 kip # ft
Ans.
5–131. An 80-mm diameter solid circular shaft is made of
an elastic-perfectly plastic material having a yield shear
stress of tY = 125 MPa. Determine (a) the maximum elastic
torque TY; and (b) the plastic torque Tp.
Maximum Elastic Torque.
TY =
=
1 3
pc tY
2
1
pa 0.043 b A 125 B a 106 b
2
= 12 566.37 N # m = 12.6 kN # m
Ans.
Plastic Torque.
TP =
=
2 3
pc tY
3
2
pa 0.043 b A 125 B a 106 b
3
= 16755.16 N # m = 16.8 kN # m
Ans.
307
g (rad)
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*5–132. The hollow shaft has the cross section shown and
is made of an elastic-perfectly plastic material having a
yield shear stress of tY. Determine the ratio of the plastic
torque Tp to the maximum elastic torque TY.
c
c
2
Maximum Elastic Torque. In this case, the torsion formula is still applicable.
tY =
TY c
J
TY =
J
t
c Y
c 4
p 4
B c - a b R tY
2
2
=
=
c
15 3
pc tY
32
Plastic Torque. Using the general equation, with t = tY,
c
TP = 2ptY
r2dr
Lc>2
c
r3
= 2ptY ¢ ≤ `
3 c>2
=
7
pc3tY
12
The ratio is
7
pc3tY
TP
12
=
= 1.24
TY
15 3
pc tY
32
Ans.
5–133. The shaft consists of two sections that are rigidly
connected. If the material is elastic plastic as shown,
determine the largest torque T that can be applied to the
shaft. Also, draw the shear-stress distribution over a radial
line for each section. Neglect the effect of stress
concentration.
T
1 in.
0.75 in.
T
0.75 in. diameter segment will be fully plastic. From Eq. 5-27 of the text:
T = Tp =
2p tY 3
(c )
3
t (ksi)
12
3
=
2p (12)(10 )
(0.3753)
3
= 1325.36 lb # in. = 110 lb # ft
Ans.
For 1 – in. diameter segment:
tmax =
1325.36(0.5)
Tc
=
p
4
J
2 (0.5)
= 6.75 ksi 6 tY
308
0.005
g (rad)
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5–134. The hollow shaft is made of an elastic-perfectly
plastic material having a shear modulus of G and a yield
shear stress of tY. Determine the applied torque Tp when the
material of the inner surface is about to yield (plastic torque).
Also, find the corresponding angle of twist and the maximum
shear strain. The shaft has a length of L.
ci
Plastic Torque. Using the general equation with t = tY,
co
TP = 2ptY
Lci
r2dr
co
r3
= 2ptY ¢ ≤ `
3 ci
=
2
pt A c 3 - ci 3 B
3 Y o
Ans.
Angle of Twist. When the material is about to yield at the inner surface, g = gY at
r = rY = ci. Also, Hooke’s Law is still valid at the inner surface.
gY =
tY
G
f =
gY
tY>G
tYL
L =
L =
rY
ci
ciG
Ans.
Shear Strain. Since the shear strain varies linearly along the radial line, Fig. a,
gmax
gY
=
co
ci
gmax = ¢
co
co tY
cotY
≤ gY = ¢ ≤ a b =
ci
ci
G
ciG
Ans.
309
c0
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5–135. The hollow shaft has inner and outer diameters of
60 mm and 80 mm, respectively. If it is made of an elasticperfectly plastic material, which has the t-g diagram shown,
determine the reactions at the fixed supports A and C.
150 mm
450 mm
B
C
15 kN⭈m
A
t (MPa)
120
Equation of Equilibrium. Refering to the free - body diagram of the shaft shown in
Fig. a,
©Mx = 0; TA + TC - 15 A 103 B = 0
(1)
Elastic Analysis. It is required that fB>A = fB>C. Thus, the compatibility equation is
fB>A = fB>C
TCLBC
TALAB
=
JG
JG
TA (0.45) = TC(0.15)
TC = 3TA
(2)
Solving Eqs. (1) and (2),
TA = 3750 N # m
TC = 11 250N # m
The maximum elastic torque and plastic torque in the shaft can be determined from
p
A 0.044 - 0.034 B
J
2
T(120) A 106 B = 8246.68 N # m
TY = tY = D
c
0.04
co
TP = 2ptY
Lci
r2dr
= 2p(120) A 106 B ¢
g (rad)
0.0016
0.04 m
r3
= 9299.11 N # m
≤`
3 0.03 m
Since TC 7 TY, the results obtained using the elastic analysis are not valid.
Plastic Analysis. Assuming that segment BC is fully plastic,
TC = TP = 9299.11N # m = 9.3kN # m
Ans.
Substituting this result into Eq. (1),
TA = 5700 N # m = 5.70 kN # m
Ans.
310
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5–135.
Continued
Since TA 6 TY, segment AB of the shaft is still linearly elastic. Here,
120 A 106 B
= 75GPa.
G =
0.0016
fB>C = fB>A =
fB>C =
gi
L ;
ci BC
5700.89(0.45)
TALAB
= 0.01244 rad
=
p
JG
A 0.044 - 0.034 B (75) A 109 B
2
0.01244 =
gi
(0.15)
0.03
gi = 0.002489 rad
Since gi 7 gY, segment BC of the shaft is indeed fully plastic.
311
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*5–136. The tubular shaft is made of a strain-hardening
material having a t-g diagram as shown. Determine the
torque T that must be applied to the shaft so that the
maximum shear strain is 0.01 rad.
T
0.5 in.
0.75 in.
t (ksi)
15
10
0.005
From the shear–strain diagram,
g
0.01
=
;
0.5
0.75
g = 0.006667 rad
From the shear stress–strain diagram,
15 - 10
t - 10
=
; t = 11.667 ksi
0.006667 - 0.005
0.01 - 0.005
15 - 11.667
t - 11.667
=
;
r - 0.5
0.75 - 0.50
t = 13.333 r + 5
co
T = 2p
tr2 dr
Lci
0.75
= 2p
(13.333r + 5) r2 dr
L0.5
0.75
= 2p
L0.5
= 2p c
(13.333r3 + 5r2) dr
13.333r4
5r3 0.75
+
d |
4
3 0.5
= 8.426 kip # in. = 702 lb # ft
Ans.
312
0.01
g (rad)
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•5–137.
The shear stress–strain diagram for a solid
50-mm-diameter shaft can be approximated as shown in
the figure. Determine the torque T required to cause a
maximum shear stress in the shaft of 125 MPa. If the shaft is
1.5 m long, what is the corresponding angle of twist?
T
1.5 m
T
t (MPa)
125
50
Strain Diagram:
rg
0.0025
=
0.025
;
0.01
0.0025
rg = 0.00625 m
Stress Diagram:
t1 =
50(106)
r = 8(109) r
0.00625
t2 - 50(106)
125(106) - 50(106)
=
r - 0.00625
0.025 - 0.00625
t2 = 4 A 109 B r + 25 A 106 B
The Ultimate Torque:
c
T = 2p
L0
t r2dr
0.00625 m
= 2p
L0
8 A 109 B r3 dr
0.025 m
+ 2p
C 4 A 10 B r + 25 A 10 B D r2dr
9
L0.00625 m
6
m
= 2p C 2 A 109 B r4 D |0.00625
0
+ 2p B 1 A 109 B r4 +
25(106)r3 0.025 m
R 2
3
0.00625 m
= 3269.30 N # m = 3.27 kN # m
Ans.
Angle of Twist:
f =
gmax
0.01
L = a
b (1.5) = 0.60 rad = 34.4°
c
0.025
Ans.
313
0.010
g (rad)
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5–138. A tube is made of elastic-perfectly plastic material,
which has the t-g diagram shown. If the radius of the
elastic core is rY = 2.25 in., determine the applied torque T.
Also, find the residual shear-stress distribution in the shaft
and the permanent angle of twist of one end relative to the
other when the torque is removed.
3 ft
3 in.
T
T
6 in.
t (ksi)
Elastic - Plastic Torque. The shear stress distribution due to T is shown in Fig. a. The
10
r = 4.444r. Thus,
linear portion of this distribution can be expressed as t =
2.25
tr = 1.5 in. = 4.444(1.5) = 6.667 ksi.
T = 2p
L
tr2dr
g (rad)
0.004
2.25 in.
= 2p
L1.5 in.
= 8.889p ¢
4.444r A r2dr B + 2p(10)
3 in.
L2.25 in.
r2dr
r4 2.25 in.
r3 3 in.
+ 20p ¢ ≤ 2
≤2
4 1.5 in.
3 2.25 in.
= 470.50 kip # in = 39.2 kip # ft
Ans.
Angle of Twist.
f =
gY
0.004
L =
(3)(12) = 0.064 rad
rY
2.25
The process of removing torque T is equivalent to the application of T¿ , which is
equal magnitude but opposite in sense to that of T. This process occurs in a linear
10
= 2.5 A 103 B ksi.
manner and G =
0.004
f¿ =
470.50(3)(2)
T¿L
=
= 0.0568 rad
p
JG
A 34 - 1.54 B (2.5) A 103 B
2
trœ = co =
470.50(3)
T¿co
=
= 11.83 ksi
p 4
4
J
2 A 3 - 1.5 B
trœ = rY =
T¿rY
470.50(2.25)
=
= 8.875 ksi
p 4
J
A 3 - 1.54 B
2
trœ = ci =
10
470.50(1.5)
T¿ci
=
= 5.917 ksi
p 4
4
J
2 A 3 - 1.5 B
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.
315
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5–139. The tube is made of elastic-perfectly plastic
material, which has the t-g diagram shown. Determine the
torque T that just causes the inner surface of the shaft to
yield. Also, find the residual shear-stress distribution in the
shaft when the torque is removed.
3 ft
3 in.
T
T
6 in.
t (ksi)
Plastic Torque. When the inner surface of the shaft is about to yield, the shaft is
about to become fully plastic.
T = 2p
L
10
tr2dr
3 in.
= 2ptY
L1.5 in.
= 2p(10)a
g (rad)
r2dr
0.004
r3 3 in.
b2
3 1.5 in.
= 494.80 kip # in. = 41.2 kip # ft
Ans.
Angle of Twist.
f =
gY
0.004
(3)(12) = 0.096 rad
L =
rY
1.5
The process of removing torque T is equivalent to the application of T¿ , which is
equal magnitude but opposite in sense to that of T. This process occurs in a linear
10
manner and G =
= 2.5 A 103 B ksi.
0.004
f¿ =
494.80(3)(12)
T¿L
=
= 0.05973 rad
p 4
JG
A 3 - 1.54 B (2.5) A 103 B
2
trœ = co =
trœ = ci =
494.80(3)
T¿co
=
= 12.44 ksi
p 4
J
4
3
1.5
A
B
2
494.80(1.5)
T¿ci
=
= 6.222 ksi
p 4
J
A 3 - 1.54 B
2
316
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5–139.
Continued
And the residual stresses are
(tr)r = co = tr = c + trœ = c = -10 + 12.44 = 2.44 ksi
Ans.
(tr)r = ci = tr = ci + trœ = ci = -10 + 6.22 = -3.78 ksi
Ans.
The shear stress distribution due to T and T¿ and the residual stress distribution are
shown in Fig. a.
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*5–140. The 2-m-long tube is made of an elastic-perfectly
plastic material as shown. Determine the applied torque T
that subjects the material at the tube’s outer edge to a shear
strain of gmax = 0.006 rad. What would be the permanent
angle of twist of the tube when this torque is removed?
Sketch the residual stress distribution in the tube.
T
35 mm
30 mm
t (MPa)
210
Plastic Torque: The tube is fully plastic if gi Ú gr = 0.003 rad.
g
0.006
=
;
0.03
0.035
0.003
g = 0.005143 rad
Therefore the tube is fully plastic.
co
TP = 2p
Lci
2p tg
=
=
3
tg r2 dr
A c3o - c3i B
2p(210)(106)
A 0.0353 - 0.033 B
3
= 6982.19 N # m = 6.98 kN # m
Ans.
Angle of Twist:
fP =
gmax
0.006
L = a
b(2) = 0.34286 rad
co
0.035
When a reverse torque of TP = 6982.19 N # m is applied,
G =
210(106)
tY
=
= 70 GPa
gY
0.003
fPœ =
6982.19(2)
TPL
= 0.18389 rad
= p
4
JG
(0.035
- 0.034)(70)(109)
2
Permanent angle of twist,
fr = fP - fPœ
= 0.34286 - 0.18389 = 0.1590 rad = 9.11°
Ans.
Residual Shear Stress:
tPœ o =
6982.19(0.035)
TP c
= 225.27 MPa
= p
4
4
J
2 (0.035 - 0.03 )
tPœ i =
TP r
6982.19(0.03)
= 193.09 MPa
= p
4
4
J
2 (0.035 - 0.03 )
(tP)o = -tg + tPœ o = -210 + 225.27 = 15.3 MPa
(tP)i = -tg + tPœ i = -210 + 193.09 = -16.9 MPa
318
g (rad)
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•5–141.
A steel alloy core is bonded firmly to the copper
alloy tube to form the shaft shown. If the materials have the
t-g diagrams shown, determine the torque resisted by the
core and the tube.
450 mm
A
100 mm
60 mm
B
15 kN⭈m
t (MPa)
Equation of Equilibrium. Refering to the free - body diagram of the cut part of the
assembly shown in Fig. a,
©Mx = 0; Tc + Tt - 15 A 103 B = 0
180
(1)
Elastic Analysis. The shear modulus of steel and copper are Gst =
36 A 106 B
and G q =
= 18 GPa. Compatibility requires that
0.002
180 A 106 B
0.0024
g (rad)
0.0024
= 75 GPa
Steel Alloy
t (MPa)
fC = ft
36
TcL
TtL
=
JcGst
JtG q
A
Tc
B
A
p
4
9
2 0.03 (75) 10
g (rad)
0.002
B
=
A
Tt
B
A
p
4
4
9
2 0.05 - 0.03 (18) 10
Copper Alloy
B
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
7610.97(0.45)
TcL
= 0.03589 rad
= p
4
JcGst
(0.03
)(75)(109)
2
0.3589 =
gi
(0.45)
0.03
gi = 0.002393 rad
Since gi 7 (gY) q = 0.002 rad, the tube is indeed fully plastic.
320
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5–142. A torque is applied to the shaft of radius r. If the
material has a shear stress–strain relation of t = kg1>6,
where k is a constant, determine the maximum shear stress
in the shaft.
r
r
r
g = gmax = gmax
c
r
T
1
gmax 6 1
b r6
t = kg = ka
r
1
6
r
T = 2p
tr2 dr
L0
1
r
= 2p
L0
gmax = a
ka
1
1
gmax 6 13
gmax 6 6
12p kg6max r3
19
b r 6 dr = 2pk a
b a b r6 =
r
r
19
19
6
19T
b
3
12p kr
1
tmax = kg6max =
19T
12p r3
Ans.
5–143. Consider a thin-walled tube of mean radius r and
thickness t. Show that the maximum shear stress in the tube
due to an applied torque T approaches the average shear
stress computed from Eq. 5–18 as r>t : q .
t
r
t
2r + t
;
ro = r + =
2
2
J =
=
t
2r - t
ri = r - =
2
2
2r - t 4
p 2r + t 4
ca
b - a
b d
2
2
2
p
p
[(2r + t)4 - (2r - t)4] =
[64 r3 t + 16 r t3]
32
32
tmax =
Tc
;
J
c = ro =
2r + t
2
T(2 r 2+ t)
T(2 r 2+ t)
= p
3
3
32 [64 r t + 16 r t ]
=
2p r t[r2 + 14t2]
2r
t
T(2r
2 + 2 r2 )
=
2p r t c rr2 + 14 rt 2 d
2
As
2
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 =
Tc
J
150(106) = p
500(0.03)
4
4
2 (0.03 - ri )
ri = 0.0293923 m = 29.3923 mm
Angle of Twist: Assume failure due to angle of twist limitation.
f =
0.08 =
TL
JG
A
500(3)
B
p
4
4
9
2 0.03 - ri (75.0)(10 )
ri = 0.0284033 m = 28.4033 mm
Choose the smallest value of ri = 28.4033 mm
t = ro - ri = 30 - 28.4033 = 1.60 mm
Ans.
322
G
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•5–145.
The A-36 steel circular tube is subjected to a
torque of 10 kN # m. Determine the shear stress at the mean
radius r = 60 mm and compute the angle of twist of the
tube if it is 4 m long and fixed at its far end. Solve the
problem using Eqs. 5–7 and 5–15 and by using Eqs. 5–18
and 5–20.
r 60 mm
4m
t 5 mm
10 kNm
Shear Stress:
Applying Eq. 5-7,
ro = 0.06 +
tr = 0.06 m =
0.005
= 0.0625 m
2
ri = 0.06 -
0.005
= 0.0575 m
2
10(103)(0.06)
Tr
= 88.27 MPa
= p
4
4
J
2 (0.0625 - 0.0575 )
Ans.
Applying Eq. 5-18,
tavg =
10(103)
T
= 88.42 MPa
=
2 t Am
29(0.005)(p)(0.062)
Ans.
Angle of Twist:
Applying Eq. 5-15,
f =
TL
JG
10(103)(4)
= p
4
2 (0.0625
- 0.05754)(75.0)(109)
= 0.0785 rad = 4.495°
Ans.
Applying Eq. 5-20,
f =
ds
TL
4A2mG L t
=
TL
ds
4A2mG t L
Where
L
ds = 2pr
2pTLr
=
4A2mG t
2p(10)(103)(4)(0.06)
=
4[(p)(0.062)]2 (75.0)(109)(0.005)
= 0.0786 rad = 4.503°
Ans.
323
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5–146. Rod AB is made of A-36 steel with an allowable
shear stress of 1tallow2st = 75 MPa, and tube BC is made of
AM1004-T61 magnesium alloy with an allowable shear
stress of 1tallow2mg = 45 MPa. The angle of twist of end C is
not allowed to exceed 0.05 rad. Determine the maximum
allowable torque T that can be applied to the assembly.
0.3 m
0.4 m
a
A
C
60 mm
Internal Loading: The internal torque developed in rod AB and tube BC are shown
in Figs. a and b, respectively.
Allowable Shear Stress: The polar moment of inertia of rod AB and tube
p
p
a0.0154 b = 25.3125(10 - 9)p m4 and JBC =
a0.034 - 0.0254 b
BC are JAB =
2
2
= 0.2096875(10 - 6)p m4. We have
A tallow B st =
TAB cAB
;
JAB
75(106) =
T(0.015)
25.3125(10 - 9)p
T = 397.61 N # m
and
A tallow B mg =
TBC cBC
;
JBC
45(106) =
T(0.03)
0.2096875(10 - 6)p
T = 988.13 N # m
Angle of Twist:
fB>A =
-T(0.7)
TAB LAB
= -0.11737(10 - 3)T = 0.11737(10 - 3)T
=
JAB Gst
25.3125(10 - 9)p(75)(109)
and
fC>B =
T(0.4)
TBC LBC
= 0.03373(10 - 3)T
=
JBC Gmg
0.2096875(10 - 6)p(18)(109)
It is required that fC>A = 0.05 rad. Thus,
fC>A = fB>A + fC>B
0.05 = 0.11737(10 - 3)T + 0.03373(10 - 3)T
T = 331 N # m A controls B
Ans.
324
50 mm
30 mm
Section a–a
T
a
B
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5–147. A shaft has the cross section shown and is made of
2014-T6 aluminum alloy having an allowable shear stress of
tallow = 125 MPa. If the angle of twist per meter length is
not allowed to exceed 0.03 rad, determine the required
minimum wall thickness t to the nearest millimeter when
the shaft is subjected to a torque of T = 15 kN # m.
30⬚ 30⬚
75 mm
Section Properties: Referring to the geometry shown in Fig. a,
Am =
C
0.075
1
1
(0.15) ¢
≤ + p A 0.0752 B = 0.01858 m2
2
tan 30°
2
ds = 2(0.15) + p(0.075) = 0.53562 m
Allowable Shear Stress:
A tavg B allow =
T
;
2tAm
125(106) =
15(103)
2t(0.01858)
t = 0.00323 m = 3.23 mm
Angle of Twist:
f =
ds
TL
2
4Am G C t
0.03 =
15(103)(1)
4(0.018582)(27)(109)
a
0.53562
b
t
t = 0.007184 m = 7.18 mm (controls)
Use t = 8 mm
Ans.
325
t
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*5–148. The motor A develops a torque at gear B of
500 lb # ft, which is applied along the axis of the 2-in.diameter A-36 steel shaft CD. This torque is to be
transmitted to the pinion gears at E and F. If these gears are
temporarily fixed, determine the maximum shear stress in
segments CB and BD of the shaft. Also, what is the angle of
twist of each of these segments? The bearings at C and D
only exert force reactions on the shaft.
B
E
F
2 ft
1.5 ft
C
D
A
Equilibrium:
TC + TD - 500 = 0
[1]
Compatibility:
fB>C = fB>D
TC(2)
TD(1.5)
=
JG
JG
TC = 0.75TD
[2]
Solving Eqs. [1] and [2] yields:
TD = 285.71 lb # ft
TC = 214.29 lb # ft
Maximum Shear Stress:
(tCB)max =
214.29(12)(1)
TCc
= 1.64 ksi
=
p
4
J
2 (1 )
Ans.
(tBD)max =
285.71(12)(1)
TDc
= 2.18 ksi
=
p
4
J
2 (1 )
Ans.
Angle of Twist:
fCB = fBD =
TD LBD
JG
285.71(12)(1.5)(12)
=
p
4
6
2 (1 )(11.0)(10 )
= 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:
©Mx = 0;
T - 550(0.13) = 0
T = 71.5 N # m
Ans.
Maximum Shear Stress:
tmax =
71.5(0.0125)
Tc
= 23.3 MPa
= p
4
J
2 (0.0125 )
Ans.
5–150. The rotating flywheel and shaft is brought to a
sudden stop at D when the bearing freezes. This causes the
flywheel to oscillate clockwise–counterclockwise, so that a
point A on the outer edge of the flywheel is displaced
through a 10-mm arc in either direction. Determine the
maximum shear stress developed in the tubular 304 stainless
steel shaft due to this oscillation. The shaft has an inner
diameter of 25 mm and an outer diameter of 35 mm. The
journal bearings at B and C allow the shaft to rotate freely.
D
2m
B
A
80 mm
Angle of Twist:
f =
TL
JG
Where
f =
10
= 0.125 rad
80
T(2)
0.125 = p
4
2 (0.0175
- 0.01254)(75.0)(109)
T = 510.82 N # m
Maximum Shear Stress:
tmax =
510.82(0.0175)
Tc
= 82.0 MPa
= p
4
4
J
2 (0.0175 - 0.0125 )
Ans.
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 =
7.85(0.15)
TL
= p
4
JG
(0.005)
(37)(109)
2
= 0.03243 rad = 1.86°
Ans.
328
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6–1. Draw the shear and moment diagrams for the shaft. The
bearings at A and B exert only vertical reactions on the shaft.
B
A
800 mm
250 mm
24 kN
6–2. Draw the shear and moment diagrams for the simply
supported beam.
4 kN
M 2 kNm
A
B
2m
329
2m
2m
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6–3. The engine crane is used to support the engine, which
has a weight of 1200 lb. Draw the shear and moment diagrams
of the boom ABC when it is in the horizontal position shown.
a + ©MA = 0;
4
F (3) - 1200(8) = 0;
5 A
+ c ©Fy = 0;
-Ay +
+ ©F = 0;
;
x
Ax -
4
(4000) - 1200 = 0;
5
3
(4000) = 0;
5
A
3 ft
5 ft
B
FA = 4000 lb
4 ft
Ay = 2000 lb
Ax = 2400 lb
*6–4. Draw the shear and moment diagrams for the cantilever beam.
2 kN/m
A
6 kNm
2m
The free-body diagram of the beam’s right segment sectioned through an arbitrary
point shown in Fig. a will be used to write the shear and moment equations of the beam.
+ c ©Fy = 0;
C
V - 2(2 - x) = 0
V = {4 - 2x} kN‚
(1)
1
a + ©M = 0; -M - 2(2 - x)c (2 - x) d - 6 = 0 M = {-x2 + 4x - 10}kN # m‚(2)
2
The shear and moment diagrams shown in Figs. b and c are plotted using Eqs. (1)
and (2), respectively. The value of the shear and moment at x = 0 is evaluated using
Eqs. (1) and (2).
Vx = 0 = 4 - 2(0) = 4 kN
Mx = 0 = C -0 + 4(0) - 10 D = -10kN # m
330
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6–5. Draw the shear and moment diagrams for the beam.
10 kN
8 kN
15 kNm
2m
3m
6–6. Draw the shear and moment diagrams for the
overhang beam.
8 kN/m
C
A
B
2m
4m
6–7. Draw the shear and moment diagrams for the
compound beam which is pin connected at B.
6 kip
8 kip
A
C
B
4 ft
331
6 ft
4 ft
4 ft
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*6–8. Draw the shear and moment diagrams for the simply
supported beam.
150 lb/ft
300 lbft
A
B
12 ft
The free-body diagram of the beam’s left segment sectioned through an arbitrary
point shown in Fig. b will be used to write the shear and moment equations. The
intensity of the triangular distributed load at the point of sectioning is
w = 150 a
x
b = 12.5x
12
Referring to Fig. b,
+ c ©Fy = 0;
a + ©M = 0; M +
275 -
1
(12.5x)(x) - V = 0
2
V = {275 - 6.25x2}lb‚ (1)
x
1
(12.5x)(x)a b - 275x = 0 M = {275x - 2.083x3}lb # ft‚(2)
2
3
The shear and moment diagrams shown in Figs. c and d are plotted using Eqs. (1)
and (2), respectively. The location where the shear is equal to zero can be obtained
by setting V = 0 in Eq. (1).
0 = 275 - 6.25x2
x = 6.633 ft
The value of the moment at x = 6.633 ft (V = 0) is evaluated using Eq. (2).
M x = 6.633 ft = 275(6.633) - 2.083(6.633)3 = 1216 lb # ft
332
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6–9. Draw the shear and moment diagrams for the beam.
Hint: The 20-kip load must be replaced by equivalent
loadings at point C on the axis of the beam.
15 kip
1 ft
A
C
4 ft
20 kip
B
4 ft
4 ft
6–10. Members ABC and BD of the counter chair are
rigidly connected at B and the smooth collar at D is allowed
to move freely along the vertical slot. Draw the shear and
moment diagrams for member ABC.
Equations of Equilibrium: Referring to the free-body diagram of the frame shown
in Fig. a,
+ c ©Fy = 0;
P 150 lb
Ay - 150 = 0
C
A
Ay = 150 lb
a + ©MA = 0;
B
1.5 ft
1.5 ft
ND(1.5) - 150(3) = 0
D
ND = 300 lb
Shear and Moment Diagram: The couple moment acting on B due to ND is
MB = 300(1.5) = 450 lb # ft. The loading acting on member ABC is shown in Fig. b
and the shear and moment diagrams are shown in Figs. c and d.
333
1.5 ft
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6–11. The overhanging beam has been fabricated with
a projected arm BD on it. Draw the shear and moment
diagrams for the beam ABC if it supports a load of 800 lb.
Hint: The loading in the supporting strut DE must be replaced
by equivalent loads at point B on the axis of the beam.
E
800 lb
B
Support Reactions:
a + ©MC = 0;
5 ft
D
2 ft
C
A
800(10) -
3
4
FDE(4) - FDE(2) = 0
5
5
6 ft
4 ft
FDE = 2000 lb
+ c ©Fy = 0;
-800 +
3
(2000) - Cy = 0
5
+ ©F = 0;
:
x
-Cx +
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;
By (a) - P(a) = 0
+ c ©Fy = 0;
Cy - P - P = 0
+ ©F = 0;
:
x
Bx = 0
By = P
a
a
a
a
Cy = 2P
From the FBD of segment AB
a + ©MA = 0;
P(2a) - P(a) - MA = 0
+ c ©Fy = 0;
P - P = 0 (equilibrium is statisfied!)
MA = Pa
6–14. The industrial robot is held in the stationary position
shown. Draw the shear and moment diagrams of the arm ABC
if it is pin connected at A and connected to a hydraulic cylinder
(two-force member) BD. Assume the arm and grip have a
uniform weight of 1.5 lbin. and support the load of 40 lb at C.
4 in.
A
10 in.
B
50 in.
120
D
335
C
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*6–16. Draw the shear and moment diagrams for the shaft
and determine the shear and moment throughout the shaft as
a function of x. The bearings at A and B exert only vertical
reactions on the shaft.
500 lb
800 lb
A
B
x
3 ft
For 0 6 x 6 3 ft
+ c ©Fy = 0.
220 - V = 0
a + ©MNA = 0.
V = 220 lb‚
Ans.
M - 220x = 0
M = (220x) lb ft‚
Ans.
For 3 ft 6 x 6 5 ft
+ c ©Fy = 0;
220 - 800 - V = 0
V = -580 lb
a + ©MNA = 0;
Ans.
M + 800(x - 3) - 220x = 0
M = {-580x + 2400} lb ft‚
Ans.
For 5 ft 6 x … 6 ft
+ c ©Fy = 0;
a + ©MNA = 0;
V - 500 = 0
V = 500 lb‚
Ans.
-M - 500(5.5 - x) - 250 = 0
M = (500x - 3000) lb ft
Ans.
336
2 ft
0.5 ft
0.5 ft
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•6–17.
Draw the shear and moment diagrams for the
cantilevered beam.
300 lb
200 lb/ft
A
6 ft
The free-body diagram of the beam’s left segment sectioned through an arbitrary
point shown in Fig. b will be used to write the shear and moment equations. The
intensity of the triangular distributed load at the point of sectioning is
x
w = 200 a b = 33.33x
6
Referring to Fig. b,
+ c ©Fy = 0;
-300 -
a + ©M = 0; M +
1
(33.33x)(x) - V = 0
2
V = {-300 - 16.67x2} lb (1)
1
x
(33.33x)(x)a b + 300x = 0 M = {-300x - 5.556x3} lb # ft (2)
2
3
The shear and moment diagrams shown in Figs. c and d are plotted using Eqs. (1)
and (2), respectively.
337
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6–18. Draw the shear and moment diagrams for the beam,
and determine the shear and moment throughout the beam
as functions of x.
2 kip/ft
10 kip
8 kip
40 kip⭈ft
Support Reactions: As shown on FBD.
Shear and Moment Function:
x
6 ft
For 0 … x 6 6 ft:
+ c ©Fy = 0;
4 ft
30.0 - 2x - V = 0
V = {30.0 - 2x} kip
Ans.
x
a + ©MNA = 0; M + 216 + 2xa b - 30.0x = 0
2
M = {-x2 + 30.0x - 216} kip # ft
Ans.
For 6 ft 6 x … 10 ft:
+ c ©Fy = 0;
a + ©MNA = 0;
V - 8 = 0
V = 8.00 kip
Ans.
-M - 8(10 - x) - 40 = 0
M = {8.00x - 120} kip # ft
Ans.
6–19. Draw the shear and moment diagrams for the beam.
2 kip/ ft
30 kip⭈ft
B
A
5 ft
338
5 ft
5 ft
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*6–20. Draw the shear and moment diagrams for the simply
supported beam.
10 kN
10 kN/m
A
B
3m
Since the area under the curved shear diagram can not be computed directly, the
value of the moment at x = 3 m will be computed using the method of sections. By
referring to the free-body diagram shown in Fig. b,
a + ©M = 0; Mx= 3 m +
1
(10)(3)(1) - 20(3) = 0
2
Mx= 3m = 45 kN # m
339
Ans.
3m
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•6–21. The beam is subjected to the uniform distributed load
shown. Draw the shear and moment diagrams for the beam.
2 kN/m
Equations of Equilibrium: Referring to the free-body diagram of the beam shown
in Fig. a,
a + ©MA = 0;
3
FBC a b (2) - 2(3)(1.5) = 0
5
B
A
1.5 m
FBC = 7.5 kN
+ c ©Fy = 0;
C
3
Ay + 7.5 a b - 2(3) = 0
5
Ay = 1.5 kN
3
Shear and Moment Diagram: The vertical component of FBC is A FBC B y = 7.5a b
5
= 4.5 kN. The shear and moment diagrams are shown in Figs. c and d.
340
2m
1m
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6–22. Draw the shear and moment diagrams for the
overhang beam.
4 kN/m
A
B
3m
Since the loading is discontinuous at support B, the shear and moment equations must
be written for regions 0 … x 6 3 m and 3 m 6 x … 6 m of the beam. The free-body
diagram of the beam’s segment sectioned through an arbitrary point within these two
regions is shown in Figs. b and c.
Region 0 … x 6 3 m, Fig. b
+ c ©Fy = 0;
1 4
a xb(x) - V = 0
2 3
2
V = e - x2 - 4 f kN
3
1 4
x
a xb(x)a b + 4x = 0
2 3
3
2
M = e - x3 - 4x f kN # m (2)
9
-4 -
a + ©M = 0; M +
(1)
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:
+ c ©Fy = 0;
V = 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;
Cy - 150.0 - 225 = 0
+ ©F = 0;
:
x
Cx = 0
Cy = 375.0 lb
Shear and Moment Diagram: The maximum positive moment occurs when V = 0.
+ c ©Fy = 0;
a + ©MNA = 0;
150.0 - 12.5x2 = 0
x = 3.464 ft
150(3.464) - 12.5 A 3.4642 B a
3.464
b - Mmax = 0
3
Mmax = 346.4 lb # ft
346
C
B
3 ft
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6–31. Draw the shear and moment diagrams for the beam and
determine the shear and moment in the beam as functions of x.
w0
Support Reactions: As shown on FBD.
A
B
x
Shear and Moment Functions:
L
–
2
For 0 … x 6 L>2
+ c ©Fy = 0;
3w0 L
-w0x - V = 0
4
V =
a + ©MNA = 0;
w0
(3L - 4x)
4
Ans.
7w0 L2
3w0 L
x
x + w0 xa b + M = 0
24
4
2
M =
w0
A -12x2 + 18Lx - 7L2)
24
Ans.
For L>2 6 x … L
+ c ©Fy = 0;
V -
1 2w0
c
(L - x) d(L - x) = 0
2 L
V =
a + ©MNA = 0;
-M -
w0
(L - x)2
L
Ans.
1 2w0
L - x
c
(L - x) d(L - x)a
b = 0
2 L
3
M = -
w0
(L - x)3
3L
Ans.
347
L
–
2
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*6–32. The smooth pin is supported by two leaves A and B
and subjected to a compressive load of 0.4 kNm caused by
bar C. Determine the intensity of the distributed load w0 of
the leaves on the pin and draw the shear and moment diagram
for the pin.
0.4 kN/m
C
A
+ c ©Fy = 0;
B
w0
1
2(w0)(20)a b - 60(0.4) = 0
2
20 mm 60 mm 20 mm
w0 = 1.2 kN>m
Ans.
•6–33.
The ski supports the 180-lb weight of the man. If
the snow loading on its bottom surface is trapezoidal as
shown, determine the intensity w, and then draw the shear
and moment diagrams for the ski.
180 lb
3 ft
w
1.5 ft
Ski:
+ c ©Fy = 0;
1
1
w(1.5) + 3w + w(1.5) - 180 = 0
2
2
Ans.
w = 40.0 lb>ft
Segment:
+ c ©Fy = 0;
30 - V = 0;
a + ©M = 0;
M - 30(0.5) = 0;
w0
V = 30.0 lb
M = 15.0 lb # ft
348
w
3 ft
1.5 ft
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6–34. Draw the shear and moment diagrams for the
compound beam.
5 kN
3 kN/m
A
B
3m
6–35. Draw the shear and moment diagrams for the beam
and determine the shear and moment as functions of x.
A
x
3m
200 - V = 0
V = 200 N
Ans.
M - 200 x = 0
M = (200 x) N # m
Ans.
For 3 m 6 x … 6 m:
200 - 200(x - 3) V = e-
1 200
c
(x - 3) d(x - 3) - V = 0
2 3
100 2
x + 500 f N
3
Ans.
Set V = 0, x = 3.873 m
a + ©MNA = 0;
M +
1 200
x - 3
c
(x - 3) d(x - 3)a
b
2 3
3
+ 200(x - 3)a
M = e-
1.5 m
B
For 0 … x 6 3 m:
+ c ©Fy = 0;
1.5 m
200 N/ m
Shear and Moment Functions:
a + ©MNA = 0;
3m
400 N/m
Support Reactions: As shown on FBD.
+ c ©Fy = 0;
D
C
x - 3
b - 200x = 0
2
100 3
x + 500x - 600 f N # m
9
Ans.
Substitute x = 3.87 m, M = 691 N # m
349
3m
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*6–36. Draw the shear and moment diagrams for the
overhang beam.
18 kN
6 kN
A
B
2m
6–37. Draw the shear and moment diagrams for the beam.
2m
M 10 kNm
2m
50 kN/m
50 kN/m
B
A
4.5 m
350
4.5 m
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6–38. The dead-weight loading along the centerline of the
airplane wing is shown. If the wing is fixed to the fuselage at
A, determine the reactions at A, and then draw the shear and
moment diagram for the wing.
3000 lb
400 lb/ft
250 lb/ft
A
8 ft
2 ft
Support Reactions:
3 ft
15 000 lb
-1.00 - 3 + 15 - 1.25 - 0.375 - Ay = 0
+ c ©Fy = 0;
Ay = 9.375 kip
Ans.
a + ©MA = 0;
1.00(7.667) + 3(5) - 15(3)
+ 1.25(2.5) + 0.375(1.667) + MA = 0
MA = 18.583 kip # ft = 18.6 kip # ft
Ans.
+ ©F = 0;
:
x
Ans.
Ax = 0
Shear and Moment Diagram:
6–39. The compound beam consists of two segments that
are pinned together at B. Draw the shear and moment
diagrams if it supports the distributed loading shown.
+ c ©Fy = 0;
2wL
1w 2
x = 0
27
2L
x =
a + ©M = 0;
w
B
2/3 L
4
L = 0.385 L
A 27
M +
C
A
1w
1
2wL
(0.385L)2 a b(0.385L) (0.385L) = 0
2L
3
27
M = 0.0190 wL2
351
1/3 L
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*6–40. Draw the shear and moment diagrams for the
simply supported beam.
10 kN
10 kN
15 kNm
A
B
2m
6–41. Draw the shear and moment diagrams for the
compound beam. The three segments are connected by
pins at B and E.
3 kN
2m
2m
3 kN
0.8 kN/m
B
E
F
A
C
2m
352
1m
1m
D
2m
1m
1m
2m
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6–42. Draw the shear and moment diagrams for the
compound beam.
5 kN/m
Support Reactions:
A
From the FBD of segment AB
B
2m
a + ©MA = 0;
By (2) - 10.0(1) = 0
By = 5.00 kN
+ c ©Fy = 0;
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.
FR =
dA =
L
LA
L0
w0
wdx =
w0
L
L L0
2
x2 dx =
w
w0 L
3
w
LA
2
x
A
w0L
w0x
= 0
12
3L2
1 1>3
x = a b L = 0.630 L
4
w0L
w0x3 1
a + ©M = 0;
(x) a xb - M = 0
12
3L2 4
M =
B
L
3
+ c ©Fy = 0;
w0
L
x3dx
L L0
3L
x =
=
=
w0 L
4
dA
3
LA
xdA
w0 2
x
L2
w0Lx
w0x4
12
12L2
Substitute x = 0.630L
M = 0.0394 w0L2
354
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6–46. Draw the shear and moment diagrams for the beam.
L
FR =
dA = w0
LA
L0
sina
w
2w0 L
p
xb dx =
p
L
w0
A
L
–
2
6–47. A member having the dimensions shown is used to
resist an internal bending moment of M = 90 kN # m.
Determine the maximum stress in the member if the moment
is applied (a) about the z axis (as shown) (b) about the y axis.
Sketch the stress distribution for each case.
200 mm
y
150 mm
The moment of inertia of the cross-section about z and y axes are
1
(0.2)(0.153) = 56.25(10 - 6) m4
12
Iy =
1
(0.15)(0.23) = 0.1(10 - 3) m4
12
M
z
x
For the bending about z axis, c = 0.075 m.
smax =
90(103) (0.075)
Mc
= 120(106)Pa = 120 MPa
=
Iz
56.25 (10 - 6)
Ans.
For the bending about y axis, C = 0.1 m.
smax =
x
B
L
–
2
Iz =
p
w w0 sin – x
L
90(103) (0.1)
Mc
= 90 (106)Pa = 90 MPa
=
Iy
0.1 (10 - 3)
Ans.
The bending stress distribution for bending about z and y axes are shown in Fig. a
and b respectively.
355
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*6–48. Determine the moment M that will produce a
maximum stress of 10 ksi on the cross section.
0.5 in.
A
3 in.
0.5 in.
0.5 in.
B
C
3 in.
M
10 in.
D
0.5 in.
Section Properties:
y =
=
©yA
©A
0.25(4)(0.5) + 2[2(3)(0.5)] + 5.5(10)(0.5)
= 3.40 in.
4(0.5) + 2[(3)(0.5)] + 10(0.5)
INA =
1
(4) A 0.53 B + 4(0.5)(3.40 - 0.25)2
12
+ 2c
1
(0.5)(33) + 0.5(3)(3.40 - 2)2 d
12
+
1
(0.5) A 103 B + 0.5(10)(5.5 - 3.40)2
12
= 91.73 in4
Maximum Bending Stress: Applying the flexure formula
smax =
10 =
Mc
I
M (10.5 - 3.4)
91.73
M = 129.2 kip # in = 10.8 kip # ft
Ans.
356
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•6–49.
Determine the maximum tensile and compressive
bending stress in the beam if it is subjected to a moment of
M = 4 kip # ft.
0.5 in.
A
0.5 in.
3 in.
0.5 in.
B
C
3 in.
M
10 in.
D
0.5 in.
Section Properties:
y =
=
©yA
©A
0.25(4)(0.5) + 2[2(3)(0.5)] + 5.5(10)(0.5)
= 3.40 in.
4(0.5) + 2[(3)(0.5)] + 10(0.5)
INA =
1
(4) A 0.53 B + 4(0.5)(3.40 - 0.25)2
12
+ 2c
1
(0.5)(33) + 0.5(3)(3.40 - 2)2 d
12
+
1
(0.5) A 103 B + 0.5(10)(5.5 - 3.40)2
12
= 91.73 in4
Maximum Bending Stress: Applying the flexure formula smax =
Mc
I
(st)max =
4(103)(12)(10.5 - 3.40)
= 3715.12 psi = 3.72 ksi
91.73
Ans.
(sc)max =
4(103)(12)(3.40)
= 1779.07 psi = 1.78 ksi
91.73
Ans.
357
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6–50. The channel strut is used as a guide rail for a trolley.
If the maximum moment in the strut is M = 30 N # m,
determine the bending stress at points A, B, and C.
50 mm
C
5 mm
5 mm
y =
B
2.5(50)(5) + 7.5(34)(5) + 2[20(5)(20)] + 2[(32.5)(12)(5)]
50(5) + 34(5) + 2[5(20)] + 2[(12)(5)]
30 mm
= 13.24 mm
A
I = c
1
(50)(53) + 50(5)(13.24 - 2.5)2 d
12
+ c
5 mm
5 mm
5 mm 7 mm 10 mm 7 mm
1
(34)(53) + 34(5)(13.24 - 7.5)2 d
12
+ 2c
1
1
(5)(203) + 5(20)(20 - 13.24)2 d + 2c (12)(53) + 12(5)(32.5 - 13.24)2 d
12
12
= 0.095883(10 - 6) m4
30(35 - 13.24)(10 - 3)
sA =
0.095883(10 - 6)
30(13.24 - 10)(10 - 3)
sB =
0.095883(10 - 6)
= 6.81 MPa
Ans.
= 1.01 MPa
Ans.
6–51. The channel strut is used as a guide rail for a
trolley. If the allowable bending stress for the material is
sallow = 175 MPa, determine the maximum bending moment
the strut will resist.
50 mm
C
5 mm
5 mm
B
-3
30(13.24)(10 )
sC =
-6
0.095883(10 )
= 4.14 MPa
©y2A
2.5(50)(5) + 7.5(34)(5) + 2[20(5)(20)] + 2[(32.5)(12)(5)]
=
= 13.24 mm
y =
©A
50(5) + 34(5) + 2[5(20)] + 2[(12)(5)]
I = c
1
1
(50)(53) + 50(5)(13.24 - 2.5)2 d + c (34)(53) + 34(5)(13.24 - 7.5)2 d
12
12
+ 2c
1
1
(5)(203) + 5(20)(20 - 13.24)2 d + 2c (12)(53) + 12(5)(32.5 - 13.24)2 d
12
12
= 0.095883(10 - 6) m4
s =
Mc
;
I
175(106) =
30 mm
Ans.
M(35 - 13.24)(10 - 3)
0.095883(10 - 6)
M = 771 N # m
Ans.
358
A
5 mm
5 mm
5 mm 7 mm 10 mm 7 mm
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*6–52. The beam is subjected to a moment M. Determine
the percentage of this moment that is resisted by the
stresses acting on both the top and bottom boards, A and
B, of the beam.
A
25 mm
M
D
Section Property:
I =
1
1
(0.2) A 0.23 B (0.15) A 0.153 B = 91.14583 A 10 - 6 B m4
12
12
150 mm
25 mm
25 mm
Bending Stress: Applying the flexure formula
B
150 mm
25 mm
My
I
s =
sE =
sD =
M(0.1)
91.14583(10 - 6)
M(0.075)
91.14583(10 - 6)
= 1097.143 M
= 822.857 M
Resultant Force and Moment: For board A or B
F = 822.857M(0.025)(0.2) +
1
(1097.143M - 822.857M)(0.025)(0.2)
2
= 4.800 M
M¿ = F(0.17619) = 4.80M(0.17619) = 0.8457 M
sc a
M¿
b = 0.8457(100%) = 84.6 %
M
Ans.
•6–53. Determine the moment M that should be applied
to the beam in order to create a compressive stress at point
D of sD = 30 MPa . Also sketch the stress distribution
acting over the cross section and compute the maximum
stress developed in the beam.
A
25 mm
Section Property:
150 mm
1
1
I =
(0.2) A 0.23 B (0.15) A 0.153 B = 91.14583 A 10 - 6 B m4
12
12
25 mm
25 mm
Bending Stress: Applying the flexure formula
s =
30 A 106 B =
My
I
M(0.075)
91.14583(10 - 6)
M = 36458 N # m = 36.5 kN # m
smax =
M
D
Ans.
36458(0.1)
Mc
= 40.0 MPa
=
I
91.14583(10 - 6)
Ans.
359
B
150 mm
25 mm
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6–54. The beam is made from three boards nailed together
as shown. If the moment acting on the cross section is
M = 600 N # m, determine the maximum bending stress in
the beam. Sketch a three-dimensional view of the stress
distribution acting over the cross section.
25 mm
150 mm
20 mm
(0.0125)(0.24)(0.025) + 2 (0.1)(0.15)(0.2)
= 0.05625 m
y =
0.24 (0.025) + 2 (0.15)(0.02)
200 mm M 600 Nm
1
(0.24)(0.0253) + (0.24)(0.025)(0.043752)
12
I =
+ 2a
20 mm
1
b (0.02)(0.153) + 2(0.15)(0.02)(0.043752)
12
= 34.53125 (10 - 6) m4
smax = sB =
Mc
I
600 (0.175 - 0.05625)
=
34.53125 (10 - 6)
= 2.06 MPa
sC =
Ans.
My
600 (0.05625)
= 0.977 MPa
=
I
34.53125 (10 - 6)
6–55. The beam is made from three boards nailed together
as shown. If the moment acting on the cross section is
M = 600 N # m, determine the resultant force the bending
stress produces on the top board.
25 mm
150 mm
(0.0125)(0.24)(0.025) + 2 (0.15)(0.1)(0.02)
= 0.05625 m
0.24 (0.025) + 2 (0.15)(0.02)
y =
20 mm
200 mm M 600 Nm
1
(0.24)(0.0253) + (0.24)(0.025)(0.043752)
12
I =
+ 2a
20 mm
1
b (0.02)(0.153) + 2(0.15)(0.02)(0.043752)
12
= 34.53125 (10 - 6) m4
s1 =
My
600(0.05625)
= 0.9774 MPa
=
I
34.53125(10 - 6)
sb =
My
600(0.05625 - 0.025)
= 0.5430 MPa
=
I
34.53125(10 - 6)
F =
1
(0.025)(0.9774 + 0.5430)(106)(0.240) = 4.56 kN
2
Ans.
360
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*6–56. The aluminum strut has a cross-sectional area in the
form of a cross. If it is subjected to the moment M = 8 kN # m,
determine the bending stress acting at points A and B, and show
the results acting on volume elements located at these points.
A
100 mm
20 mm
100 mm
B
M ⫽ 8 kN⭈m
20 mm
50 mm
50 mm
Section Property:
I =
1
1
(0.02) A 0.223 B +
(0.1) A 0.023 B = 17.8133 A 10 - 6 B m4
12
12
Bending Stress: Applying the flexure formula s =
sA =
sB =
8(103)(0.11)
17.8133(10 - 6)
8(103)(0.01)
17.8133(10 - 6)
My
I
= 49.4 MPa (C)
Ans.
= 4.49 MPa (T)
Ans.
•6–57. The aluminum strut has a cross-sectional area in the
form of a cross. If it is subjected to the moment M = 8 kN # m,
determine the maximum bending stress in the beam, and
sketch a three-dimensional view of the stress distribution
acting over the entire cross-sectional area.
A
100 mm
20 mm
100 mm
B
20 mm
M ⫽ 8 kN⭈m
50 mm
50 mm
Section Property:
I =
1
1
(0.02) A 0.223 B +
(0.1) A 0.023 B = 17.8133 A 10 - 6 B m4
12
12
Bending Stress: Applying the flexure formula smax =
smax =
8(103)(0.11)
17.8133(10 - 6)
sy = 0.01m =
My
Mc
and s =
,
I
I
= 49.4 MPa
8(103)(0.01)
17.8133(10 - 6)
Ans.
= 4.49 MPa
361
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6–58. If the beam is subjected to an internal moment
of M = 100 kip # ft, determine the maximum tensile and
compressive bending stress in the beam.
3 in.
3 in.
6 in.
M
2 in.
1.5 in.
Section Properties: The neutral axis passes through centroid C of the cross section
as shown in Fig. a. The location of C is
©yA
y =
=
©A
4(8)(6) - 2 cp A 1.52 B d
8(6) - p A 1.52 B
= 4.3454 in.
Thus, the moment of inertia of the cross section about the neutral axis is
I = ©I + Ad2
=
1
1
(6)a83 b + 6(8) A 4.3454 - 4 B 2 - B pa 1.54 b + pa 1.52 b A 4.3454 - 2 B 2 R
12
4
= 218.87 in4
Maximum Bending Stress: The maximum compressive and tensile bending stress
occurs at the top and bottom edges of the cross section.
A smax B T =
100(12)(4.3454)
Mc
=
= 23.8 ksi (T)
I
218.87
Ans.
A smax B C =
My
100(12)(8 - 4.3454)
=
= 20.0 ksi (C)
I
218.87
Ans.
362
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6–59. If the beam is made of material having an
allowable tensile and compressive stress of (sallow)t = 24 ksi
and (sallow)c = 22 ksi, respectively, determine the maximum
allowable internal moment M that can be applied to the beam.
3 in.
3 in.
6 in.
M
2 in.
1.5 in.
Section Properties: The neutral axis passes through centroid C of the cross section
as shown in Fig. a. The location of C is
©yA
y =
=
©A
4(8)(6) - 2 cp A 1.52 B d
8(6) - p A 1.52 B
= 4.3454 in.
Thus, the moment of inertia of the cross section about the neutral axis is
I = ©I + Ad2
=
1
1
(6) A 83 B + 6(8) A 4.3454 - 4 B 2 - B p A 1.54 B + p A 1.52 B A 4.3454 - 2 B 2 R
12
4
= 218.87 in4
Allowable Bending Stress: The maximum compressive and tensile bending stress
occurs at the top and bottom edges of the cross section. For the top edge,
(sallow)c =
My
;
I
22 =
M(8 - 4.3454)
218.87
M = 1317.53 kip # ina
1 ft
b = 109.79 kip # ft
12 in.
For the bottom edge,
A smax B t =
Mc
;
I
24 =
M(4.3454)
218.87
M = 1208.82 kip # ina
1 ft
b = 101 kip # ft (controls)
12 in.
363
Ans.
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*6–60. The beam is constructed from four boards as
shown. If it is subjected to a moment of Mz = 16 kip # ft,
determine the stress at points A and B. Sketch a
three-dimensional view of the stress distribution.
y
A
C
1 in. 10 in.
1 in.
10 in.
2[5(10)(1)] + 10.5(16)(1) + 16(10)(1)
y =
2(10)(1) + 16(1) + 10(1)
Mz 16 kipft
z
= 9.3043 in.
14 in.
1
1
I = 2c (1)(103) + 1(10)(9.3043 - 5)2 d +
(16)(13) + 16(1)(10.5 - 9.3043)2
12
12
+
B
1 in.
x
1 in.
1
(1)(103) + 1(10)(16 - 9.3043)2 = 1093.07 in4
12
sA =
16(12)(21 - 9.3043)
Mc
=
= 2.05 ksi
I
1093.07
Ans.
sB =
My
16(12)(9.3043)
=
= 1.63 ksi
I
1093.07
Ans.
•6–61. The beam is constructed from four boards as shown.
If it is subjected to a moment of Mz = 16 kip # ft, determine
the resultant force the stress produces on the top board C.
y
A
C
1 in. 10 in.
1 in.
10 in.
y =
2[5(10)(1)] + 10.5(16)(1) + 16(10)(1)
= 9.3043 in.
2(10)(1) + 16(1) + 10(1)
Mz 16 kipft
z
14 in.
1
1
I = 2c (1)(103) + (10)(9.3043 - 5)2 d +
(16)(13) + 16(1)(10.5 - 9.3043)2
12
12
+
1
(1)(103) + 1(10)(16 - 9.3043)2 = 1093.07 in4
12
sA =
16(12)(21 - 9.3043)
Mc
=
= 2.0544 ksi
I
1093.07
sD =
My
16(12)(11 - 9.3043)
=
= 0.2978 ksi
I
1093.07
(FR)C =
1
(2.0544 + 0.2978)(10)(1) = 11.8 kip
2
Ans.
364
1 in.
B
1 in.
x
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6–62. A box beam is constructed from four pieces of
wood, glued together as shown. If the moment acting on the
cross section is 10 kN # m, determine the stress at points A
and B and show the results acting on volume elements
located at these points.
20 mm
160 mm
25 mm
A
250 mm
25 mm
B
M 10 kNm
The moment of inertia of the cross-section about the neutral axis is
I =
1
1
(0.2)(0.33) (0.16)(0.253) = 0.2417(10 - 3) m4.
12
12
For point A, yA = C = 0.15 m.
sA =
10(103) (0.15)
MyA
= 6.207(106)Pa = 6.21 MPa (C)
=
I
0.2417(10 - 3)
Ans.
For point B, yB = 0.125 m.
sB =
MyB
10(103)(0.125)
= 5.172(106)Pa = 5.17 MPa (T)
=
I
0.2417(10 - 3)
Ans.
The state of stress at point A and B are represented by the volume element shown
in Figs. a and b respectively.
365
20 mm
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6–63. Determine the dimension a of a beam having a
square cross section in terms of the radius r of a beam with
a circular cross section if both beams are subjected to the
same internal moment which results in the same maximum
bending stress.
a
a
r
Section Properties: The moments of inertia of the square and circular cross sections
about the neutral axis are
1
a4
a A a3 B =
12
12
IS =
IC =
1 4
pr
4
Maximum Bending Stress: For the square cross section, c = a>2.
A smax B S =
M(a>2)
6M
Mc
= 3
= 4
IS
a >12
a
For the circular cross section, c = r.
A smax B c =
Mc
Mr
4M
=
Ic
1 4
pr3
pr
4
It is required that
A smax B S = A smax B C
6M
4M
=
a3
pr3
a = 1.677r
Ans.
*6–64. The steel rod having a diameter of 1 in. is subjected to
an internal moment of M = 300 lb # ft. Determine the stress
created at points A and B. Also, sketch a three-dimensional
view of the stress distribution acting over the cross section.
I =
A
B
p 4
p
r = (0.54) = 0.0490874 in4
4
4
sA =
M ⫽ 300 lb⭈ft
45⬚
300(12)(0.5)
Mc
=
= 36.7 ksi
I
0.0490874
Ans.
0.5 in.
My
300(12)(0.5 sin 45°)
sB =
=
= 25.9 ksi
I
0.0490874
Ans.
366
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•6–65.
If the moment acting on the cross section of the beam
is M = 4 kip # ft, determine the maximum bending stress in
the beam. Sketch a three-dimensional view of the stress
distribution acting over the cross section.
A
1.5 in.
12 in.
The moment of inertia of the cross-section about the neutral axis is
12 in.
1
1
(12)(153) (10.5)(123) = 1863 in4
I =
12
12
M
1.5 in.
1.5 in.
Along the top edge of the flange y = c = 7.5 in. Thus
smax =
4(103)(12)(7.5)
Mc
=
= 193 psi
I
1863
Ans.
Along the bottom edge to the flange, y = 6 in. Thus
s =
4(103)(12)(6)
My
=
= 155 psi
I
1863
6–66. If M = 4 kip # ft, determine the resultant force the
bending stress produces on the top board A of the beam.
A
1.5 in.
The moment of inertia of the cross-section about the neutral axis is
12 in.
1
1
(12)(153) (10.5)(123) = 1863 in4
12
12
I =
12 in.
M
Along the top edge of the flange y = c = 7.5 in. Thus
1.5 in.
smax =
4(103)(12)(7.5)
Mc
=
= 193.24 psi
I
1863
Along the bottom edge of the flange, y = 6 in. Thus
s =
4(103)(12)(6)
My
=
= 154.59 psi
I
1863
The resultant force acting on board A is equal to the volume of the trapezoidal
stress block shown in Fig. a.
FR =
1
(193.24 + 154.59)(1.5)(12)
2
= 3130.43 lb
= 3.13 kip
Ans.
367
1.5 in.
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6–67. The rod is supported by smooth journal bearings
at A and B that only exert vertical reactions on the shaft. If
d = 90 mm, determine the absolute maximum bending
stress in the beam, and sketch the stress distribution acting
over the cross section.
12 kN/m
d
A
B
3m
Absolute Maximum Bending Stress: The maximum moment is Mmax = 11.34 kN # m
as indicated on the moment diagram. Applying the flexure formula
smax =
Mmax c
I
11.34(103)(0.045)
=
p
4
4 (0.045 )
= 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 =
6 ft
100 lb/ft
5.75 in.
6 ft
6 ft
©yA
0 + (6.50)(0.4786)
=
= 4.6091 in.
©A
0.4786 + 0.19635
I = c
1
1
1
p(0.5)4 - p(0.3125)4 d + 0.4786(6.50 - 4.6091)2 + p(0.25)4
4
4
4
+ 0.19635(4.6091)2 = 5.9271 in4
Mmax = 300(9 - 1.5)(12) = 27 000 lb # in.
smax =
27 000(4.6091 + 0.25)
Mc
=
I
5.9271
= 22.1 ksi
Ans.
6–71. The axle of the freight car is subjected to wheel
loadings of 20 kip. If it is supported by two journal bearings at
C and D, determine the maximum bending stress developed
at the center of the axle, where the diameter is 5.5 in.
A
C
B
60 in.
10 in.
20 kip
smax =
200(2.75)
Mc
= 1
= 12.2 ksi
4
I
4 p(2.75)
Ans.
370
D
10 in.
20 kip
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*6–72. The steel beam has the cross-sectional area shown.
Determine the largest intensity of distributed load w0 that it
can support so that the maximum bending stress in the beam
does not exceed smax = 22 ksi.
w0
12 ft
12 ft
8 in.
0.30 in.
10 in.
0.3 in.
Support Reactions: As shown on FBD.
0.30 in.
Internal Moment: The maximum moment occurs at mid span. The maximum
moment is determined using the method of sections.
Section Property:
I =
1
1
(8) A 10.63 B (7.7) A 103 B = 152.344 in4
12
12
Absolute Maximum Bending Stress: The maximum moment is Mmax = 48.0w0 as
indicated on the FBD. Applying the flexure formula
smax =
22 =
Mmax c
I
48.0w0 (12)(5.30)
152.344
w0 = 1.10 kip>ft
Ans.
•6–73.
The steel beam has the cross-sectional area shown. If
w0 = 0.5 kip>ft, determine the maximum bending stress in
the beam.
w0
12 ft
12 ft
8 in.
Support Reactions: As shown on FBD.
0.3 in.
0.30 in.
Internal Moment: The maximum moment occurs at mid span. The maximum
moment is determined using the method of sections.
Section Property:
I =
1
1
(8) A 10.63 B (7.7) A 103 B = 152.344 in4
12
12
Absolute Maximum Bending Stress: The maximum moment is Mmax = 24.0 kip # ft
as indicated on the FBD. Applying the flexure formula
smax =
=
Mmax c
I
24.0(12)(5.30)
152.344
= 10.0 ksi
Ans.
371
0.30 in.
10 in.
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6–74. The boat has a weight of 2300 lb and a center of
gravity at G. If it rests on the trailer at the smooth contact A
and can be considered pinned at B, determine the absolute
maximum bending stress developed in the main strut of
the trailer. Consider the strut to be a box-beam having the
dimensions shown and pinned at C.
B
1 ft
G
C
A
3 ft
D
5 ft
4 ft
1.75 in.
1 ft
1.75 in.
3 in.
1.5 in.
Boat:
+ ©F = 0;
:
x
a + ©MB = 0;
Bx = 0
-NA(9) + 2300(5) = 0
NA = 1277.78 lb
+ c ©Fy = 0;
1277.78 - 2300 + By = 0
By = 1022.22 lb
Assembly:
a + ©MC = 0;
-ND(10) + 2300(9) = 0
ND = 2070 lb
+ c ©Fy = 0;
Cy + 2070 - 2300 = 0
Cy = 230 lb
I =
1
1
(1.75)(3)3 (1.5)(1.75)3 = 3.2676 in4
12
12
smax =
3833.3(12)(1.5)
Mc
=
= 21.1 ksi
I
3.2676
Ans.
372
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6–75. The shaft is supported by a smooth thrust bearing at
A and smooth journal bearing at D. If the shaft has the cross
section shown, determine the absolute maximum bending
stress in the shaft.
40 mm
A
B
0.75 m
D
C
1.5 m
25 mm
0.75 m
3 kN
3 kN
Shear and Moment Diagrams: As shown in Fig. a.
Maximum Moment: Due to symmetry, the maximum moment occurs in region BC
of the shaft. Referring to the free-body diagram of the segment shown in Fig. b.
Section Properties: The moment of inertia of the cross section about the neutral
axis is
I =
p
A 0.044 - 0.0254 B = 1.7038 A 10 - 6 B m4
4
Absolute Maximum Bending Stress:
2.25 A 10 B (0.04)
Mmaxc
=
= 52.8 MPa
I
1.7038 A 10 - 6 B
3
sallow =
Ans.
*6–76. Determine the moment M that must be applied to
the beam in order to create a maximum stress of 80 MPa.Also
sketch the stress distribution acting over the cross section.
300 mm
20 mm
The moment of inertia of the cross-section about the neutral axis is
I =
M
1
1
(0.3)(0.33) (0.21)(0.263) = 0.36742(10 - 3) m4
12
12
260 mm
Thus,
20 mm 30 mm
Mc
smax =
;
I
6
80(10 ) =
M(0.15)
0.36742(10 - 3)
M = 195.96 (103) N # m = 196 kN # m
The bending stress distribution over the cross-section is shown in Fig. a.
373
Ans.
30 mm
30 mm
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•6–77.
The steel beam has the cross-sectional area shown.
Determine the largest intensity of distributed load w that it
can support so that the bending stress does not exceed
smax = 22 ksi.
I =
w
1
1
(8)(10.6)3 (7.7)(103) = 152.344 in4
12
12
8 ft
w
8 ft
8 in.
Mc
smax =
I
22 =
8 ft
0.30 in.
10 in.
0.3 in.
0.30 in.
32w(12)(5.3)
152.344
w = 1.65 kip>ft
Ans.
6–78. The steel beam has the cross-sectional area shown.
If w = 5 kip>ft, determine the absolute maximum bending
stress in the beam.
w
8 ft
w
8 ft
8 ft
8 in.
0.3 in.
0.30 in.
10 in.
0.30 in.
From Prob. 6-78:
M = 32w = 32(5)(12) = 1920 kip # in.
I = 152.344 in4
smax =
1920(5.3)
Mc
=
= 66.8 ksi
I
152.344
Ans.
374
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6–79. If the beam ACB in Prob. 6–9 has a square cross
section, 6 in. by 6 in., determine the absolute maximum
bending stress in the beam.
15 kip
1 ft
A
20 kip
C
4 ft
B
4 ft
4 ft
Mmax = 46.7 kip # ft
smax =
46.7(103)(12)(3)
Mc
= 15.6 ksi
=
1
3
I
12 (6)(6 )
Ans.
*6–80. If the crane boom ABC in Prob. 6–3 has a
rectangular cross section with a base of 2.5 in., determine its
required height h to the nearest 14 in. if the allowable bending
stress is sallow = 24 ksi.
A
a + ©MA = 0;
+ c ©Fy = 0;
-Ay +
+ ©F = 0;
;
x
Ax -
4
(4000) - 1200 = 0;
5
3
(4000) = 0;
5
FB = 4000 lb
Ay = 2000 lb
Ax = 2400 lb
6000(12) A 2 B
Mc
= 24(10)3
= 1
3
I
12 (2.5)(h )
h
smax =
5 ft
B
4 ft
4
F (3) - 1200(8) = 0;
5 B
3 ft
h = 2.68 in.
Ans.
Use h = 2.75 in.
Ans.
375
C
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•6–81.
If the reaction of the ballast on the railway tie can
be assumed uniformly distributed over its length as
shown, determine the maximum bending stress developed
in the tie. The tie has the rectangular cross section with
thickness t = 6 in.
15 kip
1.5 ft
15 kip
5 ft
1.5 ft
t
w
Support Reactions: Referring to the free - body diagram of the tie shown in Fig. a,
we have
+ c ©Fy = 0;
w(8) - 2(15) = 0
w = 3.75 kip>ft
Maximum Moment: The shear and moment diagrams are shown in Figs. b and c. As
indicated on the moment diagram, the maximum moment is Mmax = 7.5 kip # ft.
Absolute Maximum Bending Stress:
smax =
12 in.
7.5(12)(3)
Mmaxc
= 1.25 ksi
=
I
1
(12)(63)
12
Ans.
376
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6–82. The reaction of the ballast on the railway tie can be
assumed uniformly distributed over its length as shown.
If the wood has an allowable bending stress of sallow =
1.5 ksi, determine the required minimum thickness t of the
rectangular cross sectional area of the tie to the nearest 18 in.
15 kip
1.5 ft
15 kip
5 ft
1.5 ft
t
w
Support Reactions: Referring to the free-body diagram of the tie shown in Fig. a, we
have
+ c ©Fy = 0;
w(8) - 2(15) = 0
w = 3.75 kip>ft
Maximum Moment: The shear and moment diagrams are shown in Figs. b and c. As
indicated on the moment diagram, the maximum moment is Mmax = 7.5 kip # ft.
Absolute Maximum Bending Stress:
t
7.5(12)a b
2
1.5 =
1
(12)t3
12
Mc
;
smax =
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
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
i
do
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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
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
i
do
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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
3
12 (b)(1.5b)
b = 0.05313 m = 53.1 mm
Ans.
380
2m
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6–86. Determine the absolute maximum bending stress
in the 2-in.-diameter shaft which is subjected to the
concentrated forces. The journal bearings at A and B only
support vertical forces.
800 lb
600 lb
A
15 in.
B
15 in.
30 in.
The FBD of the shaft is shown in Fig. a.
The shear and moment diagrams are shown in Fig. b and c, respectively. As
indicated on the moment diagram, Mmax = 15000 lb # in.
The moment of inertia of the cross-section about the neutral axis is
I =
p 4
(1 ) = 0.25 p in4
4
Here, c = 1 in. Thus
smax =
Mmax c
I
=
15000(1)
0.25 p
= 19.10(103) psi
Ans.
= 19.1 ksi
381
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6–87. Determine the smallest allowable diameter of the
shaft which is subjected to the concentrated forces. The
journal bearings at A and B only support vertical forces.
The allowable bending stress is sallow = 22 ksi.
800 lb
600 lb
A
15 in.
B
15 in.
30 in.
The FBD of the shaft is shown in Fig. a
The shear and moment diagrams are shown in Fig. b and c respectively. As
indicated on the moment diagram, Mmax = 15,000 lb # in
The moment of inertia of the cross-section about the neutral axis is
I =
p 4
p d 4
a b =
d
4 2
64
Here, c = d>2. Thus
sallow =
Mmax c
;
I
22(103) =
15000(d> 2)
pd4>64
d = 1.908 in = 2 in.
Ans.
382
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*6–88. If the beam has a square cross section of 9 in. on
each side, determine the absolute maximum bending stress
in the beam.
1200 lb
800 lb/ft
B
A
8 ft
Absolute Maximum Bending Stress: The maximum moment is Mmax = 44.8 kip # ft
as indicated on moment diagram. Applying the flexure formula
smax =
44.8(12)(4.5)
Mmax c
=
= 4.42 ksi
1
3
I
12 (9)(9)
Ans.
•6–89.
If the compound beam in Prob. 6–42 has a square
cross section, determine its dimension a if the allowable
bending stress is sallow = 150 MPa.
Allowable Bending Stress: The maximum moments is Mmax = 7.50 kN # m as
indicated on moment diagram. Applying the flexure formula
smax = sallow =
150 A 106 B =
Mmax c
I
7.50(103) A a2 B
1 4
12 a
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
Mmax c
smax =
=
I
A B
23w0 L2 h
2
216
1
3
12 bh
23w0 L2
=
Ans.
36bh2
384
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6–91. Determine the absolute maximum bending stress
in the 80-mm-diameter shaft which is subjected to the
concentrated forces. The journal bearings at A and B only
support vertical forces.
A
0.5 m
B
0.4 m
0.6 m
12 kN
20 kN
The FBD of the shaft is shown in Fig. a
The shear and moment diagrams are shown in Fig. b and c, respectively. As
indicated on the moment diagram, Mmax = 6 kN # m.
The moment of inertia of the cross-section about the neutral axis is
I =
p
(0.044) = 0.64(10 - 6)p m4
4
Here, c = 0.04 m. Thus
smax =
6(103)(0.04)
Mmax c
=
I
0.64(10 - 6)p
= 119.37(106) Pa
= 119 MPa
Ans.
385
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*6–92. Determine the smallest allowable diameter of the
shaft which is subjected to the concentrated forces. The
journal bearings at A and B only support vertical forces.
The allowable bending stress is sallow = 150 MPa.
A
0.5 m
B
0.4 m
0.6 m
12 kN
20 kN
The FBD of the shaft is shown in Fig. a.
The shear and moment diagrams are shown in Fig. b and c, respectively. As
indicated on the moment diagram, Mmax = 6 kN # m.
The moment of inertia of the cross-section about the neutral axis is
I =
pd4
p d 4
a b =
4 2
64
Here, c = d>2. Thus
sallow =
Mmax c
;
I
150(106) =
6(103)(d> 2)
pd4>64
d = 0.07413 m = 74.13 mm = 75 mm
386
Ans.
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•6–93. The man has a mass of 78 kg and stands motionless at
the end of the diving board. If the board has the cross section
shown, determine the maximum normal strain developed in
the board. The modulus of elasticity for the material is
E = 125 GPa. Assume A is a pin and B is a roller.
350 mm
30 mm
A
1.5 m
Internal Moment: The maximum moment occurs at support B. The maximum
moment is determined using the method of sections.
Section Property:
y =
©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)
I =
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 4
32 d
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 =
B
2m
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
Mmax c
I
100(103)(d)
p 4
32 d
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
0.375 4
4
4
p C A 1.25
2 B - A 2 B D = 0.11887 in
4
Mmax =
P
(4) = 2P
2
Require smax = 1.25 ksi
smax =
Mc
I
1.25 =
2P(1.25>2)
0.11887
P = 0.119 kip = 119 lb
Ans.
389
2.5 in.
0.05
P (in./ in.)
4 in.
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6–98. If the beam in Prob. 6–18 has a rectangular cross
section with a width of 8 in. and a height of 16 in., determine
the absolute maximum bending stress in the beam.
16 in.
Absolute Maximum Bending Stress: The maximum moment is Mmax = 216 kip # ft
as indicated on moment diagram. Applying the flexure formula
smax =
216(12)(8)
Mmax c
= 7.59 ksi
= 1
3
I
12 (8)(16 )
8 in.
Ans.
6–99. If the beam has a square cross section of 6 in. on
each side, determine the absolute maximum bending stress
in the beam.
400 lb/ft
B
A
6 ft
The maximum moment occurs at the fixed support A. Referring to the FBD shown
in Fig. a,
a + ©MA = 0;
Mmax - 400(6)(3) -
1
(400)(6)(8) = 0
2
Mmax = 16800 lb # ft
The moment of inertia of the about the neutral axis is I =
smax =
1
(6)(63) = 108 in4. Thus,
12
16800(12)(3)
Mc
=
I
108
= 5600 psi = 5.60 ksi
Ans.
390
6 ft
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*6–100. The steel beam has the cross-sectional area
shown. Determine the largest intensity of the distributed
load w0 that it can support so that the maximum bending
stress in the beam does not exceed sallow = 22 ksi.
w0
9 ft
9 ft
9 in.
0.25 in.
0.25 in.
12 in.
0.25 in.
Support Reactions. The FBD of the beam is shown in Fig. a.
The shear and moment diagrams are shown in Fig. a and b, respectively. As
indicated on the moment diagram, Mmax = 27wo.
The moment of inertia of the cross-section about the neutral axis is
I =
1
1
(9)(12.53) (8.75)(123)
12
12
= 204.84375 in4
Here, ¢ = 6.25 in. Thus,
sallow =
Mmax c
;
I
22(103) =
(27wo)(12)(6.25)
204.84375
wo = 2 225.46 lb>ft
= 2.23 kip>ft
Ans.
391
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•6–101.
The steel beam has the cross-sectional area
shown. If w0 = 2 kip>ft, determine the maximum bending
stress in the beam.
w0
9 ft
9 ft
9 in.
0.25 in.
0.25 in.
12 in.
0.25 in.
The FBD of the beam is shown in Fig. a
The shear and moment diagrams are shown in Fig. b and c, respectively. As
indicated on the moment diagram, Mmax = 54 kip # ft.
The moment of inertia of the I cross-section about the bending axis is
I =
1
1
(9) A 12.53 B (8.75) A 123 B
12
12
= 204.84375 in4
Here, c = 6.25 in. Thus
smax =
=
Mmax c
I
54 (12)(6.25)
204.84375
= 19.77 ksi = 19.8 ksi
Ans.
392
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6–102. The bolster or main supporting girder of a truck
body is subjected to the uniform distributed load. Determine
the bending stress at points A and B.
1.5 kip/ft
A
8 ft
B
12 ft
F2
F1
0.75 in. 6 in.
12 in.
0.5 in.
A
B
0.75 in.
Support Reactions: As shown on FBD.
Internal Moment: Using the method of sections.
+ ©MNA = 0;
M + 12.0(4) - 15.0(8) = 0
M = 72.0 kip # ft
Section Property:
I =
1
1
(6) A 13.53 B (5.5) A 123 B = 438.1875 in4
12
12
Bending Stress: Applying the flexure formula s =
My
I
sB =
72.0(12)(6.75)
= 13.3 ksi
438.1875
Ans.
sA =
72.0(12)(6)
= 11.8 ksi
438.1875
Ans.
393
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6–103. Determine the largest uniform distributed load w
that can be supported so that the bending stress in the beam
does not exceed s allow = 5 MPa .
w
The FBD of the beam is shown in Fig. a
0.5 m
The shear and moment diagrams are shown in Fig. b and c, respectively. As
indicated on the moment diagram, |Mmax| = 0.125 w.
150 mm
The moment of inertia of the cross-section is,
I =
1
(0.075) A 0.153 B = 21.09375 A 10 - 6 B m4
12
Here, c = 0.075 w. Thus,
sallow =
5 A 106 B =
Mmax c
;
I
0.125w(0.075)
21.09375 A 10 - 6 B
w = 11250 N>m = 11.25 kN>m
Ans.
394
1m
75 mm
0.5 m
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w
*6–104. If w = 10 kN>m, determine the maximum
bending stress in the beam. Sketch the stress distribution
acting over the cross section.
Support Reactions. The FBD of the beam is shown in Fig. a
0.5 m
75 mm
The shear and moment diagrams are shown in Figs. b and c, respectively. As
indicated on the moment diagram, |Mmax| = 1.25 kN # m.
150 mm
The moment of inertia of the cross-section is
I =
1
(0.075) A 0.153 B = 21.09375 A 10 - 6 B m4
12
Here, c = 0.075 m. Thus
smax =
=
Mmax c
I
1.25 A 103 B (0.075)
21.09375 A 10 - 6 B
= 4.444 A 106 B Pa
= 4.44 MPa
Ans.
The bending stress distribution over the cross section is shown in Fig. d
395
1m
0.5 m
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400 lb/ft
•6–105.
If the allowable bending stress for the wood beam
is sallow = 150 psi, determine the required dimension b to
the nearest 14 in. of its cross section. Assume the support at A
is a pin and B is a roller.
B
A
3 ft
The FBD of the beam is shown in Fig. a
The shear and moment diagrams are shown in Figs. b and c, respectively. As
indicated on the moment diagram, Mmax = 3450 lb # ft.
2b
b
The moment of inertia of the cross section is
I =
2
1
(b)(2b)3 = b4
12
3
Here, c = 2b> 2 = b. Thus,
sallow =
150 =
Mmax c
;
I
3450(12)(b)
> 3 b4
2
b = 7.453 in = 7
1
in.
2
Ans.
396
3 ft
3 ft
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400 lb/ft
6–106. The wood beam has a rectangular cross section in
the proportion shown. If b 7.5 in., determine the absolute
maximum bending stress in the beam.
B
A
The FBD of the beam is shown in Fig. a.
3 ft
The shear and moment diagrams are shown in Fig. b and c, respectively. As
indicated on the moment diagram, Mmax = 3450 lb # ft.
2b
b
The moment of inertia of the cross-section is
I =
1
(7.5) A 153 B = 2109.375 in4
12
Here, c =
15
= 7.5 in. Thus
2
smax =
3450(12)(7.5)
Mmax c
=
= 147 psi
I
2109.375
Ans.
397
3 ft
3 ft
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6–107. A beam is made of a material that has a modulus of
elasticity in compression different from that given for
tension. Determine the location c of the neutral axis, and
derive an expression for the maximum tensile stress in the
beam having the dimensions shown if it is subjected to the
bending moment M.
M
h
s
P
Ec(emax)t (h - c)
c
Ec
Location of neutral axis:
+ ©F = 0;
:
1
1
- (h - c)(smax)c (b) + (c)(smax)t (b) = 0
2
2
(h - c)(smax)c = c(smax)t
(h - c)Ec (emax)t
[1]
(h - c)
= cEt (emax)t ;
c
Ec (h - c)2 = Etc2
Taking positive root:
Ec
c
=
h - c
A Et
Ec
h
A Et
h2Ec
c =
=
Ec
2Et + 2Ec
1 +
A Et
[2] Ans.
©MNA = 0;
1
2
1
2
M = c (h - c)(smax)c (b) d a b (h - c) + c (c)(smax)t(b) d a b(c)
2
3
2
3
M =
1
1
(h - c)2 (b)(smax)c + c2b(smax)t
3
3
From Eq. [1]. (smax)c =
c
(s )
h - c max t
M =
c
1
1
(h - c)2 (b)a
b (smax)t + c2b(smax)t
3
h - c
3
M =
1
bc(smax)t (h - c + c) ;
3
(smax)t =
3M
bhc
From Eq. [2]
(smax)t =
b
Et
(emax)t (h - c)
(emax)c =
c
(smax)c = Ec(emax)c =
c
3M 2Et + 2Ec
£
≥
b h2
2Ec
Ans.
398
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*6–108. The beam has a rectangular cross section and is
subjected to a bending moment M. If the material from
which it is made has a different modulus of elasticity for
tension and compression as shown, determine the location c
of the neutral axis and the maximum compressive stress in
the beam.
M
h
s
c
b
Et
P
Ec
See the solution to Prob. 6–107
c =
h2Ec
Ans.
2Et + 2Ec
Since
(smax)c =
(smax)c =
c
(s ) =
h - c max t
2Ec
2Et
h2Ec
( 2Et + 2Ec)ch - a
h 1Ec
1Et + 1Ec
bd
(smax)t
(smax)t
(smax)c =
2Et + 2Ec
2Ec 3M
¢ 2≤¢
≤
bh
2Et
2Ec
(smax)c =
3M 2Et + 2Ec
¢
≤
bh2
2Et
Ans.
399
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•6–109.
The beam is subjected to a bending moment of
M = 20 kip # ft directed as shown. Determine the maximum
bending stress in the beam and the orientation of the
neutral axis.
y
8 in.
C
B
The y and z components of M are negative, Fig. a. Thus,
14 in.
z
My = -20 sin 45° = -14.14 kip # ft
45
16 in.
Mz = -20 cos 45° = -14.14 kip # ft.
The moments of inertia of the cross-section about the principal centroidal y and z
axes are
Iy =
1
1
(16) A 103 B (14) A 83 B = 736 in4
12
12
Iz =
1
1
(10) A 163 B (8) A 143 B = 1584 in4
12
12
My z
Mz y
Iz
+
smax = sC = -
Iy
-14.14(12)(8)
-14.14(12)( - 5)
+
1584
736
= 2.01 ksi
smax = sA = -
(T)
Ans.
-14.14(12)(-8)
-14.14(12)(5)
+
1584
736
= -2.01 ksi = 2.01 ksi (C)
Ans.
Here, u = 180° + 45° = 225°
tan a =
tan a =
Iz
Iy
D
10 in.
M
By inspection, the bending stress occurs at corners A and C are
s = -
A
tan u
1584
tan 225°
736
a = 65.1°
Ans.
The orientation of neutral axis is shown in Fig. b.
400
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6–110. Determine the maximum magnitude of the
bending moment M that can be applied to the beam so that
the bending stress in the member does not exceed 12 ksi.
y
8 in.
C
B
The y and z components of M are negative, Fig. a. Thus,
14 in.
My = -M sin 45° = -0.7071 M
z
45
16 in.
Mz = -M cos 45° = -0.7071 M
The moments of inertia of the cross-section about principal centroidal y and z
axes are
Iy =
1
1
(16) A 103 B (14) A 83 B = 736 in4
12
12
Iz =
1
1
(10) A 163 B (8) A 143 B = 1584 in4
12
12
12 = -
Myzc
Mz yc
Iz
+
D
10 in.
M
By inspection, the maximum bending stress occurs at corners A and C. Here, we
will consider corner C.
sC = sallow = -
A
Iy
-0.7071 M(12)( -5)
-0.7071 M (12)(8)
+
1584
736
M = 119.40 kip # ft = 119 kip # ft
Ans.
401
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6–111. If the resultant internal moment acting on the
cross section of the aluminum strut has a magnitude of
M = 520 N # m and is directed as shown, determine the
bending stress at points A and B. The location y of the
centroid C of the strut’s cross-sectional area must be
determined. Also, specify the orientation of the neutral axis.
y
M 520 Nm
12
20 mm
z
–y
5
13
B
C
200 mm
20 mm
20 mm
A
200 mm
Internal Moment Components:
Mz = -
12
(520) = -480 N # m
13
My =
5
(520) = 200 N # m
13
Section Properties:
0.01(0.4)(0.02) + 2[(0.110)(0.18)(0.02)]
©yA
=
©A
0.4(0.02) + 2(0.18)(0.02)
y =
= 0.057368 m = 57.4 mm
Iz =
Ans.
1
(0.4) A 0.023 B + (0.4)(0.02)(0.057368 - 0.01)2
12
+
1
(0.04) A 0.183 B + 0.04(0.18)(0.110 - 0.057368)2
12
= 57.6014 A 10 - 6 B m4
Iy =
1
1
(0.2) A 0.43 B (0.18) A 0.363 B = 0.366827 A 10 - 3 B m4
12
12
Maximum Bending Stress: Applying the flexure formula for biaxial at points A
and B
s = -
Myz
Mzy
+
Iz
Iy
200(-0.2)
-480(-0.142632)
sA = -
+
-6
57.6014(10 )
0.366827(10 - 3)
= -1.298 MPa = 1.30 MPa (C)
200(0.2)
-480(0.057368)
sB = -
Ans.
+
-6
57.6014(10 )
0.366827(10 - 3)
= 0.587 MPa (T)
Ans.
Orientation of Neutral Axis:
tan a =
tan a =
Iz
Iy
tan u
57.6014(10 - 6)
0.366827(10 - 3)
tan (-22.62°)
a = -3.74°
Ans.
402
200 mm
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*6–112. The resultant internal moment acting on the
cross section of the aluminum strut has a magnitude of
M = 520 N # m and is directed as shown. Determine
maximum bending stress in the strut. The location y of the
centroid C of the strut’s cross-sectional area must be
determined. Also, specify the orientation of the neutral axis.
y
M 520 Nm
12
20 mm
z
–y
5
13
B
C
200 mm
20 mm
20 mm
A
200 mm
Internal Moment Components:
Mz = -
12
(520) = -480 N # m
13
My =
5
(520) = 200 N # m
13
Section Properties:
0.01(0.4)(0.02) + 2[(0.110)(0.18)(0.02)]
©yA
=
©A
0.4(0.02) + 2(0.18)(0.02)
y =
= 0.057368 m = 57.4 mm
Iz =
Ans.
1
(0.4) A 0.023 B + (0.4)(0.02)(0.057368 - 0.01)2
12
1
(0.04) A 0.183 B + 0.04(0.18)(0.110 - 0.057368)2
12
+
= 57.6014 A 10 - 6 B m4
Iy =
1
1
(0.2) A 0.43 B (0.18) A 0.363 B = 0.366827 A 10 - 3 B m4
12
12
Maximum Bending Stress: By inspection, the maximum bending stress can occur at
either point A or B. Applying the flexure formula for biaxial bending at points A
and B
s = -
My z
Mz y
+
Iz
Iy
200(-0.2)
-480(-0.142632)
sA = -
+
57.6014(10 - 6)
0.366827(10 - 3)
= -1.298 MPa = 1.30 MPa (C) (Max)
200(0.2)
-480(0.057368)
sB = -
-6
57.6014(10 )
Ans.
+
0.366827(10 - 3)
= 0.587 MPa (T)
Orientation of Neutral Axis:
tan a =
tan a =
Iz
Iy
tan u
57.6014(10 - 6)
0.366827(10 - 3)
tan (-22.62°)
a = -3.74°
Ans.
403
200 mm
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6–113. Consider the general case of a prismatic beam
subjected to bending-moment components My and Mz,
as shown, when the x, y, z axes pass through the centroid
of the cross section. If the material is linear-elastic, the
normal stress in the beam is a linear function of position
such that s = a + by + cz. Using the equilibrium conditions 0 = 1A s dA, My = 1A zs dA, Mz = 1A - ys dA,
determine the constants a, b, and c, and show that the
normal stress can be determined from the equation
s = [-1MzIy + MyIyz2y + 1MyIz + MzIyz2z]>1IyIz - Iyz22,
where the moments and products of inertia are defined in
Appendix A.
y
z
My
dA
sC
y
Mz
z
Equilibrium Condition: sx = a + by + cz
0 =
LA
sx dA
0 =
LA
(a + by + cz) dA
0 = a
LA
My =
LA
z sx dA
=
LA
z(a + by + cz) dA
= a
dA + b
LA
y dA + c
LA
z dA + b
Mz =
LA
-y sx dA
=
-y(a + by + cz) dA
LA
= -a
LA
ydA - b
LA
LA
z dA
yz dA + c
y2 dA - c
LA
z2 dA
[2]
yz dA
[3]
LA
LA
[1]
Section Properties: The integrals are defined in Appendix A. Note that
LA
y dA =
LA
z dA = 0.Thus,
From Eq. [1]
Aa = 0
From Eq. [2]
My = bIyz + cIy
From Eq. [3]
Mz = -bIz - cIyz
Solving for a, b, c:
a = 0 (Since A Z 0)
b = -¢
Thus,
MzIy + My Iyz
sx = - ¢
Iy Iz - I2yz
≤
Mz Iy + My Iyz
Iy Iz - I2yz
c =
≤y + ¢
My Iz + Mz Iyz
Iy Iz - I2yz
My Iy + MzIyz
Iy Iz - I2yz
≤z
(Q.E.D.)
404
x
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6–114. The cantilevered beam is made from the Z-section
having the cross-section shown. If it supports the two
loadings, determine the bending stress at the wall in the
beam at point A. Use the result of Prob. 6–113.
50 lb
50 lb
3 ft
(My)max = 50(3) + 50(5) = 400 lb # ft = 4.80(103)lb # in.
Iy =
2 ft
0.25 in.
2 in.
1
1
(3.25)(0.25)3 + 2c (0.25)(2)3 + (0.25)(2)(1.125)2 d = 1.60319 in4
12
12
A
B
2.25 in.
1
1
Iz =
(0.25)(3.25)3 + 2 c (2)(0.25)3 + (0.25)(2)(1.5)2 d = 2.970378 in4
12
12
0.25 in.
3 in.
0.25 in.
Iyz = 2[1.5(1.125)(2)(0.25)] = 1.6875 in4
Using the equation developed in Prob. 6-113.
s = -a
sA =
Mz Iy + My Iyz
Iy Iz - I2yz
by + a
My Iz + Mz Iyz
Iy Iz - I2yz
bz
{-[0 + (4.80)(103)(1.6875)](1.625) + [(4.80)(103)(2.970378) + 0](2.125)}
[1.60319(2.970378) - (1.6875)2]
= 8.95 ksi
Ans.
6–115. The cantilevered beam is made from the Z-section
having the cross-section shown. If it supports the two
loadings, determine the bending stress at the wall in the
beam at point B. Use the result of Prob. 6–113.
50 lb
50 lb
3 ft
3
(My)max = 50(3) + 50(5) = 400 lb # ft = 4.80(10 )lb # in.
Iy =
1
1
(3.25)(0.25)3 + 2c (0.25)(2)3 + (0.25)(2)(1.125)2 d = 1.60319 in4
12
12
1
1
Iz =
(0.25)(3.25)3 + 2 c (2)(0.25)3 + (0.25)(2)(1.5)2 d = 2.970378 in4
12
12
2.25 in.
sB =
Iy Iz - I2yz
by + a
My Iz + Mz Iyz
Iy Iz - I2yz
0.25 in.
3 in.
0.25 in.
Using the equation developed in Prob. 6-113.
Mz Iy + My Iyz
A
B
Iyz = 2[1.5(1.125)(2)(0.25)] = 1.6875 in4
s = -a
2 ft
0.25 in.
2 in.
bz
-[0 + (4.80)(103)(1.6875)]( -1.625) + [(4.80)(103)(2.976378) + 0](0.125)
[(1.60319)(2.970378) - (1.6875)2]
= 7.81 ksi
Ans.
405
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*6–116. The cantilevered wide-flange steel beam is
subjected to the concentrated force P at its end. Determine
the largest magnitude of this force so that the bending stress
developed at A does not exceed sallow = 180 MPa.
10 mm
Internal Moment Components: Using method of section
y
200 mm
150 mm
10 mm
©Mz = 0;
Mz + P cos 30°(2) = 0
Mz = -1.732P
©My = 0;
My + P sin 30°(2) = 0
My = -1.00P
10 mm
A
z
Section Properties:
x
2m
30
1
1
Iz =
(0.2) A 0.173 B (0.19) A 0.153 B = 28.44583(10 - 6) m4
12
12
Iy = 2 c
P
1
1
(0.01) A 0.23 B d +
(0.15) A 0.013 B = 13.34583(10 - 6) m4
12
12
Allowable Bending Stress: By inspection, maximum bending stress occurs at points
A and B. Applying the flexure formula for biaxial bending at point A.
sA = sallow = 180 A 106 B = -
Myz
Mzy
Iz
+
Iy
(-1.732P)(0.085)
28.44583(10 - 6)
-1.00P(-0.1)
+
13.34583(10 - 6)
P = 14208 N = 14.2 kN
Ans.
•6–117.
The cantilevered wide-flange steel beam is
subjected to the concentrated force of P = 600 N at its end.
Determine the maximum bending stress developed in the
beam at section A.
10 mm
Internal Moment Components: Using method of sections
y
200 mm
150 mm
10 mm
©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
A
z
Section Properties:
x
1
1
Iz =
(0.2) A 0.173 B (0.19) A 0.153 B = 28.44583(10 - 6) m4
12
12
Iy = 2 c
Maximum Bending Stress: By inspection, maximum bending stress occurs at A and
B. Applying the flexure formula for biaxial bending at point A
s = -
Myz
Mzy
Iz
+
Iy
-600.0(-0.1)
-1039.32(0.085)
sA = -
-6
28.44583(10 )
= 7.60 MPa (T)
+
13.34583(10 - 6)
(Max)
Ans.
406
2m
30
P
1
1
(0.01) A 0.23 B d +
(0.15) A 0.013 B = 13.34583(10 - 6) m4
12
12
10 mm
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6–118. If the beam is subjected to the internal moment of
M = 1200 kN # m, determine the maximum bending stress
acting on the beam and the orientation of the neutral axis.
y
150 mm
150 mm
Internal Moment Components: The y component of M is positive since it is directed
towards the positive sense of the y axis, whereas the z component of M, which is
directed towards the negative sense of the z axis, is negative, Fig. a. Thus,
M
300 mm
30
My = 1200 sin 30° = 600 kN # m
150 mm
Mz = -1200 cos 30° = -1039.23 kN # m
z
x
150 mm
Section Properties: The location of the centroid of the cross-section is given by
©yA
0.3(0.6)(0.3) - 0.375(0.15)(0.15)
=
= 0.2893 m
©A
0.6(0.3) - 0.15(0.15)
y =
150 mm
The moments of inertia of the cross section about the principal centroidal y and z
axes are
Iy =
1
1
(0.6) A 0.33 B (0.15) A 0.153 B = 1.3078 A 10 - 3 B m4
12
12
Iz =
1
(0.3) A 0.63 B + 0.3(0.6)(0.3 - 0.2893)2
12
- c
1
(0.15) A 0.153 B + 0.15(0.15)(0.375 - 0.2893)2 d
12
= 5.2132 A 10 - 3 B m4
Bending Stress: By inspection, the maximum bending stress occurs at either corner
A or B.
s = -
Myz
Mzy
sA = -
+
Iz
Iy
c -1039.23 A 103 B d(0.2893)
5.2132 A 10 - 3 B
+
600 A 103 B (0.15)
1.3078 A 10 - 3 B
= 126 MPa (T)
sB = -
c -1039.23 A 103 B d(-0.3107)
5.2132 A 10 - 3 B
+
600 A 103 B ( -0.15)
1.3078 A 10 - 3 B
= -131 MPa = 131 MPa (C)(Max.)
Ans.
Orientation of Neutral Axis: Here, u = -30°.
tan a =
tan a =
Iz
Iy
tan u
5.2132 A 10 - 3 B
1.3078 A 10 - 3 B
tan(-30°)
a = -66.5°
Ans.
The orientation of the neutral axis is shown in Fig. b.
407
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6–119. If the beam is made from a material having
an allowable tensile and compressive stress of
(sallow)t = 125 MPa and (sallow)c = 150 MPa, respectively,
determine the maximum allowable internal moment M that
can be applied to the beam.
y
150 mm
150 mm
M
300 mm
Internal Moment Components: The y component of M is positive since it is directed
towards the positive sense of the y axis, whereas the z component of M, which is
directed towards the negative sense of the z axis, is negative, Fig. a. Thus,
30
150 mm
z
My = M sin 30° = 0.5M
x
150 mm
Mz = -M cos 30° = -0.8660M
Section Properties: The location of the centroid of the cross section is
y =
150 mm
0.3(0.6)(0.3) - 0.375(0.15)(0.15)
©yA
=
= 0.2893 m
©A
0.6(0.3) - 0.15(0.15)
The moments of inertia of the cross section about the principal centroidal y and z
axes are
Iy =
1
1
(0.6) A 0.33 B (0.15) A 0.153 B = 1.3078 A 10 - 3 B m4
12
12
Iz =
1
(0.3) A 0.63 B + 0.3(0.6)(0.3 - 0.2893)2
12
- c
1
(0.15) A 0.153 B + 0.15(0.15)(0.375 - 0.2893)2 d
12
= 5.2132 A 10 - 3 B m4
Bending Stress: By inspection, the maximum bending stress can occur at either
corner A or B. For corner A which is in tension,
sA = (sallow)t = 125 A 106 B = -
My zA
Mz yA
Iz
+
Iy
(-0.8660M)(0.2893)
5.2132 A 10
-3
B
0.5M(0.15)
+
1.3078 A 10 - 3 B
M = 1185 906.82 N # m = 1186 kN # m (controls)
Ans.
For corner B which is in compression,
sB = (sallow)c = -150 A 106 B = -
My zB
Mz yB
Iz
+
Iy
(-0.8660M)(-0.3107)
5.2132 A 10 - 3 B
0.5M(-0.15)
+
1.3078 A 10 - 3 B
M = 1376 597.12 N # m = 1377 kN # m
408
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*6–120. The shaft is supported on two journal bearings
at A and B which offer no resistance to axial loading.
Determine the required diameter d of the shaft if
the allowable bending stress for the material is
sallow = 150 MPa .
z
y
0.5 m
0.5 m
C
0.5 m
200 N
The FBD of the shaft is shown in Fig. a.
A
200 N 300 N
The shaft is subjected to two bending moment components Mz and My, Figs. b and c,
respectively.
Since all the axes through the centroid of the circular cross-section of the shaft are
principal axes, then the resultant moment M = 2My 2 + Mz 2 can be used for
design. The maximum moment occurs at D (x = 1m). Then,
Mmax = 21502 + 1752 = 230.49 N # m
Then,
sallow =
Mmax C
;
I
150(106) =
230.49(d>2)
p
4
4 (d>2)
d = 0.02501 m = 25 mm
Ans.
409
300 N
0.5 m
D
B
E
x
150 N 150 N
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•6–121.
The 30-mm-diameter shaft is subjected to the
vertical and horizontal loadings of two pulleys as shown. It
is supported on two journal bearings at A and B which offer
no resistance to axial loading. Furthermore, the coupling to
the motor at C can be assumed not to offer any support
to the shaft. Determine the maximum bending stress
developed in the shaft.
1m
1m
1m
1m
A
D
150 N
150 N
Support Reactions: As shown on FBD.
Internal Moment Components: The shaft is subjected to two bending moment
components My and Mz. The moment diagram for each component is drawn.
Maximum Bending Stress: Since all the axes through the circle’s center for circular
shaft are principal axis, then the resultant moment M = 2My 2 + Mz 2 can be used
to determine the maximum bending stress. The maximum resultant moment occurs
at E Mmax = 24002 + 1502 = 427.2 N # m.
Applying the flexure formula
Mmax c
I
427.2(0.015)
=
p
4
A 0.0154 B
= 161 MPa
Ans.
410
E
C
B
400 N
100 mm
400 N
60 mm
x
smax =
y
z
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6–122. Using the techniques outlined in Appendix A,
Example A.5 or A.6, the Z section has principal moments of
inertia of Iy = 0.060110-32 m4 and Iz = 0.471110-32 m4,
computed about the principal axes of inertia y and z,
respectively. If the section is subjected to an internal
moment of M = 250 N # m directed horizontally as shown,
determine the stress produced at point A. Solve the
problem using Eq. 6–17.
50 mm
y
A
200 mm
32.9
y¿
250 Nm
z
My = 250 cos 32.9° = 209.9 N # m
z¿
300 mm
Mz = 250 sin 32.9° = 135.8 N # m
200 mm
50 mm
B
50 mm
y = 0.15 cos 32.9° + 0.175 sin 32.9° = 0.2210 m
z = -(0.175 cos 32.9° - 0.15 sin 32.9°) = -0.06546 m
sA = -
Myz
Mzy
+
Iz
Iy
209.9(-0.06546)
-135.8(0.2210)
=
0.471(10 - 3)
+
60.0(10 - 6)
= -293 kPa = 293 kPa (C)
Ans.
6–123. Solve Prob. 6–122 using the equation developed in
Prob. 6–113.
50 mm
y
A
Internal Moment Components:
My = 250 N # m
200 mm
Mz = 0
32.9
y¿
Section Properties:
Iy =
250 Nm
1
1
(0.3) A 0.053 B + 2c (0.05) A 0.153 B + 0.05(0.15) A 0.12 B d
12
12
= 0.18125 A 10
Iz =
-3
z
z¿
300 mm
Bm
4
1
1
(0.05) A 0.33 B + 2c (0.15) A 0.053 B + 0.15(0.05) A 0.1252 B d
12
12
= 0.350(10 - 3) m4
Iyz = 0.15(0.05)(0.125)(-0.1) + 0.15(0.05)(-0.125)(0.1)
= -0.1875 A 10 - 3 B m4
Bending Stress: Using formula developed in Prob. 6-113
s =
sA =
-(Mz Iy + My Iyz)y + (My Iz + MzIyz)z
IyIz - I2yz
-[0 + 250(-0.1875)(10 - 3)](0.15) + [250(0.350)(10 - 3) + 0](-0.175)
0.18125(10 - 3)(0.350)(10 - 3) - [0.1875(10 - 3)]2
= -293 kPa = 293 kPa (C)
Ans.
411
200 mm
50 mm
B
50 mm
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*6–124. Using the techniques outlined in Appendix A,
Example A.5 or A.6, the Z section has principal moments of
inertia of Iy = 0.060110-32 m4 and Iz = 0.471110-32 m4,
computed about the principal axes of inertia y and z,
respectively. If the section is subjected to an internal
moment of M = 250 N # m directed horizontally as shown,
determine the stress produced at point B. Solve the
problem using Eq. 6–17.
50 mm
y
A
200 mm
32.9
y¿
250 Nm
z
z¿
300 mm
Internal Moment Components:
My¿ = 250 cos 32.9° = 209.9 N # m
Mz¿ = 250 sin 32.9° = 135.8 N # m
Section Property:
y¿ = 0.15 cos 32.9° + 0.175 sin 32.9° = 0.2210 m
z¿ = 0.15 sin 32.9° - 0.175 cos 32.9° = -0.06546 m
Bending Stress: Applying the flexure formula for biaxial bending
s =
sB =
My¿z¿
Mz¿y¿
Iz¿
+
Iy¿
209.9(-0.06546)
135.8(0.2210)
0.471(10 - 3)
-
0.060(10 - 3)
= 293 kPa = 293 kPa (T)
Ans.
412
200 mm
50 mm
B
50 mm
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z
•6–125. Determine the bending stress at point A of the
beam, and the orientation of the neutral axis. Using the
method in Appendix A, the principal moments of inertia of
the cross section are I¿z = 8.828 in4 and I¿y = 2.295 in4,
where z¿ and y¿ are the principal axes. Solve the problem
using Eq. 6–17.
1.183 in.
0.5 in.
z¿
A
4 in.
45
C
y
1.183 in.
0.5 in.
M 3 kip ft
y′
4 in.
Internal Moment Components: Referring to Fig. a, the y¿ and z¿ components of M
are negative since they are directed towards the negative sense of their respective
axes. Thus,
Section Properties: Referring to the geometry shown in Fig. b,
œ
= 2.817 cos 45° - 1.183 sin 45° = 1.155 in.
zA
œ
yA
= -(2.817 sin 45° + 1.183 cos 45°) = -2.828 in.
Bending Stress:
sA = -
= -
œ
My¿zA
œ
Mz¿yA
Iz¿
+
Iy¿
(-2.121)(12)(-2.828)
(-2.121)(12)(1.155)
+
8.828
2.295
= -20.97 ksi = 21.0 ksi (C)
Ans.
413
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z
6–126. Determine the bending stress at point A of the
beam using the result obtained in Prob. 6–113. The moments
of inertia of the cross sectional area about the z and y axes
are Iz = Iy = 5.561 in4 and the product of inertia of the
cross sectional area with respect to the z and y axes is
Iyz 3.267 in4. (See Appendix A)
1.183 in.
0.5 in.
z¿
A
4 in.
45
C
y
1.183 in.
0.5 in.
M 3 kip ft
y′
4 in.
Internal Moment Components: Since M is directed towards the negative sense of the y axis, its
y component is negative and it has no z component. Thus,
My = -3 kip # ft
Mz = 0
Bending Stress:
sA =
=
- A MzIy + MyIyz B yA + A MyIz + MzIyz B zA
IyIz - Iyz 2
- C 0(5.561) + (-3)(12)(-3.267) D (-1.183) + C -3(12)(5.561) + 0(-3.267) D (2.817)
5.561(5.561) - (-3.267)2
= -20.97 ksi = 21.0 ksi
Ans.
414
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6–127. The composite beam is made of 6061-T6 aluminum
(A) and C83400 red brass (B). Determine the dimension h of
the brass strip so that the neutral axis of the beam is located
at the seam of the two metals. What maximum moment will
this beam support if the allowable bending stress for the
aluminum is 1sallow2al = 128 MPa and for the brass
1sallow2br = 35 MPa?
h
B
A
150 mm
Section Properties:
n =
68.9(109)
Eal
= 0.68218
=
Ebr
101(109)
bbr = nbal = 0.68218(0.15) = 0.10233 m
y =
0.05 =
©yA
©A
0.025(0.10233)(0.05) + (0.05 + 0.5h)(0.15)h
0.10233(0.05) + (0.15)h
h = 0.04130 m = 41.3 mm
INA =
Ans.
1
(0.10233) A 0.053 B + 0.10233(0.05)(0.05 - 0.025)2
12
+
1
(0.15) A 0.041303 B + 0.15(0.04130)(0.070649 - 0.05)2
12
= 7.7851 A 10 - 6 B m4
Allowable Bending Stress: Applying the flexure formula
Assume failure of red brass
(sallow)br =
35 A 106 B =
Mc
INA
M(0.04130)
7.7851(10 - 6)
M = 6598 N # m = 6.60 kN # m (controls!)
Ans.
Assume failure of aluminium
(sallow)al = n
Mc
INA
128 A 106 B = 0.68218c
M(0.05)
7.7851(10 - 6)
d
M = 29215 N # m = 29.2 kN # m
415
50 mm
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*6–128. The composite beam is made of 6061-T6 aluminum
(A) and C83400 red brass (B). If the height h = 40 mm,
determine the maximum moment that can be applied to the
beam if the allowable bending stress for the aluminum is
1sallow2al = 128 MPa and for the brass 1sallow2br = 35 MPa.
h
B
A
Section Properties: For transformed section.
150 mm
68.9(109)
Eal
= 0.68218
=
n =
Ebr
101.0(109)
bbr = nbal = 0.68218(0.15) = 0.10233 m
y =
=
©yA
©A
0.025(0.10233)(0.05) + (0.07)(0.15)(0.04)
0.10233(0.05) + 0.15(0.04)
= 0.049289 m
INA =
1
(0.10233) A 0.053 B + 0.10233(0.05)(0.049289 - 0.025)2
12
+
1
(0.15) A 0.043 B + 0.15(0.04)(0.07 - 0.049289)2
12
= 7.45799 A 10 - 6 B m4
Allowable Bending Stress: Applying the flexure formula
Assume failure of red brass
(sallow)br =
35 A 106 B =
Mc
INA
M(0.09 - 0.049289)
7.45799(10 - 6)
M = 6412 N # m = 6.41 kN # m (controls!)
Ans.
Assume failure of aluminium
(sallow)al = n
Mc
INA
128 A 106 B = 0.68218c
M(0.049289)
7.45799(10 - 6)
d
M = 28391 N # m = 28.4 kN # m
416
50 mm
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•6–129. Segment A of the composite beam is made from
2014-T6 aluminum alloy and segment B is A-36 steel. If
w = 0.9 kip>ft, determine the absolute maximum bending
stress developed in the aluminum and steel. Sketch the
stress distribution on the cross section.
w
15 ft
A
3 in.
B
3 in.
3 in.
Maximum Moment: For the simply-supported beam subjected to the uniform
0.9 A 152 B
wL2
=
distributed load, the maximum moment in the beam is Mmax =
8
8
= 25.3125 kip # ft.
Section Properties: The cross section will be transformed into that of steel as
Eal
10.6
=
= 0.3655.
shown in Fig. a. Here, n =
Est
29
Then bst = nbal = 0.3655(3) = 1.0965 in. The location of the centroid of the
transformed section is
y =
©yA
1.5(3)(3) + 4.5(3)(1.0965)
=
= 2.3030 in.
©A
3(3) + 3(1.0965)
The moment of inertia of the transformed section about the neutral axis is
I = ©I + Ad2 =
1
(3) A 33 B + 3(3)(2.3030 - 1.5)2
12
+
1
(1.0965) A 33 B + 1.0965(3)(4.5 - 2.3030)2
12
= 30.8991 in4
Maximum Bending Stress: For the steel,
(smax)st =
25.3125(12)(2.3030)
Mmaxcst
=
= 22.6 ksi
I
30.8991
Ans.
At the seam,
ssty = 0.6970 in. =
Mmaxy
25.3125(12)(0.6970)
=
= 6.85 ksi
I
30.8991
For the aluminium,
(smax)al = n
25.3125(12)(6 - 2.3030)
Mmaxcal
= 0.3655c
d = 13.3 ksi
I
30.8991
Ans.
At the seam,
saly = 0.6970 in. = n
Mmaxy
25.3125(12)(0.6970)
= 0.3655c
d = 2.50 ksi
I
30.8991
The bending stress across the cross section of the composite beam is shown in Fig. b.
417
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6–130. Segment A of the composite beam is made from
2014-T6 aluminum alloy and segment B is A-36 steel. If
the allowable bending stress for the aluminum and steel
are (sallow)al = 15 ksi and (sallow)st = 22 ksi, determine
the maximum allowable intensity w of the uniform
distributed load.
w
15 ft
A
3 in.
B
3 in.
3 in.
Maximum Moment: For the simply-supported beam subjected to the uniform
distributed load, the maximum moment in the beam is
w A 152 B
wL2
=
= 28.125w.
Mmax =
8
8
Section Properties: The cross section will be transformed into that of steel as
Eal
10.6
=
= 0.3655.
shown in Fig. a. Here, n =
Est
29
Then bst = nbal = 0.3655(3) = 1.0965 in. The location of the centroid of the
transformed section is
y =
©yA
1.5(3)(3) + 4.5(3)(1.0965)
=
= 2.3030 in.
©A
3(3) + 3(1.0965)
The moment of inertia of the transformed section about the neutral axis is
I = ©I + Ad2 =
1
1
(3) A 33 B + 3(3)(2.3030 - 1.5)2 +
(1.0965) A 33 B
12
12
+ 1.0965 A 33 B + 1.0965(3)(4.5 - 2.3030)2
= 30.8991 in4
Bending Stress: Assuming failure of steel,
(sallow)st =
Mmax cst
;
I
22 =
(28.125w)(12)(2.3030)
30.8991
w = 0.875 kip>ft (controls)
Ans.
Assuming failure of aluminium alloy,
(sallow)al = n
Mmax cal
;
I
15 = 0.3655c
(28.125w)(12)(6 - 2.3030)
d
30.8991
w = 1.02 kip>ft
418
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6–131. The Douglas fir beam is reinforced with A-36
straps at its center and sides. Determine the maximum
stress developed in the wood and steel if the beam
is subjected to a bending moment of Mz = 7.50 kip # ft.
Sketch the stress distribution acting over the cross section.
y
0.5 in.
0.5 in.
0.5 in.
z
6 in.
2 in.
Section Properties: For the transformed section.
n =
1.90(103)
Ew
= 0.065517
=
Est
29.0(103)
bst = nbw = 0.065517(4) = 0.26207 in.
INA =
1
(1.5 + 0.26207) A 63 B = 31.7172 in4
12
Maximum Bending Stress: Applying the flexure formula
(smax)st =
7.5(12)(3)
Mc
=
= 8.51 ksi
I
31.7172
(smax)w = n
Ans.
7.5(12)(3)
Mc
= 0.065517c
d = 0.558 ksi
I
31.7172
Ans.
419
2 in.
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*6–132. The top plate is made of 2014-T6 aluminum and is
used to reinforce a Kevlar 49 plastic beam. Determine the
maximum stress in the aluminum and in the Kevlar if the
beam is subjected to a moment of M = 900 lb # ft.
6 in.
0.5 in.
0.5 in.
12 in.
M
0.5 in.
0.5 in.
Section Properties:
n =
10.6(103)
Eal
= 0.55789
=
Ek
19.0(103)
bk = n bal = 0.55789(12) = 6.6947 in.
y =
0.25(13)(0.5) + 2[(3.25)(5.5)(0.5)] + 5.75(6.6947)(0.5)
©yA
=
©A
13(0.5) + 2(5.5)(0.5) + 6.6947(0.5)
= 2.5247 in.
INA =
1
(13) A 0.53 B + 13(0.5)(2.5247 - 0.25)2
12
+
1
(1) A 5.53 B + 1(5.5)(3.25 - 2.5247)2
12
+
1
(6.6947) A 0.53 B + 6.6947(0.5)(5.75 - 2.5247)2
12
= 85.4170 in4
Maximum Bending Stress: Applying the flexure formula
(smax)al = n
(smax)k =
900(12)(6 - 2.5247)
Mc
= 0.55789 c
d = 245 psi
I
85.4170
900(12)(6 - 2.5247)
Mc
=
= 439 psi
I
85.4168
Ans.
Ans.
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•6–133.
The top plate made of 2014-T6 aluminum is used
to reinforce a Kevlar 49 plastic beam. If the allowable
bending stress for the aluminum is (sallow)al = 40 ksi and
for the Kevlar (sallow)k = 8 ksi, determine the maximum
moment M that can be applied to the beam.
6 in.
0.5 in.
0.5 in.
Section Properties:
n =
10.6(103)
Eal
= 0.55789
=
Ek
19.0(103)
12 in.
bk = n bal = 0.55789(12) = 6.6947 in.
y =
0.5 in.
© yA
0.25(13)(0.5) + 2[(3.25)(5.5)(0.5)] + 5.75(6.6947(0.5)
=
©A
13(0.5) + 2(5.5)(0.5) + 6.6947(0.5)
= 2.5247 in.
INA =
1
(13) A 0.53 B + 13(0.5)(2.5247 - 0.25)2
12
+
1
(1) A 5.53 B + 1(5.5)(3.25 - 2.5247)2
12
+
1
(6.6947) A 0.53 B + 6.6947(0.5)(5.75 - 2.5247)2
12
= 85.4170 in4
Maximum Bending Stress: Applying the flexure formula
Assume failure of aluminium
(sallow)al = n
Mc
I
40 = 0.55789 c
M(6 - 2.5247)
d
85.4170
M = 1762 kip # in = 146.9 kip # ft
Assume failure of Kevlar 49
(sallow)k =
8 =
Mc
I
M(6 - 2.5247)
85.4170
M = 196.62 kip # in
= 16.4 kip # ft
M
0.5 in.
(Controls!)
Ans.
421
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6–134. The member has a brass core bonded to a steel
casing. If a couple moment of 8 kN # m is applied at its end,
determine the maximum bending stress in the member.
Ebr = 100 GPa, Est = 200 GPa.
8 kNm
3m
20 mm
100 mm
20 mm
n =
Ebr
100
=
= 0.5
Est
200
I =
1
1
(0.14)(0.14)3 (0.05)(0.1)3 = 27.84667(10 - 6)m4
12
12
20 mm
100 mm
20 mm
Maximum stress in steel:
(sst)max =
8(103)(0.07)
Mc1
= 20.1 MPa
=
I
27.84667(10 - 6)
Ans.
(max)
Maximum stress in brass:
(sbr)max =
0.5(8)(103)(0.05)
nMc2
= 7.18 MPa
=
I
27.84667(10 - 6)
6–135. The steel channel is used to reinforce the wood
beam. Determine the maximum stress in the steel and in
the wood if the beam is subjected to a moment of
M = 850 lb # ft. Est = 29(103) ksi, Ew = 1600 ksi.
y =
4 in.
0.5 in.
(0.5)(16)(0.25) + 2(3.5)(0.5)(2.25) + (0.8276)(3.5)(2.25)
= 1.1386 in.
0.5(16) + 2(3.5)(0.5) + (0.8276)(3.5)
15 in.
M 850 lbft
0.5 in.
1
1
I =
(16)(0.53) + (16)(0.5)(0.88862) + 2 a b(0.5)(3.53) + 2(0.5)(3.5)(1.11142)
12
12
+
1
(0.8276)(3.53) + (0.8276)(3.5)(1.11142) = 20.914 in4
12
Maximum stress in steel:
(sst) =
850(12)(4 - 1.1386)
Mc
=
= 1395 psi = 1.40 ksi
I
20.914
Ans.
Maximum stress in wood:
(sw) = n(sst)max
= 0.05517(1395) = 77.0 psi
Ans.
422
0.5 in.
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*6–136. A white spruce beam is reinforced with A-36 steel
straps at its top and bottom as shown. Determine the
bending moment M it can support if (sallow)st = 22 ksi
and (sallow)w = 2.0 ksi.
y
0.5 in.
4 in.
M
0.5 in.
x
z
3 in.
Section Properties: For the transformed section.
n =
1.40(103)
Ew
= 0.048276
=
Est
29.0(103)
bst = nbw = 0.048276(3) = 0.14483 in.
INA =
1
1
(3) A 53 B (3 - 0.14483) A 43 B = 16.0224 in4
12
12
Allowable Bending Stress: Applying the flexure formula
Assume failure of steel
(sallow)st =
22 =
Mc
I
M(2.5)
16.0224
M = 141.0 kip # in
= 11.7 kip # ft (Controls !)
Ans.
Assume failure of wood
(sallow)w = n
My
I
2.0 = 0.048276 c
M(2)
d
16.0224
M = 331.9 kip # in = 27.7 kip # ft
423
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•6–137. If the beam is subjected to an internal moment of
M = 45 kN # m, determine the maximum bending stress
developed in the A-36 steel section A and the 2014-T6
aluminum alloy section B.
A
50 mm
M
15 mm
150 mm
Section Properties: The cross section will be transformed into that of steel as shown in Fig. a.
73.1 A 109 B
Eal
=
= 0.3655. Thus, bst = nbal = 0.3655(0.015) = 0.0054825 m. The
Here, n =
Est
200 A 109 B
location of the transformed section is
©yA
y =
=
©A
0.075(0.15)(0.0054825) + 0.2cp A 0.052 B d
0.15(0.0054825) + p A 0.052 B
= 0.1882 m
The moment of inertia of the transformed section about the neutral axis is
I = ©I + Ad2 =
1
(0.0054825) A 0.153 B + 0.0054825(0.15)(0.1882 - 0.075)2
12
+
1
p A 0.054 B + p A 0.052 B (0.2 - 0.1882)2
4
= 18.08 A 10 - 6 B m4
Maximum Bending Stress: For the steel,
45 A 10 B (0.06185)
Mcst
=
= 154 MPa
I
18.08 A 10 - 6 B
3
(smax)st =
Ans.
For the aluminum alloy,
45 A 10 B (0.1882)
Mcal
= 0.3655 C
S = 171 MPa
I
18.08 A 10 - 6 B
3
(smax)al = n
424
Ans.
B
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6–138. The concrete beam is reinforced with three 20-mm
diameter steel rods. Assume that the concrete cannot
support tensile stress. If the allowable compressive stress
for concrete is (sallow)con = 12.5 MPa and the allowable
tensile stress for steel is (sallow)st = 220 MPa, determine
the required dimension d so that both the concrete and steel
achieve their allowable stress simultaneously. This condition
is said to be ‘balanced’. Also, compute the corresponding
maximum allowable internal moment M that can be applied
to the beam. The moduli of elasticity for concrete and steel
are Econ = 25 GPa and Est = 200 GPa, respectively.
200 mm
M
Bending Stress: The cross section will be transformed into that of concrete as shown
Est
200
=
= 8. It is required that both concrete and steel
in Fig. a. Here, n =
Econ
25
achieve their allowable stress simultaneously. Thus,
(sallow)con =
12.5 A 106 B =
Mccon
;
I
Mccon
I
M = 12.5 A 106 B ¢
(sallow)st = n
I
≤
ccon
220 A 106 B = 8 B
Mcst
;
I
(1)
M(d - ccon)
R
I
M = 27.5 A 106 B ¢
I
≤
d - ccon
(2)
Equating Eqs. (1) and (2),
12.5 A 106 B ¢
I
I
≤ = 27.5 A 106 B ¢
≤
ccon
d - ccon
ccon = 0.3125d (3)
Section Properties: The area of the steel bars is Ast = 3c
(3)
p
A 0.022 B d = 0.3 A 10 - 3 B p m2.
4
Thus, the transformed area of concrete from steel is (Acon)t = nAs = 8 C 0.3 A 10 - 3 B p D
= 2.4 A 10 - 3 B p m2. Equating the first moment of the area of concrete above and below
the neutral axis about the neutral axis,
0.2(ccon)(ccon>2) = 2.4 A 10 - 3 B p (d - ccon)
0.1ccon 2 = 2.4 A 10 - 3 B pd - 2.4 A 10 - 3 B pccon
ccon 2 = 0.024pd - 0.024pccon
(4)
Solving Eqs. (3) and (4),
d = 0.5308 m = 531 mm
Ans.
ccon = 0.1659 m
Thus, the moment of inertia of the transformed section is
I =
1
(0.2) A 0.16593 B + 2.4 A 10 - 3 B p(0.5308 - 0.1659)2
3
425
d
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6–138.
Continued
= 1.3084 A 10 - 3 B m4
Substituting this result into Eq. (1),
M = 12.5 A 106 B C
1.3084 A 10 - 3 B
0.1659
S
= 98 594.98 N # m = 98.6 kN # m‚
Ans.
6–139. The beam is made from three types of plastic that
are identified and have the moduli of elasticity shown in the
figure. Determine the maximum bending stress in the PVC.
(bbk)1 = n1 bEs =
160
(3) = 0.6 in.
800
(bbk)2 = n2 bpvc =
450
(3) = 1.6875 in.
800
500 lb
PVC EPVC 450 ksi
Escon EE 160 ksi
Bakelite EB 800 ksi
3 ft
y =
©yA
(1)(3)(2) + 3(0.6)(2) + 4.5(1.6875)(1)
=
= 1.9346 in.
©A
3(2) + 0.6(2) + 1.6875(1)
I =
1
1
(3)(23) + 3(2)(0.93462) +
(0.6)(23) + 0.6(2)(1.06542)
12
12
+
4 ft
1 in.
2 in.
2 in.
3 in.
1
(1.6875)(13) + 1.6875(1)(2.56542) = 20.2495 in4
12
(smax)pvc = n2
500 lb
450 1500(12)(3.0654)
Mc
= a
b
I
800
20.2495
= 1.53 ksi
Ans.
426
3 ft
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*6–140. The low strength concrete floor slab is integrated
with a wide-flange A-36 steel beam using shear studs (not
shown) to form the composite beam. If the allowable
bending stress for the concrete is (sallow)con = 10 MPa, and
allowable bending stress for steel is (sallow)st = 165 MPa,
determine the maximum allowable internal moment M that
can be applied to the beam.
1m
100 mm
15 mm
400 mm
M
15 mm
15 mm
Section Properties: The beam cross section will be transformed into
Econ
22.1
that
of
steel.
Here,
Thus,
=
= 0.1105.
n =
Est
200
bst = nbcon = 0.1105(1) = 0.1105 m. The location of the transformed section is
y =
=
©yA
©A
0.0075(0.015)(0.2) + 0.2(0.37)(0.015) + 0.3925(0.015)(0.2) + 0.45(0.1)(0.1105)
0.015(0.2) + 0.37(0.015) + 0.015(0.2) + 0.1(0.1105)
= 0.3222 m
The moment of inertia of the transformed section about the neutral axis is
I = ©I + Ad2 =
1
(0.2) A 0.0153 B
12
+ 0.2(0.015)(0.3222 - 0.0075)2
+
1
(0.015) A 0.373 B + 0.015(0.37)(0.3222 - 0.2)2
12
+
1
(0.2) A 0.0153 B + 0.2(0.015)(0.3925 - 0.3222)2
12
+
1
(0.1105) A 0.13 B + 0.1105(0.1)(0.45 - 0.3222)2
12
= 647.93 A 10 - 6 B m4
Bending Stress: Assuming failure of steel,
(sallow)st =
M(0.3222)
Mcst
; 165 A 106 B =
I
647.93 A 10 - 6 B
M = 331 770.52 N # m = 332 kN # m
Assuming failure of concrete,
(sallow)con = n
Mccon
;
I
10 A 106 B = 0.1105C
M(0.5 - 0.3222)
647.93 A 10 - 6 B
S
M = 329 849.77 N # m = 330 kN # m (controls) Ans.
427
200 mm
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•6–141.
The reinforced concrete beam is used to support
the loading shown. Determine the absolute maximum
normal stress in each of the A-36 steel reinforcing rods and
the absolute maximum compressive stress in the concrete.
Assume the concrete has a high strength in compression
and yet neglect its strength in supporting tension.
10 kip
8 in.
15 in.
4 ft
8 ft
Mmax = (10 kip)(4 ft) = 40 kip # ft
Ast = 3(p)(0.5)2 = 2.3562 in2
Est = 29.0(103) ksi
Econ = 4.20(103) ksi
A¿ = nAst =
©yA = 0;
29.0(103)
4.20(103)
8(h¿)a
(2.3562) = 16.2690 in2
h¿
b - 16.2690(13 - h¿) = 0
2
h¿ 2 + 4.06724h - 52.8741 = 0
Solving for the positive root:
h¿ = 5.517 in.
I = c
1
(8)(5.517)3 + 8(5.517)(5.517>2)2 d + 16.2690(13 - 5.517)2
12
= 1358.781 in4
(scon)max =
My
40(12)(5.517)
=
= 1.95 ksi
I
1358.781
Ans.
My
29.0(103) 40(12)(13 - 5.517)
ba
b = a
b = 18.3 ksi
I
1358.781
4.20(103)
Ans.
(sst)max = na
10 kip
428
4 ft
2 in.
1 in. diameter rods
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6–142. The reinforced concrete beam is made using two
steel reinforcing rods. If the allowable tensile stress
for the steel is (sst)allow = 40 ksi and the allowable
compressive stress for the concrete is (sconc)allow = 3 ksi,
determine the maximum moment M that can be applied to
the section. Assume the concrete cannot support a tensile
stress. Est = 29(103) ksi, Econc = 3.8(103) ksi.
8 in. 6 in.
4 in.
8 in.
M
18 in.
2 in.
1-in. diameter rods
Ast = 2(p)(0.5)2 = 1.5708 in2
A¿ = nAst =
©yA = 0;
29(103)
3.8(103)
(1.5708) = 11.9877 in2
22(4)(h¿ + 2) + h¿(6)(h¿>2) - 11.9877(16 - h¿) = 0
3h2 + 99.9877h¿ - 15.8032 = 0
Solving for the positive root:
h¿ = 0.15731 in.
I = c
1
1
(22)(4)3 + 22(4)(2.15731)2 d + c (6)(0.15731)3 + 6(0.15731)(0.15731>2)2 d
12
12
+ 11.9877(16 - 0.15731)2 = 3535.69 in4
Assume concrete fails:
(scon)allow =
My
;
I
3 =
M(4.15731)
3535.69
M = 2551 kip # in.
Assume steel fails:
(sst)allow = na
My
b;
I
40 = ¢
29(103)
3
3.8(10 )
≤¢
M(16 - 0.15731)
≤
3535.69
M = 1169.7 kip # in. = 97.5 kip # ft (controls) Ans.
429
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6–143. For the curved beam in Fig. 6–40a, show that when
the radius of curvature approaches infinity, the curved-beam
formula, Eq. 6–24, reduces to the flexure formula, Eq. 6–13.
Normal Stress: Curved-beam formula
M(R - r)
s =
where A¿ =
Ar(r - R)
dA
LA r
and R =
A
dA
1A 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 y ≤ dA
r LA 1 + r
r LA 1 + r
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
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6–143. Continued
y
=
y
As
r
y
Mr
dA
r
C
¢
¢
y ≤ dA y≤ S
AI LA 1 + r
r LA 1 + r
=
: 0
LA
¢
y
r
≤ dA = 0
1 + yr
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
=
dA
1A 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)
Ans.
= 224 kPa (C)
Ans.
No, because of localized stress concentration at the wall.
Ans.
431
M
Mr
AI
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•6–145. The member has an elliptical cross section. If the
allowable bending stress is sallow = 125 MPa determine the
maximum moment M that can be applied to the member.
75 mm
150 mm
A¿
250
mm
A
100 mm
B
b = 0.0375 m
a = 0.075 m;
A = p(0.075)(0.0375) = 0.0028125 p
2p(0.0375)
dA
2pb
(0.175 - 20.1752 - 0.0752)
=
(r - 2r2 - a2) =
r
a
0.075
LA
= 0.053049301 m
R =
A
dA
1A r
=
0.0028125p
= 0.166556941 m
0.053049301
r - R = 0.175 - 0.166556941 = 8.4430586(10 - 3) m
s =
M(R - r)
Ar(r - R)
Assume tension failure.
125(106) =
M(0.166556941 - 0.1)
0.0028125p(0.1)(8.4430586)(10 - 3)
M = 14.0 kN # m (controls)
Ans.
Assume compression failure:
-125(106) =
M(0.166556941 - 0.25)
0.0028125p(0.25)(8.4430586)(10 - 3)
M = 27.9 kN # m
432
M
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6–146. Determine the greatest magnitude of the applied
forces P if the allowable bending stress is (sallow)c = 50 MPa
in compression and (sallow)t = 120 MPa in tension.
75 mm
P
10 mm
10 mm
160 mm
10 mm
P
150 mm
250 mm
Internal Moment: M = 0.160P is positive since it tends to increase the beam’s
radius of curvature.
Section Properties:
r =
=
©yA
©A
0.255(0.15)(0.01) + 0.335(0.15)(0.01) + 0.415(0.075)(0.01)
0.15(0.01) + 0.15(0.01) + 0.075(0.01)
= 0.3190 m
A = 0.15(0.01) + 0.15(0.01) + 0.075(0.01) = 0.00375 m2
©
dA
0.26
0.41
0.42
= 0.15 ln
+ 0.01 ln
+ 0.075 ln
0.25
0.26
0.41
LA r
= 0.012245 m
R =
A
© 1A dA
r
=
0.00375
= 0.306243 m
0.012245
r - R = 0.319 - 0.306243 = 0.012757 m
Allowable Normal Stress: Applying the curved-beam formula
Assume tension failure
(sallow)t =
120 A 106 B =
M(R - r)
Ar(r - R)
0.16P(0.306243 - 0.25)
0.00375(0.25)(0.012757)
P = 159482 N = 159.5 kN
Assume compression failure
(sallow)t =
-50 A 106 B =
M(R - r)
Ar(r - R)
0.16P(0.306243 - 0.42)
0.00375(0.42)(0.012757)
P = 55195 N = 55.2 kN (Controls !)
Ans.
433
150 mm
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6–147. If P = 6 kN, determine the maximum tensile and
compressive bending stresses in the beam.
75 mm
P
10 mm
10 mm
160 mm
10 mm
P
150 mm
250 mm
Internal Moment: M = 0.160(6) = 0.960 kN # m is positive since it tends to increase
the beam’s radius of curvature.
Section Properties:
r =
=
©yA
©A
0.255(0.15)(0.01) + 0.335(0.15)(0.01) + 0.415(0.075)(0.01)
0.15(0.01) + 0.15(0.01) + 0.075(0.01)
= 0.3190 m
A = 0.15(0.01) + 0.15(0.01) + 0.075(0.01) = 0.00375 m2
©
dA
0.41
0.42
0.26
= 0.15 ln
+ 0.01 ln
+ 0.075 ln
0.25
0.26
0.41
LA r
= 0.012245 m
R =
A
©1A dA
r
=
0.00375
= 0.306243 m
0.012245
r - R = 0.319 - 0.306243 = 0.012757 m
Normal Stress: Applying the curved-beam formula
(smax)t =
=
M(R - r)
Ar(r - R)
0.960(103)(0.306243 - 0.25)
0.00375(0.25)(0.012757)
= 4.51 MPa
(smax)c =
=
Ans.
M(R - r)
Ar(r - R)
0.960(103)(0.306243 - 0.42)
0.00375(0.42)(0.012757)
= -5.44 MPa
Ans.
434
150 mm
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*6–148. The curved beam is subjected to a bending
moment of M = 900 N # m as shown. Determine the stress
at points A and B, and show the stress on a volume element
located at each of these points.
A C
B
100 mm
C
A
30
20 mm
15 mm
150 mm
400 mm
B
M
Internal Moment: M = -900 N # m is negative since it tends to decrease the beam’s
radius curvature.
Section Properties:
©A = 0.15(0.015) + 0.1(0.02) = 0.00425 m2
©rA = 0.475(0.15)(0.015) + 0.56(0.1)(0.02) = 2.18875(10 - 3) m3
r =
©
2.18875 (10 - 3)
©rA
=
= 0.5150 m
©A
0.00425
dA
0.57
0.55
= 0.015 ln
+ 0.1 ln
= 8.348614(10 - 3) m
0.4
0.55
LA r
R =
A
©1A dA
r
=
0.00425
= 0.509067 m
8.348614(10 - 3)
r - R = 0.515 - 0.509067 = 5.933479(10 - 3) m
Normal Stress: Applying the curved-beam formula
sA =
M(R - rA)
-900(0.509067 - 0.57)
=
ArA (r - R)
0.00425(0.57)(5.933479)(10 - 3)
Ans.
= 3.82 MPa (T)
sB =
M(R - rB)
-900(0.509067 - 0.4)
=
ArB (r - R)
0.00425(0.4)(5.933479)(10 - 3)
= -9.73 MPa = 9.73 MPa (C)
Ans.
435
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•6–149.
The curved beam is subjected to a bending
moment of M = 900 N # m. Determine the stress at point C.
A C
B
100 mm
C
A
30
20 mm
15 mm
150 mm
400 mm
B
M
Internal Moment: M = -900 N # m is negative since it tends to decrease the beam’s
radius of curvature.
Section Properties:
©A = 0.15(0.015) + 0.1(0.02) = 0.00425 m2
©rA = 0.475(0.15)(0.015) + 0.56(0.1)(0.02) = 2.18875(10 - 3) m
r =
©
2.18875 (10 - 3)
©rA
=
= 0.5150 m
©A
0.00425
dA
0.57
0.55
= 0.015 ln
+ 0.1 ln
= 8.348614(10 - 3) m
0.4
0.55
LA r
R =
A
©1A dA
r
=
0.00425
= 0.509067 m
8.348614(10 - 3)
r - R = 0.515 - 0.509067 = 5.933479(10 - 3) m
Normal Stress: Applying the curved-beam formula
sC =
M(R - rC)
-900(0.509067 - 0.55)
=
ArC(r - R)
0.00425(0.55)(5.933479)(10 - 3)
= 2.66 MPa (T)
Ans.
436
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6–150. The elbow of the pipe has an outer radius of
0.75 in. and an inner radius of 0.63 in. If the assembly is
subjected to the moments of M = 25 lb # in., determine the
maximum stress developed at section a-a.
a
30
M 25 lbin.
1 in.
a
dA
= ©2p (r - 2r2 - c2)
LA r
= 2p(1.75 - 21.752 - 0.752) - 2p (1.75 - 21.752 - 0.632)
0.63 in.
0.75 in.
= 0.32375809 in.
A = p(0.752) - p(0.632) = 0.1656 p
R =
A
dA
1A r
=
M = 25 lbin.
0.1656 p
= 1.606902679 in.
0.32375809
r - R = 1.75 - 1.606902679 = 0.14309732 in.
(smax)t =
M(R - rA)
=
ArA(r - R)
(smax)c = =
25(1.606902679 - 1)
= 204 psi (T)
0.1656 p(1)(0.14309732)
M(R - rB)
=
ArB(r - R)
Ans.
25(1.606902679 - 2.5)
= 120 psi (C)
0.1656p(2.5)(0.14309732)
Ans.
6–151. The curved member is symmetric and is subjected
to a moment of M = 600 lb # ft. Determine the bending
stress in the member at points A and B. Show the stress
acting on volume elements located at these points.
0.5 in.
B
2 in.
A
1
A = 0.5(2) + (1)(2) = 2 in2
2
1.5 in.
8 in.
9(0.5)(2) + 8.6667 A 2 B (1)(2)
©rA
=
= 8.83333 in.
©A
2
1
r =
M
M
1(10)
dA
10
10
= 0.5 ln
+ c
cln
d - 1 d = 0.22729 in.
r
8
(10 - 8)
8
LA
R =
A
dA
1A r
=
2
= 8.7993 in.
0.22729
r - R = 8.83333 - 8.7993 = 0.03398 in.
s =
M(R - r)
Ar(r - R)
sA =
600(12)(8.7993 - 8)
= 10.6 ksi (T)
2(8)(0.03398)
Ans.
sB =
600(12)(8.7993 - 10)
= -12.7 ksi = 12.7 ksi (C)
2(10)(0.03398)
Ans.
437
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*6–152. The curved bar used on a machine has a
rectangular cross section. If the bar is subjected to a couple
as shown, determine the maximum tensile and compressive
stress acting at section a-a. Sketch the stress distribution
on the section in three dimensions.
a
75 mm
a
50 mm
162.5 mm
250 N
60
150 mm
60
250 N
75 mm
a + ©MO = 0;
M - 250 cos 60° (0.075) - 250 sin 60° (0.15) = 0
M = 41.851 N # m
r2
dA
0.2375
= b ln
= 0.05 ln
= 0.018974481 m
r
r
0.1625
1
LA
A = (0.075)(0.05) = 3.75(10 - 3) m2
R =
A
=
dA
1A r
3.75(10 - 3)
= 0.197633863 m
0.018974481
r - R = 0.2 - 0.197633863 = 0.002366137
sA =
M(R - rA)
41.851(0.197633863 - 0.2375)
=
ArA(r - R)
3.75(10 - 3)(0.2375)(0.002366137)
= -791.72 kPa
Ans.
= 792 kPa (C)
sB =
M(R - rB)
41.851 (0.197633863 - 0.1625)
=
ArB(r - R)
3.75(10 - 3)(0.1625)(0.002366137)
= 1.02 MPa (T)
438
Ans.
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•6–153.
The ceiling-suspended C-arm is used to support
the X-ray camera used in medical diagnoses. If the camera
has a mass of 150 kg, with center of mass at G, determine
the maximum bending stress at section A.
G
1.2 m
A
200 mm
100 mm
20 mm
40 mm
Section Properties:
r =
©
1.22(0.1)(0.04) + 1.25(0.2)(0.02)
©rA
=
= 1.235 m
©A
0.1(0.04) + 0.2(0.02)
dA
1.26
1.24
= 0.1 ln
+ 0.2 ln
= 6.479051 A 10 - 3 B m
r
1.20
1.24
LA
A = 0.1(0.04) + 0.2(0.02) = 0.008 m2
R =
A
dA
1A r
=
0.008
= 1.234749 m
6.479051 (10 - 3)
r - R = 1.235 - 1.234749 = 0.251183 A 10 - 3 B m
Internal Moment: The internal moment must be computed about the neutral axis as
shown on FBD. M = -1816.93 N # m is negative since it tends to decrease the
beam’s radius of curvature.
Maximum Normal Stress: Applying the curved-beam formula
sA =
M(R - rA)
ArA (r - R)
-1816.93(1.234749 - 1.26)
=
0.008(1.26)(0.251183)(10 - 3)
= 18.1 MPa (T)
sB =
M(R - rB)
ArB (r - R)
-1816.93(1.234749 - 1.20)
=
0.008(1.20)(0.251183)(10 - 3)
= -26.2 MPa = 26.2 MPa (C)
Ans.
(Max)
439
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6–154. The circular spring clamp produces a compressive
force of 3 N on the plates. Determine the maximum bending
stress produced in the spring at A. The spring has a
rectangular cross section as shown.
10 mm
20 mm
Internal Moment: As shown on FBD, M = 0.660 N # m is positive since it tends to
increase the beam’s radius of curvature.
210 mm
200 mm
A
Section Properties:
220 mm
0.200 + 0.210
r =
= 0.205 m
2
r2
dA
0.21
= 0.02 ln
= b ln
= 0.97580328 A 10 - 3 B m
r
r
0.20
1
LA
A = (0.01)(0.02) = 0.200 A 10 - 3 B m2
R =
0.200(10 - 3)
A
=
dA
1A r
0.97580328(10 - 3)
= 0.204959343 m
r - R = 0.205 - 0.204959343 = 0.040657 A 10 - 3 B m
Maximum Normal Stress: Applying the curved-beam formula
sC =
M(R - r2)
Ar2(r - R)
0.660(0.204959343 - 0.21)
=
0.200(10 - 3)(0.21)(0.040657)(10 - 3)
= -1.95MPa = 1.95 MPa (C)
st =
M(R - r1)
Ar1 (r - R)
0.660(0.204959343 - 0.2)
=
0.200(10 - 3)(0.2)(0.040657)(10 - 3)
= 2.01 MPa (T)
(Max)
Ans.
440
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6–155. Determine the maximum compressive force the
spring clamp can exert on the plates if the allowable
bending stress for the clamp is sallow = 4 MPa.
10 mm
20 mm
210 mm
200 mm
A
220 mm
Section Properties:
r =
0.200 + 0.210
= 0.205 m
2
r2
dA
0.21
= b ln
= 0.02 ln
= 0.97580328 A 10 - 3 B m
r1
0.20
LA r
A = (0.01)(0.02) = 0.200 A 10 - 3 B m2
R =
0.200(10 - 3)
A
=
dA
1A r
0.97580328(10 - 3)
= 0.204959 m
r - R = 0.205 - 0.204959343 = 0.040657 A 10 - 3 B m
Internal Moment: The internal moment must be computed about the neutral axis as
shown on FBD. Mmax = 0.424959P is positive since it tends to increase the beam’s
radius of curvature.
Allowable Normal Stress: Applying the curved-beam formula
Assume compression failure
sc = sallow =
-4 A 106 B =
M(R - r2)
Ar2(r - R)
0.424959P(0.204959 - 0.21)
0.200(10 - 3)(0.21)(0.040657)(10 - 3)
P = 3.189 N
Assume tension failure
st = sallow =
4 A 106 B =
M(R - r1)
Ar1 (r - R)
0.424959P(0.204959 - 0.2)
0.200(10 - 3)(0.2)(0.040657)(10 - 3)
P = 3.09 N (Controls !)
Ans.
441
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*6–156. While in flight, the curved rib on the jet plane is
subjected to an anticipated moment of M = 16 N # m at the
section. Determine the maximum bending stress in the rib
at this section, and sketch a two-dimensional view of the
stress distribution.
16 Nm
5 mm
20 mm
5 mm
0.6 m
5 mm
30 mm
LA
dA>r = (0.03)ln
0.625
0.630
0.605
+ (0.005)ln
+ (0.03)ln
= 0.650625(10 - 3) in.
0.6
0.605
0.625
A = 2(0.005)(0.03) + (0.02)(0.005) = 0.4(10 - 3) in2
R =
0.4(10 - 3)
A
1A dA>r
=
0.650625(10 - 3)
= 0.6147933
(sc)max =
M(R - rc)
16(0.6147933 - 0.630)
= -4.67 MPa
=
ArA(r - R)
0.4(10 3)(0.630)(0.615 - 0.6147933)
(ss)max =
M(R - rs)
16(0.6147933 - 0.6)
= 4.77 MPa
=
ArA(r - R)
0.4(10 - 3)(0.6)(0.615 - 0.6147933)
Ans.
If the radius of each notch on the plate is r = 0.5 in.,
determine the largest moment that can be applied. The
allowable bending stress for the material is sallow = 18 ksi.
•6–157.
14.5 in.
M
b =
14.5 - 12.5
= 1.0 in.
2
r
0.5
=
= 0.04
h
12.5
1
b
=
= 2.0
r
0.5
From Fig. 6-44:
K = 2.60
smax = K
Mc
I
(M)(6.25)
18(103) = 2.60c 1
d
3
12 (1)(12.5)
M = 180 288 lb # in. = 15.0 kip # ft
Ans.
442
1 in.
12.5 in.
M
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6–158. The symmetric notched plate is subjected to
bending. If the radius of each notch is r = 0.5 in. and the
applied moment is M = 10 kip # ft, determine the maximum
bending stress in the plate.
14.5 in.
M
M
12.5 in.
r
0.5
=
= 0.04
h
12.5
1
b
= 2.0
=
r
0.5
1 in.
From Fig. 6-44:
K = 2.60
smax = K
(10)(12)(6.25)
Mc
= 2.60 c 1
d = 12.0 ksi
3
I
12 (1)(12.5)
Ans.
6–159. The bar is subjected to a moment of M = 40 N # m.
Determine the smallest radius r of the fillets so that
an allowable bending stress of sallow = 124 MPa is not
exceeded.
80 mm
7 mm
20 mm
r
M
M
r
Allowable Bending Stress:
sallow = K
Mc
I
124 A 106 B = K B 1
40(0.01)
3
12 (0.007)(0.02 )
R
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 1
17.5(0.01)
3
12 (0.007)(0.02 )
R
= 54.4 MPa
Ans.
443
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•6–161. The simply supported notched bar is subjected to
two forces P. Determine the largest magnitude of P that can
be applied without causing the material to yield.The material
is A-36 steel. Each notch has a radius of r = 0.125 in.
P
P
0.5 in.
1.75 in.
1.25 in.
20 in.
b =
20 in.
20 in.
20 in.
1.75 - 1.25
= 0.25
2
0.25
b
=
= 2;
r
0.125
r
0.125
=
= 0.1
h
1.25
From Fig. 6-44. K = 1.92
sY = K
Mc
;
I
20P(0.625)
36 = 1.92 c 1
d
3
12 (0.5)(1.25)
P = 122 lb
Ans.
6–162. The simply supported notched bar is subjected to
the two loads, each having a magnitude of P = 100 lb.
Determine the maximum bending stress developed in the
bar, and sketch the bending-stress distribution acting over
the cross section at the center of the bar. Each notch has a
radius of r = 0.125 in.
P
0.5 in.
1.75 - 1.25
= 0.25
2
b
0.25
=
= 2;
r
0.125
r
0.125
=
= 0.1
h
1.25
From Fig. 6-44, K = 1.92
smax = K
1.75 in.
1.25 in.
20 in.
b =
P
2000(0.625)
Mc
= 1.92c 1
d = 29.5 ksi
3
I
12 (0.5)(1.25)
Ans.
444
20 in.
20 in.
20 in.
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6–163. Determine the length L of the center portion of
the bar so that the maximum bending stress at A, B, and C is
the same. The bar has a thickness of 10 mm.
7 mm
350 N
60 mm
A
r
7
=
= 0.175
h
40
60
w
=
= 1.5
h
40
200 mm
40 mm
7 mm
C
L
2
B
L
2
200 mm
From Fig. 6-43, K = 1.5
(sA)max = K
(35)(0.02)
MAc
d = 19.6875 MPa
= 1.5c 1
3
I
12 (0.01)(0.04 )
(sB)max = (sA)max =
19.6875(106) =
MB c
I
175(0.2 + L2 )(0.03)
1
3
12 (0.01)(0.06 )
L = 0.95 m = 950 mm
Ans.
*6–164. The stepped bar has a thickness of 15 mm.
Determine the maximum moment that can be applied to its
ends if it is made of a material having an allowable bending
stress of sallow = 200 MPa.
45 mm
30 mm
3 mm
M
M
Stress Concentration Factor:
w
30
6
r
=
= 3 and =
= 0.6, we have K = 1.2
h
10
h
10
obtained from the graph in the text.
For the smaller section with
w
45
3
r
=
= 1.5 and =
= 0.1, we have K = 1.75
h
30
h
30
obtained from the graph in the text.
For the larger section with
Allowable Bending Stress:
For the smaller section
smax = sallow = K
Mc
;
I
200 A 106 B = 1.2 B 1
M(0.005)
3
12 (0.015)(0.01 )
R
M = 41.7 N # m (Controls !)
Ans.
For the larger section
smax = sallow = K
Mc
;
I
200 A 106 B = 1.75 B 1
M(0.015)
3
12 (0.015)(0.03 )
R
M = 257 N # m
445
10 mm
6 mm
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•6–165.
The beam is made of an elastic plastic material for
which sY = 250 MPa. Determine the residual stress in the
beam at its top and bottom after the plastic moment Mp is
applied and then released.
15 mm
1
1
(0.2)(0.23)3 (0.18)(0.2)3 = 82.78333(10 - 6)m4
12
12
20 mm
200 mm
Ix =
Mp
C1 = T1 = sY (0.2)(0.015) = 0.003sY
15 mm
C2 = T2 = sY (0.1)(0.02) = 0.002sY
200 mm
Mp = 0.003sY (0.215) + 0.002sY (0.1) = 0.000845 sY
= 0.000845(250)(106) = 211.25 kN # m
s =
Mp c
211.25(103)(0.115)
=
I
82.78333(10 - 6)
y
0.115
=
;
250
293.5
= 293.5 MPa
y = 0.09796 m = 98.0 mm
stop = sbottom = 293.5 - 250 = 43.5 MPa
Ans.
6–166. The wide-flange member is made from an elasticplastic material. Determine the shape factor.
t
Plastic analysis:
T1 = C1 = sY bt;
h
T2 = C2 = sY a
MP = sY bt(h - t) + sY a
h - 2t
bt
2
t
t
h - 2t
h - 2t
b(t) a
b
2
2
b
t
= sY c bt(h - t) + (h - 2t)2 d
4
Elastic analysis:
I =
1
1
bh3 (b - t)(h - 2t)3
12
12
=
1
[bh3 - (b - t)(h - 2 t)3]
12
MY =
sy I
c
=
=
1
sY A 12
B [bh3 - (b - t)(h - 2t)3]
h
2
bh3 - (b - t)(h - 2t)3
sY
6h
Shape factor:
k =
[bt(h - t) + 4t (h - 2t)2]sY
MP
=
bh3 - (b - t)(h - 2t)3
MY
s
6h
=
Y
3h 4bt(h - t) + t(h - 2t)2
c
d
2 bh3 - (b - t)(h - 2t)3
Ans.
446
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6–167.
Determine the shape factor for the cross section.
Maximum Elastic Moment: The moment of inertia about neutral axis must be
determined first.
a
1
1
(a)(3a)3 +
(2a) A a3 B = 2.41667a4
12
12
INA =
a
a
Applying the flexure formula with s = sY, we have
sY =
MY c
I
MY =
a
a
a
sY (2.41667a4)
sYI
=
= 1.6111a3sY
c
1.5a
Plastic Moment:
MP = sY (a)(a)(2a) + sY (0.5a)(3a)(0.5a)
= 2.75a3sY
Shape Factor:
k =
MP
2.75a3sY
=
= 1.71
MY
1.6111a3sY
Ans.
*6–168. The beam is made of elastic perfectly plastic
material. Determine the maximum elastic moment and the
plastic moment that can be applied to the cross section.
Take a = 2 in. and sY = 36 ksi.
a
a
Maximum Elastic Moment: The moment of inertia about neutral axis must be
determined first.
a
1
1
(2) A 63 B +
(4) A 23 B = 38.667 in4
INA =
12
12
Applying the flexure formula with s = sY, we have
sY = =
MY =
a
MY c
I
36(38.667)
sY I
=
c
3
= 464 kip # in = 38.7 kip # ft
Ans.
Plastic Moment:
MP = 36(2)(2)(4) + 36(1)(6)(1)
= 792 kip # in = 66.0 kip # ft
Ans.
447
a
a
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•6–169.
The box beam is made of an elastic perfectly
plastic material for which sY = 250 MPa . Determine the
residual stress in the top and bottom of the beam after the
plastic moment Mp is applied and then released.
Plastic Moment:
MP = 250 A 106 B (0.2)(0.025)(0.175) + 250 A 106 B (0.075)(0.05)(0.075)
25 mm
= 289062.5 N # m
150 mm
Modulus of Rupture: The modulus of rupture sr can be determined using the flexure
formula with the application of reverse, plastic moment MP = 289062.5 N # m.
I =
25 mm
25 mm
150 mm
25 mm
1
1
(0.2) A 0.23 B (0.15) A 0.153 B
12
12
= 91.14583 A 10 - 6 B m4
sr =
289062.5 (0.1)
MP c
= 317.41 MPa
=
I
91.14583 A 10 - 6 B
Residual Bending Stress: As shown on the diagram.
œ
œ
= sbot
= sr - sY
stop
= 317.14 - 250 = 67.1 MPa
Ans.
6–170. Determine the shape factor for the wideflange beam.
15 mm
1
1
(0.2)(0.23)3 (0.18)(0.2)3 = 82.78333 A 10 - 6 B m4
12
12
Ix =
20 mm
200 mm
C1 = T1 = sY(0.2)(0.015) = 0.003sY
Mp
C2 = T2 = sY(0.1)(0.02) = 0.002sY
Mp = 0.003sY(0.215) + 0.002sY(0.1) = 0.000845 sY
15 mm
200 mm
sY =
MY =
k =
MY c
I
sY A 82.78333)10 - 6 B
0.115
Mp
MY
=
= 0.000719855 sY
0.000845sY
= 1.17
0.000719855sY
Ans.
448
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6–171. Determine the shape factor of the beam’s cross
section.
3 in.
Referring to Fig. a, the location of centroid of the cross-section is
y =
7.5(3)(6) + 3(6)(3)
©yA
=
= 5.25 in.
©A
3(6) + 6(3)
6 in.
The moment of inertia of the cross-section about the neutral axis is
I =
1
1
(3) A 63 B + 3(6)(5.25 - 3)2 +
(6) A 33 B + 6(3)(7.5 - 5.25)2
12
12
1.5 in. 3 in.
1.5 in.
4
= 249.75 in
Here smax = sY and c = y = 5.25 in. Thus
smax =
Mc
;
I
sY =
MY (5.25)
249.75
MY = 47.571sY
Referring to the stress block shown in Fig. b,
sdA = 0;
LA
T - C1 - C2 = 0
d(3)sY - (6 - d)(3)sY - 3(6)sY = 0
d = 6 in.
Since d = 6 in., c1 = 0, Fig. c. Here
T = C = 3(6) sY = 18 sY
Thus,
MP = T(4.5) = 18 sY (4.5) = 81 sY
Thus,
k =
MP
81 sY
=
= 1.70
MY
47.571 sY
Ans.
449
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*6–172. The beam is made of elastic-perfectly plastic
material. Determine the maximum elastic moment and the
plastic moment that can be applied to the cross section.
Take sY = 36 ksi.
3 in.
Referring to Fig. a, the location of centroid of the cross-section is
6 in.
7.5(3)(6) + 3(6)(3)
©yA
y =
=
= 5.25 in.
©A
3(6) + 6(3)
The moment of inertia of the cross-section about the neutral axis is
1.5 in. 3 in.
1.5 in.
I =
1
1
(3)(63) + 3(6)(5.25 - 3)2 +
(6)(33) + 6(3)(7.5 - 5.25)2
12
12
= 249.75 in4
Here, smax = sY = 36 ksi and ¢ = y = 5.25 in. Then
smax =
Mc
;
I
36 =
MY (5.25)
249.75
MY = 1712.57 kip # in = 143 kip # ft
Ans.
Referring to the stress block shown in Fig. b,
sdA = 0;
LA
T - C1 - C2 = 0
d(3) (36) - (6 - d)(3)(36) - 3(6) (36) = 0
d = 6 in.
Since d = 6 in., c1 = 0,
Here,
T = C = 3(6)(36) = 648 kip
Thus,
MP = T(4.5) = 648(4.5) = 2916 kip # in = 243 kip # ft
Ans.
450
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•6–173.
Determine the shape factor for the cross section
of the H-beam.
Ix =
1
1
(0.2)(0.023) + 2 a b(0.02)(0.23) = 26.8(10 - 6)m4
12
12
200 mm
C1 = T1 = sY(2)(0.09)(0.02) = 0.0036sy
20 mm
C2 = T2 = sY(0.01)(0.24) = 0.0024sy
Mp
20 mm
200 mm
Mp = 0.0036sY(0.11) + 0.0024sY(0.01) = 0.00042sY
20 mm
MYc
sY =
I
MY =
k =
sY(26.8)(10 - 6)
= 0.000268sY
0.1
Mp
MY
=
0.00042sY
= 1.57
0.000268sY
Ans.
6–174. The H-beam is made of an elastic-plastic material
for which sY = 250 MPa. Determine the residual stress in
the top and bottom of the beam after the plastic moment
Mp is applied and then released.
200 mm
Ix =
1
1
(0.2)(0.023) + 2 a b (0.02)(0.23) = 26.8(10 - 6)m4
12
12
20 mm
C1 = T1 = sY(2)(0.09)(0.02) = 0.0036sy
200 mm
C2 = T2 = sY(0.01)(0.24) = 0.0024sy
20 mm
Mp = 0.0036sY(0.11) + 0.0024sY(0.01) = 0.00042sY
Mp = 0.00042(250) A 106 B = 105 kN # m
s¿ =
Mp c
I
y
0.1
=
;
250
392
105(103)(0.1)
=
26.8(10 - 6)
Mp
= 392 MPa
y = 0.0638 = 63.8 mm
sT = sB = 392 - 250 = 142 MPa
Ans.
451
20 mm
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6–175.
Determine the shape factor of the cross section.
3 in.
The moment of inertia of the cross-section about the neutral axis is
I =
3 in.
1
1
(3)(93) +
(6) (33) = 195.75 in4
12
12
3 in.
Here, smax = sY and c = 4.5 in. Then
smax =
Mc
;
I
sY =
MY(4.5)
195.75
3 in.
MY = 43.5 sY
Referring to the stress block shown in Fig. a,
T1 = C1 = 3(3)sY = 9 sY
T2 = C2 = 1.5(9)sY = 13.5 sY
Thus,
MP = T1(6) + T2(1.5)
= 9sY(6) + 13.5sY(1.5) = 74.25 sY
k =
74.25 sY
MP
=
= 1.71
MY
43.5 sY
Ans.
452
3 in.
3 in.
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*6–176. The beam is made of elastic-perfectly plastic
material. Determine the maximum elastic moment and the
plastic moment that can be applied to the cross section.
Take sY = 36 ksi.
3 in.
3 in.
The moment of inertia of the cross-section about the neutral axis is
I =
3 in.
1
1
(3)(93) +
(6)(33) = 195.75 in4
12
12
Here, smax = sY = 36 ksi and c = 4.5 in. Then
Mc
;
smax =
I
3 in.
MY (4.5)
36 =
195.75
MY = 1566 kip # in = 130.5 kip # ft
Ans.
Referring to the stress block shown in Fig. a,
T1 = C1 = 3(3)(36) = 324 kip
T2 = C2 = 1.5(9)(36) = 486 kip
Thus,
MP = T1(6) + T2(1.5)
= 324(6) + 486(1.5)
= 2673 kip # in. = 222.75 kip # ft = 223 kip # ft
Ans.
453
3 in.
3 in.
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•6–177.
Determine the shape factor of the cross section
for the tube.
The moment of inertia of the tube’s cross-section about the neutral axis is
I =
5 in.
p 4
p
A r - r4i B = A 64 - 54 B = 167.75 p in4
4 o
4
6 in.
Here, smax = sY and C = ro = 6 in,
smax =
Mc
;
I
sY =
MY (6)
167.75 p
MY = 87.83 sY
The plastic Moment of the table’s cross-section can be determined by super posing
the moment of the stress block of the solid circular cross-section with radius
ro = 6 in and ri = 5 in. as shown in Figure a, Here,
T1 = C1 =
1
p(62)sY = 18psY
2
T2 = C2 =
1
p(52)sY = 12.5p sY
2
Thus,
MP = T1 b 2 c
4(6)
4(5)
d r - T2 b 2 c
dr
3p
3p
= (18psY)a
16
40
b - 12.5psY a b
p
3p
= 121.33 sY
k =
121.33 sY
MP
=
= 1.38
MY
87.83 sY
Ans.
454
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6–178. The beam is made from elastic-perfectly plastic
material. Determine the shape factor for the thick-walled tube.
ro
Maximum Elastic Moment. The moment of inertia of the cross-section about the
neutral axis is
I =
p
A r 4 - r4i B
4 o
With c = ro and smax = sY,
smax =
Mc
;
I
sY =
MY =
MY(ro)
p
A r 4 - ri 4 B
4 o
p
A r 4 - ri 4 B sY
4ro o
Plastic Moment. The plastic moment of the cross section can be determined by
superimposing the moment of the stress block of the solid beam with radius r0 and ri
as shown in Fig. a, Referring to the stress block shown in Fig. a,
T1 = c1 =
p 2
r s
2 o Y
T2 = c2 =
p 2
r s
2 i Y
MP = T1 c2 a
4ro
4ri
b d - T2 c2 a b d
3p
3p
=
8ro
8ri
p 2
p
r s a
b - ri 2sY a b
2 o Y 3p
2
3p
=
4
A r 3 - ri 3 B sY
3 o
Shape Factor.
4
A r 3 - ri 3 B sY
16ro A ro 3 - ri 3 B
MP
3 o
k =
=
=
p
MY
3p A ro 4 - ri 4 B
A ro 4 - ri 4 B sY
4ro
Ans.
455
ri
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6–179.
Determine the shape factor for the member.
Plastic analysis:
T = C =
–h
2
h
1
bh
(b)a bsY =
s
2
2
4 Y
–h
2
b h2
bh
h
MP =
sY a b =
s
4
3
12 Y
Elastic analysis:
I = 2c
1
h 3
b h3
(b)a b d =
12
2
48
b
sY A 48 B
sYI
b h2
=
s
=
h
c
24 Y
2
bh3
MY =
Shape factor:
k =
bh2
12 sY
Mp
MY
=
bh2
24 sY
= 2
Ans.
*6–180. The member is made from an elastic-plastic
material. Determine the maximum elastic moment and the
plastic moment that can be applied to the cross section.
Take b = 4 in., h = 6 in., sY = 36 ksi.
–h
2
Elastic analysis:
I = 2c
1
(4)(3)3 d = 18 in4
12
MY =
36(18)
sYI
=
= 216 kip # in. = 18 kip # ft
c
3
–h
2
Ans.
b
Plastic analysis:
T = C =
1
(4)(3)(36) = 216 kip
2
6
Mp = 2160 a b = 432 kip # in. = 36 kip # ft
3
Ans.
456
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•6–181.
The beam is made of a material that can be
assumed perfectly plastic in tension and elastic perfectly
plastic in compression. Determine the maximum bending
moment M that can be supported by the beam so that the
compressive material at the outer edge starts to yield.
h
sY
M
sdA = 0;
LA
C - T = 0
sY
a
1
s (d)(a) - sY(h - d)a = 0
2 Y
d =
M =
2
h
3
11
11a h2
2
1
sY a hb (a)a hb =
sY
2
3
18
54
Ans.
6–182. The box beam is made from an elastic-plastic
material for which sY = 25 ksi. Determine the intensity of
the distributed load w0 that will cause the moment to be
(a) the largest elastic moment and (b) the largest plastic
moment.
w0
Elastic analysis:
I =
9 ft
1
1
(8)(163) (6)(123) = 1866.67 in4
12
12
sYI
;
Mmax =
c
9 ft
8 in.
25(1866.67)
27w0(12) =
8
Ans.
w0 = 18.0 kip>ft
Plastic analysis:
16 in.
12 in.
6 in.
C1 = T1 = 25(8)(2) = 400 kip
C2 = T2 = 25(6)(2) = 300 kip
MP = 400(14) + 300(6) = 7400 kip # in.
27w0(12) = 7400
w0 = 22.8 kip>ft
Ans.
457
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6–183. The box beam is made from an elastic-plastic
material for which sY = 36 ksi. Determine the magnitude
of each concentrated force P that will cause the moment to
be (a) the largest elastic moment and (b) the largest plastic
moment.
P
From the moment diagram shown in Fig. a, Mmax = 6 P.
P
8 ft
6 ft
6 ft
The moment of inertia of the beam’s cross-section about the neutral axis is
6 in.
1
1
(6)(123) (5)(103) = 447.33 in4
I =
12
12
12 in.
10 in.
Here, smax = sY = 36 ksi and c = 6 in.
smax =
Mc
;
I
36 =
5 in.
MY (6)
447.33
MY = 2684 kip # in = 223.67 kip # ft
It is required that
Mmax = MY
6P = 223.67
P = 37.28 kip = 37.3 kip
Ans.
Referring to the stress block shown in Fig. b,
T1 = C1 = 6(1)(36) = 216 kip
T2 = C2 = 5(1)(36) = 180 kip
Thus,
MP = T1(11) + T2(5)
= 216(11) + 180(5)
= 3276 kip # in = 273 kip # ft
It is required that
Mmax = MP
6P = 273
P = 45.5 kip
Ans.
458
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*6–184. The beam is made of a polyester that has the
stress–strain curve shown. If the curve can be represented
by the equation s = [20 tan-1115P2] ksi, where tan-1115P2
is in radians, determine the magnitude of the force P that
can be applied to the beam without causing the maximum
strain in its fibers at the critical section to exceed
Pmax = 0.003 in.>in.
P
2 in.
4 in.
8 ft
s(ksi)
8 ft
s 20 tan1(15 P)
P(in./in.)
Maximum Internal Moment: The maximum internal moment M = 4.00P occurs at
the mid span as shown on FBD.
Stress–Strain Relationship: Using the stress–strain relationship. the bending stress
can be expressed in terms of y using e = 0.0015y.
s = 20 tan - 1 (15e)
= 20 tan - 1 [15(0.0015y)]
= 20 tan - 1 (0.0225y)
When emax = 0.003 in.>in., y = 2 in. and smax = 0.8994 ksi
Resultant Internal Moment: The resultant internal moment M can be evaluated
from the integal
M = 2
LA
ysdA.
ysdA
2in
= 2
LA
y C 20 tan 1 (0.0225y) D (2dy)
-
L0
2in
= 80
L0
= 80 B
y tan - 1 (0.0225y) dy
1 + (0.0225)2y2
2(0.0225)2
tan - 1 (0.0225y) -
2in.
y
R2
2(0.0225) 0
= 4.798 kip # in
Equating
M = 4.00P(12) = 4.798
P = 0.100 kip = 100 lb
Ans.
459
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•6–185.
The plexiglass bar has a stress–strain curve that
can be approximated by the straight-line segments shown.
Determine the largest moment M that can be applied to the
bar before it fails.
s (MPa)
20 mm
M
20 mm
failure
60
40
tension
0.06 0.04
P (mm/mm)
0.02
compression
80
100
Ultimate Moment:
LA
s dA = 0;
C - T2 - T1 = 0
1
1
d
1 d
sc (0.02 - d)(0.02) d - 40 A 106 B c a b(0.02) d - (60 + 40) A 106 B c(0.02) d = 0
2
2 2
2
2
s - 50s d - 3500(106)d = 0
Assume.s = 74.833 MPa; d = 0.010334 m
From the strain diagram,
0.04
e
=
0.02 - 0.010334
0.010334
e = 0.037417 mm>mm
From the stress–strain diagram,
80
s
=
0.037417
0.04
s = 74.833 MPa (OK! Close to assumed value)
Therefore,
1
C = 74.833 A 106 B c (0.02 - 0.010334)(0.02) d = 7233.59 N
2
T1 =
1
0.010334
(60 + 40) A 106 B c(0.02)a
b d = 5166.85 N
2
2
1
0.010334
b d = 2066.74 N
T2 = 40 A 106 B c (0.02)a
2
2
y1 =
2
(0.02 - 0.010334) = 0.0064442 m
3
y2 =
2 0.010334
a
b = 0.0034445 m
3
2
y3 =
0.010334
1 2(40) + 60
0.010334
+ c1 - a
bda
b = 0.0079225m
2
3
40 + 60
2
M = 7233.59(0.0064442) + 2066.74(0.0034445) + 5166.85(0.0079255)
= 94.7 N # m
Ans.
460
0.04
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6–186. The stress–strain diagram for a titanium alloy can
be approximated by the two straight lines. If a strut made of
this material is subjected to bending, determine the moment
resisted by the strut if the maximum stress reaches a value
of (a) sA and (b) sB.
3 in.
M
2 in.
s (ksi)
B
sB 180
sA 140
A
0.01
a) Maximum Elastic Moment : Since the stress is linearly related to strain up to
point A, the flexure formula can be applied.
sA =
Mc
I
M =
=
sA I
c
1
140 C 12
(2)(33) D
1.5
= 420 kip # in = 35.0 kip # ft
b)
Ans.
The Ultimate Moment :
C1 = T1 =
1
(140 + 180)(1.125)(2) = 360 kip
2
C2 = T2 =
1
(140)(0.375)(2) = 52.5 kip
2
M = 360(1.921875) + 52.5(0.5)
= 718.125 kip # in = 59.8 kip # ft
Ans.
Note: The centroid of a trapezodial area was used in calculation of moment.
461
0.04
P (in./in.)
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6–187. A beam is made from polypropylene plastic and has
a stress–strain diagram that can be approximated by the curve
shown. If the beam is subjected to a maximum tensile and
compressive strain of P = 0.02 mm>mm, determine the
maximum moment M.
M
s (Pa)
s 10(106)P1/ 4
emax = 0.02
smax = 10 A 106 B (0.02)1>4 = 3.761 MPa
M
100 mm
30 mm
P (mm/mm)
e
0.02
=
y
0.05
e = 0.4 y
s = 10 A 106 B (0.4)1>4y1>4
y(7.9527) A 106 B y1>4(0.03)dy
0.05
M =
y s dA = 2
LA
M = 0.47716 A 106 B
L0
4
y5>4dy = 0.47716 A 106 B a b(0.05)9>4
5
0.05
L0
M = 251 N # m
Ans.
*6–188. The beam has a rectangular cross section and is
made of an elastic-plastic material having a stress–strain
diagram as shown. Determine the magnitude of the
moment M that must be applied to the beam in order to
create a maximum strain in its outer fibers of P max = 0.008.
400 mm
M
200 mm
s(MPa)
200
0.004
C1 = T1 = 200 A 106 B (0.1)(0.2) = 4000 kN
C2 = T2 =
1
(200) A 106 B (0.1)(0.2) = 2000 kN
2
M = 4000(0.3) + 2000(0.1333) = 1467 kN # m = 1.47 MN # m
Ans.
462
P (mm/mm)
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s(ksi)
90
80
•6–189.
The bar is made of an aluminum alloy having a
stress–strain diagram that can be approximated by the
straight line segments shown. Assuming that this diagram is
the same for both tension and compression, determine the
moment the bar will support if the maximum strain at the
top and bottom fibers of the beam is P max = 0.03.
90 - 80
s - 80
=
;
0.03 - 0.025
0.05 - 0.025
60
4 in. M
s = 82 ksi
C1 = T1 =
1
(0.3333)(80 + 82)(3) = 81 kip
2
C2 = T2 =
1
(1.2666)(60 + 80)(3) = 266 kip
2
C3 = T3 =
1
(0.4)(60)(3) = 36 kip
2
0.006
0.025
0.05
P (in./ in.)
3 in.
M = 81(3.6680) + 266(2.1270) + 36(0.5333)
= 882.09 kip # in. = 73.5 kip # ft
Ans.
Note: The centroid of a trapezodial area was used in calculation of moment areas.
6–190. The beam is made from three boards nailed together
as shown. If the moment acting on the cross section is
M = 650 N # m, determine the resultant force the bending
stress produces on the top board.
15 mm
Section Properties:
y =
0.0075(0.29)(0.015) + 2[0.0775(0.125)(0.02)]
0.29(0.015) + 2(0.125)(0.02)
M 650 Nm
20 mm
125 mm
= 0.044933 m
20 mm
1
INA =
(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 :
20 mm
Mc
(sw)max =
I
20(106) =
M(0.0625)
0.130578(10 - 3)
;
M = 41.8 kN # m
Failure of steel :
(sst)max =
20 mm
nMc
I
130(106) =
18.182(M)(0.0625)
0.130578(10 - 3)
M = 14.9 kN # m (controls)
Ans.
465
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6–194. Solve Prob. 6–193 if the moment is applied about
the y axis instead of the z axis as shown.
y
z
125 mm
M
x
20 mm
75 mm
20 mm
n =
I =
11(109)
200(104)
= 0.055
1
1
(0.125)(0.1153) (0.118125)(0.0753) = 11.689616(10 - 6)
12
12
Failure of wood :
(sw)max =
nMc2
I
20(106) =
0.055(M)(0.0375)
11.689616(10 - 6)
;
M = 113 kN # m
Failure of steel :
(sst)max =
Mc1
I
130(106) =
M(0.0575)
11.689616(10 - 6)
M = 26.4 kN # m (controls)
Ans.
466
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6–195. A shaft is made of a polymer having a parabolic
cross section. If it resists an internal moment of
M = 125 N # m, determine the maximum bending stress
developed in the material (a) using the flexure formula and
(b) using integration. Sketch a three-dimensional view of
the stress distribution acting over the cross-sectional area.
Hint: The moment of inertia is determined using Eq. A–3 of
Appendix A.
y
100 mm
y 100 – z 2/ 25
M 125 N· m
z
Maximum Bending Stress: The moment of inertia about y axis must be
determined first in order to use Flexure Formula
I =
LA
50 mm
50 mm
y2 dA
100mm
= 2
L0
y2 (2z) dy
100mm
= 20
L0
y2 2100 - y dy
100 mm
3
5
7
3
8
16
y (100 - y)2 (100 - y)2 R 2
= 20 B - y2 (100 - y)2 2
15
105
0
= 30.4762 A 10 - 6 B mm4 = 30.4762 A 10 - 6 B m4
Thus,
smax =
125(0.1)
Mc
= 0.410 MPa
=
I
30.4762(10 - 6)
Ans.
Maximum Bending Stress: Using integration
dM = 2[y(s dA)] = 2 b yc a
M =
smax
by d(2z dy) r
100
smax 100mm 2
y 2100 - y dy
5 L0
125 A 103 B =
100 mm
smax
3
5
7
3
8
16
y(100 - y)2 (100 - y)2 R 2
B - y2(100 - y)2 5
2
15
105
0
125 A 103 B =
smax
(1.5238) A 106 B
5
smax = 0.410 N>mm2 = 0.410 MPa
Ans.
467
x
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*6–196. Determine the maximum bending stress in the
handle of the cable cutter at section a–a. A force of 45 lb is
applied to the handles. The cross-sectional area is shown in
the figure.
20
45 lb
a
5 in.
4 in.
3 in.
0.75 in.
A
a
0.50 in.
45 lb
a + ©M = 0;
M - 45(5 + 4 cos 20°) = 0
M = 394.14 lb # in.
394.14(0.375)
Mc
= 8.41 ksi
= 1
3
I
12 (0.5)(0.75 )
smax =
Ans.
M 85 Nm
•6–197.
The curved beam is subjected to a bending
moment of M = 85 N # m as shown. Determine the stress at
points A and B and show the stress on a volume element
located at these points.
100 mm
A
r2
0.57
0.59
dA
0.42
+ 0.015 ln
+ 0.1 ln
= b ln
= 0.1 ln
r1
0.40
0.42
0.57
LA r
400 mm
= 0.012908358 m
=
LA
dA
r
6.25(10 - 3)
= 0.484182418 m
0.012908358
r - R = 0.495 - 0.484182418 = 0.010817581 m
sA =
M(R - rA)
85(0.484182418 - 0.59)
=
ArA(r - R)
6.25(10 - 3)(0.59)(0.010817581)
= -225.48 kPa
sA = 225 kPa (C)
sB =
Ans.
M(R - rB)
85(0.484182418 - 0.40)
=
ArB(r - R)
150 mm
6.25(10 - 3)(0.40)(0.010817581)
20 mm
B
2
A = 2(0.1)(0.02) + (0.15)(0.015) = 6.25(10 ) m
A
20 mm
30
-3
R =
A
15 mm
B
= 265 kPa (T)
468
Ans.
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6–198. Draw the shear and moment diagrams for the
beam and determine the shear and moment in the beam as
functions of x, where 0 … x 6 6 ft.
8 kip
2 kip/ ft
50 kipft
x
6 ft
+ c ©Fy = 0;
20 - 2x - V = 0
V = 20 - 2x
c + ©MNA = 0;
4 ft
Ans.
x
20x - 166 - 2xa b - M = 0
2
M = -x2 + 20x - 166
Ans.
6–199. Draw the shear and moment diagrams for the shaft
if it is subjected to the vertical loadings of the belt, gear, and
flywheel. The bearings at A and B exert only vertical
reactions on the shaft.
300 N
450 N
A
B
200 mm
400 mm
300 mm
200 mm
150 N
469
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*6–200. A member has the triangular cross section
shown. Determine the largest internal moment M that can
be applied to the cross section without exceeding allowable
tensile and compressive stresses of (sallow)t = 22 ksi and
(sallow)c = 15 ksi, respectively.
4 in.
4 in.
M
2 in.
2 in.
y (From base) =
I =
1
242 - 22 = 1.1547 in.
3
1
(4)(242 - 22)3 = 4.6188 in4
36
Assume failure due to tensile stress :
smax =
My
;
I
22 =
M(1.1547)
4.6188
M = 88.0 kip # in. = 7.33 kip # ft
Assume failure due to compressive stress:
smax =
Mc
;
I
15 =
M(3.4641 - 1.1547)
4.6188
M = 30.0 kip # in. = 2.50 kip # ft
(controls)
Ans.
470
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•6–201.
The strut has a square cross section a by a and is
subjected to the bending moment M applied at an angle u
as shown. Determine the maximum bending stress in terms
of a, M, and u. What angle u will give the largest bending
stress in the strut? Specify the orientation of the neutral
axis for this case.
y
a
z
x
a
M
Internal Moment Components:
Mz = -M cos u
My = -M sin u
Section Property:
Iy = Iz =
1 4
a
12
Maximum Bending Stress: By Inspection, Maximum bending stress occurs at A
and B. Applying the flexure formula for biaxial bending at point A
s = -
My z
Mzy
+
Iz
Iy
-M cos u (a2)
= -
=
1 4
12 a
-Msin u ( - a2)
+
1 4
12 a
6M
(cos u + sin u)
a3
Ans.
6M
ds
= 3 (-sin u + cos u) = 0
du
a
cos u - sin u = 0
u = 45°
Ans.
Orientation of Neutral Axis:
tan a =
Iz
Iy
tan u
tan a = (1) tan(45°)
a = 45°
Ans.
471
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•7–1.
If the wide-flange beam is subjected to a shear of
V = 20 kN, determine the shear stress on the web at A.
Indicate the shear-stress components on a volume element
located at this point.
200 mm
A
20 mm
20 mm
B
V
300 mm
200 mm
The moment of inertia of the cross-section about the neutral axis is
I =
1
1
(0.2)(0.343) (0.18)(0.33) = 0.2501(10 - 3) m4
12
12
From Fig. a,
QA = y¿A¿ = 0.16 (0.02)(0.2) = 0.64(10 - 3) m3
Applying the shear formula,
VQA
20(103)[0.64(10 - 3)]
=
tA =
It
0.2501(10 - 3)(0.02)
= 2.559(106) Pa = 2.56 MPa
Ans.
The shear stress component at A is represented by the volume element shown in
Fig. b.
472
20 mm
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7–2. If the wide-flange beam is subjected to a shear of
V = 20 kN, determine the maximum shear stress in the beam.
200 mm
A
20 mm
20 mm
B
V
300 mm
200 mm
The moment of inertia of the cross-section about the neutral axis is
I =
1
1
(0.2)(0.343) (0.18)(0.33) = 0.2501(10 - 3) m4
12
12
From Fig. a.
Qmax = ©y¿A¿ = 0.16 (0.02)(0.2) + 0.075 (0.15)(0.02) = 0.865(10 - 3) m3
The maximum shear stress occurs at the points along neutral axis since Q is
maximum and thicknest t is the smallest.
tmax =
VQmax
20(103) [0.865(10 - 3)]
=
It
0.2501(10 - 3) (0.02)
= 3.459(106) Pa = 3.46 MPa
Ans.
473
20 mm
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7–3. If the wide-flange beam is subjected to a shear of
V = 20 kN, determine the shear force resisted by the web
of the beam.
200 mm
A
20 mm
20 mm
B
V
300 mm
200 mm
The moment of inertia of the cross-section about the neutral axis is
I =
1
1
(0.2)(0.343) (0.18)(0.33) = 0.2501(10 - 3) m4
12
12
For 0 … y 6 0.15 m, Fig. a, Q as a function of y is
Q = ©y¿A¿ = 0.16 (0.02)(0.2) +
1
(y + 0.15)(0.15 - y)(0.02)
2
= 0.865(10 - 3) - 0.01y2
For 0 … y 6 0.15 m, t = 0.02 m. Thus.
t =
20(103) C 0.865(10 - 3) - 0.01y2 D
VQ
=
It
0.2501(10 - 3) (0.02)
= E 3.459(106) - 39.99(106) y2 F Pa.
The sheer force resisted by the web is,
0.15 m
Vw = 2
L0
0.15 m
tdA = 2
L0
C 3.459(106) - 39.99(106) y2 D (0.02 dy)
= 18.95 (103) N = 19.0 kN
Ans.
474
20 mm
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*7–4. If the T-beam is subjected to a vertical shear of
V = 12 kip, determine the maximum shear stress in the
beam. Also, compute the shear-stress jump at the flangeweb junction AB. Sketch the variation of the shear-stress
intensity over the entire cross section.
4 in.
4 in.
3 in.
4 in.
B
6 in.
A
V ⫽ 12 kip
Section Properties:
y =
INA =
1.5(12)(3) + 6(4)(6)
©yA
=
= 3.30 in.
©A
12(3) + 4(6)
1
1
(12) A 33 B + 12(3)(3.30 - 1.5)2 +
(4) A 63 B + 4(6)(6 - 3.30)2
12
12
= 390.60 in4
Qmax = y1œ A¿ = 2.85(5.7)(4) = 64.98 in3
QAB = y2œ A¿ = 1.8(3)(12) = 64.8 in3
Shear Stress: Applying the shear formula t =
VQ
It
tmax =
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.
475
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•7–5.
If the T-beam is subjected to a vertical shear of
V = 12 kip, determine the vertical shear force resisted by
the flange.
4 in.
4 in.
3 in.
4 in.
B
6 in.
A
V ⫽ 12 kip
Section Properties:
y =
©yA
1.5(12)(3) + 6(4)(6)
=
= 3.30 in.
©A
12(3) + 4(6)
INA =
1
1
(12) A 33 B + 12(3)(3.30 - 1.5)2 +
(4) A 63 B + 6(4)(6 - 3.30)2
12
12
= 390.60 in4
Q = y¿A¿ = (1.65 + 0.5y)(3.3 - y)(12) = 65.34 - 6y2
Shear Stress: Applying the shear formula
t =
VQ
12(65.34 - 6y2)
=
It
390.60(12)
= 0.16728 - 0.01536y2
Resultant Shear Force: For the flange
Vf =
tdA
LA
3.3 in
=
L0.3 in
A 0.16728 - 0.01536y2 B (12dy)
= 3.82 kip
Ans.
476
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7–6. If the beam is subjected to a shear of V = 15 kN,
determine the web’s shear stress at A and B. Indicate the
shear-stress components on a volume element located
at these points. Show that the neutral axis is located at
y = 0.1747 m from the bottom and INA = 0.2182110-32 m4.
200 mm
A
30 mm
25 mm
V
(0.015)(0.125)(0.03) + (0.155)(0.025)(0.25) + (0.295)(0.2)(0.03)
y =
= 0.1747 m
0.125(0.03) + (0.025)(0.25) + (0.2)(0.03)
I =
1
(0.125)(0.033) + 0.125(0.03)(0.1747 - 0.015)2
12
+
1
(0.025)(0.253) + 0.25(0.025)(0.1747 - 0.155)2
12
+
1
(0.2)(0.033) + 0.2(0.03)(0.295 - 0.1747)2 = 0.218182 (10 - 3) m4
12
B
250 mm
30 mm
125 mm
œ
QA = yAA
= (0.310 - 0.015 - 0.1747)(0.2)(0.03) = 0.7219 (10 - 3) m3
QB = yABœ = (0.1747 - 0.015)(0.125)(0.03) = 0.59883 (10 - 3) m3
tA =
15(103)(0.7219)(10 - 3)
VQA
= 1.99 MPa
=
It
0.218182(10 - 3)(0.025)
Ans.
tB =
VQB
15(103)(0.59883)(10 - 3)
= 1.65 MPa
=
It
0.218182(10 - 3)0.025)
Ans.
7–7. If the wide-flange beam is subjected to a shear of
V = 30 kN, determine the maximum shear stress in the beam.
200 mm
A
30 mm
25 mm
V
B
250 mm
30 mm
Section Properties:
I =
1
1
(0.2)(0.310)3 (0.175)(0.250)3 = 268.652(10) - 6 m4
12
12
Qmax = © y¿A = 0.0625(0.125)(0.025) + 0.140(0.2)(0.030) = 1.0353(10) - 3 m3
tmax =
VQ
30(10)3(1.0353)(10) - 3
= 4.62 MPa
=
It
268.652(10) - 6 (0.025)
Ans.
477
200 mm
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*7–8. If the wide-flange beam is subjected to a shear of
V = 30 kN, determine the shear force resisted by the web
of the beam.
200 mm
A
30 mm
1
1
(0.2)(0.310)3 (0.175)(0.250)3 = 268.652(10) - 6 m4
12
12
I =
Q = a
25 mm
V
B
0.155 + y
b (0.155 - y)(0.2) = 0.1(0.024025 - y2)
2
250 mm
30(10)3(0.1)(0.024025 - y2)
tf =
268.652(10)
-6
30 mm
200 mm
(0.2)
0.155
Vf =
L
tf dA = 55.8343(10)6
L0.125
= 11.1669(10)6[ 0.024025y -
(0.024025 - y2)(0.2 dy)
1 3 0.155
y ]
2 0.125
Vf = 1.457 kN
Vw = 30 - 2(1.457) = 27.1 kN
Ans.
•7–9. Determine the largest shear force V that the member
can sustain if the allowable shear stress is tallow = 8 ksi.
3 in.
1 in.
V
3 in. 1 in.
1 in.
y =
(0.5)(1)(5) + 2 [(2)(1)(2)]
= 1.1667 in.
1 (5) + 2 (1)(2)
I =
1
(5)(13) + 5 (1)(1.1667 - 0.5)2
12
+ 2a
1
b (1)(23) + 2 (1)(2)(2 - 1.1667)2 = 6.75 in4
12
Qmax = ©y¿A¿ = 2 (0.91665)(1.8333)(1) = 3.3611 in3
tmax = tallow =
8 (103) = -
VQmax
It
V (3.3611)
6.75 (2)(1)
V = 32132 lb = 32.1 kip
Ans.
478
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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.
479
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*7–12. The beam has a rectangular cross section and is
made of wood having an allowable shear stress of tallow =
200 psi. Determine the maximum shear force V that can be
developed in the cross section of the beam. Also, plot the
shear-stress variation over the cross section.
V
12 in.
8 in.
Section Properties The moment of inertia of the cross-section about the neutral axis is
I =
1
(8) (123) = 1152 in4
12
Q as the function of y, Fig. a,
Q =
1
(y + 6)(6 - y)(8) = 4 (36 - y2)
2
Qmax occurs when y = 0. Thus,
Qmax = 4(36 - 02) = 144 in3
The maximum shear stress occurs of points along the neutral axis since Q is
maximum and the thickness t = 8 in. is constant.
tallow =
VQmax
;
It
200 =
V(144)
1152(8)
V = 12800 16 = 12.8 kip
Ans.
Thus, the shear stress distribution as a function of y is
12.8(10 ) C 4(36 - y ) D
VQ
=
It
1152 (8)
3
t =
2
= E 5.56 (36 - y2) F psi
480
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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:
60 mm
1
1
INA =
(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
20 mm
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7–15. Plot the shear-stress distribution over the cross
section of a rod that has a radius c. By what factor is the
maximum shear stress greater than the average shear stress
acting over the cross section?
c
y
V
x = 2c2 - y2 ;
p 4
c
4
I =
t = 2 x = 2 2c2 - y2
dA = 2 x dy = 22c2 - y2 dy
dQ = ydA = 2y 2c2 - y2 dy
x
Q =
Ly
2y2c2 - y2 dy = -
3 x
2
2 2
2
(c - y2)2 | y = (c2 - y2)3
3
3
3
V[23 (c2 - y2)2]
VQ
4V 2
t =
=
=
[c - y2)
p 4
2
2
It
3pc4
( c )(2 2c - y )
4
The maximum shear stress occur when y = 0
tmax =
4V
3 p c2
tavg =
V
V
=
A
p c2
4V
The faector =
tmax
4
3 pc2
= V =
tavg
3
2
pc
Ans.
482
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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
V(4h2>3a)(x2)(1 - a )
VQ
=
It
((1>36)(a)(h3))(2x)
2x
t =
t =
24V(x - a2 x2)
a2h
24V
4
dt
= 2 2 a 1 - xb = 0
a
dx
ah
At x =
y =
a
4
h
2h a
a b =
a 4
2
tmax =
24V a
2 a
a b a1 - a b b
a 4
a2h 4
tmax =
3V
ah
Ans.
No, because the shear stress is not perpendicular to the boundary. See Sec. 7-3.
483
h
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•7–17.
Determine the maximum shear stress in the strut if
it is subjected to a shear force of V = 600 kN.
30 mm
150 mm
V
100 mm
100 mm
100 mm
The moment of inertia of the cross-section about the neutral axis is
I =
1
1
(0.3)(0.213) (0.2)(0.153) = 0.175275(10 - 3) m4
12
12
From Fig. a,
Qmax = ©y¿A¿ = 0.09(0.03)(0.3) + 0.0375(0.075)(0.1)
= 1.09125(10 - 3) m3
The maximum shear stress occurs at the points along the neutral axis since Q is
maximum and thickness t = 0.1 m is the smallest.
tmax =
VQmax
600(103)[1.09125(10 - 3)]
=
It
0.175275(10 - 3) (0.1)
= 37.36(106) Pa = 37.4 MPa
Ans.
484
30 mm
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7–18. Determine the maximum shear force V that the strut
can support if the allowable shear stress for the material is
tallow = 45 MPa.
30 mm
150 mm
V
100 mm
100 mm
100 mm
The moment of inertia of the cross-section about the neutral axis is
I =
1
1
(0.3)(0.213) (0.2)(0.153) = 0.175275 (10 - 3) m4
12
12
From Fig. a
Qmax = ©y¿A¿ = 0.09(0.03)(0.3) + 0.0375 (0.075)(0.1)
= 1.09125 (10 - 3) m3
The maximum shear stress occeurs at the points along the neutral axis since Q is
maximum and thickness t = 0.1 m is the smallest.
tallow =
VQmax
;
It
45(106) =
V C 1.09125(10 - 3) D
0.175275(10 - 3)(0.1)
V = 722.78(103) N = 723 kN
Ans.
485
30 mm
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7–19. Plot the intensity of the shear stress distributed over
the cross section of the strut if it is subjected to a shear force
of V = 600 kN.
30 mm
The moment of inertia of the cross-section about the neutral axis is
I =
1
1
(0.3)(0.213) (0.2)(0.153) = 0.175275 (10 - 3) m4
12
12
For 0.075 m 6 y … 0.105 m, Fig. a, Q as a function of y is
Q = y¿A¿ =
1
(0.105 + y) (0.105 - y)(0.3) = 1.65375(10 - 3) - 0.15y2
2
For 0 … y 6 0.075 m, Fig. b, Q as a function of y is
Q = ©y¿A¿ = 0.09 (0.03)(0.3) +
1
(0.075 + y)(0.075 - y)(0.1) = 1.09125(10 - 3) - 0.05 y2
2
For 0.075 m 6 y … 0.105 m, t = 0.3 m. Thus,
t =
600 (103) C 1.65375(10 - 3) - 0.15y2 D
VQ
= (18.8703 - 1711.60y2) MPa
=
It
0.175275(10 - 3) (0.3)
At y = 0.075 m and y = 0.105 m,
t|y = 0.015 m = 9.24 MPa
ty = 0.105 m = 0
For 0 … y 6 0.075 m, t = 0.1 m. Thus,
t =
VQ
600 (103) [1.09125(10 - 3) - 0.05 y2]
= (37.3556 - 1711.60 y2) MPa
=
It
0.175275(10 - 3) (0.1)
At y = 0 and y = 0.075 m,
t|y = 0 = 37.4 MPa
ty = 0.075 m = 27.7 MPa
The plot shear stress distribution over the cross-section is shown in Fig. c.
486
150 mm
V
100 mm
100 mm
100 mm
30 mm
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*7–20. The steel rod is subjected to a shear of 30 kip.
Determine the maximum shear stress in the rod.
The moment of inertia of the ciralor cross-section about the neutral axis (x axis) is
p
p
I = r4 = (24) = 4 p in4
4
4
30 kip
dQ = ydA = y (2xdy) = 2xy dy
1
However, from the equation of the circle, x = (4 - y2)2 , Then
1
dQ = 2y(4 - y2)2 dy
Thus, Q for the area above y is
2 in
1
2y (4 - y2)2 dy
Ly
3 2 in
2
= - (4 - y2)2 y
3
=
3
2
(4 - y2)2
3
1
Here, t = 2x = 2 (4 - y2)2 . Thus
30 C 23 (4 - y2)2 D
VQ
=
t =
1
It
4p C 2(4 - y2)2 D
3
t =
5
(4 - y2) ksi
2p
By inspecting this equation, t = tmax at y = 0. Thus
¿=
tmax
A
2 in.
Q for the differential area shown shaded in Fig. a is
Q =
1 in.
20
10
= 3.18 ksi
=
p
2p
Ans.
487
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•7–21.
The steel rod is subjected to a shear of 30 kip.
Determine the shear stress at point A. Show the result on a
volume element at this point.
1 in.
A
The moment of inertia of the circular cross-section about the neutral axis (x axis) is
I =
2 in.
p 4
p
r = (24) = 4p in4
4
4
30 kip
Q for the differential area shown in Fig. a is
dQ = ydA = y (2xdy) = 2xy dy
1
However, from the equation of the circle, x = (4 - y2)2 , Then
1
dQ = 2y (4 - y2)2 dy
Thus, Q for the area above y is
2 in.
1
Q =
Ly
= -
2y (4 - y2)2 dy
2 in.
3
3
2
2
(4 - y2)2 `
= (4 - y2)2
3
3
y
1
Here t = 2x = 2 (4 - y2)2 . Thus,
30 C 23 (4 - y2)2 D
VQ
=
t =
1
It
4p C 2(4 - y2)2 D
3
t =
5
(4 - y2) ksi
2p
For point A, y = 1 in. Thus
tA =
5
(4 - 12) = 2.39 ksi
2p
Ans.
The state of shear stress at point A can be represented by the volume element
shown in Fig. b.
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7–22. Determine the shear stress at point B on the web of
the cantilevered strut at section a–a.
2 kN
250 mm
a
250 mm
4 kN
300 mm
a
20 mm
70 mm
(0.01)(0.05)(0.02) + (0.055)(0.07)(0.02)
y =
= 0.03625 m
(0.05)(0.02) + (0.07)(0.02)
I =
+
B
20 mm
50 mm
1
(0.05)(0.023) + (0.05)(0.02)(0.03625 - 0.01)2
12
1
(0.02)(0.073) + (0.02)(0.07)(0.055 - 0.03625)2 = 1.78625(10 - 6) m4
12
yBœ = 0.03625 - 0.01 = 0.02625 m
QB = (0.02)(0.05)(0.02625) = 26.25(10 - 6) m3
tB =
6(103)(26.25)(10 - 6)
VQB
=
It
1.78622(10 - 6)(0.02)
= 4.41 MPa
Ans.
7–23. Determine the maximum shear stress acting at
section a–a of the cantilevered strut.
2 kN
250 mm
a
250 mm
4 kN
300 mm
a
20 mm
70 mm
y =
(0.01)(0.05)(0.02) + (0.055)(0.07)(0.02)
= 0.03625 m
(0.05)(0.02) + (0.07)(0.02)
I =
1
(0.05)(0.023) + (0.05)(0.02)(0.03625 - 0.01)2
12
+
20 mm
50 mm
1
(0.02)(0.073) + (0.02)(0.07)(0.055 - 0.03625)2 = 1.78625(10 - 6) m4
12
Qmax = y¿A¿ = (0.026875)(0.05375)(0.02) = 28.8906(10 - 6) m3
tmax =
B
VQmax
6(103)(28.8906)(10 - 6)
=
It
1.78625(10 - 6)(0.02)
= 4.85 MPa
Ans.
489
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*7–24. Determine the maximum shear stress in the T-beam
at the critical section where the internal shear force is
maximum.
10 kN/m
A
1.5 m
3m
The shear diagram is shown in Fig. b. As indicated, Vmax = 27.5 kN
150 mm
The neutral axis passes through centroid c of the cross-section, Fig. c.
'
0.075(0.15)(0.03) + 0.165(0.03)(0.15)
© y A
=
y =
©A
0.15(0.03) + 0.03(0.15)
150 mm
1
(0.03)(0.153) + 0.03(0.15)(0.12 - 0.075)2
12
+
1
(0.15)(0.033) + 0.15(0.03)(0.165 - 0.12)2
12
= 27.0 (10 - 6) m4
From Fig. d,
Qmax = y¿A¿ = 0.06(0.12)(0.03)
= 0.216 (10 - 3) m3
The maximum shear stress occurs at points on the neutral axis since Q is maximum
and thickness t = 0.03 m is the smallest.
27.5(10 ) C 0.216(10 ) D
Vmax Qmax
=
It
27.0(10 - 6)(0.03)
3
tmax =
-3
= 7.333(106) Pa
= 7.33 MPa
Ans.
490
30 mm
30 mm
= 0.12 m
I =
B
C
The FBD of the beam is shown in Fig. a,
1.5 m
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•7–25.
Determine the maximum shear stress in the
T-beam at point C. Show the result on a volume element
at this point.
10 kN/m
A
B
C
1.5 m
3m
150 mm
150 mm
30 mm
using the method of sections,
+ c ©Fy = 0;
VC + 17.5 -
1
(5)(1.5) = 0
2
VC = -13.75 kN
The neutral axis passes through centroid C of the cross-section,
0.075 (0.15)(0.03) + 0.165(0.03)(0.15)
©yA
=
©A
0.15(0.03) + 0.03(0.15)
y =
= 0.12 m
I =
1
(0.03)(0.15) + 0.03(0.15)(0.12 - 0.075)2
12
+
1
(0.15)(0.033) + 0.15(0.03)(0.165 - 0.12)2
12
= 27.0 (10 - 6) m4
Qmax = y¿A¿ = 0.06 (0.12)(0.03)
= 0.216 (10 - 3) m3 490
The maximum shear stress occurs at points on the neutral axis since Q is maximum
and thickness t = 0.03 m is the smallest.
tmax =
30 mm
13.75(103) C 0.216(10 - 3) D
VC Qmax
=
It
27.0(10 - 6) (0.03)
= 3.667(106) Pa = 3.67 MPa
Ans.
491
1.5 m
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7–26. Determine the maximum shear stress acting in the
fiberglass beam at the section where the internal shear
force is maximum.
200 lb/ft
150 lb/ft
D
A
6 ft
6 ft
2 ft
4 in.
6 in.
0.5 in.
4 in.
Support Reactions: As shown on FBD.
Internal Shear Force: As shown on shear diagram, Vmax = 878.57 lb.
Section Properties:
INA =
1
1
(4) A 7.53 B (3.5) A 63 B = 77.625 in4
12
12
Qmax = ©y¿A¿
= 3.375(4)(0.75) + 1.5(3)(0.5) = 12.375 in3
Maximum Shear Stress: Maximum shear stress occurs at the point where the
neutral axis passes through the section.
Applying the shear formula
tmax =
=
VQmax
It
878.57(12.375)
= 280 psi
77.625(0.5)
Ans.
492
0.75 in.
0.75 in.
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7–27. Determine the shear stress at points C and D
located on the web of the beam.
3 kip/ft
D
A
C
6 ft
6 ft
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
B
6 ft
6 in.
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
6 ft
6 ft
The FBD is shown in Fig. a.
The shear diagram is shown in Fig. b, Vmax = 18.0 kip.
B
0.75 in.
6 ft
6 in.
1 in.
C
D
4 in.
4 in.
6 in.
1 in.
The moment of inertia of the beam’s cross-section about the neutral axis is
I =
1
1
(6)(103) (5.25)(83)
12
12
= 276 in4
From Fig. c
Qmax = ©y¿A¿ = 4.5 (1)(6) + 2(4)(0.75)
= 33 in3
The maximum shear stress occurs at points on the neutral axis since Q is the
maximum and thickness t = 0.75 in is the smallest
tmax =
Vmax Qmax
18.0 (33)
=
= 2.87 ksi
It
276 (0.75)
Ans.
494
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7–30. The beam has a rectangular cross section and is
subjected to a load P that is just large enough to develop a
fully plastic moment Mp = PL at the fixed support. If the
material is elastic-plastic, then at a distance x 6 L the
moment M = Px creates a region of plastic yielding with
an associated elastic core having a height 2y¿. This situation
has been described by Eq. 6–30 and the moment M is
distributed over the cross section as shown in Fig. 6–48e.
Prove that the maximum shear stress developed in the beam
is given by tmax = 321P>A¿2, where A¿ = 2y¿b, the crosssectional area of the elastic core.
P
x
Plastic region
2y¿
h
b
Elastic region
Force Equilibrium: The shaded area indicares the plastic zone. Isolate an element in
the plastic zone and write the equation of equilibrium.
; ©Fx = 0;
tlong A2 + sg A1 - sg A1 = 0
tlong = 0
This proves that the longitudinal shear stress. tlong, is equal to zero. Hence the
corresponding transverse stress, tmax, is also equal to zero in the plastic zone.
Therefore, the shear force V = P is carried by the malerial only in the elastic zone.
Section Properties:
INA =
1
2
(b)(2y¿)3 = b y¿ 3
12
3
Qmax = y¿ A¿ =
y¿
y¿ 2b
(y¿)(b) =
2
2
Maximum Shear Stress: Applying the shear formula
VA 2 B
VQmax
3P
tmax =
=
=
2
3
It
4by¿
A by¿ B (b)
y¿ 3b
3
However,
A¿ = 2by¿
tmax =
3P
‚
2A¿
hence
(Q.E.D.)
495
L
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7–31. The beam in Fig. 6–48f is subjected to a fully plastic
moment Mp . Prove that the longitudinal and transverse
shear stresses in the beam are zero. Hint: Consider an element
of the beam as shown in Fig. 7–4c.
P
x
Plastic region
2y¿
h
b
Elastic region
L
Force Equilibrium: If a fully plastic moment acts on the cross section, then an
element of the material taken from the top or bottom of the cross section is
subjected to the loading shown. For equilibrium
; ©Fx = 0;
sg A1 + tlong A2 - sg A1 = 0
tlong = 0
Thus no shear stress is developed on the longitudinal or transverse plane of the
element. (Q. E. D.)
*7–32. The beam is constructed from two boards fastened
together at the top and bottom with two rows of nails
spaced every 6 in. If each nail can support a 500-lb shear
force, determine the maximum shear force V that can be
applied to the beam.
6 in.
6 in.
2 in.
2 in.
V
6 in.
Section Properties:
I =
1
(6) A 43 B = 32.0 in4
12
Q = y¿A¿ = 1(6)(2) = 12.0 in4
Shear Flow: There are two rows of nails. Hence, the allowable shear flow
2(500)
= 166.67 lb>in.
q =
6
q =
166.67 =
VQ
I
V(12.0)
32.0
V = 444 lb
Ans.
496
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•7–33.
The beam is constructed from two boards
fastened together at the top and bottom with two rows of
nails spaced every 6 in. If an internal shear force of
V = 600 lb is applied to the boards, determine the shear
force resisted by each nail.
6 in.
6 in.
2 in.
2 in.
Section Properties:
I =
1
(6) A 43 B = 32.0 in4
12
V
6 in.
Q = y¿A¿ = 1(6)(2) = 12.0 in4
Shear Flow:
q =
VQ
600(12.0)
=
= 225 lb>in.
I
32.0
There are two rows of nails. Hence, the shear force resisted by each nail is
q
225 lb>in.
F = a bs = a
b(6 in.) = 675 lb
2
2
Ans.
7–34. The beam is constructed from two boards fastened
together with three rows of nails spaced s = 2 in. apart. If
each nail can support a 450-lb shear force, determine the
maximum shear force V that can be applied to the beam. The
allowable shear stress for the wood is tallow = 300 psi.
s
s
1.5 in.
The moment of inertia of the cross-section about the neutral axis is
I =
V
1
(6)(33) = 13.5 in4
12
6 in.
Refering to Fig. a,
QA = Qmax = y¿A¿ = 0.75(1.5)(6) = 6.75 in3
The maximum shear stress occurs at the points on the neutral axis where Q is
maximum and t = 6 in.
tallow =
VQmax
;
It
300 =
V(6.75)
13.5(6)
V = 3600 lb = 3.60 kips
Shear Flow: Since there are three rows of nails,
F
450
b = 675 lb>in.
qallow = 3 a b = 3 a
s
2
VQA
V(6.75)
;
675 =
qallow =
I
13.5
V = 1350 lb = 1.35 kip
497
Ans.
1.5 in.
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7–35. The beam is constructed from two boards fastened
together with three rows of nails. If the allowable shear
stress for the wood is tallow = 150 psi, determine the
maximum shear force V that can be applied to the beam.
Also, find the maximum spacing s of the nails if each nail
can resist 650 lb in shear.
s
s
1.5 in.
V
6 in.
The moment of inertia of the cross-section about the neutral axis is
I =
1
(6)(33) = 13.5 in4
12
Refering to Fig. a,
QA = Qmax = y¿A¿ = 0.75(1.5)(6) = 6.75 in3
The maximum shear stress occurs at the points on the neutral axis where Q is
maximum and t = 6 in.
tallow =
VQmax
;
It
150 =
V(6.75)
13.5(6)
V = 1800 lb = 1.80 kip
Since there are three rows of nails, qallow = 3 a
qallow =
VQA
;
I
Ans.
F
650
1950 lb
b = 3¢
b
≤ = a
s
s
s
in.
1800(6.75)
1950
=
s
13.5
s = 2.167 in = 2
1
in
8
Ans.
498
1.5 in.
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*7–36. The beam is fabricated from two equivalent
structural tees and two plates. Each plate has a height of
6 in. and a thickness of 0.5 in. If a shear of V = 50 kip is
applied to the cross section, determine the maximum spacing
of the bolts. Each bolt can resist a shear force of 15 kip.
0.5 in.
s
3 in.
1 in.
A
Section Properties:
INA =
V
6 in.
1
1
(3) A 93 B (2.5) A 83 B
12
12
0.5 in.
N
1
1
(0.5) A 23 B +
(1) A 63 B
12
12
3 in.
= 93.25 in4
Q = ©y¿A¿ = 2.5(3)(0.5) + 4.25(3)(0.5) = 10.125 in3
Shear Flow: Since there are two shear planes on the bolt, the allowable shear flow is
2(15)
30
.
=
q =
s
s
VQ
q =
I
50(10.125)
30
=
s
93.25
s = 5.53 in.
Ans.
•7–37.
The beam is fabricated from two equivalent
structural tees and two plates. Each plate has a height of
6 in. and a thickness of 0.5 in. If the bolts are spaced at
s = 8 in., determine the maximum shear force V that can
be applied to the cross section. Each bolt can resist a
shear force of 15 kip.
0.5 in.
s
3 in.
1 in.
A
Section Properties:
-
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
INA =
(3) A 93 B (2.5) A 83 B
12
12
VQ
I
V(10.125)
93.25
y = 34.5 kip
Ans.
499
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7–38. The beam is subjected to a shear of V = 2 kN.
Determine the average shear stress developed in each nail
if the nails are spaced 75 mm apart on each side of the
beam. Each nail has a diameter of 4 mm.
The neutral axis passes through centroid C of the cross-section as shown in Fig. a.
'
0.175(0.05)(0.2) + 0.1(0.2)(0.05)
© y A
y =
=
= 0.1375 m
©A
0.05(0.2) + 0.2(0.05)
200 mm
75 mm
50 mm 75 mm
25 mm
V
200 mm
Thus,
I =
1
(0.2)(0.053) + 0.2 (0.05)(0.175 - 0.1375)2
12
+
25 mm
1
(0.05)(0.23) + 0.05(0.2)(0.1375 - 0.1)2
12
= 63.5417(10 - 6) m4
Q for the shaded area shown in Fig. b is
Q = y¿A¿ = 0.0375 (0.05)(0.2) = 0.375(10 - 3) m3
Since there are two rows of nails q = 2a
q =
VQ
;
I
26.67 F =
F
2F
b =
= (26.67 F) N>m.
s
0.075
2000 C 0.375 (10 - 3) D
63.5417 (10 - 6)
F = 442.62 N
Thus, the shear stress developed in the nail is
tn =
F
442.62
=
= 35.22(106)Pa = 35.2 MPa
p
A
2
(0.004 )
4
Ans.
500
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7–39. A beam is constructed from three boards bolted
together as shown. Determine the shear force developed
in each bolt if the bolts are spaced s = 250 mm apart and the
applied shear is V = 35 kN.
25 mm
25 mm
100 mm 250 mm
2 (0.125)(0.25)(0.025) + 0.275 (0.35)(0.025)
y =
= 0.18676 m
2 (0.25)(0.025) + 0.35 (0.025)
I = (2)a
+
1
b(0.025)(0.253) + 2 (0.025)(0.25)(0.18676 - 0.125)2
12
V
1
(0.025)(0.35)3 + (0.025)(0.35)(0.275 - 0.18676)2
12
350 mm
s = 250 mm
= 0.270236 (10 - 3) m4
25 mm
-3
3
Q = y¿A¿ = 0.06176(0.025)(0.25) = 0.386(10 ) m
q =
35 (0.386)(10 - 3)
VQ
= 49.997 kN>m
=
I
0.270236 (10 - 3)
F = q(s) = 49.997 (0.25) = 12.5 kN
Ans.
*7–40. The double-web girder is constructed from two
plywood sheets that are secured to wood members at its top
and bottom. If each fastener can support 600 lb in single
shear, determine the required spacing s of the fasteners
needed to support the loading P = 3000 lb. Assume A is
pinned and B is a roller.
2 in.
2 in.
s
10 in.
A
4 ft
2 in.
2 in.
6 in.
0.5 in.
0.5 in.
Support Reactions: As shown on FBD.
Internal Shear Force: As shown on shear diagram, Vmax = 1500 lb.
Section Properties:
INA =
P
1
1
(7) A 183 B (6) A 103 B = 2902 in4
12
12
Q = y¿A¿ = 7(4)(6) = 168 in3
Shear Flow: Since there are two shear planes on the bolt, the allowable shear flow is
2(600)
1200
=
q =
.
s
s
VQ
q =
I
1500(168)
1200
=
s
2902
s = 13.8 in.
Ans.
501
4 ft
B
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•7–41.
The double-web girder is constructed from two
plywood sheets that are secured to wood members at its top
and bottom. The allowable bending stress for the wood is
sallow = 8 ksi and the allowable shear stress is tallow = 3 ksi.
If the fasteners are spaced s = 6 in. and each fastener can
support 600 lb in single shear, determine the maximum load
P that can be applied to the beam.
2 in.
2 in.
s
10 in.
A
4 ft
2 in.
2 in.
6 in.
0.5 in.
0.5 in.
Support Reactions: As shown on FBD.
Internal Shear Force and Moment: As shown on shear and moment diagram,
Vmax = 0.500P and Mmax = 2.00P.
Section Properties:
INA =
P
1
1
(7) A 183 B (6) A 103 B = 2902 in4
12
12
Q = y2œ A¿ = 7(4)(6) = 168 in3
Qmax = ©y¿A¿ = 7(4)(6) + 4.5(9)(1) = 208.5 in3
Shear Flow: Assume bolt failure. Since there are two shear planes on the bolt, the
2(600)
= 200 lb>in.
allowable shear flow is q =
6
VQ
q =
I
0.500P(168)
200 =
2902
P = 6910 lb = 6.91 kip (Controls !)
Ans.
Shear Stress: Assume failure due to shear stress.
VQmax
It
0.500P(208.5)
3000 =
2902(1)
tmax = tallow =
P = 22270 lb = 83.5 kip
Bending Stress: Assume failure due to bending stress.
Mc
I
2.00P(12)(9)
8(103) =
2902
smax = sallow =
P = 107 ksi
502
4 ft
B
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7–42. The T-beam is nailed together as shown. If the nails
can each support a shear force of 950 lb, determine the
maximum shear force V that the beam can support and the
corresponding maximum nail spacing s to the nearest 18 in.
The allowable shear stress for the wood is tallow = 450 psi.
2 in.
s
The neutral axis passes through the centroid c of the cross-section as shown in Fig. a.
'
13(2)(12) + 6(12)(2)
© y A
y =
=
= 9.5 in.
©A
2(12) + 12(2)
I =
1
(2)(123) + 2(12)(9.5 - 6)2
12
+
= 884 in4
Refering to Fig. a, Qmax and QA are
Qmax = y1œ A1œ = 4.75(9.5)(2) = 90.25 in3
QA = y2œ A2œ = 3.5 (2)(12) = 84 in3
The maximum shear stress occurs at the points on the neutral axis where Q is
maximum and t = 2 in.
VQmax
;
It
450 =
V (90.25)
884 (2)
V = 8815.51 lb = 8.82 kip
Here, qallow =
F
950
=
lb>in. Then
s
s
VQA
;
qallow =
I
Ans.
8815.51(84)
950
=
s
884
s = 1.134 in = 1
12 in.
V
2 in.
1
(12)(23) + 12(2)(13 - 9.5)2
12
tallow =
s
12 in.
1
in
8
Ans.
503
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7–43. Determine the average shear stress developed in the
nails within region AB of the beam. The nails are located on
each side of the beam and are spaced 100 mm apart. Each
nail has a diameter of 4 mm. Take P = 2 kN.
P
2 kN/m
A
B
C
1.5 m
The FBD is shown in Fig. a.
As indicated in Fig. b, the internal shear force on the cross-section within region AB
is constant that is VAB = 5 kN.
1.5 m
100 mm
The neutral axis passes through centroid C of the cross section as shown in Fig. c.
'
0.18(0.04)(0.2) + 0.1(0.2)(0.04)
© y A
=
y =
©A
0.04(0.2) + 0.2(0.04)
40 mm
= 0.14 m
200 mm
1
I =
(0.04)(0.23) + 0.04(0.2)(0.14 - 0.1)2
12
1
+
(0.2)(0.043) + 0.2(0.04)(0.18 - 0.14)2
12
200 mm 20 mm
20 mm
= 53.333(10 - 6) m4
Q for the shaded area shown in Fig. d is
Q = y¿A¿ = 0.04(0.04)(0.2) = 0.32(10 - 3) m3
Since there are two rows of nail, q = 2 a
q =
VAB Q
;
I
20F =
F
F
b = 2a
b = 20F N>m.
s
0.1
5(103) C 0.32(10 - 3) D
53.333(10 - 6)
F = 1500 N
Thus, the average shear stress developed in each nail is
A tnail B avg =
F
1500
=
= 119.37(106)Pa = 119 MPa
p
Anail
2
(0.004 )
4
504
Ans.
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*7–44. The nails are on both sides of the beam and each
can resist a shear of 2 kN. In addition to the distributed
loading, determine the maximum load P that can be applied
to the end of the beam. The nails are spaced 100 mm apart
and the allowable shear stress for the wood is tallow = 3 MPa.
P
2 kN/m
A
B
C
1.5 m
1.5 m
100 mm
The FBD is shown in Fig. a.
40 mm
As indicated the shear diagram, Fig. b, the maximum shear occurs in region AB of
Constant value, Vmax = (P + 3) kN.
The neutral axis passes through Centroid C of the cross-section as shown in Fig. c.
'
0.18(0.04)(0.2) + 0.1(0.2)(0.04)
© y A
=
y =
©A
0.04(0.2) + 0.2(0.04)
= 0.14 m
I =
1
(0.04)(0.23) + 0.04(0.2)(0.14 - 0.1)2
12
1
+
(0.2)(0.043) + 0.2(0.04)(0.18 - 0.142)
12
Refering to Fig. d,
Qmax = y1œ A1œ = 0.07(0.14)(0.04) = 0.392(10 - 3) m3
QA = y2œ A2œ = 0.04(0.04)(0.2) = 0.32(10 - 3) m3
The maximum shear stress occurs at the points on Neutral axis where Q is maximum
and t = 0.04 m.
Vmax Qmax
;
It
3(106) =
(P + 3)(103) C 0.392(10 - 3) D
53.333(10 - 6)(0.04)
P = 13.33 kN
Since there are two rows of nails qallow = 2 a
Vmax QA
qallow =
;
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
506
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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
509
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7–50. A shear force of V = 300 kN is applied to the box
girder. Determine the shear flow at points A and B.
90 mm
90 mm
C
A
D
200 mm
B
190 mm
V
200 mm
10 mm
180 mm
10 mm
The moment of inertia of the cross-section about the neutral axis is
I =
1
1
(0.2)(0.43) (0.18)(0.383) = 0.24359(10 - 3) m4
12
12
Refering to Fig. a Fig. b,
QA = y1œ A1œ = 0.195 (0.01)(0.19) = 0.3705 (10 - 3) m3
QB = 2yzœ A2œ + y3œ A3œ = 2 [0.1(0.2)(0.01)] + 0.195(0.01)(0.18) = 0.751(10 - 3) m3
Due to symmety, the shear flow at points A and A¿ , Fig. a, and at points B and B¿ ,
Fig. b, are the same. Thus
3
-3
1 300(10 ) C 0.3705(10 ) D
1 VQA
s
b = c
qA = a
2
I
2
0.24359(10 - 3)
= 228.15(103) N>m = 228 kN>m
qB =
Ans.
3
-3
1 VQB
1 300(10 ) C 0.751(10 ) D
s
a
b = c
2
I
2
0.24359(10 - 3)
= 462.46(103) N>m = 462 kN>m
Ans.
510
100 mm
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7–51. A shear force of V = 450 kN is applied to the box
girder. Determine the shear flow at points C and D.
90 mm
90 mm
C
A
D
200 mm
B
190 mm
V
200 mm
10 mm
180 mm
10 mm
The moment of inertia of the cross-section about the neutral axis is
I =
1
1
(0.2)(0.43) (0.18)(0.383) = 0.24359(10 - 3) m4
12
12
Refering to Fig. a, due to symmetry ACœ = 0. Thus
QC = 0
Then refering to Fig. b,
QD = y1œ A1œ + y2œ A2œ = 0.195 (0.01)(0.09) + 0.15(0.1)(0.01)
= 0.3255(10 - 3) m3
Thus,
qC =
qD =
VQC
= 0
I
Ans.
450(103) C 0.3255(10 - 3) D
VQD
=
I
0.24359(10 - 3)
= 601.33(103) N>m = 601 kN>m
Ans.
100 mm
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*7–52. A shear force of V = 18 kN is applied to the
symmetric box girder. Determine the shear flow at A and B.
10 mm
30 mm
10 mm
A
100 mm
C
B
100 mm
150 mm
10 mm
10 mm
V
150 mm
10 mm 125 mm
10 mm
Section Properties:
INA =
1
1
(0.145) A 0.33 B (0.125) A 0.283 B
12
12
+ 2c
1
(0.125) A 0.013 B + 0.125(0.01) A 0.1052 B d
12
= 125.17 A 10 - 6 B m4
QA = y2œ A¿ = 0.145(0.125)(0.01) = 0.18125 A 10 - 3 B m3
QB = y1œ A¿ = 0.105(0.125)(0.01) = 0.13125 A 10 - 3 B m3
Shear Flow:
qA =
=
1 VQA
c
d
2
I
1 18(103)(0.18125)(10 - 3)
d
c
2
125.17(10 - 6)
= 13033 N>m = 13.0 kN>m
qB =
=
Ans.
1 VQB
c
d
2
I
1 18(103)(0.13125)(10 - 3)
d
c
2
125.17(10 - 6)
= 9437 N>m = 9.44 kN>m
Ans.
512
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A shear force of V = 18 kN is applied to the box
girder. Determine the shear flow at C.
•7–53.
10 mm
30 mm
10 mm
A
100 mm
C
B
100 mm
150 mm
10 mm
10 mm
V
150 mm
10 mm 125 mm
10 mm
Section Properties:
INA =
1
1
(0.145) A 0.33 B (0.125) A 0.283 B
12
12
+2 c
1
(0.125) A 0.013 B + 0.125(0.01) A 0.1052 B d
12
= 125.17 A 10 - 6 B m4
QC = ©y¿A¿
= 0.145(0.125)(0.01) + 0.105(0.125)(0.01) + 0.075(0.15)(0.02)
= 0.5375 A 10 - 3 B m3
Shear Flow:
qC =
=
1 VQC
c
d
2
I
1 18(103)(0.5375)(10 - 3)
d
c
2
125.17(10 - 4)
= 38648 N>m = 38.6 kN>m
Ans.
513
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7–54. The aluminum strut is 10 mm thick and has the cross
section shown. If it is subjected to a shear of, V = 150 N,
determine the shear flow at points A and B.
10 mm
2[0.005(0.03)(0.01)] + 2[0.03(0.06)(0.01)] + 0.055(0.04)(0.01)
= 0.027727 m
y =
2(0.03)(0.01) + 2(0.06)(0.01) + 0.04(0.01)
I = 2c
1
(0.03)(0.01)3 + 0.03(0.01)(0.027727 - 0.005)2 d
12
+ 2c
+
40 mm
10 mm
30 mm
1
(0.01)(0.06)3 + 0.01(0.06)(0.03 - 0.027727)2 d
12
B
A
V
40 mm
10 mm
30 mm
10 mm
1
(0.04)(0.01)3 + 0.04(0.01)(0.055 - 0.027727)2 = 0.98197(10 - 6) m4
12
yB ¿ = 0.055 - 0.027727 = 0.027272 m
yA ¿ = 0.027727 - 0.005 = 0.022727 m
QA = yA ¿A¿ = 0.022727(0.04)(0.01) = 9.0909(10 - 6) m3
QB = yB ¿A¿ = 0.027272(0.03)(0.01) = 8.1818(10 - 6) m3
qA =
VQA
150(9.0909)(10 - 6)
= 1.39 kN>m
=
I
0.98197(10 - 6)
Ans.
qB =
VQB
150(8.1818)(10 - 6)
= 1.25 kN>m
=
I
0.98197(10 - 6)
Ans.
7–55. The aluminum strut is 10 mm thick and has the cross
section shown. If it is subjected to a shear of V = 150 N,
determine the maximum shear flow in the strut.
y =
2[0.005(0.03)(0.01)] + 2[0.03(0.06)(0.01)] + 0.055(0.04)(0.01)
2(0.03)(0.01) + 2(0.06)(0.01) + 0.04(0.01)
10 mm
40 mm
B
A
= 0.027727 m
I = 2c
10 mm
1
(0.03)(0.01)3 + 0.03(0.01)(0.027727 - 0.005)2 d
12
30 mm
1
+ 2 c (0.01)(0.06)3 + 0.01(0.06)(0.03 - 0.027727)2 d
12
+
10 mm
1
(0.04)(0.01)3 + 0.04(0.01)(0.055 - 0.027727)2
12
= 0.98197(10 - 6) m4
Qmax = (0.055 - 0.027727)(0.04)(0.01) + 2[(0.06 - 0.027727)(0.01)]a
0.06 - 0.0277
b
2
= 21.3(10 - 6) m3
qmax =
V
40 mm
1 150(21.3(10 - 6))
1 VQmax
b = 1.63 kN>m
a
b = a
2
I
2 0.98197(10 - 6)
Ans.
514
30 mm
10 mm
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*7–56. The beam is subjected to a shear force of
V = 5 kip. Determine the shear flow at points A and B.
0.5 in.
C
5 in.
5 in. 0.5 in.
0.5 in.
2 in.
0.25(11)(0.5) + 2[4.5(8)(0.5)] + 6.25(10)(0.5)
©yA
y =
=
= 3.70946 in.
©A
11(0.5) + 2(8)(0.5) + 10(0.5)
A
D
8 in.
1
1
(11)(0.53) + 11(0.5)(3.70946 - 0.25)2 + 2c (0.5)(83) + 0.5(8)(4.5 - 3.70946)2 d
12
12
I =
+
0.5 in.
V
B
1
(10)(0.53) + 10(0.5)(6.25 - 3.70946)2
12
= 145.98 in4
œ
= 3.70946 - 0.25 = 3.45946 in.
yA
yBœ = 6.25 - 3.70946 = 2.54054 in.
œ
QA = yA
A¿ = 3.45946(11)(0.5) = 19.02703 in3
QB = yBœ A¿ = 2.54054(10)(0.5) = 12.7027 in3
qA =
1 VQA
1 5(103)(19.02703)
a
b = a
b = 326 lb>in.
2
I
2
145.98
Ans.
qB =
1 VQB
1 5(103)(12.7027)
a
b = a
b = 218 lb>in.
2
I
2
145.98
Ans.
•7–57.
The beam is constructed from four plates and is
subjected to a shear force of V = 5 kip. Determine the
maximum shear flow in the cross section.
0.5 in.
C
5 in.
5 in. 0.5 in.
y =
©yA
0.25(11)(0.5) + 2[4.5(8)(0.5)] + 6.25(10)(0.5)
=
= 3.70946 in.
©A
11(0.5) + 2(8)(0.5) + 10(0.5)
I =
1
1
(11)(0.53) + 11(0.5)(3.45952) + 2 c (0.5)(83) + 0.5(8)(0.79052) d
12
12
0.5 in.
2 in.
A
D
8 in.
V
+
1
(10)(0.53) + 10(0.5)(2.54052)
12
= 145.98 in4
Qmax = 3.4594 (11)(0.5) + 2[(1.6047)(0.5)(3.7094 - 0.5)]
= 24.177 in3
qmax =
1 VQmax
1 5(103)(24.177)
a
b = a
b
2
I
2
145.98
= 414 lb>in.
Ans.
515
0.5 in.
B
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7–58. The channel is subjected to a shear of V = 75 kN.
Determine the shear flow developed at point A.
30 mm
400 mm
A
200 mm
30 mm
V ⫽ 75 kN
30 mm
y =
©yA
0.015(0.4)(0.03) + 2[0.13(0.2)(0.03)]
=
= 0.0725 m
©A
0.4(0.03) + 2(0.2)(0.03)
I =
1
(0.4)(0.033) + 0.4(0.03)(0.0725 - 0.015)2
12
+ 2c
1
(0.03)(0.23) + 0.03(0.2)(0.13 - 0.0725)2 d = 0.12025(10 - 3) m4
12
œ
A¿ = 0.0575(0.2)(0.03) = 0.3450(10 - 3) m3
QA = yA
q =
qA =
VQ
I
75(103)(0.3450)(10 - 3)
0.12025(10 - 3)
= 215 kN>m
Ans.
7–59. The channel is subjected to a shear of V = 75 kN.
Determine the maximum shear flow in the channel.
30 mm
400 mm
A
V ⫽ 75 kN
30 mm
y =
©yA
0.015(0.4)(0.03) + 2[0.13(0.2)(0.03)]
=
©A
0.4(0.03) + 2(0.2)(0.03)
= 0.0725 m
1
I =
(0.4)(0.033) + 0.4(0.03)(0.0725 - 0.015)2
12
1
+ 2c (0.03)(0.23) + 0.03(0.2)(0.13 - 0.0725)2 d
12
= 0.12025(10 - 3) m4
Qmax = y¿A¿ = 0.07875(0.1575)(0.03) = 0.37209(10 - 3) m3
qmax =
75(103)(0.37209)(10 - 3)
0.12025(10 - 3)
= 232 kN>m
Ans.
516
200 mm
30 mm
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*7–60. The angle is subjected to a shear of V = 2 kip.
Sketch the distribution of shear flow along the leg AB.
Indicate numerical values at all peaks.
A
5 in.
5 in.
45⬚ 45⬚
0.25 in.
Section Properties:
b =
0.25
= 0.35355 in.
sin 45°
h = 5 cos 45° = 3.53553 in.
INA = 2c
1
(0.35355) A 3.535533 B d = 2.604167 in4
12
Q = y¿A¿ = [0.25(3.53553) + 0.5y]a2.5 -
y
b(0.25)
sin 45°
= 0.55243 - 0.17678y2
Shear Flow:
VQ
I
2(103)(0.55243 - 0.17678y2)
=
2.604167
q =
= {424 - 136y2} lb>in.
At y = 0,
Ans.
q = qmax = 424 lb>in.
Ans.
517
B
V
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•7–61.
The assembly is subjected to a vertical shear of
V = 7 kip. Determine the shear flow at points A and B and
the maximum shear flow in the cross section.
A
0.5 in.
B
V
2 in.
0.5 in.
0.5 in.
6 in.
6 in.
2 in.
0.5 in.
y =
©yA
(0.25)(11)(0.5) + 2(3.25)(5.5)(0.5) + 6.25(7)(0.5)
=
= 2.8362 in.
©A
0.5(11) + 2(0.5)(5.5) + 7(0.5)
I =
1
1
(11)(0.53) + 11(0.5)(2.8362 - 0.25)2 + 2a b(0.5)(5.53) + 2(0.5)(5.5)(3.25 - 2.8362)2
12
12
+
1
(7)(0.53) + (0.5)(7)(6.25 - 2.8362)2 = 92.569 in4
12
QA = y1 ¿A1 ¿ = (2.5862)(2)(0.5) = 2.5862 in3
QB = y2 ¿A2 ¿ = (3.4138)(7)(0.5) = 11.9483 in3
Qmax = ©y¿A¿ = (3.4138)(7)(0.5) + 2(1.5819)(3.1638)(0.5) = 16.9531 in3
q =
VQ
I
7(103)(2.5862)
= 196 lb>in.
92.569
1 7(103)(11.9483)
qB = a
b = 452 lb>in.
2
92.569
1 7(103)(16.9531)
b = 641 lb>in.
qmax = a
2
92.569
qA =
Ans.
Ans.
Ans.
518
0.5 in.
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7–62. Determine the shear-stress variation over the cross
section of the thin-walled tube as a function of elevation y and
show that t max = 2V>A, where A = 2prt. Hint: Choose a
differential area element dA = Rt du. Using dQ = y dA,
formulate Q for a circular section from u to (p - u) and show
that Q = 2R2t cos u, where cos u = 2R2 - y2>R.
ds
du
y
u
t
dA = R t du
dQ = y dA = yR t du
Here y = R sin u
Therefore dQ = R2 t sin u du
p-u
Q =
p-u
R2 t sin u du = R2 t(-cos u) |
Lu
u
2
= R t [-cos (p - u) - (-cos u)] = 2R2 t cos u
dI = y2 dA = y2 R t du = R3 t sin2 u du
2p
I =
L0
2p
R3 t sin2 u du = R3 t
2p
=
t =
sin 2u
R3 t
[u ]
2
2 0
R3 t
[2p - 0] = pR3 t
2
VQ
V(2R2t cos u)
V cos u
=
=
3
It
pR t
pR t(2t)
Here cos u =
t =
=
L0
(1 - cos 2u)
du
2
2R2 - y2
R
V
2R2 - y2
pR2t
Ans.
tmax occurs at y = 0; therefore
tmax =
V
pR t
A = 2pRt; therefore
tmax =
2V
A
QED
519
R
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7–63. Determine the location e of the shear center,
point O, for the thin-walled member having the cross
section shown where b2 7 b1. The member segments have
the same thickness t.
t
h
e
b2
Section Properties:
I =
1
h 2
t h2
t h3 + 2c(b1 + b2)ta b d =
C h + 6(b1 + b2) D
12
2
12
Q1 = y¿A¿ =
h
ht
(x )t =
x
2 1
2 1
Q2 = y¿A¿ =
h
ht
(x )t =
x
2 2
2 2
Shear Flow Resultant:
P A ht2 x1 B
VQ1
6P
q1 =
= 2
=
x1
th
I
C h + 6(b1 + b2) D h C h + 6(b1 + b2) D
12
q2 =
P A ht2 x2 B
VQ2
6P
= 2
=
x2
th
I
h
h
+
6(b
h
+
6(b
+
b
)
C
C
1 + b2) D
1
2 D
12
b1
(Ff)1 =
L0
q1 dx1 =
6P
b1
h C h + 6(b1 + b2) D L0
x1 dx1
3Pb21
=
b2
(Ff)2 =
L0
q2 dx2 =
h C h + 6(b1 + b2) D
6P
b2
h C h + 6(b1 + b2) D L0
x2 dx2
3Pb22
=
h C h + 6(b1 + b2) D
Shear Center: Summing moment about point A.
Pe = A Ff B 2 h - A Ff B 1 h
Pe =
e =
3Pb22
h C h + 6(b1 + b2) D
3(b22 - b21)
h + 6(b1 + b2)
(h) -
3Pb21
h C h + 6(b1 + b2) D
(h)
Ans.
Note that if b2 = b1, e = 0 (I shape).
520
b1
O
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*7–64. Determine the location e of the shear center,
point O, for the thin-walled member having the cross
section shown. The member segments have the same
thickness t.
b
d
45⬚
O
e
Section Properties:
I =
=
t
1
a
b(2d sin 45°)3 + 2 C bt(d sin 45°)2 D
12 sin 45°
td2
(d + 3b)
3
Q = y¿A¿ = d sin 45° (xt) = (td sin 45°)x
Shear Flow Resultant:
qf =
P(td sin 45°)x
VQ
3P sin 45°
=
=
x
td2
I
d(d + 3b)
(d + 3b)
3
b
Ff =
L0
b
qfdx =
2
3P sin 45°
3b sin 45°
P
xdx =
d(d + 3b) L0
2d(d + 3b)
Shear Center: Summing moments about point A,
Pe = Ff(2d sin 45°)
Pe = c
e =
3b2 sin 45°
P d(2d sin 45°)
2d(d + 3b)
3b2
2(d + 3b)
Ans.
521
45⬚
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•7–65.
Determine the location e of the shear center,
point O, for the thin-walled member having a slit along its
side. Each element has a constant thickness t.
a
e
a
t
a
Section Properties:
I =
1
10 3
(2t)(2a)3 + 2 C at A a2 B D =
a t
12
3
Q1 = y1œ A¿ =
y
t
(yt) = y2
2
2
Q2 = ©y¿A¿ =
a
at
(at) + a(xt) =
(a + 2x)
2
2
Shear Flow Resultant:
q1 =
P A 12 y2 B
VQ1
3P 2
= 10 3 =
y
3
I
20a
a
t
3
P C at2 (a + 2x) D
VQ2
3P
=
=
(a + 2x)
q2 =
10 3
2
I
20a
a
t
3
a
(Fw)1 =
L0
a
q1 dy =
a
Ff =
L0
3P
P
y2 dy =
20
20a3 L0
a
q2 dx =
3P
3
(a + 2x)dx =
P
2
10
20a L0
Shear Center: Summing moments about point A.
Pe = 2(Fw)1 (a) + Ff(2a)
Pe = 2 a
e =
3
P
b a + a Pb 2a
20
10
7
a
10
Ans.
522
O
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7–66. Determine the location e of the shear center,
point O, for the thin-walled member having the cross
section shown.
a
60⬚
O
a
60⬚
a
e
Summing moments about A.
Pe = F2 a
I =
13
ab
2
t
1
1
1
(t)(a)3 +
a
b(a)3 = t a3
12
12 sin 30°
4
q1 =
V(a)(t)(a>4)
q2 = q1 +
F2 =
=
1
3
4ta
V
a
V(a>2)(t)(a>4)
1
3
4ta
= q1 +
V
2a
V
4V
2 V
(a) + a b (a) =
a
3 2a
3
e =
223
a
3
Ans.
523
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7–67. Determine the location e of the shear center,
point O, for the thin-walled member having the cross
section shown. The member segments have the same
thickness t.
b
t
h
2
O
e
h
2
b
Shear Flow Resultant: The shear force flows through as Indicated by F1, F2, and F3
on FBD (b). Hence, The horizontal force equilibrium is not satisfied (©Fx Z 0). In
order to satisfy this equilibrium requirement. F1 and F2 must be equal to zero.
Shear Center: Summing moments about point A.
Pe = F2(0)
e = 0
Ans.
Also,
The shear flows through the section as indicated by F1, F2, F3.
+ ©F Z 0
However, :
x
To satisfy this equation, the section must tip so that the resultant of
:
:
:
:
F1 + F2 + F3 = P
Also, due to the geometry, for calculating F1 and F3, we require F1 = F3.
Hence, e = 0
Ans.
524
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*7–68. Determine the location e of the shear center,
point O, for the beam having the cross section shown. The
thickness is t.
1
—
r
2
e
r
O
I = (2)c
1
r 2
(t)(r>2)3 + (r>2)(t)ar + b d + Isemi-circle
12
4
= 1.583333t r3 + Isemi-circle
p>2
Isemi-circle =
p>2
2
L-p>2
(r sin u) t r du = t r3
L-p>2
sin2 u du
p
Isemi-circle = t r3 a b
2
Thus,
p
I = 1.583333t r3 + t r3 a b = 3.15413t r3
2
r
r
Q = a b t a + rb +
2
4
Lu
p>2
r sin u (t r du)
Q = 0.625 t r2 + t r2 cos u
q =
VQ
P(0.625 + cos u)t r2
=
I
3.15413 t r3
Summing moments about A:
p>2
Pe =
L-p>2
(q r du)r
p>2
Pe =
e =
Pr
(0.625 + cos u)du
3.15413 L-p>2
r (1.9634 + 2)
3.15413
e = 1.26 r
Ans.
525
1
—
r
2
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•7–69.
Determine the location e of the shear center,
point O, for the thin-walled member having the cross
section shown. The member segments have the same
thickness t.
h1
h
O
e
h1
b
Summing moments about A.
Pe = F(h) + 2V(b)
h 2
1
1
(t)(h3) + 2b(t)a b +
(t)[h3 - (h - 2h1)3]
12
2
12
I =
=
(1)
t(h - 2h1)3
bth2
th3
+
6
2
12
Q1 = y¿A¿ =
t(hy - 2h1 y + y2)
1
(h - 2h1 + y)yt =
2
2
VQ
Pt(hy - 2h1 y + y2)
=
I
2I
q1 =
h
q1 dy =
1
Pt
Pt hh1 2
2
(hy - 2h1 y + y2)dy =
c
- h31 d
2I L0
2I
2
3
Q2 = ©y¿A¿ =
1
1
h
(h - h1)h1 t + (x)(t) = t[h1 (h - h1) + hx]
2
2
2
V =
L
VQ2
Pt
=
(h (h - h1) + hx)
I
2I 1
q2 =
b
F =
L
q2 dx =
Pt
Pt
hb2
[h1 (h - h1) + hx]dx =
ah1 hb - h21 b +
b
2I L0
2I
2
From Eq, (1).
Pe =
h2b2
4
Pt
[h1 h2b - h21 hb +
+ hh21 b - h31 b]
2I
2
3
I =
t
(2h3 + 6bh2 - (h - 2h1)3)
12
e =
b(6h1 h2 + 3h2b - 8h31)
t
(6h1 h2b + 3h2b2 - 8h1 3b) =
12I
2h3 + 6bh2 - (h - 2h1)3
526
Ans.
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7–70. Determine the location e of the shear center, point O,
for the thin-walled member having the cross section shown.
t
r
a
O
a
e
Summing moments about A.
Pe = r
dF
L
dA = t ds = t r du
(1)
y = r sin u
dI = y2 dA = r2 sin2 u(t r du) = r3 t sin2 udu
p+a
I = r3 t
L
sin2 u du = r3 t
Lp - a
1 - cos 2u
du
2
=
sin 2u p + a
r3 t
(u )
2
2 p - a
=
sin 2(p + a)
sin 2(p - a)
r3 t
c ap + a b - ap - a bd
2
2
2
=
r3 t
r3 t
2 (2a - 2 sin a cos a) =
(2a - sin 2a)
2
2
dQ = y dA = r sin u(t r du) = r2 t sin u du
u
Q = r2 t
q =
u
sin u du = r2 t (-cos u)|
Lp-a
= r2 t(-cos u - cos a) = -r2 t(cos u + cos a)
p-a
P(-r2t)(cos u + cos a)
-2P(cos u + cos a)
VQ
=
=
r3t
I
r(2a - sin 2a)
(2a - sin 2a)
2
L
dF =
L
q ds =
L
q r du
p+p
L
=
dF =
2P r
-2P
(cos u + cos a) du =
(2a cos a - 2 sin a)
r(2a - sin 2a) Lp - a
2a - sin 2a
4P
(sin a - a cos a)
2a - sin 2a
4P
(sin a - a cos a) d
2a - sin 2a
4r (sin a - a cos a)
e =
2a - sin 2a
From Eq. (1); P e = r c
Ans.
527
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7–71. Sketch the intensity of the shear-stress distribution
acting over the beam’s cross-sectional area, and determine
the resultant shear force acting on the segment AB. The
shear acting at the section is V = 35 kip. Show that
INA = 872.49 in4.
C
V
8 in.
B
A
6 in.
Section Properties:
y =
4(8)(8) + 11(6)(2)
©yA
=
= 5.1053 in.
©A
8(8) + 6(2)
INA =
2 in.
1
(8) A 83 B + 8(8)(5.1053 - 4)2
12
+
1
(2) A 63 B + 2(6)(11 - 5.1053)2
12
= 872.49 in4 (Q.E.D)
Q1 = y1œ A¿ = (2.55265 + 0.5y1)(5.1053 - y1)(8)
= 104.25 - 4y21
Q2 = y2œ A¿ = (4.44735 + 0.5y2)(8.8947 - y2)(2)
= 79.12 - y22
Shear Stress: Applying the shear formula t =
tCB =
VQ
,
It
VQ1
35(103)(104.25 - 4y21)
=
It
872.49(8)
= {522.77 - 20.06y21} psi
At y1 = 0,
tCB = 523 psi
At y1 = -2.8947 in.
tCB = 355 psi
tAB =
VQ2
35(103)(79.12 - y22)
=
It
872.49(2)
= {1586.88 - 20.06y22} psi
At y2 = 2.8947 in.
tAB = 1419 psi
Resultant Shear Force: For segment AB.
VAB =
L
tAB dA
0.8947 in
=
L2.8947 in
0.8947 in
=
L2.8947 in
3 in.
3 in.
A 1586.88 - 20.06y22 B (2dy)
A 3173.76 - 40.12y22 B dy
= 9957 lb = 9.96 kip
Ans.
528
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*7–72. The beam is fabricated from four boards nailed
together as shown. Determine the shear force each nail
along the sides C and the top D must resist if the nails are
uniformly spaced at s = 3 in. The beam is subjected to a
shear of V = 4.5 kip.
1 in.
1 in.
3 in.
10 in.
A
1 in.
12 in.
V
B
Section Properties:
y =
0.5(10)(1) + 2(4)(2) + 7(12)(1)
© yA
=
= 3.50 in.
©A
10(1) + 4(2) + 12(1)
INA =
1
(10) A 13 B + (10)(1)(3.50 - 0.5)2
12
+
1
(2) A 43 B + 2(4)(3.50 - 2)2
12
1
+
(1) A 123 B + 1(12)(7 - 3.50)2
12
= 410.5 in4
QC = y1œ A¿ = 1.5(4)(1) = 6.00 in2
QD = y2œ A¿ = 3.50(12)(1) = 42.0 in2
Shear Flow:
qC =
VQC
4.5(103)(6.00)
=
= 65.773 lb>in.
I
410.5
qD =
VQD
4.5(103)(42.0)
=
= 460.41 lb>in.
I
410.5
Hence, the shear force resisted by each nail is
FC = qC s = (65.773 lb>in.)(3 in.) = 197 lb
Ans.
FD = qD s = (460.41 lb>in.)(3 in.) = 1.38 kip
Ans.
529
1 in.
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•7–73.
The member is subjected to a shear force of
V = 2 kN. Determine the shear flow at points A, B, and C.
The thickness of each thin-walled segment is 15 mm.
200 mm
B
100 mm
A C
V ⫽ 2 kN
Section Properties:
y =
=
© yA
©A
0.0075(0.2)(0.015) + 0.0575(0.115)(0.03) + 0.165(0.3)(0.015)
0.2(0.015) + 0.115(0.03) + 0.3(0.015)
= 0.08798 m
1
(0.2) A 0.0153 B + 0.2(0.015)(0.08798 - 0.0075)2
12
1
+
(0.03) A 0.1153 B + 0.03(0.115)(0.08798 - 0.0575)2
12
1
+
(0.015) A 0.33 B + 0.015(0.3)(0.165 - 0.08798)2
12
INA =
= 86.93913 A 10 - 6 B m4
QA = 0
'
Q = y œ A¿ = 0.03048(0.115)(0.015) = 52.57705 A 10 - 6 B m3
B
Ans.
1
QC = ©y¿A¿
= 0.03048(0.115)(0.015) + 0.08048(0.0925)(0.015)
= 0.16424 A 10 - 3 B m3
Shear Flow:
qA =
VQA
= 0
I
Ans.
qB =
VQB
2(103)(52.57705)(10 - 6)
= 1.21 kN>m
=
I
86.93913(10 - 6)
Ans.
qC =
VQC
2(103)(0.16424)(10 - 3)
= 3.78 kN>m
=
I
86.93913(10 - 6)
Ans.
530
300 mm
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7–74. The beam is constructed from four boards glued
together at their seams. If the glue can withstand
75 lb>in., what is the maximum vertical shear V that the
beam can support?
3 in.
0.5 in.
Section Properties:
INA =
1
1
(1) A 103 B + 2c (4) A 0.53 B + 4(0.5) A 1.752 B d
12
12
3 in.
0.5 in.
= 95.667 in4
V
Q = y¿A¿ = 1.75(4)(0.5) = 3.50 in3
4 in.
Shear Flow: There are two glue joints in this case, hence the allowable shear flow is
2(75) = 150 lb>in.
q =
150 =
3 in.
0.5 in.
0.5 in.
VQ
I
V(3.50)
95.667
V = 4100 lb = 4.10 kip
Ans.
7–75. Solve Prob. 7–74 if the beam is rotated 90° from the
position shown.
3 in.
0.5 in.
3 in.
0.5 in.
V
3 in.
4 in.
0.5 in.
Section Properties:
INA =
1
1
(10) A 53 B (9) A 43 B = 56.167 in4
12
12
Q = y¿A¿ = 2.25(10)(0.5) = 11.25 in3
Shear Flow: There are two glue joints in this case, hence the allowable shear flow is
2(75) = 150 lb>in.
q =
150 =
VQ
I
V(11.25)
56.167
V = 749 lb
Ans.
531
0.5 in.
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8–1. A spherical gas tank has an inner radius of r = 1.5 m.
If it is subjected to an internal pressure of p = 300 kPa,
determine its required thickness if the maximum normal
stress is not to exceed 12 MPa.
pr
;
2t
sallow =
12(106) =
300(103)(1.5)
2t
t = 0.0188 m = 18.8 mm
Ans.
8–2. A pressurized spherical tank is to be made of
0.5-in.-thick steel. If it is subjected to an internal pressure
of p = 200 psi, determine its outer radius if the maximum
normal stress is not to exceed 15 ksi.
sallow =
pr
;
2t
15(103) =
200 ri
2(0.5)
ri = 75 in.
ro = 75 in. + 0.5 in. = 75.5 in.
Ans.
8–3. The thin-walled cylinder can be supported in one of
two ways as shown. Determine the state of stress in the wall
of the cylinder for both cases if the piston P causes the
internal pressure to be 65 psi. The wall has a thickness of
0.25 in. and the inner diameter of the cylinder is 8 in.
P
Case (a):
s1 =
pr
;
t
s1 =
65(4)
= 1.04 ksi
0.25
Ans.
s2 = 0
Ans.
Case (b):
s1 =
pr
;
t
s1 =
65(4)
= 1.04 ksi
0.25
Ans.
s2 =
pr
;
2t
s2 =
65(4)
= 520 psi
2(0.25)
Ans.
532
P
8 in.
8 in.
(a)
(b)
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*8–4. The tank of the air compressor is subjected to an
internal pressure of 90 psi. If the internal diameter of
the tank is 22 in., and the wall thickness is 0.25 in.,
determine the stress components acting at point A. Draw a
volume element of the material at this point, and show the
results on the element.
Hoop Stress for Cylindrical Vessels: Since
A
11
r
=
= 44 7 10, then thin wall
t
0.25
analysis can be used. Applying Eq. 8–1
s1 =
pr
90(11)
=
= 3960 psi = 3.96 ksi
t
0.25
Ans.
Longitudinal Stress for Cylindrical Vessels: Applying Eq. 8–2
s2 =
pr
90(11)
=
= 1980 psi = 1.98 ksi
2t
2(0.25)
Ans.
•8–5.
The spherical gas tank is fabricated by bolting together
two hemispherical thin shells of thickness 30 mm. If the gas
contained in the tank is under a gauge pressure of 2 MPa,
determine the normal stress developed in the wall of the tank
and in each of the bolts.The tank has an inner diameter of 8 m
and is sealed with 900 bolts each 25 mm in diameter.
Normal Stress: Since
4
r
=
= 133.33 7 10, thin-wall analysis is valid. For the
t
0.03
spherical tank’s wall,
s =
Referring
pr
2(4)
=
= 133 MPa
2t
2(0.03)
to
the
free-body
diagram
p 2
6
P = pA = 2 A 10 B c A 8 B d = 32p A 10 B N. Thus,
4
Ans.
shown
in
Fig.
a,
6
+ c ©Fy = 0;
32p A 106 B - 450Pb - 450Pb = 0
Pb = 35.56 A 103 B p N
The normal stress developed in each bolt is then
35.56 A 10 B p
Pb
=
= 228 MPa
p
Ab
A 0.0252 B
4
3
sb =
Ans.
533
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8–6. The spherical gas tank is fabricated by bolting
together two hemispherical thin shells. If the 8-m inner
diameter tank is to be designed to withstand a gauge pressure
of 2 MPa, determine the minimum wall thickness of the
tank and the minimum number of 25-mm diameter bolts
that must be used to seal it. The tank and the bolts are made
from material having an allowable normal stress of 150 MPa
and 250 MPa, respectively.
Normal Stress: For the spherical tank’s wall,
sallow =
pr
2t
150 A 106 B =
2 A 106 B (4)
2t
t = 0.02667 m = 26.7 mm
Since
Ans.
r
4
=
= 150 7 10, thin-wall analysis is valid.
t
0.02667
Referring
the
free-body
diagram
p
P = pA = 2 A 106 B c A 82 B d = 32p A 106 B N. Thus,
4
+ c ©Fy = 0;
to
32p A 106 B n =
shown
in
Fig.
a,
n
n
(P )
- (Pb)allow = 0
2 b allow
2
32p A 106 B
(1)
(Pb)allow
The allowable tensile force for each bolt is
(Pb)allow = sallowAb = 250 A 106 B c
p
A 0.0252 B d = 39.0625 A 103 B pN
4
Substituting this result into Eq. (1),
n =
32p A 106 B
39.0625p A 103 B
= 819.2 = 820
Ans.
534
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8–7. A boiler is constructed of 8-mm thick steel plates that
are fastened together at their ends using a butt joint
consisting of two 8-mm cover plates and rivets having a
diameter of 10 mm and spaced 50 mm apart as shown. If the
steam pressure in the boiler is 1.35 MPa, determine (a) the
circumferential stress in the boiler’s plate apart from
the seam, (b) the circumferential stress in the outer cover plate
along the rivet line a–a, and (c) the shear stress in the rivets.
a
8 mm
50 mm
a)
s1 =
pr
1.35(106)(0.75)
=
= 126.56(106) = 127 MPa
t
0.008
Ans.
126.56 (106)(0.05)(0.008) = s1 ¿(2)(0.04)(0.008)
b)
s1 ¿ = 79.1 MPa
Ans.
c) From FBD(a)
+ c ©Fy = 0;
Fb - 79.1(106)[(0.008)(0.04)] = 0
Fb = 25.3 kN
(tavg)b =
Fb
25312.5
- p
= 322 MPa
2
A
4 (0.01)
Ans.
535
0.75 m
a
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*8–8. The gas storage tank is fabricated by bolting together
two half cylindrical thin shells and two hemispherical shells
as shown. If the tank is designed to withstand a pressure
of 3 MPa, determine the required minimum thickness of
the cylindrical and hemispherical shells and the minimum
required number of longitudinal bolts per meter length at
each side of the cylindrical shell. The tank and the 25 mm
diameter bolts are made from material having an allowable
normal stress of 150 MPa and 250 MPa, respectively. The
tank has an inner diameter of 4 m.
Normal Stress: For the cylindrical portion of the tank, the hoop stress is twice as
large as the longitudinal stress.
sallow =
pr
;
t
150 A 106 B =
3 A 106 B (2)
tc
tc = 0.04 m = 40 mm
For the hemispherical cap,
pr
;
sallow =
t
150 A 10 B =
6
Ans.
3 A 106 B (2)
2ts
ts = 0.02 m = 20 mm
Since
Ans.
r
6 10, thin-wall analysis is valid.
t
Referring to the free-body diagram of the per meter length of the cylindrical
portion, Fig. a, where P = pA = 3 A 106 B [4(1)] = 12 A 106 B N, we have
+ c ©Fy = 0;
12 A 106 B - nc(Pb)allow - nc(Pb)allow = 0
nc =
6 A 106 B
(1)
(Pb)allow
The allowable tensile force for each bolt is
(Pb)allow = sallowAb = 250 A 106 B c
p
A 0.0252 B d = 122.72 A 103 B N
4
Substituting this result into Eq. (1),
nc = 48.89 = 49 bolts>meter
Ans.
536
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•8–9.
The gas storage tank is fabricated by bolting together
two half cylindrical thin shells and two hemispherical shells
as shown. If the tank is designed to withstand a pressure of
3 MPa, determine the required minimum thickness of the
cylindrical and hemispherical shells and the minimum
required number of bolts for each hemispherical cap. The
tank and the 25 mm diameter bolts are made from material
having an allowable normal stress of 150 MPa and 250 MPa,
respectively. The tank has an inner diameter of 4 m.
Normal Stress: For the cylindrical portion of the tank, the hoop stress is twice as
large as the longitudinal stress.
sallow =
pr
;
t
150 A 106 B =
3 A 106 B (2)
tc
tc = 0.04 m = 40 mm
For the hemispherical cap,
sallow =
pr
;
t
150 A 106 B =
Ans.
3 A 106 B (2)
2ts
ts = 0.02 m = 20 mm
Since
Ans.
r
6 10, thin-wall analysis is valid.
t
The allowable tensile force for each bolt is
(Pb)allow = sallowAb = 250 A 106 B c
p
A 0.0252 B d = 122.72 A 103 B N
4
Referring to the free-body diagram of the hemispherical cap, Fig. b, where
p
P = pA = 3 A 106 B c A 42 B d = 12p A 106 B N,
4
+ ©F = 0;
:
x
12p A 106 B ns =
ns
ns
(Pb)allow (Pb)allow = 0
2
2
12p A 106 B
(1)
(Pb)allow
Substituting this result into Eq. (1),
ns = 307.2 = 308 bolts
Ans.
537
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8–10. A wood pipe having an inner diameter of 3 ft is
bound together using steel hoops each having a crosssectional area of 0.2 in2. If the allowable stress for the hoops
is sallow = 12 ksi, determine their maximum spacing s along
the section of pipe so that the pipe can resist an internal
gauge pressure of 4 psi. Assume each hoop supports the
pressure loading acting along the length s of the pipe.
s
4 psi
4 psi
s
s
Equilibrium for the steel Hoop: From the FBD
+ ©F = 0;
:
x
P = 72.0s
2P - 4(36s) = 0
Hoop Stress for the Steel Hoop:
s1 = sallow =
12(103) =
P
A
72.0s
0.2
s = 33.3 in.
Ans.
8–11. The staves or vertical members of the wooden tank
are held together using semicircular hoops having a
thickness of 0.5 in. and a width of 2 in. Determine the normal
stress in hoop AB if the tank is subjected to an internal
gauge pressure of 2 psi and this loading is transmitted
directly to the hoops. Also, if 0.25-in.-diameter bolts are used
to connect each hoop together, determine the tensile stress
in each bolt at A and B. Assume hoop AB supports the
pressure loading within a 12-in. length of the tank as shown.
18 in.
6 in.
6 in.
FR = 2(36)(12) = 864 lb
©F = 0; 864 - 2F = 0; F = 432 lb
sh =
sb =
F
432
=
= 432 psi
Ah
0.5(2)
Ans.
F
432
=
= 8801 psi = 8.80 ksi
p
Ab
(0.25)2
4
Ans.
538
12 in.
A
B
12 in.
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*8–12. Two hemispheres having an inner radius of 2 ft and
wall thickness of 0.25 in. are fitted together, and the inside
gauge pressure is reduced to -10 psi. If the coefficient
of static friction is ms = 0.5 between the hemispheres,
determine (a) the torque T needed to initiate the rotation
of the top hemisphere relative to the bottom one, (b) the
vertical force needed to pull the top hemisphere off
the bottom one, and (c) the horizontal force needed to slide
the top hemisphere off the bottom one.
0.25 in.
2 ft
Normal Pressure: Vertical force equilibrium for FBD(a).
+ c ©Fy = 0;
10 C p(242) D - N = 0
N = 5760p lb
The Friction Force: Applying friction formula
Ff = ms N = 0.5(5760p) = 2880p lb
a) The Required Torque: In order to initiate rotation of the two hemispheres
relative to each other, the torque must overcome the moment produced by the
friction force about the center of the sphere.
T = Ffr = 2880p(2 + 0.125>12) = 18190 lb # ft = 18.2 kip # ft
Ans.
b) The Required Vertical Force: In order to just pull the two hemispheres apart, the
vertical force P must overcome the normal force.
P = N = 5760p = 18096 lb = 18.1 kip
Ans.
c) The Required Horizontal Force: In order to just cause the two hemispheres to
slide relative to each other, the horizontal force F must overcome the friction force.
F = Ff = 2880p = 9048 lb = 9.05 kip
Ans.
•8–13. The 304 stainless steel band initially fits snugly around
the smooth rigid cylinder. If the band is then subjected to a
nonlinear temperature drop of ¢T = 20 sin2 u °F, where u is
in radians, determine the circumferential stress in the band.
1
in.
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
1 in.
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8–14. The ring, having the dimensions shown, is placed
over a flexible membrane which is pumped up with a
pressure p. Determine the change in the internal radius of
the ring after this pressure is applied. The modulus of
elasticity for the ring is E.
ro
ri
w
p
Equilibrium for the Ring: Form the FBD
+ ©F = 0;
:
x
2P - 2pri w = 0
P = pri w
Hoop Stress and Strain for the Ring:
s1 =
pri w
pri
P
=
=
rs - ri
A
(rs - ri)w
Using Hooke’s Law
e1 =
However,
e1 =
pri
s1
=
E
E(rs - ri)
[1]
2p(ri)1 - 2pri
(ri)1 - ri
dri
=
=
.
ri
ri
2pr
Then, from Eq. [1]
pri
dri
=
ri
E(rs - ri)
dri =
pr2i
E(rs - ri)
Ans.
540
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8–15. The inner ring A has an inner radius r1 and outer
radius r2. Before heating, the outer ring B has an inner
radius r3 and an outer radius r4, and r2 7 r3. If the outer ring
is heated and then fitted over the inner ring, determine the
pressure between the two rings when ring B reaches the
temperature of the inner ring. The material has a modulus of
elasticity of E and a coefficient of thermal expansion of a.
r1
A
Equilibrium for the Ring: From the FBD
+ ©F = 0;
:
x
P = priw
2P - 2priw = 0
Hoop Stress and Strain for the Ring:
s1 =
priw
pri
P
=
=
ro - ri
A
(ro - ri)w
Using Hooke’s law
e1 =
However,
e1 =
pri
s1
=
E
E(ro - ri)
[1]
2p(ri)1 - 2pri
(ri)1 - ri
dri
=
=
.
ri
ri
2pr
Then, from Eq. [1]
pri
dri
=
ri
E(ro - ri)
dri =
pr2i
E(ro - ri)
Compatibility: The pressure between the rings requires
dr2 + dr3 = r2 - r3
[2]
From the result obtained above
dr2 =
pr22
E(r2 - r1)
dr3 =
pr23
E(r4 - r3)
Substitute into Eq. [2]
pr22
pr23
+
= r2 - r3
E(r2 - r1)
E(r4 - r3)
p =
r4
r2
E(r2 - r3)
Ans.
r22
r23
+
r2 - r1
r4 - r3
541
r3
B
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*8–16. The cylindrical tank is fabricated by welding a
strip of thin plate helically, making an angle u with the
longitudinal axis of the tank. If the strip has a width w and
thickness t, and the gas within the tank of diameter d is
pressured to p, show that the normal stress developed along
the strip is given by su = (pd>8t)(3 - cos 2u).
w
u
Normal Stress:
sh = s1 =
pr
p(d>2)
pd
=
=
t
t
2t
sl = s2 =
p(d>2)
pd
pr
=
=
2t
2t
4t
Equilibrium: We will consider the triangular element cut from the strip
shown in Fig. a. Here,
Ah = (w sin u)t
and
Thus,
Al = (w cos u)t.
pd
pwd
and
(w sin u)t =
sin u
Fh = shAh =
2t
2
pwd
pd
(w cos u)t =
cos u.
4t
4
Fl = slAl =
Writing the force equation of equilibrium along the x¿ axis,
©Fx¿ = 0;
c
pwd
pwd
sin u d sin u + c
cos u d cos u - Nu = u
2
4
Nu =
pwd
A 2 sin2 u + cos2 u B
4
However, sin2 u + cos2 u = 1. This equation becomes
Nu =
Also, sin2 u =
pwd
A sin2 u + 1 B
4
1
(1 - cos 2u), so that
2
pwd
Nu =
(3 - cos 2u)
8
Since Au = wt, then
Nu
=
su =
Au
su =
pwd
(3 - cos 2u)
8
wt
pd
(3 - cos 2u)
8t
(Q.E.D.)
542
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8–17. In order to increase the strength of the pressure vessel,
filament winding of the same material is wrapped around the
circumference of the vessel as shown. If the pretension in the
filament is T and the vessel is subjected to an internal pressure
p, determine the hoop stresses in the filament and in the wall
of the vessel. Use the free-body diagram shown, and assume
the filament winding has a thickness t and width w for a
corresponding length of the vessel.
L
w
s1
t¿
T
p
t
s1
T
Normal Stress in the Wall and Filament Before the Internal Pressure is Applied:
The entire length w of wall is subjected to pretension filament force T. Hence, from
equilibrium, the normal stress in the wall at this state is
2T - (sl ¿)w (2wt) = 0
(sl ¿)w =
T
wt
and for the filament the normal stress is
(sl ¿)fil =
T
wt¿
Normal Stress in the Wall and Filament After the Internal Pressure is Applied: The
stress in the filament becomes
sfil = sl + (sl ¿)fil =
pr
T
+
(t + t¿)
wt¿
Ans.
sw = sl - (sl ¿)w =
pr
T
(t + t¿)
wt
Ans.
And for the wall,
8–18. The vertical force P acts on the bottom of the plate
having a negligible weight. Determine the shortest distance
d to the edge of the plate at which it can be applied so that
it produces no compressive stresses on the plate at section
a–a. The plate has a thickness of 10 mm and P acts along the
center line of this thickness.
300 mm
a
a
200 mm
500 mm
sA = 0 = sa - sb
0 =
0 =
P
Mc
A
I
P
(0.2)(0.01)
d
P(0.1 - d)(0.1)
1
3
12 (0.01)(0.2 )
P
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;
N - 100 = 0
a + ©MC = 0;
100(0.1) - M = 0
A = 0.2(0.03) = 0.006 m2
I =
N = 100 kN
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;
N - 100 = 0
a + ©MC = 0;
M - 100(0.2) = 0
A = 0.2 (0.03) = 0.006 m2
I =
N = 100 kN
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 =
4(0.004)
P
Mc
40
= 123 MPa
+
= + 1
3
A
I
(0.008)(0.003)
(0.003)(0.008)
12
2(0.004)
Mc
= 62.5 MPa
= 1
3
I
12 (0.003)(0.008)
Ans.
Ans.
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 =
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)
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
w
3
sb =
w
50 mm
s due to bending:
w
30(10 )(0.05 + 2 )( 2 )
Mc
=
1
3
I
12 (0.04)(w)
4500 (103)(0.05 + w2)
=
w2
P
smax = sallow = sa + sb
w
73(106) =
4500(103)(0.05 + 2 )
750(103)
+
w
w2
73 w2 = 0.75 w + 0.225 + 2.25 w
73 w2 - 3 w - 0.225 = 0
w = 0.0797 m = 79.7 mm
Ans.
548
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8–27. The offset link has a width of w = 200 mm and a
thickness of 40 mm. If the allowable normal stress is
sallow = 75 MPa, determine the maximum load P that can
be applied to the cables.
P
A = 0.2(0.04) = 0.008 m2
I =
s =
1
(0.04)(0.2)3 = 26.6667(10 - 6) m4
12
w
50 mm
P
Mc
+
A
I
75(106) =
0.150 P(0.1)
P
+
0.008
26.6667(10 - 6)
P = 109 kN
Ans.
P
*8–28. The joint is subjected to a force of P 80 lb and
F 0. Sketch the normal-stress distribution acting over
section a–a if the member has a rectangular cross-sectional
area of width 2 in. and thickness 0.5 in.
a
B
s due to axial force:
s =
0.5 in.
P
80
=
= 80 psi
A
(0.5)(2)
2 in.
F
s due to bending:
s =
A a
100(0.25)
Mc
= 1
= 1200 psi
3
I
12 (2)(0.5)
1.25 in.
P
(smax)t = 80 + 1200 = 1280 psi = 1.28 ksi
Ans.
(smax)c = 1200 - 80 = 1120 psi = 1.12 ksi
Ans.
y
(0.5 - y)
=
1.28
1.12
y = 0.267 in.
549
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The joint is subjected to a force of P = 200 lb and
F = 150 lb. Determine the state of stress at points A and B
and sketch the results on differential elements located at
these points. The member has a rectangular cross-sectional
area of width 0.75 in. and thickness 0.5 in.
•8–29.
a
B
A a
0.5 in.
2 in.
F
1.25 in.
P
A = 0.5(0.75) = 0.375 in2
œ
QA = yA
A¿ = 0.125(0.75)(0.25) = 0.0234375 in3 ;
I =
QB = 0
1
(0.75)(0.53) = 0.0078125 in4
12
Normal Stress:
My
N
;
A
I
s =
sA =
200
+ 0 = 533 psi (T)
0.375
Ans.
sB =
50(0.25)
200
= -1067 psi = 1067 psi (C)
0.375
0.0078125
Ans.
Shear stress:
t =
VQ
It
tA =
150(0.0234375)
= 600 psi
(0.0078125)(0.75)
Ans.
tB = 0
Ans.
550
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8–30. If the 75-kg man stands in the position shown,
determine the state of stress at point A on the cross section
of the plank at section a–a. The center of gravity of the man
is at G. Assume that the contact point at C is smooth.
C
G
600 mm
A
a
50 mm
1.5 m
B
12.5 mm
30 a
600 mm
300 mm
Support Reactions: Referring to the free-body diagram of the entire plank, Fig. a,
a + ©MB = 0;
FC sin 30°(2.4) - 75(9.81) cos 30°(0.9) = 0
FC = 477.88 N
©Fx¿ = 0; Bx¿ - 75(9.81) sin 30° - 477.88 cos 30° = 0
Bx¿ = 781.73 N
©Fy¿ = 0; By¿ + 477.88 sin 30° - 75(9.81) cos 30° = 0
By¿ = 398.24 N
Internal Loadings: Consider the equilibrium of the free-body diagram of the plank’s
lower segment, Fig. b,
©Fx¿ = 0; 781.73 - N = 0
N = 781.73 N
©Fy¿ = 0; 398.24 - V = 0
V = 398.24 N
a + ©MO = 0;
M - 398.24(0.6) = 0
M = 238.94 N # m
Section Properties: The cross-sectional area and the moment of inertia about the
centroidal axis of the plank’s cross section are
A = 0.6(0.05) = 0.03 m2
I =
1
(0.6) A 0.053 B = 6.25 A 10 - 6 B m4
12
Referring to Fig. c, QA is
QA = y¿A¿ = 0.01875(0.0125)(0.6) = 0.140625 A 10 - 3 B m3
Normal Stress: The normal stress is the combination of axial and bending stress.
Thus,
s =
My
N
;
A
I
For point A, y = 0.0125 m. Then
sA =
238.94(0.0125)
-781.73
0.03
6.25 A 10 - 6 B
= -503.94 kPa = 504 kPa (C)
Ans.
551
Section a – a and b – b
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8–30.
Continued
Shear Stress: The shear stress is contributed by transverse shear stress. Thus,
VQA
tA =
=
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.
552
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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;
M - P(0.1 - d) = 0
A = 0.2 (0.02) = 0.004 m4
I =
N = P
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;
N - 80 = 0
a + ©MC = 0;
M - 80(0.05) = 0
A = 0.01(0.2) = 0.002 m2
I =
N = 80 kN
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;
M - 10(3) = 0
+
10 lb
M = 30 lb # in.
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.
555
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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.
My
28(0.1)
= 1
= 23.3 ksi (T)
3
I
12 (0.18)(0.2)
Ans.
Point E:
sE =
4 in.
tE = 0
Ans.
556
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8–35. The wide-flange beam is subjected to the loading
shown. Determine the stress components at points A and B
and show the results on a volume element at each of these
points. Use the shear formula to compute the shear stress.
500 lb
3000 lb
2500 lb
A
B
2 ft
2 ft
2 ft
4 ft
6 ft
A
0.5 in.
1
1
I =
(4)(73) (3.5)(63) = 51.33 in4
12
12
B
4 in.
2 in.
4 in.
0.5 in.
A = 2(0.5)(4) + 6(0.5) = 7 in2
QB = ©y¿A¿ = 3.25(4)(0.5) + 2(2)(0.5) = 8.5 in3
QA = 0
sA =
0.5 in.
-11500 (12)(3.5)
-Mc
=
= -9.41 ksi
I
51.33
Ans.
tA = 0
Ans.
sB =
My
11500(12)(1)
=
= 2.69 ksi
I
51.33
Ans.
tB =
VQB
2625(8.5)
=
= 0.869 ksi
It
51.33(0.5)
Ans.
557
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*8–36. The drill is jammed in the wall and is subjected to
the torque and force shown. Determine the state of stress at
point A on the cross section of drill bit at section a–a.
y
400 mm
a 20 N ·m
x
a
125 mm
y
A
z
5 mm
B
Section a – a
Internal Loadings: Consider the equilibrium of the free-body diagram of the drill’s
right cut segment, Fig. a,
4
©Fx = 0; N - 150 a b = 0
5
N = 120 N
3
©Fy = 0; 150 a b - Vy = 0
5
Vy = 90 N
T = 20 N # m
©Mx = 0; 20 - T = 0
4
3
©Mz = 0; -150 a b (0.4) + 150 a b(0.125) + Mz = 0
5
5
Mz = 21 N # m
Section Properties: The cross-sectional area, the moment of inertia about the z axis,
and the polar moment of inertia of the drill’s cross section are
A = p A 0.0052 B = 25p A 10 - 6 B m2
Iz =
p
A 0.0054 B = 0.15625p A 10 - 9 B m4
4
J =
p
A 0.0054 B = 0.3125p A 10 - 9 B m4
2
Referring to Fig. b, QA is
QA = 0
Normal Stress: The normal stress is a combination of axial and bending stress. Thus,
s =
Mzy
N
A
Iz
For point A, y = 0.005 m. Then
sA =
-120
25p A 10
-6
B
21(0.005)
-
0.15625p A 10 - 9 B
= -215.43 MPa = 215 MPa (C)
558
Ans.
3
5
4
150 N
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8–36. Continued
Shear Stress: The transverse shear stress developed at point A is
c A txy B V d
VyQA
=
A
Izt
Ans.
= 0
The torsional shear stress developed at point A is
C (txz)T D A =
20(0.005)
Tc
= 101.86 MPa
=
J
0.3125p A 10 - 9 B
Thus,
A txy B A = 0
Ans.
A txz B A = c A txz B T d = 102 MPa
Ans.
A
The state of stress at point A is represented on the element shown in Fig c.
559
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•8–37.
The drill is jammed in the wall and is subjected to
the torque and force shown. Determine the state of stress at
point B on the cross section of drill bit at section a–a.
y
400 mm
a 20 N ·m
x
a
125 mm
y
Internal Loadings: Consider the equilibrium of the free-body diagram of the drill’s
right cut segment, Fig. a,
4
©Fx = 0; N - 150 a b = 0
5
N = 120 N
3
©Fy = 0; 150 a b - Vy = 0
5
Vy = 90 N
z
Section a – a
4
3
©Mz = 0; -150 a b (0.4) + 150 a b (0.125) + Mz = 0
5
5
Mz = 21 N # m
Section Properties: The cross-sectional area, the moment of inertia about the z axis,
and the polar moment of inertia of the drill’s cross section are
A = p A 0.0052 B = 25p A 10 - 6 B m2
Iz =
p
A 0.0054 B = 0.15625p A 10 - 9 B m4
4
J =
p
A 0.0054 B = 0.3125p A 10 - 9 B m4
2
Referring to Fig. b, QB is
QB = y¿A¿ =
4(0.005) p
c A 0.0052 B d = 83.333 A 10 - 9 B m3
3p
2
Normal Stress: The normal stress is a combination of axial and bending stress. Thus,
Mzy
N
A
Iz
For point B, y = 0. Then
sB =
-120
25p A 10 - 6 B
- 0 = -1.528 MPa = 1.53 MPa(C)
560
5 mm
B
T = 20 N # m
©Mx = 0; 20 - T = 0
s =
A
Ans.
3
5
4
150 N
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8–37. Continued
Shear Stress: The transverse shear stress developed at point B is
c A txy B V d
90 c83.333 A 10 - 9 B d
VyQB
=
B
Izt
=
0.15625p A 10 - 9 B (0.01)
= 1.528 MPa
Ans.
The torsional shear stress developed at point B is
c A txy B T d
=
B
20(0.005)
TC
= 101.86 MPa
=
J
0.3125p A 10 - 9 B
Thus,
A tC B B = 0
Ans.
A txy B B = c A txy B T d - c A txy B V d
B
B
Ans.
= 101.86 - 1.528 = 100.33 MPa = 100 MPa
The state of stress at point B is represented on the element shown in Fig. d.
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;
T - N = 0
a + ©Mo = 0;
M + T(7) - 900(24) = 0
N = T
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
Support Reactions:
B
a + ©MD = 0;
150 mm
Cy = 0.6667 kN
+
: ©Fx = 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;
M - 0.6667(1) = 0
N = 4.00 kN
V = 0.6667 kN
M = 0.6667 kN # m
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 =
=
VQA
It
0.6667(103) C 0.405(10 - 3) D
82.8(10 - 6)(0.015)
= 0.217 MPa
Ans.
564
C
20 mm
15 mm
4(0.625) - Cy (3.75) = 0
A
200 mm
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
Support Reactions:
B
a + ©MD = 0;
150 mm
Cy = 0.6667 kN
+
: ©Fx = 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;
M - 0.6667(1) = 0
N = 4.00 kN
V = 0.6667 kN
M = 0.6667 kN # m
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 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
C
20 mm
15 mm
4(0.625) - Cy (3.75) = 0
A
200 mm
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
©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 =
4
5 kN
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
©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
5 kN
Referring to Fig. b,
A Qy B B = y¿A¿ = c
4(0.04) p
d c (0.042) d = 42.67(10 - 6) m3
3p
2
A Qz B B = 0
The normal stress is contributed by bending stress only. Thus,
s = -
Myz
Mzy
Iz
+
Iy
For point B, y = 0 and z = 0.04 m. Then
s = -0 +
-1.2(103)(0.04)
0.64(10 - 6)p
= -23.87(106) Pa = 23.9 MPa (C)
Ans.
The transverse shear stress developed at point B is
A txy B v =
Vy(Qy)B
Iz t
=
3(103) C 42.67(10 - 6) D
0.64(10 - 6)p (0.08)
= 0.7958(106) MPa = 0.796 MPa
568
Ans.
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8–43. Continued
A txz B v =
Vz (Qz)B
Iy t
= 0
Ans.
The state of stress for point B can be represented by the volume element shown in
Fig. c
569
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*8–44. Determine the normal stress developed at points
A and B. Neglect the weight of the block.
6 kip
3 in.
Referring to Fig. a,
12 kip
6 in.
©Fx = (FR)x;
-6 - 12 = F
©My = (MR)y;
6(1.5) - 12(1.5) = My
©Mz = (MR)z;
12(3) - 6(3) = Mz
a
F = -18.0 kip
A
My = -9.00 kip # in
Mz = 18.0 kip # in
The cross-sectional area and moment of inertia about the y and z axes of the crosssection are
A = 6(3) = 18 in2
Iy =
1
(6)(3)3 = 13.5 in4
12
Iz =
1
(3)(63) = 54.0 in4
12
The normal stress developed is the combination of axial and bending stress. Thus,
s =
My z
Mz y
F
+
A
Iz
Iy
For point A, y = 3 in. and z = -1.5 in.
sA =
18.0(3)
-9.00(-1.5)
-18.0
+
18.0
54.0
13.5
Ans.
= -1.00 ksi = 1.00 ksi (C)
For point B, y = 3 in and z = 1.5 in.
sB =
18.0(3)
-9.00(1.5)
-18.0
+
18.0
54
13.5
= -3.00 ksi = 3.00 ksi (C)
Ans.
570
B
a
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•8–45.
Sketch the normal stress distribution acting over the
cross section at section a–a. Neglect the weight of the block.
6 kip
3 in.
12 kip
6 in.
a
A
B
a
Referring to Fig. a,
©Fx = (FR)x;
-6 - 12 = F
F = -18.0 kip
©My = (MR)y;
6(1.5) - 12(1.5) = My
©Mz = (MR)z;
12(3) - 6(3) = Mz
My = -9.00 kip # in
Mz = 18.0 kip # in
The cross-sectional area and the moment of inertia about the y and z axes of the
cross-section are
A = 3 (6) = 18.0 in2
Iy =
1
(6)(33) = 13.5 in4
12
Iz =
1
(3)(63) = 54.0 in4
12
The normal stress developed is the combination of axial and bending stress. Thus,
s =
Myz
Mzy
F
+
A
Iz
Iy
For point A, y = 3 in. and z = -1.5 in.
sA =
18.0(3)
-9.00(-1.5)
-18.0
+
18.0
54.0
13.5
= -1.00 ksi = 1.00 ksi (C)
For point B, y = 3 in. and z = 1.5 in.
sB =
18.0(3)
-9.00(1.5)
-18.0
+
18.0
54.0
13.5
= -3.00 ksi = 3.00 ksi (C)
571
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8–45. Continued
For point C, y = -3 in. and z = 1.5 in.
sC =
18.0(-3)
-9.00(1.5)
-18.0
+
18.0
54.0
13.5
= -1.00 ksi = 1.00 ksi (C)
For point D, y = -3 in. and z = -1.5 in.
sD =
18.0(-3)
-9.00(-1.5)
-18.0
+
18.0
54.0
13.5
= 1.00 ksi (T)
The normal stress distribution over the cross-section is shown in Fig. b
572
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8–46. The support is subjected to the compressive load P.
Determine the absolute maximum and minimum normal
stress acting in the material.
a
—
a 2
—
2
a a
—
2 —
2
Section Properties:
w = a +x
A = a(a + x)
I =
a
1
(a) (a + x)3 =
(a + x)3
12
12
Internal Forces and Moment: As shown on FBD.
Normal Stress:
s =
=
N
Mc
;
A
I
0.5Px C 12 (a + x) D
-P
;
a
3
a(a + x)
12 (a + x)
=
P
3x
-1
;
B
R
a a+x
(a + x)2
sA = -
P
1
3x
+
B
R
a a+x
(a + x)2
= sB =
P 4x + a
B
R
a (a + x)2
[1]
P
-1
3x
+
B
R
a a+x
(a + x)2
=
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.
576
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•8–49.
If the baby has a mass of 5 kg and his center of
mass is at G, determine the normal stress at points A and B
on the cross section of the rod at section a–a. There are two
rods, one on each side of the cradle.
500 mm
15⬚
G
a
75 mm
a
6 mm
A
B
Section a–a
Section Properties: The location of the neutral surface from the center of
curvature of the rod, Fig. a, can be determined from
A
R =
©
dA
LA r
where A = p A 0.0062 B = 36p A 10 - 6 B m2
©
dA
= 2p ¢ r - 2r2 - c2 ≤ = 2p ¢ 0.081 - 20.0812 - 0.0062 ≤ = 1.398184 A 10 - 3 B m
LA r
Thus,
R =
36p A 10 - 6 B
1.398184 A 10 - 3 B
= 0.080889 m
Then
e = r - R = 0.081 - 0.080889 = 0.111264 A 10 - 3 B m
Internal Loadings: Consider the equilibrium of the free-body diagram of the
cradle’s upper segment, Fig. b,
+ c ©Fy = 0;
-5(9.81) - 2N = 0
N = -24.525 N
a + ©MO = 0;
5(9.81)(0.5+ 0.080889) - 2M = 0
M = 14.2463 N # m
Normal Stress: The normal stress is the combination of axial and bending stress.
Thus,
s =
M(R - r)
N
+
A
Aer
Here, M = -14.1747 (negative) since it tends to increase the curvature of the
rod. For point A, r = rA = 0.075 m. Then,
sA =
-24.525
36p A 10
-6
B
-14.2463(0.080889 - 0.075)
+
36p A 10 - 6 B (0.111264) A 10 - 3 B (0.075)
= -89.1 MPa = 89.1 MPa (C)
Ans.
For point B, r = rB = 0.087 m. Then,
sB =
-24.525
36p A 10
-6
B
-14.2463(0.080889 - 0.087)
+
36p A 10 - 6 B (0.111264) A 10 - 3 B (0.087)
Ans.
= 79.3 kPa (T)
dA
5
= 0.25 ln = 0.055786
4
LA r
577
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8–50. The C-clamp applies a compressive stress on the
cylindrical block of 80 psi. Determine the maximum normal
stress developed in the clamp.
1 in.
4 in.
4.5 in.
0.75 in.
R =
A
dA
1 r
=
1(0.25)
= 4.48142
0.055786
P = sbA = 80p (0.375)2 = 35.3429 lb
M = 35.3429(8.98142) = 317.4205 lb # in.
s =
M(R - r)
P
+
Ar(r - R)
A
(st)max =
317.4205(4.48142 - 4)
35.3429
+
= 8.37 ksi
(1)(0.25)(4)(4.5 - 4.48142)
(1)(0.25)
(sc)max =
317.4205(4.48142 - 5)
35.3429
+
= -6.95 ksi
1(0.25)(5)(4.5 - 4.48142)
(1)(0.25)
578
Ans.
0.25 in.
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8–51. A post having the dimensions shown is subjected to
the bearing load P. Specify the region to which this load can
be applied without causing tensile stress to be developed at
points A, B, C, and D.
x
z
a
a
A
B
Equivalent Force System: As shown on FBD.
Section Properties:
1
A = 2a(2a) + 2 B (2a)a R = 6a2
2
Iz =
1
1
1
a 2
(2a)(2a)3 + 2 B (2a) a3 + (2a) aa a + b R
12
36
2
3
= 5a4
Iy =
=
1
1
1
a 2
(2a)(2a)3 + 2 B (2a) a3 + (2a) aa b R
12
36
2
3
5 4
a
3
Normal Stress:
s =
My z
Mzy
N
+
A
Iz
Iy
=
Peyy
Pez z
-P
+ 5
2
4
4
6a
5a
3a
=
P
A -5a2 - 6eyy + 18ez z B
30a4
At point A where y = -a and z = a, we require sA 6 0.
0 7
P
C -5a2 - 6(-a) ey + 18(a) ez D
30a4
0 7 -5a + 6ey + 18ez
6ey + 18ez 6 5a
When
ez = 0,
When
ey = 0,
Ans.
5
a
6
5
ez 6
a
18
ey 6
Repeat the same procedures for point B, C and D. The region where P can be
applied without creating tensile stress at points A, B, C and D is shown shaded in the
diagram.
579
a
P
a
D
ez
ey
C
a
a
y
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*8–52. The hook is used to lift the force of 600 lb.
Determine the maximum tensile and compressive stresses
at section a–a. The cross section is circular and has a
diameter of 1 in. Use the curved-beam formula to compute
the bending stress.
300 lb
a
300 lb
2.5 in.
a
1.5 in.
Section Properties:
r = 1.5 + 0.5 = 2.00 in.
dA
= 2p A r - 2r2 - c2 B
LA r
600 lb
= 2p A 2.00 - 22.002 - 0.52 B
= 0.399035 in.
A = p A 0.52 B = 0.25p in2
A
R =
dA
1A r
=
0.25p
= 1.968246 in.
0.399035
r - R = 2 - 1.968246 = 0.031754 in.
Internal Force and Moment: As shown on FBD. The internal moment must be
computed about the neutral axis. M = 1180.95 lb # in. is positive since it tends to
increase the beam’s radius of curvature.
Normal Stress: Applying the curved-beam formula.
For tensile stress
(st)max =
=
M(R - r1)
N
+
A
Ar1 (r - R)
1180.95(1.968246 - 1.5)
600
+
0.25p
0.25p(1.5)(0.031754)
= 15546 psi = 15.5 ksi (T)
Ans.
For compressive stress,
(sc)max =
=
M(R - r2)
N
+
A
Ar2 (r - R)
1180.95(1.968246 - 2.5)
600
+
0.25p
0.25p(2.5)(0.031754)
= -9308 psi = 9.31 ksi (C)
Ans.
580
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•8–53.
The masonry pier is subjected to the 800-kN load.
Determine the equation of the line y = f1x2 along which
the load can be placed without causing a tensile stress in the
pier. Neglect the weight of the pier.
800 kN
1.5 m y
1.5 m
2.25 m
y
2.25 m
x
x
C
A
B
A = 3(4.5) = 13.5 m2
Ix =
1
(3)(4.53) = 22.78125 m4
12
Iy =
1
(4.5)(33) = 10.125 m4
12
Normal Stress: Require sA = 0
sA =
0 =
Myx
Mxy
P
+
+
A
Ix
Iy
800(103)y(2.25)
800(103)x(1.5)
-800(103)
+
+
13.5
22.78125
10.125
0 = 0.148x + 0.0988y - 0.0741
y = 0.75 - 1.5 x
Ans.
581
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8–54. The masonry pier is subjected to the 800-kN load. If
x = 0.25 m and y = 0.5 m, determine the normal stress at
each corner A, B, C, D (not shown) and plot the stress
distribution over the cross section. Neglect the weight of
the pier.
800 kN
1.5 m y
1.5 m
2.25 m
y
2.25 m
x
x
A = 3(4.5) = 13.5 m2
Ix =
1
(3)(4.53) = 22.78125 m4
12
C
1
Iy =
(4.5)(33) = 10.125 m4
12
s =
sA =
A
B
Myx
Mxy
P
+
+
A
Ix
Iy
-800(103)
400(103)(2.25)
200(103)(1.5)
+
+
13.5
22.78125
10.125
Ans.
= 9.88 kPa (T)
3
sB =
3
3
400(10 )(2.25)
200(10 )(1.5)
-800(10 )
+
13.5
22.78125
10.125
Ans.
= -49.4 kPa = 49.4 kPa (C)
sC =
-800(103)
400(103)(2.25)
200(103)(1.5)
+
13.5
22.78125
10.125
Ans.
= -128 kPa = 128 kPa (C)
sD =
-800(103)
400(103)(2.25)
200(103)(1.5)
+
13.5
22.78125
10.125
= -69.1 kPa = 69.1 kPa (C)
Ans.
582
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8–55. The bar has a diameter of 40 mm. If it is subjected to
the two force components at its end as shown, determine
the state of stress at point A and show the results on a
differential volume element located at this point.
x
100 mm
150 mm
z
A
B
500 N
y
300 N
Internal Forces and Moment:
©Fx = 0;
Nx = 0
©Fy = 0;
Vy + 300 = 0
Vy = -300 N
©Fz = 0;
Vz - 500 = 0
Vz = 500 N
©Mx = 0;
Tx = 0
©My = 0;
My - 500(0.15) = 0
My = 75.0 N # m
©Mz = 0;
Mz - 300(0.15) = 0
Mz = 45.0 N # m
Section Properties:
A = p A 0.022 B = 0.400 A 10 - 3 B p m2
p
A 0.024 B = 40.0 A 10 - 9 B p m4
4
Ix = Iy =
J =
p
A 0.024 B = 80.0 A 10 - 9 B p m4
2
(QA)z = 0
4(0.02) 1
c p A 0.022 B d = 5.333 A 10 - 6 B m3
3p
2
(QA)y =
Normal Stress:
s =
Myz
Mzy
N
+
A
Iz
Iy
sA = 0 -
75.0(0.02)
45.0(0)
-9
+
40.0(10 )p
40.0(10 - 9)p
= 11.9 MPa (T)
Ans.
Shear Stress: The tranverse shear stress in the z and y directions can be obtained
VQ
using the shear formula, tV =
.
It
(txy)A = -tVy = -
300 C 5.333(10 - 6) D
40.0(10 - 9)p (0.04)
Ans.
= -0.318 MPa
(txz)A = tVz = 0
Ans.
583
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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 2-in.-diameter rod is subjected to the loads
shown. Determine the state of stress at point A, and show the
results on a differential element located at this point.
z
B
x
A
8 in.
y
600 lb
12 in.
Consider the equilibrium of the FBD of the right cut segment, Fig. a,
©Fy = 0 ;
Ny + 800 = 0
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
©Mx = 0 ;
Mx + 600(8) = 0
Mx = -4800 lb # in
p 4
p
(1 ) = in4
4
4
A = p(12) = p in2
Ix = Iz =
J =
500 lb
800 lb
Ty = 7200 lb # in
Mz = -13600 lb # in
p 4
p
(1 ) = in4
2
2
Referring to Fig. b,
(Qz)A = y¿A¿ =
(Qx)A = 0
4(1) p 2
c (1 ) d = 0.6667 in3
3p 2
The normal stress is contributed by axial and bending stress. Thus,
s =
Mzx
Mxz
N
+
A
Ix
Iz
For point A, z = 0 and x = 1 in.
4800(0)
-13600(1)
800
p
p>4
p>4
s =
= 17.57(103) psi = 17.6 ksi (T)
Ans.
The torsional shear stress developed at point A is
(tyz)T =
TyC
J
=
7200(1)
= 4.584(103) psi = 4.584 ksi T
p>2
The transverse shear stress developed at point A is.
(tyz)g =
(txy)g =
Vz(Qz)A
Ixt
=
600(0.6667)
= 254.64 psi = 0.2546 ksi T
p
(2)
4
Vx(Qx)A
500(0)
=
= 0
p
Izt
(2)
4
585
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8–57. Continued
Combining these two shear stress components,
tyz = A tyz B T + A tyz B g
= 4.584 + 0.2546
Ans.
= 4.838 ksi = 4.84 ksi
txy = 0
Ans.
The state of stress of point A can be represented by the volume element shown in
Fig. c.
586
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8–58. The 2-in.-diameter rod is subjected to the loads
shown. Determine the state of stress at point B, and show
the results on a differential element located at this point.
z
B
x
A
8 in.
y
600 lb
12 in.
Consider the equilibrium of the FBD of the right cut segment, Fig. a,
©Fy = 0;
Ny + 800 = 0
Ny = -800 lb
©Fz = 0;
Vz + 600 = 0
Vz = -600 lb
©Fx = 0;
Vx - 500 = 0
Vx = 500 lb
©My = 0;
Ty - 600(12) = 0
©Mz = 0;
Mz + 800(12) + 500(8) = 0 Mz = -13600 lb # in
©Mx = 0;
Mx + 600(8) = 0
800 lb
Ty = 7200 lb # in
Mx = -4800 lb # in.
The cross-sectional area the moment of inertia about x and Z axes and polar
moment of inertia of the rod are
¿
A = p(12) = p in2
Ix = Iz =
p 4
p
(1 ) = in4
4
4
J =
p 4
p
(1 ) = in4
2
2
Referring to Fig. b,
(Qz)B = 0 (Qx)B = z¿A¿ =
4(1) p 2
c (1 ) d = 0.6667 in4
3p 2
The normal stress is contributed by axial and bending stress. Thus,
s =
Mzx
Mxz
N
+
A
Ix
Iz
For point B, x = 0 and z = 1 in.
s =
4800 (1)
13600 (0)
800
+
p
p>4
p>4
= 5.86 ksi (C)
The torsional shear stress developed at point B is
(txy)T =
TyC
J
=
7200(1)
= 4.584(103) psi = 4.584 ksi :
p>2
The transverse shear stress developed at point B is.
(txy)v =
(tyz)v =
Vx(Qx)B
500 (0.6667)
=
= 212.21 psi = 0.2122 ksi :
p
Izt
(2)
4
Vz(Qz)B
Ixt
=
500 lb
600 (0)
= 0
p
(2)
4
587
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8–58.
Continued
Combining these two shear stress components,
txy = (txy)T + (txy)v
= 4.584 + 0.2122
= 4.796 ksi = 4.80 ksi
Ans.
tyz = 0
Ans.
The state of stress of point B can be represented by the volume element shown in
Fig. c.
588
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8–59. If P = 60 kN, determine the maximum normal
stress developed on the cross section of the column.
2P
150 mm
150 mm
Equivalent Force System: Referring to Fig. a,
+ c ©Fx = A FR B x;
F = 180 kN
-60 - 120 = -F
15 mm
©My = (MR)y;
-60(0.075) = -My
My = 4.5kN # m
©Mz = (MR)z;
-120(0.25) = -Mz
Mz = 30kN # m
100 mm
100 mm
Section Properties: The cross-sectional area and the moment of inertia about the y
and z axes of the cross section are
A = 0.2(0.3) - 0.185(0.27) = 0.01005 m2
Iz =
1
1
(0.2) A 0.33 B (0.185) A 0.273 B = 0.14655 A 10 - 3 B m4
12
12
Iy = 2c
1
1
(0.015) A 0.23 B d +
(0.27) A 0.0153 B = 20.0759 A 10 - 6 B m4
12
12
Normal Stress: The normal stress is the combination of axial and bending stress.
Here, F is negative since it is a compressive force. Also, My and Mz are negative
since they are directed towards the negative sense of their respective axes. By
inspection, point A is subjected to a maximum normal stress. Thus,
s =
Myz
Mzy
N
+
A
Iz
Iy
smax = sA =
-180 A 103 B
0.01005
-
C -30 A 103 B D ( -0.15)
0.14655 A 10 - 3 B
15 mm
15 mm
P
+
C -4.5 A 103 B D (0.1)
20.0759 A 10 - 6 B
= -71.0 MPa = 71.0 MPa(C)
Ans.
589
75 mm
100 mm
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*8–60. Determine the maximum allowable force P, if the
column is made from material having an allowable normal
stress of sallow = 100 MPa .
2P
150 mm
150 mm
Equivalent Force System: Referring to Fig. a,
+ c ©Fx = (FR)x;
15 mm
-P - 2P = -F
100 mm
100 mm
F = 3P
©My = (MR)y;
-P(0.075) = -My
My = 0.075 P
©Mz = (MR)z;
-2P(0.25) = -Mz
Mz = 0.5P
Section Properties: The cross-sectional area and the moment of inertia about the y
and z axes of the cross section are
A = 0.2(0.3) - 0.185(0.27) = 0.01005 m2
Iz =
1
1
(0.2) A 0.33 B (0.185) A 0.273 B = 0.14655 A 10 - 3 B m4
12
12
Iy = 2 c
1
1
(0.15) A 0.23 B d +
(0.27) A 0.0153 B = 20.0759 A 10 - 6 B m4
12
12
Normal Stress: The normal stress is the combination of axial and bending stress.
Here, F is negative since it is a compressive force. Also, My and Mz are negative
since they are directed towards the negative sense of their respective axes. By
inspection, point A is subjected to a maximum normal stress, which is in
compression. Thus,
s =
Myz
Mzy
N
+
A
Iz
Iy
-100 A 106 B = -
15 mm
15 mm
P
(-0.5P)(-0.15)
-0.075P(0.1)
3P
+
0.01005
0.14655 A 10 - 3 B
20.0759 A 10 - 6 B
P = 84470.40 N = 84.5 kN
Ans.
590
75 mm
100 mm
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•8–61.
The beveled gear is subjected to the loads shown.
Determine the stress components acting on the shaft at
point A, and show the results on a volume element located
at this point. The shaft has a diameter of 1 in. and is fixed to
the wall at C.
z
y
A
200 lb
C
B
x
8 in.
3 in.
©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 = 0;
200(8) - Mx = 0;
Mx = 1600 lb # in.
©My = 0;
200(3) - Ty = 0;
Ty = 600 lb # in.
©Mz = 0;
Mz + 75(3) - 125(8) = 0;
75 lb
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 = 0;
200(8) - Mx = 0;
Mx = 1600 lb # in.
©My = 0;
200(3) - Ty = 0;
Ty = 600 lb # in.
©Mz = 0;
Mz + 75(3) - 125(8) = 0;
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 =
My z
Mz y
N
+
A
Iz
Iy
sC =
(-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 =
My z
Mzy
N
+
A
Iz
Iy
sF =
(-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
©Mx = 0; T + 50 sin 60°(12) = 0
T = -519.62 lb # in
©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
©Mx = 0; T + 50 sin 60°(12) = 0
T = -519.62 lb # in
©My = 0; My - 50 cos 60°(10) = 0
My = 250 lb # in
©Mz = 0; Mz + 50 sin 60°(10) = 0
Mz = -433.01 lb # in
Section Properties: The moment of inertia about the y and z axes and the polar
moment of inertia of the pipe are
p 4
A 1 - 0.754 B = 0.53689 in4
4
Iy = Iz =
J =
p 4
A 1 - 0.754 B = 1.07379 in4
2
Referring to Fig. b,
A Qz B B = 0
A Qy B B = y1œ A1œ - y2œ A2œ =
4(1) p 2
4(0.75) p
c A1 B d c A 0.752 B d = 0.38542 in3
3p 2
3p
2
Normal Stress: The normal stress is contributed by bending stress only. Thus,
s = -
Myz
Mzy
Iz
+
Iy
For point B, y = 0 and z = -1. Then
sB = -0 +
250(1)
= -465.64 psi = 466 psi (C)
0.53689
Ans.
Shear Stress: The torsional shear stress developed at point B is
c A txy B T d
=
B
519.62(1)
TrC
=
= 483.91 psi
J
1.07379
599
12 in.
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8–66.
Continued
The transverse shear stress developed at point B is
c A txz B V d
c A txy B V d
= 0
B
=
B
Vy A Qy B B
Izt
=
43.30(0.38542)
= 62.17 psi
0.53689(2 - 1.5)
Combining these two shear stress components,
A txy B B = c A txy B T d - c A txy B V d
B
B
Ans.
= 483.91 - 62.17 = 422 psi
A txz B B = 0
Ans.
600
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•8–67. The eccentric force P is applied at a distance ey from
the centroid on the concrete support shown. Determine the
range along the y axis where P can be applied on the cross
section so that no tensile stress is developed in the material.
x
z
P
b
2
ey
b
2
2h
3
h
3
Internal Loadings: As shown on the free - body diagram, Fig. a.
Section Properties: The cross-sectional area and moment of inertia about the z axis
of the triangular concrete support are
A =
1
bh
2
Iz =
1
bh3
36
Normal Stress: The normal stress is the combination of axial and bending stress.
Thus,
s =
Mzy
N
A
Iz
A Pey B y
-P
1
1
bh
bh3
2
36
s =
2P 2
A h + 18eyy B
bh3
s = -
Here, it is required that sA … 0 and sB … 0. For point A, y =
(1)
h
, Then. Eq. (1) gives
3
2P 2
h
ch + 18ey a b d
3
3
bh
0 Ú -
0 … h2 + 6hey
ey Ú -
For Point B, y = -
h
6
2
h. Then. Eq. (1) gives
3
0 Ú -
2P 2
2
ch + 18ey a - hb d
3
3
bh
0 … h2 - 12hey
ey …
h
12
Thus, in order that no tensile stress be developed in the concrete support, ey must be
in the range of
-
h
h
… ey …
6
12
Ans.
601
y
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*8–68. The bar has a diameter of 40 mm. If it is subjected
to a force of 800 N as shown, determine the stress components
that act at point A and show the results on a volume element
located at this point.
150 mm
200 mm
1
1
I = p r4 = (p)(0.024) = 0.1256637 (10 - 6) m4
4
4
A
A = p r2 = p(0.022) = 1.256637 (10 - 3) m2
QA = y¿A¿ = a
z
B
4 (0.02) p (0.02)2
ba
b = 5.3333 (10 - 6) m3
3p
2
y
x
30⬚
800 N
P
Mz
sA =
+
A
I
400
+ 0 = 0.318 MPa
1.256637 (10 - 3)
Ans.
692.82 (5.3333) (10 - 6)
VQA
= 0.735 MPa
=
It
0.1256637 (10 - 6)(0.04)
Ans.
=
tA =
•8–69.
Solve Prob. 8–68 for point B.
150 mm
200 mm
A
z
B
y
x
30⬚
800 N
1
1
I = p r4 = (p)(0.024) = 0.1256637 (10 - 6) m4
4
4
A = p r2 = p(0.022) = 1.256637 (10 - 3) m2
QB = 0
sB =
138.56 (0.02)
P
Mc
400
= -21.7 MPa
=
-3
A
I
1.256637 (10 )
0.1256637 (10 - 6)
tB = 0
Ans.
Ans.
602
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8–70. The 43-in.-diameter shaft is subjected to the loading
shown. Determine the stress components at point A. Sketch
the results on a volume element located at this point. The
journal bearing at C can exert only force components Cy
and Cz on the shaft, and the thrust bearing at D can exert
force components Dx , Dy , and Dz on the shaft.
D
z
125 lb
2 in.
8 in.
125 lb
2 in.
A
20 in.
8 in.
B
C
10 in.
y
20 in.
x
A =
p
(0.752) = 0.44179 in2
4
I =
p
(0.3754) = 0.015531 in4
4
QA = 0
tA = 0
sA =
Ans.
My c
I
=
-1250(0.375)
= -30.2 ksi = 30.2 ksi (C)
0.015531
Ans.
8–71. Solve Prob. 8–70 for the stress components at point B.
D
z
125 lb
2 in.
8 in.
125 lb
2 in.
A
8 in.
B
C
10 in.
x
p
A = (0.752) = 0.44179 in2
4
I =
p
(0.3754) = 0.015531 in4
4
QB = y¿A¿ =
4(0.375) 1
a b(p)(0.3752) = 0.035156 in3
3p
2
sB = 0
tB =
Ans.
VzQB
It
=
125(0.035156)
= 0.377 ksi
0.015531(0.75)
Ans.
603
20 in.
20 in.
y
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*8–72. The hook is subjected to the force of 80 lb.
Determine the state of stress at point A at section a–a. The
cross section is circular and has a diameter of 0.5 in. Use the
curved-beam formula to compute the bending stress.
80 lb
1.5 in.
45⬚
The location of the neutral surface from the center of curvature of the hook, Fig. a,
can be determined from
where A = p(0.252) = 0.0625p in2
dA
LA r
= 2p A r - 2r2 - c2 B = 2p A 1.75 - 21.752 - 0.252 B = 0.11278 in.
Thus,
R =
0.0625p
= 1.74103 in.
0.11278
Then
e = r - R = 1.75 - 1.74103 = 0.0089746 in.
Referring to Fig. b, I and QA are
I =
p
(0.254) = 0.9765625(10 - 3)p in4
4
QA = 0
Consider the equilibrium of the FBD of the hook’s cut segment, Fig. c,
+ ©F = 0;
;
x
N - 80 cos 45° = 0
N = 56.57 lb
+ c ©Fy = 0;
80 sin 45° - V = 0
V = 56.57 lb
a + ©Mo = 0;
M - 80 cos 45°(1.74103) = 0
M = 98.49 lb # in
The normal stress developed is the combination of axial and bending stress. Thus,
s =
M(R - r)
N
+
A
Ae r
Here, M = 98.49 lb # in since it tends to reduce the curvature of the hook. For point
A, r = 1.5 in. Then
s =
(98.49)(1.74103 - 1.5)
56.57
+
0.0625p
0.0625p(0.0089746)(1.5)
= 9.269(103) psi = 9.27 ksi (T)
Ans.
The shear stress in contributed by the transverse shear stress only. Thus
t =
VQA
= 0
It
Ans.
The state of strees of point A can be represented by the element shown in Fig. d.
604
A
B
B a
A
R =
dA
©
LA r
©
a
A
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•8–73.
The hook is subjected to the force of 80 lb.
Determine the state of stress at point B at section a–a. The
cross section has a diameter of 0.5 in. Use the curved-beam
formula to compute the bending stress.
80 lb
1.5 in.
45⬚
The location of the neutral surface from the center of curvature of the the hook,
Fig. a, can be determined from
R =
dA
©
LA r
dA
= 2p A r - 2r2 - c2 B = 2p A 1.75 - 21.752 - 0.252 B = 0.11278 in.
LA r
Thus,
R =
0.0625p
= 1.74103 in
0.11278
Then
e = r - R = 1.75 - 1.74103 = 0.0089746 in
Referring to Fig. b, I and QB are computed as
p
(0.254) = 0.9765625(10 - 3)p in4
4
I =
QB = y¿A¿ =
4(0.25) p
c (0.252) d = 0.0104167 in3
3p
2
Consider the equilibrium of the FBD of the hook’s cut segment, Fig. c,
+ ©F = 0;
;
x
N - 80 cos 45° = 0
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
B, r = 1.75 in. Then
s =
(98.49)(1.74103 - 1.75)
56.57
+
0.0625p
0.0625 p (0.0089746)(1.75)
= 1.62 psi (T)
Ans.
The shear stress is contributed by the transverse shear stress only. Thus,
t =
56.57 (0.0104167)
VQB
= 3.84 psi
=
It
0.9765625(10 - 3)p (0.5)
Ans.
The state of stress of point B can be represented by the element shown in Fig. d.
605
A
B
B a
A
Where A = p(0.252) = 0.0625p in2
©
a
A
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8–74. The block is subjected to the three axial loads
shown. Determine the normal stress developed at points A
and B. Neglect the weight of the block.
100 lb
250 lb
50 lb
2 in.4 in.
5 in.
2 in.
3 in.
5 in.
A
B
Mx = -250(1.5) - 100(1.5) + 50(6.5) = -200 lb # in.
My = 250(4) + 50(2) - 100(4) = 700 lb # in.
Ix =
1
1
(4)(133) + 2 a b (2)(33) = 741.33 in4
12
12
Iy =
1
1
(3)(83) + 2 a b (5)(43) = 181.33 in4
12
12
A = 4(13) + 2(2)(3) = 64 in2
s =
My x
Mx y
P
+
A
Iy
Ix
sA = -
700(4)
-200 (-1.5)
400
+
64
181.33
741.33
= -21.3 psi
sB = -
Ans.
700(2)
-200 (-6.5)
400
+
64
181.33
741.33
= -12.2 psi
Ans.
606
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8–75. The 20-kg drum is suspended from the hook
mounted on the wooden frame. Determine the state of
stress at point E on the cross section of the frame at
section a–a. Indicate the results on an element.
50 mm
25 mm
E
75 mm
Section a – a
0.5 m 0.5 m
1m
a
B
C
a
1m
30⬚
Support Reactions: Referring to the free-body diagram of member BC
shown in Fig. a,
a + ©MB = 0;
F sin 45°(1) - 20(9.81)(2) = 0
F = 554.94 N
+ ©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
b
75 mm
1m
D
F
A
25 mm
Internal Loadings: Consider the equilibrium of the free - body diagram of
the right segment shown in Fig. b.
Section b – b
+ ©F = 0;
:
x
N - 392.4 = 0
N = 392.4 N
+ c ©Fy = 0;
V - 196.2 = 0
V = 196.2 N
a + ©MC = 0;
196.2(0.5) - M = 0
M = 98.1 N # m
Section Properties: The cross -sectional area and the moment of inertia of the cross
section are
A = 0.05(0.075) = 3.75 A 10 - 3 B m2
I =
1
(0.05) A 0.0753 B = 1.7578 A 10 - 6 B m4
12
Referring to Fig. c, QE is
QE = y¿A¿ = 0.025(0.025)(0.05) = 3.125 A 10 - 6 B m3
Normal Stress: The normal stress is the combination of axial and bending stress.
Thus,
s =
My
N
;
A
I
For point A, y = 0.0375 - 0.025 = 0.0125 m. Then
sE =
392.4
3.75 A 10
-3
B
98.1(0.0125)
+
1.7578 A 10 - 6 B
= 802 kPa
Ans.
Shear Stress: The shear stress is contributed by transverse shear stress only. Thus,
tE =
196.2 C 31.25 A 10 - 6 B D
VQA
=
= 69.8 kPa
It
1.7578 A 10 - 6 B (0.05)
Ans.
The state of stress at point E is represented on the element shown in Fig. d.
607
75 mm
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8–75.
Continued
608
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*8–76. The 20-kg drum is suspended from the hook
mounted on the wooden frame. Determine the state of
stress at point F on the cross section of the frame at
section b–b. Indicate the results on an element.
50 mm
25 mm
E
75 mm
Section a – a
0.5 m 0.5 m
1m
a
B
C
a
1m
30⬚
1m
b
FBD sin 30°(3) - 20(9.81)(2) = 0
FBD = 261.6 N
+ c ©Fy = 0;
Ay - 261.6 cos 30° - 20(9.81) = 0
Ay = 422.75 N
+ ©F = 0;
:
x
Ax - 261.6 sin 30° = 0
Ax = 130.8 N
75 mm
1m
Support Reactions: Referring to the free-body diagram of the entire frame
shown in Fig. a,
a + ©MA = 0;
b
D
F
A
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,
130.8 c46.875 A 10 - 6 B d
VQA
=
= 31.0 kPa
tA =
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.
610
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•8–77. The eye is subjected to the force of 50 lb. Determine
the maximum tensile and compressive stresses at section a-a.
The cross section is circular and has a diameter of 0.25 in. Use
the curved-beam formula to compute the bending stress.
50 lb
0.25 in.
1.25 in.
a
Section Properties:
r = 1.25 +
0.25
= 1.375 in.
2
dA
= 2p A r - 2r2 - c2 B
LA r
= 2p A 1.375 - 21.3752 - 0.1252 B
= 0.035774 in.
A = p A 0.1252 B = 0.049087 in2
R =
A
dA
1A r
=
0.049087
= 1.372153 in.
0.035774
r - R = 1.375 - 1.372153 = 0.002847 in.
Internal Force and Moment: As shown on FBD. The internal moment must be
computed about the neutral axis. M = 68.608 lb # in is positive since it tends to
increase the beam’s radius of curvature.
Normal Stress: Applying the curved - beam formula, For tensile stress
(st)max =
=
M(Rr1)
N
+
A
Ar1(r - R)
68.608(1.372153 - 1.25)
50.0
+
0.049087
0.049087(1.25)(0.002847)
= 48996 psi = 49.0 ksi (T)
Ans.
For compressive stress
(sc)max =
=
M(R - r2)
N
+
A
Ar2(r - R)
68.608(1.372153 - 1.50)
50.0
+
0.049087
0.049087(1.50)(0.002847)
= -40826 psi = 40.8 ksi (C)
Ans.
611
a
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8–78. Solve Prob. 8–77 if the cross section is square, having
dimensions of 0.25 in. by 0.25 in.
50 lb
0.25 in.
1.25 in.
a
Section Properties:
r = 1.25 +
0.25
= 1.375 in.
2
r2
dA
1.5
= bln
= 0.25 ln
= 0.45580 in.
r1
1.25
LA r
A = 0.25(0.25) = 0.0625 in2
R =
0.0625
A
=
= 1.371204 in.
dA
0.045580
1A
r
r - R = 1.375 - 1.371204 = 0.003796 in.
Internal Force and Moment: As shown on FBD. The internal moment must be
computed about the neutral axis. M = 68.560 lb # in. is positive since it tends to
increase the beam’s radius of curvature.
Normal Stress: Applying the curved -beam formula, For tensile stress
(st)max =
=
M(R - r1)
N
+
A
Ar1(r - R)
68.560(1.371204 - 1.25)
50.0
+
0.0625
0.0625(1.25)(0.003796)
= 28818 psi = 28.8 ksi (T)
Ans.
For Compressive stress
(sc)max =
=
M(R - r2)
N
+
A
Ar2 (r - R)
68.560(1.371204 - 1.5)
50.0
+
0.0625
0.0625(1.5)(0.003796)
= -24011 psi = 24.0 ksi (C)
Ans.
612
a
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8–79. If the cross section of the femur at section a–a can
be approximated as a circular tube as shown, determine the
maximum normal stress developed on the cross section at
section a–a due to the load of 75 lb.
2 in.
75 lb
a
a
0.5 in.
1 in.
Section a – a
M
F
Internal Loadings: Considering the equilibrium for the free-body diagram of the
femur’s upper segment, Fig. a,
+ c ©Fy = 0;
N - 75 = 0
N = 75 lb
a + ©MO = 0;
M - 75(2) = 0
M = 150 lb # in
Section Properties: The cross-sectional area, the moment of inertia about the
centroidal axis of the femur’s cross section are
A = p A 12 - 0.52 B = 0.75p in2
I =
p 4
A 1 - 0.54 B = 0.234375p in4
4
Normal Stress: The normal stress is a combination of axial and bending stress. Thus,
s =
My
N
+
A
I
By inspection, the maximum normal stress is in compression.
smax =
150(1)
-75
= -236 psi = 236 psi (C)
0.75p
0.234375p
613
Ans.
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*8–80. The hydraulic cylinder is required to support a
force of P = 100 kN. If the cylinder has an inner diameter
of 100 mm and is made from a material having an allowable
normal stress of sallow = 150 MPa, determine the required
minimum thickness t of the wall of the cylinder.
P
t
100 mm
Equation of Equilibrium: The absolute pressure developed in the hydraulic
cylinder can be determined by considering the equilibrium of the free-body
diagram of the piston shown in Fig. a. The resultant force of the pressure on the
p
piston is F = pA = pc A 0.12 B d = 0.0025pp. Thus,
4
©Fx¿ = 0; 0.0025pp - 100 A 103 B = 0
p = 12.732 A 106 B Pa
Normal Stress: For the cylinder, the hoop stress is twice as large as the
longitudinal stress,
sallow =
pr
;
t
150 A 106 B =
12.732 A 106 B (50)
t
t = 4.24 mm
Since
Ans.
r
50
=
= 11.78 7 10, thin -wall analysis is valid.
t
4.24
614
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•8–81. The hydraulic cylinder has an inner diameter of
100 mm and wall thickness of t = 4 mm. If it is made from
a material having an allowable normal stress of sallow =
150 MPa, determine the maximum allowable force P.
P
t
100 mm
Normal Stress: For the hydraulic cylinder, the hoop stress is twice as large as the
longitudinal stress.
Since
50
r
=
= 12.5 7 10, thin-wall analysis can be used.
t
4
sallow =
pr
;
t
150 A 106 B =
p(50)
4
p = 12 A 106 B MPa
Ans.
Equation of Equilibrium: The resultant force on the piston is
F = pA = 12 A 106 B c
p
A 0.12 B d = 30 A 103 B p. Referring to the free-body diagram of
4
the piston shown in Fig. a,
©Fx¿ = 0; 30 A 103 B p - P = 0
P = 94.247 A 103 B N = 94.2 kN
Ans.
615
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8–82. The screw of the clamp exerts a compressive force
of 500 lb on the wood blocks. Determine the maximum
normal stress developed along section a-a. The cross
section there is rectangular, 0.75 in. by 0.50 in.
4 in.
Internal Force and Moment: As shown on FBD.
a
Section Properties:
A = 0.5(0.75) = 0.375 in2
I =
a
1
(0.5) A 0.753 B = 0.017578 in4
12
0.75 in.
Maximum Normal Stress: Maximum normal stress occurs at point A.
smax = sA =
=
Mc
N
+
A
I
2000(0.375)
500
+
0.375
0.017578
= 44000 psi = 44.0 ksi (T)
Ans.
8–83. Air pressure in the cylinder is increased by exerting
forces P = 2 kN on the two pistons, each having a radius
of 45 mm. If the cylinder has a wall thickness of 2 mm,
determine the state of stress in the wall of the cylinder.
p =
2(103)
P
= 314 380.13 Pa
=
A
p(0.0452)
s1 =
pr
314 380.13(0.045)
=
= 7.07 MPa
t
0.002
P
47 mm
Ans.
s2 = 0
P
Ans.
The pressure P is supported by the surface of the pistons in the longitudinal
direction.
*8–84. Determine the maximum force P that can be
exerted on each of the two pistons so that the circumferential
stress component in the cylinder does not exceed 3 MPa.
Each piston has a radius of 45 mm and the cylinder has a
wall thickness of 2 mm.
s =
pr
;
t
3(106) =
P
47 mm
p(0.045)
0.002
P = 133.3 kPa
Ans.
P = pA = 133.3 A 103 B (p)(0.045)2 = 848 N
Ans.
616
P
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•8–85.
The cap on the cylindrical tank is bolted to the tank
along the flanges. The tank has an inner diameter of 1.5 m
and a wall thickness of 18 mm. If the largest normal stress is
not to exceed 150 MPa, determine the maximum pressure
the tank can sustain. Also, compute the number of bolts
required to attach the cap to the tank if each bolt has a
diameter of 20 mm. The allowable stress for the bolts is
1sallow2b = 180 MPa.
Hoop Stress for Cylindrical Tank: Since
750
r
=
= 41.7 7 10, then thin wall
t
18
analysis can be used. Applying Eq. 8–1
s1 = sallow =
150 A 106 B =
pr
t
p(750)
18
p = 3.60 MPa
Ans.
Force Equilibrium for the Cap:
+ c ©Fy = 0;
3.60 A 106 B C p A 0.752 B D - Fb = 0
Fb = 6.3617 A 106 B N
Allowable Normal Stress for Bolts:
(sallow)b =
180 A 106 B =
P
A
6.3617(106)
n C p4 (0.022) D
n = 112.5
Use n = 113 bolts
Ans.
617
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8–86. The cap on the cylindrical tank is bolted to the tank
along the flanges. The tank has an inner diameter of 1.5 m
and a wall thickness of 18 mm. If the pressure in the tank is
p = 1.20 MPa, determine the force in each of the 16 bolts
that are used to attach the cap to the tank. Also, specify the
state of stress in the wall of the tank.
Hoop Stress for Cylindrical Tank: Since
750
r
=
= 41.7 7 10, then thin wall
t
18
analysis can be used. Applying Eq. 8–1
s1 =
pr
1.20(106)(750)
=
= 50.0 MPa
t
18
Ans.
Longitudinal Stress for Cylindrical Tank:
s2 =
pr
1.20(106)(750)
=
= 25.0 MPa
2t
2(18)
Ans.
Force Equilibrium for the Cap:
+ c ©Fy = 0;
1.20 A 106 B C p A 0.752 B D - 16Fb = 0
Fb = 132536 N = 133 kN
Ans.
618
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9–1. Prove that the sum of the normal stresses
sx + sy = sx¿ + sy¿ is constant. See Figs. 9–2a and 9–2b.
Stress Transformation Equations: Applying Eqs. 9-1 and 9-3 of the text.
sx¿ + sy¿ =
sx + sy
2
sx - sy
+
2
cos 2u + txy sin 2u
sx + sy
+
2
sx - sy
-
2
cos 2u - txy sin 2u
sx¿ + sy¿ = sx + sy
(Q.E.D.)
619
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9–2. The state of stress at a point in a member is shown on
the element. Determine the stress components acting on
the inclined plane AB. Solve the problem using the method
of equilibrium described in Sec. 9.1.
A
8 ksi
2 ksi
5 ksi
60⬚
B
Referring to Fig a, if we assume that the areas of the inclined plane AB is ¢A, then
the area of the horizontal and vertical of the triangular element are ¢A cos 60° and
¢A sin 60° respectively. The forces act acting on these two faces indicated on the
FBD of the triangular element, Fig. b.
¢Fx¿ + 2¢A sin 60° cos 60° + 5¢ A sin 60° sin 60°
+Q©Fx¿ = 0;
+ 2¢A cos 60° sin 60° - 8¢A cos 60° cos 60° = 0
¢Fx¿ = -3.482 ¢A
¢Fy¿ + 2¢A sin 60° sin 60° - 5¢ A sin 60° cos 60°
+a©Fy¿ = 0;
- 8¢A cos 60° sin 60° - 2¢A cos 60° cos 60° = 0
¢Fy¿ = 4.629 ¢A
From the definition,
sx¿ = lim¢A:0
¢Fx¿
= -3.48 ksi
¢A
tx¿y¿ = lim¢A:0
¢Fy¿
Ans.
Ans.
= 4.63 ksi
¢A
The negative sign indicates that sx¿, is a compressive stress.
620
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9–3. The state of stress at a point in a member is shown on
the element. Determine the stress components acting on
the inclined plane AB. Solve the problem using the method
of equilibrium described in Sec. 9.1.
500 psi
B
60⬚
A
Referring to Fig. a, if we assume that the area of the inclined plane AB is ¢A, then
the areas of the horizontal and vertical surfaces of the triangular element are
¢A sin 60° and ¢A cos 60° respectively. The force acting on these two faces are
indicated on the FBD of the triangular element, Fig. b
+R©Fx¿ = 0;
¢Fx¿ + 500 ¢A sin 60° sin 60° + 350¢A sin 60° cos 60°
+350¢A cos 60° sin 60° = 0
¢Fx¿ = -678.11 ¢A
+Q©Fy¿ = 0;
¢Fy¿ + 350¢A sin 60° sin 60° - 500¢A sin 60° cos 60°
-350¢A cos 60° cos 60° = 0
¢Fy¿ = 41.51 ¢A
From the definition
sx¿ = lim¢A:0
tx¿y¿ = lim¢A:0
¢Fx¿
= -6.78 psi
¢A
¢Fy¿
Ans.
Ans.
= 41.5 psi
¢A
The negative sign indicates that sx¿, is a compressive stress.
621
350 psi
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*9–4. The state of stress at a point in a member is shown
on the element. Determine the stress components acting on
the inclined plane AB. Solve the problem using the method
of equilibrium described in Sec. 9.1.
A
650 psi
¢Fx¿ - 400(¢Acos 60°)cos 60° + 650(¢ A sin 60°)cos 30° = 0
Q+ ©Fx¿ = 0
400 psi
60⬚
¢Fx¿ = -387.5¢A
¢Fy¿ - 650(¢Asin 60°)sin 30° - 400(¢ A cos 60°)sin 60° = 0
a+ ©Fy¿ = 0
B
¢Fy¿ = 455 ¢A
sx¿ = lim¢A:0
sx¿y¿ = lim¢A:0
¢Fx¿
= -388 psi
¢A
¢Fy¿
Ans.
Ans.
= 455 psi
¢A
The negative sign indicates that the sense of sx¿, is opposite to that shown on FBD.
•9–5.
Solve Prob. 9–4 using the stress-transformation
equations developed in Sec. 9.2.
sy = 400 psi
sx = -650 psi
sx¿ =
=
sx + sy
sx - sy
+
2
2
txy = 0
A
400 psi
u = 30°
650 psi
cos 2u + txy sin 2u
60⬚
-650 + 400
-650 - 400
+
cos 60° + 0 = -388 psi
2
2
Ans.
B
The negative sign indicates sx¿, is a compressive stress.
tx¿y¿ = = -a
sx - sy
2
sin 2u + txy cos 2u
-650 - 400
bsin 60° = 455 psi
2
Ans.
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9–6. The state of stress at a point in a member is shown on
the element. Determine the stress components acting on
the inclined plane AB. Solve the problem using the method
of equilibrium described in Sec. 9.1.
90 MPa
A
35 MPa
60⬚
30⬚
R+ ©Fy¿ = 0
B
50 MPa
¢Fy¿ - 50¢A sin 30° cos 30° - 35¢A sin 30° cos 60° +
90¢A cos 30° sin 30° + 35¢A cos 30° sin 60° = 0
¢Fy¿ = -34.82¢A
b+ ©Fx¿ = 0
¢Fx¿ - 50¢A sin 30° sin 30° + 35¢A sin 30° sin 60°
-90¢A cos 30° cos 30° + 35¢A cos 30° cos 60° = 0
¢Fx¿ = 49.69 ¢A
sx¿ = lim¢A:0
¢Fx¿
= 49.7 MPa
¢A
tx¿y¿ = lim¢A:0
¢Fy¿
Ans.
Ans.
= -34.8 MPa
¢A
The negative signs indicate that the sense of sx¿, and tx¿y¿ are opposite to the shown
on FBD.
9–7. Solve Prob. 9–6 using the stress-transformation
equations developed in Sec. 9.2. Show the result on a sketch.
90 MPa
A
35 MPa
60⬚
30⬚
sy = 50 MPa
sx = 90 MPa
sx¿ =
=
sx + sy
sx - sy
+
2
2
txy = -35 MPa
u = -150°
cos 2u + txy sin 2u
90 - 50
90 + 50
+
cos(-300°) + (-35) sin ( -300°)
2
2
= 49.7 MPa
tx¿y¿ = -
sx - sy
= -a
2
Ans.
sin 2u + txy cos 2u
90 - 50
bsin(-300°) + ( -35) cos ( -300°) = -34.8 MPa
2
The negative sign indicates tx¿y¿ acts in -y¿ direction.
623
Ans.
B
50 MPa
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*9–8. Determine the normal stress and shear stress acting
on the inclined plane AB. Solve the problem using the
method of equilibrium described in Sec. 9.1.
45 MPa
B
80 MPa
45⬚
A
Force Equllibrium: Referring to Fig. a, if we assume that the area of the inclined
plane AB is ¢A, then the area of the vertical and horizontal faces of the triangular
sectioned element are ¢A sin 45° and ¢A cos 45°, respectively. The forces acting on
the free-body diagram of the triangular sectioned element, Fig. b, are
©Fx¿ = 0;
¢Fx¿ + c45 A 106 B ¢A sin 45° dcos 45° + c45 A 106 B ¢A cos 45° dsin 45°
- c80 A 106 B ¢A sin 45° dcos 45° = 0
¢Fx¿ = -5 A 106 B ¢A
©Fy¿ = 0;
¢Fy¿ + c45 A 106 B ¢A cos 45° dcos 45° - c45 A 106 B ¢A sin 45° dsin 45°
- c80 A 106 B ¢ A sin 45° dsin 45° = 0
¢Fy¿ = 40 A 106 B ¢A
Normal and Shear Stress: From the definition of normal and shear stress,
sx¿ = lim¢A:0
¢Fx¿
= -5 MPa
¢A
tx¿y¿ = lim¢A:0
¢Fy¿
Ans.
Ans.
= 40 MPa
¢A
The negative sign indicates that sx¿ is a compressive stress.
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•9–9.
Determine the normal stress and shear stress acting
on the inclined plane AB. Solve the problem using the
stress transformation equations. Show the result on the
sectioned element.
45 MPa
80 MPa
45⬚
Stress Transformation Equations:
u = +135° (Fig. a)
sx = 80 MPa
sy = 0
txy = 45 MPa
we obtain,
sx¿ =
=
sx + sy
sx - sy
+
2
2
cos u + txysin 2u
80 - 0
80 + 0
+
cos 270 + 45 sin 270°
2
2
Ans.
= -5 MPa
tx¿y¿ = -
= -
sx - sy
2
B
sinu + txy cos 2u
80 - 0
sin 270° + 45 cos 270°
2
= 40 MPa
Ans.
The negative sign indicates that sx¿ is a compressive stress. These results are
indicated on the triangular element shown in Fig. b.
625
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9–10. The state of stress at a point in a member is shown
on the element. Determine the stress components acting on
the inclined plane AB. Solve the problem using the method
of equilibrium described in Sec. 9.1.
2 ksi
A
3 ksi
30⬚
4 ksi
B
Force Equllibrium: For the sectioned element,
¢Fy¿ - 3(¢A sin 30°) sin 60° + 4(¢ A sin 30°)sin 30°
a+ ©Fy¿ = 0;
-2(¢A cos 30°) sin 30° - 4(¢A cos 30°) sin 60° = 0
¢Fy¿ = 4.165 ¢A
¢Fx¿ + 3(¢A sin 30°) cos 60° + 4(¢ A sin 30°)cos 30°
Q+ ©Fx¿ = 0;
-2(¢A cos 30°) cos 30° + 4(¢A cos 30°) cos 60° = 0
¢Fx¿ = -2.714 ¢A
Normal and Shear Stress: For the inclined plane.
sx = lim¢A:0
tx¿y¿ = lim¢A:0
¢Fx¿
= -2.71 ksi
¢A
¢Fy¿
Ans.
Ans.
= 4.17 ksi
¢A
Negative sign indicates that the sense of sx¿, is opposite to that shown on FBD.
9–11. Solve Prob. 9–10 using the stress-transformation
equations developed in Sec. 9.2. Show the result on a sketch.
2 ksi
Normal and Shear Stress: In accordance with the established sign convention,
u = +60°
sx = -3 ksi
sy = 2 ksi
A
txy = -4 ksi
3 ksi
30⬚
Stress Transformation Equations: Applying Eqs. 9-1 and 9-2.
sx¿ =
=
sx + sy
sx - sy
+
2
2
B
cos 2u + txy sin 2u
-3 - 2
-3 + 2
+
cos 120° + (-4 sin 120°)
2
2
Ans.
= -2.71 ksi
tx¿y¿ = -
= -
sx - sy
2
4 ksi
sin 2u + txy cos 2u
-3 - 2
sin 120° + (-4 cos 120°)
2
= 4.17 ksi
Ans.
Negative sign indicates sx¿, is a compressive stress
626
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*9–12. Determine the equivalent state of stress on an
element if it is oriented 50° counterclockwise from the
element shown. Use the stress-transformation equations.
10 ksi
16 ksi
sy = 0
sx = -10 ksi
txy = -16 ksi
u = +50°
sx¿ =
=
sx + sy
= -a
=
2
cos 2u + txy sin 2u
-10 - 0
-10 + 0
+
cos 100° + ( -16)sin 100° = -19.9 ksi
2
2
tx¿y¿ = - a
sy¿ =
sx - sy
+
2
sx - sy
2
b sin 2u + txy cos 2u
-10 - 0
b sin 100° + (-16)cos 100° = 7.70 ksi
2
sx + sy
2
sx - sy
-
Ans.
2
Ans.
cos 2u - txy sin 2u
-10 + 0
-10 - 0
- a
bcos 100° - (-16)sin 100° = 9.89 ksi
2
2
627
Ans.
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•9–13.
Determine the equivalent state of stress on an
element if the element is oriented 60° clockwise from the
element shown. Show the result on a sketch.
350 psi
75 psi
200 psi
In accordance to the established sign covention,
u = -60° (Fig. a)
sx = 200 psi
sy = -350 psi
txy = 75 psi
Applying Eqs 9-1, 9-2 and 9-3,
sx¿ =
=
sx + sy
sx - sy
+
2
2
cos 2u + txy sin 2u
200 - ( -350)
200 + (-350)
+
cos ( -120°) + 75 sin (-120°)
2
2
= -277.45 psi = -277 psi
sy¿ =
=
sx + sy
sx - sy
-
2
2
Ans.
cos 2u - txy sin 2u
200 - ( -350)
200 + (-350)
cos ( -120°) - 75 sin ( -120°)
2
2
= 127.45 psi = 127 psi
tx¿y¿ = -
= -
sx - sy
2
Ans.
sin 2u + txy cos 2u
200 - (-350)
sin (-120°) + 75 cos (-120°)
2
= 200.66 psi = 201 psi
Ans.
Negative sign indicates that sx¿ is a compressive stress. These result, can be
represented by the element shown in Fig. b.
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9–14. The state of stress at a point is shown on the element.
Determine (a) the principal stress and (b) the maximum
in-plane shear stress and average normal stress at the point.
Specify the orientation of the element in each case. Show
the results on each element.
30 ksi
12 ksi
sx = -30 ksi
sy = 0
txy = -12 ksi
a)
sx + sy
s1, 2 =
;
2
C
a
sx - sy
2
2
b + txy 2 =
-30 + 0
-30 - 0 2
;
a
b + (-12)2
2
C
2
s1 = 4.21 ksi
Ans.
s2 = -34.2 ksi
Ans.
Orientation of principal stress:
txy
tan 2uP =
(sx - sy)>2
uP = 19.33° and
-12
= 0.8
(-30 -0)>2
=
-70.67°
Use Eq. 9-1 to determine the principal plane of s1 and s2.
sx + sy
sx¿ =
sx - sy
+
2
2
cos 2u + txy sin 2u
u = 19.33°
sx¿ =
-30 + 0
-30 - 0
+
cos 2(19.33°) + (-12)sin 2(19.33°) = -34.2 ksi
2
2
Therefore uP2 = 19.3°
Ans.
and uP1 = -70.7°
Ans.
b)
tmaxin-plane =
savg =
C
a
sx - sy
2
sx + sy
2
=
2
b + txy 2 =
-30 - 0 2
b + (-12)2 = 19.2 ksi
C
2
a
-30 + 0
= -15 ksi
2
Ans.
Ans.
Orientation of max, in - plane shear stress:
tan 2uP =
-(sx - sy)>2
=
txy
uP = -25.2°
and
-(-30 - 0)>2
= -1.25
-12
64.3°
Ans.
By observation, in order to preserve equllibrium along AB, tmax has to act in the
direction shown in the figure.
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9–15. The state of stress at a point is shown on the element.
Determine (a) the principal stress and (b) the maximum
in-plane shear stress and average normal stress at the point.
Specify the orientation of the element in each case. Show
the results on each element.
80 MPa
50 MPa
60 MPa
In accordance to the established sign convention,
sx = -60 MPa
s1, 2 =
=
sy = -80 MPa
sx + sy
;
2
C
a
sx - sy
2
txy = 50 MPa
2
b + txy 2
-60 + (-80)
-60 - (-80) 2
;
c
d + 502
2
C
2
= -70 ; 22600
s2 = -121 MPa
s1 = -19.0 MPa
txy
tan 2uP =
=
(sx - sy)>2
uP = 39.34°
Ans.
50
= 5
[-60 - (-80)]>2
and
-50.65°
Substitute u = 39.34° into Eq. 9-1,
sx¿ =
=
sx + sy
sx - sy
+
2
2
cos 2u + txy sin 2u
-60 + (-80)
-60 - ( -80)
+
cos 78.69° + 50 sin 78.69°
2
2
= -19.0 MPa = s1
Thus,
(uP)1 = 39.3°
Ans.
(uP)2 = -50.7°
The element that represents the state of principal stress is shown in Fig. a.
t max
in-plane
=
C
a
sx - sy
2
tan 2uS =
2
b + txy 2 =
-(sx - sy)>2
=
txy
-60 - (-80) 2
d + 502 = 51.0 MPa
C
2
c
-[-60 - (-80)]>2
= -0.2
50
uS = -5.65° and 84.3°
By Inspection, t max
Ans.
Ans.
has to act in the sense shown in Fig. b to maintain
in-plane
equilibrium.
savg =
sx + sy
2
=
-60 + (-80)
= -70 MPa
2
The element that represents the state of maximum in - plane shear stress is shown in
Fig. c.
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9–15. Continued
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*9–16. The state of stress at a point is shown on the
element. Determine (a) the principal stress and (b) the
maximum in-plane shear stress and average normal stress at
the point. Specify the orientation of the element in each case.
Sketch the results on each element.
60 MPa
30 MPa
45 MPa
sx = 45 MPa
sy = -60 MPa
txy = 30 MPa
a)
s1, 2 =
=
sx + sy
;
2
C
a
sx - sy
2
2
b + txy 2
45 - (-60) 2
45 - 60
a
;
b + (30)2
2
C
2
s1 = 53.0 MPa
Ans.
s2 = -68.0 MPa
Ans.
Orientation of principal stress:
tan 2uP =
txy
(sx - sy)>2
uP = 14.87,
=
30
= 0.5714
(45 - (-60))>2
-75.13
Use Eq. 9-1 to determine the principal plane of s1 and s2:
sx¿ =
=
sx + sy
sx - sy
+
2
2
cos 2u + txy sin 2u,
where u = 14.87°
45 + (-60)
45 - (-60)
+
cos 29.74° + 30 sin 29.74° = 53.0 MPa
2
2
Therefore uP1 = 14.9°
Ans.
and uP2 = -75.1°
Ans.
b)
tmaxin-plane =
savg =
C
a
sx - sy
2
sx - sy
2
=
2
b + txy 2 =
45 - (-60) 2
b + 302 = 60.5 MPa
C
2
a
45 + (-60)
= -7.50 MPa
2
Ans.
Ans.
Orientation of maximum in - plane shear stress:
tan 2uS =
-(sx - sy)>2
txy
=
-(45 - ( -60))>2
= -1.75
30
uS = -30.1°
Ans.
uS = 59.9°
Ans.
and
By observation, in order to preserve equilibrium along AB, tmax has to act in the
direction shown.
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•9–17.
Determine the equivalent state of stress on an
element at the same point which represents (a) the principal
stress, and (b) the maximum in-plane shear stress and the
associated average normal stress. Also, for each case,
determine the corresponding orientation of the element
with respect to the element shown. Sketch the results on
each element.
75 MPa
125 MPa
50 MPa
Normal and Shear Stress:
sx = 125 MPa
sy = -75 MPa
txy = -50 MPa
In - Plane Principal Stresses:
s1,2 =
=
sx - sy
;
2
B
a
sx - sy
2
2
b + txy 2
125 + (-75)
125 - (-75) 2
a
;
b + (-50)2
2
2
B
= 25; 212500
s2 = -86.8 MPa
s1 = 137 MPa
Ans.
Orientation of Principal Plane:
tan 2uP =
txy
A sx - sy B >2
-50
=
A 125 -(-75) B >2
= -0.5
up = -13.28° and 76.72°
Substitute u = -13.28° into
sx¿ =
=
sx + sy
sx - sy
+
2
2
cos 2u + txy sin 2u
125 + (-75)
125 - (-75)
+
cos(-26.57°)+(-50) sin(-26.57°)
2
2
= 137 MPa = s1
Thus,
A up B 1 = -13.3° and A up B 2 = 76.7°
Ans.
125 - (-75)>(-50)
The element that represents the state of principal stress is shown in Fig. a.
Maximum In - Plane Shear Stress:
t max
in-plane
=
C
¢
sx - sy
2
2
≤ + txy 2 =
-100 - 0 2
b + 252 = 112 MPa
2
B
a
Orientation of the Plane of Maximum In - Plane Shear Stress:
tan 2us = -
A sx - sy B >2
txy
= -
A 125 - (-75) B >2
= 2
-50
us = 31.7° and 122°
633
Ans.
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9–17.
Continued
By inspection, t max
has to act in the same sense shown in Fig. b to maintain
in-plane
equilibrium.
Average Normal Stress:
savg =
sx + sy
2
=
125 + (-75)
= 25 MPa
2
Ans.
The element that represents the state of maximum in - plane shear stress is shown in
Fig. c.
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sy
9–18. A point on a thin plate is subjected to the two
successive states of stress shown. Determine the resultant
state of stress represented on the element oriented as
shown on the right.
=
sx¿ + sy¿
sx¿ - sy¿
+
2
2
ⴙ
60⬚
Stress Transformation Equations: Applying Eqs. 9-1, 9-2, and 9-3
u = -30°, sx¿ = -200 MPa,
to
element
(a)
with
sy¿ = -350 MPa and tx¿y¿ = 0.
(sx)a =
350 MPa
cos 2u + tx¿y¿ sin 2u
-200 - (-350)
-200 + (-350)
+
cos (-60°) + 0
2
2
= -237.5 MPa
A sy B a =
=
sx¿ + sy¿
sx¿ - sy¿
-
2
2
cos 2u - tx¿y¿ sin 2u
-200 - (-350)
-200 + (-350)
cos (-60°) - 0
2
2
= -312.5 MPa
A txy B a = = -
sx¿ - sy¿
2
sin 2u + tx¿y¿ cos 2u
-200 - (-350)
sin (-60°) + 0
2
= 64.95 MPa
For element (b), u = 25°, sx¿ = sy¿ = 0 and sx¿y¿ = 58 MPa.
(sx)b =
sx¿ + sy¿
sx¿ - sy¿
+
2
2
cos 2u + tx¿y¿ sin 2u
= 0 + 0 + 58 sin 50°
= 44.43 MPa
A sy B b =
sx¿ + sy¿
sx¿ - sy¿
-
2
2
cos 2u - tx¿y¿ sin 2u
= 0 - 0 - 58 sin 50°
= -44.43 MPa
A txy B b = -
sx¿ - sy¿
2
58 MPa
200 MPa
sin 2u + tx¿y¿ cos 2u
= -0 + 58 cos 50°
= 37.28 MPa
Combining the stress components of two elements yields
ss = (sx)a + (sx)b = -237.5 + 44.43 = -193 MPa
Ans.
sy = A sy B a + A sy B b = -312.5 - 44.43 = -357 MPa
Ans.
txy = A txy B a + A txy B b = 64.95 + 37.28 = 102 MPa
Ans.
635
25⬚
ⴝ
txy
sx
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9–19. The state of stress at a point is shown on the element.
Determine (a) the principal stress and (b) the maximum
in-plane shear stress and average normal stress at the point.
Specify the orientation of the element in each case. Sketch
the results on each element.
160 MPa
120 MPa
In accordance to the established sign Convention,
sx = 0
sy = 160 MPa
s1, 2 =
=
sx + sy
;
2
B
a
txy = -120 MPa
sx - sy
2
2
b + t2xy
0 + 160
0 - 160 2
;
a
b + (-120)2
2
2
B
= 80 ; 220800
s2 = -64.2 MPa
s1 = 224 MPa
tan 2up =
txy
(sx - sy)>2
up = 28.15°
=
Ans.
-120
= 1.5
(0 - 160)>2
and -61.85°
Substitute u = 28.15° into Eq. 9-1,
sx¿ =
=
sx + sy
sx - sy
+
2
2
cos 2u + txy sin 2u
0 + 160
0 - 160
+
cos 56.31° + (-120) sin 56.31°
2
2
= -64.22 = s2
Thus,
(up)1 = -61.8°
Ans.
(up)2 = 28.2°
The element that represents the state of principal stress is shown in Fig. a
tmax
in-plane
=
B
a
sx - sy
2
tan 2us =
0 - 160 2
b + (-120)2 = 144 MPa
2
B
a
-(sx - sy)>2
us = -16.8°
By inspection,
equilibrium.
2
b + t2xy =
tmax
in-plane
savg =
txy
Ans.
-(0 - 160)>2
= -0.6667
-120
=
Ans.
and 73.2°
has to act in the sense shown in Fig. b to maintain
sx + sy
2
=
0 + 160
= 80 MPa
2
Ans.
The element that represents the state of Maximum in - plane shear stress is shown in
Fig. (c)
636
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9–19. Continued
637
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*9–20. The stress acting on two planes at a point is
indicated. Determine the normal stress sb and the principal
stresses at the point.
a
4 ksi
60⬚
45⬚
b
2 ksi
sb
a
Stress Transformation Equations: Applying Eqs. 9-2 and 9-1 with u = -135°,
sy = 3.464 ksi, txy = 2.00 ksi, tx¿y¿ = -2 ksi, and sx¿ = sb¿.,
tx¿y¿ = -
-2 = -
sx - sy
2
sin 2u + txy cos 2u
sx - 3.464
sin (-270°) + 2cos ( -270°)
2
sx = 7.464 ksi
sx¿ =
sy =
sx - sy
sx - sy
+
2
2
cos 2u + txy sin 2u
7.464 - 3.464
7.464 + 3.464
+
cos (-270°) + 2sin ( -270°)
2
2
Ans.
= 7.46 ksi
In - Plane Principal Stress: Applying Eq. 9-5.
s1, 2 =
=
sx + sy
2
;
B
a
sx - sy
2
2
b + t2xy
7.464 - 3.464 2
7.464 + 3.464
;
a
b + 22
2
2
B
= 5.464 ; 2.828
s1 = 8.29 ksi
s2 = 2.64 ksi
Ans.
638
b
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•9–21. The stress acting on two planes at a point is
indicated. Determine the shear stress on plane a–a and the
principal stresses at the point.
b
a
ta
45⬚
60 ksi
60⬚
txy = 60 cos 60° = 30 ksi
sa =
80 =
sx + sy
sx - sy
+
2
2
51.962 - sy
51.962 + sy
+
2
cos 2u + txy sin 2u
2
cos (90°) + 30 sin (90°)
sy = 48.038 ksi
ta = - a
= -a
sx - sy
2
b sin 2u + txy cos u
51.962 - 48.038
bsin (90°) + 30 cos (90°)
2
ta = -1.96 ksi
s1, 2 =
=
Ans.
sx + sy
2
;
C
a
sx - sy
2
2
b + t2xy
51.962 - 48.038 2
51.962 + 48.038
;
a
b + (30)2
2
C
2
s1 = 80.1 ksi
Ans.
s2 = 19.9 ksi
Ans.
639
90⬚
a
b
sx = 60 sin 60° = 51.962 ksi
80 ksi
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9–22. The T-beam is subjected to the distributed loading
that is applied along its centerline. Determine the principal
stress at point A and show the results on an element located
at this point.
100 kN/m
A
1m
0.5 m
200 mm
75 mm
'
©yA
0.1(0.2)(0.02) + 0.21(0.02)(0.2)
=
= 0.155 m
©A
0.2(0.02) + 0.02(0.2)
1
(0.02)(0.2 3) + 0.02(0.2)(0.155 - 0.1)2
12
I =
+
1
(0.2)(0.023) + 0.2(0.02)(0.21 - 0.155)2
12
= 37.6667(10 - 6) m4
Referring to Fig. b,
QA = y¿A¿ = 0.1175(0.075)(0.02) = 0.17625(10 - 3) m3
Using the method of sections and considering the FBD of the left cut segment of the
beam, Fig. c,
+ c ©Fy = 0;
V - 100(1) = 0
a + ©MC = 0;
100(1)(0.5) - M = 0 M = 50 kN # m
V = 100 kN
The normal stress developed is contributed by bending stress only. For point A,
y = 0.155 - 0.075 = 0.08 m. Thus
s =
My
50(103) (0.08)
= 106 MPa
=
I
37.6667(10 - 6)
The shear stress is contributed by the transverse shear stress only. Thus,
t =
100(103)[0.17625(10 - 3)]
VQA
= 23.40(106)Pa = 23.40 MPa
=
It
37.6667(10 - 6) (0.02)
The state of stress of point A can be represented by the element shown in Fig. c.
Here, sx = -106.19 MPa, sy = 0 and txy = 23.40 MPa.
s1, 2 =
=
sx + sy
2
;
B
a
sx - sy
2
2
b + txy 2
-106.19 - 0 2
-106.19 + 0
;
b + 23.402
a
2
2
B
= -53.10 ; 58.02
s1 = 4.93 MPa
20 mm
200 mm
20 mm
The location of the centroid c of the T cross-section, Fig. a, is
y =
A
s2 = -111 MPa
Ans.
640
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9–22.
Continued
tan 2up =
txy
(sx - sy)>2
up = -11.89°
=
ans
23.40
= -0.4406
( -106.19 - 0)>2
78.11°
Substitute u = -11.89°,
sx¿ =
=
sx + sy
2
sx - sy
+
2
cos 2u + txy sin 2u
-106.19 + 0
-106.19 - 0
+
cos (-23.78°) + 23.40 5m (-23.78°)
2
2
= -111.12 MPa = s2
Thus,
(up)1 = 78.1°
Ans.
(up)2 = -11.9°
The state of principal stress can be represented by the element shown in Fig. e.
641
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•9–23.
The wood beam is subjected to a load of 12 kN. If a
grain of wood in the beam at point A makes an angle of 25°
with the horizontal as shown, determine the normal and
shear stress that act perpendicular and parallel to the grain
due to the loading.
I =
12 kN
1m
2m
A
25⬚
300 mm
75 mm
1
(0.2)(0.3)3 = 0.45(10 - 3) m4
12
QA = yA¿ = 0.1125(0.2)(0.075) = 1.6875(10 - 3) m3
sA =
MyA
13.714(103)(0.075)
= 2.2857 MPa (T)
=
I
0.45(10 - 3)
tA =
6.875(103)(1.6875)(10 - 3)
VQA
= 0.1286 MPa
=
It
0.45(10 - 3)(0.2)
sx = 2.2857 MPa
sx¿ =
sx¿ =
sx + sy
sx - sy
+
2
sy = 0
2
txy = -0.1286 MPa
u = 115°
cos 2u + txy sin 2u
2.2857 - 0
2.2857 + 0
+
cos 230° + (-0.1286)sin 230°
2
2
Ans.
= 0.507 MPa
tx¿y¿ = -
sx - sy
= -a
2
sin 2u + txy cos 2u
2.2857 - 0
b sin 230° + (-0.1286)cos 230°
2
= 0.958 MPa
Ans.
642
4m
200 mm
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*9–24. The wood beam is subjected to a load of 12 kN.
Determine the principal stress at point A and specify the
orientation of the element.
12 kN
1m
2m
A
25⬚
I =
300 mm
75 mm
1
(0.2)(0.3)3 = 0.45(10 - 3) m4
12
QA = yA¿ = 0.1125(0.2)(0.075) = 1.6875(10 - 3) m3
sA =
13.714(103)(0.075)
MyA
= 2.2857 MPa (T)
=
I
0.45(10 - 3)
tA =
VQA
6.875(103)(1.6875)(10 - 3)
= 0.1286 MPa
=
It
0.45(10 - 3)(0.2)
sx = 2.2857 MPa
s1, 2 =
=
sx + sy
;
2
C
sy = 0
txy = -0.1286 MPa
sx - sy
b + t2xy
a
2
2
2.2857 - 0 2
2.2857 + 0
;
a
b + (-0.1286)2
2
C
2
s1 = 2.29 MPa
Ans.
s2 = -7.20 kPa
Ans.
tan 2up =
txy
(sx - sy)>2
=
-0.1286
(2.2857 - 0)>2
up = -3.21°
Check direction of principal stress:
sx¿ =
=
sx + sy
2
sx - sy
+
2
cos 2u + txy sin 2u
2.2857 + 0
2.2857 - 0
+
cos (-6.42°) - 0.1285 sin (-6.42)
2
2
= 2.29 MPa
643
4m
200 mm
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•9–25.
The bent rod has a diameter of 20 mm and is
subjected to the force of 400 N. Determine the principal
stress and the maximum in-plane shear stress that is
developed at point A. Show the results on a properly
oriented element located at this point.
100 mm
150 mm
400 N
400 N
250 mm
A
Using the method of sections and consider the FBD of the rod’s left cut segment,
Fig. a.
+ ©F = 0;
:
x
N - 400 = 0 N = 400 N
a + ©MC = 0;
400(0.25) - M = 0 M = 100 N # m
A = p(0.012) = 0.1(10 - 3) p m2
p
(0.014) = 2.5(10 - 9)p m4
4
I =
The normal stress developed is the combination of axial and bending stress. Thus,
My
N
;
A
I
s =
For point A, y = C = 0.01 m.
s =
100(0.01)
400
-3
0.1(10 )p
2.5(10 - 9)p
= -126.05 (106)Pa = 126.05 MPa (C)
Since no torque and transverse shear acting on the cross - section,
t = 0
The state of stress at point A can be represented by the element shown in Fig. b
Here, sx = -126.05 MPa, sy = 0 and txy = 0. Since no shear stress acting on the
element
s1 = sy = 0
s2 = sx = -126 MPa
Ans.
Thus, the state of principal stress can also be represented by the element shown in Fig. b.
tmax
in-plane
=
B
a
sx - sy
2
tan 2us = -
2
b + t2xy =
(sx - sy)>2
= -
txy
us = 45°
tx¿y¿ = -
-126.05 - 0 2
b + 02 = 63.0 MPa
2
B
a
(-126.05 - 0)>2
= q
0
and -45°
sx - sy
2
= -
= 63.0 =
sin 2u + txy cos 2u
-126.05 - 0
sin 90° + 0 cos 90°
2
tmax
in-plane
644
Ans.
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9–25. Continued
tmax
This indicates that in-plane
acts toward the positive sense of y¿ axis at the face of
element defined by us = 45°
savg =
sx + sy
2
=
-126.05 + 0
= -63.0 MPa
2
Ans.
The state of maximum In - plane shear stress can be represented by the element
shown in Fig. c
645
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9–26. The bracket is subjected to the force of 3 kip.
Determine the principal stress and maximum in-plane
shear stress at point A on the cross section at section a–a.
Specify the orientation of this state of stress and show the
results on elements.
3 kip
3 kip
a 3 in.
a
A
0.25 in.
2 in.
0.25 in.
Internal Loadings: Consider the equilibrium of the free - body diagram from the
bracket’s left cut segment, Fig. a.
B
1 in.
+ ©F = 0;
:
x
N - 3 = 0
N = 3 kip
Section a – a
M = 12 kip # in
©MO = 0; 3(4) - M = 0
Normal and Shear Stresses: The normal stress is the combination of axial and
bending stress. Thus,
s =
My
N
A
I
The cross - sectional area and the moment of inertia about the z axis of the bracket’s
cross section is
A = 1(2) - 0.75(1.5) = 0.875 in2
I =
1
1
(1) A 23 B (0.75) A 1.53 B = 0.45573 in4
12
12
For point A, y = 1 in. Then
sA =
(-12)(1)
3
= 29.76 ksi
0.875
0.45573
Since no shear force is acting on the section,
tA = 0
The state of stress at point A can be represented on the element shown in Fig. b.
In - Plane Principal Stress: sx = 29.76 ksi, sy = 0, and txy = 0. Since no shear
stress acts on the element,
s1 = sx = 29.8 ksi s2 = sy = 0
Ans.
The state of principal stresses can also be represented by the elements shown in Fig. b
Maximum In - Plane Shear Stress:
t max
in-plane
=
C
¢
sx - sy
2
2
≤ + txy 2 =
29.76 - 0 2
b + 02 = 14.9 ksi
2
B
a
Ans.
Orientation of the Plane of Maximum In - Plane Shear Stress:
tan 2us = -
A sx - sy B >2
txy
0.25 in.
= -
(29.76 - 0)>2
= -q
0
us = -45° and 45°
Ans.
646
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9–26.
Continued
Substituting u = -45° into
tx¿y¿ = -
= -
sx - sy
2
sin 2u + txy cos 2u
29.76 - 0
sin(-90°) + 0
2
= 14.9 ksi = t max
in-plane
This indicates that t max
is directed in the positive sense of the y¿ axes on the ace
in-plane
of the element defined by us = -45°.
Average Normal Stress:
savg =
sx + sy
2
=
29.76 + 0
= 14.9 ksi
2
Ans.
The state of maximum in - plane shear stress is represented by the element shown in
Fig. c.
647
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9–27. The bracket is subjected to the force of 3 kip.
Determine the principal stress and maximum in-plane
shear stress at point B on the cross section at section a–a.
Specify the orientation of this state of stress and show the
results on elements.
3 kip
3 kip
a 3 in.
a
A
0.25 in.
2 in.
0.25 in.
Internal Loadings: Consider the equilibrium of the free - body diagram of the
bracket’s left cut segment, Fig. a.
B
1 in.
+ ©F = 0;
:
x
N - 3 = 0
N = 3 kip
Section a – a
M = 12 kip # in
©MO = 0; 3(4) - M = 0
Normal and Shear Stresses: The normal stress is the combination of axial and
bending stress. Thus,
s =
My
N
A
I
The cross - sectional area and the moment of inertia about the z axis of the bracket’s
cross section is
A = 1(2) - 0.75(1.5) = 0.875 in2
I =
1
1
(1) A 23 B (0.75) A 1.53 B = 0.45573 in4
12
12
For point B, y = -1 in. Then
sB =
(-12)(-1)
3
= -22.90 ksi
0.875
0.45573
Since no shear force is acting on the section,
tB = 0
The state of stress at point A can be represented on the element shown in Fig. b.
In - Plane Principal Stress: sx = -22.90 ksi, sy = 0, and txy = 0. Since no shear
stress acts on the element,
s1 = sy = 0
s2 = sx = -22.90 ksi
Ans.
The state of principal stresses can also be represented by the elements shown in Fig. b.
Maximum In - Plane Shear Stress:
t max
in-plane
=
C
¢
sx - sy
2
2
≤ + txy 2 =
-22.90 - 0 2
b + 02 = 11.5 ksi
2
B
a
Ans.
Orientation of the Plane of Maximum In - Plane Shear Stress:
tan 2us = -
A sx - sy B >2
txy
0.25 in.
= -
(-22.9 - 0)>2
= -q
0
us = 45° and 135°
Ans.
648
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9–27. Continued
Substituting u = 45° into
tx¿y¿ = -
= -
sx - sy
2
sin 2u + txy cos 2u
-22.9 - 0
sin 90° + 0
2
= 11.5 ksi = t max
in-plane
is directed in the positive sense of the y¿ axes on the
This indicates that t max
in-plane
element defined by us = 45°.
Average Normal Stress:
savg =
sx + sy
2
=
-22.9 + 0
= -11.5 ksi
2
Ans.
The state of maximum in - plane shear stress is represented by the element shown in
Fig. c.
649
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*9–28. The wide-flange beam is subjected to the loading
shown. Determine the principal stress in the beam at point A
and at point B. These points are located at the top and
bottom of the web, respectively. Although it is not very
accurate, use the shear formula to determine the shear stress.
8 kN/m
A
B
1m
3m
B
Internal Forces and Moment: As shown on FBD(a).
200 mm
Section Properties:
A = 0.2(0.22) - 0.19(0.2) = 6.00 A 10 - 3 B m2
1
1
(0.2) A 0.223 B (0.19) A 0.22 B = 50.8 A 10 - 6 B m4
12
12
I =
QA = QB = y¿A¿ = 0.105(0.01)(0.2) = 0.210 A 10 - 3 B m3
Normal Stress:
s =
My
N
;
A
I
21.65(103)
=
73.5(103)(0.1)
;
6.00(10 - 3)
50.8(10 - 6)
sA = 3.608 + 144.685 = 148.3 MPa
sB = 3.608 - 144.685 = -141.1 MPa
VQ
.
It
Shear Stress: Applying the shear formula t =
tA = tB =
36.5(103) C 0.210(10 - 3) D
50.8(10 - 6)(0.01)
= 15.09 MPa
In - Plane Principal Stress: sx = 148.3 MPa, sy = 0, and txy = -15.09 MPa for
point A. Applying Eq. 9-5.
s1, 2 =
=
sx + sy
;
2
C
sx - sy
a
2
2
b + t2xy
148.3 - 0 2
148.3 + 0
;
a
b + (-15.09)2
2
C
2
= 74.147 ; 75.666
s1 = 150 MPa
s2 = -1.52 MPa
Ans.
sx = -141.1 MPa, sy = 0, and txy = -15.09 MPa for point B. Applying Eq. 9-5.
s1, 2 =
=
sx + sy
2
;
C
a
sx - sy
2
2
b + t2xy
( -141.1) - 0 2
-141.1 + 0
;
a
b + (-15.09)2
2
C
2
= -70.538 ; 72.134
s1 = 1.60 MPa
s2 = -143 MPa
Ans.
650
30⬚
25 kN
A
10 mm
110 mm
10 mm
200 mm
10 mm
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•9–29.
The wide-flange beam is subjected to the loading
shown. Determine the principal stress in the beam at
point A, which is located at the top of the web. Although
it is not very accurate, use the shear formula to determine
the shear stress. Show the result on an element located at
this point.
120 kN/m
30 kN
A
0.3 m
0.9 m
Using the method of sections and consider the FBD of the left cut segment of the
bean, Fig. a
V - 12 (90)(0.9) - 30 = 0
+ c ©Fy = 0;
M = 39.15 kN # m
150 mm
The moment of inertia of the cross - section about the bending axis is
I =
1
1
(0.15)(0.193) (0.13)(0.153) = 49.175(10 - 6) m4
12
12
Referring to Fig. b,
QA = y¿A¿ = 0.085 (0.02)(0.15) = 0.255 (10 - 3) m3
The normal stress developed is contributed by bending stress only. For point A,
y = 0.075 m. Thus,
s =
My
39.15(103)(0.075)
= 59.71(106)Pa = 59.71 MPa (T)
=
I
49.175(10 - 6)
The shear stress is contributed by the transverse shear stress only. Thus
70.5(10 ) C 0.255(10 ) D
VQA
= 18.28(106)Pa = 18.28 MPa
=
It
49.175(10 - 6) (0.02)
3
t =
-3
Here, sx = 59.71 MPa, sy = 0 and txy = 18.28 MPa.
s1, 2 =
=
sx + sy
;
2
C
a
sx - sy
2
2
b + txy
59.71 - 0 2
59.71 + 0
a
;
b + 18.282
2
C
2
= 29.86 ; 35.01
s2 = -5.15 MPa
s1 = 64.9 MPa
tan 2uP =
txy
(sx - sy)>2
uP = 15.74°
=
and
Ans.
18.28
= 0.6122
(59.71 - 0)>2
-74.26°
Substitute u = 15.74°,
sx¿ =
=
sx + sy
2
sx - sy
+
2
20 mm
150 mm
20 mm
V = 70.5 kN
1
2 (90)(0.9)(0.3) + 30(0.9) - M = 0
a + ©MC = 0;
A
20 mm
cos 2u + txy sin 2u
59.71 - 0
59.71 + 0
+
cos 31.48° + 18.28 sin 31.48°
2
2
= 64.9 MPa = s1
651
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9–29.
Continued
Thus,
A uP B 1 = 15.7°
A uP B 2 = -74.3°
Ans.
The state of principal stress can be represented by the element shown in Fig. d
652
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9–30. The cantilevered rectangular bar is subjected to the
force of 5 kip. Determine the principal stress at points A
and B.
1
I =
(3)(63) = 54 in4
12
1.5 in.
A
1.5 in.
A = (6)(3) = 18 in2
QA = 2.25(1.5)(3) = 10.125 in3
1.5 in.
1.5 in.
B 1 in.
3 in.
QB = 2(2)(3) = 12 in3
1 in.
Point A:
15 in.
3 in.
3 5
4
5 kip
45(1.5)
Mxz
4
P
+
=
+
= 1.472 ksi
sA =
A
I
18
54
tA =
Vz QA
=
It
3(10.125)
= 0.1875 ksi
54(3)
sx = 1.472 ksi
s1, 2 =
=
sy = 0
sx + sy
;
2
C
a
sx - sy
2
txy = 0.1875 ksi
2
b + txy 2
1.472 - 0 2
1.472 + 0
;
a
b + 0.18752
2
C
2
s1 = 1.50 ksi
Ans.
s2 = -0.0235 ksi
Ans.
Point B:
sB =
tB =
45(1)
Mxz
4
P
=
= -0.6111 ksi
A
I
18
54
Vz QB
=
It
3(12)
= 0.2222 ksi
54(3)
sx = -0.6111 ksi
s1, 2 =
=
sy = 0
sx + sy
2
;
C
a
sx - sy
2
txy = 0.2222 ksi
2
b + txy 2
-0.6111 - 0 2
-0.611 + 0
;
a
b + 0.2222
2
C
2
s1 = 0.0723 ksi
Ans.
s2 = -0.683 ksi
Ans.
653
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9–31. Determine the principal stress at point A on the
cross section of the arm at section a–a. Specify the
orientation of this state of stress and indicate the results on
an element at the point.
7.5 mm
A
50 mm
7.5 mm
Support Reactions: Referring to the free - body diagram of the entire arm shown
in Fig. a,
©MB = 0; FCD sin 30°(0.3) - 500(0.65) = 0
FCD = 2166.67 N
+ ©F = 0;
:
x
Bx - 2166.67 cos 30° = 0
Bx = 1876.39 N
+ c ©Fy = 0;
2166.67 sin 30° - 500 - By = 0
By = 583.33 N
20 mm
Section a – a
D
Internal Loadings: Consider the equilibrium of the free - body diagram of the
arm’s left segment, Fig. b.
+ ©F = 0;
:
x
1876.39 - N = 0
N = 1876.39 N
+ c ©Fy = 0;
V - 583.33 = 0
V = 583.33 N
583.33(0.15) - M = 0
M = 87.5N # m
+ ©MO = 0;
0.15 m
Referring to Fig. b,
QA = y¿A¿ = 0.02125(0.0075)(0.02) = 3.1875 A 10 - 6 B m3
Normal and Shear Stress: The normal stress is a combination of axial and bending
stress. Thus,
-1876.39
0.5625 A 10
-3
B
MyA
N
+
A
I
87.5(0.0175)
+
0.16367 A 10 - 6 B
= 6.020 MPa
The shear stress is caused by transverse shear stress.
tA =
583.33 C 3.1875 A 10 - 6 B D
VQA
=
= 1.515 MPa
It
0.16367 A 10 - 6 B (0.0075)
The share of stress at point A can be represented on the element shown in Fig. d.
In - Plane Principal Stress: sx = 6.020 MPa, sy = 0, and txy = 1.515 MPa. We have
s1,2 =
=
sx + sy
2
;
C
¢
sx - sy
2
2
≤ + txy 2
6.020 - 0 2
6.020 + 0
;
a
b + 1.5152
2
C
2
s1 = 6.38 MPa
s2 = -0.360 MPa
Ans.
654
C
0.15 m
0.35 m
500 N
1
1
(0.02) A 0.053 B (0.0125) A 0.0353 B = 0.16367 A 10 - 6 B m4
12
12
=
a
a
A = 0.02(0.05) - 0.0125(0.035) = 0.5625 A 10 - 3 B m2
sA =
60⬚
B
Section Properties: The cross - sectional area and the moment of inertia about the z
axis of the arm’s cross section are
I =
7.5 mm
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9–31. Continued
Orientation of the Principal Plane:
tan 2uP =
txy
A sx - sy B >2
=
1.515
= 0.5032
(6.020 - 0)>2
up = 13.36° and 26.71°
Substituting u = 13.36° into
sx¿ =
=
sx + sy
2
sx - sy
+
2
cos 2u + txy sin 2u
6.020 + 0
6.020 - 0
+
cos 26.71° + 1.515 sin 26.71°
2
2
= 6.38 MPa = s1
Thus, A uP B 1 = 13.4 and A uP B 2 = 26.71°
Ans.
The state of principal stresses is represented by the element shown in Fig. e.
655
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*9–32. Determine the maximum in-plane shear stress
developed at point A on the cross section of the arm at
section a–a. Specify the orientation of this state of stress and
indicate the results on an element at the point.
7.5 mm
A
50 mm
7.5 mm
20 mm
7.5 mm
Section a – a
Support Reactions: Referring to the free - body diagram of the entire arm shown
in Fig. a,
©MB = 0; FCD sin 30°(0.3) - 500(0.65) = 0
FCD = 2166.67 N
+ ©F = 0;
:
x
Bx - 2166.67 cos 30° = 0
Bx = 1876.39 N
+ c ©Fy = 0;
2166.67 sin 30° - 500 - By = 0
By = 583.33 N
D
60⬚
a
B
C
a
0.15 m
0.15 m
0.35 m
500 N
Internal Loadings: Considering the equilibrium of the free - body diagram of the
arm’s left cut segment, Fig. b,
+ ©F = 0;
:
x
1876.39 - N = 0
N = 1876.39 N
+ c ©Fy = 0;
V - 583.33 = 0
V = 583.33 N
+ ©MO = 0;
583.33(0.15) - M = 0
M = 87.5 N # m
Section Properties: The cross - sectional area and the moment of inertia about the z
axis of the arm’s cross section are
A = 0.02(0.05) - 0.0125(0.035) = 0.5625 A 10 - 3 B m2
1
1
(0.02) A 0.053 B (0.0125) A 0.0353 B = 0.16367 A 10 - 6 B m4
12
12
I =
Referring to Fig. b,
QA = y¿A¿ = 0.02125(0.0075)(0.02) = 3.1875 A 10 - 6 B m3
Normal and Shear Stress: The normal stress is a combination of axial and bending
stress. Thus,
sA =
MyA
N
+
A
I
-1876.39
=
0.5625 A 10
-3
B
87.5(0.0175)
+
0.16367 A 10 - 6 B
= 6.020 MPa
The shear stress is contributed only by transverse shear stress.
tA =
583.33 C 3.1875 A 10 - 6 B D
VQA
=
= 1.515 MPa
It
0.16367 A 10 - 6 B (0.0075)
Maximum In - Plane Shear Stress: sx = 6.020 MPa, sy = 0, and txy = 1.515 MPa.
tmax
in-plane
=
C
¢
sx - sy
2
2
≤ + txy 2 =
6.020 - 0 2
b + 1.5152 = 3.37 MPa
B
2
a
656
Ans.
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9–32.
Continued
Orientation of the Plane of Maximum In - Plane Shear Stress:
tan 2us = -
A sx - sy B >2
txy
= -
(6.020 - 0)>2
= -1.9871
1.515
us = -31.6° and 58.4°
Ans.
Substituting u = -31.6° into
tx¿y¿ = -
= -
sx - sy
2
sin 2u + txy cos 2u
6.020 - 0
sin(-63.29°) + 1.515 cos(-63.29°)
2
= 3.37 MPa = t max
in-plane
is directed in the positive sense of the y¿ axis on the face
This indicates that t max
in-plane
of the element defined by us = -31.6°.
Average Normal Stress:
savg =
sx + sy
2
=
6.020 + 0
= 3.01 MPa
2
Ans.
The state of maximum in - plane shear stress is represented on the element shown in
Fig. e.
657
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•9–33. The clamp bears down on the smooth surface at E
by tightening the bolt. If the tensile force in the bolt is 40
kN, determine the principal stress at points A and B and
show the results on elements located at each of these
points. The cross-sectional area at A and B is shown in the
adjacent figure.
300 mm
50 mm
30 mm
100 mm
B
A
Support Reactions: As shown on FBD(a).
E
Internal Forces and Moment: As shown on FBD(b).
Section Properties:
I =
1
(0.03) A 0.053 B = 0.3125 A 10 - 6 B m4
12
QA = 0
QB = y¿A¿ = 0.0125(0.025)(0.03) = 9.375 A 10 - 6 B m3
Normal Stress: Applying the flexure formula s = -
sA = -
sB = -
2.40(103)(0.025)
= -192 MPa
0.3125(10 - 6)
2.40(103)(0)
0.3125(10 - 6)
= 0
Shear Stress: Applying the shear formula t =
tA =
tB =
My
.
I
24.0(103)(0)
0.3125(10 - 6)(0.03)
VQ
It
= 0
24.0(103) C 9.375(10 - 6) D
0.3125(10 - 6)(0.03)
= 24.0 MPa
In - Plane Principal Stresses: sx = 0, sy = -192 MPa, and txy = 0 for point A.
Since no shear stress acts on the element.
s1 = sx = 0
Ans.
s2 = sy = -192 MPa
Ans.
sx = sy = 0 and txy = -24.0 MPa for point B. Applying Eq. 9-5
s1,2 =
sx + sy
2
;
C
a
sx - sy
2
2
b + t2xy
= 0 ; 20 + (-24.0)2
= 0 ; 24.0
s1 = 24.0
s2 = -24.0 MPa
Ans.
658
B
A
25 mm
100 mm
50 mm
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9–33. Continued
Orientation of Principal Plane: Applying Eq. 9-4 for point B.
tan 2up =
txy
=
-24.0
= -q
0
and
45.0°
A sx - sy B >2
up = -45.0°
Subsututing the results into Eq. 9-1 with u = -45.0° yields
sx¿ =
sx + sy
2
sx - sy
+
2
cos 2u + txy sin 2u
= 0 + 0 + [-24.0 sin (-90.0°)]
= 24.0 MPa = s1
Hence,
up1 = -45.0°
up2 = 45.0°
Ans.
659
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9–34. Determine the principal stress and the maximum inplane shear stress that are developed at point A in the
2-in.-diameter shaft. Show the results on an element located
at this point. The bearings only support vertical reactions.
300 lb
Using the method of sections and consider the FBD of shaft’s left cut segment, Fig. a,
+ ©F = 0;
:
x
N - 3000 = 0
N = 3000 lb
+ c ©Fy = 0;
75 - V = 0
V = 75 lb
a + ©MC = 0;
M - 75(24) = 0
M = 1800 lb # in
A = p(12) = p in2
I =
p 4
p
(1 ) = in4
4
4
Also,
QA = 0
The normal stress developed is the combination of axial and bending stress. Thus
My
N
;
A
I
s =
For point A, y = C = 1 in. Then
s =
1800(1)
3000
p
p>4
= -1.337 (103) psi = 1.337 ksi (c)
The shear stress developed is due to transverse shear force. Thus,
t =
VQA
= 0
It
The state of stress at point A, can be represented by the element shown in Fig. b.
Here, sx = -1.337 ksi, sy = 0 is txy = 0. Since no shear stress acting on the
element,
s1 = sy = 0
s2 = sx = -1.34 ksi
Ans.
Thus, the state of principal stress can also be represented by the element shown in Fig. b.
t
max
in-plane =
C
a
sx - sy
2
tan 2us = us = 45°
2
b + t2xy =
-1.337 - 0 2
b + 02 = 0.668 ksi - 668 psi Ans.
C
2
a
(sx - sy)>2
= -
txy
and
(-1.337 - 0)>2
= q
0
Ans.
-45°
Substitute u = 45°,
tx¿y¿ = -
= -
sx - sy
2
sin 2u + txy cos 2u
-1.337 - 0
sin 90° + 0
2
= 0.668 ksi = 668 psi =
A
3000 lb
tmax
in-plane
660
24 in.
3000 lb
12 in.
12 in.
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9–34. Continued
tmax
This indicates that in-plane
acts toward the positive sense of y¿ axis at the face of the
element defined by us = 45°.
Average Normal Stress.
The state of maximum in - plane shear stress can be represented by the element
shown in Fig. c.
661
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9–35. The square steel plate has a thickness of 10 mm and
is subjected to the edge loading shown. Determine the
maximum in-plane shear stress and the average normal
stress developed in the steel.
sx = 5 kPa
sy = -5 kPa
tmax
sx - sy
in-plane
savg =
a
b
txy = 0
200 mm
+ t2xy
C
=
5 + 5 2
b + 0 = 5 kPa
C
2
a
sx + sy
3
50 N/m
2
=
2
50 N/m
Ans.
200 mm
5 - 5
=
= 0
2
Ans.
Note:
tan 2us =
tan 2us =
-(sx - sy)>2
txy
-(5 + 5)>2
= q
0
us = 45°
*9–36. The square steel plate has a thickness of 0.5 in. and
is subjected to the edge loading shown. Determine the
principal stresses developed in the steel.
sx = 0
s1,2 =
sy = 0
txy = 32 psi
sx + sy
2
;
16 lb/in.
C
a
sx - sy
2
2
b + t2xy
= 0 ; 20 + 322
4 in.
s1 = 32 psi
Ans.
s2 = -32 psi
Ans.
Note:
tan 2up =
16 lb/in.
4 in.
txy
(sx - sy)>2
=
32
= q
0
up = 45°
662
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•9–37.
The shaft has a diameter d and is subjected to the
loadings shown. Determine the principal stress and the
maximum in-plane shear stress that is developed at point A.
The bearings only support vertical reactions.
P
F
F
A
L
2
Support Reactions: As shown on FBD(a).
Internal Forces and Moment: As shown on FBD(b).
Section Properties:
A =
p 2
d
4
p d 4
p 4
a b =
d
4 2
64
I =
QA = 0
Normal Stress:
N
Mc
;
A
I
s =
-F
= p 2 ;
4 d
sA =
A B
pL d
4
2
p 4
d
64
4
2PL
- Fb
a
2
d
pd
Shear Stress: Since QA = 0, tA = 0
In - Plane Principal Stress: sx =
4 2PL
a
- Fb.
pd2 d
sy = 0 and txy = 0 for point A. Since no shear stress acts on the element,
s1 = sx =
4
2PL
a
- Fb
d
pd2
Ans.
s2 = sy = 0
Ans.
Maximum In - Plane Shear Stress: Applying Eq. 9-7 for point A,
t max
in-plane
=
=
=
Q
£
B
a
4
2
pd
sx - sy
2
2
b + t2xy
A 2PL
d - FB - 0
2
2
≥ + 0
2PL
2
a
- Fb
d
pd2
Ans.
663
L
2
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9–38. A paper tube is formed by rolling a paper strip in
a spiral and then gluing the edges together as shown.
Determine the shear stress acting along the seam, which is
at 30° from the vertical, when the tube is subjected to an
axial force of 10 N. The paper is 1 mm thick and the tube has
an outer diameter of 30 mm.
30⬚
10 N
10 N
30 mm
10
P
= 109.76 kPa
= p
2
A
(0.03
- 0.0282)
4
s =
sx = 109.76 kPa
tx¿y¿ = -
= -
sx - sy
2
sy = 0
txy = 0
u = 30°
sin 2u + txy cos 2u
106.76 - 0
sin 60° + 0 = -47.5 kPa
2
Ans.
9–39. Solve Prob. 9–38 for the normal stress acting
perpendicular to the seam.
30⬚
10 N
10 N
30 mm
s =
sn =
=
10
P
= 109.76 kPa
= p
2
A
(0.03
- 0.0282)
4
sx + sy
2
sx - sy
+
2
cos 2u + txy sin 2u
109.76 - 0
109.76 + 0
+
cos (60°) + 0 = 82.3 kPa
2
2
Ans.
664
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*9–40. Determine the principal stresses acting at point A
of the supporting frame. Show the results on a properly
oriented element located at this point.
800 mm
B
A
300 mm
150 mm
12 mm
5
B
15 mm
130 mm
A
y =
0.065(0.13)(0.015) + 0.136(0.15)(0.012)
©yA
=
= 0.0991 m
©A
0.13(0.015) + 0.15(0.012)
I =
1
(0.015)(0.133) + 0.015(0.13)(0.0991 - 0.065)2
12
1
(0.15)(0.012 3) + 0.15(0.012)(0.136 - 0.0991)2 = 7.4862(10 - 6) m4
12
+
QA = 0
A = 0.13(0.015) + 0.15(0.012) = 3.75(10 - 3) m2
Normal stress:
s =
P
Mc
+
A
I
sA =
-3.6(103)
-3
3.75(10 )
5.2767(103)(0.0991)
-
7.4862(10 - 6)
= -70.80 MPa
Shear stress:
tA = 0
Principal stress:
s1 = 0
Ans.
s2 = -70.8 MPa
Ans.
665
4
3
6 kN
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•9–41. Determine the principal stress acting at point B,
which is located just on the web, below the horizontal
segment on the cross section. Show the results on a properly
oriented element located at this point. Although it is not very
accurate, use the shear formula to calculate the shear stress.
800 mm
B
A
300 mm
150 mm
12 mm
y =
©yA
0.065(0.13)(0.015) + 0.136(0.15)(0.012)
=
= 0.0991 m
©A
0.13(0.015) + 0.15(0.012)
I =
1
(0.015)(0.133) + 0.015(0.13)(0.0991 - 0.065)2
12
+
130 mm
A
1
(0.15)(0.0123) + 0.15(0.012)(0.136 - 0.0991)2 = 7.4862(10 - 6) m4
12
A = 0.13(0.015) + 0.15(0.012) = 3.75(10 - 3) m2
Normal stress:
s =
Mc
P
+
A
I
sB = -
3.6(103)
3.75(10 - 3)
5.2767(103)(0.130 - 0.0991)
+
7.4862(10 - 6)
= 20.834 MPa
Shear stress:
tB =
VQ
-4.8(103)(0.0369)(0.15)(0.012)
= -2.84 MPa
=
It
7.4862(10 - 6)(0.015)
Principal stress:
s1,2 = a
20.834 + 0
20.834 - 0 2
b ;
a
b + (-2.84)2
2
C
2
s1 = 21.2 MPa
Ans.
s2 = -0.380 MPa
Ans.
tan 2up =
A
-2.84
20.834 - 0
2
B
up = -7.63°
Ans.
666
5
B
15 mm
4
3
6 kN
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9–42. The drill pipe has an outer diameter of 3 in., a wall
thickness of 0.25 in., and a weight of 50 lb>ft. If it is
subjected to a torque and axial load as shown, determine
(a) the principal stress and (b) the maximum in-plane shear
stress at a point on its surface at section a.
1500 lb
800 lb⭈ft
20 ft
a
20 ft
Internal Forces and Torque: As shown on FBD(a).
Section Properties:
A =
p 2
A 3 - 2.52 B = 0.6875p in2
4
J =
p
A 1.54 - 1.254 B = 4.1172 in4
2
s =
N
-2500
=
= -1157.5 psi
A
0.6875p
Normal Stress:
Shear Stress: Applying the torsion formula.
t =
800(12)(1.5)
Tc
=
= 3497.5 psi
J
4.1172
a) In - Plane Principal Stresses: sx = 0, sy = -1157.5 psi and txy = 3497.5 psi for
any point on the shaft’s surface. Applying Eq. 9-5.
s1,2 =
=
sx + sy
2
;
C
a
sx - sy
2
2
b + t2xy
0 - (-1157.5) 2
0 + (-1157.5)
;
a
b + (3497.5)2
2
C
2
= -578.75 ; 3545.08
s1 = 2966 psi = 2.97 ksi
Ans.
s2 = -4124 psi = -4.12 ksi
Ans.
b) Maximum In - Plane Shear Stress: Applying Eq. 9-7
t max
in-plane
a
sx - sy
2
b + t2xy
=
C
=
0 - (-1157.5) 2
≤ + (3497.5)2
C
2
2
¢
= 3545 psi = 3.55 ksi
Ans.
667
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9–43. Determine the principal stress in the beam at
point A.
60 kN
50 mm
150 kN
A
A
60 mm
0.5 m
Using the method of sections and consider the FBD of the beam’s left cut segment,
Fig. a,
+ ©F = 0;
:
x
150 - N = 0
N = 150 kN
+ c ©Fy = 0;
V - 60 = 0
V = 60 kN
a + ©MC = 0;
60(0.5) - M = 0
M = 30 kN # m
A = 0.06(0.15) = 0.009 m2
1
(0.06)(0.153) = 16.875(10 - 6) m4
12
I =
Referring to Fig. b,
QA = y¿A¿ = 0.05 (0.05)(0.06) = 0.15(10 - 3) m3
The normal stress developed is the combination of axial and bending stress. Thus
My
N
;
A
I
s =
For point A, y = 0.075 - 0.05 = 0.025 m. Then
s =
30(103)(0.025)
-150(103)
0.009
16.875(10 - 6)
= -61.11(106) Pa = 61.11 MPa (c)
The shear stress developed is due to the transverse shear, Thus,
t =
60(103) C 0.15(10 - 3) D
VQA
= 8.889 MPa
=
It
16.875(10 - 6) (0.06)
Here, sx = -61.11 MPa, sy = 0 and txy = 8.889 MPa,
s1, 2 =
=
sx + sy
;
2
C
a
sx - sy
2
2
b + t2xy
-61.11 - 0 2
-61.11 + 0
;
a
b + 8.8892
2
C
2
= -30.56 ; 31.82
s2 = -62.4 MPa
s1 = 1.27 MPa
tan 2uP =
txy
(sx - sy)>2
uP = -8.11°
=
and
Ans.
8.889
= -0.2909
(-61.11 - 0)>2
81.89°
668
0.25 m
150 mm
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9–43.
Continued
Substitute u = -8.11°,
sx¿ =
=
sx + sy
2
sx - sy
+
2
cos 2u + txy sin 2u
-61.11 + 0
-61.11 - 0
+
cos (-16.22°) + 8.889 sin (-16.22°)
2
2
= -62.4 MPa = s2
Thus,
(uP)1 = 81.9°
(uP)2 = -8.11°
The state of principal stresses can be represented by the elements shown in Fig. (c)
669
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*9–44. Determine the principal stress at point A which is
located at the bottom of the web. Show the results on an
element located at this point.
150 kN/m
Using the method of sections, consider the FBD of the bean’s left cut segment, Fig. a,
+ c ©Fy = 0;
V -
1
(100)(0.6) = 0
2
V = 30 kN
a + ©MC = 0;
1
(100)(0.6)(0.2) - M = 0
2
M = 6 kN # m
I =
10 mm
A
1
1
(0.15)(0.223) (0.14)(0.23) = 39.7667(10 - 6) m4
12
12
150 mm
Referring to Fig. b
QA = y¿A¿ = 0.105 (0.01)(0.15) = 0.1575(10 - 3) m3
The normal stress developed is due to bending only. For point A, y = 0.1 m. Then
s =
My
6(103)(0.1)
=
I
= 15.09(106)Pa = 15.09 MPa (c)
39.7667(10 - 6)
The shear stress developed is due to the transverse shear. Thus,
t =
30(103) C 0.1575(10 - 3) D
VQA
= 11.88(106)Pa = 11.88 MPa
=
It
39.7667(10 - 6)(0.01)
Here, sx = -15.09 MPa, sy = 0 And txy = 11.88 MPa.
s1, 2 =
=
sx + sy
;
2
C
a
sx - sy
2
2
b + t2xy
-15.09 - 0 2
-15.09 + 0
;
a
b + 11.882
2
C
2
= -7.544 ; 14.074
s2 = -21.6 MPa
s1 = 6.53 MPa
tan 2uP =
txy
(sx - sy)>2
uP = -28.79°
=
Ans.
11.88
= -1.575
(-15.09 - 0)>2
and
61.21°
Substitute u = 61.21°,
sx¿ =
=
sx + sy
2
sx - sy
+
2
cos 2u + txy sin 2u
-15.09 + 0
-15.09 - 0
+
cos 122.42° + 11.88 sin 122.42°
2
2
= 6.53 MPa = s1
Thus,
(uP)1 = 61.2°
A
0.6 m
Ans.
(uP)2 = -28.8°
The state of principal stresses can be represented by the element shown in Fig. d.
670
0.3 m
10 mm
200 mm
10 mm
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9–44. Continued
671
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•9–45.
Determine the maximum in-plane shear stress in
the box beam at point A. Show the results on an element
located at this point.
10 kip
4 kip
A
B
Using the method of section, consider the FBD, of bean’s left cut segment, Fig. a,
8 - 10 + V = 0
+ c ©Fy = 0;
a + ©MC = 0;
M + 10(1.5) - 8(3.5) = 0
4 in.
A
M = 13 kip # ft
4 in.
The moment of inertia of the cross - section about the neutral axis is
Referring to Fig. b,
QA = 0
The normal stress developed is contributed by the bending stress only. For point A,
y = C = 3 in.
My
=
I
13(12)(3)
= 5.40 ksi (c)
86.6667
The shear stress is contributed by the transverse shear stress only. Thus
t =
VQA
= 0
It
The state of stress at point A can be represented by the element shown in Fig. c
Here, sx = -5.40 ksi, sy = 0 and txy = 0.
tmax
in-plane
=
C
a
sx - sy
2
tan 2us = -
2
b + txy 2 =
-5.40 - 0 2
b + 02 = 2.70 ksi
C
2
(sx - sy)>2
a
txy
and
-45°
us = 45°
Ans.
(-5.40 - 0)>2
= q
2
= -
Substitute u = 45°,
tx¿y¿ = -
sx - sy
= -
2
sin 2u + txy cos 2u
-5.40 - 0
sin 90° + 0
2
= 2.70 ksi =
tmax
in-plane
tmax
This indicates that in-plane
acts toward the positive sense of y¿ axis at the face of
element defined by us = 45°
savg =
sx + sy
2
=
-5.40 + 0
= -2.70 ksi
2
The state of maximum In - plane shear stress can be represented by the element
shown in Fig. d.
672
B
6 in.
1
1
I =
(6)(63) (4)(43) = 86.6667 in4
12
12
s =
1.5 ft
2 ft
V = 2 kip
2 ft
0.5 ft
3 in.
3 in.
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9–45. Continued
673
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9–46. Determine the principal stress in the box beam at
point B. Show the results on an element located at this point.
10 kip
4 kip
A
Using the method of sections, consider the FBD of bean’s left cut segment, Fig. a,
+ c ©Fy = 0;
8 - 10 + V = 0
a + ©MC = 0;
M + 10(1.5) - 8(3.5) = 0
I =
B
V = 2 kip
M = 13 kip # ft
4 in.
A
1
1
(6)(63) (4)(43) = 86.6667 in4
12
12
4 in.
Referring to Fig. b,
C
QB = 2y1œ A1œ + y2œ A2œ = 2 1(2)(1)
My
D + 2.5(1)(6) = 19 in
= 0
I
The shear stress is contributed by the transverse shear stress only. Thus
2(103)(19)
VQB
=
= 219.23 psi
It
86.6667(2)
t =
The state of stress at point B can be represented by the element shown in Fig. c
Here, sx = sy = 0 and txy = 219.23 psi.
s1, 2 =
sx + sy
;
2
C
a
sx - sy
2
2
b + txy 2
= 0 ; 20 + 219.232
s2 = -219 psi
s1 = 219 psi
tan 2uP =
txy
(sx - sy)>2
uP = 45°
and
sx + sy
sx - sy
=
Ans.
219.23
= q
0
-45°
Substitute u = 45°,
sx¿ =
2
+
2
cos 2u + txy sin 2u
= 0 + 0 + 219.23 sin 90°
= 219 psi = s1
Thus,
(uP)1 = 45°
Ans.
(uP)2 = -45°
The state of principal stress can be represented by the element shown in Fig. d.
674
B
6 in.
3
The normal stress developed is contributed by the bending stress only. For point B,
y = 0.
s =
1.5 ft
2 ft
2 ft
0.5 ft
3 in.
3 in.
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9–46. Continued
675
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9–47. The solid shaft is subjected to a torque, bending
moment, and shear force as shown. Determine the principal
stresses acting at point A.
Ix = Iy =
J =
p
(0.025)4 = 0.306796(10 - 6) m4
4
450 mm
300 N⭈m
p
(0.025)4 = 0.613592(10 - 6) m4
2
25 mm
45 N⭈m
QA = 0
800 N
60(0.025)
Mx c
= 4.889 MPa
sA =
=
I
0.306796(10 - 6)
tA =
Ty c
45(0.025)
=
J
0.613592(10 - 6)
sx = 4.889 MPa
s1, 2 =
=
*9–48.
sy = 0
sx + sy
;
2
= 1.833 MPa
C
a
txy = -1.833 MPa
sx - sy
2
2
b + txy 2
4.889 - 0 2
4.889 + 0
;
b + (-1.833)2
a
2
C
2
s1 = 5.50 MPa
Ans.
s2 = -0.611 MPa
Ans.
Solve Prob. 9–47 for point B.
Ix = Iy =
p
(0.025)4 = 0.306796(10 - 6) m4
4
450 mm
p
J = (0.025)4 = 0.613592(10 - 6) m4
2
QB = yA¿ =
300 N⭈m
4(0.025) 1
a b p (0.0252) = 10.4167(10 - 6) m3
3p
2
800 N
Ty c
VzQB
It
-
-6
800(10.4167)(10 )
=
J
0.306796(10 )(0.05)
sx = 0
s1, 2 =
sy = 0
sx + sy
2
;
C
a
sx - sy
2
45(0.025)
-
-6
0.61359(10 - 6)
= -1.290 MPa
txy = -1.290 MPa
2
b + txy 2
= 0 ; 2(0)2 + (-1.290)2
s1 = 1.29 MPa
Ans.
s2 = -1.29 MPa
Ans.
676
A
B
25 mm
45 N⭈m
sB = 0
tB =
A
B
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•9–49.
The internal loadings at a section of the beam are
shown. Determine the principal stress at point A. Also
compute the maximum in-plane shear stress at this point.
50 mm
A
200 mm
50 mm
50 mm
y
200 mm
Section Properties:
z
40 kN⭈m
30 kN⭈m
A = 0.2(0.3) - 0.15(0.2) = 0.030 m4
800 kN
1
1
Iz =
(0.2) A 0.33 B (0.15) A 0.23 B = 0.350 A 10 - 3 B m4
12
12
Iy =
1
1
(0.1) A 0.23 B +
(0.2) A 0.053 B = 68.75 A 10 - 6 B m4
12
12
(QA)y = 0
Normal Stress:
s =
Myz
Mzy
N
+
A
Iz
Iy
sA =
-30(103)(0.1)
-500(103)
40(103)(0.15)
+
3
0.030
0.350(10 )
68.75(10 - 6)
= -77.45 MPa
tA = 0.
Shear Stress: Since (QA)y = 0,
In - Plane Principal Stresses: sx = -77.45 MPa. sy = 0. and txy = 0 for point A.
Since no shear stress acts on the element.
s1 = sy = 0
Ans.
s2 = sz = -77.4 MPa
Ans.
Maximum In-Plane Shear Stress: Applying Eq. 9–7.
t max
in-plane
a
sx - sy
2
b + t2xy
=
C
=
-77.45 - 0 2
b + 0
C
2
2
a
= 38.7 MPa
Ans.
677
500 kN
x
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9–50. The internal loadings at a section of the beam
consist of an axial force of 500 N, a shear force of 800 N,
and two moment components of 30 N # m and 40 N # m.
Determine the principal stress at point A. Also calculate the
maximum in-plane shear stress at this point.
Ix =
A
1
(0.1)(0.2)3 = 66.67(10 - 6) in4
12
200 mm
40 N⭈m
B
C
50 mm
QA = 0
50 mm
100 mm
30(0.1)
Mz
P
500
= -20 kPa
=
sA =
A
Ix
(0.1)(0.2)
66.67(10 - 6)
30 N⭈m
500 N
800 N
tA = 0
Here, the principal stresses are
s1 = sy = 0
Ans.
s2 = sx = -20 kPa
Ans.
t max
in-plane=
=
C
a
sx - sy
2
2
b + txy 2
-20 - 0 2
b + 0 = 10 kPa
C
2
a
Ans.
9–51. Solve Prob. 9–4 using Mohr’s circle.
A
400 psi
650 psi
60⬚
sx + sy
2
-650 + 400
=
= -125
2
A(-650, 0)
B(400, 0)
C( -125, 0)
B
R = CA = = 650 - 125 = 525
sx¿ = -125 - 525 cos 60° = -388 psi
Ans.
tx¿y¿ = 525 sin 60° = 455 psi
Ans.
678
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*9–52.
Solve Prob. 9–6 using Mohr’s circle.
90 MPa
A
35 MPa
60⬚
30⬚
sx = 90 MPa
sx + sy
2
=
sy = 50 MPa
txy = -35 MPa
A(90, -35)
90 + 50
= 70
2
R = 2(90 - 70)2 + (35)2 = 40.311
Coordinates of point B:
f = tan - 1 a
35
b = 60.255°
20
c = 300° - 180° - 60.255° = 59.745°
sx¿ = 70 - 40.311 cos 59.745° = 49.7 MPa
Ans.
tx¿ = -40.311 sin 59.745° = -34.8 MPa
Ans.
679
B
50 MPa
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•9–53.
Solve Prob. 9–14 using Mohr’s circle.
30 ksi
12 ksi
sx + sy
2
=
-30 + 0
= -15
2
R = 2(30 - 15)2 + (12)2 = 19.21 ksi
s1 = 19.21 - 15 = 4.21 ksi
Ans.
s2 = -19.21 - 15 = -34.2 ksi
Ans.
2uP2 = tan - 1
tmax
in-plane
12
;
(30 - 15)
uP2 = 19.3°
Ans.
= R = 19.2 ksi
Ans.
savg = -15 ksi
2uP2 = tan - 1
12
+ 90°;
(30 - 15)
Ans.
us = 64.3°
Ans.
680
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9–54.
Solve Prob. 9–16 using Mohr’s circle.
350 psi
75 psi
200 psi
sx + sy
2
=
45 - 60
= -7.5 MPa
2
R = 2(45 + 7.5)2 + (30)2 = 60.467 MPa
s1 = 60.467 - 7.5 = 53.0 MPa
Ans.
s2 = -60.467 - 7.5 = -68.0 MPa
Ans.
2uP1 = tan - 1
uP1 = 14.9°
tmax
30
(45 + 7.5)
Ans.
counterclockwise
= 60.5 MPa
Ans.
savg = -7.50 MPa
Ans.
in-plane
2uP1 = 90° - tan - 1
us1 = 30.1°
30
(45 + 7.5)
Ans.
clockwise
681
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9–55.
Solve Prob. 9–12 using Mohr’s circle.
sx + sy
2
=
-10 + 0
= -5 ksi
2
10 ksi
16 ksi
R = 2(10 - 5)2 + (16)2 = 16.763 ksi
f = tan - 1
16
= 72.646°
(10 - 5)
a = 100 - 72.646 = 27.354°
sx¿ = -5 - 16.763 cos 27.354° = -19.9 ksi
Ans.
tx¿y¿ = 16.763 sin 27.354° = 7.70 ksi
Ans.
sy¿ = 16.763 cos 27.354° - 5 = 9.89 ksi
*9–56. Solve Prob. 9–11 using Mohr’s circle.
2 ksi
Construction of the Circle: In accordance with the sign convention, sx = -3 ksi,
sy = 2 ksi, and txy = -4 ksi. Hence,
savg =
sx + sy
2
=
-3 + 2
= -0.500 ksi
2
3 ksi
30⬚
The coordinates for reference point A and C are
A(-3, -4)
A
4 ksi
B
C(-0.500, 0)
The radius of circle is
R = 2(3 - 0.5)2 + 42 = 4.717 ksi
Stress on the Inclined Plane: The normal and shear stress components
A sx¿ and tx¿y¿ B are represented by the coordinates of point P on the circle.
sx¿ = -0.500 - 4.717 cos 62.01° = -2.71 ksi
Ans.
tx¿y¿ = 4.717 sin 62.01° = 4.17 ksi
Ans.
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9–57. Mohr’s circle for the state of stress in Fig. 9–15a is
shown in Fig. 9–15b. Show that finding the coordinates of
point P1sx¿ , tx¿y¿2 on the circle gives the same value as the
stress-transformation Eqs. 9–1 and 9–2.
A(sx, txy)
R =
sxœ =
C
Ca a
B(sy, -txy)
csx - a
sx + sy
2
sx + sy
+
2
C
a
sx + sy
2
2
b d + t2xy =
sx - sy
2
C
a
b, 0b
sx - sy
2
2
b + t2xy
2
b + t2xy cos u¿
(1)
u¿ = 2uP - 2u
(2)
cos (2uP - 2u) = cos 2uP cos 2u + sin 2up sin 2u
From the circle:
sx -
cos 2uP =
sin 2uP =
4A
4A
sx + sy
sx - sy
2
txy
sx - sy
2
B
2
2
(3)
+ t2xy
(4)
B 2 + t2xy
Substitute Eq. (2), (3) and into Eq. (1)
sx¿ =
tx¿y¿ =
sx + sy
C
sx - sy
2
+
a
sx - sy
2
2
cos 2u + txy sin 2u
QED
2
b + t2xy sin u¿
(5)
sin u¿ = sin (2uP - 2u)
(6)
= sin 2uP cos 2u - sin 2u cos 2uP
Substitute Eq. (3), (4), (6) into Eq. (5),
tx¿y¿ = -
sx - sy
2
sin 2u + txy cos 2u
QED
683
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9–58. Determine the equivalent state of stress if an element
is oriented 25° counterclockwise from the element shown.
550 MPa
A(0, -550)
B(0, 550)
C(0, 0)
R = CA = CB = 550
sx¿ = -550 sin 50° = -421 MPa
Ans.
tx¿y¿ = -550 cos 50° = -354 MPa
Ans.
sy¿ = 550 sin 50° = 421 MPa
Ans.
9–59. Determine the equivalent state of stress if an
element is oriented 20° clockwise from the element shown.
2 ksi
Construction of the Circle: In accordance with the sign convention, sx = 3 ksi,
sy = -2 ksi, and tx¿y¿ = -4 ksi. Hence,
savg =
sx + sy
2
=
3 + (-2)
= 0.500 ksi
2
4 ksi
The coordinates for reference points A and C are
A(3, -4)
3 ksi
C(0.500, 0)
The radius of the circle is
R = 2(3 - 0.500)2 + 42 = 4.717 ksi
Stress on the Rotated Element: The normal and shear stress components
A sx¿ and tx¿y¿ B are represented by the coordinate of point P on the circle, sy¿, can be
determined by calculating the coordinates of point Q on the circle.
sx¿ = 0.500 + 4.717 cos 17.99° = 4.99 ksi
Ans.
tx¿y¿ = -4.717 sin 17.99° = -1.46 ksi
Ans.
sy¿ = 0.500 - 4.717 cos 17.99° = -3.99 ksi
Ans.
684
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*9–60. Determine the equivalent state of stress if an
element is oriented 30° clockwise from the element shown.
Show the result on the element.
9 ksi
4 ksi
In accordance to the established sign convention, sx = -6 ksi, sy = 9 ksi and
txy = 4 ksi. Thus,
savg =
sx + sy
2
=
-6 + 9
= 1.50 ksi
2
Then, the coordinates of reference point A and C are
A(-6, 4)
C(1.5, 0)
The radius of the circle is
R = CA = 2(-6 - 1.5)2 + 42 = 8.50 ksi
Using these results, the circle shown in Fig. a can be constructed.
Referring to the geometry of the circle, Fig. a,
a = tan - 1 a
4
b = 28.07°
6 + 1.5
b = 60° - 28.07° = 31.93°
Then,
sx¿ = 1.5 - 8.50 cos 31.93° = -5.71 ksi
Ans.
tx¿y¿ = -8.5 sin 31.95° = -4.50 ksi
sy¿ = 8.71 ksi
Ans.
The results are shown in Fig. b.
685
6 ksi
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•9–61.
Determine the equivalent state of stress for an
element oriented 60° counterclockwise from the element
shown. Show the result on the element.
250 MPa
400 MPa
In accordance to the established sign convention, sx = -560 MPa, sy = 250 MPa
and txy = -400 MPa. Thus,
savg =
sx + sy
2
=
-560 + 250
= -155 MPa
2
Then, the coordinate of reference points A and C are
A(-560, -400)
C(-155, 0)
The radius of the circle is
R = CA = 3 C -560 - (-155) D 2 + (-400)2 = 569.23 MPa
Using these results, the circle shown in Fig. a can be constructed.
Referring to the geometry of the circle, Fig. a
a = tan - 1 a
400
b = 44.64°
560 - 155
b = 120° - 44.64° = 75.36°
Then,
sx¿ = -155 - 569.23 cos 75.36° = -299 MPa
Ans.
tx¿y¿ = 569.23 sin 75.36° = 551 MPa
Ans.
sy¿ = -155 + 569.23 cos 75.36° = -11.1 MPa
Ans.
The results are shown in Fig. b.
686
560 MPa
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9–62. Determine the equivalent state of stress for an
element oriented 30° clockwise from the element shown.
Show the result on the element.
5 ksi
In accordance to the established sign convention, sx = 2 ksi, sy = -5 ksi and
txy = 0. Thus,
savg =
sx + sy
2
=
2 + (-5)
= -1.50 ksi
2
Then, the coordinate of reference points A and C are
A(2, 0)
C(-1.5, 0)
The radius of the circle is
R = CA = 3 C 2 - (-1.5) D 2 + 02 = 3.50 ksi
Using these results, the circle shown in Fig. a can be constructed.
Referring to the geometry of the circle, Fig. a,
b = 60°
Then,
sx¿ = -1.50 + 3.50 cos 60° - 0.250 ksi
Ans.
tx¿y¿ = 3.50 sin 60° = 3.03 ksi
Ans.
sy¿ = -3.25 ksi
Ans.
The results are shown in Fig b.
687
2 ksi
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9–63. Determine the principal stress, the maximum in-plane
shear stress, and average normal stress. Specify the orientation
of the element in each case.
15 ksi
5 ksi
Construction of the Circle: In accordance with the sign convention, sx = 15 ksi,
sy = 0 and txy = -5 ksi. Hence,
sx + sy
savg =
=
2
15 + 0
= 7.50 ksi
2
Ans.
The coordinates for reference point A and C are
A(15, -5)
C(7.50, 0)
The radius of the circle is
R = 2(15 - 7.50)2 + 52 = 9.014 ksi
a)
In - Plane Principal Stress: The coordinates of points B and D represent s1 and
s2, respectively.
s1 = 7.50 + 9.014 = 16.5 ksi
Ans.
s2 = 7.50 - 9.014 = -1.51 ksi
Ans.
Orientation of Principal Plane: From the circle
tan 2uP1 =
5
= 0.6667
15 - 7.50
uP1 = 16.8° (Clockwise)
Ans.
b)
Maximum In - Plane Shear Stress: Represented by the coordinates of point E on
the circle.
tmax
in-plane
= -R = -9.01 ksi
Ans.
Orientation of the Plane for Maximum In - Plane Shear Stress: From the circle
tan 2us =
15 - 7.50
= 1.500
5
us = 28.2° (Counterclockwise)
Ans.
688
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*9–64. Determine the principal stress, the maximum
in-plane shear stress, and average normal stress. Specify the
orientation of the element in each case.
20 MPa
80 MPa
30 MPa
In accordance to the established sign convention, sx = 30 MPa, sy = -20 MPa and
txy = 80 MPa. Thus,
savg =
sx + sy
2
30 + ( -20)
= 5 MPa
2
=
Then, the coordinates of reference point A and the center C of the circle is
A(30, 80)
C(5, 0)
Thus, the radius of circle is given by
R = CA = 2(30 - 5)2 + (80 - 0)2 = 83.815 MPa
Using these results, the circle shown in Fig. a, can be constructed.
The coordinates of points B and D represent s1 and s2 respectively. Thus
s1 = 5 + 83.815 = 88.8 MPa
Ans.
s2 = 5 - 83.815 = -78.8 MPa
Ans.
Referring to the geometry of the circle, Fig. a
tan 2(uP)1 =
80
= 3.20
30 - 5
uP = 36.3° (Counterclockwise)
Ans.
The state of maximum in - plane shear stress is represented by the coordinate of
point E. Thus
tmax
in-plane = R = 83.8 MPa
Ans.
From the geometry of the circle, Fig. a,
tan 2us =
30 - 5
= 0.3125
80
us = 8.68° (Clockwise)
Ans.
The state of maximum in - plane shear stress is represented by the element in Fig. c
689
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9–64.
Continued
690
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•9–65. Determine the principal stress, the maximum inplane shear stress, and average normal stress. Specify the
orientation of the element in each case.
120 psi
300 psi
A(300, 120)
B(0, -120)
C(150, 0)
R = 2(300 - 150)2 + 1202 = 192.094
s1 = 150 + 192.094 = 342 psi
Ans.
s2 = 150 - 192.094 = -42.1 psi
Ans.
tan 2uP =
120
= 0.8
300 - 150
uP1 = 19.3° Counterclockwise
Ans.
savg = 150 psi
Ans.
tmax
Ans.
in-plane
= 192 psi
tan 2us =
300 - 150
= 1.25
120
us = -25.7°
Ans.
9–66. Determine the principal stress, the maximum in-plane
shear stress, and average normal stress. Specify the orientation
of the element in each case.
A(45, -50)
B(30, 50)
30 MPa
C(37.5, 0)
45 MPa
R = CA = CB = 27.52 + 502 = 50.56
50 MPa
a)
tan 2uP =
50
7.5
s1 = 37.5 + 50.56 = 88.1 MPa
Ans.
s2 = 37.5 - 50.56 = -13.1 MPa
Ans.
uP = -40.7°
2uP = 81.47°
b)
t max
in-plane
= R = 50.6 MPa
Ans.
savg = 37.5 MPa
Ans.
2us = 90 - 2uP
us = 4.27°
Ans.
691
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9–67. Determine the principal stress, the maximum in-plane
shear stress, and average normal stress. Specify the orientation
of the element in each case.
200 MPa
500 MPa
350 MPa
Construction of the Circle: In accordance with the sign convention, sx = 350 MPa,
sy = -200 MPa, and txy = 500 MPa. Hence,
savg =
sx + sy
2
=
350 + (-200)
= 75.0 MPa
2
Ans.
The coordinates for reference point A and C are
A(350, 500)
C(75.0, 0)
The radius of the circle is
R = 2(350 - 75.0)2 + 5002 = 570.64 MPa
a)
In - Plane Principal Stresses: The coordinate of points B and D represent s1 and s2
respectively.
s1 = 75.0 + 570.64 = 646 MPa
Ans.
s2 = 75.0 - 570.64 = -496 MPa
Ans.
Orientaion of Principal Plane: From the circle
tan 2uP1 =
500
= 1.82
350 - 75.0
uP1 = 30.6° (Counterclockwise)
Ans.
b)
Maximum In - Plane Shear Stress: Represented by the coordinates of point E on
the circle.
t max
in-plane
= R = 571 MPa
Ans.
Orientation of the Plane for Maximum In - Plane Shear Stress: From the circle
tan 2us =
350 - 75.0
= 0.55
500
us = 14.4° (Clockwise)
Ans.
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*9–68. Draw Mohr’s circle that describes each of the
following states of stress.
700 psi
4 ksi
40 MPa
600 psi
(a)
a) Here, sx = 600 psi, sy = 700 psi and txy = 0. Thus,
savg =
sx + sy
=
2
600 + 700
= 650 psi
2
Thus, the coordinate of reference point A and center of circle are
A(600, 0)
C(650, 0)
Then the radius of the circle is
R = CA = 650 - 600 = 50 psi
The Mohr’s circle represents this state of stress is shown in Fig. a.
b) Here, sx = 0, sy = 4 ksi and txy = 0. Thus,
savg =
sx + sy
=
2
0 + 4
= 2 ksi
2
Thus, the coordinate of reference point A and center of circle are
A(0, 0)
C(2, 0)
Then the radius of the circle is
R = CA = 2 - 0 = 2 psi
c) Here, sx = sy = 0 and txy = -40 MPa. Thus,
savg =
sx + sy
2
= 0
Thus, the coordinate of reference point A and the center of circle are
A(0, -40)
C(0, 0)
Then, the radius of the circle is
R = CA = 40 MPa
The Mohr’s circle represents this state of stress shown in Fig. c
693
(b)
(c)
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9–68.
Continued
694
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9–69. The frame supports the distributed loading of
200 N兾m. Determine the normal and shear stresses at point
D that act perpendicular and parallel, respectively, to the
grain. The grain at this point makes an angle of 30° with the
horizontal as shown.
200 N/ m
B
30⬚
1m
200 mm
75 mm
D
1.5 m
C
100 mm
4m
60⬚
E
Support Reactions: As shown on FBD(a).
50 mm
30 mm
Internal Forces and Moment: As shown on FBD(b).
1.5 m
100 mm
Section Properties:
I =
A
1
(0.1) A 0.23 B = 66.667 A 10 - 6 B m4
12
QD = y¿A¿ = 0.0625(0.075)(0.1) = 0.46875 A 10 - 3 B m3
Normal Stress: Applying the flexure formula.
sD = -
My
150(-0.025)
= 56.25 kPa
= I
66.667(10 - 6)
Shear Stress: Applying the shear formula.
tD =
50.0 C 0.46875(10 - 3) D
VQD
= 3.516 kPa
=
It
66.667(10 - 6)(0.1)
Construction of the Circle: In accordance to the established sign convention,
sx = 56.25 kPa, sy = 0 and txy = -3.516 kPa. Hence.
savg =
sx + sy
2
=
56.25 + 0
= 28.125 kPa
2
The coordinates for reference point A and C are
A(56.25, -3.516)
C(28.125, 0)
The radius of the circle is
R = 2(56.25 - 28.125)2 + 3.5162 = 28.3439 kPa
Stresses on The Rotated Element: The normal and shear stress components
A sx¿ and tx¿y¿ B are represented by the coordinates of point P on the circle. Here,
u = 60°.
sx¿ = 28.125 - 28.3439 cos 52.875° = 11.0 kPa
Ans.
tx¿y¿ = -28.3439 sin 52.875° = -22.6 kPa
Ans.
695
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9–70. The frame supports the distributed loading of
200 N兾m. Determine the normal and shear stresses at point
E that act perpendicular and parallel, respectively, to the
grain. The grain at this point makes an angle of 60° with the
horizontal as shown.
200 N/ m
B
30⬚
1m
200 mm
75 mm
D
1.5 m
C
100 mm
4m
60⬚
E
Support Reactions: As shown on FBD(a).
50 mm
30 mm
Internal Forces and Moment: As shown on FBD(b).
1.5 m
100 mm
Section Properties:
A
A = 0.1(0.05) = 5.00 A 10 - 3 B m2
Normal Stress:
sE =
N
-250
= -50.0 kPa
=
A
5.00(10 - 3)
Construction of the Circle: In accordance with the sign convention. sx = 0,
sy = -50.0 kPa, and txy = 0. Hence.
savg =
sx + sy
2
=
0 + (-50.0)
= -25.0 kPa
2
The coordinates for reference points A and C are
A(0, 0)
C(-25.0, 0)
The radius of circle is R = 25.0 - 0 = 25.0 kPa
Stress on the Rotated Element: The normal and shear stress components
A sx¿ and tx¿y¿ B are represented by coordinates of point P on the circle. Here,
u = 150°.
sx = -25.0 + 25.0 cos 60° = -12.5 kPa
Ans.
tx¿y¿ = 25.0 sin 60° = 21.7 kPa
Ans.
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9–71. The stair tread of the escalator is supported on two
of its sides by the moving pin at A and the roller at B. If a
man having a weight of 300 lb stands in the center of the
tread, determine the principal stresses developed in the
supporting truck on the cross section at point C. The stairs
move at constant velocity.
1.25 ft
30⬚
A
C 1.5 ft
30⬚
0.5 ft B
0.5 ft
2 in.
Support Reactions: As shown on FBD (a).
Internal Forces and Moment: As shown on FBD (b).
Section Properties:
A = 2(0.5) = 1.00 in2
I =
1
(0.5) A 23 B = 0.3333 in4
12
QB = y¿A¿ = 0.5(1)(0.5) = 0.250 in3
Normal Stress:
s =
My
N
;
A
I
sC =
475.48(0)
-137.26
+
= -137.26 psi
1.00
0.3333
Shear Stress: Applying the shear formula t =
tC =
VQ
.
It
79.25(0.250)
= 118.87 psi
0.3333(0.5)
Construction of the Circle: In accordance with the sign convention, sx = 0,
sy = -137.26 psi, and txy = 118.87 psi. Hence,
savg =
sx + sy
2
=
0 + (-137.26)
= -68.63 psi
2
The coordinates for reference points A and C are
A(0, 118.87)
1 in.
C
C(-68.63, 0)
The radius of the circle is
R = 2(68.63 - 0)2 + 118.872 = 137.26 psi
In - Plane Principal Stress: The coordinates of point B and D represent s1 and s2,
respectively.
s1 = -68.63 + 137.26 = 68.6 psi
Ans.
s2 = -68.63 - 137.26 = -206 psi
Ans.
697
0.5 in.
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*9–72. The thin-walled pipe has an inner diameter of
0.5 in. and a thickness of 0.025 in. If it is subjected to an
internal pressure of 500 psi and the axial tension and
torsional loadings shown, determine the principal stress at a
point on the surface of the pipe.
200 lb
200 lb
20 lb⭈ft
Section Properties:
A = p A 0.2752 - 0.252 B = 0.013125p in2
J =
p
A 0.2754 - 0.254 B = 2.84768 A 10 - 3 B in4
2
Normal Stress: Since
0.25
r
=
= 10, thin wall analysis is valid.
t
0.025
slong =
pr
500(0.25)
200
N
+
=
+
= 7.350 ksi
A
2t
0.013125p
2(0.025)
shoop =
pr
500(0.25)
=
= 5.00 ksi
t
0.025
Shear Stress: Applying the torsion formula,
t =
20(12)(0.275)
Tc
= 23.18 ksi
=
J
2.84768(10 - 3)
Construction of the Circle: In accordance with the sign convention sx = 7.350 ksi,
sy = 5.00 ksi, and txy = -23.18 ksi. Hence,
savg =
sx + sy
2
=
7.350 + 5.00
= 6.175 ksi
2
The coordinates for reference points A and C are
A(7.350, -23.18)
C(6.175, 0)
The radius of the circle is
R = 2(7.350 - 6.175)2 + 23.182 = 23.2065 ksi
In - Plane Principal Stress: The coordinates of point B and D represent s1 and s2,
respectively.
s1 = 6.175 + 23.2065 = 29.4 ksi
Ans.
s2 = 6.175 - 23.2065 = -17.0 ksi
Ans.
698
20 lb⭈ft
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•9–73. The cantilevered rectangular bar is subjected to the
force of 5 kip. Determine the principal stress at point A.
1.5 in. A
1.5 in.
1 in.
B
1.5 in.
1.5 in.
3 in.
1 in.
15 in.
3 in.
3 5
4
5 kip
Internal Forces and Moment: As shown on FBD.
Section Properties:
A = 3(6) = 18.0 in2
I =
1
(3) A 63 B = 54.0 in4
12
QA = y¿A¿ = 2.25(1.5)(3) = 10.125 in3
Normal Stress:
s =
My
N
;
A
I
sA =
45.0(1.5)
4.00
+
= 1.4722 ksi
18.0
54.0
Shear Stress: Applying the shear formula t =
tA =
VQ
.
It
3.00(10.125)
= 0.1875 ksi
54.0(3)
Construction of the Circle: In accordance with the sign convention, sx = 1.4722 ksi,
sy = 0, and txy = -0.1875 ksi. Hence,
savg =
sx + sy
2
=
1.472 + 0
= 0.7361 ksi
2
The coordinates for reference points A and C are
A(1.4722, -0.1875)
C(0.7361, 0)
The radius of the circle is
R = 2(1.4722 - 0.7361)2 + 0.18752 = 0.7596 ksi
In - Plane Principal Stress: The coordinates of point B and D represent s1 and s2,
respectively.
s1 = 0.7361 + 0.7596 = 1.50 ksi
Ans.
s2 = 0.7361 - 0.7596 = -0.0235 ksi
Ans.
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9–74. Solve Prob. 9–73 for the principal stress at point B.
1.5 in. A
1.5 in.
1 in.
B
1.5 in.
1.5 in.
3 in.
1 in.
15 in.
3 in.
3 5
4
5 kip
Internal Forces and Moment: As shown on FBD.
Section Properties:
A = 3(6) = 18.0 in2
1
(3) A 63 B = 54.0 in4
12
I =
QB = y¿A¿ = 2(2)(3) = 12.0 in3
Normal Stress:
s =
My
N
;
A
I
sB =
45.0(1)
4.00
= -0.6111 ksi
18.0
54.0
Shear Stress: Applying the shear formula t =
tB =
VQ
.
It
3.00(12.0)
= 0.2222 ksi
54.0(3)
Construction of the Circle: In accordance with the sign convention,
sx = -0.6111 ksi, sy = 0, and txy = -0.2222 ksi. Hence.
savg =
sx + sy
2
=
-0.6111 + 0
= -0.3055 ksi
2
The coordinates for reference points A and C are
A(-0.6111, -0.2222)
C(-0.3055, 0)
The radius of the circle is
R = 2(0.6111 - 0.3055)2 + 0.22222 = 0.3778 ksi
In - Plane Principal Stress: The coordinates of point B and D represent s1 and s2,
respectively.
s1 = -0.3055 + 0.3778 = 0.0723 ksi
Ans.
s2 = -0.3055 - 0.3778 = -0.683 ksi
Ans.
700
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9–75. The 2-in.-diameter drive shaft AB on the helicopter
is subjected to an axial tension of 10 000 lb and a torque
of 300 lb # ft. Determine the principal stress and the
maximum in-plane shear stress that act at a point on the
surface of the shaft.
s =
10 000
P
=
= 3.183 ksi
A
p(1)2
t =
300(12)(1)
Tc
=
= 2.292 ksi
p
4
J
2 (1)
s1, 2 =
=
t max
sx + sy
;
2
A
(
sx - sy
2
B
A
)2 + t2xy
3.183 - 0 2
3.183 + 0
; (
) + (2.292)2
2
A
2
s1 = 4.38 ksi
Ans.
s2 = -1.20 ksi
Ans.
in-plane
=
A
(
=
A
(
sx - sy
2
)2 + t2xy
3.183 - 0 2
) + (2.292)2
2
= 2.79 ksi
Ans.
701
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*9–76. The pedal crank for a bicycle has the cross section
shown. If it is fixed to the gear at B and does not rotate
while subjected to a force of 75 lb, determine the principal
stress in the material on the cross section at point C.
75 lb
B
3 in.
A
4 in.
C
0.4 in.
0.4 in.
0.2 in.
0.3 in.
Internal Forces and Moment: As shown on FBD
Section Properties:
I =
1
(0.3) A 0.83 B = 0.0128 in3
12
QC = y¿A¿ = 0.3(0.2)(0.3) = 0.0180 in3
Normal Stress: Applying the flexure formula.
sC = -
My
-300(0.2)
= = 4687.5 psi = 4.6875 ksi
I
0.0128
Shear Stress: Applying the shear formula.
tC =
VQC
75.0(0.0180)
=
= 351.6 psi = 0.3516 ksi
It
0.0128(0.3)
Construction of the Circle: In accordance with the sign convention, sx = 4.6875 ksi,
sy = 0, and txy = 0.3516 ksi. Hence,
savg =
sx + sy
2
=
4.6875 + 0
= 2.34375 ksi
2
The coordinates for reference points A and C are
A(4.6875, 0.3516)
C(2.34375, 0)
The radius of the circle is
R = 2(4.6875 - 2.34375)2 + 0.35162 = 2.3670 ksi
In - Plane Principal Stress: The coordinates of point B and D represent s1 and s2,
respectively.
s1 = 2.34375 + 2.3670 = 4.71 ksi
Ans.
s2 = 2.34375 - 2.3670 = -0.0262 ksi
Ans.
702
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•9–77.
A spherical pressure vessel has an inner radius of
5 ft and a wall thickness of 0.5 in. Draw Mohr’s circle for the
state of stress at a point on the vessel and explain the
significance of the result. The vessel is subjected to an
internal pressure of 80 psi.
Normal Stress:
s1 = s2 =
pr
80(5)(12)
=
= 4.80 ksi
2t
2(0.5)
Mohr’s circle:
A(4.80, 0)
B(4.80, 0)
C(4.80, 0)
Regardless of the orientation of the element, the shear stress is zero and the state of
stress is represented by the same two normal stress components.
9–78. The cylindrical pressure vessel has an inner radius
of 1.25 m and a wall thickness of 15 mm. It is made from
steel plates that are welded along the 45° seam. Determine
the normal and shear stress components along this seam if
the vessel is subjected to an internal pressure of 8 MPa.
sx =
45⬚
1.25 m
pr
8(1.25)
=
= 333.33 MPa
2t
2(0.015)
sy = 2sx = 666.67 MPa
A(333.33, 0)
sx¿ =
B(666.67, 0)
C(500, 0)
333.33 + 666.67
= 500 MPa
2
Ans.
tx¿y¿ = R = 666.67 - 500 = 167 MPa
Ans.
703
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•9–79.
Determine the normal and shear stresses at point
D that act perpendicular and parallel, respectively, to the
grains. The grains at this point make an angle of 30° with
the horizontal as shown. Point D is located just to the left of
the 10-kN force.
10 kN
A 100 mm
D
B
30⬚
1m
100 mm
D
100 mm
Using the method of section and consider the FBD of the left cut segment, Fig. a
+ c ©Fy = 0;
5 - V = 0
V = 5 kN
a + ©MC = 0;
M - 5(1) = 0
M = 5 kN # m
The moment of inertia of the rectangular cross - section about the neutral axis is
I =
1
(0.1)(0.33) = 0.225(10 - 3) m4
12
Referring to Fig. b,
QD = y¿A¿ = 0.1(0.1)(0.1) = 0.001 m3
The normal stress developed is contributed by bending stress only. For point D,
y = 0.05 m. Then
s =
My
5(103)(0.05)
= 1.111 MPa (T)
=
I
0.225(10 - 3)
The shear stress is contributed by the transverse shear stress only. Thus,
t =
5(103)(0.001)
VQD
= 0.2222 MPa
=
It
0.225(10 - 3)(0.1)
The state of stress at point D can be represented by the element shown in Fig. c
In accordance to the established sign convention, sx = 1.111 MPa, sy = 0 and
txy = -0.2222 MPa, Thus.
savg =
sx + sy
2
=
1.111 + 0
= 0.5556 MPa
2
Then, the coordinate of reference point A and the center C of the circle are
A(1.111, -0.2222)
C(0.5556, 0)
Thus, the radius of the circle is given by
R = 2(1.111 - 0.5556)2 + (-0.2222)2 = 0.5984 MPa
Using these results, the circle shown in Fig. d can be constructed.
Referring to the geometry of the circle, Fig. d,
a = tan - 1 a
0.2222
b = 21.80°
1.111 - 0.5556
b = 180° - (120° - 21.80°) = 81.80°
704
1m
300 mm
2m
C
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9–79. Continued
Then
sx¿ = 0.5556 - 0.5984 cos 81.80° = 0.4702 MPa = 470 kPa
Ans.
tx¿y¿ = 0.5984 sin 81.80° = 0.5922 MPa = 592 kPa
Ans.
705
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*9–80. Determine the principal stress at point D, which is
located just to the left of the 10-kN force.
10 kN
A 100 mm
D
B
30⬚
1m
100 mm
D
100 mm
Using the method of section and consider the FBD of the left cut segment, Fig. a,
+ c ©Fy = 0;
5 - V = 0
V = 5 kN
a + ©MC = 0;
M - 5(1) = 0
M = 5 kN # m
I =
1
(0.1)(0.33) = 0.225(10 - 3) m4
12
Referring to Fig. b,
QD = y¿A¿ = 0.1(0.1)(0.1) = 0.001 m3
The normal stress developed is contributed by bending stress only. For point D,
y = 0.05 m
s =
My
5(103)(0.05)
= 1.111 MPa (T)
=
I
0.225(10 - 3)
The shear stress is contributed by the transverse shear stress only. Thus,
t =
5(103)(0.001)
VQD
= 0.2222 MPa
=
It
0.225(10 - 3)(0.1)
The state of stress at point D can be represented by the element shown in Fig. c.
In accordance to the established sign convention, sx = 1.111 MPa, sy = 0, and
txy = -0.2222 MPa. Thus,
savg =
sx + sy
2
=
1.111 + 0
= 0.5556 MPa
2
Then, the coordinate of reference point A and center C of the circle are
A(1.111, -0.2222)
C(0.5556, 0)
Thus, the radius of the circle is
R = CA = 2(1.111 - 0.5556)2 + (-0.2222)2 = 0.5984 MPa
Using these results, the circle shown in Fig. d.
In-Plane Principal Stresses. The coordinates of points B and D represent s1 and s2,
respectively. Thus,
s1 = 0.5556 + 0.5984 = 1.15 MPa
Ans.
s2 = 0.5556 - 0.5984 = -0.0428 MPa
Ans.
706
1m
300 mm
2m
C
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9–80. Continued
Referring to the geometry of the circle, Fig. d,
tan (2uP)1 =
0.2222
= 0.4
1.111 - 0.5556
(uP)1 = 10.9° (Clockwise)
Ans.
The state of principal stresses is represented by the element show in Fig. e.
707
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•9–81.
Determine the principal stress at point A on the
cross section of the hanger at section a–a. Specify the
orientation of this state of stress and indicate the result on
an element at the point.
0.75 m
0.75 m
a
250 mm a
900 N
Internal Loadings: Considering the equilibrium of the free - body diagram of the
hanger’s left cut segment, Fig. a,
+ ©F = 0;
:
x
900 - N = 0
N = 900 N
+ c ©Fy = 0;
V - 900 = 0
V = 900 N
a + ©MO = 0;
900(1) - 900(0.25) - M = 0
M = 675 N # m
b
250 mm
Section Properties: The cross - sectional area and the moment of inertia about the
centroidal axis of the hanger’s cross section are
A = 0.05(0.1) - 0.04(0.09) = 1.4 A 10 - 3 B m2
1
1
(0.05) A 0.13 B (0.04) A 0.093 B = 1.7367 A 10 - 6 B m4
12
12
Referring to Fig. b,
QA = 2y1œ A1œ + y2œ A2œ = 2[0.0375(0.025)(0.005)] + 0.0475(0.005)(0.04)
= 18.875 A 10 - 6 B m3
Normal and Shear Stress: The normal stress is a combination of axial and bending
stresses. Thus,
sA =
675(0.025)
MyA
N
900
+
= +
= 9.074 MPa
-3
A
I
1.4 A 10 B
1.7367 A 10 - 6 B
The shear stress is caused by the transverse shear stress.
tA =
900 C 18.875 A 10 - 6 B D
VQA
=
= 0.9782 MPa
It
1.7367 A 10 - 6 B (0.01)
The state of stress at point A is represented by the element shown in Fig. c.
Construction of the Circle: sx = 9.074 MPa, sy = 0, and txy = 0.9782 MPa. Thus,
savg =
sx + sy
2
=
9.074 + 0
= 4.537 MPa
2
The coordinates of reference points A and the center C of the circle are
A(9.074, 0.9782)
C(4.537, 0)
Thus, the radius of the circle is
R = CA = 2(9.074 - 4.537)2 + 0.97822 = 4.641 MPa
Using these results, the circle is shown in Fig. d.
708
b
900 N
5 mm
25 mm
A
100 mm
5 mm
50 mm
I =
0.5 m
5 mm
Sections a – a
and b – b
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9–81. Continued
In - Plane Principal Stress: The coordinates of point B and D represent s1 and s2,
respectively.
s1 = 4.537 + 4.641 = 9.18 MPa
Ans.
s2 = 4.537 - 4.641 = -0.104 MPa
Ans.
Orientaion of Principal Plane: Referring to the geometry of the circle, Fig. d,
tan 2 A uP B 1 =
0.9782
= 0.2156
9.074 - 4.537
A uP B 1 = 6.08° (counterclockwise)
Ans.
The state of principal stresses is represented on the element shown in Fig. e.
709
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9–82. Determine the principal stress at point A on the
cross section of the hanger at section b–b. Specify the
orientation of the state of stress and indicate the results on
an element at the point.
0.75 m
0.75 m
a
250 mm a
900 N
Internal Loadings: Considering the equilibrium of the free - body diagram of the
hanger’s left cut segment, Fig. a,
+ c ©Fy = 0;
V - 900 - 900 = 0
V = 1800 N
900(2.25) + 900(0.25) - M = 0
M = 2250 N # m
a + ©MO = 0;
Section Properties: The cross - sectional area and the moment of inertia about the
centroidal axis of the hanger’s cross section are
A = 0.05(0.1) - 0.04(0.09) = 1.4 A 10 - 3 B m2
I =
Referring to Fig. b.
QA = 2y1œ A1œ + y2œ A2œ = 2[0.0375(0.025)(0.005)] + 0.0475(0.005)(0.04)
= 18.875 A 10 - 6 B m3
Normal and Shear Stress: The normal stress is contributed by the bending stress
only.
MyA
2250(0.025)
=
= 32.39 MPa
I
1.7367 A 10 - 6 B
The shear stress is contributed by the transverse shear stress only.
tA =
1800 C 18.875 A 10 - 6 B D
VQA
=
= 1.956 MPa
It
1.7367 A 10 - 6 B (0.01)
The state stress at point A is represented by the element shown in Fig. c.
Construction of the Circle: sx = 32.39 MPa, sy = 0, and txy = 1.956 MPa. Thus,
savg =
sx + sy
2
=
32.39 + 0
= 16.19 MPa
2
The coordinates of reference point A and the center C of the circle are
A(32.39, 1.956)
b
250 mm
C(16.19, 0)
Thus, the radius of the circle is
R = CA = 2(32.39 - 16.19)2 + 1.9562 = 16.313 MPa
Using these results, the cricle is shown in Fig. d.
710
b
900 N
5 mm
25 mm
A
100 mm
5 mm
50 mm
5 mm
Sections a – a
and b – b
1
1
(0.05) A 0.13 B (0.04) A 0.093 B = 1.7367 A 10 - 6 B m4
12
12
sA =
0.5 m
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9–82. Continued
In - Plane Principal Stresses: The coordinates of reference point B and D represent
s1 and s2, respectively.
s1 = 16.19 + 16.313 = 32.5 MPa
Ans.
s2 = 16.19 - 16.313 = -0.118 MPa
Ans.
Orientaion of Principal Plane: Referring to the geometry of the circle, Fig. d,
tan 2 A uP B 1 =
1.956
= 0.1208
32.39 - 16.19
A uP B 1 = 3.44°
(counterclockwise)
Ans.
The state of principal stresses is represented on the element shown in Fig. e.
711
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9–83. Determine the principal stresses and the maximum
in-plane shear stress that are developed at point A. Show
the results on an element located at this point. The rod has a
diameter of 40 mm.
450 N
150 mm
Using the method of sections and consider the FBD of the member’s upper cut
segment, Fig. a,
+ c ©Fy = 0;
450 - N = 0
N = 450 N
a + ©MC = 0;
450(0.1) - M = 0
M = 45 N # m
B
100 mm
A
150 mm
A = p(0.022) = 0.4(10 - 3)p m2
I =
450 N
p
(0.024) = 40(10 - 9)p m4
4
The normal stress is the combination of axial and bending stress. Thus,
s =
My
N
+
A
I
For point A, y = C = 0.02 m.
s =
45 (0.02)
450
+
= 7.520 MPa
0.4(10 - 3)p
40(10 - 9)p
Since no transverse shear and torque is acting on the cross - section
t = 0
The state of stress at point A can be represented by the element shown in Fig. b.
In accordance to the established sign convention sx = 0, sy = 7.520 MPa and
txy = 0. Thus
savg =
sx + sy
2
=
0 + 7.520
= 3.760 MPa
2
Then, the coordinates of reference point A and the center C of the circle are
A(0, 0)
C(3.760, 0)
Thus, the radius of the circle is
R = CA = 3.760 MPa
Using this results, the circle shown in Fig. c can be constructed. Since no shear stress
acts on the element,
s1 = sy = 7.52 MPa
s2 = sx = 0
Ans.
The state of principal stresses can also be represented by the element shown in Fig. b.
The state of maximum in - plane shear stress is represented by point B on the circle,
Fig. c. Thus.
tmax
in-plane
= R = 3.76 MPa
Ans.
712
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9–83. Continued
From the circle,
2us = 90°
us = 45° (counter clockwise)
Ans.
The state of maximum In - Plane shear stress can be represented by the element
shown in Fig. d.
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*9–84. Draw the three Mohr’s circles that describe each of
the following states of stress.
5 ksi
(a) Here, smin = 0, sint = 3 ksi and smax = 5 ksi. The three Mohr’s circle
of this state of stress are shown in Fig. a
3 ksi
(b) Here, smin = 0, sint = 140 MPa and smax = 180 MPa. The three
Mohr’s circle of this state of stress are shown in Fig. b
(a)
•9–85.
Draw the three Mohr’s circles that describe the
following state of stress.
180 MPa
140 MPa
(b)
300 psi
Here, smin = -300 psi, sint = 0 and smax = 400 psi. The three Mohr’s circle for this
state of stress is shown in Fig. a.
400 psi
714
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z
9–86. The stress at a point is shown on the element.
Determine the principal stress and the absolute maximum
shear stress.
y
x
80 MPa
For y – z plane:
A(0, -80)
B(90, 80)
C(45, 0)
R = 2452 + 802 = 91.79
s1 = 45 + 91.79 = 136.79 MPa
s2 = 45 - 91.79 = -46.79 MPa
Thus,
tabs
max
=
s1 = 0
Ans.
s2 = 137 MPa
Ans.
s3 = -46.8 MPa
Ans.
136.79 - (-46.79)
smax - smin
=
= 91.8 MPa
2
2
Ans.
715
90 MPa
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z
9–87. The stress at a point is shown on the element.
Determine the principal stress and the absolute maximum
shear stress.
Mohr’s circle for the element in y - 7 plane, Fig. a, will be drawn first. In accordance
to the established sign convention, sy = 30 psi, sz = 120 psi and tyz = 70 psi. Thus
savg =
sy + sz
2
=
x
y
120 psi
70 psi
30 + 120
= 75 psi
2
30 psi
Thus the coordinates of reference point A and the center C of the circle are
A(30, 70)
C(75, 0)
Thus, the radius of the circle is
R = CA = 2(75 - 30)2 + 702 = 83.217 psi
Using these results, the circle shown in Fig. b.
The coordinates of point B and D represent the principal stresses
From the results,
smax = 158 psi
smin = -8.22 psi
sint = 0 psi
Ans.
Using these results, the three Mohr’s circle are shown in Fig. c,
From the geometry of the three circles,
tabs
max
=
158.22 - ( -8.22)
smax - smin
=
= 83.22 psi
2
2
716
Ans.
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z
*9–88. The stress at a point is shown on the element.
Determine the principal stress and the absolute maximum
shear stress.
Mohr’s circle for the element in x - z plane, Fig. a, will be drawn first. In accordance
to the established sign convention, sx = -2 ksi, sz = 0 and txz = 8 ksi. Thus
savg =
sx + sz
2
=
-2 + 0
= -1 ksi
2
2 ksi
8 ksi
Thus, the coordinates of reference point A and the center C of the circle are
A( -2, 8)
C(-1, 0)
Thus, the radius of the circle is
R = CA = 2[-2 - (-1)]2 + 82 = 265 ksi
Using these results, the circle in shown in Fig. b,
The coordinates of points B and D represent s1 and s2, respectively.
s = -1 + 265 = 7.062 ksi
smax = 7.06 ksi
sint = 0
smin = -9.06 ksi
From the results obtained,
sint = 0 ksi
smax = 7.06 ksi
smin = -9.06 ksi
Ans.
Using these results, the three Mohr’s circles are shown in Fig, c.
From the geometry of the cricle,
tabs
max
=
y
x
7.06 - (-9.06)
smax - smin
=
= 8.06 ksi
2
2
Ans.
717
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•9–89.
The stress at a point is shown on the element.
Determine the principal stress and the absolute maximum
shear stress.
z
y
x
For x – y plane:
150 MPa
120 MPa
R = CA = 2(120 - 60)2 + 1502 = 161.55
s1 = 60 + 161.55 = 221.55 MPa
s2 = 60 - 161.55 = -101.55 MPa
s1 = 222 MPa
tabs
max
=
s2 = 0 MPa
s3 = -102 MPa
Ans.
221.55 - (-101.55)
smax - smin
=
= 162 MPa
2
2
Ans.
9–90. The state of stress at a point is shown on the
element. Determine the principal stress and the absolute
maximum shear stress.
z
x
For y - z plane:
A(5, -4)
B(-2.5, 4)
4 ksi
s1 = 1.25 + 5.483 = 6.733 ksi
5 ksi
s2 = 1.25 - 5.483 = -4.233 ksi
Thus,
s1 = 6.73 ksi
Ans.
s2 = 0
Ans.
s3 = -4.23 ksi
Ans.
savg =
6.73 + (-4.23)
= 1.25 ksi
2
tabs
6.73 - (-4.23)
smax - smin
=
= 5.48 ksi
2
2
=
2.5 ksi
C(1.25, 0)
R = 23.752 + 42 = 5.483
max
y
Ans.
718
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*9–92. The solid shaft is subjected to a torque, bending
moment, and shear force as shown. Determine the principal
stress acting at points A and B and the absolute maximum
shear stress.
450 mm A
B
300 N⭈m
25 mm
45 N⭈m
800 N
Internal Forces and Moment: As shown on FBD.
Section Properties:
Iz =
p
A 0.0254 B = 0.306796 A 10 - 6 B m4
4
J =
p
A 0.0254 B = 0.613592 A 10 - 6 B m4
2
(QA)x = 0
(QB)y = y¿A¿
=
4(0.025) 1
c (p) A 0.0252 B d = 10.417 A 10 - 6 B m3
3p
2
Normal stress: Applying the flexure formula.
s = -
Mzy
Iz
-60.0(0.025)
sA = -
0.306796(10 - 6)
= 4.889 MPa
-60.0(0)
sB = -
0.306796(10 - 6)
= 0
Shear Stress: Applying the torsion formula for point A,
tA =
45.0(0.025)
Tc
= 1.833 MPa
=
J
0.613592(10 - 6)
The transverse shear stress in the y direction and the torsional shear stress can be
VQ
Tr
obtained using shear formula and torsion formula. tv =
and ttwist =
,
It
J
respectively.
tB = (tv)y - ttwist
=
800 C 10.417(10 - 6) D
-6
0.306796(10 )(0.05)
45.0(0.025)
-
0.613592(10 - 6)
= -1.290 MPa
Construction of the Circle: sx = 4.889 MPa, sz = 0, and txz = -1.833 MPa for
point A. Hence,
savg =
sx + sz
2
=
4.889 + 0
= 2.445 MPa
2
The coordinates for reference points A and C are A (4.889, –1.833) and C(2.445, 0).
719
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9–92.
Continued
The radius of the circle is
R = 2(4.889 - 2.445)2 + 1.8332 = 3.056 MPa
sx = sy = 0 and txy = -1.290 MPa for point B. Hence,
savg =
sx + sz
= 0
2
The coordinates for reference points A and C are A(0. ‚–1.290) and C(0,0).
The radius of the circle is R = 1.290 MPa
In - Plane Principal Stresses: The coordinates of point B and D represent s1 and s2,
respectively. For point A
s1 = 2.445 + 3.056 = 5.50 MPa
s2 = 2.445 - 3.506 = -0.611 MPa
For point B
s1 = 0 + 1.290 = 1.29 MPa
s2 = 0 - 1.290 = -1.290 MPa
Three Mohr’s Circles: From the results obtaired above, the principal stresses for
point A are
smax = 5.50 MPa
sint = 0
smin = -0.611 MPa
Ans.
sint = 0
smin = -1.29 MPa
Ans.
And for point B
smax = 1.29 MPa
Absolute Maximum Shear Stress: For point A,
tabs
max
=
5.50 - (-0.611)
smax - smin
=
= 3.06 MPa
2
2
Ans.
=
1.29 - (-1.29)
smax - smin
=
= 1.29 MPa
2
2
Ans.
For point B,
tabs
max
720
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•9–93.
The propane gas tank has an inner diameter of
1500 mm and wall thickness of 15 mm. If the tank is
pressurized to 2 MPa, determine the absolute maximum
shear stress in the wall of the tank.
Normal Stress: Since
750
r
=
= 50 7 10, thin - wall analysis can be used. We have
t
15
s1 =
2(750)
pr
=
= 100 MPa
t
15
s2 =
2(750)
pr
=
= 50 MPa
2t
2(15)
The state of stress of any point on the wall of the tank can be represented on the
element shown in Fig. a
Construction of Three Mohr’s Circles: Referring to the element,
smax = 100 MPa
sint = 50 MPa
smin = 0
Using these results, the three Mohr’s circles are shown in Fig. b.
Absolute Maximum Shear Stress: From the geometry of three circles,
tabs
max
=
smax - smin
100 - 0
=
= 50 MPa
2
2
Ans.
721
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9–94. Determine the principal stress and absolute
maximum shear stress developed at point A on the cross
section of the bracket at section a–a.
12 in.
6 in.
5
3
a
4
a
0.5 in.
B
0.25 in.
A
0.25 in.
0.25 in.
1.5 in.1.5 in.
Section a – a
Internal Loadings: Considering the equilibrium of the free - body diagram of the
bracket’s upper cut segment, Fig. a,
+ c ©Fy = 0;
3
N - 500 a b = 0
5
N = 300 lb
+ ©F = 0;
;
x
4
V - 500 a b = 0
5
V = 400 lb
3
4
©MO = 0; M - 500 a b(12) - 500 a b(6) = 0
5
5
M = 6000 lb # in
Section Properties: The cross - sectional area and the moment of inertia of the
bracket’s cross section are
A = 0.5(3) - 0.25(2.5) = 0.875 in2
I =
1
1
(0.5) A 33 B (0.25) A 2.53 B = 0.79948 in4
12
12
Referring to Fig. b.
QA = x1œ A1œ + x2œ A2œ = 0.625(1.25)(0.25) + 1.375(0.25)(0.5) = 0.3672 in3
Normal and Shear Stress: The normal stress is
sA =
N
300
= = -342.86 psi
A
0.875
The shear stress is contributed by the transverse shear stress.
tA =
400(0.3672)
VQA
=
= 734.85 psi
It
0.79948(0.25)
The state of stress at point A is represented by the element shown in Fig. c.
Construction of the Circle: sx = 0, sy = -342.86 psi, and txy = 734.85. Thus,
savg =
sx + sy
2
=
0 + (-342.86)
= -171.43 psi
2
The coordinates of reference point A and the center C of the circle are
A(0, 734.85)
C(-171.43, 0)
Thus, the radius of the circle is
R = CA = 2[0 - (-171.43)]2 + 734.852 = 754.58 psi
722
500 lb
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9–94. Continued
Using these results, the cricle is shown in Fig. d.
In - Plane Principal Stresses: The coordinates of reference point B and D represent
s1 and s2, respectively.
s1 = -171.43 + 754.58 = 583.2 psi
s2 = -171.43 - 754.58 = -926.0 psi
Three Mohr’s Circles: Using these results,
smax = 583 psi
sint = 0 smin = -926 psi
Ans.
Absolute Maximum Shear Stress:
tabs
max
=
583.2 - (-926.0)
smax - smin
=
- 755 psi
2
2
Ans.
723
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9–95. Determine the principal stress and absolute
maximum shear stress developed at point B on the cross
section of the bracket at section a–a.
12 in.
Internal Loadings: Considering the equilibrium of the free - body diagram of the 6 in.
bracket’s upper cut segment, Fig. a,
a
+ c ©Fy = 0;
+ ©F = 0;
;
x
3
N - 500 a b = 0
5
N = 300 lb
4
V - 500 a b = 0
5
V = 400 lb
1
1
(0.5) A 33 B (0.25) A 2.53 B = 0.79948 in4
12
12
Referring to Fig. b,
QB = 0
Normal and Shear Stress: The normal stress is a combination of axial and bending
stress.
6000(1.5)
MxB
N
300
+
= +
= 10.9 ksi
A
I
0.875
0.79948
Since QB = 0, tB = 0. The state of stress at point B is represented on the element
shown in Fig. c.
In - Plane Principal Stresses: Since no shear stress acts on the element,
s2 = 0
Three Mohr’s Circles: Using these results,
smax = 10.91 ksi
sint = smin = 0
Ans.
Absolute Maximum Shear Stress:
tabs
max
=
0.25 in.
1.5 in.1.5 in.
Section a – a
M = 6000 lb # in
smax - smin
10.91 - 0
=
= 5.46 ksi
2
2
Ans.
724
500 lb
0.25 in.
A
0.25 in.
A = 0.5(3) - 0.25(2.5) = 0.875 in2
s1 = 10.91 ksi
a
0.5 in.
Section Properties: The cross - sectional area and the moment of inertia about the
centroidal axis of the bracket’s cross section are
sB =
4
B
4
3
©MO = 0; M - 500 a b(12) - 500 a b(6) = 0
5
5
I =
5
3
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*9–96. The solid propeller shaft on a ship extends outward
from the hull. During operation it turns at v = 15 rad>s
when the engine develops 900 kW of power. This causes a
thrust of F = 1.23 MN on the shaft. If the shaft has an outer
diameter of 250 mm, determine the principal stresses at any
point located on the surface of the shaft.
0.75 m
A
T
Power Transmission: Using the formula developed in Chapter 5,
P = 900 kW = 0.900 A 106 B N # m>s
0.900(106)
P
=
= 60.0 A 103 B N # m
v
15
T0 =
Internal Torque and Force: As shown on FBD.
Section Properties:
A =
p
A 0.252 B = 0.015625p m2
4
J =
p
A 0.1254 B = 0.3835 A 10 - 3 B m4
2
s =
-1.23(106)
N
=
= -25.06 MPa
A
0.015625p
Normal Stress:
Shear Stress: Applying the torsion formula,
t =
60.0(103) (0.125)
Tc
= 19.56 MPa
=
J
0.3835(10 - 3)
In - Plane Principal Stresses: sx = -25.06 MPa, sy = 0 and txy = 19.56 MPa for
any point on the shaft’s surface. Applying Eq. 9-5,
s1,2 =
=
sx + sy
2
;
C
a
sx - sy
2
2
b + t2xy
-25.06 - 0 2
-25.06 + 0
;
a
b + (19.56)2
2
C
2
= -12.53 ; 23.23
s1 = 10.7 MPa
s2 = -35.8 MPa
Ans.
725
F
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•9–97. The solid propeller shaft on a ship extends outward
from the hull. During operation it turns at v = 15 rad>s
when the engine develops 900 kW of power. This causes a
thrust of F = 1.23 MN on the shaft. If the shaft has a
diameter of 250 mm, determine the maximum in-plane shear
stress at any point located on the surface of the shaft.
0.75 m
A
T
Power Transmission: Using the formula developed in Chapter 5,
P = 900 kW = 0.900 A 106 B N # m>s
T0 =
0.900(106)
P
=
= 60.0 A 103 B N # m
v
15
Internal Torque and Force: As shown on FBD.
Section Properties:
A =
p
A 0.252 B = 0.015625p m2
4
J =
p
A 0.1254 B = 0.3835 A 10 - 3 B m4
2
s =
-1.23(106)
N
=
= -25.06 MPa
A
0.015625p
Normal Stress:
Shear Stress: Applying the torsion formula.
t =
60.0(103) (0.125)
Tc
= 19.56 MPa
=
J
0.3835 (10 - 3)
Maximum In - Plane Principal Shear Stress: sx = -25.06 MPa, sy = 0, and
txy = 19.56 MPa for any point on the shaft’s surface. Applying Eq. 9-7,
t max
in-plane
a
sx - sy
2
b + t2xy
=
C
=
-25.06 - 0 2
b + (19.56)2
C
2
2
a
= 23.2 MPa
Ans.
726
F
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9–98. The steel pipe has an inner diameter of 2.75 in. and
an outer diameter of 3 in. If it is fixed at C and subjected to
the horizontal 20-lb force acting on the handle of the pipe
wrench at its end, determine the principal stresses in the
pipe at point A, which is located on the surface of the pipe.
20 lb
12 in.
10 in.
Internal Forces, Torque and Moment: As shown on FBD.
A
Section Properties:
B
I =
p
A 1.54 - 1.3754 B = 1.1687 in4
4
J =
p
A 1.54 - 1.3754 B = 2.3374 in4
2
C
y
z
(QA)z = ©y¿A¿
x
4(1.5) 1
4(1.375) 1
=
c p A 1.52 B d c p A 1.3752 B d
3p 2
3p
2
= 0.51693 in3
Normal Stress: Applying the flexure formula s =
sA =
My z
Iy
,
200(0)
= 0
1.1687
Shear Stress: The transverse shear stress in the z direction and the torsional shear
VQ
stress can be obtained using shear formula and torsion formula, tv =
and
It
Tr
ttwist =
, respectively.
J
tA = (tv)z - ttwist
=
20.0(0.51693)
240(1.5)
1.1687(2)(0.125)
2.3374
= -118.6 psi
In - Plane Principal Stress: sx = 0, sz = 0 and txz = -118.6 psi for point A.
Applying Eq. 9-5
s1,2 =
sx + sz
2
;
C
a
sx - sz
2
2
b + t2xz
= 0 ; 20 + (-118.6)2
s1 = 119 psi
s2 = -119 psi
Ans.
727
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9–99. Solve Prob. 9–98 for point B, which is located on the
surface of the pipe.
20 lb
12 in.
10 in.
A
B
Internal Forces, Torque and Moment: As shown on FBD.
Section Properties:
C
I =
p
A 1.54 - 1.3754 B = 1.1687 in4
4
y
z
x
p
J =
A 1.54 - 1.3754 B = 2.3374 in4
2
(QB)z = 0
Normal Stress: Applying the flexure formula s =
sB =
My z
Iv
,
200(1.5)
= 256.7 psi
1.1687
Shear Stress: Torsional shear stress can be obtained using torsion formula,
Tr
.
ttwist =
J
tB = ttwist =
240(1.5)
= 154.0 psi
2.3374
In - Plane Prinicipal Stress: sx = 256.7 psi, sy = 0, and txy = -154.0 psi for point B.
Applying Eq. 9-5
s1,2 =
=
sx + sy
2
;
C
sx - sy
a
2
2
b + t2xy
256.7 - 0 2
256.7 + 0
;
a
b + ( -154.0)2
2
C
2
= 128.35 ; 200.49
s1 = 329 psi
s2 = -72.1 psi
Ans.
728
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*9–100. The clamp exerts a force of 150 lb on the boards
at G. Determine the axial force in each screw, AB and CD,
and then compute the principal stresses at points E and F.
Show the results on properly oriented elements located at
these points. The section through EF is rectangular and is
1 in. wide.
A
150 lb
C
G
0.5 in.
E
Support Reactions: FBD(a).
F
a + ©MB = 0;
FCD(3) - 150(7) = 0
FCD = 350 lb
Ans.
+ c ©Fy = 0;
350 - 150 - FAB = 0
FAB = 200 lb
Ans.
B
1.5 in. 1.5 in.
Internal Forces and Moment: As shown on FBD(b).
Section Properties:
I =
1
(1) A 1.53 B = 0.28125 in4
12
QE = 0
QF = y¿A¿ = 0.5(0.5)(1) = 0.250 in3
Normal Stress: Applying the flexure formula s = -
My
,
I
sE = -
-300(0.75)
= 800 psi
0.28125
sF = -
-300(0.25)
= 266.67 psi
0.28125
VQ
,
It
Shear Stress: Applying the shear formula t =
tE =
200(0)
= 0
0.28125(1)
tF =
200(0.250)
= 177.78 psi
0.28125(1)
In - Plane Principal Stress: sx = 800 psi, sy = 0 and txy = 0 for point E. Since no
shear stress acts upon the element.
s1 = sx = 800 psi
Ans.
s2 = sy = 0
Ans.
sx = 266.67 psi, sy = 0, and txy = 177.78 psi for point F. Applying Eq. 9-5
s1,2 =
=
sx + sy
2
;
C
sx - sy
a
2
2
b + t2xy
266.67 - 0 2
266.67 + 0
;
a
b + 177.782
2
C
2
= 133.33 ; 222.22
s1 = 356 psi
s2 = -88.9 psi
Ans.
729
150 lb
D
4 in.
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9–100.
Continued
Orientation of Principal Plane: Applying Eq. 9-4 for point F,
tan 2up =
txy
A sx - sy B >2
up = 26.57°
=
and
177.78
= 1.3333
(266.67 - 0)>2
-63.43°
Substituting the results into Eq. 9-1 with u = 26.57° yields
sx¿ =
=
sx + sy
2
sx - sy
+
2
cos 2u + txy sin 2u
266.67 - 0
266.67 + 0
+
cos 53.13° + 177.78 sin 53.13°
2
2
= 356 psi = s1
Hence,
up1 = 26.6°
up2 = -63.4°
Ans.
730
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9–101. The shaft has a diameter d and is subjected to the
loadings shown. Determine the principal stress and
the maximum in-plane shear stress that is developed
anywhere on the surface of the shaft.
F
T0
F
T0
Internal Forces and Torque: As shown on FBD(b).
Section Properties:
A =
p 2
d
4
J =
p d 4
p 4
a b =
d
2 2
32
Normal Stress:
N
-F
4F
= p 2 = - 2
A
pd
4 d
s =
Shear Stress: Applying the shear torsion formula,
T0 A 2 B
16T0
Tc
= p 4 =
J
d
pd3
32
d
t =
16T0
4F
, sy = 0, and txy = for any point on
pd2
pd3
the shaft’s surface. Applying Eq. 9-5,
In - Plane Principal Stress: sx = -
s1,2 =
sx + sy
;
2
C
a
- 4F2 + 0
pd
=
;
2
D
sx - sy
2
¢
2
b + t2xy
- 4F2 - 0
pd
2
2
≤ + a-
16T0
3
pd
b
2
=
64T20
2
-F ;
F2 +
≤
2 ¢
C
pd
d2
s1 =
64T20
2
-F +
F2 +
≤
2 ¢
C
pd
d2
Ans.
64T20
2
F +
F2 +
≤
2 ¢
C
pd
d2
Ans.
s2 = -
Maximum In - Plane Shear Stress: Applying Eq. 9-7,
t max
in-plane
=
=
=
C
a
D
¢
sx - sy
2
- 4F2 - 0
pd
2
2
b + t2xy
2
≤ + a-
b
pd3
16T0
2
64T20
2
2
F
+
pd2 C
d2
Ans.
731
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9–102. The state of stress at a point in a member is shown
on the element. Determine the stress components acting on
the plane AB.
A
50 MPa
30⬚
28 MPa
100 MPa
B
Construction of the Circle: In accordance with the sign convention, sx = -50 MPa,
sy = -100 MPa, and txy = -28 MPa. Hence,
savg =
sx + sy
2
=
-50 + (-100)
= -75.0 MPa
2
The coordinates for reference points A and C are A(–50, –28) and C(–75.0, 0).
The radius of the circle is R = 2(75.0 - 50)2 + 282 = 37.54 MPa.
Stress on the Rotated Element: The normal and shear stress components
A sx¿ and tx¿y¿ B are represented by the coordinates of point P on the circle
sx¿ = -75.0 + 37.54 cos 71.76° = -63.3 MPa
Ans.
tx¿y¿ = 37.54 sin 71.76° = 35.7 MPa
Ans.
732
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9–103. The propeller shaft of the tugboat is subjected to
the compressive force and torque shown. If the shaft has an
inner diameter of 100 mm and an outer diameter of 150 mm,
determine the principal stress at a point A located on the
outer surface.
10 kN
A
2 kN·m
Internal Loadings: Considering the equilibrium of the free - body diagram of the
propeller shaft’s right segment, Fig. a,
©Fx = 0; 10 - N = 0
N = 10 kN
©Mx = 0; T - 2 = 0
T = 2 kN # m
Section Properties: The cross - sectional area and the polar moment of inertia of the
propeller shaft’s cross section are
A = p A 0.0752 - 0.052 B = 3.125p A 10 - 3 B m2
J =
p
A 0.0754 - 0.054 B = 12.6953125p A 10 - 6 B m4
2
Normal and Shear Stress: The normal stress is a contributed by axial stress only.
sA =
10 A 103 B
N
= = -1.019 MPa
A
3.125p A 10 - 3 B
The shear stress is contributed by the torsional shear stress only.
tA =
2 A 103 B (0.075)
Tc
=
= 3.761 MPa
J
12.6953125p A 10 - 6 B
The state of stress at point A is represented by the element shown in Fig. b.
Construction of the Circle: sx = -1.019 MPa, sy = 0, and txy = -3.761 MPa.
Thus,
savg =
sx + sy
2
=
-1.019 + 0
= -0.5093 MPa
2
The coordinates of reference point A and the center C of the circle are
A(-1.019, -3.761)
C(-0.5093, 0)
Thus, the radius of the circle is
R = CA = 2[-1.019 - ( -0.5093)]2 + (-3.761)2 = 3.795 MPa
Using these results, the circle is shown is Fig. c.
In - Plane Principal Stress: The coordinates of reference points B and D represent
s1 and s2, respectively.
s1 = -0.5093 + 3.795 = 3.29 MPa
Ans.
s2 = -0.5093 - 3.795 = -4.30 MPa
Ans.
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9–103.
Continued
Orientation of the Principal Plane: Referring to the geometry of the circle, Fig. d,
tan 2 A up B 2 =
3.761
= 7.3846
1.019 - 0.5093
A up B 2 = 41.1° (clockwise)
Ans.
The state of principal stresses is represented on the element shown in Fig. d.
734
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*9–104. The box beam is subjected to the loading shown.
Determine the principal stress in the beam at points A and B.
6 in.
A
6 in. B 8 in.
8 in.
Support Reactions: As shown on FBD(a).
Internal Forces and Moment: As shown on FBD(b).
Section Properties:
I =
1
1
(8) A 83 B (6) A 63 B = 233.33 in4
12
12
QA = QB = 0
Normal Stress: Applying the flexure formula.
s = -
My
I
sA = -
-300(12)(4)
= 61.71 psi
233.33
sB = -
-300(12)(-3)
= -46.29 psi
233.33
1200 lb
800 lb
Shear Stress: Since QA = QB = 0, then tA = tB = 0.
In - Plane Principal Stress: sx = 61.71 psi, sy = 0, and txy = 0 for point A. Since
no shear stress acts on the element,
s1 = sx = 61.7 psi
Ans.
s2 = sy = 0
Ans.
sx = -46.29 psi, sy = 0, and txy = 0 for point B. Since no shear stress acts on the
element,
s1 = sy = 0
Ans.
s2 = sx = -46.3 psi
Ans.
735
A
B
3 ft
2.5 ft
2.5 ft
5 ft
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•9–105.
The wooden strut is subjected to the loading
shown. Determine the principal stresses that act at point C
and specify the orientation of the element at this point.
The strut is supported by a bolt (pin) at B and smooth
support at A.
50 N
50 N
60⬚
C
100 mm
40 N
40 N
B
A
25 mm
50 mm
200 mm
200 mm 200 mm 200 mm
100 mm 100 mm
QC = y¿A¿ = 0.025(0.05)(0.025) = 31.25(10 - 6) m3
I =
1
(0.025)(0.13) = 2.0833(10 - 6) m4
12
Normal stress: sC = 0
Shear stress:
t =
VQC
44(31.25)(10 - 6)
= 26.4 kPa
=
It
2.0833(10 - 6)(0.025)
Principal stress:
sx = sy = 0;
s1,2 =
txy = -26.4 kPa
sx + sy
;
2
C
a
sx - sy
2
2
b + t2 xy
= 0 ; 20 + (26.4)2
s1 = 26.4 kPa
s2 = -26.4 kPa
;
Ans.
Orientation of principal stress:
tan 2up =
txy
(sx - sy)
= - q
2
up = +45° and -45°
Use
Eq. 9-1 to determine the principal
sx + sy
sx - sy
+
cos 2u + txy sin 2u
sx¿ =
2
2
plane
of
s1
and
s2
u = up = -45°
sx¿ = 0 + 0 + (-26.4) sin( -90°) = 26.4 kPa
Therefore, up1 = -45°;
up2 = 45°
Ans.
736
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9–106. The wooden strut is subjected to the loading shown.
If grains of wood in the strut at point C make an angle of 60°
with the horizontal as shown, determine the normal and
shear stresses that act perpendicular and parallel to the
grains, respectively, due to the loading. The strut is supported
by a bolt (pin) at B and smooth support at A.
50 N
50 N
60⬚
C
100 mm
40 N
40 N
B
A
25 mm
50 mm
200 mm
200 mm 200 mm 200 mm
100 mm 100 mm
QC = y¿A¿ = 0.025(0.05)(0.025) = 31.25(10 - 6) m3
I =
1
(0.025)(0.13) = 2.0833(10 - 6) m4
12
Normal stress: sC = 0
Shear stress:
t =
VQC
44(31.25)(10 - 6)
= 26.4 kPa
=
It
2.0833(10 - 6)(0.025)
Stress transformation: sx = sy = 0;
sx¿ =
sx + sy
sx - sy
+
2
2
txy = -26.4 kPa;
u = 30°
cos 2u + txy sin 2u
= 0 + 0 + (-26.4) sin 60° = -22.9 kPa
tx¿y¿ = -
sx - sy
2
Ans.
sin 2u + txy cos 2u
= -0 + (-26.4) cos 60° = -13.2 kPa
Ans.
737
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10–1. Prove that the sum of the normal strains in
perpendicular directions is constant.
ex¿ =
ey¿ =
ex + ey
2
ex - ey
+
ex + ey
2
2
ex - ey
-
2
cos 2u +
cos 2u -
gxy
2
gxy
2
sin 2u
(1)
sin 2u
(2)
Adding Eq. (1) and Eq. (2) yields:
ex¿ + ey¿ = ex + ey = constant
QED
738
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10–2. The state of strain at the point has components
of Px = 200 110-62, Py = -300 110-62, and gxy = 400(10-62.
Use the strain-transformation equations to determine the
equivalent in-plane strains on an element oriented at an
angle of 30° counterclockwise from the original position.
Sketch the deformed element due to these strains within the
x–y plane.
y
x
In accordance to the established sign convention,
ex = 200(10 - 6),
ex¿ =
ex + ey
ex - ey
+
2
= c
ey = -300(10 - 6)
2
cos 2u +
gxy
2
gxy = 400(10 - 6)
u = 30°
sin 2u
200 - (-300)
200 + (-300)
400
+
cos 60° +
sin 60° d(10 - 6)
2
2
2
= 248 (10 - 6)
gx¿y¿
2
= -a
Ans.
ex - ey
2
b sin 2u +
gxy
2
cos 2u
gx¿y¿ = e - C 200 - ( -300) D sin 60° + 400 cos 60° f(10 - 6)
= -233(10 - 6)
ey¿ =
ex + ey
= c
2
Ans.
ex - ey
-
2
cos 2u -
gxy
2
sin 2u
200 - ( -300)
200 + (-300)
400
cos 60° sin 60° d(10 - 6)
2
2
2
= -348(10 - 6)
Ans.
The deformed element of this equivalent state of strain is shown in Fig. a
739
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10–3. A strain gauge is mounted on the 1-in.-diameter
A-36 steel shaft in the manner shown. When the shaft is
rotating with an angular velocity of v = 1760 rev>min, the
reading on the strain gauge is P = 800110-62. Determine
the power output of the motor. Assume the shaft is only
subjected to a torque.
v = (1760 rev>min)a
60⬚
2p rad
1 min
ba
b = 184.307 rad>s
60 sec
1 rev
ex = ey = 0
ex¿ =
ex + ey
2
ex - ey
+
2
800(10 - 6) = 0 + 0 +
cos 2u +
gxy
2
gxy
2
sin 2u
sin 120°
gxy = 1.848(10 - 3) rad
t = G gxy = 11(103)(1.848)(10 - 3) = 20.323 ksi
t =
Tc
;
J
T(0.5)
20.323 = p
;
4
2 (0.5)
T = 3.99 kip # in = 332.5 lb # ft
P = Tv = 0.332.5 (184.307) = 61.3 kips # ft>s = 111 hp
Ans.
740
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*10–4. The state of strain at a point on a wrench
has components Px = 120110-62, Py = -180110-62, gxy =
150110-62. Use the strain-transformation equations to
determine (a) the in-plane principal strains and (b) the
maximum in-plane shear strain and average normal strain.
In each case specify the orientation of the element and show
how the strains deform the element within the x–y plane.
ex = 120(10 - 6)
e1, 2 =
a)
ey = -180(10 - 6)
gxy = 150(10 - 6)
Ex - Ey 2
ex + ey
gxy 2
;
a
b + a
b
2
A
2
2
120 + (-180)
120 - ( -180) 2
150 2
-6
;
a
b + a
b d 10
2
A
2
2
e1 = 138(10 - 6);
e2 = -198(10 - 6)
= c
Ans.
Orientation of e1 and e2
gxy
150
=
= 0.5
tan 2up =
ex - ey
[120 - (-180)]
up = 13.28° and -76.72°
Use Eq. 10.5 to determine the direction of e1 and e2
ex¿ =
ex + ey
ex - ey
+
2
2
cos 2u +
gxy
2
sin 2u
u = up = 13.28°
ex¿ = c
120 + ( -180)
120 - ( -180)
150
+
cos (26.56°) +
sin 26.56° d 10 - 6
2
2
2
= 138 (10 - 6) = e1
Therefore up1 = 13.3° ;
gmax
b)
a
in-plane
=
2
A
up2 = -76.7°
ex - ey
b + a
2
gxy
b
Ans.
2
2
2
150 2
120 - ( -180) 2
-6
-6
= 2c a
b + a
b d10 = 335 (10 )
in-plane
2
2
A
gmax
eavg =
ex + ey
2
= c
120 + (-180)
d 10 - 6 = -30.0(10 - 6)
2
Ans.
Ans.
Orientation of gmax
tan 2us =
-(ex - ey)
gxy
=
-[120 - ( -180)]
= -2.0
150
us = -31.7° and 58.3°
Ans.
gmax
Use Eq. 10–6 to determine the sign of in-plane
gx¿y¿
ex - ey
gxy
= sin 2u +
cos 2u
2
2
2
u = us = -31.7°
gx¿y¿ = 2 c -
120 - (-180)
150
sin (-63.4°) +
cos (-63.4°) d10 - 6 = 335(10 - 6)
2
2
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10–5. The state of strain at the point on the arm
has components Px = 250110-62, Py = -450110-62, gxy =
-825110-62. Use the strain-transformation equations to
determine (a) the in-plane principal strains and (b) the
maximum in-plane shear strain and average normal strain.
In each case specify the orientation of the element and show
how the strains deform the element within the x–y plane.
ex = 250(10 - 6)
ey = -450(10 - 6)
y
gxy = -825(10 - 6)
x
a)
ex + ey
e1, 2 =
;
2
= c
A
ex - ey
a
2
2
b + a
gxy
2
b
2
250 - 450
250 - ( -450) 2
-825 2
-6
;
a
b + a
b d(10 )
2
A
2
2
e1 = 441(10 - 6)
Ans.
e2 = -641(10 - 6)
Ans.
Orientation of e1 and e2 :
gxy
tan 2up =
ex - ey
up = -24.84°
-825
250 - ( -450)
=
up = 65.16°
and
Use Eq. 10–5 to determine the direction of e1 and e2:
ex¿ =
ex + ey
ex - ey
+
2
2
cos 2u +
gxy
2
sin 2u
u = up = -24.84°
ex¿ = c
250 - (-450)
250 - 450
-825
+
cos (-49.69°) +
sin (-49.69°) d(10 - 6) = 441(10 - 6)
2
2
2
Therefore, up1 = -24.8°
Ans.
up2 = 65.2°
Ans.
b)
g
max
in-plane
2
=
A
a
ex - ey
max = 2c
a
in-plane
A
g
eavg =
ex + ey
2
2
b + a
2
gxy
2
b
2
250 - (-450) 2
-825 2
-6
-3
b + a
b d(10 ) = 1.08(10 )
2
2
Ans.
= a
Ans.
250 - 450
b (10 - 6) = -100(10 - 6)
2
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10–6. The state of strain at the point has components of
Px = -100110-62, Py = 400110-62, and gxy = -300110-62.
Use the strain-transformation equations to determine the
equivalent in-plane strains on an element oriented at an
angle of 60° counterclockwise from the original position.
Sketch the deformed element due to these strains within
the x–y plane.
y
x
In accordance to the established sign convention,
ex = -100(10 - 6)
ex¿ =
ex + ey
= c
ex - ey
+
2
ey = 400(10 - 6)
2
gxy
cos 2u +
2
gxy = -300(10 - 6)
u = 60°
sin 2u
-100 - 400
-300
-100 + 400
+
cos 120° +
sin 120° d(10 - 6)
2
2
2
= 145(10 - 6)
gx¿y¿
2
= -a
Ans.
ex - ey
2
b sin 2u +
gxy
2
cos 2u
gx¿y¿ = c -(-100 - 400) sin 120° + (-300) cos 120° d(10 - 6)
= 583(10 - 6)
ey¿ =
ex + ey
= c
2
Ans.
ex - ey
-
2
cos 2u -
gxy
2
sin 2u
-100 - 400
-300
-100 + 400
cos 120° sin 120° d(10 - 6)
2
2
2
= 155 (10 - 6)
Ans.
The deformed element of this equivalent state of strain is shown in Fig. a
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10–7. The state of strain at the point has components of
Px = 100110-62, Py = 300110-62, and gxy = -150110-62.
Use the strain-transformation equations to determine the
equivalent in-plane strains on an element oriented u = 30°
clockwise. Sketch the deformed element due to these
strains within the x–y plane.
y
x
In accordance to the established sign convention,
ex = 100(10 - 6)
ex¿ =
ex + ey
ex - ey
+
2
= c
ey = 300(10 - 6)
2
cos 2u +
gxy = -150(10 - 6)
gxy
2
u = -30°
sin 2u
100 - 300
-150
100 + 300
+
cos (-60°) +
sin ( -60°) d (10 - 6)
2
2
2
= 215(10 - 6)
gx¿y¿
2
= -a
Ans.
ex - ey
2
b sin 2u +
gxy
2
cos 2u
gx¿y¿ = c -(100 - 300) sin ( -60°) + ( -150) cos ( -60°) d(10 - 6)
= -248 (10 - 6)
ey¿ =
ex + ey
= c
2
Ans.
ex - ey
-
2
cos 2u -
gxy
2
sin 2u
100 - 300
-150
100 + 300
cos ( -60°) sin (-60°) d (10 - 6)
2
2
2
= 185(10 - 6)
Ans.
The deformed element of this equivalent state of strain is shown in Fig. a
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*10–8. The state of strain at the point on the bracket
has components Px = -200110-62, Py = -650110-62, gxy ⫽
-175110-62. Use the strain-transformation equations to
determine the equivalent in-plane strains on an element
oriented at an angle of u = 20° counterclockwise from the
original position. Sketch the deformed element due to these
strains within the x–y plane.
ex = -200(10 - 6)
ex¿ =
ex + ey
ex - ey
+
2
= c
ey = -650(10 - 6)
2
cos 2u +
gxy
2
y
x
gxy = -175(10 - 6)
u = 20°
sin 2u
( -200) - (-650)
(-175)
-200 + (-650)
+
cos (40°) +
sin (40°) d(10 - 6)
2
2
2
= -309(10 - 6)
ey¿ =
ex + ey
ex - ey
-
2
= c
Ans.
2
cos 2u -
gxy
2
sin 2u
-200 - ( -650)
( -175)
-200 + (-650)
cos (40°) sin (40°) d(10 - 6)
2
2
2
= -541(10 - 6)
gx¿y¿
2
ex - ey
= -
2
Ans.
sin 2u +
gxy
2
cos 2u
gx¿y¿ = [-(-200 - (-650)) sin (40°) + (-175) cos (40°)](10 - 6)
= -423(10 - 6)
Ans.
745
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10–9. The state of strain at the point has components of
Px = 180110-62, Py = -120110-62, and gxy = -100110-62.
Use the strain-transformation equations to determine (a)
the in-plane principal strains and (b) the maximum in-plane
shear strain and average normal strain. In each case specify
the orientation of the element and show how the strains
deform the element within the x–y plane.
y
x
a)
In
accordance
to
the
established
sign
convention,
ex = 180(10 - 6),
ey = -120(10 - 6) and gxy = -100(10 - 6).
ex + ey
e1, 2 =
;
2
= b
a
A
ex - ey
2
2
b + a
gxy
2
b
2
180 + (-120)
180 - ( -120) 2
-100 2
-6
;
c
d + a
b r (10 )
2
A
2
2
= A 30 ; 158.11 B (10 - 6)
e1 = 188(10 - 6)
tan 2uP =
e2 = -128(10 - 6)
gxy
Ans.
-100(10 - 6)
ex - ey
C 180 - (-120) D (10 - 6)
=
uP = -9.217°
and
= -0.3333
80.78°
Substitute u = -9.217°,
ex + ey
ex¿ =
2
= c
ex - ey
+
2
cos 2u +
gxy
2
sin 2u
180 + ( -120)
180 - ( -120)
-100
+
cos (-18.43°) +
sin (-18.43) d(10 - 6)
2
2
2
= 188(10 - 6) = e1
Thus,
(uP)1 = -9.22°
(uP)2 = 80.8°
Ans.
The deformed element is shown in Fig (a).
gmax
ex - ey 2
gxy 2
in-plane
=
b)
a
b + a
b
2
A
2
2
180 - (-120) 2
-100 2
-6
-6
d + a
b r (10 ) = 316 A 10 B
A
2
2
gmax
= b2
tan 2us = - a
ex - ey
in-plane
gxy
c
b = -c
C 180 - (-120) D (10 - 6)
us = 35.78° = 35.8° and
-100(10 - 6)
Ans.
s = 3
Ans.
-54.22° = -54.2°
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10–9.
Continued
gmax
The algebraic sign for in-plane
when u = 35.78°.
ex - ey
gxy
gx¿y¿
= -a
b sin 2u +
cos 2u
2
2
2
gx¿y¿ = e - C 180 - ( -120) D sin 71.56° + ( -100) cos 71.56° f(10 - 6)
= -316(10 - 6)
ex + ey
180 + (-120)
eavg =
= c
d(10 - 6) = 30(10 - 6)
2
2
Ans.
The deformed element for the state of maximum In-plane shear strain is shown is
shown in Fig. b
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10–10. The state of strain at the point on the bracket
has components Px = 400110-62, Py = -250110-62, gxy ⫽
310110-62. Use the strain-transformation equations to
determine the equivalent in-plane strains on an element
oriented at an angle of u = 30° clockwise from the original
position. Sketch the deformed element due to these strains
within the x–y plane.
ex = 400(10 - 6)
ex¿ =
ex + ey
= c
ex - ey
+
2
ey = -250(10 - 6)
2
cos 2u +
gxy
2
gxy = 310(10 - 6)
y
x
u = -30°
sin 2u
400 - ( -250)
400 + ( -250)
310
+
cos (-60°) + a
b sin (-60°) d(10 - 6)
2
2
2
= 103(10 - 6)
ey¿ =
ex + ey
= c
ex - ey
-
2
Ans.
2
cos 2u -
gxy
2
sin 2u
400 - (-250)
400 + (-250)
310
cos (60°) sin (-60°) d(10 - 6)
2
2
2
= 46.7(10 - 6)
gx¿y¿
2
ex - ey
= -
2
Ans.
sin 2u +
gxy
2
cos 2u
gx¿y¿ = [-(400 - (-250)) sin (-60°) + 310 cos ( -60°)](10 - 6) = 718(10 - 6)
748
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10–11. The state of strain at the point has components of
Px = -100110-62, Py = -200110-62, and gxy = 100110-62.
Use the strain-transformation equations to determine (a)
the in-plane principal strains and (b) the maximum in-plane
shear strain and average normal strain. In each case specify
the orientation of the element and show how the strains
deform the element within the x–y plane.
In accordance to the established
ey = -200(10 - 6) and gxy = 100(10 - 6).
ex + ey
e1, 2 =
;
2
= b
A
a
ex - ey
2
b + a
2
gxy
2
b
sign
y
x
convention,
ex = -100(10 - 6),
2
-100 + (-200)
100 2
-100 - (-200) 2
-6
;
c
d + a
b r (10 )
2
A
2
2
= A -150 ; 70.71 B (10 - 6)
e1 = -79.3(10 - 6)
tan 2uP =
e2 = -221(10 - 6)
gxy
100(10 - 6)
C -100 - (-200) D (10 - 6)
=
ex - ey
uP = 22.5°
and
Ans.
= 1
-67.5°
Substitute u = 22.5,
ex + ey
ex¿ =
ex - ey
cos 2u +
gxy
sin 2u
2
2
2
-100 + (-200)
-100 - (-200)
100
+
cos 45° +
sin 45° d(10 - 6)
= c
2
2
2
+
= -79.3(10 - 6) = e1
Thus,
(uP)1 = 22.5°
(uP)2 = -67.5°
Ans.
The deformed element of the state of principal strain is shown in Fig. a
gmax
ex - ey 2
gxy 2
in-plane
=
a
b + a
b
2
A
2
2
gmax
in-plane
= b2
tan 2us = - a
c
-100 - (-200) 2
100 2
-6
-6
d + a
b r (10 ) = 141(10 )
A
2
2
ex - ey
gxy
b = -c
us = -22.5°
and
The algebraic sign for
gx¿y¿
2
= -a
ex - ey
2
gmax
in-plane
b sin 2u +
C -100 - ( -200) D (10 - 6)
100(10 - 6)
Ans.
s = -1
Ans.
67.5°
when u = -22.5°.
gxy
2
cos 2u
gx¿y¿ = - C -100 - (-200) D sin ( -45°) + 100 cos (-45°)
= 141(10 - 6)
ex + ey
-100 + ( -200)
eavg =
= c
d(10 - 6) = -150(10 - 6)
2
2
Ans.
The deformed element for the state of maximum In-plane shear strain is shown in
Fig. b.
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10–11.
Continued
*10–12. The state of plane strain on an element is given by
Px = 500110-62, Py = 300110-62, and gxy = -200110-62.
Determine the equivalent state of strain on an element at
the same point oriented 45° clockwise with respect to the
original element.
y
Pydy
dy gxy
2
Strain Transformation Equations:
ex = 500 A 10 - 6 B
ey = 300 A 10 - 6 B
gxy = -200 A 10 - 6 B
u = -45°
We obtain
ex¿ =
ex + ey
+
2
= c
ex - ey
2
cos 2u +
gxy
2
sin 2u
500 - 300
-200
500 + 300
+
cos (-90°) + a
b sin (-90°) d A 10 - 6 B
2
2
2
= 500 A 10 - 6 B
gx¿y¿
2
= -a
Ans.
ex - ey
2
b sin 2u +
gxy
2
cos 2u
gx¿y¿ = [-(500 - 300) sin ( -90°) + (-200) cos ( -90°)] A 10 - 6 B
= 200 A 10 - 6 B
ey¿ =
ex + ey
= c
2
Ans.
ex - ey
-
2
cos 2u -
gxy
2
sin 2u
500 + 300
500 - 300
-200
cos ( -90°) - a
b sin (-90°) d A 10 - 6 B
2
2
2
= 300 A 10 - 6 B
Ans.
The deformed element for this state of strain is shown in Fig. a.
750
gxy
2
dx
x
Pxdx
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10–13. The state of plane strain on an element is
Px = -300110-62, Py = 0, and gxy = 150110-62. Determine
the equivalent state of strain which represents (a) the
principal strains, and (b) the maximum in-plane shear strain
and the associated average normal strain. Specify the
orientation of the corresponding elements for these states
of strain with respect to the original element.
y
gxy
dy 2
x
In-Plane Principal Strains: ex = -300 A 10 - 6 B , ey = 0, and gxy = 150 A 10 - 6 B . We
obtain
ex + ey
e1, 2 =
;
2
= C
C
¢
ex - ey
2
2
≤ + ¢
gxy
2
≤
2
-300 + 0
-300 - 0 2
150 2
;
¢
≤ + ¢
≤ S A 10 - 6 B
2
C
2
2
= ( -150 ; 167.71) A 10 - 6 B
e1 = 17.7 A 10 - 6 B
e2 = -318 A 10 - 6 B
Ans.
Orientation of Principal Strain:
tan 2up =
gxy
ex - ey
=
150 A 10 - 6 B
(-300 - 0) A 10 - 6 B
= -0.5
uP = -13.28° and 76.72°
Substituting u = -13.28° into Eq. 9-1,
ex¿ =
ex + ey
= c
ex - ey
+
2
2
cos 2u +
gxy
2
sin 2u
-300 + 0
-300 - 0
150
+
cos (-26.57°) +
sin (-26.57°) d A 10 - 6 B
2
2
2
= -318 A 10 - 6 B = e2
Thus,
A uP B 1 = 76.7° and A uP B 2 = -13.3°
Ans.
The deformed element of this state of strain is shown in Fig. a.
Maximum In-Plane Shear Strain:
gmax
ex - ey 2
gxy 2
in-plane
=
¢
≤ + ¢ ≤
2
C
2
2
gmax
in-plane
-300 - 0 2
150 2
-6
-6
b + a
b R A 10 B = 335 A 10 B
A
2
2
= B2
a
Ans.
Orientation of the Maximum In-Plane Shear Strain:
tan 2us = - ¢
ex - ey
gxy
≤ = -C
(-300 - 0) A 10 - 6 B
150 A 10 - 6 B
S = 2
us = 31.7° and 122°
Ans.
751
gxy
2
dx
Pxdx
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10–13.
Continued
The algebraic sign for
gx¿y¿
2
= -¢
ex - ey
2
gmax
in-plane
≤ sin 2u +
when u = us = 31.7° can be obtained using
gxy
2
cos 2u
gx¿y¿ = [-(-300 - 0) sin 63.43° + 150 cos 63.43°] A 10 - 6 B
= 335 A 10 - 6 B
Average Normal Strain:
eavg =
ex + ey
2
= a
-300 + 0
b A 10 - 6 B = -150 A 10 - 6 B
2
Ans.
The deformed element for this state of strain is shown in Fig. b.
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10–14. The state of strain at the point on a boom of an
hydraulic engine crane has components of Px = 250110-62,
Py = 300110-62, and gxy = -180110-62. Use the straintransformation equations to determine (a) the in-plane
principal strains and (b) the maximum in-plane shear strain
and average normal strain. In each case, specify the
orientation of the element and show how the strains deform
the element within the x–y plane.
y
a)
In-Plane Principal Strain: Applying Eq. 10–9,
ex + ey
e1, 2 =
;
2
= B
a
A
ex - ey
2
b + a
2
gxy
2
b
2
250 - 300 2
250 + 300
-180 2
-6
;
a
b + a
b R A 10 B
2
A
2
2
= 275 ; 93.41
e1 = 368 A 10 - 6 B
e2 = 182 A 10 - 6 B
Ans.
Orientation of Principal Strain: Applying Eq. 10–8,
gxy
tan 2uP =
-180(10 - 6)
ex - ey
=
(250 - 300)(10 - 6)
uP = 37.24°
and
= 3.600
-52.76°
Use Eq. 10–5 to determine which principal strain deforms the element in the x¿
direction with u = 37.24°.
ex¿ =
ex + ey
= c
2
ex - ey
+
2
cos 2u +
gxy
2
sin 2u
250 + 300
250 - 300
-180
+
cos 74.48° +
sin 74.48° d A 10 - 6 B
2
2
2
= 182 A 10 - 6 B = e2
Hence,
uP1 = -52.8°
and
uP2 = 37.2°
Ans.
b)
Maximum In-Plane Shear Strain: Applying Eq. 10–11,
g max
ex - ey 2
gxy 2
in-plane
=
a
b + a
b
2
A
2
2
g
max = 2 B
a
in-plane
A
-180 2
250 - 300 2
-6
b + a
b R A 10 B
2
2
= 187 A 10 - 6 B
Ans.
753
x
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10–14.
Continued
Orientation of the Maximum In-Plane Shear Strain: Applying Eq. 10–10,
tan 2us = -
ex - ey
us = -7.76°
2
ex - ey
= -
2
250 - 300
= -0.2778
-180
and
82.2°
g max
can be determined by substituting u = -7.76° into Eq. 10–6.
The proper sign of
gx¿y¿
= -
gxy
in-plane
sin 2u +
gxy
2
Ans.
cos 2u
gx¿y¿ = {-[250 - 300] sin (-15.52°) + (-180) cos (-15.52°)} A 10 - 6 B
= -187 A 10 - 6 B
Normal Strain and Shear strain: In accordance with the sign convention,
ex = 250 A 10 - 6 B
ey = 300 A 10 - 6 B
gxy = -180 A 10 - 6 B
Average Normal Strain: Applying Eq. 10–12,
eavg =
ex + ey
2
= c
250 + 300
d A 10 - 6 B = 275 A 10 - 6 B
2
Ans.
754
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*10–16. The state of strain at a point on a support
has components of Px = 350110-62, Py = 400110-62,
gxy = -675110-62. Use the strain-transformation equations
to determine (a) the in-plane principal strains and (b) the
maximum in-plane shear strain and average normal strain.
In each case specify the orientation of the element and show
how the strains deform the element within the x–y plane.
a)
e1, 2 =
=
ex + ey
;
2
B
a
ex -ey
2
b + a
2
gxy
2
b
2
350 - 400 2
-675 2
350 + 400
;
a
b + a
b
2
A
2
2
e1 = 713(10 - 6)
Ans.
e2 = 36.6(10 - 6)
Ans.
tan 2uP =
gxy
ex - ey
=
-675
(350 - 400)
uP = 42.9°
Ans.
b)
(gx¿y¿)max
=
2
(gx¿y¿)max
=
2
A
a
ex - ey
2
b + a
2
gxy
2
b
2
a
350 - 400 2
-675 2
b + a
b
A
2
2
(gx¿y¿)max = 677(10 - 6)
eavg =
ex + ey
tan 2us =
2
=
Ans.
350 + 400
= 375(10 - 6)
2
-(ex - ey)
gxy
=
Ans.
350 - 400
675
us = -2.12°
Ans.
755
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•10–17.
Solve part (a) of Prob. 10–4 using Mohr’s circle.
ex = 120(10 - 6)
ey = -180(10 - 6)
gxy = 150(10 - 6)
A (120, 75)(10 - 6) C (-30, 0)(10 - 6)
R = C 2[120 - (-30)]2 + (75)2 D (10 - 6)
= 167.71 (10 - 6)
e1 = (-30 + 167.71)(10 - 6) = 138(10 - 6)
Ans.
e2 = (-30 - 167.71)(10 - 6) = -198(10 - 6)
Ans.
75
tan 2uP = a
b , uP = 13.3°
30 + 120
Ans.
756
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10–18.
Solve part (b) of Prob. 10–4 using Mohr’s circle.
ex = 120(10 - 6)
ey = -180(10 - 6)
gxy = 150(10 - 6)
A (120, 75)(10 - 6) C (-30, 0)(10 - 6)
R = C 2[120 - (-30)]2 + (75)2 D (10 - 6)
= 167.71 (10 - 6)
gxy
max
2 in-plane
gxy
= R = 167.7(10 - 6)
max
in-plane
= 335(10 - 6)
Ans.
eavg = -30 (10 - 6)
tan 2us =
120 + 30
75
Ans.
us = -31.7°
Ans.
757
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10–19.
Solve Prob. 10–8 using Mohr’s circle.
ex = -200(10 - 6)
ey = -650(10 - 6)
gxy = -175(10 - 6)
gxy
2
= -87.5(10 - 6)
u = 20°, 2u = 40°
A(-200, -87.5)(10 - 6)
C(-425, 0)(10 - 6)
R = [2(-200 - (-425))2 + 87.52 ](10 - 6) = 241.41(10 - 6)
tan a =
87.5
;
-200 - (-425)
a = 21.25°
f = 40 + 21.25 = 61.25°
ex¿ = (-425 + 241.41 cos 61.25°)(10 - 6) = -309(10 - 6)
Ans.
ey¿ = (-425 - 241.41 cos 61.25°)(10 - 6) = -541(10 - 6)
Ans.
-gx¿y¿
2
= 241.41(10 - 6) sin 61.25°
gx¿y¿ = -423(10 - 6)
Ans.
758
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*10–20.
Solve Prob. 10–10 using Mohr’s circle.
ex = 400(10 - 6)
ey = -250(10 - 6)
A(400, 155)(10 - 6)
C(75, 0)(10 - 6)
gxy = 310(10 - 6)
gxy
2
= 155(10 - 6)
R = [2(400 - 75)2 + 1552 ](10 - 6) = 360.1(10 - 6)
tan a =
155
;
400 - 75
a = 25.50°
f = 60 + 25.50 = 85.5°
ex¿ = (75 + 360.1 cos 85.5°)(10 - 6) = 103(10 - 6)
Ans.
ey¿ = (75 - 360.1 cos 85.5°)(10 - 6) = 46.7(10 - 6)
Ans.
gx¿y¿
2
= (360.1 sin 85.5°)(10 - 6)
gx¿y¿ = 718(10 - 6)
Ans.
759
u = 30°
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•10–21.
Solve Prob. 10–14 using Mohr’s circle.
Construction of the Circle: In accordance with the sign convention, ex = 250 A 10 - 6 B ,
gxy
ey = 300 A 10 - 6 B , and
= -90 A 10 - 6 B . Hence,
2
eavg =
ex + ey
2
= a
250 + 300
b A 10 - 6 B = 275 A 10 - 6 B
2
Ans.
The coordinates for reference points A and C are
A(250, -90) A 10 - 6 B
C(275, 0) A 10 - 6 B
The radius of the circle is
R = a 2(275 - 250)2 + 902 b A 10 - 6 B = 93.408
In-Plane Principal Strain: The coordinates of points B and D represent e1 and e2,
respectively.
e1 = (275 + 93.408) A 10 - 6 B = 368 A 10 - 6 B
Ans.
e2 = (275 - 93.408) A 10 - 6 B = 182 A 10 - 6 B
Ans.
Orientation of Principal Strain: From the circle,
tan 2uP2 =
90
= 3.600
275 - 250
2uP2 = 74.48°
2uP1 = 180° - 2uP2
uP1 =
180° - 74.78°
= 52.8° (Clockwise)
2
Ans.
Maximum In-Plane Shear Strain: Represented by the coordinates of point E on
the circle.
g max
in-plane
2
= -R = -93.408 A 10 - 6 B
g max
in-plane =
-187 A 10 - 6 B
Ans.
Orientation of the Maximum In-Plane Shear Strain: From the circle,
tan 2us =
275 - 250
= 0.2778
90
us = 7.76° (Clockwise)
Ans.
760
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10–22. The strain at point A on the bracket has
components Px = 300110-62, Py = 550110-62, gxy =
-650110-62. Determine (a) the principal strains at A in the
x– y plane, (b) the maximum shear strain in the x–y plane,
and (c) the absolute maximum shear strain.
ex = 300(10 - 6)
ey = 550(10 - 6)
A(300, -325)10 - 6
gxy = -650(10 - 6)
y
gxy
2
= -325(10 - 6)
A
C(425, 0)10 - 6
R = C 2(425 - 300)2 + (-325)2 D 10 - 6 = 348.2(10 - 6)
a)
e1 = (425 + 348.2)(10 - 6) = 773(10 - 6)
Ans.
e2 = (425 - 348.2)(10 - 6) = 76.8(10 - 6)
Ans.
b)
g
-6
-6
max
in-plane = 2R = 2(348.2)(10 ) = 696(10 )
Ans.
c)
gabs
max
=
2
773(10 - 6)
;
2
gabs
max
= 773(10 - 6)
Ans.
10–23. The strain at point A on the leg of the angle has
components Px = -140110-62, Py = 180110-62, gxy =
-125110-62. Determine (a) the principal strains at A in the
x–y plane, (b) the maximum shear strain in the x–y plane,
and (c) the absolute maximum shear strain.
ex = -140(10 - 6)
A( -140, -62.5)10 - 6
ey = 180(10 - 6)
gxy = -125(10 - 6)
A
gxy
2
= -62.5(10 - 6)
C(20, 0)10 - 6
R = A 2(20 - ( -140))2 + (-62.5)2 B 10 - 6 = 171.77(10 - 6)
a)
e1 = (20 + 171.77)(10 - 6) = 192(10 - 6)
Ans.
e2 = (20 - 171.77)(10 - 6) = -152(10 - 6)
Ans.
(b, c)
gabs
max
gmax
= in-plane
= 2R = 2(171.77)(10 - 6) = 344(10 - 6)
Ans.
761
x
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*10–24. The strain at point A on the pressure-vessel wall
has components Px = 480110-62, Py = 720110-62, gxy =
650110-62. Determine (a) the principal strains at A, in the
x– y plane, (b) the maximum shear strain in the x–y plane,
and (c) the absolute maximum shear strain.
ex = 480(10 - 6)
ey = 720(10 - 6)
A(480, 325)10 - 6
C(600, 0)10 - 6
gxy = 650(10 - 6)
y
A
gxy
2
= 325(10 - 6)
R = (2(600 - 480)2 + 3252 )10 - 6 = 346.44(10 - 6)
a)
e1 = (600 + 346.44)10 - 6 = 946(10 - 6)
Ans.
e2 = (600 - 346.44)10 - 6 = 254(10 - 6)
Ans.
b)
g
-6
max
= 693(10 - 6)
in-plane = 2R = 2(346.44)10
Ans.
c)
gabs
max
2
=
946(10 - 6)
;
2
gabs
max
= 946(10 - 6)
Ans.
762
x
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•10–25.
The 60° strain rosette is mounted on the bracket.
The following readings are obtained for each gauge:
Pa = -100110-62, Pb = 250110-62, and Pc = 150110-62.
Determine (a) the principal strains and (b) the maximum inplane shear strain and associated average normal strain. In
each case show the deformed element due to these strains.
b
c
60⬚
60⬚
This is a 60° strain rosette Thus,
ex = ea = -100(10 - 6)
1
A 2eb + 2ec - ea B
3
ey =
1
C 2(250) + 2(150) - (-100) D (10 - 6)
3
=
= 300(10 - 6)
gxy =
2
23
(eb - ec) =
2
23
(250 - 150)(10 - 6) = 115.47(10 - 6)
In accordance to the established sign convention, ex = -100(10 - 6), ey = 300(10 - 6)
gxy
and
= 57.74(10 - 6).
2
Thus,
eavg =
ex + ey
2
= a
-100 + 300
b(10 - 6) = 100(10 - 6)
2
Ans.
Then, the coordinates of reference point A and Center C of the circle are
A( -100, 57.74)(10 - 6)
C(100, 0)(10 - 6)
Thus, the radius of the circle is
R = CA = a 2(-100 - 100)2 + 208.16b(10 - 6) = 208.17(10 - 6)
Using these result, the circle is shown in Fig. a.
The coordinates of points B and D represent e1 and e2 respectively.
e1 = (100 + 208.17)(10 - 6) = 308(10 - 6)
Ans.
e2 = (100 - 208.17)(10 - 6) = -108(10 - 6)
Ans.
Referring to the geometry of the circle,
tan 2(uP)2 =
57.74(10 - 6)
(100 + 100)(10 - 6)
= 0.2887
A uP B 2 = 8.05° (Clockwise)
Ans.
The deformed element for the state of principal strain is shown in Fig. b.
763
a
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10–25.
Continued
gmax
The coordinates for point E represent eavg and
in-plane
2
. Thus,
gmax
in-plane
2
= R = 208.17(10 - 6)
gmax
in-plane
= 416(10 - 6)
Ans.
Referring to the geometry of the circle,
tan 2us =
100 + 100
57.74
us = 36.9° (Counter Clockwise)
Ans.
The deformed element for the state of maximum In-plane shear strain is shown in
Fig. c.
764
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10–26. The 60° strain rosette is mounted on a beam.
The following readings are obtained for each gauge:
Pa = 200110-62, Pb = - 450110-62, and Pc = 250110-62.
Determine (a) the in-plane principal strains and (b) the
maximum in-plane shear strain and average normal strain.
In each case show the deformed element due to these strains.
b
a
30⬚ 30⬚
c
With ua = 60°, ub = 120° and uc = 180°,
ea = ex cos2 ua + ey sin2 ua + gxy sin ua cos ua
200(10 - 6) = ex cos2 60° + ey sin2 60° + gxy sin 60° cos 60°
0.25ex + 0.75ey + 0.4330 gxy = 200(10 - 6)
(1)
eb = ex cos2 ub + ey sin2 ub + gxy sin ub cos ub
-450(10 - 6) = ex cos2 120° + ey sin2 120° + gxy sin 120° cos 120°
0.25ex + 0.75ey - 0.4330 gxy = -450(10 - 6)
(2)
ec = ex cos2 uc + ey sin2 uc + gxy sin uc cos uc
250(10 - 6) = ex cos2 180° + ey sin2 180° + gxy sin 180° cos 180°
ex = 250(10 - 6)
Substitute this result into Eqs. (1) and (2) and solve them,
ey = -250 (10 - 6)
gxy = 750.56 (10 - 6)
In accordance to the established sign convention, ex = 250(10 - 6), ey = -250(10 - 6),
gxy
and
= 375.28(10 - 6), Thus,
2
eavg =
ex + ey
2
= c
250 + ( -250)
d(10 - 6) = 0
2
Ans.
Then, the coordinates of the reference point A and center C of the circle are
A(250, 375.28)(10 - 6)
C(0, 0)
Thus, the radius of the circle is
R = CA = A 2(250 - 0)2 + 375.282 B (10 - 6) = 450.92(10 - 6)
Using these results, the circle is shown in Fig. a.
The coordinates for points B and D represent e1 and e2, respectively. Thus,
e1 = 451(10 - 6)
e2 = -451(10 - 6)
Ans.
Referring to the geometry of the circle,
tan 2(uP)1 =
375.28
= 1.5011
250
(uP)1 = 28.2° (Counter Clockwise)
Ans.
The deformed element for the state of principal strains is shown in Fig. b.
765
60⬚
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10–26.
Continued
gmax
in-plane
The coordinates of point E represent eavg and
. Thus,
2
gmax
in-plane
gmax
= 902(10 - 6)
= R = 450.92(10 - 6)
in-plane
2
Ans.
Referring to the geometry of the circle,
tan 2us =
250
= 0.6662
375.28
us = 16.8° (Clockwise)
Ans.
766
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10–27. The 45° strain rosette is mounted on a steel shaft.
The following readings are obtained from each gauge:
Pa = 300110-62, Pb = -250110-62, and Pc = -450110-62.
Determine (a) the in-plane principal strains and (b) the
maximum in-plane shear strain and average normal strain.
In each case show the deformed element due to these strains.
b
c
45⬚
With ua = 45°, ub = 90° and uc = 135°,
ea = ex cos2 ua + ey sin2 ua + gxy sin ua cos ua
300(10 - 6) = ex cos2 45° + ey sin2 45° + gxy sin 45° cos 45°
ex + ey + gxy = 600(10 - 6)
(1)
eb = ex cos2 ub + ey sin2 ub + gxy sin ub cos ub
-250(10 - 6) = ex cos2 90° + ey sin2 90° + gxy sin 90° cos 90°
ey = -250(10 - 6)
ec = ex cos2 uc + ey sin2 uc + gxy sin uc cos uc
-450(10 - 6) = ex cos2 135° + ey sin2 135° + gxy sin 135° cos 135°
ex + ey - gxy = -900(10 - 6)
(2)
Substitute the result of ey into Eq. (1) and (2) and solve them
ex = 100(10 - 6)
gxy = 750(10 - 6)
In accordance to the established sign convention, ex = 100(10 - 6), ey = -250(10 - 6)
gxy
and
= 375(10 - 6). Thus,
2
eavg =
ex + ey
2
= c
100 + (-250)
d(10 - 6) = -75(10 - 6)
2
Ans.
Then, the coordinates of the reference point A and the center C of the circle are
A(100, 375)(10 - 6)
C(-75, 0)(10 - 6)
Thus, the radius of the circle is
R = CA = a 2 C 100 - (-75) D 2 + 3752 b(10 - 6) = 413.82(10 - 6)
Using these results, the circle is shown in Fig. a.
The Coordinates of points B and D represent e1 and e2, respectively. Thus,
e1 = A -75 + 413.82 B (10 - 6) = 339(10 - 6)
Ans.
e2 = A -75 - 413.82 B (10 - 6) = -489(10 - 6)
Ans.
Referring to the geometry of the circle
tan 2(uP)1 =
375
= 2.1429
100 + 75
(uP)1 = 32.5° (Counter Clockwise)
Ans.
767
45⬚
a
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10–27.
Continued
The deformed element for the state of principal strains is shown in Fig. b.
gmax
The coordinates of point E represent eavg and
gmax
in-plane
2
= R = 413.82(106)
in-plane
gmax
2
in-plane
. Thus
= 828(10 - 6)
Ans.
Referring to the geometry of the circle
tan 2us =
-100 + 75
= 0.4667
375
us = 12.5° (Clockwise)
Ans.
768
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*10–28. The 45° strain rosette is mounted on the link of
the backhoe. The following readings are obtained from
each gauge: Pa = 650110-62, Pb = -300110-62, Pc = 480110-62.
Determine (a) the in-plane principal strains and (b) the
maximum in-plane shear strain and associated average
normal strain.
a
45⬚
b
ea = 650(10 - 6);
ua = 180°;
eb = -300(10 - 6);
ub = 225°
c
uc = 270°
Applying Eq. 10–16, e = ex cos2 u + ey sin2 u + gxy sin u cos u
650(10 - 6) = ex cos2 (180°) + ey sin2 (180°) + gxy sin (180°) cos (180°)
ex = 650 (10 - 6)
480 (10 - 6) = ex cos2 (270°) + ey sin2 (270°) + gxy sin (270°) cos (270°)
ey = 480 (10 - 6)
-300 (10 - 6) = 650 (10 - 6) cos2 (225°) + 480 (10 - 6) sin2 (225°) + gxy sin (225°) cos (225°)
gxy = -1730 (10 - 6)
Therefore, ex = 650 (10 - 6)
gxy
2
ey = 480 (10 - 6)
gxy = -1730 (10 - 6)
= -865 (10 - 6)
Mohr’s circle:
A(650, -865) 10 - 6
C(565, 0) 10 - 6
R = CA = C 2(650 - 565)2 + 8652 D 10 - 6 = 869.17 (10 - 6)
(a)
(b)
e1 = [565 + 869.17]10 - 6 = 1434 (10 - 6)
Ans.
e2 = [565 - 869.17]10 - 6 = -304 (10 - 6)
Ans.
gmax
Ans.
in-plane
45⬚
ec = 480(10 - 6)
= 2 R = 2(869.17) (10 - 6) = 1738 (10 - 6)
eavg = 565(10 - 6)
Ans.
769
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10–30. For the case of plane stress, show that Hooke’s law
can be written as
sx =
E
E
1Px + nPy2, sy =
1Py + nPx2
11 - n22
11 - n22
Generalized Hooke’s Law: For plane stress, sz = 0. Applying Eq. 10–18,
ex =
1
A s - v sy B
E x
vEex = A sx - v sy B v
vEex = v sx - v2 sy
ey =
[1]
1
(s - v sx)
E y
E ey = -v sx + sy
[2]
Adding Eq [1] and Eq.[2] yields.
vE ex - E ey = sy - v2 sy
sy =
E
A vex + ey B
1 - v2
(Q.E.D.)
Substituting sy into Eq. [2]
E ey = -vsx +
sx =
E
A v ex + ey B
1 - v2
E A v ex + ey B
v (1 - v2)
Eey
-
v
E v ex + E ey - E ey + Eey v2
=
=
v(1 - v2)
E
(ex + v ey)
1 - v2
(Q.E.D.)
770
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10–31. Use Hooke’s law, Eq. 10–18, to develop the straintransformation equations, Eqs. 10–5 and 10–6, from the
stress-transformation equations, Eqs. 9–1 and 9–2.
Stress transformation equations:
sx + sy
sx¿ =
2
tx¿y¿ =
sy¿ =
sx - sy
+
sx - sy
2
(1)
sin 2u + txy cos 2u
2
sx + sy
sx - sy
-
2
cos 2u + txy sin 2u
2
(2)
cos 2u - txy sin 2u
(3)
Hooke’s Law:
ex =
v sy
sx
E
E
(4)
ey =
sy
-v sx
+
E
E
(5)
txy = G gxy
G =
(6)
E
2 (1 + v)
(7)
From Eqs. (4) and (5)
ex + ey =
ex - ey =
(1 - v)(sx + sy)
(8)
E
(1 + v)(sx - sy)
(9)
E
From Eqs. (6) and (7)
txy =
E
g
2 (1 + v) xy
(10)
From Eq. (4)
ex¿ =
v sy¿
sx¿
E
E
(11)
Substitute Eqs. (1) and (3) into Eq. (11)
ex¿ =
(1 - v)(sx - sy)
(1 + v)(sx - sy)
+
2E
2E
cos 2u +
(1 + v)txy sin 2u
E
(12)
By using Eqs. (8), (9) and (10) and substitute into Eq. (12),
ex¿ =
ex + ey
2
ex - ey
+
2
cos 2u +
gxy
2
QED
sin 2u
771
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10–31.
Continued
From Eq. (6).
gx¿y¿ = G gx¿y¿ =
E
g
2 (1 + v) x¿y¿
(13)
Substitute Eqs. (13), (6) and (9) into Eq. (2),
E (ex - ey)
E
E
gx¿y¿ = sin 2u +
g cos 2u
2 (1 + v)
2 (1 + v)
2 (1 + v) xy
gx¿y¿
(ex - ey)
= -
2
2
sin 2u +
gxy
2
QED
cos 2u
*10–32. A bar of copper alloy is loaded in a tension
machine and it is determined that Px = 940110-62 and
sx = 14 ksi, sy = 0, sz = 0. Determine the modulus of
elasticity, Ecu, and the dilatation, ecu, of the copper.
ncu = 0.35.
ex =
1
[s - v(sy + sz)]
E x
940(10 - 6) =
ecu =
1
[14(103) - 0.35(0 + 0)]
Ecu
Ecu = 14.9(103) ksi
Ans.
1 - 2(0.35)
1 - 2v
(14 + 0 + 0) = 0.282(10 - 3)
(sx + sy + sz) =
E
14.9(103)
Ans.
•10–33.
The principal strains at a point on the aluminum
fuselage of a jet aircraft are P1 = 780110-62 and P2 =
400110-62. Determine the associated principal stresses at
the point in the same plane. Eal = 1011032 ksi, nal = 0.33.
Hint: See Prob. 10–30.
Plane stress, s3 = 0
See Prob 10-30,
s1 =
E
(e1 + ve2)
1 - v2
10(103)
=
s2 =
1 - 0.332
Ans.
E
(e2 + ve1)
1 - v2
10(103)
=
(780(10 - 6) + 0.33(400)(10 - 6)) = 10.2 ksi
1 - 0.332
(400(10 - 6) + 0.33(780)(10 - 6)) = 7.38 ksi
Ans.
772
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10–34. The rod is made of aluminum 2014-T6. If it is
subjected to the tensile load of 700 N and has a diameter of
20 mm, determine the absolute maximum shear strain in the
rod at a point on its surface.
700 N
Normal Stress: For uniaxial loading, sy = sz = 0.
sx =
700
P
= 2.228 MPa
= p
2
A
4 (0.02 )
Normal Strain: Applying the generalized Hooke’s Law.
ex =
=
1
C s - v A sy + sz B D
E x
1
C 2.228 A 106 B - 0 D
73.1(109)
= 30.48 A 10 - 6 B
ey =
=
1
C s - v(sx + sz) D
E y
1
C 0 - 0.35 A 2.228 A 106 B + 0 B D
73.1(109)
= -10.67 A 10 - 6 B
ez =
=
1
C s - v A sx + sy B D
E z
1
C 0 - 0.35 A 2.228 A 106 B + 0 B D
73.1(109)
= -10.67 A 10 - 6 B
Therefore.
emax = 30.48 A 10 - 6 B
emin = -10.67 A 10 - 6 B
Absolute Maximum Shear Strain:
gabs
max = emax - emin
= [30.48 - (-10.67)] A 10 - 6 B = 41.1 A 10 - 6 B
Ans.
773
700 N
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10–35. The rod is made of aluminum 2014-T6. If it is
subjected to the tensile load of 700 N and has a diameter of
20 mm, determine the principal strains at a point on the
surface of the rod.
700 N
Normal Stress: For uniaxial loading, sy = sz = 0.
sx =
P
700
= 2.228 MPa
= p
2
A
4 (0.02 )
Normal Strains: Applying the generalized Hooke’s Law.
ex =
=
1
C s - v A sy + sz B D
E x
1
C 2.228 A 106 B - 0 D
73.1(109)
= 30.48 A 10 - 6 B
ey =
=
1
C s - v(sx + sz) D
E y
1
C 0 - 0.35 A 2.228 A 106 B + 0 B D
73.1(109)
= -10.67 A 10 - 6 B
ez =
=
1
C s - v A sx + sy B D
E z
1
C 0 - 0.35 A 2.228 A 106 B + 0 B D
73.1(109)
= -10.67 A 10 - 6 B
Principal Strains: From the results obtained above,
emax = 30.5 A 10 - 6 B
eint = emin = -10.7 A 10 - 6 B
Ans.
774
700 N
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*10–36. The steel shaft has a radius of 15 mm. Determine
the torque T in the shaft if the two strain gauges, attached to
the surface of the shaft, report strains of Px¿ = -80110-62
and Py¿ = 80110-62. Also, compute the strains acting in the x
and y directions. Est = 200 GPa, nst = 0.3.
ex¿ = -80(10 - 6)
y
T
ex = ey = 0
Ans.
ex¿ = ex cos2 u + ey sin2 u + gxy sin u cos u
u = 45°
-80(10 - 6) = 0 + 0 + gxy sin 45° cos 45°
gxy = -160(10 - 6)
Ans.
Also, u = 135°
80(10 - 6) = 0 + 0 + g sin 135° cos 135°
gxy = -160(10 - 6)
200(109)
E
=
= 76.923(109)
2(1 + V)
2(1 + 0.3)
t = Gg = 76.923(109)(160)(10 - 6) = 12.308(106) Pa
12.308(106) A
p
B (0.015)4
2
= 65.2 N # m
0.015
Ans.
10–37. Determine the bulk modulus for each of the
following materials: (a) rubber, Er = 0.4 ksi, nr = 0.48, and
(b) glass, Eg = 811032 ksi, ng = 0.24.
a) For rubber:
Kr =
Er
0.4
=
= 3.33 ksi
3 (1 - 2 vr)
3[1 - 2(0.48)]
Ans.
b) For glass:
Kg =
Eg
3 (1 - 2 vg)
=
x
T
Pure shear
tJ
T =
=
c
x¿
45⬚
ey¿ = 80(10 - 6)
G =
y¿
8(103)
= 5.13 (103) ksi
3[1 - 2(0.24)]
Ans.
775
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10–38. The principal stresses at a point are shown in the
figure. If the material is A-36 steel, determine the
principal strains.
12 ksi
e1 =
1
1
e 12 - 0.32 C 8 + (-20) D f = 546 (10-6)
C s - v(s2 + s3) D =
E 1
29.0(103)
e2 =
1
1
e 8 - 0.32 C 12 + (-20) D f = 364 (10-6)
C s - v(s1 + s3) D =
E 2
29.0(103)
e3 =
20 ksi
8 ksi
1
1
C s3 - v(s1 + s2) D =
C -20 - 0.32(12 + 8) D = -910 (10-6)
E
29.0(103)
emax = 546 (10 - 6)
eint = 346 (10 - 6)
emin = -910 (10 - 6)
Ans.
10–39. The spherical pressure vessel has an inner
diameter of 2 m and a thickness of 10 mm. A strain gauge
having a length of 20 mm is attached to it, and it is observed
to increase in length by 0.012 mm when the vessel
is pressurized. Determine the pressure causing this
deformation, and find the maximum in-plane shear stress,
and the absolute maximum shear stress at a point on the
outer surface of the vessel. The material is steel, for which
Est = 200 GPa and nst = 0.3.
20 mm
1000
r
=
= 100 7 10, the thin wall analysis is valid to
t
10
determine the normal stress in the wall of the spherical vessel. This is a plane stress
Normal Stresses: Since
problem where smin = 0 since there is no load acting on the outer surface of the wall.
smax = slat =
pr
p(1000)
=
= 50.0p
2t
2(10)
[1]
Normal Strains: Applying the generalized Hooke’s Law with
emax = elat =
0.012
= 0.600 A 10 - 3 B mm>mm
20
emax =
1
C s - V (slat + smin) D
E max
0.600 A 10 - 3 B =
1
[50.0p - 0.3 (50.0p + 0)]
200(104)
p = 3.4286 MPa = 3.43 MPa
Ans.
From Eq.[1] smax = slat = 50.0(3.4286) = 171.43 MPa
Maximum In-Plane Shear (Sphere’s Surface): Mohr’s circle is simply a dot. As the
result, the state of stress is the same consisting of two normal stresses with zero
shear stress regardless of the orientation of the element.
t max
Ans.
in-plane = 0
Absolute Maximum Shear Stress:
tabs
max
=
smax - smin
171.43 - 0
=
= 85.7MPa
2
2
Ans.
776
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*10–40. The strain in the x direction at point A on the
steel beam is measured and found to be Px = -100110-62.
Determine the applied load P. What is the shear strain gxy
at point A? Est = 2911032 ksi, nst = 0.3.
P
y
3 in.
A
3 ft
1
1
(5.5)(83) = 129.833 in4
(6)(9)3 12
12
Ix =
QA = (4.25)(0.5)(6) + (2.75)(0.5)(2.5) = 16.1875 in3
s = Eex = 29(103)(100)(10 - 6) = 2.90 ksi
My
,
I
s =
2.90 =
1.5P(12)(1.5)
129.833
P = 13.945 = 13.9 kip
tA =
0.5(13.945)(16.1875)
VQ
=
= 1.739 ksi
It
129.833(0.5)
G =
29(103)
E
=
= 11.154(103) ksi
2(1 + v)
2(1 + 0.3)
gxy =
txy
G
=
Ans.
1.739
= 0.156(10 - 3) rad
11.154(103)
Ans.
777
x
4 ft
7 ft
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•10–41.
The cross section of the rectangular beam is
subjected to the bending moment M. Determine an
expression for the increase in length of lines AB and CD.
The material has a modulus of elasticity E and Poisson’s
ratio is n.
C
D
B
h
For line AB,
sz = -
ey = -
A
M
12My
My
My
= 1
= 3
I
b h3
12 b h
v sz
=
E
E b h3
h
2
¢LAB =
=
b
12 v My
L0
h
ey dy =
2
12 v M
y dy
3
E b h L0
3vM
2Ebh
Ans.
For line CD,
sz = -
ex = -
M h2
Mc
6M
= - 1
= - 2
3
I
bh
b
h
12
v sz
E
=
6vM
E b h2
¢LCD = ex LCD =
=
6vM
(b)
E b h2
6vM
E h2
Ans.
10–42. The principal stresses at a point are shown in
the figure. If the material is aluminum for which
Eal = 1011032 ksi and nal = 0.33, determine the principal
strains.
26 ksi
ex =
1
1
(s - v(sy + sz)) =
(10 - 0.33(-15 - 26)) = 2.35(10 - 3)
E x
10(103)
ey =
1
1
(s - v(sx + sz)) =
(-15 - 0.33)(10 - 26)) = -0.972(10 - 3)Ans.
E y
10(103)
ez =
Ans.
1
1
(s - v(sx + sy)) =
(-26 - 0.33(10 - 15)) = -2.44(10 - 3) Ans.
E z
10(103)
778
15 ksi
10 ksi
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10–43. A single strain gauge, placed on the outer surface
and at an angle of 30° to the axis of the pipe, gives a reading
at point A of Pa = -200(10-6). Determine the horizontal
force P if the pipe has an outer diameter of 2 in. and an
inner diameter of 1 in. The pipe is made of A-36 steel.
1.5 ft
Using the method of section and consider the equilibrium of the FBD of the pipe’s
upper segment, Fig. a,
©Fz = 0;
Vz - p = 0
©Mx = 0;
Tx - p(1.5) = 0 Tx = 1.5p
©My = 0;
My - p(2.5) = 0 My = 2.5p
Vz = p
30⬚
A
The normal strees is due to bending only. For point A, z = 0. Thus
sx =
My z
Iy
= 0
The shear stress is the combination of torsional shear stress and transverse shear
stress. Here, J = p2 (14 - 0.54) = 0.46875 p in4. Thus, for point A
tt =
1.5p(12)(1)
38.4 p
Txc
=
=
p
J
0.46875p
Referring to Fig. b,
(QA)z = y1œ A1œ - y2œ A2œ =
4 (1) p 2
4(0.5) p
c (1 ) d c (0.52) d
3p 2
3p 2
= 0.5833 in3
Iy = p4 (14 - 0.54) = 0.234375 p in4
Combine these two shear stress components,
t = tt + tv =
P
2.5 ft
38.4P
2.4889P
40.8889P
+
=
p
p
p
Since no normal stress acting on point A, it is subjected to pure shear which can be
represented by the element shown in Fig. c.
For pure shear, ex = ez = 0,
ea = ex cos3 ua + ez sin2 ua + gxz sin ua cos ua
-200(10 - 6) = 0 + 0 + gxz sin 150° cos 150°
gxz = 461.88(10 - 6)
Applying the Hooke’s Law for shear,
txz = G gxz
40.8889P
= 11.0(103) C 461.88(10 - 6) D
p
P = 0.3904 kip = 390 lb
Ans.
779
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*10–44. A single strain gauge, placed in the vertical plane
on the outer surface and at an angle of 30° to the axis of the
pipe, gives a reading at point A of Pa = -200(10-6).
Determine the principal strains in the pipe at point A. The
pipe has an outer diameter of 2 in. and an inner diameter of
1 in. and is made of A-36 steel.
1.5 ft
P
2.5 ft
Using the method of sections and consider the equilibrium of the FBD of the pipe’s
upper segment, Fig. a,
©Fz = 0;
Vz - P = 0
Vz = P
©Mx = 0;
Tx - P(1.5) = 0 Tx = 1.5P
©My = 0;
My - P(2.5) = 0 My = 2.5P
By observation, no normal stress acting on point A. Thus, this is a case of pure shear.
For the case of pure shear,
ex = ez = ey = 0
ea = ex cos2 ua + ez sin2 ua + gxz sin ua cos ua
-200(10 - 6) = 0 + 0 + gxz sin 150° cos 150°
gxz = 461.88(10 - 6)
e1, 2 =
ex + ez
= B
2
+
A
a
ex - ez
2
b + a
2
gxz
2
b
2
0 - 0 2
461.88 2
0 + 0
-6
;
a
b + a
b R (10 )
2
A
2
2
e1 = 231(10 - 6)
e2 = -231(10 - 6)
Ans.
780
30⬚
A
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10–45. The cylindrical pressure vessel is fabricated using
hemispherical end caps in order to reduce the bending stress
that would occur if flat ends were used. The bending stresses
at the seam where the caps are attached can be eliminated
by proper choice of the thickness th and tc of the caps and
cylinder, respectively. This requires the radial expansion to
be the same for both the hemispheres and cylinder. Show
that this ratio is tc>th = 12 - n2>11 - n2.Assume that the
vessel is made of the same material and both the cylinder
and hemispheres have the same inner radius. If the cylinder
is to have a thickness of 0.5 in., what is the required thickness
of the hemispheres? Take n = 0.3.
tc
th
r
For cylindrical vessel:
s1 =
pr
;
tc
e1 =
1
[s - v (s2 + s3)]
E 1
=
s2 =
pr
2 tc
s3 = 0
vpr
pr
1
1 pr
a
b =
a1 - v b
E tc
2 tc
E tc
2
d r = e1 r =
p r2
1
a1 - v b
E tc
2
(1)
For hemispherical end caps:
s1 = s2 =
e1 =
=
pr
2 th
1
[s - v (s2 + s3)] ;
E 1
s3 = 0
vpr
pr
1 pr
a
b =
(1 - v)
E 2 th
2 th
2 E th
d r = e1 r =
p r2
(1 - v)
2 E th
(2)
Equate Eqs. (1) and (2):
p r2
p r2
1
a1 - vb =
(1 - v)
E tc
2
2 E th
1
2 (1 - 2 v)
tc
2 - v
=
=
th
1 - v
1 - v
th =
QED
(1 - v) tc
(1 - 0.3) (0.5)
=
= 0.206 in.
2 - v
2 - 0.3
Ans.
781
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10–46. The principal strains in a plane, measured
experimentally at a point on the aluminum fuselage of a jet
aircraft, are P1 = 630(10-6) and P2 = 350(10-6). If this is
a case of plane stress, determine the associated principal
stresses at the point in the same plane. Eal = 10(103) ksi
and nal = 0.33.
Normal Stresses: For plane stress, s3 = 0.
Normal Strains: Applying the generalized Hooke’s Law.
e1 =
1
C s - v (s2 + s3) D
E 1
630 A 10 - 6 B =
1
[s1 - 0.33(s2 + 0)]
10(103)
6.30 = s1 - 0.33s2
e2 =
[1]
1
C s - v (s1 + s3) D
E 2
350 A 10 - 6 B =
1
C s2 - 0.33(s1 + 0) D
10(103)
3.50 = s2 - 0.33s1
[2]
Solving Eqs.[1] and [2] yields:
s1 = 8.37 ksi
s2 = 6.26 ksi
Ans.
10–47. The principal stresses at a point are shown in
the figure. If the material is aluminum for which
Eal = 1011032 ksi and nal = 0.33, determine the principal
strains.
3 ksi
e1 =
1
1
e 8 - 0.33 C 3 + (-4) D f = 833 (10 - 6)
C s - v(s2 + s3) D =
E 1
10(103)
e2 =
1
1
e 3 - 0.33 C 8 + (-4) D f = 168 (10 - 6)
C s - v(s1 + s3) D =
E 2
10(103)
e3 =
1
1
C s3 - v(s1 + s2) D =
C -4 - 0.33(8 + 3) D = -763 (10 - 6)
E
10(103)
Using these results,
e1 = 833(10 - 6)
e2 = 168(10 - 6)
e3 = -763(10 - 6)
782
8 ksi
4 ksi
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*10–48. The 6061-T6 aluminum alloy plate fits snugly into
the rigid constraint. Determine the normal stresses sx and
sy developed in the plate if the temperature is increased by
¢T = 50°C. To solve, add the thermal strain a¢T to the
equations for Hooke’s Law.
y
400 mm
300 mm
x
Generalized Hooke’s Law: Since the sides of the aluminum plate are confined in the
rigid constraint along the x and y directions, ex = ey = 0. However, the plate is
allowed to have free expansion along the z direction. Thus, sz = 0. With the
additional thermal strain term, we have
ex =
0 =
1
cs - v A sy + sz B d + a¢T
E x
1
68.9 A 109 B
csx - 0.35 A sy + 0 B d + 24a 10 - 6 b(50)
sx - 0.35sy = -82.68 A 106 B
ey =
0 =
(1)
1
C s - v A sx + sz B D + a¢T
E y
1
68.9a 10 b
9
C sy - 0.35(sx + 0) D + 24 A 10 - 6 B (50)
sy - 0.35sx = -82.68 A 106 B
(2)
Solving Eqs. (1) and (2),
sx = sy = -127.2 MPa = 127.2 MPa (C)
Ans.
Since sx = sy and sy 6 sY, the above results are valid.
783
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•10–49.
Initially, gaps between the A-36 steel plate and
the rigid constraint are as shown. Determine the normal
stresses sx and sy developed in the plate if the temperature
is increased by ¢T = 100°F. To solve, add the thermal
strain a¢T to the equations for Hooke’s Law.
y
0.0015 in.
6 in.
8 in.
0.0025 in.
x
Generalized Hooke’s Law: Since there are gaps between the sides of the plate and the rigid
constraint, the plate is allowed to expand before it comes in contact with the constraint.
dy
dx
0.0025
0.0015
ey =
=
= 0.3125 A 10 - 3 B and
=
= 0.25 A 10 - 3 B .
Thus, ex =
Lx
8
Ly
6
However, the plate is allowed to have free expansion along the z direction. Thus, sz = 0.
With the additional thermal strain term, we have
ex =
1
csx - v A sy + sz B d + a¢T
E
0.3125 a 10 - 3 b =
1
29.0 a103 b
C sx - 0.32 A sy + 0 B D + 6.60 A 10 - 6 B (100)
sx - 0.32sy = -10.0775
ey =
(1)
1
C s - v A sx + sz B D + a¢T
E y
0.25 A 10 - 3 B =
1
29.0 A 103 B
C sy - 0.32(sx + 0) D + 6.60 A 10 - 6 B (100)
sy - 0.32sx = -11.89
(2)
Solving Eqs. (1) and (2),
sx = -15.5 ksi = 15.5 ksi (C)
Ans.
sy = -16.8 ksi = 16.8 ksi (C)
Ans.
Since sx 6 sY and sy 6 sY, the above results are valid.
784
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10–50. Two strain gauges a and b are attached to a plate
made from a material having a modulus of elasticity of
E = 70 GPa and Poisson’s ratio n = 0.35. If the gauges give
a reading of Pa = 450110-62 and Pb = 100110-62, determine
the intensities of the uniform distributed load wx and wy
acting on the plate. The thickness of the plate is 25 mm.
wy
b
45⬚
y
a
Normal Strain: Since no shear force acts on the plane along the x and y axes, gxy = 0.
With ua = 0 and ub = 45°, we have
2
2
ea = ex cos ua + ey sin ua + gxy sin ua cos ua
450 A 10 - 6 B = ex cos2 0° + ey sin2 0°+0
ex = 450 A 10 - 6 B
eb = ex cos2 ub + ey sin2 ub + gxy sin ub cos ub
100 A 10 - 6 B = 450 A 10 - 6 B cos2 45° + ey sin2 45° + 0
ey = -250 A 10 - 6 B
Generalized Hooke’s Law: This is a case of plane stress. Thus, sz = 0.
ex =
1
C s - v A sy + sz B D
E x
450 A 10 - 6 B =
1
70 A 109 B
C sy - 0.35 A sy + 0 B D
sx - 0.35sy = 31.5 A 106 B
ey =
(1)
1
C s - v A sx + sz B D
E y
-250 A 10-6 B =
1
70 A 109 B
C sy - 0.35 A sy + 0 B D
sy - 0.35sx = -17.5 A 106 B
(2)
Solving Eqs. (1) and (2),
sy = -7.379 A 106 B N>m2
sx = 28.917 A 106 B N>m2
Then,
wy = syt = -7.379 A 106 B (0.025) = -184 N>m
Ans.
wx = sxt = 28.917 A 106 B (0.025) = 723 N>m
Ans.
785
z
x
wx
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10–51. Two strain gauges a and b are attached to the
surface of the plate which is subjected to the uniform
distributed load wx = 700 kN>m and wy = -175 kN>m.
If the gauges give a reading of Pa = 450110-62 and
Pb = 100110-62, determine the modulus of elasticity E,
shear modulus G, and Poisson’s ratio n for the material.
wy
b
45⬚
y
Normal Stress and Strain: The normal stresses along the x, y, and z axes are
sx =
700 A 103 B
0.025
sy = -
a
= 28 A 10 B N>m
6
175 A 103 B
0.025
2
= -7 A 106 B N>m2
z
sz = 0 (plane stress)
Since no shear force acts on the plane along the x and y axes, gxy = 0. With ua = 0°
and ub = 45°, we have
ea = ex cos2 ua + ey sin2 ua + gxy sin ua cos ua
450 A 10 - 6 B = ex cos2 0° + ey sin2 0° + 0
ex = 450 A 10 - 6 B
eb = ex cos2 ub + ey sin2 ub + gxy sin ub cos ub
100 A 10 - 6 B = 450 A 10 - 6 B cos2 45°+ ey sin2 45° + 0
ey = -250 A 10 - 6 B
Generalized Hooke’s Law:
ex =
1
C s - v A sy + sz B D
E x
450 A 10 - 6 B =
1
B 28 A 106 B - v C -7 A 106 B + 0 D R
E
450 A 10 - 6 B E - 7 A 106 B v = 28 A 106 B
ey =
(1)
1
[s - v(sx + sz)]
E y
-250 A 10 - 6 B =
1
b -7 A 106 B - v C 28 A 106 B + 0 D r
E
250 A 10 - 6 B E - 28 A 106 B v = 7 A 106 B
(2)
Solving Eqs. (1) and (2),
E = 67.74 A 109 B N>m2 = 67.7 GPa
Ans.
v = 0.3548 = 0.355
Ans.
Using the above results,
G =
67.74 A 109 B
E
=
2(1 + v)
2(1 + 0.3548)
= 25.0 A 109 B N>m2 = 25.0 GPa
Ans.
786
x
wx
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*10–52. The block is fitted between the fixed supports. If the
glued joint can resist a maximum shear stress of tallow = 2 ksi,
determine the temperature rise that will cause the joint to fail.
Take E = 10 (103) ksi, n = 0.2, and Hint: Use Eq. 10–18 with
an additional strain term of a¢T (Eq. 4–4).
40⬚
Normal Strain: Since the aluminum is confined along the y direction by the rigid
frame, then ey = 0 and sx = sz = 0. Applying the generalized Hooke’s Law with
the additional thermal strain,
ey =
0 =
1
C s - v(sx + sz) D + a¢T
E y
1
C sy - 0.2(0 + 0) D + 6.0 A 10 - 6 B (¢T)
10.0(103)
sy = -0.06¢T
Construction of the Circle: In accordance with the sign convention. sx = 0,
sy = -0.06¢T and txy = 0. Hence.
savg =
sx + sy
2
=
0 + ( -0.06¢T)
= -0.03¢T
2
The coordinates for reference points A and C are A (0, 0) and C( -0.03¢T, 0).
The radius of the circle is R = 2(0 - 0.03¢T)2 + 0 = 0.03¢T
Stress on The inclined plane: The shear stress components tx¿y¿, are represented by
the coordinates of point P on the circle.
tx¿y¿ = 0.03¢T sin 80° = 0.02954¢T
Allowable Shear Stress:
tallow = tx¿y¿
2 = 0.02954¢T
¢T = 67.7 °F
Ans.
787
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z
•10–53.
The smooth rigid-body cavity is filled with liquid
6061-T6 aluminum. When cooled it is 0.012 in. from the top
of the cavity. If the top of the cavity is covered and the
temperature is increased by 200°F, determine the stress
components sx , sy , and sz in the aluminum. Hint: Use
Eqs. 10–18 with an additional strain term of a¢T (Eq. 4–4).
0.012 in.
4 in.
4 in.
6 in.
y
Normal Strains: Since the aluminum is confined at its sides by a rigid container and
0.012
allowed to expand in the z direction, ex = ey = 0; whereas ez =
= 0.002.
6
Applying the generalized Hooke’s Law with the additional thermal strain,
ex =
1
C s - v(sy + sz) D + a¢T
E x
0 =
1
C sx - 0.35 A sy + sz B D + 13.1 A 10 - 6 B (200)
10.0(103)
0 = sx - 0.35sy - 0.35sz + 26.2
[1]
ey =
1
C s - v(sx + sz) + a¢T
E y
0 =
1
C sy - 0.35(sx + sz) D + 13.1 A 10 - 6 B (200)
10.0(103)
0 = sy - 0.35sx - 0.35sz + 26.2
ez =
0.002 =
[2]
1
C s - v A sx + sy B D + a¢T
E z
1
C sz - 0.35 A sx + sy B D + 13.1 A 10 - 6 B (200)
10.0(103)
0 = sz - 0.35sx - 0.35sy + 6.20
[3]
Solving Eqs.[1], [2] and [3] yields:
sx = sy = -70.0 ksi
sz = -55.2 ksi
Ans.
788
x
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z
10–54. The smooth rigid-body cavity is filled with liquid
6061-T6 aluminum. When cooled it is 0.012 in. from the top
of the cavity. If the top of the cavity is not covered and the
temperature is increased by 200°F, determine the strain
components Px , Py , and Pz in the aluminum. Hint: Use
Eqs. 10–18 with an additional strain term of a¢T (Eq. 4–4).
0.012 in.
4 in.
4 in.
6 in.
y
Normal Strains: Since the aluminum is confined at its sides by a rigid container,
then
ex = ey = 0
Ans.
and since it is not restrained in z direction, sz = 0. Applying the generalized
Hooke’s Law with the additional thermal strain,
ex =
1
C s - v A sy + sz B D + a¢T
E x
0 =
1
C sx - 0.35 A sy + 0 B D + 13.1 A 10 - 6 B (200)
10.0(103)
0 = sx - 0.35sy + 26.2
[1]
ey =
1
C s - v(sx + sz) D + a¢T
E y
0 =
1
C sy - 0.35(sx + 0) D + 13.1 A 10 - 6 B (200)
10.0(103)
0 = sy - 0.35sx + 26.2
[2]
Solving Eqs. [1] and [2] yields:
sx = sy = -40.31 ksi
ez =
=
1
C s - v A sx + sy B D + a¢T
E z
1
{0 - 0.35[-40.31 + (-40.31)]} + 13.1 A 10 - 6 B (200)
10.0(103)
= 5.44 A 10 - 3 B
Ans.
789
x
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10–55. A thin-walled spherical pressure vessel having an
inner radius r and thickness t is subjected to an internal
pressure p. Show that the increase in the volume within
the vessel is ¢V = 12ppr4>Et211 - n2. Use a small-strain
analysis.
pr
2t
s1 = s2 =
s3 = 0
e1 = e2 =
1
(s - vs2)
E 1
e1 = e2 =
pr
(1 - v)
2t E
e3 =
1
(-v(s1 + s2))
E
e3 = -
V =
v pr
tE
4pr3
3
V + ¢V =
4p
4pr3
¢r 3
(r + ¢r)3 =
(1 +
)
r
3
3
where ¢V V V, ¢r V r
V + ¢V -
eVol =
¢r
4p r3
a1 + 3
b
r
3
¢V
¢r
= 3a b
V
r
Since e1 = e2 =
eVol = 3e1 =
2p(r + ¢r) - 2p r
¢r
=
r
2p r
3pr
(1 - v)
2t E
¢V = VeVol =
2pp r4
(1 - v)
Et
QED
790
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*10–56. A thin-walled cylindrical pressure vessel has an
inner radius r, thickness t, and length L. If it is subjected
to an internal pressure p, show that the increase in
its inner radius is dr = rP1 = pr211 - 12 n2>Et and the
increase in its length is ¢L = pLr112 - n2>Et. Using these
results, show that the change in internal volume becomes
dV = pr211 + P12211 + P22L - pr2L. Since P1 and P2 are
small quantities, show further that the change in volume
per unit volume, called volumetric strain, can be written as
dV>V = pr12.5 - 2n2>Et.
Normal stress:
pr
;
t
s1 =
s2 =
pr
2t
Normal strain: Applying Hooke’s law
e1 =
=
1
[s - v (s2 + s3)],
E 1
vpr
pr
1
1 pr
a
b =
a1 - vb
E
t
2t
Et
2
d r = et r =
e2 =
=
s3 = 0
p r2
1
a1 - v b
Et
2
1
[s - v (s1 + s3)],
E 2
QED
s3 = 0
vpr
pr 1
1 pr
a
b =
a - vb
E 2t
t
Et 2
¢L = e2 L =
pLr 1
a - vb
Et
2
V¿ = p(r + e1 r)2 (L + e2L) ;
QED
V = p r2 L
dV = V¿ - V = pr2 (1 + e1)2 (1 + e2)L - pr2 L
QED
(1 + e1)2 = 1 + 2 e1 neglect e21 term
(1 + e1)2 (1 + e2) = (1 + 2 e1)(1 + e2) = 1 + e2 + 2 e1 neglect e1 e2 term
dV
= 1 + e2 + 2 e1 - 1 = e2 + 2 e1
V
=
2pr
pr 1
1
a - vb +
a1 - v b
Et 2
Et
2
=
pr
(2.5 - 2 v)
Et
QED
791
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10–57. The rubber block is confined in the U-shape
smooth rigid block. If the rubber has a modulus of elasticity
E and Poisson’s ratio n, determine the effective modulus of
elasticity of the rubber under the confined condition.
P
Generalized Hooke’s Law: Under this confined condition, ex = 0 and sy = 0. We
have
ex =
1
C s - v A sy + sz B D
E x
0 =
1
(s - vsz)
E x
sx = vsz
(1)
ez =
1
C s - v A sx + sy B D
E z
ez =
1
[s - v(sx + 0)]
E z
ez =
1
(s - vsx)
E z
(2)
Substituting Eq. (1) into Eq. (2),
ez =
sz
E
A 1 - v2 B
The effective modulus of elasticity of the rubber block under the confined condition
can be determined by considering the rubber block as unconfined but rather
undergoing the same normal strain of ez when it is subjected to the same normal
stress sz, Thus,
sz = Eeff ez
Eeff =
sz
ez
sz
=
sz
E
A 1 - v2 B
=
E
1 - v2
Ans.
792
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z
10–58. A soft material is placed within the confines of a
rigid cylinder which rests on a rigid support. Assuming that
Px = 0 and Py = 0, determine the factor by which the
modulus of elasticity will be increased when a load is
applied if n = 0.3 for the material.
P
x
Normal Strain: Since the material is confined in a rigid cylinder. ex = ey = 0.
Applying the generalized Hooke’s Law,
ex =
1
C s - v(sy + sx) D
E z
0 = sx - v(sy + sz)
ey =
[1]
1
C s - v(sx + sz) D
E y
0 = sy - v(sx + sz)
[2]
Solving Eqs.[1] and [2] yields:
sx = sy =
v
s
1 - v z
Thus,
ez =
=
1
C s - v(sx + sy) D
E z
v
v
1
csz - v a
sz +
s bd
E
1 - v
1 - v z
sz
=
E
c1 -
2v2
d
1 - v
=
sz 1 - v - 2v2
c
d
E
1 - v
=
sz (1 + v)(1 - 2v
c
d
E
1 - v
Thus, when the material is not being confined and undergoes the same normal strain
of ez, then the requtred modulus of elasticity is
E¿ =
sz
ez
=
The increased factor is k =
1 - v
E
(1 - 2v)(1 + v)
E¿
1 - v
=
E
(1 - 2v)(1 + v)
=
1 - 0.3
[1 - 2(0.3)](1 + 0.3)
= 1.35
Ans.
793
y
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10–59. A material is subjected to plane stress. Express
the distortion-energy theory of failure in terms of sx , sy ,
and txy .
Maximum distortion energy theory:
(s21 - s1 s2 + s22) = s2Y
s1,2 =
sx + sy
;
2
Let a =
sx + sy
2
s1 = a + b;
A
a
(1)
sx - sy
2
and b =
A
a
2
2
b + txy
sx - sy
2
2
2
b + txy
s2 = a - b
s21 = a2 + b2 + 2 a b;
s22 = a2 + b2 - 2 a b
s1 s2 = a2 - b2
From Eq. (1)
(a2 + b2 + 2 a b - a2 + b2 + a2 + b2 - 2 a b) = s2y
(a2 + 3 b2) = s2Y
(sx + sy)2
4
+ 3
(sx - sy)2
4
+ 3 t2xy = s2Y
s2x + s2y - sxsy + 3 t2xy = s2Y
Ans.
*10–60. A material is subjected to plane stress. Express
the maximum-shear-stress theory of failure in terms of sx ,
sy , and txy . Assume that the principal stresses are of
different algebraic signs.
Maximum shear stress theory:
|s1 - s2| = sY
s1,2 =
(1)
sx + sy
2
` s1 - s2 ` = 2
a
sx - sy
;
A
a
sx - sy
A
2
2
2
2
b + txy
2
b + txy
2
From Eq. (1)
4 ca
sx - sy
2
2
b + t2xy d = s2Y
2
(sx - sy) + 4 t2xy = s2Y
Ans.
794
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•10–61.
An aluminum alloy 6061-T6 is to be used for
a solid drive shaft such that it transmits 40 hp at 2400
rev>min. Using a factor of safety of 2 with respect to
yielding, determine the smallest-diameter shaft that can be
selected based on the maximum-shear-stress theory.
v = a2400
T =
2p rad 1 min
rev
ba
ba
b = 80 p rad>s
min
rev
60s
40 (550) (12)
P
3300 #
=
=
lb in.
p
v
80 p
Tc
J
Applying t =
t =
A 3300
p B c
=
p 4
2 c
6600
p3 c3
The principal stresses:
s1 = t =
6600
;
p2 c3
s2 = -t =
6600
p2 c3
Maximum shear stress theory: Both principal stresses have opposite sign, hence,
` s1 - s2 ` =
2a
sY
;
F.S.
37 (103)
6600
b = `
`
2 3
2
pc
c = 0.4166 in.
d = 0.833 in.
Ans.
10–62. Solve Prob. 10–61 using the maximum-distortionenergy theory.
v = a2400
T =
2p rad 1 min
rev
ba
ba
b = 80 p rad>s
min
rev
60s
40 (550) (12)
P
3300
=
=
lb.in.
p
v
80 p
Applying t =
t =
A 3300
p B c
p 4
2 c
=
Tc
J
6600
p2 c3
The principal stresses:
s1 = t =
6600
;
p2 c3
s2 = - t = -
6600
p2 c3
The maximum distortion-energy theory:
s21 - s1 s2 + s22 = a
3B
sY 2
b
F.S.
37(103) 2
6600 2
=
a
b
R
2
p2 c3
c = 0.3971 in.
d = 0.794 in.
Ans.
795
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10–63. An aluminum alloy is to be used for a drive shaft
such that it transmits 25 hp at 1500 rev>min. Using a factor
of safety of 2.5 with respect to yielding, determine the
smallest-diameter shaft that can be selected based on the
maximum-distortion-energy theory. sY = 3.5 ksi.
1500(2p)
= 50p
60
T =
P
v
T =
25(550)(12)
3300
=
p
50p
t =
Tc
,
J
t =
3300
p c
p 4
2c
s1 =
v =
J =
=
p 4
c
2
6600
p2c3
6600
p2c3
s2 =
s21 - s1 s2 + s22 = a
3a
-6600
p2c3
sY 2
b
F.S.
3.5(103) 2
6600 2
b
b
=
a
2.5
p2c3
c = 0.9388 in.
d = 1.88 in.
Ans.
*10–64. A bar with a square cross-sectional area is made
of a material having a yield stress of sY = 120 ksi. If the bar
is subjected to a bending moment of 75 kip # in., determine
the required size of the bar according to the maximumdistortion-energy theory. Use a factor of safety of 1.5 with
respect to yielding.
Normal and Shear Stress: Applying the flexure formula,
75 A 2 B
Mc
450
= 1 4 = 3
I
a
12 a
a
s =
In-Plane Principal Stress: Since no shear stress acts on the element
s1 = sx =
450
a3
s2 = sy = 0
Maximum Distortion Energy Theory:
s21 - s1 s2 + s22 = s2allow
a
120 2
450 2
b - 0 + 0 = a
b
3
1.5
a
a = 1.78 in.
Ans.
796
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•10–65.
Solve Prob. 10–64 using the maximum-shearstress theory.
Normal and Shear Stress: Applying the flexure formula,
75 A 2 B
Mc
450
= 1 4 = 3
I
a
a
12
a
s =
In-Plane Principal Stress: Since no shear stress acts on the element.
s1 = sx =
450
a3
s2 = sx = 0
Maximum Shear Stress Theory:
|s2| = 0 6 sallow =
120
= 80.0 ksi
1.5
(O.K!)
|s1| = sallow
120
450
=
1.5
a3
a = 1.78 in.
Ans.
10–66. Derive an expression for an equivalent torque Te
that, if applied alone to a solid bar with a circular cross
section, would cause the same energy of distortion as
the combination of an applied bending moment M and
torque T.
t =
Te c
J
Principal stress:
s1 = tx ¿
ud =
s2 = -t
1 + v 2
(s1 - s1 s2 + s22)
3E
(ud)1 =
1 + v
1 + v 3 T2x c2
( 3 t2) =
a
b
3E
3E
J2
Bending moment and torsion:
s =
Mc
;
I
t =
Tc
J
Principal stress:
s1, 2 =
s1 =
s + 0
s - 0 2
2
;
a
b + t
2
A
2
s
s2
+
+ t2 ;
2
A4
s2 =
s
s2
+ t2
2
A4
797
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10–66.
Continued
Let a =
s
2
b =
s2
+ t2
A4
s21 = a2 + b2 + 2 a b
s1 s2 = a2 - b2
s22 = a2 + b2 - 2 a b
s21 - s1 s2 + s22 = 3 b2 + a2
ud =
1 + v 2
(s1 - s1 s2 + s22)
3E
(ud)2 =
=
1 + v
1 + v 3 s2
s2
(3 b2 + a2) =
a
+ 3t2 +
b
3E
3E
4
4
c2(1 + v) M2
1 + v 2
3 T2
b
(s + 3 t2) =
a 2 +
3E
3E
I
J2
(ud)1 = (ud)2
c3(1 + v) 3 Tx 2
c2(1 + v) M2
3 T2
=
a
+
b
3E
3E
J2
I2
J2
For circular shaft
p 4
J 3c
=p 4=2
I 4c
Te =
J2 M2
+ T2
A I2 3
Te =
4 2
M + T2
A3
Ans.
10–67. Derive an expression for an equivalent bending
moment Me that, if applied alone to a solid bar with a
circular cross section, would cause the same energy of
distortion as the combination of an applied bending
moment M and torque T.
Principal stresses:
s1 =
Me c
;
I
ud =
1 + v 2
(s1 - s1 s2 + s22)
3E
(ud)1 =
s2 = 0
1 + v M2e c2
a 2 b
3E
I
(1)
798
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10–67.
Continued
Principal stress:
s + 0
s - 0 2
3
;
a
b + t
2
A
2
s1, 2 =
s1 =
s
s2
+
+ t2;
2
A4
s2 =
s
s2
+ t2
2
A4
Distortion Energy:
s
s2
,b =
+ t2
2
A4
Let a =
s21 = a2 + b2 + 2 a b
s1 s2 = a2 - b2
s22 = a2 + b2 - 2 a b
s22 - s1 s2 + s22 = 3 b2 + a2
Apply s =
Mc
;
I
t =
Tc
J
(ud)2 =
3s2
1 + v
1 + v s2
(3 b2 + a2) =
a
+
+ 3 t2 b
3E
3E
4
4
=
1 + v 2
1 + v M2 c2
3 T2 c2
(s + 3 t2) =
a 2 +
b
3E
3E
I
J2
(2)
Equating Eq. (1) and (2) yields:
(1 + v) Me c2
1 + v M2 c2
3T2 c2
a 2 b =
a 2 +
b
3E
3E
I
I
J2
M2e
2
=
I
M1
3 T2
+
2
I
J2
M2e = M1 + 3 T2 a
I 2
b
J
For circular shaft
p 4
I
1
4 c
= p 4 =
J
2
c
2
1 2
Hence, M2e = M2 + 3 T2 a b
2
Me =
A
M2 +
3 2
T
4
Ans.
799
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*10–68. The short concrete cylinder having a diameter of
50 mm is subjected to a torque of 500 N # m and an axial
compressive force of 2 kN. Determine if it fails according to
the maximum-normal-stress theory. The ultimate stress of
the concrete is sult = 28 MPa.
A =
p
(0.05)2 = 1.9635(10 - 3) m2
4
J =
p
(0.025)4 = 0.61359(10 - 4) m4
2
2 kN
500 N⭈m
500 N⭈m
2 kN
3
s =
2(10 )
P
= 1.019 MPa
=
A
1.9635(10 - 3)
t =
500(0.025)
Tc
= 20.372 MPa
=
J
0.61359(10 - 6)
sx = 0
sy = -1.019 MPa
sx + sy
s1, 2 =
s1,2 =
2
;
A
a
sx - sy
2
txy = 20.372 MPa
2
b + txy
2
0 - 1.018
0 - (-1.019) 2
2
;
a
b + 20.372
2
A
2
s1 = 19.87 MPa
s2 = -20.89 MPa
Failure criteria:
|s1| 6 salt = 28 MPa
OK
|s2| 6 salt = 28 MPa
OK
No.
Ans.
•10–69.
Cast iron when tested in tension and compression
has an ultimate strength of 1sult2t = 280 MPa and
1sult2c = 420 MPa, respectively. Also, when subjected to
pure torsion it can sustain an ultimate shear stress of
tult = 168 MPa. Plot the Mohr’s circles for each case and
establish the failure envelope. If a part made of this
material is subjected to the state of plane stress shown,
determine if it fails according to Mohr’s failure criterion.
120 MPa
100 MPa
220 MPa
s1 = 50 + 197.23 = 247 MPa
s2 = 50 - 197.23 = -147 MPa
The principal stress coordinate is located at point A which is outside the shaded
region. Therefore the material fails according to Mohr’s failure criterion.
Yes.
Ans.
800
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10–69.
Continued
10–70. Derive an expression for an equivalent bending
moment Me that, if applied alone to a solid bar with a
circular cross section, would cause the same maximum
shear stress as the combination of an applied moment M
and torque T. Assume that the principal stresses are of
opposite algebraic signs.
Bending and Torsion:
Mc
Mc
4M
= p 4 =
;
I
c
p c3
4
s =
t =
Tc
Tc
2T
= p 4 =
J
c
p
c3
2
The principal stresses:
s1, 2 =
=
tabs
max
sx + sy
2
;
A
a
sx - sy
2
2
2
b + txy =
4M
4M
pc3 + 0
2
3
;
Q
¢pc
- 0
2
2
≤ + a
2T
3
pc
b
2
2
2M
;
2M2 + T2
p c3
p c3
= s1 - s2 = 2 c
2
2M2 + T2 d
p c3
(1)
Pure bending:
s1 =
tabs
max
Me c
4 Me
Mc
= p 4 =
;
I
c
p c3
4
= s1 - s2 =
s2 = 0
4 Me
(2)
p c3
Equating Eq. (1) and (2) yields:
4 Me
4
2M2 + T2 =
p c3
p c3
Me = 2M2 + T2
Ans.
801
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10–71. The components of plane stress at a critical point
on an A-36 steel shell are shown. Determine if failure
(yielding) has occurred on the basis of the maximum-shearstress theory.
60 MPa
40 MPa
In accordance to the established sign convention, sx = 70 MPa, sy = -60 MPa and
txy = 40 MPa.
s1, 2 =
=
sx + sy
2
;
A
a
sx - sy
2
70 MPa
2
b + txy
2
70 + (-60)
70 - (-60) 2
2
;
c
d + 40
2
A
2
= 5 ; 25825
s1 = 81.32 MPa
s2 = -71.32 MPa
In this case, s1 and s2 have opposite sign. Thus,
|s1 - s2| = |81.32 - (-71.32)| = 152.64 MPa 6 sy = 250 MPa
Based on this result, the steel shell does not yield according to the maximum shear
stress theory.
*10–72. The components of plane stress at a critical point
on an A-36 steel shell are shown. Determine if failure
(yielding) has occurred on the basis of the maximumdistortion-energy theory.
60 MPa
40 MPa
In accordance to the established sign convention, sx = 70 MPa, sy = -60 MPa and
txy = 40 MPa.
s1, 2 =
=
sx + sy
2
;
A
a
sx - sy
2
2
b + txy
2
70 + (-60)
70 - ( -60) 2
2
;
c
d + 40
2
A
2
= 5 ; 25825
s1 = 81.32 MPa
s2 = -71.32 MPa
s1 2 - s1 s2 + s2 2 = 81.322 - 81.32(-71.32) + (-71.32)2 = 17,500 6 sy 2 = 62500
Based on this result, the steel shell does not yield according to the maximum
distortion energy theory.
802
70 MPa
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•10–73.
If the 2-in. diameter shaft is made from brittle
material having an ultimate strength of sult = 50 ksi for
both tension and compression, determine if the shaft fails
according to the maximum-normal-stress theory. Use a
factor of safety of 1.5 against rupture.
30 kip
4 kip · ft
Normal Stress and Shear Stresses. The cross-sectional area and polar moment of
inertia of the shaft’s cross-section are
A = p A 12 B = pin2
J =
p 4
p
A 1 B = in4
2
2
The normal stress is caused by axial stress.
s =
N
30
= = -9.549 ksi
p
A
The shear stress is contributed by torsional shear stress.
t =
4(12)(1)
Tc
=
= 30.56 ksi
p
J
2
The state of stress at the points on the surface of the shaft is represented on the
element shown in Fig. a.
In-Plane Principal Stress. sx = -9.549 ksi, sy = 0 and txy = -30.56 ksi. We have
s1, 2 =
=
sx + sy
2
;
A
a
sx - sy
2
2
b + txy
2
-9.549 - 0 2
-9.549 + 0
2
;
a
b + ( -30.56)
2
A
2
= (-4.775 ; 30.929) ksi
s1 = 26.15 ksi
s2 = -35.70 ksi
Maximum Normal-Stress Theory.
sallow =
sult
50
=
= 33.33 ksi
F.S.
1.5
|s1| = 26.15 ksi 6 sallow = 33.33 ksi
(O.K.)
|s2| = 35.70 ksi 7 sallow = 33.33 ksi
(N.G.)
Based on these results, the material fails according to the maximum normal-stress
theory.
803
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10–74. If the 2-in. diameter shaft is made from cast
iron having tensile and compressive ultimate strengths
of 1sult2t = 50 ksi and 1sult2c = 75 ksi, respectively,
determine if the shaft fails in accordance with Mohr’s
failure criterion.
Normal Stress and Shear Stresses. The cross-sectional area and polar moment of
inertia of the shaft’s cross-section are
A = p A 12 B = p in2
J =
p 4
p
A 1 B = in4
2
2
The normal stress is contributed by axial stress.
s =
N
30
= = -9.549 ksi
p
A
The shear stress is contributed by torsional shear stress.
t =
4(12)(1)
Tc
=
= 30.56 ksi
p
J
2
The state of stress at the points on the surface of the shaft is represented on the
element shown in Fig. a.
In-Plane Principal Stress. sx = -9.549 ksi, sy = 0, and txy = -30.56 ksi. We have
s1, 2 =
=
sx + sy
2
;
A
a
sx - sy
2
2
b + txy
2
-9.549 - 0 2
-9.549 + 0
2
;
a
b + ( -30.56)
2
A
2
= ( -4.775 ; 30.929) ksi
s1 = 26.15 ksi
s2 = -35.70 ksi
Mohr’s Failure Criteria. As shown in Fig. b, the coordinates of point A, which
represent the principal stresses, are located inside the shaded region. Therefore, the
material does not fail according to Mohr’s failure criteria.
804
30 kip
4 kip · ft
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10–75. If the A-36 steel pipe has outer and inner
diameters of 30 mm and 20 mm, respectively, determine the
factor of safety against yielding of the material at point A
according to the maximum-shear-stress theory.
900 N
150 mm
A
100 mm
200 mm
Internal Loadings. Considering the equilibrium of the free - body diagram of the
post’s right cut segment Fig. a,
©Fy = 0; Vy + 900 - 900 = 0
Vy = 0
©Mx = 0; T + 900(0.4) = 0
T = -360 N # m
©Mz = 0; Mz + 900(0.15) - 900(0.25) = 0 Mz = 90 N # m
Section Properties. The moment of inertia about the z axis and the polar moment of
inertia of the pipe’s cross section are
Iz =
p
A 0.0154 - 0.014 B = 10.15625p A 10 - 9 B m4
4
J =
p
A 0.0154 - 0.014 B = 20.3125p A 10 - 9 B m4
2
Normal Stress and Shear Stress. The normal stress is contributed by bending stress.
Thus,
sY = -
MyA
90(0.015)
= = -42.31MPa
Iz
10.15625p A 10 - 9 B
The shear stress is contributed by torsional shear stress.
t =
360(0.015)
Tc
=
= 84.62 MPa
J
20.3125p A 10 - 9 B
The state of stress at point A is represented by the two - dimensional element shown
in Fig. b.
In - Plane Principal Stress. sx = -42.31 MPa, sz = 0 and txz = 84.62 MPa. We have
s1, 2 =
=
sx + sz
2
;
A
a
sx - sz
2
2
b + txz
2
-42.31 - 0 2
-42.31 + 0
2
;
a
b + 84.62
2
A
2
= (-21.16 ; 87.23) MPa
s1 = 66.07 MPa
s2 = -108.38 MPa
805
200 mm
900 N
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10–75.
Continued
Maximum Shear Stress Theory. s1 and s2 have opposite signs. This requires
|s1 - s2| = sallow
66.07 - (-108.38) = sallow
sallow = 174.45 MPa
The factor of safety is
F.S. =
sY
250
=
= 1.43
sallow
174.45
Ans.
806
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*10–76. If the A-36 steel pipe has an outer and inner
diameter of 30 mm and 20 mm, respectively, determine the
factor of safety against yielding of the material at point A
according to the maximum-distortion-energy theory.
900 N
150 mm
A
100 mm
200 mm
Internal Loadings: Considering the equilibrium of the free - body diagram of the
pipe’s right cut segment Fig. a,
©Fy = 0; Vy + 900 - 900 = 0
Vy = 0
©Mx = 0; T + 900(0.4) = 0
T = -360 N # m
©Mz = 0; Mz + 900(0.15) - 900(0.25) = 0 Mz = 90 N # m
Section Properties. The moment of inertia about the z axis and the polar moment of
inertia of the pipe’s cross section are
Iz =
p
A 0.0154 - 0.014 B = 10.15625p A 10 - 9 B m4
4
J =
p
A 0.0154 - 0.014 B = 20.3125p A 10 - 9 B m4
2
Normal Stress and Shear Stress. The normal stress is caused by bending stress. Thus,
sY = -
MyA
90(0.015)
= = -42.31MPa
Iz
10.15625p A 10 - 9 B
The shear stress is caused by torsional stress.
t =
360(0.015)
Tc
=
= 84.62 MPa
J
20.3125p A 10 - 9 B
The state of stress at point A is represented by the two -dimensional element shown
in Fig. b.
In - Plane Principal Stress. sx = -42.31 MPa, sz = 0 and txz = 84.62 MPa. We have
s1, 2 =
=
sx + sz
2
;
A
a
sx - sz
2
2
b + txz
2
-42.31 - 0 2
-42.31 + 0
2
;
a
b + 84.62
2
A
2
= (-21.16 ; 87.23) MPa
s1 = 66.07 MPa
s2 = -108.38 MPa
807
200 mm
900 N
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10–76.
Continued
Maximum Distortion Energy Theory.
s1 2 - s1s2 + s2 2 = sallow 2
66.072 - 66.07(-108.38) + (-108.38)2 = sallow 2
sallow = 152.55 MPa
Thus, the factor of safety is
F.S. =
sY
250
=
= 1.64
sallow
152.55
Ans.
808
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•10–77.
The element is subjected to the stresses shown. If
sY = 36 ksi, determine the factor of safety for the loading
based on the maximum-shear-stress theory.
sx = 4 ksi
s1, 2 =
=
sy = -12 ksi
sx + sy
;
2
A
a
sx - sy
2
12 ksi
txy = -8 ksi
4 ksi
2
b + txy
2
8 ksi
4 - (-12) 2
4 - 12
2
;
a
b + (-8)
2
A
2
s2 = -15.314 ksi
s1 = 7.314 ksi
tabsmax =
7.314 - (-15.314)
s1 - s2
=
= 11.314 ksi
2
2
tallow =
sY
36
=
= 18 ksi
2
2
F.S. =
tallow
18
=
= 1.59
abs
tmax
11.314
Ans.
10–78. Solve Prob. 10–77 using the maximum-distortionenergy theory.
sx = 4 ksi
s1, 2 =
=
sy = -12 ksi
sx + sy
2
;
A
a
sx - sy
txy = -8 ksi
4 ksi
2
b + txy
2
8 ksi
4 - (-12) 2
4 - 12
2
;
a
b + (-8)
2
A
2
s1 = 7.314 ksi
s2 = -15.314 ksi
s1 2 - s1 s2 + s2 2 = a
F.S. =
2
12 ksi
sY 2
b
F.S.
362
= 1.80
A (7.314)2 - (7.314)(-15.314) + (-15.314)2
Ans.
809
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