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c2014
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INSTRUCTOR'S
SOLUTION
MANUAL
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
1–1. The shaft is supported by a smooth thrust bearing at B
and a journal bearing at C. Determine the resultant internal
loadings acting on the cross section at E.
B
A
4 ft
C
E
4 ft
4 ft
D
4 ft
400 lb
800 lb
Support Reactions: We will only need to compute Cy by writing the moment
equation of equilibrium about B with reference to the free-body diagram of the
entire shaft, Fig. a.
a + ©MB = 0;
Cy(8) + 400(4) - 800(12) = 0
Cy = 1000 lb
Internal Loadings: Using the result for Cy, section DE of the shaft will be
considered. Referring to the free-body diagram, Fig. b,
+
: ©Fx = 0;
NE = 0
+ c ©Fy = 0;
VE + 1000 - 800 = 0
Ans.
Ans.
VE = - 200 lb
a + ©ME = 0; 1000(4) - 800(8) - ME = 0
ME = - 2400 lb # ft = - 2.40 kip # ft Ans.
The negative signs indicates that VE and ME act in the opposite sense to that shown
on the free-body diagram.
Ans:
NE = 0, VE = - 200 lb, ME = - 2.40 kip # ft
1
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
a
1–2. Determine the resultant internal normal and shear
force in the member at (a) section a–a and (b) section b–b,
each of which passes through point A. The 500-lb load is
applied along the centroidal axis of the member.
b
30⬚
500 lb
500 lb
A
b
a
(a)
+
: ©Fx = 0;
Na - 500 = 0
Na = 500 lb
Ans.
+ T©Fy = 0;
Va = 0
Ans.
R+ ©Fx = 0;
Nb - 500 cos 30° = 0
(b)
Ans.
Nb = 433 lb
+Q©F = 0;
y
Vb - 500 sin 30° = 0
Vb = 250 lb
Ans.
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Ans:
Na = 500 lb, Va = 0,
Nb = 433 lb, Vb = 250 lb
2
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
1–3. The beam AB is fixed to the wall and has a uniform
weight of 80 lb>ft. If the trolley supports a load of 1500 lb,
determine the resultant internal loadings acting on the cross
sections through points C and D.
5 ft
20 ft
10 ft
3 ft
A
B
C
D
1500 lb
Segment BC:
+
; ©Fx = 0;
NC = 0
+ c ©Fy = 0;
VC - 2.0 - 1.5 = 0
Ans.
VC = 3.50 kip
a + ©MC = 0;
Ans.
-MC - 2(12.5) - 1.5 (15) = 0
MC = - 47.5 kip # ft
Ans.
Segment BD:
+
; ©Fx = 0;
ND = 0
Ans.
+ c ©Fy = 0;
VD - 0.24 = 0
VD = 0.240 kip
a + ©MD = 0;
Ans.
-MD - 0.24 (1.5) = 0
MD = - 0.360 kip # ft
Ans.
Ans:
NC = 0, VC = 3.50 kip, MC = - 47.5 kip # ft,
ND = 0, VD = 0.240 kip, MD = -0.360 kip # ft
3
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*1–4. The shaft is supported by a smooth thrust bearing at A
and a smooth journal bearing at B. Determine the resultant
internal loadings acting on the cross section at C.
600 N/m
A
B
D
C
1m
1m
1m
1.5 m
1.5 m
900 N
Support Reactions: We will only need to compute By by writing the moment
equation of equilibrium about A with reference to the free-body diagram of the
entire shaft, Fig. a.
a + ©MA = 0;
By(4.5) - 600(2)(2) - 900(6) = 0
By = 1733.33 N
Internal Loadings: Using the result of By, section CD of the shaft will be
considered. Referring to the free-body diagram of this part, Fig. b,
+
Ans.
; ©Fx = 0;
NC = 0
VC = - 233 N
+ c ©Fy = 0;
VC - 600(1) + 1733.33 - 900 = 0
a + ©MC = 0;
1733.33(2.5) - 600(1)(0.5) - 900(4) - MC = 0
MC = 433 N # m
Ans.
Ans.
The negative sign indicates that VC act in the opposite sense to that shown on the
free-body diagram.
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4
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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
E
B
6 ft
4 ft
C
4 ft
Support Reactions: For member AB
a + ©MB = 0;
+
: ©Fx = 0;
+ c ©Fy = 0;
9.00(4) - Ay(12) = 0
Ay = 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.
Ans:
ND = 0, VD = 0.750 kip, MD = 13.5 kip # ft,
NE = 0, VE = - 9.00 kip, ME = - 24.0 kip # ft
5
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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
0.5 m
C
0.75 m
0.75 m
A
0.75 m
P
Support Reactions:
a + ©MA = 0;
8(2.25) - T(0.6) = 0
T = 30.0 kN
+
: ©Fx = 0;
30.0 - A x = 0
A x = 30.0 kN
+ c ©Fy = 0;
Ay - 8 = 0
A y = 8.00 kN
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.
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Negative signs indicate that NC and VC act in the opposite direction to that shown
on FBD.
Ans:
NC = - 30.0 kN, VC = - 8.00 kN,
MC = 6.00 kN # m
6
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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
0.75 m
0.75 m
A
0.75 m
P
Support Reactions:
a + ©MA = 0;
P(2.25) - 2(0.6) = 0
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.
Ans:
P = 0.533 kN, NC = - 2.00 kN, VC = - 0.533 kN,
MC = 0.400 kN # m
7
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*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
a + ©MB = 0;
- Ay(4) + 6(3.5) +
B
A
Referring to the FBD of the entire beam, Fig. a,
1
(3)(3)(2) = 0
2
Ay = 7.50 kN
C
0.5 m 0.5 m
Referring to the FBD of this segment, Fig. b,
+
: ©Fx = 0;
+ c ©Fy = 0;
a + ©MC = 0;
NC = 0
7.50 - 6 - VC = 0
Ans.
VC = 1.50 kN
MC + 6(0.5) - 7.5(1) = 0
Ans.
MC = 4.50 kN # m
Ans.
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8
D
1.5 m
1.5 m
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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
B
A
C
0.5 m 0.5 m
D
1.5 m
1.5 m
Referring to the FBD of the entire beam, Fig. a,
a + ©MA = 0;
By(4) - 6(0.5) -
1
(3)(3)(2) = 0
2
By = 3.00 kN
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) -
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
Ans.
Ans:
ND = 0, VD = - 1.875 kN,
MD = 3.94 kN # m
9
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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.
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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.
Ans:
NA = 0, VA = 450 lb, MA = - 1.125 kip # ft,
NB = 0, VB = 850 lb, MB = - 6.325 kip # ft,
VC = 0, NC = - 1.20 kip, MC = - 8.125 kip # ft
10
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
1–11. The forearm and biceps support the 2-kg load at A. If C
can be assumed as a pin support, determine the resultant
internal loadings acting on the cross section of the bone of the
forearm at E.The biceps pulls on the bone along BD.
D
75⬚
A
C E B
Support Reactions: In this case, all the support reactions will be completed.
Referring to the free-body diagram of the forearm, Fig. a,
a + ©MC = 0;
FBD sin 75°(0.07) - 2(9.81)(0.3) = 0
FBD = 87.05 N
+
: ©Fx = 0;
Cx - 87.05 cos 75° = 0
Cx = 22.53 N
+ c ©Fy = 0;
87.05 sin 75° - 2(9.81) - Cy = 0
Cy = 64.47 N
230 mm
35 mm 35 mm
Internal Loadings: Using the results of Cx and Cy, section CE of the forearm will be
considered. Referring to the free-body diagram of this part shown in Fig. b,
+
: ©Fx = 0;
NE + 22.53 = 0
NE = - 22.5 N
Ans.
+ c ©Fy = 0;
- VE - 64.47 = 0
VE = - 64.5 N
Ans.
a + ©ME = 0;
ME + 64.47(0.035) = 0
ME = - 2.26 N # m
Ans.
The negative signs indicate that NE, VE and ME act in the opposite sense to that
shown on the free-body diagram.
Ans:
NE = - 22.5 N, VE = - 64.5 N, ME = - 2.26 N # m
11
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*1–12. The serving tray T used on an airplane is supported
on each side by an arm. The tray is pin connected to the arm
at A, and at B there is a smooth pin. (The pin can move
within the slot in the arms to permit folding the tray against
the front passenger seat when not in use.) Determine the
resultant internal loadings acting on the cross section of the
arm through point C when the tray arm supports the loads
shown.
12 N
9N
15 mm
B
60⬚
500 mm
VC
C
MC
NC
b+ ©Fx = 0;
NC + 9 cos 30° + 12 cos 30° = 0;
NC = - 18.2 N
Ans.
a+ ©Fy = 0;
VC - 9 sin 30° - 12 sin 30° = 0;
VC = 10.5 N
Ans.
a + ©MC = 0; - MC - 9(0.5 cos 60° + 0.115) - 12(0.5 cos 60° + 0.265) = 0
MC = - 9.46 N # m
Ans.
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12
100 mm
A
150 mm
T
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
1–13. The blade of the hacksaw is subjected to a
pretension force of F = 100 N. Determine the resultant
internal loadings acting on section a–a that passes through
point D.
a
225 mm
30⬚ b
B
A
D
b
F
150 mm
a
F
C
Internal Loadings: Referring to the free-body diagram of the section of the hacksaw
shown in Fig. a,
+
; ©Fx = 0;
Na - a + 100 = 0
+ c ©Fy = 0;
Va - a = 0
a + ©MD = 0;
- Ma - a - 100(0.15) = 0
Na - a = - 100 N
Ans.
Ans.
Ma - a = - 15 N # m
Ans.
The negative sign indicates that Na–a and Ma–a act in the opposite sense to that
shown on the free-body diagram.
Ans:
Na - a = - 100 N, Va - a = 0, Ma - a = - 15 N # m
13
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
1–14. The blade of the hacksaw is subjected to a
pretension force of F = 100 N. Determine the resultant
internal loadings acting on section b–b that passes through
point D.
a
225 mm
30⬚ b
B
A
D
b
F
150 mm
a
F
C
Internal Loadings: Referring to the free-body diagram of the section of the hacksaw
shown in Fig. a,
©Fx¿ = 0;
Nb - b + 100 cos 30° = 0
Nb - b = - 86.6 N
Ans.
©Fy¿ = 0;
Vb - b - 100 sin 30° = 0
Vb - b = 50 N
Ans.
a + ©MD = 0;
- Mb - b - 100(0.15) = 0
Mb - b = - 15 N # m
Ans.
The negative sign indicates that Nb–b and Mb–b act in the opposite sense to that
shown on the free-body diagram.
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Ans:
Nb - b = - 86.6 N, Vb - b = 50 N,
Mb - b = - 15 N # m
14
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
1–15. A 150-lb bucket is suspended from a cable on the
wooden frame. Determine the resultant internal loadings
on the cross section at D.
1 ft 1 ft
2 ft
B
D
H
C
2 ft
30⬚
G
1 ft
E
3 ft
Support Reactions: We will only need to compute Bx, By, and FGH . Referring to the
free-body diagram of member BC, Fig. a,
a + ©MB = 0:
FGH sin 45°(2) - 150(4) = 0
+
: ©Fx = 0;
424.26 cos 45° - Bx = 0
Bx = 300 lb
+ c ©Fy = 0;
424.26 sin 45° - 150 - By = 0
By = 150 lb
I
A
FGH = 424.26 lb
Internal Loadings: Using the results of Bx and By , section BD of member BC will be
considered. Referring to the free-body diagram of this part shown in Fig. b,
+
: ©Fx = 0;
ND - 300 = 0
ND = 300 lb
Ans.
+ c ©Fy = 0;
- VD - 150 = 0
VD = - 150 lb
Ans.
a + ©MD = 0;
150(1) + MD = 0
MD = -150 lb # ft
Ans.
The negative signs indicates that VD and MD act in the opposite sense to that shown
on the free-body diagram.
Ans:
ND = 300 lb, VD = - 150 lb, MD = -150 lb # ft
15
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*1–16. A 150-lb bucket is suspended from a cable on the
wooden frame. Determine the resultant internal loadings
acting on the cross section at E.
1 ft 1 ft
2 ft
B
D
2 ft
30⬚
G
1 ft
E
3 ft
Support Reactions: We will only need to compute Ax, Ay, and FBI. Referring to the
free-body diagram of the frame, Fig. a,
a + ©MA = 0;
FBI sin 30°(6) - 150(4) = 0
FBI = 200 lb
+
: ©Fx = 0;
Ax - 200 sin 30° = 0
Ax = 100 lb
+ c ©Fy = 0;
Ay - 200 cos 30° - 150 = 0
Ay = 323.21 lb
I
Internal Loadings: Using the results of Ax and Ay, section AE of member AB will be
considered. Referring to the free-body diagram of this part shown in Fig. b,
+
: ©Fx = 0;
NE + 323.21 = 0
NE = - 323 lb
Ans.
+ c ©Fy = 0;
100 - VE = 0
VE = 100 lb
Ans.
a + ©MD = 0;
100(3) - ME = 0
ME = 300 lb # ft
Ans.
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The negative sign indicates that NE acts in the opposite sense to that shown on the
free-body diagram.
16
A
H
C
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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.
A
45
1.5 m
a
45
3m
Referring to the FBD (section b–b) in Fig. c,
+
; ©Fx = 0;
Nb - b - 5 cos 45° + 5.303 = 0 Nb - b = - 1.768 kN
= - 1.77 kN
+ c ©Fy = 0;
a + ©MC = 0;
Vb - b - 5 sin 45° = 0
Vb - b = 3.536 kN = 3.54 kN
Ans.
Ans.
5.303 sin 45° (3) - 5(1.5) - Mb - b = 0
Mb - b = 3.75 kN # m
Ans.
Ans:
Na - a = - 3.75 kN, Va - a = 1.25 kN,
Ma - a = 3.75 kN # m, Nb - b = - 1.77 kN,
Vb - b = 3.54 kN # m, Mb - b = 3.75 kN # m
17
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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;
NC = - 80 lb
Ans.
Ans.
MC + 80(6) = 0;
MC = - 480 lb # in.
Ans.
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Ans:
NC = - 80 lb, VC = 0, MC = - 480 lb # in.
18
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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
B
D
3 ft
6 ft
Referring to the FBD of the entire beam, Fig. a,
a + ©MB = 0;
1
1
(6)(6)(2) + (6)(6)(10) - Ay(12) = 0 Ay = 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.
Ans:
NC = 0, VC = 4.50 kip, MC = 31.5 kip # ft
19
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*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
6 kip/ft
A
C
3 ft
3 ft
Referring to the FBD of the entire beam, Fig. a,
a + ©MB = 0;
1
1
(6)(6)(2) + (6)(6)(10) - Ay(12) = 0 Ay = 18.0 kip
2
2
Referring to the FBD of this segment, Fig. b,
+
: ©Fx = 0;
ND = 0
+ c ©Fy = 0;
a + ©MA = 0;
18.0 -
1
(6)(6) - VD = 0
2
MD - 18.0 (2) = 0
Ans.
VD = 0
Ans.
MD = 36.0 kip # ft
Ans.
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20
B
D
6 ft
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
1–21. 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.
200 mm
F 900 N
a
30
F 900 N
A
a
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.
Ans:
Na - a = 779 N, Va - a = 450 N, Ma - a = 180 N # m:
21
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
1–22. The metal stud punch is subjected to a force of 120 N
on the handle. Determine the magnitude of the reactive force
at the pin A and in the short link BC. Also, determine the
internal resultant loadings acting on the cross section passing
through the handle arm at D.
120 N
60
50 mm 100 mm
50 mm
E
B
D
30
100 mm
300 mm
A
C
200 mm
Member:
a +©MA = 0;
FBC cos 30°(50) - 120(500) = 0
FBC = 1385.6 N = 1.39 kN
+ c ©Fy = 0;
Ans.
Ay - 1385.6 - 120 cos 30° = 0
Ay = 1489.56 N
+
; ©Fx = 0;
Ax - 120 sin 30° = 0;
Ax = 60 N
FA = 21489.562 + 602
Ans.
= 1491 N = 1.49 kN
Segment:
a+ ©Fx¿ = 0;
www.elsolucionario.org
ND - 120 = 0
ND = 120 N
Ans.
+Q©F = 0;
y¿
VD = 0
Ans.
a + ©MD = 0;
MD - 120(0.3) = 0
MD = 36.0 N # m
Ans.
Ans:
FBC = 1.39 kN, FA = 1.49 kN, ND = 120 N,
VD = 0, MD = 36.0 N # m
22
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
1–23. Solve Prob. 1–22 for the resultant internal loadings
acting on the cross section passing through the handle arm
at E and at a cross section of the short link BC.
120 N
60
50 mm 100 mm
50 mm
E
B
30
D
100 mm
300 mm
A
C
200 mm
Member:
a + ©MA = 0;
FBC cos 30°(50) - 120(500) = 0
FBC = 1385.6 N = 1.3856 kN
Segment:
+b©F = 0;
x¿
NE = 0
a+ ©Fy¿ = 0;
VE - 120 = 0;
VE = 120 N
Ans.
a + ©ME = 0;
ME - 120(0.4) = 0;
ME = 48.0 N # m
Ans.
Ans.
Short link:
+
; ©Fx = 0;
V = 0
+ c ©Fy = 0;
1.3856 - N = 0;
a + ©MH = 0;
M = 0
Ans.
N = 1.39 kN
Ans.
Ans.
Ans:
NE = 0, VE = 120 N, ME = 48.0 N # m,
Short link: V = 0, N = 1.39 kN, M = 0
23
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*1–24. Determine the resultant internal loadings acting
on the cross section of the semicircular arch at C.
C
w0
r
u
A
p
a + ©MA = 0;
By (2r) -
(w0 r du)(cos u)(r sin u)
L0
p
-
L0
(w0 r du)(sin u)r(1 - cos u) = 0
p
By (2r) - w0 r2
sin u du = 0
L0
p
By (2r) - w0 r2( -cos u)]0 = 0
By = w0 r
+
: ©Fx = 0;
NC = - w0 r sin u
+ c ©Fy = 0;
- NC - w0 r
p
2
L0
p
2
L0
cos u du = 0
Ans.
= - w0 r
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w0 r + VC - w0 r
p
2
L0
sin u du = 0
w0 r + VC - w0 r( - cos u)
a + ©M0 = 0;
p
2
L0
= 0;
V2 = 0
Ans.
w0 r(r) - MC + ( - w0 r)(r) = 0
MC = 0
Ans.
24
B
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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
Ans.
©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.
3 ft
2
7 lb/ft
6 ft
B
A
4 ft
x
y
Ans:
(VB)x = 105 lb, (VB)y = 0, (NB)z = 0,
(MB)x = 0, (MB)y = 788 lb # ft,
(TB)z = 52.5 lb # ft
25
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
z
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.
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.
www.elsolucionario.org
Ans:
(VC)x = - 250 N, (NC)y = 0, (VC)z = - 240 N,
(MC)x = - 108 N # m, (TC)y = 0,
(MC)z = - 138 N # m
26
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z
1–27. The pipe assembly is subjected to a force of 600 N
at B. Determine the resultant internal loadings acting on
the cross section at C.
600 N
B
60
30
150 mm
150 mm
A
C
500 mm
Internal Loading: Referring to the free-body diagram of the section of the pipe
shown in Fig. a,
©Fx = 0; (NC)x - 600 cos 60° sin 30° = 0
(NC)x = 150 N
Ans.
©Fy = 0; (VC)y + 600 cos 60° cos 30° = 0
(VC)y = - 260 N
Ans.
©Fz = 0; (VC)z + 600 sin 60° = 0
(VC)z = - 520 N
Ans.
x
400 mm
y
©Mx = 0; (TC)x + 600 sin 60°(0.4) - 600 cos 60° cos 30°(0.5) = 0
Ans.
(TC)x = -77.9 N # m
©My = 0; (MC)y - 600 sin 60° (0.15) - 600 cos 60° sin 30°(0.5) = 0
(MC)y = 153 N # m
Ans.
©Mz = 0; (MC)z + 600 cos 60° cos 30°(0.15) + 600 cos 60° sin 30°(0.4) = 0
(MC)z = -99.0 N # m
Ans.
The negative signs indicate that (VC)y, (VC)z, (TC)x, and (MC)z act in the opposite
sense to that shown on the free-body diagram.
Ans:
(NC)x = 150 N, (VC)y = - 260 N,
(VC)z = - 520 N, (TC)x = -77.9 N # m,
(MC)y = 153 N # m, (MC)z = -99.0 N # m
27
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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
3 in. 9 in.
Fx 30 lb
A
Fz 10 lb
9 in.
6 in.
6 in.
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.
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28
6 in.
Fy 50 lb
y
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1–29. The curved rod AD of radius r has a weight per
length of w. If it lies in the vertical plane, determine the
resultant internal loadings acting on the cross section
through point B. Hint: The distance from the centroid C of
segment AB to point O is OC = [2r sin (u>2)]>u.
u
2
A
C
B
O
u
r
D
R+ ©Fx = 0;
NB + wru cos u = 0
NB = - wru cos u
+Q©F = 0;
y
Ans.
- VB - wru sin u = 0
VB = -wru sin u
Ans.
u 2r sin (u/2)
a + ©M0 = 0; wru a cos b a
b + (NB)r + MB = 0
2
u
MB = - NBr - wr2 2 sin (u/2) cos (u/2)
MB = wr2(u cos u - sin u)
Ans.
Ans:
NB = - wru cos u, VB = -wru sin u,
MB = wr2(u cos u - sin u)
29
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M dM
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.
V dV
N dN
M V
N
T
©Fx = 0;
N cos
du
du
du
du
+ V sin
- (N + dN) cos
+ (V + dV) sin
= 0
2
2
2
2
(1)
©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
(4)
www.elsolucionario.org
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
QED
30
T dT
du
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
1–31. The supporting wheel on a scaffold is held in place
on the leg using a 4-mm-diameter pin as shown. If the wheel
is subjected to a normal force of 3 kN, determine the
average shear stress developed in the pin. Neglect friction
between the inner scaffold puller leg and the tube used on
the wheel.
3 kN
+ c ©Fy = 0;
3 kN # 2V = 0;
V = 1.5 kN
3
tavg =
1.5(10 )
V
= 119 MPa
= p
2
A
4 (0.004)
Ans.
Ans:
tavg = 119 MPa
31
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*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
= 29.5 MPa
= p 6
A
( )2
Ans.
4 1000
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32
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¿ =
savg =
P sin u
N
P
=
=
sin2 u
A
A¿
A
sin u
tavg =
P cos u
V
=
A
A¿
sin u
=
A
.
sin u
Ans.
P
P
sin u cos u =
sin 2u
A
2A
Ans.
Ans:
savg =
33
P
P
sin2 u, tavg =
sin 2u
A
2A
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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
B
A
D
6 kN
C
8 kN
6 kN E
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 =
www.elsolucionario.org
Ans:
sD = 13.3 MPa (C), sE = 70.7 MPa (T)
34
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
1–35. If the turnbuckle is subjected to an axial force of
P = 900 lb, determine the average normal stress developed
in section a–a and in each of the bolt shanks at B and C.
Each bolt shank has a diameter of 0.5 in.
a
1 in.
A
0.25 in.
P
C
B
P
a
Internal Loading: The normal force developed in section a–a of the bracket and
the bolt shank can be obtained by writing the force equations of equilibrium along
the x axis with reference to the free-body diagrams of the sections shown in Figs. a
and b, respectively.
+
: ©Fx = 0;
900 - 2Na - a = 0
Na - a = 450 lb
+
: ©Fx = 0;
900 - Nb = 0
Nb = 900 lb
Average Normal Stress: The cross-sectional areas of section a–a and the bolt shank
p
are Aa - a = (1)(0.25) = 0.25 in2 and Ab = (0.52) = 0.1963 in2, respectively.
4
We obtain
(sa - a)avg =
sb =
Na - a
450
=
= 1800 psi = 1.80 ksi
Aa - a
0.25
Ans.
Nb
900
=
= 4584 psi = 4.58 ksi
Ab
0.1963
Ans.
Ans:
(sa - a)avg = 1.80 ksi, sb = 4.58 ksi
35
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*1–36. The average normal stresses developed in section a–a
of the turnbuckle, and the bolts shanks at B and C, are not
allowed to exceed 15 ksi and 45 ksi, respectively. Determine
the maximum axial force P that can be applied to the
turnbuckle. Each bolt shank has a diameter of 0.5 in.
a
0.25 in.
P
C
B
a
Internal Loading: The normal force developed in section a–a of the bracket and the
bolt shank can be obtained by writing the force equations of equilibrium along
the x axis with reference to the free-body diagrams of the sections shown in Figs. a
and b, respectively.
+
: ©Fx = 0;
P - 2Na - a = 0
Na - a = P>2
+
: ©Fx = 0;
P - Nb = 0
Nb = P
Average Normal Stress: The cross-sectional areas of section a–a and the bolt
p
shank are Aa - a = 1(0.25) = 0.25 in2 and Ab = (0.52) = 0.1963 in2, respectively.
4
We obtain
(sa - a)allow =
Na - a
;
Aa - a
15(103) =
P>2
0.25
P = 7500 lb = 7.50 kip (controls)
sb =
Nb
;
Ab
45(103) =
1 in.
A
Ans.
P
0.1963
P = 8336 lb = 8.84 kip
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36
P
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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 (15x 1/2) MPa
30 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)x 2 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.
Ans:
P = 40 MN, d = 2.40 m
37
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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.
800 lb
30
1 in.
1 in.
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.
www.elsolucionario.org
Ans:
s = 66.7 psi, t = 115 psi
38
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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
50 mm
100 mm
100 mm
50 mm
150 mm
The cross-sectional area of the block is A = 0.6(0.3) - 0.3(0.2) = 0.12 m2.
savg =
600(103)
P
=
= 5(106) Pa = 5 MPa
A
0.12
Ans.
The average normal stress distribution over the cross section of the block and the
state of stress of a point in the block represented by a differential volume element
are shown in Fig. a
Ans:
savg = 5 MPa
39
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*1–40. Determine the average normal stress in each of the
20-mm diameter bars of the truss. Set P = 40 kN.
C
P
1.5 m
B
A
2m
Internal Loadings: The force developed in each member of the truss can
be determined by using the method of joints. First, consider the equilibrium of
joint C, Fig. a,
+
: ©Fx = 0;
4
40 - FBC a b = 0
5
FBC = 50 kN (C)
+ c ©Fy = 0;
3
50 a b - FAC = 0
5
FAC = 30 kN (T)
Subsequently, the equilibrium of joint B, Fig. b, is considered
+
: ©Fx = 0;
4
50 a b - FAB = 0
5
FAB = 40 kN (T)
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Average Normal Stress: The cross-sectional area of each of the bars is
p
A = (0.022) = 0.3142(10 - 3) m2. We obtain,
4
50(103)
FBC
(savg)BC =
= 159 MPa
=
A
0.3142(10 - 3)
Ans.
(savg)AC =
30(103)
FAC
= 95.5 MPa
=
A
0.3142(10 - 3)
Ans.
(savg)AB =
40(103)
FAB
= 127 MPa
=
A
0.3142(10 - 3)
Ans.
40
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
1–41. If the average normal stress in each of the 20-mmdiameter bars is not allowed to exceed 150 MPa, determine
the maximum force P that can be applied to joint C.
C
P
1.5 m
B
A
2m
Internal Loadings: The force developed in each member of the truss can be determined
by using the method of joints. First, consider the equilibrium of joint C, Fig. a,
+
: ©Fx = 0;
4
P - FBC a b = 0
5
FBC = 1.25P(C)
+ c ©Fy = 0;
3
1.25Pa b - FAC = 0
5
FAC = 0.75P(T)
Subsequently, the equilibrium of joint B, Fig. b, is considered
+
: ©Fx = 0;
4
1.25P a b - FAB = 0
5
FAB = P(T)
Average Normal Stress: Since the cross-sectional area and the allowable normal stress of
each bar are the same, member BC which is subjected to the maximum normal force is
p
(0.022) =
4
0.3142(10 - 3) m2. We have,
the critical member. The cross-sectional area of each of the bars is A =
(savg)allow =
FBC
;
A
150(106) =
1.25P
0.3142(10 - 3)
P = 37 699 N = 37.7 kN
Ans.
Ans:
P = 37.7 kN
41
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1–42. Determine the average shear stress developed in
pin A of the truss. A horizontal force of P = 40 kN is
applied to joint C. Each pin has a diameter of 25 mm and is
subjected to double shear.
C
P
1.5 m
B
A
2m
Internal Loadings: The forces acting on pins A and B are equal to the support
reactions at A and B. Referring to the free-body diagram of the entire truss, Fig. a,
©MA = 0;
By(2) - 40(1.5) = 0
By = 30 kN
+
: ©Fx = 0;
40 - Ax = 0
Ax = 40 kN
+ c ©Fy = 0;
30 - Ay = 0
Ay = 30 kN
Thus,
FA = 2Ax2 + Ay2 = 2402 + 302 = 50 kN
Since pin A is in double shear, Fig. b, the shear forces developed on the shear planes
of pin A are
FA
50
VA =
=
= 25 kN
2
2
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Average Shear Stress: The area of the shear plane for pin A is AA =
p
(0.0252) =
4
0.4909(10 - 3) m2. We have
(tavg)A =
25(103)
VA
= 50.9 MPa
=
AA
0.4909(10 - 3)
Ans.
Ans:
(tavg)A = 50.9 MPa
42
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
1–43. The 150-kg bucket is suspended from end E of the
frame. Determine the average normal stress in the 6-mm
diameter wire CF and the 15-mm diameter short strut BD.
0.6 m
0.6 m
D
C
E
0.6 m
30
B
1.2 m
Internal Loadings: The normal force developed in rod BD and cable CF can be
determined by writing the moment equations of equilibrium about C and A with
reference to the free-body diagram of member CE and the entire frame shown in
Figs. a and b, respectively.
a + ©MC = 0;
FBD sin 45°(0.6) - 150(9.81)(1.2) = 0
FBD = 4162.03 N
a + ©MA = 0;
FCF sin 30°(1.8) - 150(9.81)(1.2) = 0
FCF = 1962 N
F
A
Average Normal Stress: The cross-sectional areas of rod BD and cable CF are
p
p
ABD = (0.0152) = 0.1767(10 - 3) m2 and ACF = (0.0062) = 28.274(10 - 6) m2.
4
4
We have
(savg)BD =
FBD
4162.03
= 23.6 MPa
=
ABD
0.1767(10 - 3)
Ans.
(savg)CF =
FCF
1962
= 69.4 MPa
=
ACF
28.274(10 - 6)
Ans.
Ans:
(savg)BD = 23.6 MPa, (savg)CF = 69.4 MPa
43
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*1–44. The 150-kg bucket is suspended from end E of the
frame. If the diameters of the pins at A and D are 6 mm and
10 mm, respectively, determine the average shear stress
developed in these pins. Each pin is subjected to double
shear.
0.6 m
FBD sin 45°(0.6) - 150(9.81)(1.2) = 0
E
0.6 m
30
Internal Loading: The forces exerted on pins D and A are equal to the support
reaction at D and A. First, consider the free-body diagram of member CE shown
in Fig. a.
a + ©MC = 0;
0.6 m
D
C
B
1.2 m
FBD = 4162.03 N
Subsequently, the free-body diagram of the entire frame shown in Fig. b will be
considered.
a + ©MA = 0;
FCF sin 30°(1.8) - 150(9.81)(1.2) = 0
FCF = 1962 N
+
: ©Fx = 0;
Ax - 1962 sin 30° = 0
Ax = 981 N
+ c ©Fy = 0;
Ay - 1962 cos 30° - 150(9.81) = 0
Ay = 3170.64 N
F
A
Thus, the forces acting on pins D and A are
FD = FBD = 4162.03 N
FA = 2Ax2 + Ay2 = 29812 + 3170.642 = 3318.93 N
Since both pins are in double shear
VD =
FD
= 2081.02 N
2
VA =
FA
= 1659.47 N
2
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Average Shear Stress: The cross-sectional areas of the shear plane of pins D and A
p
p
(0.012) = 78.540(10 - 6) m2 and AA = (0.0062) = 28.274(10 - 6) m2.
4
4
We obtain
VA
1659.47
(tavg)A =
= 58.7 MPa
=
Ans.
AA
28.274(10 - 6)
are AD =
(tavg)D =
VD
2081.02
= 26.5 MPa
=
AD
78.540(10 - 6)
Ans.
Ans:
x = 4 in., y = 4 in., s = 9.26 psi
44
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
1–45. The pedestal has a triangular cross section as shown.
If it is subjected to a compressive force of 500 lb, specify the
x and y coordinates for the location of point P(x, y), where
the load must be applied on the cross section, so that the
average normal stress is uniform. Compute the stress and
sketch its distribution acting on the cross section at a
location removed from the point of load application.
500 lb
12 in.
y
x
A B
P(x,y)
x =
1
12
+ 12(6)(12) 12
2 (3)(12) 3
3
1
(9)(12)
2
A B
= 4 in.
Ans.
y =
1
2
1
6
2 (3)(12)(3) 3 + 2 (6)(12) 3 + 3
1
2 (9)(12)
A B
A
B
Ans.
s =
500
P
= 9.26 psi
= 1
A
2 (9)(12)
= 4 in.
Ans.
45
3 in. 6 in.
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
1–46. The 20-kg chandelier is suspended from the wall
and ceiling using rods AB and BC, which have diameters of
3 mm and 4 mm, respectively. Determine the angle u so that
the average normal stress in both rods is the same.
C
A
30⬚
B
u
Internal Loadings: The force developed in cables AB and BC can be determined by
considering the equilibrium of joint B, Fig. a,
+
: ©Fx = 0;
FBC cos u - FAB cos 30° = 0
(1)
Average Normal Stress: The cross-sectional areas of cables AB and BC are
p
p
AAB = (0.0032) = 7.069(10 - 6) m2 and ABC = (0.0042) = 12.566(10 - 6) m2.
4
4
Since the average normal stress in both cables are required to be the same, then
(savg)AB = (savg)BC
FAB
FBC
=
AAB
ABC
FAB
7.069(10 - 6)
=
FBC
12.566(10 - 6)
FAB = 0.5625FBC
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(2)
Substituting Eq. (2) into Eq. (1),
FBC(cos u - 0.5625 cos 30°) = 0
Since FBC Z 0, then
cos u - 0.5625 cos 30° = 0
u = 60.8°
Ans.
Ans:
u = 60.8°
46
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
1–47. The chandelier is suspended from the wall and ceiling
using rods AB and BC, which have diameters of 3 mm and
4 mm, respectively. If the average normal stress in both rods is
not allowed to exceed 150 MPa, determine the largest mass of
the chandelier that can be supported if u = 45.
C
A
30⬚
B
u
Internal Loadings: The force developed in cables AB and BC can be determined by
considering the equilibrium of joint B, Fig. a,
+
: ©Fx = 0;
FBC cos 45° - FAB cos 30° = 0
(1)
+ c ©Fy = 0;
FBC sin 45° + FAB sin 30° - m(9.81) = 0
(2)
Solving Eqs. (1) and (2) yields
FAB = 7.181m
FBC = 8.795m
Average Normal Stress: The cross-sectional areas of cables AB and BC are
p
p
ABC = (0.0042) = 12.566(10 - 6) m2.
AAB = (0.0032) = 7.069(10 - 6) m2 and
4
4
We have,
(savg)allow =
FAB
;
AAB
150(106) =
7.181m
7.069(10 - 6)
m = 147.64 kg = 148 kg (controls)
(savg)allow =
FBC
;
ABC
150(106) =
Ans.
8.795m
12.566(10 - 6)
m = 214.31 kg
Ans:
m = 148 kg
47
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*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
0.5 m
4P
1m
4P
1.5 m
1.5 m
2P
0.5 m
C
30⬚
B
For pins B and C:
82.5 (103)
V
tB = tC =
= p 18 2 = 324 MPa
A
4 (1000 )
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.
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48
A
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1–49. The joint is subjected to the axial member force of
6 kip. Determine the average normal stress acting on
sections AB and BC. Assume the member is smooth and is
1.5-in. thick.
6 kip
60⬚
A
C
1.5 in.
20⬚
4.5 in.
B
+ c ©Fy = 0;
-6 sin 60° + NBC cos 20° = 0
NBC = 5.530 kip
+
: ©Fx = 0;
NAB - 6 cos 60° - 5.530 sin 20° = 0
NAB = 4.891 kip
sAB =
NAB
4.891
=
= 2.17 ksi
AAB
(1.5)(1.5)
Ans.
sBC =
NBC
5.530
=
= 0.819 ksi
ABC
(1.5)(4.5)
Ans.
Ans:
sAB = 2.17 ksi, sBC = 0.819 ksi
49
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
1–50. The driver of the sports car applies his rear brakes
and causes the tires to slip. If the normal force on each rear
tire is 400 lb and the coefficient of kinetic friction between
the tires and the pavement is mk = 0.5, determine the
average shear stress developed by the friction force on the
tires. Assume the rubber of the tires is flexible and each tire
is filled with an air pressure of 32 psi.
400 lb
F = mkN = 0.5(400) = 200 lb
p =
N
;
A
tavg =
A =
400
= 12.5 in2
32
F
200
=
= 16 psi
A
12.5
Ans.
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Ans:
tavg = 16 psi
50
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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
P
P
Va - a =
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
Aa - 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.
Ans:
P = 4 kip, Ata - aB avg = 250 psi
51
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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
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 Ab = (0.0062) = 28.274(10 - 6) m2 and Ap = 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.
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52
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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 Ab = (0.0062) = 28.274(10 - 6) m2 and Ap = 0.1(0.1) = 0.01 m2, 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
Ans:
P = 9.05 kN
53
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1–54. When the hand is holding the 5-lb stone, the humerus H,
assumed to be smooth, exerts normal forces FC and FA on the
radius C and ulna A, respectively, as shown. If the smallest crosssectional area of the ligament at B is 0.30 in2, determine the
greatest average tensile stress to which it is subjected.
B
H
FB
75⬚
FC
0.8 in.
G
C
A
FA
2 in.
a + ©MO = 0;
FB sin 75°(2) - 5(14) = 0
14 in.
FB = 36.235 lb
s =
P
36.235
=
= 121 psi
A
0.30
Ans.
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Ans:
s = 121 psi
54
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
1–55. The 2-Mg concrete pipe has a center of mass at point G.
If it is suspended from cables AB and AC, determine the
average normal stress developed in the cables.The diameters of
AB and AC are 12 mm and 10 mm, respectively.
A
30⬚
45⬚
C
B
G
Internal Loadings: The normal force developed in cables AB and AC can be
determined by considering the equilibrium of the hook for which the free-body
diagram is shown in Fig. a.
FAB = 14 362.83 N (T)
©Fx¿ = 0;
2000(9.81) cos 45° - FAB cos 15° = 0
©Fy¿ = 0;
2000(9.81) sin 45° - 14 362.83 sin 15° - FAC = 0 FAC = 10 156.06 N (T)
Average Normal Stress: The cross-sectional areas of cables AB and AC are
p
p
AAB = (0.0122) = 0.1131(10 - 3) m2 and AAC = (0.012) = 78.540(10 - 6) m2.
4
4
We have,
sAB =
FAB
14 362.83
= 127 MPa
=
AAB
0.1131(10 - 3)
Ans.
sAC =
FAC
10 156.06
= 129 MPa
=
AAC
78.540(10 - 6)
Ans.
Ans:
sAB = 127 MPa, sAC = 129 MPa
55
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*1–56. The 2-Mg concrete pipe has a center of mass at
point G. If it is suspended from cables AB and AC,
determine the diameter of cable AB so that the average
normal stress developed in this cable is the same as in the
10-mm diameter cable AC.
A
30⬚
45⬚
C
B
Internal Loadings: The normal force in cables AB and AC can be determined
by considering the equilibrium of the hook for which the free-body diagram is
shown in Fig. a.
©Fx¿ = 0; 2000(9.81) cos 45° - FAB cos 15° = 0
FAB = 14 362.83 N (T)
©Fy¿ = 0; 2000(9.81) sin 45° - 14 362.83 sin 15° - FAC = 0 FAC = 10 156.06 N (T)
Average Normal Stress: The cross-sectional areas of cables AB and AC are
p
p
AAB = dAB2 and AAC = (0.012) = 78.540(10 - 6) m2.
4
4
Here, we require
sAB = sAC
FAB
FAC
=
AAB
AAC
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10 156.06
14 362.83
=
p
2
78.540(10 - 6)
4 dAB
dAB = 0.01189 m = 11.9 mm
Ans.
56
G
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1–57. If the concrete pedestal has a specific weight of g,
determine the average normal stress developed in the
pedestal as a function of z.
r0
z
h
2r0
Internal Loading: From the geometry shown in Fig. a,
h¿ + h
h¿
=
;
r0
2r0
h¿ = h
and then
r0
r
= ;
z + h
h
r =
r0
(z + h)
h
Thus, the volume of the frustrum shown in Fig. b is
V =
=
2
r0
1
1
e p c (z + h) d f(z + h) - a pr02 bh
3
h
3
pr02
3h2
c (z + h)3 - h3 d
The weight of this frustrum is
W = gV =
pr02g
3h2
c (z + h)3 - h3 d
Average Normal Stress: The cross-sectional area the frustrum as a function of z is
pr02
(z + h)2.
A = pr2 =
h2
Also, the normal force acting on this cross section is N = W, Fig. b. We have
N
savg =
=
A
pr02g
3h2
[(z + h)3 - h3]
pr02
2
h2 (z + h)
=
g (z + h)3 - h3
c
d
3
(z + h)2
Ans.
Ans:
savg =
57
g (z + h)3 - h3
c
d
3
(z + h)2
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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
3 MPa
3 MPa
B
Average Normal Stress:
4.5 MPa
For the frustum, A = 2pxL = 2p(0.025 + 0.025) A 20.05 + 0.05 B
2
= 0.02221 m
P
;
s =
A
30 mm
2
C
2
25 mm 25 mm
F1
3 A 10 B =
0.02221
6
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;
www.elsolucionario.org
P - 21.21 - 66.64 sin 45° = 0
P = 68.3 kN
Ans.
Ans:
P = 68.3 kN
58
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
1–59. The jib crane is pinned at A and supports a chain
hoist that can travel along the bottom flange of the beam,
1 ft … x … 12 ft. If the hoist is rated to support a maximum
of 1500 lb, determine the maximum average normal stress in
3
the 4 -in. diameter tie rod BC and the maximum average
shear stress in the 58 -in. -diameter pin at B.
D
B
30⬚
A
C
x
a + ©MA = 0;
TBC sin 30°(10) - 1500(x) = 0
1500 lb
10 ft
Maximum TBC occurs when x = 12 ft
TBC = 3600 lb
P
3600
= p
= 8.15 ksi
2
A
(0.75)
4
Ans.
s =
3600>2
V
= p
= 5.87 ksi
2
A
4 (5>8)
Ans.
t =
Ans:
s = 8.15 ksi, t = 5.87 ksi
59
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*1–60. If the shaft is subjected to an axial force of 5 kN,
determine the bearing stress acting on the collar A.
A
5 kN
2.5 mm
60 mm 100 mm
2.5 mm
15 mm
Bearing Stress: The bearing area on the collar, shown shaded in Fig. a, is
Ab = p a 0.052 - 0.03252 b = 4.536(10 - 3) m2. Referring to the free-body diagram of
the collar, Fig. a, and writing the force equation of equilibrium along the axis of
the shaft,
©Fy = 0;
5(103) - sb c 4.536(10 - 3) d = 0
sb = 1.10 MPa
Ans.
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60
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1–61. If the 60-mm diameter shaft is subjected to an axial
force of 5 kN, determine the average shear stress developed in
the shear plane where the collar A and shaft are connected.
A
5 kN
2.5 mm
60 mm 100 mm
2.5 mm
15 mm
Average Shear Stress: The area of the shear plane, shown shaded in Fig. a,
is A = 2p(0.03)(0.015) = 2.827(10 - 3) m2. Referring to the free-body diagram of the
shaft, Fig. a, and writing the force equation of equilibrium along the axis of the shaft,
©Fy = 0; 5(103) - tavg c 2.827(10 - 3) d = 0
tavg = 1.77 MPa
Ans.
Ans:
tavg = 1.77 MPa
61
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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
B D
Support Reactions:
5 in.
From FBD(a)
1.5 in. 2 in. 1 in.
a + ©MD = 0;
20(5) - By (1) = 0
+
: ©Fx = 0;
Bx = 0
20 lb
By = 100 lb
From FBD(b)
+
: ©Fx = 0;
Ax = 0
a + ©ME = 0;
Ay (1.5) - 100(3.5) = 0
Ay = 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.
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Ans:
(tA)avg = 3.71 ksi
62
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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
A
From FBD(a)
a + ©MD = 0;
20(5) - By (1) = 0
+
: ©Fx = 0;
Bx = 0
B D
By = 100 lb
5 in.
1.5 in. 2 in. 1 in.
20 lb
Average Shear Stress: Pin B is subjected to double shear. Hence,
By
FB
VB =
=
= 50.0 lb
2
2
(tB)avg =
VB
50.0
= p
2
AB
4 (0.2 )
= 1592 psi = 1.59 ksi
Ans.
Ans:
(tB)avg = 1.59 ksi
63
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*1–64. A vertical force of P = 1500 N is applied to the
bell crank. Determine the average normal stress developed
in the 10-mm diameter rod CD, and the average shear stress
developed in the 6-mm diameter pin B that is subjected to
double shear.
D
C
450 mm
45⬚
300 mm
Internal Loading: Referring to the free-body diagram of the bell crank shown
in Fig. a,
a + ©MB = 0;
FCD (0.3 sin 45°) - 1500(0.45) = 0
FCD = 3181.98 N
+
: ©Fx = 0;
Bx - 3181.98 = 0
Bx = 3181.98 N
+ c ©Fy = 0;
By - 1500 = 0
By = 1500 N
Thus, the force acting on pin B is
FB = 2Bx2 + By2 = 23181.982 + 15002 = 3517.81 N
Pin B is in double shear. Referring to its free-body diagram, Fig. b,
VB =
FB
3517.81
=
= 1758.91 N
2
2
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Average Normal and Shear Stress: The cross-sectional area of rod CD is
p
ACD = (0.012) = 78.540(10 - 6) m2, and the area of the shear plane of pin B is
4
p
AB = (0.0062) = 28.274(10 - 6) m2. We obtain
4
(savg)CD =
FCD
3181.98
= 40.5 MPa
=
ACD
78.540(10 - 6)
Ans.
(tavg)B =
VB
1758.91
= 62.2 MPa
=
AB
28.274(10 - 6)
Ans.
64
P
B
A
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1–65. Determine the maximum vertical force P that can
be applied to the bell crank so that the average normal
stress developed in the 10-mm diameter rod CD, and the
average shear stress developed in the 6-mm diameter
double sheared pin B not exceed 175 MPa and 75 MPa,
respectively.
D
C
450 mm
45⬚
300 mm
P
B
A
Internal Loading: Referring to the free-body diagram of the bell crank shown in Fig. a,
a + ©MB = 0;
FCD (0.3 sin 45°) - P(0.45) = 0
FCD = 2.121P
+
: ©Fx = 0;
Bx - 2.121P = 0
Bx = 2.121P
+ c ©Fy = 0;
By - P = 0
By = P
Thus, the force acting on pin B is
FB = 2Bx2 + By2 = 2(2.121P)2 + P2 = 2.345P
Pin B is in double shear. Referring to its free-body diagram, Fig. b,
VB =
FB
2.345P
=
= 1.173P
2
2
Average Normal and Shear Stress: The cross-sectional area of rod CD is
p
ACD = (0.012) = 78.540(10 - 6) m2, and the area of the shear plane of pin B
4
p
is AB = (0.0062) = 28.274(10 - 6) m2. We obtain
4
(savg)allow =
FCD
;
ACD
(tavg)allow =
VB
;
AB
175(106) =
2.121P
78.540(10 - 6)
P = 6479.20 N = 6.48 kN
75(106) =
1.173P
28.274(10 - 6)
P = 1808.43 N = 1.81 kN (controls)
Ans.
Ans:
P = 1.81 kN
65
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1–66. Determine the largest load P that can be 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
Analyze 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
Aa - a = (0.025) a
0.025
b
3>5
C
1.5 m
P
= 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)
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P = 62.5(103) N = 62.5 kN (Controls!)
Ans.
Ans:
P = 62.5 kN
66
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z
1–67. The pedestal in the shape of a frustum of a cone is made
of concrete having a specific weight of 150 lb>ft3. Determine
the average normal stress acting in the pedestal at its base.
Hint: The volume of a cone of radius r and height h is
V = 13pr2h.
1 ft
8 ft
z ⫽ 4 ft
1.5 ft
h - 8
h
=
,
1.5
1
V =
y
h = 24 ft
1
1
p (1.5)2(24) - p (1)2(16);
3
3
x
V = 39.794 ft3
W = 150(39.794) = 5.969 kip
s =
P
5.969
=
= 844 psf = 5.86 psi
A
p(1.5)2
Ans.
Ans:
s = 5.86 psi
67
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z
*1–68. The pedestal in the shape of a frustum of a cone is
made of concrete having a specific weight of 150 lb>ft3.
Determine the average normal stress acting in the pedestal
at its midheight, z = 4 ft. Hint: The volume of a cone of
1 ft
radius r and height h is V = 13pr2h.
8 ft
z ⫽ 4 ft
h
h - 8
=
,
1.5
1
1.5 ft
h = 24 ft
y
1
1
W = c p (1.25)2 20 - (p)(12)(16) d (150) = 2395.5 lb
3
3
x
+ c g Fy = 0; P - 2395.5 = 0
P = 2395.5 lb
s =
P
2395.5
=
= 488 psf = 3.39 psi
A
p(1.25)2
Ans.
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68
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1–69. 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
(38) h
12
5
h
A
h = 2.74 in.
Use h = 2
800 lb
3
in.
4
Ans.
Ans:
3
Use h = 2 in.
4
69
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1–70. 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;
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.
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Ans:
d = 5.71 mm
70
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1–71. 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
80 kN
30 mm
40 kN
40 kN
350(106)
= 140(106)
2.5
tallow = 140(106) =
20(103)
p 2
4 d
d = 0.0135 m = 13.5 mm
Ans.
Ans:
d = 13.5 mm
71
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*1–72. The truss is used to support the loading shown
Determine the required cross-sectional area of member BC
if the allowable normal stress is sallow = 24 ksi .
800 lb
400 lb
6 ft
6 ft
F
B
30⬚
A
a + © MA = 0;
45⬚
-400(6) - 800(8.485) + 2(8.485)(Dy) = 0
Dy = 541.42 lb
a + © MF = 0;
541.42(8.485) - FBC (5.379) = 0
FBC = 854.01 lb
s=
P
;
A
24000 =
854.01
A
A = 0.0356 in2
Ans.
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72
E
60⬚
6 ft
6 ft
C
D
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1–73. The steel swivel bushing in the elevator control of
an airplane is held in place using a nut and washer as shown
in Fig. (a). Failure of the washer A can cause the push rod to
separate as shown in Fig. (b). If the maximum average
normal stress for the washer is smax = 60 ksi and the
maximum average shear stress is tmax = 21 ksi , determine
the force F that must be applied to the bushing that will
1
cause this to happen. The washer is 16
in. thick.
tavg =
V
;
A
21(103) =
0.75 in.
F
F
A
(a)
(b)
F
1
2p(0.375)( 16
)
F = 3092.5 lb = 3.09 kip
Ans.
Ans:
F = 3.09 kip
73
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1–74. 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 18 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
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.
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Ans:
1
Use a = 6 in.
2
74
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1–75. The hangers support the joist uniformly, so that it is
assumed the four nails on each hanger carry an equal
portion of the load. If the joist is subjected to the loading
shown, determine the average shear stress in each nail of
the hanger at ends A and B. Each nail has a diameter of
0.25 in. The hangers only support vertical loads.
40 lb/ft
30 lb/ft
B
A
18 ft
a + ©MA = 0;
FB(18) - 540(9) - 90(12) = 0;
FB = 330 lb
+ c ©Fy = 0;
FA + 330 - 540 - 90 = 0;
FA = 300 lb
For nails at A,
FA
300
= p
tavg =
AA
4( 4 )(0.25)2
Ans.
= 1528 psi = 1.53 ksi
For nails at B,
FB
330
= p
tavg =
AB
4( 4 )(0.25)2
Ans.
= 1681 psi = 1.68 ksi
Ans:
For nails at A: tavg = 1.53 ksi
For nails at B: tavg = 1.68 ksi
75
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*1–76. The hangers support the joists uniformly, so that it
is assumed the four nails on each hanger carry an equal
portion of the load. Determine the smallest diameter of the
nails at A and at B if the allowable stress for the nails is
tallow = 4 ksi . The hangers only support vertical loads.
40 lb/ft
30 lb/ft
B
A
18 ft
a + ©MA = 0;
FB(18) - 540(9) - 90(12) = 0;
FB = 330 lb
+ c ©Fy = 0;
FA + 330 - 540 - 90 = 0;
FA = 300 lb
For nails at A,
tallow =
FA
;
AA
4(103) =
300
4(p4 )dA2
dA = 0.155 in.
Ans.
For nails at B,
tallow =
FB
;
AB
dB = 0.162 in.
4(103) =
330
4(p4 )dB2
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Ans.
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1–77. 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
(controls)
Ans.
Ans:
P = 3.26 kip
77
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1–78. The 50-kg flowerpot is suspended from wires AB
and BC. If the wires have a normal failure stress of
sfail = 350 MPa, determine the minimum diameter of each
wire. Use a factor of safety of 2.5.
45⬚
A
C
30⬚
B
Internal Loading: The normal force developed in cables AB and BC can be
determined by considering the equilibrium of joint B, Fig. a.
+
: ©Fx = 0;
FBC cos 45° - FAB cos 30° = 0
(1)
+ c ©Fy = 0;
FAB sin 30° + FBC sin 45° - 50(9.81) = 0
(2)
Solving Eqs. (1) and (2),
FAB = 359.07 N
FBC = 439.77 N
Allowable Normal Stress:
sfail
350
sallow =
=
= 140 MPa
F.S.
2.5
Using this result,
sallow =
FAB
;
AAB
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359.07
140(106) = p
2
4 dAB
dAB = 0.001807 m = 1.81 mm
sallow =
FBC
;
ABC
Ans.
439.77
140(106) = p
2
4 dBC
dBC = 0.00200 m = 2.00 mm
Ans.
Ans:
dAB = 1.81 mm, dBC = 2.00 mm
78
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1–79. The 50-kg flowerpot is suspended from wires AB and
BC which have diameters of 1.5 mm and 2 mm, respectively. If
the wires have a normal failure stress of sfail = 350 MPa,
determine the factor of safety of each wire.
45⬚
A
C
30⬚
B
Internal Loading: The normal force developed in cables AB and BC can be
determined by considering the equilibrium of joint B, Fig. a.
+
: ©Fx = 0;
FBC cos 45° - FAB cos 30° = 0
(1)
+ c ©Fy = 0;
FAB sin 30° + FBC sin 45° - 50(9.81) = 0
(2)
Solving Eqs. (1) and (2),
FAB = 359.07 N
FBC = 439.77 N
Average Normal Stress: The cross-sectional area of wires AB and BC are
p
p
AAB = (0.0015)2 = 1.767(10 - 6) m2 and ABC = (0.0022) = 3.142(10 - 6) m2.
4
4
FAB
359.07
(savg)AB =
= 203.19 MPa
=
AAB
1.767(10 - 6)
(savg)BC =
FBC
ABC
439.77
=
3.142(10 - 6)
= 139.98 MPa
We obtain,
(F.S.)AB =
sfail
350
=
= 1.72
(savg)AB
203.19
Ans.
(F.S.)BC =
sfail
350
= 2.50
=
(savg)BC
139.98
Ans.
Ans:
(F.S.)AB = 1.72, (F.S.)BC = 2.50
79
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*1–80. The thrust bearing consists of a circular collar A
fixed to the shaft B. Determine the maximum axial force P
that can be applied to the shaft so that it does not cause the
shear stress along a cylindrical surface a or b to exceed an
allowable shear stress of tallow = 170 MPa.
C
b
a
P
B
A
30 mm 58 mm
a
b
35 mm
Assume failure along a:
tallow = 170(106) =
P
p(0.03)(0.035)
P = 561 kN (controls)
Ans.
Assume failure along b:
tallow = 170(106) =
P
p(0.058)(0.02)
P = 620 kN
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80
20 mm
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1–81. The steel pipe is supported on the circular base
plate and concrete pedestal. If the normal failure stress for
the steel is (sfail)st = 350 MPa, determine the minimum
thickness t of the pipe if it supports the force of 500 kN. Use
a factor of safety against failure of 1.5. Also, find the
minimum radius r of the base plate so that the minimum
factor of safety against failure of the concrete due to
bearing is 2.5. The failure bearing stress for concrete is
(sfail)con = 25 MPa.
t
500 kN
100 mm
r
Allowable Stress:
(sallow)st =
(sfail)st
350
=
= 233.33 MPa
F.S.
1.5
(sallow)con =
(sfail)con
25
=
= 10 MPa
F.S.
2.5
The cross-sectional area of the steel pipe and the heaving area of the concrete
pedestal are Ast = p(0.12 - ri2 ) and (Acon)b = pr2. Using these results,
(sallow)st =
P
;
Ast
233.33(106) =
500(103)
p(0.12 - ri2 )
ri = 0.09653 m = 96.53 mm
Thus, the minimum required thickness of the steel pipe is
t = rO - ri = 100 - 96.53 = 3.47 mm
Ans.
The minimum required radius of the bearing area of the concrete pedestal is
(sallow)con =
P
;
(Acon)b
10(106) =
500(103)
pr2
r = 0.1262 m = 126 mm
Ans.
Ans:
t = 3.47 mm, r = 126 mm
81
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1–82. The steel pipe is supported on the circular base
plate and concrete pedestal. If the thickness of the pipe
is t = 5 mm and the base plate has a radius of
150 mm, determine the factors of safety against failure of
the steel and concrete. The applied force is 500 kN, and the
normal failure stresses for steel and concrete are
(sfail)st = 350 MPa and (sfail)con = 25 MPa, respectively.
t
500 kN
100 mm
r
Average Normal and Bearing Stress: The cross-sectional area of the steel pipe and
the bearing area of the concrete pedestal are Ast = p(0.12 - 0.0952) =
0.975(10 - 3)p m2 and (Acon)b = p(0.152) = 0.0225p m2. We have
(savg)st =
500(103)
P
= 163.24 MPa
=
Ast
0.975(10 - 3)p
(savg)con =
500(103)
P
=
= 7.074 MPa
(Acon)b
0.0225p
Thus, the factor of safety against failure of the steel pipe and concrete pedestal are
(F.S.)st =
(sfail)st
350
=
= 2.14
(savg)st
163.24
Ans.
(sfail)con
25
=
= 3.53
(savg)con
7.074
Ans.
(F.S.)con =
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Ans:
(F.S.)st = 2.14, (F.S.)con = 3.53
82
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
1–83. 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.
Ans:
P = 90 kN, A = 6.19(10 - 3) m2, Pmax = 155 kN
83
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*1–84. 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 = 2Dx 2 + Dy 2 = 25.6572 + 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
VC =
=
= 4.00 kN
VD = FD = 6.093 kN
2
2
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For pin C,
tallow =
VC
;
AC
40(106) =
4.00(103)
p
2
4 dC
dC = 0.01128 m = 11.3 mm
For pin D,
VD
;
tallow =
AD
40(106) =
Ans.
6.093(103)
p
2
4 dD
dD = 0.01393 m = 13.9 mm
Ans.
84
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1–85. The beam is made from southern pine and is
supported by base plates resting on brick work. If the
allowable bearing stresses for the materials are
(spine)allow = 2.81 ksi and (sbrick)allow = 6.70 ksi, determine
the required length of the base plates at A and B to the
nearest 14 inch in order to support the load shown. The
plates are 3 in. wide.
6 kip
200 lb/ft
A
B
5 ft
5 ft
3 ft
The design must be based on strength of the pine.
At A:
s =
P
;
A
Use lA =
2810 =
1
in.
2
3910
lA(3)
lA = 0.464 in.
Ans.
At B:
s =
P
;
A
Use lB =
2810 =
3
in.
4
4690
l(3)
lB = 0.556 in.
Ans.
Ans:
Use lA =
85
1
3
in., lB = in.
2
4
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
1–86. The two aluminum rods support the vertical force of
P = 20 kN. Determine their required diameters if the
allowable tensile stress for the aluminum is sallow = 150 MPa.
B
A
C
45⬚
P
+ c ©Fy = 0;
FAB sin 45° - 20 = 0;
FAB = 28.284 kN
+
: ©Fx = 0;
28.284 cos 45° - FAC = 0;
FAC = 20.0 kN
For rod AB:
sallow =
FAB
;
AAB
150(106) =
28.284(103)
p 2
4 dAB
dAB = 0.0155 m = 15.5 mm
Ans.
For rod AC:
sallow =
FAC
;
AAC
150(106) =
20.0(103)
p 2
4 dAC
dAC = 0.0130 m = 13.0 mm
Ans.
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Ans:
dAB = 15.5 mm, dAC = 13.0 mm
86
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1–87. The two aluminum rods AB and AC have diameters
of 10 mm and 8 mm, respectively. Determine the largest
vertical force P that can be supported. The allowable tensile
stress for the aluminum is sallow = 150 MPa.
B
C
A
45⬚
P
+ c ©Fy = 0;
FAB sin 45° - P = 0;
+
: ©Fx = 0;
FAB cos 45° - FAC = 0
P = FAB sin 45°
(1)
(2)
Assume failure of rod AB:
sallow =
FAB
;
AAB
150(106) = p
FAB
2
4 (0.01)
FAB = 11.78 kN
From Eq. (1),
P = 8.33 kN
Assume failure of rod AC:
FAC
FAC
sallow =
; 150(106) = p
2
AAC
4 (0.008)
FAC = 7.540 kN
Solving Eqs. (1) and (2) yields:
FAB = 10.66 kN ;
P = 7.54 kN
Choose the smallest value
P = 7.54 kN
Ans.
Ans:
P = 7.54 kN
87
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*1–88. 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.
2m
2m
1.5 m
C
B
FB(4.5) + 1.5(3) + 2(1.5) - FC(6) = 0
4.5 FB - 6 FC = - 7.5
(1)
From FBD (b):
a + ©MA = 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)
D
A
From FBD (a):
a + ©MD = 0;
2 kN
1.5 kN
1.5 m 1.5 m 1.5 m
3 kN
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2
2
4 (dw - 0.00611 )
dw = 0.0154 m = 15.4 mm
Ans.
88
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1–89. Determine the required minimum thickness t of
member AB and edge distance b of the frame if P = 9 kip
and the factor of safety against failure is 2. The wood has a
normal failure stress of sfail = 6 ksi, and shear failure stress
of tfail = 1.5 ksi.
P
3 in.
B
3 in.
t
A
30⬚
b
30⬚
C
Internal Loadings: The normal force developed in member AB can be determined
by considering the equilibrium of joint A. Fig. a.
+
: ©Fx = 0;
FAB cos 30° - FAC cos 30° = 0
FAC = FAB
+ c ©Fy = 0;
2FAB sin 30° - 9 = 0
FAB = 9 kip
Subsequently, the horizontal component of the force acting on joint B can be
determined by analyzing the equilibrium of member AB, Fig. b.
+
: ©Fx = 0;
(FB)x - 9 cos 30° = 0
(FB)x = 7.794 kip
Referring to the free-body diagram shown in Fig. c, the shear force developed on the
shear plane a–a is
+
: ©Fx = 0;
Va - a = 7.794 kip
Va - a - 7.794 = 0
Allowable Normal Stress:
sfail
6
=
= 3 ksi
sallow =
F.S.
2
tallow =
tfail
1.5
=
= 0.75 ksi
F.S.
2
Using these results,
sallow =
FAB
;
AAB
3(103) =
9(103)
3t
t = 1 in.
tallow =
Va - a
;
Aa - a
0.75(103) =
Ans.
7.794(103)
3b
b = 3.46 in.
Ans.
Ans:
t = 1 in., b = 3.46 in.
89
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1–90. Determine the maximum allowable load P that can
be safely supported by the frame if t = 1.25 in. and
b = 3.5 in. The wood has a normal failure stress of
sfail = 6 ksi, and shear failure stress of tfail = 1.5 ksi. Use a
factor of safety against failure of 2.
P
3 in.
B
3 in.
t
A
30⬚
b
30⬚
C
Internal Loadings: The normal force developed in member AB can be determined
by considering the equilibrium of joint A. Fig. a.
+
: ©Fx = 0;
FAB cos 30° - FAC cos 30° = 0
FAC = FAB
+ c ©Fy = 0;
2FAB sin 30° - 9 = 0
FAB = P
Subsequently, the horizontal component of the force acting on joint B can be
determined by analyzing the equilibrium of member AB, Fig. b.
+
(FB)x = 0.8660P
: ©Fx = 0;
(FB)x - P cos 30° = 0
Referring to the free-body diagram shown in Fig. c, the shear force developed on the
shear plane a–a is
+
Va - a = 0.8660P
: ©Fx = 0;
Va - a - 0.8660P = 0
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Allowable Normal and Shear Stress:
sfail
6
sallow =
=
= 3 ksi
F.S.
2
tallow =
tfail
1.5
=
= 0.75 ksi
F.S.
2
Using these results,
sallow =
FAB
;
AAB
3(103) =
P
3(1.25)
P = 11 250 lb = 11.25 kip
tallow =
Va - a
;
Aa - a
0.75(103) =
0.8660P
3(3.5)
P = 9093.27 lb = 9.09 kip (controls)
Ans.
Ans:
P = 9.09 kip
90
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1–91. 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 =
;
AA¿
1.5(106) =
P
A¿
B¿
3m
B
1.5 m
25.0(103)
a2A¿
aA¿ = 0.1291 m = 130 mm
Ans.
For plate B¿ ,
sallow =
NB
;
AB¿
1.5(106) =
135(103)
a2B¿
aB¿ = 0.300 m = 300 mm
Ans.
Ans:
aA¿ = 130 mm, aB¿ = 300 mm
91
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40 kN/m
*1–92. 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¿
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 =
;
AA¿
1.5(106) =
(75 - 0.5P)(103)
0.15(0.15)
P = 82.5 kN
For plate B¿ ,
NB
;
(sb)allow =
AB¿
1.5(106) =
(1.5P - 15)(103)
0.25(0.25)
P = 72.5 kN (Controls!)
Ans.
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B¿
3m
B
1.5 m
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1–93. The rods AB and CD are made of steel. Determine
their smallest diameter so that they can support the dead
loads shown. The beam is assumed to be pin connected at A
and C. Use the LRFD method, where the resistance factor
for steel in tension is f = 0.9, and the dead load factor is
gD = 1. 4. The failure stress is sfail = 345 MPa.
B
D
6 kN
5 kN
4 kN
A
C
2m
2m
3m
3m
Support Reactions:
a + ©MA = 0;
FCD(10) - 5(7) - 6(4) - 4(2) = 0
FCD = 6.70 kN
a + ©MC = 0;
4(8) + 6(6) + 5(3) - FAB(10) = 0
FAB = 8.30 kN
Factored Loads:
FCD = 1.4(6.70) = 9.38 kN
FAB = 1.4(8.30) = 11.62 kN
For rod AB
0.9[345(106)] p a
dAB 2
b = 11.62(103)
2
dAB = 0.00690 m = 6.90 mm
Ans.
For rod CD
0.9[345(106)] p a
dCD 2
b = 9.38(103)
2
dCD = 0.00620 m = 6.20 mm
Ans.
Ans:
dAB = 6.90 mm, dCD = 6.20 mm
93
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1–94. 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
h
8 kip
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)
h = 1.74 in.
Ans.
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Ans:
h = 1.74 in.
94
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1–95. The pin support A and roller support B of the
bridge truss are supported on concrete abutments. If the
bearing failure stress of the concrete is (sfail)b = 4 ksi,
determine the required minimum dimension of the square
1
bearing plates at C and D to the nearest 16
in. Apply a factor
of safety of 2 against failure.
150 kip
100 kip
A
B
C
D
200 kip
300 kip 300 kip 300 kip 300 kip
6 ft
6 ft
6 ft
6 ft
6 ft
6 ft
Internal Loadings: The forces acting on the bearing plates C and D can be
determined by considering the equilibrium of the free-body diagram of the truss
shown in Fig. a,
a + ©MA = 0; By(36) - 100(36) - 200(30) - 300(24) - 300(18) - 300(12) - 300(6) = 0
By = 766.67 kip
a + ©MB = 0; 150(36) + 300(30) + 300(24) + 300(18) + 300(12) + 200(6) - Ay(36) = 0
Ay = 883.33 kip
Thus, the axial forces acting on C and D are
FC = Ay = 883.33 kip
FD = By = 766.67 kip
Allowable Bearing Stress:
(sallow)b =
(sfail)b
4
=
= 2 ksi
F.S.
2
Using this result,
(sallow)b =
FD
;
AD
2(103) =
766.67(103)
aD2
aD = 19.58 in. = 19
(sallow)b =
FC
;
AC
2(103) =
5
in.
8
Ans.
1
in.
16
Ans.
883.33(103)
aC 2
aC = 21.02 in. = 21
Ans:
Use aD = 19
95
5
1
in., aC = 21
in.
8
16
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*1–96. The pin support A and roller support B of the
bridge truss are supported on the concrete abutments. If the
square bearing plates at C and D are 21 in. * 21 in., and the
bearing failure stress for concrete is (sfail)b = 4 ksi,
determine the factor of safety against bearing failure for the
concrete under each plate.
150 kip
100 kip
A
B
C
D
200 kip
300 kip 300 kip 300 kip 300 kip
6 ft
6 ft
Internal Loadings: The forces acting on the bearing plates C and D can be
determined by considering the equilibrium of the free-body diagram of the truss
shown in Fig. a,
a + ©MA = 0; By(36) - 100(36) - 200(30) - 300(24) - 300(18) - 300(12) - 300(6) = 0
By = 766.67 kips
a + ©MB = 0; 150(36) + 300(30) + 300(24) + 300(18) + 300(12) + 200(6) - Ay(36) = 0
Ay = 883.33 kips
Thus, the axial forces acting on C and D are
FC = Ay = 883.33 kips
FD = By = 766.67 kips
Allowable Bearing Stress: The bearing area on the concrete abutment is
Ab = 21(21) = 441 in2. We obtain
www.elsolucionario.org
FC
883.33
(sb)C =
=
= 2.003 ksi
Ab
441
(sb)D =
FD
766.67
=
= 1.738 ksi
Ab
441
Using these results,
(F.S.)C =
(sfail)b
4
= 2.00
=
(sb)C
2.003
Ans.
(F.S.)C =
(sfail)b
4
=
= 2.30
(sb)C
1.738
Ans.
96
6 ft
6 ft
6 ft
6 ft
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
1–97. The beam AB is pin supported at A and supported
by a cable BC. A separate cable CG is used to hold up the
frame. If AB weighs 120 lb> ft and the column FC has a
weight of 180 lb/ft, determine the resultant internal loadings
acting on cross sections located at points D and E. Neglect
the thickness of both the beam and column in the
calculation.
C
4 ft
6 ft
A
B
D
12 ft
8 ft
E
4 ft
G
F
12 ft
Segment AD:
+
: ©Fx = 0;
ND + 2.16 = 0;
ND = - 2.16 kip
Ans.
+ T ©Fy = 0;
VD + 0.72 - 0.72 = 0;
VD = 0
Ans.
a + ©MD = 0;
MD - 0.72(3) = 0;
MD = 2.16 kip # ft
Ans.
Segment FE:
+
; ©Fx = 0;
VE - 0.54 = 0;
VE = 0.540 kip
Ans.
+ T ©Fy = 0;
NE + 0.72 - 5.04 = 0;
NE = 4.32 kip
Ans.
a + ©ME = 0;
- ME + 0.54(4) = 0;
ME = 2.16 kip # ft
Ans.
Ans:
ND = - 2.16 kip, VD = 0, MD = 2.16 kip # ft,
VE = 0.540 kip, NE = 4.32 kip, ME = 2.16 kip # ft
97
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1–98. 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
= 208 MPa
= p
ss =
2
A
4 (0.007)
30 mm
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.
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Ans:
ss = 208 MPa, (tavg)a = 4.72 MPa,
(tavg)b = 45.5 MPa
98
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1
1–99. To the nearest 16
in., 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 sallow = 10 ksi.
C
1.5 in.
Referring to the FBD of member AB, Fig. a,
a + ©MA = 0;
60⬚
B
2(8)(4) - FBC sin 60° (8) = 0
+
: ©Fx = 0;
9.238 cos 60° - Ax = 0
+ c ©Fy = 0;
9.238 sin 60° - 2(8) + Ay = 0
8 ft
A
FBC = 9.238 kip
Ax = 4.619 kip
2 kip/ft
Ay = 8.00 kip
Thus, the force acting on pin A is
FA = 2A2x + A2y = 24.6192 + 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 =
ABC
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.
Ans:
Use t =
99
1
1
13
in., dA = 1 in., dB =
in.
4
8
16
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*1–100. The circular punch B exerts a force of 2 kN on the
lop of the plate A. Determine the average shear stress in the
plate due to this loading.
2 kN
B
4 mm
A
Average Shear Stress: The shear area A = p(0.004)(0.002) = 8.00(10 - 6)p m2
tavg =
2(103)
V
= 79.6 MPa
=
A
8.00(10 - 6)p
Ans.
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100
2 mm
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1–101. Determine the average punching shear stress the
circular shaft creates in the metal plate through section AC
and BD. Also, what is the bearing stress developed on the
surface of the plate under the shaft?
40 kN
50 mm
A
B
C
D
10 mm
60 mm
Average Shear and Bearing Stress: The area of the shear plane and the bearing area on the
p
punch are AV = p(0.05)(0.01) = 0.5(10 - 3)p m2 and Ab = a0.122 - 0.062 b =
4
2.7(10 - 3)p m2. We obtain
tavg =
40(103)
P
= 25.5 MPa
=
AV
0.5(10 - 3)p
Ans.
sb =
40(103)
P
= 4.72 MPa
=
Ab
2.7(10 - 3)p
Ans.
120 mm
Ans:
tavg = 25.5 MPa, sb = 4.72 MPa
101
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1–102. 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
30⬚
a
150 mm
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
Average Normal Stress And Shear Stress: The cross sectional Area at section a–a is
A = a
0.15
b (0.15) = 0.02598 m2.
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.
www.elsolucionario.org
Ans:
sa - a = 200 kPa, ta - a = 115 kPa
102
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
1–103. 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
30 mm
A
25 mm
5 kN
For the 40 – mm – dia rod:
s40 =
5 (103)
P
= p
= 3.98 MPa
2
A
4 (0.04)
Ans.
For the 30 – mm – dia rod:
s30 =
5 (103)
V
= p
= 7.07 MPa
2
A
4 (0.03)
Ans.
Average shear stress for pin A:
tavg =
2.5 (103)
P
= p
= 5.09 MPa
2
A
4 (0.025)
Ans.
Ans:
s40 = 3.98 MPa, s30 = 7.07 MPa
tavg = 5.09 MPa
103
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*1–104. 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
B
s
C
L/2
Equation of Equilibrium:
a + ©MA = 0;
Ts -
gAL L
a b = 0
2
4
T =
gAL2
8s
Average Normal Stress:
gAL2
s =
gL2
T
8s
=
=
A
A
8s
Ans.
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104
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.
P =
pd - pd0
7 - 6
=
= 0.167 in.>in.
pd0
6
Ans.
Ans:
P = 0.167 in.>in.
105
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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.)
P =
L - L0
5p - 15
=
= 0.0472 in.>in.
L0
15
Ans.
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Ans:
P = 0.0472 in.>in.
106
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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
A
B
3m
C
2m
2m
¢LBD
¢LCE
=
3
7
3 (10)
= 4.286 mm
7
¢LCE
10
=
= 0.00250 mm>mm
PCE =
L
4000
¢LBD =
PBD =
Ans.
¢LBD
4.286
=
= 0.00107 mm>mm
L
4000
Ans.
Ans:
PCE = 0.00250 mm>mm, PBD = 0.00107 mm>mm
107
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*2–4. The force applied at the handle of the rigid lever
causes the lever to rotate clockwise about the pin B through
an angle of 2°. Determine the average normal strain
developed in each wire. The wires are unstretched when the
lever is in the horizontal position.
G
200 mm
H
2°
bp rad = 0.03491 rad.
180
Since u is small, the displacements of points A, C, and D can be approximated by
Geometry: The lever arm rotates through an angle of u = a
dA = 200(0.03491) = 6.9813 mm
dC = 300(0.03491) = 10.4720 mm
dD = 500(0.03491) = 17.4533 mm
Average Normal Strain: The unstretched length of wires AH, CG, and DF are
LAH = 200 mm, LCG = 300 mm, and LDF = 300 mm. We obtain
Ans.
Ans.
www.elsolucionario.org
(Pavg)DF =
dD
17.4533
=
= 0.0582 mm>mm
LDF
300
Ans.
108
E
C
200 mm
dA
6.9813
=
= 0.0349 mm>mm
LAH
200
dC
10.4720
= 0.0349 mm>mm
=
(Pavg)CG =
LCG
300
200 mm 300 mm
300 mm
B
A
(Pavg)AH =
F
D
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
2–5. 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
mm
30⬚
30⬚
300
A
P
mm
B
œ
LAC
= 23002 + 22 - 2(300)(2) cos 150° = 301.734 mm
PAC = PAB =
œ
- LAC
LAC
301.734 - 300
=
= 0.00578 mm>mm
LAC
300
Ans.
Ans:
PAC = PAB = 0.00578 mm>mm
109
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
2–6. The rubber band of unstretched length 2r0 is forced
down the frustum of the cone. Determine the average
normal strain in the band as a function of z.
r0
z
h
2r0
Geometry: Using similar triangles shown in Fig. a,
h¿
h¿ + h
=
;
r0
2r0
h¿ = h
Subsequently, using the result of h¿
r0
r
= ;
z+h
h
r =
r0
(z + h)
h
Average Normal Strain: The length of the rubber band as a function of z is
2pr0
(z+ h). With L0 = 2r0, we have
L = 2pr =
h
L - L0
Pavg =
=
L0
2pr0
(z + h) - 2r0
h
p
= (z + h) - 1
2r0
h
Ans.
www.elsolucionario.org
Ans:
Pavg =
110
p
(z + h) - 1
h
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2–7. The pin-connected rigid rods AB and BC are inclined
at u = 30° when they are unloaded. When the force P is
applied u becomes 30.2°. Determine the average normal
strain developed in wire AC.
P
B
u
u
600 mm
A
C
Geometry: Referring to Fig. a, the unstretched and stretched lengths of wire AD are
LAC = 2(600 sin 30°) = 600 mm
LAC ¿ = 2(600 sin 30.2°) = 603.6239 mm
Average Normal Strain:
(Pavg)AC =
LAC ¿ - LAC
603.6239 - 600
=
= 6.04(10 - 3) mm>mm
LAC
600
Ans.
Ans:
(Pavg)AC = 6.04(10 - 3) mm>mm
111
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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
300 mm
A
C
400 mm
AB = 24002 + 3002 = 500 mm
AB¿ = 24002 + 3002 - 2(400)(300) cos 90.3°
= 501.255 mm
PAB =
AB¿ - AB
501.255 - 500
=
AB
500
= 0.00251 mm>mm
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Ans.
112
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u
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.
D
P
300 mm
B
300 mm
A
C
400 mm
AB = 23002 + 4002 = 500 mm
AB¿ = AB + eABAB
= 500 + 0.0035(500) = 501.75 mm
501.752 = 3002 + 4002 - 2(300)(400) cos a
a = 90.4185°
u = 90.4185° - 90° = 0.4185° =
¢ D = 600(u) = 600(
p
(0.4185) rad
180°
p
)(0.4185) = 4.38 mm
180°
Ans.
Ans:
¢ D = 4.38 mm
113
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2–10. The corners of the square plate are given the
displacements indicated. Determine the shear strain along
the edges of the plate at A and B.
y
0.2 in.
A
10 in.
D
B
x
0.3 in.
0.3 in. 10 in.
10 in.
C
10 in.
0.2 in.
At A:
9.7
u¿
= tan - 1 a
b = 43.561°
2
10.2
u¿ = 1.52056 rad
(gA)nt =
p
- 1.52056
2
= 0.0502 rad
www.elsolucionario.org
Ans.
At B:
f¿
10.2
= tan - 1 a
b = 46.439°
2
9.7
f¿ = 1.62104 rad
(gB)nt =
p
- 1.62104
2
= -0.0502 rad
Ans.
Ans:
(gA)nt = 0.0502 rad, (gB)nt = - 0.0502 rad
114
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2–11. The corners of the square plate are given the
displacements indicated. Determine the average normal
strains along side AB and diagonals AC and DB.
y
0.2 in.
A
10 in.
D
B
x
0.3 in.
0.3 in. 10 in.
10 in.
C
10 in.
0.2 in.
For AB:
A¿B¿ = 2(10.2)2 + (9.7)2 = 14.0759 in.
AB = 2(10)2 + (10)2 = 14.14214 in.
PAB =
14.0759 - 14.14214
= - 0.00469 in.>in.
14.14214
Ans.
20.4 - 20
= 0.0200 in.>in.
20
Ans.
19.4 - 20
= - 0.0300 in.>in.
20
Ans.
For AC:
PAC =
For DB:
PDB =
Ans:
PAB = - 0.00469 in.>in., PAC = 0.0200 in.>in.,
PDB = - 0.0300 in.>in.
115
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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
400 mm
A
u1 = tan u1 =
2
= 0.006667 rad
300
u2 = tan u2 =
3
= 0.0075 rad
400
gxy = u1 + u2
www.elsolucionario.org
= 0.006667 + 0.0075 = 0.0142 rad
Ans.
116
300 mm
B
2 mm
x
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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.
y
3 mm
C
D
400 mm
A
300 mm
B
2 mm
x
AD¿ = 2(400)2 + (3)2 = 400.01125 mm
f = tan - 1 a
3
b = 0.42971°
400
AB¿ = 2(300)2 + (2)2 = 300.00667
w = tan - 1 a
2
b = 0.381966°
300
a = 90° - 0.42971° - 0.381966° = 89.18832°
D¿B¿ = 2(400.01125)2 + (300.00667)2 - 2(400.01125)(300.00667) cos (89.18832°)
D¿B¿ = 496.6014 mm
DB = 2(300)2 + (400)2 = 500 mm
496.6014 - 500
= - 0.00680 mm>mm
500
400.01125 - 400
= 0.0281(10 - 3) mm>mm
PAD =
400
Ans.
PDB =
Ans.
Ans:
PDB = - 0.00680 mm>mm,
PAD = 0.0281(10 - 3) mm>mm
117
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2–14. The force P applied at joint D of the square frame
causes the frame to sway and form the dashed rhombus.
Determine the average normal strain developed in wire AC.
Assume the three rods are rigid.
D
P
200 mm
200 mm
E
C
3
400 mm
A
B
Geometry: Referring to Fig. a, the stretched length of LAC ¿ of wire AC¿ can be
determined using the cosine law.
LAC ¿ = 24002 + 4002 - 2(400)(400) cos 93° = 580.30 mm
The unstretched length of wire AC is
LAC = 24002 + 4002 = 565.69 mm
Average Normal Strain:
(Pavg)AC =
LAC ¿ - LAC
580.30 - 565.69
=
= 0.0258 mm>mm
LAC
565.69
Ans.
www.elsolucionario.org
Ans:
(Pavg)AC = 0.0258 mm>mm
118
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2–15. The force P applied at joint D of the square frame
causes the frame to sway and form the dashed rhombus.
Determine the average normal strain developed in wire
AE. Assume the three rods are rigid.
D
P
200 mm
200 mm
E
C
3
400 mm
A
B
Geometry: Referring to Fig. a, the stretched length of LAE¿ of wire AE can be
determined using the cosine law.
LAE¿ = 24002 + 2002 - 2(400)(200) cos 93° = 456.48 mm
The unstretched length of wire AE is
LAE = 24002 + 2002 = 447.21 mm
Average Normal Strain:
(Pavg)AE =
LAE ¿ - LAE
456.48 - 447.21
= 0.0207 mm>mm
=
LAE
447.21
Ans.
Ans:
(Pavg)AE = 0.0207 mm>mm
119
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*2–16. The triangular plate ABC is deformed into the
shape shown by the dashed lines. If at A, eAB = 0.0075,
PAC = 0.01 and gxy = 0.005 rad, determine the average
normal strain along edge BC.
y
C
300 mm
gxy
A
Average Normal Strain: The stretched length of sides AB and AC are
LAC¿ = (1 + ey)LAC = (1 + 0.01)(300) = 303 mm
LAB¿ = (1 + ex)LAB = (1 + 0.0075)(400) = 403 mm
Also,
u =
p
180°
- 0.005 = 1.5658 rada
b = 89.7135°
2
p rad
The unstretched length of edge BC is
LBC = 23002 + 4002 = 500 mm
and the stretched length of this edge is
LB¿C¿ = 23032 + 4032 - 2(303)(403) cos 89.7135°
= 502.9880 mm
www.elsolucionario.org
We obtain,
PBC =
LB¿C¿ - LBC
502.9880 - 500
=
= 5.98(10 - 3) mm>mm
LBC
500
120
Ans.
400 mm
B
x
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
2–17. The plate is deformed uniformly into the shape
shown by the dashed lines. If at A, gxy = 0.0075 rad., while
PAB = PAF = 0, determine the average shear strain at point
G with respect to the x¿ and y¿ axes.
y¿
y
F
E
x¿
600 mm
gxy
C
D
300 mm
c = 90° - 0.4297° = 89.5703°
G
A
180°
Geometry: Here, gxy = 0.0075 rad a
b = 0.4297°. Thus,
p rad
300 mm
600 mm
B
b = 90° + 0.4297° = 90.4297°
Subsequently, applying the cosine law to triangles AGF¿ and GBC¿, Fig. a,
LGF¿ = 26002 + 3002 - 2(600)(300) cos 89.5703° = 668.8049 mm
LGC¿ = 26002 + 3002 - 2(600)(300) cos 90.4297° = 672.8298 mm
Then, applying the sine law to the same triangles,
sin f
sin 89.5703°
=
;
600
668.8049
f = 63.7791°
sin 90.4297°
sin a
=
;
300
672.8298
a = 26.4787°
Thus,
u = 180° - f - a = 180° - 63.7791° - 26.4787°
= 89.7422° a
p rad
b = 1.5663 rad
180°
Shear Strain:
(gG)x¿y¿ =
p
p
- u =
- 1.5663 = 4.50(10 - 3) rad
2
2
Ans.
Ans:
(gG)x¿y¿ = 4.50(10 - 3) rad
121
x
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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
2 mm
D
A
x
400 mm
3 mm
Geometry: For small angles,
a = c =
2
= 0.00662252 rad
302
b = u =
2
= 0.00496278 rad
403
Shear Strain:
www.elsolucionario.org
(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.
Ans:
(gB)xy = 11.6(10 - 3) rad,
(gA)xy = 11.6(10 - 3) rad
122
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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
x
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.
Ans:
(gC)xy = 11.6(10 - 3) rad,
(gD)xy = 11.6(10 - 3) rad
123
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*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
B
4 mm
C
300 mm
2 mm
D
A
400 mm
3 mm
Geometry:
AC = DB = 24002 + 3002 = 500 mm
DB¿ = 24052 + 3042 = 506.4 mm
A¿C¿ = 24012 + 3002 = 500.8 mm
Average Normal Strain:
PAC =
www.elsolucionario.org
A¿C¿ - AC
500.8 - 500
=
AC
500
= 0.00160 mm>mm = 1.60 A 10 - 3 B mm>mm
PDB =
Ans.
DB¿ - DB
506.4 - 500
=
DB
500
= 0.0128 mm>mm = 12.8 A 10 - 3 B mm>mm
Ans.
124
x
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
2–21. The rectangular plate is deformed into the shape of a
parallelogram shown by the dashed lines. Determine the
average shear strain gxy at corners A and B.
y
5 mm
D
C
300 mm
5 mm
A
B
x
400 mm
Geometry: Referring to Fig. a and using small angle analysis,
u =
5
= 0.01667 rad
300
f =
5
= 0.0125 rad
400
Shear Strain: Referring to Fig. a,
(gA)xy = u + f = 0.01667 + 0.0125 = 0.0292 rad
Ans.
(gB)xy = u + f = 0.01667 + 0.0125 = 0.0292 rad
Ans.
Ans:
(gA)xy = 0.0292 rad, (gB)xy = 0.0292 rad
125
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
2–22. The triangular plate is fixed at its base, and its apex A is
given a horizontal displacement of 5 mm. Determine the shear
strain, gxy, at A.
y
45⬚
800 mm
45⬚
x¿
A
45⬚
A¿
5 mm
800 mm
x
L = 28002 + 52 - 2(800)(5) cos 135° = 803.54 mm
sin 135°
sin u
=
;
803.54
800
gxy =
www.elsolucionario.org
u = 44.75° = 0.7810 rad
p
p
- 2u =
- 2(0.7810)
2
2
= 0.00880 rad
Ans.
Ans:
gxy = 0.00880 rad
126
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
2–23. The triangular plate is fixed at its base, and its apex A
is given a horizontal displacement of 5 mm. Determine the
average normal strain Px along the x axis.
y
45⬚
800 mm
45⬚
x¿
A
45⬚
A¿
5 mm
800 mm
x
L = 28002 + 52 - 2(800)(5) cos 135° = 803.54 mm
Px =
803.54 - 800
= 0.00443 mm>mm
800
Ans.
Ans:
Px = 0.00443 mm>mm
127
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*2–24. The triangular plate is fixed at its base, and its apex A
is given a horizontal displacement of 5 mm. Determine the
average normal strain Px¿ along the x¿ axis.
y
45⬚
800 mm
45⬚
x¿
A
45⬚
800 mm
x
L = 800 cos 45° = 565.69 mm
Px¿ =
www.elsolucionario.org
5
= 0.00884 mm>mm
565.69
Ans.
128
A¿
5 mm
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y
2–25. The square rubber block is subjected to a shear
strain of gxy = 40(10 - 6)x + 20(10 - 6)y, where x and y are
in mm. This deformation is in the shape shown by the dashed
lines, where all the lines parallel to the y axis remain vertical
after the deformation. Determine the normal strain along
edge BC.
Shear Strain: Along edge DC, y = 400 mm. Thus, (gxy)DC = 40(10 - 6)x + 0.008.
dy
Here,
= tan (gxy)DC = tan 340(10 - 6)x + 0.0084. Then,
dx
dc
C
A
B
400 mm
300 mm
dy =
L0
D
dc = -
tan [40(10 - 6)x + 0.008]dx
L0
300 mm
e ln cos c 40(10 - 6)x + 0.008 d f `
40(10 - 6)
0
1
x
300 mm
= 4.2003 mm
Along edge AB, y = 0. Thus, (gxy)AB = 40(10 - 6)x. Here,
dy
= tan (gxy)AB =
dx
tan [40(10 - 6)x]. Then,
300 mm
dB
dy =
L0
dB = -
L0
tan [40(10 - 6)x]dx
300 mm
e ln cos c 40(10 - 6)x d f `
40(10 - 6)
0
1
= 1.8000 mm
Average Normal Strain: The stretched length of edge BC is
LB¿C¿ = 400 + 4.2003 - 1.8000 = 402.4003 mm
We obtain,
(Pavg)BC =
LB¿C¿ - LBC
402.4003 - 400
=
= 6.00(10 - 3) mm>mm
LBC
400
Ans.
Ans:
(Pavg)BC = 6.00(10 - 3) mm>mm
129
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
2–26. The square plate is deformed into the shape shown by
the dashed lines. If DC has a normal strain Px = 0.004, DA has
a normal strain Py = 0.005 and at D, gxy = 0.02 rad,
determine the average normal strain along diagonal CA.
y
y¿
x¿
600 mm
A¿
B¿
B
A
E
600 mm
D
C C¿
x
Average Normal Strain: The stretched length of sides DA and DC are
LDC¿ = (1 + Px)LDC = (1 + 0.004)(600) = 602.4 mm
LDA¿ = (1 + Py)LDA = (1 + 0.005)(600) = 603 mm
Also,
a =
p
180°
- 0.02 = 1.5508 rad a
b = 88.854°
2
p rad
Thus, the length of C¿A¿ can be determined using the cosine law with reference
to Fig. a.
LC¿A¿ = 2602.42 + 6032 - 2(602.4)(603) cos 88.854°
= 843.7807 mm
www.elsolucionario.org
The original length of diagonal CA can be determined using Pythagorean’s
theorem.
LCA = 26002 + 6002 = 848.5281 mm
Thus,
(Pavg)CA =
LC¿A¿ - LCA
843.7807 - 848.5281
= - 5.59(10 - 3) mm>mm Ans.
=
LCA
848.5281
Ans:
(Pavg)CA = - 5.59(10 - 3) mm>mm
130
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
2–27. The square plate ABCD is deformed into the shape
shown by the dashed lines. If DC has a normal strain
Px = 0.004, DA has a normal strain Py = 0.005 and at D,
gxy = 0.02 rad, determine the shear strain at point E with
respect to the x¿ and y¿ axes.
y
y¿
x¿
600 mm
A¿
B¿
B
A
600 mm
D
E
C C¿
x
Average Normal Strain: The stretched length of sides DC and BC are
LDC¿ = (1 + Px)LDC = (1 + 0.004)(600) = 602.4 mm
LB¿C¿ = (1 + Py)LBC = (1 + 0.005)(600) = 603 mm
Also,
a =
180°
p
- 0.02 = 1.5508 rada
b = 88.854°
2
p rad
f =
180°
p
+ 0.02 = 1.5908 rad a
b = 91.146°
2
p rad
Thus, the length of C¿A¿ and DB¿ can be determined using the cosine law with
reference to Fig. a.
LC¿A¿ = 2602.42 + 6032 - 2(602.4)(603) cos 88.854° = 843.7807 mm
LDB¿ = 2602.42 + 6032 - 2(602.4)(603) cos 91.146° = 860.8273 mm
Thus,
LE¿A¿ =
LC¿A¿
= 421.8903 mm
2
LE¿B¿ =
LDB¿
= 430.4137 mm
2
Using this result and applying the cosine law to the triangle A¿E¿B¿ , Fig. a,
602.42 = 421.89032 + 430.41372 - 2(421.8903)(430.4137) cos u
u = 89.9429° a
p rad
b = 1.5698 rad
180°
Shear Strain:
(gE)x¿y¿ =
p
p
- u =
- 1.5698 = 0.996(10 - 3) rad
2
2
Ans.
Ans:
(gE)x¿y¿ = 0.996(10 - 3) rad
131
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*2–28.
The wire is subjected to a normal strain that is
2
P ⫽ (x/L)e–(x/L)
2
defined by P = (x>L)e - (x>L) . If the wire has an initial
x
length L, determine the increase in its length.
x
L
L
¢L =
1
- (x>L)2
xe
dx
L L0
= -L c
=
e
- (x>L)2 L
2
d
=
0
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L
31 - (1>e)4
2
L
3e - 14
2e
Ans.
132
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2–29. The rectangular plate is deformed into the shape
shown by the dashed lines. Determine the average normal
strain along diagonal AC, and the average shear strain at
corner A.
y
6 mm
400 mm
2 mm
2 mm
6 mm
C
D
300 mm
2 mm
A
Geometry: The unstretched length of diagonal AC is
x
B
400 mm
3 mm
LAC = 2300 + 400 = 500 mm
2
2
Referring to Fig. a, the stretched length of diagonal AC is
LAC¿ = 2(400 + 6)2 + (300 + 6)2 = 508.4014 mm
Referring to Fig. a and using small angle analysis,
f =
2
= 0.006623 rad
300 + 2
a =
2
= 0.004963 rad
400 + 3
Average Normal Strain: Applying Eq. 2,
(Pavg)AC =
LAC¿ - LAC
508.4014 - 500
=
= 0.0168 mm>mm
LAC
500
Ans.
Shear Strain: Referring to Fig. a,
(gA)xy = f + a = 0.006623 + 0.004963 = 0.0116 rad
Ans.
Ans:
(Pavg)AC = 0.0168 mm>mm, (gA)xy = 0.0116 rad
133
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2–30. The rectangular plate is deformed into the shape
shown by the dashed lines. Determine the average normal
strain along diagonal BD, and the average shear strain at
corner B.
y
6 mm
400 mm
2 mm
2 mm
6 mm
C
D
300 mm
2 mm
A
Geometry: The unstretched length of diagonal BD is
x
B
400 mm
3 mm
LBD = 23002 + 4002 = 500 mm
Referring to Fig. a, the stretched length of diagonal BD is
LB¿D¿ = 2(300 + 2 - 2)2 + (400 + 3 - 2)2 = 500.8004 mm
Referring to Fig. a and using small angle analysis,
2
= 0.004963 rad
403
3
= 0.009868 rad
a =
300 + 6 - 2
f =
Average Normal Strain: Applying Eq. 2,
(Pavg)BD =
LB¿D¿ - LBD
500.8004 - 500
=
= 1.60(10 - 3) mm>mm Ans.
LBD
500
www.elsolucionario.org
Shear Strain: Referring to Fig. a,
(gB)xy = f + a = 0.004963 + 0.009868 = 0.0148 rad
Ans.
Ans:
(Pavg)BD = 1.60(10 - 3) mm>mm,
(gB)xy = 0.0148 rad
134
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2–31. The nonuniform loading causes a normal strain in
the shaft that can be expressed as Px = kx2, where k is a
constant. Determine the displacement of the end B. Also,
what is the average normal strain in the rod?
L
A
B
x
d(¢x)
= Px = kx2
dx
L
(¢x)B =
3
2
L0
kx =
kL
3
Ans.
kL3
(¢x)B
kL2
3
(Px)avg =
=
=
L
L
3
Ans.
Ans:
3
(¢x)B =
135
kL
kL2
, (Px)avg =
3
3
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*2–32 The rubber block is fixed along edge AB, and edge
CD is moved so that the vertical displacement of any point
in the block is given by v(x) = (v0>b3)x3. Determine the
shear strain gxy at points (b>2, a>2) and (b, a).
y
v (x)
A
v0
D
a
B
Shear Strain: From Fig. a,
b
dv
= tan gxy
dx
3v0 2
x = tan gxy
b3
gxy = tan - 1 a
3v0 2
x b
b3
Thus, at point (b> 2, a> 2),
gxy = tan - 1 c
3v0 b 2
a b d
b3 2
3 v0
= tan - 1 c a b d
4 b
gxy = tan - 1 c
Ans.
www.elsolucionario.org
and at point (b, a),
b3
(b2) d
= tan - 1 c 3 a
v0
bd
b
3v0
C
Ans.
136
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2–33. 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¿ .
y
B¿
vB
B
L
A
uA A¿
u
x
Geometry:
LA¿B¿ = 2(L cos u - uA)2 + (L sin u + vB)2
= 2L2 + u2A + v2B + 2L(vB sin u - uA cos u)
Average Normal Strain:
LA¿B¿ - L
PAB =
L
=
A
1 +
2(vB sin u - uA cos u)
u2A + v2B
+
- 1
L
L2
Neglecting higher terms u2A and v2B
1
PAB = B 1 +
2(vB sin u - uA cos u) 2
R - 1
L
Using the binomial theorem:
PAB = 1 +
=
2uA cos u
1 2vB sin u
¢
≤ + ... - 1
2
L
L
vB sin u
uA cos u
L
L
Ans.
Ans.
PAB =
137
vB sin u
uA cos u
L
L
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2–34. If the normal strain is defined in reference to the
final length, that is,
P¿n = 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 - Pn¿ = Pn Pn¿ .
PB =
¢S¿ - ¢S
¢S
œ
=
PB - PA
¢S¿ - ¢S
¢S¿ - ¢S
¢S
¢S¿
¢S¿ 2 - ¢S¢S¿ - ¢S¿¢S + ¢S2
¢S¢S¿
¢S¿ 2 + ¢S2 - 2¢S¿¢S
=
¢S¢S¿
=
=
(¢S¿ - ¢S)2
¢S¿ - ¢S
¢S¿ - ¢S
= ¢
≤¢
≤
¢S¢S¿
¢S
¢S¿
= PA PBœ (Q.E.D)
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138
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3–1. A tension test was performed on a steel specimen
having an original diameter of 0.503 in. and gauge length of
2.00 in. The data is listed in the table. Plot the
stress–strain diagram and determine approximately the
modulus of elasticity, the yield stress, the ultimate stress, and
the rupture stress. Use a scale of 1 in. = 20 ksi and
1 in. = 0.05 in.> in. Redraw the elastic region, using the same
stress scale but a strain scale of 1 in. = 0.001 in.> in.
A =
Load (kip) Elongation (in.)
0
1.50
4.60
8.00
11.00
11.80
11.80
12.00
16.60
20.00
21.50
19.50
18.50
1
p(0.503)2 = 0.1987 in2
4
L = 2.00 in.
s(ksi)
e(in.>in.)
0
0
7.55
0.00025
23.15
0.00075
40.26
0.00125
55.36
0.00175
59.38
0.0025
59.38
0.0040
60.39
0.010
83.54
0.020
100.65
0.050
108.20
0.140
98.13
0.200
0.230
48
= 32.0(103) ksi
Eapprox =
0.0015
0
0.0005
0.0015
0.0025
0.0035
0.0050
0.0080
0.0200
0.0400
0.1000
0.2800
0.4000
0.4600
93.10
Ans.
Ans:
(sult)approx = 110 ksi, (sR)approx = 93.1 ksi,
(sY)approx = 55 ksi, Eapprox = 32.0(103) ksi
139
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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
Ans.
Modulus of Resilience: The modulus of resilience is equal to the area under the
linear portion of the stress–strain diagram (shown shaded).
ur =
1
lb
in.
in # lb
(33.2) A 103 B ¢ 2 ≤ ¢ 0.0006
≤ = 9.96
2
in.
in
in3
Ans.
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Ans:
in # lb
E = 55.3 A 103 B ksi, ur = 9.96
in3
140
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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 =
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
1
lb
in.
(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
lb
in.
1
(12.3) A 103 B ¢ 2 ≤ (0.0004) ¢ ≤
2
in.
in
in # lb
in3
Ans.
Ans:
in # lb
(ut)approx = 85.0
in3
141
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*3–4. A tension test was performed on a steel specimen
having an original diameter of 0.503 in. and a gauge length
of 2.00 in. The data is listed in the table. Plot the
stress–strain diagram and determine approximately the
modulus of elasticity, the ultimate stress, and the rupture
stress. Use a scale of 1 in. = 15 ksi and 1 in. = 0.05 in.> in.
Redraw the linear-elastic region, using the same stress scale
but a strain scale of 1 in. = 0.001 in.
A =
Load (kip)
Elongation (in.)
0
2.50
6.50
8.50
9.20
9.80
12.0
14.0
14.5
14.0
13.2
0
0.0009
0.0025
0.0040
0.0065
0.0098
0.0400
0.1200
0.2500
0.3500
0.4700
1
p(0.503)2 = 0.19871 in2
4
L = 2.00 in.
P
s = A
(ksi)
P = ¢L
L (in.>in.)
0
0
12.58
0.00045
32.71
0.00125
42.78
0.0020
46.30
0.00325
49.32
0.0049
60.39
0.02
70.45
0.06
72.97
0.125
70.45
0.175
66.43
0.235
Eapprox =
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32.71
= 26.2(103) ksi
0.00125
Ans.
142
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3–5. A tension test was performed on a steel specimen
having an original diameter of 0.503 in. and gauge length of
2.00 in. Using the data listed in the table, plot the
stress–strain diagram and determine approximately the
modulus of toughness.
Modulus of toughness (approx)
ut = total area under the curve
(1)
= 87 (7.5) (0.025)
= 16.3
in. # kip
Load (kip)
Elongation (in.)
0
2.50
6.50
8.50
9.20
9.80
12.0
14.0
14.5
14.0
13.2
0
0.0009
0.0025
0.0040
0.0065
0.0098
0.0400
0.1200
0.2500
0.3500
0.4700
Ans.
in3
In Eq.(1), 87 is the number of squares under the curve.
P
s = A
(ksi)
P = ¢L
L (in.>in.)
0
0
12.58
0.00045
32.71
0.00125
42.78
0.0020
46.30
0.00325
49.32
0.0049
60.39
0.02
70.45
0.06
72.97
0.125
70.45
0.175
66.43
0.235
Ans:
in. # kip
ut = 16.3
in3
143
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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
¢P =
0.009
= 0.000750 in.>in.
12
Modulus of Elasticity:
E =
¢s
9.167 - 2.546
=
= 8.83 A 103 B ksi
¢P
0.000750
Ans.
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Ans:
E = 8.83 A 103 B ksi
144
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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
P =
0.02
d
=
= 0.000555 in.>in.
L
3 (12)
s = EP = 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.
Ans:
A = 0.209 in2, P = 1.62 kip
145
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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)
= 124.71 in. Thus, the wire
The unstretched length of the wire is LAB =
sin 60°
stretches
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dAB = PAB LAB = 0.6586(10 - 3)(124.71)
= 0.0821 in.
Ans.
146
B
C
Here, we are only interested in determining the force in wire AB.
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s (psi)
3–9. 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.
ur =
1
(2)(11) = 11 psi
2
Ans.
2 2.25
P (in./in.)
Ans:
E = 5.5 psi, ut = 19.25 psi, ur = 11 psi
147
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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
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.)
From the stress–strain diagram, Fig. a,
60 ksi - 0
E
=
;
1
0.002 - 0
sy = 60 ksi
E = 30.0(103) ksi
Ans.
sult = 100 ksi
Thus,
PY = sYA = 60 C p4 (0.52) D = 11.78 kip = 11.8 kip
Ans.
Pult = sult A = 100 C p4 (0.52) D = 19.63 kip = 19.6 kip
Ans.
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Ans:
E = 30.0(103) ksi, PY = 11.8 kip, Pult = 19.6 kip
148
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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
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.)
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 recovers along line AB, Fig. a,
which has a slope 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.
Ans:
Elastic Recovery = 0.003 in.>in., ¢L = 0.094 in.
149
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s (ksi)
*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.
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
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 a 0.05
b = 28.5(103)
2
in.
in
in3
Ans.
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150
P (in./in.)
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3–13. A bar having a length of 5 in. and cross-sectional
area of 0.7 in.2 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
P =
d
0.002
=
= 0.000400 in.>in.
L
5
Modulus of Elasticity:
E =
s
11.43
=
= 28.6(103) ksi
P
0.000400
Ans.
Ans:
E = 28.6(103) ksi
151
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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
4 ft
P
A
D
C
3 ft
3 ft
Here, we are only interested in determining the force in wire BD. Referring to the
FBD in Fig. a
a + ©MA = 0;
FBD A 45 B (3) - 600(6) = 0
FBD = 1500 lb
The normal stress developed in the wire is
sBD =
FBD
1500
= 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;
30.56 = 29.0(103)PBD
PBD = 1.054(10 - 3) in.>in.
The unstretched length of the wire is LBD = 232 + 42 = 5 ft = 60 in. Thus, the
wire stretches
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dBD = PBD LBD = 1.054(10 - 3)(60)
= 0.0632 in.
Ans.
Ans:
dBD = 0.0632 in.
152
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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 0.15 in.
downward.
B
4 ft
P
A
D
C
3 ft
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.
Ans:
P = 570 lb
153
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*3–16. The wire has a diameter of 5 mm and is made from
A-36 steel. If a 80-kg man is sitting on seat C, determine the
elongation of wire DE.
E
W
600 mm
D
A
B
800 mm
Equations of Equilibrium: The force developed in wire DE can be determined by
writing the moment equation of equilibrium about A with reference to the freebody diagram shown in Fig. a,
a + ©MA = 0;
3
FDE a b (0.8) - 80(9.81)(1.4) = 0
5
FDE = 2289 N
Normal Stress and Strain:
sDE =
FDE
2289
=
= 116.58 MPa
p
ADE
(0.0052)
4
Since sDE < sY , Hooke’s Law can be applied
sDE = EPDE
116.58(106) = 200(109)PDE
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PDE = 0.5829(10-3) mm>mm
The unstretched length of wire DE is LDE = 26002 + 8002 = 1000 mm. Thus, the
elongation of this wire is given by
dDE = PDELDE = 0.5829(10-3)(1000) = 0.583 mm
Ans.
154
C
600 mm
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s (ksi)
3–17. A tension test was performed on a magnesium alloy
specimen having a diameter 0.5 in. and gauge length 2 in.
The resulting stress–strain diagram is shown in the figure.
Determine the approximate modulus of elasticity and the
yield strength of the alloy using the 0.2% strain offset
method.
40
35
30
25
20
15
10
5
0
0.002
0.004
0.006
0.008
0.010
P (in./in.)
Modulus of Elasticity: From the stress–strain diagram, when P = 0.002 in.>in., its
corresponding stress is s = 13.0 ksi. Thus,
Eapprox =
13.0 - 0
= 6.50(103) ksi
0.002 - 0
Ans.
Yield Strength: The intersection point between the stress–strain diagram and the
straight line drawn parallel to the initial straight portion of the stress–strain diagram
from the offset strain of P = 0.002 in.>in. is the yield strength of the alloy. From the
stress–strain diagram,
sYS = 25.9 ksi
Ans.
Ans:
Eapprox = 6.50(103) ksi, sYS = 25.9 ksi
155
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s (ksi)
3–18. A tension test was performed on a magnesium alloy
specimen having a diameter 0.5 in. and gauge length of 2 in.
The resulting stress–strain diagram is shown in the figure. If
the specimen is stressed to 30 ksi and unloaded, determine
the permanent elongation of the specimen.
40
35
30
25
20
15
10
5
0
0.002
0.004
0.006
0.008
0.010
P (in./in.)
Permanent Elongation: From the stress–strain diagram, the strain recovered is
along the straight line BC which is parallel to the straight line OA. Since
13.0 - 0
= 6.50(103) ksi, then the permanent set for the specimen is
Eapprox =
0.002 - 0
30(103)
= 0.00318 in.>in.
PP = 0.0078 6.5(106)
Thus,
dP = PPL = 0.00318(2) = 0.00637 in.
Ans.
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Ans:
dP = 0.00637 in.
156
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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
P
P = 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(106) kPa = 2.22 GPa
dP
0.45(10 - 6)
s=0
The equation for the recovery line is s = 2.22(106)(P - 0.003).
This line intersects the stress–strain curve at sYS = 2027 kPa = 2.03 MPa
Ans.
Ans:
sYS = 2.03 MPa
157
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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-mm-long 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 P = 0.12
120(10-3) = 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 - P)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 = PL = 0.12(200) = 24 mm
Ans.
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158
P
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3–21. The two bars are made of polystyrene, which has the
stress–strain diagram shown. If the cross-sectional area of
bar AB is 1.5 in2 and BC is 4 in2, determine the largest force
P that can be supported before any member ruptures.
Assume that buckling does not occur.
P
4 ft
C
B
3 ft
A
s (ksi)
25
+ c g Fy = 0;
3
F
- P = 0;
5 AB
FAB = 1.6667 P
+
; ©Fx = 0;
4
FBC - (1.6667P) = 0;
5
FBC = 1.333 P
20
(1)
15
(2)
0
From the stress–strain diagram (sR)t = 5 ksi
FBC
;
ABC
5 =
tension
5
Assuming failure of bar BC:
s =
compression
10
FBC
;
4
0
0.20
0.40
0.60
0.80
P (in./in.)
FBC = 20.0 kip
From Eq. (2), P = 15.0 kip
Assuming failure of bar AB:
From stress–strain diagram (sR)c = 25.0 ksi
s =
FAB
;
AAB
25.0 =
FAB
;
1.5
FAB = 37.5 kip
From Eq. (1), P ⫽ 22.5 kip
Choose the smallest value
P = 15.0 kip
Ans.
Ans:
P = 15.0 kip
159
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3–22. The two bars are made of polystyrene, which has the
stress–strain diagram shown. Determine the cross-sectional
area of each bar so that the bars rupture simultaneously
when the load P = 3 kip. Assume that buckling does not
occur.
P
4 ft
C
B
3 ft
A
s (ksi)
25
+ c ©Fy = 0;
3
FBA a b - 3 = 0;
5
+
: ©Fx = 0;
4
-FBC + 5a b = 0;
5
20
FBA = 5 kip
15
FBC = 4 kip
compression
10
tension
5
For member BC:
0
(smax)t =
4 kip
FBC
; ABC =
= 0.8 in2
ABC
5 ksi
(smax)c =
FBA
;
ABA
0
0.20
0.40
0.60
0.80
P (in./in.)
Ans.
For member BA:
ABA =
5 kip
= 0.2 in2
25 ksi
Ans.
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Ans:
ABC = 0.8 in2, ABA = 0.2 in2
160
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
3–23. 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 = 30(103) 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.
Ans:
n = 2.73, k = 4.23(10 - 6)
161
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3–24. The wires AB and BC have original lengths of 2 ft
3
and 3 ft, and diameters of 18 in. and 16
in., respectively. If
these wires are made of a material that has the approximate
stress–strain diagram shown, determine the elongations of
the wires after the 1500-lb load is placed on the platform.
C
Equations of Equilibrium: The forces developed in wires AB and BC can be
determined by analyzing the equilibrium of joint B, Fig. a,
+
: ©Fx = 0;
FBC sin 30° - FAB sin 45° = 0
(1)
+ c ©Fy = 0;
FBC cos 30° + FAB cos 45° = 1500
(2)
A
3 ft
45⬚
30⬚
2 ft
B
Solving Eqs. (1) and (2),
FAB = 776.46 lb
FBC = 1098.08 lb
Normal Stress and Strain:
sAB =
FAB
776.46
=
= 63.27 ksi
p
AAB
(1>8)2
4
s (ksi)
sBC =
FBC
1098.08
=
= 39.77 ksi
p
ABC
2
(3>16)
4
80
58
The corresponding normal strain can be determined from the stress–strain diagram,
Fig. b.
39.77
58
;
=
PBC
0.002
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63.27 - 58
80 - 58
=
;
PAB - 0.002
0.01 - 0.002
PBC = 0.001371 in.>in.
0.002
PAB = 0.003917 in.>in.
Thus, the elongations of wires AB and BC are
dAB = PABLAB = 0.003917(24) = 0.0940
Ans.
dBC = PBCLBC = 0.001371(36) = 0.0494
Ans.
162
0.01
P (in./in.)
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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 =
300 N
300 N
200 mm
P
300
= 1.678 MPa
= p
2
A
(0.015)
4
Plong =
1.678(106)
s
= 0.0006288
=
E
2.70(109)
d = Plong L = 0.0006288 (200) = 0.126 mm
Ans.
Plat = - nPlong = - 0.4(0.0006288) = - 0.0002515
¢d = Platd = - 0.0002515 (15) = - 0.00377 mm
Ans.
Ans:
d = 0.126 mm, ¢d = - 0.00377 mm
163
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3–26. The thin-walled tube is subjected to an axial force of
40 kN. If the tube elongates 3 mm and its circumference
decreases 0.09 mm, determine the modulus of elasticity,
Poisson’s ratio, and the shear modulus of the tube’s
material. The material behaves elastically.
40 kN
900 mm
10 mm
40 kN
12.5 mm
Normal Stress and Strain:
s =
40(103)
P
= 226.35 MPa
=
A
p(0.01252 - 0.012)
Pa =
3
d
=
= 3.3333 (10-3) mm>mm
L
900
Applying Hooke’s law,
s = EPa;
226.35(106) = E [3.3333(10-3)]
E = 67.91(106) Pa = 67.9 GPa
Ans.
Poisson’s Ratio: The circumference of the loaded tube is 2p(12.5) - 0.09 =
78.4498 mm. Thus, the outer radius of the tube is
r =
78.4498
= 12.4857 mm
2p
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The lateral strain is
Plat =
r - r0
12.4857 - 12.5
=
= - 1.1459(10-3) mm>mm
r0
12.5
n = -
- 1.1459(10-3)
Plat
d = 0.3438 = 0.344
= -c
Pa
3.3333(10-3)
Ans.
67.91(109)
E
=
= 25.27(109) Pa = 25.3 GPa
2(1 + n)
2(1 + 0.3438)
Ans.
G =
Ans:
E = 67.9 GPa, v = 0.344, G = 25.3 GPa
164
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3–27. When the two forces are placed on the beam, the
diameter of the A-36 steel rod BC decreases from 40 mm to
39.99 mm. Determine the magnitude of each force P.
C
P
1m
A
P
1m
1m
1m
B
0.75 m
Equations of Equilibrium: The force developed in rod BC can be determined by
writing the moment equation of equilibrium about A with reference to the
free-body diagram of the beam shown in Fig. a.
4
FBC a b (3) - P(2) - P(1) = 0
5
a + ©MA = 0;
FBC = 1.25P
Normal Stress and Strain: The lateral strain of rod BC is
Plat =
d - d0
39.99 - 40
=
= - 0.25(10 - 3) mm>mm
d0
40
Plat = - nPa;
- 0.25(10-3) = - (0.32)Pa
Pa = 0.78125(10-3) mm>mm
Assuming that Hooke’s Law applies,
sBC = EPa;
sBC = 200(109)(0.78125)(10-3) = 156.25 MPa
Since s 6 sY, the assumption is correct.
sBC =
FBC
;
ABC
156.25(106) =
1.25P
p
A 0.042 B
4
P = 157.08(103)N = 157 kN
Ans.
Ans:
P = 157 kN
165
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*3–28. If P = 150 kN, determine the elastic elongation of
rod BC and the decrease in its diameter. Rod BC is made of
A-36 steel and has a diameter of 40 mm.
C
P
1m
P
1m
A
1m
1m
B
0.75 m
Equations of Equilibrium: The force developed in rod BC can be determined by
writing the moment equation of equilibrium about A with reference to the freebody diagram of the beam shown in Fig. a.
a + ©MA = 0;
4
FBC a b (3) - 150(2) - 150(1) = 0
5
FBC = 187.5 kN
Normal Stress and Strain: The lateral strain of rod BC is
sBC =
187.5(103)
FBC
=
= 149.21 MPa
p
ABC
A 0.042 B
4
Since s 6 sY, Hooke’s Law can be applied. Thus,
149.21(106) = 200(109)PBC
sBC = EPBC;
PBC = 0.7460(10-3) mm>mm
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The unstretched length of rod BC is LBC = 27502 + 10002 = 1250 mm. Thus the
elongation of this rod is given by
dBC = PBCLBC = 0.7460(10-3)(1250) = 0.933 mm
Ans.
We obtain,
Plat = - nPa ;
Plat = - (0.32)(0.7460)(10-3)
= - 0.2387(10-3) mm>mm
Thus,
dd = Plat dBC = - 0.2387(10-3)(40) = - 9.55(10-3) mm
166
Ans.
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3–29. The friction pad A is used to support the member,
which is subjected to an axial force of P = 2 kN. The pad is
made from a material having a modulus of elasticity of
E = 4 MPa and Poisson’s ratio n = 0.4. If slipping does not
occur, determine the normal and shear strains in the pad.
The width is 50 mm. Assume that the material is linearly
elastic. Also, neglect the effect of the moment acting on
the pad.
P
60⬚
25 mm
A
100 mm
Internal Loading: The normal force and shear force acting on the friction pad can be
determined by considering the equilibrium of the pin shown in Fig. a.
+
: ©Fx = 0;
V - 2 cos 60° = 0
V = 1 kN
+ c ©Fy = 0;
N - 2 sin 60° = 0
N = 1.732 kN
Normal and Shear Stress:
t =
1(103)
V
=
= 200 kPa
A
0.1(0.05)
s =
1.732(103)
N
=
= 346.41 kPa
A
0.1(0.05)
Normal and Shear Strain: The shear modulus of the friction pad is
G =
4
E
=
= 1.429 MPa
2(1 + n)
2(1 + 0.4)
Applying Hooke’s Law,
s = EP;
346.41(103) = 4(106)P
P = 0.08660 mm>mm
Ans.
t = Gg;
200(103) = 1.429(106)g
g = 0.140 rad
Ans.
Ans:
P = 0.08660 mm>mm, g = 0.140 rad
167
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3–30. The lap joint is connected together using a 1.25 in.
diameter bolt. If the bolt is made from a material having a
shear stress–strain diagram that is approximated as shown,
determine the shear strain developed in the shear plane of
the bolt when P = 75 kip.
P
2
P
2
P
t (ksi)
75
50
Internal Loadings: The shear force developed in the shear planes of the bolt can be
determined by considering the equilibrium of the free-body diagram shown in Fig. a.
+
: ©Fx = 0;
75 - 2V = 0
0.005
0.05
g (rad)
V = 37.5 kip
Shear Stress and Strain:
t =
V
37.5
=
= 30.56 ksi
p
A
A 1.252 B
4
Using this result, the corresponding shear strain can be obtained from the shear
stress–strain diagram, Fig. b.
30.56
50
=
;
g
0.005
g = 3.06(10-3) rad
Ans.
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Ans:
g = 3.06(10-3) rad
168
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3–31. The lap joint is connected together using a 1.25 in.
diameter bolt. If the bolt is made from a material having a
shear stress–strain diagram that is approximated as shown,
determine the permanent shear strain in the shear plane of
the bolt when the applied force P = 150 kip is removed.
P
2
P
2
P
t (ksi)
75
50
Internal Loadings: The shear force developed in the shear planes of the bolt can be
determined by considering the equilibrium of the free-body diagram shown in Fig. a.
+
: ©Fx = 0;
150 - 2V = 0
V = 75 kip
0.005
Shear Stress and Strain:
t =
0.05
g (rad)
V
75
=
= 61.12 ksi
p
A
A 1.252 B
4
Using this result, the corresponding shear strain can be obtained from the shear
stress–strain diagram, Fig. b.
61.12 - 50
75 - 50
=
;
g - 0.005
0.05 - 0.005
g = 0.02501 rad
When force P is removed, the shear strain recovers linearly along line BC, Fig. b,
with a slope that is the same as line OA. This slope represents the shear modulus.
G =
50
= 10(103) ksi
0.005
Thus, the elastic recovery of shear strain is
t = Ggr;
61.12 = (10)(103)gr
gr = 6.112(10-3) rad
And the permanent shear strain is
gP = g - gr = 0.02501 - 6.112(10-3) = 0.0189 rad
Ans.
Ans:
gP = 0.0189 rad
169
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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
P
Shear Stress–Strain Relationship: Applying Hooke’s law with tA =
.
2p r h
g =
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
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y = 0
0 = -
C =
Then, y =
ro
P
ln
r
2p h G
At r = ri,
y = d
d =
P
ln ro + C
2p h G
P
ln ro
2p h G
ro
P
ln
ri
2p h G
Ans.
170
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3–33. 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 =
2 in.
8 kip
8 kip
3 in.
P
8
=
= 2.667 ksi
A
(2)(1.5)
Plong =
Plat =
n =
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.
Ans:
n = 0.330, h¿ = 2.000176 in.
171
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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
a
a
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.
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Ans:
d =
172
Pa
2bhG
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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.)
s
70
=
= 11400.65 ksi
P
0.00614
Eal =
When specimen is loaded with a 9 - kip load,
s =
9
P
= 45.84 ksi
= p
2
A
4 (0.5)
45.84
s
=
= 0.0040205 in.>in.
E
11400.65
Plong =
Plat =
0.49935 - 0.5
d¿ - d
=
= - 0.0013 in.>in.
d
0.5
V = -
Gal =
Plat
- 0.0013
= 0.32334
= Plong
0.0040205
11.4(103)
Eat
=
= 4.31(103) ksi
2(1 + v)
2(1 + 0.32334)
Ans.
Ans:
Gal = 4.31(103) ksi
173
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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
P
10
s =
= p
= 50.9296 ksi
2
A
(0.5)
4
0.00614
From the stress–strain diagram
E =
70
= 11400.65 ksi
0.00614
Plong =
G =
50.9296
s
=
= 0.0044673 in.>in.
E
11400.65
E
;
2(1 + v)
3.8(103) =
11400.65
;
2(1 + v)
v = 0.500
Plat = - vPlong = - 0.500(0.0044673) = - 0.002234 in.>in.
¢d = Plat d = - 0.002234(0.5) = - 0.001117 in.
d¿ = d + ¢d = 0.5 - 0.001117 = 0.4989 in.
Ans.
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174
P (in./in.)
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3–37. 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.
80 kN
x
A
B
220 mm
210 mm
3m
a +©MA = 0;
FB(3) - 80(x) = 0;
a +©MB = 0;
- FA(3) + 80(3 - x) = 0;
FB =
80x
3
FA =
(1)
80(3 - x)
3
(2)
Since the beam is held horizontally, dA = dB
s =
P
P
;
A
s
A
=
E
E
P =
d = PL = a
P
A
E
bL =
PL
AE
80(3 - x)
(220)
3
dA = dB;
AE
=
80x
3 (210)
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
A
(0.032)
4
Plong =
sA
E
55.27(106)
= -
73.1(109)
= - 0.000756
Plat = - nPlong = - 0.35(- 0.000756) = 0.0002646
dA¿ = dA + d Plat = 30 + 30(0.0002646) = 30.008 mm
Ans.
Ans:
x = 1.53 m, dA
¿ = 30.008 mm
175
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3–38. The wires each have a diameter of 12 in., length of
2 ft, and are made from 304 stainless steel. If P = 6 kip,
determine the angle of tilt of the rigid beam AB.
D
C
2 ft
P
2 ft
A
1 ft
B
Equations of Equilibrium: Referring to the free-body diagram of beam AB shown
in Fig. a,
a +©MA = 0;
FBC(3) - 6(2) = 0
FBC = 4 kip
+ c ©MB = 0;
6(1) - FAD(3) = 0
FAD = 2 kip
Normal Stress and Strain:
sBC =
sAD =
4(103)
FBC
=
= 20.37 ksi
ABC
p 1 2
a b
4 2
2(103)
FAD
=
= 10.19 ksi
AAD
p 1 2
a b
4 2
Since sBC 6 sY and sA 6 sY, Hooke’s Law can be applied.
sBC = EPBC;
20.37 = 28.0(103)PBC
PBC = 0.7276(10-3) in.>in.
sAD = EPAD;
10.19 = 28.0(103)PAD
PAD = 0.3638(10-3) in.>in.
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Thus, the elongation of cables BC and AD are given by
dBC = PBCLBC = 0.7276(10-3)(24) = 0.017462 in.
dAD = PADLAD = 0.3638(10-3)(24) = 0.008731 in.
Referring to the geometry shown in Fig. b and using small angle analysis,
u =
dBC - dAD
0.017462 - 0.008731
180°
=
= 0.2425(10-3) rad a
b = 0.0139°
36
36
prad
Ans.
Ans:
u = 0.0139°
176
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3–39. The wires each have a diameter of 12 in., length of
2 ft, and are made from 304 stainless steel. Determine the
magnitude of force P so that the rigid beam tilts 0.015°.
D
C
2 ft
P
2 ft
A
1 ft
B
Equations of Equilibrium: Referring to the free-body diagram of beam AB shown
in Fig. a,
a +©MA = 0;
FBC(3) - P(2) = 0
FBC = 0.6667P
+ c ©MB = 0;
P(1) - FAD(3) = 0
FAD = 0.3333P
Normal Stress and Strain:
sBC =
sAD =
FBC
0.6667P
=
= 3.3953P
ABC
p 1 2
a b
4 2
FAD
0.3333P
=
= 1.6977P
AAD
p 1 2
a b
4 2
Assuming that sBC 6 sY and sAD 6 sY and applying Hooke’s Law,
sBC = EPBC;
3.3953P = 28.0(106)PBC
PBC = 0.12126(10-6)P
sAD = EPAD;
1.6977P = 28.0(106)PAD
PAD = 60.6305(10-9)P
Thus, the elongation of cables BC and AD are given by
dBC = PBCLBC = 0.12126(10-6)P(24) = 2.9103(10-6)P
dAD = PADLAD = 60.6305(10-9)P(24) = 1.4551(10-6)P
Here, the angle of the tile is u = 0.015° a
prad
b = 0.2618(10-3) rad. Using small
180°
angle analysis,
u =
dBC - dAD
;
36
0.2618(10-3) =
2.9103(10-6)P - 1.4551(10-6)P
36
P = 6476.93 lb = 6.48 kip
Ans.
Since sBC = 3.3953(6476.93) = 21.99 ksi 6 sY and sAD = 1.6977(6476.93) =
11.00 ksi 6 sY, the assumption is correct.
Ans:
P = 6.48 kip
177
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*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 3 2
A
A B
4 16
Normal Strain: Since s 6 sg, Hooke’s law is still valid.
P =
s
28.97
= 0.000999 in.>in.
=
E
29(103)
Ans.
If the nut is unscrewed, the load is zero. Therefore, the strain P = 0
Ans.
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s (ksi)
3–41. The stress–strain diagram for polyethylene, which is
used to sheath coaxial cables, is determined from testing a
specimen that has a gauge length of 10 in. If a load P on the
specimen develops a strain of P = 0.024 in.>in., determine
the approximate length of the specimen, measured between
the gauge points, when the load is removed. Assume the
specimen recovers elastically.
P
5
4
3
2
1
0
P
0
0.008
0.016
0.024
0.032
0.040
0.048
P (in./in.)
Modulus of Elasticity: From the stress–strain diagram, s = 2 ksi when
P = 0.004 in.>in.
E =
2 - 0
= 0.500(103) ksi
0.004 - 0
Elastic Recovery: From the stress–strain diagram, s = 3.70 ksi when
P = 0.024 in.>in.
Elastic recovery =
s
3.70
= 0.00740 in.>in.
=
E
0.500(103)
Permanent Set:
Permanent set = 0.024 - 0.00740 = 0.0166 in.>in.
Thus,
Permanent elongation = 0.0166(10) = 0.166 in.
L = L0 + permanent elongation
= 10 + 0.166
= 10.17 in.
Ans.
Ans:
L = 10.17 in.
179
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3–42. The pipe with two rigid caps attached to its ends is
subjected to an axial force P. If the pipe is made from a
material having a modulus of elasticity E and Poisson’s
ratio n, determine the change in volume of the material.
ri
ro
L
P
a
Section a – a
a
P
Normal Stress: The rod is subjected to uniaxial loading. Thus, slong =
P
and slat = 0.
A
dV = AdL + 2prLdr
= APlong L + 2prLPlatr
Using Poisson’s ratio and noting that AL = pr2L = V,
dV = PlongV - 2nPlongV
= Plong (1 - 2n)V
slong
=
E
(1 - 2n)V
Since slong = P>A,
dV =
=
P
(1 - 2n)AL
AE
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PL
(1 - 2n)
E
Ans.
Ans:
dV =
180
PL
(1 - 2n)
E
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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
Pb =
159.15(106)
sb
= 0.00227 mm>mm
=
Eal
70(109)
Ans.
Ps =
39.79(106)
ss
= 0.000884 mm>mm
=
Emg
45(109)
Ans.
Ans:
Pb = 0.00227 mm>mm, Ps = 0.000884 mm>mm
181
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*3–44. An acetal polymer block is fixed to the rigid plates
at its top and bottom surfaces. If the top plate displaces
2 mm horizontally when it is subjected to a horizontal force
P = 2 kN, determine the shear modulus of the polymer.
The width of the block is 100 mm. Assume that the polymer
is linearly elastic and use small angle analysis.
400 mm
P ⫽ 2 kN
200 mm
Normal and Shear Stress:
t =
2(103)
V
=
= 50 kPa
A
0.4(0.1)
Referring to the geometry of the undeformed and deformed shape of the block
shown in Fig. a,
g =
2
= 0.01 rad
200
Applying Hooke’s Law,
t = Gg;
50(103) = G(0.01)
G = 5 MPa
Ans.
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4–1. The A992 steel rod is subjected to the loading shown.
If the cross-sectional area of the rod is 60 mm2, determine
the displacement of B and A, Neglect the size of the
couplings at B, C, and D.
D
0.75 m
C
60⬚
60⬚
1.50 m
3.30 kN
3.30 kN
B
5
3
3
3
4
3
10.4(10 )(1.50)
16.116(10 )(0.75)
PL
+
dB = ©
=
-6
9
AE
60(10 )(200)(10 )
60(10 - 6)(200)(109)
2 kN
= 0.00231 m = 2.31 mm
Ans.
5
4
A
0.50 m
2 kN
8 kN
3
dA = dB +
8(10 )(0.5)
60(10 - 6)(200)(109)
= 0.00264 m = 2.64 mm
Ans.
Ans:
dB = 2.31 mm, dA = 2.64 mm
183
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*4–2. The copper shaft is subjected to the axial loads shown.
Determine the displacement of end A with respect to end D
if the diameters of each segment are dAB = 0.75 in.,
dBC = 1 in., and dCD = 0.5 in. Take Ecu = 18(103) ksi.
dA>D = ©
80 in.
150 in.
5 kip
8 kip
A
100 in.
2 kip
5 kip B
C 2 kip
6 kip
D
- 8(80)
2(150)
6(100)
PL
=
+
+
p
p 2
p
AE
(0.75)2 (18)(103)
(1) (18)(103)
(0.5)2 (18)(103)
4
4
4
= 0.111 in.
Ans.
The positive sign indicates that end A moves away from end D.
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Ans:
dA>D = 0.111 in. away from end D.
184
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4–3. The composite shaft, consisting of aluminum, copper,
and steel sections, is subjected to the loading shown.
Determine the displacement of end A with respect to end D
and the normal stress in each section. The cross-sectional
area and modulus of elasticity for each section are shown in
the figure. Neglect the size of the collars at B and C.
Aluminum
Eal = 10(103 ) ksi
AAB = 0.09 in2
1.75 kip
3.50 kip
1.50 kip
B
18 in.
PAB
2
=
= 22.2 ksi
AAB
0.09
PBC
5
=
= 41.7 ksi
sBC =
ABC
0.12
sAB =
PBC
1.5
=
= 25.0 ksi
ABC
0.06
dND = ©
Steel
Est = 29(103 ) ksi
ACD = 0.06 in2
2.00 kip
A
sCD =
Copper
Ecu = 18(103 ) ksi
ABC = 0.12 in2
(T)
Ans.
(C)
Ans.
(C)
Ans.
3.50 kip
12 in.
C
D
1.75 kip
16 in.
( -5)(12)
( -1.5)(16)
2(18)
PL
+
+
=
3
3
AE
(0.09)(10)(10 )
(0.12)(18)(10 )
(0.06)(29)(103)
= -0.00157 in.
Ans.
The negative sign indicates end A moves towards end D.
Ans:
sAB = 22.2 ksi (T), sBC = 41.7 ksi (C),
sCD = 25.0 ksi (C), dA>D = 0.00157 in.
towards end D
185
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*4–4. Determine the displacement of B with respect to C
of the composite shaft in Prob. 4–3.
Aluminum
Eal = 10(103 ) ksi
AAB = 0.09 in2
Copper
Steel
Ecu = 18(103 ) ksi
ABC = 0.12 in2
Est = 29(103 ) ksi
ACD = 0.06 in2
1.75 kip
3.50 kip
2.00 kip
1.50 kip
A
B
18 in.
dB>C =
(- 5)(12)
PL
= - 0.0278 in.
=
AE
(0.12)(18)(103)
Ans.
The negative sign indicates end B moves towards end C.
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186
3.50 kip
12 in.
C
D
1.75 kip
16 in.
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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.
Ans:
dB = 1.59 mm, dA = 6.14 mm
187
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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
8
6
AE
4
(3)(35)(10 )(4) 7
(3)(35)(10 ) L0
L0
dA = 0.0128 in.
Ans.
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Ans:
dA = 0.0128 in.
188
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P1
4–7. If P1 = 50 kip and P2 = 150 kip, determine the vertical
displacement of end A of the high strength precast concrete
column.
P1
A
6 in.
a
a
P2
P2
10 ft
6 in.
Section a-a
B
Internal Loading: The normal forces developed in segments AB and BC are shown 10 ft
on the free-body diagrams of these segments in Figs. a and b, respectively.
Displacement: The cross-sectional area of segments AB and BC are AAB = 6(6) = 36 in2
and ABC = 10(10) = 100 in2.
dA>C = ©
b
b
C
10 in.
10 in.
Section b-b
PAB LAB
PBC LBC
PL
+
=
AE
AAB Econ
ABC Econ
- 400(10)(12)
(- 100)(10)(12)
=
3
36(4.2)(10 )
+
100(4.2)(103)
= - 0.194 in.
Ans.
The negative sign indicates that end A is moving towards C.
Ans:
dA = - 0.194 in.
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*4–8. If the vertical displacements of end A of the high
strength precast concrete column relative to B and B
relative to C are 0.08 in. and 0.1 in., respectively, determine
the magnitudes of P1 and P2.
P1
P1
A
6 in.
a
a
P2
P2
10 ft
6 in.
Section a-a
B
10 ft
Internal Loading: The normal forces developed in segments AB and BC are shown
on the free-body diagrams of these segments in Figs. a and b, respectively.
Displacement: The cross-sectional area of segments AB and BC are AAB = 6(6) =
36 in2 and ABC = 10(10) = 100 in2.
dA>B =
PAB LAB
AAB Econ
- 0.08 =
- 2P1(10)(12)
36(4.2)(103)
P1 = 50.4 kip
dB>C =
-0.1 =
Ans.
PBC LBC
ABC Econ
-[2(50.4) + 2P2](10)(12)
100(4.2)(103)
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P2 = 124.6 kip = 125 kip
Ans.
The negative sign indicates that end A is moving towards C.
190
b
b
C
10 in.
10 in.
Section b-b
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4–9. The assembly consists of two 10-mm diameter red
brass C83400 copper rods AB and CD, a 15-mm diameter
304 stainless steel rod EF, and a rigid bar G. If P = 5 kN,
determine the horizontal displacement of end F of rod EF.
300 mm
A
450 mm
B
P
E
4P
F
C
DG
P
Internal Loading: The normal forces developed in rods EF, AB, and CD are shown
on the free-body diagrams in Figs. a and b.
p
Displacement: The cross-sectional areas of rods EF and AB are AEF = (0.0152) =
4
p
56.25(10 - 6)p m2 and AAB = (0.012) = 25(10 - 6)p m2.
4
dF = ©
PEF LEF
PAB LAB
PL
=
+
AE
AEF Est
AAB Ebr
20(103)(450)
=
-6
5(103)(300)
9
56.25(10 )p(193)(10 )
+
25(10 - 6)p(101)(109)
= 0.453 mm
Ans.
The positive sign indicates that end F moves away from the fixed end.
Ans:
dF = 0.453 mm
191
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4–10. The assembly consists of two 10-mm diameter red
brass C83400 copper rods AB and CD, a 15-mm diameter
304 stainless steel rod EF, and a rigid bar G. If the
horizontal displacement of end F of rod EF is 0.45 mm,
determine the magnitude of P.
300 mm
A
450 mm
B
P
E
4P
F
C
DG
P
Internal Loading: The normal forces developed in rods EF, AB, and CD are shown
on the free-body diagrams in Figs. a and b.
Displacement: The cross-sectional areas of rods EF and AB are AEF =
p
(0.0152 ) =
4
56.25(10 - 6 )p m2 and
AAB =
p
(0.012 ) = 25(10 - 6 )p m2.
4
dF = ©
PEF LEF
PAB LAB
PL
=
+
AE
AEF Est
AAB Ebr
0.45 =
P(300)
4P(450)
-6
9
56.25(10 )p(193)(10 )
P = 4967 N = 4.97 kN
+
-6
25(10 )p(101)(109)
Ans.
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Ans:
P = 49.7 kN
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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
C
1 ft
Internal Forces in the wires:
5 ft
H
D
2 ft
1.8 ft
I
FBD (b)
A
B
3 ft
a + ©MA = 0;
FBC(4) - 500(3) = 0
+ c ©Fy = 0;
FAH + 375.0 - 500 = 0
FAH = 125.0 lb
a + ©MD = 0;
FCF(3) - 125.0(1) = 0
FCF = 41.67 lb
+ c ©Fy = 0;
FDE + 41.67 - 125.0 = 0
FBC = 375.0 lb
1 ft
500 lb
FBD (a)
FDE = 83.33 lb
Displacement:
dD =
83.33(3)(12)
FDELDE
= 0.0042857 in.
=
ADEE
0.025(28.0)(106)
dC =
41.67(3)(12)
FCFLCF
= 0.0021429 in.
=
ACFE
0.025(28.0)(106)
œ
dH
0.0021429
=
;
2
3
œ
dH
= 0.0014286 in.
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
dlœ = 0.0185357 in.
dl = 0.0074286 + 0.0185357 = 0.0260 in.
Ans.
Ans:
dl = 0.0260 in.
193
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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
3 ft
C
1 ft
Internal Forces in the wires:
2 ft
1.8 ft
I
FBD (b)
A
FBG(4) - 500(3) = 0
+ c ©Fy = 0;
FAH + 375.0 - 500 = 0
FAH = 125.0 lb
a + ©MD = 0;
FCF(3) - 125.0(1) = 0
FCF = 41.67 lb
+ c ©Fy = 0;
FDE + 41.67 - 125.0 = 0
FBG = 375.0 lb
FDE = 83.33 lb
Displacement:
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
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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)
0.0247143
;
48
b = 0.0295°
Ans.
194
1 ft
500 lb
FBD (a)
dD =
B
3 ft
a + ©MA = 0;
tan b =
5 ft
H
D
dB =
G
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4–13. The rigid bar is supported by the pin-connected rod
CB that has a cross-sectional area of 14 mm2 and is made
from 6061-T6 aluminum. Determine the vertical deflection
of the bar at D when the distributed load is applied.
C
300 N/m
1.5 m
D
A
B
2m
a+ ©MA = 0;
2m
1200(2) - TCB(0.6)(2) = 0
TCB = 2000 N
dB>C =
(2000)(2.5)
PL
= 0.0051835
=
AE
14(10 - 6)(68.9)(109)
(2.5051835)2 = (1.5)2 + (2)2 - 2(1.5)(2) cos u
u = 90.248°
u = 90.248° - 90° = 0.2478° = 0.004324 rad
dD = u r = 0.004324(4000) = 17.3 mm
Ans.
Ans:
dD = 17.3 mm
195
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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
B
Equation of Equilibrium: For entire post [FBD (a)]
+ c ©Fy = 0;
F + 8.00 - 20 = 0
w
2m
F
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)
www.elsolucionario.org
= - p
2
9
4 (0.06 ) 13.1 (10 )
= - 0.8639 A 10 - 3 B m
= - 0.864 mm
Ans.
Negative sign indicates that end A moves toward end B.
Ans:
F = 12.0 kN, dA>B = - 0.864 mm
196
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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
B
Equation of Equilibrium: For entire post [FBD (a)]
+ c ©Fy = 0;
F + 3.00 - 20 = 0
w
2m
F
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
A(y)E
AE
4
L0
L0
=
2m
y3
1
a
- 20yb 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
= - 1.03 mm
Ans.
Negative sign indicates that end A moves toward end B.
Ans:
F = 17.0 kN, dA>B = - 1.03 mm
197
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*4–16. The hanger consists of three 2014-T6 aluminum
alloy rods, rigid beams AC and BD, and a spring. If the hook
supports a load of P = 60 kN, determine the vertical
displacement of F. Rods AB and CD each have a diameter
of 10 mm, and rod EF has a diameter of 15 mm. The spring
has a stiffness of k = 100 MN>m and is unstretched when
P = 0.
C
A
450 mm
E
450 mm
Internal Loading: The normal forces developed in rods EF, AB, and CD and the
spring are shown in their respective free-body diagrams shown in Figs. a, b, and c.
B
D
F
Displacements: The cross-sectional areas of the rods are
p
AEF =
(0.0152) = 56.25(10 - 6)p m2 and
4
p
AAB = ACD =
(0.012) = 25(10 - 6)p m2.
4
P
dF>E =
60(103)(450)
FEF LEF
= 2.0901 mm T
=
AEF Eal
56.25(10 - 6)p(73.1)(109)
dB>A =
30(103)(450)
FAB LAB
= 2.3514 mm T
=
AAB Eal
25(10 - 6)p(73.1)(109)
The positive signs indicate that ends F and B move away from E and A, respectively.
Applying the spring formula,
dE>B =
Fsp
k
=
- 60
= - 0.6(10 - 3) m = 0.6 mm T
100(103)
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The negative sign indicates that E moves towards B. Thus, the vertical displacement
of F is
(+ T )
dF>A = dB>A + dF>E + dE>B
= 2.3514 + 2.0901 + 0.6
= 5.04 mm T
Ans.
198
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4–17. The hanger consists of three 2014-T6 aluminum
alloy rods, rigid beams AC and BD, and a spring. If the
vertical displacement of end F is 5 mm, determine the
magnitude of the load P. Rods AB and CD each have a
diameter of 10 mm, and rod EF has a diameter of 15 mm.
The spring has a stiffness of k = 100 MN>m and is
unstretched when P = 0.
C
A
450 mm
E
450 mm
Internal Loading: The normal forces developed in rods EF, AB, and CD and the
spring are shown in their respective free-body diagrams shown in Figs. a, b, and c.
B
D
F
Displacements: The cross-sectional areas of the rods are
p
AEF = (0.0152) = 56.25(10 - 6)p m2 and
4
AAB = ACD =
P
p
(0.012) = 25(10 - 6)p m2.
4
dF>E =
P(450)
FEF LEF
= 34.836(10 - 6)P T
=
AEF Eal
56.25(10 - 6)p(73.1)(109)
dB>A =
(P>2)(450)
FAB LAB
= 39.190(10 - 6)P T
=
AAB Eal
25(10 - 6)p(73.1)(109)
The positive signs indicate that ends F and B move away from E and A, respectively.
Applying the spring formula with
k = c 100(103)
kN 1000 N
1m
da
ba
b = 100(103) N>mm.
m
1 kN
1000 mm
dE>B =
Fsp
k
=
-P
= - 10(10 - 6)P = 10(10 - 6)P T
100(103)
The negative sign indicates that E moves towards B. Thus, the vertical displacement
of F is
(+ T )
dF>A = dB>A + dF>E + dE>B
5 = 34.836(10 - 6)P + 39.190(10 - 6)P + 10(10 - 6)P
Ans.
P = 59 505.71 N = 59.5 kN
Ans:
P = 59.5 kN
199
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4–18. Collar A can slide freely along the smooth vertical
guide. If the supporting rod AB is made of 304 stainless
steel and has a diameter of 0.75 in., determine the vertical
displacement of the collar when P = 10 kip.
P
A
2 ft
B
1.5 ft
Internal Loading: The normal force developed in rod AB can be determined
by considering the equilibrium of collar A with reference to its free-body diagram,
Fig. a.
4
-FAB a b - 10 = 0 FAB = - 12.5 kip
5
+ c ©Fy = 0;
Displacements: The cross-sectional area of rod AB is
AAB =
p
(0.752) = 0.4418 in2, and the initial length of rod AB is
4
LAB = 222 + 1.52 = 2.5 ft. The axial deformation of rod AB is
dAB =
- 12.5(2.5)(12)
FAB LAB
= - 0.03032 in.
=
AAB Est
0.4418(28)(103)
The negative sign indicates that end A moves towards B. From the geometry shown
1.5
in Fig. b, we obtain u = tan - 1 a
b = 36.87°. Thus,
2
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(dA)V =
dAB
0.03032
=
= 0.0379 in. T
cos u
cos 36.87°
Ans.
Ans:
(dA)V = 0.0379 in.
200
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4–19. Collar A can slide freely along the smooth vertical
guide. If the vertical displacement of the collar is 0.035 in.
and the supporting 0.75 in. diameter rod AB is made of
304 stainless steel, determine the magnitude of P.
P
A
2 ft
B
1.5 ft
Internal Loading: The normal force developed in rod AB can be determined
by considering the equilibrium of collar A with reference to its free-body diagram,
Fig. a.
+ c ©Fy = 0;
4
-FAB a b - P = 0
5
FAB = - 1.25 P
Displacements: The cross-sectional area of rod AB is
AAB =
p
(0.752) = 0.4418 in2, and the initial length of rod AB is
4
LAB = 222 + 1.52 = 2.5 ft. The axial deformation of rod AB is
dAB =
- 1.25P(2.5)(12)
FABLAB
= - 0.003032P
=
AABEst
0.4418(28.0)(103)
The negative sign indicates that end A moves towards B. From the geometry shown
1.5
in Fig. b, we obtain u = tan - 1 a
b = 36.87°. Thus,
2
dAB = (dA)V cos u
0.003032P = 0.035 cos 36.87°
P = 9.24 kip
Ans.
Ans:
P = 9.24 kip
201
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*4–20. The A992 steel drill shaft of an oil well extends
12000 ft into the ground. Assuming that the pipe used to
drill the well is suspended freely from the derrick at A,
determine the maximum average normal stress in each pipe
segment and the elongation of its end D with respect to the
fixed end at A. The shaft consists of three different sizes of
pipe, AB, BC, and CD, each having the length, weight per
unit length, and cross-sectional area indicated.
A
= 2.50 in.2
AAB
wAB = 3.2 lb/ft
B
ABC = 1.75 in.2
wBC = 2.8 lb/ft
ACD = 1.25 in.2
wCD = 2.0 lb/ft
sA =
3.2(5000) + 18000
P
=
= 13.6 ksi
A
2.5
Ans.
sB =
2.8(5000) + 4000
P
=
= 10.3 ksi
A
1.75
Ans.
sC =
2(2000)
P
=
= 3.2 ksi
A
1.25
Ans.
dD = ©
5000 ft
2000
5000
5000
(2.8x + 4000)dx
(3.2x + 18000)dx
P(x) dx
2x dx
+
+
=
6
6
(1.25)(29)(10 )
(1.75)(29)(10 )
(2.5)(29)(106)
L0
L0
L A(x) E
L0
= 2.99 ft
Ans.
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202
5000 ft
C
2000 ft
D
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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
Internal Force in the Rods:
0.75 m
k
H
E
A
C
FBD (a)
a + ©MA = 0;
FCD (0.5) - 4(0.25) = 0
+ c ©Fy = 0;
FAB + 2.00 - 4 = 0
0.25 m 0.25 m
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 springs
dsp =
Fsp
k
=
2.00
= 0.0333333 m = 33.3333 mm
60
d = dC + dsp
Ans.
= 0.5332 + 33.3333 = 33.9 mm
Ans:
d = 33.9 mm
203
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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
Internal Force in the Rods:
0.75 m
k
H
E
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
AEFE
(0.012)
(193)(109)
4
= 34.35988(10 - 6) W
dA>B = dC>D =
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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 springs
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.
Ans:
W = 9.69 kN
204
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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.
r2
L
r(x) = r1 +
r1L + (r2 - r1)x
r2 - r1
x =
L
L
r1
p
A(x) = 2 (r1L + (r2 - r1)x)2
L
P
L
d =
PL2
dx
Pdx
=
pE L0 [r1L + (r2 - r1)x]2
L A(x)E
= -
L
1
PL2
c
dƒ
p E (r2 - r1)(r1L + (r2 - r1)x) 0
= -
=
= -
PL2
1
1
c
d
p E(r2 - r1) r1L + (r2 - r1)L
r1L
r1 - r2
1
1
PL2
PL2
c
d = c
d
p E(r2 - r1) r2L
r1L
p E(r2 - r1) r2r1L
r2 - r1
PL
PL2
c
d =
p E(r2 - r1) r2r1L
p E r2r1
QED
205
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*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
h
w = d1 +
d1 h + (d2 - d1)x
d2 - d1
x =
h
h
d1
h
P(x) dx
P
dx
=
d =
[d1h + ( d 2 - d1 )x ] t
E L0
L A(x)E
P
h
h
=
Ph
dx
E t L0 d1 h + (d2 - d1)x
=
dx
Ph
E t d1 h L0 1 + d2 - d1
h
d1 h
=
x
h
d1 h
d2 - d1
Ph
a
b c ln a1 +
xb d ƒ
E t d1 h d2 - d1
d1 h
0
=
d2 - d1
d1 + d2 - d1
Ph
Ph
c ln a 1 +
bd =
cln a
bd
E t(d2 - d1)
d1
E t(d2 - d1)
d1
=
d2
Ph
c ln d
E t(d2 - d1)
d1
Ans.
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206
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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 h = 0.5 m.
d = 2c
d2
Ph
ln d
Est t(d2 - d1) d1
= 2c
30(103) (0.5)
200(109)(0.01)(0.075 - 0.02)
ln a
0.075
bd
0.02
= 0.360(10 - 3) m = 0.360 mm
Ans.
Ans:
d = 0.360 mm
207
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r1 ⫽ 0.5 in.
4–26. Determine the elongation of the tapered A992 steel
shaft when it is subjected to an axial force of 18 kip. Hint:
Use the result of Prob. 4–23.
d = (2)
4 in.
20 in.
4 in.
18 kip
PL1
PL2
+
p E r2r1
AE
(2)(18)(4)
=
18 kip
r1 ⫽ 0.5 in.
r2 ⫽ 2 in.
3
p(29)(10 )(2)(0.5)
18(20)
+
p(2)2(29)(103)
= 0.00257 in.
Ans.
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Ans:
d = 0.00257 in.
208
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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
x
B
Displacements: The cross-sectional area of the bar as a function of x is
A(x) = pr2 = pr0 2e2ax. We have
r0
r ⫽ r0 eax
A
L
d =
L
P(x)dx
P
dx
=
pr0 2E L0 e2ax
L0 A(x)E
=
L
1
P
cd 2
2
2ax
pr0 E 2ae
0
=
P
a 1 - e - 2aL b
2apr0 2E
P
Ans.
Ans:
d =
209
P
(1 - e - 2aL)
2apr0 2E
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*4–28. Bone material has a stress–strain diagram that can
be defined by the relation s = E[P>(1 + kEP)], where k
and E are constants. Determine the compression within the
length L of the bone, where it is assumed the cross-sectional
area A of the bone is constant.
P
L
P
dx
s = ; P =
A
dx
s = Ea
P
b;
1 + kEP
P
=
A
Ea
dx
b
dx
1 + kE a
dx
b
dx
P
PkE dx
dx
P
+
a b = Ea b
A
A
dx
dx
PkE
dx
P
= aE ba b
A
A
dx
L
d
L0
dx =
L0
P dx
AE a 1 -
Pk
b
A
PL
PL
AE
=
d =
E(A - Pk)
Pk
a1 b
A
Ans.
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4–29. 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.
F
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 A(y)E
L0
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.
Ans:
p0 = 250 kN>m, d = 2.93 mm
211
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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,
determine the resisting bearing force F for equilibrium.
Take p0 = 180 kN>m. Also, find the corresponding elastic
shortening of the pile. Neglect the weight of the pile.
P
p0
12 m
F
Internal Loading: By considering the equilibrium of the pile with reference to its
entire free-body diagram shown in Fig. a. We have
+ c ©Fy = 0;
F +
1
(180)(12) - 1500 = 0
2
F = 420 kN
Ans.
Also,
p(y) =
180
y = 15y kN>m
12
The normal force developed in the pile as a function of y can be determined
by considering the equilibrium of the sectional of the pile with reference to its
free-body diagram shown in Fig. b.
www.elsolucionario.org
+ c ©Fy = 0;
1
(15y)y + 420 - P(y) = 0
2
P(y) = (7.5y2 + 420) kN
p
Displacement: The cross-sectional area of the pile is A = (0.32) = 0.0225p m2.
4
We have
12 m
P(y)dy
(7.5y2 + 420)(103)dy
=
0.0225p(29.0)(109)
L0 A(y)E
L0
L
d =
12 m
=
L0
c 3.6587(10 - 6)y2 + 0.2049(10 - 3) d dy
= c 1.2196(10 - 6)y3 + 0.2049(10 - 3)y d
12 m
0
= 4.566(10 - 3) m = 4.57 mm
Ans.
Ans:
F = 420 kN, d = 4.57 mm
212
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
4–31. The concrete column is reinforced using four steel
reinforcing rods, each having a diameter of 18 mm.
Determine the stress in the concrete and the steel if
the column is subjected to an axial load of 800 kN.
Est = 200 GPa, Ec = 25 GPa.
800 kN
300 mm
300 mm
Equilibrium:
Pst + Pcon - 800 = 0
+ c ©Fy = 0;
(1)
Compatibility:
dst = dcon
Pst(L)
p
4 a b (0.0182)(200)(109)
4
=
Pcon(L)
p
c 0.32 - 4a b (0.0182) d(25)(109)
4
Pst = 0.091513 Pcon
(2)
Solving Eqs. (1) and (2) yields:
Pst = 67.072 kN
Pcon = 732.928 kN
Average Normal Stress:
sst =
67.072(103)
p
4 a b (0.0182)
4
scon =
= 65.9 MPa
732.928(103)
p
c 0.32 - 4 a b (0.0182) d
4
Ans.
= 8.24 MPa
Ans.
Ans:
sst = 65.9 MPa, scon = 8.24 MPa
213
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*4–32. The column is constructed from high-strength
concrete and four A-36 steel reinforcing rods. If it is
subjected to an axial force of 800 kN, determine the
required diameter of each rod so that one-fourth of the load
is carried by the steel and three-fourths by the concrete.
Est = 200 GPa, Ec = 25 GPa.
Equilibrium: Require Pst =
Pcon =
800 kN
300 mm
1
(800) = 200 kN and
4
3
(800) = 600 kN.
4
Compatibility:
dcon = dst
PconL
(0.32 - Ast)(25.0)(109)
=
Ast =
PstL
Ast(200)(109)
0.09Pst
8Pcon + Pst
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0.09(200)
p
4c a b d2 d =
4
8(600) + 200
d = 0.03385 m = 33.9 mm
Ans.
214
300 mm
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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
500 mm
Pst + Pcon - 80 = 0
+ c ©Fy = 0;
(1)
dst = dcon
Pst L
p
2
2
9
4 (0.08 - 0.07 ) (200) (10 )
= p
Pcon L
2
9
4 (0.07 ) (24) (10 )
Pst = 2.5510 Pcon
(2)
Solving Eqs. (1) and (2) yields
Pst = 57.47 kN
Pcon = 22.53 kN
sst =
57.47 (103)
Pst
= 48.8 MPa
= p
2
2
Ast
4 (0.08 - 0.07 )
Ans.
scon =
22.53 (103)
Pcon
= 5.85 MPa
= p
2
Acon
4 (0.07 )
Ans.
Ans:
sst = 48.8 MPa, scon = 5.85 MPa
215
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4–34. If column AB is made from high strength pre-cast
concrete and reinforced with four 43 in. diameter A-36 steel
rods, determine the average normal stress developed in the
concrete and in each rod. Set P = 75 kip.
P
P
A
a
9 in.
a
9 in.
10 ft
Equation of Equilibrium: Referring to the free-body diagram of the cut part of the
concrete column shown in Fig. a,
Pcon + 4Pst - 2(75) = 0
+ c ©Fy = 0;
Section a-a
B
(1)
Compatibility Equation: Since the steel bars and the concrete are firmly bonded,
their deformation must be the same. Thus,
dcon = dst
Pcon(10)(12)
c (9)(9) - 4 a
2
p
3
b a b d (4.20)(103)
4
4
=
Pst(10)(12)
p 3 2
a b (29)(103)
4 4
Pcon = 25.974Pst
Solving Eqs. (1) and (2),
Pst = 5.0043 kip
(2)
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Pcon = 129.98 kip
Normal Stress: Applying Eq. (1-6),
scon =
sst =
Pcon
=
Acon
129.98
= 1.64 ksi
p 3 2
(9)(9) - 4 a b a b
4
4
Ans.
Pst
5.0043
=
= 11.3 ksi
Ast
p 3 2
a b
4 4
Ans.
Ans:
scon = 1.64 ksi, sst = 11.3 ksi
216
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4–35. If column AB is made from high strength pre-cast
concrete and reinforced with four 34 in. diameter A-36 steel
rods, determine the maximum allowable floor loadings P.
The allowable normal stress for the high strength concrete
and the steel are (sallow)con = 2.5 ksi and (sallow)st = 24 ksi,
respectively.
P
P
A
a
9 in.
a
9 in.
10 ft
Section a-a
B
Equation of Equilibrium: Referring to the free-body diagram of the cut part of the
concrete column shown in Fig. a,
+ c ©Fy = 0;
Pcon + 4Pst - 2P = 0
(1)
Compatibility Equation: Since the steel bars and the concrete are firmly bonded,
their deformation must be the same. Thus,
dcon = dst
Pcon(10)(12)
p
3 2
c (9)(9) - 4 a b a b d (4.20)(103)
4
4
=
Pst(10)(12)
p 3 2
a b (29.0)(103)
4 4
Pcon = 25.974Pst
(2)
Solving Eqs. (1) and (2),
Pst = 0.06672P
Pcon = 1.7331P
Allowable Normal Stress:
(scon)allow =
Pcon
;
Acon
2.5 =
1.7331P
p 3 2
(9)(9) - 4 a b a b
4
4
P = 114.29 kip = 114 kip (controls)
(sst)allow =
Pst
;
Ast
24 =
Ans.
0.06672P
p 3 2
a b
4 4
P = 158.91 kip
Ans:
P = 114 kip
217
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*4–36. Determine the support reactions at the rigid
supports A and C. The material has a modulus of elasticity
of E.
3 d
4
d
P
B
A
2a
Equation of Equilibrium: Referring to the free-body diagram of the assembly shown
in Fig. a,
+
: ©Fx = 0;
P - FA - FC = 0
(1)
Compatibility Equation: Using the method of superposition, Fig. b,
+)
(:
d = dP - dFC
0 =
FC =
P(2a)
a
p 2
d bE
4
- ≥
FCa
2
p 3
a db E
4 4
+
FC(2a)
a
p 2
d bE
4
¥
9
P
17
Ans.
Substituting this result into Eq. (1),
FA =
8
P
17
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Ans.
218
C
a
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
4–37. If the supports at A and C are flexible and have a
stiffness k, determine the support reactions at A and C. The
material has a modulus of elasticity of E.
3 d
4
d
P
B
A
C
a
2a
Equation of Equilibrium: Referring to the free-body diagram of the assembly
shown in Fig. a,
+
: ©Fx = 0;
P - FA - FC = 0
(1)
Compatibility Equation: Using the method of superposition, Fig. b,
+)
(:
dC = dP - dFC
FC(2a)
P(2a)
FC
FC
FCa
P
+ ¥ - ≥
+
+
= ≥
¥
2
k
k
k
p 2
p 3
p 2
a db E
a d bE
a d bE
4
4 4
4
FC = c
9(8ka + pd2E)
dP
Ans.
64ka + 9pd2E
bP
136ka + 18pd2E
Ans.
136ka + 18pd2E
Substituting this result into Eq. (1),
FA = a
Ans:
FC = c
9(8ka + pd2E)
136ka + 18pd2E
FA = a
219
dP,
64ka + 9pd2E
bP
136ka + 18pd2E
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
4–38. The load of 2800 lb is to be supported by the two
essentially vertical A-36 steel wires. If originally wire AB is
60 in. long and wire AC is 40 in. long, determine the force
developed in each wire after the load is suspended. Each
wire has a cross-sectional area of 0.02 in2.
B
C
60 in.
40 in.
A
+ c ©Fy = 0;
TAB + TAC - 2800 = 0
dAB = dAC
TAB (60)
TAC (40)
=
AE
AE
1.5TAB = TAC
Solving,
TAB = 1.12 kip
Ans.
TAC = 1.68 kip
Ans.
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Ans:
TAB = 1.12 kip, TAC = 1.68 kip
220
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4–39. The load of 2800 lb is to be supported by the two
essentially vertical A-36 steel wires. If originally wire AB is
60 in. long and wire AC is 40 in. long, determine the crosssectional area of AB if the load is to be shared equally
between both wires. Wire AC has a cross-sectional area of
0.02 in2.
B
C
60 in.
40 in.
A
TAC = TAB =
2800
= 1400 lb
2
dAC = dAB
1400(40)
1400(60)
6
(0.02)(29)(10 )
=
AAB(29)(106)
AAB = 0.03 in2
Ans.
Ans:
AAB = 0.03 in2
221
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*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.
a + ©ME = 0;
B
0.75 m
E
A
0.5 m
0.5 m
(1)
TEF - 2T = 0
+ T ©Fy = 0;
TEF = 2T
(2)
Rod EF shortens 1.5 mm causing AB (and DC) to elongate. Thus:
0.0015 = dA>B + dE>F
T(0.75)
(125)(10 - 6)(200)(109)
2T(0.75)
+
C
0.75 m
- TAB(0.5) + TCD(0.5) = 0
TAB = TCD = T
0.0015 =
D
(125)(10 - 6)(200)(109)
2.25T = 37500
T = 16,666.67 N
TAB = TCD = 16.7 kN
Ans.
TEF = 33.3 kN
Ans.
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222
F
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4–41. The 2014-T6 aluminum rod AC is reinforced with
the firmly bonded A992 steel tube BC. If the assembly fits
snugly between the rigid supports so that there is no gap at
C, determine the support reactions when the axial force of
400 kN is applied. The assembly is attached at D.
D
A
400 mm
400 kN
B
A992 steel
800 mm
50 mm
a
a
25 mm
2014–T6 aluminum alloy
Section a–a
C
Equation of Equilibrium: Referring to the free-body diagram of the assembly shown
in Fig. a,
FD + (FC)al + (FC)st - 400(103) = 0
+ c ©Fy = 0;
(1)
Compatibility Equation: Using the method of superposition, Fig. b,
(+ T )
0 = dp - dFC
0 = +
400(103)(400)
p(0.0252)(73.1)(109)
400(103) = 3(FC)al + (FC)st
- J
(FC)al(800)
p(0.0252)(73.1)(109)
+
[(FC)al + (FC)st](400)
p(0.0252)(73.1)(109)
(2)
K
Also, since the aluminum rod and steel tube of segment BC are firmly bonded, their
deformation must be the same. Thus,
(dBC)st = (dBC)al
(FC)st(800)
p(0.052 - 0.0252)(200)(109)
=
(FC)al(800)
p(0.0252)(73.1)(109)
(3)
(FC)st = 8.2079(FC)al
Solving Eqs. (1) and (2),
(FC)al = 35.689 kN
(FC)st = 292.93 kN
Substituting these results into Eq. (1),
FD = 71.4 kN
Ans.
Also,
FC = (FC)st + (FC)al
= 35.689 + 292.93
= 329 kN
Ans.
Ans:
FD = 71.4 kN, FC = 329 kN
223
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
4–42. The 2014-T6 aluminum rod AC is reinforced with
the firmly bonded A992 steel tube BC. When no load is
applied to the assembly, the gap between end C and the
rigid support is 0.5 mm. Determine the support reactions
when the axial force of 400 kN is applied.
D
A
400 mm
400 kN
B
A992 steel
800 mm
50 mm
a
a
25 mm
2014–T6 aluminum alloy
Section a–a
C
Equation of Equilibrium: Referring to the free-body diagram of the assembly
shown in Fig. a,
FD + (FC)al + (FC)st - 400(103) = 0
+ c ©Fy = 0;
(1)
Compatibility Equation: Using the method of superposition, Fig. b,
dC = dP - dFC
(+ T)
0.5 = +
400(103)(400)
2
9
p(0.025 )(73.1)(10 )
- ≥
(FC)al (800)
2
9
+
[(FC)al + (FC)st](400)
p(0.025 )(73.1)(10 )
p(0.0252)(73.1)(109)
220.585(103) = 3(FC)al + (FC)st
¥
(2)
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Also, since the aluminum rod and steel tube of segment BC are firmly bonded, their
deformation must be the same. Thus,
(dBC)st = (dBC)al
(FC)st (800)
2
2
9
=
p(0.05 - 0.025 )(200)(10 )
(FC)al (800)
p(0.0252)(73.1)(109)
(FC)st = 8.2079(FC)al
(3)
Solving Eqs. (2) and (3),
(FC)al = 19.681 kN
(FC)st = 161.54 kN
Substituting these results into Eq. (1),
FD = 218.777 kN = 219 kN
Ans.
Also,
FC = (FC)al + (FC)st
= 19.681 + 161.54
= 181 kN
Ans.
Ans:
FD = 219 kN, FC = 181 kN
224
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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
Equation of Equilibrium: Due to symmetry, FAB = FCD = F. Referring to the
free-body 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 )
d
FEF = 42 483.23 N
Substituting this result into Eq. (1),
F = 18 758.38 N
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.
Ans:
sAB = sCD = 26.5 MPa, sEF = 33.8 MPa
225
C
30 mm
40 kN
D
G
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*4–44. The assembly consists of two red brass C83400
copper rods AB and CD having a diameter of 30 mm, a
304 stainless steel rod EF having a diameter of 40 mm, and
a rigid member G. If the supports at A, C, and F each have a
stiffness of k = 200 MN>m determine the average normal
stress developed in the rods when the load is applied.
300 mm
450 mm
40 kN
A
30 mm
B
E
F
40 mm
C
30 mm
D
Equation of Equilibrium: Due to symmetry, FAB = FCD = F. Referring to the freebody diagram of the assembly shown in Fig. a,
2F + FEF - 2[40(103)] = 0
+ ©F = 0,
:
x
(1)
Compatibility Equation: Using the method of superposition, Fig. b,
+ )
(;
dF = dP - dEF
(2)
Where
dF =
FEF
FEF
=
(1000) = 5(10 - 6)FEF mm
k
200 (106)
dP =
dEF =
40(103)(300)
p
(0.032)(101)(109)
4
40(103)
+
200(106)
(1000) = 0.3681 mm
FEF(450)
(FEF>2)(300)
(FEF>2)
(1000)
+
+
p
p
200(106)
(0.042)(193)(109)
(0.032)(101)(109)
4
4
www.elsolucionario.org
= 6.4565(10 - 6)FEF mm
Thus,
5(10 - 6)FEF = 0.3681 - 6.4565(10 - 6)FEF
FEF = 32.13 kN
From Eq. (1),
2F + 32.13(103) - 2[40(103)] = 0
F = 23.93 kN
sAB = sCD =
sEF =
23.93(103)
F
=
= 33.9 MPa
p
ACD
(0.032)
4
Ans.
32.13(103)
FEF
= 25.6 MPa
=
p
AEF
(0.042)
4
Ans.
226
40 kN
G
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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
40 kN
150 mm
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
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.
Ans:
sb = 32.4 MPa, st = 34.5 MPa
227
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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 A-36 steel.
600 mm
600 mm
0.15 mm
P
A
50 mm
D
B
25 mm
C
Equation of Equilibrium: Referring to the free-body diagram of the assembly shown in Fig. a,
200(103) - FD - FA = 0
+
: ©Fx = 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 )
+ p
FD (600)
2
9
4 (0.025 )(200)(10 )
FD = 20 365.05 N = 20.4 kN
S
Ans.
Substituting this result into Eq. (1),
FA = 179 634.95 N = 180 kN
Ans.
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Ans:
FD = 20.4 kN, FA = 180 kN
228
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
4–47. The support consists of a solid red brass C83400
copper post surrounded by a 304 stainless steel tube. Before
the load is applied the gap between these two parts is 1 mm.
Given the dimensions shown, determine the greatest axial
load that can be applied to the rigid cap A without causing
yielding of any one of the materials.
P
A
1 mm
0.25 m
60 mm
80 mm
Require,
10 mm
dst = dbr + 0.001
Fst(0.25)
2
2
9
=
p[(0.05) - (0.04) ]193(10 )
Fbr(0.25)
p(0.03)2(101)(109)
+ 0.001
0.45813 Fst = 0.87544 Fbr + 106
(1)
Fst + Fbr - P = 0
+ c ©Fy = 0;
(2)
Assume brass yields, then
(Fbr)max = sg Abr = 70(106)(p)(0.03)2 = 197 920.3 N
(Pg)br = sg>E =
70.0(106)
101(109)
= 0.6931(10 - 3) mm>mm
dbr = (eg)brL = 0.6931(10 - 3)(0.25) = 0.1733 mm < 1 mm
Thus only the brass is loaded.
P = Fbr = 198 kN
Ans.
Ans:
P = 198 kN
229
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*4–48. The specimen represents a filament-reinforced
matrix system made from plastic (matrix) and glass (fiber).
If there are n fibers, each having a cross-sectional area
of Af and modulus of Ef, embedded in a matrix having
a cross-sectional area of Am and modulus of Em, determine
the stress in the matrix and each fiber when the force P is
imposed on the specimen.
P
P
+ c ©Fy = 0;
- P + Pm + Pf = 0
(1)
AmEm
P
nAfEf f
(2)
dm = df
PfL
PmL
=
;
AmEm
nAfEf
Pm =
Solving Eqs. (1) and (2) yields
Pm =
AmEm
P;
nAfEf + AmEm
sf =
Pm
=
Am
a
nAfEf + AmEm
AmEm
- Pb
nAfEf + AmEm
=
=
Am
a
Pf
nAf
nAfEf
P
www.elsolucionario.org
Normal stress:
sm =
Pf =
nAfEf
nAfEf + AmEm
nAf
Pb
Em
P
nAfEf + AmEm
Ef
=
nAfEf + AmEm
P
230
Ans.
Ans.
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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)
2(3
0.025
x)(2)(E)
L30
30
- FA
L0
60
dx
dx
+ FB
= 0
(3 - 0.025 x)
(3
0.025x)
L30
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.
Ans:
FA = 4.09 kip, FB = 2.91 kip
231
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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
greatest possible magnitude for P without exceeding an
average normal stress of sallow = 4 ksi anywhere in the
member, and determine the location x at which P would
need to be applied. 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
2(3
0.025
x)(2)(E)
2(3
0.025
x)(2)(E)
Lx
L0
x
- FA
60
dx
dx
+ FB
= 0
(3
0.025
x)
(3
0.025 x)
L0
Lx
FA(40) ln (3 - 0.025 x)|x0 - FB(40) ln (3 - 0.025x)|60
x = 0
FA ln a1 -
0.025 x
0.025x
b = - FB ln a 2 b
3
1.5
For greatest magnitude of P require,
www.elsolucionario.org
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 a 2 b
3
1.5
Solving by trial and error,
x = 28.9 in.
Ans.
Therefore,
FA = 36.4 kip
P = 60.4 kip
Ans.
Ans:
x = 28.9 in., P = 60.4 kip
232
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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
3 ft
3 ft
(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.
Ans:
TCD = 27.2 kip, TCB = 9.06 kip
233
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*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 supplied.
C
6 ft
6 kip/ft
See solution of Prob. 4-51.
A
TCD = 27.1682 kip
27.1682 245
=
= 0.1175806 ft
dDC =
0.05(31)(103)
0.05(31)(103)
3 ft
Using Eq. (2) of Prob. 4-51,
(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.
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234
D
B
TCD 245
3 ft
3 ft
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
4–53. Each of the three A-36 steel wires has the same
diameter. Determine the force in each wire needed to
support the 200-kg load.
600 mm
A
600 mm
D
C
800 mm
B
Equation of Equilibrium: Referring to the free-body diagram of joint B shown in Fig. a,
+
: ©Fx = 0,
3
3
FBC a b - FAB a b = 0
5
5
+ c ©Fy = 0;
4
2 c F a b d + FBD - 200(9.81) = 0
5
FBC = FAB = F
1.6F + FBD = 1962
(1)
Compatibility Equation: Due to symmetry, joint B will displace vertically. Referring
600
b = 36.87°. Thus,
to the geometry shown in Fig. b, u = tan - 1 a
800
dBC = dBD cos 36.87°
dBC = 0.8dBD
F(1000)
FBD(800.25)
= 0.8 ≥
¥
p
p
2
9
2
9
(0.004 )(200)(10 )
(0.004 )(200)(10 )
4
4
F = 0.6402FBD
(2)
Solving Eqs. (1) and (2),
FBD = 969 N
FAB = FBC = 620 N
Ans.
Ans:
FBD = 969 N, FAB = FBC = 620 N
235
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4–54. The 200-kg load is suspended from three A-36 steel
wires each having a diameter of 4 mm. If wire BD has a
length of 800.25 mm before the load is applied, determine
the average normal stress developed in each wire.
600 mm
600 mm
D
A
C
800 mm
B
Equation of Equilibrium: Referring to the free-body diagram of joint B shown in Fig. a,
+
: ©Fx = 0;
3
3
FBC a b - FAB a b = 0
5
5
+ c ©Fy = 0;
4
2 c Fa b d + FBD - 200(9.81) = 0
5
FBC = FAB = F
1.6F + FBD = 1962
(1)
Compatibility Equation: Due to symmetry, joint B will displace vertically. Referring
600
b = 36.87°. Thus,
to the geometry shown in Fig. b, u = tan - 1 a
800
dBC = (dBD + 0.25) cos 36.87°
www.elsolucionario.org
dBC = 0.8dBD + 0.2
F(1000)
p
(0.0042)(200)(109)
4
= 0.8 ≥
FBD(800.25)
p
(0.0042)(200)(109)
4
¥ + 0.2
F = 0.6402FBD + 502.65
(2)
Solving Eqs. (1) and (2),
FBD = 571.93 N
FAB = FBC = 868.80 N
Normal Stress:
sBD =
FBD
571.93
=
= 45.5 MPa
p
ABD
(0.0042)
4
sAB = JBC =
Ans.
F
868.80
=
= 69.1 MPa
p
ABC
(0.0042)
4
Ans.
Ans:
sBD = 45.5 MPa, sAB = 69.1 MPa
236
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
4–55. The three suspender bars are made of A992 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
C
B
80 kN
50 kN
E
D
1m
1m
1m
F
1m
Referring to the FBD of the rigid beam, Fig. a,
+ 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.
Ans:
sBE = 96.3 MPa, sAD = 79.6 MPa,
sCF = 113 MPa
237
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*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
Referring to the FBD of the rigid bar, Fig. a,
A
12
3
FBC a b (5) + FCD a b (16) - 800(10) = 0
13
5
a + ©MA = 0;
5 ft
(1)
The unstretched lengths 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 a point on
12
3
d
the rigid bar is dv =
. For points B and D, cos uB =
and cos uD = . Thus,
cos u
13
5
the vertical displacements of points B and D are
A dB B v =
cos uB
A dD B v =
cos uD
dBC
=
dCD
=
www.elsolucionario.org
FBC (13)>AE
169 FBC
=
12>13
12AE
FCD (20)>AE
100 FCD
=
3>5
3 AE
The similar triangles shown in Fig. c give
A dB B v
5
=
A dD B v
16
1 169 FBC
1 100 FCD
a
a
b =
b
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.
238
D
5 ft
6 ft
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4–56. Continued
239
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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
D
5 ft
6 ft
(1)
The unstretched lengths 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 a point on
12
3
d
the rigid bar is dv =
. For points B and D, cos uB =
and cos uD = . Thus,
cos u
13
5
the vertical displacements of points B and D are
www.elsolucionario.org
A dB B v =
cos uB
A dD B v =
cos uD
dBC
dCD
=
FBC (13)>AE
169 FBC
=
12>13
12AE
=
FCD (20)>AE
100 FCD
=
3>5
3 AE
The similar triangles shown in Fig. c gives
A dB B v
=
A dD B v
5
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 v =
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.
240
Ans:
u = 0.0633°
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
4–58. Two identical rods AB and CD each have a length L
and diameter d, and are used to support the rigid beam,
which is pinned at F. If a vertical force P is applied at the
end of the beam, determine the normal stress developed in
each rod. The rods are made of material that has a modulus
of elasticity of E.
D
P
C
F
A
a
Equation of Equilibrium: Referring to the free-body diagram of the rigid beam
shown in Fig. a,
a + ©MF = 0;
a
2a
B
FAB(a) + FCD(a) - P(3a) = 0
FAB + FCD = 3P
(1)
Compatibility Equation: Referring to the geometry of the deformation diagram of
the rods shown in Fig. b,
dAB = dCD
FABL
FCDL
=
AE
AE
FAB = FCD
(2)
Solving Eqs. (1) and (2),
FAB = FCD =
3
P
2
Normal Stress:
3
P
FCD
2
6P
=
=
sAB = sCD =
p 2
ACD
pd2
d
4
Ans.
Ans:
sAB = sCD =
241
6P
pd2
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4–59. Two identical rods AB and CD each have a length L
and diameter d, and are used to support the rigid beam,
which is pinned at F. If a vertical force P is applied at the
end of the beam, determine the angle of rotation of the
beam. The rods are made of material that has a modulus of
elasticity of E.
D
P
C
F
A
a
Equation of Equilibrium: Referring to the free-body diagram of the rigid beam
shown in Fig. a,
a + ©MF = 0;
a
2a
B
FAB(a) + FCD(a) - P(3a) = 0
FAB + FCD = 3P
(1)
Compatibility Equation: Referring to the geometry of the deformation diagram of
the rods shown in Fig. b,
dAB = dCD
FABL
FCDL
=
AE
AE
www.elsolucionario.org
FAB = FCD
(2)
Solving Eqs. (1) and (2),
FAB = FCD =
3
P
2
Displacement: Using these results,
FABLAB
dAB =
=
AE
3
a PbL
2
p
a d2 b E
4
=
6PL
pd2E
Referring to Fig. b, the angle of tilt u of the beam is
u =
dAB
a
=
6PL>pd2E
6PL
=
a
pd2Ea
Ans.
Ans:
u =
242
6PL
pd2Ea
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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 positions 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) yields
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.
243
E
F
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4–61. 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.
B
D
F
L
P
A
a + ©MA = 0;
P
2
d
2
E
d
(1)
FAB + FCD + FEF - P = 0
+ c ©Fy = 0;
d
2
d
FCD(d) + FEF(2d) - P a b = 0
2
FCD + 2FEF =
C
(2)
dC - dE
dA - dE
=
d
2d
2dC = dA + dE
2FCDL
FABL
FEFL
=
+
AE
AE
AE
2FCD - FAB - FEF = 0
(3)
www.elsolucionario.org
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 =
Ans:
sAB =
244
7P
P
P
,s
=
,s =
12A CD
3A EF
12A
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
4–62. If the 2-in. diameter supporting rods are made from
A992 steel, determine the average normal stress developed
in each rod when P = 100 kip.
P
A
2 ft
2 ft
30⬚
Equation of Equilibrium: Referring to the free-body diagram of joint A shown in Fig. a,
+
: ©Fx = 0;
FAB sin 30° - FAC sin 30° = 0
+ c ©Fy = 0;
2F cos 30° + FAD - 100 = 0
B
30⬚
D
C
FAB = FAC = F
(1)
Compatibility Equation: Due to symmetry, joint A will displace vertically. Referring
to the geometry shown in Fig. b, we have
dF = d FAD cos 30°
F(2)(12)
FAD[2 cos 30°(12)]
= e
f cos 30°
AEst
AEst
F = 0.75FAD
(2)
Solving Eqs. (1) and (2),
FAD = 43.50 kip
F = 32.62 kip
Normal Stress:
sAB = sAC =
sAD =
F
32.62
=
= 10.4 ksi
p 2
AAC
(2 )
4
Ans.
FAD
43.50
=
= 13.8 ksi
p 2
AAD
(2 )
4
Ans.
Ans:
sAB = sAC = 10.4 ksi, sAD = 13.8 ksi
245
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4–63. If the supporting rods of equal diameter are made
from A992 steel, determine the required diameter to the
nearest 18 in. of each rod when P = 100 kip. The allowable
normal stress of the steel is sallow = 24 ksi.
P
A
2 ft
2 ft
30⬚
Equation of Equilibrium: Referring to the free-body diagram of joint A shown in
Fig. a,
+
: ©Fx = 0;
FAB sin 30° - FAC sin 30° = 0
+ c ©Fy = 0;
2F cos 30° + FAD - 100 = 0
B
30⬚
D
C
FAB = FAC = F
(1)
Compatibility Equation: Due to symmetry, joint A will displace vertically. Referring
to the geometry shown in Fig. b, we have
dF = dFAD cos 30°
F(2)(12)
FAD[2 cos 30°(12)]
= e
f cos 30°
AEst
AEst
F = 0.75FAD
Solving Eqs. (1) and (2),
FAD = 43.496 kip
(2)
www.elsolucionario.org
F = 32.62 kip
Normal Stress: Since all of the rods have the same diameter and rod AD is subjected
to the greatest load, it is the critical member.
sallow =
FAD
AAD
24 =
43.496
p 2
d
4
5
Use d = 1.519 in. = 1 in.
8
Ans.
Ans:
5
Use d = 1 in.
8
246
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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
100 mm
100 mm
B
C
125 mm
800 kN/m
a+ ©MB = 0;
- FA(100) + FC(100) = 0
F A = FC = F
(1)
2F + FB - 160 = 0
+ c ©Fy = 0;
(2)
dA = dB + 0.0003
F (0.125)
-6
6
400 (10 )(70)(10 )
=
FB (0.1247)
400 (10 - 6)(70)(106)
+ 0.0003
0.125 F - 0.1247FB = 8.4
(3)
Solving Eqs. (2) and (3)
F = 75.726 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.
247
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
4–65. Initially the A-36 bolt shank fits snugly against the
rigid caps E and F on the 6061-T6 aluminum sleeve. If the
thread of the bolt shank has a lead of 1 mm, and the nut is
tightened 34 of a turn, determine the average normal stress
developed in the bolt shank and the sleeve. The diameter of
bolt shank is d = 60 mm.
E
A
450 mm
d
500 mm
15 mm
Equation of Equilibrium: Referring to the free-body diagram of the cut part of the
assembly shown in Fig. a,
+ c ©Fy = 0;
Fs - Fb = 0
B
F
(1)
250 mm
Compatibility Equation: When the nut is tightened 3>4 of a turn, the unconstrained
3
bolt will be shortened by db = (1) = 0.75 mm .
4
Referring to the initial and final position of the assembly shown in Fig. b,
db - dFb = dFs
0.75 -
Fs(450)
Fb(500)
=
p
p
(0.062)(200)(109)
(0.252 - 0.222)(68.9)(109)
4
4
1.4992Fb + Fs = 1 271 677.44
Solving Eqs. (1) and (2),
(2)
www.elsolucionario.org
Fs = Fb = 508 831.16 N
Normal Stress:
ss =
sb =
Fs
508 831.16
=
= 45.9 MPa
p
As
(0.252 - 0.222)
4
Ans.
Fb
508 831.16
=
= 180 MPa
p
Ab
(0.062)
4
Ans.
Ans:
ss = 45.9 MPa, sb = 180 MPa
248
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
4–66. Initially the A-36 bolt shank fits snugly against the
rigid caps E and F on the 6061-T6 aluminum sleeve. If the
thread of the bolt shank has a lead of 1 mm, and the nut is
tightened 34 of a turn, determine the required diameter d of
the shank and the force developed in the shank and sleeve
so that the normal stress developed in the shank is four
times that of the sleeve.
E
A
450 mm
d
500 mm
15 mm
Equation of Equilibrium: Referring to the free-body diagram of the cut part of the
assembly shown in Fig. a,
B
F
Fs - Fb = 0
+ c ©Fy = 0,
Fs = Fb = F
250 mm
Normal Stress: It is required that sb = 4ss,
sb = 4ss
F
F
= 4≥
¥
p 2
p
db
(0.252 - 0.222)
4
4
db = 0.05937 m = 59.4 mm
Ans.
Compatibility Equation: When the nut is tightened 3>4 of a turn, the unconstrained
3
bolt will be shortened by db = (1) = 0.75 mm . Referring to the initial and final
4
position of the assembly shown in Fig. b,
db - dFb = dFs
0.75 -
Fb(500)
p
(0.059372)(200)(109)
4
=
Fs(450)
p
(0.252 - 0.222)(68.9)(109)
4
Fs = Fb = F = 502 418.65 N = 502 kN
Ans.
Ans:
db = 59.4 mm, Fs = Fb = 502 kN
249
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
4–67. The assembly consists of a 6061-T6-aluminum
member and a C83400-red-brass member that rest on the
rigid plates. Determine the distance d where the vertical
load P should be placed on the plates so that the plates
remain horizontal when the materials deform. Each
member has a width of 8 in. and they are not bonded
together.
P
d
30 in.
Aluminum
Red
brass
6 in. 3 in.
+ c ©Fy = 0;
- P + Fal + Fbr = 0
a + ©MO = 0;
3 Fal + 7.5 Fbr - Pd = 0
d = dbr = dal
Fbr L
Fal L
=
Abr Ebr
Aal Eal
Fbr = Fal a
(3)(8)(14.6)(103)
Abr Ebr
b = 0.730 Fal
b = Fal a
Aal Eal
6(8)(10)(103)
Thus,
P = 1.730 Fal
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3 Fal + 7.5(0.730 Fal) = (1.730 Fal)d
d = 4.90 in.
Ans.
Ans:
d = 4.90 in.
250
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*4–68. The C83400-red-brass rod AB and 2014-T6aluminum rod BC are joined at the collar B and fixed
connected at their ends. If there is no load in the members
when T1 = 50°F, determine the average normal stress in
each member when T2 = 120°F. Also, how far will the
collar be displaced? The cross-sectional area of each
member is 1.75 in2.
3 ft
Fbr = Fal = F
©Fx = 0;
dN>C = 0
-
FalLBC
Fbr LAB
+ aB ¢T LAB + aal ¢T LBC = 0
AAB Ebr
ABCEal
F(3)(12)
-
(1.75)(14.6)(106)
F(2)(12)
-
1.75(10.6)(106)
+ 9.80(10 - 6)(120 - 50)(3)(12)
+ 12.8(10 - 6)(120 - 50)(2)(12) = 0
F = 17 093.4 lb
sbr = sal =
17 093.4
= 9.77 ksi
1.75
Ans.
and (sg)br
OK
9.77 ksi < (sg)al
dB = -
17 093.4(3)(12)
1.75(14.6)(106)
B
A
+ 9.80(10 - 6)(120 - 50)(3)(12)
dB = 0.611(10 - 3) in. :
Ans.
251
C
2 ft
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4–69. The assembly has the diameters and material makeup
indicated. If it fits securely between its fixed supports when
the temperature is T1 = 70°F, determine the average
normal stress in each material when the temperature
reaches T2 = 110°F.
2014-T6 Aluminum
C 86100 Bronze
A
12 in.
dA>D = 0;
D
8 in.
B
4 ft
©Fx = 0;
304 Stainless
steel
C
6 ft
4 in.
3 ft
FA = FB = F
F(4)(12)
-
p(6)2(10.6)(106)
F(6)(12)
-
p(4)2(15)(106)
F(3)(12)
-
p(2)2(28)(106)
+ 12.8(10 - 6)(110 - 70)(4)(12)
+ 9.60(10 - 6)(110 - 70)(6)(12)
+ 9.60(10 - 6)(110 - 70)(3)(12) = 0
F = 277.69 kip
sal =
277.69
= 2.46 ksi
p(6)2
sbr =
277.69
= 5.52 ksi
p(4)2
Ans.
sst =
277.69
= 22.1 ksi
p(2)2
Ans.
Ans.
www.elsolucionario.org
Ans:
sal = 2.46 ksi, sbr = 5.52 ksi, sst = 22.1 ksi
252
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k ⫽ 1000 lb/ in.
4–70. The rod is made of A992 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:
+)
(:
x = dT - dF
x = 6.60(10 - 6)(160 - 40)(2)(12)
1.00(x + 0.5)(2)(12)
-
p
2
3
4 (0.25 )(29.0)(10 )
x = 0.01040 in.
F = 1.00(0.01040 + 0.5) = 0.510 kip
Ans.
Ans:
F = 0.510 kip
253
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4–71. If the assembly fits snugly between two rigid
supports A and C when the temperature is at T1 , determine
the normal stress developed in both rod segments when the
temperature rises to T2. Both segments are made of the
same material, having a modulus of elasticity of E and
coefficient of thermal expansion of a.
L
2
A
L
2
d
B
1d
2
C
Compatibility Equation: When the assembly is unconstrained, it has a free
expansion of d T = a¢TL = a(T2 - T1)L. Using the method of superposition,
Fig. a,
+)
(:
0 = dT - dF
0 = a(T2 - T1)L - ≥
F =
F(L>2)
F(L>2)
p d 2
a b E
4 2
+
p
a d2 bE
4
¥
a(T2 - T1)pd2E
10
Normal Stress:
www.elsolucionario.org
a(T2 - T1)pd2E
F
10
2
sAB =
=
= a(T2 - T1)E
p 2
AAB
5
d
4
Ans.
a(T2 - T1)pd2E
F
10
8
sBC =
=
= a(T2 - T1)E
ABC
5
p d 2
a b
4 2
Ans.
Ans:
sAB =
254
8
2
a (T2 - T1)E, sBC = a (T2 - T1)E
5
5
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*4–72. If the assembly fits snugly between the two
supports A and C when the temperature is at T1, determine
the normal stress developed in both segments when the
temperature rises to T2. Both segments are made of
the same material having a modulus of elasticity of E and
coefficient of the thermal expansion of a. The flexible
supports at A and C each have a stiffness k.
L
2
A
Compatibility Equation: When the assembly is unconstrained, it has a free
expansion of dT = a¢TL = a(T2 - T1)L. Using the method of superposition,
Fig. a,
+ ) dC = dT - dF
(:
F(L>2)
F(L>2)
F
F
+
+ T
= a(T2 - T1)L - D
k
k
p 2
p d 2
a b E
a d bE
4 2
4
F =
a(T2 - T1)Lpd2Ek
10kL + 2pd2E
Normal Stress:
a(T2 - T1)Lpd2Ek
sAB =
F
=
AAB
10kL + 2pd2E
p 2
d
4
=
4Eka(T2 - T1)L
=
16Eka(T2 - T1)L
Ans.
10kL + 2pd2E
a(T2 - T1)Lpd2Ek
sBC =
F
=
ABC
10kL + 2pd2E
p d 2
a b
4 2
Ans.
10kL + 2pd2E
255
L
2
d
B
1d
2
C
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
4–73. The pipe is made of A992 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
B
8 ft
Compatibility:
L
0 = dT - dF
Where
0 = 6.60 A 10 - 6 B
8 ft
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 =
19.14 A
= 19.1 ksi
A
Ans.
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Ans:
s = 19.1 ksi
256
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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
b 140 = 200 - 17.5x
8
Compatibility:
0 = dT - dF
Where
0 = 9.60 A 10 - 6 B
8 ft
0 = 9.60 A 10 - 6 B
L0
dT = 1 a¢Tdx
2
2
3
4 (1.4 - 1 )15.0(10 )
8 ft
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.
Ans:
F = 7.60 kip
257
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d
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 required
d
to expand on each end, or d for the entire rail.
2
d = a¢TL = 6.60(10 - 6)[90 - ( - 20)](40)(12)
Ans.
= 0.34848 in. = 0.348 in.
Compatibility:
+)
(:
0.34848 = dT - dF
0.34848 = 6.60(10 - 6)[110 - ( - 20)](40)(12) -
F(40)(12)
5.10(29.0)(103)
F = 19.5 kip
Ans.
www.elsolucionario.org
Ans:
d = 0.348 in., F = 19.5 kip
258
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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
1.5 in.
B
Thermal Expansion:
A dT B CD = aal ¢TLCD = 12.8(10 - 6)(150 - 75)(1.5) = 1.44(10 - 3) in.
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. a,
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.
259
D
E
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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
B
TB
TA
0 = ¢ T - dF
(1)
However,
d¢ T = a¢ T dx = a(TA +
TB - TA
x - TA)dx
L
L
L
¢T = a
= ac
TB - TA 2
TB - TA
x dx = a c
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
Ans.
www.elsolucionario.org
Ans:
F =
260
aAE
(TB - TA)
2
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4–78. When the temperature is at 30°C, the A-36 steel
pipe fits snugly between the two fuel tanks. When fuel flows
through the pipe, the temperatures at ends A and B rise to
130°C and 80°C, respectively. If the temperature drop along
the pipe is linear, determine the average normal stress
developed in the pipe. Assume each tank provides a rigid
support at A and B.
150 mm
10 mm
Section a - a
6m
x
A
a
a
B
Temperature Gradient: Since the temperature varies linearly along the pipe, Fig. a,
the temperature gradient can be expressed as a function of x as
T(x) = 80 +
50
50
(6 - x) = a 130 x b °C
6
6
Thus, the change in temperature as a function of x is
¢T = T(x) - 30° = a 130 -
50
50
x b - 30 = a 100 xb °C
6
6
Compatibility Equation: If the pipe is unconstrained, it will have a free expansion of
6m
dT = a
L
¢Tdx = 12(10 - 6)
L0
a 100 -
50
xb dx = 0.0054 m = 5.40 mm
6
Using the method of superposition, Fig. b,
+)
(:
0 = dT - dF
0 = 5.40 -
F(6000)
p(0.162 - 0.152)(200)(109)
F = 1 753 008 N
Normal Stress:
s =
1 753 008
F
=
= 180 MPa
A
p(0.162 - 0.152)
Ans.
Ans:
s = 180 MPa
261
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4–79. When the temperature is at 30°C, the A-36 steel
pipe fits snugly between the two fuel tanks. When fuel flows
through the pipe, the temperatures at ends A and B rise
to 130°C and 80°C, respectively. If the temperature drop
along the pipe is linear, determine the average normal stress
developed in the pipe. Assume the walls of each tank act as
a spring, each having a stiffness of k = 900 MN>m .
150 mm
10 mm
Section a - a
6m
x
A
a
a
B
Temperature Gradient: Since the temperature varies linearly along the pipe, Fig. a,
the temperature gradient can be expressed as a function of x as
T(x) = 80 +
50
50
(6 - x) = a 130 x b °C
6
6
Thus, the change in temperature as a function of x is
¢T = T(x) - 30° = a 130 -
50
50
x b - 30 = a 100 xb °C
6
6
Compatibility Equation: If the pipe is unconstrained, it will have a free expansion of
6m
dT = a
L
¢Tdx = 12(10 - 6)
L0
a 100 -
50
x b dx = 0.0054 m = 5.40 mm
6
Using the method of superposition, Fig. b,
+)
(:
d = dT - dF
F
900(106)
www.elsolucionario.org
(1000) = 5.40 - C
F(6000)
p(0.162 - 0.152)(200)(109)
F
+
900(106)
(1000) S
F = 1 018 361 N
Normal Stress:
s =
1 018 361
F
= 105 MPa
=
A
p(0.162 - 0.152)
Ans.
Ans:
s = 105 MPa
262
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*4–80. When the temperature is at 30°C, the A-36 steel
pipe fits snugly between the two fuel tanks. When fuel flows
through the pipe, it causes the temperature to vary along
the pipe as T = (53x2 - 20x + 120)°C, where x is in meters.
Determine the normal stress developed in the pipe. Assume
each tank provides a rigid support at A and B.
150 mm
10 mm
Section a - a
6m
x
A
Compatibility Equation: The change in temperature as a function of x is
5
5
¢T = T - 30° = a x2 - 20x + 120 b - 30 = a x2 - 20x + 90b °C. If the pipe
3
3
is unconstrained, it will have a free expansion of
6m
dT = a
L
¢Tdx = 12(10 - 6)
L0
5
a x2 - 20x + 90b dx = 0.0036 m = 3.60 mm
3
Using the method of superposition, Fig. b,
+)
(:
0 = dT - dF
0 = 3.60 -
F(6000)
2
p(0.16 - 0.152)(200)(109)
F = 1 168 672.47 N
Normal Stress:
s =
F
1 168 672.47
=
= 120 MPa
A
p(0.162 - 0.152)
Ans.
263
a
a
B
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4–81.
The 50-mm-diameter cylinder is made from
Am 1004-T61 magnesium and is placed in the clamp when
the temperature is T1 = 20° C. If the 304-stainless-steel
carriage bolts of the clamp each have a diameter of 10 mm,
and they hold the cylinder snug with negligible force against
the rigid jaws, determine the force in the cylinder when the
temperature rises to T2 = 130° C.
+ c ©Fy = 0;
100 mm
150 mm
Fst = Fmg = F
dmg = dst
amg Lmg ¢T -
26(10 - 6)(0.1)(110) -
FmgLmg
EmgAmg
= astLst ¢T +
FstLst
EstAst
F(0.1)
F(0.150)
= 17(10 - 6)(0.150)(110) +
p
p
44.7(109) (0.05)2
193(109)(2) (0.01)2
4
4
F = 904 N
Ans.
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Ans:
F = 904 N
264
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4–82. The 50-mm-diameter cylinder is made from
Am 1004-T61 magnesium and is placed in the clamp when
the temperature is T1 = 15°C. If the two 304-stainless-steel
carriage bolts of the clamp each have a diameter of 10 mm,
and they hold the cylinder snug with negligible force against
the rigid jaws, determine the temperature at which the
average normal stress in either the magnesium or the steel
first becomes 12 MPa.
100 mm
150 mm
Fst = Fmg = F
+ c ©Fy = 0;
dmg = dst
amg Lmg ¢T -
26(10 - 6)(0.1)(¢T) -
FmgLmg
EmgAmg
= astLst ¢T +
FstLst
EstAst
F(0.1)
F(0.150)
= 17(10 - 6)(0.150)(¢T) +
p
p
44.7(109) (0.05)2
193(109)(2) (0.01)2
4
4
The steel has the smallest cross-sectional area.
p
F = sA = 12(106)(2)( )(0.01)2 = 1885.0 N
4
Thus,
¢T = 229°
T2 = 229° + 15° = 244°
Ans.
Ans:
T2 = 244°
265
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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.
C
D
B
40 in.
60 in.
45⬚
45⬚
60 in.
A
150 lb
Equations of Equilibrium:
+
: ©Fx = 0;
FAC cos 45° - FAB cos 45° = 0
FAC = FAB = F
+ c ©Fy = 0;
2F sin 45° + FAD - 150 = 0
(1)
Compatibility:
(dAC)T = 8.0(10 - 6)(80)(60) = 0.03840 in.
(dAC)T2 =
(dAC)T
0.03840
=
= 0.05431 in.
cos 45°
cos 45°
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-6
(dAD)T = 9.60(10 )(80)(40) = 0.03072 in.
d0 = (dAC)T2 - (dAD)T = 0.05431 - 0.03072 = 0.02359 in.
(dAD)F = (dAC)Fr + d0
FAD(40)
F(60)
6
=
0.0123(17.0)(10 )
0.0123(29.0)(106) cos 45°
+ 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.
Ans:
FAC = FAB = 10.0 lb, FAD = 136 lb
266
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*4–84. The center rod CD of the assembly is heated from
T1 = 30°C to T2 = 180°C using electrical resistance
heating. At the lower temperature T1 the gap between
C and the rigid bar is 0.7 mm. Determine the force in rods
AB and EF caused by the increase in temperature. Rods
AB and EF are made of steel, and each has a cross-sectional
area of 125 mm2. CD is made of aluminum and has a crosssectional area of 375 mm2. Est = 200 GPa, Eal = 70 GPa, and
aal = 23(10 - 6)>°C.
0.7 mm
B
D
dst = (dg)al - dal - 0.0007
-6
9
(125)(10 )(200)(10 )
= 23(10 - 6)(150)(0.24) -
F(0.24)
(375)(10 - 6)(70)(109)
12Fst = 128 000 - 9.1428F
+ c ©Fy = 0;
–
240 mm 300 mm
A
Fst(0.3)
F
C
- 0.0007
(1)
F - 2Fst = 0
(2)
Solving Eqs. (1) and (2) yields,
FAB = FEF = Fst = 4.23 kN
Ans.
FCD = F = 8.45 kN
267
+
E
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4–85. The center rod CD of the assembly is heated from
T1 = 30°C to T2 = 180°C using electrical resistance
heating. Also, the two end rods AB and EF are heated from
T1 = 30°C to T2 = 50°C. At the lower temperature T1 the
gap between C and the rigid bar is 0.7 mm. Determine the
force in rods AB and EF caused by the increase in
temperature. Rods AB and EF are made of steel, and each
has a cross-sectional area of 125 mm2. CD is made of
aluminum and has a cross-sectional area of 375 mm2.
Est = 200 GPa, Eal = 70 GPa , ast = 12(10 - 6)>°C and
aal = 23(10 - 6)>°C.
0.7 mm
B
F
C
–
240 mm 300 mm
D
A
+
E
dst + (dT)st = (dT)al - dal - 0.0007
Fst(0.3)
(125)(10 - 6)(200)(109)
+ 12(10 - 6)(50 - 30)(0.3)
= 23(10 - 6)(180 - 30)(0.24) -
Fal(0.24)
375(10 - 6)(70)(109)
- 0.0007
12.0Fst + 9.14286Fal = 56000
+ c ©Fy = 0;
(1)
Fal - 2Fst = 0
(2)
Solving Eqs. (1) and (2) yields:
FAB = FEF = Fst = 1.85 kN
FCD = Fal = 3.70 kN
Ans.
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Ans:
FAB = FEF = 1.85 kN
268
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4–86. The metal strap has a thickness t and width w and is
subjected to a temperature gradient T1 to T2 (T1 < T2). This
causes the modulus of elasticity for the material to vary
linearly from E1 at the top to a smaller amount E2 at the
bottom. As a result, for any vertical position y,
E = [(E2 - E1)>w] y + E1. Determine the position d
where the axial force P must be applied so that the bar
stretches uniformly over its cross section.
t
T1
P
w
d
P
T2
P = constant = P0
s
=
E
P0 =
s = P0 a
s
E2 - E1
aa
b y + E1 b
w
E2 - E1
y + E1 b
w
+
: ©Fx = 0:
P -
m
w
P =
L0
s t dy =
P = P0 t a
P0 t a
a
L0
s dA = 0
P0 a
E2 - E1
y + E1 b t dy
w
E2 - E1
E2 + E1
+ E1w b = P0 ta
bw
2
2
a + ©M0 = 0:
P0 t a
LA
P(d) -
LA
y sdA = 0
w
E2 + E1
E2 - E1 2
b wd =
P0 a a
b y + E1y b t dy
2
w
L0
E2 + E1
E2 - E1 2
E1 2
b wd = P0 ta
w +
w b
2
3
2
E2 + E1
1
b d = (2E2 + E1)w
2
6
d = a
2 E2 + E
bw
3(E2 + E1)
Ans.
Ans:
d = a
269
2E2 + E1
bw
3(E2 + E1)
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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
r ⫽ 10 mm
20 mm
For the fillet:
r
10
=
= 0.5
h
20
w
40
=
= 2
h
20
From Fig. 4–23,
K = 1.4
smax = Ksavg
= 1.4 a
8 (103)
b
0.02 (0.005)
= 112 MPa
For the hole:
2r
20
=
= 0.5
w
40
From Fig. 4–24,
K = 2.1
smax = Ksavg
= 2.1 a
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8 (103)
b
(0.04 - 0.02)(0.005)
= 168 MPa
Ans.
Ans:
smax = 168 MPa
270
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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
r ⫽ 10 mm
20 mm
Assume failure of the fillet.
r
10
=
= 0.5
h
20
w
40
=
= 2;
h
20
From Fig. 4–23.
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.
2r
20
=
= 0.5
w
40
From Fig. 4–24.
K = 2.1
sallow = smax = Ksavg
120 (104) = 2.1 a
P
b
(0.04 - 0.02) (0.005)
P = 5.71 kN (controls)
Ans.
271
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4–89. The steel bar has the dimensions shown. Determine
the maximum axial force P that can be applied so as not to
exceed an allowable tensile stress of sallow = 150 MPa.
20 mm
60 mm
30
30 mm
mm
P
P
r ⫽ 15 mm
24 mm
Assume failure occurs at the fillet:
w
60
=
= 2
h
30
From the text,
and
r
15
=
= 0.5
h
30
K = 1.4
smax = sallow = Ksavg
150 (106) = 1.4 c
P
d
0.03 (0.02)
P = 64.3 kN
Assume failure occurs at the hole:
2r
24
=
= 0.4
w
60
From the text,
K = 2.2
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smax = sallow = Ksavg
150 (106) = 2.2 c
P
d
(0.06 - 0.024) (0.02)
P = 49.1 kN (controls!)
Ans.
Ans:
P = 49.1 kN
272
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4–90. Determine the maximum axial force P that can be
applied to the steel plate. The allowable stress is
sallow = 21 ksi.
0.25 in.
1 in.
P
5 in.
0.25 in.
2.5 in.
0.25 in.
P
Assume failure at fillet
r
0.25
w
5
=
= 0.1;
=
= 2
h
2.5
h
2.5
From Fig. 4–23, K = 2.4
sallow = smax = Ksavg
21 = 2.4 c
P
d;
2.5(0.25)
P = 5.47 kip
Assume failure at hole
2r
1
=
= 0.2;
w
5
From Fig. 4–24, K = 2.45
sallow = smax = Ksavg
21 = 2.45 c
P
d
(5 - 1)(0.25)
P = 8.57 kip
P = 5.47 kip (controls)
Ans.
Ans:
P = 5.47 kip
273
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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. 4–23,
r ⫽ 0.25 in.
K = 1.75
sallow = smax = Ksavg
21 = 1.75 a
P
b
1.25 (0.125)
P = 1.875 kip
Assume failure of the hole.
2r
0.75
=
= 0.40
w
1.875
From Fig. 4–24,
K = 2.2
sallow = smax = Ksavg
21 = 2.2 a
P
b
(1.875 - 0.75)(0.125)
P = 1.34 kip (controls)
Ans.
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Ans:
P = 1.34 kip
274
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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
From Fig. 4–23,
P
w
1.875
=
= 1.5
h
1.25
0.75 in.
K = 1.75
smax = K a
P
2
d = 22.4 ksi
b = 1.75 c
A
1.25(0.125)
At hole:
2r
0.75
=
= 0.40
w
1.875
From Fig. 4–24,
K = 2.2
smax = 2.2 c
2
d = 31.3 ksi
(1.875 - 0.75)(0.125)
(controls)
275
Ans.
r ⫽ 0.25 in.
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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
From the text,
30 mm
and
P
r = 15 mm
12 mm
60
w
=
= 2
h
30
K = 1.4
smax = Ksavg = K
P
ht
= 1.4 B
8(103)
R = 74.7 MPa
(0.03)(0.005)
Maximum Normal Stress at the hole:
2r
12
=
= 0.2
w
60
From the text,
K = 2.45
smax = K savg = K
P
(w - 2r) t
= 2.45 B
8(103)
R
(0.06 - 0.012)(0.005)
= 81.7 MPa (controls)
Ans.
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Ans:
smax = 81.7 MPa
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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.
Ans:
P = 15 kip, K = 1.60
277
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4–95. The 10-mm-diameter shank of the steel bolt has a
bronze sleeve bonded to it. The outer diameter of this
sleeve is 20 mm. If the yield stress for the steel is
(sY)st = 640 MPa, and for the bronze (sY)br = 520 MPa,
determine the largest possible value of P that can be
applied to the bolt. Assume the materials to be elastic
perfectly plastic. Est = 200 GPa, Ebr = 100 GPa .
P
10 mm
20 mm
+ c ©Fy = 0:
P - Pb - Ps = 0
(1)
P
The largest possible P that can be applied is when P causes both bolt and sleeve to
yield. Hence,
p
Pb = (sst)gAb = 640(106)( )(0.012) = 50.265 kN
4
p
Ps = (sbr)gAs = 520(106)( )(0.022 - 0.012)
4
= 122.52 kN
From Eq. (1).
P = 50.265 + 122.52 = 173 kN
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Ans.
Ans:
P = 173 kN
278
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*4–96. The 10-mm-diameter shank of the steel bolt has a
bronze sleeve bonded to it. The outer diameter of this
sleeve is 20 mm. If the yield stress for the steel is
(sY)st = 640 MPa and for the bronze (sY)br = 520 MPa,
determine the magnitude of the largest elastic load P that
can be applied to the assembly. Est = 200 GPa,
Ebr = 100 GPa.
P
10 mm
20 mm
+ c ©Fy = 0;
P - Pb - Ps = 0
(1)
P
¢ b = ¢ s;
Pb(L)
p
(0.012)(200)(109)
4
=
Ps(L)
p
(0.022 - 0.012)(100)(109)
4
Pb = 0.6667 Ps
(2)
Assume yielding of the bolt:
p
Pb = (sst)gAb = 640(106) a b (0.012) = 50.265 kN
4
Using Pb = 50.265 kN and solving Eqs. (1) and (2):
Ps = 75.40 kN:
P = 125.66 kN
Assume yielding of the sleeve:
p
Ps = (sg)brAs = 520 (106) a b (0.022 - 0.012) = 122.52 kN
4
Use Ps = 122.52 kN and solving Eqs. (1) and (2):
Pb = 81.68 kN
P = 204.20 kN
P = 126 kN (controls)
Ans.
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4–97. The weight is suspended from steel and aluminum
wires, each having the same initial length of 3 m and crosssectional area of 4 mm2. If the materials can be assumed to
be elastic perfectly plastic, with (sY)st = 120 MPa and
(sY)al = 70 MPa, determine the force in each wire if the
weight is (a) 600 N and (b) 720 N. Eal = 70 GPa,
Est = 200 GPa.
Aluminum
Fal + Fst - W = 0
+ c ©Fy = 0;
Steel
(1)
Assume both wires behave elastically.
dal = dst;
FalL
FstL
=
A(70)
A(200)
Fal = 0.35 Fst
(2)
(a) When W = 600 N. solving Eqs. (1) and (2) yields:
Fst = 444.44 N = 444 N
Ans.
Fal = 155.55 N = 156 N
Ans.
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sal =
Fal
155.55
= 38.88 MPa 6 (sg)al = 70 MPa
=
Ast
4(10 - 6)
OK
sst =
Fst
444.44
= 111.11 MPa 6 (sg)st = 120 MPa
=
Ast
4(10 - 6)
OK
The elastic analysis is valid for both wires.
(b) When W = 720 N. solving Eqs. (1) and (2) yields:
Fst = 533.33 N:
Fst = 186.67 N
sal =
Fal
186.67
= 46.67 MPa 6 (sg)al = 70 MPa
=
Aal
4(10 - 6)
sst =
Fst
533.33
= 133.33 MPa 7 (sg)st = 120 MPa
=
Ast
4(10 - 6)
OK
Therefore, the steel wire yields. Hence,
Fst = (sg)stAst = 120(106)(4)(10 - 6) = 480 N
Ans.
From Eq. (1). Fal = 240 N
Ans.
sal =
240
4(10 - 6)
= 60 MPa 6 (sg)al
OK
Ans:
(a) Fst = 444 N, Fal = 156 N,
(b) Fst = 480 N, Fal = 240 N
280
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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
8 kip
B
5 ft
C 5 kip
2 ft
s (ksi)
40
20
Average Normal Stress and Strain: For segment BC
sBC =
0.001
0.021
P (in./in.)
PBC
5
=
= 10.0 ksi
ABC
0.5
10.0
20
=
;
PBC
0.001
PBC =
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
=
PAB - 0.001
0.021 - 0.001
PAB = 0.0070 in.>in.
Elongation:
dBC = PBCLBC = 0.00050(2)(12) = 0.0120 in.
dAB = PAB LAB = 0.0070(5)(12) = 0.420 in.
dTot = dBC + dAB = 0.0120 + 0.420 = 0.432 in.
Ans.
Ans:
dTot = 0.432 in.
281
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4–99. The rigid beam 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;
250 mm
150 mm
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
= 530 A 106 B a
p
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
sg
Pg =
E
530(106)
=
= 0.002650 mm>mm
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200(109)
dBE = Pg 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.
Ans:
w = 21.9 kN>m, dG = 4.24 mm
282
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*4–100. The rigid beam 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
FCDL
FBEL
= 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.
283
250 mm
150 mm
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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
D
Equation of Equilibrium. Referring to the free-body diagram of the lever shown in
Fig. a,
FAB (300) + FCD (150) - 3(103)(450) = 0
a + ©ME = 0;
2FAB + FCD = 9(103)
(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
www.elsolucionario.org
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
2
ACD
(0.004
)
4
(O.K.)
sAB =
FAB
3600
= 286.48 MPa 7 (sY)st
= p
2
AAB
(0.004
)
4
(N.G.)
Since wire AB yields, the elastic analysis is not valid. The solution must be reworked
using
FAB = (sY)st AAB = 250(106) c
p
(0.0042) d
4
Ans.
= 3141.59 N = 3.14 kN
Substituting this result into Eq. (1),
FCD = 2716.81 N = 2.72 kN
sCD =
Ans.
FCD
2716.81
= 216.20 MPa 6 (sY)st
= p
2
ACD
4 (0.004 )
(O.K.)
284
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
2
9
ACD Est
4 (0.004 )(200)(10 )
Ans.
= 0.3243 mm = 0.324 mm
From Eq. (2),
dAB = 2dCD = 2(0.3243) = 0.649 mm
Ans.
Ans:
FAB = 3.14 kN, FCD = 2.72 kN,
dCD = 0.324 mm, dAB = 0.649 mm
285
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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
E
300 mm
B
D
Equation of Equilibrium. Referring 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)
(a) Elastic Analysis. The compatibility equation can be written by referring to the
geometry of Fig. b.
dAB = a
300
bd
150 CD
dAB = 2dCD
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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.
(b) 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),
Ans.
P = 3141.59 N = 3.14 kN
Ans:
(a) P = 2.62 kN, (b) P = 3.14 kN
286
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4–103. Two steel wires, each having a cross-sectional area
of 2 mm2 are tied to a ring at C, and then stretched and tied
between the two pins A and B. The initial tension in the
wires is 50 N. If a horizontal force P is applied to the ring,
determine the force in each wire if P = 20 N. What is the
smallest force P that must be applied to the ring to reduce
the force in wire CB to zero? Take sY = 300 MPa.
Est = 200 GPa .
C
A
2m
B
P
3m
Equilibrium:
+
: ©Fx = 0:
20 + (50 - P2) - (50 + P1) = 0
P1 + P2 = 20
(1)
Compatibility Condition:
dC =
P2(3)
P1(2)
=
AE
AE
P1 = 1.5 P2
(2)
Solving Eqs. (1) and (2) yields:
P1 = 12 N,
P2 = 8 N
FAC = 50 + 12 = 62 N
Ans.
FBC = 50 - 8 = 42 N
Ans.
For FCB = 0;
50 - P2 = 0
P2 = 50 N
P1 = 1.5(50) = 75 N
P = 75 + 50 = 125 N
Ans.
FAt = 50 + 75 = 125 N
sAt =
125
2(10 - 6)
= 62.5 MPa
62.5 MPa 6 sg
OK
Ans:
FAC = 62 N, FBC = 42 N, P = 125 N
287
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*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(103) = 0
a + ©MA = 0;
FBE(400) + FCF(1200) - 230(103)(800) = 0
A
B
400 mm
400 mm
(1)
FBE + 3FCF = 460(103)
(2)
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 FADL
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
www.elsolucionario.org
FBE = 65 714.29 N FAD = 32 857.14 N
Normal Stress.
sCF =
FCF
131428.57
= 267.74 MPa 7 (sY)st
= p
2
ACF
4 (0.025 )
(N.G.)
sBE =
FBE
65714.29
= 133.87 MPa 6 (sY)st
= p
2
ABE
4 (0.025 )
(O.K.)
sAD =
FAD
32857.14
= 66.94 MPa 6 (sY)st
= p
2
AAD
4 (0.025 )
(O.K.)
Since rod CF yields, the elastic analysis is not valid. The solution must be
reworked using
FCF = (sY)st ACF = 250 (106) c
C
p
(0.0252) d = 122 718.46 N = 123 kN
4
288
Ans.
400 mm
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4–104. Continued
Substituting this result into Eq. (2),
FBE = 91844.61 N = 91.8 kN
Ans.
Substituting the result for FCF and FBE into Eq. (1),
FAD = 15436.93 N = 15.4 kN
Ans.
sBE =
FBE
91844.61
= 187.10 MPa 6 (sY)st
= p
2
ABE
4 (0.025 )
(O.K.)
sAD =
FAD
15436.93
= 31.45 MPa 6 (sY)st
= p
2
AAD
4 (0.025 )
(O.K.)
289
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4–105. The rigid beam is supported by three 25-mm
diameter A-36 steel rods. If the force of P = 230 kN is
applied on the beam and removed, determine the residual
stresses in each rod. Consider the steel to be an elastic
perfectly plastic material.
D
600 mm
P
A
B
400 mm
400 mm
Equation of Equilibrium. Referring to the free-body diagram of the beam shown in
Fig. a,
+ c ©Fy = 0;
FAD + FBE + FCF - 230(103) = 0
a + ©MA = 0;
FBE(400) + FCF(1200) - 230(103)(800) = 0
(1)
FBE + 3FCF = 460(103)
(2)
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 FAD L
1 FCF L
= a
b + a
b
AE
3 AE
3 AE
FBE =
www.elsolucionario.org
2
1
F
+ FCF
3 AD
3
(4)
Solving Eqs. (1), (2), and (4)
FCF = 131428.57 N
FBE = 65714.29 N
FAD = 32857.14 N
Normal Stress.
sCF =
FCF
131428.57
= 267.74 MPa (T) 7 (sY)st
= p
2
ACF
4 (0.025 )
(N.G.)
sBE =
FBE
65714.29
= 133.87 MPa (T) 6 (sY)st
= p
2
ABE
4 (0.025 )
(O.K.)
sAD =
FAD
32857.14
= 66.94 MPa (T) 6 (sY)st
= p
2
AAD
4 (0.025 )
(O.K.)
Since rod CF yields, the elastic analysis is not valid. The solution must be
reworked using
sCF = (sY)st = 250 MPa (T)
FCF = sCF ACF = 250(106) c
F
E
p
(0.0252) d = 122718.46 N
4
290
C
400 mm
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
4–105. Continued
Substituting this result into Eq. (2),
FBE = 91844.61 N
Substituting the result for FCF and FBE into Eq. (1),
FAD = 15436.93 N
sBE =
FBE
91844.61
= 187.10 MPa (T) 6 (sY)st
= p
2
ABE
4 (0.025 )
(O.K.)
sAD =
FAD
15436.93
= 31.45 MPa (T) 6 (sY)st
= p
2
AAD
4 (0.025 )
(O.K.)
Residual Stresses. The process of removing P can be represented by applying the
force P¿ , which has a magnitude equal to that of P but is opposite in sense. 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.
Ans:
(sCF)r = 17.7 MPa (C), (sBE)r = 53.2 MPa (T), (sAD)r = 35.5 MPa (C)
291
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4–106. A material has a stress–strain diagram that can be
described by the curve s = cP1>2. Determine the deflection
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.
s
L
A
d
1
P
s2 = c2P
s = cP2:
s2(x) = c2P(x)
However s(x) =
(1)
P(x)
;
A
e(x) =
dd
dx
From Eq. (1),
P2(x)
A2
= c2
dd
;
dx
P2(x)
dd
=
dx
A2c2
L
d =
1
1
P2(x) dx = 2 2
(gAx)2dx
A2c2 L
A c L0
g2
=
L
c L0
2
d =
x2dx =
2
gL
g2 x3 L
`
c2 3 0
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3
Ans.
3c2
Ans:
d =
292
g2L3
3c2
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
4–107. Solve Prob. 4–106 if the stress–strain diagram is
defined by s = cP3>2.
s
L
A
d
P
2
s3
3
s = cP2:
P =
(1)
2
c3
However s(x) =
P(x)
;
A
P(x) =
dd
dx
From Eq. (1),
2
1 P3
dd
= 2 2
dx
c 3 A3
d =
1
2
cA L
2
3
2
3
1
=
P 3 dx =
2
2
3
(gA)3
(cA)
L
L
1
(cA) L0
2
3
2
(gAx)3dx
2
g 3 3 5 L
2
x3 dx = a b a b x 3 `
c
5
0
L0
2
d =
3 g 3 5
a b L3
5 c
Ans.
Ans:
2
3 g 3 5
d = a b L3
5 c
293
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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 a material for which E = 70 GPa
and sY = 20 MPa. Post B has a diameter of 20 mm and
is made of a material for which E¿ = 100 GPa and
sY ¿ = 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
2m
FA = FC = Fal
©MB = 0;
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
www.elsolucionario.org
= 0.0002857 L
Fbr = 8.976 kN
sbr =
8.976(103)
p
3
4 (0.02 )
= 28.6 MPa 6 sg
OK.
From Eq. (1),
8.976 + 2(88.36) - 2P = 0
P = 92.8 kN
Ans.
(b) All the posts yield:
Fbr = (sg)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.
294
P
2m
C
2m
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 a material for which E = 70 GPa and
sY = 20 MPa. Post B is made of a material for which
E¿ = 100 GPa and sY ¿ = 590 MPa . If P = 130 kN,
determine the diameter of post B so that all three posts
are about to yield. (Do not assume that the three posts
have equal uncompressed lengths.)
+ c ©Fy = 0;
P
A
B
2m
2(Fg)al + Fbr - 260 = 0
P
2m
C
2m
2m
(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.
Ans:
dB = 17.8 mm
295
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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. Assume the
bar is rigid.
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:
FBC
720
= 58.67 ksi
= p
2
ABC
(0.125
)
4
sBC =
P (in./in.)
0.007
0.12
From the Stress–Strain diagram
58.67
70
=
;
PBC
0.007
PBC = 0.005867 in.>in.
Displacement:
dBC = PBCLBC = 0.005867(40) = 0.2347 in.
dD
80
=
dBC
50
www.elsolucionario.org
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
=
PBC - 0.007
0.12 - 0.007
PBC = 0.09997 in.>in.
Displacement:
dBC = PBCLBC = 0.09997(40) = 3.9990 in.
dBC
dD
=
;
80
50
dD =
8
(3.9990) = 6.40 in.
5
Ans.
Ans:
(a) dD = 0.375 in., (b) dD = 6.40 in.
296
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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
B
C
2 ft
3 ft
s (ksi)
20
0.001
P (in./in.)
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 = PL = (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.
Ans:
P = 126 kip, ¢d = 0.00720 in.
297
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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
B
C
2 ft
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. ;
www.elsolucionario.org
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 + 0.4(125.66) = 62.832 - 0.4(125.66) = 12.566 kip
When the supports are removed the elongation will be,
d =
12.566(5)(12)
PL
= 0.0120 in.
=
AE
p(1)2(20>0.001)
Ans.
298
P (in./in.)
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
4–113. 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
D
A
P
u
L
When all bars yield, the force in each bar is, Fg = sgA
+
: ©Fx = 0 ;
P - 2sgA cos u - sgA = 0
P = syA(2 cos u + 1)
Ans.
Bar AC will yield first followed by bars AB and AD.
dAB = dAD =
dA =
Fg(L)
sgAL
sgL
=
=
AE
AE
E
sgL
dAB
=
cos u
E cos u
Ans.
Ans:
P = sYA(2 cos u + 1), dA =
299
sgL
E cos u
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4–114. The assembly consists of two A992 steel bolts AB
and EF and an 6061-T6 aluminum rod CD. When the
temperature is at 30° C, the gap between the rod and rigid
member AE is 0.1 mm. Determine the normal stress
developed in the bolts and the rod if the temperature rises
to 130° C. Assume BF is also rigid.
A
C
0.1 mm
E
25 mm
25 mm
300 mm
400 mm
50 mm
Equation of Equilibrium: Referring to the free-body diagram of the rigid cap shown
in Fig. a,
Fr - 2Fb = 0
+ c ©Fy = 0;
B
(1)
Compatibility Equation: If the bolts and the rod are unconstrained, they will have a
free expansion of (dT)b = ast ¢TLb = 12(10 - 6)(130 - 30)(400) = 0.48 mm and
(dg)r = aal ¢TLr = 24(10 - 6)(130 - 30)(300) = 0.72 mm. Referring to the initial
and final position of the assembly shown in Fig. b,
(dT)r - dFr - 0.1 = (dT)b + dFb
0.72 -
Fr (300)
Fb(400)
- 0.1 = 0.48 +
p
p
(0.052)(68.9)(109)
(0.0252)(200)(109)
4
4
Fb + 0.5443Fr = 34361.17
(2)
Solving Eqs. (1) and (2).
Fb + 16 452.29 N
Normal Stress:
sb =
sr =
Fr = 32 904.58 N
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Fb
16 452.29
= 33.5 MPa
=
p
Ab
(0.0252)
4
Ans.
Fr
32 904.58
= 16.8 MPa
=
p
Ar
(0.052)
4
Ans.
Ans:
sb = 33.5 MPa, sr = 16.8 MPa
300
D
F
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
4–115. The assembly shown consists of two A992 steel bolts
AB and EF and an 6061-T6 aluminum rod CD. When the
temperature is at 30° C, the gap between the rod and rigid
member AE is 0.1 mm. Determine the highest temperature
to which the assembly can be raised without causing yielding
either in the rod or the bolts. Assume BF is also rigid.
A
C
0.1 mm
E
25 mm
25 mm
300 mm
400 mm
50 mm
Equation of Equilibrium: Referring to the free-body diagram of the rigid cap shown
in Fig. a,
+ c ©Fy = 0;
Fp - 2Fb = 0
B
D
F
(1)
Normal Stress: Assuming that the steel bolts yield first, then
p
Fb = (sg)stAb = 250(106) c (0.0252) d = 122 718.46 N
4
Substituting this result into Eq. (1),
Fp = 245 436.93 N
Then,
sp =
Fp
Ap
=
245 436.93
= 125 MPa 6 (sg)al
p
(0.052)
4
(O.K!)
Compatibility Equation: If the assembly is unconstrained, the bolts and the post will
have free expansion of (dT)b = ast ¢TLb = 12(10 - 6)(T - 30)(400) = 4.8(10 - 3)(T - 30)
and (dT)p = aal ¢TLp = 24(10 - 6)T - 30)(300) = 7.2(10 - 3)(T - 30). Referring to
the initial and final position of the assembly shown in Fig. b,
(dT)p - dFp - 0.1 = (dT)b + dFb
7.2(10 - 3)(T - 30) -
245 436.93(300)
122 718.46(400)
- 0.1 = 4.8(10 - 3)(T - 30) +
p
p
(0 .052)(68.9)(109)
(0.0252)(200)(109)
4
4
T = 506.78° C = 507° C
Ans.
Ans:
T = 507° C
301
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*4–116. The rods each have the same 25-mm diameter
and 600-mm length. If they are made of A992 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
cos 60°
= 2dAC .Thus,
= 2(dT)AC and dFAC ¿ = 2dFAC. Thus, this equation becomes
dFAB - A dT B AB = 2 A dT B AC - 2dAC
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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.
302
D
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4–117. Two A992 steel pipes, each having a cross- sectional
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)
0.32(29)(103)
+
0.32(29)(103)
P = 46.4 kip
sAB = sBC =
P
46.4
=
= 145 ksi
A
0.32
Ans.
Ans:
sAB = sBC = 145 ksi
303
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4–118. The force P is applied to the bar, which is composed
of an elastic perfectly plastic material. Construct a graph to
show how the force in each section AB and BC (ordinate)
varies as P (abscissa) is increased. The bar has crosssectional areas of 1 in2 in region AB and 4 in2 in region BC,
and sY = 30 ksi.
C
P
6 in.
+
: ©Fx = 0; P - FAB - FBC = 0
(1)
+
Elastic behavior: : 0 = ¢ C - dC;
0 =
P(6)
FBC(2)
FBC(6)
- c
+
d
(1)E
(4)E
(1)E
FBC = 0.9231 P
(2)
Substituting Eq. (2) into (1):
FAB = 0.07692 P
(3)
By comparison, segment BC will yield first. Hence,
www.elsolucionario.org
(FBC)g = sgA = 30(4) = 120 kip
From Eq. (1) and (3) using FBC = (FBC)g = 120 kip
P = 130 kip;
B
A
FAB = 10 kip
When segment AB yields,
(FAB)g = sgA = 30(1) = 30 kip; (FBC)g = 120 kip
From Eq. (1),
P = 150 kip
304
2 in.
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
4–119. 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 = ¢ B - ¢ T + dB
0 =
FB(13)
0.016(5)
- 12.8(10 - 6)[70° - ( - 10°)](13) +
p
p
(0.52)(10.6)(103)
(0.52)(10.6)(103)
4
4
FB = 2.1251 kip = 2.13 kip
+
: ©Fx = 0;
Ans.
2(0.008) + 2.1251 - FA = 0
FA = 2.14 kip
Ans.
Ans:
FB = 2.13 kip, FA = 2.14 kip
305
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*4–120. 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
B
P/2
P/2
5 in.
+
: = ¢ B - ¢ T + dB
0 =
P(5)
- 12.8(10 - 6)[(70)(13)] + 0
p
2
3
(0.5 )(10.6)(10 )
4
P = 4.85 kip
Ans.
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306
8 in.
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
4–121. The rigid link is supported by a pin at A and two
A-36 steel wires, each having an unstretched length of 12 in.
and cross-sectional area of 0.0125 in2. Determine the force
developed in the wires when the link supports the vertical
load of 350 lb.
12 in.
C
5 in.
B
Equations of Equilibrium:
a + ©MA = 0;
4 in.
A
- FC(9) - FB (4) + 350(6) = 0
(1)
Compatibility:
6 in.
dC
dB
=
4
9
350 lb
FC(L)
FB (L)
=
4AE
9AE
9FB - 4FC = 0‚
(2)
Solving Eqs. (1) and (2) yields:
FB = 86.6 lb
Ans.
FC = 195 lb
Ans.
Ans:
FB = 86.6 lb, FC = 195 lb
307
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4–122. The joint is made from three A992 steel plates that
are bonded together at their seams. Determine the
displacement of end A with respect to end B when the joint
is subjected to the axial loads shown. Each plate has a
thickness of 5 mm.
dA>B = ©
100mm
23 kN
46 kN
A
B
600mm
200mm
23 kN
800mm
46(103)(200)
23(103)(800)
46(103)(600)
PL
+
+
=
9
9
AE
(0.005)(0.1)(200)(10 )
3(0.005)(0.1)(200)(10 )
(0.005)(0.1)(200)(109)
= 0.491 mm
Ans.
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Ans:
dA>B = 0.491 mm
308
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5–1. 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 (T>2). 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
2T
Tr
= p 4 =
3
J
r
pr
2
a T2 b r¿
T
p(r¿)3
t = p
=
Since t =
r¿
t ;
r max
4
2 (r¿)
r¿ =
r
1
T
r¿ 2T
= a 3b
r pr
p(r¿)3
Ans.
= 0.841r
24
r
2
b)
L0
r¿
dT = 2p
r
2
r
2
L0
tr2 dr
r¿
dT = 2p
L0
L0
r
tmax r2 dr
L0 r
r¿
dT = 2p
r 2T 2
a 3 b r dr
L0 r pr
r¿
T
4T
= 4
r3 dr
2
r L0
r¿ =
r¿
1
Ans.
= 0.841r
24
Ans:
r¿ = 0.841r
309
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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-quarter of the applied torque (T>4). Solve the
problem two ways: (a) by using the torsion formula, (b) by
finding the resultant of the shear-stress distribution.
r¿
r
T
T(r)
2T
Tc
= p 4 =
J
pr3
2 (r )
a) tmax =
Since t =
t¿ =
r¿ =
2Tr¿
r¿
t
=
r max
pr4
(T4 )r¿
2Tr¿
=
p
4
pr4
2 (r¿)
T¿c¿
;
J¿
r
1
= 0.707 r
Ans.
44
b) t =
r 2T
r
2T
t
= a 3b =
r;
c max
r pr
pr4
dT = rt dA = r c
T
4
L0
dT =
4T
r L
4
r¿
4T r4
T
|;
= 4
4
r 4 0
dA = 2pr dr
4T
2T
r d (2pr dr) = 4 r3dr
4
pr
r
r¿
r3dr
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(r¿)4
1
= 4
4
r
r¿ = 0.707 r
Ans.
Ans:
r¿ = 0.707 r
310
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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
B
75 mm
4 kN⭈m
75 mm
The internal torques developed at cross-sections passing 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 =
6(103)(0.05)
TArA
= 6.036(106) Pa = 6.04 MPa.
=
J
49.70 (10 - 6)
Ans.
Ans:
tB = 6.04 MPa, tA = 6.04 MPa
311
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*5–4. 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.
A
30 N⭈m
20 N⭈m
80 N⭈m
tmax =
90(0.02)
Tmax c
= p
4
4
J
2 (0.02 - 0.0185 )
= 26.7 MPa
Ans.
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312
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5–5. 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 three torques are 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
C
450 lb⭈ft
A
350 lb⭈ft
600 lb⭈ft
tA =
250(12)(1.25)
Tc
= 3.45 ksi
= p
4
4
J
2 (1.25 - 1.15 )
Ans.
tB =
200(12)(1.25)
Tc
= 2.76 ksi
= p
4
4
J
2 (1.25 - 1.15 )
Ans.
Ans:
tA = 3.45 ksi, tB = 2.16 ksi
313
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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
A
25 lb⭈ft
40 lb⭈ft
20 lb⭈ft
35 lb⭈ft
(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
= p
= 3621 psi = 3.62 ksi
4
J
2 (0.375)
Ans.
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Ans:
(tBC)max = 5.07 ksi, (tDE)max = 3.62 ksi
314
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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
A
25 lb⭈ft
40 lb⭈ft
20 lb⭈ft
35 lb⭈ft
(tEF)max =
TEF c
= 0
J
(tCD)max =
15(12)(0.375)
TCD c
= p
4
J
2 (0.375)
Ans.
= 2173 psi = 2.17 ksi
Ans.
Ans:
(tEF)max = 0, (tCD)max = 2.17 ksi
315
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*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.
300 N⭈m
500 N⭈m
A
200 N⭈m
C
400 N⭈m
300 mm
B
400 mm
Internal Torque: As shown on torque diagram.
Maximum Shear Stress: From the torque diagram Tmax = 400 N # m. Then, applying
torsion Formula.
abs =
tmax
D
Tmax c
J
400(0.015)
= 75.5 MPa
= p
4
2 (0.015 )
Ans.
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316
500 mm
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5–9. 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 on the surface, and sketch the
shear stress on volume elements located at these points.
C
35 mm
A
B
20 mm
35 mm
300 N⭈m
800 N⭈m
tB =
TB r
800(0.02)
= 6.79 MPa
= p
4
J
2 (0.035 )
Ans.
tA =
500(0.035)
TA c
= 7.42 MPa
= p
4
J
2 (0.035 )
Ans.
Ans:
tB = 6.79 MPa, tA = 7.42 MPa
317
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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
T
n is the number of bolts and F is the shear force in each bolt.
T - nFR = 0;
F =
T
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 =
2 r3
Rd2
2T
4T
=
nRpd2
p r3
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Ans.
Ans:
n =
318
2 r3
Rd2
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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.
8 in.
15 lb
tAB =
210(0.375)
Tc
= 7.82 ksi
= p
4
4
J
2 (0.375 - 0.34 )
Ans.
tBC =
210(0.5)
Tc
= 2.36 ksi
= p
4
4
J
2 (0.5 - 0.43 )
Ans.
Ans:
tAB = 7.82 ksi, tBC = 2.36 ksi
319
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A
*5–12. The 150-mm-diameter shaft is supported by a
smooth journal bearing at E and a smooth thrust bearing at F.
Determine the maximum shear stress developed in each
segment of the shaft.
B
E
C
15 kN⭈m
30 kN⭈m
F
75 kN⭈m
30 kN⭈m
Internal Loadings: The internal torques developed in segments AB, BC, and CD of
the assembly are shown in their respective free-body diagrams shown in Figs. a,
b, and c.
Allowable Shear Stress: The polar moment of inertia of the shaft is J =
49.701(10 - 6)m4 .
p
(0.0754) =
2
(tmax)AB =
15(103)(0.075)
TAB c
= 22.6 MPa
=
J
49.701(10 - 6)
Ans.
(tmax)BC =
45(103)(0.075)
TBC c
= 67.9 MPa
=
J
49.701(10 - 6)
Ans.
(tmax)CD =
30(103)(0.075)
TCD c
= 45.3 MPa
=
J
49.701(10 - 6)
Ans.
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320
D
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5–13. If the tubular shaft is made from material having an
allowable shear stress of tallow = 85 MPa, determine the
required minimum wall thickness of the shaft to the nearest
millimeter. The shaft has an outer diameter of 150 mm.
A
B
E
C
15 kN⭈m
30 kN⭈m
F
D
75 kN⭈m
30 kN⭈m
Internal Loadings: The internal torques developed in segments AB, BC, and CD
of the assembly are shown in their respective free-body diagrams shown in
Figs. a, b, and c.
Allowable Shear Stress: Segment BC is critical since it is subjected to the greatest
p
internal torque. The polar moment of inertia of the shaft is J = (0.0754 - ci4) .
2
tallow =
TBC c
;
J
85(106) =
45(103)(0.075)
p
(0.0754 - ci4)
2
ci = 0.05022 m = 50.22 mm
Thus, the minimum wall thickness is
t = co - ci = 75 - 50.22 = 24.78 mm = 25 mm
Ans.
Ans:
Use t = 25 mm
321
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5–14. A steel tube having an outer diameter of 2.5 in. is
used to transmit 9 hp when turning at 27 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.
d
2.5 in.
v =
27(2p)
= 2.8274 rad>s
60
P = Tv
9(550) = T(2.8274)
T = 1750.7 ft # lb
tmax = tallow =
Tc
J
1750.7(12)(1.25)
p
(1.254 - c i4)
2
ci = 0.9366 in.
10(103) =
d = 1.873 in.
Use d = 1
3
in.
4
www.elsolucionario.org
Ans.
Ans:
Use d = 1
322
3
in.
4
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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 millimeter.
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
Use d = 0.03291 m = 32.91 mm = 33 mm
Ans.
Ans:
Use d = 33 mm
323
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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
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.
Since segment DE is 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.
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324
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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
A
1.5 ft
1.5 ft
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;
TA - 10(4)(2) = 0
TA = 80 lb # ft a
The polar moment of inertia of the cross section at A is J =
12 in
b = 960 lb # in.
1 ft
p
(0.54) = 0.03125p in4.
2
Thus
tmax =
960 (0.5)
TA c
=
= 4889.24 psi = 4.89 ksi
J
0.03125p
Ans.
Ans:
tmax = 4.89 ksi
325
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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
A
1.5 ft
1.5 ft
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.
1 ft
p
The polar moment of inertia of the cross-section at B is J = (0.54) = 0.03125p in4 .
2
Thus,
tmax =
1440(0.5)
TB c
=
= 7333.86 psi = 7.33 ksi
J
0.03125p
Ans.
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Ans:
tmax = 7.33 ksi
326
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5–19. The shaft consists of solid 80-mm-diameter rod
segments AB and CD, and the tubular segment BC which
has an outer diameter of 100 mm and inner diameter of
80 mm. If the material has an allowable shear stress of
tallow = 75 MPa, determine the maximum allowable torque
T that can be applied to the shaft.
T
D
C
B
T A
Internal Loadings: The internal torques developed in segments CD and BC are
shown in Figs. a and b, respectively.
Allowable Shear Stress: The polar moments of inertia of segments CD and BC are
p
p
JCD = (0.044) = 1.28(10 - 6)p m4 and JBC = (0.054 - 0.044) = 1.845(10 - 6)p m4.
2
2
tallow =
TCD cCD
;
JCD
75(106) =
T(0.04)
1.28(10 - 6)p
T = 7539.82 N # m = 7.54 kN # m (controls)
tallow =
TBC cBC
;
JBC
75(106) =
Ans.
T(0.05)
1.845(10 - 6)p
T = 8694.36 N # m = 8.69 kN # m
Ans:
T = 7.54 kN # m
327
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*5–20. The shaft consists of rod segments AB and CD, and
the tubular segment BC. If the torque T = 10 kN # m is
applied to the shaft, determine the required minimum
diameter of the rod and the maximum inner diameter of the
tube. The outer diameter of the tube is 120 mm, and the
material has an allowable shear stress of tallow = 75 MPa.
T
D
C
B
T A
Internal Loadings: The internal torques developed in segments CD and BC are
shown in Figs. a and b, respectively.
Allowable Shear Stress: The polar moments of inertia of the segments CD and BC
(dBC)i 4
p dCD 4
p
p
are
and
JCD = a
b =
dCD4
d f =
JBC = e 0.064 - c
2 2
32
2
2
pc6.48(10 - 6) -
tallow =
(dBC)4 i
d.
32
10(103)(dCD>2)
TCD cCD
; 75(106) =
p
JCD
d 4
32 CD
Ans.
dCD = 0.08790 m = 87.9 mm
www.elsolucionario.org
and
tallow =
TBC cBC
; 75(106) =
JBC
10(103)(0.06)
pc 6.48(10 - 6) -
(dBC)4 i
d
32
(dBC)i = 0.1059 m = 106 mm
Ans.
328
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5–21. If the 40-mm-diameter rod is subjected to a uniform
distributed torque of t0 = 1.5 kN # m>m, determine the
shear stress developed at point C.
300 mm
A
C
300 mm
t0
B
Internal Loadings: The internal torque developed on the cross-section of the shaft
passes through point C as shown in Fig. a.
Allowable Shear Stress: The polar moment of inertia of the shaft is J =
p
(0.024) =
2
80(10 - 9)p m4. We have
tC =
1.5(103)(0.3)(0.02)
TC c
= 35.8 MPa
=
J
80(10 - 9)p
Ans.
Ans:
tC = 35.8 MPa
329
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5–22. If the rod is subjected to a uniform distributed
torque of t0 = 1.5 kN # m>m, determine the rod’s minimum
required diameter d if the material has an allowable shear
stress of tallow = 75 MPa.
300 mm
A
C
300 mm
t0
B
Internal Loadings: The maximum internal torque developed in the shaft, which
occurs at A, is shown in Fig. a.
Allowable Shear Stress: The polar moment of inertia of the shaft is J =
p d 4
a b =
2 2
p 4
d . We have
32
tallow =
Tmax c
;
J
75(106) =
1.5(103)(0.6)(d>2)
p 4
d
32
d = 0.03939 m = 39.4 mm
Ans.
www.elsolucionario.org
Ans:
d = 39.4 mm
330
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5–23. If the 40-mm-diameter rod is made from a material
having an allowable shear stress of tallow = 75 MPa,
determine the maximum allowable intensity t0 of the
uniform distributed torque.
300 mm
A
C
300 mm
t0
B
Internal Loadings: The maximum internal torque developed in the shaft, which
occurs at A, is shown in Fig. a.
Allowable Shear Stress: The polar moment of inertia of the shaft is J =
p
(0.024) =
2
80(10 - 9)p m4 . We have
tallow =
Tmax c
;
J
75(106) =
t0(0.6)(0.02)
80(10 - 9)p
t0 = 1570.80 N # m> m = 1.57 kN # m>m
Ans.
Ans:
t0 = 1.57 kN # m>m
331
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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 lb⭈ft/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.
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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 lb⭈ft/ft
4 in.
9 in.
12 in.
Internal Torque: The maximum torque occurs at the support C.
Tmax = (125 lb # ft>ft)a
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)
= 3.59 ksi
= p
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.
Ans:
tabs = 3.59 ksi
max
333
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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
t =
T
F
r
=
=
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.
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Ans:
tmax =
334
T
2p ri 2h
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5–27. The assembly consists of the solid rod AB, tube BC,
and the lever arm. If the rod and the tube are made of
material having an allowable shear stress of tallow = 12 ksi,
determine the maximum allowable torque T that can be
applied to the end of the rod and from there the couple
forces P that can be applied to the lever arm. The diameter
of the rod is 2 in., and the outer and inner diameters of the
tube are 4 in. and 2 in., respectively.
P
12 in.
12 in.
C
B
A
P
T
Internal Loadings: The internal torques developed in rod AB and tube BC of the
shaft are shown in their respective free-body diagrams in Figs. a and b.
Allowable Shear Stress: The polar moments of inertia of rod AB and tube BC are
p
p
JAB = (14) = 0.5 p in4 and JBC = (24 - 14) = 7.5 p in4 . We will consider rod
2
2
AB first.
tallow =
TAB cAB
;
JAB
12 =
T(1)
0.5p
T = 18.85 kip # in = 1.57 kip # ft
Ans.
Using this result, TBC = 18.85 + 24P.
tallow =
TBC cBC
;
JBC
12 =
(18.85 + 24P)(2)
7.5p
P = 5.11 kip
Ans.
Ans:
T = 1.57 kip # ft, P = 5.11 kip
335
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*5–28. The assembly consists of the solid rod AB, tube BC,
and the lever arm. If a torque of T = 20 kip # n. is applied to
the rod and couple forces of P = 5 kip are applied to the
lever arm, determine the required diameter for the rod, and
the outer and inner diameters of the tube, if the ratio of the
inner diameter di, to outer diameter do, is required to be
di>do = 0.75. The rod and the tube are made of material
having an allowable shear stress of tallow = 12 ksi.
P
12 in.
12 in.
C
B
Internal Loadings: The internal torque developed in rod AB and tube BC of the
shaft are shown in their respective free-body diagrams in Figs. a and b.
A
P
Allowable Shear Stress: We will consider rod AB first. The polar moment of inertia
p dAB 4
p
of rod AB is JAB = a
b =
d 4.
2 2
32 AB
tallow =
TAB cAB
;
JAB
12 =
20(dAB > 2)
p
d 4
32 AB
dAB = 2.04 in. = 2.04 in.
Ans.
The polar moment of inertia of tube BC is JBC =
di 4
p do 4
ca b - a b d =
2
2
2
0.75do 4
p do4
c
- a
b d = 0.06711do4.
2 16
2
tallow =
TBC cBC
;
JBC
12 =
140(do >2)
0.06711do4
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do = 4.430 in. = 4.43 in.
Ans.
Thus,
di = 0.75 do = 0.75(4.430) = 3.322 in. = 3.32 in.
336
Ans.
T
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5–29. The steel shafts are connected together using a fillet
weld as shown. Determine the average shear stress in the
weld along section a–a if the torque applied to the shafts is
T = 60 N # m. Note: The critical section where the weld fails
is along section a–a.
T = 60 N⭈m
50 mm
12 mm
a
12 mm
12 mm
12 mm
45⬚
a
tavg =
(60>(0.025 + 0.003))
V
=
A
2p(0.025 + 0.003)(0.012 sin 45°)
tavg = 1.44 MPa
Ans.
Ans:
tavg = 1.44 MPa
337
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5–30. The shaft has a diameter of 80 mm. Due to friction
at its surface within the hole, it is subjected to a variable
2
torque described by the function t = (25xex ) N # m>m,
where x is in meters. Determine the minimum torque T0
needed to overcome friction and cause it to twist. Also,
determine the absolute maximum stress in the shaft.
T0
80 mm
x
2m
2
t = (25x e x ) N⭈m/m
2
2
t = 25(x e x );
T0 =
L0
2
x
25(x e x ) dx
Integrating numerically, we get
T0 = 669.98 = 670 N # m
tabs =
max
Ans.
(669.98)(0.04)
T0 c
=
= 6.66 MPa
p
4
J
2 (0.04)
Ans.
www.elsolucionario.org
Ans:
T0 = 670 N # m, tabs = 6.66 MPa
max
338
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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.
B
E
C
Ans:
(tAB)max = 1.04 MPa, (tBC)max = 3.11 MPa
339
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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 )
Ans.
= 3.44 MPa
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340
B
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5–33. The motor M is connected to the speed reducer C
by the tubular shaft and coupling. If the motor supplies
20 hp and rotates the shaft at a rate of 600 rpm, determine
the minimum inner and outer diameters di and do of the
shaft if di>do = 0.75. The shaft is made from a material
having an allowable shear stress of tallow = 12 ksi.
M
C
A
B
D
Internal Loading: The angular velocity of the shaft is
v = a 600
rev
1 min 2p rad
ba
ba
b = 20p rad>s
min
60 s
1 rev
and the power is
P = 20 hp a
550 ft # lb>s
b = 11 000 ft # lb >s
1 hp
We have
T =
P
11 000
12 in.
=
= 175.07 lb # ft a
b = 2100.84 lb # in.
v
20p
1 ft
Allowable Shear Stress: The polar moment of inertia of the shaft is
J=
di 4
0.75do 4
p do 4
p d o4
ca b - a b d =
c
- a
b d = 0.06711d o4 .
2 2
2
2 16
2
tallow =
Tc
;
J
12(103) =
2100.84(do >2)
0.06711d o4
do = 1.0926 in. = 1.09 in.
Ans.
Then
di = 0.75do = 0.75(1.0926) = 0.819 in.
Ans.
Ans:
do = 1.09 in., di = 0.819 in.
341
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5–34. The motor M is connected to the speed reducer C
by the tubular shaft and coupling. The shaft has an outer
and inner diameter of 1 in. and 0.75 in., respectively, and is
made from a material having an allowable shear stress of
tallow = 12 ksi, when the motor supplies 20 hp of power.
Determine the smallest allowable angular velocity of the
shaft.
M
C
A
Allowable Shear Stress: The polar moment of inertia of the shaft is J =
B
D
p
(0.54 2
0.3754) = 0.06711 in4 . We have
tallow =
Tc
;
J
12(103) =
T(0.5)
0.06711
T = 1610.68 lb # in a
1 ft
b = 134.22 lb # ft
12 in.
Internal Loading: The power is
P = 20 hpa
550 ft # lb>s
b = 11 000 ft # lb>s
1 hp
www.elsolucionario.org
We have
T =
P
;
v
134.22 =
11000
v
v = 82.0 rad>s
Ans.
Ans:
v = 82.0 rad>s
342
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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.
Ans:
v = 21.7 rad>s
343
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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
www.elsolucionario.org
10.7379(10-9)
P = 12 650.25 W = 12.7 kW
Ans.
344
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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 hp a
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.
Note that the shaft length is irrelevant.
Ans:
tmax = 6.02 ksi
345
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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
A
90 rev/min
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
www.elsolucionario.org
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.
Ans:
dA = 12.4 mm, dB = 16.8 mm
346
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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
D
12 kW
5 kW
25 mm
B
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.
Ans:
(tmax)CF = 12.5 MPa, (tmax)BC = 7.26 MPa
347
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*5–40. Determine the absolute maximum shear stress
developed in the shaft in Prob. 5–39.
3 kW 4 kW
A
B
D
v = 50
C
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
From the torque diagram,
Tmax = 38.2 N # m
tabs = Tc =
max
J
www.elsolucionario.org
38.2(0.0125) = 12.5 MPa
p
4
2 (0.0125 )
Ans.
348
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
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 = 2.3 mm
Ans.
Ans:
t = 2.3 mm
349
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5–42. The A-36 solid steel shaft is 2 m long and has a
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
M
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.
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Ans:
v = 17.7 rad>s
350
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5–43. The solid shaft has a linear taper from rA at one end
to rB at the other. Derive an equation that gives the
maximum shear stress in the shaft at a location x along the
shaft’s axis.
T
rB
T
B
A
x
L
rA
r = rB +
=
rBL + (rA - rB)(L - x)
rA - rB
(L - x) =
L
L
rA(L - x) + rBx
L
tmax =
=
Tr
Tc
2T
= p 4 =
J
r
p
r3
2
2T
2TL3
=
rA(L - x) + rBx 3
p[rA(L - x) + rBx]3
pc
d
L
Ans.
Ans:
tmax =
351
2TL3
p[rA(L - x) + rBx]3
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*5–44. The rod has a diameter of 0.5 in. and weight of
5 lb> ft. Determine the maximum torsional stress in the rod
at a section located at A due to the rod’s weight.
3 ft
A
1 ft
B
3 ft
1 ft
©Mx = 0;
Tx - 15(1.5) - 5(3) = 0;
Tx = 37.5 lb # ft
(tA)max =
37.5(12)(0.25)
Tc
=
p
4
J
2 (0.25)
= 18.3 ksi
Ans.
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352
1 ft
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5–45. Solve Prob. 5–44 for the maximum torsional
stress at B.
3 ft
A
1 ft
1 ft
B
3 ft
1 ft
©Mx = 0;
-15(1.5) - 5(3) + Tx = 0;
Tx = 37.5 lb # ft = 450 lb # in.
(tB)max =
450(0.25)
Tc
= 18.3 ksi
= p
4
J
2 (0.25)
Ans.
Ans:
(tB)max = 18.3 ksi
353
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5–46. A motor delivers 500 hp to the shaft, which is
tubular and has an outer diameter of 2 in. If it is rotating at
200 rad>s, determine its largest inner diameter to the
nearest 18 in. if the allowable shear stress for the material is
tallow = 25 ksi.
P = 500 hp c
T =
A
B
6 in.
550 ft # lb>s
d = 275000 ft # lb>s
1 hp
275000
P
=
= 1375 lb # ft
v
200
tmax = tallow =
Tc
J
1375(12)(1)
25(103) = p
d1 4
4
2 [1 - ( 2 ) ]
di = 1.745 in.
5
Use di = 1 in.
8
Ans.
www.elsolucionario.org
Ans:
5
Use di = 1 in.
8
354
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5–47. The propellers of a ship are connected to an 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.
Ans:
tmax = 44.3 MPa, f = 11.9°
355
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*5–48. The solid shaft of radius c is subjected to a torque T
at its ends. Show that the maximum shear strain developed
in the shaft is gmax = Tc>JG. What is the shear strain on
an element located at point A, c>2 from the center of the
shaft? Sketch the strain distortion of this element.
T
A
From the geometry:
gL = r f;
g =
Since f =
TL
, then
JG
g =
rf
L
Tr
JG
(1)
However the maximum shear strain occurs when r = c
gmax =
Tc
JG
Shear strain when r =
g =
T(c>2)
Tc
=
JG
2 JG
c/ 2
c
QED
c
is from Eq. (1),
2
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Ans.
356
L
T
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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
D
85 N⭈m
250 mm
400 mm
C
B
A
85 N⭈m
fA>D = ©
TL
JG
(85)(0.25)
2(85)(0.4)
= p
4
2 (0.015
4
9
- 0.01 )(75)(10 )
+ p
4
9
2 (0.02 )(75)(10 )
= 0.01534 rad = 0.879°
Ans.
Ans:
fA>D = 0.879°
357
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
5–50. The hydrofoil boat has an A992 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
www.elsolucionario.org
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
4
6
JG
2 (4 - 3.625 )11.0(10 )
= 0.07725 rad = 4.43°
Ans.
Ans:
tmax = 2.83 ksi, f = 4.43°
358
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5–51. The 60-mm-diameter shaft is made of 6061-T6
aluminum having an allowable shear stress of
tallow = 80 MPa. Determine the maximum allowable
torque T. Also, find the corresponding angle of twist of
disk A relative to disk C.
1.20 m
A
B
1.20 m
2T
3
T
C
1T
3
Internal Loading: The internal torques developed in segments AB and BC of the
shaft are shown in Figs. a and b, respectively.
Allowable Shear Stress: Segment AB is critical since it is subjected to a greater
p
internal torque. The polar moment of inertia of the shaft is J = (0.034) =
2
0.405(10 - 6)p m4 . We have
tallow =
TAB c
;
J
80(106) =
123 T2(0.03)
0.405(10 - 6)p
T = 5089.38 N # m = 5.09 kN # m
Ans.
Angle of Twist: The internal torques developed in segments AB and BC of the shaft
1
2
are TAB = (5089.38) = 3392.92 N # m and TBC = - (5089.38) = - 1696.46 N # m.
3
3
We have
fA>C = g
fA>C =
Ti Li
TABLAB
TBCLBC
=
+
Ji Gi
JGal
JGal
- 1696.46(1.20)
3392.92(1.20)
-6
9
0.405(10 )p(26)(10 )
+
0.405(10 - 6)p(26)(109)
= 0.06154 rad = 3.53°
Ans.
Ans:
T = 5.09 kN # m, fA>C = 3.53°
359
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*5–52. The 60-mm-diameter shaft is made of 6061-T6
aluminum. If the allowable shear stress is tallow = 80 MPa,
and the angle of twist of disk A relative to disk C is limited
so that it does not exceed 0.06 rad, determine the maximum
allowable torque T.
1.20 m
A
B
1.20 m
2T
3
T
C
1T
3
Internal Loading: The internal torques developed in segments AB and BC of the
shaft are shown in Figs. a and b, respectively.
Allowable Shear Stress: Segment AB is critical since it is subjected to a greater
internal torque. The polar moment of inertia of the shaft is J = p (0.034) =
2
0.405(10 - 6)p m4 . We have
tallow =
TAB c
;
J
80(103) =
123T2(0.03)
0.405(10 - 6)p
T = 5089.38 N # m = 5.089 kN # m
Angle of Twist: It is required that fA>C = 0.06 rad. We have
fA>C = ©
0.06 =
TiLi
TAB LAB
TBC LBC
=
+
JiGi
JGal
JGal
www.elsolucionario.org
123T2(1.2)
0.405(10 - 6)p(26)(109)
+
1- 13T2(1.2)
0.405(10 - 6)p(26)(109)
T = 4962.14 N # m = 4.96 kN # m (controls)
Ans.
360
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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
D
C
B
30 N⭈m
600 mm
200 mm
20 N⭈m
800 mm
80 N⭈m
Internal Torque: As shown on FBD.
Angle of Twist:
fB = a
TL
JG
1
[ -80.0(0.8) + (- 60.0)(0.6) + ( -90.0)(0.2)]
4
9
(0.01
)(75.0)(10
)
2
= p
= - 0.1002 rad = 5.74°
Ans.
Ans:
fB = 5.74°
361
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
5–54. The shaft is made of A992 steel with the allowable
shear stress of tallow = 75 MPa. If gear B supplies 15 kW of
power, while gears A, C, and D withdraw 6 kW, 4 kW, and
5 kW, respectively, determine the required minimum
diameter d of the shaft to the nearest millimeter. Also, find
the corresponding angle of twist of gear A relative to gear D.
The shaft is rotating at 600 rpm.
A
B
C
600 mm
D
600 mm
600 mm
Internal Loading: The angular velocity of the shaft is
v = a 600
1 min 2p rad
rev
ba
ba
b = 20p rad>s
min
60 s
1 rev
Thus, the torque exerted on gears A, C, and D are
TA =
6(103)
PA
=
= 95.49 N # m
v
20p
TC =
4(103)
PC
=
= 63.66 N # m
v
20p
TD =
5(103)
PD
=
= 79.58 N # m
v
20p
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The internal torque developed in segments AB, CD, and BC of the shaft are shown
in Figs. a, b, and c, respectively.
Allowable Shear Stress: Segment BC of the shaft is critical since it is subjected to a
greater internal torque.
d
143.24 a b
TBC c
2
tallow =
;
75(106) =
J
p d 4
a b
2 2
d = 0.02135 m = 21.35 mm
Use
Ans.
d = 22 mm
p
Angle of Twist: The polar moment of inertia of the shaft is J = (0.0114) =
2
-9
4
7.3205(10 )p m . We have
fA>D = ©
fA>D =
TiLi
TAB LAB
TBC LBC
TCD LCD
=
+
+
JiGi
JGst
JGst
JGst
0.6
( - 95.49 + 143.24 + 79.58)
7.3205(10 )p(75)(109)
-9
= 0.04429 rad = 2.54°
Ans.
Ans:
Use d = 22 mm, fA>D = 2.54°
362
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5–55. Gear B supplies 15 kW of power, while gears A, C,
and D withdraw 6 kW, 4 kW, and 5 kW, respectively. If the
shaft is made of steel with the allowable shear stress of
tallow = 75 MPa, and the relative angle of twist between
any two gears cannot exceed 0.05 rad, determine the
required minimum diameter d of the shaft to the nearest
millimeter. The shaft is rotating at 600 rpm.
A
B
C
600 mm
D
600 mm
600 mm
Internal Loading: The angular velocity of the shaft is
v = a 600
1 min 2p rad
rev
ba
ba
b = 20p rad>s
min
60 s
1 rev
Thus, the torque exerted on gears A, C, and D are
TA =
6(103)
PA
=
= 95.49 N # m
v
20p
TC =
4(103)
PC
=
= 63.66 N # m
v
20p
TD =
5(103)
PD
=
= 79.58 N # m
v
20p
The internal torque developed in segments AB, CD, and BC of the shaft are shown
in Figs. a, b, and c, respectively.
Allowable Shear Stress: Segment BC of the shaft is critical since it is subjected to a
greater internal torque.
TBC c
tallow =
;
J
75(106) =
d
143.24 a b
2
p d 4
a b
2 2
d = 0.02135 m = 21.35 mm
Angle of Twist: By observation, the relative angle of twist of gear D with respect to
gear B is the greatest.
Thus, the requirement is fD>B = 0.05 rad.
fD>B = ©
TiLi
TBC LBC
TCD LCD
=
+
= 0.05
JiGi
JGst
JGst
0.6
p d 4
9
2 1 2 2 (75)(10 )
(143.24 + 79.58) = 0.05
d = 0.02455 m = 24.55 mm = 25 mm (controls!)
Use
Ans.
d = 25 mm
Ans:
Use d = 25 mm
363
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*5–56. 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 the end B of the solid section relative to end C. 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
A
85 N⭈m
fB>C =
85(0.250)
TL
= 0.00113 rad = 0.0646°
= p
4
9
JG
2 (0.020) (75)(10 )
Ans.
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364
C
D
85 N⭈m
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
5–57. The turbine develops 150 kW of power, which is
transmitted to the gears such that C receives 70% and
D receives 30%. If the rotation of the 100-mm-diameter
A-36 steel shaft is v = 800 rev>min., determine the
absolute maximum shear stress in the shaft and the angle of
twist of end E of the shaft relative to B. The journal bearing
at E allows the shaft to turn freely about its axis.
B
v
C
D
3m
E
4m
2m
P = Tv;
150(103) W = T a 800
1 min
2p rad
rev
ba
ba
b
min 60 sec
1 rev
T = 1790.493 N # m
TC = 1790.493(0.7) = 1253.345 N # m
TD = 1790.493(0.3) = 537.148 N # m
Maximum torque is in region BC.
tmax =
1790.493(0.05)
TC
=
= 9.12 MPa
p
4
J
2 (0.05)
fE>B = © a
Ans.
TL
1
b =
[1790.493(3) + 537.148(4) + 0]
JG
JG
7520.171
= 0.0102 rad = 0.585°
4
9
2 (0.05) (75)(10 )
Ans.
= p
Ans:
tmax = 9.12 MPa, fE>B = 0.585°
365
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5–58. The turbine develops 150 kW of power, which is
transmitted to the gears such that both C and D receive
an equal amount. If the rotation of the 100-mm-diameter
A-36 steel shaft is v = 500 rev>min., determine the
absolute maximum shear stress in the shaft and the rotation
of end B of the shaft relative to E. The journal bearing at E
allows the shaft to turn freely about its axis.
B
v
C
D
3m
E
4m
2m
P = Tv;
150(103) W = T a 500
rev
1 min
2p rad
ba
ba
b
min 60 sec
1 rev
T = 2864.789 N # m
TC = TD =
T
= 1432.394 N # m
2
Maximum torque is in region BC.
tmax =
2864.789(0.05)
TC
=
= 14.6 MPa
p
4
J
2 (0.05)
fB>E = © a
Ans.
TL
1
b =
[2864.789(3) + 1432.394(4) + 0]
JG
JG
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14323.945
= 0.0195 rad = 1.11°
4
(0.05)
(75)(109)
2
= p
Ans.
Ans:
tmax = 14.6 MPa, fB>E = 1.11°
366
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5–59. The shaft is made of A992 steel. It has a diameter
of 1 in. and is supported by bearings at A and D, which
allow free rotation. Determine the angle of twist of B with
respect to D.
A
B
60 lb⭈ft
C
2 ft
60 lb⭈ft
2.5 ft
D
3 ft
The internal torques developed in segments BC and CD are shown in Figs. a and b.
The polar moment of inertia of the shaft is J =
FB>D = a
p
(0.54) = 0.03125p in4. Thus,
2
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.
Ans:
fB>D = 1.15°
367
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*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
D
3 ft
The internal torque developed in segment BC is shown in Fig. a
The polar moment of inertia of the shaft is J =
fC>B =
p
(0.54) = 0.03125p in4. Thus,
2
- 60(12)(2.5)(12)
TBC LBC
=
J Gst
(0.03125p)[11.0(106)]
= - 0.02000 rad
= 1.15°
Ans.
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368
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5–61. The two shafts are made of A992 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 lb⭈ft
A
30 in.
40 lb⭈ft
8 in.
10 in.
12 in.
4 in.
6 in.
B
Internal Torque: As shown on FBD.
Angle of Twist:
fE = a
TL
JG
= p
4
1
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
Since there is no torque applied between F and B then
fB = fF = 0.02667 rad = 1.53°
Ans.
Ans:
fB = 1.53°
369
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5–62. The two shafts are made of A992 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 lb⭈ft
A
30 in.
40 lb⭈ft
8 in.
10 in.
12 in.
4 in.
6 in.
B
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
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-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.
Ans:
fA = 1.78°
370
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
5–63. If the shaft is made of red brass C83400 copper with
an allowable shear stress of tallow = 20 MPa, determine the
maximum allowable torques T1 and T2 that can be applied
at A and B. Also, find the corresponding angle of twist of
end A. Set L = 0.75 m.
1m
a
C
80 mm
100 mm
T2
L
B
a
Section a–a
A
T1
Internal Loading: The internal torques developed in segments AB and BC of the
shaft are shown in Figs. a and b, respectively.
Allowable Shear Stress: The polar moments of inertia of segments AB and BC are
p
p
JAB = (0.14 - 0.084) = 29.52(10 - 6)p m4 and JBC = (0.14) = 50(10 - 6)p m4 . We
2
2
will consider segment AB first.
tallow =
TAB cAB
;
JAB
20(106) =
T1(0.1)
29.52(10 - 6)p
T1 = 18 547.96 N # m = 18.5 kN # m
Ans.
Using this result to consider segment BC, we have
tallow =
TBC cBC
;
JBC
20(106) =
(T2 - 18547.96)(0.1)
50(10 - 6)p
T2 = 49963.89 N # m = 50.0 kN # m
Ans.
Angle of Twist: Using the results of T1 and T2,
fA>C = ©
TiLi
TABLAB
TBCLBC
=
+
JiGi
JABGst
JBCGst
(49963.89 - 18547.96)(0.25)
-18547.96(0.75)
fA>C =
-6
9
+
29.52(10 )p(37)(10 )
50(10 - 6)p(37)(109)
Ans.
= - 0.002703 rad = 0.155°
Ans:
T1 = 18.5 kN # m, T2 = 50.0 kN # m,
fA>C = 0.155°
371
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*5–64. If the shaft is made of red brass C83400 copper and
is subjected to torques T1 = 20 kN # m and T2 = 50 kN # m,
determine the distance L so that the angle of twist at end A
is zero.
1m
a
C
80 mm
100 mm
Section a–a
Internal Loading: The internal torques developed in segments AB and BC of the
shaft are shown in Figs. a and b, respectively.
Angle of Twist: The polar moments of inertia of segments AB and BC are
p
p
JAB = (0.14 - 0.084) = 29.52(10 - 6)p m4 and JBC = (0.14) = 50(10 - 6)p m4 .
2
2
Here, it is required that. £ A>C = 0
fA>C = ©
0 =
TiLi
TABLAB
TBCLBC
=
+
JiGi
JABGst
JBCGst
- 20(103)(L)
29.52(10 - 6)p(37)(109)
30(103)(1 - L)
+
50(10 - 6)p(37)(109)
Ans.
L = 0.4697 m = 470 mm
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372
T2
B
a
A
T1
L
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5–65. The 8-mm-diameter A-36 steel bolt is screwed
tightly into a block at A. Determine the couple forces F that
should be applied to the wrench so that the maximum shear
stress in the bolt becomes 18 MPa. Also, compute the
corresponding displacement of each force F needed to
cause this stress. Assume that the wrench is rigid.
F
150 mm
150 mm
A
80 mm
F
T - F(0.3) = 0
tmax =
Tc
;
J
(1)
T(0.004)
18(106) = p
4
2 (0.004 )
T = 1.8096 N # m
From Eq. (1),
F = 6.03 N
f =
Ans.
1.8096(0.08)
TL
= 0.00480 rad
= p
4
9
JG
2 [(0.0040) ] 75(10 )
s = rf = 0.15(0.00480) = 0.00072 m = 0.720 mm
Ans.
Ans:
F = 6.03 N, s = 0.720 mm
373
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5–66. The A-36 hollow steel shaft is 2 m long and has an
outer diameter of 40 mm. When it is rotating at 80 rad>s, it
transmits 32 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 = 140 MPa and the
shaft is restricted not to twist more than 0.05 rad.
E
G
P = Tv
32(103) = T(80)
T = 400 N # m
Shear stress failure
Tc
t =
J
tallow = 140(106) = p
400(0.02)
4
4
2 (0.02 - ri )
ri = 0.01875 m
Angle of twist limitation:
f =
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TL
JG
0.05 = p
400(2)
4
9
4
2 (0.02 - ri )(75)(10 )
ri = 0.01247 m
(controls)
t = ro - ri = 0.02 - 0.01247
= 0.00753 m
= 7.53 mm
Ans.
Ans:
t = 7.53 mm
374
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5–67. The A-36 solid steel shaft is 3 m long and has a
diameter of 50 mm. It is required to transmit 35 kW of
power from the engine E to the generator G. Determine the
smallest angular velocity the shaft can have if it is restricted
not to twist more than 1°.
f =
E
G
TL
JG
1°(p)
T(3)
= p
4
180°
(0.025
)(75)(109)
2
T = 267.73 N # m
P = Tv
35(103) = 267.73(v)
v = 131 rad>s
Ans.
Ans:
v = 131 rad>s
375
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*5–68. If the shaft is subjected to a uniform distributed
torque t0, determine the angle of twist at A. The material
has a shear modulus G. The shaft is hollow for exactly half
its length.
L
C
c
2
t0
A c
Internal Loading: The internal torques developed in the hollow and solid segments
of the shaft are shown in Figs. a and b, respectively.
Angle of Twist: The polar moments of inertia of the hollow and solid segments of the
p
c 4
15p 4
p
shaft are Jh = c c4 - a b d =
c and Js = c4 , respectively. We have
2
2
32
2
fA = ©
T(x) dx
L JG
L>2
= -
L0
L>2 a tx2 +
tx1dx1
15p 4
a
c bG
32
-
L>2
= -c
32
15pc4G L0
= -c
tx12 L>2
32
a
b`
4
15pc G 2
0
= -
L0
1
tLb dx2
2
p
a c4b G
2
2
pc4G L0
L>2
a tx2 +
1
tL bdx2 d
2
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tx1dx1 +
+
L>2
tx22
1
2
a
+ tLx2 b ` d
4
2
pc G 2
0
61tL2
61tL2
=
4
60pc G
60pc4G
Ans.
376
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5–69. The A-36 steel bolt is tightened within a hole so that
the reactive torque on the shank AB can be expressed by
the equation t = (kx2) N # m>m, where x is in meters. If a
torque of T = 50 N # m is applied to the bolt head,
determine the constant k and the amount of twist in the
50-mm length of the shank. Assume the shank has a
constant radius of 4 mm.
B
t
T = 50 N⭈m
50 mm
x
A
dT = t dx
0.05
0.05 m
T =
L0
x3
kx dx = k
` = 41.667(10 - 6)k
3
2
0
50 - 41.6667(10 - 6) k = 0
k = 1.20(106) N>m2
Ans.
x
In the general position, T =
f =
L0
1.20(106)x2dx = 0.4(106)x3
0.05 m
T(x)dx
1
=
[50 - 0.4(106)x3]dx
JG L0
L JG
0.4(106)x4
1
c 50x d
=
JG
4
0.05 m
`
0
=
1.875
1.875
= p
4
JG
(0.004
)(75)(109)
2
= 0.06217 rad = 3.56°
Ans.
Ans:
k = 1.20 (106) N>m2, f = 3.56°
377
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5–70. Solve Prob. 5–69 if the distributed torque is
t = (kx2>3) N # m>m.
B
t
T = 50 N⭈m
50 mm
dT = t dx
0.05
T =
x
A
5 0.05
2
3
kx3 dx = k x3 | = (4.0716)(10-3) k
5
0
L0
50 - 4.0716(10-3) k = 0
2
k = 12.28(103) N>m(3)
Ans.
In the general position,
x
T =
L0
2
5
12.28(103)x3 dx = 7.368(103) x3
Angle of twist:
f =
T(x) dx
L
1
=
JG
JG L0
0.05 m
8
[50 - 7.3681(103)x3]dx
www.elsolucionario.org
=
1
3 8 0.05 m
c 50x - 7.3681(103) a b x3 d |
JG
8
0
1.5625
= 0.0518 rad = 2.97°
4
(0.004
)(75)(109)
2
Ans.
= p
Ans:
k = 12.28(103) N>m2>3, f = 2.97°
378
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*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
Equilibrium: Referring to the free-body diagram of segment AB shown in Fig. a,
©Mx = 0;
TAB = - 2(103)N # m
- TAB - 2(103) = 0
And the free-body diagram of segment BC, Fig. b,
©Mx = 0;
- TBC - 10(103)x - 2(103) = 0
TBC = - C 10(103)x + 2(103) D N # m
p
Angle of Twist: The polar moment of inertia of the shaft is J =
A 0.044 B =
2
-6
4
1.28(10 )p m . We have
fA = ©
LBC
TiLi
TABLAB
TBC dx
=
+
JiGi
JGal
JGal
L0
0.6 m -
- 2(103)(0.6)
=
1.28(10 - 6)p(26)(109)
+
1
= -
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.
379
A
2 kN⭈m
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
5–73. The contour of the surface of the shaft is defined by
the equation y = eax , where a is a constant. If the shaft is
subjected to a torque T at its ends, determine the angle of
twist of end A with respect to end B. The shear modulus is G.
B
y=e
x
ax
T
y
A
L
T
f =
T dx
L J(x)G
where, J(x) =
L
p ax 4
(e )
2
L
2T
dx
1
2T
=
pG L0 e4ax = pG a- 4 a e4ax b `
0
=
1
1
T
e4aL - 1
2T
+
b
ab =
a
4aL
pG
4a
2apG
4ae
e4aL
=
T
(1 - e - 4aL)
2apG
Ans.
www.elsolucionario.org
Ans:
f =
380
T
(1 - e - 4aL)
2 apG
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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
d
2
B
P
Equilibrium: Referring to the free-body diagram of the entire rod shown in Fig. a,
©Mx = 0; Pd -
1
L
(t ) a b = 0
2 0 2
t0 =
4Pd
L
Ans.
Internal Loading: The distributed torque expressed as a function of x, measured
4Pd>L
t0
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
4Pd 2
8Pd
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
TBC =
4Pd 2
x
L2
Referring to the free-body diagram shown in Fig. c,
©Mx = 0; Pd - TAB = 0
TAB = Pd
381
d
2
t0
C
A
P
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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.
www.elsolucionario.org
Ans:
t0 =
382
4Pd
4PLd
,f =
L
3pc4G
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
5–75. The A992 steel posts are “drilled” at constant angular
speed into the soil using the rotary installer. If the post has an
inner diameter of 200 mm and an outer diameter of
225 mm, determine the relative angle of twist of end A of
the post with respect to end B when the post reaches the
depth indicated. Due to soil friction, assume the torque
along the post varies linearly as shown, and a concentrated
torque of 80 kN # m acts at the bit.
B
4m
3m
15 kN⭈m/m
80 kN⭈m
A
TB - 80 -
©Mz = 0;
1
(15)(3) = 0
2
TB = 102.5 kN # m
©Mz = 0;
102.5 -
1
(5 z)(z) - T = 0
2
T = (102.5 - 2.5z2) kN # m
fA>B =
TL
T dz
+
JG L JG
3
102.5(103)(4)
= p
4
2 ((0.1125)
4
9
- (0.1) )(75)(10 )
+
(102.5 - 2.5z2)(103)dz
p
4
4
9
L0 2 ((0.1125) - (0.1) )(75)(10 )
Ans.
= 0.0980 rad = 5.62°
Ans:
fA>B = 5.62°
383
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*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
www.elsolucionario.org
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 2 - 2d
4p hG ri
ro
Ans.
384
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
5–77. The steel shaft has a diameter of 40 mm and is fixed
at its ends A and B. If it is subjected to the couple determine
the maximum shear stress in regions AC and CB of the
shaft. Gst = 75 Gpa .
3 kN
A
3 kN
C
400 mm
50 mm
50 mm
B
600 mm
Equilibrium:
(1)
TA + TB - 3000(0.1) = 0
Compatibility condition:
fC>A = fC>B
TA (400)
TB (600)
=
JG
JG
TA = 1.5 TB
(2)
Solving Eqs (1) and (2) yields:
TB = 120 N # m
TA = 180 N # m
(tAC)max =
180(0.02)
Tc
= 14.3 MPa
= p
4
J
2 (0.02 )
Ans.
(tCB)max =
120(0.02)
Tc
= 9.55 MPa
= p
4
J
2 (0.02 )
Ans.
Ans:
(tAC)max = 14.3 MPa, (tCB)max = 9.55 MPa
385
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
5–78. The A992 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
1m
1.5 m
C
A
1m
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 =
max
www.elsolucionario.org
TAC c
414.29 (0.03)
=
J
p
2
= 9.77 MPa
Ans.
4
(0.03)
Ans:
tabs = 9.77 MPa
max
386
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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 the
shaft is fixed at its ends A and B and subjected to a torque
of 500 lb # ft, determine the maximum shear stress in the
shaft. Gst = 10.8(103) ksi .
A
0.5 in.
C
D 500 lb⭈ft
5 in.
1 in.
8 in.
B
12 in.
Equilibrium:
(1)
TA + TB - 500 = 0
Compatibility condition:
fD>A = fD>B
TA(5)
p
4
2 (0.25 )G
TB(12)
= p
4
2 (0.5 )G
2 (0.5 )G
+ p
TA(8)
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 )
(max)
Ans.
Ans:
tmax = 29.3 ksi
387
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*5–80. The shaft is made of A-36 steel and is fixed at its
ends A and D. Determine the torsional reactions at these
supports.
1.5 ft
2 ft
A
1.5 ft
B
40 kip⭈ft
C
20 kip⭈ft
6 in.
Equilibrium: Referring to the free-body diagram of the shaft shown in Fig. a,
©Mx = 0;
(1)
TA + TD + 20 - 40 = 0
Compatibility Equation: Using the method of superposition, Fig. b,
fA = (fA)T - (fA)TA
0 = c
40(12)(2)(12)
20(12)(1.5)(12)
TA(12)(5)(12)
+
d JG
JG
JG
TA = 22 kip # ft
Ans.
Substituting this result into Eq. (1),
TD = - 2 kip # ft = 2 kip # ft
Ans.
The negative sign indicates that TD acts in the sense opposite to that shown on the
free-body diagram.
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388
D
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
5–81. The shaft is made of A-36 steel and is fixed at end D,
while end A is allowed to rotate 0.005 rad when the torque
is applied. Determine the torsional reactions at these
supports.
1.5 ft
2 ft
A
1.5 ft
B
40 kip⭈ft
C
20 kip⭈ft
6 in.
Equilibrium: Referring to the free-body diagram of the shaft shown in Fig. a,
(1)
©Mx = 0; TA + TD + 20 - 40 = 0
Compatibility Equation: Using the method of superposition, Fig. b,
fA = (fA)T - (fA)TA
0.005 =
40(12)(2)(12)
20(12)(1.5)(12)
TA(12)(5)(12)
+
p 4
J p (34)(11.0)(103) p (34)(11.0)(103) K
(3 )(11.0)(103)
2
2
2
TA = 12.28 kip # ft = 12.3 kip # ft
Ans.
Substituting this result into Eq. (1),
TD = 7.719 kip # ft = 7.72 kip # ft
Ans.
Ans:
TA = 12.3 kip # ft, TD = 7.72 kip # ft
389
D
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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
C
T ⫽ 50 lb⭈ft
(1)
Both the steel tube and brass core undergo the same angle of twist fC>B
fC>B =
JG
Tst (2)(12)
Tbr (2)(12)
TL
= p
4
6
2 (0.5 )(5.6)(10 )
= p
4
2 (1
- 0.54)(11.5)(106)
(2)
Tbr = 0.032464 Tst
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°
(tst)max AB =
(tst)max BC =
(gst)max =
(tbr)max =
(gbr)max =
TABc
50(12)(1)
=
J
Tst c
J
p 4
2 (1 )
Ans.
www.elsolucionario.org
= 382 psi
48.428(12)(1)
= 394.63 psi = 395 psi (Max)
= p
4
4
2 (1 - 0.5 )
(tst)max
394.63
=
G
11.5(106)
= 34.3(10 - 6) rad
Ans.
1.572(12)(0.5)
Tbr c
= 96.08 psi = 96.1 psi (Max)
=
p
4
J
2 (0.5 )
(tbr)max
96.08
=
G
5.6(106)
Ans.
= 17.2(10 - 6) rad
Ans.
Ans.
Ans:
fC = 0.116°, (tst)max = 395 psi,
(gst)max = 34.3 (10 - 6) rad, (tbr)max = 96.1 psi,
(tbr)max = 17.2 (10 - 6) rad
390
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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
450 lb·ft
F
4 ft
3 ft
C
D
A
Equilibrium:
TC + TD - 450 = 0
(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
(tBC)max =
(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 )
= 1.47 ksi
Ans.
= 1.96 ksi
Ans.
= 0.00589 rad = 0.338°
Ans.
Ans:
(tBC)max = 1.47 ksi, (tBD)max = 1.96 ksi,
fB>C = fB>D = 0.338°
391
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*5–84. The Am1004-T61 magnesium tube is bonded to
the A-36 steel rod. If the allowable shear stresses for
the magnesium and steel are (tallow)mg = 45 MPa and
(tallow)st = 75 MPa, respectively, determine the maximum
allowable torque that can be applied at A. Also, find the
corresponding angle of twist of end A.
900 mm
B
A
80 mm
T
40 mm
Equilibrium: Referring to the free-body diagram of the cut part of the assembly
shown in Fig. a,
©Mx = 0; Tmg + Tst - T = 0
(1)
Compatibility Equation: Since the steel rod is bonded firmly to the magnesium
tube, the angle of twist of the rod and the tube must be the same. Thus,
(fst)A = (fmg)A
TstL
p
(0.024)(75)(109)
2
Tmg L
=
p
(0.044 - 0.024)(18)(109)
2
Tst = 0.2778 Tmg
Solving Eqs. (1) and (2),
Tmg = 0.7826T
(2)
www.elsolucionario.org
Tst = 0.2174T
Allowable Shear Stress:
(tallow)mg =
Tmg c
J
; 45(106) =
0.7826T(0.04)
p
(0.044 - 0.024)
2
T = 5419.25 N # m
(tallow) st =
Tst c
;
J
75(106) =
0.2174T(0.02)
p
(0.024)
2
T = 4335.40 N # m = 4.34 kN # m (control!)
Ans.
Angle of Twist: Using the result of T, Tst = 942.48 N # m. We have
fA =
942.48(0.9)
Tst L
=
= 0.045 rad = 2.58°
p
JstGst
(0.024)(75)(109)
2
392
Ans.
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5–85. The Am1004-T61 magnesium tube is bonded to the
A-36 steel rod. If a torque of T = 5 kN # m is applied to
end A, determine the maximum shear stress in each
material. Sketch the shear stress distribution.
900 mm
B
A
80 mm
40 mm
Equilibrium: Referring to the free-body diagram of the cut part of the assembly
shown in Fig. a,
©Mx = 0; Tmg + Tst - 5(103) = 0
T
(1)
Compatibility Equation: Since the steel rod is bonded firmly to the magnesium
tube, the angle of twist of the rod and the tube must be the same. Thus,
(fst)A = (fmg)A
Tmg L
Tst L
=
p
(0.024)(75)(109)
2
p
(0.044 - 0.024)(18)(109)
2
Tst = 0.2778Tmg
(2)
Solving Eqs. (1) and (2),
Tmg = 3913.04 N # m
Tst = 1086.96 N # m
Maximum Shear Stress:
(tst)max =
(tmg)max =
1086.96(0.02)
Tst cst
=
= 86.5 MPa
p
Jst
4
(0.02 )
2
Tmg cmg
Jmg
(tmg )|r = 0.02 m =
=
3913.04(0.04)
= 41.5 MPa
p
4
4
(0.04 - 0.02 )
2
Tmg r
Jmg
Ans.
3913.04(0.02)
=
p
(0.044 - 0.024)
2
Ans.
= 20.8 MPa
Ans.
Ans:
(tst)max = 86.5 MPa, (tmg)max = 41.5 MPa,
(tmg)|r = 0.02 m = 20.8 MPa
393
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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 N⭈m
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
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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.
Ans:
TB = 22.2 N # m, TA = 55.6 N # m
394
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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 N⭈m
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
TB(0.75)
TA(1.5)
= 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
9
JG
2 (0.0125 )(75.0)(10 )
= 0.02897 rad = 1.66°
Ans.
Ans:
fE = 1.66°
395
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*5–88. A rod is made from two segments: AB is steel and BC
is brass. It is fixed at its ends and subjected to a torque of
T = 680 N # m. If the steel portion has a diameter of 30 mm,
determine the required diameter of the brass portion so the
reactions at the walls will be the same. Gst = 75 GPa ,
Gbr = 39 GPa .
C
1.60 m
B
680 N⭈m
A
0.75 m
Compatibility Condition:
fB>C = fB>A
T(0.75)
T(1.60)
p 4
9
2 (c )(39)(10 )
= p
4
9
2 (0.015 )(75)(10 )
c = 0.02134 m
d = 2c = 0.04269 m = 42.7 mm
Ans.
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396
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5–89. Determine the absolute maximum shear stress in
the shaft of Prob. 5–88.
C
1.60 m
B
680 N⭈m
A
0.75 m
Equilibrium,
2T = 680
T = 340 N # m
tabs occurs in the steel. See solution to Prob. 5–88.
max
tabs =
max
340(0.015)
Tc
= p
4
J
2 (0.015)
= 64.1 MPa
Ans.
Ans:
tabs = 64.1 MPa
max
397
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5–90. The composite shaft consists of a mid-section that
includes the 1-in.-diameter solid shaft and a tube that is
welded to the rigid flanges at A and B. Neglect the thickness
of the flanges and determine the angle of twist of end C of
the shaft relative to end D. The shaft is subjected to a torque
of 800 lb # ft. The material is A-36 steel.
800 lb⭈ft 1 in.
3 in.
0.25 in.
C
800 lb⭈ft
A
D
0.5 ft
0.75 ft
B
0.5 ft
Equilibrium:
800(12) - Tg - Ts = 0
Compatibility Condition:
fT = fS ;
TT(0.75)(12)
p
4
4
2 ((1.5) - (1.25) )G
=
TS(0.75)(12)
p
4
2 (0.5) G
TT = 9376.42 lb # in.
TS = 223.58 lb # in.
fC>D = ©
223.58(0.75)(12)
800(12)(1)(12)
TL
+ p
= 0.1085 rad = 6.22° Ans.
= p
4
6
4
6
JG
2 (0.5) (11.0)(10 )
2 (0.5) (11.0)(10 )
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Ans:
fC>D = 6.22°
398
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5–91. The A992 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 lb⭈in./in.
1 in.
20 in.
B
Equilibrium:
TA + TB - 60(20) = 0
(1)
Compatibility condition:
fC>B = fC>A
fC>B =
20
T(x) dx
(TB - 60x) 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
18.52TB - 74.08TA = 11112
(2)
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 )
Ans.
tabs = 5.50 ksi
max
Ans:
tabs = 5.50 ksi
max
399
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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
Section a–a
(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
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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
a 0.044 - 0.034 b
2
Ans.
400
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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.
T
2c
A
Equilibrium:
TA + TB - T = 0
B
c
[1]
L/2
Section Properties:
L/2
r(x) = c +
c
c
x =
(L + x)
L
L
J(x) =
4
p c
pc4
(L + x)4
c (L + x) d =
2 L
2L4
fT =
Tdx
=
J(x)G
L
LL2
Angle of Twist:
L
Tdx
pc4
4
2L4 (L + x) G
L
=
2TL4
dx
4
L
pc G L2 (L + x)4
= -
=
fB =
L
1
2TL4
c
d 2
4
3
L
3pc G (L + x)
2
37TL
324 pc4 G
Tdx
=
L J(x)G
L0
=
TBdx
pc4
4
2L4 (L + x) G
L
2TBL4
dx
pc4G L0 (L + x)4
= -
=
L
2TBL4
L
1
d 2
3
3pc G (L + x)
0
4
c
7TB L
12pc4G
Compatibility:
0 = fT - fB
0 =
7TBL
37TL
324pc4G
12pc4G
TB =
37
T
189
Ans.
Substituting the result into Eq. [1] yields:
TA =
152
T
189
Ans.
Ans:
TB =
401
37
152
T, TA =
T
189
189
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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
A
2t0
x
T(x) =
t0 a 1 +
L0
2
3
x
x
b dx = t0 a x +
b
L2
3L2
(1)
By superposition:
0 = f - fB
x
t0 a x + 3L
2b
3
L
0 =
L0
TB =
2
dx
-
JG
TB(L)
7t0L
=
- TB(L)
JG
12
7t0 L
12
Ans.
From Eq. (1),
T = t0 a L +
TA +
4t0 L
L3
b =
3
3L2
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7t0 L
4t0 L
= 0
12
3
TA =
3t0 L
4
Ans.
Ans:
TB =
402
7t0 L
3t0 L
, TA =
12
4
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5–95. The aluminum rod has a square cross section of
10 mm by 10 mm. If it is 8 m long, determine the torque T
that is required to rotate one end relative to the other end
by 90°. Gal = 28 GPa, (tY)al = 240 MPa .
T
8m
T
10 mm
10 mm
f =
7.10 TL
a4G
7.10T(8)
p
=
2
(0.01)4(28)(109)
T = 7.74 N # m
tmax =
Ans.
4.81T
a3
4.81(7.74)
=
0.013
= 37.2 MPa 6 tY
OK
Ans:
T = 7.74 N # m
403
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*5–96. The shafts have elliptical and circular cross sections
and are to be made from the same amount of a similar
material. Determine the percent of increase in the maximum
shear stress and the angle of twist for the elliptical shaft
compared to the circular shaft when both shafts are subjected
to the same torque and have the same length.
b
2b
c
Section Properties: Since the elliptical and circular shaft are made of the same
amount of material, their cross-sectional areas must be the same, Thus,
p(b)(2b) = pc2
c = 22b
Maximum Shear Stress: For the circular shaft,
(tmax)c =
T(22b)
Tc
22T
=
=
p
J
2pb3
(22b)4
2
For the elliptical shaft,
(tmax)e =
Thus,
T
2T
2T
=
=
2
2
pab
p(2b)(b )
pb3
% of increase in shear stress = c
(tmax)e - (tmax)c
d * 100
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= ±
22T
T
pb3
2pb3
22T
2pb3
≤ * 100
Ans.
= 41.4%
Angle of Twist: For the circular shaft,
fc =
TL
TL
=
4
p
2pb
G
( 22b)4G
2
For the elliptical shaft,
fe =
Thus,
(a2 + b2)TL
3 3
pa b G
[(2b)2 + b2]TL
=
p(2b)3b3G
% of increase in angle of twist = c
=
5TL
8pb4G
fe - fc
d * 100
fc
TL
5TL
8pb4G
2pb4G
≤ * 100
= ±
TL
2pb4G
= 25%
Ans.
404
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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
d
d
For the circular shaft:
T A d2 B
Tc
16T
(tmax)c =
=
=
p d 4
J
p
d3
A B
2 2
For the elliptical shaft:
(tmax)c =
2T
2T
16T
=
=
d kd 2
p a b2
p
k2 d3
pA B A B
2
2
16T
Factor of increase in maximum shear stress =
(tmax)c
p k2 d3
= 16T
(tmax)c
3
pd
=
1
k2
Ans.
Ans:
1
Factor of increase in max. shear stress = 2
k
405
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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 N⭈m
50 N⭈m
30 N⭈m
2m
C
50 mm
20 mm
1.5 m
B
Maximum Shear Stress:
(tBC)max =
2TBC
p a b2
2(30.0)
=
p(0.05)(0.022)
Ans.
= 0.955 MPa
(tAC)max =
2TAC
2
2(50.0)
=
pab
p(0.05)(0.022)
Ans.
= 1.59 MPa
Angle of Twist:
fB>A = a
(a2 + b2)T L
p a3b3 G
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(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.
Ans:
(tBC)max = 0.955 MPa, (tAC)max = 1.59 MPa,
fB>A = 0.207°
406
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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 N⭈m
50 N⭈m
30 N⭈m
2m
C
50 mm
20 mm
1.5 m
B
Maximum Shear Stress:
(tBC)max =
2TBC
p a b2
2(30.0)
=
p(0.05)(0.022)
Ans.
= 0.955 MPa
(tAC)max =
2TAC
2
2(50.0)
=
pab
p(0.05)(0.022)
Ans.
= 1.59 MPa
Angle of Twist:
fB>C =
(a2 + b2) TBC L
p a3 b3 G
(0.052 + 0.022)( -30.0)(1.5)
=
p(0.053)(0.023)(37.0)(109)
Ans.
= - 0.001123 rad = 0.0643°
Ans:
(tBC)max = 0.955 MPa, (tAC)max = 1.59 MPa,
fB>C = 0.0643°
407
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*5–100. If end B of the shaft, which has an equilateral
triangle cross section, is subjected to a torque of
T = 900 lb # ft, determine the maximum shear stress
developed in the shaft. Also, find the angle of twist of end B.
The shaft is made from 6061-T1 aluminum.
2 ft
A
3 in.
B
T
Maximum Shear Stress:
tmax =
20(900)(12)
20T
=
= 8000 psi = 8 ksi
3
a
33
Ans.
Angle of Twist:
f =
46TL
a4G
46(900)(12)(2)(12)
=
34(3.7)(106)
= 0.03978 rad = 2.28°
Ans.
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408
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5–101. If the shaft has an equilateral triangle cross section
and is made from an alloy that has an allowable shear stress
of tallow = 12 ksi , determine the maximum allowable
torque T that can be applied to end B. Also, find the
corresponding angle of twist of end B.
2 ft
A
3 in.
B
T
Allowable Shear Stress:
tallow =
20T
;
a3
12 =
20T
33
T = 16.2 kip # in a
1 ft
b = 1.35 kip # ft
12 in.
Ans.
Angle of Twist:
f =
46TL
a4G
46(16.2)(103)(2)(12)
=
34(3.7)(106)
= 0.05968 rad = 3.42°
Ans.
Ans:
T = 1.35 kip # ft, f = 3.42°
409
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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
B
3 ft
By superposition:
0 = f - fB
0 =
7.10(80)(2)
4
-
7.10(TB)(5)
a4 G
a G
TB = 32 lb # ft
Ans.
TA + 32 - 80 = 0
TA = 48 lb # ft
fC =
Ans.
7.10(32)(12)(3)(12)
(24)(3.8)(106)
Ans.
= 0.00161 rad = 0.0925°
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Ans:
TB = 32 lb # ft, TA = 48 lb # ft, fC = 0.0925°
410
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5–103. A torque of 2 kip # in. is applied to the tube. If the
wall thickness is 0.1 in., determine the average shear stress
in the tube.
2 in.
2 in.
1.90 in.
Am =
p(1.952)
= 2.9865 in2
4
tavg =
2(103)
T
=
= 3.35 ksi
2t Am
2(0.1)(2.9865)
Ans.
Ans:
tavg = 3.35 ksi
411
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*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
0.5 m
A
60 N·m
25 mm
Maximum Shear Stress:
tmax =
4.81Tmax
a3
4.81(80.0)
=
(0.0253)
= 24.6 MPa
Ans.
Angle of Twist:
fA>C = a
7.10(- 20.0)(1.5)
7.10(- 80.0)(0.5)
7.10TL
=
+
a4G
(0.0254)(26.0)(109)
(0.0254)(26.0)(109)
Ans.
= - 0.04894 rad = 2.80°
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412
80 N⭈m
25 mm
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5–105. If the shaft is subjected to the torque of 3 kN # m,
determine the maximum shear stress developed in the
shaft. Also, find the angle of twist of end B. The shaft is
made from A-36 steel. Set a = 50 mm.
600 mm
a
a
A
B
a
3 kN⭈m
Maximum Shear Stress:
tmax =
2(3)(103)
2T
= 61.1 MPa
=
2
pab
p(0.05)(0.0252)
Ans.
Angle of Twist:
f =
(a2 + b2)TL
pa3b3G
(0.052 + 0.0252)(3)(103)(0.6)
=
p(0.053)(0.0253)(75)(109)
= 0.01222 rad = 0.700°
Ans.
Ans:
tmax = 61.1 MPa, fB = 0.700°
413
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5–106. If the shaft is made from A-36 steel having an
allowable shear stress of tallow = 75 MPa, determine the
minimum dimension a for the cross-section to the nearest
millimeter. Also, find the corresponding angle of twist
at end B.
600 mm
a
a
A
B
a
3 kN⭈m
Allowable Shear Stress:
tallow =
2T
;
pab2
75(106) =
2(3)(103)
p(a)1a222
a = 0.04670 m
Use
a = 47 mm
Ans.
Angle of Twist:
f =
(a2 + b2)TL
pa3b3G
c 0.0472 + a
=
0.047 2
b d (3)(10 3)(0.6)
2
p(0.047 3) a
0.047 3
b (75)(10 9)
2
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= 0.01566 rad = 0.897°
Ans.
Ans:
Use a = 47 mm, fB = 0.897°
414
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5–107. If the solid shaft is made from red brass C83400
copper having an allowable shear stress of tallow = 4 ksi,
determine the maximum allowable torque T that can be
applied at B.
2 ft
A
4 in.
2 in.
2 in.
4 ft
B
T
C
Equilibrium: Referring to the free-body diagram of the square bar shown in Fig. a,
we have
TA + TC - T = 0
©Mx = 0;
(1)
Compatibility Equation: Here, it is required that
fB>A = fB>C
7.10TA(2)(12)
=
a4G
7.10TC (4)(12)
a4G
TA = 2TC
(2)
Solving Eqs. (1) and (2),
TC =
1
T
3
TA =
2
T
3
Allowable Shear Stress: Segment AB is critical since it is subjected to the greater
internal torque.
tallow =
4.81TA
a3
;
4 =
2
4.81 a Tb
3
43
T = 79.83 kip # in a
1 ft
b = 6.65 kip # ft
12 in.
Ans.
Ans:
T = 6.65 kip # ft
415
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*5–108. If the solid shaft is made from red brass C83400
copper and it is subjected to a torque T = 6 kip # ft at B,
determine the maximum shear stress developed in segments
AB and BC.
2 ft
A
4 in.
2 in.
2 in.
4 ft
B
T
C
Equilibrium: Referring to the free-body diagram of the square bar shown in Fig. a,
we have
©Mx = 0;
(1)
TA + TC - 6 = 0
Compatibility Equation: Here, it is required that
fB>A = fB>C
7.10 TA(2)(12)
a4G
=
7.10 TC (4)(12)
a4G
TA = 2TC
(2)
Solving Eqs. (1) and (2),
TC = 2 kip # ft
TA = 4 kip # ft
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Maximum Shear Stress:
(tmax)AB =
(tmax)BC =
4.81TA
a3
4.81TC
a3
4.81(4)(12)
=
43
4.81(2)(12)
=
43
= 3.61 ksi
Ans.
= 1.80 ksi
Ans.
416
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5–109. For a given maximum average stress, determine the
factor by which the torque carrying capacity is increased if
the half-circular section is reversed from the dashed-line
position to the section shown. The tube is 0.1 in. thick.
1.80 in.
0.6 in.
1.20 in.
0.5 in.
Am = (1.10)(1.75) -
p(0.552)
= 1.4498 in2
2
Am ¿ = (1.10)(1.75) +
p(0.552)
= 2.4002 in2
2
tmax =
T
2t Am
T = 2 t Am tmax
Factor =
=
2t Am ¿ tmax
2t Am tmax
Am ¿
2.4002
=
= 1.66
Am
1.4498
Ans.
Ans:
Factor of increase = 1.66
417
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5–110. For a given maximum 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
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= 2.85
Ans.
Ans:
Factor of increase = 2.85
418
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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
a
a
a
Circular tube:
qcr =
T
T
2T
=
=
2Am
2p (a>2)2
p a2
Square tube:
qsq =
T
T
=
2Am
2a2
qsq
T>(2a2)
qcr
=
2
2T>(p a )
=
p
4
Thus;
qsq =
p
q
4 cr
Ans.
Ans:
qsq =
419
p
q
4 cr
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*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?
a⫹b
2
a
b
e
2
Average Shear Stress:
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 =
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
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3
a + b 2
b
T¿ = tavg (2)c (a - b) d (p)a
4
2
tavg (2) C 4 (a - b) D (p) A 2 B
3
T¿
=
=
Factor =
a + b 2
T
4
tavg (2)(a - b)(p) A
B
3
a + b 2
2
3
Percent reduction in strength = a1 - b * 100 % = 25 %
4
420
Ans.
e
2
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5–113. Determine the constant thickness of the rectangular
tube if average 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.
2 in.
Am = 2(4) = 8 in2
tavg =
T
2t Am
12 =
20
2t (8)
t = 0.104 in.
Ans.
Ans:
t = 0.104 in.
421
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5–114. 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.
2 in.
Am = 2(4) = 8 in2
tavg =
T
2t Am ;
12 =
T
2(0.125)(8)
T = 24 kip # in. = 2 kip # ft
Ans.
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Ans:
T = 2 kip # ft
422
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5–115. The steel tube has an elliptical cross section of mean
dimensions shown and a constant thickness of t = 0.2 in. If the
allowable shear stress is tallow = 8 ksi , and the tube is to
resist a torque of T = 250 lb # ft, determine the necessary
dimension b. The mean area for the ellipse is Am = pb10.5b2.
b
0.5b
250 lb⭈ft
tavg = tallow =
8(103) =
T
2tAm
250(12)
2(0.2)(p)(b)(0.5b)
b = 0.773 in.
Ans.
Ans:
b = 0.773 in.
423
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*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 the tube is subjected to
the torque of T = 500 N # m. Show the shear stress on
volume elements located at these points. Neglect stress
concentrations at the corners.
20 mm
20 mm
A
B
30 mm
50 mm
50 mm
30 mm
1
Am = 2 c (0.04)(0.03) d + 0.1(0.04) = 0.0052 m2
2
(tavg)A = (tavg)B =
=
T
2tAm
500
2(0.005)(0.0052)
= 9.62 MPa
Ans.
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424
T
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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
0.25 m
10 mm
10 mm
0.25 m
2m
Section Properties: Referring to the geometry shown in Fig. a,
Am =
F
p
1
a 0.52 b + (1 + 0.5)(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)
=
6.1019
≤
4(1.8927 )(27)(10 ) 0.01
2
9
¢
= 7.463(10 - 3) rad = 0.428°>m
Ans.
Ans:
T = 4.73 MN # m, f = 0.428°>m
425
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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
0.25 m
10 mm
0.25 m
2m
10 mm
Section Properties: Referring to the geometry shown in Fig. a,
p
1
Am = (0.52) + (1 + 0.5)(2) = 1.8927 m2
2
2
F
ds = p(0.5) + 2 222 + 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
2
4Am G F t
4.5(106)(1)
=
6.1019
≤
4(1.8927 )(27)(10 ) 0.01
2
9
¢
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= 7.0973(10 - 3) rad = 0.407°>m
Ans.
Ans:
tavg = 119 MPa, f = 0.407°>m
426
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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 N⭈m
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.
Ans:
(tavg)A = (tavg)B = 357 kPa
427
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*5–120. The steel step shaft has an allowable shear stress
of tallow = 8 MPa . If the transition between the cross
sections has a 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
r
4
=
= 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(106) = 1.25 B p
T = 20.1 N # m
Ans.
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428
T
20 mm
T
2
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5–121. The step 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 and
the radius at the transition on the shaft is 7.5 mm.
75 mm
60 mm
v = 720
T =
rev 2p rad 1 min
a
b
= 24 p rad>s
min
1 rev
60 s
30(103)
P
=
= 397.89 N # m
v
24 p
75
D
=
= 1.25
d
60
t = K
and
r
7.5
=
= 0.125;
d
60
K = 1.29
Tc
J
t = 1.29 c
397.89(0.03)
p
4
2 (.03)
d = 12.1(10)6 = 12.1 MPa > tallow
No, it is not possible.
Ans.
Ans:
No, it is not possible.
429
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5–122. The built-up shaft is designed to rotate at 540 rpm.
If the radius at the transition on the shaft is r = 7.2 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.
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Ans:
P = 101 kW
430
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5–123. The transition at the cross sections of the step shaft
has 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
A
40 N⭈m
60 N⭈m
(tmax)CD =
100(0.025)
TCDc
= p
4
J
2 (0.025 )
= 4.07 MPa
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.
Ans:
tmax = 50.6 MPa
431
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*5–124. The steel used for the step shaft has an allowable
shear stress of tallow = 8 MPa . If the radius at the transition
between the cross sections is 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
2.25
r
=
= 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.
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432
T
2
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5–125. The step shaft is subjected to a torque of
710 lb # in. If the allowable shear stress for the material is
tallow = 12 ksi, determine the smallest radius at the junction
between the cross sections that can be used to transmit the
torque.
0.75 in.
A
710 lb⭈in.
B
1.5 in.
C
tmax = tallow = K
12(103) =
710 lb⭈ft
Tc
J
K(710)(0.375)
p
4
2 (0.375 )
K = 1.40
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
Ans:
r = 0.075 in.
433
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5–126. A solid shaft has a diameter of 40 mm and length
of 1 m. It is made from an elastic-plastic material having a
yield stress of tY = 100 MPa. Determine the maximum
elastic torque TY and the corresponding angle of twist.
What is the angle of twist if the torque is increased to
T = 1.2TY? G = 80 GPa.
Maximum elastic torque TY ,
TY c
tY =
J
100(10 )1 2 2(0.02 )
tYJ
= 1256.64 N # m = 1.26 kN # m
=
c
0.02
6
TY =
p
4
Ans.
Angle of twist:
tY
gY =
f =
G
100(106)
=
80(109)
= 0.00125 rad
gY
0.00125
L =
(1) = 0.0625 rad = 3.58°
rY
0.02
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Also,
f =
Ans.
1256.64(1)
TYL
= 0.0625 rad = 3.58°
= p
4
JG
(0.02
)(80)(109)
2
From Eq. 5–26 of the text,
T =
ptY
(4c3 - r3Y);
6
1.2(1256.64) =
p(100)(106)
[4(0.023) - r3Y]
6
rY = 0.01474 m
f¿ =
gY
0.00125
L =
(1) = 0.0848 rad = 4.86°
rY
0.01474
Ans.
Ans:
TY = 1.26 kN # m, f = 3.58°, f¿ = 4.86°
434
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5–127. Determine the torque needed to twist a short
2-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 = 50 MPa. Assume that the material
becomes fully plastic.
Fully plastic torque is applied. From Eq. 5–27,
TP =
2p
2p
t c3 =
(50)(106)(0.0013) = 0.105 N # m
3 Y
3
Ans.
Ans:
Tp = 0.105 N # m
435
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*5–128. A bar having a circular cross section of 3 in.diameter is subjected to a torque of 100 in # kip. If the
material is elastic-plastic, with tY = 16 ksi, determine the
radius of the elastic core.
Using Eq. 5–26 of the text,
T =
ptY
(4c3 - r3Y)
6
100(103) =
p(16)(103)
(4 (1.53 - rY3))
6
rY = 1.16 in.
Ans.
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436
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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
0.004
p tY
T =
A 4c3 - r3Y B
6
=
g (rad)
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
JG
(0.04
)(40)(109)
2
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
Ans:
T = 20.8 kN # m, f = 34.4°, fr = 12.2°
437
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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.
T
t (ksi)
12
From the shear–strain diagram,
rY
2
=
;
0.0006
0.0048
6
rY = 0.25 in.
0.0006
From the shear stress–strain diagram,
t1 =
0.0048
g (rad)
6
r = 24r
0.25
t2 - 6
12 - 6
=
;
r - 0.25
2 - 0.25
t2 = 3.4286 r + 5.1429
c
T = 2p
L0
t r2 dr
0.25
= 2p
L0
2
24r3 dr + 2p
= 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
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= 172.30 kip # in. = 14.4 kip # ft
Ans.
Ans:
T = 14.4 kip # ft
438
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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.
c
c
2
Maximum Elastic Torque.
TY =
=
1 3
pc tY
2
1
p a 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
p a 0.043 b A 125 B a 106 b
3
= 16755.16 N # m = 16.8 kN # m
Ans.
Ans:
TY = 12.6 kN # m, Tp = 16.8 kN # m
439
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*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
Lc>2
r2dr
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c
= 2ptY ¢
=
r3
≤`
3 c>2
7
pc3tY
12
The ratio is
7
pc3tY
TP
12
=
= 1.24
TY
15 3
pc tY
32
Ans.
440
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
5–133. If the step shaft 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
t (ksi)
12
0.75-in.-diameter segment will be fully plastic. From Eq. 5-27 of the text:
0.005
2p tY 3
(c )
T = Tp =
3
g (rad)
2p (12)(103)
(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
Ans:
T = 110 lb # ft
441
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
5–134. The solid shaft is made from an elastic-plastic
material as shown. Determine the torque T needed to form
an elastic core in the shaft having a radius of rY = 23 mm.
If the shaft is 2 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.
40 mm
␶ (MPa)
150
␳ Y = 23 mm
␥ (rad)
0.005
Use Eq. 5–26 of the text,
p(150)(106)
ptY
(4c3 - r Y3 ) =
(4(0.043) - 0.0233)
6
6
T =
= 19 151 N # m = 19.2 kN # m
f =
Ans.
gL
gYL
0.005(2)(1000)
=
= 0.4348 rad = 24.9°
=
r
rY
23
Ans.
An opposite torque T = 19 151 N # m is applied:
tr =
19 151(0.04)
Tc
= 190 MPa
= p
4
J
2 (0.04 )
G =
150(106)
= 30 GPa
0.005
fP =
19151(2)
TL
= 0.3175 rad
= p
4
JG
(0.04
)(30)(109)
2
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fr = 0.4348 - 0.3175 = 0.117 rad = 6.72°
Ans.
Ans:
T = 19.2 kN # m, f = 24.9°, fr = 6.72°
442
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
5–135. A 1.5-in.-diameter shaft is made from an elasticplastic material as shown. Determine the radius of its elastic
core if it is subjected to a torque of T = 200 lb # ft. If the
shaft is 10 in. long, determine the angle of twist.
T
10 in.
T
␶ (ksi)
3
0.006
␥ (rad)
Use Eq. 5–26 from the text:
T =
ptg
6
(4 c3 - r3g)
200(12) =
p(3)(103)
[4(0.753) - r3g]
6
rg = 0.542 in.
f =
Ans.
gY
0.006
L =
(10) = 0.111 rad = 6.34°
rY
0.542
Ans.
Ans:
rY = 0.542 in., f = 6.34°
443
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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
L0.5
(13.333r + 5) r2 dr
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0.75
= 2p
L0.5
= 2p c
(13.333r3 + 5r2) dr
5r3 0.75
13.333r4
+
d |
4
3 0.5
= 8.426 kip # in. = 702 lb # ft
Ans.
444
0.01
g (rad)
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
5–137. The shaft is made from a strain-hardening material
having a t - g diagram as shown. Determine the torque T
that must be applied to the shaft in order to create an elastic
core in the shaft having a radius of rc = 0.5 in.
T
0.6 in.
t (ksi)
15
10
10(103)
t1
=
g
0.005
0.005
t1 = 2(106)g
3
0.01
g (rad)
(1)
3
3
t2 - 10(10 )
15(10 ) - 10(10 )
=
g - 0.005
0.01 - 0.005
t2 = 1(106)g + 5(103)
0.6
(0.005) = 0.006
0.5
gmax =
g =
(2)
r
r
g
=
(0.006) = 0.01r
c max
0.6
Substituting g into Eqs. (1) and (2) yields:
t1 = 20(103)r
t2 = 10(103)r + 5(103)
c
T = 2p
L0
tr2 dr
0.5
= 2p
L0
0.6
20(103)r3 dr + 2p
L0.5
[10(103)r + 5(103)]r2 dr
= 3970 lb # in. = 331 lb # ft
Ans.
Ans:
T = 331 lb # ft
445
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
5–138. 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
Plastic Torque. When the inner surface of the shaft is about to yield, the shaft is
about to become fully plastic.
T = 2p
L
T
6 in.
t (ksi)
10
tr2dr
3 in.
= 2ptY
L1.5 in.
3
= 2p(10)a
r2dr
0.004
3 in.
r
b2
3 1.5 in.
= 494.80 kip # in. = 41.2 kip # ft
Ans.
Angle of Twist.
f =
gY
0.004
L =
(3)(12) = 0.096 rad
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
manner.
www.elsolucionario.org
trœ = co =
trœ = ci =
494.80(3)
T¿co
=
= 12.44 ksi
p 4
J
A 3 - 1.54 B
2
494.80(1.5)
T¿ci
=
= 6.222 ksi
p 4
J
A 3 - 1.54 B
2
446
g (rad)
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5–138. 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.
Ans:
T = 41.2 kip # ft, (tr)r = co = 2.44 ksi,
(tr)r = ci = -3.78 ksi
447
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␶ (MPa)
5–139. The shear stress–strain diagram for a solid
50-mm-diameter shaft can be approximated as shown in the
figure. Determine the torque required to cause a maximum
shear stress in the shaft of 125 MPa. If the shaft is 3 m long,
what is the corresponding angle of twist?
125
50
0.0025
r
g = gmax
c
0.010
␥ (rad)
gmax = 0.01
When g = 0.0025
r =
=
cg
gmax
0.025(0.0025)
= 0.00625
0.010
50(106)
t - 0
=
r - 0
0.00625
t = 8000(106)(r)
t - 50(106)
125(106) - 50(106)
=
r - 0.00625
0.025 - 0.00625
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t = 4000(106)(r) + 25(106)
c
T = 2p
L0
tr2 dr
0.00625
= 2p
8000(106)r3 dr
L0
0.025
+ 2p
L0.00625
[4000(106)r + 25(106)]r2 dr
T = 3269 N # m = 3.27 kN # m
f =
Ans.
gmax
0.01
L =
(3)
c
0.025
= 1.20 rad = 68.8°
Ans.
Ans:
T = 3.27 kN # m, f = 68.8°
448
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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)
Plastic Torque: The tube is fully plastic if gi Ú gr = 0.003 rad.
g
0.006
=
;
0.03
0.035
210
g = 0.005143 rad
0.003
Therefore the tube is fully plastic.
co
TP = 2p
Lci
tg r2 dr
=
2p tY 3
A co - c3i B
3
=
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
J
(0.035
- 0.034)
2
(tP)o = - tg + tPœ o = - 210 + 225.27 = 15.3 MPa
(tP)i = - tg + tPœ i = - 210 + 193.09 = - 16.9 MPa
449
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)
180
Equation of Equilibrium. Referring 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
(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
0.0024
Steel Alloy
= 75 GPa
t (MPa)
36
0.002
fC = ft
Copper Alloy
TcL
TtL
=
JcGst
JtG q
A
Tc
B
A
p
4
9
2 0.03 (75) 10
B
=
A
Tt
B
A
p
4
4
9
2 0.05 - 0.03 (18) 10
B
Tc = 0.6204Tt
(2)
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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
g (rad)
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.
450
g (rad)
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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.
Ans:
Tt = 7.39 kN # m, Tc = 7.61 kN # m
451
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5–142. The 2-m-long tube is made from an elastic-plastic
material as shown. Determine the applied torque T, which
subjects the material of the tube’s outer edge to a shearing
strain, of gmax = 0.008 rad. What would be the permanent
angle of twist of the tube when the torque is removed?
Sketch the residual stress distribution of the tube.
T
45 mm
40 mm
(MPa)
f =
240
0.008(2)
gmax L
=
c
0.045
0.003
f = 0.3556 rad
However,
f =
gY
L
rY
0.3556 =
0.003
(2)
rY
rY = 0.016875 m 6 0.04 m
Therefore the tube is fully plastic.
Also,
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r
0.008
=
45
40
r = 0.00711 7 0.003
Again, the tube is fully plastic,
co
TP = 2p
tY r2 dr
Lci
=
2ptY 3
(co - c3i )
3
=
2p(240)(106)
(0.0453 - 0.043)
3
= 13634.5 N # m = 13.6 kN # m
Ans.
The torque is removed and the opposite torque of TP = 13634.5 N # m is applied ,
fœ =
TP L
JG
G =
240(106)
= 80 GPa
0.003
13634.5(2)
= p
4
2 (0.045
- 0.044)(80)(106)
= 0.14085 rad
452
(rad)
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5–142. Continued
fr = f - fœ = 0.35555 - 0.14085
Ans.
= 0.215 rad = 12.3°
tpœ o =
13634.5(0.045)
TP c
= 253.5 MPa
= p
6
4
J
2 (0.045 - 0.04 )
tpœ i =
0.04
(253.5) = 225.4 MPa
0.045
Ans:
Tp = 13.6 kN # m, fr = 12.3°
453
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5–143. The shaft is made of A992 steel and has an
allowable shear stress of tallow = 75 MPa. When the shaft is
rotating at 300 rpm, the motor supplies 8 kW of power,
while gears A and B withdraw 5 kW and 3 kW, respectively.
Determine the required minimum diameter of the shaft to
the nearest millimeter. Also, find the rotation of gear A
relative to C.
300 mm
300 mm
C
B
A
Applied Torque: The angular velocity of the shaft is
v = a 300
rev
1 min 2prad
ba
ba
b = 10p rad>s
min
60 s
1 rev
Thus, the torque at C and gear A are
TC =
8(103)
PC
=
= 254.65 N # m
v
10p
TA =
5(103)
PA
=
= 159.15 N # m
v
10p
Internal Loading: The internal torque developed in segment BC and AB of the shaft
are shown in Figs. a and b, respectively.
Allowable Shear Stress: By inspection, segment BC is critical.
tallow =
TBC c
;
J
75(106) =
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254.651d2 2
p d 4
2 12 2
d = 0.02586 m
Use
d = 26 mm
Ans.
Angle of Twist: Using d = 26 mm,
fA>C = ©
Ti Li
TAB LAB
TBC LBC
=
+
Ji Gi
JG
JG
0.3
= p
4
9
2 (0.013 )(75)(10 )
(159.15 + 254.65)
= 0.03689 rad = 2.11°
Ans.
Ans:
Use d = 26 mm, fA>C = 2.11°
454
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*5–144. The shaft is made of A992 steel and has an
allowable shear stress of tallow = 75 MPa. When the shaft is
rotating at 300 rpm, the motor supplies 8 kW of power,
while gears A and B withdraw 5 kW and 3 kW, respectively.
If the angle of twist of gear A relative to C is not allowed to
exceed 0.03 rad, determine the required minimum diameter
of the shaft to the nearest millimeter.
300 mm
300 mm
C
B
A
Applied Torque: The angular velocity of the shaft is
v = a 300
rev
1 min 2prad
ba
ba
b = 10p rad>s
min
60 s
1 rev
Thus, the torque at C and gear A are
TC =
8(103)
PC
=
= 254.65 N # m
v
10p
TA =
5(103)
PA
=
= 159.15 N # m
v
10p
Internal Loading: The internal torque developed in segment BC and AB of the shaft
are shown in Figs. a and b, respectively.
Allowable Shear Stress: By inspection, segment BC is critical.
tallow =
TBC c
;
J
75(103) =
254.651d2 2
p d 4
2 12 2
d = 0.02586
Angle of Twist:
fA>C = ©
TiLi
TAB LAB
TBC LBC
=
+
JiGi
JG
JG
0.3
(159.15 + 254.65)
0.03 = p d
4
9
2 1 2 2 (75)(10 )
d = 0.02738 m = 28 mm (controls)
Ans.
455
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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
We show that two different methods give similar results:
t 5 mm
Shear Stress:
10 kNm
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
Tr
10(103)(0.06)
= 88.27 MPa
= p
4
4
J
2 (0.0625 - 0.0575 )
Applying Eq. 5-18,
tavg =
10(103)
T
= 88.42 MPa
=
2 t Am
2(0.005)(p)(0.062)
Angle of Twist:
Applying Eq. 5-15,
f =
TL
JG
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10(103)(4)
= p
4
4
9
2 (0.0625 - 0.0575 )(75.0)(10 )
= 0.0785 rad = 4.495°
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°
Rounding to three significant figures, we find
t = 88.3 MPa
Ans.
f = 4.50°
Ans.
Ans:
t = 88.3 MPa, f = 4.50°
456
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5–146. A portion of an airplane fuselage can be
approximated by the cross section shown. If the thickness of
its 2014-T6-aluminum skin is 10 mm, determine the maximum
wing torque T that can be applied if tallow = 4 MPa. Also, in a
4-m-long section, determine the angle of twist.
0.75 m
T
2m
0.75 m
tavg =
T
2tAm
4(106) =
T
2(0.01)[(p)(0.75)2 + 2(1.5)]
T = 381.37(103) = 381 kN # m
f =
f =
Ans.
TL
ds
4A2mG L t
381.37(103)(4)
c
4 + 2p(0.75)
d
0.010
4[(p(0.75) + 2(1.5)) 27(10 )]
2
2
9
f = 0.542(10 - 3) rad = 0.0310°
Ans.
Ans:
T = 381 kN # m, f = 0.0310°
457
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
5–147. The material of which each of three shafts is made
has a yield stress of tY and a shear modulus of G.
Determine which shaft geometry will resist the largest
torque without yielding. What percentage of this torque can
be carried by the other two shafts? Assume that each shaft
is made of the same amount of material and that it has the
same cross sectional area A.
A
60
A
A
60
60
For circular shaft:
c = a
A = p c2;
tmax =
Tc
,
J
tY =
3
T =
pc
t =
2 Y
p(A
p)
1
A 2
b
p
Tc
a 4
c
2
3
2
tY
2
1
2
Tcir = 0.2821 A tY
For the square shaft:
A = a2;
tmax =
a = A
4.81 T
;
a3
1
2
tY =
4.81 T
3
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A2
3
T = 0.2079 A 2 tY
For the triangular shaft:
A =
1
(a)(a sin 60°);
2
tmax =
20 T
;
a3
tY =
1
a = 1.5197A2
20 T
3
(1.5197)3A2
3
T = 0.1755A2 tY
The circular shaft will carry the largest torque
Ans.
For the square shaft:
% =
0.2079
(100%) = 73.7%
0.2821
Ans.
For the triangular shaft:
% =
0.1755
(100%) = 62.2 %
0.2821
Ans.
Ans:
The circular shaft will resist the largest torque.
For the square shaft: 73.7%,
For the triangular shaft: 62.2%
458
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*5–148. Segments AB and BC of the assembly are made
from 6061-T6 aluminum and A992 steel, respectively. If
couple forces P = 3 kip are applied to the lever arm,
determine the maximum shear stress developed in each
segment. The assembly is fixed at A and C.
P
4 ft
2.5 ft
A
2.5 ft
E
4 ft
4 in.
B
D
Equilibrium: Referring to the free-body diagram of the assembly shown in Fig. a,
©Mx = 0;
TA + TC - 3(5) = 0
(1)
Compatibility Equation: It is required that
fB>A = fB>C
TA LAB
TC LBC
=
JGal
JGst
TA L
=
3
J(3.7)(10 )
TC L
J(11)(103)
TA = 0.3364 TC
(2)
Solving Eqs. (1) and (2),
TC = 11.224 kip # ft
TA = 3.775 kip # ft
Maximum Shear Stress:
(tmax)AB =
(tmax)BC =
3.775(12)(2)
TA c
=
= 3.60 ksi
p 4
J
(2 )
2
Ans.
11.224(12)(2)
TC c
=
= 10.7 ksi
p 4
J
(2 )
2
Ans.
459
4 in.
P
C
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5–149. Segments AB and BC of the assembly are made
from 6061-T6 aluminum and A992 steel, respectively. If the
allowable shear stress for the aluminum is (tallow)al = 12 ksi
and for the steel (tallow)st = 10 ksi, determine the maximum
allowable couple forces P that can be applied to the lever
arm. The assembly is fixed at A and C.
P
4 ft
2.5 ft
A
2.5 ft
E
4 ft
4 in.
B
D
Equilibrium: Referring to the free-body diagram of the assembly shown in Fig. a,
TA + TC - P(5) = 0
©Mx = 0;
4 in.
C
P
(1)
Compatibility Equation: It is required that
fB>A = fB>C
TCLBC
TA LAB
=
JGal
JGst
TAL
J(3.7)(103)
=
TCL
J(11)(103)
TA = 0.3364 TC
(2)
Solving Eqs. (1) and (2),
TC = 3.7415P
TA = 1.259P
Allowable Shear Stress:
(tallow)AB =
TA c
;
J
12 =
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1.259P(12)(2)
p 4
(2 )
2
P = 9.98 kip
TC c
(tallow)BC =
;
J
3.7415P(12)(2)
10 =
p 4
(2 )
2
P = 2.80 kip (controls)
Ans.
Ans:
P = 2.80 kip
460
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5–150. The tapered shaft is made from 2014-T6 aluminum
alloy, and has a radius which can be described by the
function r = 0.02(1 + x3>2) m, where x is in meters.
Determine the angle of twist of its end A if it is subjected to
a torque of 450 N # m.
x
r = 0.02(1 + x3/2) m
4m
450 N⭈m
x
A
T = 450 N # m
fA =
4
4
450 dx
dx
Tdx
=
= 0.066315
3
3
p
4
4
9
JG
L0 (0.02) (1 + x2) (27)(10 )
L
L0 (1 + x2)4
2
Evaluating the integral numerically, we have
fA = 0.066315 [0.4179] rad
= 0.0277 rad = 1.59°
Ans.
Ans:
fA = 1.59°
461
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5–151. The 60-mm-diameter shaft rotates at 300 rev> min.
This motion is caused by the unequal belt tensions on the
pulley of 800 N and 450 N. Determine the power transmitted
and the maximum shear stress developed in the shaft.
300 rev/min
100 mm
450 N
v = 300
rev 2p rad 1 min
c
d
= 10 p rad>s
min 1 rev 60 s
800 N
T + 450(0.1) - 800(0.1) = 0
T = 35.0 N # m
P = Tv = 35.0(10p) = 1100 W = 1.10 kW
Ans.
35.0(0.03)
Tc
= 825 kPa
= p
4
J
2 (0.03 )
Ans.
tmax =
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Ans:
P = 1.10 kW, tmax = 825 kPa
462
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6–1. The load binder is used to support a load. If the force
applied to the handle is 50 lb, determine the tensions T1 and T2
in each end of the chain and then draw the shear and
moment diagrams for the arm ABC.
T1
A
C
B
50 lb
12 in.
3 in.
T2
a + ©MC = 0;
50(15) - T1(3) = 0
T1 = 250 lb
+ T ©Fy = 0;
Ans.
50 - 250 + T2 = 0
T2 = 200 lb
Ans.
Ans:
T1 = 250 lb, T2 = 200 lb
V (lb)
200
x
⫺50
M (lb⭈in)
x
⫺600
463
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6–2. Draw the shear and moment diagrams for the shaft.
The bearings at A and D exert only vertical reaction on the
shaft. The loading is applied to the pulleys at B and C and E.
14 in.
20 in.
15 in.
12 in.
A
E
B
C
D
35 lb
80 lb
110 lb
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Ans:
V (lb)
82.2
35
2.24
x
⫺50
⫺108
M (lb⭈in)
1151
1196
x
⫺420
464
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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
3 ft
5 ft
B
C
4 ft
a + ©MA = 0;
4
F (3) - 1200(8) = 0;
5 B
+ c ©Fy = 0;
- Ay +
+
; ©Fx = 0;
Ax -
4
(4000) - 1200 = 0;
5
3
(4000) = 0;
5
FB = 4000 lb
Ay = 2000 lb
Ax = 2400 lb
Ans:
V (lb)
1200
x
⫺2000
M (lb⭈in)
x
⫺6000
465
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*6–4. Draw the shear and moment diagrams for the beam.
2 kip
4 ft
2 kip
4 ft
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466
2 kip
4 ft
2 kip
4 ft
4 ft
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6–5. Draw the shear and moment diagrams for the beam.
10 kN
8 kN
15 kN⭈m
2m
3m
Ans:
V (kN)
18
8
x
M (kN⭈m)
x
⫺15
⫺39
⫺75
467
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6–6. Express the internal shear and moment in terms of x
and then draw the shear and moment diagrams.
w0
Support Reactions: Referring to the free-body diagram of the entire beam
shown in Fig. a,
a + ©MA = 0;
+ c ©Fy = 0;
A
B
x
L
2
1
L 5
By(L) - w0 a b a L b = 0
2
2
6
By =
5
wL
24 0
Ay +
5
1
L
w L - w0 a b = 0
24 0
2
2
Ay =
w0L
24
Shear and Moment Function: For 0 … x 6
L
2
L
, we refer to the free-body
2
diagram of the beam segment shown in Fig. b.
w0L
- V = 0
24
+ c ©Fy = 0;
V =
a + ©M = 0; M -
w0L
24
Ans.
w0L
x = 0
24
M =
w0L
x
24
Ans.
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L
6 x … L, we refer to the free-body diagram of the beam segment
2
shown in Fig. c.
For
w0L
1 w0
1
- c (2x - L) d c (2x - L) d - V = 0
24
2 L
2
+ c ©Fy = 0;
V =
a + ©M = 0; M +
Ans:
For 0 … x 6
w0
c L2 - 6(2x - L)2 d
24L
Ans.
M =
w0
L
6 x … L: V =
[L2 -6(2x - L)2],
2
24L
For
w0
1 w0
1
1
c (2x - L) d c (2x - L) d c (2x - L) d x = 0
2 L
2
6
24L
w0
c L2x - (2x - L)3 d
24L
w0L
w0L
L
:V =
,M =
x,
2
24
24
M =
w0
[L2x - (2x - L)3]
24L
V
Ans.
w0L
0.704 L
24
0
When V = 0, the shear function gives
0 = L2 - 6(2x - L)2
L
x
0.5 L
x = 0.7041L
Substituting this result into the moment equation,
⫺
5 wL
0
24
M|x = 0.7041L = 0.0265w0L2
Shear and Moment Diagrams: As shown in Figs. d and e.
M
0.0208 w0L2
0.0265 w0L2
x
0.5 L 0.704 L
468
L
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6–7. Draw the shear and moment diagrams for the
compound beam which is pin connected at B. (This
structure is not fully stable. But with the given loading, it is
balanced and will remain as shown if not disturbed.)
6 kip
8 kip
A
C
B
4 ft
6 ft
4 ft
4 ft
Ans:
V (kip)
4
x
⫺4
⫺6
M (kip⭈ft)
16
6 ft
4 ft
⫺24
469
x
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*6–8. Express the internal shear and moment in terms of x
and then draw the shear and moment diagrams for the beam.
900 lb
400 lb/ft
A
B
x
6 ft
Support Reactions: Referring to the free-body diagram of the entire beam shown
in Fig. a,
a + ©MA = 0;
By(9) - 400(6)(3) - 900(6) = 0
By = 1400 lb
+ c ©Fy = 0;
Ay + 1400 - 400(6) - 900 = 0
Ay = 1900 lb
Shear and Moment Function: For 0 … x 6 6 ft, we refer to the free-body diagram
of the beam segment shown in Fig. b.
+ c ©Fy = 0;
1900 - 400x - V = 0
V = {1900 - 400x} lb
a + ©M = 0;
Ans.
x
M + 400x a b - 1900x = 0
2
M = {1900x - 200x2} lb # ft
Ans.
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For 6 ft 6 x … 9 ft, we refer to the free-body diagram of the beam segment shown
in Fig. c.
+ c ©Fy = 0;
V + 1400 = 0
V = - 1400 lb
a + ©M = 0;
Ans.
1400(9 - x) - M = 0
M = {1400(9 - x)} lb # ft
Ans.
Shear and Moment Diagrams: As shown in Figs. d and e.
470
3 ft
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6–9. Express the internal shear and moment in terms of x
and then draw the shear and moment diagrams for the
overhanging beam.
6 kN/m
A
B
x
2m
4m
Support Reactions: Referring to the free-body diagram of the entire beam shown
in Fig. a,
a + ©MA = 0;
By(4) - 6(6)(3) = 0
By = 27 kN
+ c ©Fy = 0;
Ay + 27 - 6(6) = 0
Ay = 9 kN
Shear and Moment Function: For 0 … x 6 4 m, we refer to the free-body diagram
of the beam segment shown in Fig. b.
+ c ©Fy = 0;
9 - 6x - V = 0
V = {9 - 6x} kN
a + ©M = 0;
Ans.
x
M + 6x a b - 9x = 0
2
M = {9x - 3x2} kN # m
Ans.
For 4 m 6 x … 6 m, we refer to the free-body diagram of the beam segment shown
in Fig. c.
+ c ©Fy = 0;
V - 6(6 - x) = 0
V = {6(6 - x)} kN
a + ©M = 0;
- M - 6(6 - x) a
Ans.
6 - x
b = 0
2
Ans.
M = - {3(6 - x)2} kN # m
Ans:
For 0 … x 6 4 m: V = { 9 - 6x} kN,
M = {9x - 3x2} kN # m,
Shear and Moment Diagrams: As shown in Figs. d and e.
For 4 m 6 x … 6 m: V = { 6(6 - x)} kN # m,
M = -{3(6 - x)2} kN # m
V (kN)
12
9
0
x (m)
1.5
4
6
⫺15
M (kN⭈m)
6.75
0
4
1.5
⫺12
471
6
x (m)
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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.
P ⫽ 150 lb
C
B
A
1.5 ft
1.5 ft
1.5 ft
D
Equations of Equilibrium: Referring to the free-body diagram of the frame shown
in Fig. a,
+ c ©Fy = 0;
Ay - 150 = 0
Ay = 150 lb
a + ©MA = 0;
ND(1.5) - 150(3) = 0
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.
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Ans:
V (lb)
150
0
x (ft)
1.5
3
M (lb⭈ft)
225
0
1.5
⫺225
472
x (ft)
3
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
6–11. Draw the shear and moment diagrams for the pipe.
The end screw is subjected to a horizontal force of 5 kN.
Hint: The reactions at the pin C must be replaced by an
equivalent loading at point B on the axis of the pipe.
C
A
80 mm
5 kN
B
400 mm
Ans:
V (kN)
x
⫺1
M (kN⭈m)
x
⫺0.4
473
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*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
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474
B
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6–13. Draw the shear and moment diagrams for the rod. It
is supported by a pin at A and a smooth plate at B. The plate
slides within the groove and so it cannot support a vertical
force, although it can support a moment.
15 kN
A
B
4m
2m
Ans:
V (kN)
15
x
M (kN⭈m)
60
x
⫺30
475
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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 lb>in. and support the load of
40 lb at C.
4 in.
A
10 in.
B
50 in.
C
120⬚
D
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Ans:
V (lb)
115
9
40
x
⫺6
⫺378.8
⫺393.8
M (lb⭈in)
x
⫺12
⫺3875
476
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*6–16. 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.
P
L
–
2
A
B
a
Support Reactions: As shown on FBD.
Absolute Minimum Moment: In order to get the absolute minimum moment, the
maximum positive and maximum negative moment must be equal that is
Mmax( + ) = Mmax( - ).
For the positive moment:
a + ©MNA = 0;
Mmax( + ) - a 2P -
3PL L
ba b = 0
2a
2
Mmax( + ) = PL -
3PL2
4a
For the negative moment:
a + ©MNA = 0;
Mmax( - ) - P(L - a) = 0
Mmax( - ) = P(L - a)
Mmax( + ) = Mmax( - )
PL -
3PL2
4a
= P(L - a)
4a L - 3L2 = 4a L - 4a2
a =
P
L
–
2
23
L = 0.866L
2
Ans.
Shear and Moment Diagram:
477
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6–17. Express the internal shear and moment in the
cantilevered beam as a function of x and then draw the
shear and moment diagrams.
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) Ans.
x
1
(33.33x)(x) a b + 300x = 0 M = {-300x - 5.556x3} lb # ft (2) Ans.
2
3
The shear and moment diagrams shown in Figs. c and d are plotted using Eqs. (1)
and (2), respectively.
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Ans:
V = { -300 - 16.67x2} lb,
M = {-300x - 5.556x3} lb # ft
V (lb)
x
⫺300
⫺900
M (lb⭈ft)
x
⫺3000
478
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
6–18. Draw the shear and moment diagrams for the beam,
and determine the shear and moment throughout the beam
as functions of x.
10 kip
2 kip/ft
8 kip
40 kip⭈ft
x
6 ft
Support Reactions: As shown on FBD.
4 ft
Shear and Moment Function:
For 0 … x 6 6 ft:
+ c ©Fy = 0;
30.0 - 2x - V = 0
V = {30.0 - 2x} kip
Ans.
x
a + ©MNA = 0; M + 216 + 2x a 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.
Ans:
For 0 … x 6 6 ft: V = {30.0 - 2x} kip,
M = { -x2 + 30.0x - 216} kip # ft,
For 6 ft 6 x … 10 ft: V = 8.00 kip,
M = {8.00x - 120} kip # ft
V (kip)
30.0
M (kip⭈ft)
18.0
8.00
x (ft)
0
6
0
10
⫺72.0
⫺216
479
6
10
x (ft)
⫺40.0
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6–19. Draw the shear and moment diagrams for the beam.
2 kip/ft
30 kip⭈ft
B
A
5 ft
5 ft
5 ft
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Ans:
V (kip)
x
⫺0.5
⫺10
M (kip⭈ft)
2.5
x
⫺25
480
⫺27.5
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*6–20. Draw the shear and moment diagrams for the
overhanging beam.
3 kip/ft
A
Support Reactions: Referring to the free-body diagram of the beam shown in Fig. a,
a + ©MA = 0;
+ c ©Fy = 0;
1
(3)(12)(8) = 0
2
By = 8 kip
By(18) -
1
(3)(12) = 0
2
Ay = 10 kip
Ay + 8 -
Shear and Moment Functions: For 0 … x 6 12 ft, we refer to the free-body diagram
of the beam segment shown in Fig. b.
+ c ©Fy = 0;
10 -
1 1
a x b(x) - V = 0
2 4
V = e 10 a + ©M = 0; M +
1 2
x f kip
8
Ans.
1 1
x
a x b (x)a b - 10x = 0
2 4
3
M = e 10x -
1 3
x f kip # ft
24
Ans.
When V = 0, from the shear function,
0 = 10 -
1 2
x
8
x = 8.944 ft
Substituting this result into the moment function,
M|x = 8.944 ft = 59.6 kip # ft
For 12 ft 6 x … 18 ft, we refer to the free-body diagram of the beam segment
shown in Fig. c.
+ c ©Fy = 0;
V + 8 = 0
V = - 8 kip
Ans.
a + ©M = 0; 8(18 - x) - M = 0
M = {8(18 - x)} kip # ft
Ans.
Shear and Moment Diagrams: As shown in Figs. d and e.
481
B
12 ft
6 ft
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
6–21. The 150-lb man sits in the center of the boat, which
has a uniform width and a weight per linear foot of 3 lb>ft.
Determine the maximum bending moment exerted on the
boat. Assume that the water exerts a uniform distributed
load upward on the bottom of the boat.
7.5 ft
Mmax = 281 lb # ft
7.5 ft
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Ans.
Ans:
Mmax = 281 lb # ft
V (lb)
75
x
⫺75
M (lb⭈ft)
281
0
482
x
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
6–22. Draw the shear and moment diagrams for the
overhang beam.
4 kN/ m
A
B
3m
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 x b (x) - V = 0
2 3
2
V = e - x2 - 4 f kN
3
1 4
x
a x b (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
V冷x= 3 m - = - (32) - 4 = - 10 kN
3
V冷x=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
M冷x= 3 m = - (33) - 4(3) = - 18 kN # m
9
or
M冷x= 3 m = - 2(6 - 3)2 = - 18 kN # m
Ans:
V (kN)
12
3
⫺4
x (m)
6
⫺10
M (kN⭈m)
3
⫺18
483
6
x (m)
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
6–23. The footing supports the load transmitted by the two
columns. Draw the shear and moment diagrams for the
footing if the reaction of soil pressure on the footing is
assumed to be uniform.
14 kip
14 kip
6 ft
12 ft
6 ft
7
7
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Ans:
V (kip)
x
⫺7
⫺7
M (kip⭈ft)
21
21
x
484
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*6–24. Express the shear and moment in terms of x and
then draw the shear and moment diagrams for the simply
supported beam.
300 N/m
A
B
3m
Support Reactions: Referring to the free-body diagram of the entire beam shown
in Fig. a,
a + ©MA = 0;
By(4.5) -
1
1
(300)(3)(2) - (300)(1.5)(3.5) = 0
2
2
By = 375 N
+ c ©Fy = 0;
1
1
(300)(3) - (300)(1.5) = 0
2
2
Ay + 375 Ay = 300 N
Shear and Moment Function: For 0 … x 6 3 m, we refer to the free-body diagram
of the beam segment shown in Fig. b.
+ c ©Fy = 0;
1
(100x)x - V = 0
2
V = {300 - 50x2} N
300 -
a + ©M = 0; M +
Ans.
1
x
(100x)x a b - 300x = 0
2
3
M = e 300x -
50 3
x fN#m
3
Ans.
When V = 0, from the shear function,
0 = 300 - 50x2
x = 26 m
Substituting this result into the moment equation,
M|x = 26 m = 489.90 N # m
For 3 m 6 x … 4.5 m, we refer to the free-body diagram of the beam segment
shown in Fig. c.
+ c ©Fy = 0;
V + 375 -
1
3200(4.5 - x)4(4.5 - x) = 0
2
V = e 100(4.5 - x)2 - 375 f N
Ans.
1
4.5 - x
[200(4.5 - x)](4.5 - x) a
b - M = 0
2
3
100
(4.5 - x)3 f N # m
M = e 375(4.5 - x) Ans.
3
a + ©M = 0; 375(4.5 - x) -
Shear and Moment Diagrams: As shown in Figs. d and e.
485
1.5 m
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6–25. Draw the shear and moment diagrams for the beam
and determine the shear and moment in the beam as
functions of x, where 4 ft < x < 10 ft.
150 lb/ft
200 lb⭈ft
200 lb⭈ft
A
B
x
4 ft
+ c ©Fy = 0;
4 ft
- 150(x - 4) - V + 450 = 0
V = 1050 - 150x
a + ©M = 0;
6 ft
-200 - 150(x - 4)
Ans.
(x - 4)
- M + 450(x - 4) = 0
2
M = - 75x2 + 1050x - 3200
Ans.
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Ans:
V = 1050 - 150x
M = - 75x2 + 1050x - 3200
V (lb)
450
x
⫺450
M (lb⭈ft)
475
x
⫺200
486
⫺200
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
6–27. Draw the shear and moment diagrams for the beam.
w0
B
L
3
+ c ©Fy = 0;
A
2L
3
w0L
1 w0x
- a
b(x) = 0
4
2
L
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
Ans:
V
7w0L
36
x
⫺w0L
18
⫺w0L
4
0.707 L
M
0.0345w0L2
x
⫺0.00617w0L2
487
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*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
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488
L
3
L
3
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6–29. Draw the shear and moment diagrams for the double
overhanging beam.
8 kN/m
B
A
1.5 m
1.5 m
3m
Equations of Equilibrium: Referring to the free-body diagram shown in Fig. a,
a + ©MA = 0;
By(3) - 8(6)(1.5) = 0
By = 24 kN
+ c ©Fy = 0;
Ay + 24 - 8(6) = 0
Ay = 24 kN
Shear and Moment Diagram: As shown in Figs. b and c.
Ans:
V (kN)
12
0
1.5
12
4.5
3
⫺12
x (m)
6
⫺12
M (kN⭈m)
0
1.5
⫺9
489
3
4.5
⫺9
6
x (m)
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
6–30. The beam is bolted or pinned at A and rests on a
bearing pad at B that exerts a uniform distributed loading
on the beam over its 2-ft length. Draw the shear and
moment diagrams for the beam if it supports a uniform
loading of 2 kip>ft.
2 kip/ft
B
A
8 ft
1 ft
2 ft
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Ans:
V (kip)
8
x
⫺8
M (kip⭈ft)
24
8
8
x
490
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
6–31. The support at A allows the beam to slide freely along
the vertical guide so that it cannot support a vertical force.
Draw the shear and moment diagrams for the beam.
w
B
A
L
Equations of Equilibrium: Referring to the free-body diagram of the beam shown
in Fig. a,
a + ©MB = 0;
wL a
L
b - MA = 0
2
MA =
+ c ©Fy = 0;
wL2
2
By - wL = 0
By = wL
Shear and Moment Diagram: As shown in Figs. b and c.
Ans:
V
L
⫺wL
M
wL2
2
L
491
x
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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 kN>m 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
B
w0
w0
20 mm 60 mm 20 mm
+ c ©Fy = 0;
1
2(w0)(20)a b - 60(0.4) = 0
2
w0 = 1.2 kN>m
Ans.
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492
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
6–33. The shaft is supported by a smooth thrust bearing at
A and smooth journal bearing at B. Draw the shear and
moment diagrams for the shaft.
400 N⭈m
B
A
1m
1m
1m
900 N
Equations of Equilibrium: Referring to the free-body diagram of the shaft shown
in Fig. a,
a + ©MA = 0;
By(2) + 400 - 900(1) = 0
By = 250 N
+ c ©Fy = 0;
Ay + 250 - 900 = 0
Ay = 650 N
Shear and Moment Diagram: As shown in Figs. b and c.
Ans:
V (N)
650
1
0
2
3
x (m)
⫺250
M (N⭈m)
650
400
0
493
x (m)
1
2
3
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
6–34. Draw the shear and moment diagrams for the
cantilever beam.
2 kN
A
3 kN⭈m
1.5 m
1.5 m
Equations of Equilibrium: Referring to the free-body diagram of the beam shown in
Fig. a,
+ c ©Fy = 0;
Ay - 2 = 0
Ay = 2 kN
a + ©MA = 0;
MA - 3 - 2(3) = 0
MA = 9 kN # m
Shear and Moment Diagram: As shown in Figs. b and c.
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Ans:
V (kN)
2
0
x (m)
1.5
3
1.5
3
M (kN⭈m)
0
⫺3
⫺6
⫺9
494
x (m)
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
6–35. Draw the shear and moment diagrams for the beam
and determine the shear and moment as functions of x.
400 N/m
200 N/ m
A
B
x
3m
3m
Support Reactions: As shown on FBD.
Shear and Moment Functions:
For 0 … x 6 3 m:
200 - V = 0
+ c ©Fy = 0;
a + ©MNA = 0;
V = 200 N
Ans.
M - 200 x = 0
M = 5200 x6 N # m
Ans.
For 3 m 6 x … 6 m:
+ c ©Fy = 0;
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-
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
Ans:
For 0 … x 6 3 m: V = 200 N, M = (200x) N # m,
100 2
For 3 m 6 x … 6 m: V = e x + 500 f N,
3
100 3
x + 500x - 600 f N # m
M = e9
V (N)
200
0
3.87
6
x (m)
3
⫺700
M (N⭈m)
600 691
x (m)
0
3 3.87
495
6
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*6–36. The shaft is supported by a smooth thrust bearing at
A and a smooth journal bearing at B. Draw the shear and
moment diagrams for the shaft.
600 N⭈m
B
A
0.8 m
0.8 m
0.8 m
900 N
Equations of Equilibrium: Referring to the free-body diagram of the shaft shown in
Fig. a,
a + ©MA = 0;
By(1.6) - 600 - 900(2.4) = 0
By = 1725 N
+ c ©Fy = 0;
1725 - 900 - Ay = 0
Ay = 825 N
Shear and Moment Diagram: As shown in Figs. b and c.
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496
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
6–37. Draw the shear and moment diagrams for the beam.
50 kN/m
50 kN/m
B
A
4.5 m
4.5 m
Ans:
V (kN)
112.5
x
⫺112.5
M (kN⭈m)
169
x
497
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0.5 m
6–38. The beam is used to support a uniform load along CD
due to the 6-kN weight of the crate. If the reaction at
bearing support B can be assumed uniformly distributed
along its width, draw the shear and moment diagrams for
the beam.
0.75 m
2.75 m
2m
C
D
A
B
Equations of Equilibrium: Referring to the free-body diagram of the beam
shown in Fig. a,
a + ©MA = 0;
FB(3) - 6(5) = 0
FB = 10 kN
+ c ©Fy = 0;
10 - 6 - Ay = 0
Ay = 4 kN
Shear and Moment Diagram: The intensity of the distributed load at
FB
10
=
support B and portion CD of the beam are wB =
=
0.5
0.5
6
20 kN>m and wCD =
= 3 kN>m, Fig. b. The shear
2
and moment diagrams are shown in Figs. c and d.
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Ans:
V (kN)
2.95
2.75
6
x (m)
0
⫺4
3.25 4
6
3.25
4
6
M (kN⭈m)
2.95
2.75
0
⫺6
⫺11
498
⫺10.5
⫺11.4
x (m)
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
6–39. Draw the shear and moment diagrams for the double
overhanging beam.
400 lb
400 lb
200 lb/ft
A
B
3 ft
6 ft
3 ft
Equations of Equilibrium: Referring to the free-body diagram of the beam shown in
Fig. a,
a + ©MA = 0;
By(6) + 400(3) - 200(6)(3) - 400(9) = 0
By = 1000 lb
+ c ©Fy = 0;
Ay + 1000 - 400 - 200(6) - 400 = 0
Ay = 1000 lb
Shear and Moment Diagram: As shown in Figs. b and c.
Ans:
V (lb)
600
0
3
⫺400
400
9
6
x (ft)
12
⫺600
M (lb⭈ft)
0
3
6
9
⫺300
⫺1200
499
⫺1200
12
x (ft)
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*6–40. Draw the shear and moment diagrams for the
simply supported beam.
10 kN
10 kN
15 kN⭈m
A
B
2m
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500
2m
2m
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6–41. The compound beam is fixed at A, pin connected at
B, and supported by a roller at C. Draw the shear and
moment diagrams for the beam.
600 N
400 N/m
A
C
B
2m
2m
2m
Support Reactions: Referring to the free-body diagram of segment BC shown
in Fig. a,
a +©MB = 0;
Cy(2) - 400(2)(1) = 0
Cy = 400 N
+ c ©Fy = 0;
By + 400 - 400(2) = 0
By = 400 N
Using the result of By and referring to the free-body diagram of segment AB, Fig. b,
+ c ©Fy = 0;
Ay - 600 - 400 = 0
Ay = 1000 N
a +©MA = 0;
MA - 600(2) - 400(4) = 0
MA = 2800 N
Shear and Moment Diagrams: As shown in Figs. c and d.
Ans:
V (N)
1000
400
0
5
2
6
4
x (m)
⫺400
M (N⭈m)
0
4
⫺800
⫺2800
501
200
2
5
6
x (m)
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
5 kN/m
6–42. Draw the shear and moment diagrams for the
compound beam.
A
B
2m
D
C
1m
1m
Support Reactions:
From the FBD of segment AB
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
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
+
: ©Fx = 0;
Bx = 0
From the FBD of segment AB
+
: ©Fx = 0;
Shear and Moment Diagram:
Ax = 0
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Ans:
V (kN)
5.00
0
10.0
2
1
⫺5.00
3
5.00
x (m)
4
⫺10.0
M (kN⭈m)
2.50
0
3
1
4
2
⫺7.50
502
x (m)
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
2 kN
6–43. The compound beam is fixed at A, pin connected at
B, and supported by a roller at C. Draw the shear and
moment diagrams for the beam.
3 kN/m
C
B
A
3m
3m
Support Reactions: Referring to the free-body diagram of segment BC shown
in Fig. a,
a +©MB = 0;
Cy(3) - 3(3)(1.5) = 0
Cy = 4.5 kN
c ©Fy = 0;
By + 4.5 - 3(3) = 0
By = 4.5 kN
Using the result of By and referring to the free-body diagram of segment AB, Fig. b,
c ©Fy = 0;
Ay - 2 - 4.5 = 0
Ay = 6.5 kN
a +©MA = 0;
MA - 2(3) - 4.5(3) = 0
MA = 19.5 kN # m
Shear and Moment Diagrams: As shown in Figs. c and d.
Ans:
V (kN)
6.5
0
4.5
6
3
x (m)
4.5
⫺4.5
M (kN⭈m)
3.375
0
⫺19.5
503
x (m)
3
4.5
6
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*6–44. Draw the shear and moment diagrams for the beam.
w
8 kip/ft
1
w = – x2
8
B
A
8 ft
8
FR =
1
2
x dx = 21.33 kip
8 L0
8
1
8
x =
3
x dx
L0
= 6.0 ft
21.33
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504
x
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6–45. A short link at B is used to connect beams AB and
BC to form the compound beam shown. Draw the shear and
moment diagrams for the beam if the supports at A and B
are considered fixed and pinned, respectively.
15 kN
3 kN/m
B
A
4.5 m
C
1.5 m
1.5 m
Support Reactions: Referring to the free-body diagram of segment BC shown
in Fig. a,
a +©MC = 0;
15(1.5) - FB(3) = 0
FB = 7.5 kN
+ c ©Fy = 0;
Cy + 7.5 - 15 = 0
Cy = 7.5 kN
Using the result of FB and referring to the free-body diagram of segment AB, Fig. b,
+ c ©Fy = 0;
a +©MA = 0;
1
(3)(4.5) - 7.5 = 0
2
Ay = 14.25 kN
Ay -
1
(3)(4.5)(3) - 7.5(4.5) = 0
2
MA = 54 kN # m
MA -
Shear and Moment Diagrams: As shown in Figs. c and d.
Ans:
V (kN)
14.25
0
7.5
6
7.5
x (m)
4.5
⫺7.5
M (kN⭈m)
11.25
0
⫺54
505
x (m)
4.5
6
7.5
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
6–46. Determine the placement b of the hooks to minimize
the largest moment when the concrete member is being
hoisted. Draw the shear and moment diagrams. The
member has a square cross section of dimension a on each
side. The specific weight of concrete is g.
60⬚
60⬚
Support Reactions: The intensity of the uniform distributed load caused by its own
weight is w = ga2. Due to symmetry,
b
2Fy - ga2L = 0
+ c ©Fy = 0;
Fy =
ga2L
2
Absolute Minimum Moment: To obtain the absolute minimum moment, the
maximum positive moment must be equal to the maximum negative moment. The
maximum negative moment occurs at the supports. Referring to the free-body
diagram of the beam segment shown in Fig. b,
b
ga2b a b - Mmax( - ) = 0
2
a +©M = 0;
Mmax( - ) =
ga2b2
2
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The maximum positive moment occurs between the supports. Referring to the
free-body diagram of the beam segment shown in Fig. c,
ga2L
- ga2x = 0
2
+ c ©Fy = 0;
x =
L
2
Using this result,
a+ ©M = 0; Mmax( + ) + ga2 a
ga2L L
L L
ba b a - bb = 0
2
4
2
2
Mmax( + ) =
b
L
ga2L
(L - 4b)
8
It is required that
Mmax( + ) = Mmax( - )
ga2L
ga2b2
(L - 4b) =
8
2
4b2 + 4Lb - L2 = 0
Solving for the positive result,
b = 0.2071L = 0.207L
Ans.
Shear and Moment Diagrams: Using the result for b, Fig. d, the shear and moment
diagrams are shown in Figs. e and f.
506
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6–46. Continued
Ans:
b = 0.207L
V
0.293ga2L
0.793L
0.5L
0
0.207L
⫺0.207ga2L
M
0
x
L
⫺0.293ga2L
0.0214ga2L2
0.207L
0.793L
0.5L
⫺0.0214ga2L2
507
0.207ga2L
x
L
⫺0.0214ga2L2
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
6–47. If the A-36 steel sheet roll is supported as shown and
the allowable bending stress is 165 MPa, determine the
smallest radius r of the spool if the steel sheet has a width of
1 m and a thickness of 1.5 mm. Also, find the corresponding
maximum internal moment developed in the sheet.
r
Bending Stress-Curvature Relation:
sallow =
Ec
;
r
165(106) =
200(109)30.75(10 - 3)4
r
r = 0.9091 m = 909 mm
Ans.
Moment Curvature Relation:
1
M
=
;
r
EI
1
=
0.9091
M
200(109) c
1
(1)(0.00153) d
12
M = 61.875 N # m = 61.9 N # m
Ans.
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Ans:
r = 909 mm, M = 61.9 N # m
508
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*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 =
=
INA =
~
© 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)
1
(4)(0.53) + 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)(103) + 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.
509
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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
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)(0.53) + 4(0.5)(3.40 - 0.25)2
12
1
+ 2 c (0.5)(33) + 0.5(3)(3.40 - 2)2 d
12
1
+
(0.5)(103) + 0.5(10)(5.5 - 3.40)2
12
= 91.73 in4
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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.
Ans:
(st)max = 3.72 ksi, (sc)max = 1.78 ksi
510
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6–50. 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
Ans.
(controls)
Ans:
M = 2.50 kip # ft
511
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
6–51. A member has the triangular cross section shown. If a
moment of M = 800 lb # ft is applied to the cross section,
determine the maximum tensile and compressive bending
stresses in the member. Also, sketch a three-dimensional
view of the stress distribution action over the cross section.
4 in.
4 in.
M
2 in.
2 in.
h = 242 - 22 = 3.4641 in.
Ix =
1
(4)(3.4641)3 = 4.6188 in4
36
c =
2
(3.4641) = 2.3094 in.
3
y =
1
(3.4641) = 1.1547 in.
3
(smax)t =
800(12)(1.1547)
My
=
= 2.40 ksi
I
4.6188
Ans.
(smax)c =
800(12)(2.3094)
Mc
=
= 4.80 ksi
I
4.6188
Ans.
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Ans:
(smax)t = 2.40 ksi, (smax)t = 4.80 ksi
512
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*6–52. If the beam is subjected to an internal moment of
M = 30 kN # m, determine the maximum bending stress in
the beam. The beam is made from A992 steel. Sketch the
bending stress distribution on the cross section.
50 mm
50 mm
15 mm
A
10 mm
M
Section Properties: The moment of inertia of the cross section about the neutral
axis is
I =
1
1
(0.1)(0.153) (0.09)(0.123) = 15.165(10 - 6) m4
12
12
Maximum Bending Stress: The maximum bending stress occurs at the top and
bottom surfaces of the beam since they are located at the furthest distance from the
neutral axis. Thus, c = 75 mm = 0.075 m.
smax =
30(103)(0.075)
Mc
= 148 MPa
=
I
15.165(10 - 6)
Ans.
At y = 60 mm = 0.06 m,
s|y = 0.06 m =
My
30(103)(0.06)
= 119 MPa
=
I
15.165(10 - 6)
The bending stress distribution across the cross section is shown in Fig. a.
513
150 mm
15 mm
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
6–53. If the beam is subjected to an internal moment of
M = 30 kN # m, determine the resultant force caused by the
bending stress distribution acting on the top flange A.
50 mm
50 mm
15 mm
A
10 mm
M
150 mm
15 mm
Section Properties: The moment of inertia of the cross section about the neutral
axis is
1
1
(0.1)(0.153) (0.09)(0.123) = 15.165(10 - 6) m4
12
12
I =
Bending Stress: The distance from the neutral axis to the top and bottom surfaces of
flange A is yt = 75 mm = 0.075 m and yb = 60 mm = 0.06 m.
st =
Myt
30(103)(0.075)
= 148.37 = 148 MPa
=
I
15.165(10 - 6)
sb =
Myb
30(103)(0.06)
= 118.69 = 119 MPa
=
I
15.165(10 - 6)
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Resultant Force: The resultant force acting on flange A is equal to the volume of the
trapezoidal stress block shown in Fig. a. Thus,
FR =
1
(148.37 + 118.69)(106)(0.1)(0.015)
2
= 200 296.74 N = 200 kN
Ans.
Ans:
FR = 200 kN
514
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
6–54. If the built-up beam is subjected to an internal
moment of M = 75 kN # m, determine the maximum tensile
and compressive stress acting in the beam.
150 mm
20 mm
150 mm
10 mm
150 mm
M
10 mm
300 mm
A
Section Properties: The neutral axis passes through centroid C of the cross section
as shown in Fig. a. The location of C is
y =
0.15(0.3)(0.02) + 230.225(0.15)(0.01)] + 230.295(0.01)(0.14)4
©y~A
=
= 0.2035 m
©A
0.3(0.02) + 2(0.15)(0.01) + 2(0.01)(0.14)
Thus, the moment of inertia of the cross section about the neutral axis is
I = ©I + Ad2
=
1
(0.02)(0.33) + 0.02(0.3)(0.2035 - 0.15)2
12
1
+ 2 c (0.01)(0.153) + 0.01(0.15)(0.225 - 0.2035)2 d
12
+ 2c
1
(0.14)(0.013) + 0.14(0.01)(0.295 - 0.2035)2 d
12
= 92.6509(10 - 6) m4
Maximum Bending Stress: The maximum compressive and tensile stress occurs at
the top and bottom-most fiber of the cross section.
(smax)c =
My
75(103)(0.3 - 0.2035)
= 78.1 MPa
=
I
92.6509(10 - 6)
Ans.
(smax)t =
75(103)(0.2035)
Mc
= 165 MPa
=
I
92.6509(10 - 6)
Ans.
Ans:
(smax)c = 78.1 MPa, (smax)t = 165 MPa
515
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6–55. If the built-up beam is subjected to an internal moment of
M = 75 kN # m, determine the amount of this internal moment
resisted by plate A.
150 mm
20 mm
150 mm
10 mm
Section Properties: The neutral axis passes through centroid C of the cross section
as shown in Fig. a. The location of C is
y =
0.15(0.3)(0.02) + 230.225(0.15)(0.01)] + 230.295(0.01)(0.14)4
©y~A
=
= 0.2035 m
©A
0.3(0.02) + 2(0.15)(0.01) + 2(0.01)(0.14)
150 mm
M
10 mm
300 mm
A
Thus, the moment of inertia of the cross section about the neutral axis is
I = I + Ad2
1
(0.02)(0.33) + 0.02(0.3)(0.2035 - 0.15)2
=
12
1
+ 2 c (0.01)(0.153) + 0.01(0.15)(0.225 - 0.2035)2 d
12
+ 2c
1
(0.14)(0.013) + 0.14(0.01)(0.295 - 0.2035)2 d
12
= 92.6509(10 - 6) m4
Bending Stress: The distance from the neutral axis to the top and
bottom of plate A is yt = 0.3 - 0.2035 = 0.0965 m and
yb = 0.2035 m.
st =
Myt
75(103)(0.0965)
= 78.14 MPa (C)
=
I
92.6509(10 - 6)
sb =
Myb
75(103)(0.2035)
= 164.71 MPa (T)
=
I
92.6509(10 - 6)
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The bending stress distribution across the cross section of
plate A is shown in Fig. b. The resultant forces of the tensile
and compressive triangular stress blocks are
1
(164.71)(106)(0.2035)(0.02) = 335 144.46 N
2
1
(FR)c = (78.14)(106)(0.0965)(0.02) = 75 421.50 N
2
(FR)t =
Thus, the amount of internal moment resisted by plate A is
2
2
M = 335144.46 c (0.2035) d + 75421.50 c (0.0965) d
3
3
= 50315.65 N # m = 50.3 kN # m
Ans.
Ans:
M = 50.3 kN # m
516
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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
Section Property:
B
20 mm
1
1
I =
(0.02)(0.223) +
(0.1)(0.023) = 17.8133(10 - 6) 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)
50 mm
50 mm
My
I
= 49.4 MPa (C)
Ans.
= 4.49 MPa (T)
Ans.
517
M ⫽ 8 kN⭈m
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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
Section Property:
50 mm
1
1
I =
(0.02)(0.223) +
(0.1)(0.023) = 17.8133(10 - 6) m4
12
12
Bending Stress: Applying the flexure formula smax =
smax =
M ⫽ 8 kN⭈m
8(103)(0.11)
17.8133(10 - 6)
sy = 0.01m =
My
Mc
and s =
,
I
I
Ans.
= 49.4 MPa
8(103)(0.01)
17.8133(10 - 6)
50 mm
= 4.49 MPa
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Ans:
smax = 49.4 MPa
518
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6–58. The aluminum machine part is subjected to a moment
of M = 75 kN # m. Determine the bending stress created at
points B and C on the cross section. Sketch the results on a
volume element located at each of these points.
20 mm
20 mm
10 mm
10 mm
10 mm
10 mm
A
B
10 mm
N
C
M ⫽ 75 N⭈m
40 mm
y =
I =
0.005(0.08)(0.01) + 230.03(0.04)(0.01)4
0.08(0.01) + 2(0.04)(0.01)
= 0.0175 m
1
(0.08)(0.013) + 0.08(0.01)(0.01252)
12
+ 2c
1
(0.01)(0.043) + 0.01(0.04)(0.01252)d = 0.3633(10 - 5) m4
12
sB =
75(0.0175)
Mc
= 3.61 MPa
=
I
0.3633(10 - 6)
Ans.
sC =
75(0.0175 - 0.01)
My
= 1.55 MPa
=
I
0.3633(10 - 6)
Ans.
Ans:
sB = 3.61 MPa, sC = 1.55 MPa
519
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
6–59. The aluminum machine part is subjected to a moment
of M = 75 kN # m . Determine the maximum tensile and
compressive bending stresses in the part.
20 mm
10 mm
10 mm
A
B
y =
0.005(0.08)(0.01) + 230.03(0.04)(0.01)4
20 mm
10 mm
10 mm
10 mm
= 0.0175 m
0.08(0.01) + 2(0.04)(0.01)
1
I =
(0.08)(0.013) + 0.08(0.01)(0.01252)
12
1
+ 2 c (0.01)(0.043) + 0.01(0.04)(0.01252)d = 0.3633(10 - 5) m4
12
N
C
M ⫽ 75 N⭈m
40 mm
(smax)t =
75(0.050 - 0.0175)
Mc
= 6.71 MPa
=
I
0.3633(10 - 6)
Ans.
(smax)c =
75(0.0175)
My
= 3.61 MPa
=
I
0.3633(10 - 6)
Ans.
www.elsolucionario.org
Ans:
(smax)t = 6.71 MPa, (smax)c = 3.61 MPa
520
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*6–60. The beam is subjected to a moment of 15 kip # ft.
Determine the resultant force the bending stress produces
on the top flange A and bottom flange B. Also compute the
maximum bending stress developed in the beam.
1 in.
5 in.
8 in.
A
M ⫽ 15 kip⭈ft
1 in.
1 in.
3 in.
0.5(1)(5) + 5(8)(1) + 9.5(3)(1)
©y~A
y =
=
= 4.4375 in.
©A
1(5) + 8(1) + 3(1)
I =
1
1
(5)(13) + 5(1)(4.4375 - 0.5)2 +
(1)(83) + 8(1)(5 - 4.4375)2
12
12
1
+
(3)(13) + 3(1)(9.5 - 4.4375)2
12
= 200.27 in4
Using flexure formula s =
My
I
sA =
15(12)(4.4375 - 1)
= 3.0896 ksi
200.27
sB =
15(12)(4.4375)
= 3.9883 ksi
200.27
sC =
15(12)(9 - 4.4375)
= 4.1007 ksi
200.27
smax =
D
15(12)(10 - 4.4375)
= 4.9995 ksi = 5.00 ksi (Max)
200.27
Ans.
FA =
1
(3.0896 + 3.9883)(1)(5) = 17.7 kip
2
Ans.
FB =
1
(4.9995 + 4.1007)(1)(3) = 13.7 kip
2
Ans.
521
B
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6–61. The beam is subjected to a moment of 15 kip # ft.
Determine the percentage of this moment that is resisted by
the web D of the beam.
1 in.
5 in.
8 in.
A
M ⫽ 15 kip⭈ft
1 in.
1 in.
0.5(1)(5) + 5(8)(1) + 9.5(3)(1)
©y~A
=
= 4.4375 in.
©A
1(5) + 8(1) + 3(1)
1
1
I =
(5)(13) + 5(1)(4.4375 - 0.5)2 +
(1)(83) + 8(1)(5 - 4.4375)2
12
12
1
+
(3)(13) + 3(1)(9.5 - 4.4375)2
12
D
3 in.
y =
B
= 200.27 in4
Using flexure formula s =
My
I
sA =
15(12)(4.4375 - 1)
= 3.0896 ksi
200.27
sB =
15(12)(9 - 4.4375)
= 4.1007 ksi
200.27
FC =
1
(3.0896)(3.4375)(1) = 5.3102 kip
2
FT =
1
(4.1007)(4.5625)(1) = 9.3547 kip
2
www.elsolucionario.org
M = 5.3102(2.2917) + 9.3547(3.0417)
= 40.623 kip # in. = 3.3852 kip # ft
% of moment carried by web =
3.3852
* 100 = 22.6 %
15
Ans.
Ans:
% of moment carried by web = 22.6 %
522
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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
20 mm
25 mm
A
250 mm
25 mm
The moment of inertia of the cross-section about the neutral axis is
B
M ⫽ 10 kN⭈m
1
1
(0.2)(0.33) (0.16)(0.253) = 0.2417(10 - 3) m4.
I =
12
12
For point A, yA = C = 0.15 m.
sA =
MyA
10(103) (0.15)
= 6.207(106) Pa = 6.21 MPa (C)
=
I
0.2417(10 - 3)
Ans.
For point B, yB = 0.125 m.
sB =
10(103)(0.125)
MyB
= 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.
Ans:
sA = 6.21 MPa (C), sB = 5.17 MPa (T)
523
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
6–63. The beam is subjected to a moment of M = 30 lb # ft.
Determine the bending stress acting at point A and B. Also,
sketch a three-dimensional view of the stress distribution
acting over the entire cross-sectional area.
3 in.
A
1 in.
B
M ⫽ 30 lb⭈ft
1 in.
y =
I =
2(4)(2) - 3(12)(2)(3)
4(2) - 12(2)(3)
= 1.40 in.
1
1
1
(2)(4)3 + (4)(2)(2 - 1.40)2 - a (2)(3)3 + (2)(3)(3 - 1.40)2 b = 4.367 in4
12
36
2
sA =
(30)(12)(4 - 1.40)
My
=
= 214 psi (C)
I
4.367
Ans.
sB =
30 (12)(1.40 - 1)
My
=
= 33.0 psi (T)
I
4.367
Ans.
sC =
My
30 (12)(1.40)
=
= 115 psi (T)
I
4.367
www.elsolucionario.org
Ans:
sA = 214 psi (C), sB = 33.0 psi (T),
sC = 115 psi (T)
524
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*6–64. The axle of the freight car is subjected to wheel
loading 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.
C
A
B
60 in.
10 in.
20 kip
smax =
200(2.75)
Mc
= 12.2 ksi
= 1
4
I
4 p(2.75)
Ans.
525
D
10 in.
20 kip
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6–65. A shaft is made of a polymer having an elliptical
cross-section. If it resists an internal moment of
M = 50 N # m, determine the maximum bending stress
developed in the material (a) using the flexure formula,
where lz = 14p(0.08 m)(0.04 m)3, (b) using integration.
Sketch a three-dimensional view of the stress distribution
acting over the cross-sectional area.
y
z2
y2
⫹ ———2 ⫽ 1
———
2
(80)
(40)
80 mm
M ⫽ 50 N⭈m
z
160 mm
x
(a)
1
1
p ab3 = p(0.08)(0.04)3 = 4.021238(10 - 6) m4
4
4
50(0.04)
Mc
smax =
= 497 kPa
=
I
4.021238(10 - 6)
I =
Ans.
(b)
M =
=
smax
y2dA
c LA
smax
y22zdy
c L
z = 20.0064 - 4y2 = 2 2(0.04)2 - y2
0.04
2
0.04
2
L- 0.04
2
y 2(0.04)2 - y2 dy
L- 0.04
0.04
(0.04)4
y
1
= 4c
sin - 1 a
b - y 2(0.04)2 - y2(0.042 - 2y2) d `
8
0.04
8
- 0.0 4
y zdy = 4
=
www.elsolucionario.org
0.04
(0.04)4
y
sin - 1 a
b`
2
0.04 - 0.04
= 4.021238(10 - 6) m4
smax =
50(0.04)
4.021238(10 - 6)
= 497 kPa
Ans.
Ans:
(a) smax = 497 kPa,
(b) smax = 497 kPa
526
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
6–66. Solve Prob. 6–65 if the moment M = 50 kN # m is
applied about the y axis instead of the x axis. Here
Iy = 14 p(0.04 m)(0.08 m)3.
y
z2
y2
⫹ ———2 ⫽ 1
———
2
(80)
(40)
80 mm
M ⫽ 50 N⭈m
z
160 mm
x
(a)
1
1
p ab3 = p(0.04)(0.08)3 = 16.085(10 - 6) m4
4
4
50(0.08)
Mc
smax =
= 249 kPa
=
I
16.085(10 - 6)
I =
Ans.
(b)
M =
z(s dA) =
LA
50 = 2a
za
LA
smax
b(z)(2y)dz
0.08
0.08
1>2
smax
z2
b
z2 a 1 b
(0.04)dz
0.04 L0
(0.08)2
50 = 201.06(10 - 6)smax
smax = 249 kPa
Ans.
Ans:
(a) smax = 249 kPa,
(b) smax = 249 kPa
527
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6–67. The shaft 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
1.5 m
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.
www.elsolucionario.org
Ans:
smax = 158 MPa
528
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*6–68. The shaft 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
Allowable Bending Stress: The maximum moment is Mmax = 11.34 kN # m as
indicated on the moment diagram. Applying the flexure formula
smax = sallow =
180 A 106 B =
Mmax c
I
11.34(103) A d2 B
p
4
A d2 B 4
d = 0.08626 m = 86.3 mm
Ans.
529
1.5 m
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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
300 mm
15 mm
15 mm
(a)
Section Property:
30 mm
(b)
For section (a)
I =
1
1
(0.2) A 0.333 B (0.17)(0.3)3 = 0.21645(10 - 3) m4
12
12
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)
For section (b)
smin =
150(103)(0.18)
0.36135(10 - 3)
= 114.3 MPa
www.elsolucionario.org
Ans.
= 74.72 MPa = 74.7 MPa
Ans:
smin = 74.7 MPa
530
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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
3
1 in. 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.
Ans:
smax = 22.1 ksi
531
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
6–71. 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:
+
: ©Fx = 0;
Bx = 0
a +©MB = 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 =
www.elsolucionario.org
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.
Ans:
smax = 21.1 ksi
532
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*6–72. Determine the absolute maximum bending stress in
the 1.5-in.-diameter shaft which is subjected to the
concentrated forces. The sleeve bearings at A and B support
only vertical forces.
400 lb
A
B
12 in.
18 in.
15 in.
Mmax = 4500 lb # in.
s =
4500(0.75)
Mc
= 13.6 ksi
=
I
1
p(0.75)4
4
Ans.
533
300 lb
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
6–73. Determine the smallest allowable diameter of the
shaft which is subjected to the concentrated forces. The
sleeve bearings at A and B support only vertical forces, and
the allowable bending stress is sallow = 22 ksi.
400 lb
A
300 lb
B
12 in.
18 in.
15 in.
Mmax = 4500 lb # in.
s =
Mc
;
I
22(103) =
4500c
1 4
pc
4
c = 0.639 in.
d = 1.28 in.
Ans.
www.elsolucionario.org
Ans:
d = 1.28 in.
534
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
6–74. The pin is used to connect the three links together.
Due to wear, the load is distributed over the top and bottom
of the pin as shown on the free-body diagram. If the
diameter of the pin is 0.40 in., determine the maximum
bending stress on the cross-sectional area at the center
section a–a. For the solution it is first necessary to
determine the load intensities w1 and w2.
w2
800 lb
a
w2
w1
1 in.
1 in.
a
0.40 in.
1.5 in.
400 lb
1
w (1) = 400;
2 2
w2 = 800 lb>in.
w1(1.5) = 800;
w1 = 533 lb>in.
400 lb
M = 400(0.70833) = 283.33 lb # in
1
p(0.24) = 0.0012566 in4
4
283.33 (0.2)
Mc
smax =
=
I
0.0012566
= 45.1 ksi
I =
Ans.
Ans:
smax = 45.1 ksi
535
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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
C
1.5 m
3 kN
D
25 mm
0.75 m
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
smax =
Ans.
www.elsolucionario.org
Ans:
smax = 52.8 MPa
536
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*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
M
260 mm
20 mm 30 mm
The moment of inertia of the cross section about the neutral axis is
I =
1
1
(0.3)(0.33) (0.21)(0.263) = 0.36742(10 - 3) m4
12
12
Thus,
smax =
Mc
;
I
80(106) =
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.
537
Ans.
30 mm
30 mm
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
6–77. If the beam is subjected to an internal moment of
M = 2 kip # ft, determine the maximum tensile and
compressive stress in the beam. Also, sketch the bending
stress distribution on the cross section.
A
1 in.
4 in.
1 in.
M
Section Properties: The neutral axis passes through centroid C of the cross section
as shown in Fig. a. The location of C is
y =
3 in.
©y~A
2[1.5(3)(1)] + 3.5(8)(1) + 6(4)(1)
=
= 3.3889 in.
©A
2(3)(1) + 8(1) + 4(1)
3 in. 1 in.
3 in.
Thus, the moment of inertia of the cross section about the neutral axis is
1 in.
I = ©I + Ad2
1
1
= 2 c (1)(33) + 1(3)(3.3889 - 1.5)2 d + (8)(13) + 8(1)(3.5 + 3.3889)3
12
12
+
1
(1)(43) + (1)(4)(6 - 3.3889)2
12
= 59.278 in4
Maximum Bending Stress: The maximum compressive and tensile stress occurs at
the top and bottom-most fibers of the cross section.
(smax)c =
2(12)(8 - 3.3889)
Mc
=
= 1.87 ksi
I
59.278
Ans.
(smax)t =
2(12)(3.3889)
My
=
= 1.37 ksi
I
59.278
Ans.
www.elsolucionario.org
The bending stresses at y = 0.6111 in. and y = - 0.3889 in. are
s|y = 0.6111 in. =
My
2(12)(0.6111)
=
= 0.247 ksi (C)
I
59.278
s|y = - 0.3889 in. =
My
2(12)(0.3889)
=
= 0.157 ksi (T)
I
59.278
The bending stress distribution across the cross section is shown in
Fig. b.
Ans:
(smax)c = 1.87 ksi,
(smax)t = 1.37 ksi
538
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
6–78. If the allowable tensile and compressive stress for
the beam are (sallow)t = 2 ksi and (sallow)c = 3 ksi ,
respectively, determine the maximum allowable internal
moment M that can be applied on the cross section.
A
1 in.
4 in.
1 in.
M
Section Properties: The neutral axis passes through the centroid C of the cross
section as shown in Fig. a. The location of C is given by
y=
2[1.5(3)(1) + 3.5(8)(1) + 6(4)(1)]
©y~A
=
= 3.3889 in.
©A
2(3)(1) + 8(1) + 4(1)
3 in.
3 in. 1 in.
3 in.
1 in.
Thus, the moment of inertia of the cross section about the neutral axis is
I = ©I + Ad2
= 2c
1
1
(1)(33) + 1(3)(3.3889 - 1.5)2 d +
(8)(13) + 8(1)(3.5 - 3.3889)3
12
12
+
1
(1)(43) + (1)(4)(6 - 3.3889)2
12
= 59.278 in4
Allowable Bending Stress: The maximum compressive and tensile stress occurs at
the top and bottom-most fibers of the cross section.
(sallow)c =
Mc
;
I
3 =
M(8 - 3.3889)
59.278
M = 38.57 kip # in a
1 ft
b = 3.21 kip # ft
12 in.
For the bottom-most fiber,
(sallow)t =
My
;
I
2 =
M(3.3889)
59.278
M = 34.98 kip # in a
1 ft
b = 2.92 kip # ft (controls)
12 in.
Ans.
Ans:
M = 2.92 kip # ft
539
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
6–79. If the beam is subjected to an internal moment of
M = 2 kip # ft, determine the resultant force of the bending
stress distribution acting on the top vertical board A.
A
1 in.
4 in.
1 in.
M
Section Properties: The neutral axis passes through the centroid C of the cross
section as shown in Fig. a. The location of C is given by
~
© yA
2[1.5(3)(1) + 3.5(8)(1) + 6(4)(1)]
y =
=
= 3.3889 in.
©A
2(3)(1) + 8(1) + 4(1)
3 in.
3 in. 1 in.
3 in.
1 in.
Thus, the moment of inertia of the cross section about the neutral axis is
I = ©I + Ad2
= 2c
1
1
(1)(33) + 1(3)(3.3889 - 1.5)2 d +
(8)(13) + 8(1)(3.5 - 3.3889)3
12
12
+
1
(1)(43) + (1)(4)(6 - 3.3889)2
12
= 59.278 in4
Bending Stress: The distance from the neutral axis to the top and bottom of board A
is yt = 8 - 3.3889 = 4.6111 in. and yb = 4 - 3.3889 = 0.6111 in. We have
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st =
Myt
2(12)(4.6111)
=
= 1.8669 ksi
I
59.278
sb =
Myb
2(12)(0.6111)
=
= 0.2474 ksi
I
59.278
Resultant Force: The resultant force acting on board A is equal to the volume of the
trapezoidal stress block shown in Fig. b. Thus,
FR =
1
(1.8669 + 0.2474)(1)(4) = 4.23 kip
2
Ans.
Ans:
FR = 4.23 kip
540
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*6–80. If the beam is subjected to an internal moment of
M = 100 kN # m, determine the bending stress developed
at points A, B, and C. Sketch the bending stress distribution
on the cross section.
A
300 mm
M
30 mm
30 mm
C
150 mm
Section Properties: The neutral axis passes through centroid C of the cross section
as shown in Fig. a. The location of C is
y =
~
© yA
0.015(0.03)(0.3) + 0.18(0.3)(0.03)
=
= 0.0975 m
©A
0.03(0.3) + 0.3(0.03)
Thus, the moment of inertia of the cross section about the neutral axis is
I =
1
1
(0.3)(0.033) + 0.3(0.03)(0.0975 - 0.015)2 +
(0.03)(0.33)
12
12
+ 0.03(0.3)(0.18 - 0.0975)2
= 0.1907(10 - 3) m4
Bending Stress: The distance from the neutral axis to points A, B, and C is yA =
0.33 - 0.0975 = 0.2325 m, yB = 0.0975 m, and yC = 0.0975 - 0.03 = 0.0675 m.
sA =
100(103)(0.2325)
MyA
= 122 MPa (C)
=
I
0.1907(10 - 3)
Ans.
sB =
100(103)(0.0975)
MyB
= 51.1 MPa (T)
=
I
0.1907(10 - 3)
Ans.
sC =
MyC
100(103)(0.0675)
= 35.4 MPa (T)
=
I
0.1907(10 - 3)
Ans.
Using these results, the bending stress distribution across the cross section is shown
in Fig. b.
541
B
150 mm
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
6–81. If the beam is made of 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.
A
300 mm
M
30 mm
30 mm
B
150 mm
C
150 mm
Section Properties: The neutral axis passes through centroid C of the cross section
as shown in Fig. a. The location of C is
y =
~
0.015(0.03)(0.3) + 0.18(0.3)(0.03)
© yA
=
= 0.0975 m
©A
0.03(0.3) + 0.3(0.03)
Thus, the moment of inertia of the cross section about the neutral axis is
I =
1
1
(0.3)(0.033) + 0.3(0.03)(0.0975 - 0.015)2 +
(0.03)(0.33)
12
12
+ 0.03(0.3)(0.18 - 0.0975)2
= 0.1907(10 - 3) m4
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Allowable Bending Stress: The maximum compressive and tensile stress occurs at
the top and bottom-most fibers of the cross section. For the top-most fiber,
(sallow)c =
Mc
I
150(106) =
M(0.33 - 0.0975)
0.1907(10 - 3)
M = 123024.19 N # m = 123 kN # m (controls)
Ans.
For the bottom-most fiber,
(sallow)t =
My
I
125(106) =
M(0.0975)
0.1907(10 - 3)
M = 244 471.15 N # m = 244 kN # m
Ans:
M = 123 kN # m
542
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6–82. The shaft is supported by a smooth thrust bearing at
A and smooth journal bearing at C. If d = 3 in., determine
the absolute maximum bending stress in the shaft.
A
3 ft
d
C
B
3 ft
D
3 ft
1800 lb
3600 lb
Support Reactions: Shown on the free-body diagram of the shaft, Fig. a,
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| = 5400 lb # ft.
Section Properties: The moment of inertia of the cross section about the neutral
axis is
1
I = p(1.54) = 3.9761 in4
4
Absolute Maximum Bending Moment: Here, c =
smax =
3
= 1.5 in.
2
5400(12)(1.5)
Mmaxc
=
= 24.4 ksi
I
3.9761
Ans:
smax = 24.4 ksi
543
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
6–83. The shaft is supported by a smooth thrust bearing at
A and smooth journal bearing at C. If the material has
an allowable bending stress of sallow = 24 ksi, determine
the required minimum diameter d of the shaft to the
nearest 1>16 in.
A
3 ft
d
C
B
3 ft
D
3 ft
1800 lb
3600 lb
Support Reactions: Shown on the free-body diagram of the shaft, Fig. a.
Maximum Moment: As indicated on the moment diagram, Figs. b and c, the
maximum moment is |Mmax| = 5400 lb # ft.
Section Properties: The moment of inertia of the cross section about the neutral
axis is
I =
1 d 4
p 4
pa b =
d
4
2
64
Absolute Maximum Bending Moment:
d
5400(12) a b
2
24(103) =
p 4
d
64
d = 3.02 in.
Mc
sallow =
;
I
Use
d = 3
1
in.
16
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Ans.
Ans:
Use d = 3
544
1
in.
16
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*6–84. If the intensity of the load w = 15 kN> m, determine
the absolute maximum tensile and compressive stress in the
beam.
w
A
B
6m
300 mm
150 mm
Support Reactions: Shown on the free-body diagram of the beam, Fig. a.
Maximum Moment: The maximum moment occurs when V = 0. Referring to the
free-body diagram of the beam segment shown in Fig. b,
+ c g Fy = 0;
45 - 15x = 0
x = 3m
a + g M = 0;
3
Mmax + 15(3) a b - 45(3) = 0
2
Mmax = 67.5 kN # m
Section Properties: The moment of inertia of the cross section about the neutral
axis is
I =
1
(0.15)(0.33) = 0.1125(10 - 3) m4
36
Absolute Maximum Bending Stress: The maximum compressive and tensile stresses
occur at the top and bottom-most fibers of the cross section.
(smax)c =
67.5(103)(0.2)
Mc
= 120 MPa (C)
=
I
0.1125(10 - 3)
Ans.
(smax)t =
My
67.5(103)(0.1)
= 60 MPa (T)
=
I
0.1125(10 - 3)
Ans.
545
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6–85. If the material of the beam has an allowable bending
stress of sallow = 150 MPa, determine the maximum
allowable intensity w of the uniform distributed load.
w
A
B
6m
300 mm
150 mm
Support Reactions: As shown on the free-body diagram of the beam, Fig. a,
Maximum Moment: The maximum moment occurs when V = 0. Referring to the
free-body diagram of the beam segment shown in Fig. b,
+ c g Fy = 0;
3w - wx = 0
x = 3m
a + g M = 0;
3
Mmax + w(3) a b - 3w(3) = 0
2
Mmax =
9
w
2
Section Properties: The moment of inertia of the cross section about the neutral
axis is
I =
1
(0.15)(0.33) = 0.1125(10 - 3) m4
36
Absolute Maximum Bending Stress: Here, c =
sallow =
Mc
;
I
2
(0.3) = 0.2 m.
3
9
w(0.2)
2
150(106) =
0.1125(10 - 32
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w = 18750 N>m = 18.75 kN>m
Ans.
Ans:
w = 18.75 kN>m
546
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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
A
600 lb
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
= 19.1 ksi
Ans.
Ans:
smax = 19.1 ksi
547
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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
A
600 lb
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.
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Ans:
d = 2 in.
548
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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 the moment diagram. Applying the flexure formula
smax =
44.8(12)(4.5)
Mmax c
=
= 4.42 ksi
1
3
I
12 (9)(9)
Ans.
549
8 ft
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6–89. If the compound beam in Prob. 6–42 has a square
cross section of side length a, determine the minimum value
of a if the allowable bending stress is sallow = 150 MPa.
Allowable Bending Stress: The maximum moment 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.
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Ans:
a = 66.9 mm
550
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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 moment is Mmax =
23w0 L2
as
216
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
Ans:
smax =
551
23w0 L2
36 bh2
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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, 冷 Mmax 冷 = 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.
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Ans:
smax = 119 MPa
552
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*6–92. Determine, to the nearest millimeter, 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, 冷 Mmax 冷 = 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
553
Ans.
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6–93. The wing spar ABD of a light plane is made from
2014–T6 aluminum and has a cross-sectional area of 1.27 in.2,
a depth of 3 in., and a moment of inertia about its neutral
axis of 2.68 in4. Determine the absolute maximum bending
stress in the spar if the anticipated loading is to be as shown.
Assume A, B, and C are pins. Connection is made along the
central longitudinal axis of the spar.
smax =
Mc
;
I
smax =
80 lb/in.
2 ft
A
D
B
C
3 ft
46080(1.5)
= 25.8 ksi
2.68
Note that 25.8 ksi 6 sY = 60 ksi
6 ft
Ans.
OK
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Ans:
smax = 25.8 ksi
554
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6–94. The beam has a rectangular cross section as shown.
Determine the largest load P that can be supported on its
overhanging ends so that the bending stress does not
exceed smax = 10 MPa.
P
P
1.5 m
1.5 m
1.5 m
250 mm
150 mm
I =
1
(0.15)(0.253) = 1.953125(10 - 4) m4
12
smax =
10(106) =
Mc
I
1.5P(0.125)
1.953125(10 - 4)
P = 10.4 kN
Ans.
Ans:
P = 10.4 kN
555
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6–95. The beam has the rectangular cross section shown. If
P = 12 kN, determine the absolute maximum bending stress
in the beam. Sketch the stress distribution acting over the
cross section.
P
P
1.5 m
1.5 m
1.5 m
250 mm
150 mm
M = 1.5P = 1.5(12)(103) = 18000 N # m
I =
smax =
1
(0.15)(0.253) = 1.953125(10 - 4) m4
12
18000(0.125)
Mc
= 11.5 MPa
=
I
1.953125(10 - 4)
Ans.
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Ans:
smax = 11.5 MPa
556
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*6–96. A log that is 2 ft in diameter is to be cut into a
rectangular section for use as a simply supported beam. If
the allowable bending stress for the wood is sallow = 8 ksi,
determine the required width b and height h of the beam
that will support the largest load possible. What is this load?
h
b
2 ft
P
(24)2 = b2 + h2
8 ft
Mmax =
P
(8)(12) = 48P
2
sallow =
Mmax(h2 )
Mc
= 1
3
I
12 (b)(h)
sallow =
bh2 =
6 Mmax
bh2
6
(48 P)
8000
b(24)2 - b3 = 0.036 P
(24)2 - 3b2 = 0.036
dP
= 0
db
b = 13.856 in.
Thus, from the above equations,
b = 13.9 in.
Ans.
h = 19.6 in.
Ans.
P = 148 kip
Ans.
557
8 ft
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6–97. A log that is 2 ft in diameter is to be cut into a
rectangular section for use as a simply supported beam. If
the allowable bending stress for the wood is sallow = 8 ksi,
determine the largest load P that can be supported if the
width of the beam is b = 8 in.
h
b
2 ft
P
242 = h2 + 82
8 ft
h = 22.63 in.
Mmax =
P
(96) = 48 P
2
sallow =
Mmax c
I
8(103) = 1
8 ft
48P(22.63
2 )
3
12 (8)(22.63)
P = 114 kip
Ans.
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Ans:
P = 114 kip
558
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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.
8 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 )
Ans.
Ans:
smax = 7.59 ksi
559
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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
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 cross section about the neutral axis is I =
108 in4. Thus,
smax =
1
(6)(63) =
12
16800(12)(3)
Mc
=
I
108
= 5600 psi = 5.60 ksi
Ans.
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Ans:
smax = 5.60 ksi
560
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*6–100. If d = 450 mm, determine the absolute maximum
bending stress in the overhanging beam.
12 kN
8 kN/m
125 mm
25 mm 25 mm
75 mm
d
A
B
4m
Support Reactions: Shown on the free-body diagram of the beam, Fig. a.
Maximum Moment: The shear and moment diagrams are shown in Figs. b and c.
As indicated on the moment diagram, Mmax = 24 kN # m.
Section Properties: The moment of inertia of the cross section about the neutral
axis is
I =
1
1
(0.175)(0.453) (0.125)(0.33)
12
12
= 1.0477(10 - 3) m4
Absolute Maximum Bending Stress: Here, c =
smax =
0.45
= 0.225 m.
2
Ans.
24(103)(0.225)
Mmax c
= 5.15 MPa
=
I
1.0477(10 - 3)
561
75 mm
2m
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6–101. If wood used for the beam has an allowable
bending stress of sallow = 6 MPa, determine the minimum
dimension d of the beam’s cross-sectional area to the
nearest mm.
12 kN
8 kN/m
125 mm
25 mm 25 mm
75 mm
d
A
B
4m
75 mm
2m
Support Reactions: Shown on the free-body diagram of the beam, Fig. a.
Maximum Moment: The shear and moment diagrams are shown in Figs. b and c.
As indicated on the moment diagram, Mmax = 24 kN # m.
Section Properties: The moment of inertia of the cross section about the neutral
axis is
I =
1
1
(0.175)d3 (0.125)(d - 0.15)3
12
12
= 4.1667(10 - 3)d3 + 4.6875(10 - 3)d2 - 0.703125(10 - 3)d + 35.15625(10 - 6)
Absolute Maximum Bending Stress: Here, c =
sallow =
Mc
;
I
d
.
2
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24(103)
6(106) =
d
2
4.1667(10 - 3)d3 + 4.6875(10 - 3)d2 - 0.703125(10 - 3)d + 35.15625(10 - 6)
4.1667(10 - 3)d3 + 4.6875(10 - 3)d2 - 2.703125(10 - 3)d + 35.15625(10 - 6) = 0
Solving,
d = 0.4094 m = 410 mm
Ans.
Ans:
d = 410 mm
562
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6–102. If the concentrated force P = 2 kN is applied at the
free end of the overhanging beam, determine the absolute
maximum tensile and compressive stress developed in
the beam.
P
150 mm
25 mm
200 mm
A
B
2m
1m
25 mm 25 mm
Support Reactions: Shown on the free-body diagram of the beam, Fig. a.
Maximum Moment: The shear and moment diagrams are shown in Figs. b and c.
As indicated on the moment diagram, the maximum moment is ƒ Mmax ƒ = 2 kN # m.
Section Properties: The neutral axis passes through the centroid C of the cross
section as shown in Fig. d.
The location of C is given by
y =
g y~A
2[0.1(0.2)(0.025)] + 0.2125(0.025)(0.15)
=
= 0.13068 m
gA
2(0.2)(0.025) + 0.025(0.15)
Thus, the moment of inertia of the cross section about the neutral axis is
I = gI + Ad 2
= 2c
1
1
(0.025)(0.23) + 0.025(0.2)(0.13068 - 0.1)2 d +
(0.15)(0.0253) + 0.15(0.025)(0.2125 - 0.13068)2
12
12
= 68.0457(10 - 6) m4
Absolute Maximum Bending Stress: The maximum tensile and compressive stress
occurs at the top and bottom-most fibers of the cross section.
(smax)t =
Mmax y
2(103)(0.225 - 0.13068)
= 2.77 MPa
=
I
68.0457(10 - 6)
Ans.
(smax)c =
2(103)(0.13068)
Mmax c
= 3.84 MPa
=
I
68.0457(10 - 6)
Ans.
Ans:
(smax)t = 2.77 MPa, (smax)c = 3.84 MPa
563
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6–103. If the overhanging beam is made of wood having
the allowable tensile and compressive stresses of
(sallow )t = 4 MPa and (sallow )c = 5 MPa, determine the
maximum concentrated force P that can applied at the
free end.
P
150 mm
25 mm
200 mm
A
B
2m
1m
25 mm 25 mm
Support Reactions: Shown on the free-body diagram of the beam, Fig. a.
Maximum Moment: The shear and moment diagrams are shown in Figs. b and c.
As indicated on the moment diagram, the maximum moment is ƒ Mmax ƒ = P.
Section Properties: The neutral axis passes through the centroid C of the cross
section as shown in Fig. d. The location of C is given by
g~
yA
2[0.1(0.2)(0.025)] + 0.2125(0.025)(0.15)
y=
=
= 0.13068 m
gA
2(0.2)(0.025) + 0.025(0.15)
The moment of inertia of the cross section about the neutral axis is
I = gI + Ad 2
= 2c
1
1
(0.025)(0.23) + 0.025(0.2)(0.13068 - 0.1)2 d +
(0.15)(0.0253) + 0.15(0.025)(0.2125 - 0.13068)2
12
12
= 68.0457(10 - 6) m4
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Absolute Maximum Bending Stress: The maximum tensile and compressive stresses
occur at the top and bottom-most fibers of the cross section. For the top fiber,
(sallow )t =
Mmax y
;
I
4(106) =
P(0.225 - 0.13068)
68.0457(10 - 6)
P = 2885.79 N = 2.89 kN
For the top fiber,
(sallow )c =
P(0.13068)
Mmaxc
5(106) =
I
68.0457(10 - 6)
P = 2603.49 N = 2.60 kN (controls)
Ans.
Ans:
P = 2.60 kN
564
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z
*6–104. The member has a square cross section and is
subjected to a resultant internal bending moment of
M = 850 N # m as shown. Determine the stress at each
corner and sketch the stress distribution produced by M.
Set u = 45°.
B
250 mm
125 mm
125 mm
E
A
C
M 850 Nm
u
D
y
My = 850 cos 45° = 601.04 N # m
Mz = 850 sin 45° = 601.04 N # m
Iz = Iy =
s = -
1
(0.25)(0.25)3 = 0.3255208(10 - 3) m4
12
Myz
Mzy
sA = -
sB = -
sD = -
sE = -
Iz
+
Iy
601.04(- 0.125)
601.04( -0.125)
-3
+
0.3255208(10 )
601.04(0.125)
601.04( - 0.125)
-3
+
0.3255208(10 - 3)
+
0.3255208(10 - 3)
0.3255208(10 )
601.04(- 0.125)
601.04(0.125)
0.3255208(10 - 3)
601.04(0.125)
601.04(0.125)
-3
0.3255208(10 )
0.3255208(10 - 3)
+
0.3255208(10 - 3)
= 0
Ans.
= 462 kPa
Ans.
= - 462 kPa
Ans.
= 0
Ans.
The negative sign indicates compressive stress.
565
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6–105. The member has a square cross section and is
subjected to a resultant internal bending moment of
M = 850 N # m as shown. Determine the stress at each
corner and sketch the stress distribution produced by M.
Set u = 30°.
z
B
125 mm
250 mm
125 mm
E
A
C
M 850 Nm
u
D
y
My = 850 cos 30° = 736.12 N # m
Mz = 850 sin 30° = 425 N # m
Iz = Iy =
s = -
1
(0.25)(0.25)3 = 0.3255208(10 - 3) m4
12
Myz
Mzy
sA = -
sB = -
sD = -
sE = -
Iz
+
Iy
736.12(- 0.125)
425(- 0.125)
-3
+
0.3255208(10 )
736.12(0.125)
425(- 0.125)
-3
0.3255208(10 - 3)
+
0.3255208(10 )
0.3255208(10 - 3)
-3
+
0.3255208(10 - 3)
0.3255208(10 - 3)
+
0.3255208(10 - 3)
0.3255208(10 )
736.12(0.125)
425(0.125)
Ans.
= 446 kPa
Ans.
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736.12(- 0.125)
425(0.125)
= - 119 kPa
= - 446 kPa
Ans.
= 119 kPa
Ans.
The negative signs indicate compressive stress.
Ans:
sA = - 119 kPa, sB = 446 kPa, sD = - 446 kPa,
sE = 119 kPa
566
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y
6–106. 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
z
My
dA
sC
y
Mz
s dA
My =
z s dA,
- y s dA,
Mz =
LA
LA
LA
determine the constants a, b, and c, and show that the
normal stress can be determined from the equation
s = [- (MzIy + MyIyz)y + (MyIz + MzIyz)z]> (IyIz - Iyz2),
where the moments and products of inertia are defined in
Appendix A.
0 =
x
z
Equilibrium Condition: sx = a + by + cz
0 =
LA
sx dA
0 =
LA
(a + by + cz) dA
dA + b
y dA + c
0 = a
LA
My =
LA
z sx dA
=
LA
z(a + by + cz) dA
= a
LA
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
Ans:
≤z
(Q.E.D.)
567
a = 0; b = - ¢
MzIy + MyIyz
IyIz - I2yz
≤; c =
MyIz + MzIyz
IyIz - I2yz
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y
6–107. If the beam is subjected to the internal moment of
M = 2 kN # m, determine the maximum bending stress
developed in the beam and the orientation of the neutral axis.
100 mm
50 mm
A
100 mm
M
200 mm
60°
y
Internal Moment Components: The y and z components of M are positive since they
are directed towards the positive sense of their respective axes, Fig. a. Thus,
x
z
My = 2 sin 60° = 1.732 kN # m
B
50 mm
Mz = 2 cos 60° = 1 kN # m
Section Properties: The location of the centroid of the cross-section is
g y~A
0.025(0.05)(0.2) + 0.15(0.2)(0.05)
=
= 0.0875 m
gA
0.05(0.2) + 0.2(0.05)
y =
The moments of inertia of the cross section about the principal centroidal y and z axes are
Iy =
1
1
(0.05)(0.23) +
(0.2)(0.053) = 35.4167(10 - 6) m4
12
12
1
1
(0.2)(0.053) + 0.2(0.05)(0.0875 - 0.025)2 +
(0.05)(0.23) + 0.05(0.2)(0.15 - 0.0875)2
12
12
= 0.1135(10 - 3) m4
Iz =
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Bending Stress: By inspection, the maximum bending stress occurs at either
corner A or B.
s = -
sA = -
Myz
Mzy
Iz
+
Iy
1(103)(0.0875)
0.1135(10 - 3)
1.732(103)( -0.1)
+
35.4167(10 - 6)
= -5.66 MPa = 5.66 MPa (C) (Max.)
3
sB = -
1(10 )(- 0.1625)
0.1135(10 - 3)
Ans.
3
1.732(10 )(0.025)
+
35.4167(10 - 6)
= 2.65 MPa (T)
Orientation of Neutral Axis: Here, u = 60°.
tan a =
tan a =
Iz
Iy
tan u
0.1135(10 - 3)
35.4167(10 - 6)
tan 60°
a = 79.8°
Ans.
The orientation of the neutral axis is shown in Fig. b.
Ans:
smax = 5.66 MPa (C), a = 79.8°
568
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y
*6–108. If the wood used for the T-beam has an allowable
tensile and compressive stress of (sallow)t = 4 MPa and
(sallow)c = 6 MPa, respectively, determine the maximum
allowable internal moment M that can be applied to the
beam.
100 mm
50 mm
A
100 mm
M
200 mm
60°
y
x
z
B
Internal Moment Components: The y and z components of M are positive since they
are directed towards the positive sense of their respective axes, Fig. a. Thus,
My = M sin 60° = 0.8660M
Mz = M cos 60° = 0.5M
Section Properties: The location of the centroid of the cross section is
©y~A
0.025(0.05)(0.2) + 0.15(0.2)(0.05)
y =
=
= 0.0875 m
©A
0.05(0.2) + 0.2(0.05)
The moments of inertia of the cross section about the principal centroidal y and
z axes are
Iy =
1
1
(0.05)(0.23) +
(0.2)(0.053) = 35.417(10 - 6) m4
12
12
Iz =
1
1
(0.2)(0.053) + 0.2(0.05)(0.0875 - 0.025)2 +
(0.05)(0.23) + 0.05(0.2)(0.15 - 0.0875)2
12
12
= 0.1135(10 - 3) m4
Bending Stress: By inspection, the maximum bending stress can occur at either
corner A or B. For corner A, which is in compression,
sA = (sallow)c = - 6(106) = -
MyzA
MzyA
+
Iz
0.5M(0.0875)
0.1135(10 - 3)
Iy
0.8660M(- 0.1)
+
35.417(10 - 6)
M = 2119.71 N # m = 2.12 kN # m (controls)
Ans.
For corner B which is in tension,
sB = (sallow)t = -
4(106) = -
MyzB
MzyB
Iz
+
Iy
0.8660M(0.025)
0.5M(- 0.1625)
0.1135(10 - 3)
+
35.417(10 - 6)
M = 3014.53 N # m = 3.01 kN # m
569
50 mm
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y
6–109. The box beam is subjected to the internal moment
of M = 4 kN # m, which is directed as shown. Determine the
maximum bending stress developed in the beam and the
orientation of the neutral axis.
50 mm
25 mm
50 mm
25 mm
150 mm
50 mm
150 mm
45⬚
Internal Moment Components: The y component of M is negative since it is directed
towards the negative sense of the y axis, whereas the z component of M which is
directed towards the positive sense of the z axis is positive, Fig. a. Thus,
x
M
z
50 mm
My = - 4 sin 45° = - 2.828 kN # m
Mz = 4 cos 45° = 2.828 kN # m
Section Properties: The moments of inertia of the cross section about the principal
centroidal y and z axes are
Iy =
1
1
(0.3)(0.153) (0.2)(0.13) = 67.7083(10 - 6) m4
12
12
Iz =
1
1
(0.15)(0.33) (0.1)(0.23) = 0.2708(10 - 3) m4
12
12
Bending Stress: By inspection, the maximum bending stress occurs at corners A and D.
s = -
smax = sA = -
Myz
Mzy
Iz
+
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Iy
2.828(103)(0.15)
-3
(- 2.828)(103)(0.075)
+
67.7083(10 - 6)
0.2708(10 )
= - 4.70 MPa = 4.70 MPa (C)
3
smax = sD = -
2.828(10 )( -0.15)
-3
0.2708(10 )
Ans.
3
(- 2.828)(10 )( - 0.075)
+
67.7083(10 - 6)
Ans.
= 4.70 MPa (T)
Orientation of Neutral Axis: Here, u = - 45°.
tan a =
tan a =
Iz
Iy
tan u
0.2708(10 - 3)
67.7083(10 - 6)
tan ( -45°)
a = - 76.0°
Ans.
The orientation of the neutral axis is shown in Fig. b.
Ans:
smax = 4.70 MPa, a = - 76.0°
570
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y
6–110. If the wood used for the box beam has an
allowable bending stress of (sallow) = 6 MPa , determine
the maximum allowable internal moment M that can be
applied to the beam.
50 mm
25 mm
50 mm
25 mm
150 mm
50 mm
150 mm
45⬚
Internal Moment Components: The y component of M is negative since it is directed
towards the negative sense of the y axis, whereas the z component of M, which is
directed towards the positive sense of the z axis, is positive, Fig. a. Thus,
x
M
z
50 mm
My = - M sin 45° = - 0.7071M
Mz = M cos 45° = 0.7071M
Section Properties: The moments of inertia of the cross section about the principal
centroidal y and z axes are
Iy =
1
1
(0.3)(0.153) (0.2)(0.13) = 67.708(10 - 6) m4
12
12
Iz =
1
1
(0.15)(0.33) (0.1)(0.23) = 0.2708(10 - 3) m4
12
12
Bending Stress: By inspection, the maximum bending stress occurs at corners A and D.
Here, we will consider corner D.
sD = sallow = 6(106) = -
My zD
Mz yD
Iz
+
Iy
(- 0.7071M)(- 0.075)
0.7071M(- 0.15)
0.2708(10 - 3)
+
67.708(10 - 6)
M = 5106.88 N # m = 5.11 kN # m
Ans.
Ans:
M = 5.11 kN # m
571
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y
6–111. 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.
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
My = 1200 sin 30° = 600 kN # m
z
x
150 mm
Mz = - 1200 cos 30° = - 1039.23 kN # m
Section Properties: The location of the centroid of the cross-section is given by
y =
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)
150 mm
The moments of inertia of the cross section about the principal centroidal y and z
axes are
1
1
(0.6) A 0.33 B (0.15) A 0.153 B = 1.3078 A 10 - 3 B m4
Iy =
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.
Myz
Mzy
s = +
Iz
Iy
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sA = -
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
Ans.
= - 131 MPa = 131 MPa (C)(Max.)
Orientation of Neutral Axis: Here, u = - 30°.
Iz
tan u
tan a =
Iy
tan a =
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.
Ans:
smax = 131 MPa (C), a = - 66.5°
572
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y
*6–112. 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.
150 mm
150 mm
M
300 mm
30⬚
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,
z
x
150 mm
My = M sin 30° = 0.5M
150 mm
Mz = - M cos 30° = - 0.8660M
Section Properties: The location of the centroid of the cross section is
y =
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
573
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
6–113. The board is used as a simply supported floor joist.
If a bending moment of M = 800 lb # ft is applied 3° from the
z axis, determine the stress developed in the board at
the corner A. Compare this stress with that developed by
the same moment applied along the z axis (u = 0°).
What is the angle a for the neutral axis when u = 3°?
Comment: Normally, floor boards would be nailed to the
top of the beams so that u L 0° and the high stress due to
misalignment would not occur.
2 in.
A
M 800 lbft
6 in.
z
u 3
x
y
Mz = 800 cos 3° = 798.904 lb # ft
My = - 800 sin 3° = - 41.869 lb # ft
Iz =
1
(2)(63) = 36 in4;
12
s = -
sA = -
tan a =
1
(6)(23) = 4 in4
12
My z
Mzy
Iz
Iy =
+
Iy
798.904(12)( -3)
- 41.869(12)(- 1)
+
= 925 psi
36
4
Iz
Iy
tan u;
Ans.
www.elsolucionario.org
tan a =
36
tan (- 3°)
4
a = - 25.3°
Ans.
When u = 0°
sA =
800(12)(3)
Mc
=
= 800 psi
I
36
Ans.
Ans:
When u = 3°: sA = 925 psi, a = - 25.3°,
When u = 0°: sA = 800 psi
574
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y
6–114. The T-beam is subjected to a bending moment of
M = 150 kip # in. directed as shown. Determine the
maximum bending stress in the beam and the orientation of
the neutral axis. The location y of the centroid, C, must be
determined.
6 in.
M 150 kipin.
6 in.
2 in.
y–
z
8 in.
C
2 in.
60
My = 150 sin 60° = 129.9 kip # in.
Mz = - 150 cos 60° = - 75 kip # in.
y =
(1)(12)(2) + (6)(8)(2)
= 3 in.
12(2) + 8(2)
Iy =
1
1
(2)(123) +
(8)(23) = 293.33 in4
12
12
Iz =
1
1
(12)(23) + 12(2)(22) +
(2)(83) + 2(8)(32) = 333.33 in4
12
12
s = -
Myz
Mzy
+
Iz
Iy
sA =
-(- 75)(3)
129.9(6)
+
= 3.33 ksi
333.33
293.33
sD =
-(- 75)(- 7)
129.9(- 1)
+
= - 2.02 ksi
333.33
293.33
sB =
- (-75)(3)
129.9( - 6)
+
= - 1.982 ksi
333.33
293.33
tan a =
Iz
Iy
tan u =
Ans.
333.33
tan ( - 60°)
293.33
a = - 63.1°
Ans.
Ans:
smax = 3.33 ksi (T), a = - 63.1°
575
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y
6–115. The beam has a rectangular cross section. If it is
subjected to a bending moment of M = 3500 N # m directed
as shown, determine the maximum bending stress in the
beam and the orientation of the neutral axis.
z 150 mm
150 mm
M 3500 Nm
x
30
150 mm
My = 3500 sin 30° = 1750 N # m
Mz = - 3500 cos 30° = - 3031.09 N # m
Iy =
1
(0.3)(0.153) = 84.375(10 - 6) m4
12
Iz =
1
(0.15)(0.33) = 0.3375(10 - 3) m4
12
s = -
Myz
Mzy
Iz
sC = -
Iy
1750(0.075)
- 3031.09(0.15)
sA = -
sB = -
+
0.3375(10 - 3)
+
84.375(10 - 6)
1750( -0.075)
- 3031.09(- 0.15)
+
-3
= 2.90 MPa (max)
0.3375(10 )
84.375(10 - 6)
- 3031.09(0.15)
1750( - 0.075)
0.3375(10 - 3)
Ans.
= - 2.90 MPa (max)
Ans.
www.elsolucionario.org
+
84.375(10 - 6)
= - 0.2084 MPa
sD = 0.2084 MPa
tan a4 =
Iz
Iy
tan u =
3.375 (10 - 4)
8.4375(10 - 5)
tan ( -30°)
a = - 66.6°
Ans.
Ans:
smax = 2.90 MPa, a = - 66.6°
576
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*6–116. For the section, Iy¿ = 31.7(10 - 6) m4, Iz¿ = 114(10 - 6) m4,
Iy¿z¿ = 15.1(10 - 6) m4. Using the techniques outlined in
Appendix A, the member’s cross-sectional area has principal
moments of inertia of Iy = 29.0(10 - 6) m4 and Iz =
117(10 - 6) m4, computed about the principal axes of inertia
y and z, respectively. If the section is subjected to a moment of
M = 2500 N # m directed as shown, determine the stress
produced at point A, using Eq. 6–17.
y
60 mm
y¿
60 mm
60 mm
M 2500 Nm
z¿
10.10
z
80 mm
C
140 mm
60 mm
A
Iz = 117(10 - 6) m4
Iy = 29.0(10 - 6) m4
My = 2500 sin 10.1° = 438.42 N # m
Mz = 2500 cos 10.1° = 2461.26 N # m
y = - 0.06 sin 10.1° - 0.14 cos 10.1° = - 0.14835 m
z = 0.14 sin 10.1° - 0.06 cos 10.1° = - 0.034519 m
sA =
My z
- Mzy
Iz
+
Iy
438.42( -0.034519)
-2461.26( - 0.14835)
=
-6
117(10 )
+
29.0(10 - 6)
= 2.60 MPa (T)
577
Ans.
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6–117. Solve Prob. 6–116 using the equation developed in
Prob. 6–106.
y
60 mm
y¿
60 mm
60 mm
M 2500 Nm
z¿
10.10
z
80 mm
C
140 mm
60 mm
sA =
=
- (Mz¿Iy¿ + My¿Iy¿z¿)y¿ + (My¿Iz¿ + Mz¿Iy¿z¿)z¿
A
Iy¿Iz¿ - Iy¿z¿ 2
-32500(31.7)(10 - 6) + 04( -0.14) + 30 + 2500(15.1)(10 - 6)4( -0.06)
31.7(10 - 6)(114)(10 - 6) - 3(15.1)(10 - 6)42
= 2.60 MPa (T)
Ans.
www.elsolucionario.org
Ans:
sA = 2.60 MPa
578
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6–118. If the applied distributed loading of w = 4 kN>m can
be assumed to pass through the centroid of the beam’s cross
sectional area, determine the absolute maximum bending
stress in the joist and the orientation of the neutral axis. The
beam can be considered simply supported at A and B.
15⬚
Internal Moment Components: The uniform distributed load w can be resolved into
its y and z components as shown in Fig. a.
6m
wz = 4 sin 15° = 1.035 kN>m
wy and wz produce internal moments in the beam about the z and y axes,
respectively. For the simply supported beam subjected to the uniform distributed
wL2
. Thus,
load, the maximum moment in the beam is Mmax =
8
(Mz)max =
(My)max =
8
=
3.864(62)
= 17.387 kN # m
8
=
1.035(62)
= 4.659 kN # m
8
wzL2
8
w(6 m)
B
15 mm
100 mm
100 mm
15⬚
15⬚
wy = 4 cos 15° = 3.864 kN>m
wyL2
w
A
10 mm
15 mm
15⬚
100 mm
As shown in Fig. b, (Mz)max and (My)max are positive since they are directed towards
the positive sense of their respective axes.
Section Properties: The moment of inertia of the cross section about the principal
centroidal y and z axes are
Iy = 2 c
Iz =
1
1
(0.015)(0.1 3) d +
(0.17)(0.01 3) = 2.5142(10 - 6) m4
12
12
1
1
(0.1)(0.2 3) (0.09)(0.17 3) = 29.8192(10 - 6) m4
12
12
Bending Stress: By inspection, the maximum bending stress occurs at points A and B.
s = -
(My)max z
(Mz)max y
+
Iz
Iy
17.387(103)(- 0.1)
smax = sA = -
4.659(103)(0.05)
+
-6
29.8192 (10 )
2.5142(10 - 6)
Ans.
= 150.96 MPa = 151 MPa (T)
smax = sB = -
17.387(103)(0.1)
4.659(103)(- 0.05)
+
29.8192 (10 - 6)
2.5142(10 - 6)
= - 150.96 MPa = 151 MPa (C)
Orientation of Neutral Axis: Here, u = tan - 1 c
tan a =
tan a =
Iz
Iy
(My)max
(Mz)max
Ans.
d = tan - 1 a
4.659
b = 15°.
17.387
tan u
29.8192(10 - 6)
2.5142(10 - 6)
tan 15°
a = 72.5°
Ans.
Ans:
smax = 151 MPa, a = 72.5°
The orientation of the neutral axis is shown in Fig. c.
579
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6–119. Determine the maximum allowable intensity w of
the uniform distributed load that can be applied to the
beam. Assume w passes through the centroid of the beam’s
cross sectional area and the beam is simply supported at A
and B. The beam is made of material having an allowable
bending stress of sallow = 165 MPa .
w
A
15⬚
6m
w(6 m)
B
15 mm
100 mm
100 mm
15⬚
15⬚
10 mm
15 mm
15⬚
Internal Moment Components: The uniform distributed load w can be resolved into
its y and z components as shown in Fig. a.
100 mm
wy = w cos 15° = 0.9659w
wz = w sin 15° = 0.2588w
wy and wz produce internal moments in the beam about the z and y axes,
respectively. For the simply supported beam subjected to the a uniform distributed
wL2
load, the maximum moment in the beam is Mmax =
. Thus,
8
(Mz)max =
(My)max =
wyL2
8
=
0.9659w(62)
= 4.3476w
8
=
0.2588w(62)
= 1.1647w
8
wzL2
8
www.elsolucionario.org
As shown in Fig. b, (Mz)max and (My)max are positive since they are directed
towards the positive sense of their respective axes.
Section Properties: The moment of inertia of the cross section about the principal
centroidal y and z axes are
Iy = 2 c
Iz =
1
1
(0.015)(0.13) d +
(0.17)(0.013) = 2.5142(10 - 6) m4
12
12
1
1
(0.1)(0.23) (0.09)(0.173) = 29.8192(10 - 6) m4
12
12
Bending Stress: By inspection, the maximum bending stress occurs at points A and B.
We will consider point A.
sA = sallow = -
165(106) = -
(My)maxzA
(Mz)maxyA
+
Iz
1.1647w(0.05)
4.3467w( - 0.1)
-6
Iy
+
29.8192(10 )
2.5142(10 - 6)
w = 4372.11 N>m = 4.37 kN>m
Ans.
Ans:
w = 4.37 kN>m
580
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*6–120. The composite beam is made of steel (A) bonded
to brass (B) and has the cross section shown. If it is
subjected to a moment of M = 6.5 kN # m, determine the
maximum bending stress in the brass and steel. Also, what is
the stress in each material at the seam where they are
bonded together? Ebr = 100 GPa. Est = 200 GPa.
A
y
50 mm
200 mm
M
B
z
x
175 mm
n =
200(109)
Est
= 2
=
Ebr
100(109)
y =
(350)(50)(25) + (175)(200)(150)
= 108.33 mm
350(50) + 175(200)
I =
1
1
(0.35)(0.053) + (0.35)(0.05)(0.083332) +
(0.175)(0.23) +
12
12
(0.175)(0.2)(0.041672) = 0.3026042(10 - 3) m4
Maximum stress in brass:
(sbr)max =
6.5(103)(0.14167)
Mc1
= 3.04 MPa
=
I
0.3026042(10 - 3)
Ans.
Maximum stress in steel:
(sst)max =
(2)(6.5)(103)(0.10833)
nMc2
= 4.65 MPa
=
I
0.3026042(10 - 3)
Ans.
Stress at the junction:
sbr =
6.5(103)(0.05833)
Mr
= 1.25 MPa
=
I
0.3026042(10 - 3)
Ans.
Ans.
sst = nsbr = 2(1.25) = 2.51 MPa
581
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6–121. The composite beam is made of steel (A) bonded
to brass (B) and has the cross section shown. If the
allowable bending stress for the steel is (sallow)st = 180 MPa,
and for the brass (sallow)br = 60 MPa , determine the
maximum moment M that can be applied to the beam.
Ebr = 100 GPa, Est = 200 GPa.
A
y
50 mm
200 mm
M
B
z
x
175 mm
n =
200(109)
Est
= 2
=
Ebr
100(109)
y =
(350)(50)(25) + (175)(200)(150)
= 108.33 mm
350(50) + 175(200)
I =
1
1
(0.35)(0.053) + (0.35)(0.05)(0.083332) +
(0.175)(0.23) +
12
12
(0.175)(0.2)(0.041672) = 0.3026042(10 - 3) m4
(sst)allow =
nMc2
;
I
180(106) =
(2) M(0.10833)
0.3026042(10 - 3)
M = 251 kN # m
(sbr)allow =
Mc1
;
I
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60(106) =
M(0.14167)
0.3026042(10 - 3)
M = 128 kN # m (controls)
Ans.
Ans:
M = 128 kN # m
582
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w
6–122. 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.
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,
sst冷y = 0.6970 in. =
Mmaxy
25.3125(12)(0.6970)
=
= 6.85 ksi
I
30.8991
For the aluminum,
(smax)al = n
25.3125(12)(6 - 2.3030)
Mmaxcal
= 0.3655c
d = 13.3 ksi
I
30.8991
Ans.
At the seam,
sal冷y = 0.6970 in. = n
Mmaxy
25.3125(12)(0.6970)
= 0.3655 c
d = 2.50 ksi
I
30.8991
The bending stress across the cross section of the composite beam is shown in Fig. b.
Ans:
(smax)st = 22.6 ksi, (smax)al = 13.3 ksi
583
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w
6–123. Segment A of the composite beam is made
from 2014-T6 aluminum alloy and segment B is A-36 steel.
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.
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)
www.elsolucionario.org
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 = 0.3655 c
(28.125w)(12)(6 - 2.3030)
d
30.8991
w = 1.02 kip>ft
Ans:
w = 0.875 kip>ft
584
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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 = 0.060(10 - 3) m4 and Iz = 0.471(10 - 3) 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 N⭈m
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.
585
200 mm
50 mm
B
50 mm
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6–125. The wooden section of the beam is reinforced with
two steel plates as shown. Determine the maximum internal
moment M that the beam can support if the allowable
stresses for the wood and steel are (sallow)w = 6 MPa , and
(sallow)st = 150 MPa , respectively. Take Ew = 10 GPa and
Est = 200 GPa.
15 mm
150 mm
M
15 mm
100 mm
Section Properties: The cross section will be transformed into that of steel as shown
Ew
10
= 0.05. Thus, bst = nbw = 0.05(0.1) = 0.005 m . The
=
in Fig. a. Here, n =
Est
200
moment of inertia of the transformed section about the neutral axis is
I =
1
1
(0.1)(0.183) (0.095)(0.153) = 21.88125(10 - 6) m4
12
12
Bending Stress: Assuming failure of the steel,
(sallow)st =
Mcst
;
I
150(106) =
M(0.09)
21.88125(10 - 6)
M = 36 468.75 N # m = 36.5 kN # m
www.elsolucionario.org
Assuming failure of wood,
Mcw
(sallow)w = n
;
I
6(10 ) = 0.05 c
6
M(0.075)
21.88125(10 - 6)
d
M = 35 010 N # m = 35.0 kN # m (controls)
Ans.
Ans:
M = 35.0 kN # m
586
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6–126. The wooden section of the beam is reinforced with
two steel plates as shown. If the beam is subjected to an
internal moment of M = 30 kN # m, determine the maximum
bending stresses developed in the steel and wood. Sketch
the stress distribution over the cross section. Take
Ew = 10 GPa and Est = 200 GPa.
15 mm
150 mm
M
15 mm
100 mm
Section Properties: The cross section will be transformed into that of steel as shown
Ew
10
=
= 0.05. Thus, bst = nbw = 0.05(0.1) = 0.005 m. The
in Fig. a. Here, n =
Est
200
moment of inertia of the transformed section about the neutral axis is
I =
1
1
(0.1)(0.183) (0.095)(0.153) = 21.88125(10 - 6) m4
12
12
Maximum Bending Stress: For the steel,
(smax)st =
30(103)(0.09)
Mcst
= 123 MPa
=
I
21.88125(10 - 6)
sst|y = 0.075 m =
Ans.
My
30(103)(0.075)
= 103 MPa
=
I
21.88125(10 - 6)
For the wood,
(smax)w = n
30(103)(0.075)
Mcw
d = 5.14 MPa
= 0.05 c
I
21.88125(10 - 6)
Ans.
The bending stress distribution across the cross section is shown in Fig. b.
Ans:
(smax)st = 123 MPa, (smax)w = 5.14 MPa
587
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6–127. 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
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
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)
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Ans:
smax = 20.1 MPa
588
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*6–128. 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.
4 in.
0.5 in.
15 in.
M ⫽ 850 lb⭈ft
0.5 in.
0.5 in.
y =
(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)
I =
1
1
(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.
589
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
y
6–129. A wood beam is reinforced with steel straps at its
top and bottom as shown. Determine the maximum
bending stress developed in the wood and steel if the beam
is subjected to a bending moment of M = 5 kN # m. Sketch
the stress distribution acting over the cross section. Take
Ew = 11 GPa, Est = 200 GPa.
20 mm
300 mm
M 5 kNm
20 mm
x
z
n =
200
= 18.182
11
I =
1
1
(3.63636)(0.34)3 (3.43636)(0.3)3 = 4.17848(10 - 3) m4
12
12
200 mm
Maximum stress in steel:
(sst)max =
18.182(5)(103)(0.17)
nMc1
= 3.70 MPa
=
I
4.17848(10 - 3)
Ans.
Maximum stress in wood:
(sw)max =
5(103)(0.15)
Mc2
= 0.179 MPa
=
I
4.17848(10 - 3)
Ans.
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(sst) = n(sw)max = 18.182(0.179) = 3.26 MPa
Ans:
(sst)max = 3.70 MPa, (sw)max = 0.179 MPa
590
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6–130. 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.
500 lb
500 lb
PVC EPVC 450 ksi
Escon EE 160 ksi
Bakelite EB 800 ksi
3 ft
4 ft
3 ft
1 in.
2 in.
2 in.
3 in.
(bbk)1 = n1 bEs =
160
(3) = 0.6 in.
800
(bbk)2 = n2 bpvc =
450
(3) = 1.6875 in.
800
y =
gy A
(1)(3)(2) + 3(0.6)(2) + 4.5(1.6875)(1)
=
= 1.9346 in.
gA
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
+
1
(1.6875)(13) + 1.6875(1)(2.56542) = 20.2495 in4
12
(smax)pvc = n2
Mc
450 1500(12)(3.0654)
= a
b
I
800
20.2495
= 1.53 ksi
Ans.
Ans:
(spvc)max = 1.53 ksi
591
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
6–131. 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)con = 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. Here,
Est
200
n =
=
= 8. It is required that both concrete and steel achieve their
Econ
25
allowable stress simultaneously. Thus,
(sallow)con =
Mccon
;
I
12.5 A 106 B =
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)
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Equating Eqs. (1) and (2),
12.5 A 106 B ¢
I
I
≤ = 27.5 A 106 B ¢
≤
ccon
d - ccon
ccon = 0.3125d
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
592
d
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6–131. 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.
Ans:
d = 531 mm, M = 98.6 kN # m
593
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*6–132. The wide-flange section is reinforced with two
wooden boards as shown. If this composite beam is
subjected to an internal moment of M = 100 kN # m,
determine the maximum bending stress developed in the
steel and the wood. Take Ew = 10 GPa and Est = 200 GPa.
100 mm
10 mm
100 mm
300 mm
M
10 mm
Section Properties: The cross section will be transformed into that of steel
Ew
10
=
= 0.05. Thus, bst = 0.01 + 0.05bw =
as shown in Fig. a. Here, n =
Est
200
0.01 + 0.05(0.19) = 0.0195 m. The moment of inertia of the transformed section
about the neutral axis is
I =
1
1
(0.2)(0.33) (0.1805)(0.283) = 119.81(10 - 6) m4
12
12
Maximum Bending Stress: For the steel,
(smax)st =
100(103)(0.15)
Mcst
= 125 MPa
=
I
119.81(10 - 6)
Ans.
For the wood,
(smax)w = na
100(103)(0.14)
Mcw
d = 5.84 MPa
b = 0.05 c
I
119.81(10 - 6)
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Ans.
594
10 mm
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
6–133. The wide-flange section is reinforced with two
wooden boards as shown. If the steel and wood have an
allowable bending stress of (sallow)st = 150 MPa and
(sallow)w = 6 MPa, determine the maximum allowable
internal moment M that can be applied to the beam. Take
Ew = 10 GPa and Est = 200 GPa.
100 mm
10 mm
100 mm
300 mm
M
10 mm
Section Properties: The cross section will be transformed into that of steel
Ew
10
as shown in Fig. a. Here, n =
=
= 0.5. Thus, bst = 0.01 + nbw =
Est
200
0.01 + 0.05(0.19) = 0.0195 m. The moment of inertia of the transformed section
10 mm
about the neutral axis is
I =
1
1
(0.2)(0.33) (0.1805)(0.283) = 119.81(10 - 6) m4
12
12
Bending Stress: Assuming failure of steel,
(sallow)st =
Mcst
;
I
150(106) =
M (0.15)
119.81(10 - 6)
M = 119 805.33 N # m = 120 kN # m
Assuming failure of wood,
(sallow)w = n
Mcw
;
I
6(106) = 0.05 c
M (0.14)
119.81(10 - 6)
d
M = 102 690.29 N # m = 103 kN # m (controls)
Ans.
Ans:
M = 103 kN # m
595
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
6–134. 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
73.1 A 109 B
Eal
in Fig. a. Here, n =
=
= 0.3655. Thus, bst = nbal = 0.3655(0.015) =
Est
200 A 109 B
0.0054825 m. The location of the transformed section is
©yA
y =
=
©A
B
0.075(0.15)(0.0054825) + 0.2 cp 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
www.elsolucionario.org
= 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
S = 171 MPa
= 0.3655 C
I
18.08 A 10 - 6 B
3
(smax)al = n
Ans.
Ans:
(smax)st = 154 MPa, (smax)al = 171 MPa
596
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y
6–135. 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 = 750 kip # ft. Sketch
the stress distribution acting over the cross section.
0.5 in.
0.5 in.
0.5 in.
z
6 in.
2 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.
Ans:
(smax)st = 8.51 ksi, (smax)w = 0.558 ksi
597
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*6–136. 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
r
LA r + y
LA
=
LA
a
= A -
r - r - y
r + y
y
LA r + y
+ 1 b dA
dA
[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
=
LA
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¢
ry
r + y
y2
LA r + y
+ y - y ≤ dA - Ay
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
LA r + y
y
Mr
dA
C A - ¢A dA ≤ - y
S
AI
r
+
y
r
LA
LA + y
y
dA
Mr
C
dA - y
S
AI LA r + y
r
LA + y
598
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
6–136. Continued
y
=
As
y
r
y
dA
Mr
r
C
¢
≤ dA ¢
y≤ S
AI LA 1 + yr
r LA 1 + r
: 0
LA
Therefore,
¢
y
r
y ≤ dA = 0
1 + r
s =
and
y
y
yA
dA
¢
1A dA =
y≤ =
r LA 1 + r
r
r
yA
My
Mr
ab = AI
I
r
(Q.E.D.)
599
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
6–137. The curved member is subjected to the internal
moment of M = 50 kN # m. Determine the percentage
error introduced in the computation of maximum bending
stress using the flexure formula for straight members.
M
100 mm
M
200 mm
200 mm
Straight Member: The maximum bending stress developed in the straight member
smax =
50(103)(0.1)
Mc
=
= 75 MPa
I
1
(0.1)(0.23)
12
Curved Member: When r = 0.2 m, r = 0.3 m, rA = 0.2 m and rB = 0.4 m, Fig. a.
The location of the neutral surface from the center of curvature of the curve
member is
R =
0.1(0.2)
A
=
= 0.288539 m
dA
0.4
0.1ln
0.2
LA r
Then
e = r - R = 0.011461 m
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Here, M = 50 kN # m. Since it tends to decrease the curvature of the curved
member.
sB =
M(R - rB)
50(103)(0.288539 - 0.4)
=
AerB
0.1(0.2)(0.011461)(0.4)
= - 60.78 MPa = 60.78 MPa (C)
sA =
M(R - rA)
50(103)(0.288539 - 0.2)
=
AerA
0.1(0.2)(0.011461)(0.2)
= 96.57 MPa (T) (Max.)
Thus,
% of error = a
96.57 - 75
b 100 = 22.3%
96.57
Ans.
Ans:
% of error = 22.3%
600
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6–138. The curved member is made from material
having an allowable bending stress of sallow = 100 MPa.
Determine the maximum allowable internal moment M
that can be applied to the member.
M
100 mm
M
200 mm
200 mm
Internal Moment: M is negative since it tends to decrease the curvature of the
curved member.
Section Properties: Referring to Fig. a, the location of the neutral surface from the
center of curvature of the curve beam is
R =
0.1(0.2)
A
=
= 0.288539 m
dA
0.4
0.1ln
0.2
LA r
Then
e = r - R = 0.3 - 0.288539 = 0.011461 m
Allowable Bending Stress: The maximum stress occurs at either point A or B. For
point A which is in tension,
sallow =
M(R - rA)
;
AerA
100(106) =
M(0.288539 - 0.2)
0.1(0.2)(0.011461)(0.2)
M = 51 778.27 N # m = 51.8 kN # m (controls)
Ans.
For point B which is in compression,
sallow =
M(R - rB)
;
AerB
-100(106) =
M(0.288539 - 0.4)
0.1(0.2)(0.011461)(0.4)
M = 82 260.10 N # m = 82.3 kN # m
Ans:
M = 51.8 kN # m
601
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
6–139. The curved beam is subjected to a bending
moment of M = 40 lb # ft. Determine the maximum
bending stress in the beam. Also, sketch a two-dimensional
view of the stress distribution acting on section a–a.
2 in.
a
0.5 in.
2 in.
5 in.
a
3 in.
0.5 in.
5.5 in.
M
M
Section properties:
r =
4(2)(0.5) + 5.25(2)(0.5)
= 4.625 in.
2(0.5) + 2(0.5)
5
dA
5.5
©
= 0.5 ln + 2 ln
= 0.446033 in.
r
3
5
LA
2
A
= 4.4840 in.
=
dA
0.446033
LA r
R =
r - R = 4.625 - 4.4840 = 0.1410 in.
s =
M(R - r)
Ar(r - R)
sA =
40(12)(4.4840 - 3)
= 842 psi (T) (Max)
2(3)(0.1410)
sB =
40(12)(4.4840 - 5.5)
= - 314 psi = 314 psi (C)
2(5.5)(0.1410)
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Ans.
Ans:
smax = 842 psi (T)
602
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*6–140. The curved beam is made from material having
an allowable bending stress of sallow = 24 ksi. Determine
the maximum moment M that can be applied to the beam.
2 in.
a
0.5 in.
2 in.
5 in.
a
3 in.
0.5 in.
5.5 in.
M
r =
4(2)(0.5) + 5.25(2)(0.5)
= 4.625 in.
2(0.5) + 2(0.5)
dA
5.5
5
= 0.5 ln + 2 ln
= 0.4460 in.
©
3
5
LA r
R =
2
A
=
= 4.4840 in.
dA
0.4460
LA r
r - R = 4.625 - 4.4840 = 0.1410 in.
s =
M(R - r)
Ar(r - R)
Assume tension failure:
24 =
M(4.484 - 3)
2(3)(0.1410)
M = 13.68 kip # in. = 1.14 kip # ft (controls)
Ans.
Assume compression failure:
- 24 =
M(4.484 - 5.5)
;
2(5.5)(0.1410)
M = 36.64 kip # in. = 3.05 kip # ft
603
M
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6–141. If P = 3 kN, determine the bending stress developed
at points A, B, and C of the cross section at section a- a.
Using these results, sketch the stress distribution on
section a- a.
D
600 mm
E
a 300 mm
A
B
C
a
20 mm
50 mm
25 mm
25 mm
25 mm
Section a – a
P
Internal Moment: The internal moment developed at section a–a can be determined by
writing the moment equation of equilibrium about the neutral axis of the cross section at
a–a. Using the free-body diagram shown in Fig. a,
a+ ©MNA = 0;
3(0.6) - Ma - a = 0
Ma - a = 1.8 kN # m
Here, M a - a is considered negative since it tends to reduce the curvature of the
curved segment of the beam.
Section Properties: Referring to Fig. b, the location of the centroid of the cross
section from the center of the beam’s curvature is
r =
0.31(0.02)(0.075) + 0.345(0.05)(0.025)
©rA
=
= 0.325909 m
©A
0.02(0.075) + 0.05(0.025)
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The location of the neutral surface from the center of the beam’s curvature can be
determined from
R =
A
dA
©LA
r
where A = 0.02(0.075) + 0.05(0.025) = 2.75(10 - 3) m2
dA
0.32
0.37
©
+ 0.025 ln
= 8.46994(10 - 3) m
= 0.075 ln
r
0.3
0.32
LA
Thus,
R =
2.75(10 - 3)
8.46994(10 - 3)
= 0.324678 m
then
e = r - R = 0.325909 - 0.324678 = 1.23144(10 - 3) m
Normal Stress:
sA =
M(R - rA)
1.8(103)(0.324678 - 0.3)
= 43.7 MPa (T)
=
AerA
2.75(10 - 3)(1.23144)(10 - 3)(0.3)
Ans.
sB =
M(R - rB)
1.8(103)(0.324678 - 0.32)
= 7.77 MPa (T)
=
AerB
2.75(10 - 3)(1.23144)(10 - 3)(0.32)
Ans.
sC =
M(R - rC)
1.8(103)(0.324678 - 0.37)
= - 65.1 MPa (C)
=
AerC
2.75(10 - 3)(1.23144)(10 - 3)(0.37)
Ans.
604
Ans:
sA = 43.7 MPa (T), sB = 7.77 MPa (T),
sC = - 65.1 MPa (C)
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6–142. If the maximum bending stress at section a- a is
not allowed to exceed sallow = 150 MPa, determine the
maximum allowable force P that can be applied to the
end E.
D
600 mm
E
a 300 mm
A
B
C
a
P
20 mm
50 mm
25 mm
25 mm
25 mm
Section a – a
Internal Moment: The internal Moment developed at section a–a can be
determined by writing the moment equation of equilibrium about the neutral axis of
the cross section at a–a.
a + ©MNA = 0;
P(0.6) - Ma - a = 0
Ma - a = 0.6P
Here, Ma–a is considered positive since it tends to decrease the curvature of the
curved segment of the beam.
Section Properties: Referring to Fig. b, the location of the centroid of the cross
section from the center of the beam’s curvature is
r =
0.31(0.02)(0.075) + 0.345(0.05)(0.025)
©r~A
=
= 0.325909 m
©A
0.02(0.075) + 0.05(0.025)
The location of the neutral surface from the center of the beam’s curvature can be
determined from
A
R =
©
dA
L
A r
where A = 0.02(0.075) + 0.05(0.025) = 2.75(10 - 3) m2
©L
dA
A r
= 0.075 ln
0.37
0.32
+ 0.025 ln
= 8.46994(10 - 3) m
0.3
0.32
Thus,
R =
2.75(10 - 3)
8.46994(10 - 3)
= 0.324678 m
then
e = r - R = 0.325909 - 0.324678 = 1.23144(10 - 3) m
Allowable Normal Stress: The maximum normal stress occurs at either points A or C.
For point A which is in tension,
sallow =
M(R - rA)
;
AerA
150(106) =
0.6P(0.324678 - 0.3)
2.75(10 - 3)(1.23144)(10 - 3)(0.3)
P = 10 292.09 N = 10.3 kN
For point C which is in compression,
sallow =
M(R - rC)
;
AerC
- 150(106) =
0.6P(0.324678 - 0.37)
2.75(10 - 3)(1.23144)(10 - 3)(0.37)
P = 6911.55 N = 6.91 kN (controls)
605
Ans.
Ans:
P = 6.91 kN
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6–143. 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 lb⭈in.
1 in.
a
0.63 in.
0.75 in.
dA
= ©2p (r - 2r2 - c2)
LA r
M = 25 lb⭈in.
= 2p(1.75 - 21.752 - 0.752) - 2p (1.75 - 21.752 - 0.632)
= 0.32375809 in.
A = p(0.752) - p(0.632) = 0.1656 p
R =
A
dA
1A r
=
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 = =
M(R - rB)
ArB(r - R)
25(1.606902679 - 1)
= 204 psi (T)
0.1656 p(1)(0.14309732)
=
25(1.606902679 - 2.5)
= 120 psi (C)
0.1656p(2.5)(0.14309732)
Ans.
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Ans.
Ans:
(smax)t = 204 psi, (smax)c = 120 psi
606
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*6–144. 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
1.5 in.
8 in.
M
A = 0.5(2) +
M
1
(1)(2) = 2 in2
2
9(0.5)(2) + 8.6667 A 2 B (1)(2)
©rA
=
= 8.83333 in.
©A
2
1
r =
1(10)
10
dA
10
+ c
c ln
d - 1 d = 0.22729 in.
= 0.5 ln
r
8
(10
8)
8
LA
R =
A
1A r
dA
=
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.
607
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a
6–145. 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.
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
= 0.018974481 m
= 0.05 ln
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)
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3.75(10 - 3)(0.2375)(0.002366137)
= - 791.72 kPa
= 792 kPa (C)
sB =
M(R - rB)
ArB(r - R)
Ans.
41.851 (0.197633863 - 0.1625)
=
3.75(10 - 3)(0.1625)(0.002366137)
= 1.02 MPa (T)
Ans.
Ans:
(smax)c = 792 kPa, (smax)t = 1.02 MPa
608
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6–146. The fork is used as part of a nosewheel assembly
for an airplane. If the maximum wheel reaction at the end of
the fork is 840 lb, determine the maximum bending stress in
the curved portion of the fork at section a- a. There the
cross-sectional area is circular, having a diameter of 2 in.
a
10 in.
a
30⬚
a +©MC = 0;
M - 840(6 - 10 sin 30°) = 0
M = 840 lb # in.
6 in.
840 lb
dA
= 2p(r - 2 r- 2 - c2)
LA r
= 2p(10 - 22(102 - (1)2)
= 0.314948615 in.
A = p c2 = p (1)2 = p in2
A
R =
=
dA
L
Ar
p
= 9.974937173 in.
0.314948615
r - R = 10 - 9.974937173 = 0.025062827 in.
sA =
840(9.974937173 - 9)
M(R - rA)
=
= 1.16 ksi (T) (max)
ArA(r - R)
p(9)(0.025062827)
sB =
M(R - rB)
840(9.974937173 - 11)
=
= - 0.994 ksi (C)
ArB(r - R)
p(11)(0.025062827)
Ans.
Ans:
(smax)t = 1.16 ksi, (smax)c = 0.994 ksi
609
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6–147. If the curved member is subjected to the internal
moment of M = 600 lb # ft, determine the bending stress
developed at points A, B and C. Using these results, sketch
the stress distribution on the cross section.
C
2 in.
C
B
A
B
A
0.75 in.
1.25 in.
1 in.
6 in.
M
M
Internal Moment: M = -600 ft is negative since it tends to increase the curvature of
the curved member.
Section Properties: The location of the centroid of the cross section from the center
of the beam’s curvature, Fig. a, is
~
r =
6.625(1.25)(1) + 7.625(0.75)(2)
©rA
=
= 7.17045 in.
©A
1.25(1) + 0.75(2)
The location of the neutral surface from the center of the beam’s curvature can be
determined from
A
R =
©
L
A
dA
r
where
www.elsolucionario.org
A = 1.25(1) + 0.75(2) = 2.75 in2
dA
7.25
8
©L
= (1) ln
+ 2ln
= 0.38612 in
A r
6
7.25
Thus,
R =
2.75
= 7.122099 in.
0.38612
and
e = r - R = 0.0483559 in.
Normal Stress:
sA =
M(R - rA)
- 600(12)(7.122099 - 6)
= - 10.1 ksi = 10.1 ksi (C)
=
AerA
2.75(0.0483559)(6)
Ans.
sB =
M(R - rB)
- 600(12)(7.122099 - 7.25)
=
= 0.955 ksi (T)
AerB
2.75(0.0483559)(7.25)
Ans.
sC =
M(R - rC)
- 600(12)(7.122099 - 8)
=
= 5.94 ksi (T)
AerC
2.75(0.0483559)(8)
Ans.
The normal stress distribution across the cross section is shown in Fig. b.
Ans:
sA = 10.1 ksi (C), sB = 0.955 ksi (T),
sC = 5.94 ksi (T)
610
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*6–148. If the curved member is made from material
having an allowable bending stress of sallow = 15 ksi ,
determine the maximum allowable internal moment M
that can be applied to the member.
C
2 in.
C
B
A
B
A
1 in.
6 in.
M
M
Internal Moment: M is negative since it tends to increase the curvature of the
curved member.
Section Properties: The location of the centroid of the cross section from the center
of the beam’s curvature, Fig. a, is
©r~A
r =
=
©A
6.625(1.25)(1) + 7.625(0.75)(2)
= 7.17045 in.
1.25(1) + 0.75(2)
The location of the neutral surface from the center of the beam’s curvature can be
determined from
A
R =
©
dA
L
A r
where
A = 1.25(1) + 0.75(2) = 2.75 in2
©L
dA
A r
= (1)ln
7.25
8
+ 2ln
= 0.38612 in
6
7.25
Thus,
R =
2.75
= 7.122099 in.
0.38612
and
e = r - R = 0.0483559 in.
Normal Stress: The maximum normal stress can occur at either point A or C. For
point A which is in compression,
sallow =
M(rA - R)
;
AerA
- 15 =
M(7.122099 - 6)
2.75(0.0483559)(6)
M = - 10.67 kip # in = 889 lb # ft (controls)
For point C which is in tension,
sallow =
M(rC - R)
;
AerC
15 =
M(7.122099 - 8)
2.75(0.0483559)(8)
M = - 18.18 kip # in = 1.51 lb # ft
611
Ans.
0.75 in.
1.25 in.
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M ⫽125 N⭈m
6–149. A 100-mm-diameter circular rod is bent into an
S shape. If it is subjected to the applied moments
M = 125 N # m at its ends, determine the maximum tensile
and compressive stress developed in the rod.
400 mm
400 mm
M ⫽125 N⭈m
dA
LA r
= 2p(r - 2 r- 2 - c2)
= 2p(0.45 - 2 0.452 - 0.052) = 0.01750707495 m
A = p c2 = p (0.052) = 2.5 (10 - 3)p m2
R =
2.5(10 - 3)p
A
= 0.448606818
=
dA
0.017507495
1A r
r - R = 0.45 - 0.448606818 = 1.39318138(10 - 3) m
On the upper edge of each curve:
sA =
M(R - rA)
-125(0.448606818 - 0.4)
= - 1.39 MPa (max) Ans.
=
ArA(r - R)
2.5(10 - 3)p(0.4)(1.39318138)(10 - 3)
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125(0.448606818 - 0.5)
M(R - rD)
sD =
= - 1.17 MPa
=
ArD(r - R)
2.5(10 - 3)p(0.5)(1.39318138)(10 - 3)
On the lower edge of each curve:
sB =
M(R - rB)
- 125(0.448606818 - 0.5)
= 1.17 MPa
=
ArB(r - R)
2.5(10 - 3)p(0.5)(1.39318138)(10 - 3)
sC =
M(R - rC)
125(0.448606818 - 0.4)
= 1.39 MPa
=
ArC(r - R)
2.5(10 - 3)p(0.4)(1.39318138)(10 - 3)
Ans.
Ans:
smax = ; 1.39 MPa
612
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6–150. The bar is subjected to a moment of M = 153 N # m.
Determine the smallest radius r of the fillets so that an
allowable bending stress of sallow = 120 MPa is not exceeded.
60 mm
40 mm
r
7 mm
M
M
r
smax = K
Mc
I
120(106) = K
J 1 (0.007)(0.04)3 K
(153)(0.02)
12
K = 1.46
60
w
=
= 1.5
h
40
From Fig. 6-43,
r
= 0.2
h
r = 0.2(40) = 8.0 mm
Ans.
Ans:
r = 8.0 mm
613
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6–151. The bar is subjected to a moment of M = 17.5 N # m.
If r = 6 mm determine the maximum bending stress in the
material.
60 mm
40 mm
r
7 mm
M
M
r
w
60
=
= 1.5;
h
40
6
r
=
= 0.15
h
40
From Fig. 6–43,
K = 1.57
smax = K
17.5(0.02)
Mc
= 1.57 1
= 14.7 MPa
I
J 12(0.007)(0.04)3 K
Ans.
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Ans:
smax = 14.7 MPa
614
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80 mm
80 mm
*6–152. 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.
7 mm
7 mm
20 mm
20 mm
r
M M
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
with
r
r
w
80
=
= 4 and K = 1.45, then = 0.25.
h
20
h
r
= 0.25
20
r = 5.00 mm
Ans.
615
r
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6–153. 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
r
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
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.
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Ans:
smax = 54.4 MPa
616
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6–154. 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
b
0.25
=
= 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(103) = 1.92 c 1
d
3
12 (0.5)(1.25)
P = 122 lb
Ans.
Ans:
P = 122 lb
617
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6–155. 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 in.
1.25 in.
20 in.
b =
P
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
smax = K
2000(0.625)
Mc
d = 29.5 ksi
= 1.92 c 1
3
I
12 (0.5)(1.25)
Ans.
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Ans:
smax = 29.5 ksi
618
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*6–156. 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
200 mm
r
7
=
= 0.175
h
40
60
w
=
= 1.5
h
40
From Fig. 6-43, K = 1.5
(sA)max = K
(35)(0.02)
MAc
= 1.5 c 1
d = 19.6875 MPa
3
I
(0.01)(0.04
)
12
(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.
619
40 mm
7 mm
C
L
2
B
L
2
200 mm
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6–157. 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
10 mm
6 mm
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
r
w
45
3
=
= 1.5 and
=
= 0.1, we have K = 1.77
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
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M = 41.7 N # m (Controls !)
Ans.
For the larger section
smax = sallow = K
Mc
;
I
200 A 106 B = 1.77 B 1
M(0.015)
3
12 (0.015)(0.03 )
R
M = 254 N # m
Ans:
M = 41.7 N # m
620
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6–158.
beam.
Determine the shape factor for the wide-flange
15 mm
20 mm
200 mm
Mp
15 mm
200 mm
1
1
(0.2)(0.23)3 (0.18)(0.2)3 = 82.78333(10 - 6) m4
12
12
Ix =
C1 = T1 = sY(0.2)(0.015) = 0.003 sY
C2 = T2 = sY(0.1)(0.02) = 0.002 sY
Mp = 0.003 sY(0.215) + 0.002 sY(0.1) = 0.000845 sY
sY =
MYc
I
MY =
sY(82.78333)(10 - 6)
= 0.000719855 sY
0.115
k =
Mp
MY
=
0.000845 sY
= 1.17
0.000719855 sY
Ans.
Ans:
k = 1.17
621
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
6–159. 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
20 mm
200 mm
Mp
15 mm
200 mm
Ix =
1
1
(0.2)(0.23)3 (0.18)(0.2)3 = 82.78333(10 - 6) m4
12
12
C1 = T1 = sY(0.2)(0.015) = 0.003 sY
C2 = T2 = sY(0.1)(0.02) = 0.002 sY
Mp = 0.003 sY(0.215) + 0.002 sY(0.1) = 0.000845 sY
= 0.000845(250)(106) = 211.25 kN # m
s¿ =
Mpc
I
211.25(103)(0.115)
=
y
0.115
=
;
250
293.5
82.78333(10 - 6)
= 293.5 MPa
y = 0.09796 m = 98.0 mm
stop = sbottom = 293.5 - 250 = 43.5 MPa
Ans.
www.elsolucionario.org
Ans:
stop = sbottom = 43.5 MPa
622
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*6–160. Determine the shape factor for the cross section
of the H-beam.
200 mm
20 mm
Mp
200 mm
20 mm
1
1
(0.2)(0.023) + 2 a b (0.02)(0.23) = 26.8(10 - 6) m4
12
12
lx =
C1 = T1 = sY(2)(0.09)(0.02) = 0.0036 sY
C2 = T2 = sY(0.01)(0.24) = 0.0024 sY
Mp = 0.0036 sY(0.11) + 0.0024 sY(0.01) = 0.00042 sY
sY =
MYc
I
MY =
sY(26.8)(10 - 6)
= 0.000268 sY
0.1
k =
Mp
MY
=
0.00042 sY
= 1.57
0.000268 sY
Ans.
623
20 mm
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6–161. 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
20 mm
Mp
20 mm
200 mm
20 mm
Ix =
1
1
(0.2)(0.023) + 2 a b (0.02)(0.23) = 26.8(10 - 6) m4
12
12
C1 = T1 = sY(2)(0.09)(0.02) = 0.0036 sY
C2 = T2 = sY(0.01)(0.24) = 0.0024 sY
Mp = 0.0036 sY(0.11) + 0.0024 sY(0.01) = 0.00042 sY
Mp = 0.00042(250)(106) = 105 kN # m
s¿ =
Mpc
I
105(103)(0.1)
=
y
0.1
=
;
250
392
26.8(10 - 6)
= 392 MPa
y = 0.0638 = 63.8 mm
sT = sB = 392 - 250 = 142 MPa
Ans.
www.elsolucionario.org
Ans:
stop = sbottom = 142 MPa
624
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
6–162. 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.
25 mm
150 mm
25 mm
25 mm
150 mm
25 mm
Plastic Moment:
MP = 250 A 106 B (0.2)(0.025)(0.175) + 250 A 106 B (0.075)(0.05)(0.075)
= 289062.5 N # m
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 =
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.
Ans:
stop = sbottom = 67.1 MPa
625
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
6–163. Determine the plastic moment Mp that can be
supported by a beam having the cross section shown.
sY = 30 ksi .
2 in.
1 in.
Mp
10 in.
1 s dA = 0
1 in.
C1 + C2 - T1 = 0
p(22 - 12)(30) + (10 - d)(1)(30) - d(1)(30) = 0
3p + 10 - 2d = 0
d = 9.7124 in. 6 10 in.
2
OK
2
Mp = p(2 - 1 )(30)(2.2876)
+ (0.2876)(1)(30)(0.1438)
+ (9.7124)(1)(30)(4.8562)
= 2063 kip # in.
= 172 kip # ft
Ans.
www.elsolucionario.org
Ans:
Mp = 172 kip # ft
626
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*6–164. 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
6 in.
©yA
7.5(3)(6) + 3(6)(3)
y =
=
= 5.25 in.
©A
3(6) + 6(3)
1.5 in. 3 in.
The moment of inertia of the cross section about the neutral axis is
1.5 in.
1
1
I =
(3) A 63 B + 3(6)(5.25 - 3)2 +
(6) A 33 B + 6(3)(7.5 - 5.25)2
12
12
= 249.75 in4
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.
627
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
6–165. 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
y =
6 in.
7.5(3)(6) + 3(6)(3)
©yA
=
= 5.25 in.
©A
3(6) + 6(3)
1.5 in. 3 in.
The moment of inertia of the cross-section about the neutral axis is
1.5 in.
1
1
I =
(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,
LA
sdA = 0;
T - C1 - C2 = 0
d(3) (36) - (6 - d)(3)(36) - 3(6) (36) = 0
www.elsolucionario.org
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.
Ans:
MY = 143 kip # ft, MP = 243 kip # ft
628
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
6–166. Determine the shape factor for the cross section of
the beam.
10 mm
15 mm
200 mm
Mp
10 mm
I =
200 mm
1
1
(0.2)(0.22)3 - (0.185)(0.2)3 = 54.133(10 - 6) m4
12
12
C1 = sY(0.01)(0.2) = (0.002) sY
C2 = sY(0.1)(0.015) = (0.0015) sY
Mp = 0.002 sY(0.21) + 0.0015 sY(0.1) = 0.0005 sY
sY =
MYc
I
MY =
sY(54.133)(10 - 6)
= 0.000492 sY
0.11
k =
Mp
MY
=
0.00057 sY
= 1.16
0.000492 sY
Ans.
Ans:
k = 1.16
629
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6–167. The beam is made of an elastic-plastic material for
which sY = 200 MPa. If the largest moment in the beam
occurs within the center section a -a, determine the
magnitude of each force P that causes this moment to be
(a) the largest elastic moment and (b) the largest plastic
moment.
P
P
a
a
2m
2m
2m
2m
200 mm
100 mm
(1)
M = 2P
a) Elastic moment
I =
1
(0.1)(0.23) = 66.667(10 - 6) m4
12
sY =
MYc
I
MY =
200(106)(66.667)(10 - 6)
0.1
= 133.33 kN # m
From Eq. (1)
133.33 = 2 P
www.elsolucionario.org
Ans.
P = 66.7 kN
b) Plastic moment
b h2
sY
4
Mp =
=
0.1(0.22)
(200)(106)
4
= 200 kN # m
From Eq. (1)
200 = 2 P
P = 100 kN
Ans.
Ans:
Elastic: P = 66.7 kN,
Plastic: P = 100 kN
630
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*6–168. Determine the shape factor of the cross section.
3 in.
3 in.
3 in.
3 in.
The moment of inertia of the cross section about the neutral axis is
I =
1
1
(3)(93) +
(6) (33) = 195.75 in4
12
12
Here, smax = sY and c = 4.5 in. Then
smax =
Mc
;
I
sY =
MY(4.5)
195.75
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 =
MP
74.25 sY
=
= 1.71
MY
43.5 sY
Ans.
631
3 in.
3 in.
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*6–169. 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.
3 in.
3 in.
3 in.
3 in.
The moment of inertia of the cross section about the neutral axis is
I =
1
1
(3)(93) +
(6)(33) = 195.75 in4
12
12
Here, smax = sY = 36 ksi and c = 4.5 in. Then
smax =
Mc
;
I
36 =
MY (4.5)
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)
www.elsolucionario.org
= 324(6) + 486(1.5)
= 2673 kip # in. = 222.75 kip # ft = 223 kip # ft
Ans.
Ans:
MY = 130.5 kip # ft, Mp = 223 kip # ft
632
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6–170. 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
P
8 ft
6 ft
6 ft
6 in.
12 in.
10 in.
From the moment diagram shown in Fig. a, Mmax = 6 P.
5 in.
The moment of inertia of the beam’s cross section about the neutral axis is
I =
1
1
(6)(123) (5)(103) = 447.33 in4
12
12
Here, smax = sY = 36 ksi and c = 6 in.
smax =
Mc
;
I
36 =
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.
Ans:
Elastic: P = 37.3 kip,
Plastic: P = 45.5 kip
633
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6–171. 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 =
ri
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 c 2a
=
4ro
4ri
b d - T2 c 2 a
bd
3p
3p
8ro
8ri
p 2
p
r s a
b - ri 2sY a b
2 o Y 3p
2
3p
www.elsolucionario.org
4
= A ro 3 - ri 3 B sY
3
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 r 4 - ri 4 B sY
4ro o
Ans.
Ans:
k =
634
16ro A ro3 - r3i B
3p A ro4 - r4i B
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*6–172.
Determine the shape factor for the member.
–h
2
–h
2
Plastic analysis:
T = C =
MP =
b
h
1
bh
(b)a b sY =
s
2
2
4 Y
b h2
bh
h
sY a b =
s
4
3
12 Y
Elastic analysis:
I = 2c
h 3
1
b h3
(b)a b d =
12
2
48
sY A 48 B
sYI
b h2
MY =
=
=
s
h
c
24 Y
2
bh3
Shape Factor:
k =
Mp
MY
=
bh2
12 sY
bh2
24 sY
Ans.
= 2
635
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6–173. 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
–h
2
b
Elastic analysis:
I = 2c
1
(4)(3)3 d = 18 in4
12
MY =
36(18)
sYI
= 216 kip # in. = 18 kip # ft
=
c
3
Ans.
Plastic analysis:
T = C =
1
(4)(3)(36) = 216 kip
2
6
Mp = 2160 a b = 432 kip # in. = 36 kip # ft
3
Ans.
www.elsolucionario.org
Ans:
MY = 18 kip # ft,
Mp = 36 kip # ft
636
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w
6–174. The beam is made of an elastic-plastic material for
which sY = 30 ksi. If the largest moment in the beam
occurs at the center section a - a, determine the intensity of
the distributed load w that causes this moment to be (a) the
largest elastic moment and (b) the largest plastic moment.
a
a
10 ft
10 ft
8 in.
8 in.
M = 50 w
(1)
(a) Elastic moment
I =
1
(8)(83) = 341.33 in4
12
sY =
MYc
I
MY =
30(341.33)
4
= 2560 kip # in. = 213.33 kip # ft
From Eq. (1),
213.33 = 50 w
w = 4.27 kip>ft
Ans.
(b) Plastic moment
C = T = 30(8)(4) = 960 kip
Mp = 960 (4) = 3840 kip # in. = 320 kip # ft
From Eq. (1)
320 = 50 w
w = 6.40 kip>ft
Ans.
Ans:
Elastic: w = 4.27 kip>ft,
Plastic: w = 6.40 kip>ft
637
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
w0
6–175. 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.
Elastic analysis:
9 ft
9 ft
1
1
I =
(8)(163) (6)(123) = 1866.67 in4
12
12
sYI
;
Mmax =
c
8 in.
25(1866.67)
27w0(12) =
8
16 in.
12 in.
w0 = 18.0 kip>ft
Ans.
Plastic analysis:
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.
www.elsolucionario.org
Ans:
Elastic: w0 = 18.0 kip>ft,
Plastic: w0 = 22.8 kip>ft
638
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*6–176. The wide-flange member is made from an elasticplastic material. Determine the shape factor.
t
h
t
t
b
Plastic analysis:
T1 = C1 = sYb t ;
T2 = C2 = sY a
Mp = sY b t(h - t) + sY a
= sY[b t(h - t) +
h - 2t
bt
2
h - 2t
h - 2t
b (t) a
b
2
2
t
(h - 2t)2]
4
Elastic analysis:
I =
1
1
b h3 (b - t)(h - 2t)3
12
12
=
1
[b h3 - (b - t)(h - 2t)3]
12
MY =
1
) [bh3 - (b - t)(h - 2t)3]
sY(12
sYI
=
h
c
2
=
bh3 - (b - t)(h - 2t)3
sY
6h
Shape Factor:
k =
Mp
MY
=
=
[b t(h - t) + 14(h - 2t)2]sY
bh3 - (b - t)(h - 2t)3
sY
6h
3h 4b t(h - t) + t(h - 2t)2
c
d
2 b h3 - (b - t)(h - 2t)3
Ans.
639
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*6–177. 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-1 115P2] ksi, where tan-1 115P2
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 P = 0.0015y.
s = 20 tan - 1 (15P)
= 20 tan - 1 [15(0.0015y)]
= 20 tan - 1 (0.0225y)
When Pmax = 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.
www.elsolucionario.org
ysdA
2 in
= 2
LA
y C 20 tan 1 (0.0225y) D (2dy)
-
L0
2 in
= 80
L0
= 80 B
y tan - 1 (0.0225y) dy
1 + (0.0225)2y2
2
2(0.0225)
tan - 1 (0.0225y) -
2 in.
y
R2
2(0.0225) 0
= 4.798 kip # in
Equating
M = 4.00P(12) = 4.798
P = 0.100 kip = 100 lb
Ans.
Ans:
P = 100 lb
640
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6–178. 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
0.04
compression
80
100
Ultimate Moment:
LA
s dA = 0;
C - T2 - T1 = 0
1
d
1
1 d
s c (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
Since
P
0.04
=
, then 0.02 - d = 25Pd.
d
0.02 - d
40(106)
s
s
.
=
= 2(109), then P =
P
0.02
2(109)
And since
So then 0.02 - d =
23sd
= 1.25(10 - 8)sd.
2(109)
Substituting for 0.02 - d, then solving for s, yields s = 74.833 MPa. Then
P = 0.037417 mm>mm and d = 0.010334 m.
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.
Ans:
M = 94.7 N # m
641
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6–179. 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
0.04
P (in./in.)
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
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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 trapezoidal area was used in calculation of moment.
Ans:
Maximum elastic moment: M = 35.0 kip # ft,
Ultimate moment: M = 59.8 kip # ft
642
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*6–180. 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
100 mm
M
30 mm
P (mm/ mm)
Pmax = 0.02
smax = 10 A 106 B (0.02)1>4 = 3.761 MPa
P
0.02
=
y
0.05
P = 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 =
LA
y s dA = 2
M = 0.47716 A 106 B
L0
0.05
L0
4
y5>4dy = 0.47716 A 106 B a b (0.05)9>4
9
M = 251 N # m
Ans.
643
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± σ (ksi)
90
80
6–181. 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 Pmax = 0.03.
60
4 in. M
0.006
90 - 80
s - 80
=
;
0.03 - 0.025
0.05 - 0.025
0.025
0.05
P (in./in.)
3 in.
s = 82 ksi
C1 = T1 =
1
(0.3333)(80 + 82)(3) = 81 kip
2
C2 = T2 =
1
(1.2667)(60 + 80)(3) = 266 kip
2
C3 = T3 =
1
(0.4)(60)(3) = 36 kip
2
M = 81(3.6680) + 266(2.1270) + 36(0.5333)
Ans.
= 882.09 kip # in. = 73.5 kip # ft
Note: The centroid of a trapezoidal area was used in calculation of moment areas.
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Ans:
M = 73.5 kip # ft
644
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± σ (ksi)
90
80
6–182. 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 Pmax = 0.05.
60
4 in. M
0.006
s1 =
0.025
0.05
P (in./in.)
3 in.
60
P = 10(103)P
0.006
s2 - 60
80 - 60
=
P - 0.006
0.025 - 0.006
s2 = 1052.63P + 53.684
s3 - 80
90 - 80
=
;
P - 0.025
0.05 - 0.025
P =
s3 = 400P + 70
0.05
(y) = 0.025y
2
Substitute P into s expression:
s1 = 250y
0 … y 6 0.24 in.
s2 = 26.315y + 53.684
0.24 6 y 6 1 in.
s3 = 10y + 70
1 in. 6 y … 2 in.
dM = ys dA = ys(3 dy)
0.24
M = 2[3 10
2
1
2
2
2
250y dy + 3 10.24 (26.315y + 53.684y) dy + 3 11 (10y + 70y) dy]
= 980.588 kip # in. = 81.7 kip # ft
Ans.
Also, the solution can be obtained from stress blocks as in Prob. 6–181.
Ans:
M = 81.7 kip # ft
645
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6–183. Determine the shape factor for the wide-flange beam.
20 mm
30 mm
180 mm
Mp
20 mm
180 mm
I =
1
1
(0.18)(0.223) (0.15)(0.183)
12
12
= 86.82(10 - 6) m4
Plastic moment:
Mp = sY(0.18)(0.02)(0.2) + sY(0.09)(0.03)(0.09)
= 0.963(10 - 3)sY
Shape Factor:
MY =
k =
sY(86.82)(10 - 6)
sYI
=
= 0.789273(10 - 3)sY
c
0.11
Mp
MY
=
0.963(10 - 3)sY
0.789273(10 - 3)sY
Ans.
= 1.22
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Ans:
k = 1.22
646
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*6–184. 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.
20 mm
30 mm
180 mm
Mp
20 mm
180 mm
I =
1
1
(0.18)(0.223) (0.15)(0.183)
12
12
= 86.82(10 - 6) m4
Plastic moment:
Mp = 250(106)(0.18)(0.02)(0.2)
+ 250(106)(0.09)(0.03)(0.09)
= 240750 N # m
Applying a reverse Mp = 240750 N # m
sp =
Mpc
I
240750(0.11)
=
86.82(10 - 6)
= 305.03 MPa
¿
= 305 - 250 = 55.0 MPa
s¿top = sbottom
Ans.
647
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w
6–185. 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
A
B
2/3 L
+ c ©Fy = 0;
a + ©M = 0;
2wL
1 w 2
x = 0
27
2 L
x =
4
L = 0.385 L
A 27
M +
1w
1
2wL
(0.385L) = 0
(0.385L)2 a b (0.385L) 2L
3
27
M = 0.0190 wL2
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648
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y
6–186. 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.
z
125 mm
M
x
20 mm
75 mm
20 mm
9
n =
200(10 )
Est
= 18.182
=
Ew
11(109)
I =
1
(0.80227)(0.1253) = 0.130578(10 - 3)m4
12
Failure of wood :
(sw)max =
Mc
I
20(106) =
M(0.0625)
0.130578(10 - 3)
;
M = 41.8 kN # m
Failure of steel :
(sst)max =
nMc
I
130(106) =
18.182(M)(0.0625)
0.130578(10 - 3)
M = 14.9 kN # m (controls)
Ans.
Ans:
M = 14.9 kN # m
649
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y
6–187. Solve Prob. 6–186 if the moment is applied about
the y axis instead of the z axis as shown.
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)
Failure of steel :
(sst)max =
Mc1
I
130(106) =
;
M = 113 kN # m
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M(0.0575)
11.689616(10 - 6)
M = 26.4 kN # m (controls)
Ans.
Ans:
M = 26.4 kN # m
650
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*6–188. A shaft is made of a polymer having a parabolic
upper and lower 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
50 mm
50 mm
Maximum Bending Stress: The moment of inertia about y axis must be determined
first in order to use flexure formula
I =
LA
y2 dA
100 mm
= 2
L0
y2 (2z) dy
100 mm
= 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 106 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 y c a
M =
smax
by d (2z dy) r
100
smax 100 mm 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.
651
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6–189. 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
a
45 lb
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.
smax =
394.14(0.375)
Mc
= 8.41 ksi
= 1
3
I
12 (0.5)(0.75 )
Ans.
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Ans:
smax = 8.41 ksi
652
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M 85 Nm
6–190. 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
400 mm
A
20 mm
15 mm
B
150 mm
30
20 mm
B
r2
0.57
0.59
dA
0.42
= b ln
+ 0.015 ln
+ 0.1 ln
= 0.1 ln
r
r
0.40
0.42
0.57
1
LA
= 0.012908358 m
A = 2(0.1)(0.02) + (0.15)(0.015) = 6.25(10 - 3) m2
R =
A
=
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 =
M(R - rB )
ArB (r - R)
Ans.
85(0.484182418 - 0.40)
=
6.25(10 - 3)(0.40)(0.010817581)
= 265 kPa (T)
Ans.
Ans:
sA = 225 kPa (C), sB = 265 kPa (T)
653
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6–191. Determine the shear and moment in the beam as
functions of x, where 0 … x 6 6 ft, then draw the shear and
moment diagrams for the beam.
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 - 2x a b - M = 0
2
M = - x2 + 20x - 166
Ans.
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Ans:
V = 20 - 2x, M = - x2 + 20x - 166
654
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*6–192. A wooden beam has a square cross section as
shown. Determine which orientation of the beam provides
the greatest strength at resisting the moment M. What is the
difference in the resulting maximum stress in both cases?
a
M
M
a
a
(a)
Case (a):
smax =
M(a>2)
Mc
6M
= 1 4 = 3
I
a
12 (a)
Case (b):
I = 2c
3
1 2
1
1
1
2
1
1
2
ab a bd d = 0.08333 a4
ab c a
ab a
ab + a
ab a
a
3
2 22
36 22
22
22
22
1
ab
Ma
8.4853 M
Mc
22
=
smax =
=
I
0.08333 a4
a3
Case (a) provides higher strength since the resulting maximum stress is less for a
given M and a.
Case (a)
Ans.
¢smax =
8.4853 M
6M
M
- 3 = 2.49 a 3 b
a3
a
a
Ans.
655
a
(b)
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6–193. Draw the shear and moment diagrams for the shaft
if it is subjected to the vertical loadings of the belt, gear, and
flywheel. The bearings at A and B exert only vertical
reactions on the shaft.
300 N
450 N
A
B
200 mm
400 mm
300 mm
200 mm
150 N
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656
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6–194. 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.
Ans:
smax =
6M
(cos u + sin u),
a3
u = 45°, a = 45°
657
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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
20 mm
The moment of inertia of the cross-section about the neutral axis is
I =
1
1
(0.2)(0.343) (0.18)(0.33) = 0.2501(10 - 3) m4
12
12
From Fig. a,
QA = y¿A¿ = 0.16 (0.02)(0.2) = 0.64(10 - 3) m3
Applying the shear formula,
VQA
20(103)[0.64(10 - 3)]
=
tA =
It
0.2501(10 - 3)(0.02)
= 2.559(106) Pa = 2.56 MPa
Ans.
The shear stress component at A is represented by the volume element shown in
Fig. b.
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Ans:
tA = 2.56 MPa
658
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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 =
20 mm
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 thickness t is the smallest.
tmax =
20(103) [0.865(10 - 3)]
VQmax
=
It
0.2501(10 - 3) (0.02)
= 3.459(106) Pa = 3.46 MPa
Ans.
Ans:
tmax = 3.46 MPa
659
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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
20 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.
www.elsolucionario.org
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.
Ans:
Vw = 19.0 kN
660
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*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.
661
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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
www.elsolucionario.org
A 0.16728 - 0.01536y2 B (12dy)
= 3.82 kip
Ans.
Ans:
Vf = 3.82 kip
662
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
7–6. 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.
Ans:
Vmax = 100 kN
663
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
7–7. The shaft is supported by a smooth thrust bearing at
A and a smooth journal bearing at B. If P = 20 kN,
determine the absolute maximum shear stress in the shaft.
A
C
1m
Support Reactions: As shown on the free-body diagram of the beam, Fig. a.
B
1m
1m
P
P
Maximum Shear: The shear diagram is shown in Fig. b. As indicated, Vmax = 20 kN.
30 mm
Section Properties: The moment of inertia of the hollow circular shaft about the
neutral axis is
40 mm
I =
D
p
(0.044 - 0.034) = 0.4375(10-6)p m4
4
Qmax can be computed by taking the first moment of the shaded area in Fig. c about
the neutral axis.
Here, y¿1 =
4(0.04)
4(0.03)
4
1
=
m and y¿2 =
=
m. Thus,
3p
75p
3p
25p
Qmax = y¿1A¿1 - y¿2A¿2
=
1 p
4 p
c (0.042) d c (0.032) d = 24.667(10-6) m3
75p 2
25p 2
Shear Stress: The maximum shear stress occurs at points on the neutral axis since
Q is maximum and the thickness t = 210.04 - 0.032 = 0.02 m is the smallest.
tmax =
Vmax Qmax
20(103)(24.667)(10-6)
= 17.9 MPa
=
It
0.4375(10-6)p (0.02)
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Ans.
FPO
Ans:
tmax = 17.9 MPa
664
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*7–8. The shaft is supported by a smooth thrust bearing at A
and a smooth journal bearing at B. If the shaft is made from a
material having an allowable shear stress of tallow = 75 MPa,
determine the maximum value for P.
A
C
1m
B
1m
30 mm
Maximum Shear: The shear diagram is shown in Fig. b. As indicated, Vmax = P.
Section Properties: The moment of inertia of the hollow circular shaft about the
neutral axis is
I =
p
(0.044 - 0.034) = 0.4375(10-6)p m4
4
Qmax can be computed by taking the first moment of the shaded area in Fig. c about
the neutral axis.
Here, y¿1 =
4(0.04)
4(0.03)
4
1
=
m and y¿2 =
=
m. Thus,
3p
75p
3p
25p
Qmax = y¿1A¿1 - y¿2A¿2
=
1 p
4 p
c (0.042) d c (0.032) d = 24.667(10-6) m3
75p 2
25p 2
Shear Stress: The maximum shear stress occurs at points on the neutral axis since
Q is maximum and the thickness t = 2(0.04 - 0.03) = 0.02 m.
tallow =
Vmax Qmax
;
It
75(106) =
P(24.667)(10-6)
0.4375(10-6)p(0.02)
P = 83 581.22 N = 83.6 kN
665
Ans.
1m
P
P
Support Reactions: As shown on the free-body diagram of the shaft, Fig. a.
D
40 mm
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7–9. Determine the largest shear force V that the member
can sustain if the allowable shear stress is tallow = 8 ksi.
3 in.
1 in.
V
3 in. 1 in.
1 in.
y =
(0.5)(1)(5) + 2 [(2)(1)(2)]
= 1.1667 in.
1 (5) + 2 (1)(2)
I =
1
(5)(13) + 5 (1)(1.1667 - 0.5)2
12
+ 2a
1
b (1)(23) + 2 (1)(2)(2 - 1.1667)2 = 6.75 in4
12
Qmax = ©y¿A¿ = 2 (0.91665)(1.8333)(1) = 3.3611 in3
tmax = tallow =
8 (103) = -
VQmax
It
V (3.3611)
6.75 (2)(1)
V = 32132 lb = 32.1 kip
Ans.
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Ans:
V = 32.1 kip
666
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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.
Ans:
tmax = 4.48 ksi
667
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
w
7–11. The overhang beam is subjected to the uniform
distributed load having an intensity of w = 50 kN>m.
Determine the maximum shear stress developed in the
beam.
A
B
3m
3m
50 mm
100 mm
tmax =
VQ
150(103) N (0.025 m)(0.05 m)(0.05 m)
=
1
3
It
12 10.05 m)(0.1 m) (0.05 m)
tmax = 45.0 MPa
Ans.
Because the cross section is a rectangle, then also,
tmax = 1.5
150(103) N
V
= 1.5
= 45.0 MPa
A
(0.05 m)(0.1 m)
Ans.
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Ans:
tmax = 45.0 MPa
668
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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 lb = 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
669
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
7–13. Determine the maximum shear stress in the strut if
it is subjected to a shear force of V = 20 kN.
12 mm
60 mm
V
12 mm
80 mm
Section Properties:
20 mm
20 mm
INA =
1
1
(0.12) A 0.0843 B (0.04) A 0.063 B
12
12
= 5.20704 A 10 - 6 B m4
Qmax = ©y¿A¿
= 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
www.elsolucionario.org
20(103)(87.84)(10 - 6)
=
5.20704(10 - 6)(0.08)
= 4. 22 MPa
Ans.
Ans:
tmax = 4.22 MPa
670
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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
V
12 mm
80 mm
20 mm
20 mm
Section Properties:
INA =
1
1
(0.12) A 0.0843 B (0.04) A 0.063 B
12
12
= 5.20704 A 10 - 6 B m4
Qmax = ©y¿A¿
= 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.
Ans:
V = 190 kN
671
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7–15. The strut is subjected to a vertical shear of V = 130 kN.
Plot the intensity of the shear-stress distribution acting over
the cross-sectional area, and compute the resultant shear
force developed in the vertical segment AB.
B
150 mm
50 mm
A
V ⫽ 130 kN
I =
150 mm
1
1
(0.05)(0.353) +
(0.3)(0.053) = 0.18177083(10-3) m4
12
12
QC = y ¿A ¿ = 10.1210.05210.152 = 0.75110-32 m3
150 mm
50 mm
150 mm
QD = ©y ¿A¿ = 10.1210.05210.152 + 10.0125210.35210.0252
= 0.859375110-32 m3
t =
VQ
It
1tC2t = 0.05 m =
= 10.7 MPa
1301103210.752110-32
= 1.53 MPa
0.18177083110-3210.052
1tC2t = 0.35 m =
tD =
1301103210.752110-32
0.18177083110-3210.352
1301103210.8593752110-32
0.18177083110-3210.352
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A¿ = 10.05210.175 - y2
y¿ = y +
10.175 - y2
2
=
= 1.76 MPa
1
10.175 + y2
2
Q = y¿A¿ = 0.02510.030625 - y22
t =
=
VQ
It
13010.025210.030625 - y22
0.18177083110-3210.052
= 10951.3 - 357593.1 y2
VAB =
L
t dA
dA = 0.05 dy
0.175
=
L0.025
110951.3 - 357593.1y2210.05 dy2
0.175
=
1547.565 - 17879.66y 2 dy
2
L0.025
= 50.3 kN
Ans.
Ans:
VAB = 50.3 kN
672
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*7–16. The steel rod has a radius of 1.25 in. If it is
subjected to a shear of V = 5 kip, determine the maximum
shear stress.
1.25 in.
V= 5 kip
y¿ =
411.252
5
4r
=
=
3p
3p
3p
I =
1 4
1
pr = p11.2524 = 0.610351 p
4
4
5 p11.25 2
= 1.3020833 in3
3p
2
2
Q = y¿A¿ =
tmax =
51103211.30208332
VQ
=
= 1358 psi = 1.36 ksi
It
0.6103511p212.502
Ans.
673
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
7–17. 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. Set w = 125 mm. Show that the neutral axis
is located at y- = 0.1747 m from the bottom and
INA = 0.2182(10 - 3) m4.
200 mm
A
30 mm
25 mm
V
B
250 mm
30 mm
y =
10.015210.125210.032 + 10.155210.025210.252 + 10.295210.2210.032
0.12510.032 + 10.025210.252 + 10.2210.032
w
= 0.1747 m
1
10.125210.0332 + 0.12510.03210.1747 - 0.01522
12
1
+
10.025210.2532 + 0.2510.025210.1747 - 0.15522
12
1
+
10.2210.0332 + 0.210.03210.295 - 0.174722 = 0.218182110-32 m4
12
I =
QA = yA¿ A = 10.310 - 0.015 - 0.1747210.2210.032 = 0.7219110-32 m3
QB = yA¿ B = 10.1747 - 0.015210.125210.032 = 0.59883110-32 m3
tA =
151103210.72192110-32
VQA
= 1.99 MPa
=
It
0.218182110-320.025
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15110 210.598832110 2
VQB
= 1.65 MPa
=
It
0.218182110-320.025
3
tB =
Ans.
-3
Ans.
Ans:
tA = 1.99 MPa, tB = 1.65 MPa
674
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
7–18. If the wide-flange beam is subjected to a shear of
V = 30 kN, determine the maximum shear stress in the
beam. Set w = 200 mm.
200 mm
A
30 mm
25 mm
V
B
250 mm
30 mm
w
Section Properties:
I =
1
1
10.2210.31023 10.175210.25023 = 268.6521102-6 m4
12
12
Qmax = ©yA = 0.062510.125210.0252 + 0.14010.2210.0302 = 1.03531102-3 m3
301102 11.035321102
VQ
= 4.62 MPa
=
It
268.6521102-610.0252
3
tmax =
-3
Ans.
Ans:
tmax = 4.62 MPa
675
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
7–19. 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. Set w = 200 mm.
200 mm
A
30 mm
25 mm
V
B
250 mm
30 mm
I =
1
1
(0.2)(0.310)3 (0.175)(0.250)3 = 268.652(10) - 6 m4
12
12
Q = a
tf =
w
0.155 + y
b (0.155 - y)(0.2) = 0.1(0.024025 - y2)
2
30(10)3(0.1)(0.024025 - y2)
268.652(10)-6(0.2)
0.155
Vf =
L
tf dA = 55.8343(10)6
L0.125
(0.024025 - y2)(0.2 dy)
0.155
1
= 11.1669(10) c 0.024025y - y3 d
3
6
0.125
Vf = 1.457 kN
Vw = 30 - 2(1.457) = 27.1 kN
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Ans.
Ans:
Vw = 27.1 kN
676
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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.
2 in.
30 kip
The moment of inertia of the circular cross section about the neutral axis (x axis) is
I =
p 4
p
r = (24) = 4 p in4
4
4
Q for the differential area shown shaded 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
2y (4 - y2)2 dy
Q =
Ly
3 2 in
2
= - (4 - y2)2 冷
3
y
=
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
20
10
tmax= 2p = p = 3.18 ksi
Ans.
677
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7–21. If the beam is made from wood having an allowable
shear stress tallow = 400 psi, determine the maximum
magnitude of P. Set d = 4 in.
2P
P
A
B
2 ft
2 ft
Support Reactions: As shown on the free-body diagram of the beam, Fig. a.
2 ft
d
Maximum Shear: The shear diagram is shown in Fig. b. As indicated, Vmax = 1.667P.
2 in.
Section Properties: The moment of inertia of the the rectangular beam is
I =
1
(2)(43) = 10.667 in4
12
Qmax can be computed by taking the first moment of the shaded area in Fig. c about
the neutral axis.
Qmax = y ¿A¿ = 1(2)(2) = 4 in3
Shear Stress: The maximum shear stress occurs at points on the neutral axis since
Q is maximum and the thickness t = 2 in. is constant.
tallow =
Vmax Qmax
;
It
400 =
1.667P(4)
10.667(2)
P = 1280 lb = 1.28 kip
Ans.
www.elsolucionario.org
Ans:
P = 1.28 kip
678
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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
B
20 mm
50 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
+
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.
Ans:
tB = 4.41 MPa
679
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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
B
20 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 =
+
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
Qmax = y¿A¿ = (0.026875)(0.05375)(0.02) = 28.8906(10 - 6) m3
tmax =
6(103)(28.8906)(10 - 6)
VQmax
=
It
1.78625(10 - 6)(0.02)
= 4.85 MPa
Ans.
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Ans:
tmax = 4.85 MPa
680
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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
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
= 0.12 m
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
-3
= 7.333(106) Pa
= 7.33 MPa
Ans.
681
30 mm
30 mm
1
I =
(0.03)(0.153) + 0.03(0.15)(0.12 - 0.075)2
12
tmax =
1.5 m
3m
The shear diagram is shown in Fig. b. As indicated, Vmax = 27.5 kN
+
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 section C. Show the result on a volume element
at this point.
10 kN/m
A
B
C
1.5 m
3m
1.5 m
150 mm
150 mm
30 mm
30 mm
Using the method of sections (Fig. a),
VC + 17.5 -
+ c ©Fy = 0;
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)(0.153) + 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
www.elsolucionario.org
= 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 (Fig. b).
tmax =
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.
Ans:
tmax = 3.67 MPa
682
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
7–26. The beam has a square cross section and is made of
wood having an allowable shear stress of tallow = 1.4 ksi. If
it is subjected to a shear of V = 1.5 kip, determine the
smallest dimension a of its sides.
V = 1.5 kip
a
I =
1 4
a
12
a
a3
a a
Qmax = y ¿A¿ = a b a b a =
4 2
8
tmax = tallow =
1.4 =
1.5 a
VQmax
It
a3
b
8
1 4
(a )(a)
12
a = 1.27 in.
Ans.
Ans:
a = 1.27 in.
683
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7–27. The beam is slit longitudinally along both sides as
shown. If it is subjected to an internal shear of V = 250 kN,
compare the maximum shear stress developed in the beam
before and after the cuts were made.
25 mm
200 mm
V
100 mm
25 mm
25 mm
Section Properties: The moment of inertia of the cross section about the neutral axis is
I =
1
1
10.22(0.23) 10.1252(0.153) = 98.1771(10-6) m4
12
12
25 mm
200 mm
25 mm
Qmax is the first moment of the shaded area shown in Fig. a about the neutral axis.Thus,
Qmax = 3y¿1A¿1 + y¿2 A¿2
= 3 C 0.037510.075210.0252 D + 0.087510.025210.22
= 0.6484375(10-3) m3
Maximum Shear Stress: The maximum shear stress occurs at the points on the
neutral axis since Q is maximum and t is minimum. Before the cross section is slit,
t = 310.0252 = 0.075 m.
tmax =
250(103)(0.6484375)(10-3)
VQmax
= 22.0 MPa
=
It
98.1771(10-6)(0.075)
Ans.
www.elsolucionario.org
After the cross section is slit, t = 0.025 m.
(tmax)s =
250(103)(0.6484375)(10-3)
VQmax
= 66.0 MPa
=
It
98.1771(10-6)(0.025)
Ans.
Ans:
tmax = 22.0 MPa, (tmax)s = 66.0 MPa
684
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*7–28. The beam is to be cut longitudinally along both
sides as shown. If it is made from a material having an
allowable shear stress of tallow = 75 MPa, determine the
maximum allowable internal shear force V that can be
applied before and after the cut is made.
25 mm
200 mm
V
Section Properties: The moment of inertia of the cross section about the neutral axis is
I =
1
1
10.22(0.23) 10.1252(0.153) = 98.1771(10-6) m4
12
12
100 mm
25 mm
25 mm
Qmax is the first moment of the shaded area shown in Fig. a about the neutral axis.Thus,
25 mm
200 mm
Qmax = 3y¿1A¿1 + y¿2 A¿2
25 mm
= 310.0375210.075210.0252 + 0.087510.025210.22
= 0.6484375(10-3) m3
Shear Stress: The maximum shear stress occurs at the points on the neutral axis
since Q is maximum and thickness t is minimum. Before the cross section is slit,
t = 310.0252 = 0.075 m.
tallow =
VQmax
;
It
75(106) =
V10.64843752(10-3)
98.1771(10-6)10.0752
V = 851 656.63 N = 852 kN
Ans.
After the cross section is slit, t = 0.025 m.
tallow =
VQmax
;
It
75(106) =
Vs 10.64843752(10-3)
98.1771(10-6)10.0252
Vs = 283 885.54 N = 284 kN
Ans.
685
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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 cross-sectional 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 material only in the elastic zone.
Section Properties:
www.elsolucionario.org
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
=
=
2
3
It
4by¿
A by¿ B (b)
y¿ 2b
tmax =
3
However,
A¿ = 2by¿
tmax =
3P
‚
2A¿
hence
(Q.E.D.)
686
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.
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.)
687
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*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.
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688
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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.
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:
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
225 lb>in.
q
b (6 in.) = 675 lb
F = a bs = a
2
2
Ans.
Ans:
F = 675 lb
689
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7–34. The boards are glued together to form the built-up
beam. If the wood has an allowable shear stress of
tallow = 3 MPa, and the glue seam at B can withstand a
maximum shear stress of 1.5 MPa, determine the maximum
allowable internal shear V that can be developed in the
beam.
50 mm
50 mm
50 mm
50 mm
D
150 mm
V
50 mm
B
150 mm
The moment of inertia of the cross section about the neutral axis is
I =
1
1
10.15210.2532 10.1210.1532 = 0.167187511032 m4
12
12
Then
Qmax = 0.110.05210.152 + 10.0375210.075210.052 = 0.890625110-32 m3
QB = 0.110.05210.152 = 0.75110-32 m3
The maximum shear stress occurs at the points on the neutral axis where Q is a
maximum and t = 0.05 m is the smallest.
1tallow2w =
VQmax
;
It
311062 =
V C 0.890625110-32 D
0.1671875110-3210.052
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V = 28,157 N
For the glue seam at B,
1tallow2g =
V C 0.75110-32 D
VQB
; 1.511062 =
It
0.1671875110-3210.052
V = 16,719 N = 16.7 kN (Controls)
Ans.
Ans:
V = 16.7 kN
690
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7–35. The boards are glued together to form the built-up
beam. If the wood has an allowable shear stress of
tallow = 3 MPa, and the glue seam at D can withstand a
maximum shear stress of 1.5 MPa, determine the maximum
allowable shear V that can be developed in the beam.
50 mm
50 mm
50 mm
50 mm
D
150 mm
V
50 mm
B
150 mm
The moment of inertia of the cross section about the neutral axis is
I =
1
1
10.15210.2532 10.1210.1532 = 0.167187511032 m4
12
12
Then
Qmax = 0.110.05210.152 + 10.0375210.075210.052 = 0.890625110-32 m3
QD = 0.110.05210.052 = 0.25110-32 m3
The maximum shear stress occurs at the points on the neutral axis where Q is a
maximum and t = 0.05 m is the smallest.
1tallow2w =
VQmax
;
It
311062 =
V C 0.890625110-32 D
0.1671875110-3210.052
V = 28,158 N = 28.2 kN 1Controls2
For the glue seam at D,
1tallow2g =
Ans.
V C 0.25110-32 D
VQB
; 1.511062 =
It
0.1671875110-3210.052
V = 50,156 N
Ans:
V = 28.2 kN
691
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*7–36. Three identical boards are bolted together to form the
built-up beam. Each bolt has a shear strength of 1.5 kip and
the bolts are spaced at a distance of s = 6 in. If the wood has an
allowable shear stress of tallow = 450 psi, determine the
maximum allowable internal shear V that can act on the beam.
2 in.
2 in.
s
2 in.
s
s
V
Section Properties: The neutral axis passes through the centroid C of the cross
section as shown in Fig. a. The location of C is
1 in.
1 in.
y =
2[2(4)(1)] + 4(4)(1)
©yA
=
= 2.6667 in.
©A
2(4)(1) + 4(1)
1 in.
Thus, the moment of inertia of the cross section about the neutral axis is
I = 2c
1
1
(1)(43) + 1(4)(2.6667 - 2)2 d +
(1)(43) + 1(4)(4 - 2.6667)2
12
12
= 26.6667 in4
Here, QB can be computed by referring to Fig. b, which is
QB = y ¿1A ¿1 = 1.3333(4)(1) = 5.3333 in3
Referring to Fig. c, QD and Qmax are
QD = y¿3 A¿3 = 2.3333(2)(1) = 4.6667 in3
www.elsolucionario.org
Qmax = y¿2 A¿2 + y¿3 A¿3 = 0.6667(1.3333)(3) + 2.3333(2)(1) = 7.3333 in3
Shear Stress: The maximum shear stress occurs at either point D or points on the
neutral axis. For point D, t = 1 in.
tallow =
VQD
;
It
450 =
V(4.6667)
26.6667(1)
V = 2571.43 lb = 2571 lb
For the points on the neutral axis, t = 3(1) = 3 in. and so
tallow =
VQmax
;
It
450 =
V(7.3333)
26.6667(3)
V = 4909.09 lb
1.5 (103)
F
d =
Shear Flow: Since each bolt has two shear planes, qallow = 2 a b = 2 c
s
6
500 lb>in.
qallow =
VQB
;
I
500 =
V(5.3333)
26.6667
V = 2500 lb (Controls)
Ans.
692
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7–37. Three identical boards are bolted together to form
the built-up beam. If the wood has an allowable shear stress
of tallow = 450 psi, determine the maximum allowable
internal shear V that can act on the beam. Also, find the
corresponding average shear stress in the 38 in. diameter
bolts which are spaced equally at s = 6 in.
2 in.
2 in.
s
2 in.
s
s
V
Section Properties: The neutral axis passes through the centroid c of the cross
section as shown in Fig. a. The location of c is
1 in.
g yA
2[2(4)(1)] + 4(4)(1)
y =
=
= 2.6667 in.
gA
2(4)(1) + 4(1)
1 in.
1 in.
Thus, the moment of inertia of the cross section about the neutral axis is
I = 2c
1
1
(1)(43) + 1(4)(2.6667 - 2)2 d +
(1)(43) + 1(4)(4 - 2.6667)2
12
12
= 26.6667 in4
Here, QB can be computed by referring to Fig. b, which is
QB = y¿1 A ¿1 = 1.3333(4)(1) = 5.3333 in3
Referring to Fig. c, QD and Qmax are
QD = y¿3 A¿3 = 2.3333(2)(1) = 4.6667 in3
Qmax = y¿2 A¿2 + y¿3 A¿3 = 0.6667(1.3333)(3) + 2.3333(2)(1) = 7.3333 in3
Shear Stress: The maximum shear stress occurs at either point D or points on the
neutral axis. For point D, t = 1 in.
tallow =
VQD
;
It
450 =
V(4.6667)
26.6667(1)
V = 2571.43 lb = 2.57 kip (Controls)
Ans.
For the points on the neutral axis, t = 3(1) = 3 in, and so
tallow =
VQmax
;
It
450 =
V(7.3333)
26.6667(3)
V = 4909.09 lb
Shear Flow: Since each bolt has two shear planes, q = 2 a
q =
VQB
;
I
F
F
F
b = 2a b = .
s
6
3
2571.43(5.3333)
F
=
3
26.6667
F = 1542.86 lb
Thus, the shear stress developed in the bolt is
tb =
F
1542.86
=
= 14.0 ksi
p 3 2
Ab
A B
Ans.
4 8
Ans:
V = 2571 lb, tb = 14.0 ksi
693
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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.
200 mm
75 mm
50 mm 75 mm
25 mm
V
200 mm
25 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)
Thus,
I =
1
(0.2)(0.053) + 0.2 (0.05)(0.175 - 0.1375)2
12
+
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
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Since there are two rows of nails q = 2 a
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
(0.0042)
4
Ans.
Ans:
t = 35.2 MPa
694
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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
V
350 mm
s = 250 mm
y =
2 (0.125)(0.25)(0.025) + 0.275 (0.35)(0.025)
= 0.18676 m
2 (0.25)(0.025) + 0.35 (0.025)
I = (2) a
+
25 mm
1
b(0.025)(0.253) + 2 (0.025)(0.25)(0.18676 - 0.125)2
12
1
(0.025)(0.35)3 + (0.025)(0.35)(0.275 - 0.18676)2
12
= 0.270236 (10 - 3) m4
Q = y¿A¿ = 0.06176(0.025)(0.25) = 0.386(10 - 3) m3
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.
Ans:
F = 12.5 kN
695
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*7–40. The simply supported beam is built-up from three
boards by nailing them together as shown. The wood has an
allowable shear stress of tallow = 1.5 MPa, and an allowable
bending stress of sallow = 9 MPa . The nails are spaced at
s = 75 mm, and each has a shear strength of 1.5 kN.
Determine the maximum allowable force P that can be
applied to the beam.
P
s
A
B
1m
1m
100 mm
Support Reactions: As shown on the free-body diagram of the beam shown in Fig. a.
Maximum Shear and Moment: The shear diagram is shown in Fig. b. As indicated,
P
Vmax = .
2
25 mm
25 mm
Section Properties: The moment of inertia of the cross section about the neutral axis is
1
1
I =
(0.1)(0.253) (0.075)(0.23)
12
12
= 80.2083(10 - 6) m4
Referring to Fig. d,
QB = y 2¿ A¿2 = 0.1125(0.025)(0.1) = 0.28125(10 - 3) m3
Shear Flow: Since there is only one row of nails, qallow =
1.5(103)
F
= 20 (103) N>m.
=
s
0.075
P
c 0.28125(10 - 3) d
2
www.elsolucionario.org
VmaxQB
qallow =
;
I
20(103) =
80.2083(10 - 6)
P = 11417.41 N = 11.4 kN (controls)
Bending,
smax =
s(106) N>m2 =
Mc
I
A P2 B(0.125 m)
80.2083(10-6) m4
P = 11.550 N = 11.55 kN
Shear,
tmax =
VQ
It
Q = (0.1125)(0.025)(0.1) + (0.05)(0.1)(0.025)
= 0.40625(10-3) m3
1.5 A 106 B =
A P2 B (0.40625)(10-3) m3
80.2083(10-6) m4(0.025 m)
P = 14.808 N = 14.8 kN
696
Ans.
200 mm
25 mm
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
7–40. Continued
697
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
7–41. The simply supported beam is built-up from three
boards by nailing them together as shown. If P = 12 kN,
determine the maximum allowable spacing s of the nails to
support that load, if each nail can resist a shear force of 1.5 kN.
P
s
A
B
1m
Support Reactions: As shown on the free-body diagram of the beam shown in Fig. a.
1m
100 mm
Maximum Shear and Moment: The shear diagram is shown in Fig. b. As indicated,
P
12
Vmax =
=
= 6 kN.
2
2
25 mm
25 mm
200 mm
Section Properties: The moment of inertia of the cross section about the neutral axis is
I =
1
1
(0.1)(0.253) (0.075)(0.23)
12
12
25 mm
= 80.2083(10-6) m4
Referring to Fig. d,
QB = y¿2 A¿2 = 0.1125(0.025)(0.1) = 0.28125(10-3) m3
1.5(10 2
F
=
.
s
s
3
Shear Flow: Since there is only one row of nails, qallow =
qallow =
VmaxQB
;
I
1.5(1032
s
=
6000 C 0.28125(10-3) D
80.2083(10-62
www.elsolucionario.org
s = 0.07130 m = 71.3 mm
Ans.
Ans:
s = 71.3 mm
698
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7–42. The T-beam is constructed 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
1
8 in. The allowable shear stress for the wood is
tallow = 450 psi.
2 in.
s
12 in.
s
12 in.
V
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 =
2 in.
1
(2)(123) + 2(12)(9.5 - 6)2
12
+
1
(12)(23) + 12(2)(13 - 9.5)2
12
= 884 in4
Referring 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.
tallow =
VQmax
;
It
450 =
V (90.25)
884 (2)
V = 8815.51 lb = 8.82 kip
Here, qallow =
950
F
=
lb>in. Then
s
s
VQA
;
qallow =
I
Ans.
8815.51(84)
950
=
s
884
s = 1.134 in = 1
1
in
8
Ans.
Ans:
1
V = 8.82 kip, use s = 1 in.
8
699
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
7–43. The box beam is made from four pieces of plastic
that are glued together as shown. If the glue has an
allowable strength of 400 lb> in2, determine the maximum
shear the beam will support.
5.5 in.
0.25 in.
0.25 in.
0.25 in.
4.75 in.
V
I =
0.25 in.
1
1
(6)(5.253) (5.5)(4.753) = 23.231 in4
12
12
QB = y¿A¿ = 2.5(6)(0.25) = 3.75 in3
The beam will fail at the glue joint for board B since Q is a maximum for this board.
tallow =
VQB
;
It
400 =
V(3.75)
23.231(2)(0.25)
V = 1239 lb = 1.24 kip
Ans.
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Ans:
V = 1.24 kip
700
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*7–44. The box beam is made from four pieces of plastic that
are glued together as shown. If V = 2 kip, determine the shear
stress resisted by the seam at each of the glued joints.
5.5 in.
0.25 in.
0.25 in.
0.25 in.
4.75 in.
I =
1
1
(6)(5.253) (5.5)(4.753) = 23.231 in4
12
12
V
QB = y¿A¿ = 2.5(6)(0.25) = 3.75 in3
QA = 2.5(5.5)(0.25) = 3.4375
tB =
2(103)(3.75)
VQB
=
= 646 psi
It
23.231(2)(0.25)
Ans.
tA =
VQA
2(103)(3.4375)
=
= 592 psi
It
23.231(2)(0.25)
Ans.
701
0.25 in.
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7–45. A 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
Support Reactions: As shown on FBD.
150 mm
Internal Shear Force: As shown on shear diagram, VAB = (P + 3) kN.
Section Properties:
30 mm
1
1
INA =
(0.31) A 0.153 B (0.25) A 0.093 B
12
12
250 mm 30 mm
30 mm
= 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)
www.elsolucionario.org
P = 6.60 kN
Ans.
Ans:
P = 6.60 kN
702
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
7–46. The beam is subjected to a shear of V = 800 N.
Determine the average shear stress developed in the nails
along the sides A and B if the nails are spaced s = 100 mm
apart. Each nail has a diameter of 2 mm.
B
100 mm
100 mm
150 mm
30 mm
V
A
30 mm
250 mm
30 mm
0.015(0.03)(0.25) + 2(0.075)(0.15)(0.03)
y =
= 0.04773 m
0.03(0.25) + 2(0.15)(0.03)
I =
1
(0.25)(0.033) + (0.25)(0.03)(0.04773 - 0.015)2
12
+ (2) a
1
b (0.03)(0.153) + 2(0.03)(0.15)(0.075 - 0.04773)2
12
= 32.164773(10-6) m4
Q = y¿A¿ = 0.03273(0.25)(0.03) = 0.245475(10-3) m3
q =
VQ
800(0.245475)(10 - 3)
= 6105.44 N>m
=
I
32.164773(10 - 6)
F = qs = 6105.44(0.1) = 610.544 N
Since each side of the beam resists this shear force then
tavg =
610.544
F
= 97.2 MPa
= p
2A
2( 4 )(0.0022)
Ans.
Ans:
tavg = 97.2 MPa
703
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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
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
1
(1.5) A 103 B + (1.5)(10)(6 - 3.3548)2
+
12
INA =
= 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
www.elsolucionario.org
100
Shear Flow: The allowable shear flow at points C and D is qC =
and
s
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
Ans.
s¿ = 1.21 in.
Ans.
Ans:
s = 8.66 in., s¿ = 1.21 in.
704
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*7–48. The beam is made from three polystyrene strips
that are glued together as shown. If the glue has a shear
strength of 80 kPa, determine the maximum load P that can
be applied without causing the glue to lose its bond.
30 mm
P
1
—P
4
40 mm
1
—P
4
20 mm
60 mm
A
40 mm
Maximum shear is at the supports.
Vmax =
3P
4
I =
1
1
(0.02)(0.06)3 + 2c (0.03)(0.04)3 + (0.03)(0.04)(0.05)2 d = 6.68(10 - 6) m4
12
12
t =
VQ
;
It
80(103) =
(3P>4)(0.05)(0.04)(0.03)
6.68(10 - 6)(0.02)
P = 238 N
Ans.
705
B
0.8 m
1m
1m
0.8 m
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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
100 mm
190 mm
V
The moment of inertia of the cross section about the neutral axis is
I =
200 mm
1
1
(0.2)(0.43) (0.18)(0.383) = 0.24359(10 - 3) m4
12
12
10 mm
180 mm
Referring to Fig. a, Fig. b,
10 mm
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 symmetry, the shear flow at points A and A¿ , Fig. a, and at points B and B¿ ,
Fig. b, are the same. Thus
qA =
3
-3
1 VQA
1 300(10 ) C 0.3705(10 ) D
s
a
b = c
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)
www.elsolucionario.org
3
= 462.46(10 ) N>m = 462 kN>m
Ans
Ans:
qA = 228 kN>m, qB = 462 kN>m
706
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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
100 mm
190 mm
V
The moment of inertia of the cross section about the neutral axis is
200 mm
1
1
I =
(0.2)(0.43) (0.18)(0.383) = 0.24359(10 - 3) m4
12
12
10 mm
180 mm
10 mm
Referring to Fig. a, due to symmetry ACœ = 0. Thus
QC = 0
Then referring 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.
Ans:
qC = 0, qD = 601 kN>m
707
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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
10 mm
30 mm
10 mm
Section Properties:
INA =
10 mm 125 mm
1
(0.125) A 0.013 B + 0.125(0.01) A 0.1052 B d
12
10 mm
= 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)
www.elsolucionario.org
Ans.
= 13033 N>m = 13.0 kN>m
qB =
=
V
150 mm
1
1
(0.145) A 0.33 B (0.125) A 0.283 B
12
12
+ 2c
150 mm
1 VQB
c
d
2
I
3
-3
1 18(10 )(0.13125)(10 )
d
c
2
125.17(10 - 6)
= 9437 N>m = 9.44 kN>m
Ans.
708
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
7–53. A shear force of V = 18 kN is applied to the box
girder. Determine the shear flow at C.
10 mm
30 mm
10 mm
A
100 mm
C
B
100 mm
10 mm
30 mm
10 mm
150 mm
V
Section Properties:
150 mm
1
1
(0.145) A 0.33 B (0.125) A 0.283 B
12
12
INA =
+2c
10 mm 125 mm
1
(0.125) A 0.013 B + 0.125(0.01) A 0.1052 B d
12
10 mm
= 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
3
-3
1 18(10 )(0.5375)(10 )
d
c
4
2
125.17(10 )
= 38648 N>m = 38.6 kN>m
Ans.
Ans:
qC = 38.6 kN>m
709
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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
40 mm
B
A
10 mm
V
40 mm
30 mm
y =
2[0.005(0.03)(0.01)] + 2[0.03(0.06)(0.01)] + 0.055(0.04)(0.01)
= 0.027727 m
2(0.03)(0.01) + 2(0.06)(0.01) + 0.04(0.01)
I = 2c
30 mm
10 mm
1
(0.03)(0.01)3 + 0.03(0.01)(0.027727 - 0.005)2 d
12
+ 2c
+
10 mm
1
(0.01)(0.06)3 + 0.01(0.06)(0.03 - 0.027727)2 d
12
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 =
150(8.1818)(10 - 6)
VQB
= 1.25 kN>m
=
I
0.98197(10 - 6)
Ans.
www.elsolucionario.org
Ans:
qA = 1.39 kN>m, qB = 1.25 kN>m
710
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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.
10 mm
40 mm
10 mm
30 mm
y =
B
A
V
40 mm
10 mm
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)
30 mm
10 mm
= 0.027727 m
I = 2c
1
(0.03)(0.01)3 + 0.03(0.01)(0.027727 - 0.005)2 d
12
+ 2c
1
(0.01)(0.06)3 + 0.01(0.06)(0.03 - 0.027727)2 d
12
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 =
1 VQmax
1 150(21.3(10 - 6))
b = 1.63 kN>m
a
b = a
2
I
2 0.98197(10 - 6)
Ans.
Ans:
qmax = 1.63 kN>m
711
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*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.
A
0.5 in.
D
8 in.
V
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)(3.70946 - 0.25)2 + 2c (0.5)(83) + 0.5(8)(4.5 - 3.70946)2 d
12
12
+
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.
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712
B
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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.
0.5 in.
2 in.
A
0.5 in.
D
8 in.
V
y =
0.25(11)(0.5) + 2[4.5(8)(0.5)] + 6.25(10)(0.5)
©yA
=
= 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
+
B
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.
Ans:
qmax = 414 lb>in.
713
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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
y =
0.015(0.4)(0.03) + 2[0.13(0.2)(0.03)]
©yA
=
= 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
30 mm
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.
www.elsolucionario.org
Ans:
qA = 215 kN>m
714
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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
200 mm
30 mm
V ⫽ 75 kN
y =
0.015(0.4)(0.03) + 2[0.13(0.2)(0.03)]
©yA
=
©A
0.4(0.03) + 2(0.2)(0.03)
30 mm
= 0.0725 m
I =
1
(0.4)(0.033) + 0.4(0.03)(0.0725 - 0.015)2
12
1
+ 2 c (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.
Ans:
qmax = 232 kN>m
715
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*7–60. The built-up beam is formed by welding together
the thin plates of thickness 5 mm. Determine the location of
the shear center O.
5 mm
200 mm
O
e
100 mm
200 mm
300 mm
Shear Center. Referring to Fig. a and summing moments about point A, we have
a + ©(MR)A = ©MA;
- Pe = - (Fw)1(0.3)
e =
0.3(Fw)1
P
(1)
Section Properties: The moment of inertia of the cross section about the axis of
symmetry is
I =
1
1
(0.005)(0.43) +
(0.005)(0.23) = 30(10 - 6) m4
12
12
Referring to Fig. b, y¿ = (0.1 - s) +
s
= (0.1 - 0.5s) m. Thus, Q as a function of s is
2
Q = y¿A¿ = (0.1 - 0.5s)(0.005s) = [0.5(10 - 3) s - 2.5(10 - 3) s2] m3
Shear Flow:
q =
www.elsolucionario.org
VQ
P[0.5(10 - 3) s - 2.5(10 - 3) s2]
= P(16.6667s - 83.3333s2)
=
I
30(10 - 6)
Resultant Shear Force: The shear force resisted by the shorter web is
0.1 m
(Fw)1 = 2
L0
qds = 2
0.1 m
2
L0
P(16.6667s - 83.3333s )ds = 0.1111P
Substituting this result into Eq. (1),
e = 0.03333 m = 33.3 m
Ans.
716
100 mm
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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.
©yA
(0.25)(11)(0.5) + 2(3.25)(5.5)(0.5) + 6.25(7)(0.5)
y =
=
= 2.8362 in.
©A
0.5(11) + 2(0.5)(5.5) + 7(0.5)
6 in.
6 in.
2 in.
0.5 in.
0.5 in.
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
I =
+
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)
qmax = a
b = 641 lb>in.
2
92.569
qA =
Ans.
Ans.
Ans.
Ans:
qA = 196 lb>in., qB = 452 lb>in.,
qmax = 641 lb>in.
717
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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
R
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
= R2 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 ]冷
2
2 0
(1 - cos 2u)
du
2
www.elsolucionario.org
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
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
Ans:
t =
718
V
2R2 - y2
pR2t
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
7–63. 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.
1.80 in.
1 in.
1 in.
O
e
1 in.
1 in.
Summing moments about A,
Pe = F(4) + 2V1(1.8)
I = 2c
(1)
1
1
t (43) d t(23) + 2 C (1.80)(t)(22) D = 24.4 t in4
12
12
Q1 = y- 1 ¿A¿ = a 1 +
y
y2
b (y t) = t a y +
b
2
2
Q2 = ©y- ¿A¿ = 1.5(1)(t) + 2(x)(t) = t(1.5 + 2x)
Pt ay +
y2
b
2
Pay +
y2
b
2
q1 =
VQ1
=
I
q2 =
VQ2
P t (1.5 + 2x)
P(1.5 + 2x)
=
=
I
24.4 t
24.4
V1 =
F =
24.4 t
=
24.4
L
q1 dy =
1
y2
P
ay +
b dy = 0.0273 P
24.4 L0
2
L
q2 dy =
1.80
P
(1.5 + 2x) dx = 0.2434 P
24.4 L0
From Eq. (1),
Pe = 0.2434 P(4) + 2(0.0273)P(1.8)
e = 1.07 in.
Ans.
Ans:
e = 1.07 in.
719
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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
h1
h2
O
e
Summing moments about A,
eP = bF1
I =
q1 =
F1 =
(1)
1
1
1
(t)(h1)3 +
(t)(h2)3 =
t (h31 + h32)
12
12
12
P(h1>2)(t)(h1>4)
Ph21t
=
I
8I
Ph31t
2
q1(h1) =
3
12I
From Eq. (1),
e =
=
=
b Ph31t
a
b
P 12I
h31b
(h31 + h32)
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b
1 + (h2>h1)3
Ans.
720
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
0.5 in.
7–65. The beam supports a vertical shear of V = 7 kip.
Determine the resultant force developed in segment AB of
the beam.
10 in.
0.5 in.
A
0.5 in.
5 in.
B
V
Section Properties:
INA =
1
1
(1)(53) +
(10)(0.53) = 10.52083 in4
12
12
Q = y- ¿A¿ = (0.5y + 1.25)(2.5 - y)(0.5)
= 1.5625 - 0.25y2
Shear Flow:
q =
7(103)(1.5625 - 0.25y2)
VQ
=
I
10.52083
= {1039.60 - 166.34y2} lb 冫 in.
Resultant Shear Force: For web AB
2.5 in.
VAB =
L0.25 in.
qdy
2.5 in.
=
L0.25 in.
(1039.60 - 166.34y2) dy
= 1474 lb = 1.47 kip
Ans.
Ans:
VAB = 1.47 kip
721
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
7–66. The built-up beam is fabricated from the three thin
plates having a thickness t. Determine the location of the
shear center O.
a
2
a
a
2
a
2
Pe = Ff (a)
O
a
2
Shear Center. Referring to Fig. a and summing moments about point A, we have
a + g (MR)A = g MA;
e
(1)
Section Properties: The moment of inertia of the cross section about the axis of
symmetry is
I =
1
a 2
7
(t)(2a)3 + 2 c at a b d = a3t
12
2
6
Referring to Fig. c, y¿ =
a
. Thus, Q as a function of s is
2
Q = y¿A¿ =
a
at
(st) =
s
2
2
Shear Flow:
q =
P A at2 s B
VQ
3P
= 7 3 =
s
2
I
7a
a
t
6
www.elsolucionario.org
Resultant Shear Force: The shear force resisted by the flange is
a
Ff =
L0
qds =
a
a
3P s2 2
3
P
sds =
a b =
2
14
7a2 2 0
L0 7a
3P
Substituting this result into Eq. (1),
e =
3
a
14
Ans.
Ans:
e =
722
3
a
14
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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
h1
O
b1
h
e
h1
b1
Summing moments about A,
Pe = F2(h) - F1(h - 2h1)
I =
=
h - 2h1 2
h 2
1
(t)h3 + 2(b)(t) a b + 2(b1)(t) a
b
12
2
2
b1t(h - 2h1)2
bth2
1 3
th +
+
12
2
2
Q1 = y1 ¿A1 ¿ =
VQ1
=
q1 =
I
h - 2h1
(x1)t
2
Pt a
q1dx1 =
L0
h - 2h1
2
b x1
=
I
b1
F1 =
(1)
Q2 = y2 ¿A2 ¿ =
Pt(h - 2h1)
x1
2I
Pt(h - 2h1) b1
Ptb21(h - 2h1)
x1dx =
2I
4I
L0
h
(x )t
2 2
h
q2 =
P(2 )(x2)t
VQ2
=
I
I
b
F2 =
L
q2dx2 =
2
Pht
Phb t
x dx =
2I L0 2 2
4I
From Eq. (1),
Pe =
e =
Ptb21(h - 2h1)2
Ph2b2t
4I
4I
h2b2t - tb21 (h - 2h1)2
1
b1t(h - 2h1)
4 c 12
th3 + bth
d
2 +
2
2
2
=
3[h2b2 - (h - 2h1)2b21]
3
h + 6bh2 + 6b1(h - 2h1)2
Ans.
Ans:
e =
723
3[h2b2 - (h - 2h1)2b21]
3
h + 6bh2 + 6b1(h - 2h1)2
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*7–68. A thin plate of thickness t is bent to form the beam
having the cross section shown. Determine the location of
the shear center O.
a
a
O
e
Shear Center: Referring to Fig. a and summing moments about point A,
a + ©(MR)A = ©MA;
Pe = 2F1a
t
(1)
Section Properties: The moment of inertia of the inclined segment shown in Fig. b about
t
2
1
bh3. In this case, b =
=
t and h = 2a sin 45° = 12a.
the neutral axis is I =
12
sin 45°
12
Thus, the moment of inertia of the cross section about the axis of symmetry is
I = 2c
3
2
1
2
tb a 22a b d = a3 t
a
3
12 22
Referring to Fig. c, y¿ =
Q = y- ¿A¿ =
s
12
sin 45° =
s. Thus, Q as a function of s is
2
4
22 2
22
ts
s(st) =
4
4
Shear Flow:
VQ
3 22P 2
P A 4 ts B
=
=
s
2 3
I
8a3
3a t
22
q =
2
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Resultant Shear Force: The shear force resisted by the open-ended segment is
a
F1 =
L0
qds =
3 22P 2
3 22P s3 a
22
P
s ds =
a b` =
3
3 0
8
8a3
L0 8a
a
Substituting this result into Eq. (1),
e =
12
a
4
Ans.
724
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
7–69. A thin plate of thickness t is bent to form the beam
having the cross section shown. Determine the location of
the shear center O.
t
r
O
e
Section Properties: For the arc segment, Fig. a, y = r cos u and the area of the
differential element shown shaded is dA = t ds = tr du. Then, the moment of
inertia of the entire cross section about the axis of symmetry is
I =
1
(t)(2r)3 +
y2 dA
12
L
=
2 3
rt +
(r cos u)2 trdu
3
L0
=
2 3
r3t
rt +
(cos 2u + 1) du
3
2 L0
=
p
2 3
r3t 1
a sin 2u + u b `
rt +
3
2 2
0
=
r3t
(4 + 3p)
6
p
p
Referring to Fig. a, y¿ =
r
. Thus, Q as a function of s is
2
u
Q = y¿A¿ +
ydA =
L
r
(rt) +
r cos u(trdu)
2
L0
u
=
1 2
r t + r2t
cos udu
2
L0
=
r2 t
(1 + 2 sin u)
2
Shear Flow:
P C 2 (1 + 2 sin u) D
VQ
3P
=
(1 + 2 sin u)
=
q =
r3t
I
(4
+
3p)r
(4 + 3p)
r2t
6
Resultant Shear Force: The shear force resisted by the arc segment is
p
F =
L
qds =
L0
p
qrdu =
3P
(1 + 2 sin u)rdu
(4
+
3p)r
L0
=
p
3P
(u - 2 cos u) `
4 + 3p
0
=
3P(p + 4)
4 + 3p
Shear Center: Referring to Fig. b and summing the moments about point A,
a + ©(MR)A = ©MA;
Pe = r
L
Pe = r c
e = c
dF
3P(p + 4)
d
4 + 3p
Ans:
3(p + 4)
dr
4 + 3p
Ans.
725
e = c
3(p + 4)
dr
4 + 3p
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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
L
dF
(1)
dA = t ds = t r du
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
au b 冷
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
(2a - 2 sin a cos a) =
(2a - sin 2a)
2
2
www.elsolucionario.org
dQ = y dA = r sin u(t r du) = r2 t sin u du
u
Q = r2 t
q =
u
Lp-a
sin u du = r2 t (- cos u)|
= 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
q r du
L
p+a
-2P
- 2P r
(cos u + cos a) du =
(2a cos a - 2 sin a)
dF =
r(2a - sin 2a) Lp - a
2a - sin 2a
L
L
dF =
=
L
q ds =
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.
Ans:
4r (sin a - a cos a)
e =
2a - sin 2a
726
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7–71. 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:
1 in.
πyA
0.5(10)(1) + 2(4)(2) + 7(12)(1)
y =
=
= 3.50 in.
πA
10(1) + 4(2) + 12(1)
1
INA =
(10)(13) + (10)(1)(3.50 - 0.5)2
12
1
+
(2)(43) + 2(4)(3.50 - 2)2
12
1
+
(1)(123) + 1(12)(7 -3.50)2
12
-
= 410.5 in4
QC = y1¿A¿ = 1.5(4)(1) = 6.00 in3
QD = y2 ¿A ¿ = 3.50(12)(1) = 42.0 in3
Shear Flow:
VQC
4.5(103)(6.00)
=
= 65.773 lb 冫 in.
I
410.5
VQD
4.5(103)(42.0)
=
= 460.41 lb 冫 in.
qD =
I
410.5
qC =
Hence, the shear force resisted by each nail is
FC = qC s = (65.773 lb 冫 in.)(3 in.) = 197 lb
Ans.
FD = qD s = (460.41 lb 冫 in.)(3 in.) = 1.38 kip
Ans.
Ans:
FC = 197 lb, FD = 1.38 kip
727
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*7–72. The T-beam is subjected to a shear of V = 150 kN.
Determine the amount of this force that is supported by the
web B.
200 mm
40 mm
V = 150 kN
200 mm
B
40 mm
y =
(0.02)(0.2)(0.04) + (0.14)(0.2)(0.04)
= 0.08 m
0.2(0.04) + 0.2(0.04)
I =
1
(0.2)(0.043) + 0.2(0.04)(0.08 - 0.02)2
12
+
1
(0.04)(0.23) + 0.2(0.04)(0.14 - 0.08)2 = 85.3333(10-6) m4
12
A¿ = 0.04(0.16 - y)
y- ¿ = y +
(0.16 + y)
(0.16 - y)
=
2
2
Q = y- ¿A¿ = 0.02(0.0256 - y2)
t =
V =
150(103)(0.02)(0.0256 - y2)
VQ
= 22.5(106) - 878.9(106)y2
=
It
85.3333(10-6)(0.04)
L
t dA,
www.elsolucionario.org
dA = 0.04 dy
0.16
V =
L-0.04
(22.5(106) - 878.9(106)y2) 0.04 dy
0.16
=
L-0.04
(900(103) - 35.156(106)y2)dy
= 131 250 N = 131 kN
Ans.
728
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
7–73. The member is subject 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 in 15 mm.
200 mm
B
100 mm
A C
300 mm
V ⫽ 2 kN
Section Properties:
y =
=
πyA
πA
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
INA =
1
(0.2)(0.0153) + 0.2(0.015)(0.08798 - 0.0075)2
12
1
+
(0.03)(0.1153) + 0.03(0.115)(0.08798 - 0.0575)2
12
+
1
(0.015)(0.33) + 0.015(0.3)(0.165 - 0.08798)2
12
= 86.93913(10 - 6) m4
QA = 0
Ans.
QB = y 1¿A¿ = 0.03048(0.115)(0.015) = 52.57705(10 - 6) m - 3
QC = πy1¿A¿
= 0.03048(0.115)(0.015) + 0.08048(0.0925)(0.015)
= 0.16424(10 - 3) m3
Shear Flow:
qA =
VQA
= 0
I
qB =
2(103)(52.57705)(10 - 6)
VQB
= 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.
Ans:
qA = 0, qB = 1.21 kN>m, qC = 3.78 kN>m
729
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
7–74. Determine the shear stress at points B and C on the
web of the beam located at section a - a .
8000 lb
150 lb/ft
a
C
D
A
B
4 ft
1.5 ft
a
4 ft
1.5 ft
2 in.
0.75 in.
C
0.5 in.
B
4 in.
y =
(0.375)(4)(0.75) + (3.75)(6)(0.5) + (7.125)(2)(0.75)
= 3.075 in.
4(0.75) + 6(0.5) + 2(0.75)
I =
1
(4)(0.753) + 4(0.75)(3.075 - 0.375)2
12
+
1
(0.5)(63) + 0.5(6)(3.75 - 3.075)2
12
+
1
(2)(0.753) + 2(0.75)(7.125 - 3.075)2
12
6 in.
0.75 in.
= 57.05 in4
www.elsolucionario.org
QB = y ¿B A¿ = 2.7(4)(0.75) = 8.1 in3
QC = y C¿ A¿ = 4.05(2)(0.75) = 6.075 in3
t =
VQ
It
tB =
2800(8.1)
= 795 psi
57.05(0.5)
Ans.
tC =
2800(6.075)
= 596 psi
57.05(0.5)
Ans.
Ans:
tB = 795 psi, tC = 596 psi
730
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
7–75. Determine the maximum shear stress acting at
section a–a in the beam.
8000 lb
150 lb/ft
a
D
A
y =
C
B
4 ft
(0.375)(4)(0.75) + (3.75)(6)(0.5) + (7.125)(2)(0.75)
= 3.075 in.
4(0.75) + 6(0.5) + 2(0.75)
1.5 ft
a
4 ft
1.5 ft
2 in.
1
I =
(4)(0.753) + 4(0.75)(3.075 - 0.375)2
12
C
1
+
(0.5)(63) + 0.5(6)(3.75 - 3.075)2
12
+
0.75 in.
0.5 in.
B
4 in.
1
(2)(0.753) + 2(0.75)(7.125 - 3.075)2
12
6 in.
0.75 in.
= 57.05 in4
Qmax = © y- ¿A¿
= 2.7(4)(0.75) + 2.325(0.5)(1.1625)
= 9.4514 in3
tmax =
VQmax
2800(9.4514)
=
= 928 psi
It
57.05(0.5)
Ans.
Ans:
tmax = 928 psi
731
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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.
sallow =
pr
;
2t
12(106) =
300(103)(1.5)
2t
t = 0.0188 m = 18.8 mm
Ans.
www.elsolucionario.org
Ans:
t = 18.8 mm
732
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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.
Ans:
ro = 75.5 in.
733
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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
P
8 in.
8 in.
(a)
(b)
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.
www.elsolucionario.org
Ans:
(a) s1 = 1.04 ksi, s2 = 0,
(b) s1 = 1.04 ksi, s2 = 520 psi
734
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*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 analysis
t
0.25
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.
735
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–5. The open-ended polyvinyl chloride pipe has an inner
diameter of 4 in. and thickness of 0.2 in. If it carries flowing
water at 60 psi pressure, determine the state of stress in the
walls of the pipe.
s1 =
60(2)
pr
=
= 600 psi
t
0.2
Ans.
s2 = 0
Ans.
There is no stress component in the longitudinal direction since the pipe has
open ends.
www.elsolucionario.org
Ans:
s1 = 600 psi, s2 = 0
736
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–6. If the flow of water within the pipe in Prob. 8–5 is
stopped due to the closing of a valve, determine the state of
stress in the walls of the pipe. Neglect the weight of
the water. Assume the supports only exert vertical forces
on the pipe.
s1 =
60(2)
pr
=
= 600 psi
t
0.2
Ans.
s2 =
60(2)
pr
=
= 300 psi
2t
2(0.2)
Ans.
Ans:
s1 = 600 psi, s2 = 300 psi
737
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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)
s1 =
a
8 mm
50 mm
pr
1.35(106)(0.75)
=
= 126.56(106) = 127 MPa
t
0.008
0.75 m
a
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)
www.elsolucionario.org
Ans.
Ans:
(a) s1 = 127 MPa,
(b) s1 ¿ = 79.1 MPa, (tavg)b = 322 MPa
738
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*8–8. The steel water pipe has an inner diameter of 12 in.
and wall thickness 0.25 in. If the valve A is opened and the
flowing water is under a gauge pressure of 250 psi,
determine the longitudinal and hoop stress developed in
the wall of the pipe.
A
Normal Stress: Since the pipe has two open ends,
slong = s2 = 0
Since
Ans.
6
r
=
= 24 7 10, thin-wall analysis can be used.
t
0.25
sh = s1 =
250(6)
pr
=
= 6000 psi = 6 ksi
t
0.25
Ans.
739
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–9. The steel water pipe has an inner diameter of 12 in.
and wall thickness 0.25 in. If the valve A is closed and the
water pressure is 300 psi, determine the longitudinal and
hoop stress developed in the wall of the pipe. Draw the state
of stress on a volume element located on the wall.
A
6
r
=
= 24 > 10, thin-wall analysis can be used.
t
0.25
300(6)
pr
=
= 7200 psi = 7.20 ksi
shoop = s1 =
t
0.25
Normal Stress: Since
sloop = s2 =
pr
300(6)
=
= 3600 psi = 3.60 ksi
2t
2(0.25)
Ans.
Ans.
The state of stress on an element in the pipe wall is shown in Fig. a.
www.elsolucionario.org
Ans:
shoop = 7.20 ksi, slong = 3.60 ksi
740
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–10. The A-36-steel band is 2 in. wide and is secured
around the smooth rigid cylinder. If the bolts are tightened
so that the tension in them is 400 lb, determine the normal
stress in the band, the pressure exerted on the cylinder, and
the distance half the band stretches.
1
—
8 in.
8 in.
s1 =
400
= 1600 psi
2(1>8)(1)
s1 =
pr
;
t
1600 =
Ans.
p(8)
(1>8)
p = 25 psi
P1 =
Ans.
s1
1600
= 55.1724(10 - 6)
=
E
29(106)
d = P1L = 55.1724(10 - 6)(p) a 8 +
1
b = 0.00140 in.
16
Ans.
Ans:
s1 = 1.60 ksi, p = 25 psi, d = 0.00140 in.
741
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–11. 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.
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.
www.elsolucionario.org
F = Ff = 2880p = 9048 lb = 9.05 kip
Ans.
Ans:
(a) T = 18.2 kip # ft,
(b) P = 18.1 kip,
(c) F = 9.05 kip
742
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*8–12. A pressure-vessel head is fabricated by gluing the
circular plate to the end of the vessel as shown. If the vessel
sustains an internal pressure of 450 kPa, determine the
average shear stress in the glue and the state of stress in the
wall of the vessel.
+ c ©Fy = 0;
450 mm
10 mm
20 mm
p(0.225)2450(103) - tavg (2p)(0.225)(0.01) = 0;
tavg = 5.06 MPa
Ans.
s1 =
450(103)(0.225)
pr
=
= 5.06 MPa
t
0.02
Ans.
s2 =
pr
450(103)(0.225)
=
= 2.53 MPa
2t
2(0.02)
Ans.
743
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–13. An A-36-steel hoop has an inner diameter of
23.99 in., thickness of 0.25 in., and width of 1 in. If it and the
24-in.-diameter rigid cylinder have a temperature of 65° F,
determine the temperature to which the hoop should be
heated in order for it to just slip over the cylinder. What is
the pressure the hoop exerts on the cylinder, and the tensile
stress in the ring when it cools back down to 65° F?
24 in.
dT = a ¢ TL
p(24) - p(23.99) = 6.60(10 - 6)(T1 - 65)(p)(23.99)
T1 = 128.16° F = 128°
Ans.
Cool down:
dF = dT
FL
= a ¢ TL
AE
F(p)(24)
(1)(0.25)(29)(106)
= 6.60(10 - 6)(128.16 - 65)(p)(24)
F = 3022.21 lb
s1 =
F
;
A
s1 =
pr
;
t
s1 =
3022.21
= 12 088 psi = 12.1 ksi
(1)(0.25)
12 088 =
p(12)
(0.25)
Ans.
www.elsolucionario.org
P = 252 psi
Ans.
Ans:
T1 = 128°, s1 = 12.1 ksi, p = 252 psi
744
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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: From the FBD
+ ©F = 0;
:
x
2P - 2pri w = 0
P = pri w
Hoop Stress and Strain for the Ring:
s1 =
pri w
pri
P
=
=
ro - ri
A
(ro - ri)w
Using Hooke’s Law
P1 =
However,
P1 =
pri
s1
=
E
E(ro - ri)
(1)
2p(ri)1 - 2pri
(ri)1 - ri
dri
=
=
.
ri
ri
2pri
Then, from Eq. (1)
pri
dri
=
ri
E(ro - ri)
dri =
pr2i
E(ro - ri)
Ans.
Ans:
dri =
745
pr2i
E(ro - ri)
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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.
r4
r2
r1
A
r3
B
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
P1 =
However,
P1 =
pri
s1
=
E
E(ro - ri)
(1)
2p(ri)1 - 2pri
(ri)1 - ri
dri
=
=
.
ri
ri
2pri
www.elsolucionario.org
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 =
E(r2 - r3)
Ans.
r22
r23
+
r2 - r1
r4 - r3
Ans:
p =
746
E(r2 - r3)
r22
r23
+
r2 - r1
r4 - r3
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*8–16. A closed-ended pressure vessel is fabricated by
cross winding glass filaments over a mandrel, so that the wall
thickness t of the vessel is composed entirely of filament and
an expoxy binder as shown in the figure. Consider a segment
of the vessel of width w and wrapped at an angle u. If the
vessel is subjected to an internal pressure p, show that the
force in the segment is Fu = s0wt, where s0 is the stress in
the filaments. Also, show that the stresses in the hoop and
longitudinal directions are sh = s0 sin2 u and s1 = s0 cos2u,
respectively. At what angle u (optimum winding angle)
would the filaments have to be would so that the hoop and
longitudinal stresses are equivalent?
Fu
t
u
d
The Hoop and Longitudinal Stresses: Applying Eq. 8–1 and Eq. 8–2
pA2 B
pr
pd
=
=
t
t
2t
d
s1 =
pA2 B
pd
pr
=
=
2t
2t
4t
d
s2 =
The Hoop and Longitudinal Force for Filament:
F1 = s1A =
pd
pdw
w
a
tb =
2r sin u
2 sin u
F1 = s2A =
pd
pdw
w
a
tb =
4t cos u
4 cos u
Hence,
Fu = 2F2b + F2t
=
pdw 2
pdw 2
b + a
b
4 cos u
F 2 sin u
=
pdw
4
=
a
G
1
4
+
sin2 u
cos2 u
pdw
4 cos2 u + sin2 u
4 G
sin2 u
pdw
=
2 22 sin 2u
23 cos 2 u + 5
pdw
su =
23 cos 2 u + 5
Fu
2 22 sin 2u
=
A
wt
pd
=
2 22t
a
33 cos 2u + 5
b
sin 2u
(Q.E.D)
dsu
= 0 when su is minimum.
du
pd
dsu
2 cos 2u
3
ca 23 cos 2u + 5 b =
d = 0
du
sin2 2u
2 22r
33 cos 2u + 5
747
w
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–16.
(Continued)
3
2 cos 2u
a 33 cos 2u + 5b +
= 0
sin2 2u
3 3 cos 2u + 5
a 33 cos 2u + 5b a
a 33 cos 2u + 5b c
3
2 cos 2u
+
b = 0
3 cos 2u + 5
sin2 2u
3 cos2 2u + 10 cos 2u + 3
d = 0
sin2 2u(3 cos 2u + 5)
However, 33 cos 2u + 5 Z 0. Therefore,
3 cos2 2u + 10 cos 2u + 3
= 0
sin2 2u(3 cos 2u + 5)
3 cos2 2u + 10 cos 2u + 3 = 0
cos 2u =
- 10 ; 2102 - 4(3)(3)
2(3)
cos 2u = - 0.3333
u = 54.7°
Ans.
Force in U Direction: Consider a portion of the cylinder. For a filament wire the
cross-sectional area is A = wt, then
Fu = s0 wt
www.elsolucionario.org
(Q.E.D.)
Hoop Stress: The force in hoop direction is Fh = Fu sin u = s0 wt sin u and the
wt
area is A =
. Then due to the internal pressure p,
sin u
sh =
Fh
s0 wt sin u
=
A
wt>sin u
= s0 sin2 u
(Q. E. D.)
Longitudinal Stress: The force in the longitudinal direction is F1 = Fu cos u = su wt cos u
wt
and the area is A =
. Then due to the internal pressure p,
sin u
st =
Fh
s0 wt cos u
=
A
wt>cos u
= s0 cos2 u
Optimum Wrap Angle: This requires
(Q. E. D.)
pd>2t
sh
=
= 2. Then
sl
pd>4t
s0 sin2 u
sh
= 2
=
sl
s0 cos2 u
tan2 u = 2
u = 54.7°
Ans.
748
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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 L of
the vessel.
L
w
t¿
p
s1
T
t
s1
T
Normal Stress in the Wall and Filament Before the Internal Pressure is Applied:
The entire length L of wall is subjected to pretension filament force T. Hence, from
equilibrium, the normal stress in the wall at this state is
2T - (s¿)w (2Lt) = 0
(s¿)w =
T
Lt
and for the filament the normal stress is
(s¿)fil =
T
wt¿
Normal Stress in the Wall and Filament After the Internal Pressure is Applied: In
order to use s1 = pr>t, developed for a vessel of uniform thickness, we redistribute
the filament’s cross-section as if it were thinner and wider, to cover the vessel with
no gaps. The modified filament has width L and thickness t’w>L, still with crosssectional area wt’ subjected to tension T. Then the stress in the filament becomes
sfil = s + (s¿)fil =
pr
T
+
(t + t¿w>L)
wt¿
Ans.
pr
T
(t + t¿w>L)
Lt
Ans.
And for the wall,
sw = s - (s¿)w =
Check: 2wt¿sfil + 2Ltsw = 2rLp
OK
Ans:
pr
T
+
,
t + t¿w>L
wt¿
pr
T
sw =
t + t¿w>L
Lt
sfil =
749
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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 =
d
P
Mc
A
I
P
(0.2)(0.01)
P
P(0.1 - d)(0.1)
1
3
12 (0.01)(0.2 )
P( - 1000 + 15000 d) = 0
d = 0.0667 m = 66.7 mm
Ans.
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Ans:
d = 66.7 mm
750
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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
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)
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.
Ans:
sL = 66.7 MPa (C), sR = 33.3 MPa (T)
751
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*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
N = 100 kN
I =
M = 20 kN # m
1
(0.03)(0.23) = 20.0(10 - 6) m4
12
The normal stress developed is the combination of axial and bending stress. Thus,
s =
My
N
;
A
I
For the left edge fiber, y = C = 0.1 m. Then
sR = -
100(103)
20.0(103)(0.1)
+
0.006
20.0(10 - 6)
= 83.33(106) Pa = 83.3 MPa (T)
Ans.
www.elsolucionario.org
For the right edge fiber, y = C = 0.1 m. Thus
sR = -
20.0(103)(0.1)
100(103)
0.006
20.0(10 - 6)
= 117 MPa (C)
Ans.
752
a
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8–21. If the load has a weight of 600 lb, determine the
maximum normal stress developed on the cross section of
the supporting member at section a–a. Also, plot the normal
stress distribution over the cross section.
1.5 ft
a
Internal Loadings: Consider the equilibrium of the free-body diagram of the bottom
cut segment shown in Fig. a.
a + c ©Fy = 0;
N - 600 = 0
a + ©MC = 0;
600(1.5) - M = 0
a
1 in.
Section a – a
N = 600 lb
M = 900 lb # ft
Section Properties: The cross-sectional area and the moment of inertia about the
centroidal axis of the member are
A = p(12) = p in2
I =
p
p 4
(1 ) = in4
4
4
Normal Stress: The normal stress is the combination of axial and bending stress.
Thus,
s =
Mc
N
;
A
I
By observation, the maximum normal stress occurs at point B, Fig. b. Thus,
900(12)(1)
600
+
= 13.9 ksi (T)
p
p>4
Ans.
900(12)(1)
600
+ = - 13.6 ksi = 13.6 ksi (C)
p
p>4
Ans.
smax = sB =
For Point A,
sA =
Using these results, the normal stress distribution over the cross section is shown in
Fig. b. The location of the neutral axis can be determined from
2 - x
p
=
;
13.9
13.6
x = 1.01 in.
Ans:
smax = sL = 13.9 ksi (T), sR = 13.6 ksi (C)
753
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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
180 N
A
E
There is no 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)
Ans.
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Ans:
smax = 1.07 MPa
754
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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
B
180 N
A
E
There is no moment in this problem. Therefore, the compressive stress is produced
by axial force only.
sconst =
240
P
=
= 1.07 MPa
A
(0.015)(0.015)
Ans:
sconst = 1.07 MPa
755
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*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.
Q+ ©Fx = 0;
N - 700 cos 30° = 0;
N = 606.218 lb
a+ ©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
tA = 0
Ans.
(since QA = 0)
Ans.
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756
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.
Q+ ©Fx = 0;
N - 700 cos 30° = 0;
N = 606.218 lb
a+ ©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)
Mc
606.218
N
+
=
+ 1
3
A
I
(0.75)(0.5)
12 (0.5)(0.75)
sB = 5.35 ksi
tB = 0
Ans.
(since QB = 0)
Ans.
Ans:
sB = 5.35 ksi, tB = 0
757
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8–26. The column is built up by gluing the two identical
boards together. Determine the maximum normal stress
developed on the cross section when the eccentric force of
P = 50 kN is applied.
P
250 mm
150 mm
150 mm
75 mm
50 mm
300 mm
Section Properties: The location of the centroid of the cross section, Fig. a, is
y =
©yA
0.075(0.15)(0.3) + 0.3(0.3)(0.15)
=
= 0.1875 m
©A
0.15(0.3) + 0.3(0.15)
The cross - sectional area and the moment of inertia about the z axis of the cross
section are
A = 0.15(0.3) + 0.3(0.15) = 0.09 m2
Iz =
1
1
(0.3)(0.153) + 0.3(0.15)(0.1875 - 0.075)2 +
(0.15)(0.33) + 0.15(0.3)(0.3 - 0.1875)2
12
12
= 1.5609(10 - 3) m4
Equivalent Force System: Referring to Fig. b,
+ c ©Fx = (FR)x;
-50 = - F
F = 50 kN
©Mz = (MR)z;
- 50(0.2125) = - M
M = 10.625 kN # m
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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 occurs at points along the edge where
y = 0.45 - 0.1875 = 0.2625 m such as point A. Thus,
smax =
10.625(103)(0.2625)
-50(103)
0.09
1.5609(10 - 3)
Ans.
= - 2.342 MPa = 2.34 MPa (C)
Ans:
smax = 2.34 MPa (C)
758
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8–27. The column is built up by gluing the two identical
boards together. If the wood has an allowable normal stress
of sallow = 6 MPa, determine the maximum allowable
eccentric force P that can be applied to the column.
P
250 mm
150 mm
150 mm
75 mm
50 mm
300 mm
Section Properties: The location of the centroid c of the cross section, Fig. a, is
y =
©yA
0.075(0.15)(0.3) + 0.3(0.3)(0.15)
=
= 0.1875 m
©A
0.15(0.3) + 0.3(0.15)
The cross-sectional area and the moment of inertia about the z axis of the cross
section are
A = 0.15(0.3) + 0.3(0.15) = 0.09 m2
Iz =
1
1
(0.3)(0.153) + 0.3(0.15)(0.1875 - 0.075)2 +
(0.15)(0.33) + 0.15(0.3)(0.3 - 0.1875)2
12
12
= 1.5609(10 - 3) m4
Equivalent Force System: Referring to Fig. b,
+ c ©Fx = (FR)x;
-P = -F
F = P
©Mz = (MR)z;
- P(0.2125) = - M
M = 0.2125P
Normal Stress: The normal stress is a combination of axial and bending stress. Thus,
F =
My
N
+
A
I
By inspection, the maximum normal stress, Which is compression, occurs at points
along the edge where y = 0.45 - 0.1875 = 0.2625 m such as point A. Thus,
- 6(106) =
0.2125P(0.2625)
-P
0.09
1.5609(10 - 3)
P = 128 076.92 N = 128 kN
Ans.
Ans:
P = 128 kN
759
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*8–28. The cylindrical post, having a diameter of 40 mm, is
being pulled from the ground using a sling of negligible
thickness. If the rope is subjected to a vertical force of
P = 500 N, determine the normal stress at points A and B.
Show the results on a volume element located at each of
these points.
I =
P
1
1
p r4 = (p)(0.024) = 0.1256637(10 - 6) m4
4
4
B
A = p r2 = p(0.022) = 1.256637(10 - 3) m2
A
P
Mx
+
A
I
500
+ 0 = 0.398 MPa
=
1.256637(10 - 3)
sA =
Ans.
P
Mc
A
I
10(0.02)
500
=
-3
1.256637(10 )
0.1256637(10 - 6)
sB =
Ans.
= - 1.19 MPa
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760
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–29. Determine the maximum load P that can be applied
to the sling having a negligible thickness so that the normal
stress in the post does not exceed sallow = 30 MPa . The post
has a diameter of 50 mm.
+ T ©F = 0;
N - P = 0;
a+©M = 0;
M - P(0.025) = 0;
P
N = P
B
M = 0.025P
A
p 2
A =
d = p(0.0252) = 0.625(10 - 3)p m2
4
I =
p
p 4
r =
(0.0254) = 97.65625(10 - 9)p m4
4
4
s =
My
N
+
A
I
s = 30(106) =
P(0.025)(0.025)
P
+
0.625(10 - 3)p
97.65625(10 - 9)p
P = 11.8 kN
Ans.
Ans:
P = 11.8 kN
761
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8–30. The rib-joint pliers are used to grip the smooth pipe C.
If the force of 100 N is applied to the handles, determine the
state of stress at points A and B on the cross section of the
jaw at section a- a. Indicate the results on an element at
each point.
100 N
250 mm
25 mm
25 mm
45°
a
a
C
A
10 mm
B
20 mm
Support Reactions: Referring to the free-body diagram of the handle shown in
Fig. a,
a+ ©MD = 0;
100(0.25) - FC (0.05) = 0
7.5 mm
Section a – a
FC = 500 N
Internal Loadings: Consider the equilibrium of the free-body diagram of the
segment shown in Fig. b,
©Fy¿ = 0;
500 - V = 0
V = 500 N
a+ ©MC = 0;
M - 500(0.025) = 0
M = 12.5 N # m
Section Properties: The moment of inertia of the cross section about the centroidal
axis is
I =
1
(0.0075)(0.023) = 5(10 - 9) m4
12
Referring to Fig. c, QA and QB are
QA = 0
www.elsolucionario.org
QB = y¿A¿ = 0.005(0.01)(0.0075) = 0.375(10 - 6) m3
Normal Stress: The normal stress is contributed by bending stress only. Thus
s =
My
I
For point A, y = 0.01 m. Then
sA = -
12.5(0.01)
5(10 - 9)
= - 25 MPa = 25 MPa (C)
Ans.
For point B, y = 0. Then
sB = 0
Ans.
Shear Stress: The shear stress is contributed by transverse shear stress only. Thus,
tA =
VQA
= 0
It
Ans.
tB =
VQB
500[0.375(10 - 6)]
= 5 MPa
=
It
5(10 - 9)(0.0075)
Ans.
The state of stress of points A and B are represented by the elements shown in
Figs. d and e respectively.
762
100 N
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8–30.
Continued
Ans:
sA = 25 MPa (C), sB = 0,
tA = 0, tB = 5 MPa
763
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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)
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0 = 250 P - 7500 P (0.1 - d)
d = 0.06667 m = 66.7 mm
Ans.
Ans:
d = 66.7 mm
764
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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 =
4.00(103)(0.1)
80(103)
+
0.002
6.667(10 - 6)
= 100 (106) Pa = 100 MPa (T)
The location of the neutral axis can be determined using the similar triangles.
20.0
0.2 - d
=
d
100
20 - 100d = 20d
d =
1
m = 166.667 mm
2
765
a
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–33. The control lever is subjected to a horizontal force
of 20 lb on the handle. Determine the state of stress at
points A and B. Sketch the results on differential elements
located at each of these points. The assembly is pin
connected at C and attached to a cable at D.
20 lb
2 in.
E
1.75 in.
0.75 in.
0.5 in.
A F
A
D
F
For point B:
I =
40(0.15)
Mc
= 8.89 ksi (C)
=
I
0.675(10 - 3)
tB = 0
B
5 in.
0.3 in.
C
90
0.25 in.
1
(0.3)(0.33) = 0.675(10 - 3) in4
12
sB =
B E
Ans.
(since QB = 0)
Ans.
For point A:
I =
1
(0.25)(13) = 0.020833 in4
12
sA =
30(0.5)
Mc
=
= 720 psi (T)
I
0.020833
tA = 0
Ans.
(since QA = 0)
Ans.
www.elsolucionario.org
Ans:
sB = 8.89 ksi (C), tB = 0,
sA = 720 psi (T), tA = 0
766
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–34. The control lever is subjected to a horizontal force
of 20 lb on the handle. Determine the state of stress at
points E and F. Sketch the results on differential elements
located at each of these points. The assembly is pin
connected at C and attached to a cable at D.
20 lb
2 in.
E
1.75 in.
0.75 in.
0.5 in.
A F
A
D
F
For point E:
I =
40(0.15)
Mc
= 8.89 ksi (T)
=
I
0.675(10 - 3)
tE = 0
B
5 in.
0.3 in.
C
90
0.25 in.
1
(0.3)(0.33) = 0.675(10 - 3) in4
12
sE =
B E
Ans.
(since QE = 0)
Ans.
For point F:
I =
1
(0.25)(13) = 0.020833 in4
12
sF = 0
tF =
Ans.
VQ
40(0.25)(0.5)(0.25)
= 240 psi
=
It
1
(0.25)(1)3(0.25)
12
Ans.
Ans:
sE = 8.89 ksi (T), tE = 0, sF = 0, tF = 240 psi
767
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8–35. The tubular shaft of the soil auger is subjected to the
axial force and torque shown. If the auger is rotating at a
constant rate, determine the state of stress at points A and B
on the cross section of the shaft at section a–a.
x
1200 lb
3000 lb⭈ft
Internal Loadings: Consider the equilibrium of the free-body diagram of the
upper cut segment shown in Fig. a.
1 in.
a
a
©Fx = 0;
N - 1200 = 0
N = 1200 lb
©Mx = 0;
T - 3000 = 0
T = 3000 lb # ft
z
A
z
y
1.5 in. B
y
Section a – a
Section Properties: The cross-sectional area and the polar moment of inertia
of the shaft are
A = p(1.52 - 12) = 1.25p in2
J =
p
(1.54 - 14) = 2.03125p in4
2
Normal Stress: The normal stress is contributed by axial stress only. Thus,
sA = sB =
N
-1200
=
= - 305.58 psi = 306 psi (C)
A
1.25x
Ans.
Shear Stress: The shear stress is contributed by torsional shear stress only.
Thus,
www.elsolucionario.org
t =
Tp
J
For point A, r = 1.5 in. and the shear stress is directed along the y axis. Thus,
(txy)A =
3000(12)(1.5)
= 8462 psi = 8.46 ksi
2.03125p
Ans.
For point B, r = 1 in. and the shear stress is directed along the z axis. Thus,
(txz)B =
3000(12)(1)
= 5641 psi = 5.64 ksi
2.03125p
Ans.
The state of stress at points A and B are represented on the elements shown in
Figs. b and c, respectively.
Ans:
sA = sB = 306 psi (C),
tA = 8.46 ksi, tB = 5.64 ksi
768
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*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 - 150a 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) + 150a 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)
769
Ans.
3
5
4
150 N
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–36.
Continued
Shear Stress: The transverse shear stress developed at point A is
c A txy B V d
VyQA
=
A
Izt
= 0
The torsional shear stress developed at point A is
C (txz)T D A =
20(0.005)
Tc
= 101.86 MPa
=
J
0.3125p A 10 - 9 B
Thus,
A txy B A = 0
Ans.
A txz B A = c A txz B T d = 102 MPa
Ans.
A
The state of stress at point A is represented on the element shown in Fig. c.
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770
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y
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, in back, at section a–a.
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 - 150a b = 0
5
N = 120 N
3
©Fy = 0; 150 a b - Vy = 0
5
Vy = 90 N
z
Section a – a
3
4
©Mz = 0; -150 a b (0.4) + 150a 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,
s =
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)
771
5 mm
B
T = 20 N # m
©Mx = 0; 20 - T = 0
A
Ans.
3
5
4
150 N
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–37.
Continued
Shear Stress: The transverse shear stress developed at point B is
c A txy B V d
VyQB
=
Izt
B
=
90 c 83.333 A 10 - 9 B d
0.15625p A 10 - 9 B (0.01)
= 1.528 MPa
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 txy B B = 0
Ans.
A txy B B = c A txy B T d - c A txy B V d
B
B
= 101.86 - 1.528 = 100.33 MPa = 100 MPa
Ans.
The state of stress at point B is represented on the element shown in Fig. d.
www.elsolucionario.org
Ans:
sB = 1.53 MPa (C), tB = 100 MPa
772
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–38. The frame supports the distributed load shown.
Determine the state of stress acting at point D. Show the
results on a differential element at this point.
4 kN/m
B
A
E
D
1.5 m
1.5 m
3m
20 mm
60 mm
20 mm
D
5m
50 mm
E
3m
sD = -
My
8(103)
12(103)(0.03)
P
= - 1
3
A
I
(0.1)(0.05)
12 (0.05)(0.1)
C
sD = - 88.0 MPa
Ans.
tD = 0
Ans.
Ans:
sD = -88.0 MPa, tD = 0
773
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–39. The frame supports the distributed load shown.
Determine the state of stress acting at point E. Show the
results on a differential element at this point.
4 kN/m
B
A
E
D
1.5 m
1.5 m
3m
20 mm
60 mm
20 mm
D
5m
50 mm
E
3m
C
sE = -
tE =
My
8(103)
8.25(103)(0.03)
P
=
+ 1
= 57.8 MPa
3
A
I
(0.1)(0.05)
12 (0.05)(0.1)
VQ
4.5(103)(0.04)(0.02)(0.05)
= 864 kPa
=
1
3
It
12 (0.05)(0.1) (0.05)
Ans.
Ans.
www.elsolucionario.org
Ans:
sE = 57.8 MPa, tE = 864 kPa
774
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*8–40. The 500-kg engine is suspended from the jib crane at
the position shown. Determine the state of stress at point A
on the cross section of the boom at section a–a.
E
150 mm
20 mm
2m
C
2m
2m
a 30⬚
a
150 mm
D
0.4 m
A
300 mm
20 mm
B
20 mm
Section a – a
Support Reactions: Referring to the free-body diagram of the entire boom, Fig. a,
a+ ©MC = 0;
FDE sin 30°(6) + FDE cos 30°(0.4) - 500(9.81)(2) = 0
FDE = 2931.50 N
Internal Loadings: Considering the equilibrium of the free-body diagram of the boom’s
right cut segment, Fig. b,
+
: ©Fx = 0;
N - 2931.50 cos 30° = 0
N = 2538.75 N
+ c ©Fy = 0;
2931.50 sin 30° - V = 0
V = 1465.75 N
a + ©MO = 0;
2931.50 sin 30°(2) + 2931.50 cos 30°(0.4) - M = 0
M = 3947.00 N # m
Section Properties: The cross-sectional area and the moment of inertia about the
centroidal axis of the boom’s cross section are
A = 0.15(0.3) - 0.13(0.26) = 0.0112 m2
I =
1
1
(0.15)(0.33) (0.13)(0.263) = 0.14709(10 - 3) m4
12
12
Referring to Fig. c, QA is
QA = y¿1A¿1 + y¿1A¿2 = 0.065(0.13)(0.2) + 0.14(0.02)(0.15) = 0.589(10 - 3) 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. Then
sA =
- 2538.75
+ 0 = - 0.2267 MPa = 0.227 MPa (C)
0.0112
Ans.
Shear Stress: The shear stress is contributed by transverse shear stress only. Thus,
tA =
VQA
1465.75[0.589(10 - 3)]
= 0.293 MPa
=
It
0.14709(10 - 3)(0.02)
Ans.
The state of stress at point A is represented on the element shown in Fig. d.
775
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–40. Continued
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776
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–41. The 500-kg engine is suspended from the jib crane at
the position shown. Determine the state of stress at point B
on the cross section of the boom at section a–a. Point B is
just above the bottom flange.
E
150 mm
20 mm
2m
C
2m
2m
a 30⬚
a
150 mm
D
0.4 m
A
300 mm
20 mm
B
20 mm
Section a – a
Support Reactions: Referring to the free-body diagram of the entire boom, Fig. a,
a + ©MC = 0;
FDE sin 30°(6) + FDE cos 30°(0.4) - 500(9.81)(2) = 0
FDE = 2931.50 N
Internal Loadings: Considering the equilibrium of the free-body diagram of the
boom’s right cut segment, Fig. b,
+
: ©Fx = 0;
N - 2931.50 cos 30° = 0
N = 2538.75 N
+ c ©Fy = 0;
2931.50 sin 30° - V = 0
V = 1465.75 N
a+ ©MO = 0;
2931.50 sin 30°(2) + 2931.50 cos 30°(0.4) - M = 0
M = 3947.00 N # m
Section Properties: The cross-sectional area and the moment of inertia about the
centroidal axis of the boom’s cross section are
A = 0.15(0.3) - 0.13(0.26) = 0.0112 m2
I =
1
1
(0.15)(0.33) (0.13)(0.263) = 0.14709(10 - 3) m4
12
12
Referring to Fig. c, QB is
QB = y2A2¿ = 0.14(0.02)(0.15) = 0.42(10 - 3) m3
Normal Stress: The normal stress is the combination of axial and bending stress.
Thus,
s =
My
N
+
A
I
For point B, y = 0.13 m. Then
sB =
3947.00(0.13)
- 2538.75
= 3.26 MPa (T)
+
0.0112
0.14709(10 - 3)
Ans.
777
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–41. Continued
Shear Stress: The shear stress is contributed by transverse shear stress only. Thus,
tB =
1465.75[0.42(10 - 3)]
VQB
= 0.209 MPa
=
It
0.14709(10 - 3)(0.02)
Ans.
The state of stress at point B is represented on the element shown in Fig. d.
www.elsolucionario.org
Ans:
sB = 3.26 MPa (T), tB = 0.209 MPa
778
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–42. Determine the state of stress at point A on the cross
section of the post at section a–a. Indicate the results on a
differential element at the point.
1.5 ft
5 ft
400 lb
300 lb
Internal Loadings: Considering the equilibrium of the free-body diagram
of the post’s upper cut segment, Fig. a,
©Fy = 0;
Vy + 300 = 0
Vy = - 300 lb
©Fz = 0;
Vz + 400 = 0
Vz = - 400 lb
©Mx = 0;
T + 400(1.5) = 0
T = - 600 lb # ft
©My = 0;
My + 400(5) = 0
My = - 2000 lb # ft
©Mz = 0;
Mz - 300(5) = 0
Mz = 1500 lb # ft
a
J =
p
(2.54 - 24) = 5.765625p in4
4
p
(2.54 - 24) = 11.53125p in4
2
Referring to Fig. b,
(Qz)A = 0
(Qy)A =
4(2.5) p
4(2) p 2
c (2.52) d c (2 ) d = 5.0833 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 and z = - 2.5 in. Then
sA = - 0 +
A
2.5 in.
Section Properties: The moments of inertia about the y and z axes and the
polar moment of inertia of the post’s cross section are
Iy = Iz =
2 in.
a
- 2000(12)(- 2.5)
= 3.31 ksi (T)
5.765625p
Ans.
779
B
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8–42. Continued
Shear Stress: The torsional shear stress at points A and B are
[(txy)T]A =
600(12)(2.5)
Tc
=
= 0.4969 ksi
J
11.53125p
The transverse shear stresses at points A and B are
[(txz)V]A =
[(txy)V]A =
Vz(Qz)A
Iy t
= 0
Vy(Qy)B
Iz t
=
300(5.0833)
= 0.08419 ksi
5.765625p(5 - 4)
Combining these two shear stress components,
(txy)A = [(txy)T]A + [(txy)V]A = 0.4969 + 0.08419 = 0.581 ksi
Ans.
(txz)A = 0
Ans.
The state of stress at point A is represented on the element shown in Fig. c.
www.elsolucionario.org
Ans:
sA = 3.31 ksi (T), tA = 0.581 ksi
780
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8–43. Determine the state of stress at point B on the cross
section of the post at section a–a. Indicate the results on a
differential element at the point.
1.5 ft
5 ft
400 lb
Internal Loadings: Considering the equilibrium of the free-body diagram of the
post’s upper segment, Fig. a,
Vy + 300 = 0
Vy = - 300 lb
©Fz = 0;
Vz + 400 = 0
©Mx = 0;
T + 400(1.5) = 0
Vz = - 400 lb
T = - 600 lb # ft
©My = 0;
My + 400(5) = 0
My = - 2000 lb # ft
©Mz = 0;
Mz - 300(5) = 0
Mz = 1500 lb # ft
Section Properties: The moments of inertia about the y and z axes and the polar
moment of inertia of the post’s cross section are
J =
p
(2.54 - 24) = 5.765625p in4
4
p
(2.54 - 24) = 11.53125p in4
2
Referring to Fig. b,
(Qy)B = 0
(Qz)B =
4(2.5) p
4(2) p 2
c (2.52) d c (2 ) d = 5.0833 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 = 2 in. and z = 0. Then
sB =
2 in.
a
A
2.5 in.
©Fy = 0;
Iy = Iz =
a
300 lb
1500(12)(2)
+ 0 = - 1.987 ksi = 1.99 ksi (C)
5.765625p
Ans.
781
B
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–43. Continued
Shear Stress: The torsional shear stress at point B is
[(txz)T]B =
600(12)(2)
Tc
=
= 0.3975 ksi
J
11.53125p
The transverse shear stress at point B is
[(txy)V]B =
[(txz)V]B =
Vy(Qy)A
Iz t
= 0
Vz(Qz)B
Iy t
=
400(5.0833)
= 0.1123 ksi
5.765625p(5 - 4)
Combining these two shear stress components,
(txz)B = [(txz)T]B + [(txz)V]B = 0.3975 + 0.1123 = 0.510 ksi
(txy)B = 0
Ans.
Ans.
The state of stress at point B is represented on the element shown in Fig. c.
www.elsolucionario.org
Ans:
sB = 1.99 ksi (C), tB = 0.510 ksi
782
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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.
12 kip
Referring to Fig. a,
6 in.
a
©Fx = (FR)x;
- 6 - 12 = F
F = - 18.0 kip
©My = (MR)y;
6(1.5) - 12(1.5) = My
My = - 9.00 kip # in
©Mz = (MR)z;
12(3) - 6(3) = Mz
Mz = 18.0 kip # in
The cross-sectional area and moments of inertia about the y and z axes of the cross
section 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.
783
A
B
a
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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 =
www.elsolucionario.org
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
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
54.0
13.5
= - 3.00 ksi = 3.00 ksi (C)
Ans.
784
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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
Ans.
= - 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)
Ans.
The normal stress distribution over the cross section is shown in Fig. b
Ans:
sA = 1.00 ksi (C), sB = 3.00 ksi (C)
785
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–46. The support is subjected to the compressive load P.
Determine the absolute maximum possible and minimum
possible normal stress acting in the material, for x Ú 0.
a
—
a 2
—
2
P
a a
—
2 —
2
x
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)
=
www.elsolucionario.org
P
-1
3x
;
B
R
a a+x
(a + x)2
sA = -
3x
P
1
+
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
786
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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)
a2
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.
Ans:
smax =
787
1.33P
P
(C), smin =
(T)
a2
3a2
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–47. The bent shaft is fixed in the wall at A. If a force F is
applied at B, determine the stress components at points D
and E. Show the results on a differential element located at
each of these points. Take F = 12 lb and u = 0°.
z
y
A
x
D
6 in.
1.25 in.
E
8 in.
3 in.
©Fx = 0;
Vx - 12 = 0;
Vx = 12 lb
©My = 0;
- Ty + 12(3) = 0;
Ty = 36 lb # in.
©Mz = 0;
Mz - 12(8) = 0;
Mz = 96 lb # in.
B
u
F
A = p(0.6252) = 1.2272 in2
I =
1
p(0.6254) = 0.1198 in4
4
J =
1
p (0.6254) = 0.2397 in4
2
Point D:
4(0.625) 1
(p)(0.6252) = 0.1628 in3
3p
2
(QD)z =
sD =
Mzx
I
www.elsolucionario.org
Ans.
= 0
(tD)yx = (tD)V - (tD)twist
Ty c
Vx (QD)z
=
=
-
It
J
36(0.625)
12(0.1628)
= - 80.8 psi
0.1198(1.25)
0.2397
Ans.
Point E :
(sE)y =
Mzx
I
=
- 96(0.625)
= - 501 psi
0.1198
Ans.
(tE)yz = (tE)V - (tE)twist
= 0 -
Ty c
J
=
- 36(0.625)
0.2397
= - 93.9 psi
Ans.
Ans:
sD = 0, tD = 80.8 psi,
sE = - 501 psi, tE = 93.9 psi
788
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*8–48. The bent shaft is fixed in the wall at A. If a force F is
applied at B, determine the stress components at points D
and E. Show the results on a differential element located at
each of these points. Take F = 12 lb and u = 90°.
z
A
y
x
D
6 in.
1.25 in.
E
8 in.
3 in.
©Fy = 0;
Ny - 12 = 0;
Ny = 12 lb
©Mx = 0;
Mx - 12(3) = 0;
Mx = 36 lb # in.
F
A = p(0.6252) = 1.2272 in2
I =
1
p(0.6254) = 0.1198 in4
4
Point D :
(sD)y =
Ny
A
-
36(0.625)
Mx z
12
=
I
1.2272
0.1198
Ans.
= - 178 psi
Ans.
(tD)yz = (tD)yz = 0
Point E :
(sE)y =
Ny
A
+
B
u
Mx z
12
=
I
1.2272
= 9.78 psi
Ans.
(tE)yx = (tE)yz = 0
Ans.
789
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–49. The bent shaft is fixed in the wall at A. If a force F is
applied at B, determine the stress components at points D
and E. Show the results on a volume element located at
each of these points. Take F = 12 lb and u = 45°.
z
A
y
x
D
6 in.
1.25 in.
E
8 in.
Vx - 12 cos 45° = 0;
©Fx = 0;
Vx = 8.485 lb
©Fy = 0;
Ny - 12 sin 45° = 0;
Ny = 8.485 lb
©Mz = 0;
Mz - 12 sin 45°(3) = 0;
Mz = 25.456 lb # in.
©My = 0;
- Ty + 12 cos 45°(3) = 0;
Ty = 25.456 lb # in.
©Mz = 0;
Mz - 12 cos 45°(8) = 0;
Mz = 67.882 lb # in.
3 in.
B
u
F
A = p(0.6252) = 1.2272 in2
1
p(0.6254) = 0.1195 in4
4
1
J = p(0.6254) = 0.2397 in4
4
I =
Point D :
4(0.625) 1
(p)(0.6252) = 0.1628 in2
3p
2
Nz
25.456(0.625)
Mx z
8.485
(sD)y =
=
A
I
1.2272
0.1198
(QD)y =
www.elsolucionario.org
= - 126 psi
(tD)yx =
Ans.
Ty c
Vs (QD)z
-
It
J
(25.456)(0.625)
8.485(0.1628)
=
0.1198(1.25)
0.2397
Ans.
= - 57.2 psi
Point E :
(sE)x = 0
Ny
Mzx
(67.882)(0.625)
8.485
(sE)y =
=
A
I
1.2272
0.1198
= - 347 psi
(tE)yx =
VzQE
It
= 0 -
-
Ans.
Tc
J
(25.456)(0.625)
0.2397
= - 66.4 psi
Ans.
Ans:
sD = -126 psi, tD = 57.2 psi,
sE = -347 psi, tE = 66.4 psi
790
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–50. The coiled spring is subjected to a force P. If we
assume the shear stress caused by the shear force at any
vertical section of the coil wire to be uniform, show that the
maximum shear stress in the coil is tmax = P>A + PRr>J,
where J is the polar moment of inertia of the coil wire and A
is its cross-sectional area.
P
2r
R
Tc
PRr
= max on perimeter =
J
J
tmax =
V
Tc
P
PRr
+
=
+
A
J
A
J
QED
P
791
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
x
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.
z
a
a
A
B
a
P
a
D
ez
ey
C
a
a
y
Equivalent Force System: As shown on FBD.
Section Properties:
1
A = 2a(2a) + 2 B (2a)a R = 6a2
2
1
1
1
a 2
(2a)(2a)3 + 2 B (2a) a3 + (2a) a aa + b R
12
36
2
3
Iz =
= 5a4
1
1
1
a 2
(2a)(2a)3 + 2 B (2a) a3 + (2a) a a b R
12
36
2
3
Iy =
=
5 4
a
3
www.elsolucionario.org
Normal Stress:
s =
N
A
=
=
My z
Mzy
Iz
+
Iy
Peyy Pez z
-P
- 5
2
4
6a
5a4
3a
P
A - 5a2 - 6eyy - 18ez z B
30a4
At point B where y = - a and z = - a , we require sB 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 A, 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.
Ans:
6ey + 18ez 6 5a
792
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
P
*8–52. The vertebra of the spinal column can support a
maximum compressive stress of smax, before undergoing a
compression fracture. Determine the smallest force P that
can be applied to a vertebra, if we assume this load is
applied at an eccentric distance e from the centerline of the
bone, and the bone remains elastic. Model the vertebra as a
hollow cylinder with an inner radius ri and outer radius ro.
smax =
smax =
smax =
P =
ⴝ
Pero
P
+
p 4
A
(r - r4i )
4 0
smax = Pc
smax =
ro
4er0
1
+
d
p(r20 - r2i )
p(r40 - r4i )
4er0
P
c1 + 2
d
p(r20 - r2i )
(r0 + r2i )
P(r20 + r2i + 4er0)
p(r20 - r2i )(r20 + r2i )
P(r20 + r2i + 4er0)
p(r40 - r4i )
dmaxp(r40 - r4i )
r20 + r2i + 4er0
793
e
ri
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
y
8–53. The 1-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.
x
A
C
B
z
75 lb
8 in.
3 in.
100 lb
©Fz = 0;
Vz + 100 = 0;
Vz = - 100 lb
©Fx = 0;
Nx - 75 = 0;
Nx = 75 lb
©Fy = 0;
Vy = 80 = 0;
Vy = 80 lb
©Mz = 0;
Mz + 80(8) = 0;
Mz = - 640 lb # in.
©Mx = 0;
T2 + 80(3) = 0;
Tx = - 240 lb # in.
©My = 0;
MF + 100(8) - 75(3) = 0;
My = - 575 lb # in.
80 lb
p 2
p 2
1
d =
(1 ) = p in2
4
4
4
p
p 4
c = (0.54) = 0.03125p in4
J =
2
4
A =
(Qy)A = 0
(QA)A = y¿A =
4(0.5) 1
(p)(0.52) = 0.08333 in2
3p 2
www.elsolucionario.org
p 4
p
r = (0.54) = 0.015625p in4
4
4
Myz
Mxy
p
+
+
Normal stress: s =
A
Ix
Iy
Iy = Ix =
640(0.5)
75
+ 0 = 6.61 ksi (T)
sA = 1 +
0.0156p
4p
Ans.
Shear stress:
t =
VQ
Tc
+
It
J
(txz)A =
240(0.5)
100(0.06333)
+
0.0156p(1)
0.0312p
Ans.
= 1.39 ksi
(txy)A = 0
Ans.
Ans:
sA = 6.61 ksi (T), tA = 1.39 ksi
794
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
y
8–54. The 1-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.
x
A
C
B
z
75 lb
8 in.
3 in.
100 lb
©Fz = 0;
Vz + 100 = 0;
Vz = - 100 lb
©Fx = 0;
Nx - 75 = 0;
Nx = 75.0 lb
©Fy = 0;
Vy - 80 = 0;
Vy = 80 lb
©Mz = 0;
Mz + 80(8) = 0;
Mz = - 640 lb # in.
©Mx = 0;
Tx + 80(3) = 0;
Tx = - 240 lb # in.
©My = 0;
My + 100(8) - 75(3) = 0;
My = - 575 lb # in.
80 lb
p 2
p
p
d = (12) = in2
4
2
4
A =
p 4
p
c =
(0.54) = 0.03125p in4
2
2
4(0.5) 1 p
a b (12) = 0.08333 in2
(Qy)y =
3p 2 4
J =
Iy = Ix =
p 4
p
r = (0.54) = 0.015625p in4
4
4
Normal stress:
s =
Myz
Mxy
p
+
+
A
Ix
Iy
sA =
75
p
4
+ 0 -
575(0.5)
= - 5.76 ksi = 5.76 ksi (C)
0.015625p
Ans.
Shear stress:
t =
VQ
Tc
and t =
It
J
(txy)A =
VQ
240(0.5)
80(0.0833)
Tc
=
+
= 1.36 ksi
J
It
0.03125p
0.015625p(1)
Ans.
(txz)A = 0
Ans.
Ans:
sB = 5.76 ksi (C), tB = 1.36 ksi
795
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–55. Determine the state of stress at point A on the cross
section of the post at section a–a. Indicate the results on a
differential element at the point.
100 mm
100 mm
3 kN
4 kN
400 mm
50 mm
50 mm
A
50 mm
50 mm a
B
z
Section a – a
Internal Loadings: Considering the equilibrium of the free-body diagram of the
post’s upper segment, Fig. a,
©Fy = 0;
Vy + 3 = 0
Vy = - 3 kN
©Fz = 0;
Vz + 4 = 0
Vz = - 4 kN
©Mx = 0; T = 0
©My = 0; My + 4(0.4) = 0
My = - 1.6 kN # m
©Mz = 0; Mz - 3(0.4) = 0
Mz = - 1.2 kN # m
Section Properties: The moment of inertia about the y and z axes of the post’s cross
section is
Iy = Iz =
www.elsolucionario.org
1
(0.1)(0.13) = 8.3333(10 - 6) m4
12
Referring to Fig. b,
(Qy)A = 0
(Qz)A = 0.025(0.05)(0.1) = 0.125(10 - 3) m3
Normal Stress: The normal stress is contributed by bending stress only. Thus,
s = -
Myz
Mzy
Iz
+
Iy
For point A, y = - 0.05 m and z = 0. Then
sA =
1.2(103)( - 0.05)
8.3333(10 - 6)
+ 0 = 7.20 MPa (T)
Ans.
Shear Stress: Then transverse shear stress at point A is
[(txy)V]A =
[(txz)V]A =
Vy(Qy)A
Izt
= 0
4(103)[0.125(10 - 3)]
Vz(Qz)A
Iy t
=
8.3333(10 - 6)(0.1)
Ans.
= 0.6 MPa
Ans.
The state of stress at point A is represented on the elements shown in Figs. c and d,
respectively.
796
400 mm
a
x
y
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–55. Continued
Ans:
sA = 7.20 MPa (T), tA = 0.6 MPa
797
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*8–56. Determine the state of stress at point B on the cross
section of the post at section a–a. Indicate the results on a
differential element at the point.
100 mm
100 mm
3 kN
4 kN
400 mm
50 mm
50 mm
A
50 mm
50 mm a
B
z
Section a – a
Internal Loadings: Considering the equilibrium of the free-body diagram of the
post’s upper segment, Fig. a,
©Fy = 0;
Vy + 3 = 0
Vy = - 3 kN
©Fz = 0;
Vz + 4 = 0
Vz = - 4 kN
©Mx = 0; T = 0
©My = 0; My + 4(0.4) = 0
My = - 1.6 kN # m
©Mz = 0; Mz - 3(0.4) = 0
Mz = 1.2 kN # m
Section Properties: The moment of inertia about the y and z axes of the post’s cross
section is
Iy = Iz =
1
(0.1)(0.13) = 8.3333(10 - 6) m4
12
www.elsolucionario.org
Referring to Fig. b,
(Qz)B = 0
(Qy)B = 0.025(0.05)(0.1) = 0.125(10 - 3) m3
Normal Stress: The normal stress is contributed by bending stress only. Thus,
s =
Myz
Mzy
-
Iz
+
Iy
For point B, y = 0 and z = - 0.05 m. Then
sB = - 0 +
- 1.6(103)( - 0.05)
8.3333(10 - 6)
= 9.60 MPa (T)
Ans.
Shear Stress: Then transverse shear stress at point B is
c (txy)V d
c (txy)V d
Vz(Qz)A
=
B
Iy t
= 0
Vy(Qy)A
=
B
Iz t
3(103)[0.125(10 - 3)]
=
8.3333(10 - 6)(0.1)
Ans.
= 0.45 MPa
Ans.
The state of stress at point B is represented on the elements shown in Figs. c and d,
respectively.
798
400 mm
a
x
y
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–56. Continued
799
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z
8–57. The sign is subjected to the uniform wind loading.
Determine the stress components at points A and B on the
100-mm-diameter supporting post. Show the results on a
volume element located at each of these points.
2m
1m
1.5 kPa
3m
C
B
A
2m
D
y
x
Point A:
sA =
10.5(103)(0.05)
Mc
=
= 107 MPa (T)
p
4
I
4 (0.05)
Ans.
tA =
3(103)(0.05)
Tc
= p
= 15.279(106) = 15.3 MPa
4
J
4 (0.05)
Ans.
Point B:
sB = 0
Ans.
1
tB =
2
3000(4(0.05)/3p))(2)(p)(0.05)
VQ
Tc
= 15.279(106) p
4
J
It
4 (0.05) (0.1)
tB = 14.8 MPa
Ans.
www.elsolucionario.org
Ans:
sA = 107 MPa (T), tA = 15.3 MPa,
sB = 0, tB = 14.8 MPa
800
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
z
8–58. The sign is subjected to the uniform wind loading.
Determine the stress components at points C and D on the
100-mm-diameter supporting post. Show the results on a
volume element located at each of these points.
2m
1m
1.5 kPa
3m
C
B
A
2m
D
y
x
Point C:
sC =
10.5(103)(0.05)
Mc
=
= 107 MPa (C)
p
4
I
4 (0.05)
Ans.
tC =
3(103)(0.05)
TC
= 15.279(106) = 15.3 MPa
= p
4
J
2 (0.05)
Ans.
Point D:
sD = 0
Ans.
3
tD =
1
2
3(10 )(4(0.05)/3p)(2)(p)(0.05)
VQ
Tc
= 15.8 MPa
+
= 15.279(106) +
p
4
J
It
4 (0.05) (0.1)
Ans.
Ans:
sC = 107 MPa (C), tC = 15.3 MPa,
sD = 0, tD = 15.8 MPa
801
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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
15 mm
15 mm
P
150 mm
15 mm
75 mm
100 mm
100 mm
100 mm
Equivalent Force System: Referring to Fig. a,
+ c ©Fx = (FR)x ;
- 60 - 120 = - F
F = 180 kN
©My = (MR)y ;
- 60(0.075) = - My
My = 4.5 kN # m
©Mz = (MR)z ;
-120(0.25) = - Mz
Mz = 30 kN # m
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)(0.33) (0.185)(0.273) = 0.14655(10 - 3) m4
12
12
Iy = 2 c
1
1
(0.015)(0.23) d +
(0.27)(0.0153) = 20.0759(10 - 6) m4
12
12
www.elsolucionario.org
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 =
My z
Mzy
N
+
A
Iz
Iy
smax = sA =
[-4.5(103)](0.1)
- 180(103)
[- 30(103)]( - 0.15)
+
0.01005
0.14655(10 - 3)
20.0759(10 - 6)
= -71.0 MPa = 71.0 MPa (C)
Ans.
Ans:
smax = 71.0 MPa (C)
802
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*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
15 mm
15 mm
P
150 mm
15 mm
100 mm
100 mm
Equivalent Force System: Referring to Fig. a,
+ c ©Fx = (FR)x ;
-P - 2P = - F
F = 3P
- P(0.075) = - My
©My = (MR)y ;
My = 0.075P
- 2P(0.25) = - Mz
©Mz = (MR)z ;
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) = 0.01005m2
Iz =
1
1
(0.2)(0.33) (0.185)(0.273) = 0.14655(10 - 3) m4
12
12
Iy = 2 c
1
1
(0.15)(0.23) d +
(0.27)(0.0153) = 20.0759(10 - 6) 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 =
My z
Mzy
N
+
A
Iz
Iy
- 100(106) = -
-0.075P(0.1)
( -0.5P)( -0.15)
3P
+
0.01005
0.14655(10 - 3)
20.0759(10 - 6)
P = 84470.40 N = 84.5k N
Ans.
803
75 mm
100 mm
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–61. The C-frame is used in a riveting machine. If the
force at the ram on the clamp at D is P = 8 kN, sketch the
stress distribution acting over the section a–a.
a
a
P
D
x =
(0.005)(0.04)(0.01) + 0.04(0.06)(0.01)
©x-A
=
= 0.026 m
©A
0.04(0.01) + 0.06(0.01)
200 mm
A = 0.04(0.01) + 0.06(0.01) = 0.001 m2
I =
10 mm
40 mm
1
(0.04)(0.013) + (0.04)(0.01)(0.026 - 0.005)2
12
1
+
(0.01)(0.063) + 0.01(0.06)(0.040 - 0.026)2 = 0.4773(10 - 6) m4
12
(smax)t =
10 mm
8(103)
1.808(103)0.026
P
Mx
+
=
+
A
I
0.001
0.4773(10 - 6)
= 106.48 MPa = 106 MPa
3
(smax)c =
60 mm
Ans.
3
8(10 )
1.808(10 )(0.070 - 0.026)
P
Mc
=
A
I
0.001
0.4773(10 - 6)
Ans.
= - 158.66 MPa = - 159 MPa
x
70 - x
=
;
158.66
106.48
www.elsolucionario.org
x = 41.9 mm
Ans:
(smax)t = 106 MPa, (smax)c = - 159 MPa
804
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–62. Determine the maximum ram force P that can be
applied to the clamp at D if the allowable normal stress for
the material is sallow = 180 MPa.
a
a
P
D
x =
(0.005)(0.04)(0.01) + 0.04(0.06)(0.01)
©xA
=
= 0.026 m
©A
0.04(0.01) + 0.06(0.01)
200 mm
A = 0.04(0.01) + 0.06(0.01) = 0.001 m2
I =
s =
10 mm
40 mm
1
(0.04)(0.013) + (0.04)(0.01)(0.026 - 0.005)2
12
1
+
(0.01)(0.063) + 0.01(0.06)(0.040 - 0.026)2 = 0.4773(10 - 6) m4
12
60 mm
10 mm
P
Mx
;
A
I
Assume tension failure,
0.226P(0.026)
P
180(106) =
+
0.001
0.4773(10 - 6)
P = 13524 N = 13.5 kN
Assume compression failure,
- 180(106) =
0.226P(0.070 - 0.026)
P
0.001
0.4773(10 - 6)
P = 9076 N = 9.08 kN (controls)
Ans.
Ans:
P = 9.08 kN
805
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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
Internal Forces and Moments: As shown on FBD.
y
x
©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
Section Properties:
A = p A 32 - 2.752 B = 1.4375p in2
www.elsolucionario.org
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
Mz y
N
+
A
Iz
Iy
sC =
(- 64.8)(12)(0)
9.00(12)(2.75)
-1.50
+
1.4375p
18.6992
18.6992
= 15.6 ksi (T)
sD =
Ans.
(- 64.8)(12)(3)
9.00(12)(0)
-1.50
+
1.4375p
18.6992
18.6992
= 124 ksi (T)
Ans.
806
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8–63. 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)D = ttwist =
64.8(12)(3)
= 62.4 ksi
37.3984
Ans.
(txy)D = tVy = 0
Ans.
(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.
Ans:
sC = 15.6 ksi (T), sD = 124 ksi (T),
tD = 62.4 ksi, tC = 52.4 ksi
807
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*8–64.
Solve Prob. 8–63 for points E and F.
12 ft
B
150 lb/ft2
6 ft
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
D
C
A
z
©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
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
(QF)z = (QE)y = 0
(QF)y = (QE)z =
www.elsolucionario.org
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 =
F
E
3 ft
(- 64.8)(12)(- 3)
9.00(12)(0)
-1.50
+
1.4375p
18.6992
18.6992
= - 125 ksi = 125 ksi (C)
Ans.
808
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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.
(txz)F = tVy = 0
Ans.
809
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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
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
10 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
www.elsolucionario.org
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
810
Ans.
12 in.
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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.
Ans:
sA = 605 psi (T), tA = 327 psi
811
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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
1 in.
Section a–a
x
a
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 = 250 lb # in
©Mz = 0; Mz + 50 sin 60°(10) = 0
Mz = - 433.01 lb # in
a
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
www.elsolucionario.org
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
TrC
519.62(1)
=
= 483.91 psi
J
1.07379
812
60°
z
10 in.
©My = 0; My - 50 cos 60°(10) = 0
50 lb
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.
Ans:
sB = 466 psi (C), tB = 422 psi
813
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–67. The metal link is subjected to the axial force of
P = 7 kN. Its original cross section is to be altered by
cutting a circular groove into one side. Determine the
distance a the groove can penetrate into the cross section so
that the tensile stress does not exceed sallow = 175 MPa .
Offer a better way to remove this depth of material from
the cross section and calculate the tensile stress for this
case. Neglect the effects of stress concentration.
P
a
40 mm
P
40 mm
25 mm
M - 7(103) a 0.04 - a
a + ©MO = 0;
0.08 - a
bb = 0
2
M = 3.5(103)a
smax =
Mc
P
+
A
I
175(106) =
7(103)
3.5(103)a(0.08 - a)>2
+ 1
3
(0.025)(0.08 - a)
12 (0.025)(0.08 - a)
Set x = 0.08 - a
4375 =
21(0.08 - x)
7
+
x
x2
www.elsolucionario.org
4375x2 + 14x - 1.68 = 0
Choose positive root :
x = 0.01806
a = 0.08 - 0.01806 = 0.0619 m
a = 61.9 mm
Ans.
Remove material equally from both sides.
s =
7(103)
= 15.5 MPa
(0.025)(0.01806)
Ans.
Ans:
a = 61.9 mm. Remove material equally from
both sides, s = 15.5 MPa.
814
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*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
A
z
B
y
x
30⬚
I =
800 N
1
1
p r4 = (p)(0.024) = 0.1256637 (10 - 6) m4
4
4
A = p r2 = p(0.022) = 1.256637 (10 - 3) m2
QA = y¿A¿ = a
sA =
=
tA =
4 (0.02) p (0.02)2
ba
b = 5.3333 (10 - 6) m3
3p
2
P
Mz
+
A
I
400
+ 0 = 0.318 MPa
1.256637 (10 - 3)
Ans.
692.82 (5.3333) (10 - 6)
VQA
= 0.735 MPa
=
It
0.1256637 (10 - 6)(0.04)
Ans.
815
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8–69.
Solve Prob. 8–68 for point B.
150 mm
200 mm
A
1
1
I = p r4 = (p)(0.024) = 0.1256637 (10 - 6) m4
4
4
z
B
y
x
30⬚
A = p r2 = p(0.022) = 1.256637 (10 - 3) m2
800 N
QB = 0
sB =
138.56 (0.02)
P
Mc
400
= - 21.7 MPa
=
A
I
1.256637 (10 - 3)
0.1256637 (10 - 6)
tB = 0
Ans.
Ans.
www.elsolucionario.org
Ans:
sB = - 21.7 MPa, tB = 0
816
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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
C
20 in.
8 in.
B
10 in.
20 in.
y
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.
Ans:
tA = 0, sA = 30.2 ksi (C)
817
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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
C
A =
p
(0.752) = 0.44179 in2
4
I =
p
(0.3754) = 0.015531 in4
4
QB = y¿A¿ =
8 in.
B
10 in.
20 in.
y
x
4(0.375) 1
a b (p)(0.3752) = 0.035156 in3
3p
2
sB = 0
tB =
20 in.
Ans.
VzQB
It
=
125(0.035156)
= 0.377 ksi
0.015531(0.75)
Ans.
www.elsolucionario.org
Ans:
sB = 0, tB = 0.377 ksi
818
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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⬚
a
A
The location of the neutral surface from the center of curvature of the hook, Fig. a,
can be determined from
A
dA
©
LA r
R =
where A = p(0.252) = 0.0625p in2
©
dA
= 2p A r - 2r2 - c2 B = 2p A 1.75 - 21.752 - 0.252 B = 0.11278 in.
LA r
Thus,
R =
0.0625p
= 1.74103 in.
0.11278
Then
e = r - R = 1.75 - 1.74103 = 0.0089746 in.
Referring to Fig. b, I and QA are
I =
p
(0.254) = 0.9765625(10 - 3)p in4
4
QA = 0
Consider the equilibrium of the FBD of the hook’s cut segment, Fig. c,
+ ©F = 0;
;
x
N - 80 cos 45° = 0
N = 56.57 lb
+ c ©Fy = 0;
80 sin 45° - V = 0
V = 56.57 lb
a + ©Mo = 0;
M - 80 cos 45°(1.74103) = 0
M = 98.49 lb # in
The normal stress developed is the combination of axial and bending stress. Thus,
s =
M(R - r)
N
+
A
Ae r
Here, M = 98.49 lb # in since it tends to reduce the curvature of the hook. For point
A, r = 1.5 in. Then
s =
(98.49)(1.74103 - 1.5)
56.57
+
0.0625p
0.0625p(0.0089746)(1.5)
= 9.269(103) psi = 9.27 ksi (T)
Ans.
The shear stress in contributed by the transverse shear stress only. Thus
t =
VQA
= 0
It
Ans.
The state of strees of point A can be represented by the element shown in Fig. d.
819
A
B
B a
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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 hook, Fig. a,
can be determined from
©
A
B
B a
A
R =
a
A
dA
LA r
Where A = p(0.252) = 0.0625p in2
©
dA
= 2p A r - 2r2 - c2 B = 2p A 1.75 - 21.752 - 0.252 B = 0.11278 in.
LA r
Thus,
R =
0.0625p
= 1.74103 in
0.11278
Then
e = r - R = 1.75 - 1.74103 = 0.0089746 in
Referring to Fig. b, I and QB are computed as
p
(0.254) = 0.9765625(10 - 3)p in4
4
I =
www.elsolucionario.org
4(0.25) p
QB = y¿A¿ =
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)
= 1.48 psi (T)
Ans.
The shear stress is contributed by the transverse shear stress only. Thus,
t =
56.57 (0.0104167)
VQB
= 384 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.
820
Ans:
s = 1.48 psi (T), t = 384 psi
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–74. The eye hook has the dimensions shown. If it supports
a cable loading of 80 kN, determine the maximum normal
stress at section a-a and sketch the stress distribution acting
over the cross section.
800 lb
2.5 in.
3.75 in.
a
a
1.25 in.
L
dA
r = 2p a3.125 -
R =
A
L
dA
r
2(3.125)2 - (0.625)2 b = 0.395707
800 lb
p(0.625)2
=
= 3.09343 in.
0.396707
M = 800(3.09343) = 2.475(103)
s =
(st)max =
(sc)max =
M(R - r)
P
+
Ar(r - R)
A
2.475(103)(3.09343 - 2.5)
p(0.625)2(2.5)(3.125 - 3.09343)
+
800
= 15.8 ksi
p(0.625)2
Ans.
800
= - 10.5 ksi
p(0.625)2
Ans.
2.475(103)(3.09343 - 3.75)
p(0.625)2(3.75)(3.125 - 3.09343)
+
Ans:
(st)max = 15.8 ksi, (sc)max = - 10.5 ksi
821
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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
1m
b
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
b
75 mm
1m
D
F
A
25 mm
Section b – b
Internal Loadings: Consider the equilibrium of the free-body diagram of the right
segment shown in Fig. b.
+ ©F = 0;
:
N - 392.4 = 0
N = 392.4 N
x
+ 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
www.elsolucionario.org
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) = 31.25 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 E, 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.
822
75 mm
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8–75. Continued
Ans:
sE = 802 kPa, tE = 69.8 kPa
823
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*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
Support Reactions: Referring to the free-body diagram of the entire frame shown in
Fig. a,
C
a
1m
30⬚
a + ©MA = 0;
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
1m
b
D
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
www.elsolucionario.org
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,
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.
824
F
A
25 mm
Section b – b
+ ©F = 0;
:
x
s =
75 mm
1m
Internal Loadings: Consider the equilibrium of the free-body diagram of the lower
cut segment, Fig. b,
I =
b
75 mm
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8–76. Continued
Shear Stress: The shear stress is contributed by transverse shear stress only. Thus,
130.8c 46.875 A 10 - 6 B d
VQA
tA =
=
= 31.0 kPa
It
2.6367 A 10 - 6 B (0.075)
Ans.
The state of stress at point A is represented on the element shown in Fig. d.
825
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–77. A bar having a square cross section of 30 mm by
30 mm is 2 m long and is held upward. If it has a mass of
5 kg/m, determine the largest angle u, measured from the
vertical, at which it can be supported before it is subjected
to a tensile stress along its axis near the grip.
u
2m
A = 0.03(0.03) = 0.9(10 - 3) m2
I =
1
(0.03)(0.033) = 67.5(10 - 9) m4
12
Require sA = 0
sA = 0 =
0 =
P
Mc
+
A
I
98.1 sin u(0.015)
- 98.1 cos u
+
0.9(10 - 3)
67.5(10 - 9)
0 = - 1111.11 cos u + 222222.22 sin u
tan u = 0.005;
u = 0.286°
www.elsolucionario.org
Ans.
Ans:
u = 0.286°
826
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–78. Solve Prob. 8–77 if the bar has a circular cross
section of 30-mm diameter.
u
A =
p
(0.032) = 0.225p(10 - 3) m2
4
I =
p
(0.0154) = 12.65625p(10 - 9) m4
4
2m
Require sA = 0
sA = 0 =
0 =
Mc
P
+
A
I
98.1 sin u(0.015)
- 98.1 cos u
+
0.225p(10 - 3)
12.65625p(10 - 9)
0 = - 4444.44 cos u + 1185185.185 sin u
tan u = 0.00375
u = 0.215°
Ans.
Ans:
u = 0.215°
827
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–79. The gondola and passengers have a weight of 1500 lb
and center of gravity at G. The suspender arm AE has a
square cross-sectional area of 1.5 in. by 1.5 in., and is pin
connected at its ends A and E. Determine the largest tensile
stress developed in regions AB and DC of the arm.
1.25 ft
E
D
4 ft
1.5 in.
B
C
1.5 in.
5.5 ft
A
Segment AB:
(smax)AB =
G
G
PAB
1500
=
= 667 psi
A
(1.5)(1.5)
Ans.
Segment CD:
sa =
PCD
1500
=
= 666.67 psi
A
(1.5)(1.5)
sb =
1875(12)(0.75)
Mc
= 40 000 psi
= 1
3
I
12 (1.5)(1.5 )
(smax)CD = sa + sb = 666.67 + 40 000
www.elsolucionario.org
= 40 666.67 psi = 40.7 ksi
Ans.
Ans:
(smax)AB = 667 psi, (smax)CD = 40.7 ksi
828
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*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 piston is
p
F = pA = p c 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
829
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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 Pa
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.
www.elsolucionario.org
Ans:
P = 94.2 kN
830
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8–82. 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
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
a
0.5 in.
1 in.
Section a – a
M
Section Properties: The cross-sectional area and the moment of inertia about the
centroidal axis of the femur’s cross section are
F
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
Ans.
Ans:
smax = 236 psi (C)
831
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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
P
Ans.
s2 = 0
Ans.
The pressure P is supported by the surface of the pistons in the longitudinal
direction.
www.elsolucionario.org
Ans:
s1 = 7.07 MPa, s2 = 0
832
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*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.
P
47 mm
P
s =
pr
;
t
3(106) =
p(0.045)
0.002
P = 133.3 kPa
Ans.
P = pA = 133.3 A 103 B (p)(0.045)2 = 848 N
Ans.
833
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8–85. The wall hanger has a thickness of 0.25 in. and is
used to support the vertical reactions of the beam that is
loaded as shown. If the load is transferred uniformly to each
strap of the hanger, determine the state of stress at points C
and D on the strap at A. Assume the vertical reaction F at
this end acts in the center and on the edge of the bracket
as shown.
10 kip
2 kip/ft
A
B
2 ft
2 ft
6 ft
2 in.
2 in.
2 in.
3.75 in.
2.75 in.
3 in.
a + ©MB = 0 ;
F
D
C
1 in.
12(3) + 10(8) - FA(10) = 0
1 in.
FA = 11.60 kip
I = 2c
1
(0.25)(2)3 d = 0.333 in4
12
A = 2(0.25)(2) = 1 in2
At point C,
sC =
2(5.80)
P
=
= 11.6 ksi
A
1
tC = 0
Ans.
www.elsolucionario.org
Ans.
At point D,
sD =
2(5.80)
[2(5.80)](1)
P
Mc
=
= - 23.2 ksi
A
I
1
0.333
Ans.
tD = 0
Ans.
Ans:
sC = 11.6 ksi, tC = 0,
sD = -23.2 ksi, tD = 0
834
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–86. The wall hanger has a thickness of 0.25 in. and is
used to support the vertical reactions of the beam that is
loaded as shown. If the load is transferred uniformly to each
strap of the hanger, determine the state of stress at points C
and D on the strap at B. Assume the vertical reaction F at
this end acts in the center and on the edge of the bracket
as shown.
10 kip
2 kip/ft
A
B
2 ft
2 ft
6 ft
2 in.
2 in.
2 in.
3.75 in.
2.75 in.
3 in.
a + ©MA = 0;
I = 2c
FB(10) - 10(2) - 12(7) = 0;
1
(0.25)(2)3 d = 0.333 in4;
12
F
D
C
1 in.
FB = 10.40 kip
1 in.
A = 2(0.25)(2) = 1 in2
At point C:
sC =
2(5.20)
P
=
= 10.4 ksi
A
1
Ans.
tC = 0
Ans.
At point D:
sD =
2(5.20)
[2(5.20)](1)
Mc
P
=
= - 20.8 ksi
A
I
1
0.333
Ans.
tD = 0
Ans.
Ans:
sC = 10.4 ksi, tC = 0,
sD = -20.8 ksi, tD = 0
835
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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
sx + sy
2
cos 2u + txy sin 2u
sx - sy
-
2
cos 2u - txy sin 2u
sx¿ + sy¿ = sx + sy
(Q. E. D.)
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836
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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.
8 ksi
B
5 ksi
40⬚
A
3 ksi
¢Fx¿ + (8¢A sin 40°) cos 40° - (5¢A sin 40°) cos 50° - (3¢A cos 40°) cos 40° +
a+ ©Fx¿ = 0
(8¢A cos 40°) cos 50° = 0
¢Fx¿ = - 4.052¢A
b+ ©Fy¿ = 0
¢Fy¿ - (8¢A sin 40°) sin 40° - (5¢A sin 40°) sin 50° + (3¢A cos 40°) sin 40° +
(8¢A cos 40°) sin 50° = 0
¢Fy¿ = - 0.4044¢A
sx¿ = lim¢A : 0
tx¿y¿ = lim¢A : 0
¢Fx ¿
= - 4.05 ksi
¢A
¢Fy ¿
Ans.
Ans.
= - 0.404 ksi
¢A
The negative signs indicate that the senses of sx¿ and tx¿y¿ are opposite to that shown
on FBD.
Ans:
sx¿ = - 4.05 ksi, tx¿y¿ = - 0.404 ksi
837
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A
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.
400 psi
650 psi
60⬚
B
Q+ ©Fx¿ = 0
¢Fx¿ - 400(¢A cos 60°) cos 60° + 650(¢A sin 60°) cos 30° = 0
¢Fx¿ = - 387.5¢A
a+ ©Fy¿ = 0
¢Fy¿ - 650(¢A sin 60°) sin 30° - 400(¢A cos 60°) sin 60° = 0
¢Fy¿ = 455 ¢A
sx¿ = lim¢A : 0
¢Fx ¿
= - 388 psi
¢A
tx¿y¿ = lim¢A : 0
¢Fy ¿
Ans.
= 455 psi
Ans.
¢A
The negative sign indicates that the sense of sx¿ is opposite to that shown on FBD.
www.elsolucionario.org
Ans:
sx¿ = -388 psi, tx¿y¿ = 455 psi
838
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*9–4. 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.
15 ksi
B
A
60⬚
6 ksi
Force Equilibrium: 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 60° and ¢A cos 60°, respectively. The forces acting on
the free-body diagram of the triangular sectioned element, Fig. b, are
©Fx¿ = 0;
¢Fx¿ - (6¢A sin 60°) cos 60° - (6¢A cos 60°) sin 60° + (15¢A cos 60°) cos 60° = 0
¢Fx¿ = 1.4461¢A
©Fy¿ = 0;
¢Fy¿ + (6¢A sin 60°) sin 60° - (6¢A cos 60°) cos 60° - (15¢A cos 60°) sin 60° = 0
¢Fy¿ = 3.4952¢A
Normal and Shear Stress: From the definition of normal and shear stress,
sx¿ = lim¢A : 0
tx¿y¿ = lim¢A : 0
¢Fx ¿
= 1.45 ksi
¢A
¢Fy ¿
Ans.
Ans.
= 3.50 ksi
¢A
839
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9–5. Determine the normal stress and shear stress acting
on the inclined plane AB. Solve the problem using the stress
transformation equations. Show the results on the sectional
element.
15 ksi
B
A
60⬚
6 ksi
Stress Transformation Equations:
u = + 150° (Fig. a)
sx = 0
sy = - 15 ksi
txy = - 6 ksi
We obtain,
sx¿ =
=
sx + sy
sx - sy
+
2
2
cos 2u + txy sin 2u
0 - ( - 15)
0 + ( -15)
+
cos 300° + ( -6) sin 300°
2
2
Ans.
= 1.45 ksi
tx¿y¿ = -
= -
sx - sy
2
sin 2u + txy cos 2u
0 - ( -15)
sin 300° + (- 6) cos 300°
2
www.elsolucionario.org
= 3.50 ksi
Ans.
The results are indicated on the triangular sectioned element shown in Fig. b.
Ans:
sx¿ = 1.45 ksi, tx¿y¿ = 3.50 ksi
840
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9–6. 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¿ + c 45 A 106 B ¢A sin 45° d cos 45° + c45 A 106 B ¢A cos 45° dsin 45°
- c 80 A 106 B ¢A sin 45° d cos 45° = 0
¢Fx¿ = - 5 A 106 B ¢A
©Fy¿ = 0;
¢Fy¿ + c 45 A 106 B ¢A cos 45° d cos 45° - c45 A 106 B ¢A sin 45° dsin 45°
- c 80 A 106 B ¢ A sin 45° d sin 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.
Ans:
sx¿ = -5 MPa, tx¿y¿ = 40 MPa
841
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9–7. 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
B
sy = 0
txy = 45 MPa
A
we obtain,
sx¿ =
=
sx + sy
sx - sy
+
2
2
cos u + txysin 2u
80 - 0
80 + 0
+
cos 270 + 45 sin 270°
2
2
= - 5 MPa
tx¿y¿ = -
= -
sx - sy
2
Ans.
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.
www.elsolucionario.org
Ans:
sx¿ = - 5 MPa, tx¿y¿ = 40 MPa
842
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*9–8. Determine the equivalent state of stress on an
element at the same point oriented 30° clockwise with
respect to the element shown. Sketch the results on the
element.
75 MPa
100 MPa
Stress Transformation Equations:
u = - 30° (Fig. a)
sx = 100 MPa
sy = - 75 MPa
txy = 0
We obtain,
sx¿ =
sx + sy
sx - sy
+
2
2
100 + ( -75)
=
100 - ( - 75)
+
2
cos 2u + txy sin 2u
2
cos ( -60°) + 0 sin (- 60°)
Ans.
= 56.25 MPa
sy¿ =
=
sx + sy
sx - sy
-
2
2
cos 2u - txy sin 2u
100 - ( -75)
100 + ( -75)
cos ( -60°) - 0 sin (- 60°)
2
2
Ans.
= - 31.25 MPa
tx¿y¿ = -
= -
sx - sy
2
sin 2u + txy cos 2u
100 - ( - 75)
sin ( - 60°) + 0 cos ( -60°)
2
Ans.
= 75.8 MPa
The negative sign indicates that sy¿ is a compressive stress. These results are
indicated on the element shown in Fig. b.
843
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9–9. Determine the equivalent state of stress on an
element at the same point oriented 30° counterclockwise
with respect to the element shown. Sketch the results on the
element.
75 MPa
100 MPa
Stress Transformation Equations:
u = + 30° (Fig. a)
sy = - 75 MPa
sx = 100 MPa
We obtain,
sx¿ =
=
sx + sy
sx - sy
+
2
2
cos 2u + txy sin 2u
100 + ( - 75)
100 - ( - 75)
+
cos 60° + 0 sin 60°
2
2
Ans.
= 56.25 MPa
sy¿ =
=
sx + sy
sx - sy
-
2
2
cos 2u - txy sin 2u
100 + (- 75)
100 - ( - 75)
cos 60° - 0 sin 60°
2
2
= - 31.25 MPa
tx¿y¿ = -
= -
sx - sy
2
www.elsolucionario.org
Ans.
sin 2u + txy cos 2u
100 - ( - 75)
sin 60° + 0 cos 60°
2
Ans.
= - 75.8 MPa
The negative signs indicate that sy¿ is a compressive stress tx¿y¿ is directed towards
the negative sense of the y¿ axis. These results are indicated on the element shown
in Fig. b.
Ans:
sx¿ = 56.25 MPa, sy¿ = -31.25 MPa,
tx¿y¿ = - 75.8 MPa
844
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9–10. Determine the equivalent state of stress on an
element at the same point oriented 60° clockwise with respect
to the element shown. Sketch the results on the element.
100 MPa
75 MPa
150 MPa
Stress Transformation Equations:
u = - 60° (Fig. a)
sx = 150 MPa
sy = 100 MPa
txy = 75 MPa
We obtain,
sx¿ =
=
sx + sy
sx - sy
+
2
2
cos 2u + txy sin 2u
150 + 100
150 - 100
+
cos ( -120°) + 75 sin (-120°)
2
2
= 47.5 MPa
sy¿ =
=
Ans.
sx + sy
sx - sy
-
2
2
cos 2u - txy sin 2u
150 - 100
150 + 100
cos ( -120°) - 75 sin (-120°)
2
2
Ans.
= 202 MPa
tx¿y¿ = -
= -
sx - sy
2
sin 2u + txy cos 2u
150 - 100
sin ( - 120°) + 75 cos (- 120°)
2
= - 15.8 MPa
Ans.
The negative sign indicates that tx¿y¿ is directed towards the negative sense of the
y¿ axis. These results are indicated on the element shown in Fig. b.
Ans:
sx¿ = 47.5 MPa, sy¿ = 202 MPa,
tx¿y¿ = -15.8 MPa
845
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9–11. Determine the equivalent state of stress on an element
at the same point oriented 60° counterclockwise with respect
to the element shown. Sketch the results on the element.
100 MPa
75 MPa
150 MPa
Stress Transformation Equations:
sx = 150 MPa
u = + 60° (Fig. a)
sy = 100 MPa
txy = 75 MPa
We obtain,
sx + sy
sx¿ =
2
=
sx - sy
+
2
cos 2u + txy sin 2u
150 - 100
150 + 100
+
cos 120° + 75 sin 120°
2
2
Ans.
= 177 MPa
sy¿ =
=
sx + sy
sx - sy
-
2
2
cos 2u - txy sin 2u
150 - 100
150 + 100
cos 120° - 75 sin 120°
2
2
www.elsolucionario.org
Ans.
= 72.5 MPa
tx¿y¿ = -
= -
sx - sy
2
sin 2u + txy cos 2u
150 - 100
sin 120° + 75 cos 120°
2
Ans.
= - 59.2 MPa
The negative sign indicates that tx¿y¿ is directed towards the negative sense of the
y¿ axis. These results are indicated on the element shown in Fig. b.
Ans:
sx¿ = 177 MPa, sy¿ = 72.5 MPa,
tx¿y¿ = - 59.2 MPa
846
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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
2
= -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
+
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
847
Ans.
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9–13. Determine the equivalent state of stress on an
element if it is oriented 30° clockwise from the element
shown. Use the stress-transformation equations.
300 psi
950 psi
sx = 0
sx¿ =
=
sy = - 300 psi
sx + sy
2
= -a
=
2
u = - 30°
cos 2u + txy sin 2u
0 - (- 300)
0 - 300
+
cos ( - 60°) + 950 sin ( -60) = - 898 psi
2
2
tx¿y¿ = - a
sy¿ =
sx - sy
+
txy = 950 psi
sx - sy
2
b sin 2u + txy cos 2u
0 - ( -300)
b sin ( - 60°) + 950 cos ( - 60°) = 605 psi
2
sx + sy
2
sx - sy
-
Ans.
2
Ans.
cos 2u - txy sin 2u
www.elsolucionario.org
0 - (- 300)
0 - 300
- a
b cos ( -60°) - 950 sin ( - 60°) = 598 psi
2
2
Ans.
Ans:
sx¿ = -898 psi, tx¿y¿ = 605 psi, sy¿ = 598 psi
848
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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 2us =
- (sx - sy)>2
=
txy
us = - 25.7°
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.
Ans:
s1 = 4.21 ksi, s2 = - 34.2 ksi,
up2 = 19.3° and up1 = -70.7°,
tmax = 19.2 ksi, savg = - 15 ksi, us = - 25.7°
in-plane
and 64.3°
849
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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.
60 MPa
30 MPa
45 MPa
sx = 45 MPa
a)
s1, 2 =
=
sy = - 60 MPa
sx + sy
Aa
;
2
sx - sy
2
txy = 30 MPa
2
b + txy
2
45 - ( -60) 2
45 - 60
2
;
b + (30)
a
A
2
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°
www.elsolucionario.org
Use Eq. 9–1 to determine the principal plane of s1 and s2:
sx¿ =
=
sx + sy
sx - sy
+
2
2
where u = 14.87°
45 + (-60)
45 - ( - 60)
+
cos 29.74° + 30 sin 29.74° = 53.0 MPa
2
2
Therefore up1 = 14.9°
b) tmaxin - plane = A a
savg =
cos 2u + txy sin 2u,
sx - sy
2
sx + sy
=
2
Ans.
2
b + txy =
2
and
Aa
up2 = - 75.1°
Ans.
45 - ( - 60) 2
2
b + 30 = 60.5 MPa Ans.
2
45 + ( - 60)
= - 7.50 MPa
2
Ans.
Orientation of maximum in-plane shear stress:
tan 2uy =
- (sx - sy)>2
txy
uy = - 30.1°
=
- (45 - (- 60))>2
= - 1.75
30
Ans.
and
uy = 59.9°
Ans.
By observation, in order to preserve equilibrium along AB, tmax has to act in the
direction shown.
Ans:
s1 = 53.0 MPa, s2 = -68.0 MPa,
up1 = 14.9° and up2 = - 75.1°,
savg = - 7.50 MPa, tmax = 60.5 MPa,
in-plane
us = - 30.1° and 59.9°
850
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*9–16. Determine the equivalent state of stress on an
element at the point which represents (a) the principal
stresses 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 and sketch the results on
the element.
50 MPa
15 MPa
Normal and Shear Stress:
sx = 50 MPa
sy = 0
txy = - 15 MPa
In-Plane Principal Stresses:
s1, 2 =
=
sx + sy
2
;
Aa
sx - sy
2
2
b + txy
2
2
50 + 0
; a 50 - 0 b + ( - 15)2
2
A
2
= 25 ; 2850
s1 = 54.2 MPa
s2 = - 4.15 MPa
Ans.
Orientation of Principal Plane:
tan 2up =
txy
(sx - sy)>2
=
- 15
= - 0.6
(50 - 0)>2
up = - 15.48° and 74.52°
Substitute u = - 15.48° into
sx¿ =
=
sx + sy
2
sx - sy
+
2
cos 2u + txy sin 2u
50 + 0
50 - 0
+
cos ( - 30.96°) + ( -15) sin ( - 30.96°)
2
2
= 54.2 MPa = s1
Thus,
Ans.
(up)1 = - 15.5° and (up)2 = 74.5°
The element that represents the state of principal stress is shown in Fig. a.
Maximum In-Plane Shear Stress:
tmax
in-plane
=
Aa
sx - sy
2
2
b + txy =
2
Aa
50 - 0 2
2
b + ( -15) = 29.2 MPa
2
851
Ans.
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–16. Continued
Orientation of the Plane of Maximum In-Plane Shear Stress:
tan 2us =
-(sx - sy)>2
txy
=
- (50 - 0)>2
= 1.667
- 15
us = 29.5° and 120°
By inspection, tmax
Ans.
has to act in the same sense shown in Fig. b to maintain
in-plane
equilibrium.
Average Normal Stress:
savg =
sx + sy
2
=
50 + 0
= 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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852
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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:
sx - sy
s1,2 =
;
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
2
sx - sy
+
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
≤ + txy 2 = Ba
2
125 - (- 75)
2
2
b + 502 = 112 MPa
Orientation of the Plane of Maximum In - Plane Shear Stress:
tan 2us = -
A sx - sy B >2
txy
= -
A 125 - (- 75) B >2
- 50
= 2
us = 31.7° and 122°
853
Ans.
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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.
www.elsolucionario.org
Ans:
s1 = 137 MPa, s2 = -86.8 MPa,
up1 = - 13.3°, up2 = 76.7°, tmax = 112 MPa,
in-plane
us = -31.7° and 122°, savg = 25 MPa
854
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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.
85 MPa
60 MPa
txy
45⬚
⫹
30⬚
⫽
sx
85 MPa
For element a:
sx = sy = 85 MPa
(sx¿)a =
=
(sy¿)a =
=
sx + sy
sx - sy
+
2
txy = 0
2
u = - 45°
cos 2u + txy sin 2u
85 + 85
85 - 85
+
cos ( - 90°) + 0 = 85 MPa
2
2
sx + sy
sx - sy
+
2
2
cos 2u - txy sin 2u
85 - 85
85 + 85
cos (- 90°) - 0 = 85 MPa
2
2
(tx¿y¿)a = -
= -
sx - sy
2
sin 2u + txy cos 2u
85 - 85
sin ( - 90°) + 0 = 0
2
For element b:
sx = sy = 0
(sx¿)b =
txy = 60 MPa
sx + sy
sx - sy
+
2
2
u = - 60°
cos 2u + txy sin 2u
= 0 + 0 + 60 sin ( - 120°) = - 51.96 MPa
(sy¿)b =
sx + sy
sx - sy
-
2
2
cos 2u - txy sin 2u
= 0 - 0 - 60 sin ( - 120°) = 51.96 MPa
(tx¿y¿)b = = -
sx - sy
2
sin 2u - txy cos 2u
85 - 85
sin ( - 120°) + 60 cos ( - 120°) = - 30 MPa
2
sx = (sx¿)a + (sx¿)b = 85 + (- 51.96) = 33.0 MPa
Ans.
sy = (sy¿)a + (sy¿)b = 85 + 51.96 = 137 MPa
Ans.
txy = (tx¿y¿)a + (tx¿y¿)b = 0 + (- 30) = - 30 MPa
Ans.
Ans:
sx = 33.0 MPa, sy = 137 MPa, txy = - 30 MPa
855
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–19. 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 and sketch the
results on the element.
25 MPa
100 MPa
Normal and Shear Stress:
sx = - 100 MPa
sy = 0
txy = 25 MPa
In-Plane Principal Stresses:
sx + sy
s1, 2 =
; Aa
2
sx - sy
2
2
2
b + txy
- 100 - 0 2
- 100 + 0 ;
2
b + 25
a
A
2
2
=
= - 50 ; 23125
s1 = 5.90 MPa
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s2 = - 106 MPa
Orientation of Principal Plane:
tan 2up =
txy
(sx - sy)>2
=
Ans.
25
= - 0.5
( -100 - 0)>2
up = - 13.28° and 76.72°
Substitute u = - 13.28° into
sx¿ =
sx + sy
2
sx - sy
+
2
cos 2u + txy sin 2u
-100 + 0
- 100 - 0
+
cos ( - 26.57°) + 25 sin ( -26.57°)
2
2
=
= - 106 MPa = s2
Thus,
(up)1 = 76.7° and (up)2 = - 13.3°
Ans.
The element that represents the state of principal stress is shown in Fig. a.
Maximum In-Plane Shear Stress:
tmax
in-plane
=
A
a
sx - sy
2
b + txy2 =
2
A
a
- 100 - 0 2
2
b + 25 = 55.9 MPa
2
856
Ans.
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–19. Continued
Orientation of the Plane of Maximum In-Plane Shear Stress:
tan 2us = -
(sx - sy)>2
txy
= -
( - 100 - 0)>2
= 2
25
us = 31.7° and 122°
By inspection, tmax
Ans.
has to act in the same sense shown in Fig. b to maintain
in-plane
equilibrium.
Average Normal Stress:
savg =
sx + sy
2
=
- 100 + 0
= - 50 MPa
2
Ans.
The element that represents the state of maximum in-plane shear stress is shown
in Fig. c.
Ans:
s1 = 5.90 MPa, s2 = -106 MPa,
up1 = 76.7° and up2 = -13.3°,
tmax = 55.9 MPa, savg = - 50 MPa,
in-plane
us = 31.7° and 122°
857
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*9–20. Planes AB and BC at a point are subjected to the
stresses shown. Determine the principal stresses acting at
this point and find sBC.
A
15 ksi
5 ksi
45⬚
B
6 ksi
C
Stress Transformation Equations: Referring to Fig. a and the established sign
convention,
u = - 135°
sx = - 15 ksi
sy = sAC
txy = 5 ksi
tx¿y¿ = 6 ksi
sx¿ = sBC
We have
tx¿y¿ = -
6 = -
sx - sy
2
sin 2u + txy cos 2u
-15 - sAC
sin (-270°) + 5 cos ( -270°)
2
sAC = - 3 ksi
Using this result,
s1, 2 =
=
sx + sy
;
2
A
a
sx - sy
2
- 15 - ( -3) 2
- 15 + ( - 3)
2
d + 5
; Ac
2
2
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s1 = - 1.19 ksi
sBC = sx¿ =
=
2
b + txy2
s2 = - 16.8 ksi
sx + sy
2
sx - sy
+
2
Ans.
cos 2u - txy sin 2u
-15 + ( - 3)
-15 - ( -3)
+
cos (- 270°) - 5 sin ( -270°)
2
2
Ans.
= - 14 ksi
858
sBC
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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⬚
80 ksi
90⬚
a
b
sx = 60 sin 60° = 51.962 ksi
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
b sin (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.
Ans:
ta = - 1.96 ksi, s1 = 80.1 ksi, s2 = 19.9 ksi
859
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–22. The grains of wood in the board make an angle of
20° with the horizontal as shown. Determine the normal
and shear stress that act perpendicular and parallel to the
grains if the board is subjected to an axial load of 250 N.
sx =
300 mm
60 mm
250 N
250 N
20⬚
25 mm
250
P
=
= 166.67 kPa
A
(0.06)(0.025)
sy = 0
txy = 0
u = 70°
sx¿ =
=
sx + sy
sx - sy
+
2
2
cos 2u + txy sin 2u
166.67 - 0
166.67 + 0
+
cos 140° + 0 = 19.5 kPa
2
2
tx¿y¿ = - a
= -a
sx - sy
2
Ans.
b sin 2u + txy cos 2u
166.67 - 0
b sin 140° + 0 = - 53.6 kPa
2
Ans.
www.elsolucionario.org
Ans:
sx¿ = 19.5 kPa, tx¿y¿ = -53.6 kPa
860
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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⬚
4m
300 mm
75 mm
200 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 =
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.
Ans:
sx¿ = 0.507 MPa, tx¿y¿ = 0.958 MPa
861
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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
4m
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 =
6.875(103)(1.6875)(10 - 3)
VQA
= 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.
www.elsolucionario.org
s2 = - 7.21 kPa
tan 2up =
Ans.
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
862
200 mm
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
sy
9–25. The wooden block will fail if the shear stress acting
along the grain is 550 psi. If the normal stress sx = 400 psi ,
determine the necessary compressive stress sy that will
cause failure.
58⬚
tx¿y¿ = - a
sx - sy
550 = - a
400 - sy
2
sx ⫽ 400 psi
b sin 2u + txy cos 2u
2
b sin 296° + 0
sy = - 824 psi
Ans.
Ans:
sy = - 824 psi
863
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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
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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.
864
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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 face
in-plane
of the element defined by us = - 45°.
Average Normal Stress:
savg =
sx + sy
2
=
29.76 + 0
= 14.9 ksi
2
The state of maximum in - plane shear stress is represented by the element shown in
Fig. c.
Ans:
s1 = 29.8 ksi, s2 = 0 , tmax
us = -45° and 45°
865
in-plane
= 14.9 ksi,
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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 =
www.elsolucionario.org
( -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.
866
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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
The state of maximum in - plane shear stress is represented by the element shown in
Fig. c.
Ans:
s1 = 0, s2 = - 22.90 ksi, tmax = 11.5 ksi,
in-plane
us = 45° and 135°
867
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*9–28. The 25-mm thick rectangular bar is subjected to
the axial load of 10 kN. If the bar is joined by the weld,
which makes an angle of 60° with the horizontal, determine
the shear stress parallel to the weld and the normal stress
perpendicular to the weld.
10 kN
10 kN
60⬚
80 mm
Internal Loadings: Consider the equilibrium of the free-body diagram of the bar’s
left cut segment, Fig. a.
+ ©F = 0;
:
x
N - 10 = 0
N = 10 kN
Normal and Shear Stress: The normal stress is developed by the axial stress only.
Thus,
sx =
10(103)
N
=
= 5 MPa
A
0.025(0.08)
Since no shear force is acting on the cut section
txy = 0
The state of stress is represented by the element shown in Fig. b.
Stress Transformation Equations: The stresses acting on the plane of weld can be
determined by orienting the element in the manner shown in Fig. c. We have
www.elsolucionario.org
sx = 5 MPa
u = - 30°
sy = 0
txy = 0
We obtain
sx¿ =
=
sx + sy
sx - sy
+
2
2
cos 2u + txy sin 2u
5 + 0
5 - 0
+
cos ( - 60°) + 0 sin ( -60°)
2
2
Ans.
= 3.75 MPa
tx¿y¿ = -
= -
sx - sy
2
sin 2u + txy cos 2u
5 - 0
sin ( - 60°) + 0 cos ( -60°)
2
= 2.17 MPa
Ans.
868
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9–29. The 3-in. diameter shaft is supported by a smooth
thrust bearing at A and a smooth journal bearing at B.
Determine the principal stresses and maximum in-plane
shear stress at a point on the outer surface of the shaft at
section a–a.
A
a
500 lb⭈ft
a
B
500 lb⭈ft
3 kip
Internal Loadings: Consider the equilibrium of the free-body diagram of the shaft’s
right cut segment, Fig. a.
©Fx = 0;
N + 3000 = 0
N = - 3000 lb
©Mx = 0;
T + 500 = 0
T = - 500 lb # ft
Section Properties: The cross-sectional area and the polar moment of inertia of the
shaft’s cross section are
A = p(1.52) = 2.25p in2
J =
p
(1.54) = 2.53125p in4
2
Normal and Shear Stress: The normal stress is contributed by the axial stress only.
Thus,
s =
- 3000
N
=
= - 424.41 psi
A
2.25p
The shear stress is contributed by the torsional shear stress only.
t =
500(12)(1.5)
Tc
=
= 1131.77 psi
J
2.53125p
The state of stress at a point on the outer surface of the shaft, Fig. b, can be
represented by the element shown in Fig. c.
In-Plane Principal Stress: sx = - 424.41 psi, sy = 0, and txy = - 1131.77 psi. We
obtain,
s1,2 =
sx + sy
=
;
2
Aa
sx - sy
2
2
b + txy
2
2
-424.41 + 0
; a - 424.41 - 0 b + ( -1131.77)2
2
A
2
s1 = 939.28 psi = 0.939 ksi s2 = - 1363.70 psi = - 1.36 ksi
Ans.
Maximum In-Plane Shear Stress:
tmax
in-plane
=
A
a
sx - sy
2
b + txy2 = A a
2
-424.41 - 0 2
2
b + ( -1131.77) = 1151.49 psi
2
= 1.15 ksi
Ans.
Ans:
s1 = 0.939 ksi, s2 = -1.36 ksi, tmax
in-plane
869
= 1.15 ksi
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9–30. 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.
www.elsolucionario.org
Ans:
sx¿ = -63.3 MPa, tx¿y¿ = 35.7 MPa
870
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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
+ ©MO = 0;
583.33(0.15) - M = 0
M = 87.5N # m
0.15 m
Referring to Fig. c,
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.
583.33 C 3.1875 A 10 - 6 B D
VQA
=
tA =
= 1.515 MPa
It
0.16367 A 10 - 6 B (0.0075)
The state 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
871
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 -76.64°
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 = - 76.6°
Ans.
The state of principal stresses is represented by the element shown in Fig. e.
www.elsolucionario.org
Ans:
s1 = 6.38 MPa, s2 = -0.360 MPa,
up1 = 13.4° and up2 = - 76.6°
872
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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 = 1876.39 N
Bx - 2166.67 cos 30° = 0
D
60⬚
a
B
C
a
2166.67 sin 30° - 500 - By = 0
+ c ©Fy = 0;
By = 583.33 N
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
873
Ans.
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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.
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874
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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 MPa
s2 = - 24.0 MPa
Ans.
875
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°
Substituting 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.
www.elsolucionario.org
Ans:
Point A: s1 = 0, s2 = - 192 MPa,
up1 = 0, up2 = 90°,
Point B: s1 = 24.0 MPa, s2 = -24.0 MPa,
up1 = - 45.0°, up2 = 45.0°
876
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–34. Determine the principal stress and the maximum
in-plane 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
- 45°
Ans.
Substitute u = 45°,
tx¿y¿ = -
= -
sx - sy
2
A
3000 lb
sin 2u + txy cos 2u
- 1.337 - 0
sin 90° + 0
2
= 0.668 ksi = 668 psi = tmax
in-plane
877
24 in.
3000 lb
12 in.
12 in.
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9–34. Continued
This indicates that tmax
acts toward the positive sense of y¿ axis at the face of the
in-plane
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.
www.elsolucionario.org
Ans:
s1 = 0, s2 = -1.34 ksi, tmax
in-plane
us = 45° and -45°
878
= 668 psi,
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9–35. The square steel plate has a thickness of 10 mm and
is subjected to the edge loading shown. Determine the
maximum in-plane shear stress and the average normal
stress developed in the steel.
sx = 5 kPa
sy = - 5 kPa
tmax
sx - sy
in-plane
savg =
a
txy = 0
200 mm
b + t2xy
C
=
5 + 5 2
b + 0 = 5 kPa
C
2
a
sx + sy
3
=
50 N/m
2
=
2
50 N/m
Ans.
200 mm
5 - 5
= 0
2
Ans.
Note:
tan 2us =
tan 2us =
- (sx - sy)>2
txy
-(5 + 5)>2
= q
0
us = 45°
Ans:
tmax
in-plane
879
= 5 kPa, savg = 0
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*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°
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880
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P
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.
F
F
A
L
2
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
d
pd2
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
2
d
pd
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.
Ans:
s1 =
4
2PL
- Fb , s2 = 0,
a
d
pd2
tmax
=
in-plane
881
2PL
2
a
- Fb
d
pd2
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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.
s =
30⬚
10 N
10 N
30 mm
10
P
= 109.76 kPa
= p
2
2
A
4 (0.03 - 0.028 )
sx = 109.76 kPa
tx¿y¿ = -
= -
sx - sy
2
sy = 0
txy = 0
u = 30°
sin 2u + txy cos 2u
109.76 - 0
sin 60° + 0 = - 47.5 kPa
2
Ans.
www.elsolucionario.org
Ans:
tx¿y¿ = -47.5 kPa
882
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9–39. Solve Prob. 9–38 for the normal stress acting
perpendicular to the seam.
30⬚
10 N
10 N
30 mm
s =
10
P
= 109.76 kPa
= p
2
A
(0.03
- 0.0282)
4
sx¿ =
=
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.
Ans:
sx¿ = 82.3 kPa
883
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*9–40. The wide-flange beam is subjected to the 50-kN
force. Determine the principal stresses in the beam at point A
located on the web at the bottom of the upper flange.
Although it is not very accurate, use the shear formula to
calculate the shear stress.
50 kN
A
B
1m
3m
A
10 mm
B
200 mm
I =
1
1
(0.2)(0.274)3 (0.19)(0.25)3 = 95.451233(10 - 6) m4
12
12
QA = (0.131)(0.012)(0.2) = 0.3144(10 - 3) m3
sA =
150(103)(0.125)
My
= 196.43 MPa
=
I
95.451233(10 - 6)
tA =
VQA
50(103)(0.3144)(10 - 3)
= 16.47 MPa
=
It
95.451233(10 - 6)(0.01)
sx = 196.43 MPa
s1, 2 =
=
sx + sy
2
;
sy = 0
Aa
sx - sy
2
txy = - 16.47 MPa
2
b + txy
2
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196.43 + 0
196.43 - 0 2
2
;
b + ( - 16.47)
2
Aa
2
s1 = 198 MPa
Ans.
s2 = - 1.37 MPa
Ans.
884
12 mm
250 mm
12 mm
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–41. Solve Prob. 9–40 for point B located on the web at
the top of the bottom flange.
50 kN
A
B
1m
3m
A
12 mm
250 mm
12 mm
10 mm
B
200 mm
1
1
(0.2)(0.247)3 (0.19)(0.25)3 = 95.451233(10 - 6) m4
I =
12
12
QB = (0.131)(0.012)(0.2) = 0.3144(10 - 3)
sB = -
tB =
My
150(103)(0.125)
= = - 196.43 MPa
I
95.451233(10 - 6)
VQB
50(103)(0.3144)(10 - 3)
=
= 16.47 MPa
It
95.451233(10 - 6)(0.01)
sx = -196.43 MPa
s1, 2 =
=
sy = 0
sx + sy
;
2
A a
txy = - 16.47 MPa
sx - sy
2
2
b + txy
2
- 196.43 + 0
- 196.43 - 0 2
2
;
b + ( -16.47)
2
Aa
2
s1 = 1.37 MPa
Ans.
s2 = - 198 MPa
Ans.
Ans:
s1 = 1.37 MPa, s2 = - 198 MPa
885
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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 stresses 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
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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.
Ans:
s1 = 2.97 ksi, s2 = - 4.12 ksi, tmax = 3.55 ksi
in-plane
886
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9–43. The nose wheel of the plane is subjected to a design
load of 12 kN. Determine the principal stresses acting on
the aluminum wheel support at point A.
150 mm
A
20 mm
60⬚
30 mm
A
20 mm
300 mm
a+ ©Fy = 0;
12 cos 30° - N = 0;
N = 10.392 kN
b+ ©Fx = 0;
- 12 sin 30° + V = 0;
V = 6 kN
a + ©MA = 0;
M - (12)(0.150) = 0;
M = 1.80 kN # m
s =
10.392(103)
P
=
= 8.66 MPa (C)
A
(0.03)(0.04)
t =
VQ
6(103)(0.01)(0.03)(0.02)
= 7.50 MPa
= 1
3
It
12 (0.03)(0.04) (0.03)
s1, 2 =
=
sx + sy
;
2
A
a
sx - sy
2
12 kN
175 mm
b + txy2
2
-8.66 + 0
- 8.66 - 0 2
; a
b + (7.50)2
2
A
2
= - 4.33 ; 8.66025
s2 = - 12.990 = - 13.0 MPa
Ans.
s1 = 4.33 MPa
Ans.
Ans:
s1 = 4.33 MPa, s2 = - 13.0 MPa
887
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*9–44.
Solve Prob. 9–3 using Mohr’s circle.
sx + sy
2
=
A(- 650, 0)
- 650 + 400
= - 125
2
B(400, 0)
C( - 125, 0)
R = CA = = 650 - 125 = 525
sx¿ = - 125 - 525 cos 60° = - 388 psi
Ans.
tx¿y¿ = 525 sin 60° = 455 psi
Ans.
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888
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9–45.
Solve Prob. 9–6 using Mohr’s circle.
sx = 80 MPa
sx + sy
2
=
sy = 0
80 + 0
= 40
2
txy = 45 MPa
u = 135°
A(80, 45)
C(40, 0)
R = 2(80 - 40)2 + 452 = 60.208
f = tan - 1 a
45
b = 48.366° below the x-axis.
80 - 40
Coordinates of the rotated point: a counterclockwise rotation of 2u = 270° is the
same as a clockwise rotation of 90°, to c = 48.366° + 90° = 41.634° below the
negative x-axis.
sx¿ = 40 - 60.208 cos (41.634°) = - 5 MPa
Ans.
tx¿y¿ = 60.208 sin (41.634°) = 40 MPa
Ans.
Ans:
sx¿ = - 5 MPa, tx¿y¿ = 40 MPa
889
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9–46.
Solve Prob. 9–14 using Mohr’s circle.
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
2us = tan - 1
12
+ 90°;
(30 - 15)
Ans.
us = 64.3°
Ans.
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Ans:
s1 = 4.21 ksi, s2 = - 34.2 ksi,
up2 = 19.3° and up1 = - 70.7°,
= 19.2 ksi, savg = - 15 ksi, us = 64.3°
tmax
in-plane
890
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9–47.
Solve Prob. 9–10 using Mohr’s circle.
Construction of the
txy = 75 MPa. Thus
savg =
Circle:
u = - 60°,
sx + sy
2
=
sx = 150 MPa,
sy = 100 MPa,
150 + 100
= 125 MPa
2
The coordinates of the reference point A and center C of the circle are
A(150, 75)
C(125, 0)
Thus, the radius of the circle is
R = CA = 2(150 - 125)2 + (75)2 = 79.06 MPa
Normal and Shear Stress on Rotated Element: Here u = 60° clockwise. By rotating
the radial line CA clockwise 2u = 120°, it coincides with the radial line OP and the
coordinates of reference point P(sx¿, tx¿y¿) represent the normal and shear
stresses on the face of the element defined by u = - 60°, sy ¿ can be determined by
calculating the coordinates of point Q. From the geometry of the circle, Fig. (a).
sin a =
75
, a = 71.57°, b = 120° + 71.57° - 180° = 11.57°
79.06
sx¿ = 125 - 79.06 cos 11.57° = 47.5 MPa
Ans.
tx¿y¿ = - 79.06 sin 11.57° = - 15.8 MPa
Ans.
sy¿ = 125 + 79.06 cos 11.57° = 202 MPa
Ans.
The results are shown in Fig. (b).
Ans:
sx¿ = 47.5 MPa, tx¿ y¿ = - 15.8 MPa,
sy¿ = 202 MPa
891
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*9–48.
Solve Prob. 9–12 using Mohr’s circle.
sx + sy
2
=
- 10 + 0
= - 5 ksi
2
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
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892
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9–49.
Solve Prob. 9–16 using Mohr’s circle.
Construction of Circle: sx = 50 MPa, sy = 0, txy = -15 MPa . Thus,
savg =
sx + sy
=
2
50 + 0
= 25 MPa
2
Ans.
The coordinates of reference point A and center C of the circle are
A(50, -15)
C(25, 0)
Thus, the radius of the circle is
R = CA = 2(50 - 25)2 + (- 15)2 = 29.15 MPa
See Fig. (a).
a) Principal Stress:
s1 = 54.2 MPa,
sin 2a =
15
,
29.15
s2 = - 4.15 MPa
Ans.
a = 15.5°
Ans.
See Fig. (b).
Ans:
s1 = 54.2 MPa, s2 = - 4.15 MPa , up = - 15.5°
= 29.2 MPa, us = 29.5°
savg = 25 MPa, t max
in-plane
893
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9–50. 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 P(sx¿, tx¿y¿) 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)
c sx - a
sx + sy
2
sx + sy
+
2
C
a
sx + sy
2
2
b d + t2xy =
sx - sy
2
C
a
b , 0b
sx - sy
2
2
b + t2xy
2
b + t2xy cos u¿
(1)
u¿ = 2uP - 2u
(2)
cos (2uP - 2u) = cos 2uP cos 2u + sin 2up sin 2u
From the circle:
sx -
cos 2uP =
sin 2uP =
4A
4A
sx + sy
2
B + t2xy
(3)
B 2 + t2xy
(4)
sx - sy
2
txy
sx - sy
2
2
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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
894
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9–51. Determine the equivalent state of stress if an
element is oriented 45° clockwise from the element shown.
10 ksi
5 ksi
Construction of the Circle: sx = 0, sy = 10 ksi, and txy = - 5 ksi. Thus,
savg =
sx + sy
2
=
0 + 10
= 5 ksi
2
The coordinates of reference point A and the center C of the circle are
A(0, -5)
C(5, 0)
Thus, the radius of the circle is
R = CA = 2(0 - 5)2 + ( - 5)2 = 7.071 ksi
Using these results, the circle is shown in Fig. a.
Normal and Shear Stresses on the Rotated Element: Here, u = 45° clockwise. By
rotating the radial line CA clockwise 2u = 90°, it coincides with the radial line OP
and the coordinates of reference point P(sx¿, tx¿y¿) represent the normal and shear
stresses on the face of the element defined by u = - 45°. sy¿ can be determined by
calculating the coordinates of point Q. From the geometry of the circle,
5
a = tan-1 a b = 45°
5
b = 180° - 90° - 45° = 45°
Then
sx¿ = 5 + 7.071 cos 45° = 10 ksi
Ans.
tx¿y¿ = - 7.071 sin 45° = - 5 ksi
Ans.
sy¿ = 5 - 7.071 cos 45° = 0
Ans.
The results are indicated on the element shown in Fig. b.
Ans:
sx¿ = 10 ksi, tx¿y¿ = - 5 ksi, sy¿ = 0
895
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*9–52. Determine the equivalent state of stress if an
element is oriented 20° clockwise from the element shown.
2 ksi
Construction of the Circle: In accordance with the sign convention, sx = 3 ksi,
sy = - 2 ksi, and tx¿y¿ = - 4 ksi. Hence,
savg =
sx + sy
2
=
3 + ( -2)
= 0.500 ksi
2
4 ksi
The coordinates for reference points A and C are
A(3, - 4)
3 ksi
C(0.500, 0)
The radius of the circle is
R = 2(3 - 0.500)2 + 42 = 4.717 ksi
Stress on the Rotated Element: The normal and shear stress components
A sx¿ and tx¿y¿ B are represented by the coordinate of point P on the circle. sy¿ can be
determined by calculating the coordinates of point Q on the circle.
sx¿ = 0.500 + 4.717 cos 17.99° = 4.99 ksi
Ans.
tx¿y¿ = - 4.717 sin 17.99° = - 1.46 ksi
Ans.
sy¿ = 0.500 - 4.717 cos 17.99° = - 3.99 ksi
Ans.
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896
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9–53. Determine the equivalent state of stress if an
element is oriented 30° clockwise from the element shown.
230 MPa
350 MPa
480 MPa
A(350, - 480 )
B(230, 480)
C(290, 0)
R = 2602 + 4802 = 483.73
sx¿ = 290 + 483.73 cos 22.87° = 736 MPa
Ans.
sy¿ = 290 - 483.73 cos 22.87° = - 156 MPa
Ans.
tx¿y¿ = - 483.73 sin 22.87° = - 188 MPa
Ans.
Ans:
sx¿ = 736 MPa, sy¿ = - 156 MPa,
tx¿y¿ = - 188 MPa
897
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9–54. Determine the equivalent state of stress which
represents (a) the principal stress, and (b) the maximum
in-plane shear stress and the associated average normal stress.
For each case, determine the corresponding orientation of the
element with respect to the element shown.
120 MPa
80 MPa
40 MPa
Construction of the Circle: sx = 80 MPa, sy = - 120 MPa, and txy = - 40 MPa .
Thus,
savg =
sx + sy
2
=
80 + ( - 120)
= - 20 MPa
2
Ans.
The coordinates of reference point A and the center C of the circle are
A(80, -40)
C(-20, 0)
Thus, the radius of the circle is
R = CA = 2 [80 - ( - 20)]2 + (- 40)2 = 107.703 MPa
Using these results, the circle is shown in Fig. a.
In-Plane Principal Stress: The coordinates of reference points B and D represent
s1 and s2, respectively.
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s1 = - 20 + 107.703 = 87.7 MPa
Ans.
s2 = - 20 - 107.703 = - 128 MPa
Ans.
Orientation of the Principal Plane: Referring to the geometry of the circle,
tan 2(up)1 =
40
= 0.4
80 + 20
Ans.
(up)1 = 10.9° (clockwise)
The state of principal stress is represented on the element shown in Fig. b.
Maximum In-Plane Shear Stress: The state of maximum in-plane shear stress is
represented by the coordinates of point E. Fig. a.
tmax
= - R = - 108 MPa
Ans.
in-plane
Orientation of the Plane of Maximum In-Plane Shear Stress: From the geometry of
the circle, Fig. a,
tan 2us =
80 + 20
= 2.5
40
Ans.
us = 34.1° (counterclockwise)
Ans.
The state of in-plane maximum shear stress is represented on the element shown
in Fig. c.
Ans:
s1 = 87.7 MPa, s2 = - 128 MPa,
(up)1 = 10.9° (clockwise), savg = - 20 MPa,
tmax
= - 108 MPa ,
in-plane
us = 34.1° (counterclockwise)
898
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10 ksi
9–55. Determine the equivalent state of stress which
represents (a) the principal stress, and (b) the maximum
in-plane shear stress and the associated average normal stress.
For each case, determine the corresponding orientation of the
element with respect to the element shown.
25 ksi
Construction of the Circle: sx = - 25 ksi, sy = 0, and txy = 10 ksi. Thus,
savg =
sx + sy
2
=
- 25 + 0
= - 12.5 ksi
2
Ans.
The coordinates of reference point A and the center C of the circle are
A(- 25, 10)
C( - 12.5, 0)
Thus, the radius of the circle is
R = CA = 2[- 25 - ( - 12.5)]2 + 102 = 16.008 ksi
Using these results, the circle is shown in Fig. a.
In-Plane Principal Stress: The coordinates of reference points B and D represent
s1 and s2, respectively.
s1 = - 12.5 + 16.008 = 3.51 ksi
Ans.
s2 = - 12.5 - 16.008 = - 28.5 ksi
Ans.
Orientation of the Principal Plane: From the geometry of the circle, Fig. a,
tan 2(up)1 =
10
= 0.8
25 - 12.5
Ans.
(up)1 = 19.3° (clockwise)
The state of principal stress is represented on the element shown in Fig. b.
Maximum In-Plane Shear Stress: The state of maximum in-plane shear stress is
represented by the coordinates of point E, Fig. a.
tmax
= R = 16.0 ksi
Ans.
in-plane
Orientation of the Plane of Maximum In-Plane Shear Stress: From the geometry of
the circle,
tan 2us =
25 - 12.5
= 1.25
10
us = 25.7° (counterclockwise)
Ans.
The state of in-plane maximum shear stress is represented on the element shown
in Fig. c.
Ans:
s1 = 3.51 ksi , s2 = - 28.5 ksi ,
(up)1 = 19.3°(clockwise), t max
= 16.0 ksi ,
in-plane
savg = - 12.5 ksi, us = 25.7° (counterclockwise)
899
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*9–56. 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.
www.elsolucionario.org
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
= - R = - 9.01 ksi
Ans.
in-plane
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.
900
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9–57. Determine (a) the principal stresses and (b) the
maximum in-plane shear stress and average normal stress.
Specify the orientation of the element in each case.
50 MPa
30 MPa
A(0, -30)
B(50, 30)
C(25, 0)
R = CA = 2(25 - 0)2 + 302 = 39.05
s1 = 25 + 39.05 = 64.1 MPa
Ans.
s2 = 25 - 39.05 = - 14.1 MPa
Ans.
tan 2up =
30
= 1.2
25 - 0
up2 = 25.1°
Ans.
savg = 25.0 MPa
Ans.
tmax
= R = 39.1 MPa
Ans.
in-plane
tan 2us =
25 - 0
= 0.8333
30
us = - 19.9°
Ans.
Ans:
s1 = 64.1 MPa, s2 = - 14.1 MPa, up2 = 25.1°
savg = 25.0 MPa, tmax
= 39.1 MPa,
in-plane
us = - 19.9°
901
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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.
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Ans:
sx¿ = - 421 MPa, tx¿y¿ = - 354 MPa,
sy¿ = 421 MPa
902
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8 ksi
9–59. Determine (a) the principal stresses and (b) the
maximum in-plane shear stress and average normal stress.
Specify the orientation of the element in each case.
4 ksi
12 ksi
A(-12, 4)
B( - 8, - 4)
C( -10, 0)
R = CA = CB = 222 + 42 = 4.472
a)
s1 = - 10 + 4.472 = - 5.53 ksi
Ans.
s2 = - 10 - 4.472 = - 14.5 ksi
Ans.
tan 2up =
4
2
2up = 63.43°
up = - 31.7°
b)
tmax
= R = 4.47 ksi
Ans.
in-plane
savg = - 10 ksi
Ans.
2us = 90 - 2up
us = 13.3°
Ans.
Ans:
(a) s1 = - 5.53 ksi, s2 = - 14.5 ksi,
up = - 31.7°
(b) tmax
= 4.47 ksi, savg = - 10 ksi,
in-plane
us = 13.3°
903
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
200 MPa
*9–60. Determine the principal stresses, the maximum
in-plane shear stress, and average normal stress. Specify the
orientation of the element in each case.
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.
www.elsolucionario.org
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.
904
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9–61. Draw Mohr’s circle that describes each of the following
states of stress.
5 MPa
20 ksi
20 ksi
5 MPa
(b)
(a)
18 MPa
(c)
905
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–62. The grains of wood in the board make an angle of 20°
with the horizontal as shown. Using Mohr’s circle, determine
the normal and shear stresses that act perpendicular and
parallel to the grains if the board is subjected to an axial load
of 250 N.
sx =
300 mm
60 mm
250 N
250 N
20⬚
25 mm
250
P
=
= 166.67 kPa
A
(0.06)(0.025)
R = 83.33
Coordinates of point B:
sx¿ = 83.33 - 83.33 cos 40°
sx¿ = 19.5 kPa
Ans.
tx¿y¿ = - 83.33 sin 40° = - 53.6 kPa
Ans.
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Ans:
sx¿ = 19.5 kPa, tx¿y¿ = - 53.6 kPa
906
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z
9–63. The post has a square cross-sectional area. If it is fixed
supported at its base and a horizontal force is applied at its end
as shown, determine (a) the maximum in-plane shear stress
developed at A and (b) the principal stresses at A.
3 in.
3 in.
60 lb
y
x
18 in.
A
1 in.
1
I =
(3)(33) = 6.75 in4
12
Myx
QA = (1)(1)(3) = 3 in3
=
1080(0.5)
= - 80 psi
6.75
=
60(3)
= 8.889 psi
6.75(3)
A(-80, 8.889)
B(0, - 8.889)
sA =
tA =
tmax
I
VyQA
It
C( - 40, 0)
= R = 2402 + 8.8892 = 41.0 psi
Ans.
in-plane
s1 = - 40 + 40.9757 = 0.976 psi
Ans.
s2 = - 40 - 40.9757 = - 81.0 psi
Ans.
Ans:
tmax
in-plane
= 41.0 psi , s1 = 0.976 psi ,
s2 = - 81.0 psi
907
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*9–64. Determine the principal stress, the maximum
in-plane shear stress, and average normal stress. Specify the
orientation of the element in each case.
20 MPa
80 MPa
30 MPa
In accordance to the established sign convention, sx = 30 MPa, sy = - 20 MPa and
txy = 80 MPa. Thus,
savg =
sx + sy
30 + ( -20)
= 5 MPa
2
=
2
Then, the coordinates of reference point A and the center C of the circle is
A(30, 80)
C(5, 0)
Thus, the radius of circle is given by
R = CA = 2(30 - 5)2 + (80 - 0)2 = 83.815 MPa
Using these results, the circle shown in Fig. a, can be constructed.
The coordinates of points B and D represent s1 and s2 respectively. Thus
s1 = 5 + 83.815 = 88.8 MPa
Ans.
s2 = 5 - 83.815 = - 78.8 MPa
Ans.
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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
= R = 83.8 MPa
Ans.
in-plane
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.
908
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9–64. Continued
909
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–65. 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
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)
N
200
+
=
+
= 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)
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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.
Ans:
s1 = 29.4 ksi , s2 = - 17.0 ksi
910
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9–66. Determine the principal stress and maximum
in-plane shear stress at point A on the cross section of the
pipe at section a–a.
a 300 mm
a
300 mm
b
30 mm
A
B
200 mm
300 N
Internal Loadings: Considering the equilibrium of the free-body diagram of the
assembly’s segment, Fig. a,
©Fx = 0; N - 450 = 0
N = 450 N
©Fy = 0; Vy = 0
©Fz = 0; Vz + 300 = 0
Vz = - 300 N
©Mx = 0; T + 300(0.2) = 0
T = - 60 N # m
©My = 0; My - 450(0.3) + 300(0.3) = 0
My = 45 N # m
©Mz = 0; Mz + 450(0.2) = 0
Mz = - 90 N # m
Section Properties: The cross-sectional area, the moment of inertia about the y and
z axes, and the polar moment of inertia of the pipe’s cross section are
A = p(0.032 - 0.022) = 0.5p(10-3) m2
Iy = Iz =
J =
p
(0.034 - 0.024) = 0.1625p(10-6) m4
4
p
(0.034 - 0.024) = 0.325p(10-6) m4
2
Referring to Fig. b,
(Qy)A = 0 (Qz)A =
4(0.03) p
4(0.02) p
c (0.03)2 d c (0.022) d = 12.667(10-6) m3
3p
2
3p
2
Normal and Shear Stress: The normal stress is a combination of axial and bending
stress.
sA =
MzyA
( - 90)(- 0.03)
N
450
=
-3
A
Iz
0.5p(10 )
0.1625p(10-6)
= - 5.002 MPa
Since Vy = 0, [(txy)V]A = 0. However, the shear stress is the combination of
torsional and transverse shear stress. Thus,
(txz)A = [(txz)T]A - [(txz)V]A
=
Vz(Qz)A
60(0.03)
300[12.667(10-6)]
Tc
= 1.391 MPa
=
-6
J
Iy t
0.325p(10 )
0.1625p(10-6)(0.02)
The state of stress at point A is represented by the element shown in Fig. c.
911
450 N
20 mm
Section a – a
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9–66. Continued
Construction of the Circle: sx = - 5.002 MPa, sz = 0, and txz = 1.391 MPa. Thus,
savg =
sx + sz
2
=
- 5.002 + 0
= - 2.501 MPa
2
The coordinates of reference point A and the center C of the circle are
A(- 5.002, 1.391)
C(- 2.501, 0)
Thus, the radius of the circle is
R = CA = 2[- 5.002 - ( -2.501)]2 + 1.3912 = 2.862 MPa
Using these results, the circle is shown in Fig. d.
In-Plane Principal Stresses: The coordinates of reference points B and D represent
s1 and s2, respectively.
s1 = - 2.501 + 2.862 = 0.361 MPa
Ans.
s2 = - 2.501 - 2.862 = - 5.36 MPa
Ans.
In-Plane Maximum Shear Stress: The coordinates of point E represent the state of
maximum shear stress. Thus,
tmax
= |R| = 2.86 MPa
Ans.
in-plane
www.elsolucionario.org
Ans:
s1 = 0.361 MPa , s2 = - 5.36 MPa,
tmax
= 2.86 MPa
in-plane
912
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–67. Determine the principal stress and maximum
in-plane shear stress at point B on the cross section of the
pipe at section a–a.
a 300 mm
a
300 mm
b
30 mm
A
B
200 mm
300 N
Internal Loadings: Considering the equilibrium of the free-body diagram of the
assembly’s cut segment, Fig. a,
©Fx = 0;
N - 450 = 0
©Fy = 0;
Vy = 0
©Fz = 0;
Vz + 300 = 0
Vz = - 300 N
©Mx = 0;
T + 300(0.2) = 0
T = - 60 N # m
©My = 0;
My - 450(0.3) + 300(0.3) = 0
My = 45 N # m
©Mz = 0;
Mz + 450(0.2) = 0
Mz = - 90 N # m
N = 450 N
Section Properties: The cross-sectional area, the moment of inertia about the y and
z axes, and the polar moment of inertia of the pipe’s cross section are
A = p (0.032 - 0.022) = 0.5p (10 - 3) m2
Iy = Iz =
J =
p
(0.034 - 0.024) = 0.1625p(10 - 6) m4
4
p
(0.034 - 0.024) = 0.325p(10 - 6) m4
2
Referring to Fig. b,
(Qz)B = 0
Normal and Shear Stress: The normal stress is a combination of axial and bending
stress.
sB =
MyzB
45( -0.03)
N
450
+
=
+
-3
A
ly
0.5p(10 )
0.1625p(10 - 6)
= - 2.358 MPa
Since (Qz)B = 0, (txy)B = 0. Also Vy = 0. Then the shear stress along the y axis is
contributed by torsional shear stress only.
(txy)B = [(txy)T]B =
60(0.03)
Tc
= 1.763 MPa
=
J
0.325p(10 - 6)
The state of stress at point B is represented on the two-dimensional element shown
in Fig. c.
913
450 N
20 mm
Section a – a
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–67. Continued
Construction of the Circle: sx = - 2.358 MPa, sy = 0, and txy = - 1.763 MPa .
Thus,
sx + sy
-2.358 + 0
=
= - 1.179 MPa
savg =
2
2
The coordinates of reference point A and the center C of the circle are
A( - 2.358, - 1.763)
C( -1.179, 0)
Thus, the radius of the circle is
R = CA = 2[ -2.358 - ( -1.179)]2 + (- 1.763)2 = 2.121 MPa
Using these results, the circle is shown in Fig. d.
In-Plane Principal Stresses: The coordinates of reference points B and D represent
s1 and s2, respectively.
s1 = - 1.179 + 2.121 = 0.942 MPa
Ans.
s2 = - 1.179 - 2.121 = - 3.30 MPa
Ans.
Maximum In-Plane Shear Stress: The coordinates of point E represent the state of
maximum in-plane shear stress. Thus,
tmax
= |R| = 2.12 MPa
Ans.
in-plane
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Ans:
s1 = 0.942 MPa, s2 = - 3.30 MPa,
tmax = 2.12 MPa
in-plane
914
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*9–68. The rotor shaft of the helicopter is subjected to the
tensile force and torque shown when the rotor blades
provide the lifting force to suspend the helicopter at midair.
If the shaft has a diameter of 6 in., determine the principal
stress and maximum in-plane shear stress at a point located
on the surface of the shaft.
50 kip
10 kip⭈ft
Internal Loadings: Considering the equilibrium of the free-body diagram of the
rotor shaft’s upper segment, Fig. a,
©Fy = 0;
N - 50 = 0
N = 50 kip
©My = 0;
T - 10 = 0
T = 10 kip # ft
Section Properties: The cross-sectional area and the polar moment of inertia of the
rotor shaft’s cross section are
A = p(32) = 9p m2
J =
p 4
(3 ) = 40.5p in4
2
Normal and Shear Stress: The normal stress is contributed by axial stress only.
sA =
50
N
=
= 1.768 ksi
A
9p
The shear stress is contributed by the torsional shear stress only.
tA =
10(12)(3)
Tc
=
= 2.829 ksi
J
40.5p
The state of stress at point A is represented by the element shown in Fig. b.
915
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–68. Continued
Construction of the Circle: sx = 0, sy = 1.768 ksi, and txy = 2.829 ksi. Thus,
savg =
sx + sy
2
=
0 + 1.768
= 0.8842 ksi
2
The coordinates of reference point A and the center C of the circle are
A(0, 2.829)
C(0.8842, 0)
Thus, the radius of the circle is
R = CA = 2(0 - 0.8842)2 + 2.8292 = 2.964 ksi
Using these results, the circle is shown in Fig. c.
In-Plane Principal Stress: The coordinates of reference points B and D represent s1
and s2, respectively.
s1 = 0.8842 + 2.964 = 3.85 ksi
Ans.
s2 = 0.8842 - 2.964 = - 2.08 ksi
Ans.
Maximum In-Plane Shear Stress: The state of maximum shear stress is represented
by the coordinates of point E, Fig. a.
tmax
= R = 2.96 ksi
Ans.
in-plane
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916
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*9–69. 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 = -
- 300(0.2)
My
= = 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 = 2.370 ksi
In-Plane Principal Stress: The coordinates of point B and D represent s1 and s2,
respectively.
s1 = 2.34375 + 2.370 = 4.71 ksi
Ans.
s2 = 2.34375 - 2.370 = - 0.0262 ksi
Ans.
Ans:
s1 = 4.71 ksi, s2 = - 0.0262 ksi
917
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9–70. 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.
45⬚
1.25 m
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.
Ans.
www.elsolucionario.org
Ans:
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.
918
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–71. 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)
B(666.67, 0)
C(500, 0)
333.33 + 666.67
= 500 MPa
2
Ans.
tx¿y¿ = - R = 500 - 666.67 = - 167 MPa
Ans.
sx¿ =
Ans:
sx¿ = 500 MPa, tx¿y¿ = - 167 MPa
919
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*9–72. 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 =
www.elsolucionario.org
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 with 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°
920
1m
300 mm
2m
C
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9–72. 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.
921
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–73. 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)
www.elsolucionario.org
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 with 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 can be constructed.
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.
922
1m
300 mm
2m
C
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9–73. 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 shown in Fig. e.
Ans:
s1 = 1.15 MPa, s2 = - 0.0428 MPa
923
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–74. If the box wrench is subjected to the 50 lb force,
determine the principal stress and maximum in-plane shear
stress at point A on the cross section of the wrench at
section a–a. Specify the orientation of these states of stress
and indicate the results on elements at the point.
12 in.
50 lb
0.5 in.
2 in.
a
a
A
B
Section a – a
Internal Loadings: Considering the equilibrium of the free-body diagram of the
wrench’s segment, Fig. a,
©Fy = 0;
Vy + 50 = 0
Vy = - 50 lb
©Mx = 0;
T + 50(12) = 0
T = - 600 lb # in
©Mz = 0;
Mz - 50(2) = 0 Mz = 100 lb # in
Section Properties: The moment of inertia about the z axis and the polar moment of
inertia of the wrench’s cross section are
Iz =
p
(0.54) = 0.015625p in4
4
J =
p
(0.54) = 0.03125p in4
2
Referring to Fig. b,
(Qy)A = y¿A¿ =
www.elsolucionario.org
4(0.5) p
c (0.52) d = 0.08333 in3
3p 2
Normal and Shear Stress: The shear stress of point A along the z axis is (txz)A = 0.
However, the shear stress along the y axis is a combination of torsional and
transverse shear stress.
(txy)A = [(txy)T]A - [(txy)V]A
=
Vy(Qy)A
Tc
+
J
lzt
=
600(0.5)
-50(0.08333)
+
= 2.971 ksi
0.03125p
0.015625p(l)
The state of stress at point A is represented by the two-dimensional element shown
in Fig. c.
924
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–74. Continued
Construction of the Circle: sx = sy = 0, and txy = 2.971 ksi. Thus,
savg =
sx + sy
2
=
0 + 0
= 0
2
The coordinates of reference point A and the center C of the circle are
A(0, 2.971)
C(0, 0)
Thus, the radius of the circle is
R = CA = 2(0 - 0)2 + 2.9712 = 2.971 ksi
Using these results, the circle is shown in Fig. d.
In-Plane Principal Stress: The coordinates of reference points B and D represent s1
and s2, respectively.
s1 = 0 + 2.971 = 2.97 ksi
Ans.
s2 = 0 - 2.971 = - 2.97 ksi
Ans.
Maximum In-Plane Shear Stress: Since there is no normal stress acting on the
element,
tmax
in-plane
= (txy)A = 2.97 ksi
Ans.
Ans:
s1 = 2.97 ksi, s2 = - 2.97 ksi, up1 = 45.0°,
up2 = - 45.0°, tmax = 2.97 ksi, us = 0°
in-plane
925
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–75. If the box wrench is subjected to the 50 lb force,
determine the principal stress and maximum in-plane shear
stress at point B on the cross section of the wrench at
section a–a. Specify the orientation of these states of stress
and indicate the results on elements at the point.
12 in.
50 lb
0.5 in.
2 in.
a
a
A
B
Section a – a
Internal Loadings: Considering the equilibrium of the free-body diagram of the
wrench’s cut segment, Fig. a,
©Fy = 0;
Vy + 50 = 0
Vy = - 50 lb
©Mx = 0;
T + 50(12) = 0
T = - 600 lb # in
©Mz = 0;
Mz - 50(2) = 0 Mz = 100 lb # in
Section Properties: The moment of inertia about the z axis and the polar moment of
inertia of the wrench’s cross section are
Iz =
p
(0.54) = 0.015625p in4
4
J =
p
(0.54) = 0.03125p in4
2
www.elsolucionario.org
Referring to Fig. b,
(Qy)B = 0
Normal and Shear Stress: The normal stress is caused by the bending stress
due to Mz.
(sx)B = -
MzyB
Iz
= -
100(0.5)
= - 1.019 ksi
0.015625p
The shear stress at point B along the y axis is (txy)B = 0 since (Qy)B. However, the
shear stress along the z axis is caused by torsion.
(txz)B =
600(0.5)
Tc
=
= 3.056 ksi
J
0.03125p
The state of stress at point B is represented by the two-dimensional element shown
in Fig. c.
926
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–75. Continued
Construction of the Circle: sx = - 1.019 ksi, sz = 0, and txz = - 3.056 ksi . Thus,
savg =
sx + sy
2
=
- 1.019 + 0
= - 0.5093 ksi
2
The coordinates of reference point A and the center C of the circle are
A( - 1.019, - 3.056)
C( - 0.5093, 0)
Thus, the radius of the circle is
R = CA = 2 [- 1.019 - ( -0.5093)]2 + ( -3.056)2 = 3.0979 ksi
Using these results, the circle is shown in Fig. d.
In-Plane Principal Stress: The coordinates of reference points B and D represent
s1 and s2, respectively.
s1 = - 0.5093 + 3.0979 = 2.59 ksi
Ans.
s2 = - 0.5093 - 3.0979 = - 3.61 ksi
Ans.
Maximum In-Plane Shear Stress: The coordinates of point E represent the
maximum in-plane stress, Fig. a.
tmax
= R = 3.10 ksi
Ans.
in-plane
Ans:
s1 = 2.59 ksi, s2 = - 3.61 ksi, up1 = - 40.3°,
up2 = 49.7, tmax
= 3.10 ksi , us = 4.73°
in-plane
927
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*9–76. The ladder is supported on the rough surface at A
and by a smooth wall at B. If a man weighing 150 lb stands
upright at C, determine the principal stresses in one of the
legs at point D. Each leg is made from a 1-in.-thick board
having a rectangular cross section. Assume that the total
weight of the man is exerted vertically on the rung at C and
is shared equally by each of the ladder’s two legs. Neglect
the weight of the ladder and the forces developed by the
man’s arms.
B
C
12 ft
3 in.
1 in.
D
1 in.
3 in.
5 ft
1 in.
D
4 ft
A
A = 3(1) = 3 in2
I =
1
(1)(33) = 2.25 in4
12
QD = y¿A¿ = (1)(1)(1) = 1 in3
sD =
My
35.52(12)(0.5)
-P
- 77.55
=
= - 120.570 psi
A
I
3
2.25
tD =
8.88(1)
VQD
=
= 3.947 psi
It
2.25(1)
A( -120.57, - 3.947)
B(0, 3.947)
C( -60.285, 0)
www.elsolucionario.org
R = 2(60.285)2 + (3.947)2 = 60.412
s1 = - 60.285 + 60.4125 = 0.129 psi
Ans.
s2 = - 60.285 - 60.4125 = - 121 psi
Ans.
928
5 ft
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9–77. 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 circles
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 circles of this state of stress are shown in Fig. b
(a)
929
180 MPa
140 MPa
(b)
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9–78. Draw the three Mohr’s circles that describe the
following state of stress.
300 psi
Here, smin = - 300 psi, sint = 0 and smax = 400 psi. The three Mohr’s circles for
this state of stress are shown in Fig. a.
400 psi
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930
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z
9–79. The stress at a point is shown on the element.
Determine the principal stresses and the absolute maximum
shear stress.
x
y
90 MPa
20 MPa
60 MPa
For y – z plane:
The center of the cricle is at sAvg =
sy + sz
2
=
60 + 90
= 75 MPa
2
R = 2(75 - 60)2 + (- 20)2 = 25 MPa
s1 = 75 + 25 = 100 MPa
s2 = 75 - 25 = 50 MPa
Thus,
tabs
max
=
s1 = 100
Ans.
s2 = 50 MPa
Ans.
s3 = 0 MPa
Ans.
smax - smin
100 - 0
=
= 50 MPa
2
2
Ans.
Ans:
s1 = 100 MPa, s2 = 50 MPa, s3 = 0,
tabs = 50 MPa
max
931
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
z
*9–80. The stress at a point is shown on the element.
Determine the principal stresses and the absolute maximum
shear stress.
x
Mohr’s circle for the element in the y–z plane, Fig. a, will be drawn first. In
accordance with the established sign convention, sy = 30 psi, sz = 120 psi and
tyz = 70 psi. Thus
savg =
sy + sz
2
=
30 + 120
= 75 psi
2
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 circles are shown in Fig. c,
From the geometry of the three circles,
tabs
max
=
120 psi
70 psi
30 psi
Thus the coordinates of reference point A and the center C of the circle are
A(30, 70)
y
www.elsolucionario.org
158.22 - ( -8.22)
smax - smin
=
= 83.2 psi
2
2
932
Ans.
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z
*9–81. The stress at a point is shown on the element.
Determine the principal stresses and the absolute maximum
shear stress.
x
y
6 ksi
Mohr’s circle for the element in x–z plane, Fig. a, will be drawn first. In accordance
with the established sign convention, sx = - 1 ksi , sz = 0 and txz = 6 ksi . Thus
savg =
sx + sz
=
2
-1 - 6
= - 3.5 ksi
2
1 ksi
1 ksi
Thus, the coordinates of reference point A and the center C of the circle are
A(- 1, 1)
C( - 3.5, 0)
Thus, the radius of the circle is
R = CA = 2[- 1 - ( - 3.5)]2 + 12 = 2.69 ksi
Using these results, the circle is shown in Fig. b,
The coordinates of points B and D represent s2 and s3, respectively.
s2 = - (3.5 - 2.69) = - 0.807 ksi
s3 = - (3.5 + 2.69) = - 6.19 ksi
From the results obtained,
s2 = - 0.807 ksi
tabs
max
=
s3 = - 6.19 ksi
s1 = 0 ksi
Ans.
smax - smin
- 6.19 - 0
=
= -3.10 ksi
2
2
Ans.
Ans:
s2 = - 0.807 ksi, s3 = - 6.19 ksi, s1 = 0,
tabs = - 3.10 ksi
max
933
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
z
9–82. The stress at a point is shown on the element.
Determine the principal stresses and the absolute maximum
shear stress.
y
x
150 MPa
120 MPa
For x – z plane:
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.
www.elsolucionario.org
Ans:
s1 = 222 MPa, s2 = 0 MPa, s3 = - 102 MPa,
tabs = 162 MPa
max
934
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
z
9–83. The state of stress at a point is shown on the
element. Determine the principal stresses and the absolute
maximum shear stress.
x
For y - z plane:
A(5, - 4)
B( - 2.5, 4)
y
2.5 ksi
C(1.25, 0)
4 ksi
R = 23.752 + 42 = 5.483
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
max
=
Ans.
Ans:
s1 = 6.73 ksi, s2 = 0, s3 = - 4.23 ksi,
tabs = 5.48 ksi
max
935
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–85. The solid cylinder having a radius r is placed in a
sealed container and subjected to a pressure p. Determine
the stress components acting at point A located on the
center line of the cylinder. Draw Mohr’s circles for the
element at this point.
A
r
p
-s(dz)(2r) =
L0
p(r du) dz sin u
u
- 2s = p
p
L0
sin u du = p( - cos u)|0
s = -p
The stress in every direction is s1 = s2 = s3 = - p
Ans.
www.elsolucionario.org
Ans:
The stress in every direction is
s1 = s2 = s3 = - p
936
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–86. The plate is subjected to a tensile force P = 5 kip. If
it has the dimensions shown, determine the principal
stresses and the absolute maximum shear stress. If the
material is ductile it will fail in shear. Make a sketch of the
plate showing how this failure would appear. If the material
is brittle the plate will fail due to the principal stresses.
Show how this failure occurs.
s =
P ⫽ 5 kip
2 in.
2 in. P ⫽ 5 kip
12 in.
0.5 in.
P
5000
=
= 2500 psi = 2.50 ksi
A
(4)(0.5)
s1 = 2.50 ksi
Ans.
s2 = s3 = 0
Ans.
tabs =
max
s1
= 1.25 ksi
2
Ans.
Failure by shear:
Failure by principal stress:
Ans:
s1 = 2.50 ksi, s2 = s3 = 0, tabs = 1.25 ksi
max
937
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–87. Determine the principal stresses 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 =
www.elsolucionario.org
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 =
300
N
= = - 342.86 psi
A
0.875
The shear stress is contributed by the transverse shear stress.
tA =
VQA
400(0.3672)
=
= 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
938
500 lb
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9–87. Continued
Using these results, the circle 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.
Ans:
s1 = 583 psi, s2 = 0, s3 = - 926 psi,
tabs = 755 psi
max
939
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*9–88. Determine the principal stresses 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,
+ c ©Fy = 0;
3
N - 500 a b = 0
5
+ ©F = 0;
;
x
4
V - 500 a b = 0
5
B
V = 400 lb
A = 0.5(3) - 0.25(2.5) = 0.875 in2
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.
www.elsolucionario.org
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,
s1 = 10.91 ksi
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.
940
500 lb
0.25 in.
A
0.25 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
a
0.5 in.
N = 300 lb
4
3
©MO = 0; M - 500 a b (12) - 500 a b (6) = 0
5
5
I =
5
3
a
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–89. 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
F
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.
Ans:
s1 = 10.7 MPa, s2 = - 35.8 MPa
941
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–90. 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
F
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 =
www.elsolucionario.org
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.
Ans:
tmax
in-plane
942
= 23.2 MPa
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–91. 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
p
I =
A 1.54 - 1.3754 B = 1.1687 in4
4
J =
C
p
A 1.54 - 1.3754 B = 2.3374 in4
2
y
z
x
(QA)z = ©y¿A¿
=
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.
Ans:
s1 = 119 psi, s2 = - 119 psi
943
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*9–92. Solve Prob. 9–91 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.
C
y
Section Properties:
z
I =
p
A 1.54 - 1.3754 B = 1.1687 in4
4
J =
p
A 1.54 - 1.3754 B = 2.3374 in4
2
x
(QB)z = 0
Normal Stress: Applying the flexure formula s =
sB =
My z
Iv
,
200(1.5)
= 256.7 psi
1.1687
www.elsolucionario.org
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.
944
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9–93. Determine the equivalent state of stress if an
element is oriented 40° clockwise from the element shown.
Use Mohr’s circle.
10 ksi
6 ksi
A(6, 0)
B(- 10, 0)
C( - 2, 0)
R = CA = CB = 8
sx¿ = - 2 + 8 cos 80° = - 0.611 ksi
Ans.
tx¿y¿ = 8 sin 80° = 7.88 ksi
Ans.
sy¿ = - 2 - 8 cos 80° = - 3.39 ksi
Ans.
Ans:
sx¿ = - 0.611 ksi, tx¿y¿ = 7.88 ksi, sy¿ = - 3.39 ksi
945
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–94. The crane is used to support the 350-lb load.
Determine the principal stresses acting in the boom at
points A and B. The cross section is rectangular and has a
width of 6 in. and a thickness of 3 in. Use Mohr’s circle.
5 ft
3 in.
5 ft
B
A
45°
A = 6(3) = 18 in2
I =
45°
1
(3)(63) = 54 in4
12
QB = (1.5)(3)(3) = 13.5 in3
QA = 0
For point A:
sA = -
My
1750(12)(3)
P
597.49
=
= - 1200 psi
A
I
18
54
tA = 0
s1 = 0
Ans.
s2 = - 1200 psi = - 1.20 ksi
Ans.
For point B:
www.elsolucionario.org
P
597.49
sB = - = = - 33.19 psi
A
18
tB =
VQB
247.49(13.5)
=
= 20.62 psi
It
54(3)
A(-33.19, - 20.62)
B(0, 20.62)
C( - 16.60, 0)
R = 216.602 + 20.622 = 26.47
s1 = - 16.60 + 26.47 = 9.88 psi
Ans.
s2 = - 16.60 - 26.47 = - 43.1 psi
Ans.
Ans:
Point A: s1 = 0, s2 = - 1.20 ksi,
Point B: s1 = 9.88 psi, s2 = - 43.1 psi
946
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–95. Determine the equivalent state of stress on an
element at the same point which represents (a) the
principal stresses, 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 and sketch the
results on the element.
30 ksi
10 ksi
Normal and Shear Stress:
sx = 0
sy = - 30 ksi
txy = - 10 ksi
In-Plane Principal Stresses:
s1, 2 =
=
sx + sy
;
2
A
a
sx - sy
2
2
2
b + txy
0 + ( -30)
0 - ( - 30)2
2
;
b + (- 10)
2
Aa
2
= - 15 ; 2325
s1 = 3.03 ksi
s2 = - 33.0 ksi
Ans.
Orientation of Principal Plane:
tan 2up =
txy
(sx - sy)>2
=
- 10
= - 0.6667
[0 - (- 30)]>2
up = - 16.845° and 73.15°
Substituting u = - 16.845° into
sx + sy
sx¿ =
=
sx - sy
+
2
2
cos 2u + txy sin 2u
0 + ( - 30)
0 - ( -30)
+
cos ( - 33.69°) - 10 sin ( -33.69°)
2
2
= 3.03 ksi = s1
Thus,
(up)1 = - 16.8° and (up)2 = 73.2°
Ans.
The element that represents the state of principal stress is shown in Fig. a.
Maximum In-Plane Shear Stress:
tmax
in-plane
=
a
B
sx - sy
2
2
b + txy2 =
0 - ( - 30) 2
b + ( -10)2 = 18.0 ksi
2
B
a
947
Ans.
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9–95. Continued
Orientation of the Plane of Maximum In-Plane Shear Stress:
tan 2us = -
(sx - sy)>2
txy
=
[0 - (- 30)]>2
= 1.5
- 10
us = 28.2° and 118°
By inspection, tmax
has to act in the same sense shown in Fig. b to maintain
in-plane
equilibrium.
Average Normal Stress:
savg =
sx + sy
2
=
0 + ( - 30)
= - 15 ksi
2
Ans.
The element that represents the state of maximum in-plane shear stress is shown
in Fig. c.
www.elsolucionario.org
Ans:
s1 = 3.03 ksi, s2 = - 33.0 ksi,
up1 = - 16.8° and up2 = 73.2°,
tmax
= 18.0 ksi , savg = - 15 ksi, us = 28.2°
in-plane
948
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*9–96. 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 in 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.
949
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9–96. Continued
Orientation of the Principal Plane: Referring to the geometry of the circle, Fig. c,
tan 2 A up B 2 =
3.761
= 7.3846
1.019 - 0.5093
A up B 2 = 41.1° (clockwise)
Ans.
The state of principal stresses is represented on the element shown in Fig. d.
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950
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*9–97. The box beam is subjected to the loading shown.
Determine the principal stress in the beam at points A and B.
1200 lb
800 lb
6 in.
A
6 in. B 8 in.
8 in.
A
B
3 ft
2.5 ft
2.5 ft
5 ft
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
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.
Ans:
Point A: s1 = 61.7 psi, s2 = 0
Point B: s1 = 0, s2 = - 46.3 psi
951
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9–98. The state of stress at a point is shown on the element.
Determine (a) the principal stresses and (b) the maximum
in-plane shear stress and average normal stress at the point.
Specify the orientation of the element in each case.
60 MPa
30 MPa
45 MPa
a) sx = 45 MPa
s1, 2 =
=
txy = 30 MPa
sy = - 60 MPa
sx + sy
Aa
;
2
sx - sy
2
2
b + txy
2
45 - 60
45 - ( -60) 2
2
; a
b + 30
2
A
2
Ans.
s2 = - 68.0 MPa
s1 = 53.0 MPa
txy
tan 2up = sx - sy =
2
30
45 - ( - 60)
2
= 0.5714
up = 14.87° and - 75.13°
Use Eq. 9–1 to determine the principal plane of s1 and s2
sx + sy
sx ¿ =
sx - sy
+
2
2
www.elsolucionario.org
cos 2u + txy sin 2u
u = uy = 14.87°
sx¿ =
45 - ( -60)
45 + ( -60)
+
cos 29.74° + 30 sin 29.74° = 53.0 MPa
2
2
Therefore, up1 = 14.9° ;
b) tmax
=
A
in-plane
savg =
a
sx - sy
2
sx + sy
2
tan 2us = us = - 30.1°
=
2
=
b + txy
2
Aa
Ans.
45 - ( -60) 2
2
b + 30 = 60.5 MPa
2
45 + ( -60)
= - 7.50 MPa
2
(sx - sy)
2
txy
up2 = - 75.1°
= -
[45 - ( - 60)]
2
Ans.
30
Ans.
Ans.
= -1.75
and
59.9°
Ans.
By observation, in order to preserve equilibrium, tmax = 60.5 MPa has to act in the
direction shown in the figure.
Ans:
s1 = 53.0 MPa, s2 = - 68.0 MPa,
up1 = 14.9°, up2 = - 75.1°, tmax
= 60.5 MPa,
in-plane
savg = - 7.50 MPa, us = - 30.1° and 59.9°
952
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9–99. 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.
14 ksi
A
20 ksi
50⬚
B
+ Q©F
x¿ = 0;
sx¿ ¢A + 14 ¢A sin 50° cos 40° + 20 ¢A cos 50° cos 50° = 0
sx¿ = - 16.5 ksi
a + ©Fy¿ = 0;
Ans.
tx¿y¿ ¢A + 14 ¢A sin 50° sin 40° - 20 ¢A cos 50° sin 50° = 0
tx¿y¿ = 2.95 ksi
Ans.
Ans:
sx¿ = - 16.5 ksi, tx¿y¿ = 2.95 ksi
953
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10–1. Prove that the sum of the normal strains in
perpendicular directions is constant.
Px + Py
Px¿ =
2
Px - Py
+
Px + Py
Py¿ =
2
2
Px - Py
-
2
cos 2u +
cos 2u -
gxy
2
gxy
2
sin 2u
(1)
sin 2u
(2)
Adding Eq. (1) and Eq. (2) yields:
Px¿ + Py¿ = Px + Py = constant
(Q.E.D.)
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954
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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 with the established sign convention,
Px = 200(10 - 6),
Px + Py
Px¿ =
Px - Py
+
2
= c
Py = - 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.
Px - Py
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)
Px + Py
Py¿ =
= c
2
Ans.
Px - Py
-
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
Ans:
Px¿ = 248(10 - 6), gx¿y¿ = - 233(10 - 6),
Py¿ = - 348(10 - 6)
955
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10–3. The state of strain at a point on a wrench
Px = 120110-62,
has
components
Py = - 180110-62,
gxy = 150(10-62. 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 x–y plane.
Px = 120(10 - 6)
Py = - 180(10 - 6)
; Aa
Px + Py
a)
P1, 2 =
2
= c
Px - Py
2
2
b + a
gxy = 150(10 - 6)
gxy
2
b
2
120 + ( - 180)
120 - ( -180) 2
150 2 10 - 6
;
b
+
a
b d
a
A
2
2
2
P1 = 138(10 - 6);
P2 = - 198(10 - 6)
Ans.
Orientation of P1 and P2
tan 2up =
gxy
=
Px - Py
150
= 0.5
[120 - (- 180)]
up = 13.28° and - 76.72°
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Use Eq. 10–5 to determine the direction of P1 and P2
Px + Py
Py¿ =
2
Px - Py
+
cos 2u +
2
gxy
2
sin 2u
u = up = 13.28°
Py¿ = J
120 + ( -180)
120 - ( - 180)
150
+
cos (26.56°) +
sin 26.56° d10 - 6
2
2
2
= 138(10 - 6) = P1
Therefore up1 = 13.3°;
gmax
b)
in-plane
Aa
2
=
gmax
= 2c
in-plane
Px - Py
Aa
2
2
Ans.
gxy 2
2
b + a 2 b
2
120 - ( -180) 2
+ a 150 b d 10 - 6 = 335(10 - 6)
b
2
2
Px + Py
Pavg =
up2 = - 76.7°
= c
120 + (- 180)
d 10 - 6 = - 30.0(10 - 6)
2
956
Ans.
Ans.
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10–3. Continued
Orientation of gmax
tan 2us =
- (Px - Py)
gxy
=
us = - 31.7°
and
-[120 - ( - 180)]
= - 2.0
150
Ans.
58.3°
Use Eq. 10–11 to determine the sign of gmax
in-plane
gx¿y¿
2
Px - Py
= -
2
sin 2u +
gxy
2
cos 2u
u = us = - 31.7°
gx¿y¿ = 2 c -
120 - (- 180)
150
sin ( - 63.4°) +
cos ( -63.4°) d10 - 6 = 335(10 - 6)
2
2
Ans:
P1 = 138(10 - 6), P2 = - 198(10 - 6),
up1 = 13.3°, up2 = - 76.7°,
gmax
= 335(10 - 6), Pavg = - 30.0(10 - 6)
in-plane
us = - 31.7° and 58.3°
957
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
y
*10–4. The state of strain at the point on the gear tooth has
components
gxy =
Px = 850110-62,
Py = 480110-62,
-6
650(10 2. 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.
Px = 850(10 - 6)
Py = 480(10 - 6)
Px + Py
a)
P1, 2 =
2
= c
Aa
;
Px - Py
2
gxy = 650(10 - 6)
b + a
2
x
gxy
2
b
2
2
2
850 + 480
; a 850 - 480 b + a 650 b d(10 - 6)
2
A
2
2
P1 = 1039(10 - 6)
P2 = 291(10 - 6)
Ans.
Ans.
Orientation of P1 and P2:
gxy
tan 2uy =
=
650
850 # 480
and
120.18°
www.elsolucionario.org
Px - Py
uy = 30.18°
Use Eq. 10–5 to determine the direction of P1 and P2:
Px + Py
Px¿ =
Px - Py
+
2
2
cos 2u +
gxy
2
sin 2u
u = uy = 30.18°
Px¿ = c
850 + 480
850 - 480
650
+
cos (60.35°) +
sin (60.35°) d (10 - 6)
2
2
2
= 1039(10 - 6)
up2 = 120°
Ans.
850 - 480 2 + 650 2 d (10 - 6) = 748(10 - 6)
a
b
b
2
2
Ans.
= a
Ans.
Therefore, up1 = 30.2°
Ans.
b)
gmax
Aa
in-plane
2
gmax
in-plane
=
= 2c
2
2
Aa
Px + Py
Pavg =
Px - Py
2
b + a
gxy
2
b
2
850 + 480
b (10 - 6) = 665(10 - 6)
2
958
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
10–4. Continued
Orientation of gmax:
tan 2ut =
- (Px - Py)
gxy
=
- (850 - 480)
650
ut = - 14.8° and 75.2°
Ans.
Use Eq. 10–6 to determine the sign of gmax
:
in-plane
gx¿y¿
2
Px - Py
= -
2
sin 2u +
gxy
2
cos 2u;
u = ut = - 14.8°
gx¿y¿ = [ -(850 - 480) sin ( - 29.65°) + 650 cos ( -29.65°)](10 - 6) = 748(10 - 6)
959
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
y
10–5. The state of strain at the point on the gear tooth has
the components Px = 520110-62, Py = - 760110-62, gxy =
750(10-62. 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.
Px = 520(10 - 6)
Py = - 760(10 - 6)
Px + Py
a)
P1, 2 =
;
2
= c
Aa
Px - Py
2
x
gxy = - 750(10 - 6)
2
b + a
gxy
2
b
2
2
2
520 + ( - 760)
; a 520 - ( -760) b + a - 750 b d10 - 6
2
A
2
2
P1 = 622(10 - 6);
P2 = - 862(10 - 6)
Ans.
Orientation of P1 and P2
tan 2up =
gxy
=
Px - Py
www.elsolucionario.org
-750
= - 0.5859; up = - 15.18° and up = 74.82°
[520 - ( - 760)]
Use Eq. 10–5 to determine the direction of P1 and P2.
Px + Py
Px¿ =
Px - Py
+
2
2
cos 2u +
gxy
2
sin 2u
u = up = - 15.18°
Px¿ = J
520 + (- 760)
520 - ( - 760)
-750
+
cos ( - 30.36°) +
sin ( -30.36°) d10 - 6
2
2
2
= 622 (10 - 6) = P1
Therefore, up1 = - 15.2° and up2 = 74.8°
gmax
b)
in-plane
=
2
gmax
in-plane
gxy 2
Px - Py 2
b + a
2
2 b
A
= 2c
a
520 - ( -760) 2
- 750 2
d 10 - 6 = - 1484 (10 - 6)
b + a
A
2
2 b
Px + Py
Pavg =
2
Ans.
a
= c
520 + ( -760)
d 10 - 6 = - 120 (10 - 6)
2
Ans.
Ans.
960
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
10–5. Continued
Orientation of gmax
in-plane
tan 2us =
:
- (Px - Py)
=
gxy
-[520 - ( - 760)]
= 1.7067
- 750
us = 29.8° and us = - 60.2°
Ans.
Use Eq. 10–6 to check the sign of gmax
:
in-plane
gx¿y¿
2
Px - Py
= -
gx¿y¿ = 2 c -
2
sin 2u +
gxy
2
cos 2u;
u = us = 29.8°
520 - (- 760)
-750
sin (59.6°) +
cos (59.6°) d10 - 6 = - 1484 (10 - 6)
2
2
Ans:
P1 = 622(10 - 6), P2 = - 862(10-6),
up1 = - 15.2° and up2 = 74.8°,
gmax = - 1484(10 - 6), Pavg = - 120(10 - 6),
in-plane
us = 29.8° and -60.2°
961
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y
10–6. A differential element on the bracket is subjected
to plane strain that has the following components:
Px = 150110-62, Py = 200110-62, gxy = - 700(10-62. Use the
strain-transformation equations and determine the
equivalent in plane strains on an element oriented at an
angle of u = 60° counterclockwise from the original
position. Sketch the deformed element within the x–y plane
due to these strains.
Px = 150 (10 - 6)
Px + Py
Px¿ =
Px - Py
+
2
= c
Py = 200 (10 - 6)
2
cos 2u +
gxy
2
x
gxy = - 700 (10 - 6)
u = 60°
sin 2u
150 - 200
- 700
150 + 200
+
cos 120° + a
b sin 120° d 10 - 6
2
2
2
= - 116 (10 - 6)
Px + Py
Py¿ =
= c
Px - Py
-
2
Ans.
2
cos 2u -
gxy
2
sin 2u
www.elsolucionario.org
150 + 200
150 - 200
- 700
cos 120° - a
b sin 120° d 10 - 6
2
2
2
= 466 (10 - 6)
gx¿y¿
2
Px - Py
= -
gx¿y¿ = 2 c -
2
Ans.
sin 2u +
gxy
2
cos 2u
150 - 200
-700
sin 120° + a
b cos 120° d 10 - 6 = 393 (10 - 6)
2
2
Ans.
Ans:
Px¿ = - 116(10 - 6), Py¿ = 466(10 - 6),
gx¿y¿ = 393(10 - 6)
962
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
y
10–7. Solve Prob. 10–6 for an element oriented u = 30°
clockwise.
x
Px = 150 (10 - 6)
Px + Py
Px¿ =
Px - Py
+
2
= c
Py = 200 (10 - 6)
2
cos 2u +
gxy
2
gxy = - 700 (10 - 6)
u = - 30°
sin 2u
150 - 200
- 700
150 + 200
+
cos ( -60°) + a
b sin ( -60°) d10 - 6
2
2
2
= 466 (10 - 6)
Px + Py
Py¿ =
= c
Ans.
Px - Py
-
2
2
cos 2u -
gxy
2
sin 2u
150 - 200
- 700
150 + 200
cos ( -60°) - a
b sin ( -60°) d10 - 6
2
2
2
= - 116 (10 - 6)
gx¿y¿
2
Px - Py
= -
gx¿y¿ = 2 c -
2
sin 2u +
Ans.
gxy
2
cos 2u
- 700
150 - 200
sin ( - 60°) +
cos ( - 60°) d10 - 6
2
2
= - 393(10 - 6)
Ans.
Ans:
Px¿ = 466(10 - 6), Py¿ = - 116(10 - 6),
gx¿y¿ = - 393(10 - 6)
963
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*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.
Px = - 200(10 - 6)
Px + Py
Px¿ =
Px - Py
+
2
= c
Py = - 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)
Px + Py
Py¿ =
Px - Py
-
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
Px - Py
= -
2
Ans.
sin 2u +
gxy
2
cos 2u
www.elsolucionario.org
gx¿y¿ = [ -(- 200 - ( -650)) sin (40°) + ( - 175) cos (40°)](10 - 6)
= - 423(10 - 6)
Ans.
964
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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 with the established sign convention, Px = 180(10 - 6),
Py = - 120(10 - 6) and gxy = - 100(10 - 6).
Px + Py
P1, 2 =
;
2
= b
Aa
Px - Py
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)
P1 = 188(10 - 6)
tan 2uP =
P2 = - 128(10 - 6)
gxy
Ans.
- 100(10 - 6)
C 180 - ( -120) D (10 - 6)
=
Px - Py
uP = - 9.217°
and
= - 0.3333
80.78°
Substitute u = - 9.217°,
Px + Py
Px¿ =
2
= c
Px - Py
+
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) = P1
Thus,
(uP)1 = - 9.22°
(uP)2 = 80.8°
Ans.
The deformed element is shown in Fig (a).
gmax
Px - Py 2
gxy 2
in-plane
=
b + a
b
b)
2
Aa
2
2
gmax
= b2
tan 2us = - a
Px - Py
in-plane
gxy
A
c
180 - (- 120) 2
- 100 2
-6
-6
d + a
b r (10 ) = 316 A 10 B
2
2
b = -c
C 180 - ( - 120) D (10 - 6)
us = 35.78° = 35.8° and
- 100(10 - 6)
Ans.
s = 3
- 54.22° = - 54.2°
Ans.
965
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
10–9. Continued
gmax
The algebraic sign for in-plane
when u = 35.78°.
gx¿y¿
Px - Py
gxy
= -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)
Px + Py
180 + ( -120)
Pavg =
= c
d (10 - 6) = 30(10 - 6)
2
2
Ans.
The deformed element for the state of maximum in-plane shear strain is shown
in Fig. b
www.elsolucionario.org
Ans:
P1 = 188(10 - 6), P2 = - 128(10 - 6),
up1 = - 9.22°, up 2 = 80.8°,
gmax = 316(10 - 6), us = 35.8° and - 54.2°,
in-plane
Pavg = 30(10 - 6)
966
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
10–10. The state of strain at the point on the support
has components of Px = 350110-62, Py = 400110-62,
gxy = - 675(10-62. 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.
P
a)
Px + Py
P1, 2 =
=
;
2
A
a
Px - Py
2
2
b + a
gxy
2
b
2
350 + 400
- 675 2
350 - 400 2
;
a
b + a
b
2
A
2
2
P1 = 713 (10 - 6)
tan 2up =
P2 = 36.6 (10 - 6)
Ans.
gxy
=
Px - Py
Ans.
- 675
(350 - 400)
up1 = 133°
Ans.
b)
(gx¿y¿)max
=
2
(gx¿y¿)max
=
2
A
a
Px - Py
2
2
gxy 2
b + a 2 b
- 675 2
350 - 400 2
b + a
b
A
2
2
a
(gx¿y¿)max = 677(10 - 6)
Px + Py
Pavg =
tan 2us =
2
=
Ans.
350 + 400
= 375 (10 - 6)
2
-(Px - Py)
gxy
=
Ans.
350 - 400
675
us = - 2.12°
Ans.
Ans:
(a) P1 = 713(10-6), P2 = 36.6(10 - 6), up1 = 133°
= 677(10 - 6), Pavg = 375(10 - 6),
(b) gmax
in-plane
us = - 2.12°
967
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y
10–11. The state of strain on an element has components
Px = - 150110-62, Py = 450110-62, gxy = 200110-62. Determine
the equivalent state of strain on an element at the same point
oriented 30° counterclockwise with respect to the original
element. Sketch the results on this element.
Pydy
dy
gxy
2
gxy
2
dx
Pxdx
x
Strain Transformation Equations:
Px = - 150(10 - 6)
Py = 450(10 - 6)
gxy = 200(10 - 6)
u = 30°
We obtain
Px + Py
Px¿ =
2
= c
Px - Py
+
2
gxy
cos 2u +
2
sin 2u
- 150 - 450
200
-150 + 450
+
cos 60° +
sin 60° d(10 - 6)
2
2
2
= 86. 6(10 - 6)
gx¿y¿
2
= -a
Ans.
Px - Py
2
b sin 2u +
gxy
2
cos 2u
www.elsolucionario.org
gx¿y¿ = [ -( -150 - 450) sin 60° + 200 cos 60°](10 - 6)
= 620(10 - 6)
Px + Py
Py¿ =
= c
2
Ans.
Px - Py
-
2
cos 2u -
gxy
2
sin 2u
- 150 + 450
- 150 - 450
200
cos 60° sin 60° d(10 - 6)
2
2
2
= 213(10 - 6)
Ans.
The deformed element for this state of strain is shown in Fig. a.
Ans:
Px¿ = 86.6(10 - 6), gx¿y¿ = 620(10 - 6),
Py¿ = 213(10 - 6)
968
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y
*10–12. The state of strain on an element has components
Px = - 400110-62, Py = 0, gxy = 150110-62. Determine the
equivalent state of strain on an element at the same point
oriented 30° clockwise with respect to the original element.
Sketch the results on this element.
gxy
dy 2
x
gxy
2
dx
Strain Transformation Equations:
Px = - 400(10 - 6)
Py = 0
gxy = 150(10 - 6)
u = - 30°
We obtain
Px + Py
Px¿ =
2
= c
Px - Py
+
2
gxy
cos 2u +
2
sin 2u
- 400 - 0
150
- 400 + 0
+
cos ( -60°) +
sin (-60°) d(10 - 6)
2
2
2
= - 365(10 - 6)
gx¿y¿
= -a
2
Ans.
Px - Py
2
b sin 2u +
gxy
2
cos 2u
gx¿y¿ = [ - ( -400 - 0) sin ( -60°) + 150 cos ( -60°)](10 - 6)
= - 271(10 - 6)
Px + Py
Py¿ =
= c
2
Ans.
Px - Py
-
2
cos 2u -
gxy
2
sin 2u
- 400 + 0
- 400 - 0
150
cos ( -60°) sin ( -60°) d(10 - 6)
2
2
2
= - 35.0(10 - 6)
Ans.
The deformed element for this state of strain is shown in Fig. a.
969
Pxdx
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y
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.
gxy
dy 2
x
In-Plane Principal Strains: Px = - 300 A 10 - 6 B , Py = 0, and gxy = 150 A 10 - 6 B . We
obtain
Px + Py
C¢
;
P1, 2 =
2
= C
Px - Py
2
gxy
2
≤ + ¢ ≤
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
P1 = 17.7 A 10 - 6 B
P2 = - 318 A 10 - 6 B
Ans.
Orientation of Principal Strain:
tan 2up =
gxy
=
Px - Py
150 A 10 - 6 B
( -300 - 0) A 10 - 6 B
= - 0.5
uP = - 13.28° and 76.72°
www.elsolucionario.org
Substituting u = - 13.28° into Eq. 9-1,
Px + Py
Px¿ =
= c
Px - Py
+
2
cos 2u +
2
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 = P2
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
in-plane
2
gmax
in-plane
=
C¢
Px - Py
2
2
≤ + ¢
gxy
2
≤
2
- 300 - 0 2
150 2
-6
-6
b + a
b R A 10 B = 335 A 10 B
2
2
A
= B2
a
Ans.
Orientation of the Maximum In-Plane Shear Strain:
tan 2us = - ¢
Px - Py
gxy
≤ = -C
( - 300 - 0) A 10 - 6 B
150 A 10 - 6 B
S = 2
us = 31.7° and 122°
Ans.
970
gxy
2
dx
Pxdx
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
10–13. Continued
The algebraic sign for
gx¿y¿
2
= -¢
Px - Py
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:
Px + Py
Pavg =
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.
Ans:
P1 = 17.7(10 - 6), P2 = - 318(10 - 6),
up1 = 76.7° and up2 = - 13.3°,
gmax
= 335(10 - 6), us = 31.7° and 122°,
in-plane
Pavg = - 150(10 - 6)
971
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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,
Px + Py
P1, 2 =
;
2
= B
Aa
Px - Py
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
P1 = 368 A 10 - 6 B
P2 = 182 A 10 - 6 B
Ans.
Orientation of Principal Strain: Applying Eq. 10–8,
gxy
tan 2uP =
- 180(10 - 6)
=
Px - Py
(250 - 300)(10 - 6)
uP = 37.24°
and
= 3.600
- 52.76°
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Use Eq. 10–5 to determine which principal strain deforms the element in the x¿
direction with u = 37.24°.
Px + Py
Px¿ =
= c
2
Px - Py
+
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 = P2
Hence,
uP1 = - 52.8°
and
uP2 = 37.2°
Ans.
b)
Maximum In-Plane Shear Strain: Applying Eq. 10–11,
g max
Px - Py 2
gxy 2
in-plane
=
b + a
b
2
Aa
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.
972
x
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
10–14. Continued
Orientation of the Maximum In-Plane Shear Strain: Applying Eq. 10–10,
tan 2us = -
Px - Py
us = - 7.76°
2
Px - Py
= -
2
250 - 300
= - 0.2778
- 180
Ans.
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
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,
Px = 250 A 10 - 6 B
Py = 300 A 10 - 6 B
gxy = - 180 A 10 - 6 B
Average Normal Strain: Applying Eq. 10–12,
Px + Py
Pavg =
2
= c
250 + 300
d A 10 - 6 B = 275 A 10 - 6 B
2
Ans.
Ans:
P1 = 368(10 - 6), P2 = 182(10 - 6),
up1 = - 52.8° and up2 = 37.2°,
gmax
= 187(10 - 6), us = -7.76° and 82.2°,
in-plane
Pavg = 275 (10 - 6)
973
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
y
*10–16. The state of strain on an element has components
Px = -300 (10 - 6), Py = 100 (10 - 6), gxy = 150 (10-6). 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.
Pydy
dy
gxy
2
gxy
2
dx
In-Plane Principal Strains: Px = - 300(10 - 6), Py = 100(10 - 6), and gxy = 150(10 - 6).
We obtain
Px + Py
P1, 2 =
= c
;
2
a
A
Px - Py
2
2
b + a
gxy
2
b
2
-300 + 100
150 2
- 300 - 100 2
;
d(10 - 6)
a
b + a
2
A
2
2 b
= (- 100 ; 213.60)(10 - 6)
P1 = 114(10 - 6)
P2 = - 314(10 - 6)
Orientation of Principal Strains:
tan 2up =
www.elsolucionario.org
gxy
150(10 - 6)
=
Px - Py
Ans.
( -300 - 100)(10 - 6)
= - 0.375
up = - 10.28° and 79.72°
Substituting u = - 10.28° into
Px + Py
Px¿ =
2
= c
Px - Py
+
2
cos 2u +
gxy
2
sin 2u
- 300 + 100
- 300 - 100
150
+
cos ( - 20.56°) +
sin ( -20.56°) d(10 - 6)
2
2
2
= - 314(10 - 6) = P2
Thus,
(up)1 = 79.7° and (up)2 = - 10.3°
Ans.
The deformed element for the state of principal strain is shown in Fig. a.
Maximum In-Plane Shear Strain:
gmax
in-plane
2
gmax
in-plane
=
A
a
= c2
Px - Py
A
2
a
2
b + a
gxy
2
b
2
150 2
- 300 - 100 2
d(10 - 6) = 427(10 - 6)
b + a
2
2 b
974
Ans.
Pxdx
x
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10–16. Continued
Orientation of Maximum In-Plane Shear Strain:
tan 2us = - a
Px - Py
gxy
b = -c
(- 300 - 100)(10 - 6)
150(10 - 6)
d(10 - 6) = 2.6667
us = 34.7° and 125°
The algebraic sign for gmax
Ans.
in-plane
gx¿y¿
2
= -a
Px - Py
2
when u = us = 34.7° can be determined using
b sin 2u +
gxy
2
cos 2u
gx¿y¿ = [ - (- 300 - 100) sin 69.44° + 150 cos 69.44°](10 - 6)
= 427(10 - 6)
Average Normal Strain:
Px + Py
Pavg =
2
= a
- 300 + 100
b(10 - 6) = - 100(10 - 6)
2
Ans.
The deformed element for this state of maximum in-plane shear strain is shown
in Fig. b.
975
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
10–17.
Solve part (a) of Prob. 10–3 using Mohr’s circle.
Px = 120(10 - 6)
Py = - 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)
P1 = (- 30 + 167.71)(10 - 6) = 138(10 - 6)
Ans.
P2 = (- 30 - 167.71)(10 - 6) = - 198(10 - 6)
Ans.
tan 2uP = a
75
b , uP = 13.3°
30 + 120
Ans.
www.elsolucionario.org
Ans:
P1 = 138(10 - 6), P2 = - 198(10 - 6), up = 13.3°
976
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10–18.
Solve part (b) of Prob. 10–3 using Mohr’s circle.
Px = 120(10 - 6)
Py = - 180(10 - 6)
A (120, 75)(10 - 6)
C ( - 30, 0)(10 - 6)
gxy = 150(10 - 6)
R = C 2[120 - ( - 30)]2 + (75)2 D (10 - 6)
= 167.71 (10 - 6)
gxy
2
gmax
= R = 167.7(10 - 6)
in-plane
= 335(10 - 6)
Ans.
Pavg = - 30 (10 - 6)
tan 2us =
120 + 30
75
Ans.
us = - 31.7°
Ans.
Ans:
g max
in-plane
= 335(10 - 6), Pavg = - 30(10 - 6),
us = - 31.7°
977
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10–19.
Solve Prob. 10–4 using Mohr’s circle.
Px = 850(10 - 6)
Py = 480(10 - 6)
gxy = 650(10 - 6)
gxy
2
= 325(10 - 6)
A(850, 325)(10 - 6) C(665, 0)(10 - 6)
R = [2(850 - 665)2 + 3252](10 - 6) = 373.97(10 - 6)
P1 = (665 + 373.97)(10 - 6) = 1039(10 - 6)
Ans.
P2 = (665 - 373.97)(10 - 6) = 291(10 - 6)
Ans.
tan 2up =
325
850 - 665
(Mohr’s circle)
2up = 60.35°
up = 30.2°
Ans.
(element)
gmax
in-plane
2
gmax
in-plane
= R
www.elsolucionario.org
= 2(373.97)(10 - 6) = 748(10 - 6)
Ans.
Pavg = 665(10 - 6)
Ans.
2us = 90° - 2up = 29.65°
us = - 14.8°
Ans.
(Mohr’s circle)
(element)
Ans:
P1 = 1039(10 - 6), P2 = 291(10 - 6), up = 30.2°,
gmax
= 748(10 - 9), Pavg = 665(10 - 6),
in-plane
us = - 14.8°
978
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*10–20.
Solve Prob. 10–5 using Mohr’s circle.
a) Px = 520(10 - 6)
Py = - 760(10 - 6)
gxy = - 750(10 - 6)
gxy
2
= - 375(10 - 6)
A(520, -375); C(- 120, 0)
R = 2(520 + 120)2 + 3752 = 741.77
P1 = 741.77 - 120 = 622(10 - 6)
Ans.
P2 = - 120 - 741.77 = - 862(10 - 6)
Ans.
tan 2up1 =
375
= 0.5859
(120 + 520)
up1 = 15.2°
b)
gmax
gmax
in-plane
in-plane
Ans.
= 2R = 2(741.77)
= - 1484(10 - 6)
Ans.
Pavg = - 120(10 - 6)
tan 2us =
Ans.
(120 + 520)
= 1.7067
375
us = 29.8°
Ans.
979
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10–21.
Solve Prob. 10–7 using Mohr’s circle.
Px = 150(10 - 6)
u = - 30°
Py = 200(10 - 6)
gxy = - 700(10 - 6)
gxy
2
= - 350(10 - 6)
2u = - 60°
A(150, - 350);
C(175, 0)
R = 2(175 - 150)2 + (- 350)2 = 350.89
Coordinates of point B:
Px¿ = 350.89 cos 34.09° + 175
= 466(10 - 6)
gx¿y¿
2
Ans.
= - 350.89 sin 34.09°
gx¿y¿ = - 393(10 - 6)
Ans.
Coordinates of point D:
Py¿ = 175 - 350.89 cos 34.09°
www.elsolucionario.org
-6
= - 116(10 )
Ans.
Ans:
-6
Px¿ = 466(10 - 6), gx¿y¿ = - 393(10 ),
-6
Py¿ = - 116(10 )
980
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10–22. The strain at point A on the bracket
has
components
Px = 300110-62,
Py = 550110-62,
-6
gxy = - 650110 2, Pz = 0. 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.
Px = 300(10 - 6)
A(300, - 325)10 - 6
Py = 550(10 - 6)
gxy = - 650(10 - 6)
y
gxy
2
A
= - 325(10 - 6)
x
C(425, 0)10 - 6
R = C 2(425 - 300)2 + ( - 325)2 D 10 - 6 = 348.2(10 - 6)
a)
P1 = (425 + 348.2)(10 - 6) = 773(10 - 6)
Ans.
P2 = (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.
Ans:
(a) P1 = 773(10 - 6), P2 = 76.8(10 - 6),
(b) gmax
= 696(10 - 6),
in-plane
(c) gabs = 773(10 - 6)
max
981
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10–23. The strain at point A on a beam has components
Px = 450(10 - 6), Py = 825(10 - 6), gxy = 275(10 - 6), Pz = 0.
Determine (a) the principal strains at A, (b) the maximum
shear strain in the x–y plane, and (c) the absolute maximum
shear strain.
Px = 450(10 - 6)
Py = 825(10 - 6)
A(450, 137.5)10 - 6
gxy = 275(10 - 6)
A
gxy
2
= 137.5(10 - 6)
C(637.5, 0)10 - 6
R = [2(637.5 - 450)2 + 137.52]10 - 6 = 232.51(10 - 6)
a)
P1 = (637.5 + 232.51)(10 - 6) = 870(10 - 6)
Ans.
P2 = (637.5 - 232.51)(10 - 6) = 405(10 - 6)
Ans.
gmax
= 2R = 2(232.51)(10 - 6) = 465(10 - 6)
Ans.
870(10 - 6)
;
2
Ans.
b)
in-plane
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c)
gabs
max
2
=
gabs
max
= 870(10 - 6)
Ans:
(a) P1 = 870(10 - 6), P2 = 405(10 - 6),
= 465(10 - 6),
(b) gmax
in-plane
(c) gabs = 870(10 - 6)
max
982
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*10–24. The steel bar is subjected to the tensile load of
500 lb. If it is 0.5 in. thick determine the three principal
strains. E = 29 (103) ksi , n = 0.3.
2 in.
500 lb
500 lb
15 in.
sx =
500
= 500 psi
2(0.5)
Px =
1
1
(500) = 17.2414 (10 - 6)
(s ) =
E x
29(106)
sy = 0
sz = 0
Py = Pz = - vPx = - 0.3(17.2414)(10 - 6) = - 5.1724(10 - 6)
P1 = 17.2(10 - 6)
P2, 3 = - 5.17(10 - 6)
Ans.
983
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10–25. The 45° strain rosette is mounted on a machine
element. The following readings are obtained from each
gauge: Pa = 650(10 - 6), Pb = - 300(10 - 6), Pc = 480(10 - 6).
Determine (a) the in-plane principal strains and (b) the
maximum in-plane shear strain and associated average
normal strain. In each case show the deformed element due
to these strains.
a
b
45⬚
c
Pa = 650(10 - 6)
Pb = - 300(10 - 6)
Pc = 480(10 - 6)
ua = 90°
ub = 135°
uc = 180°
45⬚
Pa = Px cos2 ua + Py sin2 ua + gxy sin ua cos ua
650(10 - 6) = Px cos2 90° + Py sin2 90° + gxy sin 90° cos 90°
Px = 650(10 - 6)
Pc = Px cos2 uc + Py sin2 uc + gxy sin uc cos uc
480(10 - 6) = Px cos2 180° + Py sin2 180° + gxy sin 180° cos 180°
Py = 480(10 - 6)
Pb = Px cos2 ub + Py sin2 ub + gxy sin ub cos ub
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- 300(10 - 6) = 480(10 - 6) cos2 135° + 650(10 - 6) sin2 135° + gxy sin 135° cos 135°
gxy = 1730(10 - 6)
gxy
2
= 865(10 - 6)
A(480, 865)10 - 6
C(565, 0)10 - 6
R = ( 2(565 - 480)2 + 8652)10 - 6 = 869.17(10 - 6)
a)
P1 = (565 + 869.17)10 - 6 = 1434(10 - 6)
Ans.
P2 = (565 - 869.17)10 - 6 = - 304(10 - 6)
Ans.
tan 2up =
865
565 - 480
2up = 84.39°
(Mohr’s circle)
up = - 42.19°
(element)
b)
gmax
in-plane
= 2R = 2(869.17)(10 - 6) = 1738(10 - 6)
Ans.
Pavg = 565(10 - 6)
Ans.
2us = 90° - 2up = 5.61°
(Mohr’s circle)
us = 2.81°
(element)
Ans:
P1 = 1434(10 - 6), P2 = - 304(10 - 6),
gmax
= 1738(10 - 6), Pavg = 565(10 - 6)
in-plane
984
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10–26. The 60° strain rosette is attached to point A on the
surface of the support. Due to the loading the strain gauges
give a reading of Pa = 300(10 - 6), Pb = - 150 (10-6), and
Pc = - 450 (10 - 6). Use Mohr’s circle and determine (a) the
in-plane principal strains and (b) the maximum in-plane
shear strain and the associated average normal strain.
Specify the orientation of each element that has these states
of strain with respect to the x axis.
Pa = Px cos2 ua + Py sin2 ua + gxy sin ua cos ua
300(10 - 6) = Px cos2 0° + Py sin2 0° + gxy sin 0° cos 0°
Px = 300(10 - 6)
Pb = Px cos2 ub + Py sin2 ub + gxy sin ub cos ub
- 150(10 - 6) = 300(10 - 6) cos2 60° + Py sin2 60° + gxy sin 60° cos 60°
0.75Py + 0.43301gxy = - 225(10 - 6)
(1)
2
Pc = Px cos uc + Py sin uc + gxy sin uc cos uc
- 450(10 - 6) = 300(10 - 6) cos2 120° + Py sin2 120° + gxy sin 120° cos 120°
0.75Py - 0.43301gxy = - 525(10 - 6)
(2)
Solving Eqs. (1) and (2),
gxy = 346.41(10 - 6)
gxy
Construction of the Circle: Px = 300(10 - 6), Py = - 500(10 - 6), and
= 173.20(10 - 6).
2
Thus
Px + Py
Pavg =
2
= c
300 + ( - 500)
d (10 - 6) = - 100(10 - 6)
2
Ans.
The coordinates of reference point A and center of C of the circle are
A(300, 173.20)(10 - 6)
x
a
Normal and Shear Strain: With ua = 0°, ub = 60°, and uc = 120°, we have
Py = - 500(10 - 6)
60⬚
60⬚
A
2
c
b
C( - 100, 0)(10 - 6)
Thus, the radius of the circle is
R = CA = a2[300 - ( - 100)]2 + 173.202 b(10 - 6) = 435.89(10 - 6)
Using these results, the circle is shown in Fig. a.
In-Plane Principal Strains: The coordinates of reference points B and D represent
P1 and P2, respectively.
P1 = (- 100 + 435.89)(10 - 6) = 336(10 - 6)
Ans.
P2 = (- 100 - 435.89)(10 - 6) = - 536(10 - 6)
Ans.
985
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10–26. Continued
Orientation of Principal Strain: Referring to the geometry of the circle,
tan 2(up)1 =
(up)1 = 11.7°
173.20(10 - 6)
(300 + 100)(10 - 6)
= 0.43301
(counterclockwise)
Ans.
The deformed element for the state of principal strain is shown in Fig. b.
Maximum In-Plane Shear Strain: The coordinates of point E represent Pavg and
gmax
in-plane
. Thus
2
gmax
in-plane
2
gmax
in-plane
= R = (435.89)(10 - 6)
= 872(10 - 6)
Ans.
Orientation of Maximum In-Plane Shear Strain: Referring to the geometry of the
circle,
tan 2us =
300 + 100
= 2.3094
173.20
us = 33.3°
(clockwise)
Ans.
The deformed element for the state of maximum in-plane shear strain is shown in
Fig. c.
www.elsolucionario.org
Ans:
P1 = 336(10 - 6), P2 = - 536(10 - 6),
up1 = 11.7° (counterclockwise),
gmax
= 872(10 - 6), Pavg = - 100(10 - 6),
in-plane
us = 33.3° (clockwise)
986
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10–27. The strain rosette is attached at the point on the
surface of the pump. Due to the loading, the strain gauges give
a reading of Pa = - 250(10 - 6), Pb = - 300 (10-6), and
Pc = - 200 (10 - 6). Determine (a) the in-plane principal
strains, and (b) the maximum in-plane shear strain. Specify
the orientation of each element that has these states of strain
with respect to the x axis.
b
60⬚
xx
A
60⬚
c
Normal and Shear Strains: With ua = 0°, ub = 60°, and uc = - 60°, we have
Pa = Px cos2 ua + Py sin2 ua + gxy sin ua cos ua
-250(10 - 6) = Px cos2 0° + Py sin2 0° + gxy sin 0° cos 0°
Px = - 250(10 - 6)
Pb = Px cos2 ub + Py sin2 ub + gxy sin ub cos ub
300(10 - 6) = - 250(10 - 6) cos2 60° + Py sin2 60° + gxy sin 60° cos 60°
0.75Py + 0.43301gxy = 362.5(10 - 6)
(1)
Pc = Px cos2 uc + Py sin2 uc + gxy sin uc cos uc
- 200(10 - 6) = - 250(10 - 6) cos2 ( -60°) + Py sin2 ( -60°) + gxy sin ( -60°) cos ( -60°)
0.75Py - 0.43301gxy = - 137.5(10 - 6)
(2)
Solving Eqs. (1) and (2), we obtain
Py = 150(10 - 6)
gxy = 577.35(10 - 6)
gxy
Construction of the Circle: Px = - 250(10 - 6), Py = 150(10 - 6), and
= 288.68(10 - 6).
2
Thus
Px + Py
- 250 + 150
Ans.
= a
b(10 - 6) = - 50(10 - 6)
Pavg =
2
2
The coordinates of reference point A and center of C of the circle are
A( - 250, 288.68)(10 - 6)
a
C( -50, 0)(10 - 6)
Thus, the radius of the circle is
R = CA = a2[ -250 - ( - 50)]2 + 288.682 b(10 - 6) = 351.19(10 - 6)
Using these results, the circle is shown in Fig. a.
In-Plane Principal Strains: The coordinates of reference points B and D represent
P1 and P2, respectively.
P1 = (- 50 + 351.19)(10 - 6) = 301(10 - 6)
Ans.
P2 = (- 50 - 351.19)(10 - 6) = - 401(10 - 6)
Ans.
987
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10–27. Continued
Orientation of Principal Strain: Referring to the geometry of the circle,
tan 2(up)2 =
288.68
= 1.4434
250 - 50
(up)2 = 27.6° (clockwise)
Ans.
The deformed element for the state of principal strain is shown in Fig. b.
Maximum In-Plane Shear Strain: The coordinates of point E represent Pavg and
gmax
in-plane
. Thus
2
gmax
in-plane
2
gmax
in-plane
= R = 351.19(10 - 6)
= 702(10 - 6)
Ans.
Orientation of Maximum In-Plane Shear Strain: Referring to the geometry of the
circle,
tan 2us =
250 - 50
= 0.6928
288.68
us = 17.4° (counterclockwise)
Ans.
The deformed element for the state of maximum in-plane shear strain is shown in
Fig. c.
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Ans:
P1 = 301(10 - 6), P2 = - 401(10 - 6),
up2 = 27.6° (clockwise),
gmax
= 702(10 - 6), Pavg = - 50(10 - 6),
in-plane
us = 17.4° (counterclockwise)
988
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*10–28. The 60° strain rosette is mounted on a beam. The
following readings are obtained from each gauge:
Pa = 250(10 - 6), Pb = - 400 (10-6), Pc = 280(10 - 6). Determine (a) the in-plane principal strains and their orientation,
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
60⬚
c
Pa = 250(10 - 6)
Pb = - 400(10 - 6)
Pc = 280(10 - 6)
ua = 60°
ub = 120°
uc = 180°
Pc = Px cos2 uc + Pg sin2 uc + gxy si