© 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.
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.
4
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1–5. Determine the resultant internal loadings in the
beam at cross sections through points D and E. Point E is
just to the right of the 3-kip load.
3 kip
1.5 kip/ ft
A
D
6 ft
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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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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.
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) +
1
(3)(3)(2) = 0
2
Ay = 7.50 kN
+ 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
8
C
0.5 m 0.5 m
Referring to the FBD of this segment, Fig. b,
+
: ©Fx = 0;
B
A
Referring to the FBD of the entire beam, Fig. a,
Ans.
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.
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
+
: ©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
FBD = 87.05 N
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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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*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
#
Ans.
m
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.
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
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.
The negative sign indicates that NE acts in the opposite sense to that shown on the
free-body diagram.
16
I
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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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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
MC + 80(6) = 0;
Ans.
Ans.
MC = - 480 lb # in.
Ans.
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
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.
20
B
D
3 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.
1–21. 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
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 = 1.39 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;
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
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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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
z
1–25. Determine the resultant internal loadings acting on
the cross section through point B of the signpost. The post is
fixed to the ground and a uniform pressure of 7 lb>ft2 acts
perpendicular to the face of the sign.
3 ft
2 ft
©Fx = 0;
(VB)x - 105 = 0;
(VB)x = 105 lb
©Fy = 0;
(VB)y = 0
Ans.
©Fz = 0;
(NB)z = 0
Ans.
©Mx = 0;
(MB)x = 0
©My = 0;
(MB)y - 105(7.5) = 0;
©Mz = 0;
(TB)z - 105(0.5) = 0;
Ans.
3 ft
2
7 lb/ft
Ans.
(MB)y = 788 lb # ft
6 ft
Ans.
(TB)z = 52.5 lb # ft
B
Ans.
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;
©Fy = 0;
(NC)y = 0
©Fz = 0;
(VC)z + 240 = 0;
(VC)x = - 250 N
Ans.
Ans.
Ans.
(VC)z = - 240 N
(MC)x = - 108 N # m
©Mx = 0;
(MC)x + 240(0.45) = 0;
©My = 0;
(TC)y = 0
©Mz = 0;
(MC)z - 1000(0.2) + 750(0.45) = 0;
Ans.
Ans.
(MC)z = - 138 N # m Ans.
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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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
z
1–27. 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
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.
28
3 in. 9 in.
Fx 30 lb
A
Fz 10 lb
9 in.
6 in.
6 in.
6 in.
Fy 50 lb
y
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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
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
(4)
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
2
A
4 (1000 )
Ans.
32
250 mm
20 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.
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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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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 =
P
=
A
4(103)
p
2
4 (0.028
- 0.022)
= 13.3 MPa (C)
Ans.
At E:
sE =
P
=
A
8(103)
p
4
(0.0122)
= 70.7 MPa (T)
Ans.
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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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*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
36
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.
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
A¿ =
1.5 in.
800 lb
30
1 in.
1 in.
800 lb
30
1.5(1)
= 3 in2
sin 30°
s =
N
200
=
= 66.7 psi
A¿
3
Ans.
t =
V
346.41
=
= 115 psi
A¿
3
Ans.
Ans:
s = 66.7 psi, t = 115 psi
38
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
1–39. 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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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*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)
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
© 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.
1–42. 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
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
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
A B
x =
1
12
+ 12(6)(12) 12
2 (3)(12) 3
3
1
(9)(12)
2
y =
1
2
1
2 (3)(12)(3) 3 + 2 (6)(12)
1
2 (9)(12)
s =
500
P
= 9.26 psi
= 1
A
2 (9)(12)
A B
= 4 in.
A 3 + 63 B
P(x,y)
Ans.
= 4 in.
Ans.
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
(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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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*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.
48
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.
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
1.5 in.
B
+ c ©Fy = 0;
C
20⬚
4.5 in.
-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.
Ans:
tavg = 16 psi
50
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
1–51. During the tension test, the wooden specimen is
subjected to an average normal stress of 2 ksi. Determine the
axial force P applied to the specimen. Also, find the average
shear stress developed along section a–a of the specimen.
P
a
4 in.
a
2 in.
1 in.
Internal Loading: The normal force developed on the cross section of the middle
portion of the specimen can be obtained by considering the free-body diagram
shown in Fig. a.
+ c ©Fy = 0;
P
P
+
- N = 0
2
2
4 in.
N = P
Referring to the free-body diagram shown in fig. b, the shear force developed in the
shear plane a–a is
+ c ©Fy = 0;
P
- Va - a = 0
2
Va - a
P
P
=
2
Average Normal Stress and Shear Stress: The cross-sectional area of the specimen is
A = 1(2) = 2 in2. We have
savg =
N
;
A
2(103) =
P
2
P = 4(103)lb = 4 kip
Ans.
4(103)
P
=
= 2(103) lb. The area of the shear plane is
2
2
= 2(4) = 8 in2 . We obtain
Using the result of P, Va - a =
Aa - a
A ta - a B avg =
2(103)
Va - a
=
= 250 psi
Aa - a
8
Ans.
Ans:
P = 4 kip, Ata - aB avg = 250 psi
51
© 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.
*1–52. 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.
52
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
1–53. 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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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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.
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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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*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
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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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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
savg
N
=
=
A
pr02g
3h2
[(z + h)3 - h3]
pr02
h2 (z
=
+ h)2
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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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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:
For the frustum, A = 2pxL = 2p(0.025 + 0.025) A 20.05 + 0.05
2
= 0.02221 m
P
;
s =
A
3 A 10
6
B
2
4.5 MPa
B
30 mm
C
2
25 mm 25 mm
F1
=
0.02221
F1 = 66.64 kN
Average Shear Stress:
For the cylinder, A = p(0.05)(0.03) = 0.004712 m2
F2
V
;
4.5 A 106 B =
tavg =
A
0.004712
F2 = 21.21 kN
Equation of Equilibrium:
+ c ©Fy = 0;
P - 21.21 - 66.64 sin 45° = 0
P = 68.3 kN
Ans.
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
Ans.
s =
P
=
A
t =
3600>2
V
= p
= 5.87 ksi
2
A
4 (5>8)
3600
p
2
4 (0.75)
= 8.15 ksi
Ans.
Ans:
s = 8.15 ksi, t = 5.87 ksi
59
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*1–60. 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.
60
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
1–61. 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)
a + ©MD = 0;
+
: ©Fx = 0;
1.5 in. 2 in. 1 in.
20(5) - By (1) = 0
20 lb
By = 100 lb
Bx = 0
From FBD(b)
+
: ©Fx = 0;
a + ©ME = 0;
Ax = 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.
Ans:
(tA)avg = 3.71 ksi
62
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
1–63. Solve Prob. 1–62 for pin B. The pin is subjected to
double shear and has a diameter of 0.2 in.
20 lb
Support Reactions:
A
From FBD(a)
a + ©MD = 0;
+
: ©Fx = 0;
C
E
20(5) - By (1) = 0
B D
By = 100 lb
5 in.
Bx = 0
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
=
AB
p
4
50.0
(0.22)
= 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
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
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 =
(tavg)B =
C
FCD
3181.98
= 40.5 MPa
=
ACD
78.540(10 - 6)
Ans.
VB
1758.91
= 62.2 MPa
=
AB
28.274(10 - 6)
Ans.
64
P
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.
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;
+
: ©Fx = 0;
FCD (0.3 sin 45°) - P(0.45) = 0
FCD = 2.121P
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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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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
4
1.25Pa b - Va - a = 0
5
cross-sectional
area
of
section
Na - a = 0.75P
a
A
Va - a = P
a–a
is
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)
P = 62.5(103) N = 62.5 kN (Controls!)
Ans.
Ans:
P = 62.5 kN
66
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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
1–67. 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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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
z
*1–68. 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.
68
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
1–69. 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)
12
h
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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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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.
Ans:
d = 5.71 mm
70
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
1–71. 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
4
d2
d = 0.0135 m = 13.5 mm
Ans.
Ans:
d = 13.5 mm
71
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*1–72. 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;
-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.
72
E
60⬚
6 ft
45⬚
6 ft
C
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.
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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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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.
Ans:
1
Use a = 6 in.
2
74
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
1–75. 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;
+ c ©Fy = 0;
FB(18) - 540(9) - 90(12) = 0;
FB = 330 lb
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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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*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
4(103) =
330
4(p4 )dB2
dB = 0.162 in.
Ans.
76
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
1–77. 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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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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
140(106) =
359.07
p
2
4 dAB
dAB = 0.001807 m = 1.81 mm
sallow =
FBC
;
ABC
140(106) =
Ans.
439.77
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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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*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
80
20 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.
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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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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
(F.S.)con =
Ans.
(sfail)con
25
=
= 3.53
(savg)con
7.074
Ans.
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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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*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
For pin C,
tallow =
VC
;
AC
40(106) =
4.00(103)
p
4
dC 2
dC = 0.01128 m = 11.3 mm
For pin D,
VD
;
tallow =
AD
40(106) =
Ans.
6.093(103)
p
4
dD 2
dD = 0.01393 m = 13.9 mm
Ans.
84
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.
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
C
A
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.
Ans:
dAB = 15.5 mm, dAC = 13.0 mm
86
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
1–87. 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) =
FAB
p
2
4 (0.01)
FAB = 11.78 kN
From Eq. (1),
P = 8.33 kN
Assume failure of rod AC:
FAC
sallow =
; 150(106) =
AAC
FAC
p
2
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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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*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
FB(4.5) + 1.5(3) + 2(1.5) - FC(6) = 0
(1)
FB(5.5) - FC(4) - 3(2) = 0
5.5 FB - 4 FC = 6
(2)
Solving Eqs. (1) and (2) yields
FC = 4.55 kN
For bolt:
sallow = 150(106) =
4.40(103)
p
2
4 (dB)
dB = 0.00611 m
= 6.11 mm
Ans.
For washer:
sallow = 28 (104) =
4.40(103)
p
2
4 (dw
D
B
From FBD (b):
FB = 4.40 kN;
1.5 m
C
4.5 FB - 6 FC = - 7.5
a + ©MA = 0;
2m
A
From FBD (a):
a + ©MD = 0;
2 kN
1.5 kN
1.5 m 1.5 m 1.5 m
3 kN
- 0.006112)
dw = 0.0154 m = 15.4 mm
Ans.
88
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
1–89. 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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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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
Allowable Normal and Shear Stress:
sallow =
sfail
6
=
= 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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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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
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.
92
P
A¿
B¿
3m
B
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.
1–93. 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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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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.
Ans:
h = 1.74 in.
94
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
1–95. 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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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*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
A
B
C
D
200 kip
300 kip 300 kip 300 kip 300 kip
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
(sb)C =
FC
883.33
=
= 2.003 ksi
Ab
441
(sb)D =
FD
766.67
=
= 1.738 ksi
Ab
441
100 kip
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
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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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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
P
=
ss =
A
8 (103)
= 208 MPa
Ans.
(tavg)a =
8 (103)
V
=
= 4.72 MPa
A
p (0.018)(0.030)
Ans.
(tavg)b =
8 (103)
V
=
= 45.5 MPa
A
p (0.007)(0.008)
Ans.
p
4
(0.007)2
30 mm
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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*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.
100
2 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.
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
10 mm
C
D
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 =
sb =
40(103)
P
= 25.5 MPa
=
AV
0.5(10 - 3)p
120 mm
Ans.
40(103)
P
= 4.72 MPa
=
Ab
2.7(10 - 3)p
Ans.
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.
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
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 =
B
gL2
T
8s
=
=
A
A
8s
Ans.
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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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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.
Ans:
P = 0.0472 in.>in.
106
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
2–3. 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
L
4000
¢LBD =
PCE
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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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*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
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
dA
6.9813
=
= 0.0349 mm>mm
LAH
200
dC
10.4720
= 0.0349 mm>mm
=
=
LCG
300
(Pavg)AH =
Ans.
(Pavg)CG
Ans.
(Pavg)DF =
dD
17.4533
=
= 0.0582 mm>mm
LDF
300
Ans.
108
E
C
200 mm
2°
bp rad = 0.03491 rad.
180
Since u is small, the displacements of points A, C, and D can be approximated by
200 mm 300 mm
300 mm
B
A
Geometry: The lever arm rotates through an angle of u = a
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
Pavg
L - L0
=
=
L0
2pr0
(z + h) - 2r0
h
p
= (z + h) - 1
2r0
h
Ans.
Ans:
Pavg =
110
p
(z + h) - 1
h
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
2–7. 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
Ans.
112
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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
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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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*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
= 0.006667 + 0.0075 = 0.0142 rad
Ans.
116
300 mm
B
2 mm
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.
2–13. 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
400
PDB =
Ans.
PAD
Ans.
Ans:
PDB = - 0.00680 mm>mm,
PAD = 0.0281(10 - 3) mm>mm
117
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
2–14. 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.
Ans:
(Pavg)AC = 0.0258 mm>mm
118
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
2–15. 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
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
Geometry: Here, gxy
c = 90° - 0.4297° = 89.5703°
G
A
180°
= 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:
(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 =
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 =
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¿ =
5
= 0.00884 mm>mm
565.69
Ans.
128
A¿
5 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.
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
1
40(10 - 6)
e ln cos c 40(10 - 6)x + 0.008 d f `
300 mm
x
300 mm
0
= 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
1
40(10 - 6)
tan [40(10 - 6)x]dx
e ln cos c 40(10 - 6)x d f `
300 mm
0
= 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
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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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*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
=
2
L
e - (x>L) L
d = 31 - (1>e)4
2
2
0
L
3e - 14
2e
Ans.
132
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
2–29. 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
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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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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 =
(Px)avg =
3
2
L0
kx =
(¢x)B
L
kL
3
Ans.
kL3
kL2
3
=
=
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
b3
x2 = tan gxy
gxy = tan - 1 a
3v0
b3
x2 b
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
Ans.
and at point (b, a),
gxy = tan - 1 c
3v0
b3
= tan - 1 c 3 a
C
(b2) d
v0
bd
b
Ans.
136
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.
2–33. 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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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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)
138
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
3–1. 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)
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
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
93.10
Eapprox
0.230
48
=
= 32.0(103) ksi
0.0015
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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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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.
Ans:
in # lb
E = 55.3 A 103 B ksi, ur = 9.96
in3
140
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
3–3. 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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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*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 =
1
p(0.503)2 = 0.19871 in2
4
L = 2.00 in.
s =
P
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 =
32.71
= 26.2(103) ksi
0.00125
Ans.
142
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
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
3–5. 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.
s =
P
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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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
3–6. A specimen is originally 1 ft long, has a diameter of
0.5 in., and is subjected to a force of 500 lb. When the force
is increased from 500 lb to 1800 lb, the specimen elongates
0.009 in. Determine the modulus of elasticity for the
material if it remains linear elastic.
Normal Stress and Strain: Applying s =
s1 =
s2 =
¢P =
0.500
p
2
4 (0.5 )
1.80
p
2
4 (0.5 )
dL
P
and e =
.
A
L
= 2.546 ksi
= 9.167 ksi
0.009
= 0.000750 in.>in.
12
Modulus of Elasticity:
E =
¢s
9.167 - 2.546
=
= 8.83 A 103 B ksi
¢P
0.000750
Ans.
Ans:
E = 8.83 A 103 B ksi
144
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
3–7. 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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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*3–8. The strut is supported by a pin at C and an A-36 steel
guy wire AB. If the wire has a diameter of 0.2 in., determine
how much it stretches when the distributed load acts on
the strut.
A
60⬚
200 lb/ft
a + ©MC = 0;
1
FAB cos 60°(9) - (200)(9)(3) = 0
2
9 ft
FAB = 600 lb
The normal stress the wire is
sAB =
FAB
=
AAB
p
4
600
= 19.10(103) psi = 19.10 ksi
(0.22)
Since sAB 6 sy = 36 ksi, Hooke’s Law can be applied to determine the strain
in wire.
sAB = EPAB;
19.10 = 29.0(103)PAB
PAB = 0.6586(10 - 3) in>in
9(12)
= 124.71 in. Thus, the wire
The unstretched length of the wire is LAB =
sin 60°
stretches
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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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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.
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
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
150
Ans.
0
0
0.05 0.10 0.15 0.20 0.25 0.30 0.35
0.001 0.002 0.003 0.004 0.005 0.006 0.007
P (in./in.)
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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
=
ABD
p
4
1500
= 30.56(103) psi = 30.56 ksi
(0.252)
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
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
29.01(103) =
2.50 P
(0.252)
p
4
P = 569.57 lb = 570 lb
Ans.
Ans:
P = 570 lb
153
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*3–16. 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
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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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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.
Ans:
dP = 0.00637 in.
156
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
3–19. 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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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*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.
158
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.
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;
+
; ©Fx = 0;
3
F
- P = 0;
5 AB
FBC
4
- (1.6667P) = 0;
5
FAB = 1.6667 P
20
(1)
15
FBC = 1.333 P
(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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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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;
+
: ©Fx = 0;
3
FBA a b - 3 = 0;
5
-FBC
4
+ 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.
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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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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;
+ c ©Fy = 0;
FBC sin 30° - FAB sin 45° = 0
(1)
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
PBC = 0.001371 in.>in.
63.27 - 58
80 - 58
=
;
PAB - 0.002
0.01 - 0.002
PAB = 0.003917 in.>in.
0.002
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 =
P
=
A
Plong =
300
p
2
4 (0.015)
300 N
300 N
200 mm
= 1.678 MPa
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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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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
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)
G =
Ans.
67.91(109)
E
=
= 25.27(109) Pa = 25.3 GPa
2(1 + n)
2(1 + 0.3438)
Ans.
Ans:
E = 67.9 GPa, v = 0.344, G = 25.3 GPa
164
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
3–27. 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
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 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,
sBC = EPBC;
149.21(106) = 200(109)PBC
PBC = 0.7460(10-3) mm>mm
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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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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.
Ans:
g = 3.06(10-3) rad
168
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
3–31. 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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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*3–32. A shear spring is made by bonding the rubber
annulus to a rigid fixed ring and a plug. When an axial load
P is placed on the plug, show that the slope at point y in
the rubber is dy>dr = - tan g = - tan1P>12phGr22. For small
angles we can write dy>dr = - P>12phGr2. Integrate this
expression and evaluate the constant of integration using
the condition that y = 0 at r = ro. From the result compute
the deflection y = d of the plug.
P
h
ro
y
d
ri
r
y
Shear Stress–Strain Relationship: Applying Hooke’s law with tA
g =
P
=
.
2p r h
tA
P
=
G
2p h G r
dy
P
= - tan g = - tan a
b
dr
2p h G r
(Q.E.D)
If g is small, then tan g = g. Therefore,
dy
P
= dr
2p h G r
At r = ro,
y = -
dr
P
2p h G L r
y = -
P
ln r + C
2p h G
0 = -
P
ln ro + C
2p h G
y = 0
C =
Then, y =
ro
P
ln
r
2p h G
At r = ri,
y = d
d =
P
ln ro
2p h G
ro
P
ln
ri
2p h G
Ans.
170
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
3–33. 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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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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.
Ans:
d =
172
Pa
2bhG
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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 =
P
=
A
0.49935 - 0.5
d¿ - d
=
= - 0.0013 in.>in.
d
0.5
V = -
Gal =
9
= 45.84 ksi
(0.5)2
45.84
s
=
= 0.0040205 in.>in.
E
11400.65
Plong =
Plat =
p
4
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.
P
s =
=
A
70
10
= 50.9296 ksi
p
2
(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.
174
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.
3–37. 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 =
80(3 - x)
3
dA = dB;
PL
AE
(220)
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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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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.
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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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*3–40. The head H is connected to the cylinder of a
compressor using six steel bolts. If the clamping force in
each bolt is 800 lb, determine the normal strain in the
3
bolts. Each bolt has a diameter of 16
in. If sY = 40 ksi and
3
Est = 29110 2 ksi, what is the strain in each bolt when the
nut is unscrewed so that the clamping force is released?
C
L
H
Normal Stress:
s =
P
=
A
800
A B
p 3 2
4 16
= 28.97 ksi 6 sg = 40 ksi
Normal Strain: Since s 6 sg, Hooke’s law is still valid.
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
178
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.
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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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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
PL
(1 - 2n)
E
Ans.
Ans:
dV =
180
PL
(1 - 2n)
E