mechanics-of-materials-si-9th-edition-hibbeler-solutions-manual

Prévia do material em texto

StuDocu is not sponsored or endorsed by any college or university
Mechanics of Materials SI 9th Edition Hibbeler Solutions
Manual
Mechanical Engineering (Universiti Teknikal Malaysia Melaka)
StuDocu is not sponsored or endorsed by any college or university
Mechanics of Materials SI 9th Edition Hibbeler Solutions
Manual
Mechanical Engineering (Universiti Teknikal Malaysia Melaka)
Downloaded by Igor Bertoldo (igorbertoldo1001@gmail.com)
lOMoARcPSD|9428255
https://www.studocu.com/en-us?utm_campaign=shared-document&utm_source=studocu-document&utm_medium=social_sharing&utm_content=mechanics-of-materials-si-9th-edition-hibbeler-solutions-manual
https://www.studocu.com/en-us/document/universiti-teknikal-malaysia-melaka/mechanical-engineering/mechanics-of-materials-si-9th-edition-hibbeler-solutions-manual/10805978?utm_campaign=shared-document&utm_source=studocu-document&utm_medium=social_sharing&utm_content=mechanics-of-materials-si-9th-edition-hibbeler-solutions-manual
https://www.studocu.com/en-us/course/universiti-teknikal-malaysia-melaka/mechanical-engineering/3192868?utm_campaign=shared-document&utm_source=studocu-document&utm_medium=social_sharing&utm_content=mechanics-of-materials-si-9th-edition-hibbeler-solutions-manual
https://www.studocu.com/en-us?utm_campaign=shared-document&utm_source=studocu-document&utm_medium=social_sharing&utm_content=mechanics-of-materials-si-9th-edition-hibbeler-solutions-manual
https://www.studocu.com/en-us/document/universiti-teknikal-malaysia-melaka/mechanical-engineering/mechanics-of-materials-si-9th-edition-hibbeler-solutions-manual/10805978?utm_campaign=shared-document&utm_source=studocu-document&utm_medium=social_sharing&utm_content=mechanics-of-materials-si-9th-edition-hibbeler-solutions-manual
https://www.studocu.com/en-us/course/universiti-teknikal-malaysia-melaka/mechanical-engineering/3192868?utm_campaign=shared-document&utm_source=studocu-document&utm_medium=social_sharing&utm_content=mechanics-of-materials-si-9th-edition-hibbeler-solutions-manual
Mechanics of Materials SI 9th Edition Hibbeler SOLUTIONS MANUAL 
Full clear download at: 
https://testbankreal.com/download/mechanics-materials-si-9th-edition-hibbeler-
solutions-manual/ 
 
 
2-1 The center portion of th e rubber balloon has a 
diameter of d = 100 mm. If the air pressure 
within it cau ses the balloon's diameter to 
become d = 125 mm, determine the av erage 
normal strain in the rubber.
 
Given: d0 := 100mm 
 
Solution: 
d := 125mm
 
 
ε := 
 
π d − π d0 
π d0
mm ε = 0.2500 
mm 
 
Ans
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
Downloaded by Igor Bertoldo (igorbertoldo1001@gmail.com)
lOMoARcPSD|9428255
https://testbankreal.com/download/mechanics-materials-si-9th-edition-hibbeler-solutions-manual/
https://testbankreal.com/download/mechanics-materials-si-9th-edition-hibbeler-solutions-manual/
https://www.studocu.com/en-us?utm_campaign=shared-document&utm_source=studocu-document&utm_medium=social_sharing&utm_content=mechanics-of-materials-si-9th-edition-hibbeler-solutions-manual
Ans: 
 CE 0.00250 mm mm, BD 0.00107 mm mm 
 
 
105
Downloaded by Igor Bertoldo (igorbertoldo1001@gmail.com)
lOMoARcPSD|9428255
 
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.) 
 
L - L0 5p - 15 
= 
L 
= 
15 
= 0.0472 in.>in. 
0 
© Pearson Education South Asia Pte Ltd 2014. 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. 
 
 
2–2. A thin strip of rubber has an unstretched length of 
375 mm. If it is stretched around a pipe having an outer 
diameter of 125 mm, determine the average normal strain 
in the strip. 
 
L0 = 375 mm 
L = p(125 mm) 
ee 5 
 L – L0 5 125p – 375 = 0.0472 
mm/mm 
 
 
Ans.
L0 375 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
Ans: 
P = 0.0472 mm>mm 
 
Downloaded by Igor Bertoldo (igorbertoldo1001@gmail.com)
lOMoARcPSD|9428255
https://www.studocu.com/en-us?utm_campaign=shared-document&utm_source=studocu-document&utm_medium=social_sharing&utm_content=mechanics-of-materials-si-9th-edition-hibbeler-solutions-manual
106 
Downloaded by Igor Bertoldo (igorbertoldo1001@gmail.com)
lOMoARcPSD|9428255
© Pearson Education South Asia Pte Ltd 2014. 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. 
 
 
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 
D E 
be displaced 10 mm downward, determine the normal strain 
developed in wires CE and BD. 
 
4 m 
 
P 
 
A B C 
 
 
3 m 2 m 2 m 
 
 
 
 
 
 
 
 
 
 
 
 
 
¢LBD 
3 
=
 
¢ LCE 
7
 
¢LBD = 
3 (10) 
7 
= 4.286 mm
¢ LCE 10 
PCE = L 
= 
4000 
= 0.00250 mm>mm
 
Ans.
 
PBD = 
 
¢ LBD 
L 
=
 
 
4.286 
4000 
= 0.00107 mm>mm
 
 
Ans.
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
Ans: 
PCE = 0.00250 mm>mm, PBD = 0.00107 mm>mm 
Downloaded by Igor Bertoldo (igorbertoldo1001@gmail.com)
lOMoARcPSD|9428255
https://www.studocu.com/en-us?utm_campaign=shared-document&utm_source=studocu-document&utm_medium=social_sharing&utm_content=mechanics-of-materials-si-9th-edition-hibbeler-solutions-manual
 
107 
Downloaded by Igor Bertoldo (igorbertoldo1001@gmail.com)
lOMoARcPSD|9428255
© Pearson Education South Asia Pte Ltd 2014. 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. 
 
*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. 
 
 
 
 200 mm 
300 mm
 
 
A B 
 
 
G F 
 
 200 mm 
 
 
 
 
300 mm 
E
 
C D 
200 mm 
 
 H 
 
 
 
 
2° 
Geometry: The lever arm rotates through an angle of u = a 
180 
b p rad = 0.03491 rad.
 
Since u is small, the displacements of points A, C, and D can be approximated by 
 
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
 
(Pavg)AH = 
 
(Pavg)CG = 
 
 
(Pavg)DF = 
dA 
= 
LAH 
dC 
= 
LCG 
 
dD 
= 
LDF 
6.9813 
200 
= 0.0349 mm>mm
 
10.4720 
300 
= 0.0349 mm>mm
 
 
17.4533 
300 
= 0.0582 mm>mm
 
 
Ans. 
Ans. 
 
Ans.
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
108 
Downloaded by Igor Bertoldo (igorbertoldo1001@gmail.com)
lOMoARcPSD|9428255
https://www.studocu.com/en-us?utm_campaign=shared-document&utm_source=studocu-document&utm_medium=social_sharing&utm_content=mechanics-of-materials-si-9th-edition-hibbeler-solutions-manual
© Pearson Education South Asia Pte Ltd 2014. All rights reserved. This material is protected under all copyrightlaws 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. 
 
 
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 
 
 
 
 
30 
P 
30 A 
 
 
 
 
B 
 
 
 
 
 
 
 
 
 
 
 
 
 
Lœ 2 2AC = 2300 + 2 
 
Lœ
 
- 2(300)(2) cos 150° = 301.734 mm
 
PAC = PAB = 
 AC - LAC 
LAC 
301.734 - 300 
= 
300 
= 0.00578 mm>mm
 
 
Ans.
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
Ans: 
PAC = PAB = 0.00578 mm>mm 
 
Downloaded by Igor Bertoldo (igorbertoldo1001@gmail.com)
lOMoARcPSD|9428255
109 
Downloaded by Igor Bertoldo (igorbertoldo1001@gmail.com)
lOMoARcPSD|9428255
https://www.studocu.com/en-us?utm_campaign=shared-document&utm_source=studocu-document&utm_medium=social_sharing&utm_content=mechanics-of-materials-si-9th-edition-hibbeler-solutions-manual
 
r 
© Pearson Education South Asia Pte Ltd 2014. 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. 
 
 
2–6. The rubber band of unstretched length 2r
0 
is forced 
down the frustum of the cone. Determine the average r0 
normal strain in the band as a function of z. 
 
 
 
z 
 
 
 
2r0 
h 
 
Geometry: Using similar triangles shown in Fig. a,
h¿ h¿ +h 
= ; 
0 2r0 
 
h¿ = h
 
Subsequently, using the result of h¿
 r r0 
z + h 
= 
h 
;
 
r0 r = (z + h) 
h
 
Average Normal Strain: The length of the rubber band as a function of z is 
2pr0
L = 2pr = (z + h). With L0 = 2r0, we have 
h 
 
2pr0
 
Pavg = 
L - L0 
L0 
 h 
(z + h) - 
2r0 
= = 
2r0 
 
p 
(z + h) - 1 
h 
 
Ans.
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
Ans: 
 
Pavg = 
 
p 
(z + h) - 1 
h
Downloaded by Igor Bertoldo (igorbertoldo1001@gmail.com)
lOMoARcPSD|9428255
 
110 
Downloaded by Igor Bertoldo (igorbertoldo1001@gmail.com)
lOMoARcPSD|9428255
https://www.studocu.com/en-us?utm_campaign=shared-document&utm_source=studocu-document&utm_medium=social_sharing&utm_content=mechanics-of-materials-si-9th-edition-hibbeler-solutions-manual
© Pearson Education South Asia Pte Ltd 2014. 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. 
 
 
2–7. The pin-connected rigid rods AB and BC are inclined P 
at u = 30° when they are unloaded. When the force P is 
applied u becomes 30.2°. Determine the average normal B 
strain developed in wire AC. 
 
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: 
 
LAC ¿ - LAC 
 
 
603.6239 - 600 - 3
(Pavg)AC = = LAC 600 
= 6.04(10
 
) mm>mm Ans.
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
Ans: 
(Pavg)AC = 6.04(10 
- 3) mm>mm 
 
Downloaded by Igor Bertoldo (igorbertoldo1001@gmail.com)
lOMoARcPSD|9428255
111 
Downloaded by Igor Bertoldo (igorbertoldo1001@gmail.com)
lOMoARcPSD|9428255
https://www.studocu.com/en-us?utm_campaign=shared-document&utm_source=studocu-document&utm_medium=social_sharing&utm_content=mechanics-of-materials-si-9th-edition-hibbeler-solutions-manual
P 
© Pearson Education South Asia Pte Ltd 2014. 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. 
 
 
*2–8. Part of a control linkage for an airplane consists of a u 
rigid member CBD and a flexible cable AB. If a force is 
D 
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. 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 
AB¿ -AB 
 
 
501.255 - 500
PAB = 
AB 
=
 500
= 0.00251 mm>mm Ans.
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
Downloaded by Igor Bertoldo (igorbertoldo1001@gmail.com)
lOMoARcPSD|9428255
 
112 
Downloaded by Igor Bertoldo (igorbertoldo1001@gmail.com)
lOMoARcPSD|9428255
https://www.studocu.com/en-us?utm_campaign=shared-document&utm_source=studocu-document&utm_medium=social_sharing&utm_content=mechanics-of-materials-si-9th-edition-hibbeler-solutions-manual
© Pearson Education South Asia Pte Ltd 2014. 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. 
 
2–9. Part of a control linkage for an airplane consists of a 
rigid member CBD and a flexible cable AB. If a force is 
applied to the end D of the member and causes a normal 
strain in the cable of 0.0035 mm>mm, determine the 
displacement of point D. Originally the cable is unstretched. 
u 
 
D 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° 
p
u = 90.4185° - 90° = 0.4185° = 
180° 
(0.4185) rad
 
p 
¢D = 600(u) = 600(
180°
)(0.4185) = 4.38 mm
 
 
 
 
Ans.
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
Ans: 
¢D = 4.38 mm 
 
113 
Downloaded by Igor Bertoldo (igorbertoldo1001@gmail.com)
lOMoARcPSD|9428255
© Pearson Education South Asia Pte Ltd 2014. 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. 
 
y 
2–10. The corners of the square plate are given the 
displacements indicated. Determine the shear strain along 
A 
the edges of the plate at A and B. 
 
 
16 mm 
D 
 
3 mm 
B 
x 
 
 
3 mm 16 mm
 
C 
16 mm 
 
16 mm
 
 
 
 
 
 
 
 
At A: 
 
u¿ - 1 
2 
= tan
 
 
 
 9.7 
a 
10.2 
b = 43.561°
 
u¿ = 1.52056 rad 
 
p
(gA)nt = 
2 
- 1.52056
 
= 0.0502 rad 
 
Ans.
At B: 
 
f¿ 
- 1
 
2 
= tan
 
 
 
10.2 
a 
9.7 
b = 46.439°
 
f¿ = 1.62104 rad 
 
p
(gB)nt = 
2 
- 1.62104
 
= - 0.0502 rad 
 
Ans.
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
Ans:(gA)nt = 0.0502 rad, (gB)nt = - 0.0502 rad 
 
Downloaded by Igor Bertoldo (igorbertoldo1001@gmail.com)
lOMoARcPSD|9428255
https://www.studocu.com/en-us?utm_campaign=shared-document&utm_source=studocu-document&utm_medium=social_sharing&utm_content=mechanics-of-materials-si-9th-edition-hibbeler-solutions-manual
114 
Downloaded by Igor Bertoldo (igorbertoldo1001@gmail.com)
lOMoARcPSD|9428255
© Pearson Education South Asia Pte Ltd 2014. 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. 
 
 
2–11. The corners B and D of the square plate are given y 
the displacements indicated. Determine the average normal 
strains along side AB and diagonal DB. A 
 
 
16 mm 
D 
 
3 mm 
B 
x 
 
 
3 mm 16 mm
 
C 
16 mm 
 
16 mm
 
 
Referring to Fig. a, 
 
LAB = 216
2 + 162 = 2512 mm 
LAB¿ = 216
2 + 132 = 2425 mm 
LBD = 16 + 16 = 32 mm 
LB¿D¿ = 13 + 13 = 26 mm 
 
Thus, 
 
A eavg B AB = 
LAB¿ - LAB 
LAB 
 
LB¿D¿ - LBD 
2425 - 2512 
= 
2512 
 
26 - 32 
 
= - 0.0889 mm>mm 
 
Ans.
A eavg B BD = LBD 
= 
32 
= - 0.1875 mm>mm
 
Ans.
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
115 
Downloaded by Igor Bertoldo (igorbertoldo1001@gmail.com)
lOMoARcPSD|9428255
https://www.studocu.com/en-us?utm_campaign=shared-document&utm_source=studocu-document&utm_medium=social_sharing&utm_content=mechanics-of-materials-si-9th-edition-hibbeler-solutions-manual
© Pearson Education South Asia Pte Ltd 2014. 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. 
 
 
 
2–12 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
116 
Downloaded by Igor Bertoldo (igorbertoldo1001@gmail.com)
lOMoARcPSD|9428255
© Pearson Education South Asia Pte Ltd 2014. 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. 
 
 
2–13 . 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
Ans: 
PDB = PAB cos
2 
u + PCB sin
2 
u 
 
117 
Downloaded by Igor Bertoldo (igorbertoldo1001@gmail.com)
lOMoARcPSD|9428255
https://www.studocu.com/en-us?utm_campaign=shared-document&utm_source=studocu-document&utm_medium=social_sharing&utm_content=mechanics-of-materials-si-9th-edition-hibbeler-solutions-manual
© Pearson Education South Asia Pte Ltd 2014. 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. 
 
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. 
200 mm 200 mm 
P D 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 ¿ = 2400
2 + 4002 - 2(400)(400) cos 93° = 580.30 mm 
 
The unstretched length of wire AC is 
 
LAC = 2400
2 + 4002 = 565.69 mm 
Average Normal Strain: 
 
LAC ¿ - LAC 
 
 
580.30 - 565.69
(Pavg)AC = = LAC 565.69 
= 0.0258 mm>mm
 
Ans.
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
Ans: 
(Pavg)AC = 0.0258 mm>mm 
 
118 
Downloaded by Igor Bertoldo (igorbertoldo1001@gmail.com)
lOMoARcPSD|9428255
© Pearson Education South Asia Pte Ltd 2014. 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. 
 
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. 
200 mm 200 mm 
P D 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¿ = 2400
2 + 2002 - 2(400)(200) cos 93° = 456.48 mm 
 
The unstretched length of wire AE is 
 
LAE = 2400
2 
+ 200
2 
= 447.21 mm 
Average Normal Strain: 
 
(Pavg)AE = 
 
 
LAE ¿ - LAE 
LAE 
 
 
456.48 - 447.21 
= 
447.21 
= 0.0207 mm>mm
 
 
 
Ans.
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
Ans: 
(Pavg)AE = 0.0207 mm>mm 
 
119 
Downloaded by Igor Bertoldo (igorbertoldo1001@gmail.com)
lOMoARcPSD|9428255
https://www.studocu.com/en-us?utm_campaign=shared-document&utm_source=studocu-document&utm_medium=social_sharing&utm_content=mechanics-of-materials-si-9th-edition-hibbeler-solutions-manual
© Pearson Education South Asia Pte Ltd 2014. 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. 
 
 
*2–16. The triangular plate ABC is deformed into the y 
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. C 
 
300 mm 
gxy
 
 
A B 
x 
400 mm 
 
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, 
p 
 
 
180°
u = 
2 
- 0.005 = 1.5658 rad a 
p rad 
b = 89.7135°
 
 
The unstretched length of edge BC is 
 
LBC = 2300
2 
+ 400
2 
= 500 mm 
 
and the stretched length of this edge is 
 
LB¿C¿ = 2303
2 + 4032 - 2(303)(403) cos 89.7135° 
 
= 502.9880 mm 
We obtain, 
 
 
PBC = 
 
 
LB¿C¿ - LBC 
LBC 
 
 
502.9880 - 500 
= 
500 
= 5.98(10
 
 
 
 
- 3) mm>mm 
 
 
 
Ans.
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
120 
Downloaded by Igor Bertoldo (igorbertoldo1001@gmail.com)
lOMoARcPSD|9428255
© Pearson Education South Asia Pte Ltd 2014. 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. 
 
 
2–17. The plate is deformed uniformly into the shape y 
shown by the dashed lines. If at A, gxy = 0.0075 rad., while 
y¿
 
PAB = PAF = 0, determine the average shear strain at point 
G with respect to the x¿ and y¿ axes. 
 
 F E 
x¿ 
 
C
600 mm D 
gxy 
 
 
300 mm
 
 180° 
Geometry: Here, gxy= 0.0075 rad a 
p rad 
b = 0.4297°. Thus,
 
A G B 
x 
600 mm 
300 mm
c = 90° - 0.4297° = 89.5703° b = 90° + 0.4297° = 90.4297°
 
Subsequently, applying the cosine law to triangles AGF¿ and GBC¿, Fig. a, 
 
LGF¿ = 2600
2 + 3002 - 2(600)(300) cos 89.5703° = 668.8049 mm 
 
LGC¿ = 2600
2 + 3002 - 2(600)(300) cos 90.4297° = 672.8298 mm 
 
Then, applying the sine law to the same triangles, 
sin f 
600 
=
 
sin 89.5703° 
; 
668.8049 
 
f = 63.7791°
sin a 
300 
=
 
sin 90.4297° 
; 
672.8298 
 
a = 26.4787°
Thus, 
 
u = 180° - f - a = 180° - 63.7791° - 26.4787°
 
p rad 
= 89.7422° a 
180° 
b = 1.5663 rad
 
Shear Strain: 
 
(gG)x¿y¿ = 
 
 
p 
2 
- u =
 
 
 
p 
2 
- 1.5663 = 4.50(10
 
 
 
- 3) rad 
 
 
 
Ans.
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
Ans: 
(gG)x¿y¿ = 4.50(10 
- 3) rad 
Downloaded by Igor Bertoldo (igorbertoldo1001@gmail.com)
lOMoARcPSD|9428255
https://www.studocu.com/en-us?utm_campaign=shared-document&utm_source=studocu-document&utm_medium=social_sharing&utm_content=mechanics-of-materials-si-9th-edition-hibbeler-solutions-manual
 
121 
Downloaded by Igor Bertoldo (igorbertoldo1001@gmail.com)
lOMoARcPSD|9428255
© Pearson Education South Asia Pte Ltd 2014. 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. 
 
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. 
 
 
 
 
2 mm 
y 
 
2 mm 
 
 
C 
 
5 mm 
 
 
B 
 
 
 
 
4 mm
 
300 mm 
 
D A 
400 mm 
2 mm x
3 mm 
 
 
 
 
 
 
 
 
 
 
 
 
 
Geometry: For small angles, 
 
2 
a = c = 
302 
= 0.00662252 rad
 
 
2 
b = u = 
403 
= 0.00496278 rad
 
 
Shear Strain: 
(gB)xy = a + b 
 
= 0.0116 rad = 11.6 A 10 - 3 B rad 
 
(gA)xy = u + c 
 
= 0.0116 rad = 11.6 A 10 - 3 B rad 
 
 
Ans. 
 
 
 
Ans.
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
Ans: 
(gB)xy = 11.6(10 
- 3) rad, 
(gA)xy = 11.6(10 
- 3) rad 
 
122 
Downloaded by Igor Bertoldo (igorbertoldo1001@gmail.com)
lOMoARcPSD|9428255
https://www.studocu.com/en-us?utm_campaign=shared-document&utm_source=studocu-document&utm_medium=social_sharing&utm_content=mechanics-of-materials-si-9th-edition-hibbeler-solutions-manual
© Pearson Education South Asia Pte Ltd 2014. 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. 
 
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. 
 
 
 
 
2 mm 
y 
 
2 mm 
 
 
C 
 
5 mm 
 
 
B 
 
 
 
 
4 mm
 
300 mm 
 
D A 
400 mm 
2 mm x
3 mm 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
Geometry: For small angles, 
 
2 
a = c = 
403 
= 0.00496278 rad
 
2 
b = u = 
302 
= 0.00662252 rad
 
Shear Strain: 
(gC)xy = a + b 
 
= 0.0116 rad = 11.6 A 10 - 3 B rad 
 
(gD)xy = u + c 
 
= 0.0116 rad = 11.6 A 10 - 3 B rad 
 
 
Ans. 
 
 
 
Ans.
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
Ans: 
(gC)xy = 11.6(10 
- 3) rad, 
(gD)xy = 11.6(10 
- 3) rad 
 
123 
Downloaded by Igor Bertoldo (igorbertoldo1001@gmail.com)
lOMoARcPSD|9428255
© Pearson Education South Asia Pte Ltd 2014. 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. 
 
*2–20. The piece of plastic is originally rectangular. 
Determine the average normal strain that occurs along the 
diagonals AC and DB. 
 
 
 
 
2 mm 
y 
 
2 mm 
 
 
C 
 
5 mm 
 
 
B 
 
 
 
 
4 mm
 
300 mm 
 
D A 
400 mm 
2 mm x
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 AC
 
=
 
500.8 - 500 
500
= 0.00160 mm>mm = 1.60 A 10 - 3 B mm>mm Ans.
 
PDB = 
 
DB ¿ -DB DB
 
=
 
 
506.4 - 500 
500
= 0.0128 mm>mm = 12.8 A 10 - 3 B mm>mm Ans.
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
Downloaded by Igor Bertoldo (igorbertoldo1001@gmail.com)
lOMoARcPSD|9428255
https://www.studocu.com/en-us?utm_campaign=shared-document&utm_source=studocu-document&utm_medium=social_sharing&utm_content=mechanics-of-materials-si-9th-edition-hibbeler-solutions-manual
 
124 
Downloaded by Igor Bertoldo (igorbertoldo1001@gmail.com)
lOMoARcPSD|9428255
© Pearson Education South Asia Pte Ltd 2014. 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. 
 
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 
 
 
A B 
400 mm 
5 mm x
 
 
 
Geometry: Referring to Fig. a and using small angle analysis, 
5 
u = 
300 
= 0.01667 rad
 
5 
f = 
400 
= 0.0125 rad
 
Shear Strain: Referring to Fig. a, 
 
(gA)xy = u + f = 0.01667 + 0.0125 = 0.0292 rad 
 
(gB)xy = u + f = 0.01667 + 0.0125 = 0.0292 rad 
 
Ans. 
Ans.
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
Ans: 
(gA)xy = 0.0292 rad, (gB)xy = 0.0292 rad 
 
125 
Downloaded by Igor Bertoldo (igorbertoldo1001@gmail.com)
lOMoARcPSD|9428255
https://www.studocu.com/en-us?utm_campaign=shared-document&utm_source=studocu-document&utm_medium=social_sharing&utm_content=mechanics-of-materials-si-9th-edition-hibbeler-solutions-manual
© Pearson Education South Asia Pte Ltd 2014. 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. 
 
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° 
803.54 
=
 
p 
sin u 
; 
800 
p 
 
u = 44.75° = 0.7810 rad
gxy = 
2 
- 2u =
 2 
- 2(0.7810)
 
= 0.00880 rad 
 
Ans.
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
Ans: 
gxy = 0.00880 rad 
 
126 
Downloaded by Igor Bertoldo (igorbertoldo1001@gmail.com)
lOMoARcPSD|9428255
© Pearson Education South Asia Pte Ltd 2014. 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. 
 
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 
800= 0.00443 mm>mm
 
 
Ans.
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
Ans: 
Px = 0.00443 mm>mm 
 
127 
Downloaded by Igor Bertoldo (igorbertoldo1001@gmail.com)
lOMoARcPSD|9428255
https://www.studocu.com/en-us?utm_campaign=shared-document&utm_source=studocu-document&utm_medium=social_sharing&utm_content=mechanics-of-materials-si-9th-edition-hibbeler-solutions-manual
© Pearson Education South Asia Pte Ltd 2014. 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. 
 
*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 
 
A¿ 
5 mm
 
 
800 mm 
 
 
x 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
L = 800 cos 45° = 565.69 mm 
5 
Px¿ = 
565.69 
= 0.00884 mm>mm
 
 
Ans.
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
128 
Downloaded by Igor Bertoldo (igorbertoldo1001@gmail.com)
lOMoARcPSD|9428255
dy 
© Pearson Education South Asia Pte Ltd 2014. 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. 
 
 
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 D C 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 
400 mm
Here, 
dx 
= tan (gxy)DC = tan 340(10 
- 6)x + 0.0084. Then,
 
dy = 
L0 L0 
 
tan [40(10 - 6)x + 0.008]dx 
 
A B 
x
 
 1 
dc = - 
40(10 - 6) 
 
= 4.2003 mm 
 
e ln cos c 40(10 - 6)x + 0.008 d f ` 
 
300 mm 
 
0 
 
300 mm
 
 
- 6 Along edge AB, y = 0. Thus, (gxy)AB = 40(10 
tan [40(10 - 6)x]. Then, 
)x. Here, 
dx 
= tan (gxy)AB =
dB 
dy = 
L0 
300 mm 
tan [40(10 - 6)x]dx 
L0 
 
1 
 
 
 
 
300 mm
dB = - 
40(10 - 6) 
 
= 1.8000 mm 
e ln cos c 40(10 - 6)x d f ` 
0
 
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 
LBC 
 
 
402.4003 - 400 
= 
400 
= 6.00(10
 
 
 
 
- 3) mm>mm 
 
 
 
Ans.
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
Downloaded by Igor Bertoldo (igorbertoldo1001@gmail.com)
lOMoARcPSD|9428255
https://www.studocu.com/en-us?utm_campaign=shared-document&utm_source=studocu-document&utm_medium=social_sharing&utm_content=mechanics-of-materials-si-9th-edition-hibbeler-solutions-manual
Ans: 
(Pavg)BC = 6.00(10 
- 3) mm>mm 
 
129 
Downloaded by Igor Bertoldo (igorbertoldo1001@gmail.com)
lOMoARcPSD|9428255
© Pearson Education South Asia Pte Ltd 2014. 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. 
 
 
 
2– 26. 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
Ans: 
1Pavg 2CA = 
 
 
- 5.59110-32 mm > mm
 
 
 
130 
Downloaded by Igor Bertoldo (igorbertoldo1001@gmail.com)
lOMoARcPSD|9428255
https://www.studocu.com/en-us?utm_campaign=shared-document&utm_source=studocu-document&utm_medium=social_sharing&utm_content=mechanics-of-materials-si-9th-edition-hibbeler-solutions-manual
© Pearson Education South Asia Pte Ltd 2014. 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. 
 
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, 
y 
y¿ x¿ 
 
600 mm
gxy = 0.02 rad, determine the shear strain at point E with 
respect to the x¿ and y¿ axes. 
 A¿ B¿ 
A B
 
 
600 mm E 
 
 
x 
D C C ¿ 
 
 
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, 
 
p 
 
 
180°
a = 
2 
- 0.02 = 1.5508 rad a 
p rad 
b = 88.854°
 
 
p 180° 
f = 
2 
+ 0.02 = 1.5908 rad a 
p rad 
b = 91.146°
 
 
Thus, the length of C¿A¿ and DB¿ can be determined using the cosine law with 
reference to Fig. a. 
 
LC¿A¿ = 2602.4
2 + 6032 - 2(602.4)(603) cos 88.854° = 843.7807 mm 
 
LDB¿ = 2602.4
2 + 6032 - 2(602.4)(603) cos 91.146° = 860.8273 mm 
Thus, 
 
LE¿A¿ = 
 
 
LC¿A¿ 
2 
= 421.8903 mm
 
 
 
LE¿B¿ = 
 
 
LDB¿ 
2 
= 430.4137 mm
 
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 
 
p rad 
u = 89.9429° a 
180° 
b = 1.5698 rad
 
Shear Strain: 
 
(gE)x¿y¿ = 
 
 
p 
2 
- u =
 
 
 
p 
2 
- 1.5698 = 0.996(10
 
 
 
 
- 3) rad 
 
 
 
Ans.
 
 
 
 
 
 
 
 
 
 
 
 
Ans: 
(gE)x¿y¿ = 0.996(10 
- 3) rad 
 
131 
Downloaded by Igor Bertoldo (igorbertoldo1001@gmail.com)
lOMoARcPSD|9428255
0 
2
L 
= 
© Pearson Education South Asia Pte Ltd 2014. 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. 
 
*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 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 1 
¢L = 
L L 
xe
 
 
- (x>L)2
dx
 
= - L c 
e - (x>L) L 
2 
d 
0 
L 
2 
31 - (1>e)4
L 
= 
2e 
3e - 14
 
 
Ans.
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
Downloaded by Igor Bertoldo (igorbertoldo1001@gmail.com)
lOMoARcPSD|9428255
https://www.studocu.com/en-us?utm_campaign=shared-document&utm_source=studocu-document&utm_medium=social_sharing&utm_content=mechanics-of-materials-si-9th-edition-hibbeler-solutions-manual
132 
Downloaded by Igor Bertoldo (igorbertoldo1001@gmail.com)
lOMoARcPSD|9428255
© Pearson Education South Asia Pte Ltd 2014. 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. 
 
 
 
2–29. 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
Ans: 
1Pavg 2AC = 0.0168 mm >mm, 
1gA 2xy = 0.0116 rad 
 
 
133 
Downloadedby Igor Bertoldo (igorbertoldo1001@gmail.com)
lOMoARcPSD|9428255
https://www.studocu.com/en-us?utm_campaign=shared-document&utm_source=studocu-document&utm_medium=social_sharing&utm_content=mechanics-of-materials-si-9th-edition-hibbeler-solutions-manual
© Pearson Education South Asia Pte Ltd 2014. 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. 
 
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 
 
 
 
2 mm 2 mm 
 
 
 
400 mm 
 
 
6 mm 
 
 
 
 
 
6 mm
D C 
 
 
300 mm 
 
 
 
Geometry: The unstretched length of diagonal BD is 
 
LBD = 2300
2 
+ 400
2 
= 500 mm 
 
 
A B 
400 mm 
2 mm 
x 
 
3 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 
f = 
403 
= 0.004963 rad
 
3 
a = 
300 + 6 - 2 
= 0.009868 rad
 
 
Average Normal Strain: Applying Eq. 2,
 
(Pavg)BD = 
 
LB¿D¿ - LBD 
LBD 
 
500.8004 - 500 
= 
500 
= 1.60(10
 
 
- 3) mm>mm 
 
Ans.
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 
Downloaded by Igor Bertoldo (igorbertoldo1001@gmail.com)
lOMoARcPSD|9428255
L 
© Pearson Education South Asia Pte Ltd 2014. 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. 
 
 
2–31. The nonuniform loading causes a normal strain in 
the shaft that can be expressed as Px = kx
2, where k is a 
constant. Determine the displacement of the end B. Also, 
A B 
what is the average normal strain in the rod? 
x 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
d(¢x) 
2
 
dx 
= Px = kx
 
 
(¢x)B = 
 
L 
kx
2
 
L0 
 
kL
3
 
= 
3 
 
kL3 
 
 
Ans.
 
(Px)avg = 
(¢x)B 
L 
=
 
3 kL2 
L 
= 
3 
 
Ans.
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
Ans: 
 
(¢x)B = 
 
 
kL
3
 
, (Px)avg = 3 
 
 
kL2 
3
 
135 
Downloaded by Igor Bertoldo (igorbertoldo1001@gmail.com)
lOMoARcPSD|9428255
https://www.studocu.com/en-us?utm_campaign=shared-document&utm_source=studocu-document&utm_medium=social_sharing&utm_content=mechanics-of-materials-si-9th-edition-hibbeler-solutions-manual
b 3 
3 
b 
© Pearson Education South Asia Pte Ltd 2014. 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. 
 
*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>b
3)x3. Determine the 
shear strain gxy at points (b>2, a>2) and (b, a). 
y 
 
 
v (x) 
v0 
A D 
 
 
 
a 
 
 
x 
B C 
Shear Strain: From Fig. a, 
b 
dv 
dx 
= tan gxy
 
 
3v0 2
b3 
x
 
= tan gxy 
 
3v0
gxy = tan 
- 1 a x2 
b 
 
Thus, at point (b>2, a>2), 
 
3v0 b 
2
gxy = tan 
- 1 c 
b 
a 
2 
b d
 
= tan - 1 c 
3 
a 
v0 
b d 
 
Ans.
4 b 
and at point (b, a), 
 
gxy = tan 
- 1 c 
 
 
3v0 
b3 
 
 
(b2) d
 
v0 
= tan - 1 c 3 a b d 
 
Ans.
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
136 
Downloaded by Igor Bertoldo (igorbertoldo1001@gmail.com)
lOMoARcPSD|9428255
LA B A 
2
 
1
© Pearson Education South Asia Pte Ltd 2014. 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. 
 
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 
u 
x 
A uA A¿
 
 
 
Geometry: 
¿ ¿ = 2(L cos u - u ) + (L sin u + vB)
2
 
2 2
 
= 2L
2 + uA + vB + 2L(vB sin u - uA cos u) 
Average Normal Strain: 
LA¿B¿ - L 
PAB = 
L 
 
2 2
= 
A 
1 + 
uA + v B 
L2 
+
 
2(vB sin u - uA cos u) - 1 
L
 
Neglecting higher terms u
2 
and v
2 
A B 
 
PAB = B 1 + 
2(vB sin u - uA cos u) 2 
L 
R - 1
 
Using the binomial theorem: 
1 
PAB = 1 + 
2 
¢
 
2vB sin u 
L 
-
 
2uA cos u 
L 
≤ + . . . - 1
vB sin u 
= 
L 
-
 
uA cos u 
L 
 
Ans.
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
Ans. 
PAB = 
 
vB sin u 
L 
-
 
 
uA cos u 
L
 
137 
Downloaded by Igor Bertoldo (igorbertoldo1001@gmail.com)
lOMoARcPSD|9428255
https://www.studocu.com/en-us?utm_campaign=shared-document&utm_source=studocu-document&utm_medium=social_sharing&utm_content=mechanics-of-materials-si-9th-edition-hibbeler-solutions-manual
n 
A 
B 
© Pearson Education South Asia Pte Ltd 2014. 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. 
 
 
2–34. If the normal strain is defined in reference to the 
final length, that is,
 
P¿ = lim 
p : p¿ 
 
¢s¿ -¢s 
a 
¢s¿ 
b
 
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
 
PB - P
œ = 
¢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 P
œ 
(Q.E.D) 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
Downloaded by Igor Bertoldo (igorbertoldo1001@gmail.com)
lOMoARcPSD|9428255
138 
Mechanics of Materials SI 9th Edition Hibbeler SOLUTIONS 
MANUAL 
Full clear download at: 
https://testbankreal.com/download/mechanics-materials-si-9th-edition-
hibbeler-solutions-manual/ 
mechanics of materials 9th edition si pdf 
mechanics of materials 9th edition hibbeler pdf download 
hibbeler mechanics of materials pdf 
hibbeler mechanics of materials 9th edition solutions 
mechanics of materials 9th edition pdf free 
hibbeler mechanics of materials 10th edition pdf 
mechanics of materials rc hibbeler 9th edition pdf free download 
hibbeler mechanics of materials free pdf 
 
Downloaded by Igor Bertoldo (igorbertoldo1001@gmail.com)
lOMoARcPSD|9428255
https://testbankreal.com/download/mechanics-materials-si-9th-edition-hibbeler-solutions-manual/
https://testbankreal.com/download/mechanics-materials-si-9th-edition-hibbeler-solutions-manual/
https://www.google.com.vn/search?safe=active&q=mechanics+of+materials+9th+edition+si+pdf&sa=X&ved=0ahUKEwiJ0afcyOLXAhUKppQKHVdKCHkQ1QIIXigAhttps://www.google.com.vn/search?safe=active&q=mechanics+of+materials+9th+edition+hibbeler+pdf+download&sa=X&ved=0ahUKEwiJ0afcyOLXAhUKppQKHVdKCHkQ1QIIXygB
https://www.google.com.vn/search?safe=active&q=hibbeler+mechanics+of+materials+pdf&sa=X&ved=0ahUKEwiJ0afcyOLXAhUKppQKHVdKCHkQ1QIIYCgC
https://www.google.com.vn/search?safe=active&q=hibbeler+mechanics+of+materials+9th+edition+solutions&sa=X&ved=0ahUKEwiJ0afcyOLXAhUKppQKHVdKCHkQ1QIIYSgD
https://www.google.com.vn/search?safe=active&q=mechanics+of+materials+9th+edition+pdf+free&sa=X&ved=0ahUKEwiJ0afcyOLXAhUKppQKHVdKCHkQ1QIIYigE
https://www.google.com.vn/search?safe=active&q=hibbeler+mechanics+of+materials+10th+edition+pdf&sa=X&ved=0ahUKEwiJ0afcyOLXAhUKppQKHVdKCHkQ1QIIYygF
https://www.google.com.vn/search?safe=active&q=mechanics+of+materials+rc+hibbeler+9th+edition+pdf+free+download&sa=X&ved=0ahUKEwiJ0afcyOLXAhUKppQKHVdKCHkQ1QIIZCgG
https://www.google.com.vn/search?safe=active&q=hibbeler+mechanics+of+materials+free+pdf&sa=X&ved=0ahUKEwiJ0afcyOLXAhUKppQKHVdKCHkQ1QIIZSgH
https://www.studocu.com/en-us?utm_campaign=shared-document&utm_source=studocu-document&utm_medium=social_sharing&utm_content=mechanics-of-materials-si-9th-edition-hibbeler-solutions-manual