Solutions manual for mechanics of materials si 9th edition by hibbeler ibsn 9789810694364

Solutions manual for mechanics of materials si 9th edition by hibbeler ibsn 9789810694364
download: https://goo.gl/iqN3kb
People also search:
mechanics of materials 9th edition si pdf
hibbeler mechanics of materials 9th edition pdf
hibbeler mechanics of materials 10th edition pdf
hibbeler mechanics of materials 9th edition solutions
hibbeler mechanics of materials pdf
mechanics of materials rc hibbeler 9th edition pdf free download
mechanics of materials hibbeler 9th edition
mechanics of materials 9th edition solutions download

1. SOLUTIONS MANUAL for Mechanics of Materials SI 9th Edition by Hibbeler
IBSN 9789810694364
Download: https://downloadlink.org/p/solutions-manual-for-mechanics-of-materials-
si-9th-edition-by-hibbeler-ibsn-9789810694364/











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: d
0
:= 100mm

Solution:
d := 125mm


ε :=

π d − π d
0

π d
0
mm
ε = 0.2500
mm

Ans

2. Ans:

CE 0.00250 mm mm,
BD 0.00107 mm mm


105

3. 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.

L
0 = 15 in.

L = p(5 in.)

L - L
0 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

106

4. © 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













¢L
BD

3
=

¢ L
CE

7

¢L
BD =
3 (10)
7
= 4.286 mm
¢ LCE

10
P
CE =
L
=
4000
= 0.00250 mm>mm

Ans.

P
BD =

¢ L
BD

L
=


4.286
4000
= 0.00107 mm>mm


Ans.


























Ans:
P
CE = 0.00250 mm>mm, P
BD = 0.00107 mm>mm

107

5. © 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

d
A = 200(0.03491) = 6.9813 mm

d
C = 300(0.03491) = 10.4720 mm

d
D = 500(0.03491) = 17.4533 mm

Average Normal Strain: The unstretched length of wires AH, CG, and DF are

L
AH = 200 mm, L
CG = 300 mm, and L
DF = 300 mm. We obtain

(P
avg)
AH =

(P
avg)
CG =


(P
avg)
DF =
d
A

=
L
AH
d
C

=
L
CG

d
D

=
L
DF
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

6. © 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.


2–5. 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













L
œ 2 2
AC = 2300 + 2

L
œ

- 2(300)(2) cos 150° = 301.734 mm

P
AC = P
AB =
AC - LAC
L
AC
301.734 - 300
=
300
= 0.00578 mm>mm


Ans.

































Ans:
P
AC = P
AB = 0.00578 mm>mm

109

7. 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 r
0
normal strain in the band as a function of z.



z



2r
0
h


Geometry: Using similar triangles shown in Fig. a,
h¿ h¿ +h
= ;
0 2r
0

h¿ = h

Subsequently, using the result of h¿
r r
0

z + h
=
h
;

r
0

r = (z + h)
h

Average Normal Strain: The length of the rubber band as a function of z is
2pr
0
L = 2pr = (z + h). With L
0 = 2r
0, we have
h

2pr0

P
avg =
L - L
0

L
0
h
(z + h) -
2r
0

= =
2r
0

p
(z + h) - 1
h

Ans.


































Ans:

P
avg =

p
(z + h) - 1
h

110

8. © 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

L
AC = 2(600 sin 30°) = 600 mm

L
AC ¿ = 2(600 sin 30.2°) = 603.6239 mm
Average Normal Strain:

LAC ¿ - LAC



603.6239 - 600
- 3
(P
avg)
AC = =
L
AC 600
= 6.04(10

) mm>mm Ans.




































Ans:
(P
avg)
AC = 6.04(10
- 3
) mm>mm

111

9. 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 = 2400
2
+ 300
2
= 500 mm

AB¿ = 2400
2
+ 300
2
- 2(400)(300) cos 90.3°
= 501.255 mm
AB¿ -AB


501.255 - 500
P
AB =
AB
=
500
= 0.00251 mm>mm Ans.



































112

10. © 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 = 2300
2
+ 400
2
= 500 mm

AB¿ = AB + e
ABAB

= 500 + 0.0035(500) = 501.75 mm

501.75
2
= 300
2
+ 400
2
- 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

11. © 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

114

12. © 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,

L
AB = 216
2
+ 16
2
= 2512 mm
L
AB¿ = 216
2
+ 13
2
= 2425 mm
L
BD = 16 + 16 = 32 mm
L
B¿D¿ = 13 + 13 = 26 mm

Thus,

A e
avg B
AB =
L
AB¿ - L
AB
L
AB

LB¿D¿ - LBD

2425 - 2512
=
2512

26 - 32


= - 0.0889 mm>mm

Ans.
A e
avg B BD =
L
BD
=
32
= - 0.1875 mm>mm

Ans.

































115

13. © 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

14. © 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

15. © 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 L
AC ¿ of wire AC¿ can be
determined using the cosine law.

L
AC ¿ = 2400
2
+ 400
2
- 2(400)(400) cos 93° = 580.30 mm

The unstretched length of wire AC is

LAC = 2400
2
+ 400
2
= 565.69 mm
Average Normal Strain:

LAC ¿ - LAC



580.30 - 565.69
(P
avg)
AC = =
L
AC 565.69
= 0.0258 mm>mm

Ans.


































Ans:
(P
avg)
AC = 0.0258 mm>mm

118

16. © 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 L
AE¿ of wire AE can be
determined using the cosine law.

L
AE¿ = 2400
2
+ 200
2
- 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:

(P
avg)
AE =


L
AE ¿ - L
AE
LAE


456.48 - 447.21
=
447.21
= 0.0207 mm>mm



Ans.

































Ans:
(P
avg)
AE = 0.0207 mm>mm

119

17. © 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, e
AB = 0.0075,
P
AC = 0.01 and g
xy = 0.005 rad, determine the average
normal strain along edge BC.
C


300 mm
g
xy


A B
x

400 mm

Average Normal Strain: The stretched length of sides AB and AC are

L
AC¿ = (1 + e
y)L
AC = (1 + 0.01)(300) = 303 mm

L
AB¿ = (1 + e
x)L
AB = (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

L
B¿C¿ = 2303
2
+ 403
2
- 2(303)(403) cos 89.7135°

= 502.9880 mm
We obtain,


P
BC =


L
B¿C¿ - L
BC
L
BC


502.9880 - 500
=
500
= 5.98(10




- 3
) mm>mm



Ans.






























120

18. © 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
g
xy


300 mm

180°
Geometry: Here, g
xy = 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,

L
GF¿ = 2600
2
+ 300
2
- 2(600)(300) cos 89.5703° = 668.8049 mm

L
GC¿ = 2600
2
+ 300
2
- 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:

(g
G)
x¿y¿ =


p
2
- u =



p
2
- 1.5663 = 4.50(10



- 3
) rad



Ans.






















Ans:
(g
G)
x¿y¿ = 4.50(10
- 3
) rad

121

19. © 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 g
xy 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:
(g
B)
xy = a + b

= 0.0116 rad = 11.6 A 10
- 3
B rad

(g
A)
xy = u + c

= 0.0116 rad = 11.6 A 10
- 3
B rad


Ans.



Ans.





















Ans:
(g
B)
xy = 11.6(10
- 3
) rad,
(g
A)
xy = 11.6(10
- 3
) rad

122

20. © 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 g
xy 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:
(g
C)
xy = a + b

= 0.0116 rad = 11.6 A 10
- 3
B rad

(g
D)
xy = u + c

= 0.0116 rad = 11.6 A 10
- 3
B rad


Ans.



Ans.





















Ans:
(g
C)
xy = 11.6(10
- 3
) rad,
(g
D)
xy = 11.6(10
- 3
) rad

123

21. © 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 = 2400
2
+ 300
2
= 500 mm
DB¿ = 2405
2
+ 304
2
= 506.4 mm
A¿C¿ = 2401
2
+ 300
2
= 500.8 mm

Average Normal Strain:

P
AC =

A¿C¿ -AC AC

=

500.8 - 500
500
= 0.00160 mm>mm = 1.60 A 10
- 3
B mm>mm Ans.

P
DB =

DB ¿ -DB DB

=


506.4 - 500
500
= 0.0128 mm>mm = 12.8 A 10
- 3
B mm>mm Ans.

























124

22. © 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 g
xy 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,

(g
A)
xy = u + f = 0.01667 + 0.0125 = 0.0292 rad

(g
B)
xy = u + f = 0.01667 + 0.0125 = 0.0292 rad

Ans.
Ans.




































Ans:
(g
A)
xy = 0.0292 rad, (g
B)
xy = 0.0292 rad

125

23. © 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, g
xy, at A.
y



45




800 mm

45
x¿
A

45

A¿
5 mm


800 mm


x
















L = 2800
2
+ 5
2
- 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:
g
xy = 0.00880 rad

126

24. © 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 P
x along the x axis.
y



45




800 mm

45
x¿
A

45

A¿
5 mm


800 mm


x


















L = 2800
2
+ 5
2
- 2(800)(5) cos 135° = 803.54 mm

P
x =
803.54 - 800
800
= 0.00443 mm>mm


Ans.




























Ans:
P
x = 0.00443 mm>mm

127

25. © 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 P
x¿ 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
P
x¿ =
565.69
= 0.00884 mm>mm


Ans.
































128

26. 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 g
xy = 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, (g
xy)
DC = 40(10
- 6
)x + 0.008.
dy

400 mm
Here,
dx
= tan (g
xy)
DC = tan 340(10
- 6
)x + 0.0084. Then,

dy =
L0 L0

tan [40(10
- 6
)x + 0.008]dx

A B
x

1
d
c = -
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 (g
xy)
AB =
d
B

dy =
L0
300 mm
tan [40(10
- 6
)x]dx
L0

1





300 mm
d
B = -
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

L
B¿C¿ = 400 + 4.2003 - 1.8000 = 402.4003 mm
We obtain,


(P
avg)
BC =


L
B¿C¿ - L
BC
LBC


402.4003 - 400
=
400
= 6.00(10




- 3
) mm>mm



Ans.






















Ans:
(P
avg)
BC = 6.00(10
- 3
) mm>mm

129

27. © 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:
1P
avg 2
CA =


- 5.59110
-3
2 mm > mm



130

28. © 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
P
x = 0.004, DA has a normal strain P
y = 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

L
DC¿ = (1 + P
x)L
DC = (1 + 0.004)(600) = 602.4 mm

L
B¿C¿ = (1 + P
y)L
BC = (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.

L
C¿A¿ = 2602.4
2
+ 603
2
- 2(602.4)(603) cos 88.854° = 843.7807 mm

L
DB¿ = 2602.4
2
+ 603
2
- 2(602.4)(603) cos 91.146° = 860.8273 mm
Thus,

L
E¿A¿ =


L
C¿A¿

2
= 421.8903 mm



L
E¿B¿ =


L
DB¿

2
= 430.4137 mm

Using this result and applying the cosine law to the triangle A¿E¿B¿, Fig. a,

602.4
2
= 421.8903
2
+ 430.4137
2
- 2(421.8903)(430.4137) cos u

p rad
u = 89.9429° a
180°
b = 1.5698 rad

Shear Strain:

(g
E)
x¿y¿ =


p
2
- u =



p
2
- 1.5698 = 0.996(10




- 3
) rad



Ans.












Ans:
(g
E)
x¿y¿ = 0.996(10
- 3
) rad

131

29. 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.





























132

30. © 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:
1P
avg 2
AC = 0.0168 mm >mm,
1gA 2xy = 0.0116 rad


133

31. © 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

L
B¿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,

(P
avg)
BD =

L
B¿D¿ - L
BD
L
BD

500.8004 - 500
=
500
= 1.60(10


- 3
) mm>mm

Ans.
Shear Strain: Referring to Fig. a,

(g
B)
xy = f + a = 0.004963 + 0.009868 = 0.0148 rad


Ans.

























Ans:
(P
avg)
BD = 1.60(10
- 3
) mm>mm,
(g
B)
xy = 0.0148 rad

134

32. 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 P
x = 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
= P
x = kx


(¢x)
B =

L
kx
2

L0

kL
3

=
3

kL
3



Ans.

(P
x)
avg =
(¢x)
B

L
=

3 kL
2

L
=
3

Ans.






















Ans:

(¢x)
B =


kL
3

, (P
x)
avg =
3


kL
2

3

135

33. 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) = (v
0>b
3
)x
3
. Determine the
shear strain g
xy at points (b>2, a>2) and (b, a).
y


v (x)
v
0

A D



a


x
B C
Shear Strain: From Fig. a,
b
dv
dx
= tan g
xy


3v0
2
b
3
x

= tan gxy

3v
0
g
xy = tan
- 1
a x
2

b

Thus, at point (b>2, a>2),

3v
0 b
2
g
xy = tan
- 1
c
b
a
2
b d

= tan
- 1
c
3
a
v0
bd


Ans.
4 b
and at point (b, a),

g
xy = tan
- 1
c


3v
0

b
3



(b
2
) d

v
0

= tan
- 1
c 3 a bd

Ans.






























136

34. 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 u
A and v
B,
respectively, determine the normal strain in the fiber when
it is in position A¿B¿.
y

B¿
v
B
B

L
u
x
A u
A A¿



Geometry:
¿ ¿ = 2(L cos u - u ) + (L sin u + v
B)
2

2 2

= 2L
2
+ u
A + v
B + 2L(v
B sin u - u
A cos u)
Average Normal Strain:
L
A¿B¿ - L
P
AB =
L

2 2
=
A
1 +
u
A + v
B

L
2
+

2(v
B sin u - u
A cos u)
- 1
L

Neglecting higher terms u
2
and v
2
A B

P
AB = B 1 +
2(v
B sin u - u
A cos u) 2
L
R - 1

Using the binomial theorem:
1
P
AB = 1 +
2
¢

2v
B sin u
L
-

2u
A cos u
L
≤ + . . . - 1
v
B sin u
=
L
-

u
A cos u
L

Ans.
























Ans.
P
AB =

v
B sin u
L
-


u
A cos u
L

137

35. 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, P
n - P
n¿ = P
n P
n¿ .



P
B =
¢ S ¿ -¢S
¢S

P
B - P
œ
=
¢S ¿ -¢S
¢S
-

¢ S ¿ -¢S
¢S¿
¢ S ¿
2
-¢S¢S ¿ -¢S¿¢S +¢S
2

=
¢S¢S¿
¢ S ¿
2
+¢S
2
- 2¢S¿¢S
=
¢S¢S¿

(¢ S ¿ -¢S)
2

=
¢S¢S¿

¢ S¿ -¢S
= ¢
¢S

¢ S¿ -¢S
≤¢
¢S¿
≤

= PA P
œ
(Q.E.D)










































138

36. SOLUTIONS MANUAL for Mechanics of Materials SI 9th Edition by
Hibbeler IBSN 9789810694364
Download: https://downloadlink.org/p/solutions-manual-for-mechanics-of-
materials-si-9th-edition-by-hibbeler-ibsn-9789810694364/

People also search:
mechanics of materials 9th edition si pdf
hibbeler mechanics of materials 9th edition pdf
hibbeler mechanics of materials 10th edition pdf
hibbeler mechanics of materials 9th edition solutions
hibbeler mechanics of materials pdf
mechanics of materials rc hibbeler 9th edition pdf free download
mechanics of materials hibbeler 9th edition
mechanics of materials 9th edition solutions download