Ans: Full file at ... · Ans: CE 0.00250 mm mm, BD 0.00107 mm mm 105 Full file at

Ans:

CE 0.00250 mm mm, BD 0.00107 mm mm

105

Full file at https://testbank123.eu/Solutions-Manual-for-Mechanics-of-Materials-SI-9th-Edition-Hibbeler


https://testbank123.eu/Solutions-Manual-for-Mechanics-of-Materials-SI-9th-Edition-Hibbeler

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





106






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





107






*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






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





109





r


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





110






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





111





P


*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.





112






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°

pu = 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






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






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 = 2162 + 162 = 2512 mm

LAB¿ = 2162 + 132 = 2425 mm

LBD = 16 + 16 = 32 mm

LB¿D¿ = 13 + 13 = 26 mm

Thus,

A eavg B AB = LAB¿ - LAB

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






2–12

116






2–13 .

Ans:

PDB = PAB cos2 u + PCB sin2 u

117






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 ¿ = 24002 + 4002 - 2(400)(400) cos 93° = 580.30 mm

The unstretched length of wire AC is

LAC = 24002 + 4002 = 565.69 mm


LAC ¿ - LAC

580.30 - 565.69

(Pavg)AC = = LAC 565.69

= 0.0258 mm>mm

Ans.

Ans:

(Pavg)AC = 0.0258 mm>mm

118






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¿ = 24002 + 2002 - 2(400)(200) cos 93° = 456.48 mm

The unstretched length of wire AE is

LAE = 24002 + 2002 = 447.21 mm


(Pavg)AE =

LAE ¿ - LAE

LAE

456.48 - 447.21

= 447.21

= 0.0207 mm>mm

Ans.

Ans:

(Pavg)AE = 0.0207 mm>mm

119






*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 = 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

LBC

502.9880 - 500

= 500

= 5.98(10

- 3) mm>mm

Ans.

120






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¿

C600 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¿ = 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

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





121






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






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






*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


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.





124






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






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






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






*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





dy


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.





Ans:

(Pavg)BC = 6.00(10 - 3) mm>mm

129






2– 26.

Ans:

1Pavg 2CA =

- 5.59110-32 mm > mm

130






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¿ = 2602.42 + 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¿

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





0

2

L

=


*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






2–29.

Ans:

1Pavg 2AC = 0.0168 mm >mm,

1gA 2xy = 0.0116 rad

133






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 = 23002 + 4002 = 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





L


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, A B

what is the average normal strain in the rod?

x

d(¢x) 2

dx = Px = kx

(¢x)B =

L

kx2

L0

kL3

= 3

kL3

Ans.

(Px)avg = (¢x)B

L =

3 kL2

L =

3

Ans.

Ans:

(¢x)B =

kL3

, (Px)avg = 3

kL2

3

135





b 3

3

b


*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)

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 Full file at https://testbank123.eu/Solutions-Manual-for-Mechanics-of-Materials-SI-9th-Edition-Hibbeler




LA B A 2

1


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

= 2L2 + uA + vB + 2L(vB sin u - uA cos u)


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 u2 and v2 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





Documents

Ans: Full file at ... · Ans: CE 0.00250 mm mm, BD 0.00107 mm mm 105 Full file at