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2โ1. An air-filled rubber ball has a diameter of 6 in. If the air
pressure within it is increased until the ballโs 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.
2โ2. 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.
2โ3. 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
ยขLBD
ยขLCE
=
3
7
3 (10)
= 4.286 mm
7
ยขLCE
10
=
= 0.00250 mm>mm
PCE =
L
4000
ยขLBD =
PBD =
Ans.
ยขLBD
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.
*2โ4. The force applied at the handle of the rigid lever
causes the lever to rotate clockwise about the pin B through
an angle of 2ยฐ. 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
=
(Pavg)CG =
LCG
300
(Pavg)AH =
(Pavg)DF =
Ans.
Ans.
dD
17.4533
=
= 0.0582 mm>mm
LDF
300
Ans.
108
E
C
200 mm
2ยฐ
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.
2โ5. 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
30โฌ
30โฌ
300
A
P
mm
B
ล
LAC
= 23002 + 22 – 2(300)(2) cos 150ยฐ = 301.734 mm
PAC = PAB =
ล
– 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.
2โ6. 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,
hยฟ
hยฟ + h
=
;
r0
2r0
hยฟ = h
Subsequently, using the result of hยฟ
r0
r
= ;
z+h
h
r =
r0
(z + h)
h
Average Normal Strain: The length of the rubber band as a function of z is
2pr0
(z +h). With L0 = 2r0, we have
L = 2pr =
h
L – L0
Pavg =
=
L0
2pr0
(z + h) – 2r0
h
p
= (z + h) – 1
2r0
h
Ans.
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.
2โ7. The pin-connected rigid rods AB and BC are inclined
at u = 30ยฐ when they are unloaded. When the force P is
applied u becomes 30.2ยฐ. 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 30ยฐ) = 600 mm
LAC ยฟ = 2(600 sin 30.2ยฐ) = 603.6239 mm
Average Normal Strain:
(Pavg)AC =
LAC ยฟ – 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
*2โ8. 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 = 0.3ยฐ, 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
ABยฟ = 24002 + 3002 – 2(400)(300) cos 90.3ยฐ
= 501.255 mm
PAB =
ABยฟ – 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.
2โ9. 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
ABยฟ = AB + eABAB
= 500 + 0.0035(500) = 501.75 mm
501.752 = 3002 + 4002 – 2(300)(400) cos a
a = 90.4185ยฐ
u = 90.4185ยฐ – 90ยฐ = 0.4185ยฐ =
ยข D = 600(u) = 600(
p
(0.4185) rad
180ยฐ
p
)(0.4185) = 4.38 mm
180ยฐ
Ans.
Ans:
ยข 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.
2โ10. 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
uยฟ
= tan – 1 a
b = 43.561ยฐ
2
10.2
uยฟ = 1.52056 rad
(gA)nt =
p
– 1.52056
2
= 0.0502 rad
Ans.
At B:
fยฟ
10.2
= tan – 1 a
b = 46.439ยฐ
2
9.7
fยฟ = 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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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
2โ11. 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:
AยฟBยฟ = 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.
*2โ12. 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.
2โ13. 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
ADยฟ = 2(400)2 + (3)2 = 400.01125 mm
f = tan – 1 a
3
b = 0.42971ยฐ
400
ABยฟ = 2(300)2 + (2)2 = 300.00667
w = tan – 1 a
2
b = 0.381966ยฐ
300
a = 90ยฐ – 0.42971ยฐ – 0.381966ยฐ = 89.18832ยฐ
DยฟBยฟ = 2(400.01125)2 + (300.00667)2 – 2(400.01125)(300.00667) cos (89.18832ยฐ)
DยฟBยฟ = 496.6014 mm
DB = 2(300)2 + (400)2 = 500 mm
496.6014 – 500
= -0.00680 mm>mm
500
400.01125 – 400
PAD =
= 0.0281(10 – 3) mm>mm
400
Ans.
PDB =
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.
2โ14. 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 ยฟ of wire ACยฟ can be
determined using the cosine law.
LAC ยฟ = 24002 + 4002 – 2(400)(400) cos 93ยฐ = 580.30 mm
The unstretched length of wire AC is
LAC = 24002 + 4002 = 565.69 mm
Average Normal Strain:
(Pavg)AC =
LAC ยฟ – 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.
2โ15. 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 LAEยฟ of wire AE can be
determined using the cosine law.
LAEยฟ = 24002 + 2002 – 2(400)(200) cos 93ยฐ = 456.48 mm
The unstretched length of wire AE is
LAE = 24002 +2002 = 447.21 mm
Average Normal Strain:
(Pavg)AE =
LAE ยฟ – 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.
*2โ16. 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
LACยฟ = (1 + ey)LAC = (1 + 0.01)(300) = 303 mm
LABยฟ = (1 + ex)LAB = (1 + 0.0075)(400) = 403 mm
Also,
u =
p
180ยฐ
– 0.005 = 1.5658 rada
b = 89.7135ยฐ
2
p rad
The unstretched length of edge BC is
LBC = 23002 + 4002 = 500 mm
and the stretched length of this edge is
LBยฟCยฟ = 23032 + 4032 – 2(303)(403) cos 89.7135ยฐ
= 502.9880 mm
We obtain,
PBC =
LBยฟCยฟ – 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.
2โ17. 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 xยฟ and yยฟ axes.
yยฟ
y
F
E
xยฟ
600 mm
gxy
C
D
300 mm
c = 90ยฐ – 0.4297ยฐ = 89.5703ยฐ
G
A
180ยฐ
Geometry: Here, gxy = 0.0075 rad a
b = 0.4297ยฐ. Thus,
p rad
300 mm
600 mm
B
b = 90ยฐ + 0.4297ยฐ = 90.4297ยฐ
Subsequently, applying the cosine law to triangles AGFยฟ and GBCยฟ, Fig. a,
LGFยฟ = 26002 + 3002 – 2(600)(300) cos 89.5703ยฐ = 668.8049 mm
LGCยฟ = 26002 + 3002 – 2(600)(300) cos 90.4297ยฐ = 672.8298 mm
Then, applying the sine law to the same triangles,
sin f
sin 89.5703ยฐ
=
;
600
668.8049
f = 63.7791ยฐ
sin 90.4297ยฐ
sin a
=
;
300
672.8298
a = 26.4787ยฐ
Thus,
u = 180ยฐ – f – a = 180ยฐ – 63.7791ยฐ – 26.4787ยฐ
= 89.7422ยฐa
p rad
b = 1.5663 rad
180ยฐ
Shear Strain:
(gG)xยฟyยฟ =
p
p
– u =
– 1.5663 = 4.50(10 – 3) rad
2
2
Ans.
Ans:
(gG)xยฟyยฟ = 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.
2โ18. 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.
2โ19. 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.
*2โ20. 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
DBยฟ = 24052 + 3042 = 506.4 mm
AยฟCยฟ = 24012 + 3002 = 500.8 mm
Average Normal Strain:
PAC =
AยฟCยฟ – AC
500.8 – 500
=
AC
500
= 0.00160 mm>mm = 1.60 A 10 – 3 B mm>mm
PDB =
Ans.
DBยฟ – 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.
2โ21. 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.
2โ22. 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
45โฌ
800 mm
45โฌ
xยฟ
A
45โฌ
Aยฟ
5 mm
800 mm
x
L = 28002 + 52 – 2(800)(5) cos 135ยฐ = 803.54 mm
sin 135ยฐ
sin u
=
;
803.54
800
gxy =
u = 44.75ยฐ = 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.
2โ23. 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
45โฌ
800 mm
45โฌ
xยฟ
A
45โฌ
Aยฟ
5 mm
800 mm
x
L = 28002 + 52 – 2(800)(5) cos 135ยฐ = 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.
*2โ24. 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 Pxยฟ along the xยฟ axis.
y
45โฌ
800 mm
45โฌ
xยฟ
A
45โฌ
800 mm
x
L = 800 cos 45ยฐ = 565.69 mm
Pxยฟ =
5
= 0.00884 mm>mm
565.69
Ans.
128
Aยฟ
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
2โ25. The square rubber block is subjected to a shear
strain of gxy = 40(10 – 6)x + 20(10 – 6)y, where x and y are
in mm. This deformation is in the shape shown by the dashed
lines, where all the lines parallel to the y axis remain vertical
after the deformation. Determine the normal strain along
edge BC.
Shear Strain: Along edge DC, y = 400 mm. Thus, (gxy)DC = 40(10 – 6)x + 0.008.
dy
Here,
= tan (gxy)DC = tan 340(10 – 6)x + 0.0084. Then,
dx
dc
C
A
B
400 mm
300 mm
dy =
L0
D
dc = –
tan [40(10 – 6)x + 0.008]dx
L0
300 mm
e ln cos c 40(10 – 6)x + 0.008 d f `
40(10 – 6)
0
1
x
300 mm
= 4.2003 mm
Along edge AB, y = 0. Thus, (gxy)AB = 40(10 – 6)x. Here,
dy
= tan (gxy)AB =
dx
tan [40(10 – 6)x]. Then,
300 mm
dB
dy =
L0
dB = –
L0
tan [40(10 – 6)x]dx
300 mm
e ln cos c 40(10 – 6)x d f `
40(10 – 6)
0
1
= 1.8000 mm
Average Normal Strain: The stretched length of edge BC is
LBยฟCยฟ = 400 + 4.2003 – 1.8000 = 402.4003 mm
We obtain,
(Pavg)BC =
LBยฟCยฟ – 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.
2โ26. 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
yยฟ
xยฟ
600 mm
Aยฟ
Bยฟ
B
A
E
600 mm
D
C Cยฟ
x
Average Normal Strain: The stretched length of sides DA and DC are
LDCยฟ = (1 + Px)LDC = (1 + 0.004)(600) = 602.4 mm
LDAยฟ = (1 + Py)LDA = (1 + 0.005)(600) = 603 mm
Also,
a =
p
180ยฐ
– 0.02 = 1.5508 rada
b = 88.854ยฐ
2
p rad
Thus, the length of CยฟAยฟ can be determined using the cosine law with reference
to Fig. a.
LCยฟAยฟ = 2602.42 + 6032 – 2(602.4)(603) cos 88.854ยฐ
= 843.7807 mm
The original length of diagonal CA can be determined using Pythagoreanโs
theorem.
LCA = 26002 + 6002 = 848.5281 mm
Thus,
(Pavg)CA =
LCยฟAยฟ – 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.
2โ27. 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 xยฟ and yยฟ axes.
y
yยฟ
xยฟ
600 mm
Aยฟ
Bยฟ
B
A
600 mm
D
E
C Cยฟ
x
Average Normal Strain: The stretched length of sides DC and BC are
LDCยฟ = (1 + Px)LDC = (1 + 0.004)(600) = 602.4 mm
LBยฟCยฟ = (1 + Py)LBC = (1 + 0.005)(600) = 603 mm
Also,
a =
180ยฐ
p
– 0.02 = 1.5508 rada
b = 88.854ยฐ
2
p rad
f =
180ยฐ
p
+ 0.02 = 1.5908 rada
b = 91.146ยฐ
2
p rad
Thus, the length of CยฟAยฟ and DBยฟ can be determined using the cosine law with
reference to Fig. a.
LCยฟAยฟ = 2602.42 + 6032 – 2(602.4)(603) cos 88.854ยฐ = 843.7807 mm
LDBยฟ = 2602.42 + 6032 – 2(602.4)(603) cos 91.146ยฐ = 860.8273 mm
Thus,
LEยฟAยฟ =
LCยฟAยฟ
= 421.8903 mm
2
LEยฟBยฟ =
LDBยฟ
= 430.4137 mm
2
Using this result and applying the cosine law to the triangle AยฟEยฟBยฟ , Fig. a,
602.42 = 421.89032 + 430.41372 – 2(421.8903)(430.4137) cos u
u = 89.9429ยฐa
p rad
b = 1.5698 rad
180ยฐ
Shear Strain:
(gE)xยฟyยฟ =
p
p
– u =
– 1.5698 = 0.996(10 – 3) rad
2
2
Ans.
Ans:
(gE)xยฟyยฟ = 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.
*2โ28.
The wire is subjected to a normal strain that is
2
P โซฝ (x/L)eโ(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
ยขL =
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.
2โ29. 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
LACยฟ = 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 =
LACยฟ – 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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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
2โ30. 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
LBยฟDยฟ = 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 =
LBยฟDยฟ – 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.
2โ31. 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(ยขx)
= Px = kx2
dx
L
(ยขx)B =
3
2
L0
kx =
kL
3
Ans.
kL3
(ยขx)B
kL2
3
(Px)avg =
=
=
L
L
3
Ans.
Ans:
3
(ยขx)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.
*2โ32 The rubber block is fixed along edge AB, and edge
CD is moved so that the vertical displacement of any point
in the block is given by v(x) = (v0>b3)x3. Determine the
shear strain gxy at points (b>2, a>2) and (b, a).
y
v (x)
A
v0
D
a
B
Shear Strain: From Fig. a,
b
dv
= tan gxy
dx
3v0 2
x = tan gxy
b3
gxy = tan – 1 a
3v0 2
x b
b3
Thus, at point (b> 2, a> 2),
gxy = tan – 1 c
3v0 b 2
a b d
b3 2
3 v0
= tan – 1 c a b d
4 b
Ans.
and at point (b, a),
gxy = tan – 1 c
b3
(b2) d
= tan – 1 c 3 a
v0
bd
b
3v0
C
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.
2โ33. 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 AยฟBยฟ .
y
Bยฟ
vB
B
L
A
uA Aยฟ
u
x
Geometry:
LAยฟBยฟ = 2(L cos u – uA)2 + (L sin u + vB)2
= 2L2 + u2A + v2B + 2L(vB sin u – uA cos u)
Average Normal Strain:
LAยฟBยฟ – 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
ยข
โค + … – 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.
2โ34. If the normal strain is defined in reference to the
final length, that is,
Pยฟn = lim ยฟ a
p:p
ยขsยฟ – ยขs
b
ยขsยฟ
instead of in reference to the original length, Eq. 2โ2 , show
that the difference in these strains is represented as a
second-order term, namely, Pn – Pnยฟ = Pn Pnยฟ .
PB =
ยขSยฟ – ยขS
ยขS
ล
=
PB – PA
ยขSยฟ – ยขS
ยขSยฟ – ยขS
ยขS
ยขSยฟ
ยขSยฟ 2 – ยขSยขSยฟ – ยขSยฟยขS + ยขS2
ยขSยขSยฟ
ยขSยฟ 2 + ยขS2 – 2ยขSยฟยขS
=
ยขSยขSยฟ
=
=
(ยขSยฟ – ยขS)2
ยขSยฟ – ยขS
ยขSยฟ – ยขS
= ยข
โคยข
โค
ยขSยขSยฟ
ยขS
ยขSยฟ
= PA PBล (Q.E.D)
138

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