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10–1. Prove that the sum of the normal strains in
perpendicular directions is constant.
ex¿ =
ey¿ =
ex + ey
2
ex  ey
+
ex + ey
2
2
ex  ey

2
cos 2u +
cos 2u 
gxy
2
gxy
2
sin 2u
(1)
sin 2u
(2)
Adding Eq. (1) and Eq. (2) yields:
ex¿ + ey¿ = ex + ey = constant
QED
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10–2. The state of strain at the point has components
of Px = 200 11062, Py = 300 11062, and gxy = 400(1062.
Use the straintransformation equations to determine the
equivalent inplane strains on an element oriented at an
angle of 30° counterclockwise from the original position.
Sketch the deformed element due to these strains within the
x–y plane.
y
x
In accordance to the established sign convention,
ex = 200(10  6),
ex¿ =
ex + ey
ex  ey
+
2
= c
ey = 300(10  6)
2
cos 2u +
gxy
2
gxy = 400(10  6)
u = 30°
sin 2u
200  (300)
200 + (300)
400
+
cos 60° +
sin 60° d(10  6)
2
2
2
= 248 (10  6)
gx¿y¿
2
= a
Ans.
ex  ey
2
b sin 2u +
gxy
2
cos 2u
gx¿y¿ = e  C 200  ( 300) D sin 60° + 400 cos 60° f(10  6)
= 233(10  6)
ey¿ =
ex + ey
= c
2
Ans.
ex  ey

2
cos 2u 
gxy
2
sin 2u
200  ( 300)
200 + (300)
400
cos 60° sin 60° d(10  6)
2
2
2
= 348(10  6)
Ans.
The deformed element of this equivalent state of strain is shown in Fig. a
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10–3. A strain gauge is mounted on the 1in.diameter
A36 steel shaft in the manner shown. When the shaft is
rotating with an angular velocity of v = 1760 rev>min, the
reading on the strain gauge is P = 80011062. Determine
the power output of the motor. Assume the shaft is only
subjected to a torque.
v = (1760 rev>min)a
60Њ
2p rad
1 min
ba
b = 184.307 rad>s
60 sec
1 rev
ex = ey = 0
ex¿ =
ex + ey
2
ex  ey
+
2
800(10  6) = 0 + 0 +
cos 2u +
gxy
2
gxy
2
sin 2u
sin 120°
gxy = 1.848(10  3) rad
t = G gxy = 11(103)(1.848)(10  3) = 20.323 ksi
t =
Tc
;
J
20.323 =
T(0.5)
p
2
(0.5)4
;
T = 3.99 kip # in = 332.5 lb # ft
P = Tv = 0.332.5 (184.307) = 61.3 kips # ft>s = 111 hp
Ans.
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*10–4. The state of strain at a point on a wrench
has components Px = 12011062, Py = 18011062, gxy =
15011062. Use the straintransformation equations to
determine (a) the inplane principal strains and (b) the
maximum inplane shear strain and average normal strain.
In each case specify the orientation of the element and show
how the strains deform the element within the x–y plane.
ex = 120(10  6)
e1, 2 =
a)
ey = 180(10  6)
gxy = 150(10  6)
Ex  Ey 2
ex + ey
gxy 2
;
a
b + a
b
2
A
2
2
120 + (180)
120  ( 180) 2
150 2
6
;
a
b + a
b d 10
2
A
2
2
e1 = 138(10  6);
e2 = 198(10  6)
= c
Ans.
Orientation of e1 and e2
gxy
150
=
= 0.5
tan 2up =
ex  ey
[120  (180)]
up = 13.28° and 76.72°
Use Eq. 10.5 to determine the direction of e1 and e2
ex¿ =
ex + ey
ex  ey
+
2
2
cos 2u +
gxy
2
sin 2u
u = up = 13.28°
ex¿ = c
120 + ( 180)
120  ( 180)
150
+
cos (26.56°) +
sin 26.56° d 10  6
2
2
2
= 138 (10  6) = e1
Therefore up1 = 13.3° ;
gmax
b)
=
2
inplane
ex + ey
2
A
ex  ey
b + a
2
gxy
b
Ans.
2
2
2
150 2
120  ( 180) 2
6
6
= 2c a
b + a
b d10 = 335 (10 )
2
2
A
gmax
eavg =
a
inplane
up2 = 76.7°
= c
120 + (180)
d 10  6 = 30.0(10  6)
2
Ans.
Ans.
Orientation of gmax
tan 2us =
(ex  ey)
gxy
=
[120  ( 180)]
= 2.0
150
us = 31.7° and 58.3°
Ans.
gmax
Use Eq. 10–6 to determine the sign of inplane
gx¿y¿
ex  ey
gxy
= sin 2u +
cos 2u
2
2
2
u = us = 31.7°
gx¿y¿ = 2 c 
120  (180)
150
sin (63.4°) +
cos (63.4°) d10  6 = 335(10  6)
2
2
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10–5. The state of strain at the point on the arm
has components Px = 25011062, Py = 45011062, gxy =
82511062. Use the straintransformation equations to
determine (a) the inplane principal strains and (b) the
maximum inplane shear strain and average normal strain.
In each case specify the orientation of the element and show
how the strains deform the element within the x–y plane.
ex = 250(10  6)
ey = 450(10  6)
y
gxy = 825(10  6)
x
a)
ex + ey
e1, 2 =
;
2
= c
A
ex  ey
a
2
2
b + a
gxy
2
b
2
250  450
250  ( 450) 2
825 2
6
;
a
b + a
b d(10 )
2
A
2
2
e1 = 441(10  6)
Ans.
e2 = 641(10  6)
Ans.
Orientation of e1 and e2 :
gxy
tan 2up =
ex  ey
up = 24.84°
825
250  ( 450)
=
up = 65.16°
and
Use Eq. 10–5 to determine the direction of e1 and e2:
ex¿ =
ex + ey
ex  ey
+
2
2
cos 2u +
gxy
2
sin 2u
u = up = 24.84°
ex¿ = c
250  (450)
250  450
825
+
cos (49.69°) +
sin (49.69°) d(10  6) = 441(10  6)
2
2
2
Therefore, up1 = 24.8°
Ans.
up2 = 65.2°
Ans.
b)
g
max
inplane
2
g
max
inplane
eavg =
=
A
= 2c
a
ex  ey
2
2
gxy
2
b
2
250  (450) 2
825 2
6
3
b + a
b d(10 ) = 1.08(10 )
A
2
2
a
ex + ey
2
b + a
= a
250  450
b (10  6) = 100(10  6)
2
Ans.
Ans.
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10–6. The state of strain at the point has components of
Px = 10011062, Py = 40011062, and gxy = 30011062.
Use the straintransformation equations to determine the
equivalent inplane strains on an element oriented at an
angle of 60° counterclockwise from the original position.
Sketch the deformed element due to these strains within
the x–y plane.
y
x
In accordance to the established sign convention,
ex = 100(10  6)
ex¿ =
ex + ey
= c
ex  ey
+
2
ey = 400(10  6)
2
gxy
cos 2u +
2
gxy = 300(10  6)
u = 60°
sin 2u
100  400
300
100 + 400
+
cos 120° +
sin 120° d(10  6)
2
2
2
= 145(10  6)
gx¿y¿
2
= a
Ans.
ex  ey
2
b sin 2u +
gxy
2
cos 2u
gx¿y¿ = c (100  400) sin 120° + (300) cos 120° d(10  6)
= 583(10  6)
ey¿ =
ex + ey
= c
2
Ans.
ex  ey

2
cos 2u 
gxy
2
sin 2u
100  400
300
100 + 400
cos 120° sin 120° d(10  6)
2
2
2
= 155 (10  6)
Ans.
The deformed element of this equivalent state of strain is shown in Fig. a
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10–7. The state of strain at the point has components of
Px = 10011062, Py = 30011062, and gxy = 15011062.
Use the straintransformation equations to determine the
equivalent inplane strains on an element oriented u = 30°
clockwise. Sketch the deformed element due to these
strains within the x–y plane.
y
x
In accordance to the established sign convention,
ex = 100(10  6)
ex¿ =
ex + ey
ex  ey
+
2
= c
ey = 300(10  6)
2
cos 2u +
gxy = 150(10  6)
gxy
2
u = 30°
sin 2u
100  300
150
100 + 300
+
cos (60°) +
sin ( 60°) d (10  6)
2
2
2
= 215(10  6)
gx¿y¿
2
= a
Ans.
ex  ey
2
b sin 2u +
gxy
2
cos 2u
gx¿y¿ = c (100  300) sin ( 60°) + ( 150) cos ( 60°) d(10  6)
= 248 (10  6)
ey¿ =
ex + ey
= c
2
Ans.
ex  ey

2
cos 2u 
gxy
2
sin 2u
100  300
150
100 + 300
cos ( 60°) sin (60°) d (10  6)
2
2
2
= 185(10  6)
Ans.
The deformed element of this equivalent state of strain is shown in Fig. a
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*10–8. The state of strain at the point on the bracket
has components Px = 20011062, Py = 65011062, gxy ϭ
17511062. Use the straintransformation equations to
determine the equivalent inplane strains on an element
oriented at an angle of u = 20° counterclockwise from the
original position. Sketch the deformed element due to these
strains within the x–y plane.
ex = 200(10  6)
ex¿ =
ex + ey
ex  ey
+
2
= c
ey = 650(10  6)
2
cos 2u +
gxy
2
y
x
gxy = 175(10  6)
u = 20°
sin 2u
( 200)  (650)
(175)
200 + (650)
+
cos (40°) +
sin (40°) d(10  6)
2
2
2
= 309(10  6)
ey¿ =
ex + ey
ex  ey

2
= c
Ans.
2
cos 2u 
gxy
2
sin 2u
200  ( 650)
( 175)
200 + (650)
cos (40°) sin (40°) d(10  6)
2
2
2
= 541(10  6)
gx¿y¿
2
ex  ey
= 
2
Ans.
sin 2u +
gxy
2
cos 2u
gx¿y¿ = [(200  (650)) sin (40°) + (175) cos (40°)](10  6)
= 423(10  6)
Ans.
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10–9. The state of strain at the point has components of
Px = 18011062, Py = 12011062, and gxy = 10011062.
Use the straintransformation equations to determine (a)
the inplane principal strains and (b) the maximum inplane
shear strain and average normal strain. In each case specify
the orientation of the element and show how the strains
deform the element within the x–y plane.
y
x
a)
In
accordance
to
the
established
sign
convention,
ex = 180(10  6),
ey = 120(10  6) and gxy = 100(10  6).
ex + ey
e1, 2 =
;
2
= b
a
A
ex  ey
2
2
b + a
gxy
2
b
2
180 + (120)
180  ( 120) 2
100 2
6
;
c
d + a
b r (10 )
2
A
2
2
= A 30 ; 158.11 B (10  6)
e1 = 188(10  6)
tan 2uP =
e2 = 128(10  6)
gxy
Ans.
100(10  6)
ex  ey
C 180  (120) D (10  6)
=
uP = 9.217°
and
= 0.3333
80.78°
Substitute u = 9.217°,
ex + ey
ex¿ =
2
= c
ex  ey
+
2
cos 2u +
gxy
2
sin 2u
180 + ( 120)
180  ( 120)
100
+
cos (18.43°) +
sin (18.43) d(10  6)
2
2
2
= 188(10  6) = e1
Thus,
(uP)1 = 9.22°
(uP)2 = 80.8°
Ans.
The deformed element is shown in Fig (a).
gmax
ex  ey 2
gxy 2
inplane
=
b)
a
b + a
b
2
A
2
2
gmax
inplane
tan 2us =  a
= b2
180  (120) 2
100 2
6
6
d + a
b r (10 ) = 316 A 10 B
A
2
2
ex  ey
gxy
c
b = c
C 180  (120) D (10  6)
us = 35.78° = 35.8° and
100(10  6)
Ans.
s = 3
Ans.
54.22° = 54.2°
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10–9.
Continued
gmax
The algebraic sign for inplane
when u = 35.78°.
ex  ey
gxy
gx¿y¿
= a
b sin 2u +
cos 2u
2
2
2
gx¿y¿ = e  C 180  ( 120) D sin 71.56° + ( 100) cos 71.56° f(10  6)
eavg
= 316(10  6)
ex + ey
180 + (120)
=
= c
d(10  6) = 30(10  6)
2
2
Ans.
The deformed element for the state of maximum Inplane shear strain is shown is
shown in Fig. b
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Page 748
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10–10. The state of strain at the point on the bracket
has components Px = 40011062, Py = 25011062, gxy ϭ
31011062. Use the straintransformation equations to
determine the equivalent inplane strains on an element
oriented at an angle of u = 30° clockwise from the original
position. Sketch the deformed element due to these strains
within the x–y plane.
ex = 400(10  6)
ex¿ =
ex + ey
= c
ex  ey
+
2
ey = 250(10  6)
2
cos 2u +
gxy
2
gxy = 310(10  6)
y
x
u = 30°
sin 2u
400  ( 250)
400 + ( 250)
310
+
cos (60°) + a
b sin (60°) d(10  6)
2
2
2
= 103(10  6)
ey¿ =
ex + ey
= c
ex  ey

2
Ans.
2
cos 2u 
gxy
2
sin 2u
400  (250)
400 + (250)
310
cos (60°) sin (60°) d(10  6)
2
2
2
= 46.7(10  6)
gx¿y¿
2
ex  ey
= 
2
Ans.
sin 2u +
gxy
2
cos 2u
gx¿y¿ = [(400  (250)) sin (60°) + 310 cos ( 60°)](10  6) = 718(10  6)
748
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Page 749
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10–11. The state of strain at the point has components of
Px = 10011062, Py = 20011062, and gxy = 10011062.
Use the straintransformation equations to determine (a)
the inplane principal strains and (b) the maximum inplane
shear strain and average normal strain. In each case specify
the orientation of the element and show how the strains
deform the element within the x–y plane.
In accordance to the established
ey = 200(10  6) and gxy = 100(10  6).
ex + ey
e1, 2 =
;
2
= b
A
a
ex  ey
2
b + a
2
gxy
2
b
sign
y
x
convention,
ex = 100(10  6),
2
100 + (200)
100 2
100  (200) 2
6
;
c
d + a
b r (10 )
2
A
2
2
A 150 ; 70.71 B (10  6)
=
e1 = 79.3(10  6)
tan 2uP =
e2 = 221(10  6)
gxy
100(10  6)
C 100  (200) D (10  6)
=
ex  ey
uP = 22.5°
and
Ans.
= 1
67.5°
Substitute u = 22.5,
ex + ey
ex¿ =
ex  ey
cos 2u +
gxy
sin 2u
2
2
2
100 + (200)
100  (200)
100
+
cos 45° +
sin 45° d(10  6)
= c
2
2
2
+
= 79.3(10  6) = e1
Thus,
(uP)1 = 22.5°
(uP)2 = 67.5°
Ans.
The deformed element of the state of principal strain is shown in Fig. a
gmax
ex  ey 2
gxy 2
inplane
=
a
b + a
b
2
A
2
2
gmax
inplane
= b2
tan 2us =  a
c
100  (200) 2
100 2
6
6
d + a
b r (10 ) = 141(10 )
A
2
2
ex  ey
gxy
b = c
us = 22.5°
The algebraic sign for
gx¿y¿
2
= a
ex  ey
2
C 100  ( 200) D (10  6)
100(10  6)
and
gmax
inplane
b sin 2u +
Ans.
s = 1
Ans.
67.5°
when u = 22.5°.
gxy
2
cos 2u
gx¿y¿ =  C 100  (200) D sin ( 45°) + 100 cos (45°)
eavg
= 141(10  6)
ex + ey
100 + ( 200)
=
= c
d(10  6) = 150(10  6)
2
2
Ans.
The deformed element for the state of maximum Inplane shear strain is shown in
Fig. b.
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10–11.
Continued
*10–12. The state of plane strain on an element is given by
Px = 50011062, Py = 30011062, and gxy = 20011062.
Determine the equivalent state of strain on an element at
the same point oriented 45° clockwise with respect to the
original element.
y
Pydy
dy gxy
2
Strain Transformation Equations:
ex = 500 A 10  6 B
ey = 300 A 10  6 B
gxy = 200 A 10  6 B
u = 45°
We obtain
ex¿ =
ex + ey
+
2
= c
ex  ey
2
cos 2u +
gxy
2
sin 2u
500  300
200
500 + 300
+
cos (90°) + a
b sin (90°) d A 10  6 B
2
2
2
= 500 A 10  6 B
gx¿y¿
2
= a
Ans.
ex  ey
2
b sin 2u +
gxy
2
cos 2u
gx¿y¿ = [(500  300) sin ( 90°) + (200) cos ( 90°)] A 10  6 B
= 200 A 10  6 B
ey¿ =
ex + ey
= c
2
Ans.
ex  ey

2
cos 2u 
gxy
2
sin 2u
500 + 300
500  300
200
cos ( 90°)  a
b sin (90°) d A 10  6 B
2
2
2
= 300 A 10  6 B
Ans.
The deformed element for this state of strain is shown in Fig. a.
750
gxy
2
dx
x
Pxdx
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10–13. The state of plane strain on an element is
Px = 30011062, Py = 0, and gxy = 15011062. Determine
the equivalent state of strain which represents (a) the
principal strains, and (b) the maximum inplane shear strain
and the associated average normal strain. Specify the
orientation of the corresponding elements for these states
of strain with respect to the original element.
y
gxy
dy 2
x
InPlane Principal Strains: ex = 300 A 10  6 B , ey = 0, and gxy = 150 A 10  6 B . We
obtain
ex + ey
e1, 2 =
;
2
= C
C
¢
ex  ey
2
2
≤ + ¢
gxy
2
≤
2
300 + 0
300  0 2
150 2
;
¢
≤ + ¢
≤ S A 10  6 B
2
C
2
2
= ( 150 ; 167.71) A 10  6 B
e1 = 17.7 A 10  6 B
e2 = 318 A 10  6 B
Ans.
Orientation of Principal Strain:
tan 2up =
gxy
ex  ey
=
150 A 10  6 B
(300  0) A 10  6 B
= 0.5
uP = 13.28° and 76.72°
Substituting u = 13.28° into Eq. 91,
ex¿ =
ex + ey
= c
ex  ey
+
2
2
cos 2u +
gxy
2
sin 2u
300 + 0
300  0
150
+
cos (26.57°) +
sin (26.57°) d A 10  6 B
2
2
2
= 318 A 10  6 B = e2
Thus,
A uP B 1 = 76.7° and A uP B 2 = 13.3°
Ans.
The deformed element of this state of strain is shown in Fig. a.
Maximum InPlane Shear Strain:
gmax
ex  ey 2
gxy 2
inplane
=
¢
≤ + ¢ ≤
2
C
2
2
gmax
inplane
300  0 2
150 2
6
6
b + a
b R A 10 B = 335 A 10 B
A
2
2
= B2
a
Ans.
Orientation of the Maximum InPlane Shear Strain:
tan 2us =  ¢
ex  ey
gxy
≤ = C
(300  0) A 10  6 B
150 A 10  6 B
S = 2
us = 31.7° and 122°
Ans.
751
gxy
2
dx
Pxdx
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10–13.
Continued
The algebraic sign for
gx¿y¿
2
= ¢
ex  ey
2
gmax
inplane
≤ sin 2u +
when u = us = 31.7° can be obtained using
gxy
2
cos 2u
gx¿y¿ = [(300  0) sin 63.43° + 150 cos 63.43°] A 10  6 B
= 335 A 10  6 B
Average Normal Strain:
eavg =
ex + ey
2
= a
300 + 0
b A 10  6 B = 150 A 10  6 B
2
Ans.
The deformed element for this state of strain is shown in Fig. b.
752
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10–14. The state of strain at the point on a boom of an
hydraulic engine crane has components of Px = 25011062,
Py = 30011062, and gxy = 18011062. Use the straintransformation equations to determine (a) the inplane
principal strains and (b) the maximum inplane shear strain
and average normal strain. In each case, specify the
orientation of the element and show how the strains deform
the element within the x–y plane.
y
a)
InPlane Principal Strain: Applying Eq. 10–9,
ex + ey
e1, 2 =
;
2
= B
a
A
ex  ey
2
b + a
2
gxy
2
b
2
250  300 2
250 + 300
180 2
6
;
a
b + a
b R A 10 B
2
A
2
2
= 275 ; 93.41
e1 = 368 A 10  6 B
e2 = 182 A 10  6 B
Ans.
Orientation of Principal Strain: Applying Eq. 10–8,
gxy
tan 2uP =
180(10  6)
ex  ey
=
(250  300)(10  6)
uP = 37.24°
and
= 3.600
52.76°
Use Eq. 10–5 to determine which principal strain deforms the element in the x¿
direction with u = 37.24°.
ex¿ =
ex + ey
= c
2
ex  ey
+
2
cos 2u +
gxy
2
sin 2u
250 + 300
250  300
180
+
cos 74.48° +
sin 74.48° d A 10  6 B
2
2
2
= 182 A 10  6 B = e2
Hence,
uP1 = 52.8°
and
uP2 = 37.2°
Ans.
b)
Maximum InPlane Shear Strain: Applying Eq. 10–11,
g max
ex  ey 2
gxy 2
inplane
=
a
b + a
b
2
A
2
2
g
max
inplane
= 2B
180 2
250  300 2
6
b + a
b R A 10 B
A
2
2
a
= 187 A 10  6 B
Ans.
753
x
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10–14.
Continued
Orientation of the Maximum InPlane Shear Strain: Applying Eq. 10–10,
tan 2us = 
ex  ey
us = 7.76°
and
The proper sign of
gx¿y¿
2
ex  ey
= 
= 
gxy
2
g
max
inplane
250  300
= 0.2778
180
82.2°
Ans.
can be determined by substituting u = 7.76° into Eq. 10–6.
sin 2u +
gxy
2
cos 2u
gx¿y¿ = {[250  300] sin (15.52°) + (180) cos (15.52°)} A 10  6 B
= 187 A 10  6 B
Normal Strain and Shear strain: In accordance with the sign convention,
ex = 250 A 10  6 B
ey = 300 A 10  6 B
gxy = 180 A 10  6 B
Average Normal Strain: Applying Eq. 10–12,
eavg =
ex + ey
2
= c
250 + 300
d A 10  6 B = 275 A 10  6 B
2
Ans.
754
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*10–16. The state of strain at a point on a support
has components of Px = 35011062, Py = 40011062,
gxy = 67511062. Use the straintransformation equations
to determine (a) the inplane principal strains and (b) the
maximum inplane shear strain and average normal strain.
In each case specify the orientation of the element and show
how the strains deform the element within the x–y plane.
a)
e1, 2 =
=
ex + ey
;
2
B
a
ex ey
2
b + a
2
gxy
2
b
2
350  400 2
675 2
350 + 400
;
a
b + a
b
2
A
2
2
e1 = 713(10  6)
Ans.
e2 = 36.6(10  6)
Ans.
tan 2uP =
gxy
ex  ey
=
675
(350  400)
uP = 42.9°
Ans.
b)
(gx¿y¿)max
=
2
(gx¿y¿)max
=
2
A
a
ex  ey
2
b + a
2
gxy
2
b
2
a
350  400 2
675 2
b + a
b
A
2
2
(gx¿y¿)max = 677(10  6)
eavg =
ex + ey
tan 2us =
2
=
Ans.
350 + 400
= 375(10  6)
2
(ex  ey)
gxy
=
Ans.
350  400
675
us = 2.12°
Ans.
755
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•10–17.
Solve part (a) of Prob. 10–4 using Mohr’s circle.
ex = 120(10  6)
ey = 180(10  6)
gxy = 150(10  6)
A (120, 75)(10  6) C (30, 0)(10  6)
R = C 2[120  (30)]2 + (75)2 D (10  6)
= 167.71 (10  6)
e1 = (30 + 167.71)(10  6) = 138(10  6)
Ans.
e2 = (30  167.71)(10  6) = 198(10  6)
Ans.
75
tan 2uP = a
b , uP = 13.3°
30 + 120
Ans.
756
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10–18.
Solve part (b) of Prob. 10–4 using Mohr’s circle.
ex = 120(10  6)
ey = 180(10  6)
gxy = 150(10  6)
A (120, 75)(10  6) C (30, 0)(10  6)
R = C 2[120  (30)]2 + (75)2 D (10  6)
= 167.71 (10  6)
gxy
max
2 inplane
gxy
= R = 167.7(10  6)
max
inplane
= 335(10  6)
Ans.
eavg = 30 (10  6)
tan 2us =
120 + 30
75
Ans.
us = 31.7°
Ans.
757
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10–19.
Solve Prob. 10–8 using Mohr’s circle.
ex = 200(10  6)
ey = 650(10  6)
gxy = 175(10  6)
gxy
2
= 87.5(10  6)
u = 20°, 2u = 40°
A(200, 87.5)(10  6)
C(425, 0)(10  6)
R = [2(200  (425))2 + 87.52 ](10  6) = 241.41(10  6)
tan a =
87.5
;
200  (425)
a = 21.25°
f = 40 + 21.25 = 61.25°
ex¿ = (425 + 241.41 cos 61.25°)(10  6) = 309(10  6)
Ans.
ey¿ = (425  241.41 cos 61.25°)(10  6) = 541(10  6)
Ans.
gx¿y¿
2
= 241.41(10  6) sin 61.25°
gx¿y¿ = 423(10  6)
Ans.
758
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*10–20.
Solve Prob. 10–10 using Mohr’s circle.
ex = 400(10  6)
A(400, 155)(10  6)
ey = 250(10  6)
gxy = 310(10  6)
gxy
2
= 155(10  6)
C(75, 0)(10  6)
R = [2(400  75)2 + 1552 ](10  6) = 360.1(10  6)
tan a =
155
;
400  75
a = 25.50°
f = 60 + 25.50 = 85.5°
ex¿ = (75 + 360.1 cos 85.5°)(10  6) = 103(10  6)
Ans.
ey¿ = (75  360.1 cos 85.5°)(10  6) = 46.7(10  6)
Ans.
gx¿y¿
2
= (360.1 sin 85.5°)(10  6)
gx¿y¿ = 718(10  6)
Ans.
759
u = 30°
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•10–21.
Solve Prob. 10–14 using Mohr’s circle.
Construction of the Circle: In accordance with the sign convention, ex = 250 A 10  6 B ,
gxy
ey = 300 A 10  6 B , and
= 90 A 10  6 B . Hence,
2
eavg =
ex + ey
2
= a
250 + 300
b A 10  6 B = 275 A 10  6 B
2
Ans.
The coordinates for reference points A and C are
A(250, 90) A 10  6 B
C(275, 0) A 10  6 B
The radius of the circle is
R = a 2(275  250)2 + 902 b A 10  6 B = 93.408
InPlane Principal Strain: The coordinates of points B and D represent e1 and e2,
respectively.
e1 = (275 + 93.408) A 10  6 B = 368 A 10  6 B
Ans.
e2 = (275  93.408) A 10  6 B = 182 A 10  6 B
Ans.
Orientation of Principal Strain: From the circle,
tan 2uP2 =
90
= 3.600
275  250
2uP2 = 74.48°
2uP1 = 180°  2uP2
uP1 =
180°  74.78°
= 52.8° (Clockwise)
2
Ans.
Maximum InPlane Shear Strain: Represented by the coordinates of point E on
the circle.
g max
inplane
2
g
= R = 93.408 A 10  6 B
max
inplane
= 187 A 10  6 B
Ans.
Orientation of the Maximum InPlane Shear Strain: From the circle,
tan 2us =
275  250
= 0.2778
90
us = 7.76° (Clockwise)
Ans.
760
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10–22. The strain at point A on the bracket has
components Px = 30011062, Py = 55011062, gxy =
65011062. Determine (a) the principal strains at A in the
x– y plane, (b) the maximum shear strain in the x–y plane,
and (c) the absolute maximum shear strain.
ex = 300(10  6)
ey = 550(10  6)
A(300, 325)10  6
gxy = 650(10  6)
y
gxy
2
= 325(10  6)
A
C(425, 0)10  6
R = C 2(425  300)2 + (325)2 D 10  6 = 348.2(10  6)
a)
e1 = (425 + 348.2)(10  6) = 773(10  6)
Ans.
e2 = (425  348.2)(10  6) = 76.8(10  6)
Ans.
b)
g
max
inplane
= 2R = 2(348.2)(10  6) = 696(10  6)
Ans.
773(10  6)
;
2
Ans.
c)
gabs
max
=
2
gabs
max
= 773(10  6)
10–23. The strain at point A on the leg of the angle has
components Px = 14011062, Py = 18011062, gxy =
12511062. Determine (a) the principal strains at A in the
x–y plane, (b) the maximum shear strain in the x–y plane,
and (c) the absolute maximum shear strain.
ex = 140(10  6)
A( 140, 62.5)10  6
ey = 180(10  6)
gxy = 125(10  6)
A
gxy
2
= 62.5(10  6)
C(20, 0)10  6
A 2(20  ( 140))2 + (62.5)2 B 10  6 = 171.77(10  6)
R =
a)
e1 = (20 + 171.77)(10  6) = 192(10  6)
Ans.
e2 = (20  171.77)(10  6) = 152(10  6)
Ans.
(b, c)
gabs
max
=
g
max
inplane
= 2R = 2(171.77)(10  6) = 344(10  6)
Ans.
761
x
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*10–24. The strain at point A on the pressurevessel wall
has components Px = 48011062, Py = 72011062, gxy =
65011062. Determine (a) the principal strains at A, in the
x– y plane, (b) the maximum shear strain in the x–y plane,
and (c) the absolute maximum shear strain.
ex = 480(10  6)
ey = 720(10  6)
A(480, 325)10  6
C(600, 0)10  6
gxy = 650(10  6)
y
A
gxy
2
= 325(10  6)
R = (2(600  480)2 + 3252 )10  6 = 346.44(10  6)
a)
e1 = (600 + 346.44)10  6 = 946(10  6)
Ans.
e2 = (600  346.44)10  6 = 254(10  6)
Ans.
b)
g
max
inplane
= 2R = 2(346.44)10  6 = 693(10  6)
Ans.
946(10  6)
;
2
Ans.
c)
gabs
max
2
=
gabs
max
= 946(10  6)
762
x