Solution manual mechanics of materials 8th edition hibbeler chapter 10
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10 Solutions 46060
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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 = 200 110-62, Py = -300 110-62, and gxy = 400(10-62.
Use the strain-transformation equations to determine the
equivalent in-plane 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 1-in.-diameter
A-36 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 = 800110-62. 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 = 120110-62, Py = -180110-62, gxy =
150110-62. Use the strain-transformation equations to
determine (a) the in-plane principal strains and (b) the
maximum in-plane 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
in-plane
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
in-plane
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 in-plane
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 = 250110-62, Py = -450110-62, gxy =
-825110-62. Use the strain-transformation equations to
determine (a) the in-plane principal strains and (b) the
maximum in-plane 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
in-plane
2
g
max
in-plane
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 = -100110-62, Py = 400110-62, and gxy = -300110-62.
Use the strain-transformation equations to determine the
equivalent in-plane 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 = 100110-62, Py = 300110-62, and gxy = -150110-62.
Use the strain-transformation equations to determine the
equivalent in-plane 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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*10–8. The state of strain at the point on the bracket
has components Px = -200110-62, Py = -650110-62, gxy ϭ
-175110-62. Use the strain-transformation equations to
determine the equivalent in-plane 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 = 180110-62, Py = -120110-62, and gxy = -100110-62.
Use the strain-transformation equations to determine (a)
the in-plane principal strains and (b) the maximum in-plane
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
in-plane
=
b)
a
b + a
b
2
A
2
2
gmax
in-plane
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 in-plane
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 In-plane shear strain is shown is
shown in Fig. b
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10–10. The state of strain at the point on the bracket
has components Px = 400110-62, Py = -250110-62, gxy ϭ
310110-62. Use the strain-transformation equations to
determine the equivalent in-plane 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
Ans.
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10–11. The state of strain at the point has components of
Px = -100110-62, Py = -200110-62, and gxy = 100110-62.
Use the strain-transformation equations to determine (a)
the in-plane principal strains and (b) the maximum in-plane
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
in-plane
=
a
b + a
b
2
A
2
2
gmax
in-plane
= 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
in-plane
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 In-plane 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 = 500110-62, Py = 300110-62, and gxy = -200110-62.
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 = -300110-62, Py = 0, and gxy = 150110-62. Determine
the equivalent state of strain which represents (a) the
principal strains, and (b) the maximum in-plane 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
In-Plane 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. 9-1,
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 In-Plane Shear Strain:
gmax
ex - ey 2
gxy 2
in-plane
=
¢
≤ + ¢ ≤
2
C
2
2
gmax
in-plane
-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 In-Plane 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
in-plane
≤ 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 = 250110-62,
Py = 300110-62, and gxy = -180110-62. Use the straintransformation equations to determine (a) the in-plane
principal strains and (b) the maximum in-plane 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)
In-Plane 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 In-Plane Shear Strain: Applying Eq. 10–11,
g max
ex - ey 2
gxy 2
in-plane
=
a
b + a
b
2
A
2
2
g
max
in-plane
= 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 In-Plane 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
in-plane
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 = 350110-62, Py = 400110-62,
gxy = -675110-62. Use the strain-transformation equations
to determine (a) the in-plane principal strains and (b) the
maximum in-plane 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 in-plane
gxy
= R = 167.7(10 - 6)
max
in-plane
= 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
In-Plane 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 In-Plane Shear Strain: Represented by the coordinates of point E on
the circle.
g max
in-plane
2
g
= -R = -93.408 A 10 - 6 B
max
in-plane
= -187 A 10 - 6 B
Ans.
Orientation of the Maximum In-Plane 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 = 300110-62, Py = 550110-62, gxy =
-650110-62. 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
in-plane
= 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 = -140110-62, Py = 180110-62, gxy =
-125110-62. 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
in-plane
= 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 pressure-vessel wall
has components Px = 480110-62, Py = 720110-62, gxy =
650110-62. 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
in-plane
= 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
