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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.
9–57. Determine (a) the principal stresses and (b) the
maximum in-plane shear stress and average normal stress.
Specify the orientation of the element in each case.
50 MPa
30 MPa
A(0, -30)
B(50, 30)
C(25, 0)
R = CA = 2(25 - 0)2 + 302 = 39.05
s1 = 25 + 39.05 = 64.1 MPa
Ans.
s2 = 25 - 39.05 = - 14.1 MPa
Ans.
tan 2up =
30
= 1.2
25 - 0
up2 = 25.1°
Ans.
savg = 25.0 MPa
Ans.
tmax
= R = 39.1 MPa
Ans.
in-plane
tan 2us =
25 - 0
= 0.8333
30
us = - 19.9°
Ans.
Ans:
s1 = 64.1 MPa, s2 = - 14.1 MPa, up2 = 25.1°
savg = 25.0 MPa, tmax
= 39.1 MPa,
in-plane
us = - 19.9°
901
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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.
9–58. Determine the equivalent state of stress if an
element is oriented 25° counterclockwise from the element
shown.
550 MPa
A(0, -550)
B(0, 550)
C(0, 0)
R = CA = CB = 550
sx¿ = - 550 sin 50° = - 421 MPa
Ans.
tx¿y¿ = - 550 cos 50° = - 354 MPa
Ans.
sy¿ = 550 sin 50° = 421 MPa
Ans.
Ans:
sx¿ = - 421 MPa, tx¿y¿ = - 354 MPa,
sy¿ = 421 MPa
902
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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.
8 ksi
9–59. Determine (a) the principal stresses and (b) the
maximum in-plane shear stress and average normal stress.
Specify the orientation of the element in each case.
4 ksi
12 ksi
A(-12, 4)
B( - 8, - 4)
C( -10, 0)
R = CA = CB = 222 + 42 = 4.472
a)
s1 = - 10 + 4.472 = - 5.53 ksi
Ans.
s2 = - 10 - 4.472 = - 14.5 ksi
Ans.
tan 2up =
4
2
2up = 63.43°
up = - 31.7°
b)
tmax
= R = 4.47 ksi
Ans.
in-plane
savg = - 10 ksi
Ans.
2us = 90 - 2up
us = 13.3°
Ans.
Ans:
(a) s1 = - 5.53 ksi, s2 = - 14.5 ksi,
up = - 31.7°
(b) tmax
= 4.47 ksi, savg = - 10 ksi,
in-plane
us = 13.3°
903
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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.
200 MPa
*9–60. Determine the principal stresses, the maximum
in-plane shear stress, and average normal stress. Specify the
orientation of the element in each case.
500 MPa
350 MPa
Construction of the Circle: In accordance with the sign convention, sx = 350 MPa,
sy = - 200 MPa, and txy = 500 MPa. Hence,
savg =
sx + sy
2
=
350 + ( -200)
= 75.0 MPa
2
Ans.
The coordinates for reference point A and C are
A(350, 500)
C(75.0, 0)
The radius of the circle is
R = 2(350 - 75.0)2 + 5002 = 570.64 MPa
a)
In-Plane Principal Stresses: The coordinate of points B and D represent s1 and s2
respectively.
s1 = 75.0 + 570.64 = 646 MPa
Ans.
s2 = 75.0 - 570.64 = - 496 MPa
Ans.
Orientaion of Principal Plane: From the circle
tan 2uP1 =
500
= 1.82
350 - 75.0
uP1 = 30.6° (Counterclockwise)
Ans.
b)
Maximum In-Plane Shear Stress: Represented by the coordinates of point E on the
circle.
t
max
in-plane
= R = 571 MPa
Ans.
Orientation of the Plane for Maximum In-Plane Shear Stress: From the circle
tan 2us =
350 - 75.0
= 0.55
500
us = 14.4° (Clockwise)
Ans.
904
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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.
9–61. Draw Mohr’s circle that describes each of the following
states of stress.
5 MPa
20 ksi
20 ksi
5 MPa
(b)
(a)
18 MPa
(c)
905
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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.
9–62. The grains of wood in the board make an angle of 20°
with the horizontal as shown. Using Mohr’s circle, determine
the normal and shear stresses that act perpendicular and
parallel to the grains if the board is subjected to an axial load
of 250 N.
sx =
300 mm
60 mm
250 N
250 N
20⬚
25 mm
250
P
=
= 166.67 kPa
A
(0.06)(0.025)
R = 83.33
Coordinates of point B:
sx¿ = 83.33 - 83.33 cos 40°
sx¿ = 19.5 kPa
Ans.
tx¿y¿ = - 83.33 sin 40° = - 53.6 kPa
Ans.
Ans:
sx¿ = 19.5 kPa, tx¿y¿ = - 53.6 kPa
906
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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.
z
9–63. The post has a square cross-sectional area. If it is fixed
supported at its base and a horizontal force is applied at its end
as shown, determine (a) the maximum in-plane shear stress
developed at A and (b) the principal stresses at A.
3 in.
3 in.
60 lb
y
x
18 in.
A
1 in.
1
I =
(3)(33) = 6.75 in4
12
sA =
tA =
Myx
I
=
1080(0.5)
= - 80 psi
6.75
=
60(3)
= 8.889 psi
6.75(3)
VyQA
It
A(-80, 8.889)
tmax
QA = (1)(1)(3) = 3 in3
B(0, - 8.889)
C( - 40, 0)
= R = 2402 + 8.8892 = 41.0 psi
Ans.
in-plane
s1 = - 40 + 40.9757 = 0.976 psi
Ans.
s2 = - 40 - 40.9757 = - 81.0 psi
Ans.
Ans:
tmax
in-plane
= 41.0 psi , s1 = 0.976 psi ,
s2 = - 81.0 psi
907
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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.
*9–64. Determine the principal stress, the maximum
in-plane shear stress, and average normal stress. Specify the
orientation of the element in each case.
20 MPa
80 MPa
30 MPa
In accordance to the established sign convention, sx = 30 MPa, sy = - 20 MPa and
txy = 80 MPa. Thus,
savg =
sx + sy
30 + ( -20)
= 5 MPa
2
=
2
Then, the coordinates of reference point A and the center C of the circle is
A(30, 80)
C(5, 0)
Thus, the radius of circle is given by
R = CA = 2(30 - 5)2 + (80 - 0)2 = 83.815 MPa
Using these results, the circle shown in Fig. a, can be constructed.
The coordinates of points B and D represent s1 and s2 respectively. Thus
s1 = 5 + 83.815 = 88.8 MPa
Ans.
s2 = 5 - 83.815 = - 78.8 MPa
Ans.
Referring to the geometry of the circle, Fig. a
tan 2(uP)1 =
80
= 3.20
30 - 5
uP = 36.3° (Counterclockwise)
Ans.
The state of maximum in-plane shear stress is represented by the coordinate of
point E. Thus
tmax
= R = 83.8 MPa
Ans.
in-plane
From the geometry of the circle, Fig. a,
tan 2us =
30 - 5
= 0.3125
80
us = 8.68° (Clockwise)
Ans.
The state of maximum in-plane shear stress is represented by the element in Fig. c.
908
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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.
9–64.
Continued
909
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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.
9–65. The thin-walled pipe has an inner diameter of 0.5 in.
and a thickness of 0.025 in. If it is subjected to an internal
pressure of 500 psi and the axial tension and torsional
loadings shown, determine the principal stress at a point on
the surface of the pipe.
200 lb
200 lb
20 lb⭈ft
20 lb⭈ft
Section Properties:
A = p A 0.2752 - 0.252 B = 0.013125p in2
J =
p
A 0.2754 - 0.254 B = 2.84768 A 10 - 3 B in4
2
Normal Stress: Since
0.25
r
=
= 10, thin wall analysis is valid.
t
0.025
slong =
pr
500(0.25)
N
200
+
=
+
= 7.350 ksi
A
2t
0.013125p
2(0.025)
shoop =
pr
500(0.25)
=
= 5.00 ksi
t
0.025
Shear Stress: Applying the torsion formula,
t =
20(12)(0.275)
Tc
= 23.18 ksi
=
J
2.84768(10 - 3)
Construction of the Circle: In accordance with the sign convention sx = 7.350 ksi,
sy = 5.00 ksi, and txy = - 23.18 ksi. Hence,
savg =
sx + sy
2
=
7.350 + 5.00
= 6.175 ksi
2
The coordinates for reference points A and C are
A(7.350, -23.18)
C(6.175, 0)
The radius of the circle is
R = 2(7.350 - 6.175)2 + 23.182 = 23.2065 ksi
In-Plane Principal Stress: The coordinates of point B and D represent s1 and s2,
respectively.
s1 = 6.175 + 23.2065 = 29.4 ksi
Ans.
s2 = 6.175 - 23.2065 = - 17.0 ksi
Ans.
Ans:
s1 = 29.4 ksi , s2 = - 17.0 ksi
910
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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.
9–66. Determine the principal stress and maximum
in-plane shear stress at point A on the cross section of the
pipe at section a–a.
a 300 mm
a
300 mm
b
30 mm
A
B
200 mm
300 N
Internal Loadings: Considering the equilibrium of the free-body diagram of the
assembly’s segment, Fig. a,
©Fx = 0; N - 450 = 0
N = 450 N
©Fy = 0; Vy = 0
©Fz = 0; Vz + 300 = 0
Vz = - 300 N
©Mx = 0; T + 300(0.2) = 0
T = - 60 N # m
©My = 0; My - 450(0.3) + 300(0.3) = 0
My = 45 N # m
©Mz = 0; Mz + 450(0.2) = 0
Mz = - 90 N # m
Section Properties: The cross-sectional area, the moment of inertia about the y and
z axes, and the polar moment of inertia of the pipe’s cross section are
A = p(0.032 - 0.022) = 0.5p(10-3) m2
Iy = Iz =
J =
p
(0.034 - 0.024) = 0.1625p(10-6) m4
4
p
(0.034 - 0.024) = 0.325p(10-6) m4
2
Referring to Fig. b,
(Qy)A = 0 (Qz)A =
4(0.03) p
4(0.02) p
c (0.03)2 d c (0.022) d = 12.667(10-6) m3
3p
2
3p
2
Normal and Shear Stress: The normal stress is a combination of axial and bending
stress.
sA =
MzyA
( - 90)(- 0.03)
N
450
=
-3
A
Iz
0.5p(10 )
0.1625p(10-6)
= - 5.002 MPa
Since Vy = 0, [(txy)V]A = 0. However, the shear stress is the combination of
torsional and transverse shear stress. Thus,
(txz)A = [(txz)T]A - [(txz)V]A
=
Vz(Qz)A
60(0.03)
300[12.667(10-6)]
Tc
= 1.391 MPa
=
-6
J
Iy t
0.325p(10 )
0.1625p(10-6)(0.02)
The state of stress at point A is represented by the element shown in Fig. c.
911
450 N
20 mm
Section a – 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.
9–66.
Continued
Construction of the Circle: sx = - 5.002 MPa, sz = 0, and txz = 1.391 MPa. Thus,
savg =
sx + sz
2
=
- 5.002 + 0
= - 2.501 MPa
2
The coordinates of reference point A and the center C of the circle are
A(- 5.002, 1.391)
C(- 2.501, 0)
Thus, the radius of the circle is
R = CA = 2[- 5.002 - ( -2.501)]2 + 1.3912 = 2.862 MPa
Using these results, the circle is shown in Fig. d.
In-Plane Principal Stresses: The coordinates of reference points B and D represent
s1 and s2, respectively.
s1 = - 2.501 + 2.862 = 0.361 MPa
Ans.
s2 = - 2.501 - 2.862 = - 5.36 MPa
Ans.
In-Plane Maximum Shear Stress: The coordinates of point E represent the state of
maximum shear stress. Thus,
tmax
= |R| = 2.86 MPa
Ans.
in-plane
Ans:
s1 = 0.361 MPa , s2 = - 5.36 MPa,
tmax
= 2.86 MPa
in-plane
912
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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.
9–67. Determine the principal stress and maximum
in-plane shear stress at point B on the cross section of the
pipe at section a–a.
a 300 mm
a
300 mm
b
30 mm
A
B
200 mm
300 N
Internal Loadings: Considering the equilibrium of the free-body diagram of the
assembly’s cut segment, Fig. a,
©Fx = 0;
N - 450 = 0
©Fy = 0;
Vy = 0
©Fz = 0;
Vz + 300 = 0
Vz = - 300 N
©Mx = 0;
T + 300(0.2) = 0
T = - 60 N # m
©My = 0;
My - 450(0.3) + 300(0.3) = 0
My = 45 N # m
©Mz = 0;
Mz + 450(0.2) = 0
Mz = - 90 N # m
N = 450 N
Section Properties: The cross-sectional area, the moment of inertia about the y and
z axes, and the polar moment of inertia of the pipe’s cross section are
A = p (0.032 - 0.022) = 0.5p (10 - 3) m2
Iy = Iz =
J =
p
(0.034 - 0.024) = 0.1625p(10 - 6) m4
4
p
(0.034 - 0.024) = 0.325p(10 - 6) m4
2
Referring to Fig. b,
(Qz)B = 0
Normal and Shear Stress: The normal stress is a combination of axial and bending
stress.
sB =
MyzB
45( -0.03)
N
450
+
=
+
-3
A
ly
0.5p(10 )
0.1625p(10 - 6)
= - 2.358 MPa
Since (Qz)B = 0, (txy)B = 0. Also Vy = 0. Then the shear stress along the y axis is
contributed by torsional shear stress only.
(txy)B = [(txy)T]B =
60(0.03)
Tc
= 1.763 MPa
=
J
0.325p(10 - 6)
The state of stress at point B is represented on the two-dimensional element shown
in Fig. c.
913
450 N
20 mm
Section a – 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.
9–67. Continued
Construction of the Circle: sx = - 2.358 MPa, sy = 0, and txy = - 1.763 MPa .
Thus,
sx + sy
-2.358 + 0
=
= - 1.179 MPa
savg =
2
2
The coordinates of reference point A and the center C of the circle are
A( - 2.358, - 1.763)
C( -1.179, 0)
Thus, the radius of the circle is
R = CA = 2[ -2.358 - ( -1.179)]2 + (- 1.763)2 = 2.121 MPa
Using these results, the circle is shown in Fig. d.
In-Plane Principal Stresses: The coordinates of reference points B and D represent
s1 and s2, respectively.
s1 = - 1.179 + 2.121 = 0.942 MPa
Ans.
s2 = - 1.179 - 2.121 = - 3.30 MPa
Ans.
Maximum In-Plane Shear Stress: The coordinates of point E represent the state of
maximum in-plane shear stress. Thus,
tmax
= |R| = 2.12 MPa
Ans.
in-plane
Ans:
s1 = 0.942 MPa, s2 = - 3.30 MPa,
tmax = 2.12 MPa
in-plane
914
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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.
*9–68. The rotor shaft of the helicopter is subjected to the
tensile force and torque shown when the rotor blades
provide the lifting force to suspend the helicopter at midair.
If the shaft has a diameter of 6 in., determine the principal
stress and maximum in-plane shear stress at a point located
on the surface of the shaft.
50 kip
10 kip⭈ft
Internal Loadings: Considering the equilibrium of the free-body diagram of the
rotor shaft’s upper segment, Fig. a,
©Fy = 0;
N - 50 = 0
N = 50 kip
©My = 0;
T - 10 = 0
T = 10 kip # ft
Section Properties: The cross-sectional area and the polar moment of inertia of the
rotor shaft’s cross section are
A = p(32) = 9p m2
J =
p 4
(3 ) = 40.5p in4
2
Normal and Shear Stress: The normal stress is contributed by axial stress only.
sA =
50
N
=
= 1.768 ksi
A
9p
The shear stress is contributed by the torsional shear stress only.
tA =
10(12)(3)
Tc
=
= 2.829 ksi
J
40.5p
The state of stress at point A is represented by the element shown in Fig. b.
915
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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.
9–68.
Continued
Construction of the Circle: sx = 0, sy = 1.768 ksi, and txy = 2.829 ksi. Thus,
savg =
sx + sy
2
=
0 + 1.768
= 0.8842 ksi
2
The coordinates of reference point A and the center C of the circle are
A(0, 2.829)
C(0.8842, 0)
Thus, the radius of the circle is
R = CA = 2(0 - 0.8842)2 + 2.8292 = 2.964 ksi
Using these results, the circle is shown in Fig. c.
In-Plane Principal Stress: The coordinates of reference points B and D represent s1
and s2, respectively.
s1 = 0.8842 + 2.964 = 3.85 ksi
Ans.
s2 = 0.8842 - 2.964 = - 2.08 ksi
Ans.
Maximum In-Plane Shear Stress: The state of maximum shear stress is represented
by the coordinates of point E, Fig. a.
tmax
= R = 2.96 ksi
Ans.
in-plane
916
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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.
*9–69. The pedal crank for a bicycle has the cross section
shown. If it is fixed to the gear at B and does not rotate
while subjected to a force of 75 lb, determine the principal
stress in the material on the cross section at point C.
75 lb
B
3 in.
A
4 in.
C
0.4 in.
0.4 in.
0.2 in.
0.3 in.
Internal Forces and Moment: As shown on FBD
Section Properties:
I =
1
(0.3) A 0.83 B = 0.0128 in3
12
QC = y¿A¿ = 0.3(0.2)(0.3) = 0.0180 in3
Normal Stress: Applying the flexure formula.
sC = -
- 300(0.2)
My
= = 4687.5 psi = 4.6875 ksi
I
0.0128
Shear Stress: Applying the shear formula.
tC =
VQC
75.0(0.0180)
=
= 351.6 psi = 0.3516 ksi
It
0.0128(0.3)
Construction of the Circle: In accordance with the sign convention, sx = 4.6875 ksi,
sy = 0, and txy = 0.3516 ksi. Hence,
savg =
sx + sy
2
=
4.6875 + 0
= 2.34375 ksi
2
The coordinates for reference points A and C are
A(4.6875, 0.3516)
C(2.34375, 0)
The radius of the circle is
R = 2(4.6875 - 2.34375)2 + 0.35162 = 2.370 ksi
In-Plane Principal Stress: The coordinates of point B and D represent s1 and s2,
respectively.
s1 = 2.34375 + 2.370 = 4.71 ksi
Ans.
s2 = 2.34375 - 2.370 = - 0.0262 ksi
Ans.
Ans:
s1 = 4.71 ksi, s2 = - 0.0262 ksi
917
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. 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.
9–70. A spherical pressure vessel has an inner radius of
5 ft and a wall thickness of 0.5 in. Draw Mohr’s circle for
the state of stress at a point on the vessel and explain the
significance of the result. The vessel is subjected to an
internal pressure of 80 psi.
45⬚
1.25 m
Normal Stress:
s1 = s2 =
pr
80(5)(12)
=
= 4.80 ksi
2t
2(0.5)
Mohr’s circle:
A(4.80, 0)
B(4.80, 0)
C(4.80, 0)
Regardless of the orientation of the element, the shear stress is zero and the state of
stress is represented by the same two normal stress components.
Ans.
Ans:
Regardless of the orientation of the element,
the shear stress is zero and the state of stress is
represented by the same two normal stress
components.
918
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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.
9–71. The cylindrical pressure vessel has an inner radius
of 1.25 m and a wall thickness of 15 mm. It is made from
steel plates that are welded along the 45° seam. Determine
the normal and shear stress components along this seam if
the vessel is subjected to an internal pressure of 8 MPa.
sx =
45⬚
1.25 m
pr
8(1.25)
=
= 333.33 MPa
2t
2(0.015)
sy = 2sx = 666.67 MPa
A(333.33, 0)
B(666.67, 0)
C(500, 0)
333.33 + 666.67
= 500 MPa
2
Ans.
tx¿y¿ = - R = 500 - 666.67 = - 167 MPa
Ans.
sx¿ =
Ans:
sx¿ = 500 MPa, tx¿y¿ = - 167 MPa
919
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. 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.
*9–72. Determine the normal and shear stresses at point D
that act perpendicular and parallel, respectively, to the grains.
The grains at this point make an angle of 30° with the
horizontal as shown. Point D is located just to the left of the
10-kN force.
10 kN
A 100 mm
D
B
30⬚
1m
100 mm
D
100 mm
Using the method of section and consider the FBD of the left cut segment, Fig. a
+ c ©Fy = 0;
5 - V = 0
a + ©MC = 0;
V = 5 kN
M = 5 kN # m
M - 5(1) = 0
The moment of inertia of the rectangular cross - section about the neutral axis is
I =
1
(0.1)(0.33) = 0.225(10 - 3) m4
12
Referring to Fig. b,
QD = y¿A¿ = 0.1(0.1)(0.1) = 0.001 m3
The normal stress developed is contributed by bending stress only. For point D,
y = 0.05 m. Then
s =
My
5(103)(0.05)
= 1.111 MPa (T)
=
I
0.225(10 - 3)
The shear stress is contributed by the transverse shear stress only. Thus,
t =
5(103)(0.001)
VQD
= 0.2222 MPa
=
It
0.225(10 - 3)(0.1)
The state of stress at point D can be represented by the element shown in Fig. c
In accordance with the established sign convention, sx = 1.111 MPa, sy = 0 and
txy = - 0.2222 MPa, Thus.
savg =
sx + sy
2
=
1.111 + 0
= 0.5556 MPa
2
Then, the coordinate of reference point A and the center C of the circle are
A(1.111, -0.2222)
C(0.5556, 0)
Thus, the radius of the circle is given by
R = 2(1.111 - 0.5556)2 + (- 0.2222)2 = 0.5984 MPa
Using these results, the circle shown in Fig. d can be constructed.
Referring to the geometry of the circle, Fig. d,
a = tan - 1 a
0.2222
b = 21.80°
1.111 - 0.5556
b = 180° - (120° - 21.80°) = 81.80°
920
1m
300 mm
2m
C
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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.
9–72.
Continued
Then
sx¿ = 0.5556 - 0.5984 cos 81.80° = 0.4702 MPa = 470 kPa
Ans.
tx¿y¿ = 0.5984 sin 81.80° = 0.5922 MPa = 592 kPa
Ans.
921
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. 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.
9–73. Determine the principal stress at point D, which is
located just to the left of the 10-kN force.
10 kN
A 100 mm
D
B
30⬚
1m
100 mm
D
100 mm
Using the method of section and consider the FBD of the left cut segment, Fig. a,
+ c ©Fy = 0;
5 - V = 0
a + ©MC = 0;
V = 5 kN
M = 5 kN # m
M - 5(1) = 0
I =
1
(0.1)(0.33) = 0.225(10 - 3) m4
12
Referring to Fig. b,
QD = y¿A¿ = 0.1(0.1)(0.1) = 0.001 m3
The normal stress developed is contributed by bending stress only. For point D,
y = 0.05 m
s =
My
5(103)(0.05)
= 1.111 MPa (T)
=
I
0.225(10 - 3)
The shear stress is contributed by the transverse shear stress only. Thus,
t =
5(103)(0.001)
VQD
= 0.2222 MPa
=
It
0.225(10 - 3)(0.1)
The state of stress at point D can be represented by the element shown in Fig. c.
In accordance with the established sign convention, sx = 1.111 MPa, sy = 0, and
txy = - 0.2222 MPa. Thus,
savg =
sx + sy
2
=
1.111 + 0
= 0.5556 MPa
2
Then, the coordinate of reference point A and center C of the circle are
A(1.111, - 0.2222)
C(0.5556, 0)
Thus, the radius of the circle is
R = CA = 2(1.111 - 0.5556)2 + (- 0.2222)2 = 0.5984 MPa
Using these results, the circle shown in Fig. d can be constructed.
In-Plane Principal Stresses. The coordinates of points B and D represent s1 and s2,
respectively. Thus,
s1 = 0.5556 + 0.5984 = 1.15 MPa
Ans.
s2 = 0.5556 - 0.5984 = - 0.0428 MPa
Ans.
922
1m
300 mm
2m
C
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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.
9–73.
Continued
Referring to the geometry of the circle, Fig. d,
tan (2uP)1 =
0.2222
= 0.4
1.111 - 0.5556
(uP)1 = 10.9° (Clockwise)
Ans.
The state of principal stresses is represented by the element shown in Fig. e.
Ans:
s1 = 1.15 MPa, s2 = - 0.0428 MPa
923
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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.
9–74. If the box wrench is subjected to the 50 lb force,
determine the principal stress and maximum in-plane shear
stress at point A on the cross section of the wrench at
section a–a. Specify the orientation of these states of stress
and indicate the results on elements at the point.
12 in.
50 lb
0.5 in.
2 in.
a
a
A
B
Section a – a
Internal Loadings: Considering the equilibrium of the free-body diagram of the
wrench’s segment, Fig. a,
©Fy = 0;
Vy + 50 = 0
Vy = - 50 lb
©Mx = 0;
T + 50(12) = 0
T = - 600 lb # in
©Mz = 0;
Mz - 50(2) = 0 Mz = 100 lb # in
Section Properties: The moment of inertia about the z axis and the polar moment of
inertia of the wrench’s cross section are
Iz =
p
(0.54) = 0.015625p in4
4
J =
p
(0.54) = 0.03125p in4
2
Referring to Fig. b,
(Qy)A = y¿A¿ =
4(0.5) p
c (0.52) d = 0.08333 in3
3p 2
Normal and Shear Stress: The shear stress of point A along the z axis is (txz)A = 0.
However, the shear stress along the y axis is a combination of torsional and
transverse shear stress.
(txy)A = [(txy)T]A - [(txy)V]A
=
Vy(Qy)A
Tc
+
J
lzt
=
600(0.5)
-50(0.08333)
+
= 2.971 ksi
0.03125p
0.015625p(l)
The state of stress at point A is represented by the two-dimensional element shown
in Fig. c.
924
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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.
9–74.
Continued
Construction of the Circle: sx = sy = 0, and txy = 2.971 ksi. Thus,
savg =
sx + sy
2
=
0 + 0
= 0
2
The coordinates of reference point A and the center C of the circle are
A(0, 2.971)
C(0, 0)
Thus, the radius of the circle is
R = CA = 2(0 - 0)2 + 2.9712 = 2.971 ksi
Using these results, the circle is shown in Fig. d.
In-Plane Principal Stress: The coordinates of reference points B and D represent s1
and s2, respectively.
s1 = 0 + 2.971 = 2.97 ksi
Ans.
s2 = 0 - 2.971 = - 2.97 ksi
Ans.
Maximum In-Plane Shear Stress: Since there is no normal stress acting on the
element,
tmax
in-plane
= (txy)A = 2.97 ksi
Ans.
Ans:
s1 = 2.97 ksi, s2 = - 2.97 ksi, up1 = 45.0°,
up2 = - 45.0°, tmax = 2.97 ksi, us = 0°
in-plane
925
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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.
9–75. If the box wrench is subjected to the 50 lb force,
determine the principal stress and maximum in-plane shear
stress at point B on the cross section of the wrench at
section a–a. Specify the orientation of these states of stress
and indicate the results on elements at the point.
12 in.
50 lb
0.5 in.
2 in.
a
a
A
B
Section a – a
Internal Loadings: Considering the equilibrium of the free-body diagram of the
wrench’s cut segment, Fig. a,
©Fy = 0;
Vy + 50 = 0
Vy = - 50 lb
©Mx = 0;
T + 50(12) = 0
T = - 600 lb # in
©Mz = 0;
Mz - 50(2) = 0 Mz = 100 lb # in
Section Properties: The moment of inertia about the z axis and the polar moment of
inertia of the wrench’s cross section are
Iz =
p
(0.54) = 0.015625p in4
4
J =
p
(0.54) = 0.03125p in4
2
Referring to Fig. b,
(Qy)B = 0
Normal and Shear Stress: The normal stress is caused by the bending stress
due to Mz.
(sx)B = -
MzyB
Iz
= -
100(0.5)
= - 1.019 ksi
0.015625p
The shear stress at point B along the y axis is (txy)B = 0 since (Qy)B. However, the
shear stress along the z axis is caused by torsion.
(txz)B =
600(0.5)
Tc
=
= 3.056 ksi
J
0.03125p
The state of stress at point B is represented by the two-dimensional element shown
in Fig. c.
926
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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.
9–75.
Continued
Construction of the Circle: sx = - 1.019 ksi, sz = 0, and txz = - 3.056 ksi . Thus,
savg =
sx + sy
2
=
- 1.019 + 0
= - 0.5093 ksi
2
The coordinates of reference point A and the center C of the circle are
A( - 1.019, - 3.056)
C( - 0.5093, 0)
Thus, the radius of the circle is
R = CA = 2 [- 1.019 - ( -0.5093)]2 + ( -3.056)2 = 3.0979 ksi
Using these results, the circle is shown in Fig. d.
In-Plane Principal Stress: The coordinates of reference points B and D represent
s1 and s2, respectively.
s1 = - 0.5093 + 3.0979 = 2.59 ksi
Ans.
s2 = - 0.5093 - 3.0979 = - 3.61 ksi
Ans.
Maximum In-Plane Shear Stress: The coordinates of point E represent the
maximum in-plane stress, Fig. a.
tmax
= R = 3.10 ksi
Ans.
in-plane
Ans:
s1 = 2.59 ksi, s2 = - 3.61 ksi, up1 = - 40.3°,
up2 = 49.7, tmax
= 3.10 ksi , us = 4.73°
in-plane
927
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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.
*9–76. The ladder is supported on the rough surface at A
and by a smooth wall at B. If a man weighing 150 lb stands
upright at C, determine the principal stresses in one of the
legs at point D. Each leg is made from a 1-in.-thick board
having a rectangular cross section. Assume that the total
weight of the man is exerted vertically on the rung at C and
is shared equally by each of the ladder’s two legs. Neglect
the weight of the ladder and the forces developed by the
man’s arms.
B
C
12 ft
3 in.
1 in.
D
3 in.
1 in.
1 in.
5 ft
D
4 ft
A
A = 3(1) = 3 in2
I =
1
(1)(33) = 2.25 in4
12
QD = y¿A¿ = (1)(1)(1) = 1 in3
sD =
My
35.52(12)(0.5)
-P
- 77.55
=
= - 120.570 psi
A
I
3
2.25
tD =
8.88(1)
VQD
=
= 3.947 psi
It
2.25(1)
A( -120.57, - 3.947)
B(0, 3.947)
C( -60.285, 0)
R = 2(60.285)2 + (3.947)2 = 60.412
s1 = - 60.285 + 60.4125 = 0.129 psi
Ans.
s2 = - 60.285 - 60.4125 = - 121 psi
Ans.
928
5 ft
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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.
9–77. Draw the three Mohr’s circles that describe each of
the following states of stress.
5 ksi
(a) Here, smin = 0, sint = 3 ksi and smax = 5 ksi. The three Mohr’s circles
of this state of stress are shown in Fig. a
3 ksi
(b) Here, smin = 0, sint = 140 MPa and smax = 180 MPa. The three
Mohr’s circles of this state of stress are shown in Fig. b
(a)
929
180 MPa
140 MPa
(b)
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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.
9–78. Draw the three Mohr’s circles that describe the
following state of stress.
300 psi
Here, smin = - 300 psi, sint = 0 and smax = 400 psi. The three Mohr’s circles for
this state of stress are shown in Fig. a.
400 psi
930
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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.
z
9–79. The stress at a point is shown on the element.
Determine the principal stresses and the absolute maximum
shear stress.
x
y
90 MPa
20 MPa
60 MPa
For y – z plane:
The center of the cricle is at sAvg =
sy + sz
2
=
60 + 90
= 75 MPa
2
R = 2(75 - 60)2 + (- 20)2 = 25 MPa
s1 = 75 + 25 = 100 MPa
s2 = 75 - 25 = 50 MPa
Thus,
tabs
max
=
s1 = 100
Ans.
s2 = 50 MPa
Ans.
s3 = 0 MPa
Ans.
smax - smin
100 - 0
=
= 50 MPa
2
2
Ans.
Ans:
s1 = 100 MPa, s2 = 50 MPa, s3 = 0,
tabs = 50 MPa
max
931
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. 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.
z
*9–80. The stress at a point is shown on the element.
Determine the principal stresses and the absolute maximum
shear stress.
x
Mohr’s circle for the element in the y–z plane, Fig. a, will be drawn first. In
accordance with the established sign convention, sy = 30 psi, sz = 120 psi and
tyz = 70 psi. Thus
savg =
sy + sz
2
=
30 + 120
= 75 psi
2
C(75, 0)
Thus, the radius of the circle is
R = CA = 2(75 - 30)2 + 702 = 83.217 psi
Using these results, the circle shown in Fig. b.
The coordinates of point B and D represent the principal stresses
From the results,
smax = 158 psi
smin = - 8.22 psi
sint = 0 psi
Ans.
Using these results, the three Mohr’s circles are shown in Fig. c,
From the geometry of the three circles,
tabs
max
=
120 psi
70 psi
30 psi
Thus the coordinates of reference point A and the center C of the circle are
A(30, 70)
y
158.22 - ( -8.22)
smax - smin
=
= 83.2 psi
2
2
932
Ans.
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. 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.
z
*9–81. The stress at a point is shown on the element.
Determine the principal stresses and the absolute maximum
shear stress.
x
y
6 ksi
Mohr’s circle for the element in x–z plane, Fig. a, will be drawn first. In accordance
with the established sign convention, sx = - 1 ksi , sz = 0 and txz = 6 ksi . Thus
savg =
sx + sz
=
2
-1 - 6
= - 3.5 ksi
2
1 ksi
1 ksi
Thus, the coordinates of reference point A and the center C of the circle are
A(- 1, 1)
C( - 3.5, 0)
Thus, the radius of the circle is
R = CA = 2[- 1 - ( - 3.5)]2 + 12 = 2.69 ksi
Using these results, the circle is shown in Fig. b,
The coordinates of points B and D represent s2 and s3, respectively.
s2 = - (3.5 - 2.69) = - 0.807 ksi
s3 = - (3.5 + 2.69) = - 6.19 ksi
From the results obtained,
s2 = - 0.807 ksi
tabs
max
=
s3 = - 6.19 ksi
s1 = 0 ksi
Ans.
smax - smin
- 6.19 - 0
=
= -3.10 ksi
2
2
Ans.
Ans:
s2 = - 0.807 ksi, s3 = - 6.19 ksi, s1 = 0,
tabs = - 3.10 ksi
max
933
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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.
z
9–82. The stress at a point is shown on the element.
Determine the principal stresses and the absolute maximum
shear stress.
y
x
150 MPa
120 MPa
For x – z plane:
R = CA = 2(120 - 60)2 + 1502 = 161.55
s1 = 60 + 161.55 = 221.55 MPa
s2 = 60 - 161.55 = - 101.55 MPa
s1 = 222 MPa
tabs
max
=
s2 = 0 MPa
s3 = - 102 MPa
Ans.
221.55 - ( - 101.55)
smax - smin
=
= 162 MPa
2
2
Ans.
Ans:
s1 = 222 MPa, s2 = 0 MPa, s3 = - 102 MPa,
tabs = 162 MPa
max
934
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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.
z
9–83. The state of stress at a point is shown on the
element. Determine the principal stresses and the absolute
maximum shear stress.
x
For y - z plane:
A(5, - 4)
B( - 2.5, 4)
y
2.5 ksi
C(1.25, 0)
4 ksi
R = 23.752 + 42 = 5.483
s1 = 1.25 + 5.483 = 6.733 ksi
5 ksi
s2 = 1.25 - 5.483 = - 4.233 ksi
Thus,
savg =
tabs
max
=
s1 = 6.73 ksi
Ans.
s2 = 0
Ans.
s3 = - 4.23 ksi
Ans.
6.73 + (- 4.23)
= 1.25 ksi
2
6.73 - ( -4.23)
smax - smin
=
= 5.48 ksi
2
2
Ans.
Ans:
s1 = 6.73 ksi, s2 = 0, s3 = - 4.23 ksi,
tabs = 5.48 ksi
max
935
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. 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.
9–85. The solid cylinder having a radius r is placed in a
sealed container and subjected to a pressure p. Determine
the stress components acting at point A located on the
center line of the cylinder. Draw Mohr’s circles for the
element at this point.
A
r
p
-s(dz)(2r) =
L0
p(r du) dz sin u
u
- 2s = p
p
L0
sin u du = p( - cos u)|0
s = -p
The stress in every direction is s1 = s2 = s3 = - p
Ans.
Ans:
The stress in every direction is
s1 = s2 = s3 = - p
936
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. 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.
9–86. The plate is subjected to a tensile force P = 5 kip. If
it has the dimensions shown, determine the principal
stresses and the absolute maximum shear stress. If the
material is ductile it will fail in shear. Make a sketch of the
plate showing how this failure would appear. If the material
is brittle the plate will fail due to the principal stresses.
Show how this failure occurs.
s =
P ⫽ 5 kip
2 in.
2 in. P ⫽ 5 kip
12 in.
0.5 in.
P
5000
=
= 2500 psi = 2.50 ksi
A
(4)(0.5)
s1 = 2.50 ksi
Ans.
s2 = s3 = 0
Ans.
tabs =
max
s1
= 1.25 ksi
2
Ans.
Failure by shear:
Failure by principal stress:
Ans:
s1 = 2.50 ksi, s2 = s3 = 0, tabs = 1.25 ksi
max
937
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. 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.
9–87. Determine the principal stresses and absolute
maximum shear stress developed at point A on the cross
section of the bracket at section a–a.
12 in.
6 in.
5
3
a
4
a
0.5 in.
B
0.25 in.
A
0.25 in.
0.25 in.
1.5 in.1.5 in.
Section a – a
Internal Loadings: Considering the equilibrium of the free-body diagram of the
bracket’s upper cut segment, Fig. a,
+ c ©Fy = 0;
3
N - 500 a b = 0
5
N = 300 lb
+ ©F = 0;
;
x
4
V - 500 a b = 0
5
V = 400 lb
3
4
©MO = 0; M - 500 a b (12) - 500 a b (6) = 0
5
5
M = 6000 lb # in
Section Properties: The cross-sectional area and the moment of inertia of the
bracket’s cross section are
A = 0.5(3) - 0.25(2.5) = 0.875 in2
I =
1
1
(0.5) A 33 B (0.25) A 2.53 B = 0.79948 in4
12
12
Referring to Fig. b.
QA = x1œ A1œ + x2œ A2œ = 0.625(1.25)(0.25) + 1.375(0.25)(0.5) = 0.3672 in3
Normal and Shear Stress: The normal stress is
sA =
300
N
= = - 342.86 psi
A
0.875
The shear stress is contributed by the transverse shear stress.
tA =
VQA
400(0.3672)
=
= 734.85 psi
It
0.79948(0.25)
The state of stress at point A is represented by the element shown in Fig. c.
Construction of the Circle: sx = 0, sy = - 342.86 psi, and txy = 734.85. Thus,
savg =
sx + sy
2
=
0 + ( - 342.86)
= - 171.43 psi
2
The coordinates of reference point A and the center C of the circle are
A(0, 734.85)
C( - 171.43, 0)
Thus, the radius of the circle is
R = CA = 2[0 - ( -171.43)]2 + 734.852 = 754.58 psi
938
500 lb
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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.
9–87.
Continued
Using these results, the circle is shown in Fig. d.
In-Plane Principal Stresses: The coordinates of reference point B and D represent
s1 and s2, respectively.
s1 = - 171.43 + 754.58 = 583.2 psi
s2 = - 171.43 - 754.58 = - 926.0 psi
Three Mohr’s Circles: Using these results,
smax = 583 psi
sint = 0 smin = - 926 psi
Ans.
Absolute Maximum Shear Stress:
tabs
max
=
583.2 - ( - 926.0)
smax - smin
=
= 755 psi
2
2
Ans.
Ans:
s1 = 583 psi, s2 = 0, s3 = - 926 psi,
tabs = 755 psi
max
939
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. 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.
*9–88. Determine the principal stresses and absolute
maximum shear stress developed at point B on the cross
section of the bracket at section a–a.
12 in.
Internal Loadings: Considering the equilibrium of the free-body diagram of the 6 in.
bracket’s upper cut segment, Fig. a,
+ c ©Fy = 0;
3
N - 500 a b = 0
5
+ ©F = 0;
;
x
4
V - 500 a b = 0
5
B
Section Properties: The cross-sectional area and the moment of inertia about the
centroidal axis of the bracket’s cross section are
A = 0.5(3) - 0.25(2.5) = 0.875 in2
1
1
(0.5) A 33 B (0.25) A 2.53 B = 0.79948 in4
12
12
Referring to Fig. b,
QB = 0
Normal and Shear Stress: The normal stress is a combination of axial and bending
stress.
sB =
6000(1.5)
MxB
N
300
+
= +
= 10.9 ksi
A
I
0.875
0.79948
Since QB = 0, tB = 0. The state of stress at point B is represented on the element
shown in Fig. c.
In-Plane Principal Stresses: Since no shear stress acts on the element,
s1 = 10.91 ksi
s2 = 0
Three Mohr’s Circles: Using these results,
smax = 10.91 ksi
sint = smin = 0
Ans.
Absolute Maximum Shear Stress:
tabs
max
=
0.25 in.
1.5 in.1.5 in.
Section a – a
M = 6000 lb # in
smax - smin
10.91 - 0
=
= 5.46 ksi
2
2
Ans.
940
500 lb
0.25 in.
A
0.25 in.
V = 400 lb
4
3
©MO = 0; M - 500 a b (12) - 500 a b (6) = 0
5
5
I =
4
a
0.5 in.
N = 300 lb
5
3
a
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. 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.
9–89. The solid propeller shaft on a ship extends outward
from the hull. During operation it turns at v = 15 rad>s
when the engine develops 900 kW of power. This causes a
thrust of F = 1.23 MN on the shaft. If the shaft has an outer
diameter of 250 mm, determine the principal stresses at any
point located on the surface of the shaft.
0.75 m
A
T
F
Power Transmission: Using the formula developed in Chapter 5,
P = 900 kW = 0.900 A 106 B N # m>s
0.900(106)
P
=
= 60.0 A 103 B N # m
v
15
T0 =
Internal Torque and Force: As shown on FBD.
Section Properties:
A =
p
A 0.252 B = 0.015625p m2
4
J =
p
A 0.1254 B = 0.3835 A 10 - 3 B m4
2
Normal Stress:
s =
- 1.23(106)
N
=
= - 25.06 MPa
A
0.015625p
Shear Stress: Applying the torsion formula,
t =
60.0(103) (0.125)
Tc
= 19.56 MPa
=
J
0.3835(10 - 3)
In-Plane Principal Stresses: sx = - 25.06 MPa, sy = 0 and txy = 19.56 MPa for
any point on the shaft’s surface. Applying Eq. 9-5,
s1,2 =
=
sx + sy
2
;
C
a
sx - sy
2
2
b + t2xy
- 25.06 - 0 2
- 25.06 + 0
;
a
b + (19.56)2
2
C
2
= - 12.53 ; 23.23
s1 = 10.7 MPa
s2 = - 35.8 MPa
Ans.
Ans:
s1 = 10.7 MPa, s2 = - 35.8 MPa
941
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. 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.
9–90. The solid propeller shaft on a ship extends outward
from the hull. During operation it turns at v = 15 rad>s
when the engine develops 900 kW of power. This causes a
thrust of F = 1.23 MN on the shaft. If the shaft has a
diameter of 250 mm, determine the maximum in-plane shear
stress at any point located on the surface of the shaft.
0.75 m
A
T
F
Power Transmission: Using the formula developed in Chapter 5,
P = 900 kW = 0.900 A 106 B N # m>s
T0 =
0.900(106)
P
=
= 60.0 A 103 B N # m
v
15
Internal Torque and Force: As shown on FBD.
Section Properties:
A =
p
A 0.252 B = 0.015625p m2
4
J =
p
A 0.1254 B = 0.3835 A 10 - 3 B m4
2
Normal Stress:
s =
- 1.23(106)
N
=
= - 25.06 MPa
A
0.015625p
Shear Stress: Applying the torsion formula.
t =
60.0(103) (0.125)
Tc
= 19.56 MPa
=
J
0.3835 (10 - 3)
Maximum In-Plane Principal Shear Stress: sx = - 25.06 MPa, sy = 0, and
txy = 19.56 MPa for any point on the shaft’s surface. Applying Eq. 9-7,
t
max
in-plane
a
sx - sy
2
b + t2xy
=
C
=
- 25.06 - 0 2
b + (19.56)2
C
2
2
a
= 23.2 MPa
Ans.
Ans:
tmax
in-plane
942
= 23.2 MPa
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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.
9–91. The steel pipe has an inner diameter of 2.75 in. and
an outer diameter of 3 in. If it is fixed at C and subjected to
the horizontal 20-lb force acting on the handle of the pipe
wrench at its end, determine the principal stresses in the
pipe at point A, which is located on the surface of the pipe.
20 lb
12 in.
10 in.
Internal Forces, Torque and Moment: As shown on FBD.
A
Section Properties:
B
p
I =
A 1.54 - 1.3754 B = 1.1687 in4
4
J =
C
p
A 1.54 - 1.3754 B = 2.3374 in4
2
y
z
x
(QA)z = ©y¿A¿
=
4(1.5) 1
4(1.375) 1
c p A 1.52 B d c p A 1.3752 B d
3p 2
3p
2
= 0.51693 in3
Normal Stress: Applying the flexure formula s =
sA =
My z
Iy
,
200(0)
= 0
1.1687
Shear Stress: The transverse shear stress in the z direction and the torsional shear
VQ
stress can be obtained using shear formula and torsion formula, tv =
and
It
Tr
ttwist =
, respectively.
J
tA = (tv)z - ttwist
=
20.0(0.51693)
240(1.5)
1.1687(2)(0.125)
2.3374
= - 118.6 psi
In-Plane Principal Stress: sx = 0, sz = 0 and txz = - 118.6 psi for point A.
Applying Eq. 9-5
s1,2 =
sx + sz
2
;
C
a
sx - sz
2
2
b + t2xz
= 0 ; 20 + ( - 118.6)2
s1 = 119 psi
s2 = - 119 psi
Ans.
Ans:
s1 = 119 psi, s2 = - 119 psi
943
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. 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.
*9–92. Solve Prob. 9–91 for point B, which is located on
the surface of the pipe.
20 lb
12 in.
10 in.
A
B
Internal Forces, Torque and Moment: As shown on FBD.
C
y
Section Properties:
z
I =
p
A 1.54 - 1.3754 B = 1.1687 in4
4
J =
p
A 1.54 - 1.3754 B = 2.3374 in4
2
x
(QB)z = 0
Normal Stress: Applying the flexure formula s =
sB =
My z
Iv
,
200(1.5)
= 256.7 psi
1.1687
Shear Stress: Torsional shear stress can be obtained using torsion formula,
Tr
.
ttwist =
J
tB = ttwist =
240(1.5)
= 154.0 psi
2.3374
In - Plane Prinicipal Stress: sx = 256.7 psi, sy = 0, and txy = - 154.0 psi for point B.
Applying Eq. 9-5
s1,2 =
=
sx + sy
2
;
C
sx - sy
a
2
2
b + t2xy
256.7 - 0 2
256.7 + 0
a
;
b + ( -154.0)2
2
C
2
= 128.35 ; 200.49
s1 = 329 psi
s2 = - 72.1 psi
Ans.
944
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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.
9–93. Determine the equivalent state of stress if an
element is oriented 40° clockwise from the element shown.
Use Mohr’s circle.
10 ksi
6 ksi
A(6, 0)
B(- 10, 0)
C( - 2, 0)
R = CA = CB = 8
sx¿ = - 2 + 8 cos 80° = - 0.611 ksi
Ans.
tx¿y¿ = 8 sin 80° = 7.88 ksi
Ans.
sy¿ = - 2 - 8 cos 80° = - 3.39 ksi
Ans.
Ans:
sx¿ = - 0.611 ksi, tx¿y¿ = 7.88 ksi, sy¿ = - 3.39 ksi
945
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. 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.
9–94. The crane is used to support the 350-lb load.
Determine the principal stresses acting in the boom at
points A and B. The cross section is rectangular and has a
width of 6 in. and a thickness of 3 in. Use Mohr’s circle.
5 ft
3 in.
5 ft
B
A
45°
A = 6(3) = 18 in2
I =
45°
1
(3)(63) = 54 in4
12
QB = (1.5)(3)(3) = 13.5 in3
QA = 0
For point A:
sA = -
My
1750(12)(3)
P
597.49
=
= - 1200 psi
A
I
18
54
tA = 0
s1 = 0
Ans.
s2 = - 1200 psi = - 1.20 ksi
Ans.
For point B:
sB = -
tB =
P
597.49
= = - 33.19 psi
A
18
VQB
247.49(13.5)
=
= 20.62 psi
It
54(3)
A(-33.19, - 20.62)
B(0, 20.62)
C( - 16.60, 0)
R = 216.602 + 20.622 = 26.47
s1 = - 16.60 + 26.47 = 9.88 psi
Ans.
s2 = - 16.60 - 26.47 = - 43.1 psi
Ans.
Ans:
Point A: s1 = 0, s2 = - 1.20 ksi,
Point B: s1 = 9.88 psi, s2 = - 43.1 psi
946
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. 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.
9–95. Determine the equivalent state of stress on an
element at the same point which represents (a) the
principal stresses, and (b) the maximum in-plane shear
stress and the associated average normal stress. Also, for
each case, determine the corresponding orientation of the
element with respect to the element shown and sketch the
results on the element.
30 ksi
10 ksi
Normal and Shear Stress:
sx = 0
sy = - 30 ksi
txy = - 10 ksi
In-Plane Principal Stresses:
s1, 2 =
=
sx + sy
;
2
A
a
sx - sy
2
2
2
b + txy
0 + ( -30)
0 - ( - 30)2
2
;
b + (- 10)
2
Aa
2
= - 15 ; 2325
s1 = 3.03 ksi
s2 = - 33.0 ksi
Ans.
Orientation of Principal Plane:
tan 2up =
txy
(sx - sy)>2
=
- 10
= - 0.6667
[0 - (- 30)]>2
up = - 16.845° and 73.15°
Substituting u = - 16.845° into
sx + sy
sx¿ =
=
sx - sy
+
2
2
cos 2u + txy sin 2u
0 + ( - 30)
0 - ( -30)
+
cos ( - 33.69°) - 10 sin ( -33.69°)
2
2
= 3.03 ksi = s1
Thus,
(up)1 = - 16.8° and (up)2 = 73.2°
Ans.
The element that represents the state of principal stress is shown in Fig. a.
Maximum In-Plane Shear Stress:
tmax
in-plane
=
a
B
sx - sy
2
2
b + txy2 =
0 - ( - 30) 2
b + ( -10)2 = 18.0 ksi
2
B
a
947
Ans.
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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.
9–95. Continued
Orientation of the Plane of Maximum In-Plane Shear Stress:
tan 2us = -
(sx - sy)>2
txy
=
[0 - (- 30)]>2
= 1.5
- 10
us = 28.2° and 118°
By inspection, tmax
has to act in the same sense shown in Fig. b to maintain
in-plane
equilibrium.
Average Normal Stress:
savg =
sx + sy
2
=
0 + ( - 30)
= - 15 ksi
2
Ans.
The element that represents the state of maximum in-plane shear stress is shown
in Fig. c.
Ans:
s1 = 3.03 ksi, s2 = - 33.0 ksi,
up1 = - 16.8° and up2 = 73.2°,
tmax
= 18.0 ksi , savg = - 15 ksi, us = 28.2°
in-plane
948
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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.
*9–96. The propeller shaft of the tugboat is subjected to
the compressive force and torque shown. If the shaft has an
inner diameter of 100 mm and an outer diameter of 150 mm,
determine the principal stress at a point A located on the
outer surface.
10 kN
A
2 kN·m
Internal Loadings: Considering the equilibrium of the free-body diagram of the
propeller shaft’s right segment, Fig. a,
©Fx = 0; 10 - N = 0
N = 10 kN
©Mx = 0; T - 2 = 0
T = 2 kN # m
Section Properties: The cross - sectional area and the polar moment of inertia of the
propeller shaft’s cross section are
A = p A 0.0752 - 0.052 B = 3.125p A 10 - 3 B m2
J =
p
A 0.0754 - 0.054 B = 12.6953125p A 10 - 6 B m4
2
Normal and Shear Stress: The normal stress is a contributed by axial stress only.
sA =
10 A 103 B
N
= = - 1.019 MPa
A
3.125p A 10 - 3 B
The shear stress is contributed by the torsional shear stress only.
tA =
2 A 103 B (0.075)
Tc
=
= 3.761 MPa
J
12.6953125p A 10 - 6 B
The state of stress at point A is represented by the element shown in Fig. b.
Construction of the Circle: sx = - 1.019 MPa, sy = 0, and txy = - 3.761 MPa.
Thus,
savg =
sx + sy
2
=
-1.019 + 0
= - 0.5093 MPa
2
The coordinates of reference point A and the center C of the circle are
A(-1.019, - 3.761)
C( - 0.5093, 0)
Thus, the radius of the circle is
R = CA = 2[- 1.019 - ( - 0.5093)]2 + (- 3.761)2 = 3.795 MPa
Using these results, the circle is shown in Fig. c.
In-Plane Principal Stress: The coordinates of reference points B and D represent s1
and s2, respectively.
s1 = - 0.5093 + 3.795 = 3.29 MPa
Ans.
s2 = - 0.5093 - 3.795 = - 4.30 MPa
Ans.
949
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. 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.
9–96.
Continued
Orientation of the Principal Plane: Referring to the geometry of the circle, Fig. c,
tan 2 A up B 2 =
3.761
= 7.3846
1.019 - 0.5093
A up B 2 = 41.1° (clockwise)
Ans.
The state of principal stresses is represented on the element shown in Fig. d.
950
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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.
*9–97. The box beam is subjected to the loading shown.
Determine the principal stress in the beam at points A and B.
1200 lb
800 lb
6 in.
A
6 in. B 8 in.
8 in.
A
B
3 ft
2.5 ft
2.5 ft
5 ft
Support Reactions: As shown on FBD(a).
Internal Forces and Moment: As shown on FBD(b).
Section Properties:
I =
1
1
(8) A 83 B (6) A 63 B = 233.33 in4
12
12
QA = QB = 0
Normal Stress: Applying the flexure formula.
s = -
My
I
sA = -
- 300(12)(4)
= 61.71 psi
233.33
sB = -
- 300(12)(- 3)
= - 46.29 psi
233.33
Shear Stress: Since QA = QB = 0, then tA = tB = 0.
In-Plane Principal Stress: sx = 61.71 psi, sy = 0, and txy = 0 for point A. Since no
shear stress acts on the element,
s1 = sx = 61.7 psi
Ans.
s2 = sy = 0
Ans.
sx = - 46.29 psi, sy = 0, and txy = 0 for point B. Since no shear stress acts on the
element,
s1 = sy = 0
Ans.
s2 = sx = - 46.3 psi
Ans.
Ans:
Point A: s1 = 61.7 psi, s2 = 0
Point B: s1 = 0, s2 = - 46.3 psi
951
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. 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.
9–98. The state of stress at a point is shown on the element.
Determine (a) the principal stresses and (b) the maximum
in-plane shear stress and average normal stress at the point.
Specify the orientation of the element in each case.
60 MPa
30 MPa
45 MPa
a) sx = 45 MPa
s1, 2 =
=
txy = 30 MPa
sy = - 60 MPa
sx + sy
Aa
;
2
sx - sy
2
2
b + txy
2
45 - 60
45 - ( -60) 2
2
; a
b + 30
2
A
2
tan 2up =
Ans.
s2 = - 68.0 MPa
s1 = 53.0 MPa
txy
sx - sy
2
30
=
45 - ( - 60)
2
= 0.5714
up = 14.87° and - 75.13°
Use Eq. 9–1 to determine the principal plane of s1 and s2
sx + sy
sx ¿ =
sx - sy
+
2
2
cos 2u + txy sin 2u
u = uy = 14.87°
sx¿ =
45 - ( -60)
45 + ( -60)
+
cos 29.74° + 30 sin 29.74° = 53.0 MPa
2
2
Therefore, up1 = 14.9° ;
b) tmax
=
A
in-plane
savg =
a
sx - sy
2
sx + sy
2
tan 2us = us = - 30.1°
=
2
=
b + txy
2
Aa
Ans.
45 - ( -60) 2
2
b + 30 = 60.5 MPa
2
45 + ( -60)
= - 7.50 MPa
2
(sx - sy)
2
txy
up2 = - 75.1°
= -
[45 - ( - 60)]
2
Ans.
30
Ans.
Ans.
= -1.75
and
59.9°
Ans.
By observation, in order to preserve equilibrium, tmax = 60.5 MPa has to act in the
direction shown in the figure.
Ans:
s1 = 53.0 MPa, s2 = - 68.0 MPa,
up1 = 14.9°, up2 = - 75.1°, tmax
= 60.5 MPa,
in-plane
savg = - 7.50 MPa, us = - 30.1° and 59.9°
952
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. 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.
9–99. The state of stress at a point in a member is shown on
the element. Determine the stress components acting on the
inclined plane AB. Solve the problem using the method of
equilibrium described in Sec. 9.1.
14 ksi
A
20 ksi
50⬚
B
+ Q©F
x¿
= 0;
sx¿ ¢A + 14 ¢A sin 50° cos 40° + 20 ¢A cos 50° cos 50° = 0
sx¿ = - 16.5 ksi
a + ©Fy¿ = 0;
Ans.
tx¿y¿ ¢A + 14 ¢A sin 50° sin 40° - 20 ¢A cos 50° sin 50° = 0
tx¿y¿ = 2.95 ksi
Ans.
Ans:
sx¿ = - 16.5 ksi, tx¿y¿ = 2.95 ksi
953
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. 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.
10–1. Prove that the sum of the normal strains in
perpendicular directions is constant.
Px + Py
Px¿ =
2
Px - Py
+
Px + Py
Py¿ =
2
2
Px - Py
-
2
cos 2u +
cos 2u -
gxy
2
gxy
2
sin 2u
(1)
sin 2u
(2)
Adding Eq. (1) and Eq. (2) yields:
Px¿ + Py¿ = Px + Py = constant
(Q.E.D.)
954
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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.
10–2. 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 with the established sign convention,
Px = 200(10 - 6),
Px + Py
Px¿ =
Px - Py
+
2
= c
Py = - 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.
Px - Py
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)
Px + Py
Py¿ =
= c
2
Ans.
Px - Py
-
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
Ans:
Px¿ = 248(10 - 6), gx¿y¿ = - 233(10 - 6),
Py¿ = - 348(10 - 6)
955
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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.
10–3. The state of strain at a point on a wrench
Px = 120110-62,
has
components
Py = - 180110-62,
gxy = 150(10-62. 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 x–y plane.
Px = 120(10 - 6)
Py = - 180(10 - 6)
; Aa
Px + Py
a)
P1, 2 =
2
= c
Px - Py
2
2
b + a
gxy = 150(10 - 6)
gxy
2
b
2
120 + ( - 180)
120 - ( -180) 2
150 2 10 - 6
;
b
+
a
b d
a
A
2
2
2
P1 = 138(10 - 6);
P2 = - 198(10 - 6)
Ans.
Orientation of P1 and P2
tan 2up =
gxy
=
Px - Py
150
= 0.5
[120 - (- 180)]
up = 13.28° and - 76.72°
Use Eq. 10–5 to determine the direction of P1 and P2
Px + Py
Py¿ =
2
Px - Py
+
cos 2u +
2
gxy
2
sin 2u
u = up = 13.28°
Py¿ =
J
120 + ( -180)
120 - ( - 180)
150
+
cos (26.56°) +
sin 26.56° d10 - 6
2
2
2
= 138(10 - 6) = P1
Therefore up1 = 13.3°;
gmax
b)
in-plane
2
gmax
in-plane
=
Aa
= 2c
Px - Py
Aa
2
2
Ans.
gxy 2
2
b + a 2 b
2
120 - ( -180) 2
+ a 150 b d 10 - 6 = 335(10 - 6)
b
2
2
Px + Py
Pavg =
up2 = - 76.7°
= c
120 + (- 180)
d 10 - 6 = - 30.0(10 - 6)
2
956
Ans.
Ans.
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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.
10–3. Continued
Orientation of gmax
tan 2us =
- (Px - Py)
gxy
=
us = - 31.7°
and
-[120 - ( - 180)]
= - 2.0
150
Ans.
58.3°
Use Eq. 10–11 to determine the sign of gmax
in-plane
gx¿y¿
2
Px - Py
= -
2
sin 2u +
gxy
2
cos 2u
u = us = - 31.7°
gx¿y¿ = 2 c -
120 - (- 180)
150
sin ( - 63.4°) +
cos ( -63.4°) d10 - 6 = 335(10 - 6)
2
2
Ans:
P1 = 138(10 - 6), P2 = - 198(10 - 6),
up1 = 13.3°, up2 = - 76.7°,
gmax
= 335(10 - 6), Pavg = - 30.0(10 - 6)
in-plane
us = - 31.7° and 58.3°
957
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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
*10–4. The state of strain at the point on the gear tooth has
components
gxy =
Px = 850110-62,
Py = 480110-62,
-6
650(10 2. 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.
Px = 850(10 - 6)
Py = 480(10 - 6)
Px + Py
a)
P1, 2 =
2
= c
Aa
;
Px - Py
2
gxy = 650(10 - 6)
b + a
2
x
gxy
2
b
2
2
2
850 + 480
; a 850 - 480 b + a 650 b d(10 - 6)
2
A
2
2
P1 = 1039(10 - 6)
P2 = 291(10 - 6)
Ans.
Ans.
Orientation of P1 and P2:
gxy
tan 2uy =
650
850 # 480
=
Px - Py
uy = 30.18°
and
120.18°
Use Eq. 10–5 to determine the direction of P1 and P2:
Px + Py
Px¿ =
Px - Py
+
2
2
cos 2u +
gxy
2
sin 2u
u = uy = 30.18°
Px¿ = c
850 + 480
850 - 480
650
+
cos (60.35°) +
sin (60.35°) d (10 - 6)
2
2
2
= 1039(10 - 6)
up2 = 120°
Ans.
850 - 480 2 + 650 2 d (10 - 6) = 748(10 - 6)
a
b
b
2
2
Ans.
Therefore, up1 = 30.2°
Ans.
b)
gmax
Aa
in-plane
2
gmax
in-plane
=
= 2c
2
2
Aa
Px + Py
Pavg =
Px - Py
= a
2
b + a
gxy
2
b
2
850 + 480
b (10 - 6) = 665(10 - 6)
2
Ans.
958
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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.
10–4. Continued
Orientation of gmax:
tan 2ut =
- (Px - Py)
gxy
=
- (850 - 480)
650
ut = - 14.8° and 75.2°
Ans.
Use Eq. 10–6 to determine the sign of gmax
:
in-plane
gx¿y¿
2
Px - Py
= -
2
sin 2u +
gxy
2
cos 2u;
u = ut = - 14.8°
gx¿y¿ = [ -(850 - 480) sin ( - 29.65°) + 650 cos ( -29.65°)](10 - 6) = 748(10 - 6)
959
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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
10–5. The state of strain at the point on the gear tooth has
the components Px = 520110-62, Py = - 760110-62, gxy =
750(10-62. 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.
Px = 520(10 - 6)
Py = - 760(10 - 6)
Px + Py
a)
P1, 2 =
;
2
= c
Aa
Px - Py
2
x
gxy = - 750(10 - 6)
2
b + a
gxy
2
b
2
2
2
520 + ( - 760)
; a 520 - ( -760) b + a - 750 b d10 - 6
2
A
2
2
P1 = 622(10 - 6);
P2 = - 862(10 - 6)
Ans.
Orientation of P1 and P2
tan 2up =
gxy
=
Px - Py
-750
= - 0.5859; up = - 15.18° and up = 74.82°
[520 - ( - 760)]
Use Eq. 10–5 to determine the direction of P1 and P2.
Px + Py
Px¿ =
Px - Py
+
2
2
cos 2u +
gxy
2
sin 2u
u = up = - 15.18°
Px¿ =
J
520 + (- 760)
520 - ( - 760)
-750
+
cos ( - 30.36°) +
sin ( -30.36°) d10 - 6
2
2
2
= 622 (10 - 6) = P1
Therefore, up1 = - 15.2° and up2 = 74.8°
gmax
b)
in-plane
=
2
gmax
in-plane
gxy 2
Px - Py 2
b + a
2
2 b
A
= 2c
a
520 - ( -760) 2
- 750 2
d 10 - 6 = - 1484 (10 - 6)
b + a
A
2
2 b
Px + Py
Pavg =
2
Ans.
a
= c
520 + ( -760)
d 10 - 6 = - 120 (10 - 6)
2
Ans.
Ans.
960
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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.
10–5. Continued
Orientation of gmax
in-plane
tan 2us =
:
- (Px - Py)
=
gxy
-[520 - ( - 760)]
= 1.7067
- 750
us = 29.8° and us = - 60.2°
Ans.
Use Eq. 10–6 to check the sign of gmax
:
in-plane
gx¿y¿
2
Px - Py
= -
gx¿y¿ = 2 c -
2
sin 2u +
gxy
2
cos 2u;
u = us = 29.8°
520 - (- 760)
-750
sin (59.6°) +
cos (59.6°) d10 - 6 = - 1484 (10 - 6)
2
2
Ans:
P1 = 622(10 - 6), P2 = - 862(10-6),
up1 = - 15.2° and up2 = 74.8°,
gmax = - 1484(10 - 6), Pavg = - 120(10 - 6),
in-plane
us = 29.8° and -60.2°
961
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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
10–6. A differential element on the bracket is subjected
to plane strain that has the following components:
Px = 150110-62, Py = 200110-62, gxy = - 700(10-62. Use the
strain-transformation equations and determine the
equivalent in plane strains on an element oriented at an
angle of u = 60° counterclockwise from the original
position. Sketch the deformed element within the x–y plane
due to these strains.
Px = 150 (10 - 6)
Px + Py
Px¿ =
Px - Py
+
2
= c
Py = 200 (10 - 6)
2
cos 2u +
gxy
2
x
gxy = - 700 (10 - 6)
u = 60°
sin 2u
150 - 200
- 700
150 + 200
+
cos 120° + a
b sin 120° d 10 - 6
2
2
2
= - 116 (10 - 6)
Px + Py
Py¿ =
= c
Px - Py
-
2
Ans.
2
cos 2u -
gxy
2
sin 2u
150 + 200
150 - 200
- 700
cos 120° - a
b sin 120° d 10 - 6
2
2
2
= 466 (10 - 6)
gx¿y¿
2
Px - Py
= -
gx¿y¿ = 2 c -
2
Ans.
sin 2u +
gxy
2
cos 2u
150 - 200
-700
sin 120° + a
b cos 120° d 10 - 6 = 393 (10 - 6)
2
2
Ans.
Ans:
Px¿ = - 116(10 - 6), Py¿ = 466(10 - 6),
gx¿y¿ = 393(10 - 6)
962
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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
10–7. Solve Prob. 10–6 for an element oriented u = 30°
clockwise.
x
Px = 150 (10 - 6)
Px + Py
Px¿ =
Px - Py
+
2
= c
Py = 200 (10 - 6)
2
cos 2u +
gxy
2
gxy = - 700 (10 - 6)
u = - 30°
sin 2u
150 - 200
- 700
150 + 200
+
cos ( -60°) + a
b sin ( -60°) d10 - 6
2
2
2
= 466 (10 - 6)
Px + Py
Py¿ =
= c
Ans.
Px - Py
-
2
2
cos 2u -
gxy
2
sin 2u
150 - 200
- 700
150 + 200
cos ( -60°) - a
b sin ( -60°) d10 - 6
2
2
2
= - 116 (10 - 6)
gx¿y¿
2
Px - Py
= -
gx¿y¿ = 2 c -
2
sin 2u +
Ans.
gxy
2
cos 2u
- 700
150 - 200
sin ( - 60°) +
cos ( - 60°) d10 - 6
2
2
= - 393(10 - 6)
Ans.
Ans:
Px¿ = 466(10 - 6), Py¿ = - 116(10 - 6),
gx¿y¿ = - 393(10 - 6)
963
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. 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.
*10–8. The state of strain at the point on the bracket
has components Px = - 200110-62, Py = - 650110-62,
gxy ⫽ - 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.
Px = - 200(10 - 6)
Px + Py
Px¿ =
Px - Py
+
2
= c
Py = - 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)
Px + Py
Py¿ =
Px - Py
-
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
Px - Py
= -
2
Ans.
sin 2u +
gxy
2
cos 2u
gx¿y¿ = [ -(- 200 - ( -650)) sin (40°) + ( - 175) cos (40°)](10 - 6)
= - 423(10 - 6)
Ans.
964
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. 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.
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 with the established sign convention, Px = 180(10 - 6),
Py = - 120(10 - 6) and gxy = - 100(10 - 6).
Px + Py
P1, 2 =
;
2
= b
Aa
Px - Py
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)
P1 = 188(10 - 6)
tan 2uP =
P2 = - 128(10 - 6)
gxy
Ans.
- 100(10 - 6)
C 180 - ( -120) D (10 - 6)
=
Px - Py
uP = - 9.217°
and
= - 0.3333
80.78°
Substitute u = - 9.217°,
Px + Py
Px¿ =
2
= c
Px - Py
+
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) = P1
Thus,
(uP)1 = - 9.22°
(uP)2 = 80.8°
Ans.
The deformed element is shown in Fig (a).
gmax
Px - Py 2
gxy 2
in-plane
=
b + a
b
b)
2
Aa
2
2
gmax
in-plane
tan 2us = - a
= b2
A
Px - Py
gxy
c
180 - (- 120) 2
- 100 2
-6
-6
d + a
b r (10 ) = 316 A 10 B
2
2
b = -c
C 180 - ( - 120) D (10 - 6)
us = 35.78° = 35.8° and
- 100(10 - 6)
Ans.
s = 3
- 54.22° = - 54.2°
Ans.
965
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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.
10–9. Continued
gmax
The algebraic sign for in-plane
when u = 35.78°.
gx¿y¿
Px - Py
gxy
= -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)
Pavg
= - 316(10 - 6)
Px + Py
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
in Fig. b
Ans:
P1 = 188(10 - 6), P2 = - 128(10 - 6),
up1 = - 9.22°, up 2 = 80.8°,
gmax = 316(10 - 6), us = 35.8° and - 54.2°,
in-plane
Pavg = 30(10 - 6)
966
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. 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.
10–10. The state of strain at the point on the support
has components of Px = 350110-62, Py = 400110-62,
gxy = - 675(10-62. 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.
P
a)
Px + Py
P1, 2 =
=
;
2
A
a
Px - Py
2
2
b + a
gxy
2
b
2
350 + 400
- 675 2
350 - 400 2
;
a
b + a
b
2
A
2
2
P1 = 713 (10 - 6)
tan 2up =
P2 = 36.6 (10 - 6)
Ans.
gxy
=
Px - Py
Ans.
- 675
(350 - 400)
up1 = 133°
Ans.
b)
(gx¿y¿)max
=
2
(gx¿y¿)max
=
2
A
a
Px - Py
2
2
gxy 2
b + a 2 b
- 675 2
350 - 400 2
b + a
b
A
2
2
a
(gx¿y¿)max = 677(10 - 6)
Px + Py
Pavg =
tan 2us =
2
=
Ans.
350 + 400
= 375 (10 - 6)
2
-(Px - Py)
gxy
=
Ans.
350 - 400
675
us = - 2.12°
Ans.
Ans:
(a) P1 = 713(10-6), P2 = 36.6(10 - 6), up1 = 133°
= 677(10 - 6), Pavg = 375(10 - 6),
(b) gmax
in-plane
us = - 2.12°
967
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. 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
10–11. The state of strain on an element has components
Px = - 150110-62, Py = 450110-62, gxy = 200110-62. Determine
the equivalent state of strain on an element at the same point
oriented 30° counterclockwise with respect to the original
element. Sketch the results on this element.
Pydy
dy
gxy
2
gxy
2
dx
Pxdx
x
Strain Transformation Equations:
Px = - 150(10 - 6)
Py = 450(10 - 6)
gxy = 200(10 - 6)
u = 30°
We obtain
Px + Py
Px¿ =
2
= c
Px - Py
+
2
gxy
cos 2u +
2
sin 2u
- 150 - 450
200
-150 + 450
+
cos 60° +
sin 60° d(10 - 6)
2
2
2
= 86. 6(10 - 6)
gx¿y¿
2
= -a
Ans.
Px - Py
2
b sin 2u +
gxy
2
cos 2u
gx¿y¿ = [ -( -150 - 450) sin 60° + 200 cos 60°](10 - 6)
= 620(10 - 6)
Px + Py
Py¿ =
= c
2
Ans.
Px - Py
-
2
cos 2u -
gxy
2
sin 2u
- 150 + 450
- 150 - 450
200
cos 60° sin 60° d(10 - 6)
2
2
2
= 213(10 - 6)
Ans.
The deformed element for this state of strain is shown in Fig. a.
Ans:
Px¿ = 86.6(10 - 6), gx¿y¿ = 620(10 - 6),
Py¿ = 213(10 - 6)
968
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. 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
*10–12. The state of strain on an element has components
Px = - 400110-62, Py = 0, gxy = 150110-62. Determine the
equivalent state of strain on an element at the same point
oriented 30° clockwise with respect to the original element.
Sketch the results on this element.
gxy
dy 2
x
gxy
2
dx
Strain Transformation Equations:
Px = - 400(10 - 6)
Py = 0
gxy = 150(10 - 6)
u = - 30°
We obtain
Px + Py
Px¿ =
2
= c
Px - Py
+
2
gxy
cos 2u +
2
sin 2u
- 400 - 0
150
- 400 + 0
+
cos ( -60°) +
sin (-60°) d(10 - 6)
2
2
2
= - 365(10 - 6)
gx¿y¿
= -a
2
Ans.
Px - Py
2
b sin 2u +
gxy
2
cos 2u
gx¿y¿ = [ - ( -400 - 0) sin ( -60°) + 150 cos ( -60°)](10 - 6)
= - 271(10 - 6)
Px + Py
Py¿ =
= c
2
Ans.
Px - Py
-
2
cos 2u -
gxy
2
sin 2u
- 400 + 0
- 400 - 0
150
cos ( -60°) sin ( -60°) d(10 - 6)
2
2
2
= - 35.0(10 - 6)
Ans.
The deformed element for this state of strain is shown in Fig. a.
969
Pxdx
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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
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.
gxy
dy 2
x
In-Plane Principal Strains: Px = - 300 A 10 - 6 B , Py = 0, and gxy = 150 A 10 - 6 B . We
obtain
Px + Py
C¢
;
P1, 2 =
2
= C
Px - Py
2
gxy
2
≤ + ¢ ≤
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
P1 = 17.7 A 10 - 6 B
P2 = - 318 A 10 - 6 B
Ans.
Orientation of Principal Strain:
tan 2up =
gxy
=
Px - Py
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,
Px + Py
Px¿ =
= c
Px - Py
+
2
cos 2u +
2
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 = P2
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
in-plane
2
gmax
in-plane
=
C¢
Px - Py
2
2
≤ + ¢
gxy
2
≤
2
- 300 - 0 2
150 2
-6
-6
b + a
b R A 10 B = 335 A 10 B
2
2
A
= B2
a
Ans.
Orientation of the Maximum In-Plane Shear Strain:
tan 2us = - ¢
Px - Py
gxy
≤ = -C
( - 300 - 0) A 10 - 6 B
150 A 10 - 6 B
S = 2
us = 31.7° and 122°
Ans.
970
gxy
2
dx
Pxdx
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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.
10–13. Continued
The algebraic sign for
gx¿y¿
2
= -¢
Px - Py
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:
Px + Py
Pavg =
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.
Ans:
P1 = 17.7(10 - 6), P2 = - 318(10 - 6),
up1 = 76.7° and up2 = - 13.3°,
gmax
= 335(10 - 6), us = 31.7° and 122°,
in-plane
Pavg = - 150(10 - 6)
971
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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.
10–14. 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,
Px + Py
P1, 2 =
;
2
= B
Aa
Px - Py
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
P1 = 368 A 10 - 6 B
P2 = 182 A 10 - 6 B
Ans.
Orientation of Principal Strain: Applying Eq. 10–8,
gxy
tan 2uP =
- 180(10 - 6)
=
Px - Py
(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°.
Px + Py
Px¿ =
= c
2
Px - Py
+
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 = P2
Hence,
uP1 = - 52.8°
and
uP2 = 37.2°
Ans.
b)
Maximum In-Plane Shear Strain: Applying Eq. 10–11,
g max
Px - Py 2
gxy 2
in-plane
=
b + a
b
2
Aa
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.
972
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.
10–14. Continued
Orientation of the Maximum In-Plane Shear Strain: Applying Eq. 10–10,
tan 2us = -
Px - Py
us = - 7.76°
and
The proper sign of
gx¿y¿
2
Px - Py
= -
= -
gxy
2
g
max
in-plane
250 - 300
= - 0.2778
- 180
Ans.
82.2°
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,
Px = 250 A 10 - 6 B
Py = 300 A 10 - 6 B
gxy = - 180 A 10 - 6 B
Average Normal Strain: Applying Eq. 10–12,
Px + Py
Pavg =
2
= c
250 + 300
d A 10 - 6 B = 275 A 10 - 6 B
2
Ans.
Ans:
P1 = 368(10 - 6), P2 = 182(10 - 6),
up1 = - 52.8° and up2 = 37.2°,
gmax
= 187(10 - 6), us = -7.76° and 82.2°,
in-plane
Pavg = 275 (10 - 6)
973
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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
*10–16. The state of strain on an element has components
Px = -300 (10 - 6), Py = 100 (10 - 6), gxy = 150 (10-6). 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.
Pydy
dy
gxy
2
gxy
2
dx
In-Plane Principal Strains: Px = - 300(10 - 6), Py = 100(10 - 6), and gxy = 150(10 - 6).
We obtain
Px + Py
P1, 2 =
= c
;
2
a
A
Px - Py
2
2
b + a
gxy
2
b
2
-300 + 100
150 2
- 300 - 100 2
;
d(10 - 6)
a
b + a
2
A
2
2 b
= (- 100 ; 213.60)(10 - 6)
P1 = 114(10 - 6)
P2 = - 314(10 - 6)
Ans.
Orientation of Principal Strains:
tan 2up =
gxy
150(10 - 6)
=
Px - Py
( -300 - 100)(10 - 6)
= - 0.375
up = - 10.28° and 79.72°
Substituting u = - 10.28° into
Px + Py
Px¿ =
2
= c
Px - Py
+
2
cos 2u +
gxy
2
sin 2u
- 300 + 100
- 300 - 100
150
+
cos ( - 20.56°) +
sin ( -20.56°) d(10 - 6)
2
2
2
= - 314(10 - 6) = P2
Thus,
(up)1 = 79.7° and (up)2 = - 10.3°
Ans.
The deformed element for the state of principal strain is shown in Fig. a.
Maximum In-Plane Shear Strain:
gmax
in-plane
2
gmax
in-plane
=
A
a
= c2
Px - Py
A
2
a
2
b + a
gxy
2
b
2
150 2
- 300 - 100 2
d(10 - 6) = 427(10 - 6)
b + a
2
2 b
974
Ans.
Pxdx
x
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10–16. Continued
Orientation of Maximum In-Plane Shear Strain:
tan 2us = - a
Px - Py
gxy
b = -c
(- 300 - 100)(10 - 6)
150(10 - 6)
d(10 - 6) = 2.6667
us = 34.7° and 125°
The algebraic sign for gmax
Ans.
in-plane
gx¿y¿
2
= -a
Px - Py
2
when u = us = 34.7° can be determined using
b sin 2u +
gxy
2
cos 2u
gx¿y¿ = [ - (- 300 - 100) sin 69.44° + 150 cos 69.44°](10 - 6)
= 427(10 - 6)
Average Normal Strain:
Px + Py
Pavg =
2
= a
- 300 + 100
b(10 - 6) = - 100(10 - 6)
2
Ans.
The deformed element for this state of maximum in-plane shear strain is shown
in Fig. b.
975
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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.
10–17.
Solve part (a) of Prob. 10–3 using Mohr’s circle.
Px = 120(10 - 6)
Py = - 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)
P1 = (- 30 + 167.71)(10 - 6) = 138(10 - 6)
Ans.
P2 = (- 30 - 167.71)(10 - 6) = - 198(10 - 6)
Ans.
tan 2uP = a
75
b , uP = 13.3°
30 + 120
Ans.
Ans:
P1 = 138(10 - 6), P2 = - 198(10 - 6), up = 13.3°
976
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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.
10–18.
Solve part (b) of Prob. 10–3 using Mohr’s circle.
Px = 120(10 - 6)
Py = - 180(10 - 6)
A (120, 75)(10 - 6)
C ( - 30, 0)(10 - 6)
gxy = 150(10 - 6)
R = C 2[120 - ( - 30)]2 + (75)2 D (10 - 6)
= 167.71 (10 - 6)
gxy
2
gmax
= R = 167.7(10 - 6)
in-plane
= 335(10 - 6)
Ans.
Pavg = - 30 (10 - 6)
tan 2us =
120 + 30
75
Ans.
us = - 31.7°
Ans.
Ans:
g max
in-plane
= 335(10 - 6), Pavg = - 30(10 - 6),
us = - 31.7°
977
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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.
10–19.
Solve Prob. 10–4 using Mohr’s circle.
Px = 850(10 - 6)
Py = 480(10 - 6)
gxy = 650(10 - 6)
gxy
2
= 325(10 - 6)
A(850, 325)(10 - 6) C(665, 0)(10 - 6)
R = [2(850 - 665)2 + 3252](10 - 6) = 373.97(10 - 6)
P1 = (665 + 373.97)(10 - 6) = 1039(10 - 6)
Ans.
P2 = (665 - 373.97)(10 - 6) = 291(10 - 6)
Ans.
tan 2up =
325
850 - 665
(Mohr’s circle)
2up = 60.35°
up = 30.2°
Ans.
(element)
gmax
in-plane
2
gmax
in-plane
= R
= 2(373.97)(10 - 6) = 748(10 - 6)
Ans.
Pavg = 665(10 - 6)
Ans.
2us = 90° - 2up = 29.65°
us = - 14.8°
Ans.
(Mohr’s circle)
(element)
Ans:
P1 = 1039(10 - 6), P2 = 291(10 - 6), up = 30.2°,
gmax
= 748(10 - 9), Pavg = 665(10 - 6),
in-plane
us = - 14.8°
978
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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.
*10–20.
Solve Prob. 10–5 using Mohr’s circle.
a) Px = 520(10 - 6)
Py = - 760(10 - 6)
gxy = - 750(10 - 6)
gxy
2
= - 375(10 - 6)
A(520, -375); C(- 120, 0)
R = 2(520 + 120)2 + 3752 = 741.77
P1 = 741.77 - 120 = 622(10 - 6)
Ans.
P2 = - 120 - 741.77 = - 862(10 - 6)
Ans.
tan 2up1 =
375
= 0.5859
(120 + 520)
up1 = 15.2°
b)
gmax
gmax
in-plane
in-plane
Ans.
= 2R = 2(741.77)
= - 1484(10 - 6)
Ans.
Pavg = - 120(10 - 6)
tan 2us =
Ans.
(120 + 520)
= 1.7067
375
us = 29.8°
Ans.
979
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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.
10–21.
Solve Prob. 10–7 using Mohr’s circle.
Px = 150(10 - 6)
u = - 30°
Py = 200(10 - 6)
gxy = - 700(10 - 6)
gxy
2
= - 350(10 - 6)
2u = - 60°
A(150, - 350);
C(175, 0)
R = 2(175 - 150)2 + (- 350)2 = 350.89
Coordinates of point B:
Px¿ = 350.89 cos 34.09° + 175
= 466(10 - 6)
gx¿y¿
2
Ans.
= - 350.89 sin 34.09°
gx¿y¿ = - 393(10 - 6)
Ans.
Coordinates of point D:
Py¿ = 175 - 350.89 cos 34.09°
= - 116(10 - 6)
Ans.
Ans:
-6
Px¿ = 466(10 - 6), gx¿y¿ = - 393(10 ),
-6
Py¿ = - 116(10 )
980
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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.
10–22. The strain at point A on the bracket
has
components
Px = 300110-62,
Py = 550110-62,
-6
gxy = - 650110 2, Pz = 0. 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.
Px = 300(10 - 6)
A(300, - 325)10 - 6
Py = 550(10 - 6)
gxy = - 650(10 - 6)
y
gxy
2
A
= - 325(10 - 6)
x
C(425, 0)10 - 6
R = C 2(425 - 300)2 + ( - 325)2 D 10 - 6 = 348.2(10 - 6)
a)
P1 = (425 + 348.2)(10 - 6) = 773(10 - 6)
Ans.
P2 = (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)
Ans:
(a) P1 = 773(10 - 6), P2 = 76.8(10 - 6),
(b) gmax
= 696(10 - 6),
in-plane
(c) gabs = 773(10 - 6)
max
981
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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.
10–23. The strain at point A on a beam has components
Px = 450(10 - 6), Py = 825(10 - 6), gxy = 275(10 - 6), Pz = 0.
Determine (a) the principal strains at A, (b) the maximum
shear strain in the x–y plane, and (c) the absolute maximum
shear strain.
Px = 450(10 - 6)
Py = 825(10 - 6)
A(450, 137.5)10 - 6
gxy = 275(10 - 6)
A
gxy
2
= 137.5(10 - 6)
C(637.5, 0)10 - 6
R = [2(637.5 - 450)2 + 137.52]10 - 6 = 232.51(10 - 6)
a)
P1 = (637.5 + 232.51)(10 - 6) = 870(10 - 6)
Ans.
P2 = (637.5 - 232.51)(10 - 6) = 405(10 - 6)
Ans.
gmax
= 2R = 2(232.51)(10 - 6) = 465(10 - 6)
Ans.
870(10 - 6)
;
2
Ans.
b)
in-plane
c)
gabs
max
2
=
gabs
max
= 870(10 - 6)
Ans:
(a) P1 = 870(10 - 6), P2 = 405(10 - 6),
= 465(10 - 6),
(b) gmax
in-plane
(c) gabs = 870(10 - 6)
max
982
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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.
*10–24. The steel bar is subjected to the tensile load of
500 lb. If it is 0.5 in. thick determine the three principal
strains. E = 29 (103) ksi , n = 0.3.
2 in.
500 lb
500 lb
15 in.
sx =
500
= 500 psi
2(0.5)
Px =
1
1
(500) = 17.2414 (10 - 6)
(s ) =
E x
29(106)
sy = 0
sz = 0
Py = Pz = - vPx = - 0.3(17.2414)(10 - 6) = - 5.1724(10 - 6)
P1 = 17.2(10 - 6)
P2, 3 = - 5.17(10 - 6)
Ans.
983
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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.
10–25. The 45° strain rosette is mounted on a machine
element. The following readings are obtained from each
gauge: Pa = 650(10 - 6), Pb = - 300(10 - 6), Pc = 480(10 - 6).
Determine (a) the in-plane principal strains and (b) the
maximum in-plane shear strain and associated average
normal strain. In each case show the deformed element due
to these strains.
a
b
45⬚
c
Pa = 650(10 - 6)
Pb = - 300(10 - 6)
Pc = 480(10 - 6)
ua = 90°
ub = 135°
uc = 180°
45⬚
Pa = Px cos2 ua + Py sin2 ua + gxy sin ua cos ua
650(10 - 6) = Px cos2 90° + Py sin2 90° + gxy sin 90° cos 90°
Px = 650(10 - 6)
Pc = Px cos2 uc + Py sin2 uc + gxy sin uc cos uc
480(10 - 6) = Px cos2 180° + Py sin2 180° + gxy sin 180° cos 180°
Py = 480(10 - 6)
Pb = Px cos2 ub + Py sin2 ub + gxy sin ub cos ub
- 300(10 - 6) = 480(10 - 6) cos2 135° + 650(10 - 6) sin2 135° + gxy sin 135° cos 135°
gxy = 1730(10 - 6)
gxy
2
= 865(10 - 6)
A(480, 865)10 - 6
C(565, 0)10 - 6
R = ( 2(565 - 480)2 + 8652)10 - 6 = 869.17(10 - 6)
a)
P1 = (565 + 869.17)10 - 6 = 1434(10 - 6)
Ans.
P2 = (565 - 869.17)10 - 6 = - 304(10 - 6)
Ans.
tan 2up =
865
565 - 480
2up = 84.39°
(Mohr’s circle)
up = - 42.19°
(element)
b)
gmax
in-plane
= 2R = 2(869.17)(10 - 6) = 1738(10 - 6)
Ans.
Pavg = 565(10 - 6)
Ans.
2us = 90° - 2up = 5.61°
(Mohr’s circle)
us = 2.81°
(element)
Ans:
P1 = 1434(10 - 6), P2 = - 304(10 - 6),
gmax
= 1738(10 - 6), Pavg = 565(10 - 6)
in-plane
984
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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.
10–26. The 60° strain rosette is attached to point A on the
surface of the support. Due to the loading the strain gauges
give a reading of Pa = 300(10 - 6), Pb = - 150 (10-6), and
Pc = - 450 (10 - 6). Use Mohr’s circle and determine (a) the
in-plane principal strains and (b) the maximum in-plane
shear strain and the associated average normal strain.
Specify the orientation of each element that has these states
of strain with respect to the x axis.
Pa = Px cos2 ua + Py sin2 ua + gxy sin ua cos ua
300(10 - 6) = Px cos2 0° + Py sin2 0° + gxy sin 0° cos 0°
Px = 300(10 - 6)
Pb = Px cos2 ub + Py sin2 ub + gxy sin ub cos ub
- 150(10 - 6) = 300(10 - 6) cos2 60° + Py sin2 60° + gxy sin 60° cos 60°
0.75Py + 0.43301gxy = - 225(10 - 6)
(1)
2
Pc = Px cos uc + Py sin uc + gxy sin uc cos uc
- 450(10 - 6) = 300(10 - 6) cos2 120° + Py sin2 120° + gxy sin 120° cos 120°
0.75Py - 0.43301gxy = - 525(10 - 6)
(2)
Solving Eqs. (1) and (2),
gxy = 346.41(10 - 6)
gxy
Construction of the Circle: Px = 300(10 - 6), Py = - 500(10 - 6), and
= 173.20(10 - 6).
2
Thus
Px + Py
Pavg =
2
= c
300 + ( - 500)
d (10 - 6) = - 100(10 - 6)
2
Ans.
The coordinates of reference point A and center of C of the circle are
A(300, 173.20)(10 - 6)
x
a
Normal and Shear Strain: With ua = 0°, ub = 60°, and uc = 120°, we have
Py = - 500(10 - 6)
60⬚
60⬚
A
2
c
b
C( - 100, 0)(10 - 6)
Thus, the radius of the circle is
R = CA = a2[300 - ( - 100)]2 + 173.202 b(10 - 6) = 435.89(10 - 6)
Using these results, the circle is shown in Fig. a.
In-Plane Principal Strains: The coordinates of reference points B and D represent
P1 and P2, respectively.
P1 = (- 100 + 435.89)(10 - 6) = 336(10 - 6)
Ans.
P2 = (- 100 - 435.89)(10 - 6) = - 536(10 - 6)
Ans.
985
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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.
10–26. Continued
Orientation of Principal Strain: Referring to the geometry of the circle,
tan 2(up)1 =
(up)1 = 11.7°
173.20(10 - 6)
(300 + 100)(10 - 6)
= 0.43301
(counterclockwise)
Ans.
The deformed element for the state of principal strain is shown in Fig. b.
Maximum In-Plane Shear Strain: The coordinates of point E represent Pavg and
gmax
in-plane
. Thus
2
gmax
in-plane
2
gmax
in-plane
= R = (435.89)(10 - 6)
= 872(10 - 6)
Ans.
Orientation of Maximum In-Plane Shear Strain: Referring to the geometry of the
circle,
tan 2us =
us = 33.3°
300 + 100
= 2.3094
173.20
(clockwise)
Ans.
The deformed element for the state of maximum in-plane shear strain is shown in
Fig. c.
Ans:
P1 = 336(10 - 6), P2 = - 536(10 - 6),
up1 = 11.7° (counterclockwise),
gmax
= 872(10 - 6), Pavg = - 100(10 - 6),
in-plane
us = 33.3° (clockwise)
986
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. 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.
10–27. The strain rosette is attached at the point on the
surface of the pump. Due to the loading, the strain gauges give
a reading of Pa = - 250(10 - 6), Pb = - 300 (10-6), and
Pc = - 200 (10 - 6). Determine (a) the in-plane principal
strains, and (b) the maximum in-plane shear strain. Specify
the orientation of each element that has these states of strain
with respect to the x axis.
b
60⬚
xx
A
60⬚
c
Normal and Shear Strains: With ua = 0°, ub = 60°, and uc = - 60°, we have
Pa = Px cos2 ua + Py sin2 ua + gxy sin ua cos ua
-250(10 - 6) = Px cos2 0° + Py sin2 0° + gxy sin 0° cos 0°
Px = - 250(10 - 6)
Pb = Px cos2 ub + Py sin2 ub + gxy sin ub cos ub
300(10 - 6) = - 250(10 - 6) cos2 60° + Py sin2 60° + gxy sin 60° cos 60°
0.75Py + 0.43301gxy = 362.5(10 - 6)
(1)
Pc = Px cos2 uc + Py sin2 uc + gxy sin uc cos uc
- 200(10 - 6) = - 250(10 - 6) cos2 ( -60°) + Py sin2 ( -60°) + gxy sin ( -60°) cos ( -60°)
0.75Py - 0.43301gxy = - 137.5(10 - 6)
(2)
Solving Eqs. (1) and (2), we obtain
Py = 150(10 - 6)
gxy = 577.35(10 - 6)
gxy
Construction of the Circle: Px = - 250(10 - 6), Py = 150(10 - 6), and
= 288.68(10 - 6).
2
Thus
Px + Py
- 250 + 150
Ans.
= a
b(10 - 6) = - 50(10 - 6)
Pavg =
2
2
The coordinates of reference point A and center of C of the circle are
A( - 250, 288.68)(10 - 6)
a
C( -50, 0)(10 - 6)
Thus, the radius of the circle is
R = CA = a2[ -250 - ( - 50)]2 + 288.682 b(10 - 6) = 351.19(10 - 6)
Using these results, the circle is shown in Fig. a.
In-Plane Principal Strains: The coordinates of reference points B and D represent
P1 and P2, respectively.
P1 = (- 50 + 351.19)(10 - 6) = 301(10 - 6)
Ans.
P2 = (- 50 - 351.19)(10 - 6) = - 401(10 - 6)
Ans.
987
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10–27. Continued
Orientation of Principal Strain: Referring to the geometry of the circle,
tan 2(up)2 =
288.68
= 1.4434
250 - 50
(up)2 = 27.6° (clockwise)
Ans.
The deformed element for the state of principal strain is shown in Fig. b.
Maximum In-Plane Shear Strain: The coordinates of point E represent Pavg and
gmax
in-plane
. Thus
2
gmax
in-plane
2
gmax
in-plane
= R = 351.19(10 - 6)
= 702(10 - 6)
Ans.
Orientation of Maximum In-Plane Shear Strain: Referring to the geometry of the
circle,
tan 2us =
250 - 50
= 0.6928
288.68
us = 17.4° (counterclockwise)
Ans.
The deformed element for the state of maximum in-plane shear strain is shown in
Fig. c.
Ans:
P1 = 301(10 - 6), P2 = - 401(10 - 6),
up2 = 27.6° (clockwise),
gmax
= 702(10 - 6), Pavg = - 50(10 - 6),
in-plane
us = 17.4° (counterclockwise)
988
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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.
*10–28. The 60° strain rosette is mounted on a beam. The
following readings are obtained from each gauge:
Pa = 250(10 - 6), Pb = - 400 (10-6), Pc = 280(10 - 6). Determine (a) the in-plane principal strains and their orientation,
and (b) the maximum in-plane shear strain and average
normal strain. In each case show the deformed element due
to these strains.
b
a
60⬚
c
Pa = 250(10 - 6)
Pb = - 400(10 - 6)
Pc = 280(10 - 6)
ua = 60°
ub = 120°
uc = 180°
Pc = Px cos2 uc + Pg sin2 uc + gxy sin uc cos uc
280(10 - 6) = Px cos2 180° + Py sin2 180° + gxy sin 180° cos 180°
Px = 280(10 - 6)
Pa = Px cos2 ua + Py sin2 ua + gxy sin ua cos ua
250(10 - 6) = Px cos2 60° + Py sin2 60° + gxy sin 60° cos 60°
250(10 - 6) = 0.25Px + 0.75Py + 0.433gxy
(1)
Pb = Px cos 2 ub + Py sin2 ub + gxy sin ub cos ub
- 400(10 - 6) = Px cos2 120° + Py sin 2 120° + gxy sin 120° cos 120°
- 400(10 - 6) = 0.25Px + 0.75Py - 0.433gxy
(2)
Subtract Eq. (2) from Eq. (1)
650(10 - 6) = 0.866gxy
gxy = 750.56(10 - 6)
Py = - 193.33(10 - 6)
gxy
2
= 375.28(10 - 6)
A(280, 375.28)10 - 6
C(43.34, 0)10 - 6
R = ( 2(280 - 43.34)2 + 375.282)10 - 6 = 443.67(10 - 6)
a)
P1 = (43.34 + 443.67)10 - 6 = 487(10 - 6)
Ans.
P2 = (43.34 - 443.67)10 - 6 = - 400(10 - 6)
Ans.
tan 2up =
375.28
280 - 43.34
2up = 57.76°
(Mohr’s circle)
up = 28.89°
(element)
989
60⬚
60⬚
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10–28. Continued
b)
gmax
in-plane
= 2R = 2(443.67)(10 - 6) = 887(10 - 6)
Ans.
Pavg = 43.3(10 - 6)
Ans.
2us = 90° - 2uy = 32.24°
(Mohr’s circle)
us = 16.12°
(element)
990
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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.
10–30. For the case of plane stress, show that Hooke’s law
can be written as
sx =
E
E
1Px + nPy2, sy =
1Py + nPx2
11 - n22
11 - n22
Generalized Hooke’s Law: For plane stress, sz = 0. Applying Eq. 10–18,
Px =
1
A s - n sy B
E x
nEPx = A sx - n sy B n
nEPx = n sx - n2 sy
Py =
(1)
1
(s - n sx)
E y
E Py = - n sx + sy
(2)
Adding Eq. (1) and Eq.(2) yields.
nE Px + E Py = sy - n2 sy
sy =
E
A nPx + Py B
1 - n2
(Q.E.D.)
Substituting sy into Eq. (2)
E Py = - nsx +
sx =
E
A n Px + Py B
1 - n2
E A n Px + Py B
2
n (1 - n )
EPy
-
n
EnPx + EPy - EPy + EPy n2
=
=
n(1 - n2)
E
(Px + n Py)
1 - n2
(Q.E.D.)
991
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10–31. Use Hooke’s law, Eq. 10–18, to develop the straintransformation equations, Eqs. 10–5 and 10–6, from the
stress-transformation equations, Eqs. 9–1 and 9–2.
Stress Transformation Equations:
sx¿ =
sx + sy
2
tx¿y¿ = -
sy¿ =
sx - sy
+
sx - sy
2
sx + sy
2
(1)
sin 2u + txy cos 2u
sx - sy
-
2
cos 2u + txy sin 2u
2
(2)
cos 2u - txy sin 2u
(3)
Hooke’s Law:
Px =
n sy
sx
E
E
(4)
Py =
sy
- n sx
+
E
E
(5)
txy = G gxy
G =
(6)
E
2(1 + n)
(7)
From Eqs. (4) and (5)
(1 - n)(sx + sy)
Px + Py =
(8)
E
(1 + n)(sx - sy)
Px - Py =
(9)
E
From Eqs. (6) and (7)
txy =
E
g
2(1 + n) xy
(10)
From Eq. (4)
Px¿ =
n sy¿
sx¿
E
E
(11)
Substitute Eqs. (1) and (3) into Eq. (11)
(1 - n)(sx + sy)
Px¿ =
(1 + n)(sx - sy)
+
2E
2E
cos 2u +
(1 + n)txy sin 2u
E
(12)
By using Eqs. (8), (9) and (10) and substitute into Eq. (12),
Px + Py
Px¿ =
2
Px - Py
+
2
cos 2u +
gxy
2
(Q.E.D.)
sin 2u
992
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10–31. Continued
From Eq. (6).
tx¿y¿ = Ggx¿y¿ =
E
g
2(1 + n) x¿y¿
(13)
Substitute Eqs. (13), (6) and (9) into Eq. (2),
E(Px - Py)
E
E
gx¿y¿ = sin 2u +
g cos 2u
2(1 + n)
2(1 + n)
2(1 + n) xy
gx¿y¿
2
(Px - Py)
= -
2
sin 2u +
gxy
2
(Q.E.D.)
cos 2u
993
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*10–32. The principal plane stresses and associated
strains in a plane at a point are s1 = 36 ksi, s2 = 16 ksi,
P1 = 1.02(10-3), P2 = 0.180(10-3). Determine the modulus
of elasticity and Poisson’s ratio.
s3 = 0
P1 =
1
[s - v(s2 + s3)]
E 1
1.02(10 - 3) =
1
[36 - v(16)]
E
1.02(10 - 3)E = 36 - 16v
P2 =
(1)
1
[s - v(s1 + s3)]
E 2
0.180(10 - 3) =
1
[16 - v(36)]
E
0.180(10 - 3)E = 16 - 36v
(2)
Solving Eqs. (1) and (2) yields:
E = 30.7(103) ksi
Ans.
v = 0.291
Ans.
994
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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.
10–33. A rod has a radius of 10 mm. If it is subjected to an
axial load of 15 N such that the axial strain in the rod is
Px = 2.75(10-6), determine the modulus of elasticity E and
the change in its diameter. n = 0.23.
sx =
15
= 47.746 kPa,
p(0.01)2
Px =
1
[s - v(sy + sz)]
E x
2.75(10 - 6) =
sy = 0,
sz = 0
1
[47.746(103) - 0.23(0 + 0)]
E
E = 17.4 GPa
Ans.
Py = Pz = - vPx = - 0.23(2.75)(10 - 6) = - 0.632(10 - 6)
¢d = 20(- 0.632(10 - 6)) = - 12.6(10 - 6) mm
Ans.
Ans:
E = 17.4 GPa , ¢d = - 12.6(10 - 6) mm
995
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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.
10–34. The polyvinyl chloride bar is subjected to an axial
force of 900 lb. If it has the original dimensions shown
determine the change in the angle u after the load is applied.
Epvc = 800 (103) psi, npvc = 0.20.
900 lb
900 lb
3 in.
u
6 in.
sx =
900
= 300 psi
3(1)
sy = 0
Px =
=
Py =
=
1 in.
sz = 0
1
[sx - v(sy + sz)]
E
1
[300 - 0] = 0.375(10 - 3)
800(103)
1
[s - v(sx + sz)]
E y
1
[0 - 0.2(300 + 0)] = - 75(10 - 6)
800(103)
a¿ = 6 + 6(0.375)(10 - 3) = 6.00225 in.
b¿ = 3 + 3( - 75)(10 - 6) = 2.999775 in.
3
u = tan - 1 a b = 26.56505118°
6
u¿ = tan - 1 a
2.999775
b = 26.55474088°
6.00225
¢u = u¿ - u = 26.55474088° - 26.56505118° = - 0.0103°
Ans.
Ans:
¢u = - 0.0103°
996
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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.
10–35. The polyvinyl chloride bar is subjected to an axial
force of 900 lb. If it has the original dimensions shown
determine the value of Poisson’s ratio if the angle u
decreases by ¢u = 0.01° after the load is applied. Epvc =
800(103) psi.
sx =
900
= 300 psi
3(1)
Px =
1
[s - vpvc (sy + sz)]
E x
=
Py =
=
sy = 0
900 lb
900 lb
3 in.
u
6 in.
1 in.
sz = 0
1
[300 - 0] = 0.375(10 - 3)
800(103)
1
[s - vpvc (sx + sz)]
E y
1
[0 - vpvc (300 + 0)] = - 0.375(10 - 3)vpvc
800(103)
a¿ = 6 + 6(0.375)(10 - 3) = 6.00225 in.
b¿ = 3 + 3( -0.375)(10 - 3)vpvc = 3 - 1.125(10 - 3)vpvc
3
u = tan - 1 a b = 26.56505118°
6
u¿ = 26.56505118° - 0.01° = 26.55505118°
tan u¿ = 0.49978185 =
3 - 1.125(10 - 3)vpvc
6.00225
vpvc = 0.164
Ans.
Ans:
npvc = 0.164
997
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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.
*10–36. The spherical pressure vessel has an inner
diameter of 2 m and a thickness of 10 mm. A strain gauge
having a length of 20 mm is attached to it, and it is observed
to increase in length by 0.012 mm when the vessel is
pressurized. Determine the pressure causing this
deformation, and find the maximum in-plane shear stress,
and the absolute maximum shear stress at a point on the
outer surface of the vessel. The material is steel, for which
Est = 200 GPa and nst = 0.3.
20 mm
1000
r
=
= 100 7 10, the thin wall analysis is valid to
t
10
determine the normal stress in the wall of the spherical vessel. This is a plane stress
Normal Stresses: Since
problem where smin = 0 since there is no load acting on the outer surface of the wall.
smax = slat =
pr
p(1000)
=
= 50.0p
2t
2(10)
(1)
Normal Strains: Applying the generalized Hooke’s Law with
Pmax = Plat =
0.012
= 0.600 A 10 - 3 B mm>mm
20
Pmax =
1
C s - n (slat + smin) D
E max
0.600 A 10 - 3 B =
1
[50.0p - 0.3 (50.0p + 0)]
200(109)
p = 3.4286 MPa = 3.43 MPa
Ans.
From Eq. (1) smax = slat = 50.0(3.4286) = 171.43 MPa
Maximum In-Plane Shear (Sphere’s Surface): Mohr’s circle is simply a dot. As the
result, the state of stress is the same consisting of two normal stresses with zero
shear stress regardless of the orientation of the element.
tmax
= 0
Ans.
smax - smin
171.43 - 0
=
= 85.7 MPa
2
2
Ans.
in-plane
Absolute Maximum Shear Stress:
tabs
max
=
998
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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.
10–37. Determine the bulk modulus for each of the
following materials: (a) rubber, Er = 0.4 ksi, nr = 0.48, and
(b) glass, Eg = 811032 ksi, ng = 0.24.
a) For rubber:
kr =
Er
0.4
=
= 3.33 ksi
3 (1 - 2 vr)
3[1 - 2(0.48)]
Ans.
b) For glass:
kg =
Eg
3(1 - 2 vg)
=
8(103)
= 5.13 (103) ksi
3[1 - 2(0.24)]
Ans.
Ans:
(a) kr = 3.33 ksi ,
(b) kg = 5.13(103) ksi
999
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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.
10–38. The strain gauge is placed on the surface of a thinwalled steel boiler as shown. If it is 0.5 in. long, determine
the pressure in the boiler when the gauge elongates
0.2(10-3) in. The boiler has a thickness of 0.5 in. and inner
diameter of 60 in. Also, determine the maximum x, y in-plane
shear strain in the material. Est = 29(103) ksi, nst = 0.3.
P2 =
0.2(10 - 3)
= 400(10 - 6)
0.5
P2 =
1
[s - v(s1 + s3)]
E 2
s2 =
where,
1
s
2 1
400(10 - 6) =
y
x
60 in.
0.5 in.
s3 = 0
1
1
c s1 - 0.3s1 d
29(103) 2
s1 = 58 ksi
Thus,
p =
58(0.5)
s1 t
=
= 0.967 ksi
r
30
P1 =
1
[s - v(s2 + s3)]
E 1
s3 = 0 and s2 =
where,
P1 =
58
= 29 ksi
2
1
[58 - 0.3(29 + 0)] = 1700(10 - 6)
29(103)
gmax
P1 - P2
in-plane
2
gmax
Ans.
in-plane
=
2
= (1700 - 400)(10 - 6) = 1.30(10 - 3)
Ans.
Ans:
p = 0.967 ksi , gmax
in-plane
1000
= 1.30(10 - 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.
y
10–39. The strain in the x direction at point A on the A-36
structural-steel beam is measured and found to be
Px = 100(10-6). Determine the applied load P. What is the
shear strain gxy at point A?
P
2 in.
x
A
B
3 ft
4 ft
2 in.
12 in.
A
6 in.
Section Properties:
I =
1
(6)(123) = 864 in4
12
QA = y¿A¿ = 5(6)(2) = 60 in3
Normal Stress:
s = E Px = 29(103)(100)(10 - 6) = 2.90 ksi
s =
My
;
I
2.90(103) =
4P(12)(4)
864
P = 13050 lb = 13.0 kip
txy =
gxy =
Ans.
13.05(103)(60)
VQ
=
= 151.04 psi
It
864(6)
txy
G
151.04
=
11.0(106)
= - 13.7(10 - 6)
Ans.
Ans:
P = 13.0 kip, gxy = - 13.7(10 - 6)
1001
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y
*10–40. The strain in the x direction at point A on the A-36
structural-steel beam is measured and found to be
Px = 200(10-6). Determine the applied load P. What is the
shear strain gxy at point A?
P
2 in.
x
A
B
3 ft
4 ft
2 in.
A
6 in.
Section Properties:
QA = y¿A¿ = 5(6)(2) = 60 in3
I =
1
(6)(123) = 864 in4
12
Normal Stress:
s = E Px = 29(103)(200)(10 - 6) = 5.80 ksi
s =
My
;
I
5.80(103) =
4P(12)(4)
864
P = 26.1 kip
Ans.
Shear Stress and Shear Strain:
tA =
gxy =
VQ
26.1(60)
=
= 0.302 ksi
It
864(6)
txy
G
=
0.302
= - 27.5(10 - 6) rad
11.0(103)
Ans.
1002
12 in.
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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
10–41. If a load of P = 3 kip is applied to the A-36
structural-steel beam, determine the strain Px and gxy at
point A.
P
2 in.
x
A
B
3 ft
4 ft
2 in.
A
Section Properties:
12 in.
6 in.
3
QA = y¿A¿ = 2(6)(5) = 60 in
I =
1
(6)(123) = 864 in4
12
Normal Stress and Strain:
sA =
Px =
My
12.0(103)(12)(4)
=
= 666.7 psi
I
864
sx
666.7
= 23.0(10 - 6)
=
E
29(106)
Ans.
Shear Stress and Shear Strain:
tA =
gxy =
3(103)(60)
VQ
=
= 34.72 psi
It
864(6)
txy
G
34.72
=
11.0(106)
= - 3.16(10 - 6)
Ans.
Ans:
Px = 23.0(10 - 6), gxy = - 3.16(10 - 6)
1003
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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.
26 ksi
10–42. The principal stresses at a point are shown in
the figure. If the material is aluminum for which
Eal = 1011032 ksi and nal = 0.33, determine the principal
strains.
Px =
1
1
(s - v(sy + sz)) =
(10 - 0.33( - 15 - 26)) = 2.35(10 - 3)
E x
10(103)
Py =
1
1
(s - v(sx + sz)) =
( -15 - 0.33)(10 - 26)) = - 0.972(10 - 3) Ans.
E y
10(103)
Pz =
1
1
(s - v(sx + sy)) =
( -26 - 0.33(10 - 15)) = - 2.44(10 - 3)
E z
10(103)
Ans.
15 ksi
10 ksi
Ans.
Ans:
Px = 2.35(10 - 3), Py = - 0.972(10 - 3),
Pz = - 2.44(10 - 3)
1004
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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.
10–43. A strain gauge a is attached in the longitudinal
direction (x axis) on the surface of the gas tank. When the
tank is pressurized, the strain gauge gives a reading of
Pa = 100(10-6). Determine the pressure p in the tank. The
tank has an inner diameter of 1.5 m and wall thickness of
25 mm. It is made of steel having a modulus of elasticity
E = 200 GPa and Poisson’s ratio n = 13.
b
45⬚ a
x
Normal Strain: With ua = 0, we have
Pa = Px cos2 ua + Py sin2 ua + gxy sin ua cos ua
100(10 - 6) = Px cos2 0°+ Py sin2 0°+ gxy sin 0° cos 0°
Px = 100(10 - 6)
0.75
r
=
= 30 7 10, thin-wall analysis can be used.
t
0.025
Normal Stress: Since
sy = s1 =
pr
p(0.75)
=
= 30 p
t
0.025
sx = s2 =
pr
p(0.75)
=
= 15 p
2t
2(0.025)
Generalized Hooke’s Law: This is a case of plane stress. Thus, nz = 0.
Px =
1
c s - n(sy + sz) d
E x
100(10 - 6) =
1
1
c 15p - (30p + 0) d
3
200(109)
p = 4(106)N>m2 = 4 MPa
Ans.
Ans:
p = 4 MPa
1005
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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.
10–44. Strain gauge b is attached to the surface of the gas tank at
an angle of 45° with x axis as shown.When the tank is pressurized,
the strain gauge gives a reading of Pb = 250(10 - 6). Determine the
pressure in the tank.The tank has an inner diameter of 1.5 m and
wall thickness of 25 mm. It is made of steel having a modulus of
elasticity E = 200 GPa and Poisson’s ratio n = 13.
Normal Stress: Since
b
0.75
r
=
= 30 7 10, thin-wall analysis can be used.
t
0.025
sy = s1 =
pr
p(0.75)
=
= 30 p
t
0.025
sx = s2 =
pr
p(0.75)
=
= 15 p
2t
2(0.025)
Normal Strain: Since no shear force acts along the x and y axes, gxy = 0. With
ub = 45, we have
Pb = Px cos2 ub + Py sin2 ub + gxy sin ub cos ub
250(10 - 6) = Px cos2 45° + Py sin2 45° + 0
Px + Py = 500(10 - 6)
(1)
Generalized Hooke’s Law: This is a case of plane stress. Thus, nz = 0. We have
Px =
1
c s - n(sy + sz) d
E x
Px =
1
1
c 15p - (30p + 0) d
3
200(109)
Px =
200(109)
Py =
1
c s - n(sx + sz) d
E y
Py =
1
1
c 30p - (15p + 0) d
9
3
200(10 )
5p
(2)
25p
Py =
(3)
200(109)
Substituting Eqs. (2) and (3) into Eq. (1), we obtain
5p
200(109)
25p
+
200(109)
= 500(10 - 6)
p = 3.333(106) N>m2 = 3.33 MPa
Ans.
1006
45⬚ a
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.
y
10–45. A material is subjected to principal stresses sx
and sy. Determine the orientation u of a strain gauge placed
at the point so that its reading of normal strain responds
only to sy and not sx. The material constants are E and n.
u
x
sx =
sx + sy
sx - sy
+
2
cos 2u + txy sin 2u
2
Since txy = 0,
sn =
sx + sy
sx - sy
+
2
2
cos 2u
cos 2u = 2 cos2 u - 1
sn =
sy
sy
sx
sx
+
+ (sx - sy ) cos2 u +
2
2
2
2
= sy (1 - cos2 u) + sx cos2 u
= sx cos2 u + sy sin2 u
sn + 90° =
sx + sy
sx - sy
-
2
2
cos 2u
=
sx - sy
sy
sx
+
- a
b (2 cos2 u - 1)
2
2
2
=
sy
sy
sx
sx
+
- (sx - sy ) cos2 u +
2
2
2
2
= sx (1 - cos2 u) + sy cos2 u
= sx sin2 u + sy cos2 u
Pn =
=
1
(s - n sn + 90°)
E n
1
(s cos2 u + sy sin2 u - n sx sin2 u - n sy cos2 u)
E x
If Pn is to be independent of sx, then
cos2 u - n sin2 u = 0
u = tan - 1 a
1
2n
or
tan2 u = 1/n
b
Ans.
Ans:
u = tan - 1 a
1007
1
2n
b
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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.
10–46. The principal strains in a plane, measured
experimentally at a point on the aluminum fuselage of a jet
aircraft, are P1 = 630(10-6) and P2 = 350(10-6). If this is
a case of plane stress, determine the associated principal
stresses at the point in the same plane. Eal = 10(103) ksi
and nal = 0.33.
Normal Stresses: For plane stress, s3 = 0.
Normal Strains: Applying the generalized Hooke’s Law.
P1 =
1
C s - v (s2 + s3) D
E 1
630 A 10 - 6 B =
1
[s1 - 0.33(s2 + 0)]
10(103)
6.30 = s1 - 0.33s2
P2 =
[1]
1
C s2 - v (s1 + s3) D
E
350 A 10 - 6 B =
1
C s2 - 0.33(s1 + 0) D
10(103)
3.50 = s2 - 0.33s1
[2]
Solving Eqs.[1] and [2] yields:
s1 = 8.37 ksi
s2 = 6.26 ksi
Ans.
Ans:
s1 = 8.37 ksi, s2 = 6.26 ksi
1008
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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.
10–47. The principal stresses at a point are shown in the
figure. If the material is aluminum for which
Eal = 10(103) ksi and nal = 0.33, determine the principal
strains.
3 ksi
P1 =
1
1
e 8 - 0.33 C 3 + (- 4) D f = 833 (10 - 6)
C s - v(s2 + s3) D =
E 1
10(103)
P2 =
1
1
e 3 - 0.33 C 8 + (- 4) D f = 168 (10 - 6)
C s2 - v(s1 + s3) D =
E
10(103)
P3 =
1
1
C s - v(s1 + s2) D =
C - 4 - 0.33(8 + 3) D = - 763 (10 - 6)
E 3
10(103)
8 ksi
4 ksi
Using these results,
P1 = 833(10 - 6)
P2 = 168(10 - 6)
P3 = - 763(10 - 6)
Ans.
Ans:
P1 = 833(10 - 6), P2 = 168(10 - 6),
P3 = - 763(10 - 6)
1009
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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.
*10–48. The 6061-T6 aluminum alloy plate fits snugly into
the rigid constraint. Determine the normal stresses sx and
sy developed in the plate if the temperature is increased by
¢T = 50°C . To solve, add the thermal strain a¢T to the
equations for Hooke’s Law.
y
400 mm
300 mm
x
Generalized Hooke’s Law: Since the sides of the aluminum plate are confined in the
rigid constraint along the x and y directions, Px = Py = 0. However, the plate is
allowed to have free expansion along the z direction. Thus, sz = 0. With the
additional thermal strain term, we have
Px =
0 =
1
c s - v A sy + sz B d + a¢T
E x
1
68.9 A 109 B
c sx - 0.35 A sy + 0 B d + 24 a10 - 6 b(50)
sx - 0.35sy = - 82.68 A 106 B
Py =
0 =
(1)
1
C s - v A sx + sz B D + a¢T
E y
1
68.9 a 109 b
C sy - 0.35(sx + 0) D + 24 A 10 - 6 B (50)
sy - 0.35sx = - 82.68 A 106 B
(2)
Solving Eqs. (1) and (2),
sx = sy = - 127.2 MPa = 127 MPa (C)
Ans.
Since sx = sy and sy 6 sY, the above results are valid.
1010
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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.
10–49. Initially, gaps between the A-36 steel plate and the
rigid constraint are as shown. Determine the normal
stresses sx and sy developed in the plate if the temperature
is increased by ¢T = 100°F . To solve, add the thermal
strain a¢T to the equations for Hooke’s Law.
y
0.0015 in.
6 in.
8 in.
0.0025 in.
x
Generalized Hooke’s Law: Since there are gaps between the sides of the plate and the rigid
constraint, the plate is allowed to expand before it comes in contact with the constraint.
dy
dx
0.0025
0.0015
=
=
= 0.3125 A 10 - 3 B and Py =
= 0.25 A 10 - 3 B .
Thus, Px =
Lx
8
Ly
6
However, the plate is allowed to have free expansion along the z direction. Thus, sz = 0.
With the additional thermal strain term, we have
Px =
1
c s - v A sy + sz B d + a¢T
E x
0.3125a 10 - 3 b =
1
29.0 a 10 b
3
C sx - 0.32 A sy + 0 B D + 6.60 A 10 - 6 B (100)
sx - 0.32sy = - 10.0775
Py =
(1)
1
C s - v A sx + sz B D + a¢T
E y
0.25 A 10 - 3 B =
1
29.0 A 103 B
C sy - 0.32(sx + 0) D + 6.60 A 10 - 6 B (100)
sy - 0.32sx = - 11.89
(2)
Solving Eqs. (1) and (2),
sx = - 15.5 ksi = 15.5 ksi (C)
Ans.
sy = - 16.8 ksi = 16.8 ksi (C)
Ans.
Since sx 6 sY and sy 6 sY, the above results are valid.
Ans:
sx = 15.5 ksi(C), sy = 16.8 ksi(C)
1011
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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
10–50. The steel shaft has a radius of 15 mm. Determine
the torque T in the shaft if the two strain gauges, attached to
the surface of the shaft, report strains of Px ¿ = - 80(10 - 6)
and Py ¿ = 80(10 - 6). Also, compute the strains acting in the
x and y directions. Est = 200 GPa, nst = 0.3.
T
y¿
x¿
45⬚
x
T
Px¿ = - 80(10 - 6)
Py¿ = 80(10 - 6)
Pure shear Px = Py = 0
Ans.
Px ¿ = Px cos2 u + Py sin2 u + gxy sin u cos u
u = 45°
-80(10 - 6) = 0 + 0 + gxy sin 45° cos 45°
gxy = - 160(10 - 6)
Ans.
Also, u = 135°
80(10 - 6) = 0 + 0 + g sin 135° cos 135°
gxy = - 160(10 - 6)
G =
200(109)
E
=
= 76.923(109)
2(1 + n)
2(1 + 0.3)
t = Gg = 76.923(109)(160)(10 - 6) = 12.308(106) Pa
p
12.308(106) a b (0.015)4
2
tJ
=
= 65.2 N # m
T =
c
0.015
Ans.
Ans:
Px = Py = 0, gxy = - 160(10 - 6), T = 65.2 N # m
1012
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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
10–51. The shaft has a radius of 15 mm and is made of
L2 tool steel. Determine the strains in the x¿ and y¿
direction if a torque T = 2 kN # m is applied to the shaft.
T
y¿
x¿
45⬚
x
T
t =
2(103)(0.015)
Tc
= 377.26 MPa
= p
4
J
2 (0.015 )
Stress-Strain Relationship:
gxy =
txy
G
377.26(106)
= -
75.0(109)
= - 5.030(10 - 3) rad
This is a pure shear case, therefore,
Px = Py = 0
Applying Eq. 10–15,
Px ¿ = Px cos2 ua + Py sin2 ua + gxy sin ua cos ua
Here ua = 45°
Px ¿ = 0 + 0 - 5.030(10 - 3) sin 45° cos 45° = - 2.52(10 - 3)
Px ¿ = - 2.52(10 - 3)
Ans.
Py ¿ = 2.52(10 - 3)
Ans.
Ans:
Px¿ = - 2.52(10 - 3), Py¿ = 2.52(10 - 3)
1013
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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.
*10–52. The metal block is fitted between the fixed
supports. If the glued joint can resist a maximum shear stress
of tallow = 2 ksi, determine the temperature rise that will
cause the joint to fail. Take E = 10(103) ksi, n = 0.2, and
a = 6.0(10 - 6)>°F. Hint: Use Eq. 10–18 with an additional
strain term of a¢T (Eq. 4–4).
40⬚
Normal Strain: Since the aluminum is confined in the y direction by the rigid
supports, then Py = 0 and sx = sz = 0. Applying the generalized Hooke’s Law
with the additional thermal strain,
Py =
0 =
1
C s - v(sx + sz) D + a¢T
E y
1
C sy - 0.2(0 + 0) D + 6.0 A 10 - 6 B (¢T)
10.0(103)
sy = - 0.06¢T
Construction of the Circle: In accordance with the sign convention. sx = 0,
sy = - 0.06¢T and txy = 0. Hence.
savg =
sx + sy
2
=
0 + ( - 0.06¢T)
= - 0.03¢T
2
The coordinates for reference points A and C are A (0, 0) and C(-0.03¢T, 0).
The radius of the circle is R = 2(0 - 0.03¢T)2 + 0 = 0.03¢T
Stress on the inclined plane: The shear stress components tx¿y¿, are represented by
the coordinates of point P on the circle.
tx¿y¿ = 0.03¢T sin 80° = 0.02954¢T
Allowable Shear Stress:
tallow = tx¿y¿
2 = 0.02954¢T
¢T = 67.7 °F
Ans.
1014
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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.
10–53. Air is pumped into the steel thin-walled pressure
vessel at C. If the ends of the vessel are closed using two
pistons connected by a rod AB, determine the increase in
the diameter of the pressure vessel when the internal gauge
pressure is 5 MPa. Also, what is the tensile stress in rod AB
if it has a diameter of 100 mm? The inner radius of the
vessel is 400 mm, and its thickness is 10 mm. Est = 200 GPa
and nst = 0.3.
C
400 mm
A
B
Circumferential Stress:
s =
5(400)
pr
=
= 200 MPa
t
10
Note: longitudinal and radial stresses are zero.
Circumferential Strain:
P =
200(106)
s
= 1.0(10 - 3)
=
E
200(109)
¢d = P d = 1.0(10 - 3)(800) = 0.800 mm
Ans.
For rod AB:
+
; ©Fx = 0;
p
TAB - 5(106) a b (0.82 - 0.12) = 0
4
TAB = 2474 kN
sAB =
2474(103)
TAB
= 315 MPa
= p
2
AAB
4 (0.1 )
Ans.
Ans:
¢d = 0.800 mm, sAB = 315 MPa
1015
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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.
10–54. Determine the increase in the diameter of the
pressure vessel in Prob. 10–53 if the pistons are replaced by
walls connected to the ends of the vessel.
C
400 mm
A
B
Principal Stress:
s1 =
5(400)
pr
=
= 200 MPa;
t
10
s2 =
1
s = 100 MPa
2 1
s3 = 0
Circumferential Strain:
P1 =
1
1
[200(106) - 0.3{100(106) + 0}]
[s - n(s2 + s3)] =
E 1
200(109)
= 0.85(10 - 3)
¢d = P1 d = 0.85(10 - 3)(800) = 0.680 mm
Ans.
Ans:
¢d = 0.680 mm
1016
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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.
10–55. A thin-walled spherical pressure vessel having an
inner radius r and thickness t is subjected to an internal
pressure p. Show that the increase in the volume within
the vessel is ¢V = 12ppr4>Et211 - n2. Use a small-strain
analysis.
pr
2t
s1 = s2 =
s3 = 0
P1 = P2 =
1
(s - vs2)
E 1
P1 = P2 =
pr
(1 - v)
2t E
P3 =
1
(- v(s1 + s2))
E
P3 = -
V =
v pr
tE
4pr3
3
V + ¢V =
4p
4pr3
¢r 3
)
(r + ¢r)3 =
(1 +
r
3
3
where ¢V V V, ¢r V r
V + ¢V =
PVol =
¢r
4p r3
b
a1 + 3
r
3
¢r
¢V
= 3a b
V
r
Since P1 = P2 =
PVol = 3P1 =
2p(r + ¢r) - 2p r
¢r
=
r
2p r
3pr
(1 - v)
2t E
¢V = VPVol =
2pp r4
(1 - v)
Et
(Q.E.D.)
1017
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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.
*10–56. The thin-walled cylindrical pressure vessel of
inner radius r and thickness t is subjected to an internal
pressure p. If the material constants are E and n, determine
the strains in the circumferential and longitudinal
directions. Using these results, compute the increase in both
the diameter and the length of a steel pressure vessel filled
with air and having an internal gauge pressure of 15 MPa.
The vessel is 3 m long, and has an inner radius of 0.5 m and
a thickness of 10 mm. Est = 200 GPa, nst = 0.3.
3m
0.5 m
Normal Stress:
s1 =
pr
t
s2 =
pr
2t
s3 = 0
Normal Strain:
Pcir =
=
1
[s - n(s2 + s3)]
E 1
npr
pr
1 pr
a
b =
(2 - n)
E
t
2t
2Et
Plong =
=
Ans.
1
[s - n(s1 + s3)]
E 2
npr
pr
1 pr
a
b =
(1 - 2n)
E 2t
t
2Et
Ans.
Numerical Substitution:
15(106)(0.5)
Pcir =
2(200)(109)(0.01)
(2 - 0.3) = 3.1875 (10 - 3)
¢d = Pcir d = 3.1875 (10 - 3)(1000) = 3.19 mm
15(106)(0.5)
Plong =
2(200)(109)(0.01)
Ans.
(1 - 2(0.3)) = 0.75(10 - 3)
¢L = Plong L = 0.75 (10 - 3)(3000) = 2.25 mm
Ans.
1018
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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.
10–57. Estimate the increase in volume of the tank in
Prob.10–56.
3m
0.5 m
By basic principles,
¢V = p(r + ¢r)2(L + ¢L) - p r2 L = p(r2 + ¢r2 + 2 r ¢r)(L + ¢L) - p r2 L
= p(r2L + r2 ¢L + ¢r2L + ¢r2 ¢L + 2 r ¢r L + 2 r ¢r ¢L - r2 L)
= p(r2 ¢L + ¢r2 L + ¢r2 ¢L + 2 r ¢r L + 2 r ¢r ¢L)
Neglecting the second order terms,
¢V = p(r2 ¢L + 2 r ¢r L)
From Prob. 10–56,
¢L = 0.00225 m
¢r =
¢d
= 0.00159375 m
2
¢V = p[(0.52)(0.00225) + 2(0.5)(0.00159375)(3)] = 0.0168 m3
Ans.
Ans:
¢V = 0.0168 m3
1019
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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.
z
10–58. A soft material is placed within the confines of a
rigid cylinder which rests on a rigid support. Assuming that
Px = 0 and Py = 0, determine the factor by which the
modulus of elasticity will be increased when a load is
applied if n = 0.3 for the material.
P
x
y
Normal Strain: Since the material is confined in a rigid cylinder. Px = Py = 0.
Applying the generalized Hooke’s Law,
Px =
1
C s - v(sy + sz) D
E x
0 = sx - v(sy + sz)
Py =
[1]
1
C sy - v(sx + sz) D
E
0 = sy - v(sx + sz)
[2]
Solving Eqs.[1] and [2] yields:
sx = sy =
v
s
1 - v z
Thus,
Pz =
=
1
C s - v(sx + sy) D
E z
v
v
1
cs - va
s +
s bd
E z
1 - v z
1 - v z
sz
=
E
c1 -
2v2
d
1 - v
=
sz 1 - v - 2v2
c
d
E
1 - v
=
sz (1 + v)(1 - 2v)
c
d
E
1 - v
Thus, when the material is not being confined and undergoes the same normal strain
of Pz, then the requtred modulus of elasticity is
E¿ =
sz
=
Pz
The increase factor is
k =
1 - v
E
(1 - 2v)(1 + v)
E¿
1 - v
=
E
(1 - 2v)(1 + v)
=
1 - 0.3
[1 - 2(0.3)](1 + 0.3)
Ans.
= 1.35
Ans:
k = 1.35
1020
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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.
10–59. A material is subjected to plane stress. Express the
distortion-energy theory of failure in terms of sx , sy , and txy .
Maximum distortion energy theory:
(s21 - s1 s2 + s22) = s2Y
s1,2 =
sx + sy
Let a =
;
2
sx + sy
2
s1 = a + b;
A
a
(1)
sx - sy
and b =
2
A
a
2
2
b + txy
sx - sy
2
2
2
b + txy
s2 = a - b
s21 = a2 + b2 + 2 a b;
s22 = a2 + b2 - 2 a b
s1 s2 = a2 - b2
From Eq. (1)
(a2 + b2 + 2 a b - a2 + b2 + a2 + b2 - 2 a b) = sY2
(a2 + 3 b2) = s2Y
(sx + sy)2
4
+ 3
(sx - sy)2
4
+ 3 t2xy = s2Y
s2x + s2y - sxsy + 3 t2xy = s2Y
Ans.
Ans:
s2x + s2y - sxsy + 3t2xy = s2y
1021
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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.
*10–60. A material is subjected to plane stress. Express
the maximum-shear-stress theory of failure in terms of sx ,
sy , and txy . Assume that the principal stresses are of
different algebraic signs.
Maximum shear stress theory:
|s1 - s2| = sY
s1,2 =
(1)
sx + sy
;
2
` s1 - s2 ` = 2 A a
A
a
sx - sy
2
sx - sy
2
2
2
b + txy
2
b + txy
2
From Eq. (1)
4 ca
sx - sy
2
2
b + t2xy d = s2Y
2
(sx - sy) + 4 t2xy = s2Y
Ans.
1022
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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.
10–61. The yield stress for a zirconium-magnesium alloy is
sY = 15.3 ksi. If a machine part is made of this material and a
critical point in the material is subjected to in-plane principal
stresses s1 and s2 = - 0.5s1, determine the magnitude of s1
that will cause yielding according to the maximum-shear-stress
theory.
sY = 15.3 ksi
s1 - s2 = 15.3
s1 - (- 0.5 s1) = 15.3
s1 = 10.2 ksi
Ans.
Ans:
s1 = 10.2 ksi
1023
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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.
10–62. Solve Prob. 10–61 using the maximum-distortion
energy theory.
s21 - s1 s2 + s22 = sY2
s21 - s1(- 0.5s1) + (- 0.5s1)2 = sY2
1.75 s21 = (15.3)2
s1 = 11.6 ksi
Ans.
Ans:
s1 = 11.6 ksi
1024
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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.
10–63. An aluminum alloy is to be used for a drive shaft
such that it transmits 25 hp at 1500 rev>min. Using a factor
of safety of 2.5 with respect to yielding, determine the
smallest-diameter shaft that can be selected based on the
maximum-distortion-energy theory. sY = 3.5 ksi.
1500(2p)
= 50p
60
T =
P
v
T =
25(550)(12)
3300
=
p
50p
t =
Tc
,
J
t =
3300
p c
p 4
2c
s1 =
v =
J =
=
p 4
c
2
6600
p2c3
6600
p2c3
s2 =
s21 - s1 s2 + s22 = a
3a
-6600
p2c3
sY 2
b
F.S.
3.5(103) 2
6600 2
b
=
a
b
2.5
p2c3
c = 0.9388 in.
d = 1.88 in.
Ans.
Ans:
d = 1.88 in.
1025
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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.
*10–64. If a shaft is made of a material for which
sY = 50 ksi, determine the torsional shear stress required
to cause yielding using the maximum-distortion-energy
theory.
s1 = t
s2 = - t
s12 - s1 s2 + s22 = sY2
3t2 = 502
t = 28.9 ksi
Ans.
1026
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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.
10–65. Solve Prob. 10–64 using the maximum-shear-stress
theory.
s1 = t
s2 = - t
|s1 - s2| = sY
t - (- t) = 50
t = 25 ksi
Ans.
Ans:
t = 25 ksi
1027
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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.
10–66. Derive an expression for an equivalent torque Te
that, if applied alone to a solid bar with a circular cross
section, would cause the same energy of distortion as
the combination of an applied bending moment M and
torque T.
t =
Te c
J
Principal Stress:
s1 = t s2 = - t
ud =
1 + v 2
(s1 - s1 s2 + s22)
3E
1 + v
1 + v 3 T2e c2
( 3 t2) =
a
b
3E
3E
J2
(ud)1 =
Bending Moment and Torsion:
s =
Mc
;
I
t =
Tc
J
Principal Stress:
s1, 2 =
s1 =
s + 0
s - 0 2
2
;
a
b + t
A
2
2
s
s2
+
+ t2 ;
A4
2
Let a =
s
2
b =
s2 =
s
s2
+ t2
A4
2
s2
+ t2
A4
s21 = a2 + b2 + 2 a b
s1 s2 = a2 - b2
s22 = a2 + b2 - 2 a b
s21 - s1 s2 + s22 = 3 b2 + a2
ud =
1 + v 2
(s1 - s1 s2 + s22)
3E
(ud)2 =
=
1 + v
1 + v 3 s2
s2
(3 b2 + a2) =
a
+ 3t2 +
b
3E
3E
4
4
c2(1 + v) M2
1 + v 2
3 T2
(s + 3 t2) =
a 2 +
b
3E
3E
I
J2
(ud)1 = (ud)2
c2(1 + v) 3 Te2
c2(1 + v) M2
3 T2
=
+
b
a
3E
3E
J2
I2
J2
1028
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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.
10–66. Continued
For circular shaft
J
=
I
p
2
p
4
c4
c4
=2
Te =
J2 M2
+ T2
A I2 3
Te =
4 2
M + T2
A3
Ans.
Ans:
Te =
1029
4 2
M + T2
A3
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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.
10–67. Derive an expression for an equivalent bending
moment Me that, if applied alone to a solid bar with a
circular cross section, would cause the same energy of
distortion as the combination of an applied bending
moment M and torque T.
Principal Stresses:
s1 =
Me c
;
I
ud =
1 + v 2
(s1 - s1 s2 + s22)
3E
s2 = 0
1 + v M2e c2
a 2 b
3E
I
(ud)1 =
(1)
Principal Stress:
s + 0
s - 0 2
3
;
a
b + t
A
2
2
s1, 2 =
s1 =
s
s2
+
+ t2;
A4
2
s2 =
s
s2
+ t2
A4
2
Distortion Energy:
Let a =
2
s
, b = s + t2
A4
2
s21 = a2 + b2 + 2 a b
s1 s2 = a2 - b2
s22 = a2 + b2 - 2 a b
s22 - s1 s2 + s22 = 3 b2 + a2
Apply s =
(ud)2 =
=
Mc
;
I
t =
Tc
J
1 + v
1 + v s2
3s2
(3 b2 + a2) =
a
+
+ 3 t2 b
3E
3E
4
4
1 + v 2
1 + v M2 c2
3 T2 c2
(s + 3 t2) =
a 2 +
b
3E
3E
I
J2
(2)
Equating Eq. (1) and (2) yields:
(1 + v) Me c2
1 + v M2 c2
3T2 c2
a 2 b =
a 2 +
b
3E
3E
I
I
J2
M2e
I
2
=
M2
3 T2
+
2
I
J2
M2e = M2 + 3 T2 a
I 2
b
J
1030
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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.
10–67. Continued
For circular shaft
I
=
J
p
4
p
2
c4
c4
=
1
2
1 2
Hence, M2e = M2 + 3 T2 a b
2
Me =
A
M2 +
3 2
T
4
Ans.
Ans:
Me =
1031
A
M2 +
3 2
T
4
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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.
*10–68. The principal plane stresses acting on a differential
element are shown. If the material is machine steel having a
yield stress of sY = 700 MPa, determine the factor of
safety with respect to yielding if the maximum-shear-stress
theory is considered
50 MPa
80 MPa
smax = 80 MPa
smin = - 50 MPa
tabs =
80 - ( - 50)
smax - smin
=
= 65 MPa
2
2
tmax =
sY
700
=
= 350 MPa
2
2
F.S. =
tmax
350
=
= 5.38
tabs
65
max
Ans.
max
1032
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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.
10–69. The short concrete cylinder having a diameter of
50 mm is subjected to a torque of 500 N # m and an axial
compressive force of 2 kN. Determine if it fails according to
the maximum-normal-stress theory. The ultimate stress of
the concrete is sult = 28 MPa.
A =
p
(0.05)2 = 1.9635(10 - 3) m2
4
J =
p
(0.025)4 = 0.61359(10 - 4) m4
2
2 kN
500 N⭈m
500 N⭈m
2 kN
3
s =
2(10 )
P
= 1.019 MPa
=
A
1.9635(10 - 3)
t =
500(0.025)
Tc
= 20.372 MPa
=
J
0.61359(10 - 6)
sx = 0
s1, 2 =
s1,2 =
sy = - 1.019 MPa
sx + sy
2
;
Aa
sx - sy
2
txy = 20.372 MPa
2
b + txy
2
0 - 1.018
0 - ( -1.019) 2
2
;
a
b + 20.372
A
2
2
s1 = 19.87 MPa
s2 = - 20.89 MPa
Failure criteria:
|s1| 6 salt = 28 MPa
OK
|s2| 6 salt = 28 MPa
OK
No.
Ans.
Ans:
No.
1033
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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.
10–70. Derive an expression for an equivalent bending
moment Me that, if applied alone to a solid bar with a
circular cross section, would cause the same maximum
shear stress as the combination of an applied moment M
and torque T. Assume that the principal stresses are of
opposite algebraic signs.
Bending and Torsion:
Mc
Mc
4M
= p 4 =
;
I
p c3
4 c
s =
t =
Tc
Tc
2T
= p 4 =
J
p c3
2c
The principal stresses:
s1, 2 =
tabs
max
sx + sy
2
;
Aa
sx - sy
2
2
2
b + txy =
=
2M
2
;
2M2 + T2
p c3
p c3
=
s1 - s2
2
=
2M2 + T2
2
p c3
4M
pc3
4M
+ 0
2
3
;
Q
¢pc
- 0
2
2
≤ + a
b
3
pc
2T
2
(1)
Pure bending:
s1 =
tabs
max
Me c
4 Me
Mc
= p 4 =
;
I
c
p c3
4
=
s2 = 0
s1 - s2
2 Me
=
2
p c3
(2)
Equating Eq. (1) and (2) yields:
2 Me
2
2M2 + T2 =
p c3
p c3
Me = 2M2 + T2
Ans.
Ans:
Me = 2M2 + T2
1034
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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.
sy ⫽ 0.5sx
10–71. The plate is made of hard copper, which yields
at sY = 105 ksi . Using the maximum-shear-stress theory,
determine the tensile stress sx that can be applied to the
plate if a tensile stress sy = 0.5sx is also applied.
sx
s1 = sx
s2 =
1
s
2 x
|s1| = sY
sx = 105 ksi
Ans.
Ans:
sx = 105 ksi
1035
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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.
sy ⫽ 0.5sx
*10–72. Solve Prob. 10–71 using the maximum-distortion
energy theory.
sx
s1 = sx
s2 =
sx
2
s21 - s1 s2 + s22 = sY2
sx2 -
sx2
sx2
+
= (105)2
2
4
sx = 121 ksi
Ans.
1036
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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.
10–73. If the 2-in.-diameter shaft is made from brittle
material having an ultimate strength of sult = 50 ksi, for
both tension and compression, determine if the shaft fails
according to the maximum-normal-stress theory. Use a factor
of safety of 1.5 against rupture.
30 kip
4 kip · ft
Normal Stress and Shear Stresses. The cross-sectional area and polar moment of
inertia of the shaft’s cross-section are
A = p A 12 B = pin2
J =
p 4
p
A 1 B = in4
2
2
The normal stress is caused by axial stress.
s =
N
30
= = - 9.549 ksi
p
A
The shear stress is contributed by torsional shear stress.
t =
4(12)(1)
Tc
=
= 30.56 ksi
p
J
2
The state of stress at the points on the surface of the shaft is represented on the
element shown in Fig. a.
In-Plane Principal Stress. sx = - 9.549 ksi, sy = 0 and txy = - 30.56 ksi. We have
s1, 2 =
=
sx + sy
2
;
Aa
sx - sy
2
2
b + txy
2
- 9.549 - 0 2
-9.549 + 0
2
;
b + (- 30.56)
Aa
2
2
= ( -4.775 ; 30.929) ksi
s1 = 26.15 ksi
s2 = - 35.70 ksi
Maximum Normal-Stress Theory.
sallow =
sult
50
=
= 33.33 ksi
F.S.
1.5
|s1| = 26.15 ksi 6 sallow = 33.33 ksi
(O.K.)
|s2| = 35.70 ksi 7 sallow = 33.33 ksi
(N.G.)
Based on these results, the material fails according to the maximum normal-stress
theory.
Ans:
Yes.
1037
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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.
10–74. If the 2-in.-diameter shaft is made from cast
iron having tensile and compressive ultimate strengths
of 1sult2t = 50 ksi and 1sult2c = 75 ksi, respectively,
determine if the shaft fails in accordance with Mohr’s
failure criterion.
Normal Stress and Shear Stresses. The cross-sectional area and polar moment of
inertia of the shaft’s cross-section are
A = p A 12 B = p in2
J =
p 4
p
A 1 B = in4
2
2
30 kip
4 kip · ft
The normal stress is contributed by axial stress.
s =
N
30
= = - 9.549 ksi
p
A
The shear stress is contributed by torsional shear stress.
t =
4(12)(1)
Tc
=
= 30.56 ksi
p
J
2
The state of stress at the points on the surface of the shaft is represented on the
element shown in Fig. a.
In-Plane Principal Stress. sx = - 9.549 ksi, sy = 0, and txy = - 30.56 ksi. We have
s1, 2 =
=
sx + sy
2
;
Aa
sx - sy
2
2
b + txy
2
- 9.549 - 0 2
- 9.549 + 0
2
;
a
b + ( -30.56)
A
2
2
= (- 4.775 ; 30.929) ksi
s1 = 26.15 ksi
s2 = - 35.70 ksi
Mohr’s Failure Criteria. As shown in Fig. b, the coordinates of point A, which
represent the principal stresses, are located inside the shaded region. Therefore, the
material does not fail according to Mohr’s failure criteria.
Ans:
No.
1038
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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.
10–75. The components of plane stress at a critical point on
an A-36 steel shell are shown. Determine if failure (yielding)
has occurred on the basis of the maximum-shear-stress theory.
60 MPa
40 MPa
In accordance with the established sign convention, sx = 70 MPa, sy = - 60 MPa
and txy = 40 MPa.
s1, 2 =
=
sx + sy
2
;
Aa
sx - sy
2
70 MPa
2
b + txy
2
70 + (- 60)
70 - ( - 60) 2
2
;
c
d + 40
A
2
2
= 5 ; 25825
s1 = 81.32 MPa
s2 = - 71.32 MPa
In this case, s1 and s2 have opposite sign. Thus,
|s1 - s2| = |81.32 - ( -71.32)| = 152.64 MPa 6 sy = 250 MPa
Based on this result, the steel shell does not yield according to the maximum shear
stress theory.
Ans:
No.
1039
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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.
*10–76. The components of plane stress at a critical point
on an A-36 steel shell are shown. Determine if failure (yielding)
has occurred on the basis of the maximum-distortion-energy
theory.
60 MPa
40 MPa
In accordance with the established sign convention, sx = 70 MPa, sy = - 60 MPa
and txy = 40 MPa.
s1, 2 =
=
sx + sy
2
;
Aa
sx - sy
2
2
b + txy
2
70 + ( - 60)
70 - ( -60) 2
2
;
c
d + 40
A
2
2
= 5 ; 25825
s1 = 81.32 MPa
s2 = - 71.32 MPa
s1 2 - s1 s2 + s2 2 = 81.322 - 81.32( - 71.32) + ( - 71.32)2 = 17,500 6 sy 2 = 62500
Based on this result, the steel shell does not yield according to the maximum
distortion energy theory.
1040
70 MPa
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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.
10–77. If the A-36 steel pipe has outer and inner
diameters of 30 mm and 20 mm, respectively, determine the
factor of safety against yielding of the material at point A
according to the maximum-shear-stress theory.
900 N
150 mm
A
100 mm
200 mm
Internal Loadings. Considering the equilibrium of the free-body diagram of the
post’s right cut segment Fig. a,
©Fy = 0; Vy + 900 - 900 = 0
Vy = 0
T = - 360 N # m
©Mx = 0; T + 900(0.4) = 0
©Mz = 0; Mz + 900(0.15) - 900(0.25) = 0 Mz = 90 N # m
Section Properties. The moment of inertia about the z axis and the polar moment of
inertia of the pipe’s cross section are
Iz =
p
A 0.0154 - 0.014 B = 10.15625p A 10 - 9 B m4
4
J =
p
A 0.0154 - 0.014 B = 20.3125p A 10 - 9 B m4
2
Normal Stress and Shear Stress. The normal stress is contributed by bending stress.
Thus,
sY = -
MyA
90(0.015)
= = - 42.31MPa
Iz
10.15625p A 10 - 9 B
The shear stress is contributed by torsional shear stress.
t =
360(0.015)
Tc
=
= 84.62 MPa
J
20.3125p A 10 - 9 B
The state of stress at point A is represented by the two-dimensional element shown
in Fig. b.
In-Plane Principal Stress. sx = - 42.31 MPa, sz = 0 and txz = 84.62 MPa. We have
s1, 2 =
=
sx + sz
2
;
Aa
sx - sz
2
2
b + txz
2
- 42.31 - 0 2
- 42.31 + 0
2
;
a
b + 84.62
A
2
2
= ( - 21.16 ; 87.23) MPa
s1 = 66.07 MPa
s2 = - 108.38 MPa
1041
200 mm
900 N
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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.
10–77. Continued
Maximum Shear Stress Theory. s1 and s2 have opposite signs. This requires
|s1 - s2| = sallow
66.07 - (- 108.38) = sallow
sallow = 174.45 MPa
The factor of safety is
F.S. =
sY
250
=
= 1.43
sallow
174.45
Ans.
Ans:
F.S. = 1.43
1042
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. 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.
10–78. If the A-36 steel pipe has an outer and inner
diameter of 30 mm and 20 mm, respectively, determine the
factor of safety against yielding of the material at point A
according to the maximum-distortion-energy theory.
900 N
150 mm
A
100 mm
200 mm
Internal Loadings: Considering the equilibrium of the free-body diagram of the
pipe’s right cut segment Fig. a,
©Fy = 0; Vy + 900 - 900 = 0
Vy = 0
T = - 360 N # m
©Mx = 0; T + 900(0.4) = 0
©Mz = 0; Mz + 900(0.15) - 900(0.25) = 0 Mz = 90 N # m
Section Properties. The moment of inertia about the z axis and the polar moment of
inertia of the pipe’s cross section are
Iz =
p
A 0.0154 - 0.014 B = 10.15625p A 10 - 9 B m4
4
J =
p
A 0.0154 - 0.014 B = 20.3125p A 10 - 9 B m4
2
Normal Stress and Shear Stress. The normal stress is caused by bending stress. Thus,
sY = -
MyA
90(0.015)
= = - 42.31MPa
Iz
10.15625p A 10 - 9 B
The shear stress is caused by torsional stress.
t =
360(0.015)
Tc
= 84.62 MPa
=
J
20.3125p A 10 - 9 B
The state of stress at point A is represented by the two-dimensional element shown
in Fig. b.
In-Plane Principal Stress. sx = - 42.31 MPa, sz = 0 and txz = 84.62 MPa. We have
s1, 2 =
=
sx + sz
2
;
Aa
sx - sz
2
2
b + txz
2
- 42.31 - 0 2
-42.31 + 0
2
;
a
b + 84.62
A
2
2
= ( -21.16 ; 87.23) MPa
s1 = 66.07 MPa
s2 = - 108.38 MPa
1043
200 mm
900 N
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. 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.
10–78. Continued
Maximum Distortion Energy Theory.
s1 2 - s1s2 + s2 2 = sallow 2
66.072 - 66.07( - 108.38) + ( - 108.38)2 = sallow 2
sallow = 152.55 MPa
Thus, the factor of safety is
F.S. =
sY
250
=
= 1.64
sallow
152.55
Ans.
Ans:
F.S. = 1.64
1044
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. 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.
10–79. The yield stress for heat-treated beryllium copper
is sY = 130 ksi. If this material is subjected to plane stress
and elastic failure occurs when one principal stress is 145 ksi,
what is the smallest magnitude of the other principal stress?
Use the maximum-distortion-energy theory.
Maximum Distortion Energy Theory: With s1 = 145 ksi,
s21 - s1 s2 + s22 = s2Y
1452 - 145s2 + s22 = 1302
s22 - 145s2 + 4125 = 0
s2 =
-( - 145) ; 2( - 145)2 - 4(1)(4125)
2(1)
= 72.5 ; 33.634
Choose the smaller root, s2 = 38.9 ksi
Ans.
Ans:
s2 = 38.9 ksi
1045
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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.
*10–80. The yield stress for a uranium alloy is
sY = 160 MPa. If a machine part is made of this material
and a critical point in the material is subjected to plane stress,
such that the principal stresses are s1 and s2 = 0.25s1,
determine the magnitude of s1 that will cause yielding
according to the maximum-distortion energy theory.
s12 - s1 s2 + s22 = sY2
s12 - (s1)(0.25s1) + (0.25s1)2 = sY2
0.8125s12 = sY2
0.8125s12 = (160)2
s1 = 178 MPa
Ans.
1046
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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.
10–81. Solve Prob. 10–80 using the maximum-shear-stress
theory.
s1
2
tabs
=
tabs
= tallow
max
max
`
s1
` = 80;
2
tallow =
sY
160
=
= 80 MPa
2
2
Ans.
s1 = 160 MPa
Ans:
s1 = 160 MPa
1047
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. 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.
10–82. The state of stress acting at a critical point on the
seat frame of an automobile during a crash is shown in the
figure. Determine the smallest yield stress for a steel that
can be selected for the member, based on the maximumshear-stress theory.
Normal and Shear Stress: In accordance with the sign convention.
sx = 80 ksi
sy = 0
25 ksi
txy = 25 ksi
80 ksi
In-Plane Principal Stress: Applying Eq. 9-5.
s1,2 =
=
sx + sy
;
2
A
a
sx - sy
2
2
b + txy
2
80 - 0 2
80 + 0
2
;
a
b + 25
2
A
2
= 40 ; 47.170
s1 = 87.170 ksi
s2 = - 7.170 ksi
Maximum Shear Stress Theory: s1 and s2 have opposite signs so
|s1 - s2| = sY
|87.170 - ( - 7.170)| = sY
sY = 94.3 ksi
Ans.
Ans:
sY = 94.3 ksi
1048
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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.
10–83. Solve Prob. 10–82 using the maximum-distortionenergy theory.
25 ksi
Normal and Shear Stress: In accordance with the sign convention.
sx = 80 ksi
sy = 0
80 ksi
txy = 25 ksi
In-Plane Principal Stress: Applying Eq. 9-5.
s1,2 =
=
sx + sy
;
2
a
sx - s 2
2
b + txy
A
2
80 - 0 2
80 + 0
2
;
a
b + 25
2
A
2
= 40 ; 47.170
s1 = 87.170 ksi
s2 = - 7.170 ksi
Maximum Distortion Energy Theory:
s21 - s1s2 + s22 = s2Y
87.1702 - 87.170(- 7.170) + ( -7.170)2 = s2Y
sY = 91.0 ksi
Ans.
Ans:
sY = 91.0 ksi
1049
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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.
*10–84. A bar with a circular cross-sectional area is made
of SAE 1045 carbon steel having a yield stress of
sY = 150 ksi. If the bar is subjected to a torque of
30 kip # in. and a bending moment of 56 kip # in., determine
the required diameter of the bar according to the
maximum-distortion-energy theory. Use a factor of safety
of 2 with respect to yielding.
Normal and Shear Stresses: Applying the flexure and torsion formulas.
56 A d2 B
Mc
1792
=
=
p d 4
I
pd3
A B
s =
4
t =
Tc
=
J
2
30 A d2 B
A B
d 4
2
p
2
=
480
pd3
The critical state of stress is shown in Fig. (a) or (b), where
sx =
1792
pd3
sy = 0
txy =
480
pd3
In-Plane Principal Stresses: Applying Eq. 9-5,
s1,2 =
sx + sy
2
1792
3
pd
=
=
s1 =
;
A
+ 0
2
;
D
a
sx - sy
2
1792
3
pd
¢
- 0
2
2
b + txy
2
2
≤ + a
480 2
b
pd3
896
1016.47
;
pd3
pd3
1912.47
pd3
s2 = -
120.47
pd3
Maximum Distortion Energy Theory:
s21 - s1s2 + s22 = s2allow
a
1912.47
120.47
120.47 2
150 2
1912.47 2
b - a
bab + ab = a
b
3
3
3
3
2
pd
pd
pd
pd
Ans.
d = 2.03 in.
1050
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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.
10–85. The state of stress acting at a critical point on a
machine element is shown in the figure. Determine the
smallest yield stress for a steel that might be selected for the
part, based on the maximum-shear-stress theory.
10 ksi
The Principal Stresses:
4 ksi
s1,2 =
=
sx + sy
2
;
A
a
sx - sy
2
b +
2
t2xy
8 ksi
8 - ( -10) 2
8 - 10
2
;
a
b + 4
2
A
2
s1 = 8.8489 ksi
s2 = - 10.8489 ksi
Maximum Shear Stress Theory: Both principal stresses have opposite sign. hence,
|s1 - s2| = sY
8.8489 - ( - 10.8489) = sY
sY = 19.7 ksi
Ans.
Ans:
sY = 19.7 ksi
1051
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. 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.
10–86. The principal stresses acting at a point on a thinwalled cylindrical pressure vessel are s1 = pr>t, s2 = pr>2t,
and s3 = 0. If the yield stress is sY, determine the maximum
value of p based on (a) the maximum-shear-stress theory and
(b) the maximum-distortion-energy theory.
a) Maximum Shear Stress Theory: s1 and s2 have the same signs, then
|s2| = sY
2
pr
2 = sY
2t
p =
2t
s
r Y
|s1| = sY
2
pr
2 = sY
t
p =
t
s (Controls!)
r Y
Ans.
b) Maximum Distortion Energy Theory:
s21 - s1s2 + s22 = s2Y
a
pr 2
pr 2
pr pr
b - a b a b + a b = s2Y
t
t
2t
2t
p =
2t
23r
sY
Ans.
Ans:
(a) p =
(b) p =
1052
t
s ,
r y
2t
23r
sy
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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.
10–87. If a solid shaft having a diameter d is subjected to a
torque T and moment M, show that by the maximum-shearstress theory the maximum allowable shear stress is
tallow = (16>pd3) 2M2 + T2. Assume the principal stresses
to be of opposite algebraic signs.
T
M
Section properties:
I =
p d 4
pd4
a b =
;
4 2
64
J =
p d 4
pd4
a b =
2 2
32
Thus,
s =
M(d2 )
Mc
32 M
=
=
p d4
I
pd3
64
t =
Tc
=
J
T (d2 )
p d4
32
=
16 T
pd3
The principal stresses:
s1,2 =
=
sx + sy
2
;
A
a
sx - sy
2
2
b + txy
2
16 M
16 M 2
16 T 2
16 M
16
;
;
2M2 + T2
a
b + a
b =
3
3
A pd
pd
p d3
pd3
p d3
Assume s1 and s2 have opposite sign, hence,
tallow =
2 C 163 2M2 + T2 D
s1 - s2
16
pd
=
=
2M2 + T2
2
2
pd3
(Q.E.D.)
1053
T
M
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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.
*10–88. If a solid shaft having a diameter d is subjected to a
torque T and moment M, show that by the maximum-normalstress theory the maximum allowable principal stress is
sallow = 116>pd321M + 2M2 + T22.
T
M
Section properties:
I =
p d4
;
64
p d4
32
J =
Stress components:
s =
M (d2 )
Mc
32 M
= p 4 =
;
I
d
p d3
64
t =
T(d2 )
Tc
16 T
= p 4 =
J
d
p d3
32
The principal stresses:
s1,2 =
=
sx + sy
2
;
A
a
sx - sy
2
2
b + txy =
2
32 M
3
pd
32 M
+ 0
2
3
;
D
¢ pd
- 0
2
2
≤ + a
16 T 2
b
p d3
16 M
16
;
2M2 + T2
p d3
p d3
Maximum normal stress theory. Assume s1 7 s2
sallow = s1 =
=
16 M
16
+
2M2 + T2
p d3
p d3
16
[M + 2M2 + T2]
p d3
(Q.E.D.)
1054
T
M
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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.
10–89. The gas tank has an inner diameter of 1.50 m and a
wall thickness of 25 mm. If it is made from A-36 steel and the
tank is pressured to 5 MPa, determine the factor of safety
against yielding using (a) the maximum-shear-stress theory,
and (b) the maximum-distortion-energy theory.
(a) Normal Stress. Since
0.75
r
=
= 30 7 10, thin-wall analysis can be used.We have
t
0.025
s1 = sh =
5(0.75)
pr
=
= 150 MPa
t
0.025
s2 = slong =
pr
5(0.75)
=
= 75 MPa
2t
2(0.025)
Maximum Shear Stress Theory. s1 and s2 have the sign. Thus,
|s1| = sallow
sallow = 150 MPa
The factor of safety is
F.S. =
sY
250
= 1.67
=
sallow
150
Ans.
(b) Maximum Distortion Energy Theory.
s1 2 - s1s2 + s2 2 = sallow 2
1502 - 150(75) + 752 = sallow 2
sallow = 129.90 MPa
The factor of safety is
F.S. =
sY
250
=
= 1.92
sallow
129.90
Ans.
Ans:
(a) F.S. = 1.67,
(b) F.S. = 1.92
1055
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. 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.
10–90. The gas tank is made from A-36 steel and
has an inner diameter of 1.50 m. If the tank is designed
to withstand a pressure of 5 MPa, determine the required
minimum wall thickness to the nearest millimeter using
(a) the maximum-shear-stress theory, and (b) maximumdistortion-energy theory. Apply a factor of safety of
1.5 against yielding.
(a) Normal Stress. Assuming that thin-wall analysis is valid, we have
s1 = sh =
5 A 106 B (0.75)
3.75 A 106 B
pr
=
=
t
t
t
s2 = slong =
5 A 106 B (0.75)
1.875 A 106 B
pr
=
=
2t
2t
t
Maximum Shear Stress Theory.
sallow =
250 A 106 B
sY
=
= 166.67 A 106 B Pa
FS.
1.5
s1 and s2 have the same sign. Thus,
|s1| = sallow
3.75 A 106 B
= 166.67 A 106 B
t
t = 0.0225 m = 22.5 mm
Since
Ans.
0.75
r
=
= 33.3 7 10, thin-wall analysis is valid.
t
0.0225
(b) Maximum Distortion Energy Theory.
sallow =
250 A 106 B
sY
=
= 166.67 A 106 B Pa
F.S.
1.5
Thus,
s1 2 - s1s2 + s2 2 = sallow 2
C
3.75 A 106 B
t
3.2476 A 106 B
t
2
S - C
3.75 A 106 B
t
SC
1.875 A 106 B
t
S + C
1.875 A 106 B
t
2
S = c166.67 A 106 B d
= 166.67 A 106 B
t = 0.01949 m = 19.5 mm
Since
2
Ans.
0.75
r
=
= 38.5 7 10, thin-wall analysis is valid.
t
0.01949
Ans:
(a) t = 22.5 mm,
(b) t = 19.5 mm
1056
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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.
10–91. The internal loadings at a critical section along the
steel drive shaft of a ship are calculated to be a torque of
2300 lb # ft, a bending moment of 1500 lb # ft, and an axial
thrust of 2500 lb. If the yield points for tension and shear
sY = 100 ksi and tY = 50 ksi, respectively, determine the
required diameter of the shaft using the maximum-shearstress theory.
2300 lb⭈ft
A = p c2
sA =
I =
p 4
c
4
J =
p 4
c
2
1500 lb⭈ft
2500 lb
1500(12)(c)
P
Mc
2500
2500
72 000
+
b = -a
+
b
+
= -a
2
pc4
A
I
pc
p c2
p c3
4
tA =
s1,2 =
Tc
=
J
2300(12)(c)
=
p c4
2
sx + sy
= -a
2
;
A
a
55 200
p c3
sx - sy
2
2
b + txy
2
2500 c + 72 000
2500c + 72 000 2
55200 2
b ;
a
b + a
b
3
3
A
2p c
2p c
p c3
(1)
Assume s1 and s2 have opposite signs:
|s1 - s2| = sY
2500c + 72 000 2
55 200 2
3
b + a
b = 100(10 )
3
A
2p c
p c3
2
a
(2500c + 72000)2 + 1104002 = 10 000(106)p2 c6
6.25c2 + 360c + 17372.16 - 10 000p2 c6 = 0
By trial and error:
c = 0.750 57 in.
Substitute c into Eq. (1):
s1 = 22 193 psi
s2 = - 77 807 psi
s1 and s2 are of opposite signs
OK
Therefore,
d = 1.50 in.
Ans.
Ans:
d = 1.50 in.
1057
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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.
*10–92. The shaft consists of a solid segment AB and a
hollow segment BC, which are rigidly joined by the
coupling at B. If the shaft is made from A-36 steel,
determine the maximum torque T that can be applied
according to the maximum-shear-stress theory. Use a factor
of safety of 1.5 against yielding.
A
B
T
C
Shear Stress: This is a case of pure shear, and the shear stress is contributed by
p
torsion. For the hollow segment, Jh =
A 0.054 - 0.044 B = 1.845p A 10 - 6 B m4. Thus,
2
(tmax)h =
T(0.05)
Tch
=
= 8626.28T
Jh
1.845p A 10 - 6 B
For the solid segment, Js =
(tmax)s =
p
A 0.044 B = 1.28p A 10 - 6 B m4. Thus,
2
T(0.04)
Tcs
=
= 9947.18T
Js
1.28p A 10 - 6 B
By comparision, the points on the surface of the solid segment are critical and their
state of stress is represented on the element shown in Fig. a.
In-Plane Principal Stress. sx = sy = 0 and txy = 9947.18T. We have
s1,2 =
=
sx + sy
2
;
C
¢
sx - sy
2
2
≤ + t2xy
0 - 0 2
0 + 0
;
¢
≤ + (9947.18T)2
2
C
2
s1 = 9947.18T
s2 = - 9947.18T
Maximum Shear Stress Theory.
sallow =
80 mm
sY
250
=
= 166.67 MPa
F.S.
1.5
Since s1 and s2 have opposite signs,
|s1 - s2| = sallow
9947.18T - ( -9947.18T) = 166.67 A 106 B
T = 8377.58 N # m = 8.38 kN # m
Ans.
1058
80 mm
100 mm
T
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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.
10–93. The shaft consists of a solid segment AB and a
hollow segment BC, which are rigidly joined by the
coupling at B. If the shaft is made from A-36 steel,
determine the maximum torque T that can be applied
according to the maximum-distortion-energy theory. Use a
factor of safety of 1.5 against yielding.
A
B
T
C
Shear Stress. This is a case of pure shear, and the shear stress is contributed by
p
torsion. For the hollow segment, Jh =
A 0.054 - 0.044 B = 1.845p A 10 - 6 B m4. Thus,
2
(tmax)h =
80 mm
100 mm
T
T(0.05)
Tch
=
= 8626.28T
Jh
1.845p A 10 - 6 B
For the solid segment, Js =
(tmax)s =
80 mm
p
A 0.044 B = 1.28p A 10 - 6 B m4. Thus,
2
T(0.04)
Tcs
=
= 9947.18T
Js
1.28p A 10 - 6 B
By comparision, the points on the surface of the solid segment are critical and their
state of stress is represented on the element shown in Fig. a.
In-Plane Principal Stress. sx = sy = 0 and txy = 9947.18T. We have
s1,2 =
=
sx + sy
2
;
C
¢
sx - sy
2
2
≤ + t2xy
0 - 0 2
0 + 0
;
¢
≤ + (9947.18T)2
2
C
2
s1 = 9947.18T
s2 = - 9947.18T
Maximum Distortion Energy Theory.
sallow =
sY
250
=
= 166.67 MPa
F.S.
1.5
Then,
s1 2 - s1s2 + s2 2 = sallow 2
(9947.18T)2 - (9947.18T)( - 9947.18T) + ( -9947.18T)2 = C 166.67 A 106 B D 2
T = 9673.60 N # m = 9.67 kN # m
Ans.
Ans:
T = 9.67 kN # m
1059
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. 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.
10–94. In the case of plane stress, where the in-plane
principal strains are given by P1 and P2, show that the
third principal strain can be obtained from
P3 = - [n>(1 - n)](P1 + P2), where n is Poisson’s ratio for
the material.
Generalized Hooke’s Law: In the case of plane stress, s3 = 0. Thus,
P1 =
1
(s - ns2)
E 1
(1)
P2 =
1
(s - ns1)
E 2
(2)
n
(s + s2)
E 1
(3)
P3 = -
Solving for s1 and s2 using Eqs. (1) and (2), we obtain
s1 =
E(P1 + nP2)
2
1-n
s2 =
E(P2 + nP1)
1 - n2
Substituting these results into Eq. (3),
P3 = -
E(P2 + nP1)
n E(P1 + nP2)
c
+
d
E
1 - n2
1 - n2
P3 = -
(P1 +P2) + n(P1 +P2)
n
c
d
1 - n
1+n
P3 = -
(P1 +P2)(1 + n)
n
c
d
1 - n
1+n
P3 = -
n
(P +P )
1 - n 1 2
(Q.E.D.)
Ans.
1060
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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.
10–95. The plate is made of material having a modulus of
elasticity E = 200 GPa and Poisson’s ratio n = 13 . Determine
the change in width a, height b, and thickness t when it is
subjected to the uniform distributed loading shown.
a ⫽ 400 mm
2 MN/m
t ⫽ 20 mm
3 MN/m
y
b ⫽ 300 mm
Normal Stress: The normal stresses along the x, y, and z axes are
z
3(106)
sx =
= 150 MPa
0.02
sy = -
x
2(106)
= - 100 MPa
0.02
sz = 0
Generalized Hooke’s Law:
Px =
=
1
c s - n(sy + sz) d
E x
1
1
e 150(106) - c - 100(106) + 0 d f
9
3
200(10 )
= 0.9167(10 - 3)
Py =
=
1
c s - n(sx + sz) d
E y
1
1
e -100(106) - c 150(106) + 0 d f
3
200(109)
= - 0.75(10 - 3)
Pz =
=
1
c s - n(sx + sy) d
E z
1
1
e 0 - c 150(106) + (- 100)(106) d f
3
200(109)
= - 83.33(10 - 6)
Thus, the changes in dimensions of the plate are
da = Pxa = 0.9167(10 - 3)(400) = 0.367 mm
Ans.
db = Pyb = - 0.75(10 - 3)(300) = - 0.225 mm
Ans.
dt = Pzt = - 83.33(10 - 6)(20) = - 0.00167 mm
Ans.
The negative signs indicate that b and t contract.
Ans:
da = 0.367 mm, db = - 0.255 mm,
dt = - 0.00167 mm
1061
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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.
*10–96. The principal plane stresses acting at a point are
shown in the figure. If the material is machine steel having a
yield stress of sY = 500 MPa, determine the factor of
safety with respect to yielding if the maximum-shear-stress
theory is considered.
100 MPa
150 MPa
Here, the in plane principal stresses are
s1 = sy = 100 MPa
s2 = sx = - 150 MPa
Since s1 and s2 have the same sign,
F.S =
sy
|s1 - s2|
=
500
= 2
|100 - ( - 150)|
Ans.
1062
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10–97. The components of plane stress at a critical point
on a thin steel shell are shown. Determine if failure
(yielding) has occurred on the basis of the maximumdistortion-energy theory. The yield stress for the steel is
sY = 650 MPa.
340 MPa
65 MPa
55 MPa
sx = - 55 MPa
s1, 2 =
=
sy = 340 MPa
sx + sy
2
;
A
a
sx - sy
2
txy = 65 MPa
2
b + txy
2
- 55 - 340 2
- 55 + 340
2
;
a
b + 65
2
A
2
s1 = 350.42 MPa
s2 = - 65.42 MPa
(s1 2 - s1s2 + s2 ) = [350.422 - 350.42(- 65.42) + ( -65.42)2]
= 150 000 6 s2Y = 422 500
OK
No.
Ans.
Ans:
No.
1063
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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.
10–98. The 60° strain rosette is mounted on a beam. The
following readings are obtained for each gauge:
Pa = 600110-62, Pb = - 700110-62, and Pc = 350110-62.
Determine (a) the in-plane principal strains and (b) the
maximum in-plane shear strain and average normal strain.
In each case show the deformed element due to these
strains.
a
60⬚
60⬚
b
60⬚
Strain Rosettes (60º): Applying Eq. 10-15 with Px = 600 A 10 - 6 B ,
Pb = - 700 A 10 - 6 B , Pc = 350 A 10 - 6 B , ua = 150°, ub = - 150° and uc = - 90°,
c
350 A 10 - 6 B = Px cos2 (- 90°) + Py sin2 ( -90°) + gxy sin (- 90°) cos (-90°)
Py = 350 A 10 - 6 B
600 A 10 - 6 B = Px cos2 150° + 350 A 10 - 6 B sin2 150° + gxy sin 150° cos 150°
512.5 A 10 - 6 B = 0.75 Px - 0.4330 gxy
[1]
-787.5 A 10 - 6 B = 0.75Px + 0.4330 gxy
[2]
- 700 A 10 - 6 B = Px cos2 ( -150°) + 350 A 10 - 6 B sin2( -150°) + gxy sin ( -150°) cos (- 150°)
Solving Eq. [1] and [2] yields Px = - 183.33 A 10 - 6 B
gxy = - 1501.11 A 10 - 6 B
Construction of the Circle: With Px = - 183.33 A 10 - 6 B , Py = 350 A 10 - 6 B , and
gxy
= - 750.56 A 10 - 6 B .
2
Px + Py
Pavg =
2
= a
- 183.33 + 350
b A 10 - 6 B = 83.3 A 10 - 6 B
2
Ans.
The coordinates for reference points A and C are
A(- 183.33, -750.56) A 10 - 6 B
C(83.33, 0) A 10 - 6 B
The radius of the circle is
R = a 2(183.33 + 83.33)2 + 750.562 b A 10 - 6 B = 796.52 A 10 - 6 B
a)
In-Plane Principal Strain: The coordinates of points B and D represent P1 and P2,
respectively.
P1 = (83.33 + 796.52) A 10 - 6 B = 880 A 10 - 6 B
Ans.
P2 = (83.33 - 796.52) A 10 - 6 B = - 713 A 10 - 6 B
Ans.
Orientation of Principal Strain: From the circle,
tan 2uP1 =
750.56
= 2.8145
183.33 + 83.33
2uP2 = 70.44°
2uP1 = 180° - 2uP2
uP =
180° - 70.44°
= 54.8° (Clockwise)
2
Ans.
1064
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10–98. Continued
b)
Maximum In-Plane Shear Strain: Represented by the coordinates of point E on the
circle.
g max
in-plane
2
g max
in-plane
= - R = - 796.52 A 10 - 6 B
= - 1593 A 10 - 6 B
Ans.
Orientation of Maximum In-Plane Shear Strain: From the circle.
tan 2us =
183.33 + 83.33
= 0.3553
750.56
us = 9.78° (Clockwise)
Ans.
Ans:
Pavg = 83.3(10 - 6), P1 = 880(10 - 6),
P2 = - 713(10 - 6), up = 54.8° (clockwise),
gmax
= - 1593(10 - 6),
in-plane
us = 9.78° (clockwise)
1065
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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
10–99. The state of strain at the point on the bracket has components Px = 350(10 - 6), Py = - 860(10 - 6), gxy = 250(10 - 6).
Use the strain-transformation equations to determine the
equivalent in-plane strains on an element oriented at an
angle of u = 45° clockwise from the original position. Sketch
the deformed element within the x–y plane due to these
strains.
x
Px = 350(10 - 6)
Px + Py
Px ¿ =
= c
Px - Py
+
2
Px - Py
Px + Py
-
2
gxy
2
u = - 45°
sin 2u
2
cos 2u -
gxy
2
sin 2u
350 - ( -860)
350 - 860
250
cos ( -90°) sin ( -90°) d (10 - 6) = - 130(10 - 6)
2
2
2
gx¿y¿
2
2
cos 2u +
gxy = 250(10 - 6)
350 - (- 860)
350 - 860
250
+
cos ( -90°) +
sin ( -90°) d (10 - 6) = - 380(10 - 6) Ans.
2
2
2
Py¿ =
= c
Py = - 860(10 - 6)
Px - Py
= -
gx¿y¿ = 2 c- a
2
sin 2u +
Ans.
g
cos 2u
2
350 - (- 860)
250
b sin (- 90°) +
cos (- 90°)d (10 - 6) = 1.21(10 - 3) Ans.
2
2
Ans:
Px¿ = - 380(10 - 6), Py¿ = - 130(10 - 6),
gx¿y¿ = 1.21(10 - 3)
1066
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*10–100. The A-36 steel post is subjected to the forces
shown. If the strain gauges a and b at point A give readings
of Pa = 300(10 - 6) and Pb = 175(10 - 6), determine the magnitudes of P1 and P2.
P1
P2
a
+ ©F = 0;
:
x
P2 - V = 0
V = P2
+ c ©Fy = 0;
N - P1 = 0
N = P1
a + ©MO = 0;
M + P2(2) = 0
M = 2P2
Section Properties: The cross-sectional area and the moment of inertia about the
bending axis of the post’s cross-section are
A = 4(2) = 8 in2
I =
1
(2) A 43 B = 10.667 in4
12
Referring to Fig. b,
A Qy B A = x¿A¿ = 1.5(1)(2) = 3 in3
Normal and Shear Stress: The normal stress is a combination of axial and bending
stress.
sA =
2P2(12)(1)
MxA
P1
N
+
= +
= 2.25P2 - 0.125P1
A
I
8
10.667
The shear stress is caused by transverse shear stress.
tA =
P2(3)
VQA
=
= 0.140625P2
It
10.667(2)
Thus, the state of stress at point A is represented on the element shown in Fig. c.
Normal and Shear Strain: With ua = 90° and ub = 45°, we have
Pa = Px cos2ua + Py sin2ua + gxy sin ua cos ua
300 A 10 - 6 B = Px cos2 90° + Py sin2 90° + gxy sin 90° cos 90°
Py = 300 A 10 - 6 B
Pb = Px cos2ub + Py sin2 ub + gxy sin ub cos ub
175 A 10 - 6 B = Px cos2 45° + 300 A 10 - 6 B sin2 45° + gxy sin 45°cos 45°
Px + gxy = 50 A 10 - 6 B
(1)
1067
A
1 in.
b 45⬚
A
c
Internal Loadings: Considering the equilibrium of the free-body diagram of the
post’s segment, Fig. a,
2 in.
2 ft
A
1 in.
4 in.
c
Section c–c
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10–100. Continued
Since sy = sz = 0, Px = - vPy = - 0.32(300) A 10 - 6 B = - 96 A 10 - 6 B
Then Eq. (1) gives
gxy = 146 A 10 - 6 B
Stress and Strain Relation: Hooke’s Law for shear gives
tx = Ggxy
0.140625P2 = 11.0 A 103 B C 146 A 10 - 6 B D
P2 = 11.42 kip = 11.4 kip
Ans.
Since sy = sz = 0, Hooke’s Law gives
sy = EPy
2.25(11.42) - 0.125P1 = 29.0 A 103 B C 300 A 10 - 6 B D
P1 = 136 kip
Ans.
1068
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10–101. A differential element is subjected to plane strain
that has the following components; Px = 950110-62,
Py = 420110-62, gxy = - 325110-62. Use the straintransformation equations and determine (a) the principal
strains and (b) the maximum in plane shear strain and the
associated average strain. In each case specify the
orientation of the element and show how the strains deform
the element.
Aa
Px + Py
P1, 2 =
;
2
= c
Px - Py
2
2
b + gxy
2
950 - 420 2
-325 2
950 + 420
-6
;
a
b + a
b d(10 )
2
A
2
2
P1 = 996(10 - 6)
Ans.
P2 = 374(10 - 6)
Ans.
Orientation of P1 and P2 :
gxy
tan 2uP =
-325
950 - 420
=
Px - Py
uP = - 15.76°, 74.24°
Use Eq. 10.5 to determine the direction of P1 and P2.
Px + Py
Px¿ =
Px - Py
+
2
2
cos 2u +
gxy
2
sin 2u
u = uP = - 15.76°
Px¿ = b
( - 325)
950 - 420
950 + 420
+
cos (- 31.52°) +
sin ( - 31.52°) r (10 - 6) = 996(10 - 6)
2
2
2
uP1 = - 15.8°
Ans.
uP2 = 74.2°
Ans.
b)
gmax
in-plane
2
gmax
in-plane
=
Aa
= 2c
Px - Py
2
2
2
gxy
2
b
2
950 - 420 2
-325 2
-6
-6
b + a
b d (10 ) = 622(10 )
A
2
2
a
Px + Py
Pavg =
b + a
= a
Ans.
950 + 420
b (10 - 6) = 685(10 - 6)
2
Ans.
1069
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10–101. Continued
Orientation of gmax:
tan 2uP =
- (Px - Py)
=
gxy
-(950 - 420)
- 325
uP = 29.2° and uP = 119°
Ans.
Use Eq. 10.6 to determine the sign of
gx¿y¿
2
Px - Py
= -
2
sin 2u +
gxy
2
gmax
in-plane
:
cos 2u
u = us = 29.2°
gx¿y¿ = 2 c
- (950 - 420)
-325
sin (58.4°) +
cos (58.4°) d(10 - 6)
2
2
gxy = - 622(10 - 6)
Ans:
P1 = 996(10 - 6), P2 = 374(10 - 6), up1 = - 15.8,
up2 = 74.2, gmax
= 622(10 - 6),
in-plane
Pavg = 685(10 - 6), us = 29.2° and 119°
1070
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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.
10–102. The state of strain at the point on the bracket
Px = - 130110-62,
Py = 280110-62,
has
components
gxy = 75110-62. 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
Px = - 130(10 - 6)
Py = 280(10 - 6)
gxy = 75(10 - 6)
a)
Px + Py
P1, 2 =
2
= c
Aa
;
Px - Py
2
2
b + a
gxy
2
b
2
- 130 + 280
- 130 - 280 2
75 2
-6
;
a
b + a b d(10 )
2
A
2
2
P1 = 283(10 - 6)
Ans.
P2 = - 133(10 - 6)
Ans.
Orientation of P1 and P2:
tan 2up =
gxy
=
Px - Py
up = -5.18°
75
- 130 - 280
and
84.82°
Use Eq. 10–5 to determine the direction of P1 and P2:
Px + Py
Px¿ =
2
Px - Py
+
2
cos 2u +
gxy
2
sin 2u
u = up = -5.18°
Px¿ = c
- 130 + 280
- 130 - 280
75
+
cos ( -10.37°) +
sin (- 10.37°) d(10 - 6) = - 133(10 - 6)
2
2
2
Therefore up1 = 84.8°
Ans.
up2 = - 5.18°
Ans.
1071
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10–102. Continued
b)
gmax
in-plane
=
2
gmax
in-plane
A
= 2c
a
2
2
b + a
gxy
2
b
2
- 130 - 280 2
75 2
b + a b d (10 - 6) = 417(10 - 6)
A
2
2
a
Px + Py
Pavg =
Px - Py
2
= a
Ans.
- 130 + 280
b (10 - 6) = 75.0(10 - 6)
2
Ans.
Orientation of gmax:
tan 2us =
- (Px - Py)
gxy
=
- (- 130 - 280)
75
us = 39.8° and us = 130°
Ans.
Use Eq. 10–16 to determine the sign of
gx¿y¿
2
Px - Py
= -
2
sin 2u +
gxy
2
cos 2u;
gmax
in-plane
:
u = us = 39.8°
gx¿y¿ = -(- 130 - 280) sin (79.6°) + (75) cos (79.6°) = 417(10 - 6)
Ans:
P1 = 283(10 - 6), P2 = -133(10 - 6),
up1 = 84.8°, up2 = -5.18°, gmax
= 417(10 - 6),
in-plane
Pavg = 75.0(10 - 6), us = 39.8° and 130°
1072
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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
10–103. The state of plain strain on an element is
Px = 400110-62, Py = 200110-62, and gxy = - 300110-62.
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 element at the
point with respect to the original element. Sketch the results
on the element.
Pydy
dy
Construction of the Circle: Px = 400 A 10 - 6 B , Py = 200 A 10 - 6 B , and
gxy
2
= - 150 A 10 - 6 B .
Thus,
Px + Py
Pavg =
2
= a
400 + 200
b A 10 - 6 B = 300 A 10 - 6 B
2
Ans.
The coordinates for reference points A and the center C of the circle are
A(400, -150) A 10 - 6 B
C(300, 0) A 10 - 6 B
The radius of the circle is
R = CA = 2(400 - 300)2 + (- 150)2 = 180.28 A 10 - 6 B
Using these results, the circle is shown in Fig. a.
In-Plane Principal Stresses: The coordinates of points B and D represent P1 and P2,
respectively. Thus,
P1 = (300 + 180.28) A 10 - 6 B = 480 A 10 - 6 B
Ans.
P2 = (300 - 180.28) A 10 - 6 B = 120 A 10 - 6 B
Ans.
Orientation of Principal Plane: Referring to the geometry of the circle,
tan 2 A up B 1 =
150
= 1.5
400 - 300
A up B 1 = 28.2° (clockwise)
Ans.
The deformed element for the state of principal strains is shown in Fig. b.
Maximum In-Plane Shear Stress: The coordinates of point E represent Pavg and
gmax
. Thus
in-plane
gmax
= - R = - 180.28 A 10 - 6 B
in-plane
2
gmax
in-plane
= - 361 A 10 - 6 B
Ans.
Orientation of the Plane of Maximum In-plane Shear Strain: Referring to the
geometry of the circle,
tan 2us =
400 - 300
= 0.6667
150
uS = 16.8° (counterclockwise)
Ans.
The deformed element for the state of maximum In-plane shear strain is shown in
Fig. c.
1073
gxy
2
gxy
2
dx
x
Pxdx
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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.
10–103. Continued
Ans:
P1 = 480(10 - 6), P2 = 120(10 - 6),
up1 = - 28.2° (clockwise),
gmax
= - 361(10 - 6),
in-plane
us = 16.8° (counterclockwise),
Pavg = 300(10 - 6)
1074
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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.
8 kN/m
11–1. The simply supported beam is made of timber that
has an allowable bending stress of sallow = 6.5 MPa and an
allowable shear stress of tallow = 500 kPa. Determine its
dimensions if it is to be rectangular and have a height-towidth ratio of 1.25.
2m
Ix =
4m
2m
1
(b)(1.25b)3 = 0.16276b4
12
Qmax = y¿A¿ = (0.3125b)(0.625b)(b) = 0.1953125b3
Assume bending moment controls:
Mmax = 16 kN # m
sallow =
Mmax c
I
6.5(106) =
16(103)(0.625b)
0.16276b4
b = 0.21143 m = 211 mm
Ans.
h = 1.25b = 264 mm
Ans.
Check shear:
Qmax = 1.846159(10 - 3) m3
I = 0.325248(10 - 3) m4
tmax =
16(103)(1.846159)(10 - 3)
VQmax
= 429 kPa 6 500 kPa‚ OK
=
It
0.325248(10 - 3)(0.21143)
Ans:
b = 211 mm, h = 264 mm
1075
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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.
11–2. Determine the minimum width of the beam to the
nearest 14 in. that will safely support the loading of
P = 8 kip. The allowable bending stress is sallow = 24 ksi
and the allowable shear stress is tallow = 15 ksi.
P
6 ft
6 ft
6 in.
B
A
Beam design: Assume moment controls.
sallow =
Mc
;
I
24 =
48.0(12)(3)
1
3
12 (b)(6 )
b = 4 in.
Ans.
Check shear:
tmax =
8(1.5)(3)(4)
VQ
= 0.5 ksi 6 15 ksi OK
= 1
3
It
12 (4)(6) (4)
Ans:
Use b = 4 in.
1076
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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.
11–3.
Solve Prob. 11–2 if P = 10 kip.
P
6 ft
6 ft
6 in.
B
A
Beam design: Assume moment controls.
sallow =
Mc
;
I
24 =
60(12)(3)
1
3
12 (b)(6 )
b = 5 in.
Ans.
Check shear:
tmax =
VQ
10(1.5)(3)(5)
= 0.5 ksi 6 15 ksi OK
= 1
3
It
12 (5)(6) (5)
Ans:
Use b = 5 in.
1077
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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.
*11–4. The brick wall exerts a uniform distributed load
of 1.20 kip>ft on the beam. If the allowable bending
stress is sallow = 22 ksi and the allowable shear stress is
tallow = 12 ksi, select the lightest wide-flange section with
the shortest depth from Appendix B that will safely support
the load. If there are several choices of equal weight, choose
the one with the shortest height.
1.20 kip/ft
4 ft
Bending Stress: From the moment diagram, Mmax = 44.55 kip # ft. Assuming
bending controls the design and applying the flexure formula.
Sreq d =
=
Mmax
sallow
44.55 (12)
= 24.3 in3
22
Two choices of wide flange section having the weight 22 lb>ft can be made. They are
W12 * 22 and W14 * 22. However, W12 * 22 is the shortest.
Select
W12 * 22
A Sx = 25.4 in3, d = 12.31 in., tw = 0.260 in. B
V
for the W12 * 22 wide tw d
= 6.60 kip.
Shear Stress: Provide a shear stress check using t =
flange section. From the shear diagram, Vmax
tmax =
=
Vmax
tw d
6.60
0.260(12.31)
= 2.06 ksi 6 tallow = 12 ksi (O.K!)
Hence,
Use
Ans.
W12 * 22
1078
10 ft
6 ft
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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.
11–5. Select the lightest-weight steel wide-flange beam
from Appendix B that will safely support the machine loading
shown. The allowable bending stress is sallow = 24 ksi and
the allowable shear stress is tallow = 14 ksi.
5 kip
2 ft
Bending Stress: From the moment diagram, Mmax = 30.0 kip # ft.
Assume bending controls the design. Applying the flexure formula.
Sreq¿d =
=
Select
W12 * 16
5 kip
2 ft
5 kip
2 ft
5 kip
2 ft
2 ft
Mmax
sallow
30.0(12)
= 15.0 in3
24
A Sx = 17.1 in3, d = 11.99 in., tw = 0.220 in. B
V
for the W12 * 16 wide tw d
= 10.0 kip
Shear Stress: Provide a shear stress check using t =
flange section. From the shear diagram, Vmax
tmax =
=
Vmax
tw d
10.0
0.220(11.99)
= 3.79 ksi 6 tallow = 14 ksi (O.K!)
Hence,
Use
Ans.
W12 * 16
Ans:
Use W12 * 16
1079
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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.
11–6. The spreader beam AB is used to slowly lift the
3000-lb pipe that is centrally located on the straps at C and D.
If the beam is a W12 * 45, determine if it can safely support
the load. The allowable bending stress is sallow = 22 ksi and
the allowable shear stress is tallow = 12 ksi.
3000 lb
5 ft
A
12 in.
B
3 ft
6 ft
C
h =
1500
= 2700 lb
tan 29.055°
s =
M
;
S
s =
5850(12)
= 1.21 ksi 6 22 ksi
58.1
t =
V
;
A web
t =
1500
= 371 psi 6 12 ksi OK
(12.06)(0.335)
6 ft
3 ft
D
OK
Yes.
Ans.
Ans:
Yes.
1080