Principles of Foundation Engineering (MindTap Course List)
8th Edition
ISBN: 9781305081550
Author: Braja M. Das
Publisher: Cengage Learning
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The copper pipe shown in the figure has an outside diameter of 4.000 in. and a wall thickness of 0.363 in. The pipe is subjected to a
uniformly distributed torque of t = 80 lb-ft/ft along its entire length. Using a = 2.00 ft, b = 4.00 ft, and c = 7.5 ft, calculate
(a) the shear stress at A on the outer surface of the pipe.
(b) the shear stress at B on the outer surface of the pipe.
a
b
Answers:
(а) ТА
i
psi
(b) Тв-
i
psi
The copper pipe shown in the figure has an outside diameter of 3.000 in. and a wall thickness of 0.383 in. The pipe is subjected to a
uniformly distributed torque of t = 87 lb-ft/ft along its entire length. Using a = 2.00 ft, b = 4.50 ft, and c = 7.75 ft, calculate
(a) the shear stress at A on the outer surface of the pipe.
(b) the shear stress at B on the outer surface of the pipe.
Answers:
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(a) TA=
a
(b) TB =
i
b
i
psi
psi
The copper pipe shown in the figure has an outside diameter of 3.625 in. and a wall thickness of 0.208 in. The pipe is subjected to a
uniformly distributed torque of t = 82 lb-ft/ft along its entire length. Using a = 2.50 ft, b = 6.00 ft, and c = 13 ft, calculate
(a) the shear stress at A on the outer surface of the pipe.
(b) the shear stress at B on the outer surface of the pipe.
a
b
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Principles of Foundation Engineering (MindTap Course List)
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- The copper pipe shown in the figure has an outside diameter of 3.625 in. and a wall thickness of 0.373 in. The pipe is subjected to a uniformly distributed torque oft= 63lb-ft/ft along its entire length. Using a = 2.25 ft, b = 4.00 ft, and c= 8 ft, calculate (a) the shear stress at A on the outer surface of the pipe. (b) the shear stress at B on the outer surface of the pipe.arrow_forwardRequired information The tube shown has a uniform wall thickness of 12 mm. Take P = 19 kN. D HP E 28 KN 28 KN Determine the stress at points A and B. 16.11 MPa. The stress at point A is The stress at point B is MPa. 125 mm 75 mmarrow_forwardConsider a crack arbitrarily oriented with respect to a stress field, as shown below: буу I. 0 = 0 oyx X 0₂ 0 xy 0 xx The stress state in the structure's coordinate system is d c [%] b mode II stress intensity factor in MPa/m? The crack length is 2a, where a = 8 mm, and its orientation is 0 = 2.52 rad. MPa, where d = 21, b = 12, and c = 3. What is the Note: Stress intensity factor is positive and thus you have to take the absolute value of relevant shear stress.arrow_forward
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