Concept explainers
Two circular aluminum pipes of equal length L = 24 in. arc loaded by torsional moments T (sec figure). Pipe I has outside and inside diameters d2= 3 in. and L2, = 2.5 in., respectively. Pipe 2 has a constant outer diameter of d2along its entire length L and an inner diameter of d1but has an increased inner diameter of d3= 2.65 in. over the middle third.
Assume that E = 10,400 ksi, u = 0.33, and allowable shear stress ra= 6500 psi.
- Find the maximum acceptable torques that can be applied to Pipe 1; repeat for Pipe 2.
(a)
The maximum acceptable torques for pipe (1).
The maximum acceptable torques for pipe (2).
Answer to Problem 3.5.13P
Maximum acceptable torques for pipe (1) is =
Maximum acceptable torques for pipe (2) is =
Explanation of Solution
Given information:
The following figure shows the free body diagram of pipe 1:
Figure-(1) shows the diagram of two pipes:
Figure-(1)
The following figure shows the free body diagram of pipe 2:
Figure-(2)
The length of pipe is
Write the expression for polar moment of inertia for pipe (1)
Here, the polar moment of inertia is
Write the expression for maximum torque for pipe (1)
Here, acceptable shear stress is
Write the expression for polar moment of inertia for pipe (2)
Here, the polar moment of inertia is
Write the expression for maximum torque for pipe (2)
Here, acceptable shear stress is
Calculation:
Substitute,
Substitute
Substitute,
Substitute
Conclusion:
Maximum acceptable torques for pipe (1) is =
Maximum acceptable torques for pipe (2) is =
(b)
The maximum acceptable length of the middle segment.
Answer to Problem 3.5.13P
The maximum acceptable length of the middle segment.
Explanation of Solution
Given information:
Maximum twist of pipe (2) cannot exceed
Write the expression for the angle of twist for pipe (1).
Here the angle of twist is
Write the expression for total angle of twist for pipe (2)
Here, the total angle of twist is
Write the expression for the length of the segment
Write expression for the angle of twist in the section
Write expression for the angle of twist in the section
Write expression for the angle of twist in the section
Substitute
Write the expression for relation between
Calculation:
Substitute
Substitute
Substitute
Conclusion:
The maximum acceptable length of the middle segment is =
(c)
The inner diameter for the given parameters.
Answer to Problem 3.5.13P
The inner diameter for the given parameters is =
Explanation of Solution
Given information:
the maximum torque carried by pipe (2) is
Write the expression for allowable torque.
Here, the allowable torque is
Substitute
Calculation:
Substitute
Conclusion:
The inner diameter for the given parameters is
(c)
Applied torque on pipe (1)
Maximum twist of pipe (1)
Answer to Problem 3.5.13P
Applied torque on pipe (1) is =
Applied torque on pipe (2) is =
Maximum twist of pipe (1) is =
Explanation of Solution
Write the expression for maximum shear strain.
Here, maximum shear strain is
Write the expression for maximum shear stress.
Here, the maximum shear stress is
Write the expression for shear modulus of elasticity.
Write the expression for maximum torque for pipe (1)
Write the expression for maximum angle of twist for pipe (1)
Maximum angle of twist in pipe (1) is
Calculation:
Substitute
Substitute
Substitute
Substitute
Substitute
Substitute
Conclusion:
Applied torque on pipe (1) is =
Applied torque on pipe (2) is =
Maximum twist of pipe (1) is =
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Chapter 3 Solutions
Mechanics of Materials (MindTap Course List)
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