Concept explainers
(a)
Find the applied torque and the corresponding angle of twist at the onset of yield.
(a)
Answer to Problem 107P
The applied torque at the onset of yield is
The corresponding angle of twist at the onset of yield is
Explanation of Solution
Given information:
The length of the shaft (L) is 0.9 m.
The shear stress
The rigidity modulus of steel (G) is 77.2 GPa.
The inner diameter of the shaft
The inner diameter of the shaft
Calculation:
Find the inner radius
Substitute 30 mm for
Find the outer radius
Substitute 70 mm for
Find the polar moment of inertia of a shaft (J) using the relation:
Substitute 0.035 m for
At the onset of yield, the stress distribution is the elastic distribution with
Find the applied torque
Substitute
Therefore, the applied torque at the onset of yield is
Find the angle of twist
Substitute
Therefore, the corresponding angle of twist at the onset of yield is
(b)
Find the applied torque and the corresponding angle of twist of the when the plastic zone is
(b)
Answer to Problem 107P
The applied torque when the plastic zone is 10 mm deep is
The corresponding angle of twist when the plastic zone is 10 mm deep is
Explanation of Solution
Given information:
The depth of plastic zone (t) is 10 mm.
Calculation:
Find the depth of elastic portion
Substitute 0.035 m for
Find the polar moment of inertia
Substitute 0.025 m for
Find the torque
Substitute
Find the expression of torque
Integrate the Equation (9).
Substitute 180 MPa for
Find the total applied torque (T):
Substitute
Therefore, the applied torque when depth of plastic zone is 10 mm is
Find the angle of twist
Substitute 180 MPa for
Therefore, the corresponding angle of twist when plastic zone is 10 mm is
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Chapter 3 Solutions
Mechanics of Materials - With Access
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