To determine The required minimum wall thickness of the shaft. Explanation Given information: The allowable shear stress of the tubular shaft is τ allow = 85 MPa . Outer diameter of the shaft is 150 mm. Calculation: The internal torque in the region AB of the shaft is shown in Figure 1. The internal torque in the region BC of the shaft is shown in Figure 2. The internal torque in the region CD of the shaft is shown in Figure 3. Refer Figure 1. The value of torque in the region AB of the shaft is 15 kN ⋅ m . Refer Figure 2. The value of torque in the region BC of the shaft is 45 kN ⋅ m . Refer Figure 3. The value of torque in the region CD of the shaft is 30 kN ⋅ m . The segment BC is critical because it is subjected to maximum internal torque. The torsion formula for allowable shear stress in the tubular shaft ( τ allow ) is shown below: τ allow = T B C c J (1) Here, the maximum internal torque is T B C , the polar moment of inertia is J , and the outer radius of the segment is c . The polar moment of inertia for a tubular shaft of outer radius r and inner radius r ' is π 2 ( r 4 − r ' 4 ) . Outer radius of the tubular shaft is 150 mm. Substitute π 2 ( r 4 − r ' 4 ) for J in Equation (1). τ allow = T B C c π 2 ( r 4 − r ' 4 ) = 2 T B C c π ( r 4 − r ' 4 ) (2) Substitute 85 MPa for τ allow , 45 kN ⋅ m for T B C , 75 mm for c , and 75 mm for r in Equation (2)

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ISBN: 9780133254426

Author: Russell C. Hibbeler

Publisher: Prentice Hall

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Chapter 5.3, Problem 5.13P

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The required minimum wall thickness of the shaft.

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