Explanation Given information: Shaft is subjected to a torque T . Calculation: The expression to find the yield stress ( τ Y ) in the shaft is shown below: τ Y = T Y c J (1) Here, T Y is the maximum elastic torque, c is the outer radius of the shaft, and J is the polar moment of inertia of the shaft. Calculate the maximum elastic torque for the larger diameter of the segment: The polar moment of inertia of a shaft with radius r is π 2 r 4 . The larger diameter of the shaft is 80 mm. So the radius value of the shaft is 40 mm. Substitute r for c and π 2 r 4 for J in Equation (1). τ Y = T Y r π 2 r 4 (2) Substitute 40 mm for r in Equation (2). τ Y = T Y ( 40 mm ) π 2 ( 40 mm ) 4 T Y = π 2 ( 40 mm × 10 − 3 m 1 mm ) 4 τ Y ( 40 mm × 10 − 3 m 1 mm ) T Y = 1.0053 × 10 − 4 τ Y The expression for elastic-plastic torque ( T P ) occurs in the smaller diameter of the segment is shown below: T P = 2 π 3 τ Y c 3 (3) The smaller diameter of the shaft is 75 mm. So the radius value of the shaft is 37.5 mm. Substitute 37.5 mm for c in Equation (3). T P = 2 π 3 τ Y ( 37.5 mm ) 3 = 2 π 3 τ Y ( 37.5 mm × 10 − 3 m 1 mm ) 3 = 1.1045 × 10 − 4 τ Y Compare the values of elastic torque ( T Y ) and the elastic-plastic torque ( T P )

10th Edition

ISBN: 9780134319650

Author: Russell C. Hibbeler

Publisher: PEARSON

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Chapter 5.10, Problem 5.139P

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The radius of the elastic core in the smaller diameter segment of the shaft.

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Chapter 5.10, Problem 5.138P

Chapter 5.10, Problem 5.140P

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The radius of the elastic core in the smaller diameter segment of the shaft.

Solution Summary: The author explains how to determine the radius of the elastic core in the smaller diameter segment of a shaft.

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The radius of the elastic core in the smaller diameter segment of the shaft.

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