Explanation Given information: The shaft is subjected to a distributed torque of length t = ( 10 x 2 ) N ⋅ m m . Maximum stress in the shaft is 80 MPa. Calculation: Find the torque applied ( T ) to the shaft using the relation: T = ∫ t d x (1) Here, the distributed torque over the length is t . Rearrange Equation (1) for the limits 0 to x . T = ∫ 0 x t d x (2) Substitute ( 10 x 2 ) N ⋅ m m for T in Equation (2) and integrate with respect to x . T = ∫ 0 x ( 10 x 2 ) d x = 10 [ x 3 3 ] 0 x = 10 3 x 3 The torsion formula for shear stress in the solid shaft ( τ ) is shown below: τ = T c J (3) Here, the polar moment of inertia is J and the outer radius of the shaft is c . The polar moment of inertia for a solid shaft of radius ( c ) is π 2 c 4 . Substitute 80 MPa for τ , 10 3 x 3 for T , and π 2 c 4 for J in Equation (3)

10th Edition

ISBN: 9780134319650

Author: Russell C. Hibbeler

Publisher: PEARSON

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Chapter 5.3, Problem 5.29P

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The variation of radius c of the shaft for 0≤x≤3 m.

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The variation of radius c of the shaft for 0 ≤ x ≤ 3 m .

Solution Summary: The author explains how the shaft is subjected to a distributed torque of length (T).

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