Explanation Given information: The solid shaft is subjected to a torque T and a moment M . The diameter of the solid shaft is d . Calculation: The polar moment of inertia ( J ) of a solid shaft with diameter ( d ) is expressed as follows: J = π d 4 32 The moment of inertia ( I ) of a solid shaft with diameter ( d ) is expressed as follows: I = π d 4 64 Express the stress ( σ ) in the shaft using the relation: σ = M c I (1) Here, M is the moment and c is the outer radius of the shaft. Substitute d 2 for c and π d 4 64 for I in Equation (1). σ = M ( d 2 ) π d 4 64 = 32 M π d 3 Express the shear stress ( τ ) in the shaft using the relation: τ = T c J (2) Here, T is the torque applied to the shaft. Substitute d 2 for c and π d 4 32 for J in Equation (2). τ = T ( d 2 ) π d 4 32 = 16 T π d 3 Express the principal stresses ( σ 1 , σ 2 ) using the relation: σ 1 , 2 = σ x + σ y 2 ± ( σ x − σ y 2 ) 2 + τ x y 2 (3) Substitute 32 M π d 3 for σ x , 0 for σ y , and 16 T π d 3 for τ x y in Equation (3)

9th Edition

ISBN: 9780133254426

Author: Russell C. Hibbeler

Publisher: Prentice Hall

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Chapter 10.7, Problem 10.88P

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The maximum allowable shear stress is σallow=16πd3[M+M2+T2] according to maximum normal stress theory.

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Chapter 10.7, Problem 10.87P

Chapter 10.7, Problem 10.89P

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The maximum allowable shear stress is σ allow = 16 π d 3 [ M + M 2 + T 2 ] according to maximum normal stress theory.

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To show: The maximum allowable shear stress is σallow=16πd3[M+M2+T2] according to maximum normal stress theory.

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Chapter 10 Solutions

To show:

The maximum allowable shear stress is σallow=16πd3[M+M2+T2] according to maximum normal stress theory.