Explanation Given information: The load P = 26 kN . The diameter of the shaft AB is d = 32 mm . The uniaxial tension for the aluminum alloy is σ U T = 60 MPa . The uniaxial compression for the aluminum alloy is σ U C = 120 MPa . Calculation: Calculate the area of the shaft ( A ) as shown below. A = π d 2 4 Substitute 32 mm for d . A = π × 32 2 4 = 804.2477 mm 2 Calculate the axial stress ( σ x ) as shown below. σ x = P A Substitute 26 kN for P and 804.2477 mm 2 for A . σ x = 26 kN × 1 , 000 N 1 kN 804.2477 mm 2 = 32.328 N / mm 2 × 1 MPa 1 N / mm 2 = 32.328 MPa The stress along y axis σ y = 0 . Calculate the average normal stress ( σ avg ) as shown below. σ avg = σ x + σ y 2 Substitute 32.328 MPa for σ x and 0 for σ y . σ avg = 32.328 + 0 2 = 16.164 MPa Calculate the value of R as shown below. R = ( σ x − σ y 2 ) 2 + τ x y 2 Substitute 32.328 MPa for σ x and 0 for σ y . R = ( 32.328 − 0 2 ) 2 + τ x y 2 = 261.2749 + τ x y 2 Calculate principal stresses ( σ max and σ min ) as shown below. σ max , min = σ avg ± R Substitute 16.164 MPa for σ avg and 261.2749 + τ x y 2 for R . σ a , b = 16.164 ± 261.2749 + τ x y 2 σ a = 16.164 + 261.2749 + τ x y 2 σ b = 16.164 − 261.2749 + τ x y 2 Here, | σ avg | < R

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ISBN: 9780073398235

Author: Ferdinand P. Beer, E. Russell Johnston Jr., John T. DeWolf, David F. Mazurek

Publisher: McGraw-Hill Education

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Chapter 7.5, Problem 96P

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The magnitude of the torque T for which the failure should be expected using Mohr’s criterion.

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The magnitude of the torque T for which the failure should be expected using Mohr’s criterion.

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