Explanation Given information: The state of plane stress components are σ x = − 80 MPa, σ y = 0 , and τ x y = − τ 0 . The uniaxial tension for the aluminum alloy is σ U T = 75 MPa . The uniaxial compression for the aluminum alloy is σ U C = 150 MPa . Calculation: Calculate the average normal stress ( σ avg ) as shown below. σ avg = σ x + σ y 2 Substitute − 80 MPa for σ x and 0 for σ y . σ avg = − 80 + 0 2 = − 40 MPa Calculate the value of τ x y as shown below. R = ( σ x − σ y 2 ) 2 + τ x y 2 Substitute − 80 MPa for σ x , 0 for σ y , and − τ 0 for τ x y . R = ( − 80 − 0 2 ) 2 + τ 0 2 = 1 , 600 + τ 0 2 Calculate principal stresses ( σ max and σ min ) as shown below. σ max , min = σ avg ± R Substitute − 40 MPa for σ avg and 1 , 600 + τ 0 2 for R . σ a , b = − 40 ± 1 , 600 + τ 0 2 σ a = − 40 + 1 , 600 + τ 0 2 σ b = − 40 − 1 , 600 + τ 0 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 94P

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The shearing stress τ0 for which the failure should be expected using Mohr’s criterion.

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Chapter 7.5, Problem 93P

Chapter 7.5, Problem 95P

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Mechanics of Materials, 7th Edition

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The shearing stress τ 0 for which the failure should be expected using Mohr’s criterion.

Solution Summary: The author calculates the shearing stress tau_0 for which failure should be expected using Mohr’s criterion.

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The shearing stress τ0 for which the failure should be expected using Mohr’s criterion.

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