Explanation Given information: The yield strength of steel is σ Y = 45 ksi . The magnitude of load is P = 60 kips . The diameter of the shaft AB is d = 1.5 in . . Calculation: Apply maximum distortion energy criterion as shown below. σ a 2 − σ a σ b + σ b 2 = σ Y σ a 2 − σ a σ b + σ b 2 = σ Y 2 (1) Calculate the area of the shaft ( A ) as shown below. A = π d 2 4 Substitute 1.5 in . for d . A = π × 1.5 2 4 = 1.7671 in . 2 Calculate the axial stress ( σ x ) as shown below. σ x = − P A Substitute 60 kips for P and 1.7671 in . 2 for A . σ x = − 60 1.7671 = − 33.954 ksi 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 − 33.954 ksi for σ x and 0 for σ y . σ avg = − 33.954 + 0 2 = − 16.977 ksi Calculate the value of τ x y as shown below. R = ( σ x − σ y 2 ) 2 + τ x y 2 Substitute − 33.954 ksi for σ x and 0 for σ y . R = ( − 33.954 − 0 2 ) 2 + τ x y 2 = 288.2185 + τ x y 2 Calculate principal stresses ( σ max and σ min ) as shown below. σ max , min = σ avg ± R Substitute − 16.977 ksi for σ avg and 288.2185 + τ x y 2 for R . σ a , b = − 16.977 ± 288.2185 + τ x y 2 σ a = − 16.977 + 288.2185 + τ x y 2 σ b = − 16.977 − 288.2185 + τ x y 2 Calculate the value of τ x y as shown below. Substitute − 16

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ISBN: 8220100257063

Author: BEER

Publisher: YUZU

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Chapter 7.5, Problem 88P

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The magnitude of the torque T using maximum distortion energy criterion.

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Chapter 7.5, Problem 87P

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