Explanation Given information: A bar with a circular cross sectional area is made of SAE 1045 carbon steel. The yield stress of the SAE 1045 carbon steel is σ Y = 150 ksi . The bar is subjected to a torque of 30 kip ⋅ in . and a bending moment of 56 kip ⋅ in . Calculation: Find the moment of inertia for the solid bar as shown below. I = π d 4 64 Here, d is the diameter of the bar. Find the polar moment of inertia for the solid bar as shown below. J = π d 4 32 The centroid of the bar c = d 2 . Find the bending stress as shown below. σ = M c I Here, I is the moment of inertia and M is the bending moment. Substitute π d 4 64 for I , 56 kip ⋅ in . for M , and d 2 for c . σ = 56 × d 2 π d 4 64 = 28 d × 64 π d 4 = 1 , 792 π d 3 Find the shear stress as shown below. τ = T c J Here, T is the torque and J is the polar moment of inertia. Substitute π d 4 32 for J , 30 kip ⋅ in . for T , and d 2 for c . τ = 30 × d 2 π d 4 32 = 15 d × 32 π d 4 = 480 π d 3 Sketch the critical state of stress as shown in Figure 1. Refer to Figure 1. σ x = 1 , 792 π d 3 σ y = 0 τ x y = 480 π d 3 Find the principal stresses in complex system as shown below. σ 1 , 2 = σ x + σ y 2 ± ( σ x − σ y 2 ) 2 + τ x y 2 (1) Here, σ x is the normal stress in x direction, σ y is the normal stress in y direction, and τ x y is the shear stress in xy plane. Substitute 1 , 792 π d 3 for σ x , 0 for σ y , and 480 π d 3 for τ x y in Equation (1)

9th Edition

ISBN: 9780133254426

Author: Russell C. Hibbeler

Publisher: Prentice Hall

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Chapter 10.7, Problem 10.84P

To determine

The required diameter of the bar according to the maximum distortion energy theory.

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Chapter 10.7, Problem 10.83P

Chapter 10.7, Problem 10.85P

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