Explanation Given information: An aluminum alloy is to be used for a drive shaft such that it transmits 25 hp at 1 , 500 rev / min . The factor of safety with respect to yielding is 2.5. The allowable yield stress σ Y = 3.5 ksi . Calculation: Find the polar moment of inertia ( J ) of the solid bar as shown below. J = π c 4 2 Here, c is the radius of the bar. Find the angular velocity as shown below. ω = 2 π N 60 Here, N is the number of revolutions. Substitute 1 , 500 rev / min for N . ω = 2 π × 1 , 500 60 = 50 π Convert the unit of P from hp to lb-in . / sec . P = 25 hp × 550 lb-ft / sec 1 hp × 12 in . 1 ft = 165 , 000 lb-in . / sec Find the torque ( T ) as shown below. T = P ω Here, P is the power and ω is the angular velocity. Substitute 165 , 000 lb-in . / sec for P and 50 π for ω . T = 165 , 000 50 π = 3 , 300 π 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 3 , 300 π for T and π c 4 2 for J . τ = 3 , 300 π c π c 4 2 = 3 , 300 π c × 2 π c 4 = 6 , 600 π 2 c 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 0 for σ x , 0 for σ y , and 6 , 600 π 2 c 3 for τ x y in Equation (1). σ 1 , 2 = 0 + 0 2 ± ( 0 − 0 2 ) 2 + ( 6 , 600 π 2 c 3 ) 2 = ± 6 , 600 π 2 c 3 (2) Find the maximum and minimum principal stresses using Equation (2) as shown below

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ISBN: 9780133254426

Author: Russell C. Hibbeler

Publisher: Prentice Hall

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Chapter 10.7, Problem 10.63P

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The smallest diameter of the shaft that can be selected based on the maximum distortion energy theory.

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