Explanation Given information: The allowable shear stress for aluminum is 12 ksi. The allowable shear stress for steel is 10 ksi. Calculation: Sketch the free body diagram of the shaft as shown in Figure 1. Apply Equilibrium equation. Sum of moments at x is equal to 0. Σ M x = 0 T A + T C − ( P ) ( 5 ft ) = 0 T A + T C = 5 P kip ⋅ ft (1) The angle of twist of the shaft is expressed as follows: ϕ = T L J G Here, T is the torque applied, L is the length of the shaft, J is the polar moment of inertia, and G is the shear modulus of elasticity of the material. Refer the properties of 6061-T6 aluminum and A992 steel. The value of shear modulus of elasticity of 6061-T6 aluminum and A992 steel are 3.7 × 10 3 ksi and 11 × 10 3 ksi . Apply compatibility condition. The angle of twist of B with respect to end A is equal to the angle of twist of B with respect to end C . ϕ B / A = ϕ B / C T A L J G al = T C L J G st Substitute 3.7 × 10 3 ksi for G al and 11 × 10 3 ksi for G st . T A L J ( 3.7 × 10 3 ksi ) = T C L J ( 11 × 10 3 ksi ) T A = 0.3364 T C (2) Substitute 0.3364 T C for T A in Equation (1). 0.3364 T C + T C = 5 P T C = 3.7414 P kip ⋅ ft Substitute 3.7414 P kip ⋅ ft for T C in Equation (2). T A = 0.3364 ( 3.7414 P ) = 1.259 P kip ⋅ ft The outer radius of the shaft is r . The radius value of the shaft is 2 in. The polar moment of inertia for a solid shaft of radius ( r ) is π 2 r 4 . The torsion formula for allowable shear stress in the region AB of the solid shaft ( τ A B ) allow is shown below: ( τ A B ) allow = T A c J Here, c is the outer radius of the shaft. Substitute r for c and π 2 r 4 for J

Statics and Mechanics of Materials (5th Edition)

5th Edition

ISBN: 9780134382593

Author: Russell C. Hibbeler

Publisher: PEARSON

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