Explanation Given information: Diameter of the segment AC is 0.5 in. Diameter of the segment CB is 1 in. Shaft subjected to a uniform distributed torque of 60 lb ⋅ in . / in . along segment CB . Calculation: Let the torque at end A be T A . Let the torque at end B be T B . Sketch the free body diagram of the shaft as shown in Figure 1. Refer Figure 1. Apply Equilibrium condition. T A + T B − 60 ( 20 ) = 0 T A + T B = 1 , 200 (1) The angle of twist of the shaft is expressed as follows: ϕ = T L J G (2) 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. Sketch the torque at distance x from end B as shown in Figure 2. Rearrange Equation (2) by applying compatibility condition. ϕ C / B = ϕ C / A ∫ T ( x ) d x J G = T A L A C J G (3) The polar moment of inertia of a shaft of radius ( r ) is expressed as, J = π 2 ( r 4 ) Radius of segment AC is 0.25 in. and radius of segment CB is 0.5 in. Refer Figure 2. The torque at section x is T ( x ) = T B − 60 x over the limit 0 to 20 in. Rearrange Equation (3) as shown below: ∫ 0 20 in . ( T B − 60 x ) d x π 2 ( ( 0.5 in . ) 4 ) G = T A ( 5 in . ) π 2 ( ( 0.25 in . ) 4 ) G 1 π 2 ( 0.5 in . ) 4 G [ T B x − 60 x 2 2 ] 0 20 in . = T A ( 5 in . ) π 2 ( 0.25 in . ) 4 G 1 ( 0.5 in . ) 4 ( T B ( 20 ) − 60 20 2 2 ) = T A ( 5 in . ) ( 0.25 in . ) 4 320 T B − 192 , 000 = 1 , 280 T A 320 T B − 1 , 280 T A = 192 , 000 (4) Solve Equation (1) and Equation (4)

9th Edition

ISBN: 9789810694364

Author: Russell C Hibbeler

Publisher: Pearson Education

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Chapter 5.5, Problem 5.91P

To determine

The absolute maximum shear stress in the shaft.

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Chapter 5.5, Problem 5.90P

Chapter 5.5, Problem 5.92P

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