Explanation Given information: The ratio of inner diameter to the outer diameter is d i d 0 = 0.75 . The power developed by the motor is 20 hp with turning of 600 rpm . The allowable shear stress is 12 ksi. Calculation: Determine the power transmitted ( P ) by the shaft using the relation: P = T ω (1) Here, T is the applied torque and ω is the angular velocity of the shaft. The expression for angular velocity of the shaft ( ω ) is shown below: ω = 2 π f (2) Here, f is the frequency of shaft’s rotation. Substitute 600 rev min for f in Equation (2). ω = 2 π ( 600 rev min ) = 2 π ( 600 rev min ) ( 1 min 60 sec ) = 62.8319 rad sec Substitute 20 hp for P and 62.8319 rad sec for ω in Equation (1). 20 hp = T ( 62.8319 rad sec ) 20 hp ( 550 ft ⋅ lb sec 1 hp ) = T ( 62.8319 rad sec ) T = 175.0703 lb ⋅ ft T = 2 , 100.84 lb ⋅ in . The torsion formula for allowable shear stress in the shaft ( τ allow ) is shown below: τ allow = T c J (3) Here, the polar moment of inertia is J and the outer radius of the shaft is c . The ratio of inner diameter to the outer diameter is d i d 0 = 0.75 . The polar moment of inertia for a shaft of outer diameter d 0 and inner diameter d i is expressed as, J = π 2 ( ( d 0 2 ) 4 − ( d i 2 ) 4 ) = π 2 ( d 0 4 16 − ( 0

9th Edition

ISBN: 9780133254426

Author: Russell C. Hibbeler

Publisher: Prentice Hall

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Chapter 5.3, Problem 5.33P

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The inner and outer diameters of the shaft.

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Mechanics of Materials

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