Explanation Given information: Allowable shear stress is 85 MPa. The power developed by the motor is 300 W with turning of 90 rev min . Calculation: The expression for the power transmitted ( P ) by the shaft is shown below: P = T ω (1) Here, T is the applied torque and ω is the angular velocity of the shaft. Rearrange Equation (1) to find the torque at A . T A = P ω A (2) The expression for angular velocity of the shaft at A ( ω A ) is shown below: ω A = 2 π f (3) Here, f is the frequency of shaft’s rotation. Substitute 90 rev min for f in Equation (3). ω = 2 π ( 90 rev min ) ( 1 min 60 sec ) = 9.4248 rad sec Power developed by the motor is P = 300 W = 300 N ⋅ m / sec . Substitute 300 N ⋅ m / sec for P and 9.4248 rad sec for ω A in Equation (2). T A = 300 N ⋅ m / sec 9.4248 rad sec = 31.83 N ⋅ m The expression for angular velocity of the shaft at B ( ω B ) is shown below: ω B = ω A ( r A r B ) (4) Here, r A is the radius at pulley A and r B is the radius at pulley B . Substitute 9.4248 rad sec for ω A , 60 mm for r A , and 150 mm for r B in Equation (4). ω B = 9.4248 rad sec ( 60 mm 150 mm ) = 3.7699 rad sec Rearrange Equation (1) to find the torque at B . T B = P ω B (5) Substitute 300 N ⋅ m / sec for P and 3.7699 rad sec for ω B in Equation (5). T B = 300 N ⋅ m / sec 3.7699 rad sec = 79.58 N ⋅ m The torsion formula for allowable shear stress in the shaft at A ( τ allow ) is shown below: τ allow = T A c J (6) Here, J is the polar moment of inertia and the outer radius of the shaft is c

9th Edition

ISBN: 9789810694364

Author: Russell C Hibbeler

Publisher: Pearson Education

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Chapter 5.3, Problem 5.38P

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The required diameter of the steel shaft on the pulley at A.

The required diameter of the steel shaft on the pulley at B.

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Chapter 5.3, Problem 5.37P

Chapter 5.3, Problem 5.39P

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Mechanics Of Materials, Si Edition

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The required diameter of the steel shaft on the pulley at A.

The required diameter of the steel shaft on the pulley at B.