Explanation Given information: Shaft rotates at 450 rpm. Power the shaft transmits is 230 kW. The allowable shear stress is 150 MPa. Calculation: Determine the power transmitted ( P ) by the shaft using the relation: P = T ω (1) Here, the applied torque is T and the angular velocity of the shaft is ω . The expression for angular velocity of the shaft ( ω ) is shown below: ω = 2 π f (2) Here, the frequency of shaft’s rotation is f . Substitute 450 rev min for f in Equation (2). ω = 2 π ( 450 rev min ) = 2 π ( 450 rev min ) ( 1 min 60 sec ) = 47.1239 rad sec Substitute 230 kW for P and 47.1239 rad sec for ω in Equation (1). 230 kW = T ( 47.1239 rad sec ) 230 kW ( 10 3 W 1 kW ) = T ( 47.1239 rad sec ) T = 4 , 880.75 N ⋅ m Let D be the larger diameter of the built-up shaft, d be the smaller diameter of the built-up shaft, and r be the radius at transition between the cross sections. The expression to find the allowable shear stress ( τ allow ) in the built-up shaft is shown below: τ allow = K T c J (3) Here, T is the applied torque, c is the outer radius of the shaft, and J is the polar moment of inertia of the shaft. The polar moment of inertia of a shaft with radius r is π 2 r 4 . Diameter of the shaft is 60 mm. so the radius value of the shaft is 30 mm. Substitute r for c and π 2 r 4 for J in Equation (3). τ allow = K T r π 2 r 4 (4) Substitute 150 MPa for τ allow , 4 , 880

10th Edition

ISBN: 9780134319650

Author: Russell C. Hibbeler

Publisher: PEARSON

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Chapter 5.10, Problem 5.121P

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The built-up shaft is possible to rotate at 450 rpm and transmits power of 230 kW.

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Chapter 5.10, Problem 5.120P

Chapter 5.10, Problem 5.122P

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To check: The built-up shaft is possible to rotate at 450 rpm and transmits power of 230 kW.

Solution Summary: The author explains that the built-up shaft rotates at 450 rpm and transmits 230 kW. The allowable shear stress is 150 MPa.