Explanation Given information: The outer diameter of shaft is 50 mm . The transmitted power, P is 10 kW . The allowable shear stress ( τ all ) is 60 MPa . The motor speed ( N ) is 240 rpm . Calculation: Sketch the circular section as shown in Figure 1. Convert the unit of power in kW to Watts . P = 10 kW ( 10 3 W 1 kW ) = 10 , 000 W Find the frequency ( f ) using the relation: f = N 60 (1) Here, N is the number of revolutions. Substitute 240 rpm for N in Equation (1). f = 240 rpm 60 = 4 Hz Find the value of torque ( T ) for shaft ABC as follows: T = P 2 π f (2) Here, P is the power and f is the frequency. Substitute 10 , 000 W for P and 4 Hz for f in Equation (2). T = 10 , 000 2 π × 4 = 10 , 000 25.1327 = 397.887 N ⋅ m Gear A : Find the force using equilibrium equation as follows: F r A − T = 0 F = T r A (3) Here T is the torque and r A radius at point A . Substitute 397.887 N ⋅ m for T and 90 mm for r A in Equation (3). F = 397.887 90 mm ( 1 m 10 3 mm ) = 397.887 0.09 = 4 , 420.96 ≃ 4 , 421 N Calculate the bending moment at B as follows: Refer to Figure P8.25 in the textbook. M B = L A B × F (4) Here, F is the force and L A B is the length of AB . Substitute 100 mm for L A B and 4 , 421 N for F in Equation (4). M B = 100 mm ( m 10 3 mm ) × 4 , 421 = 0.1 × 4 , 421 = 442.1 N ⋅ m Find the value of outer radius ( c o ) as follows: c o = d o 2 (5) Here, d o is outer diameter. Substitute 50 mm for d o in Equation (5). c o = 50 2 = 25 mm Find the value of c using the relation: τ all = c o J M 2 + T 2 J c o = M 2 + T 2 τ all Substitute π 2 ( c o 4 − c i 4 ) c o for J c o

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ISBN: 8220100257063

Author: BEER

Publisher: YUZU

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Chapter 8.2, Problem 28P

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The largest permissible inner diameter of the shaft.

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The largest permissible inner diameter of the shaft.

Solution Summary: The author calculates the unit of power in kW and the motor speed (N) using the relation.

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