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Solid And Hollow Shafts In Torsion

A solid steel shaft will transmit 500 hp (372.8 kW) at 3600 r/min. What diameter shaft is required if the allowable stress in the shaft is 12,500 lb/in 2 (86,187.5 kPa)? What diameter hollow shaft is needed to transmit the same power if the inside diameter of the shaft is 1.0 in (2.5 cm)?

Calculation Procedure:

1. Compute the torque in the solid shaft

For any solid shaft, the torque T, lb·in = 63,000 hp/R, where R = shaft rpm. Thus, T = 63,000(500)/3600 = 8750 lb·in (988.6 N·m).

2. Compute the required shaft diameter

For any solid shaft, the required diameter d, in = 1.72 ( T/s)1/3, where s = allowable stress in shaft, lb/in 2. Thus, for this shaft, d = 1.72(8750/12,500) 1/3 = 1.526 in (3.9 cm).

3. Analyze the hollow shaft

The usual practice is to size hollow shafts such that the ratio q of the inside diameter di into the outside diameter do in is 1:2 to 1:3 or some intermediate value. With a q in this range the shaft will have sufficient thickness to prevent failure in service.

Assume q = di/do = 1/2. Then with di = 1.0 in (2.5 cm), do = di/q, or do = 1.0/0.5 = 2.0 in (5.1 cm). With q = 1/3, do = 1.0/0.33 = 3.0 in (7.6 cm).

4. Compute the stress in each hollow shaft

For the hollow shaft s = 5.1 T/d3o(1 - q4), where the symbols are as defined above. Thus, for the 2-in (5.1-cm) outside-diameter shaft, s = 5.1(8750)/[8(1 - 0.0625)] = 5950 lb/in 2 (41,028.8 kPa).

By inspection, the stress in the 3-in (7.6-cm) outside-diameter shaft will be lower because the torque is constant. Thus, s = 5.1 (8750)/[27(1 - 0.0123)] = 1672 lb/in 2 (11,528.0 kPa).

5. Choose the outside diameter of the hollow shaft

Use a trial-and-error procedure to choose the hollow shaft's outside diameter. Since the stress in the 2-in (5.1-cm) outside-diameter shaft, 5950 lb/in 2 (41,023.8 kPa), is less than half the allowable stress of 12,500 lb/in 2 (86,187.5 kPa), select a smaller outside diameter and compute the stress while holding the inside diameter constant.

Thus, with a 1.5-in (3.8-cm) shaft and the same inside diameter, s = 5.1(8750)/[3.38(1 0.197)] = 16,430 lb/in 2 (113,284.9 kPa). This exceeds the allowable stress.

Try a larger outside diameter, 1.75 in (4.4 cm), to find the effect on the stress. Or s = 5.1 (8750)/[5.35(1 - 0.107)] = 9350 lb/in 2 (64,468.3 kPa). This is lower than the allowable stress.

Since a 1.5-in (3.8-cm) shaft has a 16,430-lb/in 2 (113,284.9-kPa) stress and a 1.75-in (4.4-cm) shaft has a 9350-lb/in 2 (64,468.3-kPa) stress, a shaft of intermediate size will have a stress approaching 12,500 lb/in 2 (86,187.5 kPa). Trying 1.625 in (4.1 cm) gives s = 5.1(8750)/[4.4(1 0.143)] = 11,820 lb/in 2 (81,489.9 kPa). This is within 680 lb/in 2 (4688.6 kPa) of the allowable stress and is close enough for usual design calculations.

Related Calculations: Use this procedure to find the diameter of any solid or hollow shaft acted on only by torsional stress. Where bending and torsion occur, use the next calculation procedure. Find the allowable torsional stress for various materials in Baumeister and Marks-- Standard Handbook for Mechanical Engineers .

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