What diameter solid steel shaft should be used for a 500-hp (372.8-kW) 250-r/min application if the allowable torsional deflection is 1°, the maximum allowable stress is 10,000 lb/in 2 (68,947.6 kPa), and the modulus of rigidity is 13 X 10 6 lb/in 2 (89.6 X 10 6 kPa)? What diameter hollow steel shaft should be used if the ratio of the inside diameter to the outside diameter is 1:3, the allowable deflection is 1°, the allowable stress is 10,000 lb/in 2 (68,947.6 kPa), and the modulus of rigidity is 13 X 10 6 lb/in 2 (89.6 X 10 6 kPa)? What shaft has the greatest weight?
Calculation Procedure:
For any shaft, T = 68,000 hp/R; or for this shaft, T = 68,000(500)/250 = 126,000 lb·in (14,236.1 N·m).
For a solid metal shaft, d = (584 Tl/Ga) 1/3, where l = shaft length expressed as a number of shaft diameters, in; G = modulus of rigidity, lb/in 2; a = angle of torsion deflection, deg.
Usual specifications for noncritical applications of shafts require that the torsional deflection not exceed 1° in a shaft having a length equal to 20 diameters. Using this length gives d = [584 X 126,000 X 20/(13 X 10 6 X 1.0)] 1/3 = 4.84 in (12.3 cm). Use a 5-in (12.7-cm) diameter shaft.
Assume that the shaft has a length equal to 20 diameters. Then for a hollow shaft d = [584 Tl/Ga(1 - q4)]1/3, where q = di/do; di = inside diameter of the shaft, in; do = outside diameter of the shaft, in. For this shaft, d = {584 X 126,000 X 20/(13 X 10 6 X 1.0)[1 - (1/3) 4]}1/3 = 4.86 in (12:3 cm). Use a 5-in (12.7-cm) outside-diameter shaft. The inside diameter would be 5.0/3 = 1.667 in (4.2 cm).
Steel weighs approximately 480 lb/ft 3 (7688.9 kg/m 3). To find the weight of each shaft, compute its volume in cubic feet and multiply it by 480. Thus, for the 5-in (12.7-cm) diameter solid shaft, weight = (p 5 2/4)(5 X 20)(480)/1728 = 540 lb (244.9 kg). The 5-in (12.7-cm) outside-diameter hollow shaft weighs (p 5 2/4 - p1.667 2/4)(5 X 20)(480)/1728 = 242 lb (109.8 kg). Thus, the hollow shaft weighs less than half the solid shaft. However, it would probably be more expensive to manufacture because drilling the central hole could be costly.
Related Calculations: Use this procedure to determine the steady-load torsional deflection of any shaft of uniform cross section made of any metal--steel, bronze, brass, aluminum, Monel, etc. The assumed torsional deflection of 1° for a shaft that is 20 times as long as the shaft diameter is typical for routine applications. Special shafts may be designed for considerably less torsional deflection.