A 2-in (5.1-cm) diameter solid steel shaft has a maximum bending moment of 6000 lb·in (677.9 N·m) and an applied torque of 3000 lb·in (339.0 N·m). Is this shaft safe if the maximum allowable bending stress is 10,000 lb/in 2 (68,947.6 kPa)? What is the ideal torque for this shaft?
Calculation Procedure:
The equivalent bending moment Me lb·in for a solid shaft is Me = 0.5[M + (M 2 + T 2)0.5], where M = maximum bending moment acting on the shaft, lb·in; T = maximum torque acting on the shaft, lb·in. For this shaft, Me = 0.5[6000 + (6000 2 + 3000 2)0.5] = 6355 lb·in (718.0 N·m).
Use the flexure relation s = Mc/I, where s = stress developed in the shaft, lb/in 2; M = Me for a shaft; I = section moment of inertia of the shaft about the neutral axis; in 4; c = distance from shaft neutral axis to outside fibers, in. For a circular shaft, I = pd 4/64 = p(2) 4/64 = 0.785 in 4 (32.7 cm 4); c = d/2 = 2/2 = 1.0. Then s = Mc/I = (6855)(1.0)/0.785 = 8100 lb/in 2 (55,849.5 kPa). Thus, the actual bending stress is 1900 lb/in 2 (13,100.5 kPa) less than the maximum allowable bending stress. Therefore the shaft is safe. Alternatively, compute the maximum equivalent bending moment from Me = sI/c = (10,000)(0.785)/1.0 = 7850 lb·in (886.9 N·m). This is 7850 - 6355 = 1495 lb·in (168.9 N·m) greater than the actual equivalent bending moment. Hence, the shaft is safe.
The ideal torque Ti lb·in for a shaft is Ti = M + ( M2 + T2)0.5, where M and T are the bending and torsional moments, respectively, acting on the shaft, lb·in. For this shaft, Ti = 6000 + (6000 2 + 3000 2)0.5 = 12,710 lb·in (1436.0 N·m).
Related Calculations: Use this procedure for any shaft of uniform cross section made of metal--steel, aluminum, bronze, brass, etc.