Question

A shaft is to be designed for transmitting 100 kW power at 150 rpm. Shaft is supported in bearings 3 m apart and at 1 m from one bearing a pulley exerting a transverse load of 30 KN on shaft is mounted. Obtain the diameter of shaft if the maximum direct stress is not to exceed 100 N/mm2.

Answer 1

Ra + Rb = 30 KN

Ra = 20 KN

Rb = 10 KN

Maximum bending moment occurs at Point ‘C’.

MC = 20 × 1 = 20 KN–m ...(i)

Power transmitted by shaft P = 2πNT/60

100 × 103 = 2π.150. T/60

T = 6366.19 N–m ...(ii)

Equivalent bending moment Me = [M + (M2 + T2)1/2]

= [20,000 + (20,0002 + 63692)1/2]

Me = 20494.38 N–m ...(iii)

From bending equation Me /I = σ/y

2044.38 × 103/[ (π/64)d4] = 100/d/2

d = 127.8 mm