Calculate diameter of shaft in maximum direct stress:

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

Sol.:                                       

R_a + R_b = 30 KN

R_a = 20 KN

R_b = 10 KN

Maximum bending moment occurs at 'C'.

M_C  = 20 × 1 = 20 KN-m                                                                                ...(i)

The power transmitted by shaft P = 2  NT/60

100 × 10³  =   2  x 150. T/60

T = 6366.19 N-m                                                                                         ...(ii)

Equivalent bending moment M_e = ½[M + (M₂ + T₂)^1/2]

= ½[20,000 + (20,0002 + 63692)^1/2]

M_e  = 20494.38 N-m                                                                                      ...(iii)

From the bending equation Me/I = σ/y 2044.38 × 10³/[ ( ?/64)d⁴] 100/d/2

d = 127.8 mm