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

R_a + R_b = 30 KN 

R_a = 20 KN 

R_b = 10 KN 

Maximum bending moment occurs at Point ‘C’. 

M_C = 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 M_e = ½[M + (M² + T²)^1/2] 

= ½[20,000 + (20,000² + 6369²)^1/2] 

M_e = 20494.38 N–m ...(iii)

From bending equation M_e /I = σ/y 

2044.38 × 10³/[ (π/64)d⁴] = 100/d/2 

d = 127.8 mm