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 × 10^{3} = 2 x 150. T/60
T = 6366.19 N-m ...(ii)
Equivalent bending moment M_{e} = ½[M + (M_{2} + T_{2})^{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 × 10^{3}/[ ( ?/64)d^{4}] 100/d/2
d = 127.8 mm