Question

A solid shaft transmits power at the rate of 2000KW at the speed of 60RPM. If the safe allowable stress is 80MN/m², find the minimum diameter of the shaft.

Answer 1

P = 2000KW = 2000 × 10³ W 

N = 60RPM 

τ = 80MN/m² = 80 × 10⁶ N/m² 

d = ?

Using the relation: P = 2π.N.T_max/60 watts 

2000 X 10³ = 2π.60.T_max/60 

T_max = 318309.88 N–m

Now using the relation T_max = (π/16) τ_max.D³ 

318309.88 = (π/16) × 80 × 10⁶D³ 

D = 0.2726 m or D = 272.63 mm