Question

A motor driving a solid circular shaft transmits 30 kW at 500 r.p.m. What is the torque activity on the shaft, if allowable shear stress is 42 MPa?
1. 427 Nm
2. 573 Nm
3. 180 Nm
4. 219 Nm

Answer 1

Correct Answer - Option 2 : 573 Nm

Concept:

Power (P)

P = T × W

Torsion equation

\(\frac{T}{J} = \frac{{G\theta }}{L} = \frac{\tau }{r}\)

Calculation:

Given:

P = 30 kW, N = 500 rpm, D = 40 mm

T = ?

As, P = T × W

\(30 \times {10^3} = {\rm{T}} \times \frac{{2{\rm{\pi }} \times 500}}{{60}} \)

T = 572.9 kN ≈ 573 N-m