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?

Question

Question

This question was previously asked in

ESE Civil 2021 Official Paper

Answer 1

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

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