Question

A motor driving a solid circular steel shaft transmits 40 kW of power at 500 rpm. If the diameter of the shaft is 40 mm, the maximum shear stress in the shaft is ________ MPa.

Answer 1

Concept:

Power (P)

P = T × W

Torsion equation

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

Calculation:

Given:

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

As, P = T × W

\(40 \times {10^3} = {\rm{T}} \times \frac{{2{\rm{\pi }} \times 500}}{{60}} ⇒ {\rm{T}} = 763.9{\rm{\;Nm}} = 763.9 \times {10^3}{\rm{Nmm}}\)

Now,

\(\frac{T}{J} = \frac{\tau }{R} ⇒ \tau = \frac{{TR}}{J} = \frac{{763.9 \times {{10}^3} \times 20 \times 32}}{{\pi \times {{40}^4}}}\)

τ_max. = 60.79 N/mm² = 60.79 MPa