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

$T=\frac{P\times 60\times {10}^6}{2\pi N}=\frac{40\times 60\times {10}^6}{2\pi \left(500\right)}$
= 763943.7268 N-mm
${\tau }_{max}=\frac{16T}{\pi d^3}=\frac{16\times 763943.7268}{\pi (40{)}^3}$
= 60.792 MPa