Question

Calculate a suitable diameter for a solid shaft to transmit power of 200kW when

running at 180rev/min if the maximum shear stress in the shaft must not exceed

85N/mm2

Answer 1

Power $P = 200KW = 200,000 Nm$

RPM N = 180

$T = \frac{P \times 60}{2\pi N} = \frac{200000 \times 60}{2\pi 180} = 10.61 KNm$

Diameter of shaft $d = \sqrt[3]{\frac{16T}{\pi\times \tau }} = \sqrt[3]{\frac{16\times10.61\times10^3}{\pi\times 85 }}=8.598mm$