Question

A solid steel shaft has to transmit 50kw at 410rp.m.Taking allowable sheer stress as 90mn/m. Find the suitable diameter of the shaft,if the maximum torque transmitted on each revolution exceeds the mean by 80 percent

Answer 1

$$P=\frac{2\pi NT}{60}\\50000=\frac{2\pi (410)T}{60}\\T=1165\ Nm=1165\cdot10^3\ Nmm$$

$$T_{max}=\frac{\pi}{16}\tau d^3\\(1.8)1165\cdot10^3=\frac{\pi}{16}(90) d^3\\d=49\ mm$$