a. The necessary diameter for solid circular shaft. b. The necessary diameter for hollow circular shaft, if inside diameter being $3/4^{th}$ of outside diameter. Take shear stress as 75 Mpa.

Given

$P = 450 \hspace{0.05cm} kw = 450\hspace{0.05cm}\times\hspace{0.05cm}10^3\hspace{0.05cm}watt\\ N = 100\hspace{0.05cm}rpm\\ \tau = 75\hspace{0.05cm}MPa$

To find

1) solid shaft; D = ?

2) Hollow shaft; D = ? If $d=\frac{3}{4}D$

Solution

Torque $P = \frac{2\pi NT}{60}\\ 450\hspace{0.05cm}\times\hspace{0.05cm}10^3 = \frac{2\pi\hspace{0.05cm}\times\hspace{0.05cm}100}{60}\\ T = 42.97\hspace{0.05cm}\times\hspace{0.05cm}10^3\hspace{0.05cm}Nmm = 42.97\hspace{0.05cm}\times10^6\hspace{0.05cm}Nmm$

1) For solid shaft

$\frac{T}{J} = {\tau}{R}\\ \frac{42.97\hspace{0.05cm}\times\hspace{0.05cm}10^6}{\frac{\pi}{32}\hspace{0.05cm}\times\hspace{0.05cm}D^4} = \frac{75}{\frac{D}{2}}\\ D = 142.8\hspace{0.05cm}mm$

2) For Hollow Shaft

$d = \frac{3}{4}D = 0.75D\\ J = \frac{\pi}{32}[D64 - d^4] = \frac{\pi}{32}[D^4 - (0.75D)^4] = 0.067D^4\\ \textit{for hollow shaft}\hspace{0.25cm}R = \frac{D}{2}\\ \frac{T}{J} = \frac{\tau}{R}\\ \frac{42.97\hspace{0.05cm}\times\hspace{0.05cm}10^6}{0.067D^4} = \frac{75}{\frac{D}{2}}\\ D = 162.22\hspace{0.05cm}mm\\ d = \frac{3}{4}D = 121.665\hspace{0.05cm}mm$