b) If the shaft in (a) is hollow with inside diameter equal to $3/4^{th}$ of the outside diameter of the shaft for the same condition.

Given

$P = 50\hspace{0.05cm}KW = 50\hspace{0.05cm}\times\hspace{0.05cm}10^3\hspace{0.05cm}Watt\\ N = 120\hspace{0.05cm}rpm$

Solution

$\omega = \frac{2\pi N}{60} = 12.50\hspace{0.05cm}rad/sec\\ \rho = \frac{2\pi NT}{60}\\ \hspace{0.25cm} = \omega .T\\ T = \frac{P}{\omega} = \frac{50\hspace{0.05cm}\times\hspace{0.05cm}10^3}{12.50} = 3.98\hspace{0.05cm}\times\hspace{0.05cm}10^3\hspace{0.05cm}Nm = 3.98\hspace{0.05cm}\times\hspace{0.05cm}10^6\hspace{0.05cm}Nmm$

If $\theta = 0.5^\circ = 8.72\hspace{0.05cm}\times\hspace{0.05cm}10^{-3}\hspace{0.05cm}rad$

A] For solid shaft, L = 1000 mm, $J = \frac{\pi}{32}D^4$

$\textit{For}\hspace{0.05cm} \theta \lt 0.5\\ \frac{T}{J} = \frac{G\theta}{L}\\ \frac{3.98\hspace{0.05cm}\times\hspace{0.05cm}10^6}{\frac{\pi}{32}\hspace{0.05cm}\times\hspace{0.05cm}D^4} = \frac{80\hspace{0.05cm}\times\hspace{0.05cm}10^3\hspace{0.05cm}\times\hspace{0.05cm}8.72\hspace{0.05cm}\times\hspace{0.05cm}10^{-3}}{1000}\\ D = 87.31\hspace{0.05cm}mm$

For $\tau \lt 230 MPa\\ \frac{T}{J} = \frac{\tau}{R}\\ \frac{3.98\hspace{0.05cm}\times\hspace{0.05cm}10^6}{\frac{\pi}{32}D^4} = \frac{230}{\frac{D}{2}}\\ D = 44.50\hspace{0.05cm}mm$

Accordig to the life of shaft selecting D = 87.31\hspace{0.05cm}mm as the shaft diameter.

B] For Hollow shaft

$d = \frac{3}{4}D = 0.75 D\\ J = \frac{\pi}{32}[D^4 - d^4] = \frac{\pi}{32}[D^4 - (0.75D)^4]\\ J = 0.067D^4$

for $\theta = 0.5^\circ = 8.72\hspace{0.05cm}\times\hspace{0.05cm}10^{-3}\\ \frac{T}{J} = \frac{G\theta}{L}\\ \frac{3.98\hspace{0.05cm}\times\hspace{0.05cm}10^6}{0.067 D^4} = \frac{80\hspace{0.05cm}\times\hspace{0.05cm}10^3\hspace{0.05cm}\times\hspace{0.05cm}8.72\hspace{0.05cm}\times\hspace{0.05cm}10^{-3}}{1000}\\ D = 96.06\hspace{0.05cm}mm\\ d = \frac{3}{4}D = 72.03\hspace{0.05cm}mm$

For $\tau = 230\hspace{0.05cm}MPa\\ \frac{T}{J} = \frac{\tau}{R}\\ \frac{3.98\hspace{0.05cm}\times\hspace{0.05cm}10^6}{0.067 D^4} = \frac{230}{\frac{D}{2}}\\ D = 50.54\hspace{0.05cm}mm\\ d = \frac{3}{4}D = 37.94\hspace{0.05cm}mm$