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} = …