Solution :
Given:
Power, P = 200 hp
= 110000 lbf ft/s
Speed, N = 3600 rpm
Power , [tex]$P=\frac{2 \pi NT}{60}$[/tex]
∴ Torque, [tex]$T = \frac{60 P}{2 \pi N}$[/tex]
[tex]$T = \frac{60 \times 110000}{2 \pi \times 3600}$[/tex]
T = 291.93 lb -ft
[tex]$T_{allow} = 20,250 \ psi$[/tex] ---- allowable shear stress
Therefore, torque , T = 391.93 lb-ft
T = 291.93 x 12 lb-in
∴ T 3503.16 lb-in
Now from the Torsion formula,
[tex]$\frac{T}{J}=\frac{T_{allow}}{ r}$[/tex]
Here, [tex]$r=\frac{d}{2} ; \ \ \ J=\frac{\pi d^4}{32}$[/tex] ---- polar moment of inertia.
∴ [tex]$\frac{T}{\frac{\pi d^4}{32}}=\frac{T_{allow}}{ d/2}$[/tex]
[tex]$T_{allow}=\frac{16 T}{\pi d^3}$[/tex]
[tex]$\pi d^3=\frac{16 \times 3503.16}{20520}$[/tex]
[tex]$\pi d^3 = 2.7315$[/tex]
[tex]$d^3 = 0.8699$[/tex]
or [tex]$d= (0.8699)^{1/3}$[/tex]
d = 0.9546 in
Therefore the diameter of the shaft is 0.9546 in.
