Determine suitable diameter of a solid shaft transmit 1 MW power rotating at 220 RPM if the following working conditions are to be satisfied- a) The shaft must not twist more than 1 degree on length of 12 times the shaft diameter and b) The shear stress must not exceed 60 N/mm^2

$Data: Power P= 1M \hspace{1mm}W $ $N= 220 \hspace{1mm} RPM (revolution \hspace{1mm} per \hspace{1mm} minute) $ $\theta = 1 ^ \circ (angle \hspace{1mm} of \hspace{1mm} twist)$ $Length (l) \hspace{1mm} 12\hspace{1mm} times \hspace{1mm} the \hspace{1mm} shaft$ $diameter (d) $

$i.e l=12d$

$Shear \hspace{1mm} stress \hspace{1mm} \tau=60N/mm^2$

$G=84 KN/mm^2$

$G=84*10^3 N/mm^2 (modulus \hspace{1mm} of \hspace{1mm} rigidity)$

1. Power transmitted by shaft is,

$P= \frac{2 \pi NT}{60}$

$1*10^6=\frac{2 \pi *220*T}{60}$

$T=43405.89 Nm = 43405.89*10^3 Nmm$

1. For strength:

$\frac{T}{J_p}=\frac{\tau}{r}$

$\frac{43405.89*10^3}{\frac{\pi d^4}{32}} = \frac{60}{\frac{d}{2}}$

1. For stiffness:

$\frac{T}{J_p}=\frac{G\theta}{l}$

$\frac{43405.89*10^3}{\frac{\pi d^4}{32}} = \frac{84*10^3*(1*\frac{\pi}{180})}{12d}$

$d=153.52mm$

Safe diameter=greater of two values =154.45mm

Note:

The angle twist (θ) always in radian.

N in RPM.

Torque (T) in Power formula always be in Nm.