Question 9.12: A solid shaft of diameter d is used in power transmission. D...

Step I Torque transmitted by solid shaft
From Eq. (9.3),

\tau=\frac{16 M_{t}}{\pi d^{3}}                 (9.3).

M_{t}=\left(\frac{\pi d^{3}}{16}\right) \tau                   (a).

Step II Torque transmitted by hollow shaft

M_{t}=\left(\frac{J}{r}\right) \tau \quad \text { where } \quad J=\frac{\pi\left(d_{o}^{4}-d_{i}^{4}\right)}{32} .

∴            M_{t}=\frac{\pi\left(d_{o}^{4}-d_{i}^{4}\right)}{16 d_{o}} \tau           (b).

Step III Outer diameter of hollow shaft
The solid and hollow shafts are equally strong in torsion. Equating (a) and (b),

d^{3}=\frac{d_{o}^{4}-d_{i}^{4}}{d_{o}} .

or            d_{i}^{4}=d_{o}^{4}-d^{3} d_{o}                 (c).

Since the weight of hollow shaft per metre length is half of the solid shaft,

\frac{\pi\left(d_{o}^{2}-d_{i}^{2}\right)}{4}=\frac{1}{2} \times \frac{\pi d^{2}}{4} .

or          d_{i}^{2}=d_{o}^{2}-\frac{d^{2}}{2} .

∴                d_{i}^{4}=\left(d_{o}^{2}-\frac{d^{2}}{2}\right)^{2} .

\text { Eliminating the term }\left(d_{i}^{4}\right) \text { from Eqs (c) and (d) } ,

d_{o}^{4}-d^{3} d_{o}=\left(d_{o}^{2}-\frac{d^{2}}{2}\right)^{2} .

\therefore \quad d_{o}^{4}-d^{3} d_{o}=d_{o}^{4}-d_{o}^{2} d^{2}+\frac{d^{4}}{4} .

or            d_{o}^{2} d^{2}-d_{o} d^{3}-\frac{d^{4}}{4}=0             (e).

Equation (e) is a quadratic equation and the roots are given by,

d_{o}=\frac{d^{3} \pm \sqrt{d^{6}-4\left(d^{2}\right)\left(-\frac{d^{4}}{4}\right)}}{2 d^{2}}=\frac{d \pm \sqrt{2} d}{2} .

Taking positive root,

d_{o}=\left(\frac{d+\sqrt{2} d}{2}\right) .