Question

A solid shaft has length and diameter ${}^{\prime }{L}_s^{\prime }$ and D respectively. A hollow shaft of length $L_h$, external diameter D, and internal diameter 'd' repectively. Both are of the same material. The ratio of torsional stiffness of hollow shaft to that of solid shaft is

• A. $[1+{\left(\frac{d}{D}\right)}^4].\frac{L_s}{L_h}$
• B. $[1-{\left(\frac{d}{D}\right)}^4].\frac{L_h}{L_s}$
• C. $[1-{\left(\frac{d}{D}\right)}^4].\frac{L_s}{L_h}$
• D. $[1+{\left(\frac{D}{d}\right)}^4].\frac{L_s}{L_h}$

Answer 1

The correct option is C $[1-{\left(\frac{d}{D}\right)}^4].\frac{L_s}{L_h}$

The torsional stiffness of a bar is the torque required to produce a unit angle of rotation. It is denoted by k.

$\therefore \frac{T}{\theta }=k=\frac{G{I}_P}{L}$

$\Rightarrow \frac{k_h}{k_s}=\frac{G{I}_{ph}/L_h}{G{I}_{PS}/L_s}[\because materialissame]$

$\Rightarrow \frac{k_h}{k_s}=\frac{I_{ph}}{I_{ps}}\times \frac{L_s}{L_h}$

$\Rightarrow \frac{k_h}{k_s}=\left[\frac{\frac{\pi }{32}\times (D^4-d^4)}{\frac{pi}{32}\times D^4}\right]\times \frac{L_s}{L_h}$

$\Rightarrow \frac{k_h}{k_s}=\left[\frac{D^4-d^4}{D^4}\right]\times \frac{L_s}{L_h}$

$\Rightarrow \frac{k_h}{h_s}=[1-{\left(\frac{d}{D}\right)}^4]\times \frac{L_s}{L_h}$