Question

A solid shaft and a hollow shaft of the same material have same length and outer radius R. The inner radius of the hollow shaft is 0.7R. a) If both the shafts are subjected to the same torque, compare their shear stresses, angle of twist and mass. b) Find the strength to weight ratio for both the shafts.

Answer 1

Answer with Explanation:

By the equation or Torque we have

$\frac{T}{I_{p}}=\frac{\tau }{r}=\frac{G\theta }{L}$

where

T is the torque applied on the shaft

$I_{p}$ is the polar moment of inertia of the shaft

$\tau$ is the shear stress developed at a distance 'r' from the center of the shaft

$\theta$ is the angle of twist of the shaft

'G' is the modulus of rigidity of the shaft

We know that for solid shaft $I_{p}=\frac{\pi R^4}{2}$

For a hollow shaft $I_{p}=\frac{\pi (R_o^4-R_i^4)}{2}$

Since the two shafts are subjected to same torque from the relation of Torque we have

1) For solid shaft

$\frac{2T}{\pi R^4}\times r=\tau _{solid}$

2) For hollow shaft we have

$\tau _{hollow}=\frac{2T}{\pi (R^4-0.7R^4)}\times r=\frac{2T}{\pi 0.76R^4}$

Comparing the above 2 relations we see

$\frac{\tau _{solid}}{\tau _{hollow}}=0.76$

Similarly for angle of twist we can see

$\frac{\theta _{solid}}{\theta _{hollow}}=\frac{\frac{LT}{I_{solid}}}{\frac{LT}{I_{hollow}}}=\frac{I_{hollow}}{I_{solid}}=1.316$

Part b)

Strength of solid shaft = $\tau _{max}=\frac{T\times R}{I_{solid}}$

Weight of solid shaft =$\rho \times \pi R^2\times L$

Strength per unit weight of solid shaft = $\frac{\tau _{max}}{W}=\frac{T\times R}{I_{solid}}\times \frac{1}{\rho \times \pi R^2\times L}=\frac{2T}{\rho \pi ^2R^5L}$

Strength of hollow shaft = $\tau '_{max}=\frac{T\times R}{I_{hollow}}$

Weight of hollow shaft =$\rho \times \pi (R^2-0.7R^2)\times L$

Strength per unit weight of hollow shaft = $\frac{\tau _{max}}{W}=\frac{T\times R}{I_{hollow}}\times \frac{1}{\rho \times \pi (R^2-0.7^2)\times L}=\frac{5.16T}{\rho \pi ^2R^5L}$

Thus $\frac{Strength/Weight _{hollow}}{Strength/Weight _{Solid}}=5.16$