Topic: Strength of Materials
A solid shaft of diameter 'D' undergoes a twisting moment that develops maximum shear stress '$\tau$'. If the shaft is replaced by a hollow one of outside diameter 'D' and inside diameter D/2, then the maximum shear stress will be
- A. $1.067\tau$
- B. $1.143\tau$
- C. $1.33\tau$
- D. $2\tau$
Answer: A. $1.067\tau$
Explanation: For a solid shaft, $\tau = \frac{16T}{\pi D^3}$.
For a hollow shaft with $D_o = D$ and $D_i = D/2$, polar moment of inertia $J_h = \frac{\pi}{32}[D^4 - (D/2)^4] = \frac{\pi}{32}D^4(1 - 1/16) = \frac{15\pi D^4}{512}$.
Maximum shear stress in hollow shaft, $\tau_h = \frac{T (D/2)}{J_h} = \frac{T D / 2}{\frac{15\pi D^4}{512}} = \frac{256T}{15\pi D^3}$.
Ratio $\frac{\tau_h}{\tau} = \frac{\frac{256T}{15\pi D^3}}{\frac{16T}{\pi D^3}} = \frac{256}{15 \times 16} = \frac{16}{15} \approx 1.067$.
So, $\tau_h = 1.067\tau$.