Question:easy

A solid shaft of diameter D carries a twisting moment that develops a maximum shear stress $\tau$. If the solid shaft is replaced by a hollow shaft having outside diameter D and inside diameter D/2; then the maximum shear stress in the hollow shaft will be:

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Removing the center of a shaft (making it hollow) has a surprisingly small impact on its strength. [cite: 15] By removing 50
Updated On: Jul 1, 2026
  • $\frac{16}{15} \tau$
  • $\frac{8}{7} \tau$
  • $\frac{4}{3} \tau$
  • $2 \tau$
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The Correct Option is A

Solution and Explanation

1. Solid Shaft Stress: [cite: 23, 25] For a solid shaft of diameter $D$, the polar moment of inertia is $J_s = \frac{\pi D^4}{32}$. [cite: 23, 25] The maximum shear stress $\tau$ at the outer surface ($r = D/2$) is: $$\tau = \frac{T \cdot (D/2)}{\pi D^4 / 32} = \frac{16T}{\pi D^3}$$ [cite: 23, 25]

2. Hollow Shaft Stress: [cite: 23, 25] For the hollow shaft with $D_o = D$ and $D_i = D/2$: $$J_h = \frac{\pi}{32} (D^4 - (D/2)^4) = \frac{\pi}{32} (D^4 - \frac{D^4}{16}) = \frac{\pi D^4}{32} (\frac{15}{16})$$ [cite: 23, 25] The new maximum shear stress $\tau'$ at the same outer radius $r = D/2$ is: $$\tau' = \frac{T \cdot (D/2)}{J_h} = \frac{T \cdot (D/2)}{\frac{\pi D^4}{32} \cdot \frac{15}{16}} = \frac{16T}{\pi D^3} \cdot \frac{16}{15}$$ [cite: 23, 25]

3. Final Comparison: [cite: 23, 25] Substituting the original $\tau$ value: $$\tau' = \frac{16}{15} \tau$$ [cite: 23, 25]
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