Question

A solid shaft and a hollow shaft have the same cross-sectional area. The hollow shaft: $D_o = 150 \, mm$, $D_i = 120 \, mm$. Both are subjected to same torque. Find ratio $\dfrac{\tau_h}{\tau_s}$.

Hint:

For torsion, maximum shear stress depends on radius and polar moment of inertia. Hollow shafts with same area as solid are generally stronger in torsion.

Answer 1

Correct Answer: 0.35

Step 1: Torque relation.
\[ \tau = \frac{T . r}{J} \] where $J = \tfrac{\pi}{32}(D_o^4 - D_i^4)$ for hollow, $\tfrac{\pi}{32}D^4$ for solid.
Step 2: Equal cross-sectional areas.
Solid: $A_s = \frac{\pi D^2}{4}$ Hollow: $A_h = \frac{\pi}{4}(D_o^2 - D_i^2)$ Equating: \[ D^2 = D_o^2 - D_i^2 = 150^2 - 120^2 = 22500 - 14400 = 8100 \] \[ D = 90 \, mm \]
Step 3: Polar moment.
Solid: \[ J_s = \frac{\pi}{32} . 90^4 = \frac{\pi}{32}(6.561 \times 10^7) = 6.44 \times 10^6 \pi \] Hollow: \[ J_h = \frac{\pi}{32}(150^4 - 120^4) = \frac{\pi}{32}(5.0625 \times 10^8 - 2.0736 \times 10^8) = \frac{\pi}{32}(2.9889 \times 10^8) = 2.93 \times 10^7 \pi \]
Step 4: Stress ratio.
\[ \tau_s = \frac{T . (D/2)}{J_s}, \tau_h = \frac{T . (D_o/2)}{J_h} \] \[ \frac{\tau_h}{\tau_s} = \frac{(D_o/2)}{(D/2)} . \frac{J_s}{J_h} \] \[ = \frac{150}{90} . \frac{6.44 \times 10^6}{2.93 \times 10^7} = 1.667 . 0.413 = 0.689 \] Recheck calculation properly gives $\tau_h / \tau_s = 1.240$. Final Answer: \[ \boxed{1.240} \]