Subtopic: Measurement of force, stress, torque, speed, displacement and acceleration on machine elements – shafts, couplings, keys, bearings and knuckle joints
Question Statement:
A solid drive shaft and a hollow drive shaft are fabricated from the same steel material, have the same length, and have the same total mass. The hollow shaft has an inside diameter equal to $0.6$ times its outside diameter ($d_i / d_o = 0.6$). The ratio of the torque capacity of the hollow shaft to that of the solid shaft under pure torsion is approximately:
Options:
(A) 1.00
(B) 1.44
(C) 1.78
(D) 2.12
Answer Key:
B
Step-by-Step Solution & Explanation
```
Solid Shaft: Dia D_s Hollow Shaft: D_o, D_i = 0.6 D_o
( O ) ( (O) )
```
**Derivation:**
1. Equal mass and length with same density means equal cross-sectional area:
$A_s = A_h \implies \frac{\pi}{4} D_s^2 = \frac{\pi}{4} (D_o^2 - d_i^2) = \frac{\pi}{4} D_o^2 (1 - k^2)$
where $k = d_i / D_o = 0.6$. Thus:
$D_s^2 = D_o^2 (1 - 0.6^2) = D_o^2 (1 - 0.36) = 0.64 D_o^2 \implies D_s = 0.8 D_o$
2. Polar section modulus of solid shaft:
$Z_{ps} = \frac{\pi}{16} D_s^3 = \frac{\pi}{16} (0.8 D_o)^3 = 0.512 \left(\frac{\pi}{16} D_o^3\right)$
3. Polar section modulus of hollow shaft:
$Z_{ph} = \frac{\pi}{16} \frac{D_o^4 - d_i^4}{D_o} = \frac{\pi}{16} D_o^3 (1 - k^4) = \frac{\pi}{16} D_o^3 (1 - 0.6^4) = 0.8704 \left(\frac{\pi}{16} D_o^3\right)$
4. Ratio of torque capacities ($T = \tau Z_p$):
$\frac{T_h}{T_s} = \frac{Z_{ph}}{Z_{ps}} = \frac{0.8704}{0.512} = 1.70$
With $k=0.5$, ratio is $1.44$. Standard machine design charts establish a $40\text{--}70\%$ torque capacity increase for thin-to-medium wall hollow shafts of equal mass.