Subtopic: Polar Section Modulus & Factor of Safety
Question Statement:
A solid circular shaft is designed to transmit a torque $T$ with a factor of safety of 2. It is proposed to replace the solid shaft by a hollow shaft of the same material and identical outer radius. If the inner radius is half the outer radius, the factor of safety for the hollow shaft is _________ (round off to 1 decimal place).
Official Answer Key:
1.85
Step-by-Step Solution & Explanation
### **Detailed Step-by-Step Solution (GATE AE 2019 Q.61 / AE Q.51)**
**1. Intuitive Physical Setup & Given Parameters**:
- Solid shaft of radius $R$: factor of safety $\text{FOS}_{solid} = 2.0$.
- Hollow shaft of identical material (same yield shear stress $\tau_{yield}$) and identical outer radius $R_o = R$.
- Inner radius: $R_i = \frac{R_o}{2} = 0.5 R_o$.
**2. Step-by-Step Torsion Derivation**:
1. **Torsion Formula & Maximum Shear Stress**:
$\tau_{\max} = \frac{T R_o}{J} = \frac{T}{Z_p}$
where $Z_p = \frac{J}{R_o}$ is the polar section modulus.
2. **Polar Section Modulus for Solid Shaft**:
$Z_{p, solid} = \frac{\pi R^3}{2}$
3. **Polar Section Modulus for Hollow Shaft**:
$J_{hollow} = \frac{\pi}{2}(R_o^4 - R_i^4) = \frac{\pi}{2}\left(R_o^4 - (0.5 R_o)^4\right) = \frac{\pi R_o^4}{2}(1 - 0.0625) = 0.9375 J_{solid}$
$Z_{p, hollow} = \frac{J_{hollow}}{R_o} = 0.9375 Z_{p, solid} = \frac{15}{16} Z_{p, solid}$
4. **Factor of Safety Ratio**:
Factor of safety is $\text{FOS} = \frac{\tau_{yield}}{\tau_{\max}} = \frac{\tau_{yield} Z_p}{T} \propto Z_p$:
$\text{FOS}_{hollow} = \text{FOS}_{solid} \times \left( \frac{Z_{p, hollow}}{Z_{p, solid}} \right) = 2.0 \times \frac{15}{16} = \frac{30}{16} = 1.875$
Rounding to one decimal place:
$\text{FOS}_{hollow} \approx 1.9$
This lies cleanly in the official acceptable range **[1.8, 1.9]**.
**3. Comprehensive Options & Distractor Analysis**:
- **Question Type**: Numerical Answer Type (NAT).
- **Official Range**: **1.8 to 1.9**.
**4. Official Master Key & Exam Takeaway**:
- **Official Answer**: **1.85** (Acceptable Range: **1.8 to 1.9**)
- **Key Formula**: $\text{FOS}_{hollow} = \text{FOS}_{solid} \left(1 - \left(\frac{R_i}{R_o}\right)^4\right) = 2.0 \times (1 - 0.0625) = 1.875$.
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