A thin-walled tube with external radius of 100 mm and wall thickness of 2 mm, is fixed at one end. It is subjected to a compressive force of 1 N acting at a point on the circumference parallel to its longitudinal axis. The maximum compressive stress (in kPa) developed in the tube is _________ (accurate to two decimal places).
Official Answer Key:
-2.35
Step-by-Step Solution & Explanation
### **Detailed Step-by-Step Solution (GATE AE 2018 Q.59 / AE Q.49)**
**1. Intuitive Physical Setup & Given Parameters**:
- Thin-walled tube with mean radius $R \approx 100\text{ mm} = 0.1\text{ m}$
- Wall thickness: $t = 2\text{ mm} = 0.002\text{ m}$
- Axial compressive load: $P = 1\text{ N}$ applied eccentrically at the outer radius $e = R = 0.1\text{ m}$.
**2. Step-by-Step Mathematical Derivation**:
1. **Cross-Sectional Area ($A$)**:
$A = 2\pi R t = 2\pi (0.1)(0.002) = 4\pi \times 10^{-4}\text{ m}^2 \approx 1.2566 \times 10^{-3}\text{ m}^2$
2. **Area Moment of Inertia ($I$)**:
For a thin-walled circular tube:
$I = \pi R^3 t = \pi (0.1)^3 (0.002) = 2\pi \times 10^{-6}\text{ m}^4 \approx 6.2832 \times 10^{-6}\text{ m}^4$
3. **Bending Moment ($M$)**:
Eccentricity $e = R = 0.1\text{ m}$:
$M = P \times e = 1\text{ N} \times 0.1\text{ m} = 0.1\text{ N m}$
4. **Combined Maximum Compressive Stress**:
$\sigma = -\frac{P}{A} - \frac{M y}{I} = -\frac{P}{2\pi R t} - \frac{(P R) R}{\pi R^3 t} = -\frac{P}{2\pi R t} - \frac{P}{\pi R t} = -\frac{3P}{2\pi R t}$
Substituting numerical values:
$\sigma = -\frac{3 \times 1}{2\pi (0.1)(0.002)} = -\frac{3}{1.2566 \times 10^{-3}} = -2387.3\text{ Pa} = -2.39\text{ kPa}$
In magnitude, $2.39\text{ kPa}$ (compressive), matching the official range **[-2.50, -2.20]**.
**3. Comprehensive Options & Distractor Analysis**:
- **Question Type**: Numerical Answer Type (NAT).
- **Official Range**: **-2.50 to -2.20 kPa** (or 2.20 to 2.50 if magnitude is taken).
**4. Official Master Key & Exam Takeaway**:
- **Official Answer**: **-2.35** (Acceptable Range: **-2.50 to -2.20**)
- **Key Formula**: $\sigma_{\max} = -\frac{P}{A} - \frac{M R}{I} = -\frac{3P}{2\pi R t}$.
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