Subtopic: Cantilever Beam Deflection under Distributed Load
Question Statement:
An aircraft wing is idealized as a cantilever beam of constant width and length $L = 5$ m, carrying a uniformly distributed aerodynamic lift load of $w = 2$ kN/m. If the flexural rigidity of the wing box is $EI = 10^7$ N-m$^2$, the tip deflection (in mm) of the wing is _________ (round off to the nearest integer).
Official Answer Key:
16.0
Step-by-Step Solution & Explanation
### **Detailed Step-by-Step Solution (GATE AE 2017 Q.51)**
**1. Intuitive Physical Setup & Given Parameters**:
- Beam span: $L = 5\text{ m}$
- Uniformly distributed load: $w = 2\text{ kN/m} = 2000\text{ N/m}$
- Flexural rigidity: $EI = 10^7\text{ N m}^2$
**2. Step-by-Step Structural Derivation**:
1. **Euler-Bernoulli Beam Tip Deflection Formula**:
For a cantilever beam subjected to a uniform distributed load $w$ over its entire span:
$\delta_{tip} = \frac{w L^4}{8 EI}$
2. **Substituting Given Numbers**:
$L^4 = 5^4 = 625\text{ m}^4$
$\delta_{tip} = \frac{2000 \times 625}{8 \times 10^7} = \frac{1250000}{8 \times 10^7} = \frac{1.25 \times 10^6}{8 \times 10^7} = \frac{1.25}{80} = 0.015625\text{ m}$
3. **Converting to Millimeters**:
$\delta_{tip} = 0.015625 \times 1000 = 15.625\text{ mm}$
Rounding to the nearest integer:
$\delta_{tip} \approx 16\text{ mm}$
This matches the official acceptable interval **[15, 17] mm**.
**3. Comprehensive Options & Distractor Analysis**:
- **Question Type**: Numerical Answer Type (NAT).
- **Exact Analytical Evaluation**: **15.625 mm** (Acceptable Range: [15, 17]).
**4. Official Master Key & Exam Takeaway**:
- **Official Answer**: **16.0**
- **Key Formula**: $\delta_{tip} = \frac{w L^4}{8 EI}$.
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