A 1 m long massless cantilever beam oscillates at 2 Hz, while a 60 kg mass is attached at the tip of it. The flexural rigidity of the beam (in kN-m$^2$) is _________________ (accurate to two decimal places).
Official Answer Key:
3.15
Step-by-Step Solution & Explanation
### **Detailed Step-by-Step Solution (GATE AE 2018 Q.60 / AE Q.50)**
**1. Intuitive Physical Setup & Given Parameters**:
- Length of cantilever beam: $L = 1\text{ m}$
- Tip mass: $m = 60\text{ kg}$
- Natural frequency: $f = 2\text{ Hz} = 2\text{ cycles/s}$
- Angular natural frequency: $\omega_n = 2\pi f = 2\pi (2) = 4\pi\text{ rad/s} \approx 12.5664\text{ rad/s}$
**2. Step-by-Step Mathematical Derivation**:
1. **Equivalent Tip Stiffness of Cantilever Beam ($k$)**:
From Euler-Bernoulli beam theory:
$k = \frac{3EI}{L^3}$
2. **Natural Frequency of Single Degree of Freedom System**:
$\omega_n = \sqrt{\frac{k}{m}} = \sqrt{\frac{3EI}{m L^3}}$
3. **Solving for Flexural Rigidity ($EI$)**:
$\omega_n^2 = \frac{3EI}{m L^3} \implies EI = \frac{m L^3 \omega_n^2}{3} = \frac{m L^3 (4\pi)^2}{3}$
4. **Numerical Evaluation**:
$EI = \frac{60 \times (1)^3 \times 16 \pi^2}{3} = 20 \times 16 \pi^2 = 320 \pi^2\text{ N m}^2$
Using $\pi^2 \approx 9.8696$:
$EI = 320 \times 9.8696 = 3158.27\text{ N m}^2 = 3.158\text{ kN m}^2$
Rounding to two decimal places:
$EI \approx 3.16\text{ kN m}^2$
This lies squarely within the official acceptable range **[3.10, 3.20]**.
**3. Comprehensive Options & Distractor Analysis**:
- **Question Type**: Numerical Answer Type (NAT).
- **Official Range**: **3.10 to 3.20 kN-m$^2$**.
**4. Official Master Key & Exam Takeaway**:
- **Official Answer**: **3.15** (Acceptable Range: **3.10 to 3.20**)
- **Key Formula**: $EI = \frac{m L^3 (2\pi f)^2}{3}$.
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