Subtopic: Euler Buckling of Columns & Energy Methods
Question Statement:
A uniform symmetric cross-section cantilever beam of length $L$ is subjected to a transverse force $P$ at the free end, as shown in the figure. The Young's modulus of the material is $E$ and the moment of inertia is $I$. Ignoring the contributions due to transverse shear, the strain energy stored in the beam is:
Scientific schematic diagram reproduced from official IIT paper
Options:
(A) $\frac{P^2L^3}{6EI}$
(B) $\frac{PL^3}{3EI}$
(C) $\frac{PL^3}{6EI}$
(D) $\frac{P^2L^3}{3EI}$
Official Answer Key:
A
Step-by-Step Solution & Explanation
### **Detailed Step-by-Step Solution (GATE AE 2025 Q.29)**
**1. Intuitive Physical Setup & Given Parameters**:
- Cantilever beam of length $L$ fixed at $x = 0$ and free at $x = L$.
- Uniform symmetric cross-section with flexural rigidity $EI$.
- Transverse concentrated point force $P$ applied at the free end ($x = L$).
- Transverse shear deformation contributions are ignored (pure Euler-Bernoulli bending).
We need to determine the total strain energy $U$ stored in the beam.
**2. Step-by-Step Structural Mechanics Derivation**:
1. **Bending Moment Distribution**:
Taking coordinate $x$ from the fixed support ($0 \le x \le L$), or coordinate $\xi$ from the free tip ($0 \le \xi \le L$ where $\xi = L - x$):
M(\xi) = P \xi
2. **Strain Energy Formula in Pure Bending**:
The elastic bending strain energy stored in a beam is:
U = \int_0^L \frac{M^2}{2EI} d\xi
3. **Evaluate the Integral**:
U = \int_0^L \frac{(P\xi)^2}{2EI} d\xi = \frac{P^2}{2EI} \int_0^L \xi^2 d\xi
U = \frac{P^2}{2EI} \left[ \frac{\xi^3}{3} \right]_0^L = \frac{P^2}{2EI} \left(\frac{L^3}{3}\right) = \frac{P^2 L^3}{6EI}
4. **Alternative Work-Energy Check**:
By Clapeyron's theorem, the external work done by load $P$ slowly applied is:
W_{\text{ext}} = \frac{1}{2} P \delta_{\text{tip}}
For a cantilever beam with tip load $P$, tip deflection is $\delta_{\text{tip}} = \frac{P L^3}{3EI}$.
U = W_{\text{ext}} = \frac{1}{2} P \left(\frac{P L^3}{3EI}\right) = \frac{P^2 L^3}{6EI}
**3. Comprehensive Options & Distractor Analysis**:
- **(A) P^2 L^3 / (6EI)**: ✅ **CORRECT**. Exactly matches $\int \frac{M^2}{2EI}dx = \frac{1}{2}P\delta_{\text{tip}}$.
- **(B) P L^3 / (3EI)**: ❌ **INCORRECT**. This is the tip deflection $\delta_{\text{tip}}$, having dimensions of length, not energy.
- **(C) P L^3 / (6EI)**: ❌ **INCORRECT**. Dimensional error ($P$ instead of $P^2$).
- **(D) P^2 L^3 / (3EI)**: ❌ **INCORRECT**. Forgets the factor of $1/2$ in strain energy ($P \delta_{\text{tip}}$ instead of $\frac{1}{2}P\delta_{\text{tip}}$).
**4. Official Master Key & Exam Takeaway**:
- **Official Answer Key**: **A**
- **Exam Takeaway**: Elastic strain energy is $U = \frac{1}{2} P \delta = \frac{P^2 L^3}{6EI}$. Always check dimensional consistency: $[P^2 L^3 / EI] = \text{Force} \times \text{Length} = \text{Energy}$.
