Two cantilever beams (Beam 1 and Beam 2) are made of the same homogeneous isotropic material and have identical length $L$. Beam 1 has a solid circular cross-section of radius $R$, while Beam 2 has a hollow circular cross-section with outer radius $R$ and inner radius $R/2$. If identical tip loads $P$ are applied, the ratio of the tip deflection of Beam 2 to that of Beam 1 ($\delta_2 / \delta_1$) is:
Options:
(A) 16/15
(B) 15/16
(C) 4/3
(D) 2
Official Answer Key:
A
Step-by-Step Solution & Explanation
### **Detailed Step-by-Step Solution (GATE AE 2021 Q.20 / AE Q.10)**
**1. Intuitive Physical Setup & Given Parameters**:
- Both beams: cantilever, length $L$, modulus $E$, tip load $P$.
- Beam 1: Solid circular cross-section, radius $R$.
- Beam 2: Hollow circular cross-section, outer radius $R_o = R$, inner radius $R_i = R/2$.
**2. Step-by-Step Structural Mechanics Derivation**:
1. **Tip Deflection of Cantilever Beam**:
\delta = \frac{P L^3}{3 E I} \implies \delta \propto \frac{1}{I}
\frac{\delta_2}{\delta_1} = \frac{I_1}{I_2}
2. **Area Moment of Inertia for Beam 1 (Solid)**:
I_1 = \frac{\pi R^4}{4}
3. **Area Moment of Inertia for Beam 2 (Hollow)**:
I_2 = \frac{\pi}{4}(R_o^4 - R_i^4) = \frac{\pi}{4}\left(R^4 - \left(\frac{R}{2}\right)^4\right) = \frac{\pi}{4} R^4 \left(1 - \frac{1}{16}\right) = \frac{15}{16} \left(\frac{\pi R^4}{4}\right) = \frac{15}{16} I_1
4. **Evaluating Deflection Ratio**:
\frac{\delta_2}{\delta_1} = \frac{I_1}{I_2} = \frac{I_1}{\frac{15}{16} I_1} = \frac{16}{15}
**3. Comprehensive Options & Distractor Analysis**:
- **(A)** 16/15: ✅ **CORRECT**. Exactly equals $1 / (15/16) = 16/15 \approx 1.067$.
- **(B)** 15/16: ❌ Incorrect. This is the ratio of stiffnesses $I_2/I_1$, not deflections.
- **(C)** 4/3: ❌ Incorrect.
- **(D)** 2: ❌ Incorrect.
**4. Official Master Key & Exam Takeaway**:
- **Official Answer**: **A**
- **Exam Rule**: Deflection is inversely proportional to moment of inertia: $\frac{\delta_{hollow}}{\delta_{solid}} = \frac{I_{solid}}{I_{hollow}} = \frac{1}{1 - (R_i/R_o)^4}$.