Subtopic: Stress and Strain Tensors & Generalized Hooke's Law
Question Statement:
A $10\text{ kN}$ axial load is applied eccentrically on a rod of square cross-section ($1\text{ cm} \times 1\text{ cm}$) as shown in the figure. The strains measured by the two strain gages attached to the top and bottom surfaces at a distance of $0.5\text{ m}$ from the tip are $\epsilon_1 = 0.0016$ and $\epsilon_2 = 0.0004$, respectively. The eccentricity in loading, $e$, is ______ $\text{mm}$.
Official Answer Key:
1.0
Step-by-Step Solution & Explanation
### **Detailed Step-by-Step Solution (GATE AE 2022 Q.60)**
**1. Combined Axial and Bending Strain**:
$$\epsilon_1 = \frac{P}{AE} + \frac{P e (c)}{I E}, \quad \epsilon_2 = \frac{P}{AE} - \frac{P e (c)}{I E}$$
Sum and difference:
$$\epsilon_1 - \epsilon_2 = \frac{2 P e c}{I E}, \quad \epsilon_1 + \epsilon_2 = \frac{2P}{AE}$$
Dividing the two relations:
$$\frac{\epsilon_1 - \epsilon_2}{\epsilon_1 + \epsilon_2} = \frac{e c A}{I}$$
For square section of side $a = 1\text{ cm} = 10\text{ mm}$:
$$c = \frac{a}{2}, \quad I = \frac{a^4}{12}, \quad A = a^2 \implies \frac{c A}{I} = \frac{a/2 \cdot a^2}{a^4/12} = \frac{6}{a}$$
$$\frac{0.0016 - 0.0004}{0.0016 + 0.0004} = \frac{0.0012}{0.0020} = 0.6 = \frac{6 e}{10\text{ mm}} \implies e = 1.0\text{ mm}$$
- **Official Answer Key**: **1.0**