For the state of stress shown in the figure, the normal stress, $\sigma_n$, on a plane inclined at 45 degrees to the principal planes with $\sigma_1 = 300$ MPa and $\sigma_2 = 200$ MPa is _________ MPa (round off to the nearest integer).
Official Answer Key:
250.0
Step-by-Step Solution & Explanation
### **Detailed Step-by-Step Solution (GATE AE 2019 Q.34 / AE Q.24)**
**1. Intuitive Physical Setup & Given Parameters**:
- Principal stresses: $\sigma_1 = 300\text{ MPa}$, $\sigma_2 = 200\text{ MPa}$.
- Inclination angle to the principal plane: $\theta = 45^\circ$.
**2. Step-by-Step Mathematical Derivation**:
1. **Stress Transformation Equation**:
The normal stress on an inclined plane at angle $\theta$ is:
$\sigma_n = \frac{\sigma_1 + \sigma_2}{2} + \frac{\sigma_1 - \sigma_2}{2}\cos(2\theta)$
2. **Evaluating at $\theta = 45^\circ$**:
Since $\theta = 45^\circ$, $2\theta = 90^\circ$, and $\cos(90^\circ) = 0$:
$\sigma_n = \frac{\sigma_1 + \sigma_2}{2} + 0 = \frac{300 + 200}{2} = \frac{500}{2} = 250\text{ MPa}$
3. On the plane of maximum shear stress (which occurs at $45^\circ$ to principal planes), the normal stress is always equal to the average normal stress:
$\sigma_{avg} = \frac{\sigma_1 + \sigma_2}{2} = 250\text{ MPa}$
This falls right in the middle of the official acceptable range **[247, 255]**.
**3. Comprehensive Options & Distractor Analysis**:
- **Question Type**: Numerical Answer Type (NAT).
- **Official Range**: **247 to 255 MPa**.
**4. Official Master Key & Exam Takeaway**:
- **Official Answer**: **250.0** (Range: **247 to 255**)
- **Key Formula**: On maximum shear planes ($\theta = 45^\circ$), $\sigma_n = \frac{\sigma_1 + \sigma_2}{2}$.
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