A batch of aluminium alloy yields in uniaxial tension at the stress $\sigma_y = 330$ MPa. Using the von Mises yield criterion, the yield stress in pure shear (in MPa) is _________ (round off to 1 decimal place).
Official Answer Key:
190.5
Step-by-Step Solution & Explanation
### **Detailed Step-by-Step Solution (GATE AE 2017 Q.41)**
**1. Intuitive Physical Setup & Given Parameters**:
- Uniaxial yield stress under tension: $\sigma_y = 330\text{ MPa}$.
- Yield criterion: Distortion Energy Theory (von Mises criterion).
- Target: Yield stress under pure shear $\tau_y$.
**2. Step-by-Step Mathematical Derivation**:
1. **von Mises Yield Criterion Formulation**:
Under a general 3D stress state, yielding occurs when the equivalent stress $\sigma_{vm}$ reaches the uniaxial yield strength $\sigma_y$:
$\sigma_{vm} = \sqrt{\frac{1}{2}\left((\sigma_1 - \sigma_2)^2 + (\sigma_2 - \sigma_3)^2 + (\sigma_3 - \sigma_1)^2\right)} = \sigma_y$
2. **State of Pure Shear**:
In pure shear of magnitude $\tau$, the principal stresses are:
$\sigma_1 = \tau, \quad \sigma_2 = -\tau, \quad \sigma_3 = 0$
3. **Evaluating Equivalent Stress in Pure Shear**:
$\sigma_{vm} = \sqrt{\frac{1}{2}\left((\tau - (-\tau))^2 + (-\tau - 0)^2 + (0 - \tau)^2\right)} = \sqrt{\frac{1}{2}\left((2\tau)^2 + \tau^2 + \tau^2\right)} = \sqrt{\frac{1}{2}(4\tau^2 + 2\tau^2)} = \sqrt{3\tau^2} = \sqrt{3}\tau$
4. **Solving for Shear Yield Stress $\tau_y$**:
$\sqrt{3}\tau_y = \sigma_y \implies \tau_y = \frac{\sigma_y}{\sqrt{3}}$
5. **Numerical Substitution**:
$\tau_y = \frac{330}{\sqrt{3}} = \frac{330}{1.73205} \approx 190.526\text{ MPa}$
Rounding to one decimal place:
$\tau_y \approx 190.5\text{ MPa}$
Official acceptable range: **[189.0, 192.0] MPa**.
**3. Comprehensive Options & Distractor Analysis**:
- **Question Type**: Numerical Answer Type (NAT).
- **Exact Analytical Result**: **$330/\sqrt{3} \approx 190.5$ MPa** (Acceptable Range: [189.0, 192.0]).
**4. Official Master Key & Exam Takeaway**:
- **Official Answer**: **190.5**
- **Exam Rule**: According to von Mises: $\tau_y = \frac{\sigma_y}{\sqrt{3}} \approx 0.577 \sigma_y$; according to Tresca: $\tau_y = 0.5 \sigma_y$.
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