GATE AE 2019 • Q62 NAT • 2 Marks Aerospace Structures

Aerospace Structures

Strength of Materials & Elasticity

Subtopic: Stress and Strain Tensors & Generalized Hooke's Law

Question Statement:

In the structure shown in the figure, bars AB and BC are made of identical material and have circular cross-sections of 10 mm radii. The yield stress of the material under uniaxial tension is 280 MPa. Using the von Mises yield criterion, the maximum load (in N) along the z-direction (perpendicular to the plane of paper) that can be applied at C, such that AB does not yield is _________ N (round off to the nearest integer). (Take $AB = 0.55$ m, $BC = 0.50$ m)

Official Answer Key:

315.0

Step-by-Step Solution & Explanation

### **Detailed Step-by-Step Solution (GATE AE 2019 Q.62 / AE Q.52)** **1. Intuitive Physical Setup & Given Parameters**: - Circular bar AB: radius $R = 10\text{ mm} = 0.01\text{ m}$. - Bar lengths: $L_{AB} = 0.55\text{ m}$, $L_{BC} = 0.50\text{ m}$. - Yield strength: $\sigma_y = 280\text{ MPa} = 280 \times 10^6\text{ N/m}^2$. - Force $P$ applied at tip $C$ along $z$-axis. **2. Step-by-Step Structural Mechanics Derivation**: 1. **Load Transfer to Critical Section of Bar AB (at Fixed Support A)**: - Bending Moment: $M = P \times L_{AB} = 0.55 P$ - Torsional Moment: $T = P \times L_{BC} = 0.50 P$ 2. **Maximum Normal and Shear Stresses in Bar AB**: For a solid circular shaft of radius $R$: \sigma_b = \frac{4M}{\pi R^3} = \frac{4(0.55 P)}{\pi R^3} = \frac{2.20 P}{\pi R^3} \tau = \frac{2T}{\pi R^3} = \frac{2(0.50 P)}{\pi R^3} = \frac{1.00 P}{\pi R^3} 3. **von Mises Equivalent Stress ($\sigma_{vm}$)**: \sigma_{vm} = \sqrt{\sigma_b^2 + 3\tau^2} = \frac{P}{\pi R^3} \sqrt{(2.20)^2 + 3(1.00)^2} = \frac{P}{\pi R^3} \sqrt{4.84 + 3.0} = \frac{P}{\pi R^3} \sqrt{7.84} = 2.80 \frac{P}{\pi R^3} 4. **Setting $\sigma_{vm} = \sigma_y$ at Yield Limit**: 2.80 \frac{P}{\pi R^3} = \sigma_y \implies P = \frac{\pi R^3 \sigma_y}{2.80} Substituting numerical values ($R = 0.01\text{ m}$, $\sigma_y = 280 \times 10^6\text{ Pa}$): P = \frac{\pi \times (10^{-6}) \times (280 \times 10^6)}{2.80} = \frac{280\pi}{2.80} = 100\pi \approx 314.16\text{ N} Rounding to the nearest integer: P \approx 314\text{ N} This falls perfectly in the center of the official acceptable interval **[310, 320] N**. **3. Comprehensive Options & Distractor Analysis**: - **Question Type**: Numerical Answer Type (NAT). - **Exact Analytical Result**: **$100\pi \approx 314.16$ N** (Official Range: [310, 320]). **4. Official Master Key & Exam Takeaway**: - **Official Answer**: **315.0** (Acceptable Range: **310 to 320**) - **Key Formula**: $\sigma_{vm} = \sqrt{\sigma_b^2 + 3\tau^2} \le \sigma_y$.
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