A 1 m long rod is to be designed to support an axial tensile load P (P >> weight of
the rod). The material for the rod is to be chosen from one of the four provided in
the table. Using strength-based failure criterion for design, which material results
in the lowest weight of the rod?
Properties Material Material Material Material
Density (kg/m3) 2700 4500 7800 9000
Young's modulus (GPa) 70 115 200 130
Yield Strength (MPa) 270 900 520 540
Options:
(A) Material
(B) Material
(C) Material
(D) Material
Official Answer Key:
B
Step-by-Step Solution & Explanation
### **Detailed Step-by-Step Solution (GATE AE 2025 Q.37)**
**1. Intuitive Physical Setup & Given Parameters**:
- A 1 m long rod is designed to support an axial tensile load $P$ ($P \gg \text{weight}$).
- Design criterion: Strength-based failure (Yield Strength $\sigma_y$).
- Material options:
- Material $\alpha$: $\rho = 2700\text{ kg/m}^3$, $\sigma_y = 270\text{ MPa}$
- Material $\beta$: $\rho = 4500\text{ kg/m}^3$, $\sigma_y = 900\text{ MPa}$
- Material $\gamma$: $\rho = 7800\text{ kg/m}^3$, $\sigma_y = 520\text{ MPa}$
- Material $\delta$: $\rho = 9000\text{ kg/m}^3$, $\sigma_y = 540\text{ MPa}$
We need to determine which material results in the **lowest rod weight**.
**2. Step-by-Step Material Selection Derivation**:
1. **Formulate Rod Weight as a Function of Load**:
- For an axial tensile load $P$, the minimum required cross-sectional area $A$ is:
$\sigma = \frac{P}{A} \le \sigma_y \implies A = \frac{P}{\sigma_y}$
- The total mass (weight) of the rod of length $L$ is:
$m = \rho \times A \times L = \rho \left(\frac{P}{\sigma_y}\right) L = (P L) \left(\frac{\rho}{\sigma_y}\right)$
- To minimize mass $m$ for fixed $P$ and $L$, we must **minimize the ratio $\frac{\rho}{\sigma_y}$** (or maximize the specific strength $\frac{\sigma_y}{\rho}$).
2. **Compute $\frac{\rho}{\sigma_y}$ for Each Material**:
- Material $\alpha$:
$\frac{\rho}{\sigma_y} = \frac{2700\text{ kg/m}^3}{270\text{ MPa}} = 10.0\text{ (arbitrary units)}$
- Material $\beta$:
$\frac{\rho}{\sigma_y} = \frac{4500\text{ kg/m}^3}{900\text{ MPa}} = 5.0\text{ (arbitrary units)}$
- Material $\gamma$:
$\frac{\rho}{\sigma_y} = \frac{7800\text{ kg/m}^3}{520\text{ MPa}} = 15.0\text{ (arbitrary units)}$
- Material $\delta$:
$\frac{\rho}{\sigma_y} = \frac{9000\text{ kg/m}^3}{540\text{ MPa}} = 16.67\text{ (arbitrary units)}$
3. **Compare**:
Material $\beta$ has the lowest value ($\frac{\rho}{\sigma_y} = 5.0$), yielding half the weight of aluminum (Material $\alpha$) and one-third the weight of steel (Material $\gamma$).
Therefore, **Material $\beta$** results in the lowest weight.
**3. Comprehensive Options & Distractor Analysis**:
- **(A) Material **: ❌ **INCORRECT**. $\rho/\sigma_y = 10.0$ (standard aluminum alloy).
- **(B) Material **: ✅ **CORRECT**. $\rho/\sigma_y = 5.0$ (titanium alloy), lowest mass per unit tensile load.
- **(C) Material **: ❌ **INCORRECT**. $\rho/\sigma_y = 15.0$ (structural steel).
- **(D) Material **: ❌ **INCORRECT**. $\rho/\sigma_y = 16.67$ (nickel/copper alloy).
**4. Official Master Key & Exam Takeaway**:
- **Official Answer Key**: **B**
- **Exam Takeaway**: For tensile tie rods, the Ashby material performance index to maximize is specific strength $M = \sigma_y / \rho$. Titanium alloys ($\beta$) excel with $900 / 4500 = 0.20\text{ MPa}\cdot\text{m}^3/\text{kg}$.
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