A solid round bar supports an axial load of 150 kN . The yield stress of the material is \( 540 \mathrm{MN} / \mathrm{m}^{2} \) Calculate the minimum diameter of the bar to support the load if a safety of factor of 2 is to be considered. (A) 26.6 mm (B) 3.5 mm (C) 15.9 mm (D) 23.2 mm

To determine the minimum diameter of the solid round bar that can support an axial load of 150 kN with a safety factor of 2, we'll follow these steps: ### **Given:** - **Axial Load (P):** 150 kN = 150,000 N - **Yield Stress (σ_yield):** \( 540 \, \text{MN/m}^2 = 540 \times 10^6 \, \text{N/m}^2 \) - **Safety Factor (SF):** 2 ### **Steps:** 1. **Calculate the Allowable Stress:** \[ \sigma_{\text{allow}} = \frac{\sigma_{\text{yield}}}{\text{SF}} = \frac{540 \times 10^6 \, \text{N/m}^2}{2} = 270 \times 10^6 \, \text{N/m}^2 \] 2. **Determine the Required Cross-Sectional Area (A):** The stress (\(\sigma\)) is related to the load and the area by: \[ \sigma = \frac{P}{A} \leq \sigma_{\text{allow}} \] Solving for A: \[ A \geq \frac{P}{\sigma_{\text{allow}}} = \frac{150,000 \, \text{N}}{270 \times 10^6 \, \text{N/m}^2} \approx 5.56 \times 10^{-4} \, \text{m}^2 \] 3. **Relate Area to Diameter:** For a circular cross-section: \[ A = \frac{\pi d^2}{4} \] Solving for \( d \): \[ d \geq \sqrt{\frac{4A}{\pi}} = \sqrt{\frac{4 \times 5.56 \times 10^{-4}}{\pi}} \approx 0.0266 \, \text{m} = 26.6 \, \text{mm} \] ### **Conclusion:** The minimum diameter required to safely support the axial load is **26.6 mm**. **Answer:** 26.6 mm