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 calculate the minimum diameter, we start by determining the maximum allowable stress that the material can withstand, considering the safety factor. The yield stress is \( 540 \, \text{MN/m}^2 \), and with a safety factor of 2, the maximum allowable stress (σ_allow) is: \[ \sigma_{\text{allow}} = \frac{\text{yield stress}}{\text{safety factor}} = \frac{540 \, \text{MN/m}^2}{2} = 270 \, \text{MN/m}^2 \] Next, we use σ_allow to calculate the cross-sectional area (A) required to support the axial load (F) of 150 kN: \[ A = \frac{F}{\sigma_{\text{allow}}} = \frac{150 \times 10^3 \, \text{N}}{270 \times 10^6 \, \text{N/m}^2} \approx 0.0005556 \, \text{m}^2 \] Now, since the area A of a circular cross-section is given by \( A = \frac{\pi d^2}{4} \), we can solve for the diameter (d): \[ d^2 = \frac{4A}{\pi} = \frac{4 \times 0.0005556}{\pi} \approx 0.000707 \, \text{m}^2 \] \[ d \approx \sqrt{0.000707} \approx 0.0266 \, \text{m} \approx 26.6 \, \text{mm} \] Therefore, the minimum diameter of the bar to support the load, considering the safety factor, is (A) 26.6 mm.