Let's break down the question shown in the image: **Question:** A 1.5-meter-long brass round bar is designed to absorb the energy generated by impact loading at a tension of 50 Nm. Brass developed 200 MPa of stress. Calculate the diameter of the bar. Given that the elastic modulus of brass is 115 GPa. It is assumed that there is no yield in the round rod. Let's go step by step: ### Given Data: - Length of bar, \( L = 1.5 \) m - Energy absorbed, \( U = 50 \) Nm (Joules) - Stress developed, \( \sigma = 200 \) MPa \( = 200 \times 10^6 \) Pa - Elastic modulus of brass, \( E = 115 \) GPa \( = 115 \times 10^9 \) Pa ### Find: Diameter of the bar (\( d \)) --- ### Step 1: Strain Energy in a Bar The energy stored in a bar under axial load (strain energy) is given by: \[ U = \frac{\sigma^2}{2E} \times \text{Volume} \] The volume of the bar \( V = A \times L \) where \( A \) is the cross-sectional area. \[ U = \frac{\sigma^2}{2E} \times A \times L \] --- ### Step 2: Rearrange to Solve for Area \[ A = \frac{2EU}{\sigma^2 L} \] --- ### Step 3: Substitute Values - \( U = 50 \) Nm \( = 50 \) Joules - \( E = 115 \times 10^9 \) Pa - \( \sigma = 200 \times 10^6 \) Pa - \( L = 1.5 \) m \[ A = \frac{2 \times 115 \times 10^9 \times 50}{(200 \times 10^6)^2 \times 1.5} \] --- ### Step 4: Calculate First, let's calculate the denominator: \[ (200 \times 10^6)^2 = 4 \times 10^{16} \] \[ 4 \times 10^{16} \times 1.5 = 6 \times 10^{16} \] Now, the numerator: \[ 2 \times 115 \times 10^9 \times 50 = 11,500 \times 10^9 = 1.15 \times 10^{13} \] So, \[ A = \frac{1.15 \times 10^{13}}{6 \times 10^{16}} \] \[ A = 1.9167 \times 10^{-4} \text{ m}^2 \] --- ### Step 5: Area to Diameter Area of a circle \( A = \frac{\pi d^2}{4} \): \[ \frac{\pi d^2}{4} = 1.9167 \times 10^{-4} \] \[ d^2 = \frac{4 \times 1.9167 \times 10^{-4}}{\pi} \] \[ d^2 = \frac{7.6668 \times 10^{-4}}{3.1416} \] \[ d^2 = 2.44 \times 10^{-4} \] \[ d = \sqrt{2.44 \times 10^{-4}} \] \[ d \approx 0.0156 \text{ m} = 15.6 \text{ mm} \] --- ### Step 6: Match Closest Option The closest option is **14.20 mm**. --- ### **Final Answer:** \[ \boxed{14.20 \text{ mm}} \]