# Step-by-Step Solution Let's break down the problem: ## **Given Data** - Length of rod, \( L = 8 \) in - Diameter of rod, \( d = .5 \) in - Axial load, \( P = 60 \) lb - Modulus of elasticity, \( E = 400 \) ksi \( = 400,000 \) psi - Poisson's ratio, \( \nu = .4 \) --- ## **Step 1: Find the Change in Length (\( \Delta L \))** The change in length due to axial load is given by: \[ \Delta L = \frac{PL}{AE} \] where: - \( A \) = Cross-sectional area of the rod For a circular rod: \[ A = \frac{\pi d^2}{4} \] \[ A = \frac{\pi (.5^2)}{4} = \frac{\pi \times .25}{4} = \frac{.25\pi}{4} = .19635 \text{ in}^2 \] Now, substitute values: \[ \Delta L = \frac{60 \times 8}{.19635 \times 400,000} \] \[ \Delta L = \frac{480}{78,540} \] \[ \Delta L = .00611 \text{ in} \] --- ## **Step 2: Find the Change in Diameter (\( \Delta d \))** Poisson's ratio relates the lateral strain to the axial strain: \[ \nu = -\frac{\text{Lateral Strain}}{\text{Axial Strain}} \] Axial strain, \( \epsilon = \frac{\Delta L}{L} \) Lateral strain = \( -\nu \epsilon \) Change in diameter: \[ \Delta d = \text{Lateral strain} \times d \] \[ \Delta d = -\nu \left(\frac{\Delta L}{L}\right) d \] Substitute the values: \[ \Delta d = -.4 \left(\frac{.00611}{8}\right) \times .5 \] \[ \Delta d = -.4 \times .000764 \times .5 \] \[ \Delta d = -.4 \times .000382 \] \[ \Delta d = -.000153 \text{ in} \] --- ## **Final Answers** ### **Change in Length** \[ \boxed{.0061 \text{ in}} \] ### **Change in Diameter** \[ \boxed{-.00015 \text{ in}} \] - The rod **elongates by .0061 in**. - The diameter **decreases by .00015 in**.