Let's break down the problem step by step using the given data: ### **Given Data** - **Length, \( L \)** = 50 in - **Diameter, \( d \)** = 1.3 in - **Yield Stress, \( \sigma_Y \)** = 50 ksi - **Elastic Modulus, \( E \)** - **Post-yield stiffness** = 2.5% of \( E \) - **Yield Strain** = \( \epsilon_Y = \frac{\sigma_Y}{E} = 0.00172 \) (from the graph) - **Cross-sectional Area, \( A \) = \(\frac{\pi d^2}{4}\)** - **1 kip = 1000 lb** Let's solve each part: --- ## **(a) Elongation under \( P = 40 \) kips** Find stress: \[ \sigma = \frac{P}{A} \] \[ A = \frac{\pi}{4} d^2 = \frac{\pi}{4} (1.3)^2 = 1.327 \, \text{in}^2 \] \[ \sigma = \frac{40,000}{1.327} = 30,146 \, \text{psi} = 30.15 \, \text{ksi} \] Since \( 30.15 \) ksi < \( \sigma_Y \), the rod is in the elastic regime. \[ \epsilon = \frac{\sigma}{E} \] From the graph, \( \epsilon_Y = 0.00172 \) at \( \sigma_Y = 50 \) ksi: \[ E = \frac{50}{0.00172} = 29,070 \, \text{ksi} \] \[ \epsilon = \frac{30.15}{29,070} = 0.001037 \] Elongation: \[ \delta = \epsilon L = 0.001037 \times 50 = 0.0518 \, \text{in} \] --- ## **(b) Permanent elongation after loading to 90 kips and unloading** First, calculate the stress at 90 kips: \[ \sigma = \frac{90,000}{1.327} = 67,844 \, \text{psi} = 67.84 \, \text{ksi} \] This exceeds yield, so the rod deforms plastically. ### **Step 1: Find total strain at 67.84 ksi:** - The elastic region goes up to \( \sigma_Y = 50 \) ksi, \( \epsilon_Y = 0.00172 \). - Beyond that, use the post-yield slope (\( E_p = 0.025E \)): \[ E_p = 0.025 \times 29,070 = 726.75 \, \text{ksi} \] \[ \epsilon = \epsilon_Y + \frac{\sigma - \sigma_Y}{E_p} \] \[ \epsilon = 0.00172 + \frac{67.84-50}{726.75} = 0.00172 + \frac{17.84}{726.75} = 0.00172 + 0.02455 = 0.02627 \] Total elongation at max load: \[ \delta_{total} = \epsilon \times L = 0.02627 \times 50 = 1.314 \, \text{in} \] ### **Step 2: When unloaded, only elastic strain remains:** Elastic strain at yield: \[ \epsilon_{elastic} = \frac{\sigma_{unload}}{E} = \frac{0}{E} = 0 \] But the *residual* elastic strain will be that corresponding to the yield stress \( \sigma_Y \), because when unloaded, the rod follows the original elastic modulus back down to zero stress, from the maximum strain. \[ \epsilon_{residual} = \epsilon_{max} - \frac{\sigma_Y}{E} \] \[ \epsilon_{plastic} = \epsilon_{max} - \frac{\sigma_Y}{E} \] \[ = 0.02627 - 0.00172 = 0.02455 \] Permanent elongation: \[ \delta_{perm} = \epsilon_{plastic} \cdot L = 0.02455 \times 50 = 1.227 \, \text{in} \] --- ## **(c) Permanent elongation after elongating to 0.15 in and unloading** Maximum total strain: \[ \delta_{max} = 0.15 \, \text{in} \] \[ \epsilon_{max} = \frac{\delta_{max}}{L} = \frac{0.15}{50} = 0.003 \] This is above yield (\( \epsilon_Y = 0.00172 \)), so it enters the plastic region. ### **Step 1: Find the max stress** \[ \epsilon_{max} = \epsilon_Y + \frac{\sigma_{max} - \sigma_Y}{E_p} \] \[ 0.003 = 0.00172 + \frac{\sigma_{max} - 50}{726.75} \] \[ 0.003 - 0.00172 = \frac{\sigma_{max} - 50}{726.75} \] \[ 0.00128 \times 726.75 = \sigma_{max} - 50 \] \[ 0.930 = \sigma_{max} - 50 \] \[ \sigma_{max} = 50.93 \, \text{ksi} \] ### **Step 2: Permanent strain after unloading** Unloading follows the original elastic slope \( E \): \[ \epsilon_{perm} = \epsilon_{max} - \frac{\sigma_{max}}{E} \] \[ = 0.003 - \frac{50.93}{29,070} = 0.003 - 0.001752 = 0.00125 \] Permanent elongation: \[ \delta_{perm} = \epsilon_{perm} \times L = 0.00125 \times 50 = 0.0625 \, \text{in} \] --- ## **Summary of Answers** **(a)** Elongation under 40 kip load: \[ \boxed{0.0518 \, \text{in}} \] **(b)** Permanent elongation after 90 kip load and unload: \[ \boxed{1.23 \, \text{in}} \] **(c)** Permanent elongation after 0.15 in elongation and unload: \[ \boxed{0.0625 \, \text{in}} \] Let me know if you want detailed step-by-step calculations for any part!