Let's break down the problem step by step using the information provided and the figure. --- ## **Given Data** - **Original diameter, \( d_ \):** 14. mm - **Original gage length, \( L_ \):** 66.8 mm - **Poisson's ratio, \( \nu \):** .35 - **Maximum stress, \( \sigma \):** 560 MPa (from the graph) - **Maximum strain, \( \epsilon \):** .008 mm/mm (from the graph) --- ## **Step 1: Find Young's Modulus \( E \)** The slope of the linear (elastic) portion of the stress-strain curve is the **Young's modulus**: \[ E = \frac{\sigma}{\epsilon} \] Plug in the values: \[ E = \frac{560 \, \text{MPa}}{.008} \] \[ E = \frac{560}{.008} = 70,000 \, \text{MPa} = 70 \, \text{GPa} \] --- ## **Step 2: Find the Change in Length (\( \Delta L \)) at Maximum Stress** Strain is defined as: \[ \epsilon = \frac{\Delta L}{L_} \] So, \[ \Delta L = \epsilon \cdot L_ \] \[ \Delta L = .008 \times 66.8 \, \text{mm} \] \[ \Delta L = .5344 \, \text{mm} \] --- ## **Step 3: Find the Lateral Strain and Change in Diameter (\( \Delta d \))** Lateral strain is related to axial strain by Poisson's ratio: \[ \text{Lateral strain} = -\nu \times \text{Axial strain} \] \[ \text{Lateral strain} = -.35 \times .008 = -.0028 \] Change in diameter: \[ \text{Lateral strain} = \frac{\Delta d}{d_} \] \[ \Delta d = \text{Lateral strain} \cdot d_ \] \[ \Delta d = -.0028 \times 14. \, \text{mm} \] \[ \Delta d = -.0392 \, \text{mm} \] So, the diameter **decreases** by .0392 mm. --- ## **Step 4: Final Answers** ### **(a) Young's Modulus** \[ \boxed{E = 70\,\text{GPa}} \] ### **(b) Change in gage length at 560 MPa** \[ \boxed{\Delta L = .534\,\text{mm}} \] ### **(c) Change in diameter at 560 MPa** \[ \boxed{\Delta d = -.039 \,\text{mm}} \] --- ### **Summary Table** | Quantity | Value | |-----------------------|--------------| | Young's Modulus (E) | 70 GPa | | Change in Length | .534 mm | | Change in Diameter | -.039 mm | If you need further breakdown or additional calculations (like final area or volume), let me know!

# Step-by-Step Solution for Stress-Strain Problem ## **Given Data** - **Original diameter, \( d \):** 14.0 mm - **Original gage length, \( L \):** 66.8 mm - **Poisson's ratio, \( \nu \):** 0.35 - **Maximum stress, \( \sigma \):** 560 MPa (from the graph) - **Maximum strain, \( \epsilon \):** 0.008 (mm/mm from the graph) --- ## **Step 1: Calculate Young's Modulus \( E \)** Young's modulus is calculated using the formula: \[ E = \frac{\sigma}{\epsilon} \] Substituting the values: \[ E = \frac{560 \, \text{MPa}}{0.008} = 70,000 \, \text{MPa} = 70 \, \text{GPa} \] --- ## **Step 2: Calculate Change in Length (\( \Delta L \))** Using the definition of strain: \[ \epsilon = \frac{\Delta L}{L} \] Rearranging gives: \[ \Delta L = \epsilon \cdot L \] Substituting the values: \[ \Delta L = 0.008 \times 66.8 \, \text{mm} = 0.5344 \, \text{mm} \] --- ## **Step 3: Calculate Lateral Strain and Change in Diameter (\( \Delta d \))** Lateral strain is calculated as: \[ \text{Lateral strain} = -\nu \cdot \text{Axial strain} \] Substituting the values: \[ \text{Lateral strain} = -0.35 \times 0.008 = -0.0028 \] Next, calculate the change in diameter: \[ \Delta d = \text{Lateral strain} \cdot d \] \[ \Delta d = -0.0028 \times 14.0 \, \text{mm} = -0.0392 \, \text{mm} \] --- ## **Final Answers** ### **(a) Young's Modulus** \[ \boxed{E = 70\,\text{GPa}} \] ### **(b) Change in Gage Length at 560 MPa** \[ \boxed{\Delta L = 0.534\,\text{mm}} \] ### **(c) Change in Diameter at 560 MPa** \[ \boxed{\Delta d = -0.039\,\text{mm}} \] --- ## **Summary Table** | Quantity | Value | |-----------------------|--------------| | Young's Modulus (E) | 70 GPa | | Change in Length | 0.534 mm | | Change in Diameter | -0.039 mm | This concludes the calculations based on the provided stress-strain diagram for the aluminum alloy specimen. If you have any further questions or need additional calculations, feel free to ask!