Let's go step by step to solve for the main physical properties requested in a typical tensile test analysis: --- ## **Given Data** - **Diameter, \( d \) = 12 mm** - **Initial Length, \( L_ \) = 50 mm** - **Maximum Load, \( F_{max} \) = 55.2 kN** - **Fracture Load, \( F_{fracture} \) = 50.8 kN** --- ## **1. Calculate Initial Cross-sectional Area (\( A_ \))** \[ A_ = \frac{\pi d^2}{4} \] \[ A_ = \frac{\pi (12\ \text{mm})^2}{4} = \frac{3.1416 \times 144}{4} = \frac{452.39}{4} = 113.1\ \text{mm}^2 \] --- ## **2. Engineering Stress and Strain** ### **Engineering Stress (\( \sigma \))** \[ \sigma = \frac{F}{A_} \] - \( F \) in Newtons (1 kN = 100 N) - \( A_ \) in mm² (but convert to m² if you want the answer in Pascals, otherwise leave in MPa) ### **Engineering Strain (\( \epsilon \))** \[ \epsilon = \frac{\Delta L}{L_} \] --- ## **3. Young's Modulus (E)** - Use the *linear* region (first 2-3 data points): \[ E = \frac{\sigma}{\epsilon} \] --- ## **Step-by-Step Calculations** ### **A. Fill Out a Table** Let's use the first three data points to calculate stress, strain, and estimate Young's Modulus. #### **Sample Calculations for Data Point 2:** - \( \Delta L = .042 \) mm - \( F = 13.4 \) kN = 13,400 N #### **Strain:** \[ \epsilon = \frac{.042}{50} = .00084 \] #### **Stress:** \[ \sigma = \frac{13,400}{113.1} = 118.6\ \text{MPa} \] #### **Repeat for Data Point 3:** - \( \Delta L = .085 \) mm - \( F = 26.7 \) kN = 26,700 N \[ \epsilon = \frac{.085}{50} = .0017 \] \[ \sigma = \frac{26,700}{113.1} = 236.1\ \text{MPa} \] --- ### **B. Young's Modulus \( E \) (Slope):** Using data points 2 and 3: \[ E = \frac{\Delta \sigma}{\Delta \epsilon} \] \[ \Delta \sigma = 236.1 - 118.6 = 117.5\ \text{MPa} \] \[ \Delta \epsilon = .0017 - .00084 = .00086 \] \[ E = \frac{117.5}{.00086} \approx 136,630\ \text{MPa} = 136.6\ \text{GPa} \] --- ### **C. Ultimate Tensile Strength (UTS):** UTS is the maximum engineering stress: - Max load = 55.2 kN = 55,200 N \[ UTS = \frac{55,200}{113.1} = 487.9\ \text{MPa} \] --- ### **D. Strain at Fracture:** - \( \Delta L_{fracture} = 25.908 \) mm \[ \epsilon_{fracture} = \frac{25.908}{50} = .518 \] --- ### **E. Summary Table** | Data # | ΔL (mm) | Load (kN) | Strain (ε) | Stress (MPa) | |--------|---------|-----------|------------|--------------| | 1 | .000 | . | .000 | . | | 2 | .042 | 13.4 | .00084 | 118.6 | | 3 | .085 | 26.7 | .00170 | 236.1 | | 4 | .106 | 33.4 | .00212 | 295.4 | | 5 | .229 | 40.1 | .00458 | 354.5 | | ... | ... | ... | ... | ... | | 8 | 12.700 | 55.2 | .254 | 487.9 | | 9 | 25.908 | 50.8 | .518 | 449.2 | --- ## **Summary of Results** - **Young’s Modulus (E):** ~137 GPa - **Ultimate Tensile Strength (UTS):** ~488 MPa - **Strain at Fracture:** .518 (51.8%) - **Initial Cross-sectional Area:** 113.1 mm² --- ### **If you need to plot or do further calculations (like true stress/strain), let me know!** If you want a spreadsheet template, just ask!