Question: A 2.8-kN tensile load ($P$) is applied to a test coupon made from 1.6-mm flat steel plate ($E = 200$ GPa, $\nu = 0.30$). $P \leftarrow$ top-left $P \leftarrow$ top-right 50 mm A B 12 mm Determine the resulting change in the 50-mm gage length. The resulting change in the 50-mm gage length is ____ $\times 10^{-6}$ m .

1. **State the problem:** A tensile load $P = 2.8$ kN is applied to a flat steel plate with thickness $t = 1.6$ mm, Young's modulus $E = 200$ GPa, and Poisson's ratio $\nu = 0.30$. The original gage length is $L_0 = 50$ mm. We need to find the change in length $\Delta L$ of the 50-mm gage length. 2. **Relevant formula:** The axial stress $\sigma$ is given by: $$\sigma = \frac{P}{A}$$ where $A$ is the cross-sectional area. The axial strain $\epsilon$ is: $$\epsilon = \frac{\Delta L}{L_0} = \frac{\sigma}{E}$$ Rearranged to find $\Delta L$: $$\Delta L = \epsilon L_0 = \frac{\sigma}{E} L_0$$ 3. **Calculate cross-sectional area $A$:** Width $w = 12$ mm, thickness $t = 1.6$ mm $$A = w \times t = 12 \text{ mm} \times 1.6 \text{ mm} = 19.2 \text{ mm}^2$$ Convert to $m^2$: $$19.2 \text{ mm}^2 = 19.2 \times 10^{-6} \text{ m}^2$$ 4. **Calculate axial stress $\sigma$:** Load $P = 2.8$ kN = $2800$ N $$\sigma = \frac{2800}{19.2 \times 10^{-6}} = 145833333.3 \text{ Pa} = 145.83 \text{ MPa}$$ 5. **Calculate axial strain $\epsilon$:** $$\epsilon = \frac{\sigma}{E} = \frac{145.83 \times 10^{6}}{200 \times 10^{9}} = 0.00072917$$ 6. **Calculate change in length $\Delta L$:** Original length $L_0 = 50$ mm = $0.05$ m $$\Delta L = \epsilon L_0 = 0.00072917 \times 0.05 = 3.6458 \times 10^{-5} \text{ m}$$ 7. **Express in $\times 10^{-6}$ m:** $$\Delta L = 36.458 \times 10^{-6} \text{ m}$$ **Final answer:** The resulting change in the 50-mm gage length is approximately $36.5 \times 10^{-6}$ m.