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, $v = 0.30$).
$P$
$P$
$A$
$B$
50$ mm
$12$ mm
Determine the resulting change in the width of portion $AB$ of the test coupon. (Include a minus sign if necessary. Round the final answer to three decimal places.)
The resulting change in the width of portion $AB$ is ______ ×10$^{-6}$ m.
1. **Problem Statement:**
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 $v = 0.30$. The width of portion $AB$ is $w = 50$ mm. We need to find the change in width $\Delta w$ of portion $AB$ due to the tensile load.
2. **Relevant Formulas and Concepts:**
- The tensile stress $\sigma$ is given by:
$$\sigma = \frac{P}{A}$$
where $A$ is the cross-sectional area.
- The cross-sectional area $A$ is:
$$A = w_{narrow} \times t$$
where $w_{narrow} = 12$ mm is the narrow width where the load is applied.
- The longitudinal strain $\varepsilon_x$ is:
$$\varepsilon_x = \frac{\sigma}{E}$$
- The lateral strain $\varepsilon_y$ (change in width direction) is related to longitudinal strain by Poisson's ratio:
$$\varepsilon_y = -v \varepsilon_x$$
- The change in width $\Delta w$ is:
$$\Delta w = \varepsilon_y \times w$$
3. **Calculate Cross-Sectional Area:**
$$A = 12\ \text{mm} \times 1.6\ \text{mm} = 19.2\ \text{mm}^2$$
Convert to $m^2$:
$$A = 19.2 \times 10^{-6} \ m^2$$
4. **Calculate Tensile Stress $\sigma$:**
Convert $P$ to Newtons:
$$P = 2.8 \times 10^3 \ N$$
Calculate $\sigma$:
$$\sigma = \frac{2.8 \times 10^3}{19.2 \times 10^{-6}} = 145.833 \times 10^6 \ Pa = 145.833 \ MPa$$
5. **Calculate Longitudinal Strain $\varepsilon_x$:**
$$\varepsilon_x = \frac{\sigma}{E} = \frac{145.833 \times 10^6}{200 \times 10^9} = 7.2917 \times 10^{-4}$$
6. **Calculate Lateral Strain $\varepsilon_y$:**
$$\varepsilon_y = -v \varepsilon_x = -0.30 \times 7.2917 \times 10^{-4} = -2.1875 \times 10^{-4}$$
7. **Calculate Change in Width $\Delta w$:**
Convert $w$ to meters:
$$w = 50 \ mm = 0.05 \ m$$
$$\Delta w = \varepsilon_y \times w = -2.1875 \times 10^{-4} \times 0.05 = -1.09375 \times 10^{-5} \ m$$
Express in $\times 10^{-6} m$:
$$\Delta w = -10.938 \times 10^{-6} \ m$$
8. **Final Answer:**
The resulting change in the width of portion $AB$ is $-10.938 \times 10^{-6}$ m.
**Rounded to three decimal places:**
$$\boxed{-10.938 \times 10^{-6} \ m}$$