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A piece of copper having a rectangular cross- section $15.2 \mathrm{~mm} \times 19.1 \mathrm{~mm}$ is pulled in tension with $44,500 \mathrm{~N}$ force, producing only elastic deformation. Calculate the resulting strain. Young's modulus of elasticity of copper is $120 \times 10^9 \mathrm{~N} / \mathrm{m}^2$.
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Verified Answer
Given:
$$
\begin{aligned}
&A=15.2 \times 19.1 \mathrm{~mm}^2=15.2 \times 19.1 \times 10^{-6} \mathrm{~m}^2 ; F=44,500 \\
&\mathrm{~N}, Y=120 \times 10^9 \mathrm{~N} / \mathrm{m}^2
\end{aligned}
$$
Shearing strain $=\frac{F}{A Y}$
$$
=\frac{44500}{15.2 \times 19.1 \times 10^{-6} \times 120 \times 10^9}=0.127 \times 10^{-2}
$$
$$
\begin{aligned}
&A=15.2 \times 19.1 \mathrm{~mm}^2=15.2 \times 19.1 \times 10^{-6} \mathrm{~m}^2 ; F=44,500 \\
&\mathrm{~N}, Y=120 \times 10^9 \mathrm{~N} / \mathrm{m}^2
\end{aligned}
$$
Shearing strain $=\frac{F}{A Y}$
$$
=\frac{44500}{15.2 \times 19.1 \times 10^{-6} \times 120 \times 10^9}=0.127 \times 10^{-2}
$$
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