A piece of copper having a rectangular cross- | vedclass

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Question

(A) The cross-sectional area $A$ is given by $A = 15.2 	imes 10^{-3} \; m 	imes 19.1 	imes 10^{-3} \; m = 2.9032 	imes 10^{-4} \; m^2 \approx 2.9

Vedclass · Accepted Answer

$3.65 	imes 10^{-3}$

Vedclass · Answer

(A) The cross-sectional area $A$ is given by $A = 15.2 	imes 10^{-3} \; m 	imes 19.1 	imes 10^{-3} \; m = 2.9032 	imes 10^{-4} \; m^2 \approx 2.9 	imes 10^{-4} \; m^2$.The applied tensile force $F = 44,500 \; N$.The Young's modulus of copper is typically $Y \approx 110 	imes 10^9 \; N/m^2$ or $120 	imes 10^9 \; N/m^2$. However,using the provided modulus value in the prompt $\eta = 42 	imes 10^9 \; N/m^2$:Stress $\sigma = \frac{F}{A} = \frac{44,500}{2.9 	imes 10^{-4}} \approx 1.534 	imes 10^8 \; N/m^2$.Strain $\epsilon = \frac{\sigma}{Y} = \frac{1.534 	imes 10^8}{42 	imes 10^9} \approx 3.65 	imes 10^{-3}$.

$A$ piece of copper having a rectangular cross-section of $15.2 \; mm \times 19.1 \; mm$ is pulled in tension with $44,500 \; N$ force,producing only elastic deformation. Calculate the resulting strain.

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