A specimen of copper having a rectangular cross section \(15.2 \mathrm{~mm} \times 19.1 \mathrm{~mm}(0.60 \mathrm{in} . \times 0.75 \mathrm{in} .)\) is pulled in tension with \(44,500 \mathrm{~N}\left(10,000 \mathrm{lb}_{\mathrm{f}}\right)\) force, producing only elastic deformation. Calculate the resulting strain.

In an elastic material, the stress and the strain are linearly related by the Young's modulus (Y). The relationship can be expressed as: $$ \sigma = Y \cdot \epsilon $$ Where \(\sigma\) is the stress, \(\epsilon\) is the strain, and \(Y\) is the Young's modulus.

To calculate the stress experienced by the specimen, we first need to find the cross-sectional area it. After that, we can use the formula: $$ \sigma = \frac{F}{A} $$ Where \(F\) is the force applied on the specimen and \(A\) is its cross-sectional area. Given the dimensions of the copper specimen in meters: \(15.2 \mathrm{~mm} \times 19.1 \mathrm{~mm} = 0.0152 \mathrm{~m} \times 0.0191 \mathrm{~m}\) The cross-sectional area is: $$ A = 0.0152 \mathrm{~m} \times 0.0191 \mathrm{~m} = 2.9 \times 10^{-4} \mathrm{m^2} $$ Now, given that the tensile force is \(F = 44,500 \mathrm{~N}\), we can calculate the stress: $$ \sigma = \frac{44,500 \mathrm{~N}}{2.9 \times 10^{-4} \mathrm{m^2}} = 1.53 \times 10^8 \mathrm{~Pa} $$

Now that we have the stress, we can find the strain using the stress-strain relationship. The Young's modulus for copper is \(Y = 11 \times 10^{10} \mathrm{~Pa}\). We can rearrange the formula to solve for the strain: $$ \epsilon = \frac{\sigma}{Y} $$ Calculating the strain using the given values: $$ \epsilon = \frac{1.53 \times 10^8 \mathrm{~Pa}}{11 \times 10^{10} \mathrm{~Pa}} = 1.39 \times 10^{-3} $$ The resulting strain of the copper specimen is \(1.39 \times 10^{-3}\).