A steel alloy bar 100 mm long with a rectangular cross section of 10 mm X 40 mm is subjected to tension with a load of 80 kN and experiences an increase in length of 0.1 mm. If the increase in length is entirely elastic, calculate the modulus of elasticity of the steel alloy.

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Step 1Concept: Since it has been given that the material is linearly elastic within applied load, the modulus of elasticity can be determined as the ratio of stress to strain. (Please refer images below for solution) Step 2Given:- length =100 \mathrm{~mm}Cross-section;\begin{aligned}\text { Dection; width } & =40 \mathrm{~mm} \\\text { Depth } & =10 \mathrm{~mm} \\\text { Area of cross-section } & =40 \times 10 \\& =400 \mathrm{~mm}^{2}\end{aligned}\begin{array}{l}\text { load applied }=80 \mathrm{kN} \text { (tension) } \\\text { increase in length }=0.1 \mathrm{~mm} \text {. }\end{array}Solution:\begin{aligned}\text { Stress } & =\frac{\text { load }}{\text { Area of cross-section }}=\frac{80 \times 10^{3}}{400} \mathrm{~N} / \mathrm{mm}^{2} \\& =200 \mathrm{~N} / \mathrm{mm}^{2}\end{aligned}\begin{aligned}\text { strain } & =\frac{\text { Change in Dimension }(\text { length })}{\text { Orignal Jimensin (lensth) }}=\frac{0.1}{100} \\& =10^{-3}\end{aligned}\begin{array}{l}\text { modulus of Elasticity }=\frac{\text { stres }}{\text { strain }}=\frac{200}{10^{-3}} \\=2 \times 10^{5} \mathrm{~N} / \mathrm{mm}^{2} \\\end{array}Answer ...