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Q8 A piece of copper having a rectangular cross-section of 15.2 mm × 19.1 mm is pulled in tension with 44,500 N force, producing only elastic deformation. Calculate the resulting strain?

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Length of the copper piece, l = 19.1 mm

The breadth of the copper piece, b = 15.2 mm

Force acting, F = 44500 N

Modulus of Elasticity of copper, \eta =42\times 10^{9}Nm^{-2} 

\\Strain=\frac{F}{A\eta }\\ =\frac{F}{lb\eta }\\ =\frac{44500}{15.2\times 10^{-3}\times 19.1\times 10^{-3}\times 42\times 10^{9}}\\ =3.65\times 10^{-3}

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