Question

A piece of copper having a rectangular cross section of `15.2 xx 19.1 mm` is pulled in tension with 45,500N, force producing only elastic deformation. Calculate the resulting strain. Shear modulus of elasticity of copper is `42 xx 10^(9)Nm^(-2)`.

Answer 1

Length of the piece of copper, l = 19.1 mm = 19.1 `xx` `10^(-3)` m
Breadth of the piece of copper, b = 15.2 mm = 15.2 `xx``10^(-3)` m
Area of the copper piece: A = l`xx`b
= 19.1`xx``10^(-3)``xx` 15.2 `xx``10^(-3)`
=2.9`xx``10^(-4)``"m"^(2)`
Tension force applied on the piece of copper, F = 44500 N
Modulus of elasticity of copper, `eta=42xx10^(9)"N"//"m"^(2)`
Modulus of elasticity, `eta=("Stress")/("Strain")=((F)/(A))/("Strain")`
`therefore"Strain"=(F)/(Aeta)` = `(44500)/(2.9xx10^(-4)xx42xx10^(9))` = 3.65`xx` `10^(-3)`