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)`.

Explanation

Here,
`A = 15.2mm xx 19.1mm = 2.90xx10^(2) mm^(2)`
`= 2.90xx10^(-4) m^(2)`
`F = 44,500N, eta = 42GPa = 42xx10^(9) N//m^(2)`
stress `= (F)/(A) = (44500N)/(2.90xx10^(-4) m^(2)) = 1.53xx10^(8) N//m^(2)`
As `eta = ("shear stress")/("shear strain")`
Shear strain `= ("shear stress")/(eta) = (1.53xx10^(8))/(42xx10^(9))`
`= 0.036xx10^(-1) = 3.6xx10^(-3)`