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?

#### Solution 1

Length of the piece of copper, l = 19.1 mm = 19.1 × 10^{–3} m

Breadth of the piece of copper, b = 15.2 mm = 15.2 × 10^{–3} m

Area of the copper piece:

A = l × b

= 19.1 × 10^{–3} × 15.2 × 10^{–3}

= 2.9 × 10^{–4} m^{2}

Tension force applied on the piece of copper, F = 44500 N

Modulus of elasticity of copper, η = 42 × 10^{9} N/m^{2}

Modulus of elasticity, η = `"Stress"/"Strain" = (F/A)/"Strain"`

`:. "Strain" = F/(Aeta)`

`= 44500/(2.9xx10^(-4)xx42xx10^(9))`

`= 3.65 xx 10^(-3)`

#### Solution 2

A = 15.2 x 19.2 x 10^{-6} m^{2}; F = 44500 N; η = 42 x 10^{9} Nm^{-2}

Strain = `"Stress"/"modulus of elasticity" = "F/A"/eta`

`=F/(Aeta) = 44500/((15.2xx19.2xx10^(-6))xx42xx10^9) = 3.65 xx 10^(-3)`