Question

Derive an expression for the elastic energy stored per unit volume of a wire.

Answer 1

When a body is stretched, work is done against the restoring force (internal force). This work done is stored in the body in the form of elastic energy.

Consider a wire whose un-stretch length is L and the area of cross-section is A. Let a force produce an extension 1 and further assume that the elastic limit of the wire has not been exceeded and there is no loss in energy. Then, the work done by the force F is equal to the energy gained by the wire.

The work done in stretching the wire by dl, dW = Fdl

The total work done in stretching the wire from 0 to l is

W = `int_0^"l" "Fdl"` .............(1)

From Young’s modulus of elasticity,

Y = `"F"/"A" xx "L"/"l"⇒ F = "YAl"/"L"` .........(2)

Substituting equation (2) in equation (1), we get

W = `int_0^"l" "YAl"/"L" "dl"`

Since l is the dummy variable in the integration, we can change l to l’ (not in limits), therefore

W = `int_0^"l" "YAl’"/"L" "dl’" = "YA"/"L"("l’"^2/2)_0^"l" = "YA"/"L" "l"^2/2 = 1/2 ("YAl"/"L")"l" = 1/2"Fl"`

W = `1/2` Fl = Elastic potential energy

Energy per unit volume is called energy density

u = `"Elastic potential energy"/"Volume" = (1/2 "Fl")/("AL")`

= `1/2 "F"/"A" "l"/"L" = 1/2 xx "Stress" xx "Strain"`