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

Explanation

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 area of cross section is A. Let a force produce an extension l 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 = F dl`
The total work done in stretching the wire from 0 to l is
`W = int_(0)^(l) F dl " " ....(1)`
From Young.s modulus of elasticity,
`Y = F/A xx L/l implies F = (YAl)/L " " ....(2)`
Substituting equation (2) in equation (1), we get
`W = int_(0)^(l) (YA l)/(L) dl`
Since, l is the dummy variable in the intergration, 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 ("Stress" xx "Strain")`.