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

1.	Derive an expression for the elastic energy stored per unit volume of a wire.
Answer» SOLUTION :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")`.

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