Q.If the length of a wire is increased by applying an external force, prove that the work done per unit volume of the wire = 1/2 x stress x strain.

Integrating the Hooke's-law restoring force over the extension shows the elastic work stored per unit volume of a stretched wire is exactly (1/2) * stress * strain.

Consider a wire of natural length L, cross-sectional area A, and Young's modulus Y, stretched by an external force. When it has been extended by an amount x (0 <= x <= l, where l is the final extension), Hooke's law gives the relation:

Y = stress/strain = (F/A)/(x/L)

which gives F = (Y*A/L)*x

This restoring force increases linearly with x, starting from 0 (at x=0) up to F_max = YAl/L (at x=l) - so it behaves exactly like a spring of effective constant k = Y*A/L.

The work done in stretching the wire by a further small amount dx, at extension x, is:

dW = Fdx = (YA/L)xdx

Total work done in stretching the wire from x=0 to x=l:

W = integral from 0 to l of (YA/L)xdx = (YA/L)(l^2/2) = (YAl^2)/(2L)

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