Question

show that when a wire is stretched within elastic limit, the potential energy per unit volume stored in it is half the product of stress & strain.​

Answer 1

Answer:

Let the wire be of unstretched length L and let a force F produce an extension e. (Assume that the elastic limit of the wire has not been exceeded and that no energy is lost as heat.)

Consider Figure 1(a). The work done a the force is Fs but in this case the force varies from 0 at the start to F at the end when the wire is stretched by an amount e. Therefore:

Work done on the wire during stretching = average force x extension = ½ Fe

But the work done by F is equal to the energy gained by the wire.

Therefore:

work done = average force x extension = ½ Fe

Therefore:

work done = energy stored = ½ Fe = ½ EAe2/L

And this energy is the shaded area of the graph.

If the extension is increased from e1 to e2 then the extra energy stored is given by:

Energy stored = ½ F[e2 – e1] = ½ EA[e22 – e12]/L

This is the shaded area on the graph in Figure 1(b), and in general the energy stored in an extension is the area below the line in the force-extension graph. It can also be shown that:

Energy stored per unit volume of a specimen = ½ stress x strain

Explanation: