Hey student’s we know that when load is applied to free end of wire, extension is produced in the wire. Work is to be done during extension of wire due to external applied force. This work is stored in the form of potential energy in the wire known as strain energy or elastic energy.

**Let’s derive the formula for strain energy…..!**

Suppose that the wire of length L and area of cross section A is elongated by gradually increasing the load from 0 to ∆l the magnitude F as shown in fig.

Let ∆l be the extension produced in the wire for final loading, and then the tensile stress is given as,

Tensile strain = F/A

Also tensile strain is given by,

Tensile strain = ∆l/L

By definition for Young’s modulus is given as the ratio of tensile stress to tensile strain then we can write,

During the procedure of loading, let at any instance be the stretching force f and x be the corresponding extension, then

Let the wire is extended by infinitesimal small value ‘dx’ against the force ‘f’ then the amount of work done during this will be,

∴ dW = f dx

Equation (3) gives the work done during extension of wire for infinitesimal extension of ‘dx’ The total work done during the extension of wire from l=0 to l=△l is can be obtained by integrating the above equation

This work is stored in the form of potential energy here it is called strain energy or elastic energy.

∴ Strain energy = 1/2 × F × l ………….(5)

Multiplying and dividing the above equation by quantity volume, V=AL

∴ Strain energy =1/2 × F/A × l/L × AL

∴Strain energy =1/2 × stress × strain × volume

Dividing both sides by the volume ‘V’

∴Strain energy/volume =1/2 × stress × strain …………………..(6)

∴Strain Energy per unit volume = ½ × stress × strain

We have, Y = stress/strain

∴ Strain energy/unit volume = ½ × strain^{2} × Y …………………..(7)

**Some important points to note……!**

- Strain energy is directly proportional to applied force and extension produced.
- Strain energy per unit volume depends on Young’s modulus.

**Let’s go in more detail with some numerical….!**

Ex: 1) Wire of length 2 m and area 0.225 cm^{2}, extends by 0.2 cm when load of 2000 g is applied at its free end. Find the energy per unit volume of wire.

∴ Strain Energy per unit volume = 435.55 J