# Strain Energy Formula

The strain energy is defined as the energy stored in any object which is loaded within its elastic limits. In other words, the strain energy is the energy stored in anybody due to its deformation. The strain energy is also known as Resilience. The unit of strain energy is N-m or Joules. Here, the derivation of the strain energy formula, its unit, equations, etc are discussed thoroughly. The concepts are explained with the help of relevant diagrams and symbols.

Conditions for Strain Energy are:

• The material should be loaded in its elastic limit.

• The material shouldn’t undergo permanent deformation.

• The amount of strain energy depends upon the amount of load and the type of load loaded.

## Strain Energy Equation:

Consider a rod of length l and let the load be applied on either side of the rod gradually. Due to applied load, the rod deforms by a length $\delta l$. From the definition of strain energy, we know that it is the energy stored in a body when work is done on it to deform within the elastic limit.

Then, the elastic strain energy formula is given by,

$\Rightarrow$ Work Done = Elastic strain energy

$\Rightarrow U = W = F dx$ ……(1)

Here the force is the applied load and since the load was applied gradually we will consider the average of the load and the displacement is the deformation. Therefore equation (1) becomes,

$\Rightarrow U = F dx = \left(0 + \dfrac{F}{2}\right)\delta l$

$\Rightarrow U = \dfrac{F \delta l}{2}$ ……….(2)

Equation (2) is known as Strain Energy Formula or Strain Energy Equation.

### Different Forms of Strain Energy or Strain Potential Energy Formula:

We know that from Hooke’s law,

#### $\Rightarrow E = \dfrac{\text{Stress}}{\text{Strain}}$

Where $E$ is Young’s modulus.

$\Rightarrow E = \left(\dfrac{F}{A}\right) \left(\dfrac{l}{\delta l}\right)$ …….(1)

On rearranging the above equation for $\delta l$ we get,

#### $\Rightarrow \delta l = \dfrac{Fl}{E}$ …….(2)

Substituting, the value of deformation in the expression of strain energy,

#### $\Rightarrow U = \dfrac{F^2A}{2EA}$……(3)

Multiply and divide by A in the equation (3), we get:

#### $U = \dfrac{F^2lA}{2EA^2}$

We know that $l \times A$ is the volume of the rod and $\dfrac{F}{A}$ is the stress(\sigma) applied to the material, substituting the same in the above expression we get:

#### $\Rightarrow U = \dfrac{\sigma V \varepsilon}{2} \left(\therefore E = \dfrac{\text{stress}}{\text{strain}} = \dfrac{\sigma}{\varepsilon}\right)$

Where,

$\sigma$ - Stress in the material.

$V$ - Volume of the material.

$\varepsilon$ - Strain in the material

Equations (3) and (4) are the two different forms of strain energy that we use while studying the strength of any given material.

### Strain Energy Per Unit Volume Formula:

Strain energy per unit volume is also known as the strain energy density formula is given by:

We know that the stress-energy formula is,

#### $\Rightarrow U = \dfrac{\sigma V \varepsilon}{2}$

Where,

$\sigma$ - Stress in the material.

$V$ - Volume of the material.

$\varepsilon$ - Strain in the material.

Then, strain energy per unit volume is,

$\Rightarrow \dfrac{U}{V} = \dfrac{\sigma \varepsilon V}{2V}$

#### $\Rightarrow$ Stress energy per unit volume $= \dfrac{\sigma \varepsilon}{2}$

This is the required elastic energy density formula.

### Examples:

1. Calculate the Work Done in Stretching a Wire of Length 5m and its Cross-Sectional Area is Given by 1mm2 is Deformed by the Length of 1mm if Young’s Modulus is Given of the Wire is Given by $2 \times 10^{11}N/m^2$

Ans:

Given,

Length of the wire $= l = 5m$

Area of the wire $= A= 1mm^2=10^{-6}m^2$

Deformation due to stretching $= \delta l = 1mm = 10^{-3}mA$

Young’s modulus of the wire $= E = 2 \times 10^{11} N/m^2$

Young’s Modulus of Elasticity is given by,

$E = \dfrac{F l}{A \delta l}$

$\therefore F = EA \dfrac{\delta l}{l} = \dfrac{(2 \times 10^{11}N/m^2)(10^{-6}m^2)(10^{-3}m)}{5m}$

$\Rightarrow F = 40N$

Now work done in stretching the wire is given by,

$\Rightarrow U = \dfrac{F \delta l}{2}$

$\Rightarrow U = \dfrac{40 \times 10^{-3}}{2}$

$\Rightarrow U = 0.02J$

Therefore, $0.02J$ of work is done in stretching $5m$ long wire.

The Strain Energy Formula has applications in several concepts. Hence it is important to learn and understand its derivation and features. The strain energy formula is derived and strain energy can be written in the form of the volumetric term, young’s modulus, etc.. depending upon the need of calculation and data available.