Strain Energy: Strain energy is defined as the energy stored in any material due to deformation. When a deformable structure, such as rubber, spring, metals, etc., stretches, then it stores a type of energy called strain energy. In many such cases, we can turn back this energy into kinetic energy relatively in an easy way. Students must also follow the strain energy density formula.

If a spring attached to a block placed on a smooth surface is compressed, it has been given strain energy. As the spring returns to its original length, its strain energy is transferred to the block in the form of kinetic energy. Strain energy stored in a material for any particular stress can be fully transformed into kinetic energy only if the stress is under the elastic limit. In this article, we will learn more about the strain energy formula with examples.

Stress and Strain

When we apply force on any material, it will generate equal resistive force. This resistive force generated inside the material per unit area is called stress. The elastic strain energy formula will be available in the coming sections.

Stress \((\sigma)=\frac{F}{A}\)

Here \(F\) is the applied force, and \(A\) is the cross-section area.

When force is applied to a material, there will be deformation in the material. The linear deformation (Change in length) per unit length is called longitudinal Strain.

Strain \((\varepsilon ) = \frac{{{\rm{ Change\, in\, length }}}}{{{\rm{ Orignal\, length }}}} = \frac{{\Delta L}}{L}\)

Study Strain in Detail Here

Relation Between Stress and Strain

Hooke’s law states that for small deformation of elastic material, the Strain will be directly proportional to stress. Hooke’s law is only valid for small deformation (up to the proportional limit) in an elastic material.

According to Hooke’s law,

Stress \(\propto\) Strain

Or

Stress \(=E \times\) Strain

Here \(E\) is the proportionality constant and is known as the elastic strain energy formula.

Stress vs Strain Graph

When a gradually increasing force is applied to a material and the stress applied is plotted for the corresponding strain, then we will get the stress vs strain graph for that particular material. The figure given below shows a Stress vs Strain graph.

In the below graph,

1. Point \(A\) is the proportional limit.
2. Point \(B\) is the elastic limit.
3. Point \(C\) is the yield point.
4. Point \(D\) represents ultimate strength.
5. Point \(E\) represents the fracture point.
6. The area under the curve \(O\) to \(B\) represents the elastic region, and the area under the curve \(B\) to \(E\) is the plastic region.

From the graph, we can see that in the region between \(O\) to \(A\), the curve is linear (straight line), so we can say that the stress is directly proportional to Strain and obeys Hooke’s law. On unloading, the material regains its original dimensions, and all the stored potential energy (strain energy) energy is released. In this region, the material behaves as an elastic body. The slope of the line \(OA\) gives Young’s modulus and is denoted by the symbol \(Y\). It has the same units as stress. i.e. \(\mathrm{N} / \mathrm{m}^{2}\) or \(Pa\)

\({\rm{Y}} = {\rm{Slope}} = \frac{{{\rm{ Stress }}}}{{{\rm{ Strain }}}}\)

In the region from \(A\) to \(B\) stress and Strain are not proportional. Between \(A\) and \(B\) the body still returns to its original dimension when the load is removed. Point \(B\) in the above curve is known as yield point (also known as elastic limit), and the corresponding stress is known as the yield strength of the material. If the stress is increased more than point \(B\), plastic deformation will occur, and the material will not come to its original shape and size after the release of the load.

Derivation of Strain Energy Formula

When we apply force to the material, it will deform. The external force will do work on the material, which will be stored in the material as strain energy. Under the elastic limit, the work done by external force will be equal to the strain energy stored (Work-energy theorem). It can also be calculated by calculating the area under the curve of the Stress vs Strain graph up to the elastic limit. The strain energy up to the elastic limit is also known as Resilience. The unit of strain energy is \(\rm{N-m}\) or Joules. 

Before going for the derivation of the strain energy formula, we will have to make certain assumptions for the ideal condition.

The assumption for Strain Energy Formula

1. The material should be elastic.
2. Stress developed should be within the proportional limit.
3. Load should be gradually applied.

Let at any instant, the material having original length \(L\) and having area \(A\) is elongated by \(x\) unit.

From Hooke’s law,

\({\rm{Stress = E \times Strain}}\)

\(\Rightarrow \frac{F}{A}=E \frac{x}{L}\)

\(\Rightarrow F=\frac{EAx}{L}\)

So, the force is variable and gradually increasing with deformation. Let for small deformation \(dx\) differential work done will be \(dW.\) We know that the work done by the variable force is given by,

\(dW = \vec F \cdot d\vec s = F\;dx\) (Both force and displacement have the same direction)

\(W=\int_{0}^{\Delta L} F d x\)

\(W=\int_{0}^{\Delta L} \frac{EA x}{L} d x\)

\(W=\left[\frac{E A x^{2}}{2 L}\right]_{0}^{\Delta L}=\frac{E A(\Delta L)^{2}}{2 L}\)

Thus the total strain energy \((U)\) for small deformation will be,

\(U=\frac{E A(\Delta L)^{2}}{2 L}\)

\(=\frac{1}{2} \times\left(E \times \frac{\Delta L}{L}\right) \times A L \times \frac{\Delta L}{L}\)

\(= = \frac{1}{2} \times (E \times {\rm{strain}}) \times AL \times {\rm{strain}}\)

\({\rm{ = }}\frac{{\rm{1}}}{{\rm{2}}}{\rm{ \times Stress \times Strain \times volume\, of \,material}}\)

The formula of strain energy can also be written as,

\(U = \frac{1}{2} \times \frac{{{{{\rm{(Stress)}}}^{\rm{2}}}}}{{\rm{E}}}{\rm{ \times volume\, of\, material}}\)

Strain Energy Density

If strain energy is distributed inside the material uniformly, then the strain energy per unit volume is known as the strain energy density. Its value is given by,

\({\rm{u = }}\frac{{{\rm{ Total\, strain\, energy }}}}{{{\rm{ Volume\, of\, the\, material }}}}\)

\({\rm{ = }}\frac{{\rm{1}}}{{\rm{2}}}{\rm{ \times Stress \times Strain}}\)

Or

\( = \frac{1}{2} \times \frac{{{{({\rm{ Stress }})}^2}}}{E}\)

Sample Problems on Strain Energy Formula

Q.1.  A bar having an area of \(90 \mathrm{~mm}^{2}\) has a length of \(3\,\rm{m}\). Determine the strain energy if the stress of \(300\,\rm{MPa}\) is applied when stretched. Young’s modulus is given as \(200\,\rm{GPa.}\)
Ans: Given:
Area of the bar, \(\mathrm{A}=90 \mathrm{~mm}^{2}\)
Length of the bar, \(L=3 \mathrm{~m}\)
Stress applied in the bar, \(\sigma=300 \,\mathrm{MPa}\)
Young’s modulus, \(\mathrm{E}=200 \,\mathrm{GPa}\)
Now the volume of the bar is given by,
\({\rm{V = area \times length}}\)
\(=\left(90 \times 10^{-6}\right) \times 3\)
\(=270 \times 10^{-6} \mathrm{~m}^{3}\)
Now the strain energy formula is given as,
\({\rm{U = }}\frac{{\rm{1}}}{{\rm{2}}}{\rm{ \times }}\frac{{{{{\rm{(Stress )}}}^{\rm{2}}}}}{{\rm{E}}}{\rm{ \times volume\, of\, material}}\)
\(=\frac{1}{2} \times \frac{\left(300 \times 10^{6}\right)^{2}}{200 \times 10^{9}} \times 270 \times 10^{-6}\)
\(=60.75 \mathrm{~J}\)
Therefore, the strain energy stored inside the rod is \(60.75 \mathrm{~J} .\)

Summary

When a deformable structure, such as a spring, rubber or metal stretches, then it stores a type of energy known as strain energy. Thus the total strain energy \((U),\) for small deformation is given by \(U = \frac{{EA{{(\Delta L)}^2}}}{{2L}}\). It can also be written in the form of applied stress and produced Strain. It is given as, 

\({\rm{U = }}\frac{{\rm{1}}}{{\rm{2}}}{\rm{ \times Stress \times Strain \times volume\, of\, material}}\) or \({\rm{U = }}\frac{{\rm{1}}}{{\rm{2}}}{\rm{ \times }}\frac{{{{{\rm{( Stress )}}}^{\rm{2}}}}}{{\rm{E}}}{\rm{ \times volume\, of\, material}}{\rm{.}}\)

During derivation of the above formula, Hooke’s law is used. So, the above formula is valid for deformation under elastic material up to the proportional limit. The strain energy per unit volume is known as the strain energy density. It is given by, 

\({\rm{u = }}\frac{{{\rm{ Total\, strain\, energy }}}}{{{\rm{ Volume\, of\, the\, material }}}}{\rm{ = }}\frac{{\rm{1}}}{{\rm{2}}}{\rm{ \times Stress \times Strain}}\)

FAQs

Q.1. What is the SI unit of Stress and Strain?
Ans: The SI unit of stress is Newton per square meter. Or we can express in Pascal. \(1\) Pascal \(=1 \mathrm{~Pa}=1 \mathrm{~N} \mathrm{~m}^{-2}\)
While Strain is a dimensionless quantity, this is because it is the ratio of change of length to the original length.

Q.2. What does the area under stress vs strain graph represent?
Ans: Area under stress vs strain graph represents the work required to stretch the material. The area under the curve up to the elastic limit represents the stored elastic strain energy.

Q.3. What is the SI unit of strain energy?
Ans: Strain energy has the unit of energy or work. Its SI unit is \(\rm{J}\) or \(\rm{Nm}.\) The dimensional formula of strain energy is \(\mathrm{ML}^{2} \mathrm{~T}^{-2}\).

Q.4. If the value of stress is doubled, what will be the effect on strain energy?
Ans: We know that the formula of strain energy Is given by,
\({\rm{U = }}\frac{{\rm{1}}}{{\rm{2}}}{\rm{ \times }}\frac{{{{{\rm{(Stress)}}}^{\rm{2}}}}}{{\rm{E}}}{\rm{ \times volume\, of\, material}}\)
So, when the stress is doubled, then the strain energy will become four times.

Q.5. For the same stress, does the strain energy will depend on the type of material?
Ans: The formula of strain energy can be written as,
\({\rm{U = }}\frac{{\rm{1}}}{{\rm{2}}}{\rm{ \times }}\frac{{{{{\rm{(Stress)}}}^{\rm{2}}}}}{{\rm{E}}}{\rm{ \times volume\, of\, material}}\)
Thus from the above formula, we can observe that the strain energy for a particular magnitude of stress depends on the elastic moduli of the material. We know that the value of elastic moduli will be different for different materials. Thus we can say that the value of strain energy for any particular stress depends on the type of material.

We hope you find this article on ‘Strain Energy Formula‘ helpful. In case of any queries, you can reach back to us in the comments section, and we will try to solve them.