---
author: Akbar Rahman
date: \today
title: MMME2053 // Strain Energy
tags: [ strain_energy ]
uuid: 9cbe6fab-c733-45d0-b8e7-b3d949b3eeb7
lecture_slides: [ ./lecture_slides/MMME2053 SE L1 Slides.pdf, ./lecture_slides/MMME2053 SE L2 Slides.pdf, ./lecture_slides/MMME2053 SE L3 Slides.pdf ]
lecture_notes: [ ./lecture_notes/Strain Energy Methods Notes.pdf ]
worked_examples: [ ./worked_examples/MMME2053 SE WE Slides.pdf ]
exercise_sheets: [ ./exercise_sheets/Strain Energy Methods Exercise Sheet.pdf, ./exercise_sheets/Strain Energy Methods Exercise Sheet Solutions.pdf ]
---

# Strain Energy Definition

This section refers to the first two slide sets.

Strain energy in a body is equal to the work done on the body by the applies loads:

$$U = \int_0^u P\mathrm du$$

![](./images/vimscrot-2023-02-16T14:19:48,580021845+00:00.png)

## Bending

\begin{equation}
U = \int_0^\Phi M\mathrm d\Phi = \int^L_0\frac{M^2}{2EI}\delta s
\label{eqn:bendingeqn1}
\end{equation}

![](./images/vimscrot-2023-02-16T14:22:41,358964129+00:00.png)

If this material represents an element of a larger beam of length $L$ and curvature of radius $R$

![](./images/vimscrot-2023-02-16T14:41:32,526245370+00:00.png)

The strain energy within this element will be:

$$\delta U = \frac12 M\delta \Phi$$

From the elastic beam bending equation we know:

$$\frac MI = \frac ER$$

and as the angle subtended by the element is equal to the change in slope, the expression for the
arc created by the element is:

$$\delta s = R\delta \Phi$$

Eliminating $R$ from the above two equations and rearranging gives

$$\delta \Phi = \frac{M}{EI}\delta s$$

This can be substituted into equation \ref{eqn:bendingeqn1} (left and middle) to get

$$\delta U = \frac{M^2}{2EI}\delta s$$

Integrate over full length of beam to get total strain energy:

$$U = \int^L_0\frac{M^2}{2EI}\delta s$$

## Torsion

$$U = \int_0^\theta T\mathrm d\theta = \int^L_0 \frac{T^2}{2GJ} \delta s$$

Derivations for the equation above is analogous to those for bending and axial loads, and can be
found in the second lecture slide set (p9-p11).

![](./images/vimscrot-2023-02-16T14:23:06,415583742+00:00.png)

## Elastic Axial Loading

\begin{equation}
U = \frac12 Pu = \int^L_0\frac{P^2}{2EA}\delta s
\label{eqn:elasticaxialloading1}
\end{equation}

![](./images/vimscrot-2023-02-16T14:26:48,809335151+00:00.png)

If this material represents an element, of length $\delta s$, of a larger beam of length $L$, and
the change in length of this element due to the applied load $P$ is $\delta u$ ten the strain energy
within the element is

$$\delta U = \frac12 P\delta u$$


![](./images/vimscrot-2023-02-16T14:40:43,750030500+00:00.png)

> Note that there are transverse strains/displacements due to Poisson's effects but there are no
> transverse stresses/loads, thus there is no work done in the transverse direction.

Axial strain is:

$$\epsilon = \frac{\delta u}{\delta s}$$

Equating this to Hooke's law yields:

$$\delta u = \frac{P}{EA}\delta s$$

which can be substituted into equation \ref{eqn:elasticaxialloading1} (left and middle) to get

$$\delta U = \frac{P^2}{2EA}\delta s$$

Integrate over full length of beam to get total strain energy:

$$U = \int^L_0\frac{P^2}{2EA}\delta s$$

# Castigliano's Theorem

This section refers to the third slide set.