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+++ b/uni/mmme/2053_mechanics_of_solids/strain_energy.md
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+---
+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 \label{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.
+
+