notes/uni/mmme/2053_mechanics_of_solids/strain_energy.md

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

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}

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

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).

Elastic Axial Loading

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

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

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.