# elongation

## Axial Deformation

In the linear portion of the stress-strain diagram, the tress is proportional to strain and is given by

$\sigma = E \varepsilon$

since $\sigma = P / A$ and $\varepsilon = \delta / L$, then $\dfrac{P}{A} = E \dfrac{\delta}{L}$

$\delta = \dfrac{PL}{AE} = \dfrac{\sigma L}{E}$

To use this formula, the load must be axial, the bar must have a uniform cross-sectional area, and the stress must not exceed the proportional limit.