Airy's Stress Functions
============================
>Airy's stress functions pose a useful technique for solving two dimensional linear elasticity problems in equilibrium.

Given a two dimensional continuous and isotropic elastic body, subjected to external tractions $t_i(x_1,x_2)$

```{image} ../LEFM/Gen_2D.png
:alt: general 2D shape
:width: 400px
:align: center
```
and assuming that the body forces can be written as a derivative of a scalar function of positions such that :

$$
\rho F_1 = \frac{\partial \Omega}{\partial x_1} \quad ; \quad \rho F_2 = \frac{\partial \Omega}{\partial x_2} \quad ; \quad \rho F_3 = 0 \text{   (!2D)}
$$

there exist a function $\Phi(x_1,x_2)$ for which :

$$
\nabla^2\nabla^2\Phi = \frac{\partial^4{\Phi}}{\partial x^4} +2\frac{\partial^4{\Phi}}{\partial x^2 \partial y^2} +\frac{\partial^4{\Phi}}{\partial y^4} =\left ( \frac{\partial F_1}{\partial x_1} + \frac{\partial F_2}{\partial x_2}\right)C(\nu) 
$$

where

$$
C(\nu) = \begin{cases}
\frac{1}{1-\nu},& \text{plane stress}\\
\\
\frac{1-\nu}{1-2\nu},& \text{plane strain}
\end{cases}
$$


For the case on no body forces $\Phi$ satisfied the bi-harmonic equation 

$$
\nabla^2\nabla^2\Phi = \frac{\partial^4{\Phi}}{\partial x^4} +2\frac{\partial^4{\Phi}}{\partial x^2 \partial y^2} +\frac{\partial^4{\Phi}}{\partial y^4} =0 
$$

After demanding that $\Phi$ satisfies the boundary conditions, we can use $\Phi$ to derive the stresses:

$$ \sigma_{xx} = \frac{\partial^2{\Phi}}{\partial{y}^2} \quad ; \quad
 \sigma_{yy} = \frac{\partial^2{\Phi}}{\partial{x}^2} \quad ; \quad
 \tau_{xy} = \frac{\partial^2{\Phi}}{\partial{x}\partial{y}} $$

 $\Phi$ is guaranteed to satisfy both $\color{red}{\text{equilibrium}}$ and $\color{green}{\text{compatability}}$ equations automatically:

$$
\begin{align*}
& \color{red}{\frac{\partial \sigma_{xx}}{\partial x} +\frac{\partial \tau_{xy}}{\partial y} =0} \quad  ; \quad \color{red}{\frac{\partial \sigma_{yy}}{\partial y} +\frac{\partial \tau_{xy}}{\partial x} =0 } \\
\\ 
 & \color{green}{\nabla^2\left ( \sigma_{xx} + \sigma_{yy}\right ) = 0 }
\end{align*}
$$

We can use the the constitutive relation to extract the strains from the stresses and if the displacement filed is sought after, the strains can be integrated. 

## Airy stress functions in polar coordinates

In polar coordinates, where $u_{\theta}$ and $u_r$ represent the tangential and radial displacements, the strains are given by:

$$ \epsilon_{rr} = \frac{\partial u_r}{\partial_r} \quad ; \quad \epsilon_{\theta \theta} = \frac{1}{r}\frac{\partial u_{\theta}}{\partial_{\theta}} + \frac{u_r}{r} \quad ; \quad \epsilon_{r \theta} = \frac{1}{2} \left ( \frac{1}{r}\frac{\partial u_{r}}{\partial_{\theta}} +\frac{\partial u_{\theta}}{\partial_{r}} - \frac{u_{\theta}}{r} \right) $$

and the $\color{red}{\text{equilibrium}}$ and $\color{green}{\text{compatability}}$ equations are: 

$$ 
\begin{align*}
 &\color{red}{\frac{\partial \sigma_{rr}}{\partial r} +\frac{1}{r}\frac{\partial \tau_{r \theta}}{\partial \theta} + \frac{\sigma_{rr} - \sigma_{\theta \theta}}{r} =0} \quad ; \quad \color{red}{\frac{1}{r}\frac{\partial \sigma_{\theta \theta}}{\partial \theta} +\frac{\partial \tau_{r \theta}}{\partial r} + \frac{2\tau_{r \theta}}{r} =0} \\
\\
&\color{green}{\nabla^2\left ( \sigma_{r r} + \sigma_{\theta \theta} \right ) = 0 }
\end{align*}
$$

In polar coordinates, stresses are expressed as:

$$
\begin {align*}
\sigma_{rr}& = \frac{1}{r^2} \frac{\partial^2 \Phi}{\partial \theta ^2} + \frac{1}{r}\frac{\partial \Phi}{\partial r}\\
\\
\tau_{r \theta}& = \frac{1}{r^2} \frac{\partial^2 \Phi}{\partial \theta \partial r} + \frac{1}{r^2}\frac{\partial \Phi}{\partial \theta} \\
\\
\sigma_{\theta \theta}& = \frac{\partial ^2 \Phi}{\partial r^2}
\end{align*}
$$


Next, we will see how to obtain the stress field around a circular hole in an infinite plate subjected to arbitrary remote tractions using Airy's stress function.