# Airy stress function

## Definition

The Airy stress function (${\displaystyle \varphi }$ ):

• Scalar potential function that can be used to find the stress.
• Satisfies equilibrium in the absence of body forces.
• Only for two-dimensional problems (plane stress/plane strain).

### Airy stress function in rectangular Cartesian coordinates

If the coordinate basis is rectangular Cartesian ${\displaystyle (\mathbf {e} _{1},~\mathbf {e} _{2})}$  with coordinates denoted by ${\displaystyle (x_{1},~x_{2})}$  then the Airy stress function ${\displaystyle (\varphi )}$  is related to the components of the Cauchy stress tensor ${\displaystyle ({\boldsymbol {\sigma }})}$  by

{\displaystyle {\begin{aligned}\sigma _{11}&=\varphi _{,22}={\cfrac {\partial ^{2}\varphi }{\partial x_{2}^{2}}}\\\sigma _{22}&=\varphi _{,11}={\cfrac {\partial ^{2}\varphi }{\partial x_{1}^{2}}}\\\sigma _{12}&=-\varphi _{,12}=-{\cfrac {\partial ^{2}\varphi }{\partial x_{1}\partial x_{2}}}\end{aligned}}}

Alternatively, if we write the basis as ${\displaystyle (\mathbf {e} _{x},\mathbf {e} _{y})}$  and the coordinates as ${\displaystyle (x,y)\,}$ , then the Cauchy stress components are related to the Airy stress function by

{\displaystyle {\begin{aligned}\sigma _{xx}&={\cfrac {\partial ^{2}\varphi }{\partial y^{2}}}\\\sigma _{yy}&={\cfrac {\partial ^{2}\varphi }{\partial x^{2}}}\\\sigma _{xy}&=-{\cfrac {\partial ^{2}\varphi }{\partial x\partial y}}\end{aligned}}}

### Airy stress function in polar coordinates

In polar basis ${\displaystyle (\mathbf {e} _{r},\mathbf {e} _{\theta })}$  with co-ordinates ${\displaystyle (r,\theta )\,}$ , the Airy stress function is related to the components of the Cauchy stress via

{\displaystyle {\begin{aligned}\sigma _{rr}&={\cfrac {1}{r}}{\cfrac {\partial \varphi }{\partial r}}+{\cfrac {1}{r^{2}}}{\cfrac {\partial ^{2}\varphi }{\partial \theta ^{2}}}\\\sigma _{\theta \theta }&={\cfrac {\partial ^{2}\varphi }{\partial r^{2}}}\\\sigma _{r\theta }&=-{\cfrac {\partial }{\partial r}}\left({\cfrac {1}{r}}{\cfrac {\partial \varphi }{\partial \theta }}\right)\end{aligned}}}
 Something to think about ... Do you think the Airy stress function can be extended to three dimensions?

## Stress equation of compatibility in 2-D

In the absence of body forces,

${\displaystyle \nabla ^{2}{(\sigma _{11}+\sigma _{22})}=0}$

or,

${\displaystyle \sigma _{11,11}+\sigma _{11,22}+\sigma _{22,11}+\sigma _{22,22}=0\,}$
• Note that the stress field is independent of material properties in the absence of body forces (or homogeneous body forces).
• Therefore, the plane strain and plane stress solutions are the same if the boundary conditions are expressed as traction BCS.

In terms of the Airy stress function

${\displaystyle \varphi _{,1122}+\varphi _{,2222}+\varphi _{,1111}+\varphi _{,1122}=0\,}$

or,

${\displaystyle {\cfrac {\partial ^{4}\varphi }{\partial x_{1}^{4}}}+2{\cfrac {\partial ^{4}\varphi }{\partial x_{1}^{2}\partial x_{2}^{2}}}+{\cfrac {\partial ^{4}\varphi }{\partial x_{2}^{4}}}=0}$

or,

${\displaystyle \nabla ^{4}\varphi =0\,}$
• The stress function ${\displaystyle (\varphi )}$  is biharmonic.
• Any polynomial in ${\displaystyle x_{1}}$  and ${\displaystyle x_{2}}$  of degree less than four is biharmonic.
• Stress fields that are derived from an Airy stress function which satisfies the biharmonic equation will satisfy equilibrium and correspond to compatible strain fields.

## Some biharmonic Airy stress functions

In cylindrical co-ordinates, some biharmonic functions that may be used as Airy stress functions are

{\displaystyle {\begin{aligned}\varphi &=C\theta \\\varphi &=Cr^{2}\theta \\\varphi &=Cr\theta \cos \theta \\\varphi &=Cr\theta \sin \theta \\\varphi &=f_{n}(r)\cos(n\theta )\\\varphi &=f_{n}(r)\sin(n\theta )\\\end{aligned}}}

where

{\displaystyle {\begin{aligned}f_{0}(r)&=a_{0}r^{2}+b_{0}r^{2}\ln r+c_{0}+d_{0}\ln r\\f_{1}(r)&=a_{1}r^{3}+b_{1}r+c_{1}r\ln r+d_{1}r^{-1}\\f_{n}(r)&=a_{n}r^{n+2}+b_{n}r^{n}+c_{n}r^{-n+2}+d_{n}r^{-n}~,~~n>1\end{aligned}}}

## Displacements in terms of scalar potentials

If the body force is negligible, then the displacements components in 2-D can be expressed as

${\displaystyle 2\mu u_{1}=-\varphi _{,1}+\alpha \psi _{,2}~,~~2\mu u_{2}=-\varphi _{,2}+\alpha \psi _{,1}}$

where,

${\displaystyle \alpha ={\begin{cases}1-\nu &{\text{for plane strain}}\\{\cfrac {1}{1+\nu }}&{\text{for plane stress}}\end{cases}}}$

and ${\displaystyle \psi (x_{1},x_{2})}$  is a scalar displacement potential function that satisfies the conditions

${\displaystyle \nabla ^{2}{\psi }=0~,~~\psi _{,12}=\nabla ^{2}{\varphi }}$

To prove the above, you have to use the plane strain/stress constitutive relations

${\displaystyle {\begin{matrix}\sigma _{\alpha \beta }&=2\mu \left[\varepsilon _{\alpha \beta }+\left({\cfrac {1-\alpha }{2\alpha -1}}\right)\varepsilon _{\gamma \gamma }\delta _{\alpha \beta }\right]\\\varepsilon _{\alpha \beta }&={\cfrac {1}{2\mu }}\left[\sigma _{\alpha \beta }+\left(1-\alpha \right)\sigma _{\gamma \gamma }\delta _{\alpha \beta }\right]\end{matrix}}}$

Note also that the plane stress/strain compatibility equations can be written as

${\displaystyle \nabla ^{2}{\sigma _{\gamma \gamma }}=-{\cfrac {1}{\alpha }}f_{\gamma ,\gamma }}$