Airy stress function

Definition

The Airy stress function ($\phi$):

• 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 $({𝐞}_1,{𝐞}_2)$ with coordinates denoted by $(x_1,x_2)$ then the Airy stress function $(\phi )$ is related to the components of the Cauchy stress tensor $(\mathit{\sigma })$ by $\begin{array}{cc}{\sigma }_{11}& ={\phi }_{,22}=\frac{{\partial }^2\phi }{\partial x_2^2}\\ {\sigma }_{22}& ={\phi }_{,11}=\frac{{\partial }^2\phi }{\partial x_1^2}\\ {\sigma }_{12}& =-{\phi }_{,12}=-\frac{{\partial }^2\phi }{\partial x_1\partial x_2}\end{array}$ Alternatively, if we write the basis as $({𝐞}_x,{𝐞}_y)$ and the coordinates as $(x,y)$, then the Cauchy stress components are related to the Airy stress function by $\begin{array}{cc}{\sigma }_{xx}& =\frac{{\partial }^2\phi }{\partial y^2}\\ {\sigma }_{yy}& =\frac{{\partial }^2\phi }{\partial x^2}\\ {\sigma }_{xy}& =-\frac{{\partial }^2\phi }{\partial x\partial y}\end{array}$

Airy stress function in polar coordinates In polar basis $({𝐞}_r,{𝐞}_{\theta })$ with co-ordinates $(r,\theta )$, the Airy stress function is related to the components of the Cauchy stress via $\begin{array}{cc}{\sigma }_{rr}& =\frac{1}{r}\frac{\partial \phi }{\partial r}+\frac{1}{r^2}\frac{{\partial }^2\phi }{\partial {\theta }^2}\\ {\sigma }_{\theta \theta }& =\frac{{\partial }^2\phi }{\partial r^2}\\ {\sigma }_{r\theta }& =-\frac{\partial }{\partial r}\left(\frac{1}{r}\frac{\partial \phi }{\partial \theta }\right)\end{array}$

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, Template:Center top${\mathrm{\nabla }}^2({\sigma }_{11}+{\sigma }_{22})=0$Template:Center bottom or, Template:Center top${\sigma }_{11,11}+{\sigma }_{11,22}+{\sigma }_{22,11}+{\sigma }_{22,22}=0$Template:Center bottom

• 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

 Template:Center top${\phi }_{,1122}+{\phi }_{,2222}+{\phi }_{,1111}+{\phi }_{,1122}=0$Template:Center bottom

or,

 Template:Center top$\frac{{\partial }^4\phi }{\partial x_1^4}+2\frac{{\partial }^4\phi }{\partial x_1^2\partial x_2^2}+\frac{{\partial }^4\phi }{\partial x_2^4}=0$Template:Center bottom

or,

 Template:Center top${\mathrm{\nabla }}^4\phi =0$Template:Center bottom

• The stress function $(\phi )$ is biharmonic.
• Any polynomial in $x_1$ and $x_2$ 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

$\begin{array}{cc}\phi & =C\theta \\ \phi & =C{r}^2\theta \\ \phi & =Cr\theta \cos \theta \\ \phi & =Cr\theta \sin \theta \\ \phi & =f_n(r)\cos (n\theta )\\ \phi & =f_n(r)\sin (n\theta )\\ \end{array}$

where

$\begin{array}{cc}{f}_0(r)& =a_0{r}^2+b_0{r}^2\ln r+c_0+d_0\ln r\\ f_1(r)& =a_1{r}^3+b_{1}r+c_{1}r\ln r+d_1{r}^{-1}\\ f_n(r)& =a_n{r}^{n+2}+b_n{r}^n+c_n{r}^{-n+2}+d_n{r}^{-n},n>1\end{array}$

Displacements in terms of scalar potentials

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

 Template:Center top$2\mu u_1=-{\phi }_{,1}+\alpha {\psi }_{,2},2\mu u_2=-{\phi }_{,2}+\alpha {\psi }_{,1}$Template:Center bottom

where,

 Template:Center top$\alpha =\{\begin{array}{cc}1-\nu & \text{for plane strain}\\ \frac{1}{1+\nu }& \text{for plane stress}\end{array}$Template:Center bottom

and $\psi (x_1,x_2)$ is a scalar displacement potential function that satisfies the conditions

 Template:Center top${\mathrm{\nabla }}^2\psi =0,{\psi }_{,12}={\mathrm{\nabla }}^2\phi$Template:Center bottom

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

 Template:Center top$\begin{array}{cc}{\sigma }_{\alpha \beta }& =2\mu [{\epsilon }_{\alpha \beta }+\left(\frac{1-\alpha }{2\alpha -1}\right){\epsilon }_{\gamma \gamma }{\delta }_{\alpha \beta }]\\ {\epsilon }_{\alpha \beta }& =\frac{1}{2\mu }[{\sigma }_{\alpha \beta }+(1-\alpha ){\sigma }_{\gamma \gamma }{\delta }_{\alpha \beta }]\end{array}$Template:Center bottom

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

 Template:Center top${\mathrm{\nabla }}^2{\sigma }_{\gamma \gamma }=-\frac{1}{\alpha }{f}_{\gamma ,\gamma }$Template:Center bottom

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