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The curious fact about Airy stress functions is that their use typically leads one to obtain a solution first, and then the next step is to determine what is the actual problem to which the solution applies. It's as if the answer comes before the question.

The governing differential equation for equilibrium expresses \(\sum {\bf F} = m \, {\bf a}\) in terms of derivatives of the stress tensor as

\[ \nabla \cdot \boldsymbol{\sigma} + \rho \, {\bf f} = \rho \, {\bf a} \]

where:

\({\boldsymbol \sigma} \; \) is the stress tensor\(\rho \; \, \) is density

\({\bf f} \; \, \) is the body force vector

\({\bf a} \; \) is the acceleration vector

A derivation of the equilibrium equation is available here. The complete set of equations is

\[ {\partial \sigma_{11} \over \partial x_1} + {\partial \sigma_{12} \over \partial x_2} + {\partial \sigma_{13} \over \partial x_3} + \rho f_x = \rho \, a_x \] \[ {\partial \sigma_{21} \over \partial x_1} + {\partial \sigma_{22} \over \partial x_2} + {\partial \sigma_{23} \over \partial x_3} + \rho f_y = \rho \, a_y \] \[ {\partial \sigma_{31} \over \partial x_1} + {\partial \sigma_{32} \over \partial x_2} + {\partial \sigma_{33} \over \partial x_3} + \rho f_z = \rho \, a_z \]

First, note that in equilibrium (\({\bf a} = 0\)), and in the absence of body forces (\({\bf f} = 0\)), the equilbrium equation, \(\nabla \boldsymbol{\sigma} + \rho \, {\bf f} = \rho \, {\bf a}\), reduces to \(\nabla \boldsymbol{\sigma} = 0\). And in 2-D, the two component equations are simply

\[ {\partial \sigma_{xx} \over \partial x} + {\partial \tau_{xy} \over \partial y} = 0 \quad \qquad \text{and} \qquad \quad {\partial \sigma_{yy} \over \partial y} + {\partial \tau_{xy} \over \partial x} = 0 \]

Next, propose that a scalar function, \(\phi\), exists (this is the Airy stress function) and is related to the 2-D stress components by the following cleverly chosen relationships.

\[ \sigma_{xx} = {\partial^2 \phi \over \partial y^2} \qquad \sigma_{yy} = {\partial^2 \phi \over \partial x^2} \qquad \tau_{xy} = - {\partial^2 \phi \over \partial x \partial y} \]

Then, substituting the above \(\phi\) relationships into the equilibrium equations gives a remarkable result.

\[ {\partial \over \partial x} \left( {\partial^2 \phi \over \partial y^2} \right) - {\partial \over \partial y} \left( {\partial^2 \phi \over \partial x \partial y} \right) = 0 \] \[ {\partial \over \partial y} \left( {\partial^2 \phi \over \partial x^2} \right) - {\partial \over \partial x} \left( {\partial^2 \phi \over \partial x \partial y} \right) = 0 \]

The remarkable result here is that the equilibrium equations are

Take for example, \(\phi = {1 \over 2} A y^2\). This is the solution to something. But what? To find out, take the partial derivatives to determine the stress fields. This leads to

\[ \sigma_{xx} \; = \; {\partial^2 \over \partial y^2} \left( {1 \over 2} A y^2 \right) \; = \; A \]

Therefore, this is easily recognized as a simple case of uniaxial tension in the \(x\) direction. Likewise, letting \(\phi = -B x y\) leads to a state of uniform pure shear in which \(\tau_{xy} = B\).

Nevertheless, nothing is quite THAT easy. There is one limitation on the choice of \(\phi\) that results from the facts that the solutions are restricted to isotropic materials, the strains are related to stresses through Hooke's Law, and they must make physical sense. For example, the strains cannot be so negative that the material folds back on itself. The limitation is that \(\phi\) must satisfy the

\[ {\partial^4 \phi \over \partial x^4} + 2 {\partial^4 \phi \over \partial x^2 \partial y^2} + {\partial^4 \phi \over \partial y^4} = 0 \]

and is abbreviated \(\nabla^4 \phi = 0\). It is not at all intuitive why the restrictions lead to the biharmonic equation, and there is a great deal of tedious algebra required to show it, but it is indeed the case. Any \(\phi\) function satisfying \(\nabla^4 \phi = 0\) is guaranteed to produce stress and strain fields that are in equilibrium for an isotropic solid not subjected to body forces.

Note that any polynomial of degree 3 or less in \(x\) and \(y\) is automatically a solution of the biharmonic equation because the equation contains 4th order derivatives.

\[ \left( {1 \over r} {\partial \over \partial r} \left( r \, {\partial \over \partial r} \right) + {1 \over r^2} {\partial^2 \over \partial \theta^2} \right)^2 \phi = 0 \]

and the relationships for the stress components are

\[ \sigma_{rr} = {1 \over r} {\partial \phi \over \partial r} + {1 \over r^2} {\partial^2 \phi \over \partial \theta^2} \qquad \qquad \sigma_{\theta \theta} = {\partial^2 \phi \over \partial r^2} \qquad \qquad \tau_{r \theta} = - {\partial \over \partial r} \left( {1 \over r} {\partial \phi \over \partial \theta} \right) \]

\[ \phi = - {P' \over \pi} r \, \theta \cos \theta \]

The function can be inserted in the biharmonic equation to verify that it is indeed a solution. The stress components obtained from differentiating the stress function are therefore a valid solution to a particular problem. But which one? To determine that, first evaluate the stresses.

\[ \begin{eqnarray} \sigma_{rr} & = & {1 \over r} {\partial \phi \over \partial r} + {1 \over r^2} {\partial^2 \phi \over \partial \theta^2} \\ \\ & = & - {2 \, P' \over \pi \, r} \cos \theta \\ \\ \\ \sigma_{\theta \theta} & = & {\partial^2 \phi \over \partial r^2} \; = \; 0 \\ \\ \\ \tau_{r \theta} & = & - {\partial \over \partial r} \left( {1 \over r} {\partial \phi \over \partial \theta} \right) \; = \; 0 \end{eqnarray} \]

This stress field results from a distributed line load of zero width. This can be varified by computing the net vertical force due to the radial stress using

\[ \begin{eqnarray} \text{Vertical Load / Length} & = & - \int_{-\pi/2}^{\pi/2} \sigma_{rr} \cos \theta \, r d \theta \qquad \qquad \quad \end {eqnarray} \]

where the \(\cos \theta\) term gives the vertical component of force due to the radial stress. Substituting the expression for \(\sigma_{rr}\) into the equation and integrating gives

\[ \begin{eqnarray} \text{Vertical Load / Length} & = & - \int_{-\pi/2}^{\pi/2} \left( - {2 \, P' \over \pi \, r} \cos \theta \right) \cos \theta \, r d \theta \\ \\ & = & {2\, P' \over \pi} \int_{-\pi/2}^{\pi/2} \cos^2 \theta \; d \theta \\ \\ & = & P' \end{eqnarray} \]