Introduction
The use of
Airy Stress Functions is a powerful technique for solving 2-D
equilibrium problems. They are covered here because the approach was
used by several researchers in the mid 1900's to develop analytical solutions
to linear elastic problems involving cracks.
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.
Equilibrium
The story of Airy stress functions begins with the concept of
equilibrium. Everyone knows that \(\sum {\bf F} = m \, {\bf a}\), and if the object
is in equilibrium, then \({\bf a} = 0\) and \(\sum {\bf F} = 0\).
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
\[
{\partial \sigma_{11} \over \partial x_1} +
{\partial \sigma_{12} \over \partial x_2} +
{\partial \sigma_{13} \over \partial x_3} + \rho f_x = \rho \, a_x
\]
\[
{\partial \sigma_{21} \over \partial x_1} +
{\partial \sigma_{22} \over \partial x_2} +
{\partial \sigma_{23} \over \partial x_3} + \rho f_y = \rho \, a_y
\]
\[
{\partial \sigma_{31} \over \partial x_1} +
{\partial \sigma_{32} \over \partial x_2} +
{\partial \sigma_{33} \over \partial x_3} + \rho f_z = \rho \, a_z
\]
Airy Stress Functions
Airy stress functions are used to solve 2-D equilibrium problems.
The approach will be presented here for the special case of no body forces.
First, note that in equilibrium (\({\bf a} = 0\)),
and in the absence of body forces (\({\bf f} = 0\)),
the
equilibrium 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
always satisfied
regardless of the choice of \(\phi\). So any choice of \(\phi\) is the solution to
a problem (well almost, more on this in a moment).
But which problem? Indeed, when one works with Airy stress functions, one
can find oneself with a solution, but not know what problem it is a solution to!
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
Biharmonic Equation. It is
\[
{\partial^4 \phi \over \partial x^4} + 2 {\partial^4 \phi \over \partial x^2 \partial y^2} +
{\partial^4 \phi \over \partial y^4} = 0
\]
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.
Polar Coordinates
In polar coordinates, the biharmonic equation is
\[
\left( {1 \over r} {\partial \over \partial r} \left( r \, {\partial \over \partial r} \right) +
{1 \over r^2} {\partial^2 \over \partial \theta^2} \right)^2 \phi = 0
\]
and the relationships for the stress components are
\[
\sigma_{rr} = {1 \over r} {\partial \phi \over \partial r} + {1 \over r^2} {\partial^2 \phi \over \partial \theta^2}
\qquad \qquad
\sigma_{\theta \theta} = {\partial^2 \phi \over \partial r^2}
\qquad \qquad
\tau_{r \theta} = - {\partial \over \partial r} \left( {1 \over r} {\partial \phi \over \partial \theta} \right)
\]
Line Load Example
The case of a distributed linear load \(P'\) on an infinite solid can be solved with Airy
stress functions in polar coordinates. The stress function in this case is
\[
\phi = - {P' \over \pi} r \, \theta \sin \theta
\]
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
verified 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
\[
\begin{eqnarray}
\text{Vertical Load / Length} & = & - \int_{-\pi/2}^{\pi/2} \left( - {2 \, P' \over \pi \, r} \cos \theta \right) \cos \theta \, r d \theta \\
\\
& = & {2\, P' \over \pi} \int_{-\pi/2}^{\pi/2} \cos^2 \theta \; d \theta \\
\\
& = & P'
\end{eqnarray}
\]
