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The one caveat to note is that Westergaard's solution applies to an infinite plate in equalbiaxial, not uniaxial, tension. Nevertheless, it does provide much additional insight into the stress fields surrounding cracks.

There is little need to continue with this page if you are not familiar with these two subjects. This page will present a very high level summary of each. The following webpages go into more detail: (i) Airy stress functions, and (ii) calculus of complex numbers.

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

The Airy stress function, \(\phi\), is related to the 2-D stress components by the following cleverly chosen relationships.

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

Substituting these \(\phi\) relationships into the equilibrium equations gives the following 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 is that the equilibrium equations are

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

and is abbreviated \(\nabla^4 \phi = 0\). 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. Also, the strain fields are never so negative that they describe the material as folding back on itself, a physical impossibility.

The Cauchy-Riemann equations are relationships between the many different types of derivatives of complex functions. They are

\[ \begin{eqnarray} \text{Re} \, { dZ \over dz } & = & {\partial \, \text{Re}Z \over \partial x} & = & \;\;\;\;{\partial \, \text{Im}Z \over \partial y} \\ \\ \text{Im} \, { dZ \over dz } & = & {\partial \, \text{Im}Z \over \partial x} & = & - {\partial \, \text{Re}Z \over \partial y} \end{eqnarray} \]

The equations lead to some interesting, if not intuitive, relationships among the derivatives. For example, just as any complex function, \(Z(z)\), can be separated into real and imaginary parts like \(Z(z) = \text{Re}Z + i\,\text{Im}Z\), so can its derivative. This looks like

\[ { dZ \over dz } = \text{Re} \, { dZ \over dz } + i \, \text{Im} \, { dZ \over dz } \]

But thanks to the Cauchy-Riemann equations, the derivative can also be expressed as

\[ \begin{eqnarray} { dZ \over dz } & = & \text{Re} \, { dZ \over dz } + i \, \text{Im} \, { dZ \over dz } \\ \\ & = & {\partial \, \text{Re}Z \over \partial x} + i \, {\partial \, \text{Im}Z \over \partial x} \\ \\ & = & {\partial \, \text{Im}Z \over \partial y} - i \, {\partial \, \text{Re}Z \over \partial y} \\ \end{eqnarray} \]

Go to complex.html for a detailed explanation and proof of the Cauchy-Riemann equations.

\[ { d \overline{\overline{Z}} \over dz } = \overline{Z} \qquad \qquad { d \overline{Z} \over dz } = Z \qquad \qquad { dZ \over dz } = Z' \]

The notation is useful because, as we will see, the Airy stress function, \(\phi\), will be in terms of \(\overline{\overline{Z}}\), but the stresses are all functions of 2nd derivatives of \(\phi\). So it is not necessary to know explicitly what \(\overline{Z}\) and \(\overline{\overline{Z}}\) are; the notation is only needed to keep track of the integrals.

Westergaard's choice for the Airy stress function, \(\phi\), was

\[ \phi = \text{Re} \, \overline{\overline{Z}} + y \, \text{Im} \, \overline{Z} \]

with

\[ Z(z) = {\sigma_\infty \over \sqrt{1 - \left( a \over z \right)^2}} \]

where \(a\) is crack length.

The first step following selection of an Airy stress function is to confirm that it satisfies \(\nabla^4 \phi = 0\). Trust me, this one does, but we'll skip the proof because of all the tedious steps required to show it. Nevertheless, there is one fascinating fact to note regarding the proof. It is that the differential equation is satified by \(\phi = \text{Re} \, \overline{\overline{Z}} + y \, \text{Im} \, \overline{Z}\) regardless of the choice of \(Z(z)\). This made it possible for Westergaard to select many different functions for \(Z(z)\), each being the solution to a different problem. The function for \(Z(z)\) listed above prooved to be the solution for a crack in an infinite plate. To see this, take the derivatives of \(\phi\) to get expressions for the stress components.

The expression for \(\sigma_{xx}\) is obtained by taking the derivatives of \(\phi\) with respect to \(y\). The first derivative is

\[ \begin{eqnarray} {\partial \phi \over \partial y} & = & - \text{Im}\,\overline{Z} + \text{Im}\,\overline{Z} + y \, \text{Re} \, Z \\ \\ & = & y \, \text{Re} \, Z \end{eqnarray} \]

Note that the product rule of differentiation was used here, and will be used again to find the 2nd derivative.

The 2nd derivative gives \(\sigma_{xx}\).

\[ \sigma_{xx} \; = \; {\partial^2 \phi \over \partial y^2} \; = \; \text{Re}\,Z - y \, \text{Im} \, Z' \]

In a similar manner, the expression for \(\sigma_{yy}\) is obtained by taking the derivatives of \(\phi\) with respect to \(x\).

\[ {\partial \phi \over \partial x} \; = \; \text{Re}\,\overline{Z} + y \, \text{Im} \, Z \]

And taking the 2nd derivative gives \(\sigma_{yy}\).

\[ \sigma_{yy} \; = \; {\partial^2 \phi \over \partial x^2} \; = \; \text{Re}\,Z + y \, \text{Im} \, Z' \]

The result for \(\tau_{xy}\) is

\[ \tau_{xy} \; = \; - {\partial^2 \phi \over \partial x \partial y} \; = \; - y \, \text{Re} \, Z' \]

The complete set of equations for the stress field is

\[ \begin{eqnarray} \sigma_{xx} & = & &{\partial^2 \phi \over \partial y^2} & = & \text{Re}\,Z - y \, \text{Im} \, Z' \\ \\ \sigma_{yy} & = & &{\partial^2 \phi \over \partial x^2} & = & \text{Re}\,Z + y \, \text{Im} \, Z' \\ \\ \tau_{xy} & = & - & {\partial^2 \phi \over \partial x \partial y} & = & - y \, \text{Re} \, Z' \end{eqnarray} \]

And recall that

\[ Z(z) = {\sigma_\infty \over \sqrt{1 - \left( a \over z \right)^2}} \]

where \(a\) is the crack length and \(z = x + i y\).

Fortunately, a series expansion can be avoided along the plane of the crack where \(y = 0\), and this permits easy insight into the stress field. When \(y = 0\), the equations reduce to

\[ \begin{eqnarray} \sigma_{xx}|_{y=0} & = & \text{Re}\,Z \\ \\ \sigma_{yy}|_{y=0} & = & \text{Re}\,Z \\ \\ \tau_{xy}|_{y=0} & = & 0 \end{eqnarray} \]

And the square root function simplifies because \(z = x\) when \(y = 0\).

\[ Z(x) = {\sigma_\infty \over \sqrt{1 - \left( a \over x \right)^2}} \]

At this point, it becomes clear that the solution is indeed that of a cracked plate in tension. Recall that \(\sigma_{yy} = \text{Re}\,Z\), and \(Z(x)\) is real only when \(x \gt a\). In contrast, when \(x \lt a\), then \(Z(x)\) is imaginary (because the argument of the square-root function is negative) and therefore has no real part. So \(\sigma_{yy} = 0\). This is the stress state in the crack plane (\(y = 0\)): zero stress on the crack's face, infinite stress at the crack tip, and then decreasing back to \(\sigma_\infty\) with increasing \(x\).

The \(\sigma_{xx}\) and \(\sigma_{yy}\) stresses along \(y = 0\) and \(x \gt a\) are

\[ \sigma_{xx} \; = \; \sigma_{yy} \; = \; { \sigma_\infty \over \sqrt{1 - \left( a \over x \right)^2} } \]

A plot is shown below. It shows the stress value quickly dropping from infinity at the crack tip, \(x = a\), to the far-field value of \(\sigma_\infty\). Keep in mind that this is along the crack plane, \(y=0\). As reflected by the equation, the entire stress field is proportional to \(\sigma_\infty\).

The equation and its graph are the key results of Westergaard's solution that are shared by most authors, and with good reason. Computing the stress at any other position near the crack tip requires a Taylor series expansion in order to partition the function into its real and imaginary parts... a great deal of work.

- Westergaard, H.M., "Bearing Pressures and Cracks,"
*Journal of Applied Mechanics,*Vol. 6, pp. A49-53, 1939. - Inglis, C.E., "Stresses in Plates Due to the Presence of Cracks and Sharp Corners,"
*Transactions of the Institute of Naval Architects,*Vol. 55, pp. 219-241, 1913. - http://en.wikipedia.org/wiki/Cauchy-Riemann_equations