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Westergaard's Solution for Cracks

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Introduction

In 1939, Harold M. Westergaard developed a solution for the stress field surrounding a crack [1] that has two advantages over Inglis's solution [2]. First, Westergaard's solution applies directly to cracks, not to an ellipse that approaches a crack in the limit. Second, the solution is expressed in rectangular coordinates rather than elliptical coordinates. Granted, Westergaard chose to express the rectangular coordinates as complex numbers, $$z = x + i y$$. Nevertheless, the resulting expressions are much easier to interpret than Inglis's equations in elliptical coordinates. A copy of Westergaard's paper is available here.

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.

Prerequisites

There are two subjects that are essential to know in order to follow the discussion on this page. They are (i) Airy stress functions, and the (ii) calculus of complex numbers. Neither topic is difficult; they are simply not well publicized by most engineering curriculums.

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.

Airy Stress Functions

The use of Airy Stress Functions is a powerful technique for solving 2-D equilibrium elasticity problems. The component equations of equilibrium for 2-D problems without body forces are

${\partial \sigma_{xx} \over \partial x} + {\partial \tau_{xy} \over \partial y} = 0 \quad \qquad \text{and} \qquad \quad {\partial \sigma_{yy} \over \partial y} + {\partial \tau_{xy} \over \partial x} = 0$
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 always satisfied regardless of the choice of $$\phi$$. So any choice of $$\phi$$ is the solution to a problem (well almost). Actually, the function must satisfy the biharmonic equation, which 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 $$\nabla^4 \phi = 0$$. 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.

Complex Numbers

Westergaard chose a function of complex numbers, $$z = x + i y$$, as the Airy stress function for a crack in tension. This permits stress to be expressed as a function of $$x$$ and $$y$$, $$\boldsymbol{\sigma} = f(x,y)$$, but more compactly by using complex numbers. And since stress is the second derivative of an Airy stress function (see above), we need to review the calculus of complex numbers, specifically the Cauchy-Riemann Equations [3], in order to understand Westergaard's solution.

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.

Westergaard's Solution

Westergaard found an Airy stress function of complex numbers that is the solution for the stress field in an infinite plate containing a crack. But before getting to the function, a little more notation is needed. Fortunately, it is simple: The integral of $$Z$$ is represented by a bar, $$\overline{Z}$$, and the integral of $$\overline{Z}$$ is represented by two bars, $$\overline{\overline{Z}}$$. Finally, the derivative of $$Z$$ is represented by $$Z'$$. In summary,

${ d \overline{\overline{Z}} \over dz } = \overline{Z} \qquad \qquad { d \overline{Z} \over dz } = Z \qquad \qquad { dZ \over dz } = Z'$
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 $$z = x + i y$$.

Analysis of the Solution

Westergaard's solution containing complex numbers is amazingly compact compared to Inglis's result involving elliptical coordinates. The only challenge is the necessity of partitioning the square-root function into real and imaginary components. The process requires a Taylor series expansion.

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.

Two decades later, Irwin showed that the solution could be simplified in the area immediately surrounding the crack tip, and invented the stress intensity factor in the process. We will cover this next.

References

1. Westergaard, H.M., "Bearing Pressures and Cracks," Journal of Applied Mechanics, Vol. 6, pp. A49-53, 1939.
2. 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.
3. http://en.wikipedia.org/wiki/Cauchy-Riemann_equations