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Crack Tip Displacements

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Introduction

The objective of this page is to develop expressions for the displacement field at a crack. This is accomplished by first inserting Westergaard's complex-valued expressions for stress into Hooke's Law to obtain strains. Then the strain equations are integrated to obtain displacements. Finally, we apply Irwin's near crack tip approximation to the displacement equations to obtain expressions in terms of the stress intensity factor, \(K\), and polar coordinates, \(r\) and \(\theta\).

This is a very long page containing a great deal of calculus of complex-valued functions. You can review complex numbers and the Cauchy-Riemann relationships here, complex.html, and then review this page, westergaard.html, to see how Westergaard used the complex functions to solve for the stress field. You may also want to review Irwin's near crack tip approximation that was originally applied to Westergaard's solution for stress here, sif.html.

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Westergaard's Solution for Stress

Recall that Westergaard found an Airy stress function of complex numbers that is the solution for the stresses in an infinite plate containing a crack [1]. See westergaard.html for details.

The complete set of equations for the stress field is

\[ \begin{eqnarray} \sigma_{xx} & = & \text{Re}\,Z - y \, \text{Im} \, Z' \\ \\ \sigma_{yy} & = & \text{Re}\,Z + y \, \text{Im} \, Z' \\ \\ \tau_{xy} & = & - y \, \text{Re} \, Z' \end{eqnarray} \]
where \(Z\) and its derivative, \(Z'\), are

\[ Z(z) = {\sigma_\infty \over \sqrt{1 - \left( a \over z \right)^2}} \quad \qquad \text{and} \qquad \quad Z'(z) = {- \sigma_\infty \, a^2 \over z^3 \left[ 1 - \left( a \over z \right)^2 \right]^{3/2} } \]
and \(a\) is crack length, and \(z\) equals \(x + i y\).

Also, the integral of \(Z(z)\) is

\[ \overline{Z} \, = \int Z(z) \, dz \; = \; \sigma_\infty \sqrt{z^2 - a^2} \]
This equation for \(\overline{Z}\) will be needed when the expressions for \(\epsilon_{xx}\) and \(\epsilon_{yy}\) are integrated to obtain displacements, \(u_x\) and \(u_y\).

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Strains

Westergaard's expressions for stress are inserted into Hooke's Law to obtain strains. Hooke's Law equations for the x-y plane are

\[ \begin{eqnarray} \epsilon_{xx} & = & {1 \over E} \left[ \sigma_{xx} - \nu \; (\sigma_{yy} + \sigma_{zz}) \right] \\ \\ \epsilon_{yy} & = & {1 \over E} \left[ \sigma_{yy} - \nu \; (\sigma_{xx} + \sigma_{zz}) \right] \\ \\ \gamma_{xy} & = & \tau_{xy} \, / \, G \end{eqnarray} \]
At this point, the analysis splits into two parallel paths, one for plane stress conditions where \(\sigma_{zz} = 0\), and the second for plane strain conditions where \(\epsilon_{zz} = 0\) and \(\sigma_{zz} = \nu (\sigma_{xx} + \sigma_{yy})\).

Plane Stress

We'll start with plane stress because it's the easier of the two. Inserting Westergaard's expressions for stress into Hooke's Law gives

\[ \begin{eqnarray} \epsilon_{xx} & = & {1 \over E} \left[ ( \text{Re}\,Z - y \, \text{Im} \, Z') - \nu \; ( \text{Re}\,Z + y \, \text{Im} \, Z' ) \right] \\ \\ \epsilon_{yy} & = & {1 \over E} \left[ ( \text{Re}\,Z + y \, \text{Im} \, Z') - \nu \; ( \text{Re}\,Z - y \, \text{Im} \, Z' ) \right] \\ \\ \gamma_{xy} & = & - y \, \text{Re} \, Z' \, / \, G \end{eqnarray} \]
and simplifying yields

\[ \begin{eqnarray} \epsilon_{xx} & = & {1 \over E} \left[ (1 - \nu) \, \text{Re} \, Z - (1 + \nu) \, y \, \text{Im} \, Z' \right] \\ \\ \epsilon_{yy} & = & {1 \over E} \left[ (1 - \nu) \, \text{Re} \, Z + (1 + \nu) \, y \, \text{Im} \, Z' \right] \\ \\ \gamma_{xy} & = & - y \, \text{Re} \, Z' \, / \, G \end{eqnarray} \]

Plane Strain

Plane strain conditions require that \(\sigma_{zz}\) first be determined before inserting the stresses into Hooke's Law. The expression for \(\sigma_{zz}\) is

\[ \begin{eqnarray} \sigma_{zz} & = & \nu \, (\sigma_{xx} + \sigma_{yy}) \\ \\ & = & \nu \, \left[ ( \text{Re} \, Z - y \, \text{Im} \, Z') + ( \text{Re} \, Z + y \, \text{Im} \, Z') \right] \\ \\ & = & 2 \, \nu \, \text{Re} \, Z \end{eqnarray} \]
Inserting this and the remaining stresses into Hooke's Law gives

\[ \begin{eqnarray} \epsilon_{xx} & = & {1 \over E} \left[ ( \text{Re}\,Z - y \, \text{Im} \, Z') - \nu \; ( \text{Re}\,Z + y \, \text{Im} \, Z' ) - 2 \, \nu^2 \, \text{Re} \, Z \right] \\ \\ \epsilon_{yy} & = & {1 \over E} \left[ ( \text{Re}\,Z + y \, \text{Im} \, Z') - \nu \; ( \text{Re}\,Z - y \, \text{Im} \, Z' ) - 2 \, \nu^2 \, \text{Re} \, Z \right] \\ \\ \gamma_{xy} & = & - y \, \text{Re} \, Z' \, / \, G \end{eqnarray} \]
and simplifying produces

\[ \begin{eqnarray} \epsilon_{xx} & = & {(1 + \nu) \over E} \left[ (1 - 2 \nu) \, \text{Re} \, Z - y \, \text{Im} \, Z' \right] \\ \\ \epsilon_{yy} & = & {(1 + \nu) \over E} \left[ (1 - 2 \nu) \, \text{Re} \, Z + y \, \text{Im} \, Z' \right] \\ \\ \gamma_{xy} & = & - y \, \text{Re} \, Z' \, / \, G \end{eqnarray} \]
This completes the solution of the strains at a crack. One could next use Irwin's near crack tip approximation to obtain expressions in terms of \(r\) and \(\theta\) similar to those for stress [2]. However, there is little interest in strains compared to stresses and displacements. Therefore, we will skip Irwin's approximation and continue on to solve for displacements by integrating the strain expressions. We will then apply Irwin's approximation to the displacement results.

Kolosov Constant

In summary, the strains at a crack for the case of plane stress are

\[ \begin{eqnarray} \epsilon_{xx} & = & {1 \over E} \left[ (1 - \nu) \, \text{Re} \, Z - (1 + \nu) \, y \, \text{Im} \, Z' \right] \\ \\ \epsilon_{yy} & = & {1 \over E} \left[ (1 - \nu) \, \text{Re} \, Z + (1 + \nu) \, y \, \text{Im} \, Z' \right] \\ \\ \gamma_{xy} & = & - y \, \text{Re} \, Z' \, / \, G \end{eqnarray} \]
and for plane strain, the strains are

\[ \begin{eqnarray} \epsilon_{xx} & = & {(1 + \nu) \over E} \left[ (1 - 2 \nu) \, \text{Re} \, Z - y \, \text{Im} \, Z' \right] \\ \\ \epsilon_{yy} & = & {(1 + \nu) \over E} \left[ (1 - 2 \nu) \, \text{Re} \, Z + y \, \text{Im} \, Z' \right] \\ \\ \gamma_{xy} & = & - y \, \text{Re} \, Z' \, / \, G \end{eqnarray} \]
The only difference between the two sets of equations for \(\epsilon_{xx}\) and \(\epsilon_{yy}\) is the dependence on Poisson's Ratio, \(\nu\). Remarkably, both sets of equations can be generalized into one master set by introducing the Kolosov constant, \(\kappa\), defined as

\[ \kappa \; = \; \left\{ \begin{eqnarray} 3 - 4 \nu & \qquad \text{Plane Strain} \\ \\ {3 - \nu \over 1 + \nu} & \qquad \text{Plane Stress} \end{eqnarray} \right. \]
* It appears to be standard practice to list the plane strain relationship first, followed by the plane stress expression. I don't know why.

The strain fields can now be described by one set of equations applicable to both plane stress and plane strain conditions with only the value of \(\kappa\) differing between the two cases.

\[ \begin{eqnarray} \epsilon_{xx} & = & {1 \over 2G} \left[ \left( {\kappa - 1 \over 2 } \right) \, \text{Re} \, Z - y \, \text{Im} \, Z' \right] \\ \\ \epsilon_{yy} & = & {1 \over 2G} \left[ \left( {\kappa - 1 \over 2 } \right) \, \text{Re} \, Z + y \, \text{Im} \, Z' \right] \\ \\ \gamma_{xy} & = & - y \, \text{Re} \, Z' \, / \, G \end{eqnarray} \]
For metals where Poisson's Ratio is approximately 1/3, the values of \(\kappa\) are

\[ \kappa \; = \; \left\{ \begin{eqnarray} 5 / 3 & \qquad \text{Plane Strain} \\ \\ 2 \;\; & \qquad \text{Plane Stress} \end{eqnarray} \right. \]
The different values of \(\kappa\) reflect the fact that normal strains under plane stress conditions are greater than under plane strain conditions at equal far field stress, \(\sigma_\infty\).

Do not confuse this difference with the fact that a thin object deforms more than a thick object under the same load. Keep in mind that we are not looking at equal load, but instead at equal stress.

The reduced strain under plane strain conditions is due to the elevated normal stress, \(\sigma_{zz}\), present in plane strain loading, but absent in plane stress loading. Poisson effects tied to \(\sigma_{zz}\) work to reduce the normal strains.

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Displacements

The relationships between the strain components and displacements are

\[ \epsilon_{xx} = {\partial u_x \over \partial x} \qquad \qquad \epsilon_{yy} = {\partial u_y \over \partial y} \qquad \qquad \gamma_{xy} = {\partial u_x \over \partial y} + {\partial u_y \over \partial x} \]
These equations permit us to integrate \(\epsilon_{xx}\) with respect to \(x\) to obtain \(u_x\) and integrate \(\epsilon_{yy}\) with respect to \(y\) to obtain \(u_y\). And that is just what we are going to do, except... it's not trivial because we're dealing with complex numbers.

Forturnately, we have the Cauchy-Riemann equations to help us (see complex.html for more information). They are

\[ \begin{eqnarray} \text{Re} \, { dZ \over dz } \; & \; = \; & \; {\partial \, \text{Re}\, Z \over \partial x} & = & \;\;\;{\partial \, \text{Im}\,Z \over \partial y} \\ \\ \text{Im} \, { dZ \over dz } \; & \; = \; & \; {\partial \, \text{Im}\,Z \over \partial x} & = & - {\partial \, \text{Re}\, Z \over \partial y} \end{eqnarray} \]
Both normal strain equations contain \(\text{Re} Z\) and \(y \, \text{Im} Z'\), and both of these terms must be integrated with respect to \(x\) and \(y\).

For the case of integrating \(\text{Re} Z\) with respect to \(x\), the relevant part of the Cauchy-Riemann equations is

\[ \text{Re} \, { dZ \over dz } \; = \; {\partial \, \text{Re}\, Z \over \partial x} \]
And therefore, integration by \(x\) produces

\[ \int \text{Re} \, Z \, dx = \text{Re} \left( \int Z \, dz \right) = \text{Re} \, \overline{Z} \]
Likewise, integration of the imaginary term with respect to \(x\) gives

\[ \int y \; \text{Im}\,Z' \, dx = y \; \text{Im} \left( \int Z' \, dz \right) = y \; \text{Im} \, Z \]
Note that \(y\) is treated as a constant here because the integration is with respect to \(x\). For the \(u_y\) displacement field, we must integrate \(\text{Re} Z\) and \(y\,\text{Im} Z'\) with respect to \(y\). Integration of the first term gives

\[ \int \text{Re} \, Z \, dy = \text{Im} \left( \int Z \, dz \right) = \text{Im} \, \overline{Z} \]
Integration of the second term requires some trial and error (more like wild guess and error, actually). It is rather tricky due to the presence of \(y\) in the \(y \, \text{Im}\,Z'\) term.

\[ \int y \, \text{Im}\,Z' \, dy = \text{Im} \, \overline{Z} - y \, \text{Re} \, Z \] The integration identities lead to the following equations for the displacements.

\[ \begin{eqnarray} u_x & = & {1 \over 2G} \left[ \left( {\kappa - 1 \over 2 } \right) \, \text{Re} \, \overline{Z} - y \, \text{Im} \, Z \right] \\ \\ u_y & = & {1 \over 2G} \left[ \left( {\kappa - 1 \over 2 } \right) \, \text{Im} \, \overline{Z} + \text{Im} \, \overline{Z} - y \, \text{Re} \, Z \right] \\ \\ & = & {1 \over 2G} \left[ \left( {\kappa + 1 \over 2 } \right) \, \text{Im} \, \overline{Z} - y \, \text{Re} \, Z \right] \end{eqnarray} \]

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Crack Opening Profile

As is typical with complex-valued functions, it is practically impossible to interpret what the equations for \(u_x\) and \(u_y\) are telling us. Does the displacement field vary linearly, quadratically, logarithmically, etc, in any particular direction? It is just not possible to determine from simple observation.

However, the crack opening profile is an exception to this complexity. It is very easy to compute. All we need is \(u_y\) at \(y = 0\) and \(|x| \le a\).

Setting \(y = 0\) leaves \(z = x\) and \(u_y\) reduces to

\[ u_y = {1 \over 2G} \left( {\kappa + 1 \over 2 } \right) \, \text{Im} \, \overline{Z} \]
With \(z = x\), \(\; \overline{Z}\) reduces to \( \; \sigma_\infty \sqrt{x^2 - a^2}\).  And since \(|x| \le a\), the imaginary part of \(\overline{Z}\) is

\[ \text{Im} \, \overline{Z}_{\begin{matrix} y = 0 \\ |x| \le a \end{matrix}} \; = \; \text{Im} \left( \sigma_\infty \sqrt{x^2 - a^2} \right) \; = \; \sigma_\infty \sqrt{a^2 - x^2} \]
and the full expression for \(u_y\) is

\[ u_y = {\sigma_\infty \over 2G} \left( {\kappa + 1 \over 2 } \right) \sqrt{a^2 - x^2} \]
This result reveals a great deal about the crack's deformed shape. First, the derivative of its displacement is infinite at the crack tip, \(x = a\). But this should not be a surprise because it is already known that the stresses and strains are infinite at the tip. Also, the crack forms an ellipse when deformed. Finally, the equation can be written as

\[ u_y = {\sigma_\infty \over 2G} \left( {\kappa + 1 \over 2 } \right) a \; \sqrt{1 - \left( x \over a \right)^2} \]
which shows that the entire displacement profile can be expressed in terms of the dimensionless parameter, \(x/a\), and is proportional to the crack's length, \(a\) (or \(2a\) if you prefer).

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Irwin's Near Crack Tip Approximation

Although the crack opening profile is relatively easy to compute, it does not tell us about the displacements in the remaining area around the crack tip. However, Irwin's near crack tip approximation can be used to gain insight into this.

Irwin's approximation can be used to obtain displacement expressions near the crack tip just as was done with the original stress expressions. This will reveal the behavior near the crack tip in terms of distance, \(r\), and direction, \(\theta\), from it.

Recall the process begins by expressing \(z\) as the sum of two terms.

\[ z = a + r e^{i \theta} \]
This equation is inserted into the two expressions for \(\overline{Z}\) and \(Z\) and negligible terms will be dropped just as Irwin did with the original stress equations. The process for \(Z\) has already been presented in detail here: sif.html.

It led to

\[ Z = \sigma_\infty \sqrt{ a \over 2 \, r } \; \left( \cos {\theta \over 2} - i \, \sin {\theta \over 2} \right) \]
We will repeat the process for \(\overline{Z}\) next. Substituting \(z = a + r e^{i \theta}\) into \(\overline{Z}\) gives

\[ \begin{eqnarray} \overline{Z} & = & \sigma_\infty \sqrt{z^2 - a^2} \\ \\ & = & \sigma_\infty \sqrt{ (a + r e^{i \theta})^2 - a^2} \\ \\ & = & \sigma_\infty \sqrt{ 2 a r e^{i \theta} + r^2 e^{i 2 \theta}} \end{eqnarray} \]
Recognize that \(r \lt \lt a\) near the crack tip, and therefore \(r^2 \lt \lt ar\). This permits the 2nd term to be neglected, leaving

\[ \overline{Z} = \sigma_\infty \sqrt{ 2 \, a \, r} \, e^{i {\theta \over 2} } \]
Apply Euler's identity, \(e^{i \phi} = \cos \phi + i \, \sin \phi\), to obtain the final relationship for \(\overline{Z}\).

\[ \overline{Z} = \sigma_\infty \sqrt{2 \, a \, r } \; \left( \cos {\theta \over 2} + i \, \sin {\theta \over 2} \right) \]
Now insert the real and imaginary parts of \(\overline{Z}\) and \(Z\) into the equations for \(u_x\) and \(u_y\).

\[ \begin{eqnarray} u_x & = & {1 \over 2G} \left[ \left( {\kappa - 1 \over 2 } \right) \, \text{Re} \, \overline{Z} - y \, \text{Im} \, Z \right] \\ \\ & = & {1 \over 2G} \left[ \left( {\kappa - 1 \over 2 } \right) \, \left( \sigma_\infty \sqrt{2 \, a \, r } \; \cos {\theta \over 2} \right) - y \, \left( - \sigma_\infty \sqrt{ a \over 2 \, r } \; \sin {\theta \over 2} \right) \right] \end{eqnarray} \]
Substitute \(r \sin \theta\) for \(y\), and then substitute \(2 \sin {\theta \over 2} \cos {\theta \over 2}\) for \(\sin \theta\).

\[ u_x = {1 \over 2G} \left[ \left( {\kappa - 1 \over 2 } \right) \left( \sigma_\infty \sqrt{2 \, a \, r } \; \cos {\theta \over 2} \right) - \left( 2 \, r \sin {\theta \over 2} \cos {\theta \over 2} \right) \left( - \sigma_\infty \sqrt{ a \over 2 \, r } \; \sin {\theta \over 2} \right) \right] \]
After much clean-up, one arrives at

\[ u_x = {\sigma_\infty \sqrt{a} \over 2G} \sqrt{ r \over 2} \cos {\theta \over 2} \left[ \kappa - 1 + 2 \sin^2 {\theta \over 2} \right] \]
And as is common, multiply by \(\sqrt{\pi / \pi}\)

\[ u_x = {\sigma_\infty \sqrt{\pi a} \over 2G} \sqrt{ r \over 2 \pi} \cos {\theta \over 2} \left[ \kappa - 1 + 2 \sin^2 {\theta \over 2} \right] \]
Finally, recognize that \(\sigma_\infty \sqrt{\pi a}\) is the stress intensity factor, \(K\). So the expression becomes

\[ u_x = {K \over 2G} \sqrt{ r \over 2 \pi} \cos {\theta \over 2} \left[ \kappa - 1 + 2 \sin^2 {\theta \over 2} \right] \]
This is the final result for \(u_x\). As was the case for stress, we see that the entire displacement field is proportional to the stress intensity factor, \(K\). Also, the dependences on \(K\), \(r\), and \(\theta\) are all completely independent from each other. The key difference between the stress field and displacement field is that one depends on \(\sqrt{1/r}\) while the other depends on \(\sqrt{r}\). This should not in fact be very surprising. After all, if the displacements depend on \(\sqrt{r}\), then the strains will depend on its derivative, which involves a \(\sqrt{1/r}\) term. And if the strains depend on \(\sqrt{1/r}\), then the stresses will as well because the two are linearly related.

There is no need to repeat all the steps required to obtain \(u_y\). They are the same as for \(u_x\). The result is

\[ u_y = {K \over 2G} \sqrt{ r \over 2 \pi} \sin {\theta \over 2} \left[ \kappa + 1 - 2 \cos^2 {\theta \over 2} \right] \]
and to keep all the important results together, recall the Kolosov constant is

\[ \kappa \; = \; \left\{ \begin{eqnarray} 3 - 4 \nu & \qquad \text{Plane Strain} \\ \\ {3 - \nu \over 1 + \nu} & \qquad \text{Plane Stress} \end{eqnarray} \right. \]

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Summary

Westergaard's exact solution for displacements at the crack shown here is

\[ \begin{eqnarray} u_x & = & {1 \over 2G} \left[ \left( {\kappa - 1 \over 2 } \right) \, \text{Re} \, \overline{Z} - y \, \text{Im} \, Z \right] \\ \\ u_y & = & {1 \over 2G} \left[ \left( {\kappa + 1 \over 2 } \right) \, \text{Im} \, \overline{Z} - y \, \text{Re} \, Z \right] \end{eqnarray} \]
where

\[ Z(z) = {\sigma_\infty \over \sqrt{1 - \left( a \over z \right)^2}} \quad \qquad \text{and} \qquad \quad \overline{Z} = \; \sigma_\infty \sqrt{z^2 - a^2} \]
Kolosov's constant is

\[ \kappa \; = \; \left\{ \begin{eqnarray} 3 - 4 \nu & \qquad \text{Plane Strain} \\ \\ {3 - \nu \over 1 + \nu} & \qquad \text{Plane Stress} \end{eqnarray} \right. \]
Displacement of the crack profile is

\[ u_y = {\sigma_\infty \over 2G} \left( {\kappa + 1 \over 2 } \right) a \; \sqrt{1 - \left( x \over a \right)^2} \]
The total crack opening displacement is \(2 u_y\).

\[ d = 2 \, u_y = {\sigma_\infty \over 2G} ( \kappa + 1 ) \; a \; \sqrt{1 - \left( x \over a \right)^2} \]
Irwin's crack tip approximation is

\[ \begin{eqnarray} u_x & = & {K \over 2G} \sqrt{ r \over 2 \pi} \cos {\theta \over 2} \left[ \kappa - 1 + 2 \sin^2 {\theta \over 2} \right] \\ \\ u_y & = & {K \over 2G} \sqrt{ r \over 2 \pi} \sin {\theta \over 2} \left[ \kappa + 1 - 2 \cos^2 {\theta \over 2} \right] \end{eqnarray} \]
Irwin's approximation of the crack profile is

\[ u_y = {K \over 2G} \sqrt{ r \over 2 \pi} \; ( \kappa + 1 ) \]
Irwin's approximation of the cracking opening displacement is

\[ d = 2 \, u_y = {K \over G} \sqrt{ r \over 2 \pi} \; ( \kappa + 1 ) \]

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References

1. Westergaard, H.M., "Bearing Pressures and Cracks," Journal of Applied Mechanics, Vol. 6, pp. A49-53, 1939.
2. Irwin, G.R., "Analysis of Stresses and Strains Near the End of a Crack Traversing a Plate," Journal of Applied Mechanics, Vol. 24, pp. 361-364, 1957.