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## Introduction

So far, we have dealt with a very specific loading condition in which a tensile force is applied perpendicular to the crack, e.g., the force is vertical while the crack is horizontal. This is the natural orientation to study in the beginning. But it is not the only one. On this page, we will generalize the force to be at any orientation relative to the crack.

In such a scenario, the general stress state can be partitioned into three components. They consist of a tensile stress (the same one we've studied so far) and two shear stresses, one in-plane and the second out-of-plane. They correspond to the three key crack loading modes and are introduced next.

In general, an object can be loaded in any direction relative to the crack. The sketch at the right shows a force vector at such a random orientation. It is mostly perpendicular to the crack, but also contains components that produce in-plane and out-of-plane shear.

When this occurs, the logical thing to do is partition the force into its fundamental components. This process leads to the three loading modes shown below.

 Mode I loading occurs most often and produces the most damage. Because of this, it naturally receives the most attention in research, structural design, failure analysis, etc. It is commonly called the Opening Mode. Mode II corresponds to shearing of the crack face due to in-plane shear stresses. Mode II loading influences the crack growth direction in a way that minimizes further Mode II loading while maximizing Mode I. Mode III is the Tearing Mode for obvious reasons. It is driven by out-of-plane shear stresses, and does not seem to occur as often as the other two.

The remainder of this page will introduce the stress, strain, and displacement fields for the three loading modes. The discussions will be at a high level because the Mode I case has already been covered in great detail, and derivation of the Mode II and Mode III relationships follows those of Mode I very closely. So it is not necessary to retrace all the steps in detail.

This page effectively serves as a reference document. For more details about the derivation steps, see these pages: (i) Westergaard's Solution for Cracks, (ii) Stress Intensity Factor, and (iii) Crack Tip Displacements.

## Mode I

Recall that in 1939, Westergaard found an Airy stress function, $$\phi$$, that is the solution for the stress field in an infinite plate containing a crack .

$\phi = \text{Re} \, \overline{\overline{Z}} + y \, \text{Im} \, \overline{Z}$
The equations for the stress field are

$\begin{eqnarray} \sigma_{xx} & = & &{\partial^2 \phi \over \partial y^2} & = & \text{Re}\,Z - y \, \text{Im} \, Z' \\ \\ \sigma_{yy} & = & &{\partial^2 \phi \over \partial x^2} & = & \text{Re}\,Z + y \, \text{Im} \, Z' \\ \\ \tau_{xy} & = & - & {\partial^2 \phi \over \partial x \partial y} & = & - y \, \text{Re} \, Z' \end{eqnarray}$
where $$\overline{Z}$$, $$Z$$, and its derivative, $$Z'$$, are

$\overline{Z}(z) = \sigma_\infty \sqrt{z^2 - a^2} \quad \qquad \qquad \quad Z(z) = {\sigma_\infty \over \sqrt{1 - \left( a \over z \right)^2}} \quad \qquad \qquad \quad Z'(z) = {- \sigma_\infty \, a^2 \over z^3 \left[ 1 - \left( a \over z \right)^2 \right]^{3/2} }$
and $$a$$ is crack length, and $$z$$ equals $$x + i y$$. Fortunately, $$\overline{\overline{Z}}$$, the integral of $$\overline{Z}$$, is not needed.

Two decades later, Irwin showed that Westergaard's result could be simplified in the area immediately surrounding the crack tip by expressing $$z$$ as the sum of two terms .

$z = a + r e^{i \theta}$
The key feature of this expression is that $$r = 0$$ at the crack tip ($$x=a$$). Irwin's expression leads to the following equations for the Mode I stress state near the crack tip in terms of polar coordinates, $$r$$ and $$\theta$$.

$\sigma_{xx} = {\sigma_\infty \sqrt{\pi a} \over \sqrt{ 2 \pi r }} \cos {\theta \over 2} \left( 1 - \sin {\theta \over 2} \sin {3 \theta \over 2} \right)$
$\sigma_{yy} = {\sigma_\infty \sqrt{\pi a} \over \sqrt{ 2 \pi r }} \cos {\theta \over 2} \left( 1 + \sin {\theta \over 2} \sin {3 \theta \over 2} \right)$
$\tau_{xy} = {\sigma_\infty \sqrt{\pi a} \over \sqrt{ 2 \pi r }} \cos {\theta \over 2} \sin {\theta \over 2} \cos {3 \theta \over 2}$
An expression for $$\sigma_{yy}$$ along the crack path, is found by setting $$y = 0$$, leaving $$z = x$$.

$\sigma_{yy} = { \sigma_\infty \over \sqrt{1 - \left( a \over x \right)^2} }$
and Irwin's approximation (with $$\theta = 0$$) is

$\sigma_{yy} = { \sigma_\infty \sqrt{ \pi a} \over \sqrt{2 \pi r} }$
The two equations are shown in the graph below. It is clear that both are very close at the crack tip and diverge as the distance from the tip increases. The region of close agreement is approximately $$r \le a/10$$. Beyond this, the approximate expression continues to decrease toward zero because $$1/\sqrt{r}$$ always decreases as $$r$$ increases. In contrast, the exact solution levels out at $$\sigma_\infty$$.

The strain fields are obtained by inserting Westergaard's expressions for stress into into Hooke's Law. The result is

$\begin{eqnarray} \epsilon_{xx} & = & {1 \over 2G} \left[ \left( {\kappa - 1 \over 2 } \right) \, \text{Re} \, Z - y \, \text{Im} \, Z' \right] \\ \\ \epsilon_{yy} & = & {1 \over 2G} \left[ \left( {\kappa - 1 \over 2 } \right) \, \text{Re} \, Z + y \, \text{Im} \, Z' \right] \\ \\ \gamma_{xy} & = & - y \, \text{Re} \, Z' \, / \, G \end{eqnarray}$
The Kolosov constant, $$\kappa$$, in the strain equations is defined as

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

For metals where Poisson's Ratio is approximately 1/3, the values of $$\kappa$$ are

$\kappa \; = \; \left\{ \begin{eqnarray} 5 / 3 & \qquad \text{Plane Strain} \\ \\ 2 \;\; & \qquad \text{Plane Stress} \end{eqnarray} \right.$
Westergaard's exact solution for the displacement field is obtained by integrating the strain equations.

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

$u_y = {\sigma_\infty \over 2G} \left( {\kappa + 1 \over 2 } \right) a \; \sqrt{1 - \left( x \over a \right)^2}$

Irwin's approximation of the Mode I displacement field near the crack tip 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

$u_y = {K \over 2G} \sqrt{ r \over 2 \pi} \; ( \kappa + 1 )$

## Mode II

Recall that Mode II loading consists of in-plane shear stress applied to an infinite plate containing a crack.

Development of expressions for the stress, strain, and displacement fields for Mode II loading is practically identical to that of Mode I. The only difference is to propose a new complex function in which $$i \, \tau_\infty$$ is used for Mode II loading in place of $$\sigma_\infty$$.

$Z(z) = {i \, \tau_\infty \over \sqrt{1 - \left( a \over z \right)^2}}$
The same Airy stress function, $$\phi$$, is used as the solution for the stress field.

$\phi = \text{Re} \, \overline{\overline{Z}} + y \, \text{Im} \, \overline{Z}$
The equations for the stress field remain unchanged

$\begin{eqnarray} \sigma_{xx} & = & &{\partial^2 \phi \over \partial y^2} & = & \text{Re}\,Z - y \, \text{Im} \, Z' \\ \\ \sigma_{yy} & = & &{\partial^2 \phi \over \partial x^2} & = & \text{Re}\,Z + y \, \text{Im} \, Z' \\ \\ \tau_{xy} & = & - & {\partial^2 \phi \over \partial x \partial y} & = & - y \, \text{Re} \, Z' \end{eqnarray}$
but $$\overline{Z}$$, $$Z$$, and its derivative, $$Z'$$, are subtly different

$\overline{Z}(z) = i \, \tau_\infty \sqrt{z^2 - a^2} \quad \qquad \qquad \quad Z(z) = {i \, \tau_\infty \over \sqrt{1 - \left( a \over z \right)^2}} \quad \qquad \qquad \quad Z'(z) = {- i \, \tau_\infty \, a^2 \over z^3 \left[ 1 - \left( a \over z \right)^2 \right]^{3/2} }$
and $$a$$ is crack length, and $$z$$ equals $$x + i y$$. Fortunately, $$\overline{\overline{Z}}$$, the integral of $$\overline{Z}$$, is not needed.

Irwin's approximation for the stress field near the crack tip is

$\sigma_{xx} = {\sigma_\infty \sqrt{\pi a} \over \sqrt{ 2 \pi r }} \cos {\theta \over 2} \left( 1 - \sin {\theta \over 2} \sin {3 \theta \over 2} \right)$
$\sigma_{yy} = {\sigma_\infty \sqrt{\pi a} \over \sqrt{ 2 \pi r }} \cos {\theta \over 2} \left( 1 + \sin {\theta \over 2} \sin {3 \theta \over 2} \right)$
$\tau_{xy} = {\sigma_\infty \sqrt{\pi a} \over \sqrt{ 2 \pi r }} \cos {\theta \over 2} \sin {\theta \over 2} \cos {3 \theta \over 2}$
An expression for $$\sigma_{yy}$$ along the crack path, is found by setting $$y = 0$$, leaving $$z = x$$.

$\sigma_{yy} = { \sigma_\infty \over \sqrt{1 - \left( a \over x \right)^2} }$
and Irwin's approximation (with $$\theta = 0$$) is

$\sigma_{yy} = { \sigma_\infty \sqrt{ \pi a} \over \sqrt{2 \pi r} }$
The two equations are shown in the graph below. It is clear that both are very close at the crack tip and diverge as the distance from the tip increases. The region of close agreement is approximately $$r \le a/10$$. Beyond this, the approximate expression continues to decrease toward zero because $$1/\sqrt{r}$$ always decreases as $$r$$ increases. In contrast, the exact solution levels out at $$\sigma_\infty$$.

The strain fields are obtained by inserting Westergaard's expressions for stress into into Hooke's Law. The result is

$\begin{eqnarray} \epsilon_{xx} & = & {1 \over 2G} \left[ \left( {\kappa - 1 \over 2 } \right) \, \text{Re} \, Z - y \, \text{Im} \, Z' \right] \\ \\ \epsilon_{yy} & = & {1 \over 2G} \left[ \left( {\kappa - 1 \over 2 } \right) \, \text{Re} \, Z + y \, \text{Im} \, Z' \right] \\ \\ \gamma_{xy} & = & - y \, \text{Re} \, Z' \, / \, G \end{eqnarray}$
The Kolosov constant, $$\kappa$$, in the strain equations is defined as

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

For metals where Poisson's Ratio is approximately 1/3, the values of $$\kappa$$ are

$\kappa \; = \; \left\{ \begin{eqnarray} 5 / 3 & \qquad \text{Plane Strain} \\ \\ 2 \;\; & \qquad \text{Plane Stress} \end{eqnarray} \right.$
Westergaard's exact solution for the displacement field is

$\begin{eqnarray} u_x & = & {1 \over 2G} \left[ \left( {\kappa - 1 \over 2 } \right) \, \text{Re} \, \overline{Z} - y \, \text{Im} \, Z \right] \\ \\ u_y & = & {1 \over 2G} \left[ \left( {\kappa + 1 \over 2 } \right) \, \text{Im} \, \overline{Z} - y \, \text{Re} \, Z \right] \end{eqnarray}$
Irwin's crack tip approximation is

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

$u_y = {\sigma_\infty \over 2G} \left( {\kappa + 1 \over 2 } \right) a \; \sqrt{1 - \left( x \over a \right)^2}$

Irwin's approximation of the crack profile is

$u_y = {K \over 2G} \sqrt{ r \over 2 \pi} \; ( \kappa + 1 )$