Mixed Mode Fracture

Mixed Mode Fracture#

Motivation

So far we have discussed mostly mode I cracks (tensile) however real life scenarios will often involve not only tensile tractions ths leading to mixed mode interactions

Mixed-mode scenarios will often be found in heterogenous structures (multi-phase materials, weldements, coatings, composites etc.).

We can use superposition and obtain the crack tip fields as

\begin{array}{r} σ_{11} = \frac{K_{I}}{\sqrt{2 π r}} [\cos \frac{θ}{2} (1 - \sin \frac{θ}{2} \sin \frac{2 θ}{2})] + \frac{K_{I I}}{\sqrt{2 π r}} [- \sin \frac{θ}{2} (2 + \cos \frac{θ}{2} \cos \frac{2 θ}{2})] \\ σ_{22} = \frac{K_{I}}{\sqrt{2 π r}} [\cos \frac{θ}{2} (1 + \sin \frac{θ}{2} \sin \frac{2 θ}{2})] + \frac{K_{I I}}{\sqrt{2 π r}} [\sin \frac{θ}{2} \cos \frac{θ}{2} \cos \frac{2 θ}{2}] \\ σ_{12} = \frac{K_{I}}{\sqrt{2 π r}} [\cos \frac{θ}{2} \sin \frac{θ}{2} \sin \frac{2 θ}{2}] + \frac{K_{I I}}{\sqrt{2 π r}} [\cos \frac{θ}{2} (1 - \sin \frac{θ}{2} \sin \frac{2 θ}{2})] \end{array}

And the displacements:

\begin{array}{r} u_{1} = \frac{K_{I}}{2 μ} \sqrt{\frac{r}{2 π}} [\cos \frac{θ}{2} (κ - \cos θ)] + \frac{K_{I I}}{2 μ} \sqrt{\frac{r}{2 π}} [\sin \frac{θ}{2} (κ - + \cos θ)] \\ u_{2} = \frac{K_{I}}{2 μ} \sqrt{\frac{r}{2 π}} [\sin \frac{θ}{2} (κ - \cos θ)] + \frac{K_{I I}}{2 μ} \sqrt{\frac{r}{2 π}} [- \cos \frac{θ}{2} (κ - 2 + \cos θ)] \end{array}

Energy release rate#

We can write the total energy release write to be

\begin{array}{r} G = \frac{K_{I}^{2}}{E^{'}} + \frac{K_{I I}^{2}}{E^{'}} + \frac{(1 + ν) K_{I I I}^{2}}{E} \\ E^{'} = E / (1 - ν^{2}) for plane strain \\ E^{'} = E for plane stress \end{array}

Assuming we already measured the critical value of $G_{I} = G_{I C} = \frac{K_{I C}^{2}}{E^{'}}$

We can see that for mixed mode loading we obtain

K_{I C}^{2} = K_{I}^{2} + K_{I I}^{2} + \frac{E^{'} (1 + ν)}{E} K_{I I I}^{2}

Question

What is the meaning of the expression we just obtained?

For the 2D problem sketched above we can write the stresses as

\begin{array}{r} σ_{x x} = σ \cos^{2} β \\ σ_{y y} = σ \sin^{2} β \\ τ_{x y} = σ \cos β \sin β \end{array}

leading to

\begin{array}{r} K_{I} = σ_{y y} \sqrt{π a} = σ \sqrt{π a} \sin^{2} β \\ K_{I I} = τ_{x y} \sqrt{π a} = σ \sqrt{π a} \sin β \cos β \end{array}

and then

K_{I C}^{2} = K_{I}^{2} + K_{I I}^{2} \to K_{I C} = σ_{f} \sqrt{π a} \sin β

Example

Consider a plate with a $10 m m$ crack at angle $β$ subjected to an applied stress of $130 M P a$ . Assuming that the $K_{I C}$ is known and equal $25 M P a \sqrt{m}$ find out if the plate will fail for any value of $β$

Using the above equations we can find the values of $K_{I}$ and $K_{I I}$ as a function of the angle to be

\begin{array}{r} K_{I} = 130 M P a \sqrt{π 10 e - 3} \sin^{2} β = 23 \sin^{2} β M P a \sqrt{m} \\ K_{I I} = 130 M P a \sqrt{π 10 e - 3} \sin β \cos β = \sin β \cos β M P a \sqrt{m} \end{array}

and then

\sqrt{K_{I}^{2} + K_{I I}^{2}} = 23 * \sqrt{\sin^{4} β + (\sin β \cos β)^{2}}

This expression will always yield a result smaller than $25 M P a \sqrt{m}$

Under mixed mode conditions, cracks may propagate at an inclined angle ( $θ_{c}$ ) to their original direction ( $β$ )

Writing $θ_{c} = \frac{π}{2} - β$ and using the $K_{I C}$ expression we obtain

K_{I C} = σ_{f} \sqrt{π a} \cos θ_{C}

I the crack will grow aligned with the $x$ direction, this will lead to $K = K_{I}$ and $K_{I I} \to 0$ .

For some materials, it was observed that the relation

\K_{I C} = \sqrt{\frac{2}{3}}

is a better approximation to the mixed-mode conditions leading to

K_{I}^{2} + \frac{2}{3} K_{I I}^{2} = K_{I C}^{2}

Which leads to

K_{I C} = σ_{f} \sqrt{\frac{π a}{3} (3 - \cos^{2} β)} \sin β

Principle stress#

The principle stress criterion requires that the crack will grow in the direction perpendicular to the maximum principal stress.

\begin{array}{r} \frac{\partial σ_{θ θ}}{\partial θ} = 0 for θ = θ_{c} \\ \frac{\partial^{2} σ_{θ θ}}{\partial θ^{2}} < 0 f o r σ_{θ θ} > 0 \end{array}

Considering the same 2D problem as before, and setting $τ_{r θ} = 0$ (so that $σ_{θ θ}$ will become a principal stress) we can rewrite the expressions for the stress intensity factors :

\begin{array}{r} K_{I} \sin θ_{c} + K_{I I} (3 \cos θ_{c} - 1) = 0 \end{array}

we define $K_{r} = \frac{K_{I}}{K_{I I}}$ to obtain

θ_{c} = - \arccos \frac{1}{3} [1 - \frac{K_{r} (K_{r} - 3 \sqrt{K_{r}^{2} + 8}}{K_{R}^{2} + 9}] for θ_{c} < \frac{p i}{2}

This gives us a way of finding the crack growth direction if we know $K_{r}$

For a pure mode II loading we will obtain $θ_{c} = - {70.53}^{o}$

The maximum principle stress can be derived to be (some trgo which we skip) :

σ_{θ} (θ = θ_{c}) = \frac{1}{\sqrt{2 π r}} \cos {\frac{θ_{c}}{2}}^{2} [K_{I} \cos \frac{θ_{c}}{2} - 3 K_{I I} \sin \frac{θ_{c}}{2}]

and using the definition of $K_{I C}$ for a pure mode I we obtain

K_{I C} = K_{I} \cos^{3} \frac{θ_{c}}{2} - 3 K_{I I} \sin \frac{θ_{c}}{2} \cos^{2} \frac{θ_{c}}{2}

Looking at the angle for a pure mode II ( $θ_{c} = - 70.53$ ) we can see that under this criterion

K_{I I C} = 1.15 K_{I C}

Strain energy density criterion#

Sih proposed that the strain energy density $S$ dictates the crack growth direction such that crack growth will start in the direction for which :

\begin{array}{r} \frac{\partial S}{\partial θ} = 0 \\ \frac{\partial^{2} S}{\partial θ^{2}} > 0 \end{array}

To use this criterion we thus have to define $W_{c}$ at $θ = θ_{c}$ .

The strain energy can be derived using the stress fields and the definition of work done on a body to obtain

\begin{aligned} S = & K_{I}^{2} \frac{1}{16 G} [(1 + \cos θ) (κ - \cos θ)] \\ + & K_{I I}^{2} \frac{1}{16 G} [(κ + 1) (1 - \cos θ) + (1 + \cos θ) (3 \cos θ - 1)] \\ + & K_{I} K_{I I} \frac{1}{16 G} \sin θ [2 \cos θ) - (κ - 1)] \end{aligned}

Excersice

Consider the same plate as before being subjected to a mixed mode loading under plane strain conditions. We are given with the following inputs:

The plate fractures if $σ_{y y} = 150 M P a$ or $τ_{x y} = 90 M P a$
The crack length is $a = 30 m m$
$ν = 0.33$ and $E = \1 95 G P a$

Use the strain energy density criteria to calculate:

$θ_{c}$
$β$
$K_{I} C$ and $K_{I I C}$

Compare the results to what you would have obtained using the maximum principle stress criterion.

Guidance principla stress

Find the stress intensity factors.
Find $K_{r}$
Use the equation we derived for $θ_{c}$ based on $K_{r}$
FInd $β$
Calculate $K_{I C}$ and $K_{I I C}$

strain energy density

calculate $κ$
use the relation between $S$ and the stress intensity factors to find $θ_{c}$ by taking the derivative with respect to $θ$ as $0$ and setting $θ = θ_{c}$ 3.Calculate $K_{I C}$ and $K_{I I C}$
Find $S_{C}$