Mixed Mode Fracture ============================ ```{admonition} 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.). ```{image} ../LEFM/mix1.png :alt: mix plate :width: 500px :align: center ``` We can use superposition and obtain the crack tip fields as $$ \sigma_{11} = \frac{K_I}{\sqrt{ 2 \pi r}} \left [ \cos \frac{\theta}{2} \left ( 1- \sin \frac{\theta}{2} \sin \frac{2\theta}{2} \right ) \right ] +\frac{K_{II}}{\sqrt{ 2 \pi r}} \left [ -\sin \frac{\theta}{2} \left ( 2+ \cos \frac{\theta}{2} \cos \frac{2\theta}{2} \right ) \right ] \\ \sigma_{22} = \frac{K_I}{\sqrt{ 2 \pi r}} \left [ \cos \frac{\theta}{2} \left ( 1+ \sin \frac{\theta}{2} \sin \frac{2\theta}{2} \right ) \right ] +\frac{K_{II}}{\sqrt{ 2 \pi r}} \left [ \sin \frac{\theta}{2} \cos \frac{\theta}{2} \cos \frac{2\theta}{2} \right ] \\ \sigma_{12} = \frac{K_I}{\sqrt{ 2 \pi r}} \left [ \cos \frac{\theta}{2} \sin \frac{\theta}{2} \sin \frac{2\theta}{2} \right ] +\frac{K_{II}}{\sqrt{ 2 \pi r}} \left [ \cos \frac{\theta}{2} \left ( 1- \sin \frac{\theta}{2} \sin \frac{2\theta}{2} \right ) \right ] $$ And the displacements: $$ u_1 = \frac{K_I}{2 \mu} \sqrt{ \frac{r}{2 \pi}} \left [ \cos \frac{\theta}{2} \left ( \kappa - \cos \theta \right ) \right ] +\frac{K_{II}}{2 \mu} \sqrt{ \frac{r}{2 \pi}} \left [ \sin \frac{\theta}{2} \left ( \kappa - + \cos \theta \right ) \right ] \\ u_2 = \frac{K_I}{2 \mu} \sqrt{ \frac{r}{2 \pi}} \left [ \sin \frac{\theta}{2} \left ( \kappa - \cos \theta \right ) \right ] +\frac{K_{II}}{2 \mu} \sqrt{ \frac{r}{2 \pi}} \left [ - \cos \frac{\theta}{2} \left ( \kappa -2 + \cos \theta \right ) \right ] $$ ## Energy release rate We can write the total energy release write to be $$ G = \frac{K_I^2}{E'}+\frac{K_{II}^2}{E'} + \frac{(1+\nu)K_{III}^2}{E} \\ E' = E/(1-\nu^2) \ \ \text{for plane strain} \\ E' = E \ \ \text{for plane stress} $$ Assuming we already measured the critical value of $G_I=G_{IC}=\frac{K_{IC}^2}{E'}$ We can see that for mixed mode loading we obtain $$ K_{IC}^2=K_I^2+K_{II}^2+\frac{E'(1+\nu)}{E}K_{III}^2 $$ ```{admonition} Question What is the meaning of the expression we just obtained? ``` For the 2D problem sketched above we can write the stresses as $$ \sigma_{xx} = \sigma \cos^2\beta \\ \sigma_{yy} = \sigma \sin^2\beta \\ \tau_{xy} = \sigma \cos\beta\sin\beta \\ $$ leading to $$ K_I = \sigma_{yy}\sqrt{\pi a} = \sigma \sqrt{\pi a} \sin^2 \beta\\ K_{II} = \tau_{xy}\sqrt{\pi a} = \sigma \sqrt{\pi a} \sin \beta\cos\beta $$ and then $$ K_{IC}^2 = K_I^2 + K_{II}^2 \rightarrow K_{IC}=\sigma_f \sqrt{\pi a} \sin \beta $$ ```{admonition} Example Consider a plate with a $10mm$ crack at angle $\beta$ subjected to an applied stress of $130 MPa$. Assuming that the $K_{IC}$ is known and equal $25 MPa\sqrt{m}$ find out if the plate will fail for any value of $\beta$ Using the above equations we can find the values of $K_I$ and $K_{II}$ as a function of the angle to be $$ K_I = 130MPa\sqrt{\pi 10e-3}\sin^2\beta = 23\sin^2\beta MPa\sqrt{m}\\ K_{II} = 130MPa\sqrt{\pi 10e-3}\sin\beta\cos\beta =\sin\beta\cos\beta MPa\sqrt{m}\\ $$ and then $$ \sqrt{K_I^2+K_{II}^2} = 23*\sqrt{\sin^4\beta + (\sin\beta\cos\beta)^2} $$ This expression will always yield a result smaller than $25MPa\sqrt{m}$ ``` Under mixed mode conditions, cracks may propagate at an inclined angle ($\theta_c$) to their original direction ($\beta$) Writing $\theta_c=\frac{\pi}{2}-\beta$ and using the $K_{IC}$ expression we obtain $$ K_{IC} = \sigma_f \sqrt{\pi a} \cos\theta_C $$ I the crack will grow aligned with the $x$ direction, this will lead to $K=K_I$ and $K_{II} \to 0$. For some materials, it was observed that the relation $$ \K_{IC} = \sqrt{\frac{2}{3}} $$ is a better approximation to the mixed-mode conditions leading to $$ K_I^2 + \frac{2}{3}K_{II}^2 = K_{IC}^2 $$ Which leads to $$ K_{IC} = \sigma_f \sqrt{\frac{\pi a}{3} (3-\cos^2\beta)}\sin\beta $$ ## Principle stress The principle stress criterion requires that the crack will grow in the direction perpendicular to the maximum principal stress. $$ \frac{\partial \sigma_{\theta \theta}}{\partial \theta} = 0 \ \ \text{for} \theta=\theta_c \\ \frac{\partial^2 \sigma_{\theta \theta}}{\partial \theta^2} < 0 \ \ for \sigma_{\theta \theta}>0 $$ Considering the same 2D problem as before, and setting $\tau_{r \theta}=0$ (so that $\sigma_{\theta \theta}$ will become a principal stress) we can rewrite the expressions for the stress intensity factors : $$ K_I \sin \theta_c + K_{II}(3\cos\theta_c -1)=0 \\ $$ we define $K_r = \frac{K_I}{K_{II}}$ to obtain $$ \theta_c = -\arccos \frac{1}{3}\left [ 1 - \frac{K_r (K_r -3 \sqrt{K_r^2+8} }{K_R^2+9}\right ] \ \ \text{for} \ \ \theta_c<\frac{pi}{2} $$ 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 $\theta_c = -70.53^o$ The maximum principle stress can be derived to be (some trgo which we skip) : $$ \sigma_{\theta}(\theta=\theta_c) = \frac{1}{\sqrt{2\pi r}}\cos\frac{\theta_c}{2}^2 \left [ K_I\cos\frac{\theta_c}{2} - 3K_{II} \sin \frac{\theta_c}{2} \right] $$ and using the definition of $K_{IC}$ for a pure mode I we obtain $$ K_{IC} = K_I\cos^3\frac{\theta_c}{2} - 3K_{II} \sin \frac{\theta_c}{2} \cos^2 \frac{\theta_c}{2} $$ Looking at the angle for a pure mode II ($\theta_c = -70.53$) we can see that under this criterion $$ K_{IIC} = 1.15K_{IC} $$ ## 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 : $$ \frac{\partial S}{\partial \theta} = 0 \\ \frac{\partial^2 S}{\partial \theta^2} > 0 \\ $$ To use this criterion we thus have to define $W_c$ at $\theta=\theta_c$. The strain energy can be derived using the stress fields and the definition of work done on a body to obtain ```{math} S = &K_I^2 \frac{1}{16G} [ (1+\cos \theta)(\kappa -\cos \theta)] \\ + &K_{II}^2 \frac{1}{16G} [(\kappa +1) (1-\cos \theta) + (1+\cos \theta)(3\cos \theta -1)] \\ + &K_I K_{II} \frac{1}{16G}\sin \theta [ 2\cos \theta) -(\kappa -1)] ``` ```{admonition} 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 $\sigma_{yy} = 150MPa$ or $\tau_{xy}=90MPa$ - The crack length is $a=30 mm$ - $\nu = 0.33$ and $E=\195GPa$ Use the strain energy density criteria to calculate: 1. $\theta_c$ 2. $\beta$ 3. $K_IC$ and $K_{IIC}$ Compare the results to what you would have obtained using the maximum principle stress criterion. **Guidance** *principla stress* 1. Find the stress intensity factors. 2. Find $K_r$ 3. Use the equation we derived for $\theta_c$ based on $K_r$ 4. FInd $\beta$ 5. Calculate $K_{IC}$ and $K_{IIC}$ *strain energy density* 1. calculate $\kappa$ 2. use the relation between $S$ and the stress intensity factors to find $\theta_c$ by taking the derivative with respect to $\theta$ as $0$ and setting $\theta= \theta_c$ 3.Calculate $K_{IC}$ and $K_{IIC}$ 4. Find $S_C$ ```