Contents

# Shape optimization

## Crack path problem

Here, we use the symbols used in Section 2.2.1 . The simplest problem for finding a crack path is to find the direction in the kinked crack such as $D=(-1,1)^2$, \begin{eqnarray*} \Sigma _0&=&\{(x_1,x_2):\, x_2=0,-1\le x_1\le 0\} \\ \Sigma ^{\alpha }(t)&=&\Sigma _0\cup \{(x_1,x_2):\, x_1=s\cos \alpha , x_2=s\sin \alpha , 0\le s\le t\} \end{eqnarray*} In this case, $\mathcal{G}(f,\Sigma ^{\alpha }(\cdot )) = \lim _{t→+0}t^{-1}\left [\mathcal{E}(u;f,\Omega )-\mathcal{E}(u^{\alpha }(t);f,\Omega ^{\alpha }(t))\right ]$ where $u^{\alpha }(t)$ is the solution of the elastic plane defined on $\Omega ^{\alpha }(t)=D\setminus \Sigma ^{\alpha }(t)$. Maximum energy release rate criterion. Find $\alpha$ such that $\alpha ^*$ $$\mathcal{G}(f,\Sigma ^{\alpha ^*}(\cdot ))= \max _{-\pi \lt \alpha \lt \pi }\mathcal{G}(f,\Sigma ^{\alpha }(\cdot )) \tag{3.21}$$ Putting $\omega (t)=\{x:\, |x-\gamma _{\Sigma }^{\alpha }(t)| \lt t/2\} , \gamma _{\Sigma }^{\alpha }(t)=(t\cos \alpha ,t\sin \alpha )$, we have from (2.20) \begin{eqnarray*} -\frac{d}{ds}\mathcal{E}(u^{\alpha }(t+s);f,\Omega ^{\alpha }(t+s))|_{s=0} &=&J_{\omega (t)}(u^{\alpha }(t),\mu _\alpha |\partial \omega (t))\\ &=&\frac{\kappa +1}{8G}(K_1(\gamma _{\Sigma }^{\alpha }(t))^2+K_2(\gamma _{\Sigma }^{\alpha }(t))^2) \end{eqnarray*} for each $t$, where $\mu _{\alpha }=(\cos \alpha ,\sin \alpha )$ and $K_i(\gamma _{\Sigma }^{\alpha }(t)),i=1,2$ are the stress intensity factors at $\gamma _{\Sigma }^{\alpha }(t)$ (see (2.18a) ). Therefore, we arrive $$\mathcal{G}(f,\Sigma ^{\alpha }(\cdot )) =\frac{\kappa +1}{8G}(K_1^{\alpha }(\gamma _{\Sigma })^2+K_2^{\alpha }(\gamma _{\Sigma })^2) \tag{3.22}$$ where $K_i^{\alpha }(\gamma _{\Sigma })=\lim _{t→+0}K_i(\gamma _{\Sigma }^{\alpha }(t)), i=1,2$. Maximum energy release rate criterion becomes to find $\alpha ^*$ such that $$\mathcal{G}(f,\Sigma ^{\alpha ^*}(\cdot ))= \max _{-\pi \lt \alpha \lt \pi }\frac{\kappa +1}{8G}\{K_1^{\alpha }(\gamma _{\Sigma })^2+K_2^{\alpha }(\gamma _{\Sigma })^2\} \tag{3.23}$$ From (3.23) , we have the following if $\alpha \neq 0$, \begin{eqnarray} \mathcal{G}(f,\Sigma ^{\alpha }(\cdot ))&\neq & J_{\omega }(u(0),\mu _{\alpha }|\partial \omega )\\ J_{\omega }(u(0),\mu _{\alpha }|\partial \omega )&=&J_{\omega }(u(0),e_1)\cos \alpha +J_{\omega }(u(0),e_2|\partial \omega )\sin \alpha \notag \tag{3.24} \end{eqnarray} where $\omega$ is a domain containing $\gamma _{\Sigma }$. We must notice that $J_{\omega }(u(0),e_2|\partial \omega )$ has vertical sensitivity to $\Sigma _0$ which does not show the sensitivity by crack propagation. Therefore, from (3.24) it is quite difficult to use the H1-gradient method finding the direction of crack propagation. For the crack path, see [Oh02, Sumi] for more.

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