表面力vectorから3次元のMohr円を求める

記号の準備

$ \begin{dcases}\sigma&=n_0^2\sigma_{00}+n_1^2\sigma_{11}+n_2^2\sigma_{22}\\\tau^2&=n_0^2\sigma_{00}^2+n_1^2\sigma_{11}^2+n_2^2\sigma_{22}^2-\sigma^2\\1&=n_0^2+n_1^2+n_2^2\end{dcases}

$ \iff\begin{pmatrix}1&1&1\\\sigma_{00}&\sigma_{11}&\sigma_{22}\\{\sigma_{00}}^2&{\sigma_{11}}^2&{\sigma_{22}}^2\end{pmatrix}\begin{pmatrix}{n_0}^2\\{n_1}^2\\{n_2}^2\end{pmatrix}=\begin{pmatrix}1\\\sigma\\\sigma^2+\tau^2\end{pmatrix}

$ \iff\begin{pmatrix}{n_0}^2\\{n_1}^2\\{n_2}^2\end{pmatrix}=\begin{pmatrix}1&1&1\\\sigma_{00}&\sigma_{11}&\sigma_{22}\\{\sigma_{00}}^2&{\sigma_{11}}^2&{\sigma_{22}}^2\end{pmatrix}^{-1}\begin{pmatrix}1\\\sigma\\\sigma^2+\tau^2\end{pmatrix}

$ = -\frac1{4q_{01}q_{12}q_{20}}\begin{pmatrix}q_{12}\sigma_{11}\sigma_{22}-2q_{12}p_{12}\sigma+q_{12}(\sigma^2+\tau^2)\\q_{20}\sigma_{22}\sigma_{00}-2q_{20}p_{20}\sigma+q_{20}(\sigma^2+\tau^2)\\q_{01}\sigma_{00}\sigma_{11}-2q_{01}p_{01}\sigma+q_{01}(\sigma^2+\tau^2)\end{pmatrix}

$ \because\begin{pmatrix}1&1&1\\a&b&c\\a^2&b^2&c^2\end{pmatrix}^{-1}=\frac{-1}{(a-b)(b-c)(c-a)}\begin{pmatrix}bc(b-c)&-(b-c)(b+c)&b-c\\ca(c-a)&-(c-a)(c+a)&c-a\\ab(a-b)&-(a-b)(a+b)&a-b\end{pmatrix}

$ = -\frac1{4q_{01}q_{12}q_{20}}\begin{pmatrix}q_{12}\tau^2+q_{12}(\sigma-p_{12})^2-q_{12}{p_{12}}^2+q_{12}\sigma_{11}\sigma_{22}\\q_{20}\tau^2+q_{20}(\sigma-p_{20})^2-q_{20}{p_{20}}^2+q_{20}\sigma_{22}\sigma_{00}\\q_{01}\tau^2+q_{01}(\sigma-p_{01})^2-q_{01}{p_{01}}^2+q_{01}\sigma_{00}\sigma_{11}\end{pmatrix}

$ = -\frac1{4q_{01}q_{12}q_{20}}\begin{pmatrix}q_{12}(\sigma-p_{12})^2+q_{12}\tau^2-{q_{12}}^3\\q_{20}(\sigma-p_{20})^2+q_{20}\tau^2-{q_{20}}^3\\q_{01}(\sigma-p_{01})^2+q_{01}\tau^2-{q_{01}}^3\end{pmatrix}

$ \underline{\iff \begin{dcases}(\sigma-p_{12})^2+\tau^2&=4{n_0}^2q_{20}q_{01}+{q_{12}}^2\\(\sigma-p_{20})^2+\tau^2&=4{n_1}^2q_{01}q_{12}+{q_{20}}^2\\(\sigma-p_{01})^2+\tau^2&=4{n_2}^2q_{12}q_{20}+{q_{01}}^2\end{dcases}\quad}_\blacksquare

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