3次元のMohr円

これについて言及している資料は探しにくい

見つけたものを列挙しておく

中瀬さんが書いている章

応力円の話はないが、3次元の応力tensorの図的な解説が冒頭にある

角度周りの話がくわしい

その他、関連ページリストにもある

導出

他の方法も探りたい

この説明はあやしい

本当に不要なのか？

$ \bm\sigma={\cal\pmb D}:\bm\sigma+\frac13({\rm tr}\bm\sigma)\bm I

$ \sigma_m:=\frac13({\rm tr}\bm\sigma)=\frac13J_1

$ \sigma_0-\sigma_m=\frac13(\sigma_0-\sigma_1)+\frac13(\sigma_0-\sigma_2)

$ \sigma_1-\sigma_m=\frac13(\sigma_1-\sigma_2)+\frac13(\sigma_1-\sigma_0)

$ \sigma_2-\sigma_m=\frac13(\sigma_2-\sigma_0)+\frac13(\sigma_2-\sigma_1)

$ \sigma_0-\sigma_m=\frac23q_{01}+\frac23q_{01}+\frac23q_{12}=\frac23(2q_{01}+q_{12})=\frac23(-q_{01}+q_{12})+q_{01}=\frac13(q_{01}+2q_{12})+q_{01}

$ \sigma_1-\sigma_m=\frac23q_{12}-\frac23q_{01}=\frac23(-q_{01}+q_{12})=\frac13(q_{01}+2q_{12})-q_{01}=\frac23(-q_{01}+q_{12})-q_{12}=\frac13(-2q_{01}-q_{12})+q_{12}

$ \sigma_2-\sigma_m=-\frac23q_{01}-\frac23q_{12}-\frac23q_{12}=\frac23(-q_{01}-2q_{12})=\frac23(-q_{01}+q_{12})-q_{12}=\frac13(-2q_{01}-q_{12})-q_{12}

$ [\bm\sigma]^{\sf EE}=\begin{pmatrix}\sigma_{00}&\sigma_{01}&\sigma_{02}\\\sigma_{10}&\sigma_{11}&\sigma_{12}\\\sigma_{20}&\sigma_{21}&\sigma_{22}\end{pmatrix}=\begin{pmatrix}\sigma_{00}&\sigma_{01}&\sigma_{02}\\\sigma_{01}&\sigma_{11}&\sigma_{12}\\\sigma_{02}&\sigma_{12}&\sigma_{22}\end{pmatrix}

$ = \begin{pmatrix}\frac23q_{01}&\sigma_{01}&0\\\sigma_{01}&-\frac23q_{01}&0\\0&0&0\end{pmatrix}+\begin{pmatrix}0&0&0\\0&\frac23q_{12}&\sigma_{12}\\0&\sigma_{12}&-\frac23q_{12}\end{pmatrix}+\begin{pmatrix}\frac23(q_{01}+q_{12})&0&\sigma_{02}\\0&0&0\\\sigma_{02}&0&-\frac23(q_{01}+q_{12})\end{pmatrix}+\sigma_m\begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix}

$ \sigma_{00}-\sigma_m=\frac23q_{01}+\frac23(q_{01}+q_{12})

$ \sigma_{11}-\sigma_m=\frac23q_{12}-\frac23q_{01}

$ \sigma_{22}-\sigma_m=-\frac23(q_{01}+q_{12})-\frac23q_{12}

$ [\bm I]^{\sf FE}_{ij}=\bm f_i\cdot\bm e_j=\cos\theta\delta_{ij}+(1-\cos\theta)n_in_j+\sin\theta\epsilon_{jki}n_k=[\bm R(\bm n,\theta)]^{\sf EE}_{ji}=[\bm R(\bm n,\theta)^\top]^{\sf EE}_{ij}

$ \because\bm f_i=\bm R(\bm n,\theta)\cdot\bm e_i=\bm e_i\cos\theta+(1-\cos\theta )n_i\hat{\bm n}+\sin\theta\hat{\bm n}\times\bm e_i

$ \pmb R(\pmb n,\theta):=\pmb I\cos\theta+(1-\cos\theta)\hat{\pmb n}\hat{\pmb n}-(\sin\theta)\hat{\pmb n}\cdot{\Large\pmb\epsilon}

$ [\bm\sigma]^{\sf FF}=[\bm I]^{\sf FE}[\bm\sigma]^{\sf EE}[\bm I]^{\sf EF}

$ =[\bm R(\bm n,\theta)^\top]^{\sf EE}[\bm\sigma]^{\sf EE}[\bm R(\bm n,\theta)]^{\sf EE}

一つづつ計算

等方成分はそのまま相殺される

$ \begin{pmatrix}\frac23q_{01}&\sigma_{01}&0\\\sigma_{01}&-\frac23q_{01}&0\\0&0&0\end{pmatrix}

$ \cos\theta\begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix}+(1-\cos\theta)\begin{pmatrix}n_0n_0&n_0n_1&n_0n_2\\n_1n_0&n_1n_1&n_1n_2\\n_2n_0&n_2n_1&n_2n_2\end{pmatrix}-(\sin\theta)\begin{pmatrix}0&-n_2&n_1\\n_2&0&-n_0\\-n_1&n_0&0\end{pmatrix}

$ [\bm I]^{\sf EE}\begin{pmatrix}1&0&0\\0&-1&0\\0&0&0\end{pmatrix}=\begin{pmatrix}1&0&0\\0&-1&0\\0&0&0\end{pmatrix}

$ \begin{pmatrix}1&0&0\\0&-1&0\\0&0&0\end{pmatrix}[\bm I]^{\sf EE}=\begin{pmatrix}1&0&0\\0&-1&0\\0&0&0\end{pmatrix}

$ [\hat{\bm n}\hat{\bm n}]^{\sf EE}\begin{pmatrix}1&0&0\\0&-1&0\\0&0&0\end{pmatrix}=\begin{pmatrix}n_0^2&-n_0n_1&0\\n_1n_0&-n_1^2&0\\n_2n_0&-n_2n_1&0\end{pmatrix}

うーん、一般的に解こうとするのは相当難しいな

行列計算が尋常じゃない

$ \begin{pmatrix}n_0^2(1-\cos\theta)+\cos\theta&n_0n_1(1-\cos\theta)+n_2\sin\theta&n_0n_2(1-\cos\theta)-n_1\sin\theta\\n_0n_1(1-\cos\theta)-n_2\sin\theta&n_1^2(1-\cos\theta)+\cos\theta&n_1n_2(1-\cos\theta)+n_0\sin\theta\\n_0n_2(1-\cos\theta)+n_1\sin\theta&n_1n_2(1-\cos\theta)-n_0\sin\theta&n_2^2(1-\cos\theta)+\cos\theta\end{pmatrix}\begin{pmatrix}1&0&0\\0&-1&0\\0&0&1\end{pmatrix}\begin{pmatrix}n_0^2(1-\cos\theta)+\cos\theta&n_0n_1(1-\cos\theta)-n_2\sin\theta&n_0n_2(1-\cos\theta)+n_1\sin\theta\\n_0n_1(1-\cos\theta)+n_2\sin\theta&n_1^2(1-\cos\theta)+\cos\theta&n_1n_2(1-\cos\theta)-n_0\sin\theta\\n_0n_2(1-\cos\theta)-n_1\sin\theta&n_1n_2(1-\cos\theta)+n_0\sin\theta&n_2^2(1-\cos\theta)+\cos\theta\end{pmatrix}

$ =\begin{pmatrix}n_0^2(1-\cos\theta)+\cos\theta&-n_0n_1(1-\cos\theta)+n_2\sin\theta&n_0n_2(1-\cos\theta)-n_1\sin\theta\\n_0n_1(1-\cos\theta)-n_2\sin\theta&-n_1^2(1-\cos\theta)+\cos\theta&n_1n_2(1-\cos\theta)+n_0\sin\theta\\n_0n_2(1-\cos\theta)+n_1\sin\theta&-n_1n_2(1-\cos\theta)-n_0\sin\theta&n_2^2(1-\cos\theta)+\cos\theta\end{pmatrix}\begin{pmatrix}n_0^2(1-\cos\theta)+\cos\theta&n_0n_1(1-\cos\theta)-n_2\sin\theta&n_0n_2(1-\cos\theta)+n_1\sin\theta\\n_0n_1(1-\cos\theta)+n_2\sin\theta&n_1^2(1-\cos\theta)+\cos\theta&n_1n_2(1-\cos\theta)-n_0\sin\theta\\n_0n_2(1-\cos\theta)-n_1\sin\theta&n_1n_2(1-\cos\theta)+n_0\sin\theta&n_2^2(1-\cos\theta)+\cos\theta\end{pmatrix}

$ =\begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix}-2 \begin{pmatrix}0&n_0n_1(1-\cos\theta)+n_2\sin\theta&0\\0&n_1^2(1-\cos\theta)+\cos\theta&0\\0&n_1n_2(1-\cos\theta)-n_0\sin\theta&0\end{pmatrix}\begin{pmatrix}n_0^2(1-\cos\theta)+\cos\theta&n_0n_1(1-\cos\theta)-n_2\sin\theta&n_0n_2(1-\cos\theta)+n_1\sin\theta\\n_0n_1(1-\cos\theta)+n_2\sin\theta&n_1^2(1-\cos\theta)+\cos\theta&n_1n_2(1-\cos\theta)-n_0\sin\theta\\n_0n_2(1-\cos\theta)-n_1\sin\theta&n_1n_2(1-\cos\theta)+n_0\sin\theta&n_2^2(1-\cos\theta)+\cos\theta\end{pmatrix}

(計算途中)

$ =\begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix}-2 \begin{pmatrix}(n_0n_1(1-\cos\theta)+n_2\sin\theta)^2&(n_0n_1(1-\cos\theta)+n_2\sin\theta)(n_1^2(1-\cos\theta)+\cos\theta)&(n_0n_1(1-\cos\theta)+n_2\sin\theta)(n_1n_2(1-\cos\theta)-n_0\sin\theta)\\0&n_1^2(1-\cos\theta)+\cos\theta&0\\0&n_1n_2(1-\cos\theta)-n_0\sin\theta&0\end{pmatrix}

いやー、これはちょっと無理です