2階tensorから任意方向成分を取り出す

from 表面力vectorから3次元のMohr円を求める

記号の準備

$ \bm\sigma：Cauchy応力tensor

数学上は任意の3次元対称tensorで成立する

$ \bm n：求めたい表面力vectorの作用する面の法線vector

基底を正規直交基底$ \sf Eで固定し成分表示する

表面力vector$ \bm\sigma\cdot\bm nの$ \bm n方向成分を$ \sigma、それに直交する成分の2乗を$ \tau^2とする

任意次元で計算すると

$ \sigma=(\bm\sigma\cdot\bm n)\cdot\bm n

$ = \bm\sigma:\bm n\bm n

$ \tau^2=|\bm\sigma\cdot\bm n-(\bm\sigma:\bm n\bm n)\bm n|^2

$ = |(\bm\sigma\cdot\bm n)\cdot(\bm I-\bm n\bm n)|^2

$ =\bm n\cdot\bm\sigma\cdot(\bm I-\bm n\bm n)^2\cdot\bm\sigma\cdot\bm n

$ =\bm n\cdot\bm\sigma\cdot(\bm I-2\bm n\bm n+\bm n\bm n)\cdot\bm\sigma\cdot\bm n

$ =\bm n\cdot\bm\sigma\cdot(\bm I-\bm n\bm n)\cdot\bm\sigma\cdot\bm n

$ =\bm\sigma^2:\bm n\bm n-\sigma^2

以下、特定の次元にて$ \sigma,\tauを成分で表す

3次元2階tensorのとき(対称tensor)

$ \sigma= n_0^2\sigma_{00}+2n_0n_1\sigma_{01}+n_1^2\sigma_{11}+2n_1n_2\sigma_{12}+n_2^2\sigma_{22}+2n_2n_0\sigma_{20}

$ \tau^2=n_0^2(\sigma_{00}^2+\sigma_{01}^2+\sigma_{02}^2)+2n_0n_1(\sigma_{00}\sigma_{01}+\sigma_{01}\sigma_{11}+\sigma_{02}\sigma_{21})

$ +n_1^2(\sigma_{11}^2+\sigma_{12}^2+\sigma_{10}^2)+2n_1n_2(\sigma_{10}\sigma_{02}+\sigma_{11}\sigma_{12}+\sigma_{12}\sigma_{22})

$ +n_2^2(\sigma_{22}^2+\sigma_{20}^2+\sigma_{21}^2)+2n_2n_0(\sigma_{20}\sigma_{00}+\sigma_{21}\sigma_{10}+\sigma_{22}\sigma_{20})

$ -\sigma^2

$ \sf Eを$ \bm\sigmaの主軸に合わせて簡略化する

$ \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

$ =-2{n_0}^2{n_1}^2\sigma_{00}\sigma_{11}-2{n_1}^2{n_2}^2\sigma_{11}\sigma_{22}-2{n_2}^2{n_0}^2\sigma_{22}\sigma_{00}-{n_0}^4\sigma_{00}^2-{n_1}^4\sigma_{11}^2-{n_2}^4\sigma_{22}^2

$ +n_0^2\sigma_{00}^2+n_1^2\sigma_{11}^2+n_2^2\sigma_{22}^2

$ ={n_0}^2{n_1}^2(\sigma_{00}-\sigma_{11})^2+{n_1}^2{n_2}^2(\sigma_{11}-\sigma_{22})^2+{n_2}^2{n_0}^2(\sigma_{22}-\sigma_{00})^2

$ -{n_0}^2({n_1}^2+{n_2}^2)\sigma_{00}^2-{n_1}^2({n_2}^2+{n_0}^2)\sigma_{11}^2-{n_2}^2({n_0}^2+{n_1}^2)\sigma_{22}^2

$ -{n_0}^4\sigma_{00}^2-{n_1}^4\sigma_{11}^2-{n_2}^4\sigma_{22}^2

$ +n_0^2\sigma_{00}^2+n_1^2\sigma_{11}^2+n_2^2\sigma_{22}^2

$ ={n_0}^2{n_1}^2(\sigma_{00}-\sigma_{11})^2+{n_1}^2{n_2}^2(\sigma_{11}-\sigma_{22})^2+{n_2}^2{n_0}^2(\sigma_{22}-\sigma_{00})^2

$ -n_0^2\sigma_{00}^2-n_1^2\sigma_{11}^2-n_2^2\sigma_{22}^2

$ +n_0^2\sigma_{00}^2+n_1^2\sigma_{11}^2+n_2^2\sigma_{22}^2

$ ={n_0}^2{n_1}^2(\sigma_{00}-\sigma_{11})^2+{n_1}^2{n_2}^2(\sigma_{11}-\sigma_{22})^2+{n_2}^2{n_0}^2(\sigma_{22}-\sigma_{00})^2

$ \bm n=\frac1{\sqrt3}\begin{pmatrix}1\\1\\1\end{pmatrix}(正八面体の法線vector)のとき

$ \sigma=\frac13(\sigma_{00}+\sigma_{11}+\sigma_{22})=\sigma_{oct}=p

$ p：平均応力

$ \sigma_{oct}：八面体垂直応力

$ \tau^2=\frac19((\sigma_{00}-\sigma_{11})^2+(\sigma_{11}-\sigma_{22})^2+(\sigma_{22}-\sigma_{00})^2)={\tau_{oct}}^2

$ \tau_{oct}：八面体剪断応力

八面体剪断成分

2次元2階tensorのとき(対称tensor)

$ \sigma= {n_0}^2\sigma_{00}+2n_0n_1\sigma_{01}+{n_1}^2\sigma_{11}

$ \tau^2={n_0}^2(\sigma_{00}^2+\sigma_{01}^2)+2n_0n_1(\sigma_{00}\sigma_{01}+\sigma_{01}\sigma_{11})+{n_1}^2(\sigma_{01}^2+\sigma_{11}^2) -\sigma^2

$ \sf Eを$ \bm\sigmaの主軸に合わせて簡略化する

$ \sigma=n_0^2\sigma_{00}+n_1^2\sigma_{11}

$ \tau^2=n_0^2\sigma_{00}^2+n_1^2\sigma_{11}^2-\sigma^2

$ = n_0^2\sigma_{00}^2+n_1^2\sigma_{11}^2-n_0^4\sigma_{00}^2-n_1^4\sigma_{11}^2-2n_0^2n_1^2\sigma_{00}\sigma_{11}

$ = n_0^2(n_0^2+n_1^2)\sigma_{00}^2+n_1^2(n_0^2+n_1^2)\sigma_{11}^2-n_0^4\sigma_{00}^2-n_1^4\sigma_{11}^2-2n_0^2n_1^2\sigma_{00}\sigma_{11}

$ = {n_0}^2{n_1}^2(\sigma_{00}-\sigma_{11})^2

$ \bm n=\frac1{\sqrt2}\begin{pmatrix}1\\1\end{pmatrix}($ x=yに並行な単位vector)のとき

$ \sigma=\frac12(\sigma_{00}+\sigma_{11})=p

$ p：平均応力

$ \tau^2=\left(\frac12(\sigma_{00}-\sigma_{11})\right)^2=q^2

$ q:=\frac12(\sigma_{00}-\sigma_{11}：2次元偏差応力

Reference

『連続体の力学序説』 p.30

#2024-02-04 11:47:37

#2024-01-08 13:58:26