応力の対称性

証明

$ \frac{\partial}{\partial t}\int_{B(t)}\bm x\times\rho\bm v\mathrm dv=\int_{\partial B(t)}\bm x\times(\bm\sigma\cdot\mathrm d\bm s)+\int_{B(t)}\bm x\times\rho\bm g\mathrm dv\quad\text{.for }\forall B

$ \iff \int_{B(t)}\rho\frac{\mathrm D(\bm x\times\bm v)}{\mathrm Dt}\mathrm dv=\int_{\partial B(t)}\bm x\times(\bm\sigma\cdot\mathrm d\bm s)+\int_{B(t)}\bm x\times\rho\bm g\mathrm dv

$ ={\Large\bm\epsilon}:\int_{\partial B(t)}\bm x\bm\sigma\cdot\mathrm d\bm s+\int_{B(t)}\bm x\times\rho\bm g\mathrm dv

$ ={\Large\bm\epsilon}:\int_{B(t)}(\bm x\bm\sigma)\cdot\overleftarrow{\bm\nabla}\mathrm dv+\int_{B(t)}\bm x\times\rho\bm g\mathrm dv

$ ={\Large\bm\epsilon}:\int_{B(t)}\left(\bm I\cdot\bm\sigma^\top+\bm x\bm\sigma\cdot\overleftarrow{\bm\nabla}\right)\mathrm dv+\int_{B(t)}\bm x\times\rho\bm g\mathrm dv

$ =\int_{B(t)}\left({\Large\bm\epsilon}:\bm\sigma^\top+\bm x\times(\nabla\cdot\bm\sigma^\top+\rho\bm g)\right)\mathrm dv

$ =\int_{B(t)}\left({\Large\bm\epsilon}:\bm\sigma^\top+\bm x\times\rho\frac{\mathrm D\bm v}{\mathrm Dt}\right)\mathrm dv\quad\text{.for }\forall B

$ \iff\rho\frac{\mathrm D(\bm x\times\bm v)}{\mathrm Dt}={\Large\bm\epsilon}:\bm\sigma^\top+\bm x\times\rho\frac{\mathrm D\bm v}{\mathrm Dt}

$ \iff\rho\bm x\times\frac{\mathrm D\bm v}{\mathrm Dt}={\Large\bm\epsilon}:\bm\sigma^\top+\bm x\times\rho\frac{\mathrm D\bm v}{\mathrm Dt}

$ \iff {\Large\bm\epsilon}:\bm\sigma^\top=\bm 0

$ \iff {\Large\bm\epsilon}:\bm\sigma=-\bm0=\bm0

$ \iff {\Large\bm\epsilon}\cdot{\Large\bm\epsilon}:\bm\sigma=\bm0

$ \iff{\cal\bm W}:\bm\sigma=\bm 0

$ \iff\bm\sigma=\bm\sigma^\top