初等梁の変位勾配tensor
$ \pmb\varepsilon=\varepsilon_{xx}\pmb e_x\pmb e_x
これを$ \pmb\varepsilon=\frac12(\pmb\nabla\pmb u+(\pmb\nabla\pmb u)^\top)に代入する
$ \frac{\partial u_x}{\partial x}=\varepsilon_{xx}ー①
$ \frac{\partial u_y}{\partial y}=\frac{\partial u_z}{\partial z}=0ー②
$ \frac{\partial u_x}{\partial y}=-\frac{\partial u_y}{\partial x}ー③
$ \frac{\partial u_y}{\partial z}=-\frac{\partial u_z}{\partial y}ー④
$ \frac{\partial u_z}{\partial x}=-\frac{\partial u_x}{\partial z}ー⑤
微分する
$ ①\implies\frac{\partial^2u_x}{{\partial x}^2}=\frac{\partial\varepsilon_{xx}}{\partial x}
$ ③\implies\frac{\partial^2u_x}{\partial y\partial x}=-\frac{\partial^2u_y}{{\partial x}^2}
$ ④\implies\frac{\partial^2u_x}{\partial z\partial x}=-\frac{\partial^2u_z}{{\partial x}^2}
$ ③\implies\frac{\partial^2u_x}{\partial x\partial y}=-\frac{\partial^2u_y}{{\partial x}^2}
$ ③\land②\implies\frac{\partial^2u_x}{{\partial y}^2}=0
$ ③\land④\land⑤\implies\frac{\partial^2u_x}{\partial z\partial y}=-\frac{\partial^2u_y}{\partial x\partial z}=\frac{\partial^2u_z}{\partial y\partial x}=-\frac{\partial^2u_x}{\partial z\partial y}
$ \implies \frac{\partial^2u_x}{\partial y\partial z}=\frac{\partial^2u_z}{\partial x\partial y}=\frac{\partial^2u_y}{\partial z\partial x}=0ー⑥
よって、任意の函数$ \theta(x)を用いて
$ \frac{\partial u_x}{\partial y}=\theta(x)ー⑦
$ ⑤\implies\frac{\partial^2u_x}{\partial x\partial z}=-\frac{\partial^2u_z}{{\partial x}^2}
$ ⑥\implies\frac{\partial^2u_x}{\partial y\partial z}=0
$ ⑤\land②\implies\frac{\partial^2u_x}{{\partial z}^2}=-\frac{\partial^2u_z}{\partial x\partial z}=0
よって、任意の函数$ \phi(x)を用いて
$ \frac{\partial u_x}{\partial z}=-\phi(x)ー⑧
以上より$ u_xが求まる
$ \mathrm{d}u_x=\varepsilon_{xx}\mathrm{d}x+\mathrm{d}(\theta(x)y-\phi(x)z)ー⑨
$ ⑦\land③\implies\frac{\partial u_y}{\partial x}=-\theta(x)
$ ②\implies\frac{\partial u_y}{\partial y}=0
$ ⑥\implies\frac{\partial^2u_y}{\partial x\partial z}=0
$ ②\implies\frac{\partial^2u_y}{\partial y\partial z}=0
$ ②\land④\implies\frac{\partial^2u_y}{{\partial z}^2}=-\frac{\partial^2u_z}{\partial y\partial z}=0
$ \therefore\frac{\partial u_y}{\partial z}=-\frac{\partial u_z}{\partial y}=\varphiー⑩
$ \varphiは任意定数
$ \therefore\mathrm{d}u_y=-\theta(x)\mathrm{d}x+\varphi\mathrm{d}zー⑪
$ ⑧\land⑤\implies\frac{\partial u_z}{\partial x}=\phi(x)
$ ⑩\implies\frac{\partial u_z}{\partial y}=0
$ ②\implies\frac{\partial u_z}{\partial z}=0
$ \therefore\mathrm{d}u_z=\phi(x)\mathrm{d}x-\varphi\mathrm dyー⑫
⑨, ⑪, ⑫より、$ \pmb\nabla\pmb uが求まる
$ \pmb\nabla\pmb u=\begin{pmatrix}\varepsilon_{xx}&\theta(x)&-\phi(x)\\-\theta(x)&0&\varphi\\\phi(x)&-\varphi&0\end{pmatrix}
断面が剛体回転することを表す
あとは$ \varepsilon_{xx}の関係式が求まればいいのだが
$ \frac{\partial\varepsilon_{xx}}{\partial y}=-\frac{\partial^2u_x}{\partial y\partial x}=\frac{\partial^2u_y}{{\partial x}^2}=-\theta'(x)
$ \frac{\partial\varepsilon_{xx}}{\partial z}=-\frac{\partial^2u_x}{\partial z\partial x}=\frac{\partial^2u_z}{{\partial x}^2}=\phi'(x)
$ \therefore \varepsilon_{xx}=f(x)-\theta'(x)y+\phi'(x)z
求まっちゃった