すっごく丁寧に剪断力と曲げmomentの関係式を求める

https://kakeru.app/3dcdce31c87bf84ce0a3c9368acf296e https://i.kakeru.app/3dcdce31c87bf84ce0a3c9368acf296e.svg

記号の定義

O点まわりの力のmomentの釣合いと力の釣合いを立てる

$ \begin{dcases}H=\color{red}{\iint_S\sigma_{xx}(y,z)\mathrm{d}y\mathrm{d}z}\\V=\textcolor{orange}{\iint_Tq_o\mathrm{d}z\mathrm{d}x}+\color{skyblue}{\iint_S\sigma_{xy}(y,z)\mathrm{d}y\mathrm{d}z}\\0=\iint_S(a\sigma_{xy}(y,z)-y\sigma_{xx}(y,z))\mathrm{d}y\mathrm{d}z+\color{orange}{\iint_Txq_0\mathrm{d}z\mathrm{d}x}\end{dcases}

色分けは図の色と対応している

$ \iff\begin{dcases}H=N\\V=q_0ab+Q\\0=\iint_Sa\sigma_{xy}(y,z)\mathrm{d}y\mathrm{d}z-M+\frac{1}{2}q_0a^2b\end{dcases}

$ \iff\begin{dcases}H=N\\V=q_0ab+Q\\0=a{\color{skyblue}{\iint_S\sigma_{xy}(y,z)\mathrm{d}y\mathrm{d}z}}-M+\frac{1}{2}q_0a^2b\end{dcases}

$ \iff\begin{dcases}H=N\\V=q_0ab+Q\\0=aQ-M+\frac{1}{2}q_0a^2b\end{dcases}

$ \implies \mathrm{d}M=(Q+bq_0a)\mathrm{d}a

数式だけだとこんがらがってくる

図を追加しよう

あと応力tensorの成分の従属変数を明示すべき

$ \bm\sigma\cdot\overleftarrow{\bm\nabla}+\bm b=\bm 0

$ \implies\bm r\bm\sigma\cdot\overleftarrow{\bm\nabla}+\bm r\bm b=\bm 0

$ \implies(\bm r\bm\sigma)\cdot\overleftarrow{\bm\nabla}-\bm r\overleftarrow{\bm\nabla}:\bm\sigma+\bm r\bm b=\bm 0

$ \iff(\bm r\bm\sigma)\cdot\overleftarrow{\bm\nabla}-\bm\sigma+\bm r\bm b=\bm 0

$ \because\bm r\overleftarrow{\bm\nabla}=\bm I

$ \implies\int_{\partial B}\bm r\bm\sigma\cdot\mathrm d\bm a-\int_B\bm\sigma\mathrm dv+\int_B\bm r\bm b\mathrm dv=\bm 0

$ \implies\int_{\partial B}\bm r\times\bm\sigma\cdot\mathrm d\bm a+0+\int_B\bm r\times\bm b\mathrm dv=\bm 0

ここからz方向成分だけ見ると

$ \int_{\partial B}(x\sigma_{yx}\mathrm da_x+x\sigma_{yy}\mathrm da_y+x\sigma_{yz}\mathrm da_z-y\sigma_{xx}\mathrm da_x-y\sigma_{xy}\mathrm da_y-y\sigma_{xz}\mathrm da_z)+\int_B(xb_y-yb_x)\mathrm dv=0

$ \int_{A_1\cup A_2}(x\sigma_{yx}\mathrm da_x-y\sigma_{xx}\mathrm da_x)+\int_{S}(x\sigma_{yy}\mathrm da_y+x\sigma_{yz}\mathrm da_z-y\sigma_{xy}\mathrm da_y-y\sigma_{xz}\mathrm da_z)+\int_B(xb_y-yb_x)\mathrm dv=0

$ \int_{A_1\cup A_2}(x\sigma_{yx}\mathrm da_x-y\sigma_{xx}\mathrm da_x)+\int_{S}x\sigma_{yy}\mathrm da_y+\int_B(xb_y-yb_x)\mathrm dv=0

$ \int_{A_1\cup A_2}(x\sigma_{yx}\mathrm da_x-y\sigma_{xx}\mathrm da_x)+\int_B(xb_y-yb_x)\mathrm dv=0

$ \int_{A_1\cup A_2}x\sigma_{yx}\mathrm da_x-M(x)+M(x_0)+\int_B(xb_y-yb_x)\mathrm dv=0

$ \iff x_0\int_{A_1}\sigma_{yx}\mathrm da_x+x\int_{A_2}\sigma_{yx}\mathrm da_x-M(x)+M(x_0)+\int_B(xb_y-yb_x)\mathrm dv=0

$ \iff (xQ(x)-x_0Q(x_0))-M(x)+M(x_0)+\int_B(xb_y-yb_x)\mathrm dv=0

$ \implies Q(x)+xQ'(x)-M'(x)+\frac{\partial}{\partial x}\int_B(xb_y-yb_x)\mathrm dv=0

$ \iff M'(x)=Q(x)+xQ'(x)+\frac{\partial}{\partial x}\int_B(xb_y-yb_x)\mathrm dv

$ \int_{\partial B}\bm\sigma\cdot\mathrm d\bm a+\int_B\bm b\mathrm dv=\bm 0

$ \implies\int_{A_1\cup A_1}\sigma_{yx}\mathrm da_x+\int_{S}(\sigma_{yy}\mathrm da_y+\sigma_{yz}\mathrm da_z)+\int_Bb_y\mathrm dv=0

$ \iff\int_{A_1\cup A_1}\sigma_{yx}\mathrm da_x+\int_Bb_y\mathrm dv=0

$ \implies Q'(x)+\frac{\partial}{\partial x}\int_Bb_y\mathrm dv=0

$ \iff Q'(x)+p_y(x)=0

$ \iff N'(x)+p_x(x)=0

$ \int_B(xb_y-yb_x)\mathrm dv=\int_{x_0}^x\int_{A(x)}(xb_y-yb_x)\mathrm da_x\mathrm dx

$ =\int_{x_0}^x\int_{A(x)}(xb_y-yb_x)\mathrm da_x\mathrm dx

$ =\int_{x_0}^xx\int_{A(x)}b_y\mathrm da_x\mathrm dx-\int_{x_0}^x\int_{A(x)}yb_x\mathrm da_x\mathrm dx

$ =\int_{x_0}^xsp_y(s)\mathrm ds-\int_{x_0}^x\int_{A(x)}yb_x\mathrm da_x\mathrm dx

$ \therefore M'(x)=Q(x)+xQ'(x)+xp_y(x)-\int_{A(x)}yb_x\mathrm da_x

$ = Q(x)-\int_{A(x)}yb_x\mathrm da_x