薄肉閉断面部材のSaint-Venantのねじり

記号は変更してある

https://kakeru.app/cec15119684d6023ab032f489e53a157 https://i.kakeru.app/cec15119684d6023ab032f489e53a157.svg

仮定

軸方向変形なし

$ u_x=\rm const.

$ \pmb\varepsilon-\pmb\varepsilon\cdot\pmb e_x\pmb e_x-\pmb e_x\pmb e_x\cdot\pmb\varepsilon+\pmb e_x\pmb e_x\cdot\pmb\varepsilon\cdot\pmb e_x\pmb e_x=\pmb 0

$ \because\pmb\varepsilon-\pmb\varepsilon\cdot\pmb e_x\pmb e_x-\pmb e_x\pmb e_x\cdot\pmb\varepsilon+\pmb e_x\pmb e_x\cdot\pmb\varepsilon\cdot\pmb e_x\pmb e_x=\pmb\varepsilon\cdot(\pmb I-\pmb e_x\pmb e_x)-\pmb e_x\pmb e_x\cdot\pmb\varepsilon\cdot(\pmb I-\pmb e_x\pmb e_x)

$ =(\pmb I-\pmb e_x\pmb e_x)\cdot\pmb\varepsilon\cdot(\pmb I-\pmb e_x\pmb e_x)

$ \begin{pmatrix}\varepsilon_{xx}&\varepsilon_{xr}&\varepsilon_{x\theta}\\\varepsilon_{rx}&\varepsilon_{rr}&\varepsilon_{r\theta}\\\varepsilon_{\theta x}&\varepsilon_{\theta r}&\varepsilon_{\theta\theta}\end{pmatrix}=\begin{pmatrix}\frac{\partial u_x}{\partial x}&\frac12\left(\frac{\partial u_r}{\partial x}+\frac{\partial u_x}{\partial r}\right)&\frac12\left(\frac{\partial u_\theta}{\partial x}+\frac1r\frac{\partial u_x}{\partial\theta}\right)\\\varepsilon_{rx}&\frac{\partial u_r}{\partial r}&\frac12\left(\frac{\partial u_\theta}{\partial r}-\frac{u_\theta}{r}+\frac1r\frac{\partial u_r}{\partial\theta}\right)\\\varepsilon_{\theta x}&\varepsilon_{\theta r}&\frac{u_r}{r}+\frac1r\frac{\partial u_\theta}{\partial\theta}\end{pmatrix}

$ \begin{pmatrix}\varepsilon_{xx}&\varepsilon_{xr}&\varepsilon_{x\theta}\\\varepsilon_{rx}&\varepsilon_{rr}&\varepsilon_{r\theta}\\\varepsilon_{\theta x}&\varepsilon_{\theta r}&\varepsilon_{\theta\theta}\end{pmatrix}=\begin{pmatrix}0&\frac12\left(\frac{\partial u_r}{\partial x}+0\right)&\frac12\left(\frac{\partial u_\theta}{\partial x}+0\right)\\\varepsilon_{rx}&\frac{\partial u_r}{\partial r}&\frac12\left(\frac{\partial u_\theta}{\partial r}-\frac{u_\theta}{r}+\frac1r\frac{\partial u_r}{\partial\theta}\right)\\\varepsilon_{\theta x}&\varepsilon_{\theta r}&\frac{u_r}{r}+\frac1r\frac{\partial u_\theta}{\partial\theta}\end{pmatrix}

$ =\begin{pmatrix}0&\frac12\frac{\partial u_r}{\partial x}&\frac12\frac{\partial u_\theta}{\partial x}\\\frac12\frac{\partial u_r}{\partial x}&\frac{\partial u_r}{\partial r}&\frac12\left(\frac{\partial u_\theta}{\partial r}-\frac{u_\theta}{r}+\frac1r\frac{\partial u_r}{\partial\theta}\right)\\\frac12\frac{\partial u_\theta}{\partial x}&\varepsilon_{\theta r}&\frac{u_r}{r}+\frac1r\frac{\partial u_\theta}{\partial\theta}\end{pmatrix}

$ \varepsilon_{xr}=\frac12\frac{\partial u_r}{\partial x}

$ \varepsilon_{x\theta}=\frac12\frac{\partial u_\theta}{\partial x}

$ \varepsilon_{rr}=\frac{\partial u_r}{\partial r}=0

$ \varepsilon_{r\theta}=\frac12\left(\frac{\partial u_\theta}{\partial r}-\frac{u_\theta}{r}+\frac1r\frac{\partial u_r}{\partial\theta}\right)=0

$ \varepsilon_{\theta\theta}=\frac{u_r}{r}+\frac1r\frac{\partial u_\theta}{\partial\theta}=0

$ u_r=-\frac{\partial u_\theta}{\partial\theta}=f(x,\theta)

$ u_\theta=\frac{\partial u_\theta}{\partial r}r+\frac{\partial u_r}{\partial\theta}

$ u_\theta=\frac{\partial u_\theta}{\partial r}r-\frac{\partial^2 u_\theta}{{\partial\theta}^2}

$ \frac{\partial u_\theta}{\partial r}=\frac{\partial u_\theta}{\partial r}+\frac{\partial^2 u_\theta}{{\partial r}^2}+0

$ \therefore \frac{\partial u_\theta}{\partial r}=g(x,\theta)

うーん、うまく行かない

$ y=r\cos\theta,z=r\sin\theta

$ u_y\pmb e_y+u_z\pmb e_z=u_r\pmb e_r+u_\theta\pmb e_\theta

$ u_\theta=(u_y\pmb e_y+u_z\pmb e_z)\cdot\pmb e_\theta

$ =-u_y\sin\theta+u_z\cos\theta

$ = -u_{0y}(x)\sin\theta-f(x)r\sin\theta\sin\theta+u_{0z}(x)\cos\theta-f(x)r\cos\theta\cos\theta

$ = -u_{0y}(x)\sin\theta+u_{0z}(x)\cos\theta-f(x)r

$ u_r=(u_y\pmb e_y+u_z\pmb e_z)\cdot\pmb e_r

$ = u_y\cos\theta+u_z\sin\theta

$ =u_{0y}(x)\cos\theta+u_{0z}(x)\sin\theta+f(x)r(\sin\theta\cos\theta-\cos\theta\sin\theta)

$ = u_{0y}(x)\cos\theta+u_{0z}(x)\sin\theta

ようやく求まった

第1,2項が断面の平行移動を表し、第3項が回転を表す

難しく考えすぎた。基本は常微分方程式と同じなのだ。

テキストだと断面形状不変の仮定にひっくるめてしまっている

仕切り直し

$ u_x=0

$ u_r=0

$ u_\theta=\omega_{yz}r

となる

$ \sigma_{x\theta}=2G\varepsilon_{x\theta}=G\frac{\partial u_\theta}{\partial x}=Gr\frac{\partial\omega_{yz}}{\partial x}=:Gr\gamma

$ T=\int_S r\sigma_{x\theta}\mathrm dS

$ = \int_0^{2\pi}\int_0^{r_0}rGr\gamma r\mathrm dr\mathrm d\theta

$ = G\gamma\frac12\pi{r_0}^4

$ T=GJ\frac{\partial\omega_{yz}}{\partial x}

応力は丸棒の場合と同じ

$ J=2\pi\int_{r_c-\frac12t}^{r_c+\frac12t}r^3\mathrm dr

$ = \frac12\pi\left(\left(r_c+\frac12t\right)^2+\left(r_c-\frac12t\right)^2\right)\cdot2r_c\cdot t

$ =\pi\left(2{r_c}^2+\frac12t^2\right)r_ct

$ =2\pi{r_c}^3t\left(1+\left(\frac{t}{2r_c}\right)^2\right)

$ \simeq2\pi{r_c}^3t

$ J\simeq2\pi{r_c}^3t

$ = 2A_cr_ct

$ \because A_c=\pi{r_c}^2

$ = \frac{4{A_c}^2}{\frac1t2\pi r_c}

$ = \frac{4{A_c}^2}{\int_{\partial S_c}\frac1t\mathrm dl}

この形式だと任意の薄肉閉断面部材に適用できるようになる

導出をまだ理解できていない

テキストの方法とは違い、極座標で求めている

$ T=\int_Sr\sigma_{x\theta}\mathrm dS

$ = \int_0^{2\pi}\int_{r_c-\frac12t}^{r_c+\frac12t}r\sigma_{x\theta}r\mathrm dr\mathrm d\theta

$ = q\int_0^{2\pi}\int_{r_c-\frac12t}^{r_c+\frac12t}\frac{r^2}{t}\mathrm dr\mathrm d\theta

$ = q\int_0^{2\pi}\frac13\frac1t\left(\left(r_c+\frac12t\right)^3-\left(r_c-\frac12t\right)^3\right)\mathrm d\theta

$ = q\int_0^{2\pi}\frac13\frac1t\cdot t\left(\left(r_c+\frac12t+r_c-\frac12t\right)^2-{r_c}^2+\frac14t^2\right)\mathrm d\theta

$ =q\int_0^{2\pi}\frac13\frac1t\cdot t\left(3{r_c}^2+\frac14t^2\right)\mathrm d\theta

$ \simeq q\int_0^{2\pi}{r_c}^2\mathrm d\theta

$ = 2q\int_0^{2\pi}\int_0^{r_c}r\mathrm dr\mathrm d\theta

$ =2qA_c

$ \therefore\sigma_{x\theta}=\frac{T}{2A_ct}

ねじり率との関係を求める

$ \therefore T=2A_ctGr\gamma

$ = G\cdot2A_ctr\cdot\gamma

$ =:GJ\gamma

$ J=2A_ctr

$ = \frac{4{A_c}^2}{\frac{2A_c}{tr}}