Non-Euclidean Euler Buckling
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Non-Euclidean Euler Buckling
In the classical Euler buckling, a beam lies on a plane. However, when a rod lies on a curved surfaced, what will happen? Exactly, what is the effect of gaussian curvature on the buckling critical loading.
Frame transformation
We consider two frames in this system, the material frame of the rod $\mathbf{d}={\mathbf{d}_1,\mathbf{d}_2,\mathbf{d}_3}$, and the surface frame ${\mathbf{n}\times\mathbf{d}_3,\mathbf{n},\mathbf{d}_3}$ where $\mathbf{n}$ is the normal vector of the surface. For convenience, we take $\mathbf{e}_3=\mathbf{d}_3$, $\mathbf{e}_1=\mathbf{n}\times\mathbf{d}_3$ and $\mathbf{e}_2=\mathbf{n}$ for the surface frame $\mathbf{e}={\mathbf{e}_1,\mathbf{e}_2,\mathbf{e}_3}$. The motion equation of the frame can be written as:
\[\partial_s \mathbf{d}=\mathbf{\Omega}\mathbf{d},\] \[\partial_s \mathbf{e}=\mathbf{\Gamma}\mathbf{e},\]where $\mathbf{\Omega}$ and $\mathbf{\Gamma}$ are the rotation tensor for the two frames respectively. In detail they can be expressed as:
\[\mathbf{\Omega}=\begin{pmatrix}0& \omega_3 & -\omega_2\\ -\omega_3 & 0 & \omega_1 \\ \omega_2 & -\omega_1 & 0 \end{pmatrix}\]and
\[\mathbf{\Gamma}=\begin{pmatrix}0 & \tau_g & -\kappa_g \\ -\tau_g & 0 & -\kappa_n \\ \kappa_g & \kappa_n & 0\end{pmatrix}\]We assume the material frame can be generated by rotating the surface frame along the $\mathbf{d}_3$ with angle $\phi$, then we have $\mathbf{d}=\mathbf{R}\mathbf{e}$ and finally $\mathbf{\Omega}$ can be expressed by $\mathbf{\Gamma}$ and $\mathbf{R}$:
\[\mathbf{\Omega}=\mathbf{R}_s\mathbf{R^T}+\mathbf{R}\mathbf{\Gamma}\mathbf{R^T},\]where $\mathbf{R}$ is the rotation matrix descrbing the rotation along $\mathbf{d}_3$ with $\phi$, $\mathbf{R}_s=\partial_s\mathbf{R}$ is the partial derivative of the $\mathbf{R}$ and $\mathbf{R^T}$ is the transpose of $\mathbf{R}$.
Space transformation
We have describe the internal motion in the last section, now we consider the motion of the space curve in absolute frame ${\mathbf{i},\mathbf{j},\mathbf{k}}$. We assume the transformation from absolute frame to material frame is described by the quaterion $\mathbf{q}={q_0(s),q_1(s),q_2(s),q_3(s)}^T$, then the rotation brings a constrain for $\mathbf{q}$:
\[\mathbf{q}_s=\mathbf{Q}\mathbf{q},\]where $\mathbf{Q}=\begin{pmatrix}0 & -\mathbf{\omega} \ \mathbf{\omega}^T & \mathbf{\Omega} \end{pmatrix}$ and $\mathbf{\omega} = \mathrm{ax}(\mathbf{\Omega}) = {\omega_1,\omega_2,\omega_3} $ is the axis vector. Note that the deformed configuration $\mathbf{r}(s)=x(s)\mathbf{i}+y(s)\mathbf{j}+z(s)\mathbf{k}={x(s),y(s),z(s)}^T$ can be expressed as:
\[\mathbf{r}_s=\mathbf{O}_3\mathbf{^T},\]where $\mathbf{O}_3$ is the third row of the matrix $\mathbf{O}$ and $\mathbf{O}$ can be expressed by the quaternion $\mathbf{q}$.
\[\mathbf{O} = 2 \begin{pmatrix} q_{0}^{2} + q_{1}^{2} - 1/2 & q_{1}q_{2} + q_{0}q_{3} & q_{1}q_{3} - q_{0}q_{2} \\ q_{1}q_{2} - q_{0}q_{3} & q_{0}^{2} + q_{2}^{2} - 1/2 & q_{2}q_{3} + q_{0}q_{1} \\ q_{1}q_{3} + q_{0}q_{2} & q_{2}q_{3} - q_{0}q_{1} & q_{0}^{2} + q_{3}^{2} - 1/2 \end{pmatrix}\]Surface constrains
We consider the constrain of the curved surface in this section. We assume the surface is parameterized with $(u,v)$ and the deformed configuration can be expressed as: $\mathbf{r}=\mathbf{r}(s)=\mathbf{r}(u(s),v(s))$, finally the normal vector can be written as:
\[\mathbf{e}_2=\frac{\mathbf{r}_u\times\mathbf{r}_v}{||\mathbf{r}_u\times\mathbf{r}_v||},\]where $\mathbf{r}_u=\partial_u \mathbf{r}$ and $\mathbf{r}_v=\partial_v\mathbf{r}$. The rotation angle can be expressed as $\cos\phi=\mathbf{e}_2\cdot\mathbf{d}_2$ and $\mathbf{e}_2$ has the relationship with $\mathbf{O}$:
\[\mathbf{e}_2=\mathbf{O}_2\mathbf{^T}\]The $\kappa_g$, $\kappa_n$ and $\tau_g$ can also be expressed with the surface parameters:
\[\begin{aligned} \kappa_{g}(s) = &\sqrt{\frac{E G-F^{2}}{\left(E u_s^{2}+2 F u_s v_s+G v_s^{2}\right)^{3}}}\times \left[ \Gamma_{11}^{2} u_s^{3} - \Gamma_{22}^{1} v_s^{3} + \left( 2\Gamma_{12}^{2} - \Gamma_{11}^{1} \right) u_s^2 v_s - \left( 2\Gamma_{12}^{1} - \Gamma_{22}^{2} \right) u_s v_s^2 + u_s v_{ss} - u_{ss} v_s \right] \end{aligned}\]where $(\cdot){ss}=\partial_s\partial_s(\cdot)$ and $\Gamma{\alpha\beta}^{\gamma}=\frac{1}{2}g^{\gamma\theta}(\frac{\partial g_{\theta \alpha}}{\partial x^\beta}+\frac{\partial g_{\theta \beta}}{\partial x^\alpha}-\frac{\partial g_{\alpha\beta}}{\partial x^{\theta}})$ is the Christoffel symbol.
\[\kappa_{n}(s) = \frac{L u_s^{2} + 2M u_s v_s + N v_s^{2}}{E u_s^{2} + 2F u_s v_s + G v_s^{2}},\] \[\tau_{g}(s) = \frac{(M E - LF) u_s^{2} + (NE - LG) u_s v_s + (NF - MG) v_s^{2}}{\left(E u_s^{2} + 2F u_s v_s + G v_s^{2}\right) \sqrt{EG - F^{2}}},\]where $E,F,G$ is the first fundamental form and $L,M,N$ is the second fundamental form of the surface.
Kirchhoff rod
The equilibrium equation of the Kirchhoff rod can be written as:
\[\begin{aligned} \mathbf{F}_s+\mathbf{f}&=0\\ \mathbf{M}_s + \mathbf{d}_3\times\mathbf{F}+\mathbf{m} &=0\end{aligned}\]We expand the $\mathbf{F}$ and $\mathbf{M}$ in surface frame $\mathbf{e}$, the components equations can be finally written as:
\[\begin{aligned} \mathbf{\bar{F}}_s+\mathbf{\bar{F}}\mathbf{\Gamma}+\mathbf{\bar{f}}&=0 \\ \mathbf{\bar{M}}_s+\mathbf{\bar{M}}\mathbf{\Gamma}+\mathbf{\bar{F}}\mathbf{\Lambda}+\mathbf{\bar{m}}&=0\end{aligned}\]where $\mathbf{\Lambda}=\begin{pmatrix} 0 & 1 &0 \ -1 & 0 & 0 \ 0 & 0 & 0\end{pmatrix}$ is the cross time matrix. The variable $(\cdot)$ with $\bar{(\cdot)}$ is the components form in $\mathbf{e}$, for example $\mathbf{\bar{F}}={F_1,F_2,F_3} \left( \begin{array}{c} \mathbf{e}_1 \ \mathbf{e}_2 \ \mathbf{e}_3\end{array}\right)$.
Constitutive law
For kirchhoff rod, we only consider bending deformation and the constitutive law can be written as:
\[\mathbf{M}=(\mathbf{\omega}-\mathbf{\omega}_0)\mathbf{S}\mathbf{d}=(\mathbf{\omega}-\mathbf{\omega}_0)\mathbf{S}\mathbf{R^T}\mathbf{e}\]where $\mathbf{S}= \operatorname{diag}({\mathrm{EI}_1,\mathrm{EI}_2,\mathrm{GJ}})$, $\mathrm{EI}_1$ is the bending stiffness along $\mathbf{d}_1$, $\mathrm{EI}_2$ is the bending stiffness along $\mathbf{d}_2$ and $\mathrm{GJ}$ is the twisting stiffness.
The constitutive law can be finally written as:
\[\mathbf{\bar{M}}=(\mathbf{\omega}-\mathbf{\omega}_0)\mathbf{S}\mathbf{R^T}\]Summary
We summary the control equation for the rod lies on the surface:
\[\begin{cases} \mathbf{\bar{F}}_s+\mathbf{\bar{F}}\mathbf{\Gamma}+\mathbf{\bar{f}}=0 \\ \mathbf{\bar{M}}_s+\mathbf{\bar{M}}\mathbf{\Gamma}+\mathbf{\bar{F}}\mathbf{\Lambda}+\mathbf{\bar{m}}=0 \\ \mathbf{\bar{M}}=(\operatorname{ax}(\mathbf{\Omega})-\operatorname{ax}(\mathbf{\Omega}_0))\mathbf{S}\mathbf{R^T} \\ \mathbf{q}_s=\mathbf{Q}\mathbf{q}\\ \mathbf{r}_s=\mathbf{O}_3\mathbf{^T}\\ \mathbf{e}_2=\frac{\mathbf{r}_u\times\mathbf{r}_v}{||\mathbf{r}_u\times\mathbf{r}_v||}=\mathbf{O}_2\mathbf{^T}\\ \mathbf{\Omega}=\mathbf{R}_s\mathbf{R^T}+\mathbf{R}\mathbf{\Gamma}\mathbf{R^T}\\ \mathbf{\Gamma}=\mathbf{\Gamma}(u(s),v(s))\\ \end{cases}\]Component form for numerical solution
The component form can be written as:
\[\begin{cases}(F_1)_s+F_3\omega_2-F_2\omega_3=0\\ (F_2)_s+F_1\omega_3-F_3\omega_1+p=0\\ (F_3)_s+F_2\omega_1 -F_1\omega_2=0\\ (M_1)_s+M_3\omega_2-M_2\omega_3-F_2=0\\ (M_2)_s+M_1\omega_3-M_3\omega_1+F_1=0\\ (M_3)_s+M_2\omega_1 -M_1\omega_2=0\\ \omega_3=\tau_g+(\phi)_s\\\omega_2=\kappa_g\cos\phi+\kappa_n\sin\phi\\ \omega_1=\kappa_g\sin\phi-\kappa_n\cos\phi\\ M_1=\mathrm{EI}_1(\omega _1-\bar{\omega}_1)\\ M_2=\mathrm{EI}_2(\omega _2-\bar{\omega}_2)\\ M_3=\mathrm{GJ}(\omega _3-\bar{\omega}_3) \\ \left( q_0 \right)_s=-\frac12\left( q_1\omega_{1}+q_2\omega_{2}+q_3\omega_{3} \right)\\ \left( q_1 \right)_s=\frac12\left( q_0\omega_{1}-q_3\omega_{2}+q_2\omega_{3} \right)\\ \left( q_2 \right)_s=\frac12\left( q_3\omega_{1}+q_0\omega_{2}-q_1\omega_{3} \right)\\ \left( q_3 \right)_s=\frac12\left( q_1\omega_{2}-q_2\omega_{1}+q_0\omega_{3} \right)\\ \mathbf{r}_u (u(s))_s+\mathbf{r}_v (v(s))_s=\mathbf{O}_3\mathbf{^T}=2\{ q_{1}q_{3} + q_{0}q_{2}, q_{2}q_{3} - q_{0}q_{1}, q_{0}^{2} + q_{3}^{2} - 1/2 \}^T\\ \kappa_{g}(s) = \sqrt{\frac{E G-F^{2}}{\left(E u_s^{2}+2 F u_s v_s+G v_s^{2}\right)^{3}}}\times \left[ \Gamma_{11}^{2} u_s^{3} - \Gamma_{22}^{1} v_s^{3} + \left( 2\Gamma_{12}^{2} - \Gamma_{11}^{1} \right) u_s^2 v_s - \left( 2\Gamma_{12}^{1} - \Gamma_{22}^{2} \right) u_s v_s^2 + u_s v_{ss} - u_{ss} v_s \right]\\ \kappa_{n}(s) = \frac{L u_s^{2} + 2M u_s v_s + N v_s^{2}}{E u_s^{2} + 2F u_s v_s + G v_s^{2}}\\ \tau_{g}(s) = \frac{(M E - LF) u_s^{2} + (NE - LG) u_s v_s + (NF - MG) v_s^{2}}{\left(E u_s^{2} + 2F u_s v_s + G v_s^{2}\right) \sqrt{EG - F^{2}}}\\ \end{cases}\]The component equations can be simplified according to the constrains for the arc-length $E u_s^2+2Fu_sv_s+Gv_s^2=1$, we have:
\[\begin{cases}(F_1)_s+F_3\omega _2-F_2\omega _3=0 \\ (F_3)_s+F_2\omega _1-F_1\omega _2=0 \\ (M_2)_s+M_1\omega _3-M_3\omega _1+F_1=0 \\ (M_3)_s+M_2\omega _1-M_1\omega _2=0 \\ (\phi )_s=\tau _g-\omega _3 \\ (\kappa_n)_s=(\omega_2\sin\phi-\omega_1\cos\phi)_s \\ \left( q_0 \right) _s=-\frac{1}{2}\left( q_1\omega _1+q_2\omega _2+q_3\omega _3 \right) \\ \left( q_1 \right) _s=\frac{1}{2}\left( q_0\omega _1-q_3\omega _2+q_2\omega _3 \right) \\ \left( q_2 \right) _s=\frac{1}{2}\left( q_3\omega _1+q_0\omega _2-q_1\omega _3 \right) \\ \left( q_3 \right) _s=\frac{1}{2}\left( q_1\omega _2-q_2\omega _1+q_0\omega _3 \right) \\ \mathbf{r}_u(u(s))_s+\mathbf{r}_v(v(s))_s={\mathbf{O}_3}^{\mathbf{T}}=2\{q_1q_3+q_0q_2,q_2q_3-q_0q_1,q_{0}^{2}+q_{3}^{2}-1/2\}^T\end{cases}\]The variables needed to be solved is ${F_1,F_3,\phi,\omega_1,\omega_2,\omega_3,q_0,q_1,q_2,q_3,u,v}$
Other variables can be expressed as:
\[\begin{cases}F_2=(M_1)_s+M_3\omega _2-M_2\omega _3 \\ M_1=\mathrm{EI}_1(\omega _1-\bar{\omega}_1) \\ M_2=\mathrm{EI}_2(\omega _2-\bar{\omega}_2) \\ M_3=\mathrm{GJ(}\omega _3-\bar{\omega}_3) \\ \kappa _g(s)=\sqrt{EG-F^2}\times \left[ \Gamma _{11}^{2}u_{s}^{3}-\Gamma _{22}^{1}v_{s}^{3}+\left( 2\Gamma _{12}^{2}-\Gamma _{11}^{1} \right) u_{s}^{2}v_s-\left( 2\Gamma _{12}^{1}-\Gamma _{22}^{2} \right) u_sv_{s}^{2}+u_sv_{ss}-u_{ss}v_s \right] \\ \kappa _n(s)=Lu_{s}^{2}+2Mu_sv_s+Nv_{s}^{2} \\ \tau _g(s)=\frac{(ME-LF)u_{s}^{2}+(NE-LG)u_sv_s+(NF-MG)v_{s}^{2}}{\sqrt{EG-F^2}} \\ p=F_3\omega _1-(F_2)_s-F_1\omega _3 \\ \kappa _g=\omega _2\cos \phi +\omega_1\sin \phi\end{cases}\]always $0$. The control equation can be simplified as: