Simplified Forms of Kirchhoff Rod Theory
Published:
How does the Kirchhoff rod reduce to some classical theories?
All assumed to be constant circular cross-section elastic rods
Static Elastic Rod
\[\begin{array}{c} \frac{\partial \mathbf{F}}{\partial s}+\mathbf{q}=0 \\[6pt] \frac{\partial \mathbf{M}}{\partial s}+\frac{\partial \mathbf{r}}{\partial s} \times \mathbf{F}+\mathbf{m}=0 \\[6pt] \dot{\mathbf{r}}=0 \\[6pt] \mathbf{d}_{1}=\frac{\partial u}{\partial s} \\[6pt] \mathbf{\Gamma}=\mathbf{R}_{s} \mathbf{R}^{\mathbf{T}} \\[6pt] \mathbf{\Omega}=0 \\[6pt] \mathbf{M}=\mathrm{EI}(\mathbf{\Gamma}) \end{array}\]The above simplifies to:
\[\mathrm{F}_s+\left( \mathrm{FR}_{\mathrm{s}}+\mathrm{q} \right) \mathrm{R}^{\mathrm{T}}=0\] \[\mathrm{M}_s+\left( \mathrm{MR}_{\mathrm{s}}+\mathrm{m} \right) \mathrm{R}^{\mathrm{T}}+\mathrm{F}\Lambda =0\] \[\Lambda =\begin{pmatrix} 0 & 0 & 0\\ 0 & 0 & -1\\ 0 & 1 & 0 \end{pmatrix}\] \[\Gamma =\begin{pmatrix} 0 & \omega_3 & -\omega_2\\ -\omega_3 & 0 & \omega_1\\ \omega_2 & -\omega_1 & 0 \end{pmatrix} =\mathrm{R}_s \mathrm{R}^{\mathrm{T}}\] \[\mathrm{M}=\begin{pmatrix} \mathrm{GI}\omega_1 & -\mathrm{EI}\omega_2 & \mathrm{EI}\omega_3 \end{pmatrix}\]Two cases are discussed:
① Rotation expressed in Euler angles (3-2-3 rotation)
The rotation matrix is:
\[\mathrm{R}=\begin{pmatrix} c_{\alpha}c_{\beta}c_{\gamma}-s_{\alpha}s_{\gamma} & -c_{\alpha}c_{\beta}s_{\gamma}-c_{\gamma}s_{\alpha} & c_{\alpha}s_{\beta}\\ c_{\beta}c_{\gamma}s_{\alpha}+c_{\alpha}s_{\gamma} & c_{\alpha}c_{\gamma}-c_{\beta}s_{\alpha}s_{\gamma} & s_{\alpha}s_{\beta}\\ -c_{\gamma}s_{\beta} & s_{\beta}s_{\gamma} & c_{\beta} \end{pmatrix}\]The Frenet matrix is:
\[\Gamma =\begin{pmatrix} 0 & -\alpha'(s)-\cos(\beta(s))\gamma'(s) & \cos(\alpha(s))\beta'(s)+\sin(\alpha(s))\sin(\beta(s))\gamma'(s)\\ \alpha'(s)+\cos(\beta(s))\gamma'(s) & 0 & \sin(\alpha(s))\beta'(s)-\cos(\alpha(s))\sin(\beta(s))\gamma'(s)\\ -\cos(\alpha(s))\beta'(s)-\sin(\alpha(s))\sin(\beta(s))\gamma'(s) & \cos(\alpha(s))\sin(\beta(s))\gamma'(s)-\sin(\alpha(s))\beta'(s) & 0 \end{pmatrix}\]Combining (2), (3), and (4) in component form:
\[\begin{pmatrix} \mathrm{F}_1(s)\\ \mathrm{F}_2(s)\\ \mathrm{F}_3(s) \end{pmatrix}'- \begin{pmatrix} 0 & \omega_3(s) & -\omega_2(s)\\ -\omega_3(s) & 0 & \omega_1(s)\\ \omega_2(s) & -\omega_1(s) & 0 \end{pmatrix} \begin{pmatrix} \mathrm{F}_1(s)\\ \mathrm{F}_2(s)\\ \mathrm{F}_3(s) \end{pmatrix} + \begin{pmatrix} \cos(\alpha(s))\cos(\beta(s))\cos(\gamma(s))-\sin(\alpha(s))\sin(\gamma(s)) & -\sin(\alpha(s))\cos(\gamma(s))-\cos(\alpha(s))\cos(\beta(s))\sin(\gamma(s)) & \cos(\alpha(s))\sin(\beta(s))\\ \sin(\alpha(s))\cos(\beta(s))\cos(\gamma(s))+\cos(\alpha(s))\sin(\gamma(s)) & \cos(\alpha(s))\cos(\gamma(s))-\sin(\alpha(s))\cos(\beta(s))\sin(\gamma(s)) & \sin(\alpha(s))\sin(\beta(s))\\ -\sin(\beta(s))\cos(\gamma(s)) & \sin(\beta(s))\sin(\gamma(s)) & \cos(\beta(s)) \end{pmatrix} \begin{pmatrix} \mathrm{q}_x(s)\\ \mathrm{q}_y(s)\\ \mathrm{q}_z(s) \end{pmatrix} =0\] \[\begin{pmatrix} \mathrm{M}_1(s)\\ \mathrm{M}_2(s)\\ \mathrm{M}_3(s) \end{pmatrix}'- \begin{pmatrix} 0 & \omega_3(s) & -\omega_2(s)\\ -\omega_3(s) & 0 & \omega_1(s)\\ \omega_2(s) & -\omega_1(s) & 0 \end{pmatrix} \begin{pmatrix} \mathrm{M}_1(s)\\ \mathrm{M}_2(s)\\ \mathrm{M}_3(s) \end{pmatrix} + \begin{pmatrix} \cos(\alpha(s))\cos(\beta(s))\cos(\gamma(s))-\sin(\alpha(s))\sin(\gamma(s)) & -\sin(\alpha(s))\cos(\gamma(s))-\cos(\alpha(s))\cos(\beta(s))\sin(\gamma(s)) & \cos(\alpha(s))\sin(\beta(s))\\ \sin(\alpha(s))\cos(\beta(s))\cos(\gamma(s))+\cos(\alpha(s))\sin(\gamma(s)) & \cos(\alpha(s))\cos(\gamma(s))-\sin(\alpha(s))\cos(\beta(s))\sin(\gamma(s)) & \sin(\alpha(s))\sin(\beta(s))\\ -\sin(\beta(s))\cos(\gamma(s)) & \sin(\beta(s))\sin(\gamma(s)) & \cos(\beta(s)) \end{pmatrix} \begin{pmatrix} \mathrm{m}_x(s)\\ \mathrm{m}_y(s)\\ \mathrm{m}_z(s) \end{pmatrix} + \begin{pmatrix} 0\\ \mathrm{F}_3(s)\\ -\mathrm{F}_2(s) \end{pmatrix} =0\] \[\begin{pmatrix} \mathrm{M}_1(s)\\ \mathrm{M}_2(s)\\ \mathrm{M}_3(s) \end{pmatrix} = \begin{pmatrix} \mathrm{GI} & 0 & 0\\ 0 & -\mathrm{EI} & 0\\ 0 & 0 & \mathrm{EI} \end{pmatrix} \begin{pmatrix} \omega_1(s)\\ \omega_2(s)\\ \omega_3(s) \end{pmatrix}\] \[\begin{pmatrix} \omega_1(s)\\ \omega_2(s)\\ \omega_3(s) \end{pmatrix} = \begin{pmatrix} \sin(\alpha(s))\beta'(s)-\cos(\alpha(s))\sin(\beta(s))\gamma'(s)\\ -\cos(\alpha(s))\beta'(s)-\sin(\alpha(s))\sin(\beta(s))\gamma'(s)\\ -\alpha'(s)-\cos(\beta(s))\gamma'(s) \end{pmatrix}\]There are 12 sets of equations and 12 variables ($\omega_i; \mathrm{M}_i; \mathrm{F}_i; \alpha; \beta; \gamma$). The system is closed.
② Rotation expressed in quaternion form
The rotation matrix is:
\[\mathrm{R}=\begin{pmatrix} -2q_2(s)^2-2q_3(s)^2+1 & 2(q_0(s)q_3(s)+q_1(s)q_2(s)) & 2q_1(s)q_3(s)-2q_0(s)q_2(s)\\ 2q_1(s)q_2(s)-2q_0(s)q_3(s) & -2q_1(s)^2-2q_3(s)^2+1 & 2(q_0(s)q_1(s)+q_2(s)q_3(s))\\ 2(q_0(s)q_2(s)+q_1(s)q_3(s)) & 2q_2(s)q_3(s)-2q_0(s)q_1(s) & -2q_1(s)^2-2q_2(s)^2+1 \end{pmatrix}\]The Frenet matrix is:
\[\Gamma=\begin{pmatrix} 0 & 2\left( -q_3(s)q_0'(s)+q_0(s)q_3'(s)+q_2(s)q_1'(s)-q_1(s)q_2'(s) \right) & 2\left( q_2(s)q_0'(s)-q_0(s)q_2'(s)+q_3(s)q_1'(s)-q_1(s)q_3'(s) \right)\\ 2\left( q_3(s)q_0'(s)-q_0(s)q_3'(s)-q_2(s)q_1'(s)+q_1(s)q_2'(s) \right) & 0 & 2\left( -q_1(s)q_0'(s)+q_0(s)q_1'(s)+q_3(s)q_2'(s)-q_2(s)q_3'(s) \right)\\ 2\left( -q_2(s)q_0'(s)+q_0(s)q_2'(s)-q_3(s)q_1'(s)+q_1(s)q_3'(s) \right) & 2\left( q_1(s)q_0'(s)-q_0(s)q_1'(s)-q_3(s)q_2'(s)+q_2(s)q_3'(s) \right) & 0 \end{pmatrix}\]Plus the constraint:
\[q_0(s)^2+q_1(s)^2+q_2(s)^2+q_3(s)^2=1\]Combining (2), (6), (7), and (8) in component form:
\[\begin{pmatrix} \mathrm{F}_1(s)\\ \mathrm{F}_2(s)\\ \mathrm{F}_3(s) \end{pmatrix}'- \begin{pmatrix} 0 & \omega_3(s) & -\omega_2(s)\\ -\omega_3(s) & 0 & \omega_1(s)\\ \omega_2(s) & -\omega_1(s) & 0 \end{pmatrix} \begin{pmatrix} \mathrm{F}_1(s)\\ \mathrm{F}_2(s)\\ \mathrm{F}_3(s) \end{pmatrix} + \begin{pmatrix} -2q_2(s)^2-2q_3(s)^2+1 & 2(q_0(s)q_3(s)+q_1(s)q_2(s)) & 2q_1(s)q_3(s)-2q_0(s)q_2(s)\\ 2q_1(s)q_2(s)-2q_0(s)q_3(s) & -2q_1(s)^2-2q_3(s)^2+1 & 2(q_0(s)q_1(s)+q_2(s)q_3(s))\\ 2(q_0(s)q_2(s)+q_1(s)q_3(s)) & 2q_2(s)q_3(s)-2q_0(s)q_1(s) & -2q_1(s)^2-2q_2(s)^2+1 \end{pmatrix} \begin{pmatrix} q_x(s)\\ q_y(s)\\ q_z(s) \end{pmatrix} =0\] \[\begin{pmatrix} \mathrm{M}_1(s)\\ \mathrm{M}_2(s)\\ \mathrm{M}_3(s) \end{pmatrix}'- \begin{pmatrix} 0 & \omega_3(s) & -\omega_2(s)\\ -\omega_3(s) & 0 & \omega_1(s)\\ \omega_2(s) & -\omega_1(s) & 0 \end{pmatrix} \begin{pmatrix} \mathrm{M}_1(s)\\ \mathrm{M}_2(s)\\ \mathrm{M}_3(s) \end{pmatrix} + \begin{pmatrix} -2q_2(s)^2-2q_3(s)^2+1 & 2(q_0(s)q_3(s)+q_1(s)q_2(s)) & 2q_1(s)q_3(s)-2q_0(s)q_2(s)\\ 2q_1(s)q_2(s)-2q_0(s)q_3(s) & -2q_1(s)^2-2q_3(s)^2+1 & 2(q_0(s)q_1(s)+q_2(s)q_3(s))\\ 2(q_0(s)q_2(s)+q_1(s)q_3(s)) & 2q_2(s)q_3(s)-2q_0(s)q_1(s) & -2q_1(s)^2-2q_2(s)^2+1 \end{pmatrix} \begin{pmatrix} \mathrm{m}_x(s)\\ \mathrm{m}_y(s)\\ \mathrm{m}_z(s) \end{pmatrix} + \begin{pmatrix} 0\\ \mathrm{F}_3(s)\\ -\mathrm{F}_2(s) \end{pmatrix} =0\] \[\begin{pmatrix} \mathrm{M}_1(s)\\ \mathrm{M}_2(s)\\ \mathrm{M}_3(s) \end{pmatrix} = \begin{pmatrix} \mathrm{GI} & 0 & 0\\ 0 & -\mathrm{EI} & 0\\ 0 & 0 & \mathrm{EI} \end{pmatrix} \begin{pmatrix} \omega_1(s)\\ \omega_2(s)\\ \omega_3(s) \end{pmatrix}\] \[\begin{pmatrix} \omega_1(s)\\ \omega_2(s)\\ \omega_3(s) \end{pmatrix} = \begin{pmatrix} 2\left( -q_1(s)q_0'(s)+q_0(s)q_1'(s)+q_3(s)q_2'(s)-q_2(s)q_3'(s) \right)\\ 2\left( -q_2(s)q_0'(s)+q_0(s)q_2'(s)-q_3(s)q_1'(s)+q_1(s)q_3'(s) \right)\\ 2\left( -q_3(s)q_0'(s)+q_0(s)q_3'(s)+q_2(s)q_1'(s)-q_1(s)q_2'(s) \right) \end{pmatrix}\]The choice of rotation representation only affects equation (5). When external loads are present, it differs from the Euler angle solution; when external loads are zero, the local frame and global frame are decoupled.
Simplified Discussion
If the external load is zero, the Euler angles decouple from the above equations, giving:
\[\begin{pmatrix} \mathrm{F}_1(s)\\ \mathrm{F}_2(s)\\ \mathrm{F}_3(s) \end{pmatrix}'- \begin{pmatrix} 0 & \omega_3(s) & -\omega_2(s)\\ -\omega_3(s) & 0 & \omega_1(s)\\ \omega_2(s) & -\omega_1(s) & 0 \end{pmatrix} \begin{pmatrix} \mathrm{F}_1(s)\\ \mathrm{F}_2(s)\\ \mathrm{F}_3(s) \end{pmatrix} =0\] \[\begin{pmatrix} \mathrm{M}_1(s)\\ \mathrm{M}_2(s)\\ \mathrm{M}_3(s) \end{pmatrix}'- \begin{pmatrix} 0 & \omega_3(s) & -\omega_2(s)\\ -\omega_3(s) & 0 & \omega_1(s)\\ \omega_2(s) & -\omega_1(s) & 0 \end{pmatrix} \begin{pmatrix} \mathrm{M}_1(s)\\ \mathrm{M}_2(s)\\ \mathrm{M}_3(s) \end{pmatrix} + \begin{pmatrix} 0\\ \mathrm{F}_3(s)\\ -\mathrm{F}_2(s) \end{pmatrix} =0\] \[\begin{pmatrix} \mathrm{M}_1(s)\\ \mathrm{M}_2(s)\\ \mathrm{M}_3(s) \end{pmatrix} = \begin{pmatrix} \mathrm{GI} & 0 & 0\\ 0 & -\mathrm{EI} & 0\\ 0 & 0 & \mathrm{EI} \end{pmatrix} \begin{pmatrix} \omega_1(s)\\ \omega_2(s)\\ \omega_3(s) \end{pmatrix}\] \[\begin{pmatrix} \omega_1(s)\\ \omega_2(s)\\ \omega_3(s) \end{pmatrix} = \begin{pmatrix} \sin(\alpha(s))\beta'(s)-\cos(\alpha(s))\sin(\beta(s))\gamma'(s)\\ -\cos(\alpha(s))\beta'(s)-\sin(\alpha(s))\sin(\beta(s))\gamma'(s)\\ -\alpha'(s)-\cos(\beta(s))\gamma'(s) \end{pmatrix} = \begin{pmatrix} 2\left( -q_1(s)q_0'(s)+q_0(s)q_1'(s)+q_3(s)q_2'(s)-q_2(s)q_3'(s) \right)\\ 2\left( -q_2(s)q_0'(s)+q_0(s)q_2'(s)-q_3(s)q_1'(s)+q_1(s)q_3'(s) \right)\\ 2\left( -q_3(s)q_0'(s)+q_0(s)q_3'(s)+q_2(s)q_1'(s)-q_1(s)q_2'(s) \right) \end{pmatrix}\]Simplifying:
\[\begin{cases} \mathrm{EI}\left( \left( \omega_1\left( \omega_3^2-\omega_2^2 \right) \right)'+\omega_3(s)\omega_2''(s)-2\omega_1(s)^2\omega_2(s)\omega_3(s)+\omega_2(s)\omega_3''(s) \right) -\mathrm{GI}\left( \left( \omega_2(s)^2+\omega_3(s)^2 \right) \omega_1'(s)+\frac{\omega_1(s)}{2}\left( \omega_2(s)^2+\omega_3(s)^2 \right)' \right) =0\\ \mathrm{EI}\left( \frac{3\omega_1'(s)\omega_3'(s)}{\omega_2(s)}+\omega_3(s)\left( \frac{\omega_1'(s)}{\omega_2(s)} \right)'+2\omega_1(s)\left( \frac{\omega_3'(s)}{\omega_2(s)} \right)'-\left( \omega_1(s)^2+\frac{1}{2}\left( \omega_3(s)^2-\omega_2(s)^2 \right) \right)'+2\omega_1(s)\omega_2(s)\omega_3(s)+\left( \frac{\omega_2''(s)}{\omega_2(s)} \right)' \right) +\mathrm{GI}\left( -\omega_3(s)\left( \frac{\omega_1'(s)}{\omega_2(s)} \right)'-\omega_3'(s)\left( \frac{\omega_1(s)}{\omega_2(s)} \right)'-\frac{\left( \omega_1(s)\omega_3'(s) \right)'}{\omega_2(s)}-\left( \omega_1(s)^2 \right)' \right) =0\\ \mathrm{EI}\left( \frac{3\omega_1'(s)\omega_2'(s)}{\omega_3(s)}-\omega_2(s)\left( \frac{\omega_1'(s)}{\omega_3(s)} \right)'+2\omega_1(s)\left( \frac{\omega_2'(s)}{\omega_3(s)} \right)'+\left( \omega_1(s)^2+\frac{1}{2}\left( \omega_2(s)^2-\omega_3(s)^2 \right) \right)'+2\omega_1(s)\omega_2(s)\omega_3(s)-\left( \frac{\omega_3''(s)}{\omega_3(s)} \right)' \right) +\mathrm{GI}\left( \omega_1(s)\left( \frac{\omega_2'(s)}{\omega_3(s)} \right)'+\omega_1'(s)\left( \frac{\omega_2(s)}{\omega_3(s)} \right)'+\frac{\left( \omega_2(s)\omega_1'(s) \right)'}{\omega_3(s)}-\left( \omega_1(s)^2 \right)' \right) =0\\ \end{cases}\]Even for a Kirchhoff rod without external loads, the governing equations are already quite complex. Now consider choosing the Frenet frame, i.e., let $\omega_3=\kappa;\omega_2=0;\omega_1=\tau$. The above simplifies to:
\[\begin{cases} \frac{\tau'}{\tau}+\frac{\left( 2\mathrm{EI}-\mathrm{GI} \right)}{2\left( \mathrm{EI}-\mathrm{GI} \right)}\frac{\left( \kappa^2 \right)'}{\kappa^2}=0\\ \left( \frac{\kappa''}{\kappa} \right)'+\left( \frac{\mathrm{GI}}{\mathrm{EI}}-1 \right) \left( \tau^2 \right)'+\frac{1}{2}\left( \kappa^2 \right)'=0\\ \end{cases}\]Define $\frac{\mathrm{EI}}{\mathrm{GI}}=\nu+1$, $\nu$ is Poisson’s ratio. The governing equations become:
\[\begin{cases} \frac{\tau'}{\tau}+\left( 1+\frac{1}{2\nu} \right) \frac{\left( \kappa^2 \right)'}{\kappa^2}=0\\ \left( \frac{\kappa''}{\kappa} \right)'-\frac{\nu}{\nu+1}\left( \tau^2 \right)'+\frac{1}{2}\left( \kappa^2 \right)'=0\\ \end{cases}\]Integrating (9):
\[\begin{cases} \tau \kappa^{2+\frac{1}{\nu}}=\mathrm{c}_0\\ \kappa''-\frac{\nu}{\nu+1}\tau^2\kappa +\frac{1}{2}\kappa^3+\mathrm{c}_1=0\\ \end{cases}\]The governing equation for the unloaded static elastic rod with torsion is:
\[\kappa''+\frac{1}{2}\kappa^3+\mathrm{c}_1\kappa -\frac{\nu}{\nu+1}\mathrm{c}_0^2\frac{1}{\kappa^{3+\frac{2}{\nu}}}=0\]If torsional energy is neglected, i.e., $\mathrm{GI}\rightarrow0;\nu\rightarrow\infty$, (11) reduces to the Euler’s Elastica Rod equation obtained via the energy functional in The Elastic Rod Problem and Its Numerical Solution.
Steady-State Elastic Rod (Moving Along the Rope)
The second derivative of the local frame is:
\[\ddot{\mathbf{d}}=\left( v_t\Gamma +v^2\left( \Gamma_s+\Gamma\Gamma \right) \right) \mathbf{d}\] \[\ddot{r}=r_{ss}v^2+r_sv_t=v_t\mathbf{d}_1+v^2\left( m_1\mathbf{d}_2+m_2\mathbf{d}_3 \right)\]The elastic rod equations:
\[\frac{\partial \mathbf{F}}{\partial s}+\mathbf{q}=\rho \mathbf{A}\ddot{\mathbf{r}}=\rho \mathbf{A}(v_t\mathbf{d}_1+v^2(m_1\mathbf{d}_2+m_2\mathbf{d}_3))\] \[\frac{\partial \mathbf{M}}{\partial s}+\frac{\partial \mathbf{r}}{\partial s}\times \mathbf{F}+\mathbf{m}=\rho \left( \mathrm{I}_3\mathbf{d}_2\times \ddot{\mathbf{d}}_2+\mathrm{I}_2\mathbf{d}_3\times \ddot{\mathbf{d}}_3 \right)\] \[=\rho I_0\left( \left( 2v^2m'+2mv_t \right) \mathbf{d}_1+\left( -m_2v_t-v^2m_2'+v^2mm_1 \right) \mathbf{d}_2+\left( m_1v_t+v^2m_1'+v^2mm_2 \right) \mathbf{d}_3 \right)\] \[\dot{\mathbf{r}}=\frac{\partial u}{\partial t}=\frac{\partial u}{\partial s}v=v\mathbf{d}_1\] \[\mathbf{d}_1=\frac{\partial u}{\partial s}\] \[\mathbf{\Gamma}=\mathbf{R}_s\mathbf{R}^{\mathbf{T}}\] \[\mathbf{\Omega}=v\mathbf{R}_s\mathbf{R}^{\mathbf{T}}=v\mathbf{\Gamma}\] \[\mathbf{M}=\mathrm{EI}(\mathbf{\Gamma})\]Two-Dimensional Theory
Two-Dimensional Static Theory
Starting from the static theory equations:
\[\frac{\partial \mathbf{F}}{\partial s}+\mathbf{q}=0\] \[\frac{\partial \mathbf{M}}{\partial s}+\frac{\partial \mathbf{r}}{\partial s}\times \mathbf{F}+\mathbf{m}=0\] \[\dot{\mathbf{r}}=0\] \[\mathbf{d}_1=\frac{\partial u}{\partial s}\] \[\mathbf{\Gamma}=\mathbf{R}_s\mathbf{R}^{\mathbf{T}}\] \[\mathbf{\Omega}=0\] \[\mathbf{M}=\mathrm{EI}(\mathbf{\Gamma})\]Simplification conditions:
\[\mathbf{F}=\mathrm{F}_1\mathbf{d}_1+\mathrm{F}_2\mathbf{d}_2\] \[\mathbf{M}=\mathrm{M}\mathbf{d}_3\] \[\tau =0\] \[\dot{\mathbf{d}}_3=0\] \[\dot{\mathbf{d}}_1=\kappa \mathbf{d}_2\] \[\dot{\mathbf{d}}_2=-\kappa \mathbf{d}_1\] \[\mathrm{M}=\mathrm{EI}\kappa\] \[\begin{pmatrix} \mathbf{d}_1\\ \mathbf{d}_2 \end{pmatrix} = \begin{pmatrix} \cos\theta(s) & -\sin\theta(s)\\ \sin\theta(s) & \cos\theta(s) \end{pmatrix} \begin{pmatrix} \mathbf{x}\\ \mathbf{y} \end{pmatrix}\] \[\begin{pmatrix} 0 & \kappa\\ -\kappa & 0 \end{pmatrix} = \begin{pmatrix} 0 & -\theta'(s)\\ \theta'(s) & 0 \end{pmatrix}\] \[\mathbf{m}=m(s)\mathbf{d}_3\] \[\mathbf{q}=q_x(s)\mathbf{i}+q_y(s)\mathbf{j} = \begin{pmatrix} q_x(s) & q_y(s) \end{pmatrix} \begin{pmatrix} \mathbf{i}\\ \mathbf{j} \end{pmatrix} = \begin{pmatrix} q_x(s) & q_y(s) \end{pmatrix} \begin{pmatrix} \cos\theta(s) & \sin\theta(s)\\ -\sin\theta(s) & \cos\theta(s) \end{pmatrix} \begin{pmatrix} \mathbf{d}_1\\ \mathbf{d}_2 \end{pmatrix}\]The vector equations simplify to:
\[\begin{cases} \mathrm{F}_1'(s)-\mathrm{F}_2(s)\kappa(s)+\cos\theta(s) q_x(s) -\sin\theta(s) q_y(s) =0\\ \mathrm{F}_1(s)\kappa(s)+\mathrm{F}_2'(s)+\cos\theta(s) q_y(s) +\sin\theta(s) q_x(s) =0\\ \mathrm{M}'(s)+\mathrm{F}_2(s) +m(s) =0\\ \kappa(s) =-\theta'(s) \end{cases}\]Two cases are discussed:
① Small angle case: $\theta(s)\approx0$, the frame coincides with the fixed coordinate system. Equation (6) simplifies to:
\[\begin{cases} \mathrm{F}_1'(s)-\mathrm{F}_2(s)\kappa(s)+q_x(s) =0\\ \mathrm{F}_1(s)\kappa(s)+\mathrm{F}_2'(s)+q_y(s) =0\\ \mathrm{M}'(s)+\mathrm{F}_2(s) +m(s) =0\\ \kappa(s) =-\theta'(s)\\ \mathrm{M}(s) =\mathrm{EI}\kappa(s) \end{cases}\]For an inextensible beam, the axial force is constant. For a cantilever beam with zero axial force at the right end, $\mathrm{F}_1 =0$.
\[\begin{cases} \mathrm{F}_2(s)\kappa(s)+q_x(s) =0\\ \mathrm{F}_2'(s)+q_y(s) =0\\ \mathrm{M}'(s)+\mathrm{F}_2(s) +m(s) =0\\ \kappa(s) =-\theta'(s)\\ \mathrm{M}(s) =\mathrm{EI}\kappa(s) \end{cases}\]From equation (8): $q_y(s) =\mathrm{EI}\kappa’‘(s) +m’(s)$.
For a beam under constant bending moment load, the above reduces to the classical Euler beam formula:
\[q_y(s) =(\mathrm{EI}\kappa(s))''=\frac{\partial^2}{\partial s^2}(\mathrm{EI}\frac{\partial^2u(s)}{\partial s^2})\]② Case with zero external load: $q_x(s)=0, q_y(s)=0$.
Equation (6) simplifies to:
\[\begin{cases} \mathrm{F}_1'(s)-\mathrm{F}_2(s)\kappa(s)=0\\ \mathrm{F}_1(s)\kappa(s)+\mathrm{F}_2'(s)=0\\ \mathrm{M}'(s)+\mathrm{F}_2(s) +m(s) =0\\ \kappa(s) =-\theta'(s)\\ \mathrm{M}(s) =\mathrm{EI}\kappa(s) \end{cases}\]Equation (10) simplifies to:
\[\begin{cases} \mathrm{m}(s) \kappa(s) +\left( \frac{\mathrm{m}'(s)}{\kappa(s)} \right)'+\mathrm{EI}\left( \frac{\kappa(s)^2}{2}+\frac{\kappa''(s)}{\kappa(s)} \right)'=0\\ \mathrm{F}_2(s) \kappa(s) +\left( \frac{\mathrm{F}_2'(s)}{\kappa(s)} \right)'=0\\ \mathrm{m}(s) +\mathrm{EI}\kappa'(s) +\mathrm{F}_2(s) =0 \end{cases}\]If the external moment load is zero, $\mathrm{m}(s)=0$. We obtain: $\frac{\kappa(s)^2}{2}+\frac{\kappa’‘(s)}{\kappa(s)}=c$, which expands to the Euler Elastica equation (already obtained via the energy method in The Elastic Rod Problem and Its Numerical Solution):
\[\kappa''(s) +\frac{\kappa(s)^3}{2}+\mathrm{c}\kappa(s) =0\]Two-Dimensional Steady-State Theory
The two-dimensional steady-state theory can be simplified to:
\[\frac{\partial \mathbf{F}}{\partial s}+\mathbf{q}=\rho\mathrm{A}\left( v_t\mathbf{d}_1+v^2\kappa(s) \mathbf{d}_2 \right)\] \[\frac{\partial \mathbf{M}}{\partial s}+\frac{\partial \mathbf{r}}{\partial s}\times \mathbf{F}+\mathbf{m}=\rho\mathrm{I}_0\left( \kappa(s) v_t+v^2\kappa'(s) \right) \mathbf{d}_3\] \[\dot{\mathbf{r}}=v\mathbf{d}_1\] \[\mathbf{d}_1=\frac{\partial u}{\partial s}\] \[\mathbf{\Gamma}=\mathbf{R}_s\mathbf{R}^{\mathbf{T}}\] \[\mathbf{\Omega}=v\mathbf{R}_s\mathbf{R}^{\mathbf{T}}\] \[\mathbf{M}=\mathrm{EI}(\mathbf{\Gamma})\]Simplification conditions:
\[\mathbf{F}=\mathrm{F}_1\mathbf{d}_1+\mathrm{F}_2\mathbf{d}_2\] \[\mathbf{M}=\mathrm{M}\mathbf{d}_3\] \[\tau =0\] \[\mathbf{d}'_3=0\] \[\mathbf{d}'_1=\kappa \mathbf{d}_2\] \[\mathbf{d}'_2=-\kappa \mathbf{d}_1\] \[\mathrm{M}=\mathrm{EI}\kappa\] \[\begin{pmatrix} \mathbf{d}_1\\ \mathbf{d}_2 \end{pmatrix} = \begin{pmatrix} \cos\theta(s) & -\sin\theta(s)\\ \sin\theta(s) & \cos\theta(s) \end{pmatrix} \begin{pmatrix} \mathbf{x}\\ \mathbf{y} \end{pmatrix}\] \[\begin{pmatrix} 0 & \kappa\\ -\kappa & 0 \end{pmatrix} = \begin{pmatrix} 0 & -\theta'(s)\\ \theta'(s) & 0 \end{pmatrix}\] \[\mathbf{m}=m(s)\mathbf{d}_3\] \[\mathbf{q}=q_x(s)\mathbf{i}+q_y(s)\mathbf{j} = \begin{pmatrix} q_x(s) & q_y(s) \end{pmatrix} \begin{pmatrix} \mathbf{i}\\ \mathbf{j} \end{pmatrix} = \begin{pmatrix} q_x(s) & q_y(s) \end{pmatrix} \begin{pmatrix} \cos\theta(s) & \sin\theta(s)\\ -\sin\theta(s) & \cos\theta(s) \end{pmatrix} \begin{pmatrix} \mathbf{d}_1\\ \mathbf{d}_2 \end{pmatrix}\]The vector equations simplify to:
\[\begin{cases} \mathrm{F}_1'(s)-\mathrm{F}_2(s)\kappa(s)+\cos\theta(s) q_x(s) -\sin\theta(s) q_y(s) =\rho\mathrm{A}v_t\\ \mathrm{F}_1(s)\kappa(s)+\mathrm{F}_2'(s)+\cos\theta(s) q_y(s) +\sin\theta(s) q_x(s) =\rho\mathrm{A}v^2\kappa(s)\\ \mathrm{M}'(s)+\mathrm{F}_2(s) +m(s) =\rho I_0\left( \kappa(s) v_t+v^2\kappa'(s) \right)\\ \kappa(s) =-\theta'(s) \end{cases}\]The above simplifies to:
\[\begin{cases} \mathrm{m}(s) \kappa(s) +\left( \frac{\mathrm{m}'(s)}{\kappa(s)} \right)'+\mathrm{EI}\left( \frac{\kappa(s)^2}{2}+\frac{\kappa''(s)}{\kappa(s)} \right)'-\sin(\theta(s))\left( \left( \frac{q_x(s)}{\kappa(s)} \right)'+2q_y(s) \right) +\cos(\theta(s))\left( 2q_x(s)\frac{q_y(s)}{\kappa(s)}-\left( \frac{q_y(s)}{\kappa(s)} \right)' \right)=v_t\left( \rho\mathrm{A}+\rho\mathrm{I}_0\kappa(s)^2+\rho\mathrm{I}_0\left( \frac{\kappa'(s)}{\kappa(s)} \right)' \right) +v^2\rho\mathrm{I}_0\left( \left( \frac{\kappa(s)^2}{2} \right)'+\left( \frac{\kappa''(s)}{\kappa(s)} \right)' \right)\\ -\sin(\theta(s))\left( \left( \frac{q_x(s)}{\kappa(s)} \right)'+2q_y(s) \right) +\cos(\theta(s))\left( 2q_x(s)-\left( \frac{q_y(s)}{\kappa(s)} \right)' \right) =\rho\mathrm{A}v_t\\ \mathrm{F}_2(s) \kappa(s) +\left( \frac{\mathrm{F}_2'(s)}{\kappa(s)} \right)'=0\\ \mathrm{m}(s) +\mathrm{EI}\kappa'(s) +\mathrm{F}_2(s) =\rho\mathrm{I}_0\left( v_t\kappa(s) +v^2\kappa'(s) \right) \end{cases}\]