## 2 Elastodynamic Model of Parallel Mechanism with Tube Structure
### Elastodynamic Equation of the PM  
As shown in Equations (14)-(16), the elastic energy \(V\), kinematic energy \(T\), and external force virtual work \(\delta W\) of all elements can be expressed by all independent node displacements \(\mathbf{u}=\left[\mathbf{u}_{1};\ \mathbf{u}_{2};\ \cdots;\ \mathbf{u}_{n}\right]\) and node velocities \(\dot{\mathbf{u}}=\left[\dot{\mathbf{u}}_{1};\ \dot{\mathbf{u}}_{2};\ \cdots;\ \dot{\mathbf{u}}_{n}\right]\). \[V=\sum_{i=1}^{m}\frac{1}{2}\mathbf{u}^{T}K_{i}\mathbf{u} \tag{14}\]  
\[T=\sum_{i=1}^{m}\frac{1}{2}\dot{\mathbf{u}}^{T}M_{i}\dot{u} \tag{15}\]  
\[\delta W=\mathbf{F}^{T}\delta\mathbf{u} \tag{16}\]  
Then, according to the Lagrange formulation in Equation (17), the elastodynamic equation of the PM is expressed in Equation (18). Then, we add displacement boundary conditions to solve the equation. We divide the independent node displacement \(\mathbf{u}\) into two parts \(\mathbf{u}=\left[\mathbf{u}_{1};\mathbf{u}_{2}\right]\mathbf{u}_{2}\) contains the known boundary condition \(\mathbf{u}_{2}=\mathbf{u}_{2}^{0}\). Then, the mass matrix \(\mathbf{M}\), stiffness matrix \(\mathbf{K}\), and external force matrix \(\mathbf{F}\) are also divided into corresponding block matrices, as shown below. \[\frac{d}{dt}\frac{\partial T}{\partial\dot{\mathbf{u}}}+\frac{\partial V}{\partial \mathbf{u}}=\mathbf{F} \tag{17}\]  
\[\mathbf{M}\ddot{\mathbf{u}}+\mathbf{K}\mathbf{u}=\mathbf{F}\]  
\[\mathbf{M}=\sum_{i=1}^{m}\mathbf{M}_{i}= \begin{bmatrix}\mathbf{M}_{11}&\mathbf{M}_{12}\\ \mathbf{M}_{21}&\mathbf{M}_{22}\end{bmatrix} ,\mathbf{K}=\sum_{i=1}^{m}\mathbf{K}_{i}= \begin{bmatrix} \mathbf{K}_{11}&\mathbf{K}_{12}\\ \mathbf{K}_{21}&\mathbf{K}_{22}\end{bmatrix}  \tag{18}\]  
\[\mathbf{u}= \begin{bmatrix}\mathbf{u}_{1}\\ \mathbf{u}_{2}^{0}\end{bmatrix} ,\ \mathbf{F}= \begin{bmatrix}\mathbf{F}_{1}\\ \mathbf{F}_{2}\end{bmatrix} \]  
After simplification with the block matrix, Equation (18) becomes the equivalent Equations (19) and (20). The unknown variables \(\mathbf{u}_{1}\) and \(\mathbf{F}_{2}\) can be obtained by solving these ordinary differential equations. \[\mathbf{M}_{11}\ddot{\mathbf{u}}_{1}+\mathbf{K}_{11}\mathbf{u}_{1}=\mathbf{F}_{1}-\mathbf{K}_{12}\mathbf{ u}_{2}^{0}-\mathbf{M}_{12}\ddot{\mathbf{u}}_{2}^{0} \tag{19}\]  
\[\mathbf{F}_{2}=\mathbf{M}_{21}\ddot{\mathbf{u}}_{1}+\mathbf{K}_{21}\mathbf{u}_{1}+\mathbf{K}_{22}\mathbf{ u}_{2}^{0}+\mathbf{M}_{22}\ddot{\mathbf{u}}_{2}^{0} \tag{20}\]  
Then, the independent node displacement \(\mathbf{u}=\left[\mathbf{u}_{1};\mathbf{u}_{2}^{0}\right]\) is derived. The node displacement \(\left[\mathbf{u}_{1}^{1};\mathbf{u}_{1}^{2}\right]\) at two sides of any beam element can also be obtained in the global coordinate. Coordinate transformation can be used to derive the node displacement \(\left[\overline{\mathbf{u}}_{1}^{1};\overline{\mathbf{u}}_{1}^{2}\right]\) in the beam local coordinate, as shown in Equation (21). \(\mathbf{R}\) is the rotation matrix of the local coordinate expressed in the global coordinate. \[ \begin{bmatrix}\overline{\mathbf{u}}_{1}^{1}\\ \overline{\mathbf{u}}_{1}^{2}\end{bmatrix} =\widetilde{\mathbf{R}}^{T} \begin{bmatrix} \mathbf{u}_{1}^{1}\\ \mathbf{u}_{1}^{1}\end{bmatrix}  \tag{21}\]  
\[\widetilde{\mathbf{R}}=\text{diag}(\left[\mathbf{R},\mathbf{R},\mathbf{R},\mathbf{R}\right])\]  
According to elastic theory [35], for any interpolation point in a beam element, the strain field \(\varepsilon\), the stress field \(\sigma\), and the stress tensor \(\mathbf{Q}\) can be obtained as follows. The details of Equations (22) and (23) are shown in Appendix C.  
\[\varepsilon=\mathbf{D}(\mathbf{S}) \begin{bmatrix}\overline{\mathbf{u}}_{1}^{1}\\ \overline{\mathbf{u}}_{2}^{2}\end{bmatrix}  \tag{22}\]  
\[\sigma=\mathbf{f}(\varepsilon) \tag{23}\]\[