# Equivalent Analysis of Thermo-Dynamic Blow-Off Impulse under X-ray Irradiation
## 2 Finite Element Calculation Software
### Basic Equation  
As shown in Figure 1, if the X-ray is perpendicular to an axial irradiated cylinder, and the states of each section are assumed to be the same, any physical quantity is independent of the axial coordinate \(Z\), then the problem is simplified to be a two-dimensional plane one [27; 28]. For a two-dimensional problem,  
\[\epsilon_{yz}=0,\epsilon_{xz}=0,\epsilon_{zz}=0,\;\sigma_{xz}=0,\sigma_{yz}=0 \tag{10}\]  
The symmetric stress tensor and the deformation velocity tensor are as follows [30],  
Figure 1: Two-dimensional plane problem. \[\sigma=\left( \begin{array}{ccc}\sigma_{xx}&\sigma_{xy}&0\\ \sigma_{xy}&\sigma_{yy}&0\\ 0&0&\sigma_{zz}\end{array} \right)\text{ and }\dot{\epsilon}^{\prime}=\left( \begin{array}{ccc}\dot{\epsilon}^{\prime}_{xx}&\dot{\epsilon}^{\prime}_{xy}&0 \\ \dot{\epsilon}^{\prime}_{xy}&\dot{\epsilon}^{\prime}_{yy}&0\\ 0&0&0\end{array} \right)=\left( \begin{array}{ccc}\frac{\partial u}{\partial x }&\frac{1}{2}\Big{(}\frac{\partial u}{\partial y}+\frac{\partial v}{\partial x }\Big{)}&0\\ \frac{1}{2}\Big{(}\frac{\partial u}{\partial y}+\frac{\partial v}{\partial x} \Big{)}&\frac{\partial v}{\partial y}&0\\ 0&0&0\end{array} \right) \tag{11}\]  
The two-dimensional strain irradiation problem satisfies the following conserved elastoplastic hydrodynamic equations in the Euler coordinate system, in the case of ignoring the force and heat conduction [31],  
\[\frac{\partial\mathbf{U}}{\partial t}+\frac{\partial\mathbf{F}(\mathbf{U})}{\partial x}+ \frac{\partial\mathbf{G}(\mathbf{U})}{\partial y}=\mathbf{W} \tag{12}\]  
\[\mathbf{U}=\left( \begin{array}{c}\rho\\ \rho u\\ \rho v\\ \rho e_{k}\end{array} \right)\text{, }F\text{=}\left( \begin{array}{c}\rho u \\ \rho u^{2}-\sigma_{xx}\\ \rho uv-\sigma_{xy}\\ u(\rho e_{k}-\sigma_{xx})-v\sigma_{xy}\end{array} \right)\text{, }\mathbf{G}=\left( \begin{array}{c}\rho v\\ \rho uv-\sigma_{xy}\\ \rho v^{2}-\sigma_{yy}\\ v(\rho e_{k}-\sigma_{yy})-u\sigma_{xy}\end{array} \right)\text{, }\mathbf{W}=\left( \begin{array}{c}0\\ 0\\ 0\\ \rho Q\end{array} \right) \tag{13}\]  
\[e_{k}=[E+(u^{2}+v^{2})/2]\text{, and }\left\{ \begin{array}{c}\sigma_{xx}=S_{xx}-p+q \\ \sigma_{yy}=S_{yy}-p+q\\ \sigma_{zz}=S_{zz}-p\\ \sigma_{xy}=S_{xy}\end{array} \right. \tag{14}\]  
where \(q\) is the artificial viscosity. The Lagrange method is used in numerical simulation, and the motion equations is,  
\[\left\{ \begin{array}{c}\rho\frac{\partial u}{\partial t}=\frac{\partial \sigma_{xx}}{\partial x}+\frac{\partial\sigma_{xy}}{\partial y}\\ \rho\frac{\partial v}{\partial t}=\frac{\partial\sigma_{yy}}{\partial x}+ \frac{\partial\sigma_{yy}}{\partial y}\end{array} \right. \tag{15}\]  
where \(x\) and \(y\) are the Euler space coordinates of the trajectory of Lagrange particles. When used, the material element will rotate in the plane dynamic stress state and Lagrange coordinates. However, the above equations of motion are given in a fixed coordinate system, so the influence of such rotation should be considered in the numerical calculation. Since the stress-strain relation must be independent of the rigid body motion, the coordinate correction of the stress-strain relation must be carried out in the coordinate system to give the stress at the new position in the coordinate system. The stress caused by rotation is modified to,  
\[\left\{ \begin{array}{c}\Delta S_{xx}=S_{xx}\cos^{2}\Delta\alpha+S_{yy}\sin^ {2}\Delta\alpha+2S_{xy}\sin\Delta\alpha\cos\Delta\alpha-S_{xx}\\ \Delta S_{yy}=S_{xx}\sin^{2}\Delta\alpha+S_{yy}\cos^{2}\Delta\alpha-2S_{xy}\sin \Delta\alpha\cos\Delta\alpha-S_{yy}\\ \Delta S_{xy}=S_{xy}(\cos^{2}\Delta\alpha-\sin^{2}\Delta\alpha)+(S_{yy}-S_{xx}) \sin\Delta\alpha\cos\Delta\alpha-S_{xy}\end{array} \right. \tag{16}\]  
Or,  
\[\left\{ \begin{array}{c}\Delta S_{xx}=\frac{S_{xx}-S_{yy}}{2}(\cos 2\Delta \alpha-1)+S_{xy}\sin 2\Delta\alpha\\ \Delta S_{yy}=\frac{S_{xx}-S_{yy}}{2}(1-\cos 2\Delta\alpha)-S_{xy}\sin 2 \Delta\alpha\\ \Delta S_{xy}=S_{xy}(\cos 2\Delta\alpha-1)+\frac{S_{yy}-S_{xx}}{2}\sin 2 \Delta\alpha\end{array} \right.