## 5 Numerical Examples  
### Solid Plate Numerical Example  
A numerical example is proposed to validate the method: the dynamic performance of a solid plate. The solid plate is composed of two independent parts of the same size: \(1000\ \mathrm{mm}\times 500\ \mathrm{mm}\times 30\ \mathrm{mm}\). One side of the solid plate is fixed, and on the opposite side, there is a loading point and a response point, as shown in Figure 2. The two parts are directly connected on the interface. The external force acting on the loading point lasts for 0.1 s:  
\[\mathbf{f}=\left\{ \begin{array}{cc}40t&0\leq t\leq 0.025\\ 1&0.025\leq t\leq 0.050\\ 3-40t&0.050\leq t\leq 0.075\\ 0&0.075\leq t\leq 0.1\end{array} \right. \tag{78}\]  
The FE model is shown in Figure 3. The two FE substructures with coincident interface nodes are tied by a multi-point constraint (MPC). First, to implement the interpolation function, the assembly of the IBS method is discussed. To implement the IBS method, the IRFs matrix is obtained using FEM (shown in Figure 4). The IRF matrix of Substructure 1 is obtained under the free boundary condition, and Substructure 2 is one-side fixed. Figure 4: External force against time. Figure 3: Two FE substructures with coincident interface nodes. Figure 2: A solid plate assembled with plane interface. To validate the method, the support nodes on the interface are purposely chosen differently from each grid. In Substructure 1, only 18 points are extracted on the interface, and there are 34 interface points in Substructure 2. Each point has three translational DOFs. The interface has a size of 500 mm \(\times\) 80 mm and the distribution of the interface support nodes of the substructures is shown in Figure 5, yellow circles represent for interface node. In this configuration, we set the intermediate surface nodes to coincide with Interface 2. Then, \(\mathbf{G}_{u}^{(2)}=\mathbf{I}\), \(\mathbf{G}_{f}^{(2)}=\mathbf{I}\) and  
\[\left\{ \begin{array}{l}\mathbf{u}_{b}^{(2)}=\mathbf{G}_{u}^{(1)}\mathbf{u}_{b}^{(1)}\\ \mathbf{g}^{(2)}=\mathbf{B}^{(2)\intercal}\mathbf{\lambda}\\ \mathbf{g}^{(1)}=-\mathbf{G}_{f}^{(1)}\mathbf{B}^{(1)\intercal}\mathbf{\lambda}\end{array} \right. \tag{79}\]  
where \(\mathbf{G}_{f}^{(1)}=\mathbf{G}_{u}^{(1)\intercal}\), considering the conservation of energy, and \(\mathbf{B}^{(1)}=\mathbf{B}^{(2)}=\mathbf{I}\) since there is only one intermediate surface in this configuration. In order to obtain \(\mathbf{G}_{u}^{(1)}\), the moving Kriging shape function matrix is employed [31]:  
\[\mathbf{R}\big{(}\mathbf{x}_{i},\mathbf{x}_{j}\big{)}=\left\{ \begin{array}{l}\exp(-(\alpha h /h_{c})^{2})\hskip 42.679134 pth\leq h_{c}\\ 0\hskip 42.679134 pth>h_{c}\end{array} \right. \tag{80}\]  
where \(h\) denotes the Euclidean distance between the given interpolation nodes \(\mathbf{x}_{i}\) and \(\mathbf{x}_{j}\), \(h_{c}\) is the radius of the compact support domain, and \(\alpha\) is the correlation coefficient to fit the model, which should be chosen properly. Using different correlation coefficients \(\alpha\) will obtain results with different precision levels. While using the moving least squares interpolation to obtain \(\mathbf{G}_{u}^{(1)}\), we employ a weight function model with the same expression as in Equation (71) [33]. First, we set the radius of the compact support domain \(h_{c}=85\) and \(\alpha=1\). On the interface, displacement compatibility is naturally satisfied in Equation (15). For both the moving Kriging and moving least squares interpolation schemes, the sum of interface forces of direction Y on the two sides is in the order of \(10^{-14}\) (equal to zero, the force equilibrium satisfied), as shown in Figure 6.  
Figure 5: Interface node distribution of two solid substructures. Figure 6: The sum of interface forces of direction Y. The displacement, velocity and acceleration of the response point are used to illustrate the performance of the method.