## 4 Closed Solutions  
Consider an FG-GRC plate with four sides simply supported, as shown in Figure 1. The transversal load \(q\) acts on the surface of the FG-GRC plate. The Navier solution is used to solve the closed solution of Equations (27)-(31). According to the Navier solution, the solution form is assumed to be as follows:  
\[ \begin{array}{l}u(x,y,t)=\sum\limits_{r=1}^{R}\sum\limits_{s=1}^{S}(U_{rs}e^ {i\omega t}\cos\alpha x\sin\beta y)\\ v(x,y,t)=\sum\limits_{r=1}^{R}\sum\limits_{s=1}^{S}\big{(}V_{rs}e^{i\omega t} \sin\alpha x\cos\beta y)\\ w_{b}(x,y,t)=\sum\limits_{r=1}^{R}\sum\limits_{s=1}^{S}(W_{bsr}e^{i\omega t }\sin\alpha x\sin\beta y)\\ w_{s}(x,y,t)=\sum\limits_{r=1}^{R}\sum\limits_{s=1}^{S}\big{(}W_{ssr}e^{i \omega t}\sin\alpha x\sin\beta y)\\ w_{z}(x,y,t)=\sum\limits_{r=1}^{R}\sum\limits_{s=1}^{S}\big{(}W_{ssr}e^{i \omega t}\sin\alpha x\sin\beta y\big{)}\end{array}  \tag{37}\]  
where \(U_{rs}\), \(V_{rs}\), \(W_{bsr}\), \(W_{ssr}\), and \(W_{2 rs}\) are unknown coefficients, while \(r\) and \(s\) are the half-wave numbers in the \(x\) and \(y\) directions, respectively. \(\omega\) is the natural frequency. In addition,  
\[i=\sqrt{-1},\alpha=\frac{r\pi}{a},\beta=\frac{s\pi}{b} \tag{38}\]  
The form of transversal load \(q\) is assumed to be as follows:  
\[q(x,y)=\sum\limits_{r=1}^{R}\sum\limits_{s=1}^{S}Q_{rs}\sin\alpha x\sin\beta y \tag{39}\]  
For the bending analysis in the following, it is necessary to consider a uniformly distributed transverse load \(q(x,y)=F_{0}\), in which case [43],  
\[Q_{rs}=4\lambda_{rs}F_{0}/(E_{\text{M}}rs\pi^{2}) \tag{40}\]  
where  
\[\lambda_{rs}=1-(-1)^{r}-(-1)^{s}+(-1)^{r+s} \tag{41}\]  
Substituting Equations (37) and (39) into Equations (27)-(31), the closed solutions can be derived from the following equation. \[\left(\left[ \begin{array}{ccccc}j_{11}&j_{12}&j_{13}&j_{14}&j_{15}\\ j_{12}&j_{22}&j_{23}&j_{24}&j_{25}\\ j_{13}&j_{23}&j_{33}&j_{34}&j_{35}\\ j_{14}&j_{24}&j_{34}&j_{44}&j_{45}\\ j_{15}&j_{25}&j_{35}&j_{45}&j_{55}\end{array} \right]-\omega^{2}\left[ \begin{array} []{ccccc}m_{11}&0&m_{13}&m_{14}&0\\ 0&m_{22}&m_{23}&m_{24}&0\\ m_{31}&m_{32}&m_{33}&m_{34}&m_{35}\\ m_{41}&m_{42}&m_{43}&m_{44}&m_{45}\\ 0&0&m_{53}&m_{54}&m_{55}\end{array} \right]\right)\left\{ \begin{array}{c}U_{ rs}\\ V_{rs}\\ W_{bsr}\\ W_{zrs}\end{array} \right\}=\left\{ \begin{array}{c}0\\ 0\\ Q_{rs}\\ 0\end{array} \right\} \tag{42}\]where  
\[ \begin{array}{l}j_{11}=A_{11}\alpha^{2}+A_{66}\beta^{2},j_{12}=(A_{12}+A_{66}) \alpha\beta,j_{13}=-B_{11}\alpha^{3}-(B_{12}+2B_{66})\alpha\beta^{2}\\ j_{14}=-B_{11}^{5}\alpha^{3}-\big{(}B_{12}^{5}+2B_{66}^{6}\big{)}\alpha\beta^{2},j_{15}=-X_{13}\alpha,j_{22}=A_{66}\alpha^{2}+A_{22}\beta^{2}\\ j_{23}=-B_{2}\beta^{3}-(B_{12}+2B_{66})\alpha^{2}\beta,j_{24}=-B_{22}^{5}\beta^ {3}-(B_{12}^{5}+2B_{66}^{6})\alpha^{2}\beta,j_{25}=-X_{23}\beta\\ j_{33}=D_{11}\alpha^{4}+2(D_{12}+2D_{66})\alpha^{2}\beta^{2}+D_{22}^{5}\beta^ {4},j_{45}=(Y_{13}^{5}+A_{55}^{2})\alpha^{2}+(Y_{23}^{5}+A_{44}^{5})\beta^{2} \\ j_{34}=D_{11}^{5}\alpha^{4}+2\big{(}D_{12}^{5}+2D_{66}^{6}\big{)}\alpha^{2} \beta^{2}+D_{22}^{5}\beta^{4},j_{55}=A_{55}^{5}\alpha^{2}+A_{44}^{5}\beta^{2}+ Z_{33}\\ j_{35}=Y_{13}\alpha^{2}+Y_{23}\beta^{2},j_{44}=H_{11}^{5}\alpha^{4}+2\big{(}H_{1 2}^{5}+2H_{66}^{5}\big{)}\alpha^{2}\beta^{2}+H_{22}^{5}\beta^{4}+A_{55}^{2} \alpha^{2}+A_{44}^{5}\beta^{2}\\ m_{11}=I_{0},m_{13}=-\alpha I_{1},m_{14}=-\alpha I_{1},m_{22}=I_{0},m_{23}=- \beta I_{1},m_{24}=-\beta I_{1}\\ m_{33}=I_{0}+I_{2}\big{(}\alpha^{2}+\beta^{2}\big{)},m_{34}=I_{0}+J_{1}\big{(} \alpha^{2}+\beta^{2}\big{)},m_{35}=J_{0},m_{44}=I_{0}+K_{2}\big{(}\alpha^{2}+ \beta^{2}\big{)}\\ m_{44}=I_{0}+K_{2}\big{(}\alpha^{2}+\beta^{2}\big{)},m_{45}=J_{0},m_{55}=K_{0 }\end{array} \]  
To solve for the bending deflection, just take \(\omega=0\), and we can find \(U_{rs}\), \(V_{rs}\), \(W_{brs}\), \(W_{srs}\), and \(W_{rs}\), and then the deflection value using Equation (42). The free vibration frequency can be found by simply substituting Equation (42) for \(Q_{rs}=0\).