## 2 Basic Equations
\begin{array}{ll}\left[\left( \frac{\delta_{m}^{2}}{4}+\frac{\lambda_{n}^{2}}{4}\right)^{2}-\frac{\rho h \omega^{2}}{D}\right]w_{mn}\\ -\frac{2}{b}(-1)^{\frac{n-1}{2}}\bigg{[}\mu\left(\frac{\delta_{m}}{2}\right)^{ 2}+\left(\frac{\lambda_{n}}{2}\right)^{2}\bigg{]}h_{m}+\frac{2}{b}\frac{\lambda _{n}}{2}i_{m}\\ -\frac{2}{a}(-1)^{\frac{m-1}{2}}\bigg{[}\left(\frac{\delta_{m}}{2}\right)^{ 2}+\mu\left(\frac{\lambda_{n}}{2}\right)^{2}\bigg{]}d_{n}+\frac{2}{a}\frac{ \delta_{m}}{2}e_{n}\end{array} \right.\right\}\sin\frac{\alpha_{m}x}{2}\sin \frac{\beta_{n}y}{2}\;=0 \tag{17}\]After the simple derivation, one can obtain the expression of \(w_{mn}\) with \(d_{n}\), \(e_{n}\), \(h_{m}\) and \(i_{m}\) to be determined. \[w_{mn}=\xi_{mn}\left[\frac{2}{b}\chi_{2 mn}h_{m}-\frac{2}{b}\frac{\lambda_{n}}{2} i_{m}+\frac{2}{a}\chi_{1 mn}d_{n}-\frac{2}{a}\frac{\delta_{m}}{2}e_{n}\right] \tag{18}\]  
where  
\[\xi_{mn}=\frac{1}{\left(\frac{\delta_{m}^{2}}{4}+\frac{\lambda_{n}^{2}}{4} \right)^{2}-\frac{\rho h\omega^{2}}{D}}\,\ \chi_{1 mn}=(-1)^{\frac{m-1}{2}}\left[\frac{\delta_{m}^{2}}{4}+\mu \frac{\lambda_{n}^{2}}{4}\right]\,\ \chi_{2 mn}=(-1)^{\frac{m-1}{2}}\left[\mu\frac{\delta_{m}^{2}}{4}+\frac{ \lambda_{n}^{2}}{4}\right] \tag{19}\]  
Through the above solving procedure, it is clear that the preselected trial function satisfies the boundaries in Equation (4) automatically. The assumed deflection is still required to satisfy the boundaries in Equation (5), which can determine the unknown constants in \(w_{mn}\). The boundary condition at the rotationally restrained edge will be satisfied by substituting the Equation (16) and the slopes in Equations (8) and (9) into Equation (5), which produces Equations (20) and (22). Similarly, utilizing the second-order derivatives of \(w(x,y)\) in Equations (8) and (9) to acquire the bending moments at free edges and letting them to be equal to zero, which can satisfy the remain bending moments requirements and produce Equations (21) and (23). \[-D\sum\limits_{n=1,3}^{\infty}e_{n}\sin\frac{\lambda_{n}y}{2}=-R_{x0}\sum \limits_{n=1,3}^{\infty}\sum\limits_{m=1,3}^{\infty}\frac{\delta_{m}}{2}w_{mn }\sin\frac{\lambda_{n}y}{2} \tag{20}\]  
\[-D\sum\limits_{n=1,3}^{\infty}\sum\limits_{m=1,3}^{\infty}(-1)^{\frac{m-1}{2} }\left\{ \begin{array}{l}\frac{2}{a}(-1)^{\frac{m-1}{2}}d_{n}+\frac{2}{b} \mu(-1)^{\frac{m-1}{2}}h_{m}\\ -\left(\frac{\delta_{m}^{2}}{4}+\mu\frac{\lambda_{n}^{2}}{4}\right)w_{mn} \end{array} \right\}\sin\frac{\lambda_{n}y}{2}=0 \tag{21}\]  
\[-D\sum\limits_{m=1,3}^{\infty}i_{m}\sin\frac{\delta_{m}x}{2}=-R_{y0}\sum \limits_{n=1,3}^{\infty}\sum\limits_{m=1,3}^{\infty}\frac{\lambda_{n}}{2}w_{ mn}\sin\frac{\delta_{m}x}{2} \tag{22}\]  
\[-D\sum\limits_{n=1,3}^{\infty}\sum\limits_{m=1,3}^{\infty}(-1)^{\frac{m-1}{2} }\left\{ \begin{array}{l}\frac{2}{a}\mu(-1)^{\frac{m-1}{2}}d_{n}+\frac{2}{b} (-1)^{\frac{m-1}{2}}h_{m}\\ -\left(\mu\frac{\delta_{m}^{2}}{4}+\frac{\lambda_{n}^{2}}{4}\right)w_{mn} \end{array} \right\}\sin\frac{\delta_{m}x}{2}=0 \tag{23}\]  
Multiplying each side of Equations (20) and (21) by \(\sin\frac{\lambda_{n}y}{2}dy\) and integrating from 0 to b gives Equations (24) and (25) while multiplying each side of Equations (22) and (23) by \(\sin\frac{\delta_{m}x}{2}dx\) and integrating from 0 to a gives Equations (26) and (27), which are shown as follows:  
\[ \begin{array}{l}\sum\limits_{m=1,3}^{\infty}\frac{2}{b}S_{1 mn}\chi_{2 mn}h_{m }-\sum\limits_{m=1,3}^{\infty}\frac{2}{b}S_{1 nm}\frac{\chi_{n}}{2}i_{m}+\sum \limits_{m=1,3}^{\infty}\frac{2}{a}S_{1 nm}\chi_{1 nm}d_{n}\\ -\left[2a(1-r_{x0})+\sum\limits_{m=1,3}^{\infty}\frac{2}{a}S_{1 nm}\frac{\delta _{m}}{2}\right]e_{n}=0\ \ \ \ n=1,3,5\cdots\end{array}  \tag{24}\]  
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