## 4 Special Case Using the Navier Solution  
The special case assumes that the boundary conditions from all edges are simply supported. The analytic solution of the bending deflection through the plate thickness is completed using the Navier solution of classical laminate plate theory. Recall Equation (5):  
Figure 9: Loading and the boundary condition of Levy solution.  
\[\mathrm{D}_{11}*\left(\frac{\partial^{4}w_{o}}{\partial x^{4}}\right)+2*(\mathrm{D}_ {12}+2*\mathrm{D}_{66})*\left(\frac{\partial^{4}w_{o}}{\partial x^{2}\partial y^{ 2}}\right)+\mathrm{D}_{22}*\left(\frac{\partial^{4}w_{o}}{\partial y^{4}} \right)+\rho_{o}*\frac{\partial^{2}w_{o}}{\partial t^{2}}+\mathrm{N}_{xx}* \frac{\partial^{2}w_{o}}{\partial x^{2}}=0 \tag{12}\]  
The solution of the above equation is below:  
\[w_{o}(x,y,t)=\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}w_{mn}*\sin(\alpha_{m}*x)* \sin(\beta_{n}*y)*\sin(\omega_{mn}*t) \tag{13}\]  
where  
\[\frac{\partial^{4}w_{o}}{\partial x^{4}}=\sum_{m=1}^{\infty}\sum_{n=1}^{ \infty}\alpha_{m}^{4}*w_{mn}*\sin(\alpha_{m}*x)*\sin(\beta_{n}*y)*\sin(\omega _{mn}*t)\]  
\[\frac{\partial^{4}w_{o}}{\partial y^{4}}=\sum_{m=1}^{\infty}\sum_{n=1}^{ \infty}\beta_{n}^{4}*w_{mn}*\sin(\alpha_{m}*x)*\sin(\beta_{n}*y)*\sin(\omega _{mn}*t)\]  
\[\frac{\partial^{4}w_{o}}{\partial x^{2}\partial y^{2}}=\sum_{m=1}^{\infty} \sum_{n=1}^{\infty}\alpha_{m}^{2}*\beta_{n}^{2}*w_{mn}*\sin(\alpha_{m}*x)*\sin (\beta_{n}*y)*\sin(\omega_{mn}*t)\]  
\[\frac{\partial^{2}w_{o}}{\partial t^{2}}=-\omega_{mn}^{2}\sum_{m=1}^{\infty} \sum_{n=1}^{\infty}w_{mn}*\sin(\alpha_{m}*x)*\sin(\beta_{n}*y)*\sin(\omega_{ mn}*t)\]  
and  
\[\alpha_{m}=\frac{m*\pi}{a}\]  
\[\beta_{n}=\frac{n*\pi}{b}\]  
Substitute the expressions (\(\frac{\partial^{4}w_{o}}{\partial x^{4}}\), \(\frac{\partial^{4}w_{o}}{\partial y^{4}}\) and \(\frac{\partial^{4}w_{o}}{\partial x^{2}\partial y^{2}}\)) into Equation (12) and solve the plate equation for (N\({}_{xx}\)):  
\[\mathrm{N}_{xx}=\frac{\pi^{2}*a^{2}}{m^{2}}\bigg{(}\mathrm{D}_{11}*\frac{m^{4 }}{a^{4}}+2*(\mathrm{D}_{12}+2*\mathrm{D}_{66})*\frac{m^{2}*n^{2}}{a^{2}*b^{2 }}+\mathrm{D}_{22}*\frac{n^{4}}{b^{4}}-\frac{\rho_{o}*w_{mn}^{2}}{\pi^{4}} \bigg{)} \tag{14}\]  
By putting the buckling mode (n = 1) while Equation (14) is derived with respect to the buckling mode (m), to calculate the critical buckling load is as below:  
\[\frac{\mathrm{d}N_{xx}}{\mathrm{d}m}=0 \tag{15}\]  
The critical buckling load is:  
\[\mathrm{N}_{xx}=\frac{\pi^{2}}{b^{2}}\sqrt{\mathrm{D}_{11}\mathrm{D}_{22}} \bigg{(}\sqrt{\frac{\mathrm{D}_{22}-\rho_{o}*w_{mn}*b^{2}}{\mathrm{D}_{22}}}+ \frac{2*(\mathrm{D}_{22}+2*\mathrm{D}_{66})}{\sqrt{\mathrm{D}_{11}\mathrm{D}_ {22}}}+\sqrt{\frac{\mathrm{D}_{22}}{\mathrm{D}_{22}-\rho_{o}*w_{mn}^{2}*b^{2}} }-\frac{\rho_{o}*w_{mn}^{2}*b^{2}}{\pi^{4}}*\sqrt{\frac{1}{\mathrm{D}_{22}^{2} -\mathrm{D}_{22}+\rho_{o}*w_{mn}^{2}*b^{2}}}\bigg{)} \tag{16}\]  
and  
\[\frac{a}{b}=m*\left(\frac{\mathrm{D}_{11}}{\mathrm{D}_{22}-\rho_{o}*\omega_{mn }^{2}*b^{2}}\right)^{\frac{1}{4}} \tag{17}\]  
The buckling factor is  
\[\mathrm{k}_{1}=\frac{m^{2}}{c^{2}}\sqrt{\frac{\mathrm{D}_{11}}{\mathrm{D}_{22} }}+\frac{2*(\mathrm{D}_{12}+2*\mathrm{D}_{66})}{\sqrt{\mathrm{D}_{11}\mathrm{D }_{22}}}+\frac{c^{2}}{m^{2}}\sqrt{\frac{\mathrm{D}_{22}}{\mathrm{D}_{11}}}- \frac{\rho_{o}*\omega_{mn}^{2}*c^{2}*b^{2}}{m^{2}*\pi^{4}}*\frac{1}{\sqrt{ \mathrm{D}_{11}\mathrm{D}_{22}}} \tag{18}\]