# Closed Form Solutions for Thermal Buckling of Functionally Graded Rectangular Thin Plates
## 2 Governing Equation of Thermal Buckling
However, the coupling can be eliminated by choosing the neutral surface instead of the geometrical midplane as a reference surface in which there are no in-plane displacements [22] (p. 1091). The in-plane displacements of the midplane can be expressed as  
\[u_{0}=z_{0}\frac{\partial w}{\partial x},\hskip 14.226378 ptv_{0}=z_{0}\frac{ \partial w}{\partial y} \tag{7}\]  
where \(z_{0}\) is the distance between the geometrical midplane and the neutral surface. Substituting Equation (7) into Equation (5) yields  
\[\left[ \begin{array}{c}N_{xx}\\ M_{xx}\end{array} \right]=\left[ \begin{array}{c}\int_{-h/2}^{h/2}\left[\frac {E(z_{0}-z)}{1-\
u^{2}}\left(\frac{\partial^{2}w}{\partial x^{2}}+\
u\frac{ \partial^{2}w}{\partial y^{2}}\right)\right]\mathrm{d}z-N_{T}\\ \int_{-h/2}^{h/2}\left[\frac{Ez(z_{0}-z)}{1-\
u^{2}}\left(\frac{\partial^{2}w}{ \partial x^{2}}+\
u\frac{\partial^{2}w}{\partial y^{2}}\right)\right]\mathrm{ d}z-M_{T}\end{array} \right]\]  
\[\left[ \begin{array}{c}N_{yy}\\ M_{yy}\end{array} \right]=\left[ \begin{array}{c}\int_{-h/2}^{h/2}\left[\frac {E(z_{0}-z)}{1-\
u^{2}}\left(\frac{\partial^{2}w}{\partial y^{2}}+\
u\frac{ \partial^{2}w}{\partial x^{2}}\right)\right]\mathrm{d}z-N_{T}\\ \int_{-h/2}^{h/2}\left[\frac{Ez(z_{0}-z)}{1-\
u^{2}}\left(\frac{\partial^{2}w}{ \partial y^{2}}+\
u\frac{\partial^{2}w}{\partial x^{2}}\right)\right]\mathrm{ d}z-M_{T}\end{array} \right] \tag{8}\]  
\[\left[ \begin{array}{c}N_{xy}\\ M_{xy}\end{array} \right]=\left[ \begin{array}{c}\int_{-h/2}^{h/2}\left(\frac {E(z_{0}-z)}{1+\
u}\frac{\partial^{2}w}{\partial x\partial y}\right)\mathrm{ d}z\\ \int_{-h/2}^{h/2}\left(\frac{Ez(z_{0}-z)}{1+\
u}\frac{\partial^{2}w}{\partial x \partial y}\right)\mathrm{d}z\end{array} \right]\]  
From Equation (8), it can be seen that the in-plane loading and bending of the plate have been decoupled. The total potential energy of a plate in a thermal environment is [23; 24]  
\[U=\frac{1}{2}\int_{V}\left\{\sigma_{xx}(\varepsilon_{xx}-\alpha\Delta T)+ \sigma_{yy}\big{(}\varepsilon_{yy}-\alpha\Delta T\big{)}+\sigma_{xy}\varepsilon_ {xy}\right\}\mathrm{d}V \tag{9}\]Using Equation (8) and employing the Euler equations to minimize the total potential energy function yields the equilibrium equations of an FGM plate  
\[ \begin{split}&\frac{\partial N_{xx}}{\partial x}+\frac{\partial N_{ xy}}{\partial y}=0\\ &\frac{\partial N_{xy}}{\partial x}+\frac{\partial N_{yy}}{ \partial y}=0\\ &\frac{\partial^{2}M_{xx}}{\partial x^{2}}+\frac{\partial^{2}M_{ yy}}{\partial y^{2}}+2\frac{\partial^{2}M_{xy}}{\partial x\partial y}+N_{xx} \frac{\partial^{2}w}{\partial x^{2}}+2N_{xy}\frac{\partial^{2}w}{\partial x \partial y}+N_{yy}\frac{\partial^{2}w}{\partial y^{2}}=0\end{split}  \tag{10}\]  
Substituting Equation (8) into Equation (10) determines the distance \(z_{0}\) and yields the thermal buckling that governs the equation of an FGM thin rectangular plate as  
\[z_{0}=\frac{B_{11}+G_{1}}{A_{11}+G_{0}} \tag{11}\]  
\[\left(D_{11}+G_{2}-\frac{\left(B_{11}+G_{1}\right)^{2}}{A_{11}+G_{0}}\right) \left(\frac{\partial^{4}w}{\partial x^{4}}+2\frac{\partial^{4}w}{\partial x^{ 2}\partial y^{2}}+\frac{\partial^{4}w}{\partial y^{4}}\right)-G_{0}\left( \frac{\partial^{2}w}{\partial x^{2}}+\frac{\partial^{2}w}{\partial y^{2}} \right)=0 \tag{12}\]  
or  
\[\left(\frac{\partial^{4}w}{\partial x^{4}}+2\frac{\partial^{4}w}{\partial x^{ 2}\partial y^{2}}+\frac{\partial^{4}w}{\partial y^{4}}\right)+\hat{G}_{0} \left(\frac{\partial^{2}w}{\partial x^{2}}+\frac{\partial^{2}w}{\partial y^{2 }}\right)=0 \tag{13}\]  
where  
\[ \begin{split}& l\hat{G}_{0}=\frac{-G_{0}}{D_{11}+G_{2}-\frac{ \left(B_{11}+G_{1}\right)^{2}}{A_{11}+G_{0}}}\\ & G_{i}=-\int\frac{\frac{h}{2}}{\frac{h}{2}}\,z^{i}\frac{\alpha E }{1-\
u}(T-T_{0})\mathrm{d}z(i=0,1,2)\\ &\left(A_{ij},B_{ij},D_{ij}\right)=\int\frac{\frac{h}{2}}{\frac{ h}{2}}\,Q_{ij}\Big{(}1,z,z^{2}\Big{)}\mathrm{d}z(i,j=1,2,6)\end{split}  \tag{14}\]  
\[ \begin{split}& Q_{11}=Q_{22}=\frac{E}{1-\
u^{2}}\\ & Q_{12}=Q_{21}=\frac{\
u E}{1-\
u^{2}}\\ & Q_{66}=\frac{E}{2(1+\
u)}\end{split}