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1. 中北大学理学院, 太原 030051;
2. 同济大学航空航天与力学学院, 上海 200092

Analytical Solutions for Bending Problems of Functionally Graded Rectangular Plates With Specific Material Gradient
CHEN Shuping1, ZHAO Hongxiao2, WU Jinwen1, GENG Shaobo1
1. College of Science, North University of China, Taiyuan 030051;
2. Schoolof Aerospace Engineering and Applied Mechanics, Tongji University, Shanghai 200092
Abstract: Bending problems of functionally graded rectangular plates were analyzed. In terms of Plevako function, assuming that the Yang's modulus of the material was distributed through a power function in the thickness direction, an exact solution of bending problem of simply supported functionally graded rectangular plates was obtained by variable separation method. Through numerical examples, the variations of displacement and stress components with different graded material distributions were analyzed and different graded parameters were found to have significant effects on the structural behavior. The research serves as theoretical basis for the development of numerical methods and simplified theories of functionally graded material structure.
Key words: Functionally graded material     Rectangular plate     Simply supported     Variable separation method     Analytical solution
0 引言

1 问题的描述和基本方程

 图 1 功能梯度材料矩形板示意图 Figure 1 schematic of a functionally graded rectangular plate

 $\begin{array}{l} E\left( z \right){\nabla ^2}u + \frac{{E\left( z \right)}}{{1 - 2\nu }}\frac{{\partial e}}{{\partial x}} + \left( {\frac{{\partial u}}{{\partial z}} + \frac{{\partial w}}{{\partial x}}} \right)\frac{{{\rm{d}}E\left( z \right)}}{{{\rm{d}}z}} = 0\\ E\left( z \right){\nabla ^2}v + \frac{{E\left( z \right)}}{{1 - 2\nu }}\frac{{\partial e}}{{\partial y}} + \left( {\frac{{\partial v}}{{\partial z}} + \frac{{\partial w}}{{\partial y}}} \right)\frac{{{\rm{d}}E\left( z \right)}}{{{\rm{d}}z}} = 0\\ E\left( z \right){\nabla ^2}w + \frac{{E\left( z \right)}}{{1 - 2\nu }}\frac{{\partial e}}{{\partial z}} + e\frac{{\rm{d}}}{{{\rm{d}}z}}\left[ {\frac{{2E\left( z \right)\nu }}{{1 - 2\nu }}} \right] + 2\frac{{\partial w}}{{\partial x}}\frac{{{\rm{d}}E\left( z \right)}}{{{\rm{d}}z}} = 0 \end{array}$ (1)

 $\begin{array}{l} z = 0,{\sigma _z} = {Z_0}\left( {x,y} \right),{\tau _{xz}} = {X_0}\left( {x,y} \right),{\tau _{yz}} = {Y_0}\left( {x,y} \right)\\ z = h,{\sigma _z} = {Z_1}\left( {x,y} \right),{\tau _{xz}} = {X_1}\left( {x,y} \right),{\tau _{yz}} = {Y_1}\left( {x,y} \right) \end{array}$ (2)

 $\begin{array}{l} {\sigma _x} = v = w = 0\;\;\;\;\;\left( {x = 0,a} \right)\\ {\sigma _y} = u = w = 0\;\;\;\;\;\left( {y = 0,b} \right) \end{array}$ (3)

2 问题的求解

 $\begin{array}{l} u = - \frac{{1 + v}}{E}\left( {\nu {\nabla ^2} - \frac{{{\partial ^2}}}{{\partial {z^2}}}} \right)\frac{{\partial L}}{{\partial x}} + \frac{{\partial N}}{{\partial y}}\\ v = - \frac{{1 + v}}{E}\left( {\nu {\nabla ^2} - \frac{{{\partial ^2}}}{{\partial {z^2}}}} \right)\frac{{\partial L}}{{\partial y}} - \frac{{\partial N}}{{\partial x}}\\ w = - 2\frac{{1 + v}}{E}\left( {{\nabla ^2} - \frac{{{\partial ^2}}}{{\partial {z^2}}}} \right)\frac{{\partial L}}{{\partial z}} + \\ \frac{\partial }{{\partial z}}\left[ {\frac{{1 + v}}{E}\left( {\nu {\nabla ^2} - \frac{{{\partial ^2}}}{{\partial {z^2}}}} \right)L} \right] \end{array}$ (4)

 ${\nabla ^2}\left( {\frac{1}{E}{\nabla ^2}L} \right) - \frac{1}{{1 - v}}\left( {{\nabla ^2} - \frac{{{\partial ^2}}}{{\partial {z^2}}}} \right)L\frac{{{{\rm{d}}^2}}}{{{\rm{d}}{z^2}}}\left( {\frac{1}{E}} \right) = 0$ (5)
 ${\nabla ^2}N + \frac{{E'}}{E}\frac{{\partial N}}{{\partial z}} = 0$ (6)

 ${\sigma _x} = \left( {\nu \frac{{{\partial ^2}}}{{\partial {y^2}}}{\nabla ^2} - \frac{{{\partial ^4}}}{{\partial {x^2}\partial {z^2}}}} \right)L + \frac{E}{{1 + \nu }}\frac{{{\partial ^2}N}}{{\partial x\partial y}}$
 ${\sigma _y} = \left( {\nu \frac{{{\partial ^2}}}{{\partial {x^2}}}{\nabla ^2} + \frac{{{\partial ^4}}}{{\partial {y^2}\partial {z^2}}}} \right)L - \frac{E}{{1 + \nu }}\frac{{{\partial ^2}N}}{{\partial x\partial y}}$
 ${\sigma _z} = {\left( {\frac{{{\partial ^2}}}{{\partial {x^2}}} + \frac{{{\partial ^2}}}{{\partial {y^2}}}} \right)^2}L$
 $\begin{array}{l} {\tau _{yz}} = - \left( {\frac{{{\partial ^2}}}{{\partial {x^2}}} + \frac{{{\partial ^2}}}{{\partial {y^2}}}} \right)\frac{{{\partial ^2}L}}{{\partial y\partial z}} - \frac{E}{{2\left( {1 + \nu } \right)}}\frac{{{\partial ^2}N}}{{\partial x\partial z}}\\ {\tau _{xz}} = - \left( {\frac{{{\partial ^2}}}{{\partial {x^2}}} + \frac{{{\partial ^2}}}{{\partial {y^2}}}} \right)\frac{{{\partial ^2}L}}{{\partial x\partial z}} + \frac{E}{{2\left( {1 + \nu } \right)}}\frac{{{\partial ^2}N}}{{\partial y\partial z}}\\ {\tau _{xy}} = - \left[ {\nu {\nabla ^2} - \frac{{{\partial ^2}}}{{\partial {z^2}}}} \right]\frac{{{\partial ^2}L}}{{\partial x\partial y}} - \frac{E}{{2\left( {1 + \nu } \right)}}\left( {\frac{{{\partial ^2}}}{{\partial {x^2}}} - \frac{{{\partial ^2}}}{{\partial {y^2}}}} \right)N \end{array}$ (7)

L=L(x, y, z)和N=N(x, y, z)写成分离变量的形式：

 $\begin{array}{l} L\left( {x,y,z} \right) = \sum\limits_{m,n = 1}^\infty {{\psi _{mn}}\left( z \right)\sin \left( {{\alpha _m}x} \right)\sin \left( {{\beta _n}y} \right)} \\ N\left( {x,y,z} \right) = \sum\limits_{m,n = 1}^\infty {{\varphi _{mn}}\left( z \right)\cos \left( {{\alpha _m}x} \right)\cos \left( {{\beta _n}y} \right)} \end{array}$ (8)

 ${\alpha _m} = \frac{{m{\rm{ \mathit{ π} }}}}{{{L_1}}},{\beta _n} = \frac{{n{\rm{ \mathit{ π} }}}}{{{L_2}}}$ (9)

 $\begin{array}{l} \frac{{{{\rm{d}}^4}{\psi _{mn}}}}{{{\rm{d}}{z^4}}} - 2\frac{{E'}}{E}\frac{{{{\rm{d}}^3}{\psi _{mn}}}}{{{\rm{d}}{z^3}}} + \left[ {2{{\left( {\frac{{E'}}{E}} \right)}^2} - \frac{{E''}}{E} - 2\lambda } \right]\frac{{{{\rm{d}}^2}{\psi _{mn}}}}{{{\rm{d}}{z^2}}} + \\ 2\frac{{E'}}{E}\lambda \frac{{{\rm{d}}{\psi _{mn}}}}{{{\rm{d}}z}} + \left\{ {\lambda + \frac{\nu }{{1 - \nu }}\left[ {2{{\left( {\frac{{E'}}{E}} \right)}^2} - \frac{{E''}}{E}} \right]} \right\}\lambda {\psi _{mn}} = 0 \end{array}$ (10)
 $\frac{{{{\rm{d}}^2}{\varphi _{mn}}}}{{{\rm{d}}{z^2}}} + \frac{{E'}}{E}\frac{{{\rm{d}}{\varphi _{mn}}}}{{{\rm{d}}z}} - \lambda {\varphi _{mn}} = 0$ (11)

 $\lambda = {\lambda _{mn}} = \alpha _m^2 + \beta _n^2$ (12)

 $\begin{array}{l} {Z_0}\left( {x,y} \right) = \sum\limits_{m,n = 1}^\infty {{{\bar Z}_{0mn}}\sin \left( {{\alpha _m}x} \right)\sin \left( {{\beta _n}y} \right)} \\ {Z_1}\left( {x,y} \right) = \sum\limits_{m,n = 1}^\infty {{{\bar Z}_{1mn}}\sin \left( {{\alpha _m}x} \right)\sin \left( {{\beta _n}y} \right)} \\ {X_0}\left( {x,y} \right) = \sum\limits_{m,n = 1}^\infty {{{\bar X}_{0mn}}\cos \left( {{\alpha _m}x} \right)\sin \left( {{\beta _n}y} \right)} \\ {X_1}\left( {x,y} \right) = \sum\limits_{m,n = 1}^\infty {{{\bar X}_{1mn}}\cos \left( {{\alpha _m}x} \right)\sin \left( {{\beta _n}y} \right)} \\ {Y_0}\left( {x,y} \right) = \sum\limits_{m,n = 1}^\infty {{{\bar Y}_{0mn}}\sin \left( {{\alpha _m}x} \right)\cos \left( {{\beta _n}y} \right)} \\ {Y_1}\left( {x,y} \right) = \sum\limits_{m,n = 1}^\infty {{{\bar Y}_{1mn}}\sin \left( {{\alpha _m}x} \right)\cos \left( {{\beta _n}y} \right)} \end{array}$ (13)

 $\begin{array}{l} {{\bar Z}_{0mn}} = \frac{1}{{4ab}}\int_0^a {\int_0^b {{Z_0}\left( {x,y} \right)\sin \left( {{\alpha _m}x} \right)\sin \left( {{\beta _n}y} \right){\rm{d}}x{\rm{d}}y} } \\ {{\bar Z}_{1mn}} = \frac{1}{{4ab}}\int_0^a {\int_0^b {{Z_1}\left( {x,y} \right)\sin \left( {{\alpha _m}x} \right)\sin \left( {{\beta _n}y} \right){\rm{d}}x{\rm{d}}y} } \\ {{\bar X}_{0mn}} = \frac{1}{{4ab}}\int_0^a {\int_0^b {{X_0}\left( {x,y} \right)\cos \left( {{\alpha _m}x} \right)\sin \left( {{\beta _n}y} \right){\rm{d}}x{\rm{d}}y} } \\ {{\bar X}_{1mn}} = \frac{1}{{4ab}}\int_0^a {\int_0^b {{X_1}\left( {x,y} \right)\cos \left( {{\alpha _m}x} \right)\sin \left( {{\beta _n}y} \right){\rm{d}}x{\rm{d}}y} } \\ {{\bar Y}_{0mn}} = \frac{1}{{4ab}}\int_0^a {\int_0^b {{Y_0}\left( {x,y} \right)\sin \left( {{\alpha _m}x} \right)\cos \left( {{\beta _n}y} \right){\rm{d}}x{\rm{d}}y} } \\ {{\bar Y}_{1mn}} = \frac{1}{{4ab}}\int_0^a {\int_0^b {{Y_1}\left( {x,y} \right)\sin \left( {{\alpha _m}x} \right)\cos \left( {{\beta _n}y} \right){\rm{d}}x{\rm{d}}y} } \end{array}$ (14)

 $\begin{array}{l} {\psi _{mn}}\left( 0 \right) = \frac{{{{\bar Z}_{0mn}}}}{{\lambda _{mn}^2}},\;\;\;{\psi _{mn}}\left( h \right) = \frac{{{{\bar Z}_{1mn}}}}{{\lambda _{mn}^2}}\\ \frac{{{\rm{d}}{\psi _{mn}}}}{{{\rm{d}}z}}\left| {_{z = 0}} \right. = \frac{1}{{\lambda _{mn}^2}}\left( {{{\bar X}_{0mn}}{\alpha _m} + {{\bar Y}_{0mn}}{\beta _n}} \right)\\ \frac{{{\rm{d}}{\psi _{mn}}}}{{{\rm{d}}z}}\left| {_{z = h}} \right. = \frac{1}{{\lambda _{mn}^2}}\left( {{{\bar X}_{1mn}}{\alpha _m} + {{\bar Y}_{1mn}}{\beta _n}} \right) \end{array}$ (15)
 $\begin{array}{l} \frac{{{\rm{d}}{\varphi _{mn}}}}{{{\rm{d}}z}}\left| {_{z = 0}} \right. = - \frac{{2\left( {1 + \nu } \right)}}{{E{\lambda _{mn}}}}\left( {{{\bar X}_{0mn}}{\beta _n} - {{\bar Y}_{0mn}}{\alpha _m}} \right)\\ \frac{{{\rm{d}}{\varphi _{mn}}}}{{{\rm{d}}z}}\left| {_{z = h}} \right. = - \frac{{2\left( {1 + \nu } \right)}}{{E{\lambda _{mn}}}}\left( {{{\bar X}_{1mn}}{\beta _n} - {{\bar Y}_{1mn}}{\alpha _m}} \right) \end{array}$ (16)

3 幂函数梯度分布

 $E\left( z \right) = {E_0}{\left( {1 + cz} \right)^p}$ (17)

 $\begin{array}{l} \frac{{{{\rm{d}}^4}{\psi _{mn}}}}{{{\rm{d}}{z^4}}} - \frac{{2cp}}{{1 + cz}}\frac{{{{\rm{d}}^3}{\psi _{mn}}}}{{{\rm{d}}{z^3}}} + \left[ {\frac{{p{c^2}\left( {p + 1} \right)}}{{{{\left( {1 + cz} \right)}^2}}} - 2\lambda } \right]\frac{{{{\rm{d}}^2}{\psi _{mn}}}}{{{\rm{d}}{z^2}}} + \\ \frac{{2pc\lambda }}{{1 + cz}}\frac{{{\rm{d}}{\psi _{mn}}}}{{{\rm{d}}z}} + \left[ {\lambda + \frac{{v\left( {p + 1} \right){c^2}p}}{{\left( {1 - v} \right){{\left( {1 + cz} \right)}^2}}}} \right]\lambda {\psi _{mn}} = 0 \end{array}$ (18)
 $\frac{{{{\rm{d}}^2}{\varphi _{mn}}}}{{{\rm{d}}{z^2}}} + \frac{{pc}}{{1 + cz}}\frac{{{\rm{d}}{\varphi _{mn}}}}{{{\rm{d}}z}} - \lambda {\varphi _{mn}} = 0$ (19)

 $\begin{array}{*{20}{c}} {{\psi _{mn}} = {z_1}^{g - 1/2}\left[ {{A_1}{W_{\chi ,g}}\left( {2\eta {z_1}} \right) + {A_2}{W_{ - \chi ,g}}\left( {2\eta {z_1}} \right) + } \right.}\\ {\left. {{A_3}{W_{\chi ,g}}\left( { - 2\eta {z_1}} \right) + {A_4}{W_{ - \chi ,g}}\left( { - 2\eta {z_1}} \right)} \right]} \end{array}$ (20)

 $\begin{array}{l} {\psi _{mn}} = {z_1}^g\left\{ {{A_1}{I_g}\left( {\eta {z_1}} \right) + {A_2}{K_g}\left( {\eta {z_1}} \right) + } \right.\\ {A_3}\left[ {{I_g}\left( {\eta {z_1}} \right)\int {K_g^2\left( {\eta {z_1}} \right)d{z_1}} - } \right.\\ \left. {{K_g}\left( {\eta {z_1}} \right)\int {{I_g}\left( {\eta {z_1}} \right){K_g}\left( {\eta {z_1}} \right)d{z_1}} } \right] + \\ {A_4}\left[ {{I_g}\left( {\eta {z_1}} \right)\int {{I_g}\left( {\eta {z_1}} \right){K_g}\left( {\eta {z_1}} \right)d{z_1}} - } \right.\\ \left. {\left. {{K_g}\left( {\eta {z_1}} \right)\int {I_g^2\left( {\eta {z_1}} \right)d{z_1}} } \right]} \right\} \end{array}$ (21)

 ${\varphi _{mn}} = {z_1}^{ - g + \frac{3}{2}}\left[ {{B_1}{I_{ - g + 3/2}}\left( {\eta {z_1}} \right) + {B_2}{K_{ - g + 3/2}}\left( {\eta {z_1}} \right)} \right]$ (22)

 $\chi = \frac{1}{2}\sqrt {\left( {p + 1} \right)\left( {1 - \frac{{vp}}{{1 - v}}} \right)}$ (23)

4 算例与讨论

 $\begin{array}{l} {Z_1}\left( {x,y} \right) = - q\sin \left( {\frac{{\rm{ \mathit{ π} }}}{a}x} \right)\sin \left( {\frac{{\rm{ \mathit{ π} }}}{b}y} \right),q = 1{\rm{Pa}}\\ {X_0}\left( {x,y} \right) = {Y_0}\left( {x,y} \right) = {Z_0}\left( {x,y} \right) = {X_1}\left( {x,y} \right) = {Y_1}\left( {x,y} \right) = 0 \end{array}$

E(h)/E(0)=10，选择不同的非均匀系数p，弹性模量沿着板厚方向的变化如图 2所示。

 图 2 弹性模量沿着板厚方向的变化示意图 Figure 2 Variation of Young's modulus along thickness of plate

 图 3 对于薄板h/L1=0.1，在点(x/L1=0.25, y/L2=0.25)处，各个物理量沿着板厚度方向的变化图 Figure 3 Variation of physical quantities with z-coordinate at a chosen location (x/a=0.25, y/b=0.25) for a thin plate (h/a=0.1)
 图 4 对于厚板h/L1=0.4，在点(x/L1=0.25，y/L2=0.25)处，各个物理量沿着板厚度方向的变化图 Figure 4 Variation of physical quantities with z-coordinate at a chosen location (x/a=0.25, y/b=0.25) for a thick plate (h/a=0.4)

5 结论

http://dx.doi.org/10.12044/j.issn.1007-2330.2018.01.004

0

#### 文章信息

CHEN Shuping, ZHAO Hongxiao, WU Jinwen, GENG Shaobo

Analytical Solutions for Bending Problems of Functionally Graded Rectangular Plates With Specific Material Gradient

Aerospace Materials & Technology, 2018, 48(1): 16-21.
http://dx.doi.org/10.12044/j.issn.1007-2330.2018.01.004