## Symplectic Superposition Solution for the Bending of a Corner Point-Supported and the Other Opposite Edge Clamped Orthotropic Rectangular Thin Plate

Kou Tianjiao,1, Eburilitu ,1, Alatancang ,2

 基金资助: 国家自然科学基金.  11862019国家自然科学基金.  11761029内蒙古自然科学基金.  2020ZD01

 Fund supported: the NSFC.  11862019the NSFC.  11761029the NSF of Inner Mongolia.  2020ZD01

Abstract

The bending problem of a uniformly loaded orthotropic rectangular thin plate with a corner point-supported and the other opposite edge clamped is studied. First, we divide the bending problem into two sub-problems with two opposite edges slidingly clamped and another sub-problem with one edge slidingly clamped and its opposite edge simply supported. Then we obtain the eigenvalues and eigenfunctions of the Hamiltonian operator corresponding to the three sub-problems, respectively. Then the solutions of the above three sub-problems are solved by the method of symplectic eigenfunction expansion respectively. Finally, we obtain the symplectic superposition solution of the original bending problem by superposition of the solutions of the three sub-problems. In addition, we calculate the deflection and bending moment values at some points of the bending problems of isotropic rectangular plate and orthotropic rectangular plate by using the symplectic superposition solution obtained in this paper.

Keywords： Orthotropic rectangular thin plate ; Hamiltonian operator ; Eigenfunction ; Corner point-supported ; Symplectic superposition solution

Kou Tianjiao, Eburilitu , Alatancang . Symplectic Superposition Solution for the Bending of a Corner Point-Supported and the Other Opposite Edge Clamped Orthotropic Rectangular Thin Plate. Acta Mathematica Scientia[J], 2021, 41(3): 797-810 doi:

## 1 引言

1991年, 文献[7]提出了辛弹性力学方法, 该方法不需要事先人为设定试验函数且求解过程具有统一性, 但是应用该方法研究部分复杂边界条件问题时, 如研究对边固支条件下的矩形薄板弯曲问题[8]时, 应用辛弹性力学方法只能得到原方程的数值解而得不到其解析解. 为此, 2010年李锐等学者[9]将辛弹性力学方法与经典叠加方法相结合, 提出了辛叠加方法使得这一问题得到部分解决.

## 2 Hamilton正则方程

$$$D_{11}\frac{\partial^4w}{\partial x^{4}}+2H\frac{\partial^4w}{\partial x^2\partial y^{2}}+D_{22}\frac{\partial^4w}{\partial y^{4}} = q,$$$

$$$D_{11}\frac{{\rm d}^{4}X_1(x)}{{\rm d}x^{4}}+2H\mu^{2}\frac{{\rm d}^{2}X_1(x)}{{\rm d}x^{2}}+D_{22}\mu^{4}X_1(x) = 0.$$$

$X_1(x) = e^{\lambda x}$得其解为

$$$X_1(x) = c_1e^{\lambda_1 x}+c_2e^{-\lambda_1 x}+c_3e^{\lambda_2 x}+c_4e^{-\lambda_2 x},$$$

$$$\lambda_1 = \sqrt \frac{-H\mu^{2}+\sqrt{\mu^{4}(H^{2}-D_{11}D_{22})}}{D_{11}}, \lambda_2 = \sqrt \frac{-H\mu^{2}-\sqrt{\mu^{4}(H^{2}-D_{11}D_{22})}}{D_{11}}.$$$

#### 3.1 对边滑支矩形薄板所对应本征值和本征函数

$$$\frac{\partial w}{\partial x}|_{x = 0, a} = V_x|_{x = 0, a} = 0.$$$

$$$\lambda_1 = \lambda_2 = -\alpha_{n}\text{i}, (n = 1, 2, 3, \cdots),$$$

$$$\mu_{1} = \pm \sqrt{\frac{\alpha_{n}^{2}H}{D_{22}}+\sqrt{\frac{\alpha_{n}^{4}(H^{2}-D_{11}D_{22})}{D_{22}^{2}}}}, \mu_{2} = \pm \sqrt{\frac{\alpha_{n}^{2}H}{D_{22}}-\sqrt{\frac{\alpha_{n}^{4}(H^{2}-D_{11}D_{22})}{D_{22}^{2}}}},$$$

#### 3.1.1 零本征值的情形

$n = 0$时, 根据(3.11)式可得4重零本征值$\mu_{0}$. 由(3.4)式可得$\mu_{0}$相应的本征函数为

$n\neq0 $$H^{2}-D_{11}D_{22} = 0 时, 根据(3.11)式可得2重根的非零本征值 $$\mu_{n} = \alpha_{n} \sqrt \frac{H}{D_{22}}, \mu_{-n} = -\mu_{n} , n = 1, 2, 3, \cdots.$$ 由(3.4)式可得 \mu_{n} 相应的本征函数为 根据 {\bf HX}_{n}^{1}(x) = \mu_{n} {\bf X}_{n}^{1}(x)+{\bf X}_{n}^{0}(x) [17], 得到 \mu_{n} 对应的一阶Jordan型本征函数 通过计算, 我们还可得到 \mu_{-n} 对应的本征函数以及一阶Jordan型本征函数, 分别为 #### 3.1.3 非零本征值为单根的情形 n\neq0$$ H^{2}-D_{11}D_{22}\neq0$时, 根据(3.11)式可得单重本征值

$\begin{eqnarray} && \mu_{n1} = \sqrt{\frac{\alpha_{n}^{2}H}{D_{22}}+\sqrt{\frac{\alpha_{n}^{4}(H^{2}-D_{11}D_{22})}{D_{22}^{2}}}} , \mu_{n2} = -\mu_{n1} , {} \\ &&\mu_{n3} = \sqrt{\frac{\alpha_{n}^{2}H}{D_{22}}-\sqrt{\frac{\alpha_{n}^{4}(H^{2}-D_{11}D_{22})}{D_{22}^{2}}}} , \mu_{n4} = -\mu_{n3} \ (n = 1, 2, 3, \cdots). \end{eqnarray}$

$$${\bf X}_{ni}(x) = \left(1, \mu_{ni}, \mu_{ni}^{3}D_{22}-\alpha_{n}^{2}(D_{12}+4D_{66})\mu_{ni}, \alpha_{n}^{2}D_{12}-\mu _{ni}^{2}D_{22}\right)^{T}\cos[\alpha_{n}x],$$$

##### 3.2 一边滑支对边简支矩形薄板所对应本征值和本征函数

$$$\frac{\partial w}{\partial y}|_{y = 0} = V_y|_{y = 0} = 0, w|_{y = b} = M_y|_{y = b} = 0 .$$$

$$$\xi_1 = \xi_2 = -\beta_{m}\text{i} , (m = 1, 3, 5, \cdots),$$$

$$$\xi_{1} = \pm \sqrt{\frac{\beta_{m}^{2}H}{D_{11}}+\sqrt{\frac{\beta_{m}^{4}(H^{2}-D_{11}D_{22})}{D_{11}^{2}}}}, \xi_{2} = \pm \sqrt{\frac{\beta_{m}^{2}H}{D_{11}}-\sqrt{\frac{\beta_{m}^{4}(H^{2}-D_{11}D_{22})}{D_{11}^{2}}}},$$$

##### 3.2.1 非零本征值为重根的情形

$H^{2}-D_{11}D_{22} = 0$, 根据(3.17)式可得2重根的本征值

$$$\xi_{m} = \beta_{m} \sqrt \frac{H}{D_{11}} , \xi_{-m} = -\xi_{m} , m = 1, 3, 5, \cdots.$$$

##### 3.2.2 非零本征值为单根的情形

$H^{2}-D_{11}D_{22}\neq0$, 根据(3.17)式可得单重本征值

$\begin{eqnarray} && \xi_{m1} = \sqrt{\frac{\beta_{m}^{2}H}{D_{11}}+\sqrt{\frac{\beta_{m}^{4}(H^{2}-D_{11}D_{22})}{D_{11}^{2}}}}, \xi_{m2} = -\xi_{m1}, {}\\ &&\xi_{m3} = \sqrt{\frac{\beta_{m}^{2}H}{D_{11}}-\sqrt{\frac{\beta_{m}^{4}(H^{2}-D_{11}D_{22})}{D_{11}^{2}}}}, \xi_{m4} = -\xi_{m3}\ (m = 1, 3, 5, \cdots). \end{eqnarray}$

$$${\bf Y}_{mi}(y) = \left(1, \xi_{mi}, \xi_{mi}^{3}D_{11}-\beta_{m}^{2}\xi_{mi}(D_{12}+4D_{66}), \beta_{m}^{2}D_{12}-\xi _{mi}^{2}D_{11}\right)^{T}\cos[\beta_{m}y],$$$

### 图 1

(ⅰ) 一边简支三边滑支正交各向异性矩形薄板分别受到均匀荷载以及在点$(a, 0)$处受到集中荷载$P$的弯曲问题, 在$x = 0 $$x = a 边滑支, 在滑支边 y = 0 和简支边 y = b 满足条件 $$\frac{\partial w}{\partial y}|_{y = 0} = V_{y}|_{y = 0} = 0, w|_{y = b} = M_{y}|_{y = b} = 0;$$ (ⅱ) 在 x = 0$$ x = a$边滑支, 在滑支边$y = 0$和简支边$y = b$满足条件

$\begin{eqnarray} &&V_{y}|_{y = 0} = 0, \theta_{1} = \frac{\partial w}{\partial y}|_{y = 0} = \sum\limits_{n = 0}^{\infty}A_{n}\cos[\alpha_{n}x], {}\\ &&w|_{y = b} = 0, M_{y}|_{y = b} = \sum\limits_{n = 0}^{\infty}F_{n}\cos[\alpha_{n}x]; \end{eqnarray}$

(ⅲ) 在$y = 0$边滑支, 在$y = b$边简支, 在滑支边$x = 0 $$x = a 满足条件 \begin{eqnarray} &&V_{x}|_{x = 0} = 0, \varphi_{1} = \frac{\partial w}{\partial x}|_{x = 0} = \sum\limits_{m = 1}^{\infty}G_{m}\cos[\beta_{m}y], {}\\ & &V_{x}|_{x = a} = 0, \varphi_{2} = \frac{\partial w}{\partial x}|_{x = a} = \sum\limits_{m = 1}^{\infty}H_{m}\cos[\beta_{m}y]. \end{eqnarray} 将上述三个子问题的解进行叠加后可得到均匀荷载作用下一角点支撑另一对边固支的正交各向异性矩形薄板弯曲问题的辛叠加解. ### 4.1 非零本征值为重根情形下的辛叠加解 H^{2}-D_{11}D_{22} = 0 时, 我们先来求解子问题(ⅰ), 此时需要求解无穷维Hamilton正则方程(2.5), 根据本征函数系的完备性, 可设非齐次项 \begin{eqnarray} {\bf f}& = &a_{1}{\bf X}_{0}^{0}(x)+b_{1}{\bf X}_{0}^{1}(x)+c_{1}{\bf X}_{0}^{2}(x)+d_{1}{\bf X}_{0}^{3}(x){}\\ &&+ \sum^{\infty}_{n = 1}(a_{n}{\bf X}_{n}^{0}(x)+b_{n}{\bf X}_{n}^{1}(x)+c_{n}{\bf X}_{-n}^{0}(x)+d_{n}{\bf X}_{-n}^{1}(x)). \end{eqnarray} 根据本征函数系的辛正交性, 可得系数 \begin{eqnarray} &&a_{1} = 0, b_{1} = 0, c_{1} = 0, d_{1} = \frac{\int_0^a q(x, y){\text d}x}{aD_{22}}, {}\\ &&a_{n} = -\frac{a \int_0^a \frac{ (1+\mu_{n}) q(x, y) \cos[\alpha_{n}x] }{\mu_{n}} \, {\text d}x}{2 n^2 \pi ^2 H}, b_{n} = \frac{a \int_0^a q(x, y) \cos[\alpha_{n}x] \, {\text d}x}{2 n^2 \pi ^2 H}, \\ & &c_{n} = -\frac{a \int_0^a q(x, y) \cos[\alpha_{n}x]\, {\text d}x}{2 n^2 \pi ^2 H}, d_{n} = \frac{a \int_0^a q(x, y) \cos[\alpha_{n}x]\, {\text d}x}{2 n^2 \pi ^2 H}. {} \end{eqnarray} 根据本征函数系的完备性, 我们假设在边界条件(3.9)和(3.15)下Hamilton正则方程(2.5)的解为 \begin{eqnarray} {\bf U}(x, y)& = &Y_{0}^{0}(y){\bf X}_{0}^{0}(x)+Y_{0}^{1}(y){\bf X}_{0}^{1}(x)+Y_{0}^{2}(y){\bf X}_{0}^{2}(x)+Y_{0}^{3}(y){\bf X}_{0}^{3}(x)\\ &&+ \sum^{\infty}_{n = 1}(Y_{n}^{0}(y){\bf X}_{n}^{0}(x)+Y_{n}^{1}(y){\bf X}_{n}^{1}(x)+Y_{-n}^{0}(y){\bf X}_{-n}^{0}(x)+Y_{-n}^{1}(y){\bf X}_{-n}^{1}(x)). \end{eqnarray} {\bf U}(x, y) 的第一分量, 经计算可得 \begin{eqnarray} w_{1}(x, y)& = &C_{1}+yC_{2}+\frac{y^{2}C_{3}}{2}+\frac{y^{3}C_{4}}{6}+\frac{qy^{4}}{24D_{22}}\\ &&+\sum^{\infty}_{n = 1}(\cos[\alpha_{n}x](1+\frac{1}{\mu_{n}})e^{-\mu_{n}y}C_{n4}+ \cos[\alpha_{n}x]e^{\mu_{n}y}C_{n2}\\ &&+\cos[\alpha_{n}x](e^{-\mu_{n}y}(C_{n3}+yC_{n4}))+\cos[\alpha_{n}x](e^{\mu_{n}y}(C_{n1}+yC_{n2}))), \end{eqnarray} 其中 C_{1}$$ C_{4} $$C_{n1}$$ C_{n4}$为待定常数.

$\begin{eqnarray} w_{1}(x, y)& = &\frac{q(5b^{4}-6b^{2}y^{2}+y^{4})}{24D_{22}}+\frac{P(y^{3}+b(2b^{2}-3y^{2}))}{6aD_{22}}{}\\ &&+\sum^{\infty}_{n = 1}\frac{aP}{2n^{2}\pi^{2}(D_{12}+2D_{66})} \cos[n\pi]\cos[\alpha_{n}x](-e^{-y\mu_{n}}+e^{y\mu_{n}}{}\\ &&-\frac{e^{y\mu_{n}}(-1+e^{2b\mu_{n}})}{1+e^{2b\mu_{n}}} +e^{-y\mu_{n}}y+e^{y\mu_{n}}y\\ &&+\frac{e^{-y\mu_{n}}(1+e^{2b\mu_{n}}(1-2b-y)+e^{4b\mu_{n}}(-1+y)-y+e^{2b\mu_{n}}(-1-2b+y))}{(1+e^{2b\mu_{n}})^{2}}\\ &&-e^{y\mu_{n}}(1+\frac{1}{\mu_{n}})+\frac{e^{-2y\mu_{n}}(-e^{y\mu_{n}}+e^{(2b+y)\mu_{n}}+e^{y\mu_{n}}(1+e^{2b\mu_{n}}))(1+\mu_{n})}{(1+e^{2b\mu_{n}})\mu_{n}}\\ &&+\frac{1}{(1+e^{2b\mu_{n}})^{2} \mu_{n}}e^{y\mu_{n}}(-1+e^{4b\mu_{n}}+(-1+e^{4b\mu_{n}}(1-y) \\ &&+e^{2b\mu_{n}}(1-2b-y)+y+e^{2b\mu_{n}}(-1-2b+y))\mu_{n})). \end{eqnarray}$

$\begin{eqnarray} w_{2}(x, y)& = &\frac{(b-y)(-2A_{0}D_{22}+(b+y)F_{0})}{2D_{22}}{}\\ &&+\sum^{\infty}_{n = 1}\frac{1}{4a^{2}D_{22}\mu_{n}^{3}}\cos[\alpha_{n}x]\text{sech}[b\mu_{n}]^{2}(2a^{2} (b\cosh[y\mu_{n}]\sinh[b\mu_{n}] \\ &&-y\cosh[b\mu_{n}]\sinh[y\mu_{n}])F_{n}\mu_{n}^{2}+A_{n}(2n^{2}\pi^{2}\cosh[b\mu_{n}]\sinh[(b-y) \mu_{n}](D_{12}+4D_{66})\\ &&+n^{2}\pi^{2}(y\cosh[(2b-y)\mu_{n}]+(-2b+y)\cosh[y\mu_{n}])(D_{12}+4D_{66})\mu_{n} \\ & &+3a^{2}(-\sinh[(2b-y)\mu_{n}]+\sinh[y\mu_{n}])D_{22}\mu_{n}^{2}-a^{2}( y\cosh[(2b-y)\mu_{n}]{}\\ &&+(-2b+y)\cosh[y\mu_{n}])D_{22}\mu_{n}^{3})). \end{eqnarray}$

$\begin{eqnarray} w_{3}(x, y)& = &\sum^{\infty}_{m = 1}\frac{1}{8b^{2}(-1+e^{2a\xi_{m}})^{2}D_{11}\xi_{m}^{3}}e^{(a-x)\xi_{m}} {}\\ &&\times \cos[\beta_{m}y](m^{2}\pi^{2}D_{12}(H_{m}((1-e^{2a\xi_{m}}) (1+e^{2x\xi_{m}})+(-a(1+e^{2a\xi_{m}})(1+e^{2x\xi_{m}}){}\\ &&+(-1+e^{2a\xi_{m}})(-1+e^{2x\xi_{m}})x)\xi_{m}) +2e^{(a+x)\xi_{m}}G_{m}(\sinh[(2a-x)\xi_{m}]{}\\ &&+\sinh[x\xi_{m}]+(x\cosh[(2a-x)\xi_{m}]+(2a-x)\cosh[x\xi_{m}])\xi_{m}))\\ &&+4(m^{2}\pi^{2}D_{66}(H_{m}((1-e^{2a\xi_{m}})(1+e^{2x\xi_{m}})+(-a(1+e^{2a\xi_{m}})(1+e^{2x\xi_{m}})\\ &&+(-1+e^{2a\xi_{m}})(-1+e^{2x\xi_{m}})x)\xi_{m})+2e^{(a+x)\xi_{m}}G_{m}(2\cosh[(a-x)\xi_{m}] \sinh[a\xi_{m}]\\ &&+(x\cosh[(2a-x)\xi_{m}]+(2a-x)\cosh[x\xi_{m}])\xi_{m})) \\ &&+b^{2}D_{11}\xi_{m}^{2}(H_{m}(3(-1+e^{2a\xi_{m}})(1+e^{2x\xi_{m}})+(a(1+e^{2a\xi_{m}})(1+e^{2x\xi_{m}}) \\ &&-(-1+e^{2a\xi_{m}}) (-1+e^{2x\xi_{m}})x)\xi_{m})-2e^{(a+x)\xi_{m}}G_{m}(3(\sinh[(2a-x)\xi_{m}]{}\\ &&+\sinh[x\xi_{m}])+(x\cosh[(2a-x)\xi_{m}] +(2a-x)\cosh[x\xi_{m}])\xi_{m})))), \end{eqnarray}$

$\begin{eqnarray} \frac{b^{2}q}{2}+\frac{bP}{a}+F_{0} = 0 \end{eqnarray}$

$\begin{eqnarray} &&\frac{2e^{2b\mu_{i}}P\cos[i\pi](i^{2}\pi^{2}D_{12}(\sinh[2b\mu_{i}]-2b\mu_{i})+a^{2}D_{22}\mu_{i}^{2}(\sinh[2b\mu_{i}]+2b\mu_{i}))} {a(1+e^{2b\mu_{i}})^{2}i^{2}\pi^{2}(D_{12}+2D_{66})\mu_{i}} \\ &&+\frac{1}{4a^{4}D_{22}\mu_{i}^{3}}\text{sech}[b\mu_{i}]^{2}(2a^{2}F_{i}\mu_{i}^{2}(bi^{2}\pi^{2} \sinh[b\mu_{i}]D_{12}{}\\ &&+a^{2}D_{22}\mu_{i}(2\cosh[b\mu_{i}]-b\sinh[b\mu_{i}]\mu_{i}))+ A_{i}(i^{4}\pi^{4} D_{12}^{2}(\sinh[2b\mu_{i}]-2b\mu_{i}){}\\ &&+2i^{2}\pi^{2}D_{12}(\sinh[2b\mu_{i}]-2b\mu_{i})(2i^{2}\pi^{2}D_{66}-a^{2}D_{22}\mu_{i}^{2}){}\\ &&+a^{2}D_{22}\mu_{i}^{2}(a^{2}D_{22}\mu_{i}^{2}(\sinh[2b\mu_{i}]-2b\mu_{i}) +4i^{2}\pi^{2}D_{66}(\sinh[2b\mu_{i}]+2b\mu_{i})))){}\\ && +\sum\limits_{m = 1}^{\infty}\frac{1}{8b^{4}D_{11}(i^{2}\pi^{2}+a^{2}\xi_{m}^{2})^{2}} (a(G_{m}-\cos[i\pi]H_{m})(4b^{2}i^{2}m^{2}\pi^{4}D_{12}^{2}{}\\ &&+D_{12}(m^{2}\pi^{4}(a^{2}m^{2}D_{22} +16b^{2}i^{2}D_{66})+16a^{2}b^{4}D_{11}\xi_{m}^{4})\\ &&+4m^{2}\pi^{2}D_{22}(a^{2}m^{2}\pi^{2}D_{66}-b^{2}D_{11}(i^{2}\pi^{2}+2a^{2}\xi_{m}^{2})))) = 0. \end{eqnarray}$

$\begin{eqnarray} &&-\frac{b^{2}P}{2aD_{22}}-\frac{b^{3}q}{3D_{22}}+A_{0}-\frac{bF_{0}}{D_{22}}\\ &&+\sum\limits_{m = 1}^{\infty} (-\frac{m\pi(G_{m}-H_{m})\sin[\frac{m\pi}{2}](m^{2}\pi^{2}(D_{12}+4D_{66})-8b^{2}D_{11}\xi_{m}^{2})}{8ab^{3}D_{11}\xi_{m}^{4}}) = 0 \end{eqnarray}$

$\begin{eqnarray} &&-\frac{abP\cos[i\pi]\text{sech}[b\mu_{i}]\tanh[b\mu_{i}]\mu_{i}}{i^{2}\pi^{2}(D_{12}+2D_{66})}-\frac{1}{4a^{2}D_ {22}\mu_{i}}\text{sech}[b\mu_{i}]^{2}(2b i^{2}\pi^{2}A_{i}\sinh[b\mu_{i}](D_{12}+4D_{66})\\ & &+a^{2} (F_{i}(\sinh[2b\mu_{i}]+2b\mu_{i})-2A_{i}D_{22}\mu_{i}(2\cosh[b\mu_{i}]+b\sinh[b\mu_{i}]\mu_{i})))\\ &&+\sum\limits_{m = 1}^{\infty}\frac{1}{4b^{3}D_{11}(i^{2}\pi^{2}+a^{2}\xi_{m}^{2})^{2}} (-am\pi(G_{m}-\cos[i\pi]H_{m})\sin[\frac{m\pi}{2}](a^{2}\pi^{2}m^{2}(D_{12}+4D_{66})\\ &&-4b^{2}D_{11}(i^{2}\pi^{2}+2a^{2}\xi_{m}^{2}))) = 0. \end{eqnarray}$

$\begin{eqnarray} &&-\frac{8b(-mP\pi-2abq\sin[\frac{m\pi}{2}])D_{12}}{am^{3}\pi^{3}D_{22}}+\sum\limits_{n = 1}^{\infty}\frac{8bP\cos[n\pi] \mu_{n}(8b^{2}n^{2}\pi^{2}D_{11}\mu_{n}+2a^{2}m^{2}\pi^{2}D_{12}\mu_{n})} {an^{2}\pi^{2}(D_{12}+2D_{66})(m^{2}\pi^{2}+4b^{2}\mu_{n}^{2})^{2}}\\ &&+\frac{1}{8b^{4}(-1+e^{2a\xi_{m}})^{2}D_{11}\xi_{m}^{3}}(e^{2a\xi_{m}}(m^{4}\pi^{4}(\cosh[a\xi_{m}]G_{m} -H_{m})\sinh[a\xi_{m}]D_{12}(D_{12}+4D_{66})\\ &&+am^{4}\pi^{4}(G_{m}-\cosh[a\xi_{m}]H_{m})D_{12}(D_{12}+4D_{66})\xi_{m} -8b^{2}m^{2}\pi^{2}(\cosh[a\xi_{m}] G_{m}-H_{m})\\ &&\times\sinh[a\xi_{m}]D_{11}(D_{12}-2D_{66})\xi_{m}^{2} -8ab^{2}m^{2}\pi^{2}(G_{m} -\cosh[a\xi_{m}]H_{m})D_{11}(D_{12}+2D_{66})\xi_{m}^{3} \\ &&+16b^{4}(\cosh[a\xi_{m}]G_{m}-H_{m})\sinh[a\xi_{m}]D_{11}^{2}\xi_{m}^{4}+16ab^{4}(G_{m}-\cosh[a\xi_{m}]H_{m})D_{11}^{2}\xi_{m}^{5})) \\ &&+\frac{4\sin[\frac{m\pi}{2}]D_{12}F_{0}}{m\pi D_{22}} +\sum\limits_{n = 1}^{\infty}\frac{1}{a^{4}D_{22}\mu_{n}(m^{2}\pi^{2}+4b^{2}\mu_{n}^{2})^{2}}((4a^{2}m\pi^{3}F_{n} \sin[\frac{m\pi}{2}](4b^{2}n^{2}D_{11}\\ &&+a^{2}m^{2}D_{12})\mu_{n}+8bA_{n} \mu_{n}(n^{2}\pi^{2}D_{11}(\pi^{2}(-a^{2}m^{2}D_{22}+4b^{2}n^{2}(D_{12}+4D_{66})) \\ &&-8a^{2}b^{2}D_{22}\mu_{n}^{2})+a^{2}D_{12} (m^{2}n^{2}\pi^{4}(D_{12}+4D_{66})+4a^{2}b^{2}D_{22}\mu_{n}^{4})))) = 0. \end{eqnarray}$

$\begin{eqnarray} &&-\frac{8b(-mP\pi-2abq\sin[\frac{m\pi}{2}])D_{12}}{am^{3}\pi^{3}D_{22}}{}\\ &&+\sum\limits_{n = 1}^{\infty}\frac{8bP\cos[n\pi]^{2}\mu_{n}(8b^{2}n^{2}\pi^{2}D_{11}\mu_{n}+2a^{2}m^{2}\pi^{2}D_{12}\mu_{n})} {an^{2}\pi^{2}(D_{12}+2D_{66})(m^{2}\pi^{2}+4b^{2}\mu_{n}^{2})^{2}}\\ &&-\frac{1}{8b^{4}(-1+e^{2a\xi_{m}})^{2}D_{11}\xi_{m}^{3}}(e^{2a\xi_{m}}(-m^{4}\pi^{4}(G_{m} -\cosh[a\xi_{m}]H_{m})\\ &&\times\sinh[a\xi_{m}]D_{12}(D_{12}+4D_{66})+am^{4}\pi^{4}(-\cosh[a\xi_{m}]G_{m}+H_{m})D_{12}(D_{12} +4D_{66})\xi_{m}\\ &&+8b^{2}m^{2}\pi^{2}(G_{m} -\cosh[a\xi_{m}]H_{m})\sinh[a\xi_{m}]D_{11}(D_{12}-2D_{66})\xi_{m}^{2} \\ &&+8ab^{2}m^{2}\pi^{2} (\cosh[a\xi_{m}]G_{m}-H_{m})D_{11}(D_{12}+2D_{66})\xi_{m}^{3}{}\\ &&-16b^{4}(G_{m}-\cosh[a\xi_{m}]H_{m})\sinh[a\xi_{m}]D_{11}^{2}\xi_{m}^{4}{}\\ &&+16ab^{4}(-\cosh[a\xi_{m}]G_{m}+H_{m})D_{11}^{2} \xi_{m}^{5}))+\frac{4F_{0} \sin[\frac{m\pi}{2}]D_{12}}{m\pi D_{22}}\\ &&+\sum\limits_{n = 1}^{\infty}\frac{1}{a^{4}D_{22}\mu_{n}(m^{2}\pi^{2}+4b^{2}\mu_{n}^{2})^{2}}(4\cos[n\pi](a^{2}m\pi^{3}F_{n}\sin[\frac{m\pi}{2}](4b^{2}n^{2} D_{11}\\ &&+a^{2}m^{2}D_{12})\mu_{n}+2bA_{n}\mu_{n}(n^{2}\pi^{2}D_{11} (\pi^{2}(-a^{2}m^{2}D_{22}+4b^{2}n^{2}(D_{12}+4D_{66})) \\ &&-8a^{2}b^{2}D_{22}\mu_{n}^{2})+a^{2}D_{12}(m^{2}n^{2}\pi^{4}(D_{12} +4D_{66})+4a^{2}b^{2}D_{22}\mu_{n}^{4})))) = 0. \end{eqnarray}$

$\begin{eqnarray} &&\frac{5b^{4}q}{24D_{22}}+\frac{b^{3}P}{3aD_{22}}+\frac{1}{2}b(-2A_{0}+\frac{bF_{0}}{D_{22}}){}\\ &&+\sum\limits_{n = 1}^{\infty} \frac{aP\cos[n\pi]^{2}\text{sech}[b\mu_{n}]^{2}(\sinh[2b\mu_{n}]-2b\mu_{n})}{2n^{2}\pi^{2}(D_{12}+2D_{66})\mu_{n}}\\ &&+\sum\limits_{n = 1}^{\infty}\frac{1}{4a^{2}D_{22}\mu_{n}^{3}}\cos[n\pi]\text{sech}[b\mu_{n}]^{2}(2a^{2}bF_{n}\sinh[b\mu_{n}]\mu_{n}^{2} {}\\ &&+A_{n}(n^{2}\pi^{2} D_{12}(\sinh[2b\mu_{n}]-2b\mu_{n})+4n^{2}\pi^{2}D_{66}(\sinh[2b\mu_{n}]-2b\mu_{n}){}\\ &&+a^{2}D_{22}\mu_{n}^{2}(-3\sinh[2b\mu_{n}]+2b\mu_{n}))) \\ &&+\sum\limits_{m = 1}^{\infty}\frac{1}{4b^{2}(-1+e^{2a\xi_{m}})^{2}D_{11}\xi_{m}^{3}}(e^{2a\xi_{m}} (2m^{2}\pi^{2}(G_{m}-\cosh[a\xi_{m}]H_{m})\\ &&\times\sinh[a\xi_{m}](D_{12}+4D_{66})+2am^{2}\pi^{2}(\cosh[a\xi_{m}]G_{m}-H_{m})(D_{12}+4D_{66})\xi_{m}{}\\ &&-24b^{2}(G_{m} -\cosh[a\xi_{m}]H_{m})\sinh[a\xi_{m}] D_{11}\xi_{m}^{2}{}\\ &&+8ab^{2}(-\cosh[a\xi_{m}]G_{m}+H_{m})D_{11}\xi_{m}^{3})) = 0. \end{eqnarray}$

$\begin{eqnarray} w(x, y) = w_{1}(x, y)+w_{2}(x, y)+w_{3}(x, y). \end{eqnarray}$

### 4.2 非零本征值为单根情形下的辛叠加解

$\begin{eqnarray} w_{1}(x, y)& = &\frac{q(5b^{4}-6b^{2}y^{2}+y^{4})}{24D_{22}}+\frac{P(2b^{3}-3by^{2}+y^{3})}{6aD_{22}}{}\\ &&+\sum^{\infty}_{n = 1}\frac{1}{2}aP\cos[n\pi]\cos[\alpha_{n}x](-(-1 +e^{2b\mu_{n1}})(e^{-y\mu_{n1}}+e^{y\mu_{n1}}){}\\ &&\times\frac{1}{(1+e^{2b\mu_{n1}})\mu_{n1} (-n^{2}\pi^{2}D_{12}-2n^{2}\pi^{2}D_{66}+a^{2}D_{22}\mu_{n1}^{2})}\\ &&-\frac{2e^{-y\mu_{n1}}} {-2n^{2}\pi^{2}(D_{12}+2D_{66})\mu_{n1}+2a^{2}D_{22}\mu_{n1}^{3}}{}\\ &&+\frac{2e^{y\mu_{n1}}} {-2n^{2}\pi^{2}(D_{12}+2D_{66})\mu_{n1}+2a^{2}D_{22}\mu_{n1}^{3}}\\ &&+\frac{e^{-y\mu_{n3}}(1-e^{2b\mu_{n3}})(1+e^{2y\mu_{n3}})} {(1+e^{2b\mu_{n3}})\mu_{n3}(-n^{2}\pi^{2}D_{12}-2n^{2}\pi^{2}D_{66}+a^{2}D_{22}\mu_{n3}^{2})})\\ &&-\frac{2e^{-y\mu_{n3}}} {-2n^{2}\pi^{2}(D_{12}+2D_{66})\mu_{n3}+2a^{2}D_{22}\mu_{n3}^{3}}{}\\ &&+\frac{2e^{y\mu_{n3}}} {-2n^{2}\pi^{2}(D_{12}+2D_{66})\mu_{n3}+2a^{2}D_{22}\mu_{n3}^{3}}, \end{eqnarray}$

$\begin{eqnarray} w_{2}(x, y)& = &\frac{(b-y)((b+y)F_{0}-2A_{0}D_{22})}{2D_{22}}+\sum^{\infty}_{n = 1}(\frac{1}{a^{2}(1+e^{2b\mu_{n3}})D_{22}\mu_{n3}(-\mu_{n1}^{2}+\mu_{n3}^{2})}\\ &&\times(e^{y\mu_{n3}}\cos[\alpha_{n}x](A_{n}(n^{2}\pi^{2}(D_{12}+4D_{66})-a^{2}D_{22}\mu_{n1}^{2})-a^{2}e^{b\mu_{n3}}F_{n}\mu_{n3}))\\ &&-\frac{1}{a^{2}(1+e^{2b\mu_{n3}})D_{22}\mu_{n3}(-\mu_{n1}^{2}+\mu_{n3}^{2})}(e^{-y\mu_{n3}}\cos[\alpha_{n}x] \\ &&\times (e^{2b\mu_{n3}}A_{n}(n^{2}\pi^{2}(D_{12}+4D_{66})-a^{2}D_{22}\mu_{n1}^{2})+a^{2}e^{b\mu_{n3}}F_{n}\mu_{n3}))){}\\ &&+\frac{1}{a^{2}(1+e^{2b\mu_{n1}})D_{22}\mu_{n1}(\mu_{n1}^{2}-\mu_{n3}^{2})}(e^{y\mu_{n1}}\cos[\alpha_{n}x](-a^{2}e^{b\mu_{n1}}F_{n}\mu_{n1}\\ &&+A_{n}(n^{2}\pi^{2}(D_{12}+4D_{66}) -a^{2}D_{22}\mu_{n3}^{2})))\\ &&-\frac{1}{a^{2}(1+e^{2b\mu_{n1}})D_{22}\mu_{n1}(\mu_{n1}^{2}-\mu_{n3}^{2})}(e^{-y\mu_{n1}}\cos[\alpha_{n}x] (a^{2}e^{b\mu_{n1}}F_{n}\mu_{n1}\\ &&+e^{2b\mu_{n1}}A_{n}(n^{2}\pi^{2}(D_{12}+4D_{66})-a^{2}D_{22}\mu_{n3}^{2}))), \end{eqnarray}$

$\begin{eqnarray} w_{3}(x, y)& = &\sum^{\infty}_{m = 1}\frac{1}{8b^{2}\xi_{m1}^{2}\xi_{m3}^{2}} \cos[\beta_{m}y](\frac{1}{D_{11}(\xi_{m1}^{2}-\xi_{m3}^{2})} 2\xi_{m1}\xi_{m3}(\text{csch}[a\xi_{m3}] \\ &&\times(\cosh[(a-x)\xi_{m3}]G_{m}-\cosh[x\xi_{m3}]H_{m})\xi_{m1}(m^{2}\pi^{2}(D_{12}+4D_{66})-4b^{2}D_{11}\xi_{m1}^{2}) \\ &&-m^{2}\pi^{2}\text{csch}[a\xi_{m1}](\cosh[(a-x)\xi_{m1}]G_{m}-\cosh[x\xi_{m1}]H_{m})(D_{12}+4D_{66})\xi_{m3} \\ &&+4b^{2}\text{csch}[a\xi_{m1}](\cosh[(a-x)\xi_{m1}]G_{m}-\cosh[x\xi_{m1}]H_{m})D_{11}\xi_{m3}^{3}){}\\ &&+\frac{64b^{4}q\sin[\frac{m\pi}{2}] }{m\pi} (\frac{\xi_{m3}^{2}}{m^{2}\pi^{2}(D_{12}+2D_{66})-4b^{2}D_{11}\xi_{m1}^{2}}{}\\ &&+\frac{\xi_{m1}^{2}}{m^{2}\pi^{2}(D_{12}+2D_{66}) -4b^{2}D_{11}\xi_{m3}^{2}})). \end{eqnarray}$

$\begin{eqnarray} w(x, y) = w_{1}(x, y)+w_{2}(x, y)+w_{3}(x, y). \end{eqnarray}$

## 5 算例

 $y = b/2$ $b/a$ $x = 0$ $x = a/4$ $x = a/2$ $x = 3a/4$ $x = a$ $1.0$ $Dw/(qa^{4})$ present 0.02943 0.02436 0.01842 0.01115 0.002789 Ref.[10] 0.02949 0.02442 0.01847 0.01118 0.002783 $M_y/(qa^{2})$ present -0.01137 -0.008667 0.002233 0.01683 0.03146 Ref.[10] -0.01161 -0.008872 0.002031 0.01663 0.03130 $1.2$ $Dw/(qa^{4})$ present 0.05251 0.04247 0.03168 0.01956 0.006469 $M_y/(qa^{2})$ present 0.01263 0.013099 0.022244 0.03556 0.05058 $1.4$ $Dw/(qa^{4})$ present 0.08433 0.06754 0.05032 0.03202 0.013093 $M_y/(qa^{2})$ present 0.04384 0.041989 0.048879 0.06048 0.07534 $1.6$ $Dw/(qa^{4})$ present 0.12619 0.10093 0.07574 0.04992 0.023990 $M_y/(qa^{2})$ present 0.08095 0.076966 0.081643 0.09153 0.10597 $1.8$ $Dw/(qa^{4})$ present 0.17962 0.14421 0.10954 0.07485 0.040737 $M_y/(qa^{2})$ present 0.12328 0.117456 0.120189 0.12860 0.14255

 $y = b/2$ $b/a$ $x = 0$ $x = a/4$ $x = a/2$ $x = 3a/4$ $x = a$ $1.0$ $100E_{T}h^{3}Dw/(qa^{4})$ present 4.50209 4.33322 3.83644 2.59816 0.53881 $M_y/(qa^{2})$ present -0.269027 -0.247932 -0.206159 -0.101362 0.121649 $1.2$ $100E_{T}h^{3}Dw/(qa^{4})$ present 9.67375 8.79869 7.35414 4.77077 1.11120 $M_y/(qa^{2})$ present -0.382724 -0.341621 -0.264498 -0.106729 0.173482 $1.4$ $100E_{T}h^{3}Dw/(qa^{4})$ present 18.21916 15.81380 12.62913 7.97018 2.02218 $M_y/(qa^{2})$ present -0.499650 -0.437193 -0.320878 -0.104608 0.231455 $1.6$ $100E_{T}h^{3}Dw/(qa^{4})$ present 31.03660 26.01927 20.08335 12.44326 3.35009 $M_y/(qa^{2})$ present -0.612517 -0.528390 -0.372671 -0.096795 0.293216 $1.8$ $100E_{T}h^{3}Dw/(qa^{4})$ present 48.96215 40.04606 30.15277 18.44113 5.16290 $M_y/(qa^{2})$ present -0.716491 -0.609928 -0.416522 -0.084249 0.356611

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