数学物理学报, 2021, 41(3): 797-810 doi:

论文

一角点支撑另一对边固支正交各向异性矩形薄板弯曲的辛叠加解

寇天娇,1, 阿拉坦仓,2

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

通讯作者: 额布日力吐, E-mail: ebu@imu.edu.cn

收稿日期: 2020-05-14  

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

Received: 2020-05-14  

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

作者简介 About authors

寇天娇,E-mail:1849208969@qq.com , E-mail:1849208969@qq.com

阿拉坦仓,E-mail:altanca@imu.edu.cn , E-mail:altanca@imu.edu.cn

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

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本文引用格式

寇天娇, 阿拉坦仓. 一角点支撑另一对边固支正交各向异性矩形薄板弯曲的辛叠加解. 数学物理学报[J], 2021, 41(3): 797-810 doi:

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

矩形薄板广泛运用于钢筋混凝土房屋建造, 桥梁建筑以及船舶构造等工程领域中, 因此诸多学者对于矩形薄板的求解进行了深入的研究, 其中数值解法较为常见, 如有限元法[1]、有限差分法[2]和谱方法[3]等. 与数值方法相比, 研究矩形薄板问题的解析方法相对较少. 现有的解析法有级数法[4], 复变函数法[5]和半逆叠加法[6]等, 但是这些解析法求解具体实际问题时都需要事先选定好恰当的试验函数, 其求解方法不具有规律性.

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.

辛叠加方法已解决了大量不同边界条件下各向同性板和一些正交各向异性矩形薄板的弯曲[10-11]与振动[12-13]问题的解析求解, 同时也解决了若干壳体的问题[14-16]等. 但是到目前为止我们未见到应用辛叠加法研究角点支撑的正交各向异性矩形薄板弯曲问题的相关结论, 因此本文应用辛叠加法研究了均匀荷载下一角点支撑另一对边固支正交各向异性矩形薄板弯曲问题. 首先, 根据边界条件将原问题分解为三个子问题, 其中两个子问题为对边滑支问题, 另外一个子问题为一边滑支对边简支子问题. 然后应用辛本征函数展开法分别得到上述三个子问题所对应的Hamilton正则方程的通解, 进而得到这三个子问题挠度形式的解, 再将以上三个子问题挠度形式的解进行叠加, 求出均匀荷载下一角点支撑另一对边固支正交各向异性矩形薄板弯曲问题的辛叠加解. 最后我们应用所得辛叠加解分别计算了各向同性和正交各向异性矩形薄板一些点处的挠度值与弯矩值, 特别将各向同性方形板的计算数值与已有文献[10] 的数值结果进行比较, 验证了本文所得辛叠加解的正确性.

2 Hamilton正则方程

考虑正交各向异性矩形薄板的基本方程

$ \begin{equation} 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, \end{equation} $

定义区域为$ \{(x, y, z)|0\leq x\leq a, 0\leq y\leq b, -\frac{h}{2}\leq z\leq \frac{h}{2}\} $, 其中$ w $是板的挠度, $ q $是分布的横向载荷; $ D_{11} = \frac{E_{1}h^{3}}{12(1-\upsilon_{12}\upsilon_{21})} $$ D_{22} = \frac{E_{2}h^{3}}{12(1-\upsilon_{12}\upsilon_{21})} $分别为板关于$ x $轴和$ y $轴方向的弯曲刚度, $ H = D_{12}+2D_{66} $是板的有效扭转刚度, $ D_{66} = \frac{G_{12}h^{3}}{12} $是板的扭转刚度, $ E_{1} $, $ E_{2} $, $ G_{12} $是相互独立的弹性常数, $ v_{12} $$ v_{21} $是Poisson比, $ h $为板的厚度, 其中$ D_{12} = \upsilon_{12}D_{22} = \upsilon_{21}D_{11} $.

板内弯矩$ M_{x} $$ M_{y} $、扭矩$ M_{xy} $、剪力$ Q_{x} $$ Q_{y} $以及等效剪力$ V_{x} $$ V_{y} $分别表示如下:

$ \begin{equation} M_{x} = -(D_{11}\frac{\partial^{2} w}{\partial x^{2}}+D_{12}\frac{\partial^{2} w}{\partial y^{2}}), M_{y} = -(D_{22}\frac{\partial^{2} w}{\partial y^{2}}+D_{12}\frac{\partial^{2} w}{\partial x^{2}}), M_{xy} = -2D_{66}\frac{\partial^{2} w}{\partial x\partial y}; \end{equation} $

$ \begin{equation} Q_{x} = -\frac{\partial}{\partial x}(D_{11}\frac{\partial^{2} w}{\partial x^{2}}+H\frac{\partial^{2} w}{\partial y^{2}}), Q_{y} = -\frac{\partial}{\partial y}(D_{22}\frac{\partial^{2} w}{\partial y^{2}}+H\frac{\partial^{2} w}{\partial x^{2}}); \end{equation} $

$ \begin{equation} V_x = -(D_{11}\frac{\partial^{3}w}{\partial x^{3}}+(D_{12}+4D_{66})\frac{\partial^{3}w}{\partial x\partial^{2}y}), V_y = -(D_{22}\frac{\partial^{3}w}{\partial y^{3}}+(D_{12}+4D_{66})\frac{\partial^{3}w}{\partial x^{2}\partial y}). \end{equation} $

$ \partial w/\partial y = \theta $, 则由方程(2.1)–(2.4)可得到Hamilton正则方程

$ \begin{equation} \frac{\partial {\bf U}}{\partial y} = {\bf H}{\bf U}+{\bf f}, \end{equation} $

其中

直接计算可以验证算子矩阵$ {\bf H} $满足$ {\bf H}^{T} = {\bf JHJ} $, 因此$ {\bf H} $是Hamilton算子矩阵, 从而(2.5)式为矩形薄板方程(2.1)的Hamilton正则方程.

3 本征值和本征函数

为了求解方程(2.5), 我们首先求解对应的齐次方程

$ \begin{equation} \frac{\partial {\bf U}}{\partial y} = {\bf H}{\bf U}. \end{equation} $

利用分离变量法求解方程(3.1), 令

$ \begin{equation} {\bf U} = {\bf X}(x)Y(y). \end{equation} $

将(3.2)式代入方程(3.1)可得

$ \begin{equation} \frac{Y^{\prime}(y)}{Y(y)}\ = \frac{{\bf H}{\bf X}(x)}{{\bf X}(x)} = \mu , \end{equation} $

由上式可得

$ \begin{equation} {\bf H}{\bf X}(x) = \mu {\bf X}(x), \end{equation} $

其中$ \mu $为本征值, $ {\bf X}(x) $为相应的本征函数. (3.4)式可写为

$ \begin{equation} ({\bf H}-\mu{\bf I}){\bf X}(x) = 0, \end{equation} $

其中$ {\bf I} $$ 4\times4 $的单位矩阵. $ {\bf H} $代入(3.5)式整理可得

$ \begin{equation} 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. \end{equation} $

$ X_1(x) = e^{\lambda x} $得其解为

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

其中

$ \begin{equation} \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}}. \end{equation} $

本文研究的均匀荷载下一角点支撑另一对边固支正交各向异性矩形薄板弯曲问题通过边界条件的分析可分解为三个子问题, 其中两个子问题为对边滑支问题, 另外一个子问题为一边滑支对边简支子问题, 以下分别研究这两类子问题对应的本征值和本征函数.

3.1 对边滑支矩形薄板所对应本征值和本征函数

对边滑支的边界条件为

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

将(3.7)式代入(3.9)式中得到

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

其中$ \alpha_{n} = \frac{n\pi}{a} $, $ \text{i} $为虚数单位. 由(3.8)式和(3.10)式计算可得

$ \begin{equation} \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}}}}, \end{equation} $

其中$ n = 1, 2, 3, \cdots $.

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

根据$ {\bf H}{\bf X}_{0}^{i}(x) = {\bf X}_{0}^{i-1}(x) $[17]$ (i\geq1) $, 得到$ \mu_{0} $对应的$ i (i = 1, 2, 3) $阶Jordan型本征函数

3.1.2 非零本征值为重根的情形

$ 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

$ \begin{equation} \mu_{n} = \alpha_{n} \sqrt \frac{H}{D_{22}}, \mu_{-n} = -\mu_{n} , n = 1, 2, 3, \cdots. \end{equation} $

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} $

对应的本征函数系为

$ \begin{equation} {\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], \end{equation} $

其中$ n = 1, 2, 3, \cdots, i = 1, 2, 3, 4 $.

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

一边滑支对边简支的边界条件为

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

仿照对边滑支情形的计算过程, 由(3.15)式计算可得

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

其中$ \beta_{m} = \frac{m\pi}{2b} $. 又由(3.16)式计算得到

$ \begin{equation} \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}}}}, \end{equation} $

其中$ m = 1, 3, 5, \cdots. $

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

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

由(3.4)式可得$ \xi_{m} $相应的本征函数和一阶Jordan型本征函数为

通过计算, 我们还可得到$ \xi_{-m} $对应的本征函数以及一阶Jordan型本征函数, 分别为

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} $

对应的本征函数系为

$ \begin{equation} {\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], \end{equation} $

其中$ m = 1, 3, 5, \cdots, i = 1, 2, 3, 4. $

3.3 辛正交性与完备性

设空间$ X = L^{2}[0, a]\times L^{2}[0, a]\times L^{2}[0, a]\times L^{2}[0, a] $, Hamilton算子$ {\bf H} $的本征函数系$ {\bf X}_{n}^{i}(x)( i = 0, 1;n = \pm1, \pm2, \pm3, \cdots) $以及$ {\bf X}_{ni}(x)( i = 1, 2, 3, 4; n = 1, 2, 3, \cdots) $在空间$ X $中具有辛正交性与完备性, 具体证明过程参见文献[18].

4 子问题的解析解

图 1

图 1   均匀荷载下一角点支撑另一对边固支矩形薄板辛叠加解结构图


为了研究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, 我们考虑如下三个子问题[10]:

(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

$ \begin{equation} \frac{\partial w}{\partial y}|_{y = 0} = V_{y}|_{y = 0} = 0, w|_{y = b} = M_{y}|_{y = b} = 0; \end{equation} $

(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.

将(4.7)式代入边界条件(4.1)中, 得到子问题(ⅰ)的解

$ \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} $

类似可得到子问题(ⅱ)的通解

$ \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} $

还可得子问题(ⅲ)的通解

$ \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} $

其中$ \xi_{m} = \frac{m\pi}{2b} \sqrt \frac{H}{D_{11}} , m = 1, 3, 5, \cdots. $

在边$ y = 0 $处, 三个子问题的弯矩之和应为零, 即满足$ M_{y}|_{y = 0} = 0 $, 计算得到

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

以及

$ \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} $

在边$ y = b $处, 三个子问题的转角之和应为零, 即满足$ \frac{\partial w}{\partial y}|_{y = b} = 0 $, 计算得到

$ \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} $

以及

$ \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} $

在边$ x = 0 $处, 三个子问题的弯矩之和应为零, 即满足$ M_{x}|_{x = 0} = 0 $, 计算得到

$ \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} $

在边$ x = a $处, 三个子问题的弯矩之和应为零, 即满足$ M_{x}|_{x = a} = 0 $, 计算得到

$ \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} $

在点$ (a, 0) $处, 挠度之和为零, 计算得到

$ \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} $

通过求解方程组(4.11)–(4.17), 可得到对应的系数$ P $, $ A_0 $, $ F_0 $, $ A_n $, $ F_n $, $ G_m $$ H_m $$ (n = 1, 2, 3, \cdots, m = 1, 3, 5, \cdots) $, 将这些系数分别代入解(4.8)–(4.10), 我们便得到辛叠加解

$ \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

非零本征值为单根时的求解过程类似于非零本征值为重根时的计算过程, 可得子问题(ⅰ), (ⅱ)和(ⅲ)的解分别为:

$ \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} $

类似于本征值为重根的求解过程, 我们得到在不同边界条件下的七个无穷方程, 通过计算这七个无穷方程解出对应的系数, 再将这些系数分别代入解(4.19)–(4.21)中得到本征值为单根情形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

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

5 算例

这里我们分别计算了均匀荷载下一角点支撑另一对边固支各向同性矩形薄板和正交各向异性矩形薄板一些点处的挠度值与弯矩值.

例1    计算在均匀荷载下一角点支撑另一对边固支各向同性矩形薄板的挠度与弯矩值, 此时原矩形薄板方程(2.1)中的对应参数取为

其中泊松比$ \upsilon = 0.3 $. 应用辛叠加解(4.18)的前100项计算了一些点处的挠度与弯矩值, 具体结果列于表 1.

表 1   均匀荷载下一角点支撑另一对边固支各向同性矩形薄板的挠度和弯矩

$y = b/2$
$b/a$$x = 0$$x = a/4$$x = a/2$$x = 3a/4$$x = a$
$1.0$$Dw/(qa^{4})$present0.029430.024360.018420.011150.002789
Ref.[10]0.029490.024420.018470.011180.002783
$M_y/(qa^{2})$present-0.01137-0.0086670.0022330.016830.03146
Ref.[10]-0.01161-0.0088720.0020310.016630.03130
$1.2$$Dw/(qa^{4})$present0.052510.042470.031680.019560.006469
$M_y/(qa^{2})$present0.012630.0130990.0222440.035560.05058
$1.4$$Dw/(qa^{4})$present0.084330.067540.050320.032020.013093
$M_y/(qa^{2})$present0.043840.0419890.0488790.060480.07534
$1.6$$Dw/(qa^{4})$present0.126190.100930.075740.049920.023990
$M_y/(qa^{2})$present0.080950.0769660.0816430.091530.10597
$1.8$$Dw/(qa^{4})$present0.179620.144210.109540.074850.040737
$M_y/(qa^{2})$present0.123280.1174560.1201890.128600.14255

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例2    计算了一角点支撑另一对边固支的正交各向异性矩形薄板的挠度与弯矩值, 取原矩形薄板方程(2.1)中的对应参数分别为

应用辛叠加解(4.22)的前30项计算了一些点处的挠度与弯矩值, 具体结果列于表 2.

表 2   均匀荷载下一角点支撑另一对边固支正交各向异性矩形薄板的挠度和弯矩

$y = b/2$
$b/a$$x = 0$$x = a/4$$x = a/2$$x = 3a/4$$x = a$
$1.0$$100E_{T}h^{3}Dw/(qa^{4})$present4.502094.333223.836442.598160.53881
$M_y/(qa^{2})$present-0.269027-0.247932-0.206159-0.1013620.121649
$1.2$$100E_{T}h^{3}Dw/(qa^{4})$present9.673758.798697.354144.770771.11120
$M_y/(qa^{2})$present-0.382724-0.341621-0.264498-0.1067290.173482
$1.4$$100E_{T}h^{3}Dw/(qa^{4})$present18.2191615.8138012.629137.970182.02218
$M_y/(qa^{2})$present-0.499650-0.437193-0.320878-0.1046080.231455
$1.6$$100E_{T}h^{3}Dw/(qa^{4})$present31.0366026.0192720.0833512.443263.35009
$M_y/(qa^{2})$present-0.612517-0.528390-0.372671-0.0967950.293216
$1.8$$100E_{T}h^{3}Dw/(qa^{4})$present48.9621540.0460630.1527718.441135.16290
$M_y/(qa^{2})$present-0.716491-0.609928-0.416522-0.0842490.356611

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6 结论

本文利用辛叠加方法推导出了均匀荷载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解析解. 由于在求解辛叠加解的过程中, 子问题(ⅱ)和(ⅲ)的通解中均含有四组无穷多个待定常数, 而这些待定常数分别由对应另外两边的边界条件所确定. 因此应用辛叠加方法还可以研究正交各向异性矩形薄板更为复杂边界条件下的弯曲以及振动问题, 同时该方法也可以研究任意荷载下正交各向异性矩形薄板的弯曲以及振动问题.

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