THE SYMPLECTIC METHOD FOR THE BUCKLING/POST-BUCKLING PROBLEMS OF RECTANGULAR PLATES ON A TENSIONLESS ELASTIC FOUNDATION
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摘要: 无拉力弹性地基上矩形薄板的屈曲/后屈曲问题是板壳力学中一类重要课题, 在工程中有着大量应用. 因涉及接触非线性, 目前主要采用数值方法对该类问题进行求解, 发展具有重要基准价值的解析方法是当前面临的一项挑战. 针对上述问题, 本文将板划分为若干包含强制边界条件的板, 形成子问题, 在辛空间下利用分离变量与辛本征展开对子问题进行解析求解, 通过子问题边界处的连续条件确定板与地基的接触状态; 通过迭代求解上述过程, 获得子问题划分的收敛结果, 并得到最终屈曲载荷及模态. 结果表明, 无拉力弹性地基与Winkler地基上板的屈曲行为存在显著差异, 且无拉力弹性地基的刚度对板的屈曲载荷与屈曲模态均有重要影响. 在此基础上, 结合Koiter摄动法与辛方法, 对无拉力弹性地基上矩形板的后屈曲问题进行求解, 获得板的后屈曲平衡路径. 所得到的屈曲与后屈曲分析结果均与有限元计算结果吻合良好, 确认了本文结果的正确性. 由于本文方法数学推导严格, 求解效率高, 因此不仅为研究无拉力弹性地基上矩形薄板的屈曲/后屈曲行为提供了一种有价值的理论工具, 更有望拓展至更多复杂板壳力学问题的求解.Abstract: The buckling/post-buckling problems of rectangular thin plates on a tensionless elastic foundation constitute an important class of topics in mechanics of plates and shells, with extensive applications in engineering. Due to involving contact nonlinearity, this kind of problems have been primarily solved using numerical methods, while the development of analytical methods with significant benchmark value is currently a challenge. To address the aforementioned issue, a plate is divided into several subproblems in this paper, each containing enforced boundary conditions. The subproblems are solved analytically using the separation of variables and the symplectic eigen expansion in the symplectic space. The contact state between the plate and the foundation is determined by the continuity conditions at the boundaries of the subproblems. By iteratively solving the above process, the convergent division of the subproblems is obtained, along with the buckling load and buckling mode shape of the plate. The results indicate that there are significant differences in the buckling behavior between a plate on a tensionless elastic foundation and that on a Winkler foundation. The stiffness of the tensionless elastic foundation has a significant influence on both the buckling loads and buckling mode shapes. Based on this, the post-buckling problem of a rectangular plate on a tensionless elastic foundation is solved by combining the Koiter perturbation method with the symplectic method, yielding the post-buckling equilibrium path of the plate. The obtained buckling and post-buckling results both agree well with those by the finite element method, which confirms the correctness of the present results. Due to the rigorous mathematical derivation and high computational efficiency of the method proposed in this paper, it not only provides a valuable theoretical tool for the study of buckling/post-buckling behaviors of rectangular thin plates on a tensionless elastic foundation, but also can be extended to solve more complex mechanical problems of plates and shells.
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Keywords:
- symplectic method /
- buckling /
- post-buckling /
- tensionless elastic foundation /
- plate
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引 言
弹性地基板模型常用于描述道路桥梁刚性路面、建筑物承载基础和微电子器件薄膜−基底结构等重要工程结构的力学行为. 由于弹性地基板在承受面内载荷时容易发生屈曲/后屈曲失效, 因此对该类问题的研究能够为相关结构的分析和设计提供重要依据.
就弹性地基本身的数学表征而言, 已有大量的研究工作. Winkler[1]在1867年将弹性地基的反作用力通过离散的弹簧来模拟, 弹簧的刚度是均匀的, 反力与结构和地基的接触面积成正比, 这即是至今仍广泛使用的Winkler地基模型. 例如, Zhang等[2]基于改进的Ritz法研究了Winkler地基上功能梯度碳纳米管增强复合材料板的屈曲问题. Kiani等[3]解析求解了局部Winkler地基上圆板在热环境下的屈曲问题. Pasternak地基模型是在Winkler地基模型基础上改进的典型模型, 由于引入了剪切刚度, 可以更好地描述结构在地基上的响应, 因此同样得到了广泛应用. Kiani等[4]利用一阶剪切理论研究了力−热载荷作用下Pasternak地基上夹芯板的屈曲问题. Parida等[5]研究了Pasternak地基上功能梯度板的振动及屈曲问题. Sobhy[6]研究了湿热环境下功能梯度夹芯板在Pasternak地基上的振动与屈曲问题. 此外, 还有大量其他弹性地基模型应用于各类问题当中[7-10].
传统的弹性地基模型在拉、压工况下均可产生反力, 相较于工程中真实存在的无拉力弹性地基存在显著差异, 但无拉力弹性地基需要考虑接触非线性, 计算较为复杂. Ma等[11-15]针对无拉力弹性地基板的屈曲问题开展了研究, 对不同的结构以及载荷形式推导了理论解, 并开展了相关试验研究. Shen等[16]基于二次摄动方法求解了无拉力弹性地基上的层合板在力−热载荷作用下的后屈曲问题. Zhong等[17]基于微分求积方法、Newmark方法以及Newton迭代法研究了无拉力弹性地基梁的非线性瞬态热响应. Wang等[18]基于有限样条法求解了无拉力弹性地基上管道的屈曲问题.
对于屈曲和后屈曲问题的求解, 其本质是高阶偏微分方程复杂边值问题的求解, 而各类数值方法无疑是解决上述问题的重要手段. Thom等[19]基于有限元方法分析了弹性地基上功能梯度裂纹板的屈曲问题. Chaabani等[20]基于改进的有限元方法求解了弹性地基上多孔功能梯度夹芯板的屈曲问题. Yang等[21]采用微分求积方法求解了弹性地基上板的热屈曲及后屈曲问题. Kumar等[22]基于伽辽金法研究了弹性地基以及多孔性对功能梯度板屈曲的影响. Lopatin等[7]采用Ritz法求解了弹性地基上正交各向异性复合材料矩形板的屈曲问题. Liang等[23-25]结合Koiter方法与弧长法, 在整个后屈曲平衡路径上使用渐进展开式, 实现了不同缺陷对屈曲载荷的影响分析.
尽管数值方法应用广泛, 但并不能动摇解析方法的地位, 因为解析解可作为检验数值解精度的基准, 且由于能够精确描述各参量之间的关系而常用于快速优化设计, 因而具有重要的理论和应用价值. Shen等[26-27]基于二次摄动方法对弹性地基上板壳结构的后屈曲问题开展了研究. Khorasani等[28]研究了弹性地基上蜂窝夹芯板的屈曲问题, 并利用Navier法对控制方程进行求解. Yaghoobi等[29]基于Navier法研究了不同类型弹性地基上石墨烯增强夹芯板的振动、屈曲及弯曲问题. Ruocco等[30]采用Lévy法求解了Pasternak地基上纳米板的屈曲问题. Xing等[31-33]提出了分离变量法用于求解矩形板的屈曲、振动等问题. 近年来, 基于钟万勰院士和姚伟岸等开创的辛弹性力学求解体系[34], 李锐等提出了辛叠加方法, 获得了一系列复杂约束下板壳结构的振动[35]、屈曲[36-39]和弯曲[40-41]解析解.
针对无拉力弹性地基上矩形板的屈曲问题, 本文基于辛方法, 将板划分为若干包含强制边界条件的板, 形成子问题, 并在辛空间下利用分离变量与辛本征展开对子问题进行求解, 通过子问题边界处的连续条件获得板的屈曲模态, 进而确定板与地基的接触状态; 通过迭代求解上述过程, 获得子问题划分及临界屈曲载荷的收敛结果. 在此基础上, 本文结合辛方法与Koiter摄动法, 对无拉力弹性地基上矩形板的后屈曲问题进行求解, 给出了不同弹性地基刚度下板的后屈曲平衡路径. 所得屈曲与后屈曲解均与有限元结果吻合良好. 本文所发展的方法求解过程严格, 无需假定解的形式, 不仅为其他求解方法提供了对比基准, 还可望为相关工程结构的分析与设计提供理论指导.
1. 无拉力弹性地基上矩形板的屈曲分析
1.1 屈曲问题数学模型
如图1所示, 本文的研究对象为单向均布载荷作用下无拉力弹性地基上的简支矩形板(“S”代表简支边界). 沿x方向长度为a, 沿y方向宽度为b, 沿z方向厚度为h. 坐标系原点位于矩形板的角点. 地基仅在与板互相接触时产生反力.
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图 1 无拉力弹性地基板示意图Figure 1. Schematic diagram of a plate on a tensionless elastic foundation图1对应的无拉力弹性地基上矩形板屈曲问题的控制方程为
$$ \left. \begin{split} & D{\nabla ^4}w + {N_x}\frac{{{\partial ^2}w}}{{\partial {x^2}}} = q \\ & q = \left\{ \begin{aligned} & { - kw,{\text{ }}w \leqslant 0} \\ & {0,{\text{ }}w > 0} \end{aligned} \right. \end{split} \right\} $$ (1) 其中$ w $为板中面沿着z方向的挠度; $ {N_x} $为板中面沿x方向的均布内力; k为弹性地基刚度; 弯曲刚度$ D $可表达为
$$ D = \frac{{E{h^3}}}{{12\left( {1 - {\nu ^2}} \right)}} $$ (2) 其中$E$为弹性模量, $\nu $为泊松比.
上述问题涉及接触非线性, 相较于传统的Winkler地基板, 其求解的关键难点在于获取板与地基的接触关系, 区分接触区域与非接触区域. 由于当前研究对象边界以及外载荷的对称性, 使得接触/非接触区域分界线垂直于x轴, 因此沿加载方向将整块板按分界线划分为若干矩形板的拼接[12]. 将挠度向上的部分视为无地基板, 挠度向下的部分则视为Winkler地基板, 不同子问题间通过强制边界条件约束. 基于此思路, 本文拟通过迭代确定接触关系, 期间对于屈曲问题的解析求解通过辛方法实现. 具体求解流程如图2所示: 首先求解Winkler地基上矩形板的屈曲问题, 获得初始的接触状态(屈曲模态); 根据初始接触状态将板划分为Winkler地基板及无地基板的子问题, 向上变形的部分为无地基板, 向下变形的则为Winkler地基板; 对各子问题施加强制边界条件, 利用辛方法求得各子问题的挠度表达式, 再通过各子问题边界处的连续条件求得屈曲载荷及屈曲模态; 由上述屈曲模态确定接触状态, 若相邻两次迭代对应的临界屈曲载荷及接触/非接触区域长度(即每个子问题板的长度)的变化率小于0.01%, 则判定求解收敛, 否则根据最新的屈曲模态结果重新划分子问题, 继续计算, 直至迭代收敛.
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图 2 无拉力弹性地基板屈曲问题求解流程图Figure 2. Solution flowchart for the buckling of a rectangular plate on a tensionless elastic foundation1.2 导入Hamilton体系
将需要考虑接触状态的原问题简化为若干Winkler地基板及无地基板的子问题, 各子问题是否为Winkler地基板取决于迭代过程的具体求解结果, 图3为子问题示意图.
对于Winkler地基板在单向面内载荷作用下的屈曲问题, 其控制方程为高阶偏微分方程, 传统的半逆法需要预先假定解的形式, 这使得其适用范围受限. 近年来发展起来的辛方法基于严格的理论推导, 无需假定解的形式, 可解析求解此类问题. 以下利用辛方法推导获取子问题的屈曲解答.
弹性地基上矩形薄板屈曲问题的平衡方程如下式所示, 其中令$ k = 0 $即可获得无地基板的控制方程
$$ \left. \begin{split} & \frac{{\partial {M_x}}}{{\partial x}} + \frac{{\partial {M_{xy}}}}{{\partial y}} - {Q_x} = 0 \\ & \frac{{\partial {M_y}}}{{\partial y}} + \frac{{\partial {M_{xy}}}}{{\partial x}} - {Q_y} = 0 \\ & \frac{{\partial {Q_x}}}{{\partial x}} + \frac{{\partial {Q_y}}}{{\partial y}} + {N_x}\frac{{{\partial ^2}w}}{{\partial {x^2}}} + kw = 0 \end{split} \right\} $$ (3) 这里板内弯矩$ {M_x} $和$ {M_y} $, 扭矩$ {M_{xy}} $, 剪力$ {Q_x} $和$ {Q_y} $以及等效剪力$ {V_x} $和$ {V_y} $的表达式为
$$ \left.\begin{aligned} & {M_x} = - D\left( {\frac{{{\partial ^2}w}}{{\partial {x^2}}} + \nu \frac{{{\partial ^2}w}}{{\partial {y^2}}}} \right) \\ & {M_y} = - D\left( {\frac{{{\partial ^2}w}}{{\partial {y^2}}} + \nu \frac{{{\partial ^2}w}}{{\partial {x^2}}}} \right) \\ & {M_{xy}} = - D\left( {1 - \nu } \right)\frac{{{\partial ^2}w}}{{\partial x\partial y}} \end{aligned} \right\} $$ (4) $$ \left.\begin{aligned} & {Q_x} = - D\frac{\partial }{{\partial x}}{\nabla ^2}w \\ & {Q_y} = - D\frac{\partial }{{\partial y}}{\nabla ^2}w \end{aligned} \right\} $$ (5) $$ \left. \begin{aligned} & {V_x} = {Q_x} + \frac{{\partial {M_{xy}}}}{{\partial y}} + {N_x}\frac{{\partial w}}{{\partial x}} \\ & {V_y} = {Q_y} + \frac{{\partial {M_{xy}}}}{{\partial x}} \end{aligned} \right\} $$ (6) 由文献[42]的推导方法可将控制方程导入Hamilton体系, 形成如下的矩阵形式
$$ \frac{{\partial {\boldsymbol Z}}}{{\partial x}} = {\bf{H}}{\boldsymbol{Z}} $$ (7) 其中状态向量
$$ {{\boldsymbol{Z}}} = {\left[ {w,{\theta _x}, - {V_x},{M_x}} \right]^{\text{T}}} $$ (8) $$ {{\bf{H}}} = \left[ {\begin{array}{*{20}{c}} 0 & 1 & 0 & 0 \\ { - \nu \dfrac{{{\partial ^2}}}{{\partial {y^2}}}} & 0 & 0 & { - \dfrac{1}{D}} \\ {D\left( {{\nu ^2} - 1} \right)\dfrac{{{\partial ^4}}}{{\partial {y^4}}} - k} & 0 & 0 &{\nu \dfrac{{{\partial ^2}}}{{\partial {y^2}}}} \\ 0 & {2D\left( {1 - \nu } \right)\dfrac{{{\partial ^2}}}{{\partial {y^2}}} - {N_x}} & { - 1} & 0 \end{array}} \right] $$ (9) 是Hamilton算子矩阵, 满足${{{\bf{H}}}^{\text{T}}} = {{\boldsymbol{J}}{\bf{H}}{\boldsymbol{J}}}$, ${{\boldsymbol{J}}} = \left[ {\begin{array}{*{20}{c}} 0&{{{{\boldsymbol{I}}}_2}} \\ { - {{{\boldsymbol{I}}}_2}}&0 \end{array}} \right]$是4阶单位辛矩阵, $ {{{\boldsymbol{I}}}_2} $是2阶单位矩阵; $ {\theta _x} = {{\partial w} \mathord{\left/ {\vphantom {{\partial w} {\partial x}}} \right. } {\partial x}} $.
通过以上步骤, 完成了Winkler地基板屈曲问题向Hamilton体系的导入.
1.3 方程求解
利用Hamilton体系下的分离变量法求解式(7), 令
$$ {{\boldsymbol{Z}}} = X\left( x \right){{\boldsymbol{Y}}}\left( y \right) $$ (10) 其中${{\boldsymbol{Y}}}\left( y \right) = {\left[ {w\left( y \right),\theta \left( y \right), - {V_x}\left( y \right),{M_x}\left( y \right)} \right]^{{\rm{T}}}}$为H的本征向量, 仅与y有关. 将式(10)代入式(7)可得本征方程
$$ {\bf{H}}{{\boldsymbol{Y}}}\left( y \right) = {\mu _{in}}{{\boldsymbol{Y}}}\left( y \right) $$ (11) 及方程
$$ \frac{{{\text{d}}X\left( x \right)}}{{{\text{d}}x}} = {\mu _{in}}X\left( x \right) $$ (12) 其中$ {\mu _{in}} $为H的本征值. 利用下述简支边界条件求解式(11)
$$ \left. \begin{split} & {\left. {w\left( y \right)} \right|_{y = 0,{\text{ }}b}} = 0 \\ & {\left. {{M_y}\left( y \right)} \right|_{y = 0,{\text{ }}b}} = 0 \end{split} \right\} $$ (13) 可求得本征值
$$ \left.\begin{split} & {\mu _{1n}} = \sqrt {R + \beta _n^2 - \sqrt {R\left( {2\beta _n^2 + R} \right) - \frac{k}{D}} } \\ & {\mu _{2n}} = - \sqrt {R + \beta _n^2 - \sqrt {R\left( {2\beta _n^2 + R} \right) - \frac{k}{D}} } \\ & {\mu _{3n}} = \sqrt {R + \beta _n^2 + \sqrt {R\left( {2\beta _n^2 + R} \right) - \frac{k}{D}} } \\ & {\mu _{4n}} = - \sqrt {R + \beta _n^2 + \sqrt {R\left( {2\beta _n^2 + R} \right) - \frac{k}{D}} } \end{split} \right\} $$ (14) 其中$R = {{{N_x}} \mathord{\left/ {\vphantom {{{N_x}} {\left( {2 D} \right)}}} \right. } {\left( {2 D} \right)}}$, $ {\beta _n} = {{n\text{π} } \mathord{\left/ {\vphantom {{n\text{π} } b}} \right. } b} $, $ n = 1,2,3, \cdots $. 对应的本征向量为
$$ {{{\boldsymbol{Y}}}_{in}} = \sin \left( {{\beta _n}y} \right)\left[ {\begin{array}{*{20}{c}} 1 \\ {{\mu _{in}}} \\ {{\mu _{in}}\left[ {D\mu _{in}^2 + \left( {\nu - 2} \right)D\beta _n^2 - {N_x}} \right]} \\ {D\left( {\nu \beta _n^2 - \mu _{in}^2} \right)} \end{array}} \right]{\text{ }} $$ (15) 其中$ i = 1,2,3,4 $. 式(12)的解答为
$$ {X_{ni}}(y) = {C_{ni}}{{\rm{e}}^{{\mu _{in}}x}} $$ (16) 由式(15)与式(16), 状态向量${{\boldsymbol{Z}}}$可展开为
$$ {{\boldsymbol{Z}}} = \sum\limits_{n = 1}^\infty {\sum\limits_{i = 1}^4 {{C_{ni}}{{\rm{e}}^{{\mu _{in}}x}}{{{\boldsymbol{Y}}}_{in}}} } $$ (17) 其中待定系数$ {C_{ni}} $由另一个方向的边界条件确定. 首先有
$$ \left. \begin{split} & {\left. w \right|_{x = 0,{\text{ }}a}} = 0 \\ & {\left. {{M_x}} \right|_{x = 0,{\text{ }}a}} = 0 \end{split} \right\} $$ (18) 此外, 如图3所示, 原问题被划分为若干包含强制边界条件的子问题, 在第i条边界上的强制边界条件可表示为
$$ \left. \begin{split} & {\left. {{\theta _x}} \right|_{x = {a_i}}} = \sum\limits_{n = 1,2,3, \cdots }^\infty {{E_{ni}}\sin \left( {{\beta _n}y} \right)} \\ & {\left. {{V_x}} \right|_{x = {a_i}}} = \sum\limits_{n = 1,2,3, \cdots }^\infty {{F_{ni}}\sin \left( {{\beta _n}y} \right)} \end{split} \right\} $$ (19) 其中$ {E_{ni}} $与$ {F_{ni}} $为级数展开系数. 由式(18)和式(19)可将式(17)中的待定系数$ {C_{ni}} $求出, 需注意的是此时$ {E_{ni}} $与$ {F_{ni}} $仍为未知数. 将所得带有$ {E_{ni}} $与$ {F_{ni}} $的式(17)代入以下挠度解的连续条件
$$ \left.\begin{split} & {\left. {{w_i}} \right|_{x = {a_i}}} = {\left. {{w_{i + 1}}} \right|_{x = {a_i}}} \\ & {\left. {M_x^i} \right|_{x = {a_i}}} = {\left. {M_x^{i + 1}} \right|_{x = {a_i}}} \end{split} \right\} $$ (20) 其中$ {w_i} $及$ M_x^i $分别为第i个子问题的挠度及弯矩表达式, 由此得到一组关于$ {E_{ni}} $与$ {F_{ni}} $的联立线性方程. 为了获得非零解, 令方程的系数矩阵行列式为零, 从而求出相应的屈曲载荷, 同时也可获得整体屈曲模态. 进一步, 通过模态结果判断接触关系, 即可获得新的子问题分区, 继而按照图2所示步骤开展迭代求解, 直至满足收敛条件, 从而获得无拉力弹性地基上矩形板的屈曲解答.
1.4 屈曲数值结果与讨论
通过试算发现, 弹性地基刚度越大, 达到收敛所需迭代次数就越多. 同时, 存在多个半波的情况才需要考虑接触问题, 而对于仅存在单个半波的情况, 其临界屈曲模态趋向于不与地基发生接触, 最终结果与无地基板相同. 以弹性地基刚度${k_w} = {{k{b^4}} \mathord{\left/ {\vphantom {{k{b^4}} D}} \right. } D}$ = 2000, 长宽比$ {a \mathord{\left/ {\vphantom {a b}} \right. } b} = 2 $、弯曲刚度$D = 1$的板作为对象, 验证本文方法迭代求解的收敛性. 迭代过程的屈曲模态结果如图4所示, 其中曲线Iter.0代表初始的Winkler地基板计算结果, 用于确定初始分区情况. 由图4中局部放大图可知, 曲线Iter.6与Iter.7几乎重合, 表明求解在7次迭代后达到收敛. 在迭代过程中, 整体屈曲模态由Winkler地基板的4个半波逐渐变为无拉力弹性地基板的3个半波, 可见无拉力弹性地基对矩形板的屈曲行为具有显著影响.
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图 4 屈曲模态迭代的收敛性验证(kw = 2000, a/b = 2)Figure 4. Iterative convergence verification of buckling mode shapes (kw = 2000, a/b = 2)为验证本文方法求解的正确性, 基于有限元(FEM)分析软件Abaqus开展屈曲分析, 计算模型中采用S4单元, 网格尺寸为宽度的1/100. 由于线性屈曲分析中无法实现无拉力弹性地基设置, 因此需按照图2所示流程进行迭代计算. 与本文解析方法的思路一致, 通过程序实现Abaqus的参数化建模: 首先通过Winkler地基板确定初始的子问题分区, 并根据不同分区的接触状态设定弹性地基刚度, 进而进行线性屈曲Buckle分析, 根据分析结果重新划分子问题, 并计算新的屈曲载荷及模态, 反复迭代直至屈曲载荷及子问题划分结果收敛, 收敛准则与解析方法一致. 结果表明, 有限元与本文解析方法的迭代结果高度一致, 如图4所示(其中圆圈代表有限元计算得到的迭代结果).
本文方法与有限元所得屈曲模态结果如图5所示, 两者高度吻合.
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图 5 无拉力弹性地基上矩形板的临界屈曲模态(kw = 2000, a/b = 2)Figure 5. Critical buckling mode shape of a rectangular plate on a tensionless elastic foundation (kw = 2000, a/b = 2)为了进一步验证本文方法, 同时获得更为普适性的计算结果, 表1给出了不同长宽比及弹性地基刚度下板的无量纲临界屈曲载荷$ - {{{N_{{\text{cr}}}}{b^2}} \mathord{\left/ {\vphantom {{{N_{{\text{cr}}}}{b^2}} D}} \right. } D}$. 所有临界屈曲载荷解均与有限元计算结果吻合良好, 进一步验证了本文结果的准确性.
表 1 不同长宽比及弹性地基刚度下矩形板的临界屈曲载荷$ - {{{N_{{\text{cr}}}}{b^2}} \mathord{\left/ {\vphantom {{{N_{{\text{cr}}}}{b^2}} D}} \right. } D}$Table 1. Critical buckling load factors, $ - {{{N_{{\text{cr}}}}{b^2}} \mathord{\left/ {\vphantom {{{N_{{\text{cr}}}}{b^2}} D}} \right. } D}$, of rectangular plates with different aspect ratios and elastic foundation stiffnesses${k_w}$ Method a/b = 1.5 a/b = 2 a/b = 2.5 a/b = 3 1 FEM 44.228 39.502 44.046 44.105 present 44.258 39.529 44.075 44.027 10 FEM 43.038 39.923 42.683 42.265 present 43.067 39.949 42.711 42.292 100 FEM 43.687 42.172 42.119 41.436 present 43.716 42.199 42.146 41.462 200 FEM 43.038 42.998 40.998 39.770 present 43.068 43.026 41.023 39.794 2000 FEM 42.835 44.134 40.804 39.488 present 42.865 44.163 40.829 39.512 图6对比分析了弹性地基刚度对屈曲模态的影响. 在图6(a)所示含有两个半波的算例中, 随着弹性地基刚度的增大, 半波的幅值差异越发明显, 与地基发生接触的区域板的变形相对较小. 在图6(b)所示a/b = 2.0的算例中, 弹性地基刚度较小时板的屈曲模态呈现出两个半波, 而系数为2000时则为3个半波, 可见, 更高弹性地基刚度下的临界屈曲模态趋向于呈现更多半波数. 如图6(c)所示, 在弹性地基刚度较小时, 屈曲模态呈现出对称性特点, 但当系数增大后, 转变为非对称的屈曲模态. 该现象与文献中结论一致: 即使结构与边界条件均对称, 屈曲模态也可能呈现非对称性[15]. 图6(d)再次表明, 在不同的弹性地基刚度下, 屈曲模态的主要变化在于接触区域板的变形逐渐减小, 而与图6(a)对比可见, 随着板的长宽比增大, 半波数呈现出增多的趋势.
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图 6 不同长宽比矩形板的临界屈曲模态Figure 6. Critical buckling mode shape of rectangular plate with different aspect ratio2. 无拉力弹性地基上矩形板的后屈曲分析
针对无拉力弹性地基上的矩形板, 其后屈曲分析更加困难. 以下基于Koiter后屈曲摄动展开理论与辛方法, 研究无拉力弹性地基上含缺陷薄板的后屈曲问题. 摄动法常用于求解非线性方程, 其主要思想是将非线性问题的解用某个小量的渐近形式来表示.
2.1 后屈曲问题控制方程与求解
基于非线性几何方程式
$$ \left. \begin{split} & {\varepsilon _x} = \frac{{\partial u}}{{\partial x}} + \frac{1}{2}{\left( {\frac{{\partial w}}{{\partial x}}} \right)^2}, {\kappa _x} = - \frac{{{\partial ^2}w}}{{\partial {x^2}}} \\ & {\varepsilon _y} = \frac{{\partial v}}{{\partial y}} + \frac{1}{2}{\left( {\frac{{\partial w}}{{\partial y}}} \right)^2}, {\kappa _y} = - \frac{{{\partial ^2}w}}{{\partial {y^2}}} \\ & {\gamma _{xy}} = \frac{{\partial u}}{{\partial y}} + \frac{{\partial v}}{{\partial x}} + \frac{{\partial w}}{{\partial x}}\frac{{\partial w}}{{\partial y}}, {\kappa _{xy}} = - 2\frac{{{\partial ^2}w}}{{\partial x\partial y}} \end{split} \right\} $$ (21) 内力−应变关系式(4)和下式
$$ \left[ {\begin{array}{*{20}{c}} {{N_x}} \\ {{N_y}} \\ {{N_{xy}}} \end{array}} \right] = \frac{{Eh}}{{1 - {\nu ^2}}}\left[ {\begin{array}{*{20}{c}} 1&\nu &0 \\ \nu &1&0 \\ 0&0&{\dfrac{{1 - \nu }}{2}} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{\varepsilon _x}} \\ {{\varepsilon _y}} \\ {{\gamma _{xy}}} \end{array}} \right] $$ (22) 以及平衡方程式(3), 并引入应力函数F
$$ {N_x} = \frac{{{\partial ^2}F}}{{\partial {y^2}}},{N_y} = \frac{{{\partial ^2}F}}{{\partial {x^2}}},{N_{xy}} = - \frac{{{\partial ^2}F}}{{\partial x\partial y}} $$ (23) 可得薄板后屈曲问题的非线性控制方程为
$$ \left. \begin{split} & D{\nabla ^4}w = \frac{{{\partial ^2}F}}{{\partial {y^2}}}\frac{{{\partial ^2}\left( {w + \bar w} \right)}}{{\partial {x^2}}} + \frac{{{\partial ^2}F}}{{\partial {x^2}}}\frac{{{\partial ^2}\left( {w + \bar w} \right)}}{{\partial {y^2}}} -\\ &\qquad 2\frac{{{\partial ^2}\left( {w + \bar w} \right)}}{{\partial x\partial y}}\frac{{{\partial ^2}F}}{{\partial x\partial y}} - kw \\ & {\nabla ^4}F + \frac{{Eh}}{2}\left\{ \frac{{{\partial ^2}\left( {w + \bar w} \right)}}{{\partial {x^2}}}\frac{{{\partial ^2}w}}{{\partial {y^2}}} + \frac{{{\partial ^2}w}}{{\partial {x^2}}}\frac{{{\partial ^2}\left( {w + \bar w} \right)}}{{\partial {y^2}}} - \right.\\ &\qquad \left.2{{\left[ {\frac{{{\partial ^2}\left( {w + \bar w} \right)}}{{\partial x\partial y}}} \right]}^2} \right\} = 0 \end{split} \right\} $$ (24) 其中$\bar w$为初始几何缺陷. 需指出, 这里的中面内力${N_x}$、${N_y}$及${N_{xy}}$通过严格推导确定, 不同于屈曲问题中的均布内力分布设定. 利用摄动方法, 将解展开为
$$ \left\{ \begin{gathered} w \\ F \\ \end{gathered} \right\} = \left\{ \begin{gathered} 0 \\ {F_0} \\ \end{gathered} \right\} + \xi \left\{ \begin{gathered} {w_1} \\ {F_1} \\ \end{gathered} \right\} + {\xi ^2}\left\{ \begin{gathered} {w_2} \\ {F_2} \\ \end{gathered} \right\} + \cdots $$ (25) 其中$\xi $为摄动小参数. 单轴压外载荷${P_x}$可展开为
$$ {P_x} = {P_c} + {a_e}{P_c}\xi + {b_e}{P_c}{\xi ^2} $$ (26) 其中$ {P_{{c}}} $为临界屈曲载荷, $ {a_e} $及$ {b_e} $为后屈曲展开系数. 令初始几何缺陷为板的一阶屈曲模态
$$ \bar w = \xi \mu {w_1} $$ (27) 其中$\mu = {\lambda \mathord{\left/ {\vphantom {\lambda \xi }} \right. } \xi }$, $\lambda $是缺陷的幅值. 将式(25)代入式(24), 有方程的零阶控制方程
$$ {\nabla ^4}{F_0} = 0 $$ (28) 利用边界条件可得其解答为[16]
$$ {F_0} = - \frac{{{{\bar P}_x}{y^2}}}{2} $$ (29) 其中$ {\bar P_x} = {{{P_x}} \mathord{\left/ {\vphantom {{{P_x}} b}} \right. } b} $, ${P_x} = \displaystyle\int_0^b {{N_x}{\text{d}}y}$为外载荷.
由式(24)、式(25)及式(29)可得一阶控制方程
$$ D{\nabla ^4}{w_1} + \left( {1 + \mu } \right){\bar P_x}\frac{{{\partial ^2}{w_1}}}{{\partial {x^2}}} + kw = 0 $$ (30) 一阶方程表达形式与式(1)一致, 替换式(17)中$ \left( {1 + \mu } \right){\bar P_x} = {N_x} $, 即可获得一阶控制方程的解
$$ {w_1} = \sum\limits_{i = 1}^4 {{c_i}{{\rm{e}}^{{{\bar \mu }_{in}}x}}\sin \left( {{\beta _n}y} \right)} $$ (31) 其中$ {\bar \mu _{in}} $为本征值, 也可由式(14)替换获得, 此处不再赘述.
同理, 2阶控制方程为
$$ {\nabla ^4}{F_2} + Eh\left( {1 + 2\mu } \right)\left[ {\frac{{{\partial ^2}{w_1}}}{{\partial {x^2}}}\frac{{{\partial ^2}{w_1}}}{{\partial {y^2}}} - {{\left( {\frac{{{\partial ^2}{w_1}}}{{\partial x\partial y}}} \right)}^2}} \right] = 0 $$ (32) 将其导入Hamilton体系, 有
$$ \frac{{\partial {{\boldsymbol{Z}}}}}{{\partial x}} = {{\bf{H}}}{\boldsymbol{Z}} + {{\boldsymbol{f}}} $$ (33) 构造状态向量
$$ {{\boldsymbol{Z}}} = {\left[ {{F_2},\frac{{\partial {F_2}}}{{\partial x}}, - \left( {\frac{{{\partial ^3}{F_2}}}{{\partial {x^3}}} + 2\frac{{{\partial ^3}{F_2}}}{{\partial x\partial {y^2}}}} \right),\frac{{{\partial ^2}{F_2}}}{{\partial {x^2}}}} \right]^{\text{T}}} $$ (34) Hamilton算子矩阵H为
$$ {{\bf{H}}} = \left[ {\begin{array}{*{20}{c}} 0&1&0&0 \\ 0&0&0&1 \\ {\dfrac{{{\partial ^4}}}{{\partial {y^4}}}}&0&0&0 \\ 0&{ - 2\dfrac{{{\partial ^2}}}{{\partial {y^2}}}}&{ - 1}&0 \end{array}} \right] $$ (35) ${{\boldsymbol{f}}} = {\left[ {0,0,q\left( {x,y} \right),0} \right]^{\text{T}}}$, 其中由一阶解可得非齐次项表达式
$$ \begin{split} & q\left( {x,y} \right) = Eh\left( {1 + 2\mu } \right)\sum\limits_{j = 1}^4 {\sum\limits_{i = 1}^4 {{c_i}{c_j}{{\bar \mu }_{jn}}\alpha _n^2{{\text{e}}^{x\left( {{{\bar \mu }_{in}} + {{\bar \mu }_{jn}}} \right)}}} } \cdot \\ &\qquad {\text{ }}\frac{1}{2}\left\{ {{{\bar \mu }_{in}}\left[ {1 + \cos \left( {2{\beta _n}y} \right)} \right] + {{\bar \mu }_{jn}}\left[ {1 - \cos \left( {2{\beta _n}y} \right)} \right]} \right\} \end{split} $$ (36) 利用分离变量, 令
$$ {{\boldsymbol{Z}}} = X\left( x \right){{\boldsymbol{Y}}}\left( y \right) $$ (37) 将式(37)代入式(33)可得
$$\qquad\qquad\qquad {{\bf{H}}}{\boldsymbol{Y}}\left( y \right) = \gamma {{\boldsymbol{Y}}}\left( y \right) $$ (38) $$\qquad\qquad\qquad \frac{{{\text{d}}X\left( x \right)}}{{{\text{d}}x}} = \gamma X\left( x \right) $$ (39) 类似于上文求解思路, 由边界条件
$$ {\left. {\frac{{\partial F}}{{\partial y}}} \right|_{y = 0,b}} = {\left. {\frac{{{\partial ^3}F}}{{\partial {y^3}}}} \right|_{y = 0,b}} = 0 $$ (40) 可得式(38)中的本征值
$$ \left.\begin{split} & {\gamma _{1m}} = {\gamma _{2m}} = \frac{{m\text{π} }}{b} = {\beta _m} \\ & {\gamma _{3m}} = {\gamma _{4m}} = - {\beta _m} \\ &\qquad {m = 0,1,2,\cdots} \end{split} \right\} $$ (41) 对应的本征向量及其一阶Jordan型为
$$ \left.\begin{split} &{{{\boldsymbol{Y}}}}_{1m} = \mathrm{cos}\left({\beta }_{m}y\right){\left[1,{\beta }_{m},{\beta }_{m}^{3},{\beta }_{m}^{2}\right]}^{\text{T}}\\ &{{{\boldsymbol{Y}}}}_{3m} = \mathrm{cos}\left({\beta }_{m}y\right){\left[1,-{\beta }_{m},-{\beta }_{m}^{3},{\beta }_{m}^{2}\right]}^{\text{T}}\\ &{{{\boldsymbol{Y}}}}_{2m} = \mathrm{cos}\left({\beta }_{m}y\right){\left[1,1 + {\beta }_{m},{\beta }_{m}^{2}\left({\beta }_{m}-1\right),{\beta }_{m}\left(2 + {\beta }_{m}\right)\right]}^{\text{T}}\\ &{{{\boldsymbol{Y}}}}_{4m} = \mathrm{cos}\left({\beta }_{m}y\right){\left[1 + \frac{1}{{\beta }_{m}}\text{, } - {\beta }_{m}, - {\beta }_{m}^{2}\left(2 + {\beta }_{m}\right),{\beta }_{m}\left({\beta }_{m} - 1\right)\right]}^{\text{T}}\end{split}\right\} $$ (42) 当$ m = 0 $时, 对应的本征向量及其满足边界条件的Jordan型为
$$ \left. \begin{split} & {{{\boldsymbol{Y}}}_{00}} = {\left[ {1,0,0,0} \right]^{\text{T}}} \\ & {{{\boldsymbol{Y}}}_{10}} = {\left[ {0,1,0,0} \right]^{\text{T}}} \\ & {{{\boldsymbol{Y}}}_{20}} = {\left[ {0,0,0,1} \right]^{\text{T}}} \\ & {{{\boldsymbol{Y}}}_{30}} = {\left[ {0,0, - 1,0} \right]^{\text{T}}} \end{split} \right\} $$ (43) ${{\boldsymbol{f}}}$可按照辛本征向量展开为
$$ {{\boldsymbol{f}}} = {{\boldsymbol{Y}}}\left( y \right){{\boldsymbol{G}}} $$ (44) 其中${{\boldsymbol{G}}} = {\left[ {{g_{00}},{g_{01}},{g_{02}},{g_{03}}, \cdots {g_{1 m}},{g_{3 m}}, \cdots ,{g_{4 m}},{g_{2 m}}, \cdots } \right]^{\text{T}}}$. 由式(33)、式(37)、式(38)及式(44)可得
$$ \frac{{{\text{d}}{{\boldsymbol{X}}}\left( x \right)}}{{{\text{d}}x}} - {{\boldsymbol{MX}}}\left( x \right) = {{\boldsymbol{G}}} $$ (45) 其中${{\boldsymbol{M}}} = {\rm{diag}}\left( {{{{\boldsymbol{M}}}_4}, \cdots , {{{\boldsymbol{M}}}_m}, \cdots , {{{\bar {\boldsymbol{M}}}}_m}, \cdots } \right)$, ${{{\boldsymbol{M}}}_4} = \left[ {\begin{array}{*{20}{c}} 0 & 1& 0 & 0 \\ 0 & 0& 1 & 0 \\ 0 & 0& 0 & 1 \\ 0 & 0& 0 & 0 \end{array}} \right]$, ${{{\boldsymbol{M}}}_m} = \left[ {\begin{array}{*{20}{c}} {{\beta _m}}&1 \\ 0&{{\beta _m}} \end{array}} \right]$, ${{\bar {\boldsymbol{M}}}_m} = \left[ {\begin{array}{*{20}{c}} { - {\beta _m}}&0 \\ 1&{ - {\beta _m}} \end{array}} \right]$. 由于三角级数的正交性, 当$m = 2 n$时有${g_{im}} \ne 0$, 展开式(45)有一组联立的常微分方程
$$ \left.\begin{split} & {g_{00}} = {g_{10}} = {g_{20}} = 0 \\ & {g_{30}} = \frac{{Eh\left( {1 + 2\mu } \right)\beta _n^2}}{2}. \\ &\qquad {\text{ }}\sum\limits_{j = 1}^4 {\sum\limits_{i = 1}^4 {{c_i}{c_j}{{\bar \mu }_{jn}}\left( {{{\bar \mu }_{in}} + {{\bar \mu }_{jn}}} \right){{\text{e}}^{x\left( {{{\bar \mu }_{in}} + {{\bar \mu }_{jn}}} \right)}}} } \\ & {g_{1n}} = \frac{{Eh\left( {b + 2n\text{π} } \right)\left( {1 + 2\mu } \right)}}{{64n\text{π} }}. \\ &\qquad {\text{ }}\sum\limits_{j = 1}^4 {\sum\limits_{i = 1}^4 {{c_i}{c_j}{{\bar \mu }_{jn}}\left( {{{\bar \mu }_{jn}} - {{\bar \mu }_{in}}} \right){{\text{e}}^{x\left( {{{\bar \mu }_{in}} + {{\bar \mu }_{jn}}} \right)}}} } \\ & {g_{2n}} = - {g_{3n}} = - {g_{4n}} = \frac{{Eh\left( {1 + 2\mu } \right)}}{{32}}. \\ &\qquad {\text{ }}\sum\limits_{j = 1}^4 {\sum\limits_{i = 1}^4 {{c_i}{c_j}{{\bar \mu }_{jn}}\left( {{{\bar \mu }_{jn}} - {{\bar \mu }_{in}}} \right){{\text{e}}^{x\left( {{{\bar \mu }_{in}} + {{\bar \mu }_{jn}}} \right)}}} } \end{split} \right\} $$ (46) 由式(46)的展开系数, 可得式(45)的解为
$$ \begin{split} & {X_{00}}\left( x \right) = {C_{00}} + {C_{10}}x + \frac{{{C_{20}}}}{2}{x^2} + \frac{{{C_{30}}}}{6}{x^3} + \\ &\qquad {\text{ }}\sum\limits_{j = 1}^4 {\sum\limits_{i = 1,i \ne j - {{\left( { - 1} \right)}^j}}^4 {\frac{{{c_i}{c_j}Eh\left( {1 + 2\mu } \right){{\bar \mu }_{jn}}\beta _n^2{{\text{e}}^{x\left( {{{\bar \mu }_{in}} + {{\bar \mu }_{jn}}} \right)}}}}{{2{{\left( {{{\bar \mu }_{in}} + {{\bar \mu }_{jn}}} \right)}^3}}}} } \end{split} $$ (47) $$ \begin{split} & {X_{10}}\left( x \right) = {C_{10}} + {C_{20}}x + \frac{{{C_{30}}}}{2}{x^2} + \\ &\qquad {\text{ }}\sum\limits_{j = 1}^4 {\sum\limits_{i = 1,i \ne j - {{\left( { - 1} \right)}^j}}^4 {\frac{{{c_i}{c_j}Eh\left( {1 + 2\mu } \right){{\bar \mu }_{jn}}\beta _n^2{{\text{e}}^{x\left( {{{\bar \mu }_{in}} + {{\bar \mu }_{jn}}} \right)}}}}{{2{{\left( {{{\bar \mu }_{in}} + {{\bar \mu }_{jn}}} \right)}^2}}}} } \end{split} $$ (48) $$ \begin{split} & {X_{20}}\left( x \right) = {C_{20}} + x{C_{30}} + \\ &\qquad {\text{ }}\sum\limits_{j = 1}^4 {\sum\limits_{i = 1,i \ne j - {{\left( { - 1} \right)}^j}}^4 {\frac{{{c_i}{c_j}Eh{{\bar \mu }_{jn}}\beta _n^2\left( {1 + 2\mu } \right){{\text{e}}^{x\left( {{{\bar \mu }_{in}} + {{\bar \mu }_{jn}}} \right)}}}}{{2\left( {{{\bar \mu }_{in}} + {{\bar \mu }_{jn}}} \right)}}} } \end{split} $$ (49) $$ \begin{split} & {X_{30}}\left( x \right) = {C_{30}} + \frac{{Eh\left( {1 + 2\mu } \right)}}{2}. \\ &\qquad {\text{ }}\sum\limits_{j = 1}^4 {\sum\limits_{i = 1,i \ne j - {{\left( { - 1} \right)}^j}}^4 {{c_i}{c_j}{{\bar \mu }_{jn}}\beta _n^2{{\text{e}}^{x\left( {{{\bar \mu }_{in}} + {{\bar \mu }_{jn}}} \right)}}} }\end{split} $$ (50) $$ \begin{split} & {X_{1n}}\left( x \right) = \left( {{C_{1n}} + x{C_{3n}}} \right){{\text{e}}^{2x{\beta _n}}} + \\ &\qquad {\text{ }}\sum\limits_{j = 1}^4 {\sum\limits_{i = 1}^4 {\frac{{{c_i}{c_j}Eh{{\bar \mu }_{in}}\left( {1 + 2\mu } \right)\left( {{{\bar \mu }_{jn}} - {{\bar \mu }_{in}}} \right)}}{{64{\beta _n}{{\left( {{{\bar \mu }_{in}} + {{\bar \mu }_{jn}} - 2{\beta _n}} \right)}^2}}}. } } \\ &\qquad {\text{ }}\left[ {\left( {{{\bar \mu }_{in}} + {{\bar \mu }_{jn}}} \right)\left( {1 + 2{\beta _n}} \right) - 4{\beta _n}\left( {1 + {\beta _n}} \right)} \right]{{\text{e}}^{x\left( {{{\bar \mu }_{in}} + {{\bar \mu }_{jn}} - 2{\beta _n}} \right)}} \end{split} $$ (51) $$ \begin{split} & {X_{2n}}\left( x \right) = \left( {{C_{2n}} + x{C_{4n}}} \right){{\text{e}}^{ - 2x{\beta _n}}} + \\ &\qquad {\text{ }}\sum\limits_{j = 1}^4 {\sum\limits_{i = 1}^4 {\frac{{{c_i}{c_j}Eh{{\bar \mu }_{jn}}\left( {1 + 2\mu } \right)}}{{32{{\left( {2{\beta _n} + {{\bar \mu }_{in}} + {{\bar \mu }_{jn}}} \right)}^2}}}} } . \\ &\qquad {\text{ }}\left( {{{\bar \mu }_{jn}} - {{\bar \mu }_{in}}} \right)\left( {2{\beta _n} + {{\bar \mu }_{in}} + {{\bar \mu }_{jn}} - 1} \right){{\text{e}}^{x\left( {2{\beta _n} + {{\bar \mu }_{in}} + {{\bar \mu }_{jn}}} \right)}} \end{split} $$ (52) $$ \begin{split} & {X_{3n}}\left( x \right) = {C_{3n}}{{\text{e}}^{2x{\beta _n}}} + \sum\limits_{j = 1}^4 {\sum\limits_{i = 1}^4 {\frac{{{c_i}{c_j}Eh{{\bar \mu }_{jn}}\left( {1 + 2\mu } \right)}}{{32\left( {{{\bar \mu }_{in}} + {{\bar \mu }_{jn}} - 2{\beta _n}} \right)}}} } . \\ &\qquad {\text{ }}\left( {{{\bar \mu }_{in}} - {{\bar \mu }_{jn}}} \right){{\text{e}}^{x\left( {{{\bar \mu }_{in}} + {{\bar \mu }_{jn}} - 2{\beta _n}} \right)}} \end{split} $$ (53) $$ \begin{split} & {X_{4n}}\left( x \right) = {C_{4n}}{{\text{e}}^{ - 2x{\beta _n}}} + \sum\limits_{j = 1}^4 {\sum\limits_{i = 1}^4 {\frac{{{c_i}{c_j}Eh{{\bar \mu }_{jn}}\left( {1 + 2\mu } \right)}}{{32\left( {2{\beta _n} + {{\bar \mu }_{in}} + {{\bar \mu }_{jn}}} \right)}}} } . \\ &\qquad {\text{ }}\left( {{{\bar \mu }_{in}} - {{\bar \mu }_{jn}}} \right){{\text{e}}^{x\left( {2{\beta _n} + {{\bar \mu }_{in}} + {{\bar \mu }_{jn}}} \right)}} \end{split} $$ (54) 式中待定系数需通过边界条件确定, 由式(42)、式(43)及式(47) ~ 式(54)可得2阶方程的解答为
$$ \begin{split} & {F_2} = {C_{00}} + {C_{10}}x + \frac{{{C_{20}}}}{2}{x^2} + \frac{{{C_{30}}}}{6}{x^3} + \\ &\qquad \sum\limits_{j = 1}^4 {\sum\limits_{i = 1,i \ne j - {{\left( { - 1} \right)}^j}}^4 {\frac{{{c_i}{c_j}Eh\left( {1 + 2\mu } \right)\beta _n^2{{\bar \mu }_{jn}}{{\text{e}}^{x\left( {{{\bar \mu }_{in}} + {{\bar \mu }_{jn}}} \right)}}}}{{2{{\left( {{{\bar \mu }_{in}} + {{\bar \mu }_{jn}}} \right)}^3}}}} } + \\ &\qquad \cos \left( {2{\beta _n}y} \right)\Biggr\{ {\left[ {{C_n} + \left( {1 + x} \right){C_{3n}}} \right]{{\text{e}}^{2x{\beta _n}}}} + \\ &\qquad \frac{{\left\{ {{C_{4n}} + 2{\beta _n}\left[ {{C_{2n}} + \left( {1 + x} \right){C_{4n}}} \right]} \right\}{{\text{e}}^{ - 2x{\beta _n}}}}}{{2{\beta _n}}} + \\ &\qquad {\frac{{Eh\left( {1 + 2\mu } \right){{\bar \mu }_{jn}}\left( {{{\bar \mu }_{in}} - {{\bar \mu }_{jn}}} \right)\beta _n^2}}{{2{{\left[ {{{\left( {{{\bar \mu }_{in}} + {{\bar \mu }_{jn}}} \right)}^2} - 4\beta _n^2} \right]}^2}}}{{\text{e}}^{x\left( {{{\bar \mu }_{in}} + {{\bar \mu }_{jn}}} \right)}}} \Biggr\} \end{split} $$ (55) 具体地, 对各子问题用于求解2阶解的边界条件为
$$ {\left. {\frac{{{\partial ^2}F}}{{\partial x\partial y}}} \right|_{x = 0,a}} = {\left. {\frac{{{\partial ^3}F}}{{\partial {x^3}}}} \right|_{x = 0,a}} = 0 $$ (56) 将各阶方程的解代入式(25), 可得挠度及应力函数的最终解答. 令线性屈曲的最大挠度为${w^*}\left( {{x_0},{y_0}} \right) = h$. 由于分叉屈曲明显是对称的, 因此$ {a_e} = 0 $, 对矩形板面内区域$\varOmega $进行积分, 可得后屈曲展开系数$ {b_e} $[43]
$$ {b_e} = \frac{{\displaystyle\iint {\left[ {\left( {\frac{{{\partial ^2}{F_2}}}{{\partial {y^2}}}} \right){{\left( {\frac{{\partial {w_1}}}{{\partial x}}} \right)}^2} + \left( {\frac{{{\partial ^2}{F_2}}}{{\partial {x^2}}}} \right){{\left( {\frac{{\partial {w_1}}}{{\partial y}}} \right)}^2}} \right]{\text{d}}\varOmega }}}{{\displaystyle\iint {{{\bar P}_x}{{\left( {\frac{{\partial {w_1}}}{{\partial y}}} \right)}^2}{\text{d}}\varOmega }}} $$ (57) 因此, 载荷可表示为
$$ \frac{{{{\bar P}_x}}}{{{N_{{\text{cr}}}}}} = 1 + {b_e}{\xi ^2} $$ (58) 改变摄动小参数$\xi $的值即可获得相应的变形及载荷, 从而获得后屈曲平衡路径.
2.2 后屈曲数值结果与讨论
本节展示无拉力弹性地基上无缺陷及有缺陷矩形板的后屈曲平衡路径. 以弯曲刚度$D = 1$、长宽比a/b = 1.5和厚度h = 0.01的简支矩形板作为算例, 分别展示不同弹性地基刚度下的后屈曲平衡路径, 并与有限元计算结果进行对比. 有限元模型基于Riks算法, 采用S4单元进行模拟, 单元尺寸为宽度的1/100, 加载边采用耦合约束; 同时改变网格节点位置引入一阶线性屈曲模态缺陷, 缺陷幅值$\lambda $为0.05及0.1倍板的厚度. 无拉力弹性地基则通过施加如图7所表征的非线性弹簧单元实现[12], 其一端固定、另一端与矩形板连接, 在板内施加密集的连接单元从而模拟无拉力弹性地基.
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图 7 无拉力弹性地基的力−位移曲线示意图Figure 7. Schematic diagram of the force–displacement relation of a tensionless elastic foundation如图8所示, 简支矩形板的后屈曲平衡路径与有限元模拟结果在小变形阶段(即${{{w^*}} \mathord{\left/ {\vphantom {{{w^*}} h}} \right. } h} < 1$时)吻合良好. 随着变形逐渐增大, 有限元与本文方法之间的差异有所增加, 但总体误差相对较小, 主要原因在于小参数摄动法仅适用于弱非线性问题, 而随着变形的逐渐增大, 问题的非线性程度增加, 从而导致了一定误差, 但两者之间整体趋势保持一致, 验证了本文方法在后屈曲分析时的有效性.
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图 8 无缺陷和含缺陷简支板的后屈曲平衡路径 (续)Figure 8. The post-buckling equilibrium paths of perfect and imperfect simply supported plates (continued)3. 结论
在无拉力弹性地基的作用下, 矩形板的屈曲/后屈曲行为受到显著影响. 本文首先求解了无拉力弹性地基上矩形板的屈曲问题: 针对接触非线性难题, 将原问题按照接触状态进行分区, 划分为若干Winkler地基板与无地基板的子问题, 利用连续条件获得屈曲载荷及屈曲模态, 进而对子问题重新划分, 迭代计算屈曲载荷及屈曲模态, 直至结果收敛. 进一步, 基于Koiter后屈曲摄动展开理论与辛方法, 本文求解了无拉力弹性地基板的后屈曲问题. 对于屈曲和后屈曲问题, 均通过算例对比, 证明了本文求解的准确性, 并得到以下结论: (1) 随着弹性地基刚度的增大, 临界屈曲模态趋向于呈现更多半波数, 接触区域板的变形逐渐减小; (2) 在弹性地基刚度较小时, 屈曲模态呈现出对称性特点, 但当系数增大后, 转变为非对称的屈曲模态; (3) 随着板的长宽比增大, 半波数呈现出增多的趋势. 本文方法具有理论推导严格、计算效率高及无需进行复杂建模等优点, 可望推广至更多复杂屈曲/后屈曲问题的求解.
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图 1 无拉力弹性地基板示意图
Figure 1. Schematic diagram of a plate on a tensionless elastic foundation
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图 2 无拉力弹性地基板屈曲问题求解流程图
Figure 2. Solution flowchart for the buckling of a rectangular plate on a tensionless elastic foundation
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图 4 屈曲模态迭代的收敛性验证(kw = 2000, a/b = 2)
Figure 4. Iterative convergence verification of buckling mode shapes (kw = 2000, a/b = 2)
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图 5 无拉力弹性地基上矩形板的临界屈曲模态(kw = 2000, a/b = 2)
Figure 5. Critical buckling mode shape of a rectangular plate on a tensionless elastic foundation (kw = 2000, a/b = 2)
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图 6 不同长宽比矩形板的临界屈曲模态
Figure 6. Critical buckling mode shape of rectangular plate with different aspect ratio
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图 7 无拉力弹性地基的力−位移曲线示意图
Figure 7. Schematic diagram of the force–displacement relation of a tensionless elastic foundation
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图 8 无缺陷和含缺陷简支板的后屈曲平衡路径 (续)
Figure 8. The post-buckling equilibrium paths of perfect and imperfect simply supported plates (continued)
表 1 不同长宽比及弹性地基刚度下矩形板的临界屈曲载荷$ - {{{N_{{\text{cr}}}}{b^2}} \mathord{\left/ {\vphantom {{{N_{{\text{cr}}}}{b^2}} D}} \right. } D}$
Table 1 Critical buckling load factors, $ - {{{N_{{\text{cr}}}}{b^2}} \mathord{\left/ {\vphantom {{{N_{{\text{cr}}}}{b^2}} D}} \right. } D}$, of rectangular plates with different aspect ratios and elastic foundation stiffnesses
${k_w}$ Method a/b = 1.5 a/b = 2 a/b = 2.5 a/b = 3 1 FEM 44.228 39.502 44.046 44.105 present 44.258 39.529 44.075 44.027 10 FEM 43.038 39.923 42.683 42.265 present 43.067 39.949 42.711 42.292 100 FEM 43.687 42.172 42.119 41.436 present 43.716 42.199 42.146 41.462 200 FEM 43.038 42.998 40.998 39.770 present 43.068 43.026 41.023 39.794 2000 FEM 42.835 44.134 40.804 39.488 present 42.865 44.163 40.829 39.512 
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