力学学报, 2020, 52(1): 73-81 DOI: 10.6052/0459-1879-19-283

固体力学

应用边界积分法求圆形夹杂问题的解析解 1)

郭树起,2)

石家庄铁道大学省部共建交通工程结构力学行为与系统安全国家重点实验室, 石家庄 050043;石家庄铁道大学工程力学系, 石家庄 050043

EXACT SOLUTION OF CIRCULAR INCLUSION PROBLEMS BY A BOUNDARY INTEGRAL METHOD 1)

Guo Shuqi,2)

State Key Laboratory of Mechanical Behavior and System Safety of Traffic Engineering Structures, Shijiazhuang Tiedao University, Shijiazhuang 050043, China; Engineering Mechanics Department, Shijiazhuang Tiedao University, Shijiazhuang 050043, China

通讯作者: 2) 郭树起, 教授, 主要研究方向: 弹性动力学, 振动力学. E-mail:guoshuqi@stdu.edu.cn

收稿日期: 2019-10-14   接受日期: 2019-12-24   网络出版日期: 2020-01-18

基金资助: 1) 国家自然科学基金资助项目.  11272219

Received: 2019-10-14   Accepted: 2019-12-24   Online: 2020-01-18

作者简介 About authors

摘要

边界元方法作为一种数值方法, 在各种科学工程问题中得到了广泛的应用.本文参考了边界元法的求解思路, 从Somigliana等式出发, 利用格林函数性质,得到了一种边界积分法, 使之可以用来寻求弹性问题的解析解.此边界积分法也可以从Betti互易定理得到. 应用此新方法, 求解了圆形夹杂问题.首先设定夹杂与基体之间完美连接, 将界面处的位移与应力按照傅里叶级数展开,根据问题的对称性与三角函数的正交性来简化假设, 减少待定系数的个数.其次选择合适的试函数(试函数满足位移单值条件以及无体力的线弹性力学问题的控制方程),应用边界积分法, 求得界面处的位移与应力的值. 然后再求解域内位移与应力.得到了问题的精确解析解, 当夹杂弹性模量为零或趋向于无穷大时,退化为圆孔或刚性夹杂问题的解析解. 求解过程表明,若问题的求解区域包含无穷远处时, 所取的试函数应满足无穷远处的边界条件.若求解区域包含坐标原点, 试函数在原点处位移与应力应是有限的.结果表明了此方法的有效性.

关键词: 边界积分法 ; 边界元 ; 圆形夹杂 ; Somigliana等式 ; Betti互易定理

Abstract

As an excellent numerical method, boundary element method (BEM) has been widely applied in various scientific and engineering problems. In this paper, a new boundary integral method is obtained based on Somigliana's equation and the properties of Green's function by referring to the idea of boundary element method. It can be used to find the analytic solution of linear elastic problems. The boundary integral method can also be obtained from Betti's reciprocity theorem. By using this new method, the classical problem of elastic circular inclusion under a uniform tensile field at infinity is solved. Firstly, the perfect bonding between inclusion and matrix is set up, and the displacement and stress at interface are expanded according to Fourier series. According to the symmetry of the problem and the orthogonality of trigonometric function, the hypothesis is simplified and the number of undetermined coefficients is reduced. Secondly, the appropriate trial functions are selected (these trial functions satisfy the condition of displacement single value and the control equation of linear elasticity without body force). And the boundary integral method is used to calculate the displacement and stress at the interface. Then the displacement and stress in the domain are solved using similar tricks. The exact analytical solution of the problem is obtained, which is exactly the same with the results in literatures. When the elastic modulus of the inclusion is zero or tends to infinity, it degenerates to the analytical solution of the problem of circular hole or rigid inclusion. The solution process shows that if the problem has boundary conditions at infinity, trial functions should meet the boundary condition at infinity. If the domain of the problem contains the coordinate origin, the displacement and stress of trial functions at the origin should be limited. The results show that the method is effective.

Keywords: boundary integral method ; boundary element method ; elastic circular inclusion ; Somigliana equation ; Betti's reciprocal theorem

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本文引用格式

郭树起. 应用边界积分法求圆形夹杂问题的解析解 1). 力学学报[J], 2020, 52(1): 73-81 DOI:10.6052/0459-1879-19-283

Guo Shuqi. EXACT SOLUTION OF CIRCULAR INCLUSION PROBLEMS BY A BOUNDARY INTEGRAL METHOD 1). Chinese Journal of Theoretical and Applied Mechanics[J], 2020, 52(1): 73-81 DOI:10.6052/0459-1879-19-283

引言

边界元方法[1-6]作为一种非常成功的数值方法, 在力、热、电、磁等多个物理分支都得到了广泛的应用[7-16]. 对于难于求得解析解的问题, 可以采用边界元法进行数值求解[17-19]. 边界元法通常导致边界积分方程, 也称边界积分方程方法.

本文则稍稍变化一下边界元法的求解思路, 从Somigliana等式开始推导, 利用了线弹性算子零空间的不平凡性, 即格林函数的不唯一性, 导出了不同于边界积分方程方法的另一种方法, 使线弹性问题在求解解析解时提供了一个新的选择. 首先给出此边界积分法的基本理论与基本步骤. 然后为检验此法的有效性, 求解了一个典型的弹性问题, 圆形夹杂问题.

1 边界积分法

可以导出边界元法的边界积分方程的Somigl- iana等式为

$ \begin{eqnarray} \label{eq1} &&u_i(P)=\int_{\varOmega} u_{ij}^\ast (P,Q)f_j (Q){\rm d}\varOmega (Q) +\int_s \big[u_{ij}^\ast (P,q)t_j (q)-\\&& t_{ij}^\ast (P,q)u_j (q) \big]{\rm d}S(q) \end{eqnarray} $

其中, $u_i(P)$是在$P$点的位移, $u_{ij}^\ast (P,Q)$是在任意一点$P$沿$x_i$方向作用单位集中力时, 在任意一点$Q$处引起的$x_j$方向的位移分量, $f_j $是体力, $u_j (q)$是边界位移, $t_j (q)$是边界面力, $t_{ij}^\ast(P,q)$是相应于边界位移$u_{ij}^\ast (P,q)$的边界面力. $u_j^\ast $是在$P$处沿着任意方向$n_m $的单位集中力作用下, 在$Q$处的引起的位移. $u_j^\ast $与$u_{ij}^\ast $的关系为

$\begin{eqnarray} \label{eq2} u_j^\ast =u_{mj}^\ast n_m \end{eqnarray} $

满足方程

$ \begin{eqnarray} \mathcal{L}_{ik} u_{mk}^\ast =-\delta (P,Q)\delta _{im} \end{eqnarray} $

将式(3a)等号两边同时乘以单位方向$n_m $, 得$\mathcal{L}_{ik} u_{mk}^\ast n_m \mbox{=}-\delta \left( {P,Q} \right)\delta _{im} n_m $即$u_j^\ast $满足方程

$ \begin{eqnarray} \mathcal{L}_{ik} u_k^\ast =-\delta (P,Q)n_i \end{eqnarray} $

其中$\mathcal{L}_{ik} $是线弹性算子

$ \begin{eqnarray} \label{eq4} \mathcal{L}_{ik} =C_{ijkl} \frac{\partial ^2}{\partial x_l \partial x_j } \end{eqnarray} $

其中, $C_{ijkl} $是线弹性常数. 线弹性算子$\mathcal{L}_{ik} $与其共轭算子$\mathcal{L}_{ik}^\ast$相等. 因此式(3a)和式(3b)可以重写为

$ \begin{eqnarray*} &&C_{ijkl} \frac{\partial ^2}{\partial x_l \partial x_j }u_{mk}^\ast =-\delta \left( {P,Q} \right)\delta _{im}\\ &&C_{ijkl} \frac{\partial ^2}{\partial x_l \partial x_j }u_k^\ast =-\delta \left( {P,Q} \right)n_i \end{eqnarray*} $

对于各向同性弹性体, 则进一步化为

$ \begin{eqnarray} \mathcal{L}_{ik} u_{mk}^\ast =(\lambda +G)u_{mj,ji}^\ast +\mu u_{mi,jj}^\ast =-\delta (P,Q)\delta _{im} \end{eqnarray} $

$ \begin{eqnarray} \mathcal{L}_{ik} u_k^\ast =(\lambda +G)u_{j,ji}^\ast +\mu u_{i,jj}^\ast =-\delta (P,Q)n_i \end{eqnarray} $

其中, $\lambda $和$G$是Lamé常数. $u_j^\ast$, $u_{mj}^\ast $称格林函数[20].

由于线弹性算子$\mathcal{L}_{ik} $的零空间是非平凡的, 因此格林函数不唯一. 利用此不唯一性, 记两个任意的格林函数为$u_{mj}^{\ast 1}$, $u_{mj}^{\ast 2}$, 均满足方程(3a), 将其代入到Somigliana等式(1)中[4,20],将得到的两个方程相减得到

$ \begin{eqnarray} \label{eq6} \int_\varOmega w_{ij}^\ast f_j {\rm d}\varOmega +\int_S \left(w_{ij}^\ast t_j-t_{ij}^\ast u_j\right){\rm d}S =0 \end{eqnarray} $

或者

$ \begin{eqnarray} \label{eq7} \int_\varOmega w_j^\ast f_j {\rm d}\varOmega +\int_S \left( w_j^\ast t_j -t_j^\ast u_j \right){\rm d}S =0 \end{eqnarray} $

其中, $w_{ij}^\ast =u_{ij}^{\ast 1} -u_{ij}^{\ast 2} $, $t_{ij}^\ast$为相应于$w_{ij}^\ast $的在边界上的面力; $w_j^\ast =u_j^{\ast 1} -u_j^{\ast 2} =u_{ij}^{\ast 1} n_i -u_{ij}^{\ast 2} n_i $, $t_j^\ast $为相应于$w_j^\ast$的在边界上的面力. 其中$w_j^\ast $满足齐次方程

$ \begin{eqnarray} \label{eq8} \mathcal{L}_{ik} w_k^\ast =C_{ijkl} \frac{\partial ^2}{\partial x_l \partial x_j }w_k^\ast =0 \end{eqnarray} $

各向同性线弹性体则满足

$ \begin{eqnarray} \label{eq9} \mathcal{L}_{ik} w_k^\ast =(\lambda +G)w_{j,ji}^\ast +\mu w_{i,jj}^\ast =0 \end{eqnarray} $

方程(7)与式(10)所示的Betti互易定理[21-22]形式上类似.

$ \begin{eqnarray} \label{eq10} &&\int_\varOmega \left( u_j^{(2)} f_j^{(1)}-u_j^{(1)} f_j^{(2)} \right){\rm d}\varOmega +\\&&\int_S \left(u_j^{(2)}t_j^{(1)} -u_j^{(1)} t_j^{(2)} \right){\rm d}S =0 \end{eqnarray} $

其中, $(\ )^{(1)}$和$(\ )^{(2)}$分别表示同一物体的线弹性边值问题I, II的位移$u_j $, 面力$t_j $与体力$f_j $. 但$w_j^\ast $并不是某个力学边值问题的解, 而是需要满足无体力的齐次方程(8)或(9). Somigliana等式可以从Betti互易定理推导而来[4], 即把格林函数定义式(5)代入式(10). 式(7)也可以由Betti互易定理得到. 在式(10)中令$f_i^{(2)} =0$, 所得方程即与式(7)含义相同, 只是记号不同. 将$f_i^{(2)} =0$代入应力平衡方程$\sigma_{ji,j}^{(2)} +f_i^{(2)} =0$中. 即得无体力的应力平衡方程, $\sigma _{ji,j}^{(2)} =0$. 在式(8)和式(9)中代入本构方程与几何方程, 同样得无体力的应力平衡方程.

无体力时, 式(7)变为式程(11)

$ \begin{eqnarray} \label{eq11} \int_S \left(w_j^\ast t_j -t_j^\ast u_j \right){\rm d}S =0 \end{eqnarray} $

方程(11)是本文计算的主要理论依据. 为简单起见, 下面本文研究无体力的平面弹性力学问题. 这时边界面$S$进一步化为边界曲线$L$

$ \begin{eqnarray} \label{eq12} \int_L \left( w_j^\ast t_j -t_j^\ast u_j \right){\rm d}s =0 \end{eqnarray} $

利用式(12)求得未知的边界位移与面力. 对于无体力平面问题来说, 位移函数$w_j^\ast$可以采用相应于Michell解的位移函数[23-25]. 一般来说, 边界位移与面力假设中一般包含不止一个待定常数, 因此需要符合题意的多个位移函数$w_j^\ast$以及面力函数$t_j^\ast $. 这样的函数在此称为{\bf 试函数}. 边界位移与边界面力全部已知后, 边界元法用Somigliana等式计算域内位移. 本文则继续用式(12)求域内位移, 避免使用格林函数带来奇异积分问题. 基本步骤是作一条虚拟边界$L^\ast$经过需要计算位移与应力的点, 即式(12)变为

$ \begin{eqnarray} \label{eq13} \int_L \left( w_j^\ast t_j -t_j^\ast u_j \right){\rm d}s+ \int_{L^\ast}\left( w_j^\ast t_j -t_j^\ast u_j \right){\rm d}s =0 \end{eqnarray} $

这类似于矩阵传递法. 因此方程(13)预期可以用于多相介质问题的分析. 表1简叙述了边界元法与此新方法的主要思想与各自特点.

表 1   边界元法与本文方法比较

Table 1  Comparison between BEM and the proposed method

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2 圆形夹杂问题

圆形夹杂问题是弹性力学的一个经典问题, 通常采用Airy应力函数法与复变函数法求解[26-30]. 理论结果被实验所验证[31]. 在此用边界积分法求其解析解. 如图1所示, 无限大平板上有半径$r=a$的圆形夹杂, 夹杂内的弹性常数与基体不同.基体的弹性常数用$E$, $\nu $, $\mu $分别来表示杨氏模量、泊松比、剪切模量, 夹杂内的弹性常数用下标1以示区别, 如$E_1$, $\nu _1$, $\mu _1 $. 夹杂所在区域为$\varOmega^{\rm f}$, 其他不含夹杂的区域为$\varOmega $.

图 1

图 1   圆形夹杂示意图

Fig. 1   Schematic diagram of a circular inclusion


2.1 界面位移与面力

界面处认为是完美连接的, 位移、应力连续

$ \begin{eqnarray} \label{eq16} r=a:\left\{ {\begin{array}{l} \left( {u_r ,u_\theta } \right)=\left( {u_r^{\rm f} ,u_\theta ^{\rm f} } \right) \\ \left( {\sigma _r ,\sigma _\theta ,\sigma _{r\theta } } \right)=\left( {\sigma _r^{\rm f} ,\sigma _\theta ^{\rm f} ,\sigma _{r\theta }^{\rm f} }\right) \\ \end{array}} \right. \end{eqnarray}$

其中, 上标$(\ )^{\rm f}$表示与夹杂有关的量. 平板在无穷远处有均布应力$\sigma _x =q$, 其他应力分量为零. 相应的极坐标下位移$u_r^0$和$u_\theta ^0 $如式(15a)所示; 径向应力$\sigma _r^0 $、周向应力$\sigma _\theta ^0 $与切应力$\sigma _{r\theta }^0$如式(15b)所示

$ \begin{eqnarray} && \left. \begin{array}{l} u_r^0 =\dfrac{qr}{E}\left( {\dfrac{1-\nu }{2}+\dfrac{1+\nu }{2}\cos 2\theta }\right) \\ u_\theta ^0 =-\dfrac{qr}{E}\dfrac{1+\nu }{2}\sin 2\theta \\ \end{array} \right\} \end{eqnarray} $
$ \begin{eqnarray} \left. \begin{array}{l} \sigma _r^0 =\dfrac{q}{\mbox{2}}\left( {1+\cos 2\theta } \right) \\ \sigma _\theta ^0 =\dfrac{q}{\mbox{2}}\left( {1-\cos 2\theta }\right) \\ \sigma _{r\theta }^0 =-\dfrac{q}{2}\sin 2\theta \\ \end{array} \right\} \end{eqnarray} $

先研究基体部分, 即夹杂之外的区域$\varOmega $. 把位移与应力分解为均匀场部分与扰动部分[33], $u_i =u_i +u_i^0 ,\sigma _{ij} =\tau _{ij} +\sigma _{ij}^0 $, 并考虑式(15a)和式(15b), 有

$ \left. \begin{array}{l} u_r =u_r -u_r^0 =u_r -\dfrac{qr}{E}\left( {\dfrac{1-\nu }{2}+\dfrac{1+\nu }{2}\cos 2\theta } \right) \\ u_\theta =u_\theta -u_\theta ^0 =u_\theta +\dfrac{qr}{E}\dfrac{1+\nu }{2}\sin 2\theta \\ \tau _r =\sigma _r -\sigma _r^0 =\sigma _r -\dfrac{q}{2}\left( {1+\cos 2\theta } \right) \\ \tau _{r\theta } =\sigma _{r\theta } -\sigma _{r\theta }^0 =\sigma _{r\theta } +\dfrac{q}{2}\sin 2\theta \end{array} \right\} $

这时, 无穷远处单向均匀应力边界条件化为了位移$u_i $与应力$\tau _{ij} $在无穷远处为零, 即

$ \begin{eqnarray} \label{eq17} u_i =0, \ \ \tau _{ij} =0, \ \ {\rm when}\ \ r\to \infty \end{eqnarray}$

采用极坐标系, 式(12)在本节中改为

$ \begin{eqnarray} \label{eq18} \int_L \left( u_r t_r^\ast +u_\theta t_\theta ^\ast \right){\rm d}s =\int_L\left( w_r^\ast t_r +w_\theta ^\ast t_\theta \right){\rm d}s \end{eqnarray}$

考虑边界$L$即为$r=a$, 以及${\rm d}s=r{\rm d}\theta =a{\rm d}\theta $, 并将式(16)代入式(18)得到

$ \begin{eqnarray} \label{eq19} &&\int_0^{2\pi} \left\{ {\left[ {u_r -\dfrac{qr}{E}\left( {\dfrac{1-\nu }{2}+\dfrac{1+\nu }{2}\cos 2\theta } \right)} \right]t_r^\ast } \right\}_{r=a} {\rm d}\theta + \\&& \int_0^{2\pi} \left[ {\left( {u_\theta +\dfrac{qr}{E}\dfrac{1+\nu }{2}\sin 2\theta } \right)t_\theta ^\ast } \right]_{r=a} {\rm d}\theta + \\&& \int_0^{2\pi} \left\{ {w_r^\ast \left[ {\dfrac{q}{\mbox{2}}\left( {1+\cos 2\theta } \right)-\sigma _r } \right]} \right\}_{r=a} {\rm d}\theta + \\&& \int_0^{2\pi} \left[ {w_\theta ^\ast \left( {-\dfrac{q}{2}\sin 2\theta -\sigma _{r\theta } } \right)} \right]_{r=a} {\rm d}\theta =0 \end{eqnarray}$

其中, $w_r^\ast $, $w_\theta ^\ast $采用Michell解的位移部分[24]. 由于Michell的位移解以Fourier级数形式出现, $w=\sum\limits_{n=0}^\infty {\left[ {a_n \left( r \right)\cos n\theta +b_n \left( r \right)\sin n\theta } \right]} $. 根据三角级数的正交性, 式(19)中位移$w_r^\ast $, $w_\theta ^\ast$只能出现常数项与$n=2$的项. 所以$t_r $, $t_\theta $, $u_r $, $u_\theta$也只能出现$n=0$与$n=2$的项. 再根据对称性, $r$为常数时, $u_r \left( \theta \right)=u_r\left( {-\theta } \right)$, $u_r \left( \theta \right)=u_r \left( {\pi -\theta } \right)$, $u_\theta \left( {-\theta } \right)=-u_\theta \left(\theta \right)$, $u_\theta \left( \theta \right)= -u_\theta \left({\pi -\theta } \right)$. 因此位移$u_r $, $u_\theta $与面力$t_r $, $t_\theta $ 在$r=a$的位移设为

$ \begin{eqnarray} \label{eq20} \left. {\begin{array}{l} \left( {u_r } \right)_{r=a} =a_0 +a_2 \cos 2\theta \\ \left( {u_\theta } \right)_{r=a} =b_2 \sin 2\theta \\ t_r =\left( {\sigma _r } \right)_{r=a} =c_0 +c_2 \cos 2\theta \\ t_\theta =\left( {\sigma _{r\theta } } \right)_{r=a} =d_2 \sin 2\theta \\ \end{array}} \right\} \end{eqnarray}$

其中, $a_0 $, $a_2 $, $b_2 $, $c_0 $, $c_2 $, $d_2 $是待定常数.

所选位移试函数$w_r^\ast $, $w_\theta ^\ast $与相应的应力函数$\varphi $ 如表2所示. 函数$w_r^\ast $, $w_\theta ^\ast $, $t_r^\ast $, $t_\theta ^\ast$需要满足无穷远处趋于零的边界条件, 另外还需满足位移单值条件. 表2的$\varphi$相应于位移$w_r^\ast $、$w_\theta ^\ast $的Airy应力函数[31, 34], 其中, $\mu =E/(2+2\nu )$. 平面应力, $\kappa =(3-\nu )/(1+\nu)$; 平面应变$\kappa =3-4\nu$, 相应的边界上的面力为

$ \begin{eqnarray} \label{eq21} \left. {\begin{array}{l} t_r^\ast =(\sigma _r )_{r=a} =\left(\dfrac{1}{r}\dfrac{\partial \varphi }{\partial r}\right)_{r=a} \\ t_\theta ^\ast =(\sigma _{r\theta } )_{r=a} =-\left( {\dfrac{\partial ^2\varphi }{r\partial r\partial \theta }} \right)_{r=a} \end{array}} \right\} \end{eqnarray}$

表 2   位移$w_r^\ast $和$w_\theta ^\ast $与应力函数$\varphi $

Table 2  Displacements $w_r^\ast$, $w_\theta ^\ast $ and stress function $\varphi $

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这样在夹杂与基体的交界处$r=a$处的位移$u$与面力$t$为

$ \begin{eqnarray} \label{eq22} &&\left. {\begin{array}{l} \left( {u_r } \right)_{r=a} =a_0 +a_2 \cos 2\theta -\dfrac{qa}{E}\left( {\dfrac{1-\nu }{2}+\dfrac{1+\nu }{2}\cos 2\theta } \right) \\ \left( {u_\theta } \right)_{r=a} =b_2 \sin 2\theta +\dfrac{qa}{E}\dfrac{1+\nu }{2}\sin 2\theta \\ t_r =c_0 +c_2 \cos 2\theta -\dfrac{q}{2}\left( {1+\cos 2\theta } \right) \\ t_\theta =d_2 \sin 2\theta +\dfrac{q}{2}\sin 2\theta \\ \end{array}} \right\}\\&& \end{eqnarray}$

将表达式(22)代入式(18), 得到式(23). 然后将阶数$n=0$与$n=2$阶的位移试函数与面力$w_r^\ast $, $w_\theta ^\ast$, $t_r^\ast $, $t_\theta ^\ast $代入, 得到关于待定系数方程式(24).

$ \begin{eqnarray} \label{eq23} &&\!\!\!\int_0^{2\pi} {\left[ {a_0 -\dfrac{qa}{E}\dfrac{1-\nu }{2}+\left( {a_2 -\dfrac{qa}{E}\dfrac{1+\nu }{2}\cos 2\theta } \right)\left( {t_r^\ast } \right)_{r=a} } \right]{\rm d}\theta } + \\ &&\!\!\!\qquad \int_0^{2\pi} {\left( {b_2 +\dfrac{qa}{E}\dfrac{1+\nu }{2}} \right)\sin 2\theta \left( {t_\theta ^\ast } \right)_{r=a} {\rm d}\theta } + \\ &&\!\!\!\qquad \int_0^{2\pi} {\left[ {\dfrac{q}{\mbox{2}}\left( {1+\cos 2\theta } \right)-c_0 -c_2 \cos 2\theta } \right]\left( {w_r^\ast } \right)_{r=a} {\rm d}\theta }- \\ &&\!\!\!\qquad \int_0^{2\pi} {\left( {\dfrac{q}{2}+d_2 } \right)\sin 2\theta \left( {w_\theta ^\ast } \right)_{r=a} {\rm d}\theta } =0 \end{eqnarray}$
$ \begin{eqnarray} \label{eq24} \left. {\begin{array}{l} Ea_0 +a\left( {1+\nu } \right)c_0 -aq=0 \\ 2a_2 +b_2 +\left({\nu d_2 -2q+2c_2 -d_2 } \right)\dfrac{a}{E}=0 \\ a\left(1+\nu\right)\left(c_2 +d_2\right)+3E\left( {a_2+b_2} \right)=0 \\ \end{array}} \right\} \end{eqnarray}$

从式(24)解出待定常数$a_0 $, $a_2 $, $b_2 $

$ \begin{eqnarray} \label{eq25} \left. {\begin{array}{l} a_0 =-\dfrac{a\left( {1+\nu } \right)c_0 }{E}+\dfrac{qa}{E} \\ a_2 =\dfrac{a}{3E} \left( {\nu -5} \right)c_2 +\dfrac{a}{3E}\left( {4-2\nu } \right)d_2 +\dfrac{2qa}{E} \\ b_2 =-\dfrac{a}{3E}\left( {2\nu -4} \right)c_2 -\dfrac{a}{3E}\left( {5-\nu } \right)d_2 -\dfrac{2qa}{E} \\ \end{array}} \right\} \end{eqnarray}$

若$c_0 =c_2 =d_2 =0$, 得$a_0 ={qa}/{E}$, $a_2 ={2qa}/{E}$, $b_2=-{2qa}/{E}$, 将其代入式(20), 即得夹杂退化为圆孔时, 圆孔边界的位移、面力

$ \begin{eqnarray} \label{eq26} \left.\begin{array}{l} \left( {u_r } \right)_{r=a} =\dfrac{qa}{E}+\dfrac{2qa}{E}\cos 2\theta \\ \left( {u_\theta } \right)_{r=a} =-\dfrac{2qa}{E}\sin 2\theta \\ t_r =\left( {\sigma _r } \right)_{r=a} =0 \\ t_\theta =\left( {\sigma _{r\theta } } \right)_{r=a} =0 \\ \end{array} \right\} \end{eqnarray}$

式(24)中只有3个方程, 少于式(20)中待定系数的个数. 为此需要单独考虑圆形夹杂的情况. 研究$r\leqslant a$的夹杂区域$\varOmega ^{\rm f}$, 由于此时所考虑区域不含无穷远处, 因此不再将位移、面力分解为均匀场部分与扰动部分, 此时边界积分方程(18)更改为

$ \begin{eqnarray} \label{eq27} \int_{r=a} {\left( {u_r t_r^\ast +u_\theta t_\theta ^\ast } \right){\rm d}s} =\int_{r=a} {\left( {w_r^\ast t_r +w_\theta ^\ast t_\theta } \right){\rm d}s} \end{eqnarray}$

边界位移与面力仍如式(20)所示. 圆形夹杂的弹性常数与基体不同, 所取位移$w_r^\ast $, $w_\theta ^\ast $的弹性常数需要相应改变. 另外由于$r\leqslant a$区域包含原点, 所取位移试函数$w_r^\ast $, $w_\theta ^\ast$在原点的位移值以及相应的应力值都应是有限的. 所取位移试函数$w_r^\ast$, $w_\theta ^\ast $如表3所示, 其中$\mu_{1}=E_{1}/(2+2\nu_{1})$. 平面应力, $\kappa_{1}=(3-\nu _{1})/(1+\nu_{1})$; 平面应变$\kappa_{1}=3-4\nu_{1}$. 所得方程为式(28), 解出待定常数$a_0 $, $a_2 $与$b_2 $, 如式(29)所示.

表 3   位移$w_r^\ast $和$w_\theta ^\ast $与应力函数$\varphi $

Table 3  Displacements $w_r^\ast $, $w_\theta ^\ast $ and stress function $\varphi $

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$ \begin{eqnarray} \label{eq28} &&\left.\begin{array}{l} E_1 a_0 +a\left( {\nu _1 -1} \right)c_0 =0 \\ 3b_2 E_1 +2a\nu _1 c_2 -ad_2 \nu _1 -3ad_2 =0 \\ \left( {b_2 -a_2 } \right)E_1 +a\nu _1 c_2 -ad_2 \nu _1 +ac_2 -ad_2 =0 \end{array} \right\} \end{eqnarray}$
$ \begin{eqnarray} \label{eq29} \left. {\begin{array}{l} a_0 =\dfrac{ac_0 }{E_1 }\left( {1-\nu _1 } \right) \\ a_2 =\dfrac{a}{3E_1 }\left( {c_2 \nu _1 -2d_2 \nu _1 +3c_2 } \right) \\ b_2 =-\dfrac{a}{3E_1 }\left( {2c_2 \nu _1 -d_2 \nu _1 -3d_2 } \right) \end{array}} \right\} \end{eqnarray}$

根据位移与应力连续性条件, 两式(25)和式(29)中的$a_0 $, $a_2$, $b_2 $, $c_0$, $c_2 $, $d_2$不仅记号相同、数值也相等. 联立式(24)和式(27)得

$ \begin{eqnarray} \label{eq30} \left.\begin{array}{l} c_0 =-\dfrac{qE_1 }{E\nu _1 -\nu E_1 -E-E_1 } \\ c_2 =\dfrac{2qE_1 }{E\nu _1 -\nu E_1 +E+3E_1 } \\ d_2 =-\dfrac{2qE_1 }{E\nu _1 -\nu E_1 +E+3E_1 } \\ \end{array} \right\} \end{eqnarray}$

由式(30)知道$d_2 =-c_2 $, 代入式(25)和式(29)中可以进一步简化为式(31)和式(32)

$ \begin{eqnarray} \label{eq31} &&\left. {\begin{array}{l} a_0 =-\dfrac{a\left( {1+\nu } \right)c_0 }{E}+\dfrac{qa}{E} \\ a_2 =\dfrac{a}{E} \left( {\nu -3} \right)c_2 +2\dfrac{qa}{E} \\ b_2 =\dfrac{a}{3E}\left( {3-\nu } \right)c_2 -2\dfrac{qa}{E} \\ \end{array}} \right\} \end{eqnarray}$
$ \begin{eqnarray} && \left. \begin{array}{l} a_0 =\dfrac{a\left( {1-\nu _1 } \right)}{E_1 }c_0 \\ a_2 =\dfrac{a\left( {1+\nu _1 } \right)}{E_1 }c_2 \\ b_2 =-\dfrac{a\left( {1+\nu _1 } \right)}{E_1 }c_2 \\ \end{array} \right\} \end{eqnarray}$

由式(30)知道, 当$E_1 =0$, $c_0 =c_2 =d_2 =0$, 界面面力为零, 简化为圆孔解答. 当$E_1 =+\infty $, 表示刚性夹杂, 式(30)化为$c_0 ={q}/({1+\nu })$, $c_2={2q}/({3-\nu })$, $d_2 =-{2q}/({3-\nu })$, 将其代入式(25)中得到$a_0 =a_2=b_2 =0$, 这表示界面位移等于0; 将其代入式(20)得刚性夹杂的$r=a$处的面力

$ \begin{eqnarray} \label{eq33} \left.\begin{array}{l} t_r =\dfrac{q}{1+\nu }+\dfrac{2q}{3-\nu }\cos 2\theta\\ t_\theta =-\dfrac{2q}{3-\nu }\sin 2\theta\\ \end{array}\right\} \end{eqnarray}$

2.2 域内位移与应力

2.2.1 基体位移与应力($r\geqslant a)$

有了位移与应力在界面处的值, 就可以计算域内位移场与应力场. 先研究夹杂之外的区域$\varOmega $, 虚设一条边界$r=\rho >a$. 研究由$r=a$与$r=\rho $围成的环形区域, 所考虑区域不含无穷远处, 因此不再将位移、面力分解为均匀场部分与扰动部分, 此时边界积分方程式(18)变为

$ \begin{eqnarray} \label{eq34} && a\int_0^{2\pi} {\left( {u_r t_r^\ast +u_\theta t_\theta ^\ast } \right)_{r=a} {\rm d}\theta } -\rho \int_0^{2\pi} {\left( {u_r t_r^\ast +u_\theta t_\theta ^\ast } \right)_{r=\rho } {\rm d}\theta } = \\ && a\int_0^{2\pi} {\left( {w_r^\ast t_r +w_\theta ^\ast t_\theta } \right)_{r=a} {\rm d}\theta } -\\&&\rho \int_0^{2\pi} {\left( {w_r^\ast t_r +w_\theta ^\ast t_\theta } \right)_{r=\rho } {\rm d}\theta } \end{eqnarray}$

在$r=a$处, 位移$u_r $, $u_\theta $与面力$t_r $, $t_\theta $如式(20)所示, 其中待定系数如式(25)和式(30)所示. 在$r=\rho $处, 假设位移$u_r $, $u_\theta$与面力$t_r $, $t_\theta $设为式(33), 即类似式(20)的形式

$ \begin{eqnarray} \label{eq35} \left.\begin{array}{l} \left( {u_r } \right)_{r=\rho } =A_0 +A_2 \cos 2\theta \\ \left( {u_\theta } \right)_{r=\rho } =B_2 \sin 2\theta \\ t_r =\left( {\sigma _r } \right)_{r=\rho } =C_0 +C_2 \cos 2\theta \\ t_\theta =\left( {\sigma _{r\theta } } \right)_{r=\rho } =D_2 \sin 2\theta\\ \end{array} \right\} \end{eqnarray}$

将式(20)和式(35)代入式(30), 并选取适当的位移试函数$w_r^\ast $, $w_\theta ^\ast$与面力$t_r^\ast $, $t_\theta ^\ast $. 由于此环形区域既不包含原点也没有无穷远处做边界, 因此只满足位移单值条件即可. 所选取的$w_r^\ast $, $w_\theta ^\ast $、应力函数$\varphi $如表2表3所示, 此时表3中的$\mu _1 $, $\kappa _1 $应替换为$\mu $, $\kappa $. 面力$t_r^\ast $, $t_\theta ^\ast $由式(21)求出或者直接查表得到[33]. 并利用式(25)、$c_2 +d_2 =0$与$a_2 +b_2 =0$, 所得方程如下

\begin{eqnarray*} \left. {\begin{array}{l} -\dfrac{C_0 \left( {1+\nu } \right)}{E}-\dfrac{A_0 }{\rho }+\dfrac{q}{E}=0 \\ -\dfrac{2\rho ^2(\nu -1)}{E}C_0 +\dfrac{2(q-2c_0 )a^2}{E}-2\rho A_0 =0 \\ \dfrac{2C_2 }{E}+\dfrac{2A_2 }{\rho }+\dfrac{B_2 }{\rho }+\dfrac{\nu D_2 -2q-D_2 }{E}=0 \\ \rho (1+\nu )(C_2 +D_2 )+3E(A_2 +B_2 )=0 \\ 12a^4c_2 -6a^4q+\rho ^4(3+\nu )D_2 -2\rho ^4\nu C_2 -3E\rho ^3B_2 =0 \\ \dfrac{\rho ^2\left( {1+\nu } \right)}{E}(D_2 -C_2 )+\rho (A_2 -B_2 )+\dfrac{4a^2}{E}(2c_2 -q)=0 \\ \end{array}} \right\} \end{eqnarray*}

联立此6个方程, 得待定系数$A_0 $, $A_2 $, $B_2 $, $C_0 $, $C_2 $与$D_2 $, 考虑式(30), 并用弹性常数$\kappa $与$\mu $来替换$E$与$\nu $, 用$\kappa _1$与$\mu _1 $ 来替换$E_1 $与$\nu _1 $, 结果如式(36)所示

$ \begin{eqnarray} \label{eq36} \left. {\begin{array}{l} A_0 =\dfrac{q}{2}\left( {-\dfrac{\beta }{2\mu }\dfrac{a^2}{\rho }+\dfrac{1}{4}\dfrac{\kappa -1}{\mu }\rho } \right) \\ A_2 =\dfrac{q}{4}\left( {-\dfrac{\alpha }{\mu }\dfrac{a^4}{\rho ^3}+\dfrac{\alpha \left( {\kappa +1} \right)}{\mu }\dfrac{a^2}{\rho }+\dfrac{\rho }{\mu }} \right) \\ B_2 =\dfrac{q}{4}\left( {-\dfrac{\alpha }{\mu }\dfrac{a^4}{\rho ^3}-\dfrac{\alpha \left( {\kappa -1} \right)}{\mu }\dfrac{a^2}{\rho }-\dfrac{\rho }{\mu }} \right) \\ C_0 =\dfrac{q}{2}\left( {1+\beta \dfrac{a^2}{\rho }} \right) \\ C_2 =\dfrac{q}{2}\left( {1-4\alpha \dfrac{a^2}{\rho ^2}+3\alpha \dfrac{a^4}{\rho ^4}} \right) \\ D_2 =\dfrac{q}{2}\left( {-1-2\alpha \dfrac{a^2}{\rho ^2}+3\alpha \dfrac{a^4}{\rho ^4}} \right) \\ \end{array}} \right\} \end{eqnarray}$

其中, $\alpha $, $\beta $是弹性常数的组合, 如式(37)所示

$ \begin{eqnarray} \label{eq37} \alpha =\dfrac{\mu -\mu _1 }{\kappa \mu _1 +\mu },\ \ \beta =\dfrac{\kappa \mu _1 -\mu \kappa _1 +\mu -\mu _1 }{\mu \kappa _1 -\mu +2\mu _1 } \end{eqnarray}$

将式(36)代入式(35)得到位移场$u_r $, $u_\theta$与除$\sigma _\theta $之外的应力场. $\sigma _\theta $可以由位移$u_r $, $u_\theta $代入几何方程, 再代入本构方程得出. 并将$\rho $仍写为$r$, 结果如式(38) 所示

$ \begin{eqnarray} \label{eq38} \left. {\begin{array}{l} u_r =\dfrac{q}{2}\left( {-\dfrac{\beta }{2\mu }\dfrac{a^2}{r}+\dfrac{1}{4}\dfrac{\kappa -1}{\mu }r} \right)+ \\ \qquad\dfrac{q}{4}\left( {-\dfrac{\alpha }{\mu }\dfrac{a^4}{r^3}+\dfrac{\alpha \left( {\kappa +1} \right)}{\mu }\dfrac{a^2}{r}+\dfrac{r}{\mu }} \right)\cos 2\theta \\ u_\theta =\dfrac{q}{4}\left( {-\dfrac{\alpha }{\mu }\dfrac{a^4}{r^3}-\dfrac{\alpha \left( {\kappa -1} \right)}{\mu }\dfrac{a^2}{r}-\dfrac{r}{\mu }} \right)\sin 2\theta \\ \sigma _r =\dfrac{q}{2}\left( {1+\beta \dfrac{a^2}{r^2}} \right)+ \\ \qquad\dfrac{q}{2}\left( {1-4\alpha \dfrac{a^2}{r^2}+3\alpha \dfrac{a^4}{r^4}} \right)\cos 2\theta \\ \sigma _{r\theta } =\dfrac{q}{2}\left( {-1-2\alpha \dfrac{a^2}{r^2}+3\alpha \dfrac{a^4}{r^4}} \right)\sin 2\theta \\ \sigma _\theta =\dfrac{q}{2}\left( {1-\beta \dfrac{a^2}{r^2}} \right)-\dfrac{q}{2}\left( {1+3\alpha \dfrac{a^4}{r^4}} \right)\cos 2\theta \\ \end{array}} \right\} \end{eqnarray}$

2.2.2 夹杂中位移与应力($r<a)$

研究夹杂区域$\varOmega ^{\rm f}$, 虚设一条边界$r=\rho <a$. 研究由$r<a$与$r=\rho$围成的环形区域, 此时式(18)变为

$ \begin{eqnarray} \label{eq39} &&a\int_0^{2\pi} {\left( {u_r t_r^\ast +u_\theta t_\theta ^\ast } \right)_{r=a} {\rm d}\theta } -\rho \int_0^{2\pi} {\left( {u_r t_r^\ast +u_\theta t_\theta ^\ast } \right)_{r=\rho } {\rm d}\theta } = \\&& a\int_0^{2\pi} {\left( {w_r^\ast t_r +w_\theta ^\ast t_\theta } \right)_{r=a} {\rm d}\theta } -\\&&\rho \int_0^{2\pi} {\left( {w_r^\ast t_r +w_\theta ^\ast t_\theta } \right)_{r=\rho } {\rm d}\theta } \end{eqnarray}$

式(39)形式上与式(34)完全相同, 除了$r=\rho <a$. 在$r=a$处, 位移$u_r $, $u_\theta $与面力$t_r $, $t_\theta $仍如式(20)所示, 其中系数如式(30)和式(32)所示. 在$r=\rho $处, 仍设位移$u_r $, $u_\theta $与面力$t_r $, $t_\theta $设为式(40), 与式(35)形式相同, 只是待定系数的记号不同.

$ \begin{eqnarray} \label{eq40} \left. {\begin{array}{l} \left( {u_r } \right)_{r=\rho } =A_0^{\rm f} +A_2^{\rm f} \cos 2\theta \\ \left( {u_\theta } \right)_{r=\rho } =B_2^{\rm f} \sin 2\theta \\ t_r =\left( {\sigma _r } \right)_{r=\rho } =C_0^{\rm f} +C_2^{\rm f} \cos 2\theta \\ t_\theta =\left( {\sigma _{r\theta } } \right)_{r=\rho } =D_2^{\rm f} \sin 2\theta \end{array}} \right\} \end{eqnarray}$

将式(20)和式(40)代入式(39), 并选取适当的位移$w_r^\ast $, $w_\theta ^\ast $与面力$t_r^\ast $, $t_\theta ^\ast $. 由于此环形区域既不包含原点也没有无穷远处做边界, 因此只满足位移单值条件即可. 因此预期可以得到6个方程, 解出6个未知系数. 所选取的位移试函数$w_r^\ast $, $w_\theta ^\ast $ 可以分为两类, 一类如表3所示, 一类如表2所示. 先考虑表3中的$w_r^\ast $, $w_\theta ^\ast $, 这时有式(27)成立, 将其代入式(39)得到

$ \begin{eqnarray} \label{eq41} \int_{r=\rho } {\left( {u_r t_r^\ast +u_\theta t_\theta ^\ast } \right){\rm d}s} =\int_{r=\rho } {\left( {w_r^\ast t_r +w_\theta ^\ast t_\theta } \right){\rm d}s} \end{eqnarray}$

式(41)与式(27)形式完全类似, 从数学上看只是做了符号替换, $r=a$换成了$r=\rho $、式(35)换成了式(40), 弹性常数也相同. 结果上也完全类似, 式(29)替换相应的符号得式(40)

$ \begin{eqnarray} \label{eq42} \left. {\begin{array}{l} A_0^{\rm f} =\dfrac{\rho }{E_1 }\left( {1-\nu _1 } \right)C_0^{\rm f} \\ A_2^{\rm f} =\dfrac{\rho }{3E_1 }\left( {C_2^{\rm f} \nu _1 -2D_2^{\rm f} \nu _1 +3C_2^{\rm f} } \right) \\ B_2^{\rm f} =-\dfrac{\rho }{3E_1 }\left( {2C_2^{\rm f} \nu _1 -D_2^{\rm f} \nu _1 -3D_2^{\rm f} } \right) \\ \end{array}} \right\} \end{eqnarray}$

再考虑表2中的位移$w_r^\ast $, $w_\theta ^\ast $, 替换弹性常数(用$E_1 $, $\nu _1 $, $\mu _1 $替换$E$, $\nu $, $\mu )$, 再代入式(39), 面力$t_r^\ast $和$t_\theta ^\ast $由式(21)求出或者直接查表得到[33]代入式(39), 界面位移与界面面力式(20)和式(40)代入式(39), 积分整理后, 得到式(41)

$ \begin{eqnarray} \label{eq43} \left. {\begin{array}{l} \dfrac{c_0 -C_0^{\rm f} }{2\mu _1 }+\dfrac{a_0 }{a}-\dfrac{A_0^{\rm f} }{\rho }=0 \\ (C_2^{\rm f} +D_2^{\rm f} )\rho +6\mu _1 (A_2^{\rm f} +B_2^{\rm f} )=0 \\ \dfrac{(1+\kappa _1 )C_2^{\rm f} -(\kappa _1 -1)D_2^{\rm f} -2\kappa _1 c_2 }{4\mu _1 }+\\ \qquad \dfrac{2A_2^{\rm f} +B_2^{\rm f} }{\rho }=\dfrac{a_2 }{a} \\ \end{array}} \right\} \end{eqnarray}$

从式(43)中求得$C_0^{\rm f} $, $C_2^{\rm f} $, $D_2^{\rm f} $, 并利用式(30)和 式(32)得

$ \begin{eqnarray} \label{eq44} \left. {\begin{array}{l} C_0^{\rm f} =c_0 =\dfrac{q}{2}\dfrac{(1+\kappa )\mu _1 }{\mu \kappa _1 -\mu +2\mu _1 } \\ C_2^{\rm f} =c_2 =\dfrac{q}{2}\dfrac{\mu _1 (1+\kappa )}{\kappa \mu _1 +\mu } \\ D_2^{\rm f} =d_2 =-\dfrac{q}{2}\dfrac{\mu _1 (1+\kappa )}{\kappa \mu _1 +\mu } \\ \end{array}} \right\} \end{eqnarray}$

从式(44)得知, 夹杂内部的应力与界面处相同, 不随极径变化. 利用式(44)简化式(42), 并用$r$替换$\rho $, 得到

$ \begin{eqnarray} \label{eq45} A_0^{\rm f} =\dfrac{\left( {\kappa _1 -1} \right)r}{4\mu _1 }c_0 ,\ \ A_2^{\rm f} =\dfrac{r}{2\mu _1 }c_2 ,\ \ B_2^{\rm f} =-\dfrac{r}{2\mu _1 }c_2 \end{eqnarray}$

式(45)代入式(40), 并利用式(30), 得到夹杂内的位移与应力

$ \left. {\begin{array}{l} u_r =\dfrac{\left( {\kappa _1 -1} \right)r}{4\mu _1 }\beta ^{\rm f}\dfrac{q}{2}+\dfrac{r}{2\mu _1 }\alpha ^{\rm f}\dfrac{q}{2}\cos 2\theta \\ u_\theta =-\dfrac{r}{2\mu _1 }\alpha ^{\rm f}\dfrac{q}{2}\sin 2\theta \\ \sigma _r =\beta ^{\rm f}\dfrac{q}{2}+\alpha ^{\rm f}\dfrac{q}{2}\cos 2\theta \\ \sigma _{r\theta } =-\alpha ^{\rm f}\dfrac{q}{2}\sin 2\theta \\ \sigma _\theta =\beta ^{\rm f}\dfrac{q}{2}-\alpha ^{\rm f}\dfrac{q}{2}\cos 2\theta \end{array}} \right\} $

其中

\begin{eqnarray*} &&\beta ^f=\dfrac{\left( {1+\kappa } \right)\mu _1 }{\mu \kappa _1 -\mu +2\mu _1 },\ \ \alpha ^f=\dfrac{\mu _1 \left( {1+\kappa } \right)}{\kappa \mu _1 +\mu } \end{eqnarray*}

结果与文献[17]相同. 同边界元法一样, 此边界积分法也可以用于求解振动问题[35].

3 结论

本文参考了边界元的思路, 从Somigliana等式出发, 利用格林函数的不唯一性, 得到了适用于本文的边界积分方程. 可以用此方程求解弹性力学问题的解析解. 求解步骤也与边界元类似, 先求出未知边界位移与面力, 再求域内位移与面力. 对于含有有限边界的平面弹性问题来说, 第一步可以将边界上位移、面力进行傅里叶级数展开, 利用三角函数的正交性, 求解边界上未知位移与面力. 第二步类似矩阵传递法. 知道边界上的位移与面力后, 虚设边界通过域内的点, 再次利用第一步的方法求解域内位移与应力.

为了验证本方法, 具体说明本方法求解思路. 求解了单向均布应力下平板含圆形夹杂问题, 得到了精确解. 在特殊情况下可以化为含圆孔或刚性夹杂的解析解. 结果说明了此方法的有效性.

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姚振汉, 王海涛 . 边界元法. 北京: 高等教育出版社, 2010

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