磁-电-弹性半空间在轴对称热载荷作用下的三维问题研究 1)

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胡克强^,^*^,²⁾, 高存法^*, 仲政^?, Chen Zengtao^**

^*南京航空航天大学机械结构强度与振动国家重点实验室, 南京 210016

^?哈尔滨工业大学(深圳)理学院, 深圳 518055

^**Department of Mechanical Engineering, University of Alberta, Edmonton, AB, T6G 2G8, Canada

THREE-DIMENSIONAL ANALYSIS OF A MAGNETOELECTROELASTIC HALF-SPACE UNDER AXISYMMETRIC TEMPERATURE LOADING ¹⁾

Hu Keqiang^,^*^,²⁾, Gao Cunfa^*, Zhong Zheng^?, Chen Zengtao^**

^*State Key Laboratory of Mechanics and Control of Mechanical Structures, Nanjing University of Aeronautics and Astronautics, Nanjing 210016, China

^?School of Science, Harbin Institute of Technology, Shenzhen 518055, China

^**Department of Mechanical Engineering, University of Alberta, Edmonton, AB, T6G 2G8, Canada

通讯作者: 2)胡克强, 教授, 主要研究方向: 多场耦合复合材料断裂理论. E-mail:keqiang@nuaa.edu.cn

收稿日期:2020-04-20接受日期:2020-04-20网络出版日期:2020-09-18

基金资助:

1)国家自然科学基金项目.11872203
创新研究群体项目.51921003

Received:2020-04-20Accepted:2020-04-20Online:2020-09-18
作者简介 About authors

摘要
考虑力-电-磁-热等多场耦合作用, 基于线性理论给出了磁-电-弹性半空间在表面轴对称温度载荷作用下的热-磁-电-弹性分析, 并得到了问题的解析解. 利用Hankel 积分变换法求解了磁-电-弹性材料中的热传导及控制方程, 讨论了在磁-电-弹性半空间在边界表面上作用局部热载荷时的混合边值问题, 利用积分变换和积分方程技术, 通过在边界表面上施加应力自由及磁-电开路条件, 推导得到了磁-电-弹性半空间中位移、电势及磁势的积分形式的表达式. 获得了磁-电-弹性半空间中温度场的解析表达式并且给出了应力, 电位移和磁通量的解析解. 数值计算结果表明温度载荷对磁-电-弹性场的分布有显著影响. 当温度载荷作用的圆域半径增大时, 最大正应力发生的位置会远离半无限大体的边界; 反之当温度载荷作用的圆域半径减小时, 最大应力发生的位置会靠近半无限大体的边界. 电场和磁场在温度载荷作用的圆域内在边界表面附近有明显的强化, 而磁-电-弹性场强化区域的强化程度跟温度载荷的大小和作用区域大小相关. 本研究的相关结果对智能材料和结构在热载荷作用下的设计和制造具有指导意义.
关键词： 磁-电-弹性材料;轴对称温度载荷;Hankel变换;热传导;磁-电-弹性场

Abstract
Considering the coupling effects between mechanical, electrical, magnetic and thermal fields, we have presented an analytical solution for the thermo-magneto-electro-elastic problem of a magnetoelectroelastic half-space under axisymmetric thermal loading based on the linear theory. Integral transform method and integral equation technique are applied to analytically solve the heat conduction equation, the governing equations of the magnetoelectroelastic material, and the mixed boundary value problem on the boundary of the magnetoelectroelastic half-space. A general closed-form solution is presented for the complementary and particular parts of the components of the displacement, electric potential and magnetic potential. Traction-free and open circuit electro-magnetic conditions are applied on the boundary surface and an integral form solution for the displacement, electric and magnetic potentials in the magnetoelectroelastic half-space has been successfully obtained. Temperature field in the half-space has been obtained analytically and the expression of the stresses, electric displacements and magnetic induction due to the temperature change applied on the surface are derived and given in an explicit closed form. Numerical results show that the temperature loading has much effect on the field distribution of the mechanical, electric and magnetic fields in the magnetoelectroelastic half-space. As the radius of the constant temperature loading increases, the distance from the region of the maximum normal stress to the free boundary will become larger, and the normal stress becomes much smaller in the regions outside of the circular region. The maximum shear stress appears just below the boundary surface at the edge of the circular region. The electric field is found to be intensified near to the boundary surface within the circular region, and similarly, intensities of the positive and negative magnetic fields are observed at different locations in the half-space under the temperature loading applied on the boundary. The results of this study are helpful for the design and manufacturing of smart materials/structures under thermal loading.
Keywords：magnetoelectroelastic material;axisymmetric temperature loading;Hankel transform;heat conduction;electric and magnetic field

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本文引用格式
胡克强, 高存法, 仲政, Chen Zengtao. 磁-电-弹性半空间在轴对称热载荷作用下的三维问题研究 ¹⁾. 力学学报[J], 2020, 52(5): 1235-1244 DOI:10.6052/0459-1879-20-127
Hu Keqiang, Gao Cunfa, Zhong Zheng, Chen Zengtao. THREE-DIMENSIONAL ANALYSIS OF A MAGNETOELECTROELASTIC HALF-SPACE UNDER AXISYMMETRIC TEMPERATURE LOADING ¹⁾. Chinese Journal of Theoretical and Applied Mechanics[J], 2020, 52(5): 1235-1244 DOI:10.6052/0459-1879-20-127

引言

磁-电-弹性材料是一种同时具有压电压磁磁电耦合效应的复合材料, 这种材料在工程结构中, 尤其是在智能材料和结构系统中, 有越来越广泛的应用. 磁-电-弹性材料所固有的力电磁耦合特性使它们成为智能结构中传感与执行元件的首选材料, 广泛应用于传感器、执行器、滤波器、换能器和其他智能器件, 在能量转换系统中有着非常可观的潜在的应用前景^[1-8]. 由于这些智能器件通常在复杂的力、电、磁和热耦合载荷环境下工作, 因此对磁-电-弹性材料在多场耦合环境下的响应问题研究具有重要的意义.

近年来, 关于磁-电-弹性材料力学问题的研究备受关注. Wang等^[9]推出了各向异性磁-电-弹性半空间在圆周方向载荷作用下的解析解. 利用推广的Stroh公式和坐标变换技术, Qin^[10]求得了磁-电-弹性介质中含有任意方向半平面或双材料界面时的Green函数解. 利用积分变换和奇异积分方程方法, Hu等^[11]研究了磁-电-弹性和正交各向异性半空间中的界面裂纹问题, 发现裂纹尖端的振荡奇异性或非振荡奇异性取决于双材料的特定材料属性组合. 段淑敏等^[12]用奇异积分方程法求解了磁-电-弹性材料与加层间的界面裂纹在反平面剪切冲击载荷和面内电磁冲击载荷作用下的动态响应问题, 讨论了载荷、材料及几何参数对能量释放率的影响. Ma等^[13]用扩展有限元法分析了磁-电-弹性双材料界面裂纹的静态断裂和多场耦合效应. Yang和Li^[14]通过考虑表面效应并利用Kirchhoff薄板理论, 获得了纳米尺度圆盘状磁-电-弹性薄板弯曲和自由振动的解析解.

鉴于磁-电-弹性材料和结构通常会用在磁场、电场和温度场等多场耦合的载荷环境中, 而其性能通常会受到温度载荷显著的影响, 因此有必要对其热效应进行深入的研究^[15]. Hou等^[16]求得了各向同性磁-电-弹性材料中的二维形式的通解和基本解. 通过引入5个调和函数, Chen等^[17]推导出了横观各向同性磁-电-热弹性体的一般解并得到了各耦合场的表达式. Carman等^[18]提出了一个含有保守非线性的磁-电-弹性材料中的微观机械模型. 利用微观机械方法可以对完全耦合的磁-电-热弹性多相复合材料的性能进行系统的分析^[19-20]. Ke和Wang^[21]基于非局部理论和Kirchhoff板理论研究了磁-电-弹性纳米板的自由振动, 发现其固有频率对于力、电和磁载荷很敏感但对热载荷不敏感. 田晓耕和沈亚鹏^[22]讨论并综述了磁-电多场耦合的广义电磁热弹性耦合问题方面的研究以及计及扩散效应和黏弹性效应的广义热弹性理论的发展. He等^[23]研究了无限长空心圆柱体中的广义电磁热弹性问题. Ootao和Tanigama^[24]研究了多层磁-电-热弹性板条在非稳定及非均匀热载荷作用下的动态响应. Karimi和Shahidi^[25]利用非局部理论研究了磁-电-弹性纳米板在外加磁势、面内剪切和热载荷作用时的自由振动及表面效应. Gao等^[26]根据广义Stroh公式研究了完全耦合的磁-电-弹性介质中的共线电渗透型裂纹问题并给出了在远场均匀热流作用时磁-电-弹性场强度因子的简洁表达式. 利用积分方程法, Niraula和Wang^[27]求得了磁-电-弹性介质中的圆币型裂纹在均匀热流作用下的精确解. Li等^[28]求得了具有横观各向同性性质的无限大热-磁-电-弹性介质中的圆币型裂纹的三维基本解. Zhao等^[29-30]用扩展的位移间断边界积分方程法分析了三维横观各向同性热-磁-电-弹性双材料中任意形状的界面裂纹问题并讨论了电-磁场边界条件对各个场强度因子的影响.

从检索到的相关文献资料来看, 现有的工作主要集中在求解无限大磁-电-弹性体中的力学问题或者相应的热载荷是作用在全部边界上的情形, 而考虑边界的影响以及局部范围内分布载荷的作用则更具有实际意义. 本文利用积分变换方法推导并求解了磁-电-弹性半空间在边界上作用轴对称温度载荷时的热传导和控制方程, 得到了磁-电-弹性半空间中的温度场、应力、电位移和磁通量的解析解. 本文结果对磁-电-弹性材料在热环境中的应用具有重要的指导意义.

1 基本公式

考虑一个横观各向同性的磁-电-弹性材料, 其极化方向沿$z$轴方向, 各向同性平面为$xy$平面, 如图1所示. 对于轴对称问题, 线性磁-电-弹性介质的本构方程为

(1) $ \begin{eqnarray} \label{eq1} && \left\{ {{\begin{array}{*{20}c} {\sigma _{rr} } \\ {\sigma _{\theta \theta } } \\ {\sigma _{zz} } \\ {D_z } \\ {B_z } \\ \end{array} }} \right\}=\left[ {{\begin{array}{*{20}c} {c_{11} } & {c_{12} } & {c_{13} } & {e_{31} } & {h_{31} } \\ {c_{12} } & {c_{11} } & {c_{13} } & {e_{31} } & {h_{31} } \\ {c_{13} } & {c_{13} } & {c_{33} } & {e_{33} } & {h_{33} } \\ {e_{31} } & {e_{31} } & {e_{33} } & {-\varepsilon _{33} } & {-d_{33} } \\ {h_{31} } & {h_{31} } & {h_{33} } & {-d_{33} } & {-\mu _{33} } \\ \end{array} }} \right]\left\{ {{\begin{array}{*{20}c} {u_{r,r} } \\ {{u_r } / r} \\ {u_{z,z} } \\ {\phi _{,z} } \\ {\varphi _{,z} } \\ \end{array} }} \right\}- \\&&\qquad \left[ {{\begin{array}{*{20}c} {\beta _1 } & {\beta _2 } & {\beta _3 } & {\beta _4 } & {\beta _5 } \\ \end{array} }} \right]'\Delta T \end{eqnarray}$

(2) $ \begin{eqnarray}\label{eq2} \left\{ {{\begin{array}{*{20}c} {\sigma _{rz} } \\ {D_r } \\ {B_r } \\ \end{array} }} \right\}=\left[ {{\begin{array}{*{20}c} {c_{44} } & {e_{15} } & {h_{15} } \\ {e_{15} } & {-\varepsilon _{11} } & {-d_{11} } \\ {h_{15} } & {-d_{11} } & {-\mu _{11} } \\ \end{array} }} \right]\left\{ {{\begin{array}{*{20}c} {u_{z,r} +u_{r,z} } \\ {\phi _{,r} } \\ {\varphi _{,r} } \\ \end{array} }} \right\} \end{eqnarray}$

图1

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图1轴对称热载荷作用下的磁-电-弹性半空间

Fig.1A magnetoelectroelastic half-space under axisymmetric thermal loading

其中, 场变量为$r$和的$z$函数, 与角度$\theta $无关; $u_r $和$u_z $分别是径向和轴向弹性位移的分量; $\phi $和$\varphi $分别是电势和磁势; $\sigma _{rr}$, $\sigma _{\theta \theta}$, $\sigma _{zz}$, $\sigma _{rz}$为应力张量的分量; $D_r $和$D_z $为电位移的分量; $B_r $和$B_z$为磁通量的分量; $c_{11}$, $c_{12}$, $c_{13}$, $c_{33}$, $c_{44}$为弹性模量; $e_{15}$, $e_{31}$, $e_{33} $ 为压电常数; $\varepsilon _{11}$, $\varepsilon _{33} $ 为介电常数; $d_{11}$, $d_{33} $ 为电磁常数; $\mu _{11}$, $\mu _{33} $ 为磁渗透系数; $\beta _j$ $(j=1,2,...,5)$是热应力系数; $\Delta T$是温度变化量; $[\ ]'$表示矩阵的转置; $(\ )$表示对相应坐标的求偏导.

电场$E_j (j=r,z)$和磁场$H_j (j=r,z)$可以分别由电势和磁势的偏导数表示为

(3) $ \left. \begin{array}{ll} E_r =-\phi _{,r} ,\ \ E_z =-\phi _{,z} \\ H_r =-\varphi _{,r} ,\ \ H_z =-\varphi _{,z} \end{array} \right\} $

若不计体力及自由电荷, 平衡方程可以表示为

(4) $ \left. \begin{array}{ll} \sigma _{rr,r} +\sigma _{rz,z} +{(\sigma _{rr} -\sigma _{\theta \theta } )}/ r=0 \\ \sigma _{rz,r} +\sigma _{zz,z} +{\sigma _{rz} } / r=0 \\ D_{r,r} +D_{z,z} +{D_r } / r=0 \\ B_{r,r} +B_{z,z} +{B_r } / r=0 \\ \end{array} \right\} $

将方程(1)和(2)代入以上方程可以得到关于位移$u_r $, $u_z $, 电势$\phi$和磁势$\varphi $的控制方程如下

(5) $ \left. \begin{array}{ll} c_{11} \left( {u_{r,rr} +\dfrac{u_{r,r} }{r}-\dfrac{u_r }{r^2}} \right)+c_{44} u_{r,zz} +(c_{13} +c_{44} )u_{z,rz} + \\[2mm]\qquad (e_{15} +e_{31} )\phi _{,rz} +(h_{15} +h_{31} )\varphi _{,rz} =\beta _1 \Delta T_{,r} \\[2mm] c_{44} \left( {u_{z,rr} +\dfrac{u_{z,r} }{r}} \right)+(c_{13} +c_{44} )\left( {u_{r,rz} +\dfrac{u_{r,z} }{r}} \right)+\\[2mm]\qquad c_{33} u_{z,zz} + e_{15} \left( {\phi _{,rr} +\dfrac{\phi _{,r} }{r}} \right)+ e_{33} \phi _{,zz} +\\[2mm]\qquad h_{15} \left( {\varphi _{,rr} +\dfrac{\varphi _{,r} }{r}} \right)+ h_{33} \varphi _{,zz} =\beta _3 \Delta T_{,z} \\[2mm] (e_{15} +e_{31} )\left( {u_{r,rz} +\dfrac{u_{r,z} }{r}} \right)+e_{15} \left( {u_{z,rr} +\dfrac{u_{z,r} }{r}} \right)+\\[2mm]\qquad e_{33} u_{z,zz} -\varepsilon _{11} \left( {\phi _{,rr} +\dfrac{\phi _{,r} }{r}} \right)-\varepsilon _{33} \phi _{,zz} -\\[2mm]\qquad d_{11} \left( {\varphi _{,rr} +\dfrac{\varphi _{,r} }{r}} \right)- d_{33} \varphi _{,zz} =\beta _4 \Delta T_{,z} \\[2mm] (h_{15} +h_{31} )\left( {u_{r,rz} +\dfrac{u_{r,z} }{r}} \right)+h_{15} \left( {u_{z,rr} +\dfrac{u_{z,r} }{r}} \right)+\\[2mm]\qquad h_{33} u_{z,zz} - d_{11} \left( {\phi _{,rr} +\dfrac{\phi _{,r} }{r}} \right)-d_{33} \phi _{,zz} -\\[2mm]\qquad \mu _{11} \left( {\varphi _{,rr} +\dfrac{\varphi _{,r} }{r}} \right)- \mu _{33} \varphi _{,zz}=\beta _5 \Delta T_{,z} \\ \end{array} \right\} $

2 问题求解

考虑一个在其边界表面上作用有轴对称热载荷的磁-电-弹性半空间, 如图1所示. 采用柱坐标($r$, $\theta$, $z)$, 并假定极化方向沿$z$轴方向; $T_B(r)$是一个定义在圆域($r\leqslant a, z=0)$内的轴对称函数.

横观各向同性材料在轴对称条件下的稳态热传导方程为

(6) $ \begin{eqnarray} \label{eq6} T_{,rr} +{T_{,r} } / r+{k_z T_{,zz} } / {k_r }=0 \end{eqnarray}$

其中, $T=T(r,z)$是介质中的温度分布函数, $k_r $和$k_z $ 分别为$r$ 和 $z$方向的热传导系数. 应用Hankel 变换可以得到温度的表达式为

(7) $ \begin{eqnarray} \label{eq7} T(r,z)=A_0 +\int_0^\infty {A(\xi )\exp \left( {-\lambda \xi z} \right)} {\rm J}_0 (\xi r){\rm d}\xi \end{eqnarray}$

其中, $A(\xi )$为待求函数, $A_0 $是一个可以由远场边界条件确定的常数, J$_0 (\ )$ 是第一类零阶Bessel 函数, 系数$\lambda $定义为

(8) $ \begin{eqnarray} \label{eq8} \lambda =\sqrt {{k_r } / {k_z }} \end{eqnarray}$

假设无穷远处的温度为一有限值$T_I $并且作用于磁-电-弹性介质边界上有界区域内的温度为$T_B $, 即

(9) $ \begin{eqnarray} \label{eq9} T(r,\infty )=T_I,\ \ T(r,0)=T_B (r) \end{eqnarray}$

则$A_0 $和$A(\xi )$可以表示为

(10) $ \begin{eqnarray} \begin{array}{ll} A_0 =T_I \\ A(\xi )=\xi \int_0^\infty {r\left[ {T_B (r)-T_I } \right]{\rm J}_0 (\xi r){\rm d}r} =\\\qquad \xi \int_0^\infty {rT_0 (r){\rm J}_0 (\xi r){\rm d}r} \\ \end{array} \end{eqnarray}$

将方程(10)代入方程(7) 可以得到温度场的表达式为

(11) $ \begin{eqnarray} \label{eq10} &&T(r,z)=T_I+\int_0^\infty \xi \exp ( -\lambda \xi z){\rm J}_0 (\xi r)\cdot \\&&\qquad \int_0^\infty \eta T_0 (\eta ){\rm J}_0 (\xi \eta ){\rm d}\eta {\rm d}\xi \end{eqnarray}$

温度变化可以表示为

(12) $ \begin{eqnarray} \label{eq11} &&\Delta T=T(r,z)-T_I=\int_0^\infty \xi \exp \left( {-\lambda \xi z} \right){\rm J}_0 (\xi r)\cdot \\&&\qquad\int_0^\infty \eta T_0 (\eta ){\rm J}_0 (\xi \eta ){\rm d}\eta {\rm d}\xi \end{eqnarray}$

对方程(5)做Hankel变换, 可以得到弹性位移, 电势和磁势的通解如下

(13) $ u_r (r,z)=\int_0^\infty {\rm J}_1 (\xi r)\Bigg[B(\xi )\exp (-\lambda \xi z)+ \\ \qquad\sum\limits_{j=1}^4 {a_j A_j (\xi )\exp (-\gamma _j \xi z)}\Bigg]{\rm d}\xi $

(14) $ u_z (r,z)=\int_0^\infty {\rm J}_0 (\xi r)\Bigg[ C(\xi )\exp (-\lambda \xi z) + \\ \qquad\sum\limits_{j=1}^4 {\frac{A_j (\xi )}{\gamma _j }\exp (-\gamma _j \xi z)}\Bigg]{\rm d}\xi $

(15) $ \phi (r,z)=\int_0^\infty {\rm J}_0 (\xi r)\Bigg[ D(\xi )\exp (-\lambda \xi z) - \\ \qquad\sum\limits_{j=1}^4 {\frac{b_j A_j (\xi )}{\gamma _j }\exp (-\gamma _j \xi z)}\Bigg]{\rm d}\xi $

(16) $ \varphi (r,z)=\int_0^\infty {\rm J}_0 (\xi r)\Bigg[ E(\xi )\exp (-\lambda \xi z)- \\ \qquad\sum\limits_{j=1}^4 {\frac{d_j A_j (\xi )}{\gamma _j }\exp (-\gamma _j \xi z)}\Bigg]{\rm d}\xi $

其中, $A_j (\xi )$ $(j=1,2,3,4)$, $B(\xi )$, $C(\xi )$, $D(\xi )$, $E(\xi)$为待求的未知函数, $a_j $, $b_j $, $d_j $, ($j=1,2,3,4)$ 是与材料系数相关的已知的系数, 见附录; ${\rm J}_1 (\ )$ 是第一类的一阶Bessel函数, $\gamma _j $ ($j=1,2,3,4)$ 为以下特征方程的特征根

(17) $ \begin{eqnarray} \label{eq17} \left| {{\begin{array}{*{20}c} {c_{11} -c_{44} \gamma ^2} & {(c_{13} +c_{44} )\gamma } & {(e_{31} +e_{15} )\gamma } & {(h_{31} +h_{15} )\gamma } \\ {(c_{13} +c_{44} )\gamma } & {c_{33} \gamma ^2-c_{44} } & {e_{33} \gamma ^2-e_{15} } & {h_{33} \gamma ^2-h_{15} } \\ {(e_{31} +e_{15} )\gamma } & {e_{33} \gamma ^2-e_{15} } & {\varepsilon _{11} -\varepsilon _{33} \gamma ^2} & {d_{11} -d_{33} \gamma ^2} \\ {(h_{31} +h_{15} )\gamma } & {h_{33} \gamma ^2-h_{15} } & {d_{11} -d_{33} \gamma ^2} & {\mu _{11} -\mu _{33} \gamma ^2} \\ \end{array} }} \right| =0 \\ \end{eqnarray}$

其中, $\left| {M} \right|$ 表示矩阵${M}$的行列式值. 这里需要指出的是, 八阶特征方程(17)有8个特征根成对出现, 每一对大小相同符号相反, 复根以共轭形式成对出现. 在式(13)$\sim$式(16)中, 取特征根$\gamma _j $ ($j=1,2,3,4)$ 的实部为正, 以确保稳定系统的内能正定^[31-32].

利用本构方程, 可以得到应力, 电位移及磁通量的各个分量为

(18) $ \begin{eqnarray} &&\sigma _{zz} (r,z) =\int_0^\infty {\rm J}_0 (\xi r)\Bigg[ K_{10} (\xi )\exp (-\lambda \xi z)+ \\&&\qquad\sum\limits_{j=1}^4 {K_{1j} (\xi )\exp (-\gamma _j \xi z)}\Bigg]{\rm d}\xi -\beta _3 \Delta T \\ \end{eqnarray}$

(19) $ \begin{eqnarray} &&\sigma _{rz} (r,z) =\int_0^\infty {\rm J}_1 (\xi r)\Bigg[ K_{20} (\xi )\exp (-\lambda \xi z)+ \\&&\qquad\sum\limits_{j=1}^4 {K_{2j} (\xi )\exp (-\gamma _j \xi z)}\Bigg]{\rm d}\xi \\ \end{eqnarray}$

(20) $ \begin{eqnarray} && D_z (r,z)=\int_0^\infty {\rm J}_0 (\xi r)\Bigg[ K_{30} (\xi )\exp (-\lambda \xi z)+ \\&&\qquad\sum\limits_{j=1}^4 {K_{3j} (\xi )\exp (-\gamma _j \xi z)}\Bigg]{\rm d}\xi -\beta _4 \Delta T\\ \end{eqnarray}$

(21) $ \begin{eqnarray} && B_z (r,z)=\int_0^\infty {\rm J}_0 (\xi r)\Bigg[ K_{40} (\xi )\exp (-\lambda \xi z)+ \\&&\qquad\sum\limits_{j=1}^4 {K_{4j} (\xi )\exp (-\gamma _j \xi z)}\Bigg]{\rm d}\xi -\beta _5 \Delta T \end{eqnarray}$

其中

(22a) $ \left. \begin{array}{ll} K_{10} (\xi )=\xi \lt\{c_{13} B(\xi )-\lambda \lt[c_{33} C(\xi )+e_{33} D(\xi )+ \\ h_{33} E(\xi ) ] \} \\ K_{1j} (\xi )=\xi \left[ {c_{13} a_j -c_{33} +e_{33} b_j +h_{33} d_j }\right]A_j (\xi ) \\ \end{array} \right\} $

(22b) $ \left. \begin{array}{ll} K_{20} (\xi )=-\xi \lt[c_{44} \lambda B(\xi )+c_{44} C(\xi )+e_{15} D(\xi )+ \\ h_{15} E(\xi ) ] \\ K_{2j} (\xi )=\xi \bigg[ -c_{44} \left( {a_j \gamma _j +\dfrac{1}{\gamma _j }} \right) +e_{15} \dfrac{b_j }{\gamma _j }+\\\qquad h_{15} \dfrac{d_j }{\gamma _j } \bigg]A_j (\xi ) \\ \end{array} \right\} $

(22c) $ \left. \begin{array}{ll} K_{30} (\xi )=\xi \bigg\{ e_{31} B(\xi )+\lambda \bigg[ \varepsilon _{33} D(\xi ) +h_{33} E(\xi )-\\ \qquad e_{33} C(\xi ) \bigg] \bigg\}\\ K_{3j} (\xi )=\xi \left[e_{31} a_j -e_{33} -\varepsilon _{33} b_j -h_{33} d_j \right]A_j (\xi ) \\ \end{array} \right\} $

(22d) $ \left. \begin{array}{ll} K_{40} (\xi )=\xi \lt\{ h_{31} B(\xi )+\lambda \lt[d_{33} D(\xi ) +\mu _{33} E(\xi )- \\ h_{33} C(\xi ) ] \} \\ K_{4j} (\xi )=\xi \left[ h_{31} a_j -h_{33} -d_{33} b_j -\mu _{33} d_j \right]A_j (\xi ) \\ \end{array} \right\} $

当在半无限大体的表面只作用有热载荷作用时, 磁-电-弹性半空间表面上电、磁和机械边界条件为

(23) $ \left. \begin{array}{ll} D_z (r,0)=0 \\ B_z (r,0)=0 \\ \sigma _{rz} (r,0)=0 \\ \sigma _{zz} (r,0)=0 \\ \end{array} \right\}$

将方程(18)$\sim$(21)代入方程(23), 可以得到以下关系式

(24a) $ \left. \begin{array}{ll} \sum\limits_{j=1}^4 {X_{1j} A_j (\xi )} =Y_1 \\ Y_1 =-c_{13} B(\xi )+\lambda \left[ c_{33} C(\xi )+e_{33} D(\xi )+ \\ \qquad h_{33} E(\xi )\right]+\beta _3 \dfrac{A(\xi )}{\xi } \\ \end{array} \right\} $

(24b) $ \left. \begin{array}{ll} \sum\limits_{j=1}^4 {X_{2j} A_j (\xi )} =Y_2 \\ Y_2 =c_{44} \lambda B(\xi )+c_{44} C(\xi )+e_{15} D(\xi )+h_{15} E(\xi ) \\ \end{array} \right\} $

(24c) $ \left. \begin{array}{ll} \sum\limits_{j=1}^4 {X_{3j} A_j (\xi )} =Y_3 \\ Y_3 =-e_{31} B(\xi )+\lambda \left[ e_{33} C(\xi )-\varepsilon _{33} D(\xi )- \\ \qquad h_{33} E(\xi )\right]+\beta _4 \dfrac{A(\xi )}{\xi } \\ \end{array} \right\} $

(24d) $ \left. \begin{array}{ll} \sum\limits_{j=1}^4 {X_{4j} A_j (\xi )} =Y_4 \\ Y_4 =-h_{31} B(\xi )+\lambda \left[ h_{33} C(\xi ) -d_{33} D(\xi )- \\ \qquad \mu _{33} E(\xi )\right]+\beta _5 \dfrac{A(\xi )}{\xi } \\ \end{array} \right\} $

其中

(25a) $ X_{1j} =c_{13} a_j -c_{33} +e_{33} b_j +h_{33} d_j $

(25b) $ X_{2j} (\xi )=-c_{44} \left( {a_j \gamma _j +\frac{1}{\gamma _j }} \right) +e_{15} \frac{b_j }{\gamma _j }+h_{15} \frac{d_j }{\gamma _j } $

(25c) $ X_{3j} =e_{31} a_j -e_{33} -\varepsilon _{33} b_j -h_{33} d_j $

(25d) $ X_{4j} =h_{31} a_j -h_{33} -d_{33} b_j -\mu _{33} d_j $

未知函数$B(\xi )$, $C(\xi )$, $D(\xi )$, $E(\xi )$可以由函数$A(\xi )$表示为

(26) $ \begin{eqnarray} \label{eq26} \left\{ {{\begin{array}{*{20}c} {B(\xi )} \\ {C(\xi )} \\ {D(\xi )} \\ {E(\xi )} \\ \end{array} }} \right\}=\left\{ {{\begin{array}{*{20}c} {\varDelta _1 } \\ {\varDelta _2 } \\ {\varDelta _3 } \\ {\varDelta _4 } \\ \end{array} }} \right\}\frac{A(\xi )}{\xi } \end{eqnarray}$

其中, $\Delta _j $ ($j=1,2,3,4)$ 的表达式在附录中给出.

$Y_j $ ($j=1,2,3,4)$ 可以表示为$A(\xi )$的函数

(27) $ \begin{eqnarray} \label{eq27} \left\{ {{\begin{array}{*{20}c} {Y_1 } \\ {Y_2 } \\ {Y_3 } \\ {Y_4 } \\ \end{array} }} \right\}=\left\{ {{\begin{array}{*{20}c} {\delta _1 } \\ {\delta _2 } \\ {\delta _3 } \\ {\delta _4 } \\ \end{array} }} \right\}\frac{A(\xi )}{\xi } \end{eqnarray}$

(28a) $ \delta _1 =-c_{13} \varDelta _1 +c_{33} \lambda \varDelta _2 +e_{33} \lambda \varDelta _3 +h_{33} \lambda \varDelta _4 +\beta _3 $

(28b) $ \delta _2 =c_{44} \lambda \varDelta _1 +c_{44} \varDelta _2 +e_{15} \varDelta _3 +h_{15} \varDelta _4 $

(28c) $ \delta _3 =-e_{31} \varDelta _1 +e_{33} \lambda \varDelta _2 -\varepsilon _{33} \lambda \varDelta _3 -h_{33} \lambda \varDelta _4 +\beta _4 $

(28d) $ \delta _4 =-h_{31} \varDelta _1 +h_{33} \lambda \varDelta _2 -d_{33} \lambda \varDelta _3 -\mu _{33} \lambda \varDelta _4 +\beta _5 $

$A_j (\xi )\ (j=1,2,3,4)$与$A(\xi )$的关系为

(29) $ \begin{eqnarray} \label{eq28} \left[ {{\begin{array}{*{20}c} {X_{11} } & {X_{12} } & {X_{13} } & {X_{14} }\\ {X_{21} } & {X_{22} } & {X_{23} } & {X_{24} }\\ {X_{31} } & {X_{32} } & {X_{33} } & {X_{34} }\\ {X_{41} } & {X_{42} } & {X_{43} } & {X_{44} }\\ \end{array} }} \right]\left\{ {{\begin{array}{*{20}c} {A_1 (\xi )} \\ {A_2 (\xi )} \\ {A_3 (\xi )} \\ {A_4 (\xi )} \\ \end{array} }} \right\}=\left\{ {{\begin{array}{*{20}c} {\delta _1 } \\ {\delta _2 } \\ {\delta _3 } \\ {\delta _4 } \\ \end{array} }} \right\}\frac{A(\xi )}{\xi } \end{eqnarray}$

求解以上方程可得

(30) $ \begin{eqnarray} \label{eq29} \left\{ {{\begin{array}{*{20}c} {A_1 (\xi )} \\ {A_2 (\xi )} \\ {A_3 (\xi )} \\ {A_4 (\xi )} \\ \end{array} }} \right\}=\left\{ {{\begin{array}{*{20}c} {\varOmega _1 } \\ {\varOmega _2 } \\ {\varOmega _3 } \\ {\varOmega _4 } \\ \end{array} }} \right\}\frac{A(\xi )}{\xi } \end{eqnarray}$

其中, $\varOmega _j $ ($j=1,2,3,4)$ 在附录中给出.

本文考虑如下定常相对温度作用在磁-电-弹性半空间表面上一个圆域上的情形

(31) $ \begin{eqnarray} \label{eq30} T_0 (r)=\left\{ {{\begin{array}{ll} T_0,\ \ & 0\leqslant r\leqslant a \\ 0,\ \ & r>a \\ \end{array} }} \right. \end{eqnarray}$

需要指出的是, $T_0 $是作用在圆域上的温度与圆域外的温度的相对值, 而所考查的半空间中的磁-电-弹性场的变化与$T_0 $有关.

将方程(31)代入方程(10)可以得到

(32) $ \begin{eqnarray} \label{eq32} A(\xi )=T_0 a{\rm J}_1 (\xi a) \end{eqnarray}$

温度场的积分表达式如方程(7)所示, 利用关系式(26)和式(30), 可以得到位移、电势、磁势、应力、电位移和磁通量的解析形式的表达式. 在此略去具体的细节, 应力、电位移和磁通量的分量可以表示为

(33) $ \sigma _{zz} (r,z)=T_0 a\left[ {(V_1 -\beta _3 )I_1 (\lambda )+\sum\limits_{j=1}^4 {E_j I_1 (\gamma _j )} } \right] $

(34) $ D_z (r,z)=T_0 a\left[ {(V_2 -\beta _4 )I_1 (\lambda )+\sum\limits_{j=1}^4 {F_j I_1 (\gamma _j )} } \right] $

(35) $ B_z (r,z)=T_0 a\left[ {(V_3 -\beta _5 )I_1 (\lambda )+\sum\limits_{j=1}^4 {G_j I_1 (\gamma _j )} } \right] $

(36) $ \sigma _{rz} (r,z)=T_0 a\left[ {V_4 I_2 (\lambda )+\sum\limits_{j=1}^4 {H_j I_2 (\gamma _j )} } \right] $

(37) $ D_r (r,z)=T_0 a\left[ {V_5 I_2 (\lambda )+\sum\limits_{j=1}^4 {K_j I_2 (\gamma _j )} } \right] $

(38) $ B_r (r,z)=T_0 a\left[ {V_6 I_2 (\lambda )+\sum\limits_{j=1}^4 {L_j I_2 (\gamma _j )} } \right] $

其中, 系数$V_k$ $(k=1,2,\dots, 6)$, $E_j$, $F_j$, $G_j $, $H_j$, $K_j $, $L_j$ ($j=1,2, 3,4)$在附录中给出, 积分算子$I_j \left( \lambda \right)$ $(j=1,2)$定义为

(39) $ \left. \begin{array}{ll} I_1 (\lambda )=\int_0^\infty {{\rm J}_1 (\xi a){\rm J}_0 (\xi r)\exp (-\lambda \xi z){\rm d}\xi } =\\[2mm]\qquad \dfrac{a}{2}\dfrac{\lambda z}{\left[ {r^2+\left( {\lambda z} \right)^2} \right]^{3 / 2}} \\[4mm] I_2 (\lambda )=\int_0^\infty {{\rm J}_1 (\xi a){\rm J}_1 (\xi r)\exp (-\lambda \xi z){\rm d}\xi } =\\[2mm]\qquad \dfrac{ar}{2}\dfrac{1}{\left[{r^2+\left( {\lambda z} \right)^2} \right]^{3 / 2}} \\ \end{array} \right\}$

常数$\beta _j$ $(j=1,2,3,4,5)$可表示为

(40) $ \left. \begin{array}{ll} \beta _1 =c_{11} \alpha _r +c_{12} \alpha _\theta +c_{13} \alpha _z \\ \beta _2 =c_{12} \alpha _r +c_{11} \alpha _\theta +c_{13} \alpha _z \\ \beta _3 =c_{13} \alpha _r +c_{13} \alpha _\theta +c_{33} \alpha _z \\ \beta _4 =e_{31} \alpha _r +e_{31} \alpha _\theta +e_{33} \alpha _z \\ \beta _5 =h_{31} \alpha _r +h_{31} \alpha _\theta +h_{33} \alpha _z \\ \end{array} \right\} $

其中, $\alpha _i$ $(i=r,\theta ,z)$为热膨胀系数.

由以上所得到的磁、电、热弹性场的解析表达式可知, 对于在任意区域作用任意温度载荷的情形, 本文的结果可以作为一个基本解, 在线弹性和小变形范围内利用叠加原理可以得到一般温度载荷情形的解. 需要指出的是, 由于解的表达式中含有无限积分项, 在不能获得精确积分结果的情况下, 数值积分可能会产生一定的误差.

3 数值计算及分析

在数值计算中, 不失一般性, 我们取一具有完全的机械、电和磁场耦合作用的横观各向同性性质的磁-电-弹性材料-复合材料BaTiO$_{3}$-CoFe$_{2}$O$_{4}$, 其材料常数为^[31]

(41) $ \left. \begin{array}{ll} c_{11} =22.6\times 10^{10}~{\rm N} / {{\rm m}^{2}},\\ c_{13} =12.4\times 10^{10}~{\rm N} / {{\rm m}^{2}} \\ c_{33} =21.6\times 10^{10}~{\rm N} / {{\rm m}^{2}},\\ c_{44} =4.4\times 10^{10}~{\rm N} / {{\rm m}^{2}} \\ e_{15} =5.8~{\rm C} / {{\rm m}^{2}},\ \ e_{31} =-2.2~{\rm C} / {{\rm m}^{2}} \\ e_{33} =9.3~{\rm C} / {{\rm m}^{2}},\ \ h_{15} =275~{\rm N} /{{\rm Am}} \\ h_{31} =290.2~{\rm N} / {{\rm Am}},\ \ h_{33} =350~{\rm N} / {{\rm Am}} \\ \lambda _{11} =5.6\times 10^{-9}~{{\rm C}^{2}} / {{\rm Nm}^{2}},\\ \lambda _{33} =6.3\times 10^{-9}~{{\rm C}^{2}} / {{\rm Nm}^{2}} \\ \mu _{11} =29.7\times 10^{-5}~{{\rm Ns}^{2}} / {{\rm C}^{2}},\\ \mu _{33} =8.3\times 10^{-5}~{{\rm Ns}^{2}} / {{\rm C}^{2}} \\ d_{11} =4.0\times 10^{-9}~{{\rm Ns}} / {{\rm VC}},\\ d_{33} =4.7\times 10^{-9}~{{\rm Ns}} / {{\rm VC}} \\ \end{array} \right\}$

对于以$z$轴为极化方向的横观各向同性材料, 径向和切向的热膨胀系数相同, 即$\alpha _r =\alpha _\theta $. 因此, 由方程(40)可知$\beta _1 =\beta _2 $. 在以下的数值计算中, 热膨胀系数假定为$\alpha _r =\alpha _z =7.0\times 10^{-6}$ K$^{-1}$, 圆域内的相对温度值假设为$T_0 =100$ K.

图2给出了磁-电-弹性半空间中的无量纲化的正应力的分布. 图中显示最大应力发生在沿-轴方向距离半空间边界约$z\approx 1.3a$的位置. 可以推断当温度载荷作用的圆域半径增大时($a$增大), 最大应力发生的位置会远离半无限大体的边界, 即最大应力会发生在更深处; 反之当温度载荷作用的圆域半径减小时($a$减小), 最大应力发生的位置会靠近半无限大体的边界, 即最大应力会发生在靠近边界的位置. 需要指出的是, 在圆域外($r>a)$的正应力很小.

图2

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图2正应力${\sigma _{zz} } / {\beta _3 }T_0 $的分布

Fig.2Distributions of the normalized stress ${\sigma _{zz} } / {\beta _3}T_0 $

图3显示了在圆域温度载荷作用下磁-电-弹性半空间中最大剪应力${\sigma _{rz} } /{\beta _3 }T_0 $的分布. 最大剪应力发生在靠近边界表面大约在圆形区域的边上($r\approx a)$. 在远离温度载荷作用区域($r>2a$, $z>a)$时, 剪应力变得相当小.

图3

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图3剪应力${\sigma _{rz} } / {\beta _3 }\Delta T$的分布

Fig.3Distributions of the shear stress ${\sigma _{rz} } / {\beta _3 }\Delta T$

图4给出了无量纲化的电场的分布, 可以看出, 电场在温度载荷作用的圆域内在边界表面附近有明显的强化. 这一结果预示在高温载荷情形时, 高温度区域可能会出现电击穿的情形. 图5显示了半无限大空间中磁场的分布, 正的和负的磁场在不同的区域都得到了强化: 负的磁场在$z$轴附近约的区域强化显著, 而正的磁场在$z\approx 0.1a$, $r=1.2a$的区域得到了强化.

图4

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图4电场${e_{33} E_z } / {\beta _3 }\Delta T$的分布

Fig.4Distributions of the normalized electric field ${e_{33} E_z } /{\beta _3 }\Delta T$

图5

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图5磁场${h_{33} H_z } / {\beta _3 }\Delta T$的分布

Fig.5Distributions of the normalized magnetic field ${h_{33} H_z } / {\beta _3 }\Delta T$

4 结论

本文给出了磁-电-弹性半空间在轴对称温度载荷作用下的磁-电-热-弹性多场耦合的解析解. 利用变换法成功求解了磁-电-弹性介质中的热传导方程和控制方程, 得到了温度场分布的解析表达式, 并进一步获得了应力, 电位移和磁通量在半空间中的场分布. 数值结果表明在温度载荷作用下应力, 电位移和磁通量在各自特定的区域内会得到强化, 磁-电-弹性场强化区域的强化程度跟温度载荷的大小和作用区域大小相关. 本文结果对磁-电-弹性材料和结构的设计和应用具有重要的理论指导意义.

附录

方程(13), 方程(15)和(16)中的常数 $a_j$, $b_j$, $d_j $ ($j=1,2,3,4)$ 定义为

(A1)$\begin{eqnarray}\%A1 &&\left\{ {{\begin{array}{*{20}c} {a_j } \\ {b_j } \\ {d_j } \\ \end{array} }} \right\}=\left[ {{\begin{array}{*{20}c} {C_{11} -C_{44} \gamma _j^2 } & {e_{31} +e_{15} } & {h_{31} +h_{15} } \\ {(C_{13} +C_{44} )\gamma _j^2 } & {e_{33} \gamma _j^2 -e_{15} } & {h_{33} \gamma _j^2 -h_{15} } \\ {(e_{31} +e_{15} )\gamma _j^2 } & {\lambda _{11} -\lambda _{33} \gamma _j^2 } & {d_{11} -d_{33} \gamma _j^2 } \\ \end{array} }} \right]^{-1}\cdot \\&&\qquad \left\{ {{\begin{array}{*{20}c} {C_{13} +C_{44} } \\ {C_{33} \gamma _j^2 -C_{44} } \\ {e_{33} \gamma _j^2 -e_{15} } \\ \end{array} }} \right\}\ \ (j=1,2,3,4) \end{eqnarray}$

其中, $\gamma _j $ ($j=1,2,3)$ 为特征方程(17)的特征根. 在方程(26)中$\Delta _j$ ($j=1,2,3,4)$ 定义为

(A2a)$\begin{eqnarray} && \varDelta _1 = \frac{1}{\varDelta }\cdot\left| {{\begin{array}{*{20}c} {-\beta _1 } & {(C_{13} +C_{44} )\lambda } & {(e_{31} +e_{15} )\lambda } & {(h_{31} +h_{15} )\lambda } \\[1.5mm] {\beta _3 \lambda } & {C_{44} -C_{33} \lambda ^2} & {e_{15} -e_{33} \lambda ^2} & {h_{15} -h_{33} \lambda ^2} \\[1.5mm] {\beta _4 \lambda } & {e_{15} -e_{33} \lambda ^2} & {\varepsilon _{33} \lambda ^2-\varepsilon _{11} } & {d_{33} \lambda ^2-d_{11} } \\[1.5mm] {\beta _5 \lambda } & {h_{15} -h_{33} \lambda ^2} & {d_{33} \lambda ^2-d_{11} } & {\mu _{33} \lambda ^2-\mu _{11} } \\[1.5mm] \end{array} }} \right|\\[1.5mm] \end{eqnarray}$

(A2b)$\begin{eqnarray} && \varDelta _2 = \frac{1}{\varDelta }\cdot\left| {{\begin{array}{*{20}c} {C_{44} \lambda ^2-C_{11} } & {-\beta _1 } & {(e_{31} +e_{15} )\lambda } & {(h_{31} +h_{15} )\lambda } \\[1.5mm] {(C_{13} +C_{44} )\lambda } & {\beta _3 \lambda } & {e_{15} -e_{33} \lambda ^2} & {h_{15} -h_{33} \lambda ^2} \\[1.5mm] {(e_{31} +e_{15} )\lambda } & {\beta _4 \lambda } & {\varepsilon _{33} \lambda ^2-\varepsilon _{11} } & {d_{33} \lambda ^2-d_{11} } \\[1.5mm] {(h_{31} +h_{15} )\lambda } & {\beta _5 \lambda } & {d_{33} \lambda ^2-d_{11} } & {\mu _{33} \lambda ^2-\mu _{11} } \\[1.5mm] \end{array} }} \right|\\[1.5mm] \end{eqnarray}$

(A2c)$\begin{eqnarray} && \varDelta _3 = \frac{1}{\varDelta }\cdot\left| {{\begin{array}{*{20}c} {C_{44} \lambda ^2-C_{11} } & {(C_{13} +C_{44} )\lambda } & {-\beta _1 } & {(h_{31} +h_{15} )\lambda } \\[1.5mm] {(C_{13} +C_{44} )\lambda } & {C_{44} -C_{33} \lambda ^2} & {\beta _3 \lambda } & {h_{15} -h_{33} \lambda ^2} \\[1.5mm] {(e_{31} +e_{15} )\lambda } & {e_{15} -e_{33} \lambda ^2} & {\beta _4 \lambda } & {d_{33} \lambda ^2-d_{11} } \\[1.5mm] {(h_{31} +h_{15} )\lambda } & {h_{15} -h_{33} \lambda ^2} & {\beta _5 \lambda } & {\mu _{33} \lambda ^2-\mu _{11} } \\[1.5mm] \end{array} }} \right|\\[1.5mm] \end{eqnarray}$

(A2d)$\begin{eqnarray} && \varDelta _4 = \frac{1}{\varDelta }\cdot\left| {{\begin{array}{*{20}c} {C_{44} \lambda ^2-C_{11} } & {(C_{13} +C_{44} )\lambda } & {(e_{31} +e_{15} )\lambda } & {-\beta _1 } \\[1.5mm] {(C_{13} +C_{44} )\lambda } & {C_{44} -C_{33} \lambda ^2} & {e_{15} -e_{33} \lambda ^2} & {\beta _3 \lambda } \\[1.5mm] {(e_{31} +e_{15} )\lambda } & {e_{15} -e_{33} \lambda ^2} & {\varepsilon _{33} \lambda ^2-\varepsilon _{11} } & {\beta _4 \lambda } \\[1.5mm] {(h_{31} +h_{15} )\lambda } & {h_{15} -h_{33} \lambda ^2} & {d_{33} \lambda ^2-d_{11} } & {\beta _5 \lambda } \\[1.5mm] \end{array} }} \right|\\[1.5mm] \end{eqnarray}$

(A2e)$\left. \begin{array}{ll} \varDelta =\left| { \varDelta } \right| \\ { \varDelta }(1,1)=C_{44} \lambda ^2-C_{11}, { \varDelta }(1,2)={ \varDelta }(2,1)=(C_{13} +C_{44} )\lambda \\ { \varDelta }(1,3)={ \varDelta }(3,1)=(e_{31} +e_{15} )\lambda \\ { \varDelta }(1,4)={ \varDelta }(4,1)=(h_{31} +h_{15} )\lambda \\ { \varDelta }(2,2)=C_{44} -C_{33} \lambda ^2, \ \ { \varDelta }(2,3)={ \varDelta }(3,2)=e_{15} -e_{33} \lambda ^2 \\ { \varDelta }(2,4)={ \varDelta }(4,2)=h_{15} -h_{33} \lambda ^2 \\ { \varDelta }(3,3)=\varepsilon _{33} \lambda ^2-\varepsilon _{11}, { \varDelta }(3,4)={ \varDelta }(4,3)=d_{33} \lambda ^2-d_{11} \\ { \varDelta }(4,4)=\mu _{33} \lambda ^2-\mu _{11} \\ \end{array} \right\} $

在方程(30)中$\varOmega _j (j=1,2,3,4)$定义为

(A3a)$\begin{eqnarray} && \varOmega _1 =\frac{1}{\varDelta _X }\left| {{\begin{array}{c@{\quad }c@{\quad }c@{\quad }c} {\delta _1 } & {X_{12} } & {X_{13} } & {X_{14} } \\[1.5mm] {\delta _2 } & {X_{22} } & {X_{23} } & {X_{24} } \\[1.5mm] {\delta _3 } & {X_{32} } & {X_{33} } & {X_{34} } \\[1.5mm] {\delta _4 } & {X_{42} } & {X_{43} } & {X_{44} } \\[1.5mm] \end{array} }} \right|\\[1.5mm] \end{eqnarray}$

(A3b)$\begin{eqnarray} && \varOmega _2 =\frac{1}{\varDelta _X }\left| {{\begin{array}{c@{\quad }c@{\quad }c@{\quad }c} {X_{11} } & {\delta _1 } & {X_{13} } & {X_{14} } \\[1.5mm] {X_{21} } & {\delta _2 } & {X_{23} } & {X_{24} } \\[1.5mm] {X_{31} } & {\delta _3 } & {X_{33} } & {X_{34} } \\[1.5mm] {X_{41} } & {\delta _4 } & {X_{43} } & {X_{44} } \\[1.5mm] \end{array} }} \right|\\[1.5mm] \end{eqnarray}$

(A3c)$\begin{eqnarray} && \varOmega _3 =\frac{1}{\varDelta _X }\left| {{\begin{array}{c@{\quad }c@{\quad }c@{\quad }c} {X_{11} } & {X_{12} } & {\delta _1 } & {X_{14} } \\[1.5mm] {X_{21} } & {X_{22} } & {\delta _2 } & {X_{24} } \\[1.5mm] {X_{31} } & {X_{32} } & {\delta _3 } & {X_{34} } \\[1.5mm] {X_{41} } & {X_{42} } & {\delta _4 } & {X_{44} } \\[1.5mm] \end{array} }} \right|\\[1.5mm] \end{eqnarray}$

(A3d)$\begin{eqnarray} && \varOmega _4 =\frac{1}{\varDelta _X }\left| {{\begin{array}{c@{\quad }c@{\quad }c@{\quad }c} {X_{11} } & {X_{12} } & {X_{13} } & {\delta _1 } \\[1.5mm] {X_{21} } & {X_{22} } & {X_{23} } & {\delta _2 } \\[1.5mm] {X_{31} } & {X_{32} } & {X_{33} } & {\delta _3 } \\[1.5mm] {X_{41} } & {X_{42} } & {X_{43} } & {\delta _4 } \\[1.5mm] \end{array} }} \right|\\[1.5mm] \end{eqnarray}$

(A3e)$\begin{eqnarray} && \varDelta _X =\left| {{\begin{array}{c@{\quad }c@{\quad }c@{\quad }c} {X_{11} } & {X_{12} } & {X_{13} } & {X_{14} } \\[1.5mm] {X_{21} } & {X_{22} } & {X_{23} } & {X_{24} } \\[1.5mm] {X_{31} } & {X_{32} } & {X_{33} } & {X_{34} } \\[1.5mm] {X_{41} } & {X_{42} } & {X_{43} } & {X_{44} } \\[1.5mm] \end{array} }} \right| \end{eqnarray} $

其中分量$X_{ij}$ $(i,j=1,2,3,4)$在方程(25)中定义.

方程(33)$\sim$(38)中的系数$V_k$ $(k=1,2,\dots, 6)$定义为

(A4)$\begin{eqnarray}\%A4 \left\{ {{\begin{array}{*{20}c} {V_1 } \\[.5mm] {V_2 } \\[.5mm] {V_3 } \\[.5mm] {V_4 } \\[.5mm] {V_5 } \\[.5mm] {V_6 } \\[.5mm] \end{array} }} \right\}=\left[ {{\begin{array}{*{20}c} {c_{13} } & {-\lambda c_{33} } & {-\lambda e_{33} } & {-\lambda h_{33} } \\[.5mm] {e_{31} } & {-\lambda e_{33} } & {\lambda \varepsilon _{33} } & {\lambda h_{33} } \\[.5mm] {h_{31} } & {-\lambda h_{33} } & {\lambda d_{33} } & {\lambda \mu _{33} } \\[.5mm] {-\lambda c_{44} } & {-c_{44} } & {-e_{15} } & {-h_{15} } \\[.5mm] {-\lambda e_{15} } & {-e_{15} } & {\varepsilon _{11} } & {d_{11} } \\[.5mm] {-\lambda h_{15} } & {-h_{15} } & {d_{11} } & {\mu _{11} } \\[.5mm] \end{array} }} \right]\left\{ {{\begin{array}{*{20}c} { \varDelta _1 } \\[.5mm] { \varDelta _2 } \\[.5mm] { \varDelta _3 } \\[.5mm] { \varDelta _4 } \\[.5mm] \end{array} }} \right\} \end{eqnarray}$

$E_j$, $F_j$, $G_j$, $H_j$, $K_j$, $L_j$ ($j=1,2,3,4)$定义为

(A5)$\left. \begin{array}{ll} E_j =X_{1j} \varOmega _j,\ \ F_j =X_{3j} \varOmega _j \\[.5mm] G_j =X_{4j} \varOmega _j,\ \ H_j =X_{2j} \varOmega _j \\[.5mm] K_j =X_{5j} \varOmega _j,\ \ L_j =X_{6j} \varOmega _j \\[.5mm] \end{array} \right\}$

(A6a)$\begin{eqnarray} && X_{5j} (\xi )=-e_{15} \left( {a_j \gamma _j +\frac{1}{\gamma _j }} \right) -\varepsilon _{11} \frac{b_j }{\gamma _j }-d_{11} \frac{d_j }{\gamma _j }\end{eqnarray}$

(A6b)$\begin{eqnarray} X_{6j} (\xi )=-h_{15} \left( {a_j \gamma _j +\frac{1}{\gamma _j }} \right) -d_{11} \frac{b_j }{\gamma _j }-\mu _{11} \frac{d_j }{\gamma _j} \end{eqnarray} $

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