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 Journal of Zhejiang University SCIENCE A 2005 Vol.6 No.2 P.126~131 http://doi.org/10.1631/jzus.2005.A0126

The analytical solutions for orthotropic cantilever beams (I): Subjected to surface forces*

 Author(s):  Ai-min Jiang1,2, Hao-jiang Ding1 Affiliation(s):  1. Department of Civil Engineering, Zhejiang University, Hangzhou 310027, China; more Corresponding email(s):   jam@vip.sina.com Key Words:  General solution, Orthotropic media, Cantilever beams, Analytical solutions Share this article to： More <<< Previous Article|Next Article >>>

JIANG Ai-min, DING Hao-jiang. The analytical solutions for orthotropic cantilever beams (I): Subjected to surface forces[J]. Journal of Zhejiang University Science A, 2005, 6(2): 126~131.

@article{title="The analytical solutions for orthotropic cantilever beams (I): Subjected to surface forces",
author="JIANG Ai-min, DING Hao-jiang",
journal="Journal of Zhejiang University Science A",
volume="6",
number="2",
pages="126~131",
year="2005",
publisher="Zhejiang University Press & Springer",
doi="10.1631/jzus.2005.A0126"
}

%0 Journal Article
%T The analytical solutions for orthotropic cantilever beams (I): Subjected to surface forces
%A JIANG Ai-min
%A DING Hao-jiang
%J Journal of Zhejiang University SCIENCE A
%V 6
%N 2
%P 126~131
%@ 1673-565X
%D 2005
%I Zhejiang University Press & Springer
%DOI 10.1631/jzus.2005.A0126

TY - JOUR
T1 - The analytical solutions for orthotropic cantilever beams (I): Subjected to surface forces
A1 - JIANG Ai-min
A1 - DING Hao-jiang
J0 - Journal of Zhejiang University Science A
VL - 6
IS - 2
SP - 126
EP - 131
%@ 1673-565X
Y1 - 2005
PB - Zhejiang University Press & Springer
ER -
DOI - 10.1631/jzus.2005.A0126

Abstract:
This paper first gives the general solution of two-dimensional orthotropic media expressed with two harmonic displacement functions by using the governing equations. Then, based on the general solution in the case of distinct eigenvalues, a series of beam problems, including the problem of cantilever beam under uniform loads, cantilever beam with axial load and bending moment at the free end, cantilever beam under the first, second, third and fourth power of x tangential loads, is solved by the superposition principle and the trial-and-error methods.

## .  INTRODUCTION

The problem of cantilever beams subjected to uniform loads is a classic one in elasticity studies. Timoshenko and Goodier () presented a solution for an isotropic cantilever beam subjected to uniform load and cross load at free end. Lekhnitskii () obtained analytical solutions for an orthotropic cantilever beam subjected to cross load at free end and uniform load on the upper surface. The solutions for constant body force cases were also presented in the above two books. To the authors’ knowledge, no literature about the corresponding solution of orthotropic cantilever beam with variable body forces had been published yet. The problems of density functionally graded media can be transformed into those ones with variable body forces. In order to solve the problems of variable body forces, we should first analyze the solution for cantilever beam with axial load and bending moment at free end, and under the normal and tangential loads on the upper and bottom surfaces.

In this paper, we will consider the orthotropic plane problems. The general solution of two-dimensional orthotropic media expressed with two harmonic displacement functions is given at first by use of the governing equations. Then, based on the general solution in the case of distinct eigenvalues, a series of beam problems, including cantilever beam under uniform loads, cantilever beam with axial load and bending moment at the free end, cantilever beam under the first, second, third and fourth power of x tangential loads, is solved by the trial-and-error methods.

Analytical solutions for various problems are obtained by the superposition principle.

## .  GENERAL SOLUTION FOR THE PLANE PROBLEM OF ORTHOTROPIC SOLID

For the plane problems of orthotropic media, the displacements ui are assumed to be independent of y for the plane-strain case. The basic equations for two-dimensional orthotropic solid in xoz coordinates can be simplified as follows: $$\frac{{\partial {\sigma _x}}}{{\partial x}} + \frac{{\partial {\tau _{xz}}}}{{\partial z}} + {f_x} = 0,$$ $$\frac{{\partial {\tau _{xz}}}}{{\partial x}} + \frac{{\partial {\sigma _z}}}{{\partial z}} + {f_z} = 0$$$${\sigma _x} = {c_{11}}\frac{{\partial u}}{{\partial x}} + {c_{13}}\frac{{\partial w}}{{\partial z}},$$ $${\tau _{xz}} = {c_{55}}\left( {\frac{{\partial u}}{{\partial z}} + \frac{{\partial w}}{{\partial x}}} \right),$$ $${\sigma _z} = {c_{13}}\frac{{\partial u}}{{\partial x}} + {c_{33}}\frac{{\partial w}}{{\partial z}}$$ where σx (σz,τxz) and u(w) are the components of stress and displacement, respectively; fx and fz are body force; cij are the elastic constants.

Governing Eq.(1) can be expressed in terms of u and w by virtue of Eq.(2) as follows $$\left( {{c_{11}}\frac{{{\partial ^2}}}{{\partial {x^2}}} + {c_{55}}\frac{{{\partial ^2}}}{{\partial {z^2}}}} \right)u + ({c_{13}} + {c_{55}})\frac{{{\partial ^2}w}}{{\partial x\partial z}} + {f_x} = 0$$ (3) $$({c_{13}} + {c_{55}})\frac{{{\partial ^2}u}}{{\partial x\partial z}} + \left( {{c_{55}}\frac{{{\partial ^2}}}{{\partial {x^2}}} + {c_{33}}\frac{{{\partial ^2}}}{{\partial {z^2}}}} \right)w + {f_z} = 0$$

Ding et al.(; ) derived the general solution for piezoelectric plane problem without body forces, in which all physical quantities are expressed in three harmonic functions. With the method and the strict differential operator theorem presented in Ding et al.(; ), the general solution of two-dimensional orthotropic media without body forces in the case of distinct eigenvalues can be easily derived and expressed in two harmonic functions as follows $$u = \sum\limits_{j = 1}^2 {\frac{{\partial {\psi _j}}}{{\partial x}},}$$ $$w = \sum\limits_{j = 1}^2 {{s_j}{k_j}\frac{{\partial {\psi _j}}}{{\partial {z_j}}}} ,$$ $${\sigma _x} = \sum\limits_{j = 1}^2 {{\omega _{2j}}\frac{{{\partial ^2}{\psi _j}}}{{\partial z_j^2}},}$$ $${\sigma _z} = \sum\limits_{j = 1}^2 {{\omega _{1j}}\frac{{{\partial ^2}{\psi _j}}}{{\partial z_j^2}}} ,$$ $${\tau _{xz}} = \sum\limits_{j = 1}^2 {{s_j}{\omega _{1j}}\frac{{{\partial ^2}{\psi _j}}}{{\partial x\partial {z_j}}}}$$ where the functions ψj satisfy the following equations: $$\left( {\frac{{{\partial ^2}}}{{\partial x_{}^2}} + \frac{{{\partial ^2}}}{{\partial z_j^2}}} \right){\psi _j} = 0,{\text{ (}}j{\text{ = 1,2)}}$$ where zj=sjz (j=1,2) and $$s_j^2$$ are the two roots of the equation [we take Re(sj)>0] $${a_1}{s^4} - {a_2}{s^2} + {a_3} = 0$$ where $${a_1} = {c_{33}}{c_{44}},$$ $${a_2} = {c_{11}}{c_{33}} + c_{55}^2 - {({c_{13}} + {c_{55}})^2},$$ $${a_3} = {c_{11}}{c_{55}}$$ $${k_j} = \frac{{ - {c_{11}} + {c_{55}}s_j^2}}{{ - ({c_{13}} + {c_{55}})s_j^2}},$$ $${\omega _{1j}} = {c_{33}}s_j^2{k_j} - {c_{13}},$$ $${\omega _{2j}} = - s_j^2{\omega _{1j}},\left( {j = {\text{1}},{\text{2}}} \right)$$

The polynomials listed in Appendix A can be chosen as harmonic functions ψj simply by replacing z

with zj. In the next sections, we will consider three loads cases of cantilever beam shown in Fig.1, and derive the analytical solutions by using the general solution (5).

Fig.1
The geometry and coordinate system of a cantilever beam

## .  THREE SOLUTIONS FOR CANTILEVER BEAM WITHOUT BODY FORCES

### .  Cantilever beam under uniform loads on the upper and bottom surfaces

We introduce the displacement function as follows $$\begin{matrix} {\psi _j} = ({x^2} - z_j^2){A_{2j}} + ({x^2}{z_j} - \frac{1}{3}z_j^3){B_{3j}} \\ + {B_{5j}}({x^4}{z_j} - 2{x^2}z_j^3 + \frac{1}{5}z_j^5) \\ \end{matrix}$$ where A2j, B3j and B5j (j=1,2) are unknown constants to be determined .

Substituting Eq.(9) into Eq.(5) leads to $$u = \sum\limits_{j = 1}^2 {\left[ {2x{A_{2j}} + 2x{z_j}{B_{3j}} + (4{x^3}{z_j} - 4xz_j^3){B_{5j}}} \right]}$$ $$\begin{matrix} w = \sum\limits_{j = 1}^2 {{s_j}{k_j}[ - 2{z_j}{A_{2j}} + } ({x^2} - z_j^2){B_{3j}} \\ + ({x^4} - 6{x^2}z_j^2 + z_j^4){B_{5j}}] \\ \end{matrix}$$ (10b)$${\sigma _z} = \sum\limits_{j = 1}^2 {{\omega _{1j}}[ - 2{A_{2j}} - 2{z_j}{B_{3j}} + ( - 12{x^2}{z_j} + 4z_j^3){B_{5j}}]}$$ $${\tau _{xz}} = \sum\limits_{j = 1}^2 {{s_j}{\omega _{1j}}[2x{B_{3j}} + (4{x^3} - 12xz_j^2){B_{5j}}]}$$ (10d) $${\sigma _x} = \sum\limits_{j = 1}^2 {{\omega _{2j}}[ - 2{A_{2j}}} - 2{z_j}{B_{3j}} + ( - 12{x^2}{z_j} + 4z_j^3){B_{5j}}]$$

The boundary conditions are $$z = \pm h/2:{\sigma _z} = {\beta _1} \pm {C_1},{\text{ }}{\tau _{xz}} = 0$$ $$x = 0:\int_{ - h/2}^{h/2} {{\sigma _x}{\text{d}}z = 0} ,\int_{ - h/2}^{h/2} {{\sigma _x}z{\text{d}}z = 0} ,\int_{ - h/2}^{ + h/2} {{\tau _{xz}}{\text{d}}z = 0}$$ $$(x = L,z = 0):u = 0,w = 0,\partial w/\partial x = 0$$

Substituting Eqs.(10c), (10d) and (10e) into Eqs.(11a) and (11b) , we arrive at $$\sum\limits_{j = 1}^2 {{\omega _{1j}}{A_{2j}} = - {\beta _1}/2} ,$$ $$\sum\limits_{j = 1}^2 {{\omega _{1j}}( - h{s_j}{B_{3j}} + \frac{1}{2}{h^3}s_j^3{B_{5j}}) = {C_1}}$$ $$\sum\limits_{j = 1}^2 {{s_j}{\omega _{1j}}{B_{5j}} = 0} , {\text{ }}\sum\limits_{j = 1}^2 {{\omega _{2j}}{A_{2j}} = 0}$$ $$\sum\limits_{j = 1}^2 {{s_j}{\omega _{1j}}(2{B_{3j}} - 3{h^2}s_j^2{B_{5j}}) = 0}$$ $$\sum\limits_{j = 1}^2 {{s_j}{\omega _{2j}}( - 10{B_{3j}} + 3{h^2}s_j^2{B_{5j}}) = 0}$$

Then, the unknown constants A2j, B3j and B5j (j=1,2) can be determined from Eqs.(12)−(15). To satisfy the boundary conditions Eq.(11c), the solution above should be superposed on the rigid body displacements solutions as follows $${u_1} = {u_0} + {\omega _0}z, {w_1} = {w_0} - {\omega _0}x$$ (16) where $${u_0} = - 2L\sum\limits_{j = 1}^2 {{A_{2j}}} ,{\text{ }}{\omega _0} = 2L\sum\limits_{j = 1}^2 {{s_j}{k_j}({B_{3j}} + 2{L^2}{B_{5j}})}$$ (17a) $${w_0} = {L^2}\sum\limits_{j = 1}^2 {{s_j}{k_j}({B_{3j}} + 3{L^2}{B_{5j}})}$$

### .  Cantilever beam with axial force N and bending moment M at the free end

We constitute the displacement function as follows $${\psi _j} = ({x^2} - z_j^2){A_{2j}} + ({x^2}{z_j} - \frac{1}{3}z_j^3){B_{3j}},{\text{ }}(j = 1,2)$$

Substituting Eq.(18) into Eq.(5) leads to $$\begin{matrix} u = \sum\limits_{j = 1}^2 {(2x{A_{2j}} + 2x{z_j}{B_{3j}})} , \\ w = \sum\limits_{j = 1}^2 {{s_j}{k_j}[ - 2{z_j}{A_{2j}} + ({x^2} - z_j^2){B_{3j}}]} {\text{ }} \\ \end{matrix}$$ (19a)$${\sigma _z} = \sum\limits_{j = 1}^2 {{\omega _{1j}}( - 2{A_{2j}} - 2{z_j}{B_{3j}})} ,{\text{ }}{\tau _{xz}} = 2x\sum\limits_{j = 1}^2 {{s_j}{\omega _{1j}}{B_{3j}}}$$ $${\sigma _x} = \sum\limits_{j = 1}^2 {{\omega _{2j}}( - 2{A_{2j}} - 2{z_j}{B_{3j}})}$$

The boundary conditions are $$z = \pm h/2:{\sigma _z} = 0,{\text{ }}{\tau _{xz}} = 0$$ $$x = 0:\int_{ - h/2}^{h/2} {{\sigma _x}{\text{d}}z = N} ,{\text{ }}\int_{ - h/2}^{h/2} {{\sigma _x}z{\text{d}}z = M} ,{\text{ }}\int_{ - h/2}^{h/2} {{\tau _{xz}}{\text{d}}z = 0}$$ $$(x = L,z = 0):u = 0,{\text{ }}w = 0,{\text{ }}\partial w/\partial x = 0$$

Substituting Eqs.(19b) and (19c) into Eqs.(20a) and (20b), we have $$\sum\limits_{j = 1}^2 {{\omega _{1j}}{A_{2j}} = 0} ,{\text{ }} \sum\limits_{j = 1}^2 {{s_j}{\omega _{1j}}{B_{3j}} = 0}$$ $$- \frac{{{h^3}}}{6}\sum\limits_{j = 1}^2 {{s_j}{\omega _{2j}}{B_{3j}} = M} , {\text{ }} - 2h\sum\limits_{j = 1}^2 {{\omega _{2j}}{A_{2j}} = N}$$

Then, the constants A2j and B3j can be determined from Eqs.(21) and (22). To satisfy the boundary conditions Eq.(20c), the solution above should be superposed on the rigid body displacement solutions as follows $${u_1} = {u_0} + {\omega _0}z, {\text{ }}{w_1} = {w_0} - {\omega _0}x$$ where $${u_0} = - 2L\sum\limits_{j = 1}^2 {{A_{2j}}} ,{\omega _0} = 2L\sum\limits_{j = 1}^2 {{s_j}{k_j}{B_{3j}}} ,{w_0} = {L^2}\sum\limits_{j = 1}^2 {{s_j}{k_j}{B_{3j}}}$$

### .  Cantilever beam with the nth power of x tangential loads on the upper and bottom surfaces

The boundary conditions are taken as $$z = \pm h/2:{\sigma _z} = 0,{\text{ }}{\tau _{xz}} = {T_n}{x^n}$$ $$x = 0:\int_{ - h/2}^{h/2} {{\sigma _x}{\text{d}}z = 0} ,{\text{ }}\int_{ - h/2}^{h/2} {{\sigma _x}z{\text{d}}z = 0} ,{\text{ }}\int_{ - h/2}^{h/2} {{\tau _{xz}}{\text{d}}z = 0}$$ $$(x = L,z = 0):u = 0,{\text{ }}w = 0,{\text{ }}\partial w/\partial x = 0$$

We introduce the displacement function as follows $$\begin{matrix} {\psi _j} = {B_{2j}}\varphi _2^1(x,{z_j}) + {B_{4j}}\varphi _4^1(x,{z_j}) + \cdots + {B_{n + 4,j}}\varphi _{n + 4}^1(x,{z_j}) \\ (j = 1,2; n = 2,4,6, \cdots ) \\ \end{matrix}$$ $$\begin{matrix} {\psi _j} = {B_{3j}}\varphi _3^1(x,{z_j}) + {B_{5j}}\varphi _5^1(x,{z_j}) + \cdots + {B_{n + 4,j}}\varphi _{n + 4}^1(x,{z_j}) \\ (j = 1,2; n = 1,3,5, \cdots ) \\ \end{matrix}$$ where Bmj are undetermined constants, and $$\varphi _m^1(x,{z_j})$$ are taken from Appendix A.

Substituting Eq.(26) into Eq.(5) leads to the expressions of displacements and stresses. When n is an even number, we have $$\begin{matrix} u = \sum\limits_{j = 1}^2 {[{z_j}{B_{2j}} + (3{x^2}{z_j} - z_j^3){B_{4j}} + (5{x^4}{z_j} - 10{x^2}z_j^3} \\ + z_j^5){B_{6j}} + (7{x^6}{z_j} - 35{x^4}z_j^3 + 21{x^2}z_j^5 - z_j^7){B_{8j}} + \cdots ] \\ \end{matrix}$$ (27a) $$\begin{matrix} w = \sum\limits_{j = 1}^2 {{s_j}{k_j}[x{B_{2j}} + ({x^3} - 3xz_j^2){B_{4j}} + ({x^5} - 10{x^3}z_j^2} \\ + 5xz_j^4){B_{6j}} + ({x^7} - 21{x^5}z_j^2 + 35{x^3}z_j^4 - 7xz_j^6){B_{8j}} + \cdots ] \\ \end{matrix}$$ $$\begin{matrix} {\sigma _z} = \sum\limits_{j = 1}^2 {{\omega _{1j}}[ - 6x{z_j}{B_{4j}} + 20x{z_j}(z_j^2 - {x^2}){B_{6j}} + ( - 42{x^5}{z_j}} \\ + 140{x^3}z_j^3 - 42xz_j^5){B_{8j}} + \cdots ] \\ \end{matrix}$$ $$\begin{matrix} {\tau _{xz}} = \sum\limits_{j = 1}^2 {{s_j}{\omega _{1j}}[{B_{2j}} + (3{x^2} - 3z_j^2){B_{4j}} + (5{x^4} - 30{x^2}z_j^2} \\ + 5z_j^4){B_{6j}} + (7{x^6} - 105{x^4}z_j^2 + 105{x^2}z_j^4 - 7z_j^6){B_{8j}} + \cdots ] \\ \end{matrix}$$ $$\begin{matrix} {\sigma _x} = \sum\limits_{j = 1}^2 {{\omega _{2j}}[ - 6x{z_j}{B_{4j}} + 20x{z_j}(z_j^2 - {x^2}){B_{6j}} + ( - 42{x^5}{z_j}} \\ + 140{x^3}z_j^3 - 42xz_j^5){B_{8j}} + \cdots ] \\ \end{matrix}$$ (27e)

When n is an odd number, we have $$\begin{matrix} u = \sum\limits_{j = 1}^2 {[2x{z_j}{B_{3j}} + (4{x^3}{z_j} - 4xz_j^3){B_{5j}}} \\ + (6{x^5}{z_j} - 20{x^3}z_j^3 + 6xz_j^5){B_{7j}} + \cdots ] \\ \end{matrix}$$ $$\begin{matrix} w = \sum\limits_{j = 1}^2 {{s_j}{k_j}[({x^2} - z_j^2){B_{3j}} + ({x^4} - 6{x^2}z_j^2 + z_j^4){B_{5j}} + ({x^6}} \\ - 15{x^4}z_j^2 + 15{x^2}z_j^4 - z_j^6){B_{7j}} + \cdots ] \\ \end{matrix}$$ $$\begin{matrix} {\sigma _z} = \sum\limits_{j = 1}^2 {{\omega _{1j}}[ - 2{z_j}{B_{3j}} + ( - 12{x^2}{z_j} + 4z_j^3){B_{5j}}} \\ + ( - 30{x^4}{z_j} + 60{x^2}z_j^3 - 6z_j^5){B_{7j}} + \cdots ] \\ \end{matrix}$$ $$\begin{matrix} {\sigma _x} = \sum\limits_{j = 1}^2 {{\omega _{2j}}[ - 2{z_j}{B_{3j}} + ( - 12{x^2}{z_j} + 4z_j^3){B_{5j}}} \\ + ( - 30{x^4}{z_j} + 60{x^2}z_j^3 - 6z_j^5){B_{7j}} + \cdots ] \\ \end{matrix}$$ $$\begin{matrix} {\tau _{xz}} = \sum\limits_{j = 1}^2 {{s_j}{\omega _{1j}}[2x{B_{3j}} + 4({x^3} - 3xz_j^2){B_{5j}} + 6({x^5}} \\ - 10{x^3}z_j^2 + 5xz_j^4){B_{7j}} + \cdots ] \\ \end{matrix}$$

When n=1, we substitute Eqs.(28c), (28d) and (28e) into Eqs.(25a) and (25b) and have $$\sum\limits_{j = 1}^2 {{\omega _{1j}}( - h{s_j}{B_{3j}} + \frac{1}{2}{h^3}s_j^3{B_{5j}}) = 0} ,\sum\limits_{j = 1}^2 {{s_j}{\omega _{1j}}{B_{5j}} = 0}$$ $$\sum\limits_{j = 1}^2 {{s_j}{\omega _{1j}}(2{B_{3j}} - 3{h^2}s_j^2{B_{5j}}) = {T_1}} ,$$ $$\sum\limits_{j = 1}^2 {{s_j}{\omega _{2j}}( - 10{B_{3j}} + 3{h^2}s_j^2{B_{5j}}) = 0}$$

Then, the unknown constants B3j and B5j can be determined from Eqs.(29) and (30). To satisfy the boundary conditions in Eq.(25c), the solution above should be superposed on the rigid body displacements solutions as follows $${u_1} = {\omega _0}z, {\text{ }}{w_1} = {w_0} - {\omega _0}x$$ where $$\begin{matrix} {\omega _0} = 2L\sum\limits_{j = 1}^2 {{s_j}{k_j}({B_{3j}} + 2{L^2}{B_{5j}})} , \\ {w_0} = {L^2}\sum\limits_{j = 1}^2 {{s_j}{k_j}({B_{3j}} + 3{L^2}{B_{5j}})} \\ \end{matrix}$$

When n=3, we substitute Eqs.(28c), (28d) and (28e) into Eqs.(25a) and (25b) and have $$\sum\limits_{j = 1}^2 {{\omega _{1j}}\left( { - 6h{s_j}{B_{5j}} + \frac{{15}}{2}{h^3}s_j^3{B_{7j}}} \right) = 0} , {\text{ }}\sum\limits_{j = 1}^2 {{s_j}{\omega _{1j}}{B_{7j}} = 0}$$ $$\sum\limits_{j = 1}^2 {{\omega _{1j}}\left( { - {s_j}h{B_{3j}} + \frac{1}{2}s_j^3{h^3}{B_{5j}} - \frac{3}{{16}}{h^5}s_j^5{B_{7j}}} \right) = 0}$$ (34) $$\sum\limits_{j = 1}^2 {{s_j}{\omega _{1j}}(4{B_{5j}} - 15s_j^2{h^2}{B_{7j}})} = {T_3}$$ $$\sum\limits_{j = 1}^2 {{s_j}{\omega _{1j}}\left( {2{B_{3j}} - 3{h^2}s_j^2{B_{5j}} + \frac{{15}}{8}s_j^4{h^4}{B_{7j}}} \right) = 0}$$ (36) $$\sum\limits_{j = 1}^2 {{s_j}{\omega _{2j}}\left( { - \frac{1}{3}{B_{3j}} + \frac{1}{{10}}s_j^2{h^2}{B_{5j}} - \frac{3}{{112}}s_j^4{h^4}{B_{7j}}} \right) = 0}$$

Then, B3j, B5j and B7j can be determined from Eqs.(33)−(37). To satisfy the left boundary conditions in Eq.(25c), the solution above should be superposed on the rigid body displacement solutions as follows $${u_1} = {\omega _0}z,{\text{ }} {w_1} = {w_0} - {\omega _0}x$$ where $${\omega _0} = 2L\sum\limits_{j = 1}^2 {{s_j}{k_j}({B_{3j}} + 2{L^2}{B_{5j}} + 3{L^4}{B_{7j}})}$$ (39a) $${w_0} = {L^2}\sum\limits_{j = 1}^2 {{s_j}{k_j}({B_{3j}} + 3{L^2}{B_{5j}} + 5{L^4}{B_{7j}})}$$

When n=2, we substitute Eqs.(27c), (27d) and (27e) into Eqs.(25a) and (25b) and have $$\sum\limits_{j = 1}^2 {{\omega _{1j}}\left( { - 3h{s_j}{B_{4j}} + \frac{5}{2}{h^3}s_j^3{B_{6j}}} \right) = 0} , {\text{ }}\sum\limits_{j = 1}^2 {{s_j}{\omega _{1j}}{B_{6j}} = 0}$$ $$\sum\limits_{j = 1}^2 {{s_j}{\omega _{1j}}\left( {3{B_{4j}} - \frac{{15}}{2}{h^2}s_j^2{B_{6j}}} \right) = {T_2}}$$ $$\sum\limits_{j = 1}^2 {{s_j}{\omega _{1j}}\left( {{B_{2j}} - \frac{3}{4}{h^2}s_j^2{B_{4j}} + \frac{5}{{16}}s_j^4{h^4}{B_{6j}}} \right) = 0}$$ (42) $$\sum\limits_{j = 1}^2 {{s_j}{\omega _{1j}}\left( {{B_{2j}} - \frac{1}{4}s_j^2{h^2}{B_{4j}} + \frac{1}{{16}}s_j^4{h^4}{B_{6j}}} \right) = 0}$$

Substituting Eq.(27b) into the third of Eq.(25c), we have $$\sum\limits_{j = 1}^2 {{s_j}{k_j}({B_{2j}} + 3{L^2}{B_{4j}} + 5{L^4}{B_{6j}}) = 0}$$

Then, the constants B2j, B4j and B6j can be determined from Eqs.(40)−(44). To satisfy the left boundary conditions of Eq.(25c), the solution above should be superposed on the rigid body displacements solutions as follows $${w_1} = {w_0} = - L\sum\limits_{j = 1}^2 {{s_j}{k_j}({B_{2j}} + {L^2}{B_{4j}} + {L^4}{B_{6j}})}$$

When n=4, we substitute Eqs.(27c), (27d) and (27e) into Eqs.(25a) and (25b) and have $$\sum\limits_{j = 1}^2 {{s_j}{\omega _{1j}}{B_{8j}} = 0} , {\text{ }}\sum\limits_{j = 1}^2 {{\omega _{1j}}\left( { - 10{s_j}h{B_{6j}} + \frac{{35}}{2}s_j^3{h^3}{B_{8j}}} \right) = 0}$$ $$\sum\limits_{j = 1}^2 {{\omega _{1j}}\left( { - 3{s_j}h{B_{4j}} + \frac{5}{2}s_j^3{h^3}{B_{6j}} - \frac{{21}}{{16}}s_j^5{h^5}{B_{8j}}} \right) = 0}$$ $$\sum\limits_{j = 1}^2 {{s_j}{\omega _{1j}}\left( {5{B_{6j}} - \frac{{105}}{4}s_j^2{h^2}{B_{8j}}} \right) = {T_4}}$$ $$\sum\limits_{j = 1}^2 {{s_j}{\omega _{1j}}\left( {3{B_{4j}} - \frac{{15}}{2}s_j^2{h^2}{B_{6j}} + \frac{{105}}{{16}}s_j^4{h^4}{B_{8j}}} \right) = 0}$$ $$\sum\limits_{j = 1}^2 {{s_j}{\omega _{1j}}\left( {{B_{2j}} - \frac{3}{4}s_j^2{h^2}{B_{4j}} + \frac{5}{{16}}s_j^4{h^4}{B_{6j}} - \frac{7}{{64}}s_j^6{h^6}{B_{8j}}} \right)} = 0$$ $$\sum\limits_{j = 1}^2 {{s_j}{\omega _{1j}}\left( {{B_{2j}} - \frac{1}{4}s_j^2{h^2}{B_{4j}} + \frac{1}{{16}}s_j^4{h^4}{B_{6j}} - \frac{1}{{64}}s_j^6{h^6}{B_{8j}}} \right)} = 0$$

From Eq.(27b) and Eq.(25c), we have $$\sum\limits_{j = 1}^2 {{s_j}{k_j}({B_{2j}} + 3{L^2}{B_{4j}} + 5{L^4}{B_{6j}} + 7{L^6}{B_{8j}}) = 0}$$

Then, B2j, B4j, B6j and B8j can be determined from Eqs.

(46)−(52). To satisfy the left boundary conditions in Eq.(25c), the solution above should be superposed on the rigid body displacement solutions as follows $${w_1} = {w_0} = - L\sum\limits_{j = 1}^2 {{s_j}{k_j}({B_{2j}} + {L^2}{B_{4j}} + {L^4}{B_{6j}} + {L^6}{B_{8j}})}$$

APPENDIX A

Harmonic polynomials for the plane problems can be written in the following form: $$\begin{matrix} \varphi _n^m(x,z) = {x^{n - m}}{z^m} + \sum\limits_{i = 1}^{[(n - m)/2]} {{{( - 1)}^i}} \frac{{(n - m)(n - m - 1) \cdot \cdot \cdot (n - m - 2i + 1)}}{{(2i + m){\text{!}}}}{x^{n - 2i - m}}{z^{2i + m}} \\ (m = 0,1; n = 1,2,...) \\ \end{matrix}$$ where [(n(m)/2] denotes the largest integer ((n(m)/2. From Eq.(A1),the first seventeen harmonic polynomials can be written as follows: $$\varphi _0^0(x,z) = 1,$$ $$\varphi _1^0(x,z) = x$$ , $$\varphi _1^1(x,z) = z$$ $$\varphi _2^0(x,z) = {x^2} - {z^2}$$ , $$\varphi _2^1(x,z) = xz$$ $$\varphi _3^0(x,z) = {x^3} - 3x{z^2}$$ , $$\varphi _3^1(x,z) = {x^2}z - \frac{1}{3}{z^3}$$ $$\varphi _4^0(x,z) = {x^4} - 6{x^2}{z^2} + {z^4}$$ $$\varphi _4^1(x,z) = {x^3}z - x{z^3}$$ $$\varphi _5^0(x,z) = {x^5} - 10{x^3}{z^2} + 5x{z^4}$$ $$\varphi _5^1(x,z) = {x^4}z - 2{x^2}{z^3} + \frac{1}{5}{z^5}$$ $$\varphi _6^0(x,z) = {x^6} - 15{x^4}{z^2} + 15{x^2}{z^4} - {z^6}$$ $$\varphi _6^1(x,z) = {x^5}z - \frac{{10}}{3}{x^3}{z^3} + x{z^5}$$ $$\varphi _7^0(x,z) = {x^7} - 21{x^5}{z^2} + 35{x^3}{z^4} - 7x{z^6}$$ $$\varphi _7^1(x,z) = {x^6}z - 5{x^4}{z^3} + 3{x^2}{z^5} - \frac{1}{7}{z^7}$$ $$\varphi _8^0(x,z) = {x^8} - 28{x^6}{z^2} + 70{x^4}{z^4} - 28{x^2}{z^6} + {z^8}$$ $$\varphi _8^1(x,z) = {x^7}z - 7{x^5}{z^3} + 7{x^3}{z^5} - x{z^7}$$

* Project (Nos. 10432030, 10472102) supported by the National Natural Science Foundation of China

## References

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[2] Ding, H.J., Wang, G.Q., Chen, W.Q., 1997. Green’s functions for a two-phase infinite piezoelectric plane. Proceedings of Royal Society of London (A), 453:2241-57.

[3] Lekhnitskii, S.G., 1969. Anisotropic Plate. , Gordon and Breach, London, :

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