# Symplectic Analytical Solutions for the Magnetoelectroelastic Solids Plane Problem in Rectangular Domain

Authors: Xiao-Chuan Li; Wei-An Yao
Journal: Journal of Applied Mathematics
 (2011)

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## Abstract

The transversely isotropic magnetoelectroelastic solids plane problem in rectangular domain is derived to Hamiltonian system. In symplectic geometry space with the origin variables—displacements, electric potential, and magnetic potential, as well as their duality variables—lengthways stress, electric displacement, and magnetic induction, on the basis of the obtained eigensolutions of zero-eigenvalue, the eigensolutions of nonzero-eigenvalues are also obtained. The former are the basic solutions of Saint-Venant problem, and the latter are the solutions which have the local effect, decay drastically with respect to distance, and are covered in the Saint-Venant principle. So the complete solution of the problem is given out by the symplectic eigensolutions expansion. Finally, a few examples are selected and their analytical solutions are presented.


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## Body

## 1. Introduction

Magnetoelectroelastic solids are a kind of the emerging functional composite material. Due to possessing mechanical, electric and magnetic field coupling capacity, these materials show better foreground in many high-tech areas (see [1]). They not only convert energy from one form to the other (among magnetic, electric, and mechanical energies), but also possess some new properties of magnetoelectric effect, which are not found in single-phase piezoelectric or piezomagnetic materials. Over the years, a large amount of studies have been done in mechanics, materials science, and physics fields (see [2–6]), and it has become a new cross subject.Due to multifields  coupling, the magnetoelectroelastic solids problem is solved more difficultly than elasticity one. Zhong et al. (see [7, 8]) introduced a symplectic dual method based on the conservative Hamiltonian system to solve the elastic problem which is different from the traditional semi-inverse solution method. The complete solutions space can be obtained and a satisfactory solution can be obtained under the boundary conditions (see [9–11]). The symplectic dual method has been developed for studying piezoelectric effects (see [12]) and magnetoelectroelastic problems (see [13]).With the symplectic approach, the plane problem of magnetoelectroelastic solids in rectangular domain is derived into the Hamiltonian system by means of the generalized variable principle of the magnetoelectroelastic solids. In symplectic geometry space with the origin variables—displacements, electric potential, and magnetic potential, as well as their duality variables—lengthways stress, electric displacement, and magnetic induction, symplectic dual equations are employed. Yao and Li have obtained all the eigensolutions of zero-eigenvalue, which have their specific physical interpretation and are the basic solutions of plane Saint-Venant problem in [13]. This paper gets the eigensolutions of nonzero-eigenvalues, which are the solutions that have the local effect, decay drastically with respect to distance, and are covered in the Saint-Venant principle. So the complete solution of the problem is given out by the symplectic eigensolutions expansion. Finally, a few examples are selected and their analytical solutions are presented.


## 2. Functional Equations and Boundary Conditions

The transversely isotropic magnetoelectroelastic solids are studied here, with thez-axis being the polar direction. If geometry size, load, and so forth, in y direction satisfy the given condition, the problem can be simplified as a plane problem in the xoz plane. The functional equations of the plane problem are as follows (see [14]).(1)
Governing equations:(2.1)∂σx∂x+∂τxz∂z+fx=0,∂τxz∂x+∂σz∂z+fz=0,∂Dx∂x+∂Dz∂z=ρ,∂Bx∂x+∂Bz∂z=0.(2)
Gradient equations:(2.2)εx=∂u∂x,εz=∂w∂z,γxz=∂w∂x+∂u∂z,Ex=-∂ϕ∂x,Ez=-∂ϕ∂z,Hx=-∂ψ∂x,Hz=-∂ψ∂z.(3)
Constitutive equations:(2.3)σx=c11εx+c13εz-e31Ez-q31Hz,σz=c13εx+c33εz-e33Ez-q33Hz,τxz=c44γxz-e15Ex-q15Hx,Dx=e15γxz+κ11Ex+α11Hx,Dz=e31εx+e33εz+κ33Ez+α33Hz,Bx=q15γxz+α11Ex+μ11Hx,Bz=q31εx+q33εz+α33Ez+μ33Hz,where u,w are displacement components in the x,z direction, respectively; σx, σz, and τxz are stress components, respectively; Dx, Dz, and ϕ are electric  displacement components and electric potential, respectively; Bx,Bz, and ψ are magnetic induction components and magnetic potential, respectively; fx,fz, and ρ are body force components and density of free charges in region V, respectively. cij,κij, and μij are elastic modus, dielectric constant, and magnetic constants, respectively; eij, qij, and αij are piezoelectric, piezomagnetic, and electromagnetic constants, respectively. Equation (2.3) can also be rewritten as follows:
(2.4)εx=s11σx+s13σz+d31Dz+b31Bz,εz=s13σx+s33σz+d33Dz+b33Bz,γxz=s44τxz+d15Dx+b15Bx,Ex=-d15τxz+λ11Dx+β11Bx,Ez=-d31σx-d33σz+λ33Dz+β33Bz,Hx=-b15τxz+β11Dx+ν11Bx,Hz=-b31σx-b33σz+β33Dz+ν33Bz.The rectangle domain as showing in Figure1 is studied in this paper
(2.5)V:0≤z≤l,-h≤x≤h.Figure 1
The rectangular domain problems on magnetoelectroelastic solids.And the boundary conditions are expressed as (see [15])
(2.6)σx=F̅x1(z),τxz=F̅z1(z),Dx=D̅x1(z),Bx=B̅x1(z)onx=-h,σx=F̅x2(z),τxz=F̅z2(z),Dx=D̅x2(z),Bx=B̅x2(z)onx=h.
In addition, there are body force and density of free charges in region V.Onz=0,l, the boundary conditions are (see [15])
(2.7)w=w̅,u=u̅,ϕ=ϕ̅,ψ=ψ̅onz=0orl.
Or
(2.8)σz=σ̅z,τxz=τ̅xz,Dz=D̅z,Bz=B̅zonz=0orl.


## 3. Duality Equation in Symplectic Geometry Space and Separation of Variables (See [13])

Consider rectangle domain as showing in Figure1. At first, the coordinate z here is employed to simulate the time variable in the Hamiltonian system, and a symbol “̇” in the following derivation will be used denoting the differential with respect to z, that is, (̇)=∂/∂z. For the sake of simplicity, the notations σ,τ,D, and B are introduced to represent σz,τxz,Dz, and Bz, respectively. If we omit body forces and density of free charges, a dual equation with the full state function vector is given as
(3.1)v̇=Hv,
and H is the operator  matrix in [13]
(3.2)H=[0C10∂∂x00C110C14C15-∂∂x0-C21∂∂x-C22∂∂x0C17000C18∂∂x00C140-C12-C160C19∂∂x00C150-C16-C1300000-∂∂x000-C20∂2∂x200C10∂∂x0C18∂∂xC19∂∂x00C23∂2∂x2-C24∂2∂x20-C21∂∂x0000-C24∂2∂x2C25∂2∂x20-C22∂∂x00],
where the constants C10~C25 are listed in the appendix.Consider the following boundary conditions on two sides in [13]:
(3.3)1s11(∂u∂x-s13σ-d31D-b31B)=0,τ=0,1Δ(-ν11∂ϕ∂x+β11∂ψ∂x+C1τ)=0,1Δ(β11∂ϕ∂x-λ11∂ψ∂x+C2τ)=0,onx=±h,
where constant Δ,C1, and C2 are listed in the appendix. Equation (3.1) can usually be solved by using the method of partition of variables. Let
(3.4)v(x,z)=ξ(z)Y(x),
and substitute (3.4) into (3.1); the following equations can be obtained:
(3.5)HY(x)=μY(x),ξ(z)=eμz,
where μ is the eigenvalue and Y(x) is the eigenfunction vector, which satisfies the boundary condition (3.3) on x=±h.The full state vectorsv form a symplectic space according to the following definition of symplectic inner product:
(3.6)〈v(1),v(2)〉=def∫-hhv(1)TJv(2)dx,J=[0I4-I40].If onlyv(1) and v(2) meet the requirement of (3.3), the following invariant is obtained as
(3.7)〈v(1),Hv(2)〉≡〈v(2),Hv(1)〉.
Therefore, the operator matrix H is the Hamiltonian operator matrix in the symplectic geometry space. So its eigenvalues have some characteristics, that is, if μ is an eigenvalue, then -μ must also be one, and the eigenfunction vectors satisfy the symplectic adjoint orthogonal relationship. After eigenvalues and eigenfunction vectors of H are given, the origin problem can be solved by the method of eigenfunction expansion (see [7, 8]).


## 4. Eigenfunction Vectors of Eigenvalue Zero (See [13])

With the free boundary condition at both sides(x=±h), the magnetoelectroelastic solids plane problem in rectangular domain exists eigenvalue zero.  The eigenvalue zero is a special eigenvalue of the Hamiltonian operator matrix, which eigenfunction vectors are not only the fundamental solution with the special physical significance but also a nondecaying solution. Solve the following eigenequation with conditions (3.3):
(4.1)HY(x)=0.
After obtaining the fundamental eigensolutions and eigensolutions in Jordan form, the solutions of (3.1) can be given as
(4.2)v0f(0)=(10000000)T,v0s(0)=(01000000)T,v0t(0)=(00100000)T,v0r(0)=(00010000)T,v0f(1)=(zb1x00a10a2a3)T,v0s(1)=(-xz000000)T,v0t(1)=(0b2xz0a20a4a5)T,v0r(1)=(0b3x0za30a5a6)T,v0s(2)=(-xz0.5(z2-b1x2)00-a1x0-a2x-a3x)T.
The component forms of v0s(3) are
(4.3)w=a8x3+b5x-12xz2,u=-12b1x2z+16z3,ϕ=a9(x3-3h2x),ψ=a10(x3-3h2x),σ=-a1xz,τ=12a1(x2-h2),D=-a2xz,B=-a3xz.
The constants in (4.2)-(4.3) are listed in [13]. The eigensolutions of eigenvalue zero just are the fundamental solution of Saint Venant problem. But the solutions having the local effect are constituted by the eigensolutions of eigenvalue nonzero of (3.1), which are obtained in the next section.


## 5. Eigenfunction Vectors of Eigenvalue Nonzero

The eigenvalue equation (3.5) is a system of ordinary differential equations with respect to x which can be solved by first determining the eigenvalue λ̃ with respect to the x direction.  The corresponding equation is
(5.1)det⁡⁡[-μC10λ̃00C110C14C15-λ̃-μ-C21λ̃-C22λ̃0C17000C18λ̃-μ0C140-C12-C160C19λ̃0-μC150-C16-C130000-μ-λ̃000-C20λ̃200C10λ̃-μC18λ̃C19λ̃00C23λ̃2-C24λ̃20-C21λ̃-μ000-C24λ̃2C25λ̃20-C22λ̃0-μ]=0,
where the constants C10~C25 are listed in the appendix. Expanding the determinant yields the eigenvalue equation
(5.2)c1λ̃8+c2λ̃6μ2+c3λ̃4μ4+c4λ̃2μ6+μ8=0.
Apparently, where c1~c4 constituted by C10~C25,  (5.2) has eight roots
(5.3)λ̃i=λiμ,λ̃4+i=-λiμ(i=1,2,3,4).Only discuss the general case that there are eight different rootsλ̃i. Obviously, -λ̃i must be an eigenvalue if λ̃i is an eigenvalue, so the general solution of (5.2) can be expressed as
(5.4)ŵ=∑i=14A1icosh⁡⁡(λiμx)+∑i=14D1isinh⁡⁡(λiμx),û=∑i=14A2isinh⁡⁡(λiμx)+∑i=14D2icosh⁡⁡(λiμx),ϕ̂=∑i=14A3icosh⁡⁡(λiμx)+∑i=14D3isinh⁡⁡(λiμx),ψ̂=∑i=14A4icosh⁡⁡(λiμx)+∑i=14D4isinh⁡⁡(λiμx),σ̂=∑i=14A5icosh⁡⁡(λiμx)+∑i=14D5isinh⁡⁡(λiμx),τ̂=∑i=14A6isinh⁡⁡(λiμx)+∑i=14D6icosh⁡⁡(λiμx),D̂=∑i=14A7icosh⁡⁡(λiμx)+∑i=14D7isinh⁡⁡(λiμx),B̂=∑i=14A8icosh⁡⁡(λiμx)+∑i=14D8isinh⁡⁡(λiμx).It shows that the partial solutions relevant toA are the solutions of symmetric deformation with the z-axis while the partial solutions relevant to D are the solutions of antisymmetric deformation with the z-axis.Firstly discuss the general solution of symmetric deformation(5.5)ŵ=∑i=14A1icosh⁡⁡(λiμx),û=∑i=14A2isinh⁡⁡(λiμx),ϕ̂=∑i=14A3icosh⁡⁡(λiμx),ψ̂=∑i=14A4icosh⁡⁡(λiμx),σ̂=∑i=14A5icosh⁡⁡(λiμx),τ̂=∑i=14A6isinh⁡⁡(λiμx),D̂=∑i=14A7icosh⁡⁡(λiμx),B̂=∑i=14A8icosh⁡⁡(λiμx),
where constants Aji(j=1,2,…,8;i=1,2,3,4) are not all independent. Substitute (5.5) into (3.5); Aji can be expressed  by the independent constants A6i(i=1,2,3,4)(5.6)Aji=fNj(λi)fDj(λi)μA6i(j=1,2,3,4),A5i=-λiA6i,Aji=fNj(λi)fDj(λi)A6i(j=7,8),
where fNj(λi) and fDj(λi) are expressions of λi.So far, onlyA6i(i=1,2,3,4) are unknown constants in the general solution. Substituting (5.5) and (5.6) into the boundary conditions (3.3) yields
(5.7)∑i=14(λifN2(λi)fD2(λi)+s13λi-d31fN7(λi)fD7(λi)-b31fN8(λi)fD8(λi))cosh⁡⁡(λiμh)A6i=0,∑i=14sinh⁡⁡(λiμh)A6i=0,∑i=14(C1-ν11λifN3(λi)fD3(λi)+β11λifN4(λi)fD4(λi))sinh⁡⁡(λiμh)A6i=0,∑i=14(C2+β11λifN3(λi)fD3(λi)-λ11λifN4(λi)fD4(λi))sinh⁡⁡(λiμh)A6i=0.
For the sake of simplicity, (5.7) is denoted as
(5.8){Bji}{A6i}={0}(i,j=1,2,3,4).
For nontrivial solution to exist, the determinant of coefficient matrix vanishes. 						(5.9)|{Bji}|=0.
Equation (5.9) can be solved by numerical methods. If μn(n=1,2,…) is roots of (5.9), each μn in reality has its symplectic adjoint eigenvalue -μn and their complex conjugate eigenvalues. After obtaining A6i(i,j=1,2,3,4) by substituting μn into (5.8), the corresponding eigenvector function is
(5.10)Yn=(ŵnûnϕ̂nψ̂nσ̂nτ̂nD̂nB̂n)T,
where
(5.11)ŵn=∑i=14fN1(λi)fD1(λi)μA6icosh⁡⁡(λiμnx),ûn=∑i=14fN2(λi)fD2(λi)μA6isinh⁡⁡(λiμnx),ϕ̂n=∑i=14fN3(λi)fD3(λi)μA6icosh⁡⁡(λiμnx),ψ̂n=∑i=14fN4(λi)fD4(λi)μA6icosh⁡⁡(λiμnx),σ̂n=∑i=14-λiA6icosh⁡⁡(λiμnx),τ̂n=∑i=14A6isinh⁡⁡(λiμnx),D̂n=∑i=14fN7(λi)fD7(λi)A6icosh⁡⁡(λiμnx),B̂n=∑i=14fN8(λi)fD8(λi)A6icosh⁡⁡(λiμnx).The corresponding solution of (3.1) in symmetric deformation is
(5.12)vn=eμnzYn.Likewise, the corresponding eigenvector function of eigenvaluesμn′(n=1,2,…) in antisymmetric deformation is
(5.13)Yn′=(ŵn′ûn′ϕ̂n′ψ̂n′σ̂n′τ̂n′D̂n′B̂n′)T,
where
(5.14)ŵn′=∑i=14D1isinh⁡⁡(λiμn′x),ûn′=∑i=14D2icosh⁡⁡(λiμn′x),ϕ̂n′=∑i=14D3isinh⁡⁡(λiμn′x),ψ̂n′=∑i=14D4isinh⁡⁡(λiμn′x),σ̂n′=∑i=14D5isinh⁡⁡(λiμn′x),τ̂n′=∑i=14D6icosh⁡⁡(λiμn′x),D̂n′=∑i=14D7isinh⁡⁡(λiμn′x),B̂n′=∑i=14D8isinh⁡⁡(λiμn′x).The corresponding solution of (3.1) in antisymmetric deformation is
(5.15)vn′=eμn′zYn′.Thus, all eigensolutions of nonzero eigenvalues are obtained. These solutions are covered in the Saint-Venant principle and decay with distance depending on the characteristics of eigenvalues (see [7, 8]). Together with the eigensolutions of zero eigenvalue, they constitute a complete adjoint symplectic orthonormal basis and the expansion theorem is then applicable. Further discussion concerns solutions of magnetoelectroelastic solids plane problems in rectangular domain.


## 6. Solutions of Generalized Plane Problems in Rectangular Domain

The solution of homogeneous (3.1) by the method of separation variables has been discussed in the previous several sections. The analytical expressions of eigensolutions of zero eigenvalue and of nonzero eigenvalues have also been presented. Based on expansion theorems, the general solution of homogeneous equation (3.1) for magnetoelectroelastic solids plane problems in rectangular domain is
(6.1)v=∑i=01(a0f(i)v0f(i)+a0t(i)v0t(i)+a0r(i)v0r(i))+∑i=03a0s(i)v0s(i)+∑i=1∞ãivi+∑i=1∞b̃ivi′,
where a0f(i), a0s(i), a0t(i), a0r(i), ãi, and b̃i are undetermined constants. Determining these constants needs the variational equations corresponding to the two-point boundary conditions obtained by applying the variational principle.If the end boundary conditions for specified generalized displacements are (2.7), it can be expressed as
(6.2)q0=q̅0(x)=[w̅0(x)u̅0(x)ϕ̅0(x)ψ̅0(x)]Tatz=0,ql=q̅l(x)=[w̅l(x)u̅l(x)ϕ̅l(x)ψ̅l(x)]Tatz=l,
where q0 and ql denote the values of variable q at z=0 and z=l, respectively.If the end boundary conditions for specified generalized forces are (2.8), it can be expressed as
(6.3)p0=p̅0(x)=[σ̅0(x)τ̅0(x)D̅0(x)B̅0(x)]Tz=0,pl=p̅l(x)=[σ̅l(x)τ̅l(x)D̅l(x)B̅l(x)]Tz=l,
where p0 and pl denote the values of variable p at z=0 and z=l, respectively.The boundary conditions at both ends (z=0orl) can also be mixed boundary conditions.According to approach derived by Zhong et al.  in [7, 8] in the Hamiltonian system, the process of obtaining the undetermined constants of the general solution is presented here. For the sake of simplicity, the general solution is given as
(6.4)v=∑i=1nAivi,
where n is the number of eigensolutions, vi are all solutions that relate to zero eigenvalue and nonzero eigenvalues, and Ai are constants determined by a linear system of equations resulted from the Hamiltonian variational principle. If we specify generalized displacement q̅0 at z=0 and generalized force p̅l at z=l, the variational equation is obtained as
(6.5)∫-hh(∑i=1nδAip0iT)(∑j=1nAjq0j-q̅0)dx+∫-hh(∑i=1nδAiqliT)(∑j=1nAjplj-p̅l)dx=0.
Equation (6.5) can be expressed as
(6.6)∑i=1n∑j=1n[∫-hh[Aj(p0iTq0j+qliTplj)-p0iTq̅0-qliTp̅l]dx]δAi=0.Ai can be determined by the following equations:
(6.7)[c̃11c̃12⋯c̃1nc̃21c̃22⋯c̃2n⋮⋮⋱⋮c̃n1c̃n2⋯c̃nn]{A1A2⋮An}={d̃1d̃2⋮d̃n},
where
(6.8)c̃ij=∫-hh(p0iTq0j+qliTplj)dx,d̃i=∫-hh(p0iTq̅0+qliTp̅l)dx(i,j=1,2,…,n).


## 7. Numerical Examples

The Saint-Venant principle is applicable to the problem forl≫h in a rectangular domain. The influence of self-equilibrium forces at both ends (z=0orl) is only confined to the vicinity of the region. It is then appropriate to neglect the solutions of nonzero eigenvalues and, therefore, to apply only the solutions of zero eigenvalue in the expansion theorem.Example 7.1.
Consider a magnetoelectroelastic rectangle domain, under uniform axial tension, electrical displacement, or magnetic induction, respectively. Three load cases are considered and the boundary conditions are given byz=l:
(a)
σz=P0,τxz=Dz=Bz=0, under the uniform tension;(b)
Dz=D0,σz=τxz=Bz=0, under the uniform electric displacement;(c)
Bz=B0,σz=τxz=Dz=0, under the uniform magnetic induction.z=0:
τxz=w=ϕ=ψ=0;x=±h:σx=τxz=Dx=Bx=0.
The problem is treated as a symmetric deformation one. The solution is formed from (4.2). For load (a),(7.1)v1=(s33v0f(1)+d33v0t(1)+b33v0r(1))P0.For load (b),(7.2)v2=(d33v0f(1)-λ33v0t(1)-β33v0r(1))D0.For load (c),(7.3)v3=(b33v0f(1)-β33v0r(1)-ν33v0r(1))B0.
For numerical calculation, the composite materials BaTiO3-CoFe2O4 are specified, which material constants are given in [5]. Take σz=10N/m2, Dz=10-9C/m2, Bz=10-7N/Am, h=0.03m, and  l=0.1m. The numerical results at point (h,l) are given in Table 1, where the values in the parenthesis are the exact ones which are calculated with the formulas presented by Ding and Jiang in [5]. The results show that the numerical solutions by the method have higher accuracy. The eigensolutions of zero eigenvalue correspond to the solutions of Saint-Venant problems. On the other hand, the solutions of the portion covered by the Saint-Venant principle correspond to the eigensolutions with nonzero eigenvalues. Subsequently, a very simple example is presented here.Table 1
The numerical results in the rectangular domain problem under three-load case.
Caseu (10-13 m)w (10-12 m)ϕ (10-3 V)ψ (10-4 A)σz (N/m2)Dz (10-9 C/m2)Bz (10-7 N/Am)a−9.49985.68349.49552.139110.00000.00000.0000(−9.4998)(5.6834)(9.4955)(2.1391)(10.0000)(0.0000)(0.0000)b−2.10790.94955−6.28940.256690.00001.00000.0000(−2.1079)(0.94955)(−6.2894)(0.25669)(0.0000)(1.0000)(0.0000)c5.07672.13912.5669−7.52130.00000.00001.0000(5.0767)(2.1391)(2.5669)(−7.5213)(0.0000)(0.0000)(1.0000)Example 7.2.
Consider a simple tension problem of  composite materials BaTiO3-CoFe2O4 semi-infinite strip fixed at z=0, and take -h≤x≤h, h=0.1m. There is only the tension stress σ∞ for z→∞, and τzx=Dz=Bz=0. The stress distribution at the fixed end is determined here.
Firstly, by utilizing (5.2) and (5.3), λi(i=1,2,…,4) are calculated and listed in Table 2 for the composite materials BaTiO3-CoFe2O4.
Secondly, obtain nonzero eigenvalues. With reference to the problem, there is only the tension stressσ∞ for z→∞ and the deformation is symmetric with respect to z-axis. Solve (5.9) by numerical methods. Table 3 lists the first five eigenvalues in the first quadrant. Each μn in reality has its symplectic adjoint eigenvalue -μn and their complex conjugate eigenvalue.
Thirdly, the general solution is then formed from the eigensolution of the zero eigenvalue (4.2) and (4.3) and the eigensolution of the nonzero eigenvalues of the symmetric deformation  (5.12). Consider the deformation is symmetric with respect to z-axis and rigid body translation, constant electric potential or constant magnetic potential has no effect on stress, only v0f(1), v0t(1), and v0r(1) are selected in (4.2). Their coefficients are determined by the boundary conditions at z→∞. So the general solution can be expressed as
(7.4)v=σ∞(s33v0f(1)+d33v0t(1)+b33v0r(1))+∑i=1nAivi,
where n is the number of the eigensolutions of nonzero eigenvalues of the symmetric deformation. For the above general solution, only eigensolutions of α-set in (5.12) are adopted, that is,  Reμi<0. Thus it consistent with the far-end boundary condition at z→∞.
Finally, Substituting (7.4) into the equation of variation (6.5) and adopting a total of ten eigensolutions of nonzero eigenvalues in the calculation yield σz/σ∞ at the fixed end z=0 as illustrated in Figure 2. The figure shows that there is stress singularity at the edge corner. Fluctuation of stress observed is common when truncated finite terms are adopted in the expansion. Such phenomenon has been observed, for instance, when finite terms in a Fourier series are assumed. Figure 3 shows that the normal stress distribution approximates uniformity at z=2h.Table 2
The eigenvalues with respect to thex-direction.
i1234λi0.58880i1.0711i0.60197 + 1.1734i0.60197 − 1.1734iTable 3
Nonzero eigenvalues for symmetric deformation.
n12345μnh1.8209 + 1.2193i2.83493.89723.9331 + 2.1502i5.8470Figure 2
Normal stress analysis with fixed end.Figure 3
Normal stress distribution in the partial region of the left end.


## 8. Conclusions

In this paper, the transversely isotropic magnetoelectroelastic solids plane problem in rectangular domain is considered from a symplectic approach. The eigensolutions of nonzero-eigenvalues obtained decay drastically with respect to distance can express the end effects and corner stresses. There is no requirement of experience in the symplectic approach for solving the problem since it is a rational, analytical approach to satisfy the boundary conditions in a straightforward manner.



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