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Research article

Design of robust fuzzy iterative learning control for nonlinear batch processes


  • In this paper, a two-dimensional (2D) composite fuzzy iterative learning control (ILC) scheme for nonlinear batch processes is proposed. By employing the local-sector nonlinearity method, the nonlinear batch process is represented by a 2D uncertain T-S fuzzy model with non-repetitive disturbances. Then, the feedback control is integrated with the ILC scheme to be investigated under the constructed model. Sufficient conditions for robust asymptotic stability and 2D H performance requirements of the resulting closed-loop fuzzy system are established based on Lyapunov functions and some matrix transformation techniques. Furthermore, the corresponding controller gains can be derived from a set of linear matrix inequalities (LMIs). Finally, simulations on the three-tank system and the highly nonlinear continuous stirred tank reactor (CSTR) are carried out to prove the feasibility and efficiency of the proposed approach.

    Citation: Wei Zou, Yanxia Shen, Lei Wang. Design of robust fuzzy iterative learning control for nonlinear batch processes[J]. Mathematical Biosciences and Engineering, 2023, 20(11): 20274-20294. doi: 10.3934/mbe.2023897

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  • In this paper, a two-dimensional (2D) composite fuzzy iterative learning control (ILC) scheme for nonlinear batch processes is proposed. By employing the local-sector nonlinearity method, the nonlinear batch process is represented by a 2D uncertain T-S fuzzy model with non-repetitive disturbances. Then, the feedback control is integrated with the ILC scheme to be investigated under the constructed model. Sufficient conditions for robust asymptotic stability and 2D H performance requirements of the resulting closed-loop fuzzy system are established based on Lyapunov functions and some matrix transformation techniques. Furthermore, the corresponding controller gains can be derived from a set of linear matrix inequalities (LMIs). Finally, simulations on the three-tank system and the highly nonlinear continuous stirred tank reactor (CSTR) are carried out to prove the feasibility and efficiency of the proposed approach.



    The moderately thick plate is an important material that can adapt to high pressure, high temperature and strong radiation environments, often used in construction engineering, machinery manufacturing, container manufacturing and so on [1,2]. In recent years, many achievements have been made in numerical methods of moderately thick plate problems. There are some typical numerical methods, such as the meshless local radial point interpolation method [3], discrete singular convolution methods [4], spectral element method [5], meshless local Petrov-Galerkin method [6], differential quadrature element method [7], finite integral transform method [8] and chaotic radial basis function method [9]. Compared with numerical methods, the analytical method seems more accurate, but it is difficult to construct. Through the efforts of many scholars, the analytical method has made great progress. The common analytical methods are integral transformation [10,11], semi-inverse method [12], infinite series method [13] and Green's functions [14]. Traditionally, some analytical solutions always need to eliminate the unknown functions by various methods in the range of a class of variables so as to obtain a partial differential equation with higher order, which will make separating variables and other mathematical methods impossible. In order to avoid this phenomenon, academician Zhong put forward the SEA in the 1990s. The essence of SEA is the method of separation of variables based on Hamiltonian system, which does not need to assume the form of the solution in advance. The main feature of the approach is that the solution procedure is conducted in the symplectic space with dual variables rather than in the Euclidean space with one kind of variables, which is rational to solve the equations of elasticity [15,16]. Within the framework of the Hamiltonian system, the analytical solutions of the considered problem can be obtained by using the complete expansion of symplectic eigenvectors without any prior assumptions of the solution forms, which shows the distinctive advantage of the SEA.

    SEA needs to transform the equations of elasticity into a proper Hamiltonian system and can uniformly solve the state vector, including as many unknown functions, such as the physical displacements and forces, as possible. Nowadays, the SEA has been extensively studied by some scholars. Xu et al. [17] presented the fracture analysis of fractional two-dimensional viscoelastic media by SEA. Zhao et al. [18] utilized SEA on the wave propagation problem for periodic structures with uncertainty. Qiao et al. [19] considered the plane elasticity problems of two-dimensional octagonal quasicrystals by SEA. Zhou et al. [20] studied new buckling solutions of moderately thick rectangular plates by the symplectic superposition method within the Hamiltonian system framework. Li et al. [21] explored new analytic free vibration solutions of doubly curved shallow shells. Su et al. [22] obtained an analytical free vibration solution of fully free orthotropic rectangular thin plates on two-parameter elastic foundations by the symplectic superposition method. In the last thirty years, the application of SEA has been extended continuously in the electromagnetic elastic solid problem [23], non-Lévy-type functionally graded rectangular plates [24], special mechanical problems in mine engineering [25], buckling of regular and auxetic honeycombs [26], free vibration of non-Lévy-type porous FGM rectangular plates [27], the free vibration of edge-cracked thick rectangular plates [28], the buckling problem of CNT reinforced composite rectangular plates [29] and so on.

    In this paper, based on the idea of [30], the mechanical problems of rectangular moderately thick plates are analyzed and a unified framework for a class of rectangular moderately thick plates under the simply supported boundary conditions at opposite sides is given. The model considered in this paper contains free parameters, including the bending, buckling and free vibration problems of the plates. We start from the basic equations of the rectangular moderately thick plates, guide the equations to the Hamiltonian system by introducing appropriate state vector and solve the eigenvalue problem of the Hamiltonian operators. More importantly, we prove the completeness of the eigenfunction system, which lays a theoretical foundation for our general solution. For the applications of the model, based on the obtained general solutions, rectangular moderately thick plates under Winkler type foundation and Pasternak type two-parameter foundation are numerically simulated, which have not been solved by SEA so far. The free vibration problem of moderately thick plates with fully simply supported is also discussed. The satisfactory agreement further confirms the stability of the present method.

    For the problems of rectangular moderately thick plates, we propose the following general model:

    Mxx+MxyyQx=q1(x,y), (2.1a)
    Myy+MxyxQy=q2(x,y), (2.1b)
    Qxx+Qyy+α12wx2+β12wy2+γw=q3(x,y), (2.1c)

    where q1, q2, q3 are loads, α1, β1, γ are free parameters, Mx, My, Mxy, Qx, Qy, w are the bending moments, torsional moment, shear forces and the transverse modal displacement, respectively.

    The internal forces of the plates can be presented as

    Mx=D(ψxx+νψyy), (2.2a)
    My=D(ψyy+νψxx), (2.2b)
    Mxy=D(1ν)2(ψxy+ψyx), (2.2c)
    Qx=C(wxψx), (2.2d)
    Qy=C(wyψy), (2.2e)

    where C=5Eh12(1+ν), D=Eh312(1ν2) are the elastic constants, in which E denotes the modulus of elasticity, h represents the thickness of the plate and ν is Poisson's ratio. From (2.1a)–(2.2e), the model can be transformed into the following form:

    (D2x2D(1ν)22y2+C)ψxD(1+ν)22xyψyCxw=q1(x,y), (2.3a)
    (D2y2D(1ν)22x2+C)ψyD(1+ν)22xyψxCyw=q2(x,y), (2.3b)
    CxψxCyψy+(α2x2+β2y2+γ)w=q3(x,y), (2.3c)

    where α=α1+C, β=β1+C.

    Setting Vx=αwxCψx and combing (2.2a)–(2.3c), we can obtain the following matrix equation:

    U(x,y)x=HU(x,y)+f(x,y). (2.4)

    It is easy to verify that H=(0TS0) satisfies H=JHJ in which J=(0I3I30), I3 is a 3×3 unit matrix. So H is a Hamiltonian operator matrix, in which

     T=(1Dνy0νyD(ν21)2y2+CCy0Cyβ2y2γ),S=(C2αCyCαy2D(ν1)0Cα01α),

    f(x,y)=(0,q2(x,y),q3(x,y),q1(x,y),0,0)T, U(x,y)=(ψx,Mxy,Vx,Mx,ψy,w)T.

    Next, we continue to use the SEA to solve the Hamiltonian system (2.4).

    To solve the Hamiltonian system, the homogeneous system is first considered:

    U(x,y)x=HU(x,y). (2.5)

    Using the separation of variables method, assuming U(x,y)=X(x)Y(y), we have

    dX(x)dx=μX(x), (2.6)
    HY(y)=μY(y), (2.7)

    with eigenvalue μ and eigenvector Y(y).

    The considered model is solved in the rectangular region {(x,y)0xa,0yb} and the boundary conditions are simply supported at y=0 and y=b as

    w=ψx=My=0. (2.8)

    The eigenvalue problems (2.6) and (2.7) are actually quite complicated to compute. For the sake of brevity below, let ζn=nπb and

    Pn=C42αC3+C2(α2+2Dζ2n(αβ)+2γD)+2αCD(ζ2n(αβ)+γ)+D2(ζ2n(αβ)+γ)2D2,

    nN, where N stands for the set of positive integers.

    With the help of symbolic software, the corresponding results may be divided into the following three cases.

    Case 1. μ1=2CD(ν1),μ2=μ1 are eigenvalues of H and the corresponding eigenvectors are

    Y1=(0,C,0,0,μ1,0)T,Y2=(0,C,0,0,μ2,0)T,

    where the superscript T signifies the transpose of a vector.

    Case 2. When Pn0, H has six simple eigenvalues, which can be expressed as

    μn1=D(ν1)ζ2n2CD(ν1),μn2=μn1;
    μn3=C2+D(ζ2n((α+β))+Pn+γ)αC2αD,μn4=μn3

    and

    μn5=C2+D(ζ2n(α+β)+Pnγ)+αC2αD,μn6=μn5 (nN).

    The corresponding eigenvectors with respect to μni can be rewritten as

    Yn,i=(sin(yζn)D(ν1)cos(yζn)(C2(ζ2n+μ2ni)C(γβζ2n+αμ2ni)+D(νζ2n+μ2ni)(γβζ2n+αμ2ni))ζn(2C2+D(ν+1)(γβζ2n+αμ2ni))Csin(yζn)(2C2+2αCγD(ν+1)+Dζ2n(α(ν)+α+βν+β)2αDμ2ni)2C2+D(ν+1)(γβζ2n+αμ2ni)Dsin(yζn)(2C2(ν1)μ2ni2Cν(γβζ2n+αμ2ni)+D(ν1)(νζ2n+μ2ni)(γβζ2n+αμ2ni))μni(2C2+D(ν+1)(γβζ2n+αμ2ni))cos(yζn)(2C2μ2ni2C(γβζ2n+αμ2ni)+D((ν1)ζ2n+2μ2ni)(γβζ2n+αμ2ni))ζnμni(2C2+D(ν+1)(γβζ2n+αμ2ni))Csin(yζn)(2C+D(ν1)(μ2niζ2n))μni(2C2+D(ν+1)(γβζ2n+αμ2ni))),

    (i=1,2,,6; nN).

    Case 3. When Pn=0, H contains two single eigenvalues:

    μn1=D(ν1)ζ2n2CD(ν1),μn2=μn1,

    and two double eigenvalues:

    μn3=C(αC)+Dζ2n(α+β)γD2αD,μn4=μn3(nN).

    In this case, the basic eigenvectors corresponding to μn1(i=1,2,3,4;nN) are

    Y0n,i=(sin(yζn)D(ν1)cos(yζn)(C2(ζ2n+μ2ni)C(γβζ2n+αμ2ni)+D(νζ2n+μ2ni)(γβζ2n+αμ2ni))ζn(2C2+D(ν+1)(γβζ2n+αμ2ni))Csin(yζn)(2C2+2αCγD(ν+1)+Dζ2n(α(ν)+α+βν+β)2αDμ2ni)2C2+D(ν+1)(γβζ2n+αμ2ni)Dsin(yζn)(2C2(ν1)μ2ni2Cν(γβζ2n+αμ2ni)+D(ν1)(νζ2n+μ2ni)(γβζ2n+αμ2ni))μni(2C2+D(ν+1)(γβζ2n+αμ2ni))cos(yζn)(2C2μ2ni2C(γβζ2n+αμ2ni)+D((ν1)ζ2n+2μ2ni)(γβζ2n+αμ2ni))ζnμni(2C2+D(ν+1)(γβζ2n+αμ2ni))Csin(yζn)(2C+D(ν1)(μ2niζ2n))μni(2C2+D(ν+1)(γβζ2n+αμ2ni))).

    As for the situation of double eigenvalues, there exist first-Jordan form eigenvectors. Solving γ from Pn=0 yields that

    γ=γn1=2αC32C2αCαDζ2n+βDζ2nD

    and

    γ=γn2=2αC32C2αCαDζ2n+βDζ2nD

    for further simplification. According to HY1n,i(y)=μnY1n,i(y)+Y0n,i(y), (i=3,4,nN). The first-order Jordan eigenvectors are simplified as follows.

    Case 3.1. If γ=γn1=2αC32C2αCαDζ2n+βDζ2nD, then

    Y1n,3=(sin(yζn)(C(2α+C)2αDζ2n)C32C32α+C+Dζ2nDD(ν1)ζncos(yζn)(C32+2αC+2αDζ2n)C32C32α+C+Dζ2nDsin(yζn)(C32+αC+2αDζ2n)CC32α+C+Dζ2nD2αDsin(yζn)(C+(DDν)ζ2n)C322αDζncos(yζn)C320),Y1n,4=(sin(yζn)(C32+2αC+2αDζ2n)C32C32α+C+Dζ2nDD(ν1)ζncos(yζn)(C32+2αC+2αDζ2n)C32C32α+C+Dζ2nDsin(yζn)(C(α+C)2αDζ2n)CC32α+C+Dζ2nD2αDsin(yζn)(C+(DDν)ζ2n)C322αDζncos(yζn)C320).

    Case 3.2. If γ=γn2=2αC32C2αCαDζ2n+βDζ2nD, we have

    Y1n,3=(sin(yζn)(C(2α+C)+2αDζ2n)C32C32α+C+Dζ2nDD(1ν)ζncos(yζn)(C32+2αC+2αDζ2n)C32C32α+C+Dζ2nDsin(yζn)(C32αC2αDζ2n)CC32α+C+Dζ2nD2αDsin(yζn)(C+(DDν)ζ2n)C322αDζncos(yζn)C320),Y1n,4=(sin(yζn)(C322αC2αDζ2n)C32C32α+C+Dζ2ndD(ν1)ζncos(yζn)(C32+2αC+2αDζ2n)C32C32α+C+Dζ2nDsin(yζn)(C(α+C)+2αDζ2n)CC32α+C+Dζ2nD2αDsin(yζn)(C+(DDν)ζ2n)C322αDζncos(yζn)C320).

    After calculating the eigenvalues and eigenvectors, we continue to study the properties of the eigenvectors. And for the analytical solution of this kind of moderately thick rectangular plates, the choice of space is very important, and here we choose the Hilbert space Y:=L2(0,b)L2(0,b)L2(0,b)L2(0,b)L2(0,b)L2(0,b), where L2(0,b) is the space of all square integrable functions. For L(y),R(y)Y, the symplectic inner product is defined as

    L(y),R(y)=b0L(y)TJR(y)dy,

    where J is the 6×6 symplectic unit matrix.

    Symplectic orthogonality of eigenvectors lays a foundation for the expansion of eigenvectors. The following lemma can be obtained via straightforward calculation.

    Lemma 1. The system of eigenvectors {Y1(y),Y2(y)}{Yn,i(y)nN,i=1,2,,6} of H satisfies the following symplectic orthogonality relations:

    Y1(y),Y2(y)={22b(C)32D(ν1),n=m,0,nm;
    Yn,3(y),Ym,4(y)={A34,n,n=m,0,nm;
    Yn,5(y),Ym,6(y)={A56,n,n=m,0,nm;
    Yn,1(y),Ym,2(y)={bCD(ν1)ζ2n2CD(ν1)ζ2n,n=m,0,nm,

    where the constants A34,n and A56,n are presented in Appendix A. Other eigenvectors satisfy the following symplectic orthogonal relation:

    Yn,1(y),Ym,3(y)=Yn,1(y),Ym,4(y)=Yn,1(y),Ym,5(y)=Yn,1(y),Ym,6(y)=Yn,2(y),Ym,3(y)=Yn,2(y),Ym,4(y)=Yn,2(y),Ym,5(y)=Yn,2(y),Ym,6(y)=Yn,3(y),Ym,5(y)=Yn,3(y),Ym,6(y)=Yn,4(y),Ym,5(y)=Yn,4(y),Ym,6(y)=Yn,i(y),Ym,i(y)=0,n,mN.

    Based on the conjugate symplectic orthogonality of eigenvectors, we can obtain the following completeness theorem of eigenvectors, which provides the theoretical basis for the expression of general solutions.

    Theorem 1. The system of eigenvectors {Y1(y),Y2(y)}{Yn,i(y)}+n=1 (i=1,2,,6) of H is complete in Y.

    Proof. For any F(y)=(f1(y),f2(y),f3(y),f4(y),f5(y),f6(y))TY, it is sufficient to prove that there exist constant sequences {Cn,i}+n=1, (i=1,2,,6), C1 and C2 such that

    F(y)=C1Y1+C2Y2++n=1(6i=1Cn,iYn,i(y)). (2.9)

    Applying Lemma 1, we can take

    Cn,1=Yn,2(y),F(y)Yn,2(y),Yn,1(y),Cn,2=Yn,1(y),F(y)Yn,1(y),Yn,2(y),Cn,3=Yn,4(y),F(y)Yn,4(y),Yn,3(y),Cn,4=Yn,3(y),F(y)Yn,3(y),Yn,4(y),Cn,5=Yn,6(y),F(y)Yn,6(y),Yn,5(y),Cn,6=Yn,5(y),F(y)Yn,5(y),Yn,6(y),C1=Y2(y),F(y)Y2(y),Y1(y),C2=Y1(y),F(y)Y1(y),Y2(y), (2.10)

    and then

    C1Y1+C2Y2++n=1(6i=1Cn,iYn,i(y))=(01bb0f2(y)dy001bb0f5(y)dy0)++n=1(2b(b0f1(y)sinnπybdy)sinnπyb2b(b0f2(y)cosnπybdy)cosnπyb2b(b0f3(y)sinnπybdy)sinnπyb2b(b0f4(y)sinnπybdy)sinnπyb2b(b0f5(y)cosnπybdy)cosnπyb2b(b0f6(y)sinnπybdy)sinnπyb).

    Clearly, \, +n=12b(b0fi(y)sinnπybdy)sinnπyb (i=1,3,4,6) and

    1bb0fi(y)dy++n=12b(b0fi(y)cosnπybdy)cosnπyb(i=2,5)

    are the corresponding Fourier series expansions of fi(y) based on the orthogonal systems {sinnπyb}+n=1 and {cosnπyb}+n=0 in L2(0,b), respectively. Thus, F(y) can be expressed by (2.9) and the proof is completed.

    Similar to the case where the eigenvalues are single, we first discuss the orthogonal conjugations of eigenvectors. These relations play a crucial role in proving the symplectic eigenvectors expansion theorem.

    Lemma 2. The system of eigenvectors {Y1(y),Y2(y)}{Y0n,i(y),Y1n,j(y)nN;i=1,2,3,4;j=3,4} of H have the following symplectic orthogonality relations:

    Y0n,1(y),Y0m,3(y)=Y0n,1(y),Y1m,3(y)=Y0n,1(y),Y0m,4(y)=Y0n,1(y),Y1m,4(y)=Y0n,2(y),Y0m,3(y)=Y0n,2(y),Y1m,3(y)=Y0n,2(y),Y0m,4(y)=Y0n,2(y),Y1m,4(y)=Y0n,3(y),Y1m,3(y)=Y0n,3(y),Y0m,4(y)=Y0n,4(y),Y1m,4(y)=Y0n,i(y),Y0m,i(y)=0,n,mN.

    When γ=γn1=2αC32C2αCαDζ2n+βDζ2nD, we have

    Y1(y),Y2(y)={22b(C)32D(ν1),n=m,0,nm;
    Y0n,3(y),Y1m,4(y)={2αbDC,n=m,0,nm;
    Y1n,3(y),Y0m,4(y)={2αbDC,n=m,0,nm;
    Y0n,1(y),Y0m,2(y)={bCD(ν1)ζ2n2CD(ν1)ζ2n,n=m,0,nm;
    Y1n,3(y),Y1m,4(y)={2b(αC32D+2αCD+2αD2ζ2n)C2C32α+C+Dζ2nD,n=m,0,nm.

    If γ=γn2=2αC32C2αCαDζ2n+βDζ2nD, then

    Y1(y),Y2(y)={22b(C)32D(ν1),n=m,0,nm;
    Y0n,3(y),Y1m,4(y)={2αbDC,n=m,0,nm;
    Y1n,3(y),Y0m,4(y)={2αbDC,n=m,0,nm;
    Y0n,1(y),Y0m,2(y)={bCD(ν1)ζ2n2CD(ν1)ζ2n,n=m,0,nm;
    Y1n,3(y),Y1m,4(y)={2b(αC32D2αCD2αD2ζ2n)C2C32α+C+Dζ2nD,n=m,0,nm.

    Based on Lemma 2, the following completeness theorem is obtained.

    Theorem 2. The system of eigenvectors {Y1(y),Y2(y)}{Y0n,i(y)}{Y1n,j(y)}, (i=1,2,3,4;j=3,4;nN) of H is complete in Y.

    Proof. It is similar to the proof of Theorem 1 that we just have to take the sequences:

    C0n,1=Y0n,2(y),F(y)Y0n,2(y),Y0n,1(y),C0n,2=Y0n,1(y),F(y)Y0n,1(y),Y0n,2(y),C1n,3=Y0n,4(y),F(y)Y0n,4(y),Y1n,3(y),C1n,4=Y0n,3(y),F(y)Y0n,3(y),Y1n,4(y),C1=Y2(y),F(y)Y2(y),Y1(y),C2=Y1(y),F(y)Y1(y),Y2(y),C0n,3=Y1n,4(y),F(y)C1n,3Y1n,4(y),Y1n,3(y)Y1n,4(y),Y0n,3(y),C0n,4=Y1n,3(y),F(y)C1n,4Y1n,3(y),Y1n,4(y)Y1n,3(y),Y0n,4(y). (2.11)

    It is easy to calculate that

    C1Y1+C2Y2++n=1(4i=1C0n,iY0n,i(y)+4j=3C1n,jY1n,j(y))=(01bb0f2(y)dy001bb0f5(y)dy0)++n=1(2b(b0f1(y)sinnπybdy)sinnπyb2b(b0f2(y)cosnπybdy)cosnπyb2b(b0f3(y)sinnπybdy)sinnπyb2b(b0f4(y)sinnπybdy)sinnπyb2b(b0f5(y)cosnπybdy)cosnπyb2b(b0f6(y)sinnπybdy)sinnπyb).

    As a result,

    F(y)=C1Y1+C2Y2++n=1(4i=1C0n,iY0n,i(y)+4j=3C1n,jY1n,j(y)). (2.12)

    This section will give the general solutions of the Hamiltonian system under simply supported boundary conditions at y=0 and y=b based on the proven completeness theorems.

    The general solution of (2.5) can be represented in the form:

    U(x,y)=X1(x)Y1(y)+X2(x)Y2(y)++n=1(6i=1(Xn,i(x)Yn,i(y))). (2.13)

    The nonhomogeneous term f(x,y) expands to

    f(x,y)=θ1(x)Y1(y)+θ2(x)Y2(y)++n=1(6i=1(θn,i(x)Yn,i(y))). (2.14)

    From the above formulas and (2.6), we obtain

    X1(x)=Ψ1eμ1x+x0θ1(t)eμ1(xt)dt,X2(x)=Ψ2eμ2x+x0θ2(t)eμ2(xt)dt,Xn,i(x)=Ψnieμnix+x0θn,i(t)eμni(xt)dt, (2.15)

    where Ψ1,Ψ2,Ψni are undetermined constants, θ1,θ2,θn,i can be expressed in a form similar to (2.10) via the symplectic inner product of f(x,y) and Y. In particular, if qk(x,y) (k=1,2,3) is the uniformly distributed load acting on the plate, which means that it is a constant independent of x and of y. If qk(x,y) (k=1,2,3) represents a concentrated load of strength B acting on the plate, then it can be expressed as Bδ(xx0,yy0) in which δ(xx0,yy0) denotes the δ function.

    The general solutions of ψy and w can be obtained by taking the fifth and sixth components of (2.13), respectively.

    ψy=[(Ψ1eμ1x+x0y0q2(x,y)2bCeμ1(xt)dydt)(Ψ2eμ2x+x0y0q2(x,y)2bCeμ2(xt)dydt)]μ1++n=1{[(x0eμn1(xt)θn,1(t)dt+Ψn1eμn1x)(x0eμn1(xt)θn,2(t)dt+Ψn2eμn1x)]cos(yζn)[2C2μ2n12C(αμ2n1+γβζ2n)+D(2μ2n1+(ν1)ζ2n)(αμ2n1+γβζ2n)]μn1ζn(2C2+D(ν+1)(αμ2n1+γβζ2n))+[(x0eμn3(xt)θn,3(t)dt+eμn3xΨn3)(x0eμn3(xt)θn,4(t)dt+Ψn4eμn3x)]cos(yζn)[2C2μ2n32C(αμ2n3+γβζ2n)+D(2μ2n3+(ν1)ζ2n)(αμ2n3+γβζ2n)]μn3ζn(2C2+D(ν+1)(αμ2n3+γβζ2n))+[(x0eμn5(xt)θn,5(t)dt+Ψn5eμn5x)(x0eμn5(xt)θn,6(t)dt+Ψn6eμn5x)]cos(yζn)[2C2μ2n52C(αμ2n5+γβζ2n)+D(2μ2n5+(ν1)ζ2n)(αμ2n5+γβζ2n)]μn5ζn(2C2+D(ν+1)(αμ2n5+γβζ2n))}, (2.16)
    w=+n=1{[(x0eμn3(xt)Ψn,3(t)dt+Ψn3eμn3x)(x0eμn3(xt)θn,4(t)dt+Ψn4eμn3x)]Csin(yζn)(2C+D(ν1)(μ2n3ζ2n))μn3(2C2+D(ν+1)(αμ2n3+γβζ2n))+Csin(yζn)(2C+D(ν1)(μ2n5ζ2n))μn5(2C2+D(ν+1)(αμ2n5+γβζ2n))×[(x0eμn5(xt)θn,5(t)dt+Ψn5eμn5x)(x0eμn5(xt)θn,6(t)dt+Ψn6eμn5x)]}. (2.17)

    Compared to the case where the eigenvalues are simple, we combine (2.6) with (2.12) to obtain:

    X1(x)=Q1eμ1x+x0Θ1(t)eμ1(xt)dt,X2(x)=Q2eμ2x+x0Θ2(t)eμ2(xt)dt,X0n,1(x)=Q0n,1eμn1(x)+x0Θ0n,1(t)eμn1(xt)dt,X0n,2(x)=Q0n,2eμn2(x)+x0Θ0n,2(t)eμn2(xt)dt,X1n,3(x)=Q1n,3eμn3(x)+x0Θ0n,3(t)eμn3(xt)dt,X1n,4(x)=Q1n,4eμn4(x)+x0Θ0n,4(t)eμn4(xt)dt,X0n,3(x)=(Q0n,3+Q1n,3x)eμn3(x)+x0Θ0n,3(t)eμn3(xt)dt+x0t0Θ1n,3(t)eμn3(xτ)dτdt,X0n,4(x)=(Q0n,4+Q1n,4x)eμn4(x)+x0Θ0n,4(t)eμn4(xt)dt+x0t0Θ1n,4(t)eμn4(xτ)dτdt,

    where Q1,Q2,Q0n,i,Q1n,j (i=1,2,3,4; j=3,4) are undetermined constants, Θ1,2,Θ0n,i,Θ1n,j can be computed similarly as (2.11) by symplectic inner product between f(x,y) and the eigenvectors.

    Then, U(x,y) can be given by

    U(x,y)=X1(x)Y1(y)+X2(x)Y2(y)++n=1(4i=1(X0n,i(x)Y0n,i(y))+4j=3(X1n,j(x)Y1n,j(y))). (2.18)

    As for the case of double eigenvalues, in order to obtain a simple form of ψy and w, a classification discussion is carried out.

    (i) When γ=γn1=2αC32C2αCαDζ2n+βDζ2nD, we have

    ψy=[(Q1eμ1x+x0Θ1(t)eμ1(xt)dt)(Q2eμ2x+x0Θ2(t)eμ1(xt)dt)]2CD(ν1)++n=1{[eμn1xQ0n,1+x0eμn1(xt)Θ0n,1(t)dt(eμn2xQ0n,2+x0eμn2(xt)Θ0n,2(t)dt)](cos(yζn)D(ν1)ζ2n2CD(ν1)ζn)+[(eμn3x(xQ1n,3+Q0n,3)+x0eμn3(xt)Θ0n,3(t)dt+x0t0eμn3(xτ)Θ1n,3(t)dτdt)(eμn4x(xQ1n,4+Q0n,4)+x0t0eμn4(xτ)Θ1n,4(t)dτdt+x0eμn4(xt)Θ9n,4(t)dt)](ζncos(yζn)C32α+C+Dζ2nD)+[(eμn3xQ1n,3+x0eμn3(xt)Θ0n,3(t)dt)+(eμn4xQ1n,4+x0eμn4(xt)Θ0n,4(t)dt)](2αDζncos(yζn)C32)},
    w=+n=1{[eμn3x(xQ1n,3+Q0n,3)+x0t0eμn3(xτ)Θ1n,3(t)dτdt+x0eμn3(xt)Θ0n,3(t)dt][eμn4x(xQ1n,4+Q0n,4)+x0t0eμn4(xτ)Θ1n,4(t)dτdt+x0eμn4(xt)Θ0n,4(t)dt]×(Csin(yζn)αC32α+C+Dζ2nD)}.

    (ii) When \gamma = \gamma_{n2} = \frac{2 \sqrt{\alpha } C^{\frac{3}{2}}-C^2-\alpha C-\alpha D \zeta _n^2+\beta D \zeta _n^2}{D}, we can see that

    \begin{align*} \psi_{y} = &\left[\left(Q_1 e^{\mu_1 x}+\int_0^x \Theta _1(t) e^{\mu_1 (x-t)} \, dt\right)-\left(Q_2 e^{\mu_2 x}+\int_0^x \Theta _2(t) e^{\mu_2 (x-t)} \, dt\right)\right]\frac{\sqrt{-2 C}}{\sqrt{D (\nu -1)}}\nonumber\\ &+\sum\limits_{n = 1}^{+\infty}\left\{\left[e^{\mu_{n1} x} Q_{n,1}^0+\int_0^x e^{\mu_{n1} (x-t)} \Theta _{n,1}^0(t) \, {\mathrm{d}}t-\left(e^{\mu_{n2} x} Q_{n,2}^0+\int_0^x e^{\mu_{n2} (x-t)} \Theta _{n,2}^0(t) \, {\mathrm{d}}t\right)\right]\right.\nonumber\\ &\left(\frac{\cos \left(y \zeta _n\right) \sqrt{D (\nu -1) \zeta _n^2-2 C}}{\sqrt{D (\nu -1)} \zeta _n}\right)+\left[\left(e^{\mu_{n3} x} \left(x Q_{n,3}^1+Q_{n,3}^0\right)+\int_0^x e^{\mu_{n3} (x-t)} \Theta _{n,3}^0(t) \, {\mathrm{d}}t\right.\right.\nonumber\\ &\left.+\int _0^x\int _0^te^{\mu_{n3} (x-\tau )} \Theta _{n,3}^1(t){\mathrm{d}}\tau {\mathrm{d}}t\right)-\left(e^{\mu_{n4} x} \left(x Q_{n,4}^1+Q_{n,4}^0\right)+\int_0^x e^{\mu_{n4} (x-t)} \Theta _{n,4}^0(t) \, {\mathrm{d}}t\right.\nonumber\\ &\left.\left.+\int _0^x\int _0^te^{\mu_{n4} (x-\tau )} \Theta _{n,4}^1(t){\mathrm{d}}\tau {\mathrm{d}}t\right)\right]\left(\frac{\zeta _n \cos \left(y \zeta _n\right)}{\sqrt{\frac{-\frac{C^{\frac{3}{2}}}{\sqrt{\alpha }}+C+D \zeta _n^2}{D}}}\right)+\left(\frac{2 \sqrt{\alpha } D \zeta _n \cos \left(y \zeta _n\right)}{C^{\frac{3}{2}}}\right)\\ &\times\nonumber\left.\left[\left(e^{\mu_{n3} x} Q_{n,3}^1+\int_0^x e^{\mu_{n3} (x-t)} \Theta _{n,3}^0(t) \, {\mathrm{d}}t\right)+\left(e^{\mu_{n4} x} Q_{n,4}^1+\int_0^x e^{\mu_{n4} (x-t)} \Theta _{n,4}^0(t) \, {\mathrm{d}}t\right)\right]\right\}, \end{align*}
    \begin{align*} w = &\sum\limits_{n = 1}^{+\infty}\left\{\left[\left(e^{\mu_{n3} x} \left(x Q_{n,3}^1+Q_{n,3}^0\right)+\int _0^x\int _0^te^{\mu_{n3} (x-\tau )} \Theta _{n,3}^1(t)d\tau dt+\int_0^x e^{\mu_{n3} (x-t)} \Theta _{n,3}^0(t) \, {\mathrm{d}}t\right)\right.\right.\nonumber\\ &\left.-\left(e^{\mu_{n4} x} \left(x Q_{n,4}^1+Q_{n,4}^0\right)+\int _0^x\int _0^te^{\mu_{n4} (x-\tau )} \Theta _{n,4}^1(t){\mathrm{d}}\tau {\mathrm{d}}t+\int_0^x e^{\mu_{n4} (x-t)} \Theta _{n,4}^0(t) \, {\mathrm{d}}t\right)\right]\\ &\times\nonumber\left.\left(\frac{\sqrt{C} \sin \left(y \zeta _n\right)}{\sqrt{\alpha } \sqrt{\frac{-\frac{C^{\frac{3}{2}}}{\sqrt{\alpha }}+C+D \zeta _n^2}{D}}}\right)\right\}. \end{align*}

    The general solution of a kind of rectangular moderately thick plates has been obtained, which can be applied to some specific mechanical problems, such as bending, buckling, free vibration, etc. For example, when \alpha = \beta = C, \gamma = q_1(x, y) = q_2(x, y) = 0, q_3(x, y) = -q , the model reduces to the bending problem of moderately thick plates with two opposite edges simply supported [31]; when \alpha = C+N_{x}, \beta = C+N_{y}, \gamma = q_1(x, y) = q_2(x, y) = q_3(x, y) = 0 , it represents the document [20] in discussing the buckling problem of the moderately thick rectangular plates; the reference [32] discusses a free vibration problem of rectangular moderately thick plates, corresponding to \alpha = \beta = C, \, \gamma = \rho h \varpi^2 ( \varpi represents circular frequency) and q_1(x, y) = q_2(x, y) = q_3(x, y) = 0 . For these specific problems, although they may correspond to different boundary conditions, most of them can be obtained by the general solution in this paper.

    So far, SEA has not been applied to the problems of moderately thick rectangular plates with elastic foundations. In this section, the solutions to two elastic foundation problems are discussed first based on the general solutions of the considered model and the numerical results are compared with those in the literature.

    Example 1. Consider a uniformly loaded and simply supported square plate resting on a Winkler-type elastic foundation with side length a . In this case, q_{1}(x, y) = q_{2}(x, y) = 0 , q_{3}(x, y) = q , \alpha_1 = \beta_1 = 0 and \gamma = k , where q denotes the uniform load and k is the subgrade reaction operator for Winkler-type foundation [33,34,35], the corresponding equilibrium has the following form,

    \begin{align} \left\{ \begin{array}{cc} \frac{\partial M_{x}}{\partial x}+\frac{\partial M_{xy}}{\partial y}-Q_{x} = 0,\\ \frac{\partial M_{y}}{\partial y}+\frac{\partial M_{xy}}{\partial x}-Q_{y} = 0,\\ \frac{\partial Q_{x}}{\partial x}+\frac{\partial Q_{y}}{\partial y}+kw = q. \end{array}\right. \end{align} (3.1)

    According to (2.17), the form of w is given in detail:

    \begin{array}{l} w = \lim\limits_{N\rightarrow +\infty}\sum\limits_{n = 1}^{N}\left[\frac{e^{\mu_{n3} (-x)} \left(\left(e^{\mu_{n3} x}-1\right)^2 \theta _{n,3}+\mu_{n3}\Psi_{n3} e^{2 \mu_{n3} x}-\mu_{n3}\Psi_{n4}\right)}{\mu_{n3}}\right.\\\times\frac{\sqrt{2} \sin \left(y \zeta _n\right) \left(4 C^2-D (\nu -1) (k+P_1)\right)}{\left(4 C^2+D (\nu +1) (k-P_1)\right) \sqrt{\frac{2 C \zeta _n^2-k-P_1}{C}}}\nonumber\\ \left.+\frac{e^{\mu_{n5} (-x)} \left(\left(e^{\mu_{n5} x}-1\right)^2 \theta _{n,5}+\mu_{n5} \Psi_{n5} e^{2 \mu_{n5} x}-\mu_{n5}\Psi_{n6}\right)}{\mu_{n5}} \\\times \frac{\sqrt{2} \sin \left(y \zeta _n\right) \left(4 C^2-D (\nu -1) (k-P_1)\right)}{\left(4 C^2+D (\nu +1) (k+P_1)\right) \sqrt{\frac{2 C \zeta _n^2-k+P_1}{C}}}\right], \end{array}

    where \mu_{n3} = \sqrt{-\frac{-2 C \zeta _n^2+k+P_1}{2C}}, \mu_{n5} = \sqrt{\frac{2 C \zeta _n^2-k+P_1}{2C}}, P_1 = \sqrt{k \left(\frac{4 C^2}{D}+k\right)} ,

    \begin{align*} \theta_{n,3} = &-\left[2 C^2 q \zeta _n \left(\cos \left(b \zeta _n\right)-1\right) \left(4 C^2-D (\nu -1) (k+P_1)\right) \left(4 C^2+D (\nu +1) (k-P_1)\right)\right]/\nonumber\\ &\left\{bD\left[32 C^5 (k-P_1) \zeta _n^2+16 C^4 k^2 (\nu -1)-8 C^3 D \zeta _n^2 \left(k^2 \left(\nu ^2-2\right)+2 k P_1+(1-2 \nu ) P_1^2\right)\right.\right.\nonumber\\ &\left.\left.+8 C^2 D k(\nu -1) \left(k^2-P_1^2\right)+D^2 (\nu -1) \left(k^2-P_1^2\right)^2-2 C D^2 \left(\nu ^2-1\right) (k-P_1)^2 (k+P_1) \zeta _n^2\right]\right\},\nonumber\\ \theta_{n,5} = &-\left[2 C^2 q \zeta _n \left(\cos \left(b \zeta _n\right)-1\right) \left(4 C^2-D (\nu -1) (k-P_1)\right) \left(4 C^2+D (\nu +1) (k+P_1)\right)\right]/\nonumber\\ &\left\{b D\left[32 C^5 (k+P_1) \zeta _n^2+16 C^4k^2 (\nu -1)-8 C^3 D \zeta _n^2 \left(k^2 \left(\nu ^2-2\right)-2 k P_1+(1-2 \nu ) P_1^2\right)\right.\right.\nonumber\\ &\left.\left.+8 C^2 D k (\nu -1) \left(k^2-P_1^2\right)+D^2 (\nu -1) \left(k^2-P_1^2\right)^2-2 C D^2 \left(\nu ^2-1\right) (k-P_1) (k+P_1)^2 \zeta _n^2\right]\right\}. \end{align*}

    Herein, \Psi_{n, i} are determined by the simply supported conditions at x = 0 and x = a . The constants E = 2.06\times10^{11}pa, \ \nu = 0.3, \ q = 6.89\times10^{7} pa, \ a = 1.016m are taken from [33,34,35]. The corresponding results are shown in Table 1.

    Table 1.  Central deflection w of SEA and comparison with others.
    k h/a w (mm)
    (10^{5}N/m^{2}) [33] [34] [35] Present
    54.25 0.5 0.2758
    0.3 0.8153 0.8154 0.8588 0.8157
    0.2 2.2896 2.2896 2.3546 2.2858
    0.1 15.9992 15.987 16.1275 15.8836
    0.05 117.463 118.32 117.94 118.2312
    542.5 0.5 0.2759
    0.3 0.8125 0.8129 0.8555 0.8162
    0.2 2.2670 2.2688 2.3307 2.2696
    0.1 14.9535 15.023 15.0655 15.0431
    0.05 77.6112 78.535 77.999 78.0131
    5425 0.5 0.2763
    0.3 0.7846 0.7858 0.8247 0.8195
    0.2 2.0625 2.0615 2.1150 2.0843
    0.1 9.0424 9.0472 9.0833 9.0525
    0.05 17.668 17.921 17.703 17.8997

     | Show Table
    DownLoad: CSV

    In Table 1, we only take the first 20 items of the series to achieve a better fitting effect with the data in the reference [33,34,35], which indicates that it is satisfactory to use the SEA to solve this moderately thick rectangular plate with single parameter foundation. Table 2 presents the data convergence analysis for the central deflection w when \frac{h}{a} = 0.3 with the different parameters k . It can be seen that convergence of the series is satisfactory, and it confirms the completeness theorem of the eigenvector system discussed above in this paper.

    Table 2.  Convergence analysis of central deflection w when h/a = 0.3 .
    k w (mm)
    (10^{5}N/m^{2}) N 1 5 8 10 15 20 25 30
    54.25 0.8362 0.8169 0.8155 0.8162 0.8159 0.8157 0.8157 0.8157
    542.5 0.8362 0.8173 0.8158 0.8165 0.8162 0.8162 0.8162 0.8162
    5425 0.8397 0.8205 0.8191 0.8197 0.8195 0.8195 0.8195 0.8195

     | Show Table
    DownLoad: CSV

    Example 2. The equilibrium equations of the bending problem of rectangular moderately thick plates with Pasternak type two-parameter elastic foundation are[36,37]:

    \begin{align*} \left\{ \begin{array}{cc} \frac{\partial M_{x}}{\partial x}+\frac{\partial M_{xy}}{\partial y}-Q_{x} = 0,\\ \frac{\partial M_{y}}{\partial y}+\frac{\partial M_{xy}}{\partial x}-Q_{y} = 0,\\ \frac{\partial Q_{x}}{\partial x}+\frac{\partial Q_{y}}{\partial y}+G_f(\frac{\partial^2 w}{\partial x^2}+\frac{\partial^2 w}{\partial y^2})-k_fw = -q. \end{array}\right. \end{align*}

    For comparison with the data in [36,37], a square plate with the side length of a is considered under fully simply supported conditions. This is the special case of our model when q_{1}(x, y) = q_{2}(x, y) = 0 , q_{3}(x, y) = -q , \alpha = \beta = C+G_f , \gamma = -k_f . Furthermore, we take the general solution of the fifth and sixth components of {\textbf{U}}(x, y) , and the specific forms of the deflection w and the bending moment M_x are as follows:

    \begin{align*} w = -\sum\limits_{n = 1}^{+\infty}&\left[\frac{e^{\mu_{n3} (-x)} \left(\left(e^{\mu_{n3} x}-1\right)^2 \theta _{n,3}+\mu_{n3} \Psi_{n3} e^{2 \mu_{n3} x}-\mu_{n3} \Psi_{n4}\right)}{\mu_{n3}}\right.\times\\ &\frac{C \sin \left(y \zeta _n\right) \left(4 C^2+C (\nu +3) G_f+D (\nu -1) \left(k_f-P_{2,n}\right)\right)}{\left(C+G_f\right) \left(4 C^2+C (\nu +1) G_f-D (\nu +1) \left(k_f+P_{2,n}\right)\right) \sqrt{\frac{C G_f+D k_f-D P_{2,n}}{2 C D+2 D G_f}+\zeta _n^2}}\\ &+\frac{e^{\mu_{n5} (-x)} \left(\left(e^{\mu_{n5} x}-1\right)^2 \theta _{n,5}+\mu_{n5} \Psi_{n5} e^{2 \mu_{n5} x}-\mu_{n5} \Psi_{n6}\right)}{\mu_{n5}}\times\\ &\left.\frac{C \sin \left(y \zeta _n\right) \left(4 C^2+C (\nu +3) G_f+D (\nu -1) \left(k_f+P_{2,n}\right)\right)}{\left(C+G_f\right) \left(4 C^2+C (\nu +1) G_f-D (\nu +1) \left(k_f-P_{2,n}\right)\right) \sqrt{\frac{C G_f+D k_f+D P_{2,n}}{2 C D+2 D G_f}+\zeta _n^2}}\right], \end{align*}
    \begin{align*} M_x = -\sum\limits_{n = 1}^{+\infty}&\left\{\frac{\sin \left(y \zeta _n\right)(\Psi_{n1} D e^{\mu_{n1} x}-\Psi_{n2} D e^{-\mu_{n1}x})}{\mu_{n1} \left(2 C^2+D (\nu +1) \left(\left(C+G_f\right) \left(-\left(\zeta _n^2-\mu_{n1}^2\right)\right)-k_f\right)\right)}\times\left[2 C^2 \mu_{n1}^2 (\nu -1)+2C\nu\times\right.\right.\\ &\left.\left(\left(C+G_f\right)(\zeta _n^2-\mu_{n1}^2)-k_f\right)+D (\nu -1) \left(\mu_{n1}^2+\nu \zeta _n^2\right) \left(\left(C+G_f\right)\left(-\left(\zeta _n^2-\mu_{n1}^2\right)\right)-k_f\right)\right]\\ &+\frac{\sin \left(y \zeta _n\right)D e^{-\mu_{n3} x}\left(\left(e^{\mu_{n3} x}-1\right)^2 \theta_{n,3}+\mu_{n3}\left(\Psi_{n3} e^{2 \mu_{n3} x}-\Psi_{n4}\right)\right)}{\mu_{n3}^2 \left(2 C^2+D (\nu +1) \left(\left(C+G_f\right) \left(-\left(\zeta _n^2-\mu_{n3}^2\right)\right)-k_f\right)\right)}\times\left[2 C^2\mu_{n3}^2 (\nu -1)\right.\\ &\left.-2 C \nu \left(\left(C+G_f\right) \left(-\left(\zeta _n^2-\mu_{n3}^2\right)\right)-k_f\right)-D (\nu -1)\left(\mu_{n3}^2+\nu \zeta _n^2\right)\left(\left(C+G_f\right)\left(\zeta _n^2-\mu_{n3}^2\right)\right.\right.\\ &\left.\left.+k_f\right)\right]+\frac{\sin \left(y \zeta _n\right) D e^{-\mu_{n5}x} \left(\left(e^{\mu_{n5} x}-1\right)^2 \theta _{n,5}+\mu_{n5} \left(\Psi_{n5} e^{2 \mu_{n5} x}-\Psi_{n6}\right)\right)}{\mu_{n5}^2 \left(2 C^2+D (\nu +1) \left(\left(C+G_f\right) \left(-\left(\zeta _n^2-\mu_{n5}^2\right)\right)-k_f\right)\right)}\times\left[2 C^2\mu_{n5}^2\times \right.\\ &(\nu -1)-2 C \nu \left(\left(C+G_f\right) \left(-\left(\zeta _n^2-\mu_{n5}^2\right)\right)-k_f\right)-D (\nu -1) \left(\mu_{n5}^2+\nu \zeta _n^2\right)\left(\left(C+G_f\right)\times\right.\\ &\left.\left.\left(\left(\zeta _n^2-\mu_{n5}^2\right)+k_f\right)\right]\right\}, \end{align*}

    in which

    \begin{align*} &P_{2,n} = \sqrt{\frac{C^2 \left(G_f^2-4 D k_f\right)-2 C D G_f k_f+D^2 k_f^2}{D^2}},\\ &\mu_{n3} = \sqrt{\frac{C G_f+D k_f-D P_{2,n}}{2 C D+2 D G_f}+\zeta _n^2},\\ &\mu_{n5} = \sqrt{\frac{C G_f+D k_f+D P_{2,n}}{2 C D+2 D G_f}+\zeta _n^2}, \end{align*}

    \theta_{n, 3} and \theta_{n, 5} are displayed in Appendix B.

    Taking the first 20 terms of the above formula, we calculate the center deflection w and bending moment M_x corresponding to different values of h/a and G_f . The relevant results are listed in Table 3 and compared with the data in the references. The reference [36] uses the grid method of 19 \times 19 , and the reference [37] uses the analytic trigonometric series method. Compared with them, it is shown that the SEA in the presented paper is effective.

    Table 3.  The center deflection w and bending moment M_x of fully simply supported square plate under two-parameter foundation, where k_f = \frac{200D}{a^{4}} , \nu = 0.25 .
    G_f=\frac{5D}{a^{2}} G_f=\frac{20D}{a^{2}}
    h/a w \times(\frac{10^3 D }{q a^{4}}) M_x \times(\frac{10^2 }{q a^{2}}) w \times(\frac{10^3 D }{q a^{4}}) M_x \times(\frac{10^2 }{q a^{2}})
    0.005 Present 2.264014 2.418051 1.567611 1.612293
    [36] 2.264013 2.417782 1.567607 1.612873
    [37] 2.263888 2.417870 1.567556 1.612893
    0.100 Present 2.313681 2.361804 1.587336 1.569424
    [36] 2.313650 2.361704 1.587351 1.569177
    [37] 2.303082 2.352117 1.581634 1.563309
    0.200 Present 2.449645 2.207744 1.641966 1.449790
    [36] 2.449532 2.207587 1.641869 1.450101
    [37] 2.440886 2.198786 1.637296 1.444638

     | Show Table
    DownLoad: CSV

    Example 3. For the free vibration of fully simply supported moderately thick rectangular plates, its governing equations are as follows[38,39]:

    \begin{align*} \left\{ \begin{array}{cc} \frac{\partial M_{x}}{\partial x}+\frac{\partial M_{xy}}{\partial y}-Q_{x} = 0,\\ \frac{\partial M_{y}}{\partial y}+\frac{\partial M_{xy}}{\partial x}-Q_{y} = 0,\\ \frac{\partial Q_{x}}{\partial x}+\frac{\partial Q_{y}}{\partial y}+\rho \omega ^2 w = 0, \end{array}\right. \end{align*}

    where \rho is the mass per unit area of the plate and \omega is the natural frequency of the plates.

    By analyzing the boundary conditions and letting the determinant of the coefficients be equal to zero, the frequency equation of the free vibration problem is given by

    \begin{align} &\frac{1}{4 \mu_{n5}^2}\left(C^2 D^2 \left(e^{2 a \mu_{n1}}-1\right) \left(e^{2 a \mu_{n3}}-1\right) \left(e^{2 a \mu_{n5}}-1\right) \left(\mu_{n1}^2-\mu_{n3}^2\right)^2 e^{-a (\mu_{n1}+\mu_{n3}+\mu_{n5})}\right.\\ &\left.\left(D (\nu -1) \zeta _n^2-2 C\right){}^2 \left(2 C \mu_{n5}^2+D (\nu -1) \zeta _n^2 \left(\zeta _n^2-\mu_{n5}^2\right)\right)\right) = 0. \end{align} (3.2)

    To justify the frequency equation, we give the numerical results for the lowest six natural frequency parameters \bar{k} = \frac{\rho \omega^2 b^4}{D \pi^4} in Table 4. Compared with [38] and [39], the satisfactory agreement further confirms the stability of the present model and method.

    Table 4.  The lowest six natural frequency parameters \bar{k} of fully simply supported moderately thick rectangular plates, where \delta_b = \frac{ \pi^2 D}{b^2 C} , \vartheta = \frac{b}{a} and \nu = 0.3.
    \vartheta 0.2 0.6 1 2
    \delta_b Modes [38] [39] Present [38] [39] Present [38] [39] Present [38] [39] Present
    0.05 1 1.028 1.028 1.0281 1.732 1.732 1.7318 3.636 3.636 3.6364 20.00 20.00 20.0000
    2 1.272 1.272 1.2718 5.306 5.306 5.3062 20.00 20.00 20.0000 45.71 45.71 45.7143
    3 1.732 1.732 1.7318 14.83 14.83 14.8330 45.71 45.71 45.7143 102.4 102.4 102.4240
    4 - - 2.4858 - - 15.6072 - - 66.6667 - - 156.2160
    5 - - 3.6364 - - 23.2654 - - 102.4240 - - 200.0000
    6 - - 5.3062 - - 34.1537 - - 156.2160 - - 277.7780
    0.10 1 0.980 0.980 0.9797 1.628 1.628 1.6282 3.333 3.333 3.3333 16.67 16.67 16.6667
    2 1.206 1.206 1.2057 4.786 4.786 4.7858 16.67 16.67 16.6667 35.56 35.56 35.5556
    3 1.628 1.628 1.6282 12.63 12.63 12.6247 35.56 35.56 35.5556 73.48 73.48 73.4783
    4 - - 2.3107 - - 13.2379 - - 50.0000 - - 107.0370
    5 - - 3.3333 - - 19.1668 - - 73.4783 - - 133.3330
    6 - - 4.7859 - - 27.2657 - - 107.0370 - - 178.5710
    0.20 1 0.895 0.895 0.89536 1.454 1.454 1.4541 2.857 2.857 2.8571 12.50 12.50 12.5000
    2 1.092 1.092 1.0922 4.001 4.001 4.0011 12.50 12.50 12.5000 24.62 24.62 24.6154
    3 1.454 1.454 1.4541 9.728 9.728 9.7281 24.62 24.62 24.6154 46.94 46.94 46.9444
    4 - - 2.0253 - - 10.1547 - - 33.3333 - - 65.6818
    5 - - 2.8571 - - 14.1732 - - 46.9444 - - 80.0000
    6 - - 4.0011 - - 19.4293 - - 65.6818 - - 104.1670
    0.40 1 0.764 0.764 0.7638 1.198 1.198 1.1979 2.222 2.222 2.2222 8.333 8.333 8.3333
    2 0.919 0.919 0.9191 3.013 3.013 3.0130 8.333 8.333 8.3333 15.24 15.24 15.2381
    3 1.198 1.198 1.1979 6.668 6.668 6.6683 15.24 15.24 15.2381 27.26 27.26 27.2581
    4 - - 1.6242 - - 6.9277 - - 20.0000 - - 37.0513
    5 - - 2.2222 - - 9.3179 - - 27.2581 - - 44.4444
    6 - - 3.0130 - - 12.3374 - - 37.0513 - - 56.8182
    0.60 1 0.666 0.666 0.6660 1.019 1.019 1.0185 1.818 1.818 1.8182 6.250 6.250 6.2500
    2 0.793 0.793 0.7934 2.416 2,416 2.4162 6.250 6.250 6.2500 11.03 11.03 11.0345
    3 1.019 1.019 1.0185 5.073 5.073 5.0727 11.03 11.03 11.0345 19.21 19.21 19.2045
    4 - - 1.3557 - - 5.2571 - - 14.2857 - - 25.8036
    5 - - 1.8182 - - 6.9403 - - 19.2045 - - 30.7692
    6 - - 2.4162 - - 9.0383 - - 25.8036 - - 39.0625
    0.80 1 0.590 0.590 0.5904 0.886 0.886 0.8858 1.538 1.538 1.5385 5.000 5.000 5.0000
    2 0.698 0.698 0.6979 2.017 2.017 2.0168 5.000 5.000 5.0000 8.649 8.649 8.6487
    3 0.886 0.886 0.8858 4.093 4.093 4.0933 8.649 8.649 8.6487 14.83 14.83 14.8246
    4 - - 1.1633 - - 4.2357 - - 11.1111 - - 19.7945
    5 - - 1.5385 - - 5.5295 - - 14.8246 - - 23.5294
    6 - - 2.0168 - - 7.1313 - - 19.7945 - - 29.7619
    1.00 1 0.530 0.530 0.5302 0.784 0.784 0.7837 1.333 1.333 1.3333 4.167 4.167 4.1667
    2 0.623 0.623 0.6230 1.731 1.731 1.7307 4.167 4.167 4.1667 7.111 7.111 7.1111
    3 0.784 0.784 0.7837 3.431 3.431 3.4308 7.111 7.111 7.1111 12.07 12.07 12.0714
    4 - - 1.0188 - - 3.5466 - - 9.0909 - - 16.0556
    5 - - 1.3333 - - 4.5953 - - 12.0710 - - 19.0476
    6 - - 1.7307 - - 5.8889 - - 16.0560 - - 24.0385

     | Show Table
    DownLoad: CSV

    In this paper, the model for a class of moderately thick rectangular plates is considered, which includes the bending, free vibration and buckling problems of plates and is expected to be widely used in functionally graded material problems. Based on the equilibrium equations of the model, the relationship between force and displacement, the Hamiltonian system and the corresponding Hamiltonian operator are derived, which provides the possibility for the applications of SEA and the symplectic superposition method. Based on the calculation of eigenvalues and the analysis of the properties of symplectic eigenvectors, the general solution under the simply supported boundary conditions of opposite edges is given by using the proved completeness theorem. In addition, the general solution is applied to two kinds of elastic foundation problems of moderately thick rectangular plates, which have not been solved by the SEA before. At the same time, the higher order and more accurate frequency parameters are solved for the free vibration problem of moderately thick rectangular plates with fully simply supported. The satisfactory numerical results confirm the effectiveness and universality of our model. It should be noted that the buckling problem of the plates is solved similarly to the free vibration problem except that the buckling factor is included in the values of \alpha_{1} and \beta_{1} in the model.

    The authors declare they have not used Artificial Intelligence (AI) tools in the creation of this article.

    This work was supported by the Natural Science Foundation of Inner Mongolia (No.2021MS01004)

    All authors declare no conflicts of interest in this paper.

    \begin{align*} A_{34,n} = &-\left(\left(b\left((1-\nu) \left(C^4-2 \alpha C^3+C^2 \left(\alpha ^2+2 \gamma D+D \zeta _n^2 (\alpha (\nu -1)-2 \beta )\right)+C D\alpha (2 \gamma+ \right.\right.\right.\right.\\ &\left.\left.\zeta _n^2 (\alpha (-\nu )+\alpha -2 \beta )\right)-D^2 \left(-\gamma +(\beta -\alpha ) \zeta _n^2+P_n\right) \left(\gamma +\zeta _n^2 (-(\alpha \nu +\beta ))+P_{n}\right)\right)\\ &\left(C^4-2 \alpha C^3+C^2 \left(\alpha ^2+2 \gamma D+2 D \zeta _n^2 (\alpha (\nu -2)-\beta )\right)+2 \alpha C D \left(\gamma -\zeta _n^2 (\alpha \nu +\beta )\right)\right.\\ &\left.-D^2 \left(-\gamma +(\beta -\alpha ) \zeta _n^2+P_{n}\right) \left(\gamma +\zeta _n^2 (-(2 \alpha \nu +\alpha +\beta ))+P_{n}\right)\right)+\alpha D \zeta _n^2\left(C^2 (\nu- \right.\\ &3)\left.+D (\nu +1) \left(-\gamma +(\beta -\alpha ) \zeta _n^2+P_{n}\right)-\alpha C (\nu +1)\right)\left(3 C^4 (\nu -1)+\alpha C^3 (2-6 \nu )\right.\\ &+D^2 (\nu -1) \left(-\gamma +(\beta -\alpha ) \zeta _n^2+P_{n}\right) \left(-\gamma +\zeta _n^2 (2 \alpha \nu +\alpha +\beta )-P_{n}\right)-2 \alpha C D(P_{n}\\ &\left.-2 \gamma \nu +\zeta _n^2 (\alpha ((\nu -2) \nu -1)+2 \beta \nu )+\nu P_{n}\right)+C^2 \left(\alpha ^2 (3 \nu +1)+2 D (\nu -1)(2 \gamma +\right.\\ &\left.\left.\zeta _n^2 (\alpha (\nu -1)-2 \beta )+P_{n}\right)\right)+4 C^2 D \alpha \zeta _n^2 \left(C^2 (\nu -1)-\alpha C (\nu +3)+D (\nu -1)(\gamma + \right.\\ &\left.\left.\left.(\alpha -\beta ) \zeta _n^2+P_{n}\right))(C^2-\alpha C+D \left(\nu \left(\gamma +(\alpha -\beta ) \zeta _n^2\right)-P_{n}\right)\right)\right)\left/\left(\sqrt{2} \alpha ^2 D \zeta _n^2\left(C^2\right.\right.\right.\\ &\left.(\nu -3)-\alpha C (\nu +1)+D (\nu +1) \left(-\gamma +(\beta -\alpha ) \zeta _n^2+P_{n}\right)\right)^{2}(C (\alpha -C)-D(P_{n}+\\ &\left.\alpha +C (\nu +1)+D (\nu +1) \left(\gamma +(\alpha -\beta ) \zeta _n^2+P_{n}\right)\right)^{2}, \end{align*}
    \begin{align*} A_{56,n} = &\left(b(\left(\nu -1)\left(-C^4+2 \alpha C^3+C^2 \left(-\alpha ^2-2 \gamma D+2 D \zeta _n^2 (\beta -\alpha (\nu -2))\right)\right.+2 \alpha C D\right.\right.\\ &\left.\left(\zeta _n^2 (\alpha \nu +\beta )-\gamma \right)+D^2 \left(\gamma +(\alpha -\beta ) \zeta _n^2+P_{n}\right) \left(-\gamma +\zeta _n^2 (2 \alpha \nu +\alpha +\beta )+P_{n}\right)\right)\\ &\left(-C^4+2 \alpha C^3-C^2 \left(\alpha ^2+2 \gamma D+D \zeta _n^2 (\alpha (\nu -1)-2 \beta )\right)+D^2 \left(\gamma +(\alpha -\beta ) \zeta _n^2\right.\right.\\ &\left.+P_{n})(-\gamma +\zeta _n^2 (\alpha \nu +\beta )+P_{n})+\alpha C D \left(\zeta _n^2 (\alpha (\nu -1)+2 \beta )-2 \gamma \right)\right)-4 C^2 \alpha \left(C^2\right.\\ &\left.(\nu -1)-\alpha C (\nu +3)-D (\nu -1) \left(-\gamma +(\beta -\alpha ) \zeta _n^2+P_{n}\right)\right)\left(C^2-\alpha C+D\right.\\ &\left.(\nu \left(\gamma +(\alpha -\beta ) \zeta _n^2\right)+P_{n})\right)+\alpha \left(-C^2 (\nu -3)+\alpha C (\nu +1)+D(\nu +1)(P_{n}+\gamma +\right.\\ &\left.\left.(\alpha -\beta ) \zeta _n^2\right)\right)\left(3 C^4 (\nu -1)+\alpha C^3 (2-6 \nu )-D^2 (\nu -1) \left(\gamma +(\alpha -\beta ) \zeta _n^2+P_{n}\right)\right.\\ &(-\gamma +\zeta _n^2 (2 \alpha \nu +\alpha +\beta )+P_{n})+2 C D \alpha \left(2 \gamma \nu +\zeta _n^2 (\alpha (1-(\nu -2) \nu )-2 \beta \nu )+P_{n}\right.\\ &\left.\left.\left.\nu +P_{n})+C^2 \left(\alpha ^2 (3 \nu +1)-2 D (\nu -1) \left(-2 \gamma +\zeta _n^2 (\alpha (-\nu )+\alpha +2 \beta )+P_{n}\right)\right)\right)\right)\right))\\ &\left/\left(\sqrt{\frac{-C^2+\alpha C+D \left(-\gamma +(\alpha +\beta ) \zeta _n^2+P_{n}\right)}{\alpha D}}\sqrt{2}D\zeta _n^2\alpha ^2\left(C (\nu +1)-C^2 (\nu -3)\right.\right.\right.\\ &\left.\alpha +C (\nu +1)+D (\nu +1) \left(\gamma +(\alpha -\beta ) \zeta _n^2+P_{n}\right)\right)^{2}. \end{align*}
    \begin{align*} \theta_{n,3} = &-\left(C D(C+G_f)q(4 C^2+C G_f(\nu +1)-D (\nu +1) (k_f+P_{2,n}))\left(4 C^2+D \left(k_f-P_{2,n}\right)\right.\right.\\ &\left.\left.C (\nu +3) G_f+(\nu -1)\right)\zeta _n\sqrt{\frac{2 \left(C G_f+D \left(k_f-P_{2,n}\right)\right)}{D \left(C+G_f\right)}+4 \zeta _n^2} (\cos(b \zeta _n)-1)\right)/ \\ &\left(b\sqrt{\frac{C G_f+D k_f-D P_{2,n}}{2 C D+2 D G_f}+\zeta _n^2}\left(4 C^2 D\left(4 C^2+C (\nu +3) G_f+D (\nu -1) \left(k_f-P_{2,n}\right)\right)\right.\right.\\ &\left(C+G_f\right)\zeta _n^2 \left(C G_f+D \left(\nu k_f+P_{2,n}\right)\right)+(1-\nu)\left(-4 C^2 D k_f+C^2 G_f^2-2 C D G_f k_f+\right.\\ &D^2 (k_f^2-P_{2,n}^2)\left.+D \zeta _n^2 \left(C+G_f\right) \left(-4 C^2-C (\nu +1) G_f+D (\nu +1) \left(k_f+P_{2,n}\right)\right)\right)\left(C^2 G_f^2\right.\\ &+D^2 (k_f^2- P_{2,n}^2)-4 C^2 D k_f-2 C D G_f k_f+2 D (C+G_f)\left(-4 C^2-C (\nu +1) G_f+ D\times\right.\\ &\left.\left.(\nu +1)\left(k_f+P_{2,n}\right)\right)\zeta _n^2\right)+D \zeta _n^2 \left(C+G_f\right) \left(-4 C^2-C (\nu +1) G_f+D (\nu +1) \left(k_f+P_{2,n}\right)\right)\\ &\left(C^3 (2-6 \nu ) \left(C+G_f\right)+3 C^4 (\nu -1)+D^2(k_f+P_{2,n})(\nu -1)\left(2 \left(C+G_f\right)+k_f-P_{2,n}\times\right.\right.\\ &\left.(\nu +1) \zeta _n^2\right)-2 C D \left(C+G_f\right) \left(\left(\nu ^2-1\right) \zeta _n^2 \left(C+G_f\right)+2 \nu k_f+\nu P_{2,n}+P_{2,n}\right)+\left(\left(C+G_f\right){}^2\right.\\ &\left.\left.\left.\left.2 D (\nu -1) \left((\nu -3) \zeta _n^2 \left(C+G_f\right)-2 k_f+P_{2,n}\right)+(3 \nu +1)\right)C^2\right)\right)\right), \end{align*}
    \begin{align*} \theta_{n,5} = &\left(C (C+G_f)q(4 C^2+C G_f(\nu +1)-D (\nu +1) (k_f+P_{2,n}))\left(4 C^2+D \left(k_f+P_{2,n}\right)\right.\right.\\ &\left.\left.C (\nu +3) G_f+(\nu -1)\right)\zeta _n\sqrt{\frac{2 \left(C G_f+D \left(k_f+P_{2,n}\right)\right)}{D \left(C+G_f\right)}+4 \zeta _n^2} (\cos(b \zeta _n)-1)\right)/ \\ &\left(b\zeta _n\sqrt{\frac{C G_f+D k_f+D P_{2,n}}{2 C D+2 D G_f}+\zeta _n^2}\left(-4 C^2 \left(C+G_f\right)(C G_f+D \nu k_f-D P_{2,n})\right.\right.\left(4 C^2+\right.\\ +1 &\left.C (\nu +3) G_f+D (\nu -1) \left(k_f+P_{2,n}\right)\right)+\frac{1}{D \zeta _n^2}(\nu -1)\left(4 C^2 D k_f-C^2 G_f^2+2 C D G_f k_f\right.\\ &+D^2P_{2,n}^2-D^2k_f^2\left.+D \zeta _n^2 \left(C+G_f\right) \left(4 C^2+C (\nu +1) G_f-D (\nu +1) \left(k_f-P_{2,n}\right)\right)\right)\times\\ &\left(-C^2 G_f^2+4 C^2 D k_f+2 C D G_f k_f-D^2 k_f^2+D^2 P_{2,n}^2+2 D (C+G_f)\left(C (\nu +1) G_f\right.\right.+\\ &4 C^2-\left.\left.D (\nu +1) \left(k_f-P_{2,n}\right)\right)\zeta _n^2\right)+\left(C+G_f\right) \left(4 C^2+C (\nu) G_f-D (\nu +1)\left(k_f-P_{2,n}\right)\right) \\ &\left(4 C^3 \left(G_f-2 D (\nu -1) \zeta _n^2\right)+D^2 (\nu -1) \left(k_f-P_{2,n}\right) \left(2 (\nu +1) G_f \zeta _n^2+k_f+P_{2,n}\right)+C^2 \right.\\ &\left(-2 D (\nu -1) (\nu +5) G_f \zeta _n^2+4 D \left(-2 \nu k_f+k_f+P_{2,n}\right)+(3 \nu +1) G_f^2\right)+2 C DG_f \times\\ &\left.\left.\left.\left(-2 \nu k_f+\nu P_{2,n}+P_{2,n}\right)-\left(\nu ^2-1\right) \zeta _n^2 \left(D \left(P_{2,n}-k_f\right)+G_f^2\right)\right)\right)\right). \end{align*}


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