SEMIGROUP WELL-POSEDNESS OF MULTILAYER MEAD-MARKUS PLATE WITH SHEAR DAMPING

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1 SEMIGROUP WELL-POSEDNESS OF MULTILAYER MEAD-MARKUS PLATE WITH SHEAR DAMPING Scott W. Hansen Department of Mathematics Iowa State University Ames, IA Abstract A multilayer plate model consisting of m + 1 relatively stiff plate layers bonded together by m compliant plate layers is described. In the case of only three layers, the model reduces to a Mead-Markus sandwich plate with viscous damping in the central layer. Existence and uniqueness of solutions are established using semigroup theory. In particular, the homogeneous problem may be written in the form x (t) = Ax(t) where A is the generator of a C 0 contraction semigroup. Keywords: Multilayer plate, Mead-Markus plate, sandwich plate, shear damping Introduction The classical sandwich plate is a three-layer plate model consisting of two relatively stiff outer layers and a more compliant inner layer. Various sandwich plate and constrained layer models have been proposed and analyzed, see e.g., DiTaranto [], Mead and Markus [9], Rao and Nakra [10]. For a review and some comparisons of these models see e.g., Sun and Lu [11]or Mead [8]. In Hansen [3], a multilayer generalization of the Mead-Markus model (and Rao-Nakra model) was derived. The multilayer models consist of alternating stiff and compliant layers. The stiff layers do not allow shear and are modelled as Kirchhoff plates, while the compliant layers allow shear and may be modelled in various ways. For the multilayer Mead-Markus model, the in-plane inertia is ignored in each layer and bending stiffness are ignored in the compliant layers. In the undamped This research was supported in part by the National Science Foundation under grant DMS

2 case, existence and uniqueness of solutions for the multilayer systems in [3]was established on the natural energy spaces using the standard variational theory (e.g., Lions and Dautray [1]). However, in the case when shear damping is included in the multilayer Mead-Markus model, the variational approach approach in ([3]) could not be applied (at least directly) due a lack of coerciveness in the bilinear form associated with the highest order time-derivatives. We overcome this problem in this article for the special case in which all of the stiff layers have the same Poisson s ratio. In this case it is possible to solve for the shear in each compliant layer and write the homogeneous problem in the standard form y = Ay. We then show that A is the generator of a strongly continuous semigroup of contractions on the natural energy space. Analogous models to the one of this article have been investigated in the case of a layered beam. For example, an analogous three-layer beam model was investigated in [4], where it was found (for either clamped or simply-supported boundary conditions) that the associated semigroup is in fact exponentially stable (moreover is analytic, if rotational moment of inertia is neglected). Similar results were found in the multilayer beam model in [5]for the case of simply supported boundary conditions. 1. Multilayer Mead-Markus Model We first describe the multilayer Mead-Markus of [3]. (For a more careful derivation see [3].) The multilayer sandwich plate is assumed to consist of n = m + 1 layers of alternating stiff and compliant plates, that occupy the region Ω (0, h) at equilibrium, where Ω is a smooth bounded domain in the plane. The layers are indexed from 1 to n, from bottom to top, with odd indices for stiff layers and even indices for compliant layers. We use coordinates x = {x 1, x } to denote points in Ω and use x 3 as the transverse independent variable. As is typical in plate theories, it is assumed that the transverse displacement is independent of x 3, i.e., we may use the scalar w(x) to denote the transverse displacement at the point x Ω. We let v i = {v1 i, vi } i = 1,,... n denote the in-plane displacements along the midplane of the ith layer. It is assumed that all the layers are bonded to one another so that no slip occurs. The Kirchhoff hypothesis applies to the stiff layers, (i.e., no shear) while the compliant layers allow shear and deform linearly with respect to the transverse variable. Under these assumptions any displacement is completely determined by specification of the state variables v i,, and w.

3 Semigroup well-posedness of multilayer Mead-Markus plate with shear damping3 If θ and ξ are matrices in R lm, by θ : ξ we mean the scalar product in R lm. We also denote (θ, ξ) Ω = θ : ξ dx, (θ, ξ) Γ = θ : ξ dγ. Ω Define the form l for functions θ(x) = {θ 1 (x), θ (x)} by l(θ; ˆθ) = ( θ1, ˆθ ) 1 + ( + θ Ω x, ˆθ ) ( x ν θ 1, ˆθ ) (( x + 1 ν Ω Γ ( + ν θ Ω x, ˆθ 1 ) ( x ) 1 ( )Ω θ1 x + θ, ˆθ1 x + ˆθ ))Ω. where ν is the Poisson s ratio (0 < ν < 1/). It is assumed that the in-plane kinetic energy is negliglble and bending potential energy of the even layers are negligible in comparison to those of the surrounding odd layers. The energy is the sum of the kinetic and potential energies of each layer E i = K i + P i, where K i = P i = { 1 1 { 1 1 Ω ρ ih i (w ) + α i w dx Ω ρ ih i (w ) dx i even Ω Kl(h3 i D i w; w) + 1l(h i D i v i ; v i ) dx Ω G ih i ϕ i dx i even In the above primes (e.g., w ) denote differentiation with respect to time, D i = E i /1(1 ν ), α i = ρ i h 3 i /1 where E i > 0 denotes the in-plane Young s modulus, h i the thickness, ρ i the volume density, all for the ith layer. In addition, G i is the transverse shear modulus. The variable ϕ i is the shear of the ith layer defined in (1) below. Define the following n by n matrices: h = diag (h 1, h,..., h n ) D = diag (D 1, D,..., D n ) p = diag (ρ 1, ρ,..., ρ n ) G = diag (G 1, G,..., G n ). In addition, we let e.g., h O and h E denote the diagonal matrices of oddindexed and even-indexed thicknesses h i, respectively. Actually, we will only need to refer to h O, h E, D O, G E, p O. Also define 1 E and 1 O as the column vectors of m and m + 1 ones, respectively. Let v O denote the (m+1) matrix with rows v i, i = 1, 3, 5,... m+1. Likewise, let ϕ O and ϕ E denote the matrix with rows ϕ i, and even, respectively. Since no shear occurs in the odd layers, ϕ O = 0. The shear ϕ E can be expressed in terms of v O, w as follows: h E ϕ E = Bv O + h E N w; N = h E Ah O 1 O + 1 E (1)

4 4 where A = (a ij ), B = (b ij ) are the m (m + 1) matrices defined by a ij = { 1/ if j = i or i otherwise b ij = { () i+j+1 if j = i or i otherwise. Also define l O (v O, ˆv O ) = n l(v i ; ˆv i ) Collecting the energies one finds that the total potential and kinetic energy may be expressed as K(t) = c(v O, w ; v O, w )/ P(t) = a(v O, w; v O, w)/ where c( ; ) and a( ; ) denote the bilinear forms c(v O, w; ˆv O, ŵ) = (mw, ŵ) Ω + (α w, ŵ) Ω a(v O, w; ˆv O, ŵ) = l O (K 1 O w; 1 O ŵ) + 1l O (h O D O v O ; ˆv O ) +(G E h E ϕ E, ˆϕ E ) Ω where ϕ and ˆϕ satisfy (1), and n m = h i ρ i, α = 1 n ρ i h 3 i, 1 i=1 K = D i h 3 i. Let us assume the plate is clamped on a portion Γ 0 of the boundary Γ and subject to applied forces on the complementary portion Γ 1 and distributed forces on Ω. The equations of motion are determined by Hamilton s principal from the energy. The variational differential equation one obtains is the following: (mw, ŵ) Ω + (α w, ŵ) Ω + l(k O w, ŵ) +1l O (h O D O v O ; ˆv O ) + (G E h E ϕ E, h E Bˆv O + N ŵ) = W ({ˆv O, ŵ}) = Ω ŵf 3 + ˆv O f O dx + Γ 1 ŵg 3 + ˆv O g O ŵ n M n ds. In the above, f 3 is the (scalar) transverse applied force in Ω and f O is the net in-plane force acting on the odd layers in Ω. (f O has rows f i = {f1 i, f i }, i = 1, 3, 5,... m + 1.) The boundary forces acting on Γ 1 are: the transverse force g 3 (scalar valued), the in-plane force (dimensions matching f O ) and the bending moment M n (scalar valued). (For precise description of forces see [3].) The test functions ŵ, ˆv O satisfy the clamped boundary conditions on Γ 0 and are sufficiently regular.

5 Semigroup well-posedness of multilayer Mead-Markus plate with shear damping5 Inclusion of shear damping. Damping may be introduced into any of the plate layers by replacing the standard stress-strain relation for transversely isotropic materials by an appropriate dissipative constitutive law. In the case of strain-rate shear damping, the stress-strain relation for transverse shear: σ 13 = Gɛ 13 is replaced by a dissipative stress-strain relation of the form: σ 13 = (G + G d dt )ɛ 13. The equations of motion are then modified by the correspondence G (G + G t ). In our situation, shear damping occurs only in the even layers. Let G E be the diagonal matrix with diagonal elements G i, i even. We obtain the following variational differential equation when shear damping is included in the even layers: (mw, ŵ) Ω + (α w, ŵ) Ω + l(k w, ŵ) + 1l O (h O D O v O ; ˆv O ) +(G E h E ϕ E + G E h E ϕ E, h E Bˆv O + N ŵ) = W ({ˆv O, ŵ}). () 1.1 Boundary value problem We first need to define some operators and boundary operators. Define the operator L by Lφ = L{φ 1, φ } = {L 1 (φ), L (φ)} where L 1 φ = [( φ 1 + ν φ x )] + x [( 1 ν )( φ 1 x + φ )] L φ = x [( φ x + ν φ 1 )] + [( 1 ν )( φ + φ 1 x )]. The associated boundary operator Bφ = {B 1 (φ 1, φ ), (B (φ 1, φ )} is defined by [( ) ( ) ( ) ] B 1 (φ 1, φ ) = φ1 n 1 + ν φ x n ν φ1 x + φ n [( ) ( ) ( ) ] B (φ 1, φ ) = φ x n + ν φ 1 n + 1 ν φ + φ 1 x n 1. where n = (n 1, n ) denotes the outward unit normal to Γ. (For later reference: also define τ = ( n, n 1 ) as the unit tangent vector to Γ.) The following Green s formula is valid for all sufficiently smooth ˆφ, φ: l(φ, ˆφ) = (Bφ, ˆφ) Γ (Lφ, ˆφ) Ω. (3) For ξ = (ξj i ) (i = 1,,..., n, j = 1, ) define the matrices Lξ and Bξ by (Lξ) ij = (L j ξ i ), (Bξ) ij = (B j ξ i ), i = 1,,..., n, j = 1,. Furthermore we define the operators L O, L E, B O, B E from L and B based upon the convention that O and E subscripts refer to the parts of the operators that act upon odd and even rows respectively.

6 6 Similar to (3) the following Green s formula is valid for all sufficiently smooth ξ, ˆξ: l O (ξ, ˆξ) = (B O ξ, ˆξ) Γ (L O ξ, ˆξ) Ω. (4) The equations of motion can be found from () using the Green s formulas (3), (4) and further integrations by parts: mw α w + K w div N T h E (G E ϕ E + G E ϕ E ) = f } 3 h O D O L O v O + B T (G E ϕ E + Gϕ E ) = f (5) O where ϕ E = h E Bv O + N w αw n K( τ ( B w) τ) K( w) n + N T h E (G E ϕ E + G E ϕ E ) n = g 3 K(B w) n = M n 1D O B O h O v O = g O } w = w n = 0 v O = 0 where (5) holds on Q = Ω (0, ), (6) holds on Σ 1 = Γ 1 (0, ), and (7) holds on Σ 0 = Γ 0 (0, ). Appropriate initial conditions are of the form (6) (7) w(0) = w 0, w (0) = w 1, ϕ E (0) = ϕ 0 E (8). State variable formulation There is a difficulty in proving existence and uniqueness in the formulation above since it is not an easy task to solve for ϕ E in (5), which is needed to in order to solve for the generator of a semigroup. However, this can be accomplished in the present situation (when all the Poisson ratios in the stiff layers are same) due to the decomposition described in this section. Let W = span 1 O, and V = span D O h O 1 O. Clearly these spaces are not orthogonal. Therefore there is a unique decomposition of R m+1 (or C m+1 ) into a part in W and a part in V. Since the kernel of B is easily seen to be W, any vector in W can be written as an element in the image of B T. Thus at each point in Ω and Γ 1, respectively we have the decomposition f O = B T f E + f O, g O = B T g E + g O where f O and g O at each point belong to V V and f E and g E at each point belong to R m.

7 Semigroup well-posedness of multilayer Mead-Markus plate with shear damping7 We return to (5) (7) but focus on the terms coupled to v O. relevant part of the system is The h O D O L O v O + B T (G E ϕ E + G E ϕ E) = f O + B T f E in Ω (9) 1D O h O B O v O = g O + B T g E on Γ 1 (10) v = 0 on Γ 0. (11) We multiply (9) and (10) on the left by 1 T O and obtain L( D i h i v i ) = f i O in Ω (1) 1B( D i h i v i ) = g O i in Γ 1. (13) If we define v O = C T v O = D i h i v i, then (1) (13) can be written as f = f i O, ḡ = g i O (14) L v O = f in Ω 1B v O = ḡ on Γ 1 v O = 0 on Γ 0. (15) When the Poisson s ratios are constant we can write BL O v O = L E Bv O = L E h E (ϕ E N w). (16) Define the matrix P by P = 1 1 BD O h O BT. (17) We multiply (9) and (10) by 1 1 Bh O D O and obtain BL O v O + P G E ϕ E = P f E in Ω. (18) Thus, after substitution of (16) into the above, we obtain the system L E h E (ϕ E N w) + P (G E ϕ E + G E ϕ E) = P f E in Ω (19) B E h E (ϕ E N w) = P g E on Γ 1 (0) v O = 0 on Γ O. (1)

8 8 Lemma 1 The matrix P in (17) is positive, i.e., z T P z > 0 for all z R m. Moreover, every element p ij of P is positive. Proof: We can write P = BΛB T where Λ = diag (λ 1, λ,..., λ m+1 ), with λ i > 0, all i. Moreover B can be written as the sum of partitioned matrices B = ( 0. I m ) (I m. 0), where I m is the identity on R m, and 0 is a column vector of zeros. Performing the block multiplications in (17), one obtains P = Λ 11 + Λ m+1,m+1 Λ 1,m+1 Λ m+1,1 where Λ ij is the minor for the ij-th spot in Λ. Thus the diagonal of P has the elements p ii = λ i + λ i+1, for i = 1,,... m. The superdiagonal and subdiagonal sequence (beginning in the upper left) are each λ, λ 3,..., λ m. All other elements are zero. We therefore have p 11 = λ 1 + λ > p mm = λ m + λ m+1 > m p 1,k = λ k= m p m,k = λ m k=1 p kk = λ k + λ k+1 p 1,j = λ k + λ k+1. j k Thus P is diagonally dominant (with strict inequality in the first and last rows). Furthermore it is easily verified that P is irreducible. It follows from the theory of M-matrices (see e.g., [7],Theorem 3, p. 531) that P is a nonnegative matrix (every element nonnegative). This completes the proof. Since P is invertible, one can solve for G E ϕ E + G E ϕ E in (19) (1) and substitute into (5) (7) to obtain the following system: mw α w + K w div N T {h E P L E Bv O } = f 3 div N T h E f E G E ϕ E + G E ϕ E P L E Bv O = f E Where α(w n ) τ ( τ KB w) + K( w) n + N T h E (G E ϕ E + G E ϕ E ) n = g 3 KB w n = M n P B E Bv O = g E in Q () on Σ 1 (3) ϕ E = 0, w = 0, w n = 0 on Σ 0 (4) Bv O = h E (ϕ E N w), C T v O = v O, (5)

9 Semigroup well-posedness of multilayer Mead-Markus plate with shear damping9 where v O is the solution to the stationary Lamé system (15). Initial conditions are specified for w, w and ϕ E. Suppose for the moment that the system () (4) above together with appropriate initial conditions uniquely determines ϕ E and v O. Then v O is also determined: Indeed, define B C to be the partitioned (square) matrix defined by BC T = (BT. C), ( C is defined in (14)). Then (since B C is invertible from our earlier discussion) v O can be obtained from the following: ( ) h B C v O = E ϕ E h EN w. (6) v O 3. Semigroup formulation of homogeneous problem We consider the homogeneous problem mw α w + K w div N T {h E P L E Bv O } = 0 in Q G E ϕ E + G Eϕ E P L E Bv O = 0 in Q } (7) For simplicity, we restrict our interest here to the case of case of hinged boundary conditions: } B w n = 0 on Σ (8) B E Bv O = 0 w = 0 on Σ. (9) Here Σ = Ω (0, ). For the variable v O = C T v O we assumme fixed boundary conditions, so that v O is the solution of L v O = 0 in Ω v O = 0 on Γ 0. (30) The system (30) has the unique solution v O = 0 since the form l is known to be coercive on H 1 0 (Ω) ([6]). Therefore we eliminate v O and henceforth may assume C T v O = 0. Define the spaces L O (Ω) = {v O = (vj i), i = 1, 3, 5,... n, j = 1, : vi j L (Ω)} HO,Γ 1 = {v O L O (Ω) : vi j H1 0 (Ω)} L E (Ω) = {φ E = (φ i j ), i =, 4,... m, j = 1, : φi j L (Ω)} HE 1 = {φ E L E (Ω) : φi j H1 (Ω)}.

10 10 The energy space is {w, w, v O V : C T v O = 0} where V = H (Ω) H 1 0 (Ω) H 1 0 (Ω) H 1 O,Γ. Let φ = Ry denote the solution to the following elliptic problem mφ α φ = y, φ = 0 on Γ. (31) Then R maps H (Ω) into H0 1 (Ω) continuously. Let y T = (y 1, y, y 3 ) = (w, w, ϕ E ) then y = Ay = Define matrix S by y R( K y 1 + div N T {h E P L E h E (y 3 N y 1 )} G E G Ey 3 + G E P L E h E (y 3 N y 1 ) B C y = It is simple to verify that a 0 (3) y = S a (33) BSy = y y R m (34) SBu = u u R m+1 : C T u = 0 (35) Along with the state variables y, define v y = Sh E (y 3 N y 1 ). The energy inner product is defined by (y, z) E = (my, z ) Ω + α( y, z ) Ω + Kl O ( 1 0 y 1, 1 0 z 1 ) Define H by +(G E h E y 3, z 3 ) Ω + 1l 0 (h 0 D O v y, v z ). H = {w, w, ϕ E : w H (Ω) H 1 0 (Ω), w H 1 0 (Ω), ϕ E H 1 E}. (36) Then one can verify that A : D(A) H where D(A) = {y H : y 1 H 3 (Ω), y H (Ω), y 3 H 3 E, + BC s} (37) where +BC s means the boundary conditions (B y 1 ) n = 0, B E (Bv y ) = 0 are imposed on Γ. Theorem The operator A in (3) is the generator of strongly continuous semigroup of contractions on H. Moreover for any y D(A), Re (y, Ay) E = ( G E h EU, U) Ω ; U = P BL O v y G E y 3. (38)

11 Semigroup well-posedness of multilayer Mead-Markus plate with shear damping11 Proof: To show that A is the generator of a contraction semigroup, By the Lumer-Phillips theorem it is enough to demonstrate that A is dissipative and satisfies the range condition: i.e., (I A) maps D(A) onto H. First we show dissipativity. Let z = Ay, then (y, Ay) E = (my, z ) Ω + α( y, z ) Ω + Kl( y 1, z 1 ) +(G E h E y 3, z 3 ) Ω + 1l 0 (h 0 D O v y, v z ) = (my, R[ K y 1 + div N T {h E P L E h E (y 3 N y 1 )}]) Ω +α( y, R[ K y 1 + div N T {h E P L E h E (y 3 N y 1 )}]) Ω +Kl( y 1, y ) + (G E h E y 3, G E G Ey 3 + G E P L E h E (y 3 N y 1 )) Ω +1l O (h O D O v y, Sh E ( G E G Ey 3 + G E P L E h E (y 3 N y 1 ) N y )) = {(y, K y 1 ) Ω + Kl( y 1, y )} + {(y, div N T h E P L E Bv y ) Ω +1l O (h O D O v y, Sh en y )} + {terms with G E } =: {T 1 } + {T } + {T 3 }. The real part of the first bracketed term T 1 is easily shown to vanish using integrations by parts. For the second term T we integrate by parts, apply (16) and use the boundary conditions to obtain T = (y, div N T h E P L E Bv y ) Ω + 1(h O D O B O v y, Sh E N y ) Γ +1(h O D O L O v y, Sh E N y ) Ω = (y, div N T h E P BL O v y ) Ω (div N T h E S T L O h O D O v y, y ) Ω + 1(S T L O h O D O v y n, h E Ny ) Γ = (y, div N T h E P BL O v y ) Ω One has that (div N T h E P BD O h O BT S T h O D O L O v y, y ) Ω (B BD O h O BT S T h O D O ) = BD O h O (I BT S T )h O D O = 0 where we have applied the transposition of formula (35) to obtain the last equality. Thus the real part of T vanishes. Hence T 3 = (G E h E y 3, G G E P L E Bv y ) Ω E G Ey 3 ) Ω + (G E h E y 3, +1l O (h O D O v y, Sh E ( G E G Ey 3 + G E P L E Bv y )) = (G E h E y 3, G E G Ey 3 ) Ω + (G E h E y 3, +1( G E ST h O D O L O v y, h E G E y 3 ) Ω (h O D O L O v y, Sh G E E P BL O v y )) Ω. G E P BL O v y ) Ω

12 1 One can easily show using (35) and the definition of P that 1S T h O D O = P B. (39) Applying (39) to the present calculation one finds that Re (y, Ay) E = ( G +( ( = ( E h EG E y 3, G E y 3 ) Ω + Re {( G G E h EG E y 3, P BL O v y ) Ω E h EP BL O v y, G E y 3 ) Ω } G E h EP BL O v y, P BL O v y ) Ω G E h E(P BL O v y G E y 3 ), (P BL O v y G E y 3 )) Ω Since G h E is positive definite, A is dissipative. Next we check the range condition. In what follows we use the notation w s = w H s (Ω), ϕ E E,s = m i even j=1, ϕ i j s. Assume that Ay + y = z where z H. This is the same as y + y 1 = z 1 (40) R( K y 1 + div N T h E P L E h E (y 3 N y 1 )) + y = z (41) G E G Ey 3 G E P L E h E (y 3 N y 1 ) + y 3 = z 3 (4) Solving for y 1 and L E y 3 in (41) and (4) gives K y 1 α y 1 + my 1 = div N T GE h E (y 3 z 3 ) + (m α )(z 1 + z )(43) G E y 3 P L E h E y 3 + G E y 3 = P L E h E N y1 + G E z 3.(44) The left hand side of (43) is associated with a coercive bilinear symmetric form on H (Ω) H0 1 (Ω) (see [3]) and hence for functions y D(A) one can show that y 1 4 C K y 1 α y 1 + my 1 0. Thus, considering the right hand side of (43), y 1 4 C 1 ( z 3 E,1 + y 3 E,1 + z 1 + z ). (45) Estimation of (43) with R 1/ applied to both sides leads to y 1 3 C ( z 3 E,1 + y 3 E,1 + z z 1 ). (46) Using that P is symmetric, positive definite, we also have that y 3 E, C G E y 3 P L E h E y 3 + G E y 3 0 C 1 z 3 E,0 + y 1 3. (47)

13 Semigroup well-posedness of multilayer Mead-Markus plate with shear damping13 Hence from (46) From (40) we also have y 3 E, C z 3 E,1 + y 3 E,1 + z z 1 ) (48) y C( y 1 + z 1 ). (49) Since A has is dissipative, I A is injective and hence any solutions to (40) (4) are unique. Applying the usual compactness/uniqueness argument to the estimates (46), (48), (49) yields y y + y 3 E, C( z 1 + z 3 E,1 + z 1 ). (50) Thus y D(A) and I A is surjective from D(A) to H. This completes the proof. References [1] R. Dautray, M. Lions, J.-L. (with collaboration of Artola, M. Cessenat, and H. Lanchon). Mathematical Analysis and Numerical Methods for Science and Technology; Evolution Problems I, volume 5. Springer-Verlag, 199. [] R.A. DiTaranto. Theory of vibratory bending for elastic and viscoelastic layered finite-length beams. J. Appl. Mech., 3: , [3] S.W. Hansen. Several related models for multilayer sandwich plates, 003. submitted. [4] S.W. Hansen and I. Lasiecka. Analyticity, hyperbolicity and uniform stability of semigroups arising in models of composite beams. Math. Models Methods Appl. Sci., 10: , 000. [5] S.W. Hansen and Z. Liu. Analyticity of semigroup associated with laminated composite beam, in: Control of Distributed Parameter and Sochastic Systems; eds. S. Chen, X. Li, J. Yong, X.Y. Zhou pp , (Kluwer Acad. Publ.). [6] J.E. Lagnese and J.-L. Lions. Modelling, Analysis and Control of Thin Plates (collection: Recherches en Mathématiques Appliquées ), RMA 6. Springer- Verlag, [7] P. Lancaster and M Tismenetsky. The Theory of Matrices. series: Computer Science and Applied Mathematics: Academic Press, [8] D.J. Mead. A comparison of some equations for the flexural vibration of damped sandwich beams. J. Sound Vib., 83: , 198. [9] D.J. Mead and S. Markus. The forced vibration of a three-layer, damped sandwich beam with arbitrary boundary conditions. J. Sound Vibr., 10: , [10] Y.V.K.S. Rao and B.C. Nakra. Vibrations of unsymmetrical sandwich beams and plates with viscoelastic cores. J. Sound Vibr., 34:309 36, [11] C.T. Sun and Y.P. Lu. Vibration Damping of Structural Elements. Prentice Hall, 1995.