ANALYSIS OF MIXED FINITE ELEMENTS FOR

LAMINATED COMPOSITE PLATES

F. Auricchio

Dipartimento di Meccanica Strutturale

Universita di Pavia, Italy

C. Lovadina∗

Dipartimento di Ingegneria Meccanica e Strutturale

Universita di Trento, Italy

E. Sacco

Dipartimento di Meccanica, Strutture, A. & T

Universita di Cassino, Italy

Abstract

The approximation to the solution of a laminated plate problem is considered. The mainresult consists in proving that a performant scheme can be obtained by simply coupling a stan-dard membrane element for the in-plane displacements with a robust method for plate bendingproblem, both elements related to the case of homogeneous material. A rigorous convergenceanalysis is developed for the case of clamped boundary conditions. Numerical results are pre-sented, showing the accordance between the theoretical results and the actual computations.

1 Introduction

Laminated composite plates are nowadays widely used in engineering practice, especially in space,automobile and civil applications [18]. The main motivation for this growing interest is related tothe improved ratio between performances and weight associated with laminated plates with respect

∗Corresponding author. Address:

Carlo Lovadina

Dipartimento di Ingegneria Meccanica e Strutturale, Universita di Trento

Via Mesiano 77, I-38050, Italy.

E-mail: lovadina@ing.unitn.it

1

to the case of homogeneous ones. However, the behavior of laminated plates is more complex,because

• they are anisotropic materials;

• they present significant shear deformation in the thickness direction;

• they exhibit extension-bending coupling.

In literature, many different models have been presented. The simplest one is surely the ClassicalLaminated Plate Theory (CLPT), which is based on the Kirchhoff hypotheses (cf. e.g. [22] for moredetails). However, due to the fact that shear deformations are not taken into account, the modelapproximation is quite poor. Moreover, in the framework of finite element procedures (cf. [7], [15]),CLPT models require C1-conforming schemes, which are quite expensive from the computationalpoint of view.

According to these considerations, models arising from the Reissner-Mindlin assumptions (calledFirst-order Shear-Deformation Theory: FSDT) are often preferred. The main FSDT featuresare related to the possibility of using C0-conforming methods and the ability to capture sheardeformations, which makes it possible to consider also the case of moderately thick plates.

In the present paper we develop an analysis of mixed finite element schemes relative to aFSDT model. Our main result is that a performant method can be obtained by simply taking astandard membrane element for the in-plane displacements and a robust plate bending element,both elements related to homogeneous plates (cf. [2], [3] and [10], for example). We remark thatthis conclusion is not straightforward, since the laminated structure exhibits an extension-bendingcoupling (i.e. in-plane displacements arise even for vertical loads).

The paper is organized as follows. In Section 2 we briefly introduce the laminated plate model,together with the variational formulation that will be used as the starting point of the discretizationprocedure. We establish an existence and uniqueness result for the continuous solution, mainly byrecognizing that the variational formulation of the problem is well-posed. In Section 3 we developour error bounds for mixed finite element methods in an abstract framework, as we want to give ananalysis covering as many schemes as possible. In Section 4 we present two examples of applicationsof our theory. Section 5 is devoted to present numerical experiments, showing that the predictedbehaviors are in fact met by actual computations.

In what follows, we will use standard notations (cf. [9] and [13]). Moreover, C will be a constantindependent of t and h, not necessarily the same in each occurrence. Finally, for the continuoussolution and the applied loads, we will suppose the necessary regularity.

2

2 The laminated composite plate model and the mixed formula-

tion

Let us denote with A = Ω×(−t/2, t/2) the region in R3 occupied by an undeformed elastic plate ofthickness t > 0 and midplane Ω. We suppose that the plate is formed by n superposed orthotropiclayers of equal thickness t

nand with material axes arbitrarily oriented in the x-y plane. Thus, the

k-th layer (1 ≤ k ≤ n) is positioned in Ω× (zk−1, zk), where z0 = − t2 and zk = − t

2 + k tn. Moreover

we will work in the framework of the First-Order Shear-Deformation Theory (shortly: FSDT (cf.[22])). Restricting ourselves to the case of a clamped plate subjected to a vertical load f , thefunctional to be considered reads as follows

Π =1

2

(∫

ΩAnε(u∗) : ε(u∗) dΩ + 2

∫

ΩBnε(u∗) : ε(θ∗) dΩ +

∫

ΩDnε(θ∗) : ε(θ∗) dΩ

)

+

+

∫

Ωs∗ · (θ∗ −∇w∗) dΩ −

1

2

∫

ΩH−1

n s∗ · s∗ dΩ −

∫

Ωfw∗ dΩ . (1)

Above, u∗ are the in-plane displacements, θ∗ are the rotations, w∗ represents the vertical dis-placement field, and s∗ are the resultant shear stresses. Moreover, An, Bn and Dn are fourth ordertensors defined by

Anτ =n∑

k=1

(zk − zk−1)Ckτ , (2)

Bnτ =1

2

n∑

k=1

(z2k − z2

k−1)Ckτ , (3)

Dnτ =1

3

n∑

k=1

(z3k − z3

k−1)Ckτ , (4)

where τ is a second order tensor (which in our case will always be symmetric). Moreover, Ck is afourth order symmetric and positive definite tensor.

Finally, Hn is a second order tensor, defined by

Hnτ = κn∑

k=1

(zk − zk−1)Skτ , (5)

where τ is a given vector and Sk is a second order symmetric and positive definite tensor.

Remark 2.1 The coefficient κ in (5) is called shear correction factor, and, in the case of homoge-neous plates, it is known a priori and it is taken as 5/6. For a laminated structure this value is not

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the optimal one, in general, and it is not known a priori. However, in the following we suppose toknow the value of the shear correction factor κ, also for the composite plate.

Remark 2.2 The functional (1) reveals an interesting feature of laminated composites: the bendingbehaviour and the membrane one are coupled through the term involving the tensor Bn. Thus, eventhough the plate is acted upon a vertical load, in-plane displacements can occur.

2.1 Analysis of the model

We first begin by setting (cf. (2)–(5))

Aτ := t−1Anτ = t−1n∑

k=1

(zk − zk−1)Ckτ , Bτ := t−2Bnτ =t−2

2

n∑

k=1

(z2k − z2

k−1)Ckτ

Dτ := t−3Dnτ =t−3

3

n∑

k=1

(z3k − z3

k−1)Ckτ , Hτ := t−1Hnτ = t−1κn∑

k=1

(zk − zk−1)Skτ . (6)

Hence, the functional to be considered is easily recognized to be (cf. (1))

Πt =1

2

(

t

∫

ΩAε(u∗) : ε(u∗) dΩ + 2t2

∫

ΩBε(u∗) : ε(θ∗) dΩ + t3

∫

ΩDε(θ∗) : ε(θ∗) dΩ

)

+

+

∫

Ωs∗ · (θ∗ −∇w∗) dΩ −

t−1

2

∫

ΩH−1s∗ · s∗ dΩ −

∫

Ωfw∗ dΩ . (7)

In order to study the behaviour of the solution when t is small, we decide to scale the unknownsand the load in a proper way. We thus set

u = t−1u∗ , θ = θ∗ , w = w∗ , s = t−3s∗ , (8)

and we choose the load as

f = t3g . (9)

Inserting these expressions in (7) and dividing by t3, we finally obtain the functional

Πt(u, θ, w, s) =1

2

(∫

ΩAε(u) : ε(u) dΩ + 2

∫

ΩBε(u) : ε(θ) dΩ +

∫

ΩDε(θ) : ε(θ) dΩ

)

+

+

∫

Ωs · (θ −∇w) dΩ −

t2

2

∫

ΩH−1s · s dΩ −

∫

Ωgw dΩ . (10)

4

Setting U = H10 (Ω)2, Θ = H1

0 (Ω)2, W = H10 (Ω), and S = L2(Ω)2, we are thus led to consider

the variational problem

Problem 2.1 Find (u, θ, w; s) ∈ U × Θ × W × S, solution of the following system

(Aε(u), ε(v)) + (Bε(v), ε(θ)) = 0 ∀v ∈ U

(Bε(u), ε(η)) + (Dε(θ), ε(η)) + (s, η −∇ζ) = (g, ζ) ∀(η, ζ) ∈ Θ × W

(r, θ −∇w) − t2(H−1s, r) = 0 ∀r ∈ S

. (11)

We will show that Problem 2.1 admits a unique solution in U × Θ × W × S. To this end, weneed to establish the following lemma.

Lemma 2.1 There exist positive constants C1 and C2, independent of t, such that

C1

(

||τ ||20 + ||σ||20

)

≤ (Aτ , τ) + 2(Bτ , σ) + (Dσ, σ) ≤ C2

(

||τ ||20 + ||σ||20

)

, (12)

for every τ , σ ∈ L2(Ω)4s, L2(Ω)4s denoting the space of symmetric tensors with components in L2(Ω).

Proof: We first introduce the linear operator

L : L2(Ω)4s × L2(Ω)4s −→ L2(Ω)4s × L2(Ω)4s

defined by

L[τ , σ] = [Aτ + Bσ,Bτ + Dσ] , (13)

for each [τ , σ] ∈ L2(Ω)4s × L2(Ω)4s. Notice that we have

(L[τ , σ], [τ , σ]) = (Aτ , τ) + 2(Bτ , σ) + (Dσ, σ) , (14)

because of the symmetry of the operator B. Moreover, if one defines, for 1 ≤ k ≤ n,

Lk[τ , σ] = [t−1(zk − zk−1)Ckτ +t−2

2(z2

k − z2k−1)Ckσ,

t−2

2(z2

k − z2k−1)Ckτ +

t−3

3(z3

k − z3k−1)Ckσ] , (15)

then it is easily recognized that it holds (cf. (6))

L[τ , σ] =n∑

k=1

Lk[τ , σ] , (16)

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for each [τ , σ] ∈ L2(Ω)4s × L2(Ω)4s. We will show that each Lk is indeed a continuous and coercivelinear operator with continuity and coerciveness constants independent of t. Hence, (12) will be astraightforward consequence of (14) and (16). We proceed by splitting Lk as

Lk = L(2)k L

(1)k , (17)

where

L(1)k [τ , σ] := [Ckτ , Ckσ] ∀[τ , σ] ∈ L2(Ω)4s × L2(Ω)4s , (18)

and

L(2)k [τ ′, σ′] := [t−1(zk − zk−1)τ

′ +t−2

2(z2

k − z2k−1)σ

′,t−2

2(z2

k − z2k−1)τ

′ +t−3

3(z3

k − z3k−1)σ

′] , (19)

for each [τ ′, σ′] ∈ L2(Ω)4s × L2(Ω)4s. It is straightforward to check that both L(1)k and L

(2)k are

symmetric operators and they commute. Moreover, L(1)k is clearly continuous and coercive with

continuity and coerciveness constants independent of t. To study the features of L(2)k we only need

to consider the real matrix A = A(k) (cf. (19)), where A(k) is defined by

A(k) =

t−1(zk − zk−1)t−2(z2

k−z2

k−1)

2

t−2(z2

k−z2

k−1)

2t−3

3 (z3k − z3

k−1)

. (20)

In fact, an easy computation shows that L(2)k is coercive on L2(Ω)4s × L2(Ω)4s if and only if A(k) is

positive-definite. Thus, the determinant ∆(k) of A(k) is

∆(k) =t−4

3(z3

k − z3k−1)(zk − zk−1) −

t−4

4(z2

k − z2k−1)

2 =t−4

12(zk − zk−1)

4 =1

12n4> 0 , (21)

since zk − zk−1 = tn. Moreover, its trace is

Tr(k) =1

n+

t−2

3n(z2

k + zkzk−1 + z2k−1) . (22)

It follows that the eigenvalues are given by

λm(k) =Tr(k)

2

(

1 −

√

1 −4∆(k)

Tr(k)2

)

λM (k) =Tr(k)

2

(

1 +

√

1 −4∆(k)

Tr(k)2

)

. (23)

6

Noting that (cf. (22))

1

n≤ Tr(k) ≤

5

4n, (24)

and recalling (21), we get

λm(k) ≥ C1(k)

λM (k) ≤ C2(k) , (25)

for some positive constants C1(k) and C2(k), independent of t.

Hence, L(2)k is coercive and continuous on L2(Ω)4s × L2(Ω)4s, with continuity and coerciveness

constants independent of t. It follows that the composition Lk = L(2)k L

(1)k is coercive and continuous

on L2(Ω)4s × L2(Ω)4s, with continuity and coerciveness constants independent of t. Recalling also(14) and (16) we finally obtain

C1

(

||τ ||20 + ||σ||20

)

≤ (Aτ , τ) + 2(Bτ , σ) + (Dσ, σ) ≤ C2

(

||τ ||20 + ||σ||20

)

, (26)

for every τ , σ ∈ L2(Ω)4s.

Remark 2.3 We remark that the constants C1 and C2 above, even though they do not depend ont, do depend on the number n of layers in the laminated plate. The analysis of the model behaviorat fixed t and n → +∞, although interesting, will not be treated in this paper.

We are now ready to prove the following Proposition.

Proposition 2.1 For each t > 0 fixed, variational Problem 2.1 has a unique solution (u, θ, w; s) inU × Θ × W × S.

Proof: We proceed in two steps.

1– Uniqueness. Suppose f = 0. Choosing in (11) v = u, η = θ, ζ = w and r = −s, we get, summingthe equations of (11)

(Aε(u), ε(u)) + 2(Bε(u), ε(θ)) + (Dε(θ), ε(θ)) + t2(H−1s, s) = 0 . (27)

We first remark that there exists C > 0 such that (cf. (6))

t2(H−1s, s) ≥ Ct2||s||20 . (28)

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To continue, by Lemma 2.1 and Korn’s inequality, we have

(Aε(u), ε(u)) + 2(Bε(u), ε(θ)) + (Dε(θ), ε(θ)) ≥ C(

||u||21 + ||θ||21

)

. (29)

From (27)–(29) we infer

u = 0 , θ = 0 , s = 0 . (30)

Finally, using (30), from the third equation of (11) we get

(r,∇w) = 0 ∀r ∈ S , (31)

which means w = 0, due to Poincare’s inequality. We conclude that for variational Problem 2.1the solution, if it exists, is unique.

2– Existence. As far as existence is concerned, consider

Problem 2.2 Find (u, θ, w) ∈ U × Θ × W , solution of the following system

(Aε(u), ε(v)) + (Bε(v), ε(θ)) = 0 ∀v ∈ U

(Bε(u), ε(η)) + (Dε(θ), ε(η)) + t−2(H(θ −∇w), η −∇ζ) = (g, ζ) ∀(η, ζ) ∈ Θ × W. (32)

By (29) and a little algebra, we can apply Lax-Milgram Lemma to conclude that Problem 2.2has a unique solution (u, θ, w) ∈ U × Θ × W . Setting

s = t−2H(θ −∇w) ∈ S , (33)

we easily see that (u, θ, w, s) ∈ U × Θ × W × S is a solution of Problem 2.1. The proof is nowcomplete.

Furthermore, for Problem 2.1 we have, using the technique of [12] and Lemma 2.1

Proposition 2.2 There exist positive constants C1 and C2, independent of t, such that

C1 ≤ ||u||1 + ||θ||1 + ||w||1 + ||s||H−1(div) ≤ C2 , (34)

where

H−1(div) =

r : r ∈ H−1(Ω)2 , div r ∈ H−1(Ω)

,

8

equipped with the obvious norm.Moreover, as t tends to zero, (u, w) converges to (u0, w0) ∈ H1

0 (Ω)2 × H20 (Ω), solution of

(Aε(u0), ε(v)) + (Bε(v), ε(∇w0)) = 0 ∀v ∈ H10 (Ω)2

(Bε(u0), ε(∇ζ)) + (Dε(∇w0), ε(∇ζ)) = (g, ζ) ∀ζ ∈ H20 (Ω)

. (35)

Remark 2.4 Notice that Problem (35) is a Kirchhoff-type laminated composite plate problem(CLPT)- cf. [22]. In a distributional sense, (u0, w0) is the solution of the system

div(

Aε(u0) + Bε(∇w0))

= 0 in Ω

div div(

Bε(u0) + Dε(∇w0))

= g in Ω. (36)

Remark 2.5 In the sequel, we will use the variational formulation (11) as the starting point forour discretization procedure by means of finite elements.

3 The discretized problem and error analysis

The aim of this section is essentially to show that a performant discretization procedure can be ob-tained by coupling a good element for homogeneous plate problem with a good membrane element.

3.1 The discretized scheme

To begin, we select finite element spaces Uh ⊂ U , Θh ⊂ Θ, Wh ⊂ W and Sh ⊂ S (cf. [13]), h > 0denoting the meshsize. We then consider the discretized problem

Problem 3.1 Find (uh, θh, wh; sh) ∈ Uh × Θh × Wh × Sh, solution of the following system

(Aε(uh), ε(v)) + (Bε(v), ε(θh)) = 0 ∀v ∈ Uh

(Bε(uh), ε(η)) + (Dε(θh), ε(η)) + (sh, Rhη −∇ζ) = Gh(η, ζ) ∀(η, ζ) ∈ Θh × Wh

(r, Rhθh −∇wh) − t2(H−1sh, r) = 0 ∀r ∈ Sh

. (37)

Above, Rh is an interpolation or projection operator acting on a suitable space and valued inSh.

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Remark 3.1 The loading term Gh(η, ζ) appearing in (37) should be a proper approximation ofG(η, ζ) =: (g, ζ) (cf. (10)). In many cases, but not always, it holds Gh(η, ζ) = G(η, ζ) ∀(η, ζ) ∈Θh ×Wh. Moreover, we will suppose that (G−Gh)(η, ζ) = (G−Gh)(η), i.e. the consistency errordepends only on the rotational field; this is the case of schemes based on the Linked InterpolationTechnique.

In the sequel, we suppose that Uh ⊂ U is a good finite element space for standard second-orderelliptic problems, i.e. for every u ∈ U it holds

infv∈Uh

||u − v||1 = O(h) . (38)

Moreover, we assume that Θh, Wh and Sh, together with the reduction operator Rh, provide agood scheme for homogeneous plate problem. This means that we assume the following

HYPOTHESISConsider the problem

Problem 3.2 Find (θ1, w1; s1) ∈ Θ × W × S, solution of the following system

(D1ε(θ1), ε(η)) + (s1, η −∇ ζ) = (m, η) + (g, ζ) ∀(η, ζ) ∈ Θ × W

(r, θ1 −∇w1) − t2(H−11 s1, r) = 0 ∀r ∈ S

, (39)

where D1 and H1 are defined by (4)–(5).

We suppose that for the discretized problem

Problem 3.3 Find (θ1h, w1

h; s1h) ∈ Θh × Wh × Sh, solution of the following system

(D1ε(θ1h), ε(η)) + (s1h, Rhη −∇ ζ) = (m, η) + Gh(η, ζ) ∀(η, ζ) ∈ Θh × Wh

(r, Rhθ1h −∇w1

h) − t2(H−11 s1h, r) = 0 ∀r ∈ Sh

, (40)

we have the error estimate

||θ1 − θ1h||1 + ||w1 − w1

h||1 + t||s1 − s1h||0 ≤ Ch . (41)

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3.2 Error estimates

We begin by setting, for notational simplicity,

E(u, θ; v, η) =: (Aε(u), ε(v)) + (Bε(v), ε(θ)) + (Bε(u), ε(η)) + (Dε(θ), ε(η)) . (42)

We remark that, by Lemma 2.1 and Korn’s inequality, it holds

E(v, η; v, η) ≥ C(

||v||21 + ||η||21

)

∀(v, η) ∈ U × Θ , (43)

with C independent of t.We have the following error estimate, whose proof is very similar to that developed in [14].

Proposition 3.1 Suppose that ∇Wh ⊂ Sh and that

||Rhη − η||0 ≤ Ch||η||1 ∀η ∈ Θ . (44)

Let uI ∈ Uh, θI ∈ Θh and wI ∈ Wh. Set

sI = t−2H(RhθI −∇wI) . (45)

Then it holds

||u − uh||1 + ||θ − θh||1 + t||s − sh||0

≤ C(

||uI − u||1 + ||θI − θ||1 + t||sI − s||0 + h||s||0 + ||G − Gh||−1

)

. (46)

Proof: Subtracting the first two equations of (37) from the first two of (11) we have the errorequation

E(u − uh, θ − θh; v, η) + (s − sh, Rhη −∇ ζ) = (G − Gh)(η, ζ) + (s, Rhη − η) , (47)

for every v ∈ Uh, η ∈ Θh and ζ ∈ Wh.Hence we get, recalling also that (G − Gh)(η, ζ) = (G − Gh)(η) (cf. Remark 3.1)

E(uI − uh, θI − θh; v, η) + (sI − sh, Rhη −∇ ζ) = E(uI − u, θI − θ; v, η)

+ (sI − s, Rhη −∇ ζ) + (G − Gh)(η) + (s, Rhη − η) . (48)

11

Choosing v = uI − uh, η = θI − θh and ζ = wI − wh, we have

Rhη −∇ ζ = t2H−1(sI − sh) . (49)

Inserting (49) in (48), using coercivity (cf. (43)) we get

||uI − uh||21 + ||θI − θh||

21 + t2||sI − sh||

20 ≤ C

(

E(uI − u, θI − θ; uI − uh, θI − θh)

+ t2(sI − s,H−1(sI − sh)) + (G − Gh)(θI − θh) + (s, Rh[θI − θh] − [θI − θh]))

. (50)

We thus easily obtain

||uI − uh||1 + ||θI − θh||1 + t||sI − sh||0

≤ C(

||uI − u||1 + ||θI − θ||1 + t||sI − s||0 + h||s||0 + ||G − Gh||−1

)

. (51)

The triangle inequality now leads to the following estimate

||u − uh||1 + ||θ − θh||1 + t||s − sh||0

≤ C(

||uI − u||1 + ||θI − θ||1 + t||sI − s||0 + h||s||0 + ||G − Gh||−1

)

. (52)

The proof is complete.

We now choose suitable interpolants uI ∈ Uh, θI ∈ Θh and wI ∈ Wh. For uI we take theusual Clement interpolant of u. For the other variables, we take advantage from that fact that wehave selected a performant scheme for homogeneous plate problems. More precisely, fix (θ, w; s) ∈Θ × W × S, part of the solution of Problem 2.1. In particular, we have

s = t−2H(θ −∇w) . (53)

Consider now the following homogeneous-type problem.

Problem 3.4 Find (θ, w; s) ∈ Θ × W × S, solution of the following system

(Dε(θ), ε(η)) + (s, η −∇ζ) = (Dε(θ), ε(η)) + (s, η −∇ζ) ∀(η, ζ) ∈ Θ × W

(r, θ −∇ w) − t2(H−1s, r) = 0 ∀r ∈ S. (54)

12

Clearly, Problem 3.4 has the unique solution (θ, w; s) = (θ, w; s). For the discretized problem

Problem 3.5 Find (θI , wI ; sI) ∈ Θh × Wh × Sh, solution of the following system

(Dε(θI), ε(η)) + (sI , Rhη −∇ζ) = (Dε(θ), ε(η)) + (s, η −∇ζ) ∀(η, ζ) ∈ Θh × Wh

(r, RhθI −∇ wI) − t2(H−1sI , r) = 0 ∀r ∈ Sh

, (55)

we thus have the error estimate (cf. (41))

||θ − θI ||1 + ||w − wI ||1 + t||s − sI ||0 ≤ Ch . (56)

If we now choose θI = θI and wI = wI , we see that (cf. (45))

sI = t−2H(RhθI −∇wI) = sI . (57)

Hence, from Proposition 3.1, (56), (57) and standard interpolation estimates (cf. [13]), we get

||u − uh||1 + ||θ − θh||1 + t||s − sh||0 ≤ C(

h + ||G − Gh||−1,h

)

. (58)

If the consistency error ||G − Gh||−1 is O(h), we obtain

||u − uh||1 + ||θ − θh||1 + t||s − sh||0 ≤ Ch . (59)

The estimate above provides an error bound for all the variables, but the vertical displacementfields. We notice that

s = t−2H(θ −∇w) , sh = t−2H(Rhθh −∇wh) , (60)

so that

∇ (w − wh) = t2H−1(sh − s) + (θ − Rhθh) . (61)

Hence we obtain

||∇ (w − wh)||0 ≤ C(

t2||sh − s||0 + ||θ − Rhθ||0 + ||Rh(θ − θh)||0)

. (62)

By Poicare’s inequality, Proposition 3.1 and (59), we get

||w − wh||1 ≤ Ch . (63)

We can summarize what we have developed so far in the following

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Proposition 3.2 Let (u, θ, w; s) be the solution of Problem 2.1, and (uh, θh, wh; sh) be the solutionof Problem 3.1. If

• Uh is chosen so that

infv∈Uh

||u − v||1 = O(h) ; (64)

• Θh, Wh, Sh and Rh provides a good aproximation scheme for homogeneous plate bendingproblem (in particular ∇Wh ⊂ Sh, (41) and (44) hold);

• the consistency error ||G − Gh||−1 is O(h),

then the following error bound holds true:

||u − uh||1 + ||θ − θh||1 + ||w − wh||1 + t||s − sh||0 ≤ Ch . (65)

4 Applications

In this section we will present two examples of laminated composite plate element, showing thatthey meet the analysis developed above. We will suppose that our computational domain Ω is arectangle, and we will partition it into uniform meshes Th of rectangles K, whose diameters arebounded by h. For the in-plane displacement field, we choose standard bilinear and continuousfunctions for both the schemes, i.e. we set

Uh = uh ∈ Θ : uh|K ∈ Q1(K)2 ∀K ∈ Th , (66)

where Q1(T ) is the space of the functions defined on K and bilinear with respect to the usual localcoordinates ξ and η. Hence, from standard approximation theory we have

infv∈Uh

||u − v||1 = O(h) . (67)

14

4.1 A Linked Intepolation Scheme

Our first example arises from an element based on the Linked Interpolation Technique, which hasalready been analysed in the framework of homogeneous plate problems (cf. [6]). We first set

Sh = sh ∈ S : sh|K = (a + bη, c + dξ) ∀K ∈ Th . (68)

We then select

Θh = θh ∈ Θ : θh|K ∈ Q1(K)2 ⊕ Shb4 ∀K ∈ Th , (69)

where b4 = (1 − ξ2)(1 − η2). Finally, we take

Wh = wh ∈ W : wh|K ∈ Q1(K) ∀K ∈ Th . (70)

Notice that in this case ∇Wh ⊂ Sh. The reduction operator Rh is defined through the relation

Rhη = Ph(Id −∇L)η , (71)

where Ph is the L2-projection operator on Sh, Id is the identity operator and L is the so-calledLinking operator, which we define below. Let us first introduce for each K ∈ Th the functions

ϕi = λjλkλm . (72)

In (72) λi1≤i≤4 are the equations of the sides of K and the indices (i, j, k, m) form a permu-tation of the set (1, 2, 3, 4). The function ϕi is an edge bubble relatively to the edge ei of K. Letus now set

EB(K) = Span ϕi1≤i≤4 . (73)

The operator L is locally defined as

L|Kηh

=4∑

i=1

αiϕi ∈ EB(K), (74)

by requiring that

(ηh−∇Lη

h) · ti = constant along each ei, (75)

where ti is the tangent vector along the side ei. Notice that with the definition above, L is definedonly on Θh, but a natural extension to Θ is provided by setting

Lη := LηP

∀η ∈ Θ , (76)

15

where ηP

is the H1-projection of η over Θh. Thus, also Rh is defined on the whole Θ. Next, wedefine Gh by

Gh(η, ζ) = (g, ζ + Lη) , (77)

so that

(G − Gh)(η, ζ) = −(g, Lη) . (78)

It has been proved that for the operator L it holds (cf. [6])

||Lη||1 ≤ Ch||η||1 ∀η ∈ Θh , (79)

so that (cf. (76))

||Lη||1 = ||LηP||1 ≤ Ch||η

P||1 ≤ Ch||η||1 ∀η ∈ Θ . (80)

It follows that

• ||G − Gh||−1 ≤ Ch (cf. (78));

• for the operator Rh we have (cf. (71))

||η−Rhη||0 = ||η−Phη+Ph∇Lη||0 ≤ ||η−Phη||0 + ||Ph∇Lη||0 ≤ Ch||η||1 ∀η ∈ Θ . (81)

Moreover, the scheme under consideration is a robust and first order convergent when appliedto homogeneous plate bending problems (cf. [6]). We thus conclude that our method is first orderconvergent also when used in the case of a laminated composite structure (cf. Proposition 3.2).

4.2 A Mixed Interpolated with Tensorial Components Scheme

The second example we provide comes from the well-known MITC4 plate element (cf. [8] and [10]),which has been successfully used for plate bending problems in the case of homogeneous materials.For this scheme we set

Sh = sh ∈ H0(rot; Ω) : sh|K = (a + bη, c + dξ) ∀K ∈ Th , (82)

where (cf. [9], for instance)

H0(rot; Ω) =

r ∈ L2(Ω)2 : rot r ∈ L2(Ω) , r · t = 0 on ∂Ω

.

For the approximated space of rotations we select

16

Θh = θh ∈ Θ : θh|K ∈ Q1(K)2 ∀K ∈ Th . (83)

Finally, we take

Wh = wh ∈ W : wh|K ∈ Q1(K) ∀K ∈ Th . (84)

Notice that also in this case ∇Wh ⊂ Sh. The reduction operator Rh is defined through therelation (cf. [10])

Rhη ∈ Sh ,

∫

eRhη · te =

∫

eη · te , (85)

for η smooth enough, and for each side e of the rectangles in the mesh.As for Gh, we simply choose

Gh(η, ζ) = (g, ζ) , (86)

so that

(G − Gh)(η, ζ) = 0 . (87)

It is well-known (cf. [8] and [14]) that if the refinement of the mesh is obtained by splittingeach rectangle into sixteen equal subrectangles, such a scheme meets all the features required byProposition 3.2 (in particular, estimate (41) holds). Hence, we can conclude that in this case ourmethod is first order convergent when used for a laminated composite structure. We remark thatsome numerical tests relative to this element have been already presented in [1].

Remark 4.1 The analysis of [8] and [14] can be easily extended to the case of a refinement ob-tained by splitting each rectangle into four equal subrectangles (cf. also the results in [16] and [20],concerning the analysis of the Q1−P0 element for the Stokes problem, to which the MITC4 elementis strictly linked).

5 Numerical examples

The aim of this Section is to investigate the numerical performances of the interpolating schemespreviously described. To reach this goal the schemes have been implemented into the Finite ElementAnalysis Program (FEAP) [26] and the convergence rate of the proposed schemes is checked on amodel problem, for which the exact solution can be computed.

We consider a composite made with two layers of an high modulus graphite/epoxy material,whose properties are set as follows

EL = 25 MPa , ET = 1 MPa , νTT = 0.25 , GLT = 0.5 MPa , GTT = 0.2 MPa

17

where the indeces L and T indicate the longitudinal and the transversal directions, while E indicatesa Young’s modulus, ν a Poisson ratio, G a shear modulus. In particular, we observe that

EL

ET

= 25 ,GLT

ET

= 0.5 ,GTT

ET

= 0.2 ,

We study the case of an unsymmetric 0/90 cross-ply laminate. Accordingly, recalling equations(2)–(5) and expressing all the entries in MPa, we have

C1 =

250.627 .250627 0.250627 1.00251 0

0 0 .5

, C2 =

1.00251 .250627 0.250627 250.627 0

0 0 .5

S1 =

[

.5 00 .2

]

, S2 =

[

.2 00 .5

]

The plate midplane is assumed to be the square Ω = [0, a]× [0, a] and subjected to the followingsinusoidal load

f(x, y) = t3g(x, y) = t3q0 sin

(

πx

a

)

sin

(

πy

a

)

(88)

with q0 measuring the maximum load intensity. Finally, we consider the following simply supportedboundary conditions

at x = 0 and x = a ⇒ u2 = 0, w = 0, θ1 = 0at y = 0 and y = a ⇒ u1 = 0, w = 0, θ2 = 0

and the shear factor is set equal to 5/6.Following References [21] and [22], the exact solution has the form

u1(x, y) = u1,0 cos

(

πx

a

)

sin

(

πy

a

)

(89)

u2(x, y) = u2,0 sin

(

πx

a

)

cos

(

πy

a

)

(90)

w(x, y) = w0 sin

(

πx

a

)

sin

(

πy

a

)

(91)

θ1(x, y) = θ1,0 sin

(

πx

a

)

cos

(

πy

a

)

(92)

θ2(x, y) = θ2,0 cos

(

πx

a

)

sin

(

πy

a

)

(93)

where u1,0, u2,0, w0, θ1,0 and θ2,0 are factors that can be easily computed. To do so, we start fromfunctional (10), we take the variation with respect to s and we require a strong satisfaction of thecorresponding equation. It follows that (u, θ, w) minimizes the potential energy functional

18

Πt =1

2

(∫

ΩAε(u) : ε(u) dΩ + 2

∫

ΩBε(u) : ε(θ) dΩ +

∫

ΩDε(θ) : ε(θ) dΩ

)

+

+t−2

2

∫

ΩH(θ −∇w) · (θ −∇w) dΩ −

∫

Ωgw dΩ . (94)

Hence, using positions (88)-(93) and requiring the potential energy stationarity, we obtain asystem of five equations, which can be solved in terms of the five quantities u1,0, u2,0, w0, θ1,0 andθ2,0. The solution clearly depends on the specific form of the fourth order tensors A,B,D,H, henceon the specific lamination sequence as well as on the plate geometry.

Although the numerical computations have been performed for a simply supported boundarycondition (while the theoretical analysis has been developed for the case of a clampled plate),we nonetheless believe that the presented tests are significant of the actual performances of theschemes.

The error of a discrete solution is measured through the relative errors Eu, Ew, Eθ and Etot,defined as

E2u =

∑

Ni

[

(uh1(Ni) − u1(Ni))2 + (uh2(Ni) − u2(Ni))

2]

∑

Ni

[

(u1(Ni))2 + (u2(Ni))

2] (95)

E2w =

∑

Ni(wh(Ni) − w(Ni))

2

∑

Niw(Ni)2

, (96)

E2θ =

∑

Ni

[

(θh1(Ni) − θ1(Ni))2 + (θh2(Ni) − θ2(Ni))

2]

∑

Ni

[

(θ1(Ni))2 + (θ2(Ni))

2] (97)

E2tot = E2

u + E2w + E2

θ (98)

For simplicity, the summations are performed on all the nodes Ni relative to global interpolationparameters (that is, the internal parameters associated with bubble functions are neglected). Theabove error measures can also be seen as discrete L2-type errors and a h2 convergence rate in L2

norm actually means a h convergence rate in the H1 energy-type norm.The analyses are performed using uniform meshes and discretizing only one quarter of the plate,

due to symmetry considerations. Moreover, the refinement procedure is obtained by splitting eachsquare of a mesh into four equal subsquares. The plate side length is set equal to 1, while twodifferent values of thickness are considered, i.e. t ∈ 10−1, 10−3, indicated in the following as thickand thin case, respectively.

Figures 1-4 show the relative errors versus the number of nodes per side for the interpolationschemes considered.

It is interesting to observe that:

19

• Both the methods show the appropriate convergence rate for both rotations and verticaldisplacements.

• Both the methods are almost insensitive to the thickness, in such a way that the error graphsfor two different choices of t are very close to each other. As a consequence, all the proposedelements are actually locking-free and they can be used for both thick and thin plate problems.

References

[1] G. Alfano, F. Auricchio, L. Rosati and E. Sacco, MITC finite elements for laminated compositeplates, to appear in Int. J. Numer. Methods Eng.

[2] D.N. Arnold, Innovative finite element methods for plates, Math. Applic. Comp., V.10 (1991)77-88.

[3] D.N. Arnold and R.S. Falk, A uniformly accurate finite element method for the Reissner-Mindlin plate model, SIAM J. Numer. Anal., 26(6) (1989) 1276-1290.

[4] F. Auricchio and E. Sacco, A mixed-enhanced finite element for the analysis of laminatedcomposite plates, Int. J. Numer. Methods Eng., 44 (1999) 1481-1504.

[5] F. Auricchio and R.L. Taylor, A shear deformable plate element with an exact thin limit,Comput. Methods Appl. Mech. Engrg., 118 (1994) 393-412.

[6] F. Auricchio and C. Lovadina, Analysis of kinematic linked interpolation methods for Reissner–Mindlin plate problems, to appear in Comput. Methods Appl. Mech. Engrg.

[7] K.-J. Bathe, Finite Element Procedures, (Englewood Cliffs, NJ, 1995).

[8] K.-J. Bathe and F. Brezzi, On the convergence of a four-node plate bending element based onMindlin-Reissner plate theory and a mixed interpolation, MAFELAP V (J.R. Whiteman, ed.),London (1985) 491-503.

[9] F. Brezzi and M. Fortin, Mixed and Hybrid Finite Element Methods, (Springer-Verlag,1991).

[10] F. Brezzi, K.-J. Bathe and M. Fortin, Mixed-Interpolated elements for Reissner-Mindlin plates,Int. J. Numer. Methods Eng., 28 (1989) 1787-1801.

[11] F. Brezzi, M. Fortin and R. Stenberg, Error analysis of mixed-interpolated elements forReissner-Mindlin plates, Math. Models and Methods in Appl. Sci., 1 (1991) 125-151.

20

[12] D. Chenais and J.C. Paumier, On the locking phenomenon for a class of elliptic problems,Numer. Math. 67 (1994), 427-440.

[13] P.G. Ciarlet, The Finite Element Method for Elliptic Problems, (North Holland, 1978).

[14] R.G. Duran and E. Liberman, On mixed finite element methods for the Reissner-Mindlin platemodel, Math. Comp., 58(198) (1992) 561-573.

[15] T.J.R. Hughes, The Finite Element Method, (Englewood Cliffs NY, 1987).

[16] C. Johnson and J. Pitkaranta, Analysis of Some Mixed Finite Element Methods Related toReduced Integration, Math. Comp., 38(158) (1982) 375-400.

[17] C. Lovadina, Analysis of a mixed finite element method for Reissner-Mindlin plate problem,Comput. Methods Appl. Mech. Engrg., 163 (1998) 71-85.

[18] T. Massard (Editor), Proceedings of 12th International Conference on CompositeMaterials, Paris, July 5-9, (1999).

[19] J. Pitkaranta, Analysis of some low-order finite element schemes for Mindlin-Reissner andKirchhoff plates, Numer. Math., 53 (1988) 237-254.

[20] J. Pitkaranta and R. Stenberg, Error bounds for the approximation of the Stokes problem usingbilinear/constant elements on irregular quadrilateral meshes, MAFELAP V (J.R. Whiteman,ed.), London (1985) 325-334.

[21] J.N. Reddy, Energy and Variational Methods in Applied Mechanics, (John Wiley &Sons, 1984).

[22] J.N. Reddy, Mechanics of Laminated Composite Plates- Theory and Analysis, (CRCPress, 1997).

[23] R.L. Taylor and F. Auricchio, Linked interpolation for Reissner-Mindlin plate elements: PartII- A simple triangle, Int. J. Numer. Methods Eng., 36 (1993) 3057-3066.

[24] Z. Xu, A simple and efficient triangular finite element for plate bending, Acta Mech. Sin., 2(1986) 185-192.

[25] Z. Xu, A thick-thin triangular plate element, Int. J. Numer. Methods Eng., 33 (1992) 963-973.

[26] O.C. Zienkiewicz and R.L. Taylor, The Finite Element Methods, (McGraw-Hill, NewYork, NY, 1989).

21

[27] O.C. Zienkiewicz, Z. Xu, L.F. Zeng, A. Samuelsson and N.-E. Wiberg, Linked interpolationfor Reissner-Mindlin plate elements: Part I- A simple quadrilateral, Int. J. Numer. MethodsEng., 36 (1993) 3043-3056.

22

100

101

102

10−6

10−5

10−4

10−3

10−2

10−1

Number of nodes

Rel

ativ

e er

rors

ETot

E

U

EW

Eθ

Figure 1: Linked interpolation scheme: thick case (t = 10−1). Relative errors (Etot, Eu, Ew, Eθ)versus number of nodes per side for different values of thickness. It can be observed the attainmentof the h2 convergence rate in the L2 error norm, corresponding to a h convergence rate in the H1

energy-type norm.

23

100

101

102

10−6

10−5

10−4

10−3

10−2

10−1

Number of nodes

Rel

ativ

e er

rors

ETot

E

U

EW

Eθ

Figure 2: Linked interpolation scheme: thin case (t = 10−3). Relative errors (Etot, Eu, Ew, Eθ)versus number of nodes per side for different values of thickness. It can be observed the attainmentof the h2 convergence rate in the L2 error norm, corresponding to a h convergence rate in the H1

energy-type norm.

24

100

101

102

10−6

10−5

10−4

10−3

10−2

10−1

Number of nodes

Rel

ativ

e er

rors

ETot

E

U

EW

Eθ

Figure 3: MITC4 interpolation scheme: thick case (t = 10−1). Relative errors (Etot, Eu, Ew, Eθ)versus number of nodes per side for different values of thickness. It can be observed the attainmentof the h2 convergence rate in the L2 error norm, corresponding to a h convergence rate in the H1

energy-type norm.

25

100

101

102

10−6

10−5

10−4

10−3

10−2

10−1

Number of nodes

Rel

ativ

e er

rors

ETot

E

U

EW

Eθ

Figure 4: MITC4 interpolation scheme: thin case (t = 10−3). Relative errors (Etot, Eu, Ew, Eθ)versus number of nodes per side for different values of thickness. It can be observed the attainmentof the h2 convergence rate in the L2 error norm, corresponding to a h convergence rate in the H1

energy-type norm.

26