Anisotropic inhomogeneous rectangular thin-walled beams - cvgmt
Anisotropic inhomogeneous rectangular thin-walled beams Lorenzo Freddi
∗
Franc ¸ ois Murat†
Roberto Paroni‡
Abstract This paper is devoted to the asymptotic analysis of the problem of linear elasticity for an anisotropic and inhomogeneous body occupying, in its reference configuration, a cylindrical domain with a rectangular cross section with sides proportional to ε and ε2 and clamped on one of its bases. The sequence of solutions uε of the equilibrium problem is shown to converge in an appropriate topology, as ε goes to zero, to the solution of a problem for a beam in which the extensional, flexural, and torsional effects are all coupled together. Keywords: asymptotic analysis, calculus of variations, thin-walled beams, dimension reduction, variational convergence, linear elasticity
1
Introduction
Geometrically, a thin-walled beam is a slender structural element whose length is much larger than the diameter of the cross section which, on its hand, is larger than the thickness of the thin wall. These kinds of beams have been used for a long time in civil and mechanical engineering and, most of all, in flight vehicle structures because of their high ratio between maximum strength and weight. More recently, their importance has increased because of the introduction of fiber-reinforced composite materials in structural components. These materials are finding more and more applications for their high resistance to corrosion and high strength. Composite beams are usually made up by fiber-reinforced laminates and, hence, are anisotropic and inhomogeneous, even in cross-section planes. These peculiarities make classical thin-walled beam theories not applicable. The problem though has attracted the interest of several researchers ∗ Dipartimento di Matematica e Informatica, Universit` a di Udine, via delle Scienze 206, 33100 Udine, Italy, email: [email protected] † Laboratoire Jacques-Louis Lions, Universit´ e Paris VI, Boˆıte courrier 187, 75252 Paris Cedex 05, France, email: [email protected] ‡ Dipartimento di Architettura e Pianificazione, Universit` a degli Studi di Sassari, Palazzo del Pou Salit, Piazza Duomo, 07041 Alghero, Italy, email: [email protected]
1
ANISOTROPIC INHOMOGENEOUS THIN-WALLED BEAMS
2
and by now a huge number of articles can be found on the subject; see, for instance, [11] and the references therein. This is strongly remarked also in the first sentences of the abstract of [12]: “There is no lack of composite beam theories. Quite to the contrary, there might be too many of them. Different approaches, notations, etc., are used by the authors of those theories, so it is not always straightforward to compare the assumptions made and to assess the quantitative consequences of those assumptions.” The problem under study has a huge technological interest. One very suggestive, mentioned in [2], concerns the rotor blades of helicopters. The blades are composite beams and, hence, anisotropic and inhomogeneous. The anisotropy and the inhomogeneity introduce, as we shall also deduce, structural couplings between bending, extension, and twisting behaviors. It has been observed experimentally that these couplings have a powerful influence on blade dynamics including vibrations and the aeroelastic stability; see [2]. If a model of composite beams that accurately describes the structural couplings was at our disposal, then we could try to vary the anisotropy and inhomogeneity so as to minimize undesired effects like, for instance, vibrations. Through the control of lamination parameters (ply orientation and stacking sequence), it would then be possible for industry to minimize the undesired effects. Our aim here is to deduce a “rigorous” model for a composite thin-walled beam, that is, a inhomogeneous and anisotropic beam. We shall achieve our goal by means of well-established asymptotic methods starting from the threedimensional linear theory of elasticity. This paper is devoted to the asymptotic analysis of the linearized system of equilibrium equations of a body which occupies, in its reference configuration, a cylindrical domain with a rectangular cross section with sides proportional to ε and ε2 and clamped on one of its two bases. In particular, we study the compactness properties of the sequence of solutions uε of the equilibrium problems and, letting ε go to zero, we are concerned with the identification of the limit problem. The same problem has been studied from the point of view of Γconvergence: in [4] in the simpler setting of homogeneous and isotropic material and in [3] in the case of an anisotropic material which is inhomogeneous only along the longitudinal axis and subject to residual stress. Trabucho and Viano [9] also studied the same problem by superimposing two asymptotic analyses where the lengths of the two sides of the cross section go to zero independently. Besides the material properties of the body, our treatment differs from the preceding works also in the topology used in the passage from the threedimensional problem to the one-dimensional: the one used in the present paper delivers much more information on the deformation of the beam. The approach is close to the one developed in a recent paper of Murat and Sili [8] for a thin cylinder of radius ε. The lack of isotropy or homogeneity assumptions leads to a limit problem where the extensional, flexural, and torsional effects are coupled together. In fact, we prove that the limit problem can be written as a system of five equations in a 5-tuple of unknowns (u, v, w, p, q) (see Theorem 6.1) and that uε − (u + εv + ε2 w + ε3 p + ε4 q) converges strongly to zero in H 1 (Ω), under some regularity assumptions on v, w, p, and q (see Corollary 6.1). We also
ANISOTROPIC INHOMOGENEOUS THIN-WALLED BEAMS
3
derive the set of Euler equations of the variational limit problem, that is, the system of equilibrium differential equations, in the fully general case. Then we show that a strong simplification and a partial decoupling occurs when the material is homogeneous, and a complete decoupling is obtained for a homogeneous orthotropic material. Notation. Throughout this paper Ω1 , Ω2 , and Ω3 will denote the following three intervals: Ωα := (−aα /2, +aα /2) for α = 1, 2 and Ω3 := (0, `), where a1 , a2 , and ` are three positive real numbers. Also, for i, j = 1, 2, 3 we set Ωij := Ωi × Ωj and Ω := Ω1 × Ω2 × Ω3 . Unless otherwise specified, we use the Einstein summation convention. Moreover, we use the following convention for indexing vector and tensor components: Greek indices α and β take their values in the set {1, 2} and Latin indices i, j, and k in the set {1, 2, 3}. With a little abuse of notation, and because this is a common practice and does not give rise to any mistakes, we call “sequences” even those families indicized by a continuous parameter ε ∈ (0, 1). The component k of a vector v will be denoted either with (v)k or vk , and an analogous notation will be used to denote tensor components. Eαβ denotes the Ricci’s symbol, that is, E11 = E22 = 0, E12 = 1, and E21 = −1. Since usually x = (x1 , x2 , x3 ), we shall then denote by x0 := (x1 , x2 ). A wide use will be made of vector valued distributions and Sobolev spaces; for a brief account of which and for the current notation we refer the reader to the book of Le Dret [5]. Throughout this paper C will denote a constant which may change line by line.
2
The three-dimensional problem
We consider a body which occupies, in its reference configuration, the region Ωε := ε2 Ω1 × εΩ2 × Ω3 ⊂ R3 . We denote by E(u) the strain of the displacement u, whose components are µ ¶ 1 ∂ui ∂uj Eij (u) = + . 2 ∂xj ∂xi The elasticity tensor, with respect to the reference configuration Ωε , of the material will be denoted by Cε . We assume it to be essentially bounded, Cε ∈ L∞ (Ωε ; R3×3×3×3 ); to have the minor symmetries, Cεijkl = Cεjikl = Cεijlk ;
ANISOTROPIC INHOMOGENEOUS THIN-WALLED BEAMS
4
and to be positive definite. That is, there exists a constant c > 0 such that Cε A · A ≥ c|A|2 ,
(1)
for all three by three symmetric matrices A and for all ε. We consider the body clamped on Γεb := ∂Ωε ∩ {x3 = 0} and we denote by © ª 1 Hdn (Ωε ; R3 ) := ϕ ∈ H 1 (Ωε ; R3 ) : ϕ = 0 on Γεb . The weak form of the equilibrium problem can be written as ε 1 ˜ ∈ Hdn (Ωε ; R3 ), u Z Z ε ε 1 C E(˜ u ) · E(ϕ) dv = F˜ ε · E(ϕ) dv ∀ϕ ∈ Hdn (Ωε ; R3 ), Ωε
(2)
Ωε
where the matrix field F˜ ε , which takes into account the presence of external forces, is assumed to be an element of L2 (Ωε ; R3×3 sym ). If F˜ ε is not just in L2 (Ωε ; R3×3 ) but in sym 2 3 H(div, Ωε ) := {T ∈ L2 (Ωε ; R3×3 sym ) : div T ∈ L (Ωε ; R )},
then the previous problem can be seen as the weak form of the following problem: ε uε ) + ˜bε = 0 in Ωε , div C E(˜ (3) Cε E(˜ uε )nε = c˜ε on Γεc , ε ε u ˜ =0 on Γb , where Γεc := ∂Ωε \Γεb , and nε denotes the unit outward normal vector to Ωε , while the body loads ˜bε and the contact loads c˜ε are simply given by ˜bε := −divF˜ ε
in Ωε ,
c˜ε := F˜ ε nε
in Γεc .
(4)
Note that given ˜bε ∈ L2 (Ωε ; R3 ) and c˜ε ∈ H −1/2 (∂Ωε ; R3 ) it is always possible to find an F˜ ε ∈ H(div, Ωε ) which satisfies (4).
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The rescaled problem
It is convenient to work with the domain Ω instead of the domain Ωε . We therefore rescale the problem by means of the scaling map sε : Ω → Ωε , sε (x1 , x2 , x3 ) = (ε2 x1 , εx2 , x3 ). Let u ˜ε be the solution of (2); then we define the “rescaled solution” uε by uε1 := ε2 u ˜ε1 ◦ sε ,
uε2 := ε˜ uε2 ◦ sε ,
uε3 := u ˜ε3 ◦ sε .
ANISOTROPIC INHOMOGENEOUS THIN-WALLED BEAMS Let E ε be the “rescaled strain” defined by 1 1 ε4 E11 (ϕ) ε3 E12 (ϕ) E ε (ϕ) := ε13 E21 (ϕ) ε12 E22 (ϕ) 1 ε2 E31 (ϕ)
1 ε E32 (ϕ)
1 ε2 E13 (ϕ) 1 ε E23 (ϕ)
5
.
(5)
E33 (ϕ)
It follows that E ε (uε ) = E(˜ uε ) ◦ sε . We further assume that there exists a C ∈ L∞ (Ω; R3×3×3×3 ) such that C ε = C ◦ sε , ε 2 3×3 and we denote with F ε = F˜ ε ◦ s−1 ε ∈ L (Ω; Rsym ). With this notation u turns out to be the unique solution of ε 1 3 u ∈ Hdn (Ω; R ), Z Z (6) ε ε ε 1 CE F ε · E ε (ϕ) dx ∀ϕ ∈ Hdn (Ω; R3 ), (u ) · E (ϕ) dx = Ω
Ω
where Γb := ∂Ω ∩ {x3 = 0} and n o 1 Hdn (Ω; R3 ) := ϕ ∈ H 1 (Ω; R3 ) : ϕ = 0 on Γb . By taking ϕ = uε and using (1), we find ckE ε (uε )kL2 (Ω) ≤ kF ε kL2 (Ω) .
(7)
Thus a uniform bound on kF ε kL2 (Ω) would lead to rescaled strains uniformly bounded in ε. We augment this requirement by assuming Fε → F
in L2 (Ω; R3×3 sym ).
(8)
Remark 3.1 Let S ε := diag(1/ε2 , 1/ε, 1); then E ε (ϕ) = S ε E(ϕ)S ε . If we assume F ε ∈ H(div, Ω), then we can write Z Z ε ε F · E (ϕ) dx = S ε F ε S ε · E(ϕ) dx Ω
Ω
Z div(S ε F ε S ε ) · ϕ dx + hS ε F ε S ε n, ϕi∂Ω ,
=− Ω
1 (Ω; R3 ), and conclude that instead of considering F ε we could for all ϕ ∈ Hdn have been using the following body and contact forces
bε := −div(S ε F ε S ε ) in Ω −1
−1
and cε := S ε F ε S ε n
in Γc .
Note that if F ε = S ε F (0) S ε for some F (0) ∈ H(div, Ω) the body and −1 the contact forces would be independent of ε. Since S ε = diag(ε2 , ε, 1), −1 −1 the sequence {S ε F (0) S ε }, with F (0) ∈ H(div, Ω), strongly converges in
ANISOTROPIC INHOMOGENEOUS THIN-WALLED BEAMS
6
L2 (Ω; R3×3 sym ) and hence satisfies assumption (8). Assumption (8) though allows −1 −1 us to consider “stronger” forces than F ε = S ε F (0) S ε , like F ε = F (0) or, more generally, F ε = F (0) + εF (1) + ε2 F (2) + ε3 F (3) + ε4 F (4) , (4)
with F (i) ∈ H(div, Ω) for i = 0, 1, . . . , 4, where, for instance, the term ε4 F11 would lead to the definition of body and contact forces independent of ε.
4
Partial Korn’s inequalities
In this section we state and prove several Korn’s inequalities. The proofs of Theorems 4.2 and 4.3 follow some of the lines of that of Theorem 4.4, which is due to Monneau, Murat, and Sili [7]. Theorem 4.1 There exists a constant C such that Z ° ° ° ∂u ° ° ° ° 1° ≤C° °u1 − − u1 dx1 ° 2 ° ∂x1 L2 (Ω) L (Ω) Ω1 for every u1 ∈ H 1 (Ω1 ; L2 (Ω23 )). 1 ¯ Proof. By density we may restrict ourselves to considering u1 ∈ C R (Ω). For every x2 and x3 there exists a ξ = ξ(x2 , x3 ) such that u1 (ξ, x2 , x3 ) = −Ω1 u1 (s, x2 , x3 ) ds and Z x1 ∂u1 u1 (x1 , ·, ·) − u1 (ξ, ·, ·) = (s, ·, ·) ds. ∂x1 ξ
Taking squares and applying Jensen’s inequality we conclude the proof.
2
Theorem 4.2 There exists a constant C such that ° µZ ¶° Z ° ° °u2 − − u2 dx1 − x1 ∂ − u1 dx1 ° ° ° −1 ∂x2 Ω1 Ω1 H (Ω2 ;L2 (Ω13 )) ° ° ° ¡° ¢ ≤ C °E11 (u)°L2 (Ω) + °E12 (u)°L2 (Ω) for every u ∈ H 1 (Ω; R2 ). Proof. Let Z u ¯1 := u1 − − u1 dx1 , Ω1
Z Z ∂ u ¯2 := u2 − − u2 dx1 + x1 − u1 dx1 , ∂x2 Ω1 Ω1
ANISOTROPIC INHOMOGENEOUS THIN-WALLED BEAMS
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R and note that −Ω1 u ¯2 dx1 = 0 and u ¯2 ∈ H 1 (Ω1 ; L2 (Ω23 )). Let ψ ∈ H01 (Ω2 ); then ¶ Z ∂u2 ∂ ψu ¯2 dx2 = ψ + − u1 dx1 dx2 ∂x1 ∂x2 Ω1 Ω2 Ω2 µ ¶ Z ∂u ¯1 = dx2 ψ 2E12 (u) − ∂x2 Ω2 Z dψ = 2ψE12 (u) + u ¯1 dx2 . dx 2 Ω2 R R R Since Ω2 ψ¯ u2 dx2 ∈ H 1 (Ω1 ; L2 (Ω23 )) and Ω1 Ω2 ψ¯ u2 dx2 dx1 = 0, by Theorem 4.1 and the above equation we deduce ° °Z ° ° Z ° ° ° ∂ ° ° ° ° ≤C° ψ¯ u2 dx2 ° ψ¯ u2 dx2 ° ° ° 2 ∂x1 Ω2 Ω2 L2 (Ω) L (Ω) ∂ ∂x1
Z
Z
µ
≤ C kψkH 1 (Ω2 ) (kE12 (u)kL2 (Ω) + k¯ u1 kL2 (Ω) ) ≤ C kψkH 1 (Ω2 ) (kE12 (u)kL2 (Ω) + kE11 (u)kL2 (Ω) ). Let ϕ ∈ L2 (Ω13 ); then ¯Z ¯ °Z ¯ ¯ ° ¯ ¯ 2 ϕψ u ¯2 dx¯ ≤ kϕkL (Ω13 ) ° ¯ ° Ω
Ω2
° ° ψu ¯2 dx2 ° °
L2 (Ω13 )
≤ C kϕkL2 (Ω13 ) kψkH 1 (Ω2 ) (kE11 (u)kL2 (Ω13 ) + kE12 (u)kL2 (Ω) ). A density argument concludes the proof. Indeed, let {ϕn } be an orthonormal basis of L2 (Ω13 ) and for any v ∈ H01 (Ω2 ; L2 (Ω13 )), let Z ψn (x2 ) := ϕn (x1 , x3 )v(x) dx1 dx3 ∈ H01 (Ω2 ). Ω13
PN
Then for vN := n=1 ψn ϕn we have ¯Z ¯ ¯ ¯ ¯ vN u ¯2 dx¯¯ ≤ kvN kH01 (Ω2 ;L2 (Ω13 )) (kE11 (u)kL2 (Ω) + kE12 (u)kL2 (Ω) ), ¯ Ω
and letting N go to infinity we conclude the proof.
2
Remark 4.1 In spite of the fact that the left-hand side belongs to L2 (Ω), the inequality of Theorem 4.2 does not hold true if one replaces the norm H −1 with the norm of L2 , because of the following counterexample, which is inspired by an example contained in [7] in a quite similar framework. ¯ α ) with ϕ1 satisfying Consider two scalar smooth functions ϕα ∈ C ∞ (Ω Z Z ∂ϕ1 ϕ1 dx1 = dx1 = 0. Ω1 Ω1 ∂x1
ANISOTROPIC INHOMOGENEOUS THIN-WALLED BEAMS Define u1 := −ϕ2
∂ϕ1 , ∂x1
u2 := ϕ1
∂ϕ2 , ∂x2
8
u3 := 0.
Then u ∈ H 1 (Ω; R3 ), and the inequality of Theorem 4.2 reduces to ° ° ° ° ° ∂ϕ2 ° ° ∂ 2 ϕ1 ° °ϕ1 ° ° ° , ≤ C °ϕ2 ° ∂x2 ° −1 ∂x21 °L2 (Ω) H (Ω2 ;L2 (Ω1 )) which cannot be true if we replace H −1 by L2 because in such a case, taking kϕ1 kL2 (Ω1 ) = k∂ 2 ϕ1 /∂x21 kL2 (Ω1 ) would imply that ° ° ° ∂ϕ2 ° ° ° ≤ Ckϕ2 kL2 (Ω) ° ∂x2 ° 2 L (Ω) ¯ 2 ), which is clearly impossible. for any ϕ2 ∈ C ∞ (Ω Define (using the summation convention) rd2 = {r ∈ L2 (Ω12 ; R2 ) : ∃ c ∈ R, d ∈ R2 such that rα (y) = Eβα xβ d + cα }, where E denotes the Ricci’s symbol. The elements of rd2 are the infinitesimal rigid displacements on Ω12 . It is easy to see that rd2 ⊂ H 1 (Ω12 ; R2 ); moreover, being finite-dimensional, it is closed in L2 (Ω12 ; R2 ). Thus, the orthogonal projection operator of L2 (Ω12 ; R2 ) on rd2 , which will be denoted by ℘, is well defined. Given a vector function v ∈ L2 (Ω12 , Rm ) with m ≥ 2, we define Z Z 1 0 (9) ϑ(v) := (x1 v2 − x2 v1 ) dx , where IO = (x21 + x22 ) dx0 . IO Ω12 Ω12 If v ∈ L2 (Ω12 ; R2 ), then the components of ℘ turn out to be Z ℘α (v) = Eβα xβ ϑ(v) + − vα dx0 .
(10)
Ω12
Furthermore, the two-dimensional Korn’s inequality can be written as X kEαβ (v)kL2 (Ω12 ;R2×2 ) kv − ℘(v)kH 1 (Ω12 ;R2 ) ≤ C
(11)
α,β
for all v ∈ H 1 (Ω12 ; R2 ) with a constant C which is independent of v. Similarly, given a vector function u ∈ L2 (Ω3 ; L2 (Ω12 , Rm )) with m ≥ 2, we analogously define ϑ(u) ∈ L2 (Ω3 ) and ℘α (u). The operator ℘ associates to any u ∈ L2 (Ω3 ; L2 (Ω12 , Rm )) a function ℘(u) ∈ L2 (Ω3 ; L2 (Ω12 , R2 )) which is an infinitesimal rigid displacement on Ω12 for almost every x3 ∈ Ω3 . Let us observe that the orthogonal complement, with respect to the L2 (Ω12 ; R2 ) inner product, of rd2 in H 1 (Ω12 ; R2 ) can be then characterized as 1 2 1 2 rd⊥ 2 = {v ∈ H (Ω12 ; R ) : ℘(v) = 0} = {v ∈ Hm (Ω12 ; R ) : ϑ(v) = 0}.
ANISOTROPIC INHOMOGENEOUS THIN-WALLED BEAMS
9
Moreover, we denote by 1 RD2⊥ (Ω) = {v ∈ L2 (Ω3 ; Hm (Ω12 ; R2 )) : ϑ(v) = 0 a.e. x3 ∈ Ω3 }.
(12)
Hereafter, for any u ∈ L2 (Ω3 ; H 1 (Ω12 ; Rm )), m ≥ 2, we set u ˜α := uα − ℘α (u). (13) R Of course u ˜ ∈ L2 (Ω3 ; H 1 (Ω12 ; R2 )) and −Ω12 u ˜ dx1 dx2 = 0 and ϑ(˜ u) = 0 a.e. in Ω3 , where the latter follows from the linearity of ϑ and the fact that ϑ(u) = ϑ(℘(u)). Thus u ˜ ∈ RD2⊥ (Ω). Lemma 4.1 There exists a constant C such that X k˜ ukL2 (Ω3 ;H 1 (Ω12 ;R2 )) ≤ C kEαβ (u)kL2 (Ω3 ;L2 (Ω12 )) α,β
for every u ∈ L2 (Ω3 ; H 1 (Ω12 ; Rm )), m ≥ 2. Proof. Since uα (·, ·, x3 ) ∈ L2 (Ω12 ) for almost every x3 ∈ Ω3 , from (10) and (11) we have that the relations Z 1 ℘α (u) = Eβα xβ ϑ(u) + uα dx0 , (14) |Ω12 | Ω12 X k˜ ukH 1 (Ω12 ;R2 ) ≤ C kEαβ (u)kL2 (Ω12 ) (15) α,β
hold for almost every x3 in Ω3 , and the claimed inequality follows by integration. 2 A different proof of the lemma above can be found in Le Dret [6]. Proposition 4.1 RD2⊥ (Ω) is a Hilbert space with the norm kvkRD2⊥ (Ω) :=
µX
¶1/2 kEαβ (v)k2L2 (Ω)
.
α,β
Proof. We have only to prove that kvkRD2⊥ (Ω) is equivalent to the norm induced on RD2⊥ (Ω) by L2 (Ω3 ; H 1 (Ω12 ; R2 )) since the former space is a closed subspace of the latter. For any v ∈ RD2⊥ (Ω), recalling that ℘(v) = 0, so that v = v˜, and using Lemma 4.1, we then have kvkL2 (Ω3 ;H 1 (Ω12 ;R2 )) = k˜ v kL2 (Ω3 ;H 1 (Ω12 ;R2 )) ≤ CkvkRD2⊥ (Ω) while the opposite inequality is trivially satisfied.
2
ANISOTROPIC INHOMOGENEOUS THIN-WALLED BEAMS
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Theorem 4.3 There exists a constant C such that ° µ ¶° Z Z ° ° °u3 − x1 x2 dϑ(u) − x1 d − u1 dx0 + − u3 dx1 ° ° ° −1 dx3 dx3 Ω12 Ω1 H (Ω3 ;L2 (Ω12 )) µX ¶ ≤C kEαβ (u)kL2 (Ω) + kE13 (u)kL2 (Ω) α,β
for every u ∈ H 1 (Ω; R3 ). Proof. Let u ˜α := uα − ℘α (u) as in (13) and µ ¶ Z Z dϑ(u) d u ˜3 := u3 − x1 x2 − x1 − u1 dx0 + − u3 dx1 . dx3 dx3 Ω12 Ω1 Since E13 (u) = E13 (˜ u),
∂u ˜3 ∂u ˜1 = 2E13 (u) − . ∂x1 ∂x3
Let ψ ∈ H01 (Ω3 ); then Z ∂ ψ˜ u3 dx3 ∂x1 Ω3
Z =
³ ∂u ˜1 ´ ψ 2E13 (u) − dx3 ∂x3 Ω3
Z
³ dψ ´ 2 ψE13 (u) + u ˜1 dx3 . dx3 Ω3 R R R Since Ω3 ψ˜ u3 dx3 ∈ H 1 (Ω) and Ω1 Ω3 ψ u ˜3 dx3 dx1 = 0, by Theorem 4.1, Lemma 4.1, and the previous equality we deduce °Z ° ° ° Z ` ° ° ° ° ∂ ° ° ° ° ψ˜ u dx ψ u ˜ dx ≤ C 3 3° 3 3° ° ° ∂x1 0 Ω3 L2 (Ω) L2 (Ω) =
≤ ≤ Let ϕ ∈ L2 (Ω12 ); ¯Z ¯ ¯ ¯ ¯ ϕψ u ¯ = ˜ dx 3 ¯ ¯ Ω
≤
then ¯Z ¯ ¯ ¯
Ω12
CkψkH 1 (Ω3 ) (kE13 (u)kL2 (Ω) + k˜ u1 kL2 (Ω) ) P CkψkH 1 (Ω3 ) (kE13 (u)kL2 (Ω) + α,β kEαβ (u)kL2 (Ω) ). Z
ϕ Ω3
¯ °Z ¯ ° ψu ˜3 dx3 dx0 ¯¯ ≤ kϕkL2 (Ω12 ) ° °
Ω3
° ° ψ˜ u3 dx3 ° °
L2 (Ω12 )
¶ µ X kEαβ (u)kL2 (Ω) . CkϕkL2 (Ω12 ) kψkH 1 (Ω3 ) kE13 (u)kL2 (Ω) + α,β
Arguing as in the proof of Theorem 4.2, we conclude the proof.
2
ANISOTROPIC INHOMOGENEOUS THIN-WALLED BEAMS
11
Remark 4.2 As in Remark 4.1, in spite of the fact that the left-hand side belongs to L2 (Ω), the inequality of Theorem 4.3 does not hold true if one replaces the norm H −1 with the norm of L2 , because of the following counterexample. ¯ i ) with ϕα satisfying Consider three scalar smooth functions ϕi ∈ C ∞ (Ω Z Z ∂ϕα ϕα dxα = dxα = 0. Ωα Ωα ∂xα Define u1 := −ϕ2
∂ϕ1 ϕ3 , ∂x1
∂ϕ2 ϕ3 , ∂x2
u2 := −ϕ1
u3 := +ϕ1 ϕ2
∂ϕ3 . ∂x3
Then u ∈ H 1 (Ω; R3 ), and the inequality of Theorem 4.3 reduces to ° ° ° µ 2 ¶° 2 ° ° ° ° °ϕ3 ∂ ϕ1 ϕ2 + ϕ1 ∂ ϕ2 ° °ϕ1 ϕ2 ∂ϕ3 ° ≤ C , 2 2 ° ° ° ∂x3 H −1 (Ω2 ;L2 (Ω1 )) ∂x1 ∂x2 °L2 (Ω) which cannot be true if we replace H −1 by L2 , because in such a case taking kϕα kL2 (Ωα ) = k∂ 2 ϕα /∂xα 1 kL2 (Ωα ) would imply that ° ° ° ∂ϕ3 ° ° ° ≤ Ckϕ3 kL2 (Ω) ° ∂x3 ° 2 L (Ω) ¯ 3 ), which is clearly impossible. for any ϕ3 ∈ C ∞ (Ω The next partial Korn’s inequality is proved in Monneau, Murat, and Sili [7]. Theorem 4.4 There exists a constant C such that ° ¶° µZ Z ° ° °u3 − − u3 dx0 − xα d − uα dx0 ° ° ° −1 dx3 Ω12 Ω12 H (Ω3 ;L2 (Ω12 )) µX ¶ X kEαβ (u)kL2 (Ω) + ≤C kEα3 (u)kL2 (Ω) α
αβ 1 for every u ∈ Hdn (Ω; R3 ).
5
Limit strain characterization
Let 1 Hm (Ω12 ) := {z ∈ H 1 (Ω12 ) :
R
z dx0 = 0}, R 1 Hm (Ω1 ; L2 (Ω23 )) := {z ∈ H 1 (Ω1 ; L2 (Ω23 )) : Ω1 z dx1 = 0 a.e. in Ω23 }, 1 Ω12
−1 Hm (Ω3 ; L2 (Ω12 )) := {v ∈ H −1 (Ω3 ; L2 (Ω12 )) : hz3 , ϕi = 0 1
∀ ϕ ∈ H01 (Ω3 ; L2 (Ω2 ))},
ANISOTROPIC INHOMOGENEOUS THIN-WALLED BEAMS
12
where the bracket in the last definition has to be understood in the sense of the duality H −1 (Ω3 ; L2 (Ω12 )) × H01 (Ω3 ; L2 (Ω12 )). Let (uε ) be the sequence of solutions to problems (6). From (7) and assumption (8) it follows that sup kE ε (uε )kL2 (Ω) < +∞.
(16)
ε
Hence, possibly passing to a subsequence, we have that E ε (uε ) * E
in L2 (Ω; R3×3 sym )
for some E ∈ L2 (Ω; R3×3 sym ). In this section we characterize the limit strain E. For clarity we state several lemmas. Lemma 5.1 (component 33) There exists a function u ¯ in the set of the socalled Bernoulli–Navier displacements 1 U := {u ∈ Hdn (Ω; R3 ) : Eαi (u) = 0}
such that
∂u ¯3 = E33 (¯ u). ∂x3 Proof. From (16), the structure of E ε , and Korn’s inequality, we have that E33 =
C ≥ kE ε (uε )kL2 (Ω) ≥ kE(uε )kL2 (Ω) ≥ CK kuε kH 1 (Ω) , where C is a constant independent of ε and CK is Korn’s constant. Hence, up 1 to a subsequence uε * u ¯ in H 1 (Ω; R3 ), for some u ¯ ∈ Hdn (Ω; R3 ). The claim ε ε ε 2 follows by noticing that kEαi (u )kL (Ω) ≤ Cε, and E33 (u ) = ∂uε3 /∂x3 . In fact 1 it follows that u ¯ ∈ {u ∈ Hdn (Ω; R3 ) : Eαi (u) = 0}. 2 Remark 5.1 (representation of the space U ) It is well known (see, for instance, Le Dret [5]) that the space of Bernoulli–Navier displacements admits the following representation: ½ 2 1 2 1 U := u ∈ Hdn (Ω)2 × Hdn (Ω) : exists ζ ∈ Hdn (Ω3 )2 × Hdn (Ω3 ) such that u1 = ζ1 , u2 = ζ2 , u3 = ζ3 − x1 Moreover, U is a Hilbert space with the norm kukU := kE33 (u)kL2 (Ω) , 1 (Ω; R3 ) (see [8]). which is equivalent to that induced by Hdn
dζ1 dζ2 − x2 dx3 dx3
¾ .
ANISOTROPIC INHOMOGENEOUS THIN-WALLED BEAMS
13
Lemma 5.2 (component 23) There exists a function v¯ in the space ½ 1 1 (Ω)2 × L2 (Ω3 ; Hm (Ω12 )) : exist ϑ ∈ H 1 (Ω3 ) such that ϑ(0) = 0 V := v ∈ Hdn and % ∈ L2 (Ω3 ; H 1 (Ω2 )) such that v1 (x) = −x2 ϑ(x3 ), v2 (x) = x1 ϑ(x3 ), ¾ dϑ (x3 ) + %(x2 , x3 ) v3 (x) = x1 x2 dx3 such that
E23 = E23 (¯ v ). Proof. Let viε := 1ε uεi and ϑε := ϑ(v ε ) (see (9)). First of all, by adapting an argument of [4] and Lemmas 4.4 and 4.5, we prove that there exists ϑ ∈ H 1 (Ω3 ) with ϑ(0) = 0, such that, up to subsequences, ϑε → ϑ in L2 (Ω3 ). Applying Lemma 4.1, there exists a constant C such that X k˜ v ε kL2 (Ω3 ;H 1 (Ω12 ;R2 )) ≤ C kEαβ (v ε )kL2 (Ω3 ;L2 (Ω12 )) .
(17)
(18)
α,β
Since, furthermore, ε ε ε (uε ), (uε ), E22 (v ε ) = εE22 (uε ), E12 (v ε ) = ε2 E12 E11 (v ε )11 = ε3 E11
and using (16) we have ε kEαβ (v ε )kL2 (Ω) ≤ εkEαβ (uε )kL2 (Ω) ≤ C ε.
From (18) we get hence Now let η ∈
Cc∞ (Ω12 )
(19)
k˜ v ε kL2 (Ω3 ;H 1 (Ω12 ;R2 )) ≤ Cε;
(20)
v˜ε → 0 in L2 (Ω3 ; H 1 (Ω12 ; R2 )).
(21)
be such that Z η dx0 = − Ω12
IO . 2
ANISOTROPIC INHOMOGENEOUS THIN-WALLED BEAMS Then, taking into account (14), we have Z Z IO ϑε = −2ϑε η dx0 = −ϑε Ω12
14
ηDα xα dx0
Ω12
Z
Z
= ϑε
Dα η xα dx0 = ϑε
Eαγ Eβγ Dα η xβ dx0
Ω12
Ω12
Z Eαγ Dα ηEβγ xβ ϑε dx0
= Ω12
µ Eαγ Dα η ℘γ (v ε ) −
Z = Ω12
1 |Ω12 |
Z Ω12
¶ vγε dx0 dx0
Z Eαγ Dα η ℘γ (v ε ) dx0
= Ω12
Z
Z
= Ω12
Eαγ Dα η vγε dx0 −
Hence, denoting by
1 ϑ˜ε = IO
¡ ¢ Eαγ Dα η v ε − ℘γ (v ε ) dx0 .
Ω12
Z Ω12
Eαγ Dα η wγε dx0
and recalling (21), we find ϑε − ϑ˜ε → 0 in L2 (Ω3 ).
(22)
We now show that D3 ϑ˜ε is bounded in L2 . Since Eαγ Dα Dγ η = 0 everywhere in Ω3 and Dα η = 0 on ∂Ω3 , we have Z Z IO D3 ϑ˜ε = Eαγ Dα η D3 vγε dx0 = 2 Eαγ Dα η Eγ3 (v ε ) dx0 , Ω12
Ω12
ε ε (uε ) and E23 (v ε ) = E23 (uε ), and therefore D3 ϑ˜ε is bounded but E13 (v ε ) = εE13 2 ε ε ˜ ˜ in L (Ω3 ). Since ϑ (0) = 0, ϑ is then bounded in H 1 (Ω3 ) so that there exists ϑ ∈ H 1 (Ω3 ) with ϑ(0) = 0, such that, up to subsequences,
ϑ˜ε * ϑ in H 1 (Ω3 ). Thus, from (22) we obtain (17). Let us now set Z v¯1ε := v1ε − − v1ε dx0 ,
Z v¯2ε := v2ε − −
Ω12
and v¯3ε
:=
v3ε
µZ − − Ω12
v3ε
Ω12
v2ε dx0 ,
¶ Z d ε 0 − v dx . dx − xα dx3 Ω12 α 0
ANISOTROPIC INHOMOGENEOUS THIN-WALLED BEAMS
15
Observing that, by the definitions, v¯1ε = v˜1ε − x2 ϑε ,
v¯2ε = v˜2ε + x1 ϑε ,
from (17) and (21) we have that v¯1ε → −x2 ϑ,
v¯2ε → +x1 ϑ
in L2 (Ω3 ; H 1 (Ω12 )).
By Theorem 4.4, we have that k¯ v3ε kH −1 (Ω3 ;L2 (Ω12 )) ≤ C, and hence there exists v¯3 ∈ H −1 (Ω3 ; L2 (Ω12 )) such that, up to subsequences, v¯3ε * : v¯3 in H −1 (Ω3 ; L2 (Ω12 )).
(23)
Moreover, let us set v¯1 = −x2 ϑ,
v¯2 = +x1 ϑ,
and check that the vector field v¯ so defined satisfies the properties claimed in the statement of Lemma 4.1. A simple computation shows that E13 (¯ v ε ) = E13 (v ε ) and
E23 (¯ v ε ) = E23 (v ε ).
ε (uε ) → 0 in L2 (Ω) and E13 (¯ v ε ) → E13 (¯ v ) in Noticing that E13 (v ε ) = εE13 0 D (Ω), we obtain that E13 (¯ v ) = 0. Hence,
∂¯ v3 ∂¯ v1 dϑ =− = x2 ∂x1 ∂x3 dx3 and, integrating with respect to x1 , v¯3 = x1 x2
dϑ + %(x2 , x3 ) dx3
for some function % ∈ L2 (Ω3 ; H 1 (Ω2 )). Moreover, v¯3 ∈ L2 (Ω3 ; H 1 (Ω R 12 )) and, from (23) and the fact that v¯3 , v¯3ε ∈ L2 (Ω), we then obtain easily that −Ω12 v¯3 dx0 = 0, which concludes the proof. 2 Lemma 5.3 (characterization of the space V ) The space V admits the following characterization: 1 1 V = {v ∈ Hdn (Ω)2 × L2 (Ω3 ; Hm (Ω12 )) : Eαβ (v) = 0, E13 (v) = 0, R E23 (v) ∈ L2 (Ω) and Ω12 vα dx0 = 0 a.e.}.
Moreover, it is a Hilbert space with the norm kvkV := kW13 (v)kL2 (Ω) + kE23 (v)kL2 (Ω) , where W13 (v) =
µ ¶ ∂v3 1 ∂v1 − . 2 ∂x3 ∂x1
(24)
ANISOTROPIC INHOMOGENEOUS THIN-WALLED BEAMS
16
Proof. Let us call V the space at the right-hand side of equality (24) and let V be as in the statement of Lemma 5.2. It is trivial to check that V ⊆ V . Let 1 us prove the opposite inclusion. Let v ∈ V . Since vα ∈ Hdn (Ω) and Eαβ (v) = 0, 2 by integration there exists ϑ ∈ L (Ω3 ) such that v1 = −x2 ϑ(x3 ) + a1 (x3 ),
v2 = x1 ϑ(x3 ) + a2 (x3 ),
R
and since Ω12 vα dx0 = 0, we have a1 = a2 = 0 a.e.. From the resulting 1 expression of vα it follows that ϑ ∈ Hdn (Ω) and ϑ(0) = 0. Since E13 (v) = 0 we obtain that ∂v3 /∂x1 = x2 ϑ0 (x3 ). Then there exists % ∈ L2 (Ω23 ) such that v3 = x1 x2 ϑ0 (x3 ) + %(x2 , x3 ), 1 from which it follows also that % ∈ L2 (Ω3 ; Hm (Ω2 )) and that E23 (v) ∈ L2 (Ω). 1 The last part of the claim follows from the fact that L2 (Ω3 ; Hm (Ω12 ) is a Hilbert space with the scalar product Z 1 (Ω )) = 1 (Ω ) dx3 , hu3 , v3 iL2 (Ω3 ;Hm hu3 (x3 ), v3 (x3 )iHm 12 12 Ω3
1 1 and V is a closed subspace of Hdn (Ω)2 × L2 (Ω3 ; Hm (Ω12 )) which is Hilbert with the product norm 1 (Ω) + kv2 kH 1 (Ω) + kv3 kL2 (Ω ;H 1 (Ω )) kv1 kHdn 3 12 m dn
1 1 induced by Hdn (Ω)2 × L2 (Ω3 ; Hm (Ω12 )). The proof that this norm is equivalent to kvkV is an easy consequence of the Poincar´e inequality, the representation lemma, Lemma 5.2, and the characterization of the space V proved above. 2
Lemma 5.4 (components 22 and 13) There exists a function w ¯ in the space −1 W = {w ∈ RD2⊥ (Ω) × Hm (Ω3 ; L2 (Ω12 )) : 1
E11 (w) = E12 (w) = 0, E13 (w) ∈ L2 (Ω)}
such that E22 =
∂w ¯2 = E22 (w), ¯ ∂x2
E13 =
1 2
µ
∂w ¯3 ∂w ¯1 + ∂x3 ∂x1
¶ = E13 (w). ¯
Moreover, W is a Hilbert space with the norm −1 kwkW := kE22 (w)kL2 (Ω) + kE13 (w)kL2 (Ω) + kw3 kHm (Ω3 ;L2 (Ω12 )) . 1 Proof. Let 1 wε := 2 uε ; ε
ANISOTROPIC INHOMOGENEOUS THIN-WALLED BEAMS then
kE11 (wε )kL2 (Ω) ≤ Cε2 ,
kE12 (wε )kL2 (Ω) ≤ Cε,
kE22 (wε )kL2 (Ω) ≤ C,
kE13 (wε )kL2 (Ω) ≤ C.
17
(25)
Let us recall that w ˜αε := wαε − ℘α (wε ). By Lemma 4.1 and using (25), we have X kw ˜ ε kL2 (Ω3 ;H 1 (Ω12 ;R2 )) ≤ C kEαβ (wε )kL2 (Ω) ≤ C, α,β
R −Ω12 w ˜ ε dx0 = 0, and ϑ(w ˜ ε ) = 0. Thus, up to a subsequence, w ˜αε * w ¯α moreover, a.e.,
in L2 (Ω3 ; H 1 (Ω12 )) for α = 1, 2;
Z −
w ¯α dx0 = 0, ϑ(w) ¯ = 0.
(26)
(27)
Ω12
Hence (w ¯1 , w ¯2 ) ∈ RD2⊥ (Ω) (see (12)). Using (25) we also obtain E11 (w) ¯ = E12 (w) ¯ = 0. Moreover, ε E22 (uε ) = E22 (wε ) = E22 (w ˜ ε ) * E22 (w) ¯ in L2 (Ω).
Let w ˜3ε
:=
w3ε
µ ¶ Z Z dϑ(wε ) d ε 0 ε − x1 x2 − x1 − w dx + − w3 dx1 . dx3 dx3 Ω12 1 Ω1
By Theorem 4.3 we have that X kEαβ (wε )kL2 (Ω) + kE13 (wε )kL2 (Ω) ≤ C. kw ˜3ε kH −1 (Ω3 ;L2 (Ω12 )) ≤ C α,β
Hence, up to a subsequence, w ˜3ε * w3
in H −1 (Ω3 ; L2 (Ω12 )).
Taking into account that (see (16)) Z ℘1 (wε ) = − w1ε dx0 − x2 ϑ(wε ), Ω12
Z ℘2 (wε ) = − Ω12
w2ε dx0 + x1 ϑ(wε ),
ε we easily deduce that E13 (uε ) = E13 (wε ) = E13 (w ˜ ε ) and hence that µ ¶ 1 ∂w ¯1 ∂w ¯3 E13 = + . 2 ∂x3 ∂x1 R Finally, since −Ω1 w ˜3ε dx1 = 0, we have that
hw ¯3 , ϕiH −1 (Ω3 ;L2 (Ω12 ))×H01 (Ω3 ;L2 (Ω12 )) = 0
ANISOTROPIC INHOMOGENEOUS THIN-WALLED BEAMS
18
for all ϕ ∈ H01 (Ω3 ; L2 (Ω2 )). The last part of the claim follows from the fact that W is a closed subspace of −1 {z ∈ RD2⊥ (Ω) × Hm (Ω3 ; L2 (Ω12 )) : E13 (z) ∈ L2 (Ω)} 1 which, in turn, is a Hilbert space under the scalar product Z −1 hz, ζi := hz, ζiRD⊥ (Ω)×Hm + E13 (z)E13 (ζ) dx (Ω3 ;L2 (Ω12 )) 2
1
Ω
and from an application of Proposition 4.1.
2
Lemma 5.5 (representation of the space W ) The space W admits the following representation: ½ −1 W = w ∈ L2 (Ω3 ; H 1 (Ω12 ))2 × Hm (Ω3 ; L2 (Ω12 )) : E13 (w) ∈ L2 (Ω), 1 there exists η1 ∈ L2 (Ω3 ; H 2 (Ω2 )) and η2 ∈ L2 (Ω3 ; H 1 (Ω2 )) such that ∂η1 (x2 , x3 ) + η2 (x2 , x3 ), ∂x2 ¶ ¾ Z µ 2 a1 ∂η1 + x2 η1 dx2 = 0 . 12 ∂x2 Ω2
w1 (x) = η1 (x2 , x3 ), Z ηα dx2 = 0, Ω2
w2 (x) = −x1
(28) Proof. Let us call W the set on the right-hand side of equality (28), and let W be as in the statement of Lemma 5.4. Then it is trivial to check that W ⊆ W . Let us prove the converse inequality. Let w ∈ W as in Lemma 5.4. Since E11 (w) = E12 (w) = 0, by integration we deduce that there exist η1 ∈ L2 (Ω3 ; H 2 (Ω2 )) and η2 ∈ L2 (Ω3 ; H 1 (Ω2 )) such that w1 (x) = η1 (x2 , x3 ), Since
Z −
w2 (x) = −x1
wα dx0 = 0,
∂η1 (x2 , x3 ) + η2 (x2 , x3 ). ∂x2
ϑ(w) = 0,
(29)
Ω12
we have that
Z ηα dx2 = 0
a.e. for α = 1, 2,
Ω2
and
Z Ω2
a.e..
µ
¶ a21 ∂η1 + x2 η1 dx2 = 0, 12 ∂x2 2
ANISOTROPIC INHOMOGENEOUS THIN-WALLED BEAMS
19
Lemma 5.6 (component 12) There exists a vector function p¯ in the set 1 P = {0} × Hm (Ω1 ; L2 (Ω23 )) × {0} 1
such that E12 =
1 ∂ p¯2 = E12 (¯ p). 2 ∂x1
(30)
Moreover, P is a Hilbert space with the norm kpkP := kE12 (p)k. Proof. Let pεα :=
1 ε u ε3 α
for α = 1, 2,
and p¯ε1
:=
pε1
Z − − pε1 dx1 ,
p¯ε2
Ω1
:=
Z Z ∂ ε − − p2 dx1 + x1 − pε dx1 . ∂x2 Ω1 1 Ω1
pε2
ε (uε ); hence, by Theorem 4.1 we have Then E11 (¯ pε ) = E11 (pε ) = εE11
p¯ε1 → 0,
∂ p¯ε1 →0 ∂x1
in L2 (Ω).
ε (uε ), by Theorem 4.2 we also have that, up to Since E12 (¯ pε ) = E12 (pε ) = E12 subsequences, p¯ε2 * p¯2 in H −1 (Ω2 ; L2 (Ω13 )).
Setting p¯ := (0, p¯2 , 0), we have E12 = E12 (¯ p) =
1 ∂ p¯2 , 2 ∂x1
R that is, (30). It remains then to prove that p¯2 ∈ H 1 (Ω1 ; L2 (Ω23 )) and that −Ω1 p¯2 dx1 = 0. Since p¯2 ∈ H −1 (Ω2 ; L2 (Ω13 )), for any ϕ ∈ H01 (Ω2 ) and any ψ ∈ L2 (Ω13 ) the product ϕ ⊗ ψ belongs to H01 (Ω2 ; L2 (Ω13 )), and the linear map Pϕ : L2 (Ω13 ) → R,
hPϕ , ψi := h¯ p2 , ϕ ⊗ ψi
satisfies the estimate |hPϕ , ψi| ≤ k¯ p2 kH −1 kϕkH01 kψkL2 . Thus Pϕ ∈ L2 (Ω13 ). Moreover, from the definition of Pϕ and the fact that ∂Pϕ ∂ p¯2 2 2 ∂x1 ∈ L (Ω), we obtain that ∂x1 ∈ L (Ω13 ) and also that ∂Pϕ = ∂x1
Z Ω2
∂ p¯2 ϕ dx2 . ∂x1
(31)
ANISOTROPIC INHOMOGENEOUS THIN-WALLED BEAMS
20
R Since −Ω1 p¯ε2 dx1 = 0 it follows from the definitions that Z − Pϕ (s, x3 ) ds = h¯ p2 , ϕi = 0
(32)
Ω1
for almost every x3 ∈ Ω3 ; using this fact, the following Poincar´e inequality holds: ° ° ° ∂Pϕ ° ° , kPϕ (·, x3 )kL2 (Ω1 ) ≤ C ° (·, x ) 3 ° ° ∂x1 L2 (Ω1 ) where the constant C depends only on the domain Ω1 and is therefore independent of x3 . By substituting (31) inside the Poincar´e inequality, we obtain that ° ° ° ∂ p¯2 ° ° kPϕ kL2 (Ω13 ) ≤ C ° (33) ° ∂x1 ° 2 kϕkL2 (Ω2 ) . L (Ω) Using the density of Cc∞ (Ω2 ) ⊗ Cc∞ (Ω13 ) in L2 (Ω) (see, for instance, Treves [10, Theorem 39.2 and subsequent Corollary 3]), the fact that Pϕ ∈ L2 (Ω13 ), and inequality (33), we have that k¯ p2 kL2 (Ω)
=
|h¯ p2 , ϕψi| ψ∈Cc∞ (Ω13 ) kϕkL2 (Ω2 ) kψkL2 (Ω13 )
sup ϕ∈Cc∞ (Ω2 ),
|hPϕ , ψi| ϕ∈Cc∞ (Ω2 ), ψ∈Cc∞ (Ω13 ) kϕkL2 (Ω2 ) kψkL2 (Ω13 ) ° ° ° ∂ p¯2 ° kPϕ kL2 (Ω13 ) ° ° ≤ sup ≤ C° ; ∂x1 °L2 (Ω) ϕ∈Cc∞ (Ω2 ) kϕkL2 (Ω2 ) =
sup
R hence p¯2 ∈ L2 (Ω). Thus p¯2 ∈ H 1 (Ω1 ; L2 (Ω23 )), and (32) implies −Ω1 p¯2 dx1 = 0. 1 The last part of the claim follows from the fact that Hm (Ω1 ; L2 (Ω23 )) is a 1 Hilbert space with the norm ° ° ° ∂p2 ° ° ° 1 (Ω ;L2 (Ω )) := kpkHm 1 23 ° ∂x1 ° 2 , 1 L (Ω) which, by Theorem 4.1, turns out to be equivalent to the canonical one. Lemma 5.7 (component 11) There exists a function q¯ in the space 1 Q := Hm (Ω1 ; L2 (Ω23 )) × {0}2 1
such that E11 =
∂ q¯1 = E11 (¯ q ). ∂x1
Moreover, Q is a Hilbert space with the norm kqkQ := kE11 (q)k.
2
ANISOTROPIC INHOMOGENEOUS THIN-WALLED BEAMS Proof. Let q1ε := then
21
µ ¶ Z 1 ε ε ; u − − u dx 1 1 ε4 1 Ω1
° ε° ° ∂q1 ° ε ° sup ° = sup kE11 (uε )kL2 (Ω) ≤ C, ° ∂x1 ° 2 ε ε L (Ω)
and by Theorem 4.1 we have supε kq1ε kL2 (Ω) ≤ C. Then, up to a subsequence, ε (uε ) = ∂ q¯1ε /∂x1 * ∂ q¯1 /∂x1 in L2 (Ω) for some q1ε * q¯1 in L2 (Ω), and E11 2 q¯1 ∈ L (Ω). 1 The last part of the claim follows from the fact that Hm (Ω1 ; L2 (Ω23 )) is a 1 Hilbert space with the norm 1 kqkHm
1
(Ω1 ;L2 (Ω23 ))
:= kq1,1 kL2 (Ω) ,
which, by Theorem 4.1, turns out to be equivalent to the canonical one.
6
2
The limit problem
Let us consider the space A := U × V × W × P × Q. According to the notation and the results proved in the previous section, A is a Hilbert space when endowed with the product norm k(u, v, w, p, q)kA := kukU + kvkV + kwkW + kpkP + kqkQ . Given a 5-tuple of vector valued distributions (u, v, w, p, q) ∈ D0 (Ω; R3 )5 , let us define E11 (q) E12 (p) E13 (w) E22 (w) E23 (v) . E(u, v, w, p, q) := (34) Sym. E33 (u) We are now in a position to state the main result of this paper. Theorem 6.1 Let C be a positive definite fourth order tensor field on Ω with the minor symmetries, i.e., Cijkl = Cjikl = Cijlk . Let F ε be a second order symmetric tensor field which belongs to L2 (Ω; R3×3 ). Then problem (6), that is, ε 1 3 u ∈ Hdn (Ω; R ), Z Z (35) 1 CE ε (uε ) · E ε (ϕ) dx = F ε · E ε (ϕ) dx ∀ϕ ∈ Hdn (Ω; R3 ), Ω
Ω
admits a unique solution uε . Moreover, if F ε → F in L2 (Ω; R3×3 ), then we have the following:
ANISOTROPIC INHOMOGENEOUS THIN-WALLED BEAMS
22
1. the problem Z Z CE(¯ u, v¯, w, ¯ p¯, q¯) · E(u, v, w, p, q) dx = F · E(u, v, w, p, q) dx Ω
Ω
∀ (u, v, w, p, q) ∈ A (36) admits a unique solution (¯ u, v¯, w, ¯ p¯, q¯) ∈ A ; 2. uε * u ¯ in H 1 (Ω; R3 ); 3. E ε (uε ) → E(¯ u, v¯, w, ¯ p¯, q¯) in L2 (Ω; R3×3 ). The following corollary can be seen as a corrector result. Corollary 6.1 If the solution (¯ u, v¯, w, ¯ p¯, q¯) of problem (36) is such that ∂¯ v3 ∂w ¯3 ∂ q¯1 ∂ q¯1 ∂ p¯2 ∂ p¯2 , E23 (w), ¯ , , , , ∈ L2 (Ω), ∂x3 ∂x3 ∂x2 ∂x3 ∂x2 ∂x3 then where Proof. Since
(37)
kE ε (uε ) − E ε (¯ uε )kL2 (Ω;R3×3 ) → 0, u ¯ε := u ¯ + ε¯ v + ε2 w ¯ + ε3 p¯ + ε4 q¯.
E12 (¯ q) 0 0 0 E13 (¯ q) E22 (¯ p) E23 (w) ¯ +ε2 0 E23 (¯ p) E ε (¯ uε ) = E(¯ u, v¯, w, ¯ p¯, q¯)+ε Sym. E33 (¯ v) Sym. E33 (w) ¯ 0
the additional regularity assumptions imply that E ε (¯ uε ) ∈ L2 (Ω; R3×3 ) and kE ε (¯ uε ) − E(¯ u, v¯, w, ¯ p¯, q¯)kL2 (Ω;R3×3 ) → 0. Then the claim follows from step 3 of Theorem 6.1.
2
1 In order to prove Theorem 6.1, we introduce the subspaces of Hdn (Ω; R3 ), 1 Uˆ := {u ∈ Hdn (Ω; R3 ) : Eαβ (u) = Eα3 (u) = 0}, 1 Vˆ := {v ∈ Hdn (Ω; R3 ) : Eαβ (v) = E13 (v) = 0}, 1 Wˆ := {w ∈ Hdn (Ω; R3 ) : E1β (w) = 0}, 1 Pˆ := {p ∈ Hdn (Ω; R3 ) : E11 (p) = 0}, 1 Qˆ := Hdn (Ω; R3 ),
ˆ and define Aˆ := Uˆ × Vˆ × Wˆ × Pˆ × Q. ˆ Let us note that U = U , but similar equalities are not true for the spaces V , W , P, and Q. Nevertheless, for such spaces we can prove the following approximation lemma.
ANISOTROPIC INHOMOGENEOUS THIN-WALLED BEAMS
23
Lemma 6.1 For every v ∈ V , w ∈ W , p ∈ P, and q ∈ Q there exist sequences ˆ and (ˆ (ˆ v n ) in Vˆ, (w ˆ n ) in Wˆ , (ˆ pn ) in P, q n ) in Qˆ such that the following 2 convergences hold in the norm of L (Ω): (i) E23 (ˆ v n ) → E23 (v), (ii) E13 (w ˆ n ) → E13 (w) and E22 (w ˆ n ) → E22 (w), pn ) → E12 (p), (iii) E12 (ˆ (iv) E11 (ˆ q n ) → E11 (q). Proof. Let us prove (i). Since v ∈ V (see Lemma 5.2), there exist ϑ ∈ H 1 (Ω3 ) with ϑ(0) = 0 and % ∈ L2 (Ω3 ; H 1 (Ω2 )) such that v1 (x) = −x2 ϑ(x3 ), v3 (x) = x1 x2
v2 (x) = x1 ϑ(x3 ),
dϑ (x3 ) + %(x2 , x3 ). dx3
2 1 Then there exist sequences ϑn ∈ Hdn (Ω3 ) and %n ∈ Hdn (Ω23 ) such that
ϑn → ϑ in H 1 (Ω3 ), Setting
%n → % in L2 (Ω3 ; H 1 (Ω2 )).
vˆ1n (x) = −x2 ϑn (x3 ), vˆ3n (x) = x1 x2
vˆ2n (x) = x1 ϑn (x3 ),
dϑn (x3 ) + %n (x2 , x3 ), dx3
we obtain the claim. Let us prove (ii). As w ∈ W (see Lemma 5.4), there exist η1 ∈ L2 (Ω3 ; H 2 (Ω2 )) and η2 ∈ L2 (Ω3 ; H 1 (Ω2 )) such that w1 (x) = η1 (x2 , x3 ),
w2 (x) = −x1
∂η1 (x2 , x3 ) + η2 (x2 , x3 ). ∂x2
2 (Ω) with ∂η1n /∂x1 = 0 such that Then, there exist a sequence η1n ∈ Hdn
η1n → η1
in L2 (Ω3 ; H 2 (Ω2 ))
1 (Ω) with ∂η2n /∂x1 = 0 such that and a sequence η2n ∈ Hdn
η2n → η2
in L2 (Ω3 ; H 1 (Ω2 )).
Since ∂w3 /∂x1 = 2E13 (w)−∂η1 /∂x3 and E13 (w) ∈ L2 (Ω), by integration, there exists G13 ∈ H 1 (Ω1 ; L2 (Ω23 )) such that ∂G13 = E13 (w) ∂x1
and
w3 = 2G13 − x1
∂η1 , ∂x3
where we have also used the fact that η1 does not depend on x1 .
ANISOTROPIC INHOMOGENEOUS THIN-WALLED BEAMS
24
1 We may also find a sequence Gn13 ∈ Hdn (Ω) such that
Gn13 → G13
in H 1 (Ω1 ; L2 (Ω23 ))
and define w ˆ1n := η1n ,
w ˆ2n := −x1
∂η1n + η2n , ∂x2
w ˆ3n := Gn13 − x1
∂η1n . ∂x3
1 Then w ˆ n ∈ Hdn (Ω), E11 (w ˆ n ) = ∂η1n /∂x1 = 0, and E12 (w ˆ n ) = 0 so that w ˆn ∈ Wˆ . Moreover,
E13 (w ˆn) =
∂Gn13 ∂G13 → = E13 (w) ∂x1 ∂x1
in L2 (Ω),
and E22 (w ˆn) =
∂η n ∂w ˆ2n ∂ 2 η1n ∂w ˆ2 + 2 → = −x1 = E22 (w) ˆ 2 ∂x2 ∂x2 ∂x2 ∂x2
in L2 (Ω).
To prove (iii) it is enough to consider, for a given p = (0, p2 , 0) ∈ P, 1 (Ω), which converges to p2 in the norm of H 1 (Ω1 ; L2 (Ω23 )), a sequence pˆn2 ∈ Hdn and set pˆn 1 n = (0, pˆ2 , 0). Finally, claim (iv) simply follows from the density of Hdn (Ω) 1 in Hm (Ω1 ; L2 (Ω23 )). 2 1 Proof of Theorem 6.1. The existence and uniqueness of the solution of problem (36) follows from an application of the Lax–Milgram lemma to the symmetric bilinear form defined on A by Z a[(u, v, w, p, q), (ˆ u, vˆ, w, ˆ pˆ, qˆ)] := CE(u, v, w, p, q) · E(ˆ u, vˆ, w, ˆ pˆ, qˆ) dx, Ω
which is continuous and coercive with respect to the Hilbertian norm on A defined at the beginning of this section. Part 2 of the statement of Theorem 6.1 is actually a consequence of step 3. Let us now prove part 3. According to the results proved in the previous section, we have E ε (uε ) * E in L2 (Ω; R3×3 ), (38) and there exists a (¯ u, v¯, w, ¯ p¯, q¯) ∈ A such that E11 (¯ q ) E12 (¯ p) E13 (w) ¯ E22 (w) ¯ E23 (¯ v ) = E(¯ E= u, v¯, w, ¯ p¯, q¯). Sym. E33 (¯ u) The result will be achieved in two steps: (i) we prove that (¯ u, v¯, w, ¯ p¯, q¯) satisfies equality (36) and therefore coincides with the unique solution of the variational problem, and (ii) we show that the convergence in (38) is indeed strong.
ANISOTROPIC INHOMOGENEOUS THIN-WALLED BEAMS
25
Let (ˆ u, vˆ, w, ˆ pˆ, qˆ) ∈ Aˆ and set ϕˆε := u ˆ + εˆ v + ε2 w ˆ + ε3 pˆ + ε4 qˆ; 1 then ϕˆε ∈ Hdn (Ω; R3 ) and an easy computation shows that, as ε → 0, E11 (ˆ q ) E12 (ˆ p) E13 (w) ˆ E22 (w) ˆ E23 (ˆ v ) = E(ˆ E ε (ϕˆε ) → u, vˆ, w, ˆ pˆ, qˆ) Sym. E33 (ˆ u)
(39)
in the norm convergence of L2 (Ω; R3×3 ). Taking ϕ = ϕˆε in (35) and passing to the limit we find Z Z CE(¯ u, v¯, w, ¯ p¯, q¯) · E(ˆ u, vˆ, w, ˆ pˆ, qˆ) dx = F · E(ˆ u, vˆ, w, ˆ pˆ, qˆ) dx, Ω
Ω
for every (ˆ u, vˆ, w, ˆ pˆ, qˆ) ∈ Aˆ. This equality holds in fact for any (u, v, w, p, q) ∈ A in place of (ˆ u, vˆ, w, ˆ pˆ, qˆ) ∈ Aˆ because of the approximation Lemma 6.1 which ensures that there exists a sequence (u, vˆn , w ˆ n , pˆn , qˆn ) ∈ Aˆ such that k(u, vˆn , w ˆ n , pˆn , qˆn ) − (u, v, w, p, q)kA → 0. To show that the convergence in (38) is indeed strong, it suffices to prove that limε→0 kE ε (uε )kL2 (Ω;R3×3 ) = kE(¯ u, v¯, w, ¯ p¯, q¯)kL2 (Ω;R3×3 ) or, equivalently, that Z Z lim CE ε (uε ) · E ε (uε ) dx = lim F ε · E ε (uε ) dx ε→0
ε→0
Ω
Ω
Z =
F · E(¯ u, v¯, w, ¯ p¯, q¯) dx Ω
Z =
CE(¯ u, v¯, w, ¯ p¯, q¯) · E(¯ u, v¯, w, ¯ p¯, q¯) dx, Ω
where we passed to the limit thanks to the strong convergence of F ε .
7
2
Equilibrium differential equations
In this last section we derive the differential formulation of the limit problem. For simplicity we assume here that the elasticity tensor also satisfies the major symmetries; that is, Cijkl = Cklij for any i, j, k, l. Nevertheless, the same computation can be performed also in the general case. To make (36) more explicit and to keep the notation compact, in writing the elasticity tensor components Cijkl we associate to a pair of components ij a single component s following the rule 11 7→ 1, 22 7→ 2, 33 7→ 3, 23 7→ 4, 13 7→ 5, 12 7→ 6, and we write, for instance, c14 for C1123 ; see Auld [1] for more details
ANISOTROPIC INHOMOGENEOUS THIN-WALLED BEAMS
26
on the notation used. Clearly cij = cji . Still, for brevity, define e¯1 = E11 (¯ q ), e¯2 = E22 (w), ¯ e¯3 = E33 (¯ u), e¯4 = 2E23 (¯ v ), e¯5 = 2E13 (w), ¯ e¯6 = 2E12 (¯ p). Letting Z L (u, v, w, p, q) := F · E(u, v, w, p, q) dx, Ω
we can then rewrite (36) as Z X 6 £ c1j e¯j E11 (q) + c2j e¯j E22 (w) + c3j e¯j E33 (u) + 2c4j e¯j E23 (v) Ω j=1
¤ + 2c5j e¯j E13 (w) + 2c6j e¯j E12 (p) dx = L (u, v, w, p, q),
for every (u, v, w, p, q) ∈ A . Thus Z X 6
c1j e¯j E11 (q) dx = L (0, 0, 0, 0, q),
(40)
2c6j e¯j E12 (p) dx = L (0, 0, 0, p, 0),
(41)
Ω j=1
Z X 6 Ω j=1
Z X 6 £ ¤ c2j e¯j E22 (w) + 2c5j e¯j E13 (w) dx = L (0, 0, w, 0, 0),
(42)
Ω j=1
Z X 6
2c4j e¯j E23 (v) dx = L (0, v, 0, 0, 0),
(43)
c3j e¯j E33 (u) dx = L (u, 0, 0, 0, 0).
(44)
Ω j=1
Z X 6 Ω j=1
In this section, for simplicity, we assume Z L (0, 0, ·, ·, ·) = 0,
`
L (0, v, 0, 0, 0) = Z
L (u, 0, 0, 0, 0) =
m(x3 )ϑ(x3 ) dx3 ,
Z
0
b · u dx + Ω
(45) 2
s · u dH , Γ`
where Γ` = ∂Ω ∩ {x3 = `}; u ∈ U ; b ∈ L2 (Ω); s ∈ L2 (Γ` ); v ∈ V ; ϑ ∈ H 1 (Ω3 ), ϑ(0) = 0, is related to v as in Lemma 5.2; and m ∈ L2 (Ω3 ). Such assumptions are quite often satisfied in engineering applications. We now derive the equilibrium equations in differential form. Let ψ ∈ L2 (Ω) and define Z x1 Z Z x1 q1 := ψ(s, ·, ·) ds − − ψ(s, ·, ·) ds dx1 . −a1 /2
Ω1
−a1 /2
ANISOTROPIC INHOMOGENEOUS THIN-WALLED BEAMS
27
Then q := (q1 , 0, 0) ∈ Q and E11 (q) = ψ; hence, from (40) and (45) we deduce 6 X
c1j e¯j = 0 a.e..
(46)
j=1
With the same argument it follows from (41) that 6 X
c6j e¯j = 0 a.e..
(47)
j=1
From (46) and (47) we deduce, since c11 c66 − c216 > 0, that e¯1 = E11 (¯ q) = −
5 X c66 c1j − c16 c6j e¯j c11 c66 − c216 j=2
e¯6 = 2E12 (¯ p) = −
5 X c11 c6j − c16 c1j e¯j c11 c66 − c216 j=2
a.e.,
(48)
a.e..
(49)
Using (45), (48), and (49) we can rewrite (42), after setting c˜ij = cij − ci1
c11 c6j − c16 c1j c66 c1j − c16 c6j − ci6 , c11 c66 − c216 c11 c66 − c216
for i, j = 2, . . . , 5, as Z X 5 £ ¤ c˜2j e¯j E22 (w) + 2˜ c5j e¯j E13 (w) dx = L (0, 0, w, 0, 0) = 0.
(50)
Ω j=2
Since w ∈ W , it then admits the representation given in Lemma 5.5 in terms of functions η1 and η2 . Choosing η1 = η2 = 0, so that E22 (w) = 0, and w3 like it has been chosen q1 previously, we find from (50) that 5 X
c˜5j e¯j = 0 a.e..
(51)
j=2
Let ψ ∈ L2 (Ω23 ). Taking η1 = w3 = 0 and Z x2 Z Z η2 := ψ(s, ·) ds − − −a2 /2
Ω2
x2
ψ(s, ·) ds dx2
−a2 /2
so that E22 (w) = ψ, we find from (50) that 5 Z X j=2
Ω1
c˜2j e¯j dx1 = 0
a.e..
(52)
ANISOTROPIC INHOMOGENEOUS THIN-WALLED BEAMS Taking instead η2 = w3 = 0 and Z x2 Z t η1 := −a2 /2
28
ψ(s, ·) ds dt − K1 x2 − K2 ,
−a2 /2
where the constants K1 and K2 are chosen in order to satisfy the mean integral conditions required on η1 by (28), we have E22 (w) = −x1 ψ, and hence, from (50) we deduce 5 Z X x1 c˜2j e¯j dx1 = 0 a.e.. (53) Ω1
j=2
From (51), and observing that the positive definiteness of the elastic tensor implies c˜55 > 0, we find e¯5 = 2E13 (w) ¯ =−
4 X c˜5j e¯j c˜ j=2 55
a.e..
(54)
To solve (52) and (53) we need to write explicitly e¯2 = E22 (w). ¯ Since w ¯ ∈W, by Lemma 5.5, we can write w ¯1 (x) = η¯1 (x2 , x3 ),
w ¯2 (x) = −x1
∂ η¯1 (x2 , x3 ) + η¯2 (x2 , x3 ), ∂x2
where η¯1 and η¯2 belong to the appropriate spaces. Since e¯2 = −x1
∂ η¯2 ∂ 2 η¯1 + 2 ∂x2 ∂x2
(55)
and using (54), we can rewrite (52) and (53) as ∂ η¯2 ∂ 2 η¯1 − s0;22 s1;22 2 ∂x2 ∂x2 s2;22
∂ η¯2 ∂ 2 η¯1 − s1;22 ∂x22 ∂x2
where we have set cˆij :=
c˜55 c˜ij − c˜i5 c˜j5 , c˜55
4 Z X = − cˆ2j e¯j dx1 , j=3 Ω1
=
4 Z X − x1 cˆ2j e¯j dx1 , j=3 Ω1
Z sk;ij := − xk1 cˆij dx1 ,
(56)
Ω1
for i, j = 2, 3, 4 and k = 0, 1, 2. From these equations we find µ ¶ 4 Z 4 Z X X ∂ 2 η¯1 1 = s − x c ˆ e ¯ dx − s − c ˆ e ¯ dx , 0;22 1 2j j 1 1;22 2j j 1 ∂x22 s0;22 s2;22 − s21;22 j=3 Ω1 j=3 Ω1 µ ¶ 4 Z 4 Z X X 1 ∂ η¯2 = s − x c ˆ e ¯ dx − s − c ˆ e ¯ dx 1;22 1 2j j 1 2;22 2j j 1 , ∂x2 s0;22 s2;22 − s21;22 j=3 Ω1 j=3 Ω1
ANISOTROPIC INHOMOGENEOUS THIN-WALLED BEAMS
29
and then by integration η¯1 and η¯2 (the fact that s0;22 s2;22 − s21;22 > 0 can be checked, for instance, by using H¨older’s inequality; see Wheeden and Zygmund [13, Chapter 8, Exercise 4]). According to Remark 5.1 and Lemma 5.2, we let 0
0
00
00
e¯4 = 2x1 ϑ¯ +
e¯3 = ζ¯3 − x1 ζ¯1 − x2 ζ¯2 , Setting ijkl := Smpqr
∂ %¯ . ∂x2
(57)
si;2j sk;2l − sm;2p sq;2r , s0;22 s2;22 − s21;22
we then have ∂ 2 η¯1 ∂x22
¡ 00 ¢ ¯ 0213 ¯0 0223 ¯00 0224 ¯0 0214 ∂ % = S0312 ζ3 − x2 ζ¯2 − S1213 ζ1 + 2S1214 ϑ + S0412 , ∂x2
∂ η¯2 ∂x2
¡ 00 ¢ ¯ 1213 ¯0 1223 ¯00 1224 ¯0 1214 ∂ % = S0322 ζ3 − x2 ζ¯2 − S2213 ζ1 + 2S1422 ϑ + S0422 , ∂x2
and taking into account the relations (48), (49), (54), (55), and (57), we find 6 X
¢¡ 0 ¡ 00 ¢ 1213 0213 + cˆi2 S0322 ζ¯3 − x2 ζ¯2 cij e¯j = cˆi3 − x1 cˆi2 S0312
j=1
¡ ¢ 0223 1223 ¯00 − x1 cˆi3 − x1 cˆi2 S1213 + cˆi2 S2213 ζ1 ¡ ¢ 0224 1224 ¯0 +2 x1 cˆi4 − x1 cˆi2 S1214 + cˆi2 S1422 ϑ
(58)
¢ ¯ ¡ 1214 ∂ % 0214 + cˆi2 S0422 + cˆi4 − x1 cˆi2 S0412 , ∂x2 for i = 3, 4. Now let v ∈ V and ϑ and % be as in Lemma 5.2. With ψ ∈ L2 (Ω23 ) and ϑ = 0 and Z x2 Z Z x2 % := ψ(s, ·) ds − − ψ(s, ·) ds dx2 , −a2 /2
we find from (43) that
Ω2
−a2 /2
Z X 6 − c4j e¯j dx1 = 0. Ω1 j=1
It then follows that ¡ ¢¡ 0 00 ¢ 0213 1213 0 = s0;43 − s1;42 S0312 + s0;42 S0322 ζ¯3 − x2 ζ¯2 ¡ ¢ 0223 1223 ¯00 − s1;43 − s1;42 S1213 + s0;42 S2213 ζ1 ¡ ¢ 0224 1224 ¯0 +2 s1;44 − s1;42 S1214 + s0;42 S1422 ϑ ¡ ¢ ¯ 0214 1214 ∂ % , + s0;44 − s1;42 S0412 + s0;42 S0422 ∂x2
ANISOTROPIC INHOMOGENEOUS THIN-WALLED BEAMS
30
0214 1214 and provided that the last coefficient s0;44 − s1;42 S0412 + s0;42 S0422 6= 0, one finds ∂ %¯ −1 = 0214 1214 ∂x2 s0;44 − s1;42 S0412 + s0;42 S0422 h¡ ¢¡ 0 00 ¢ 0213 1213 · s0;43 − s1;42 S0312 + s0;42 S0322 ζ¯3 − x2 ζ¯2 ¡ ¢ 0223 1223 ¯00 − s1;43 − s1;42 S1213 + s0;42 S2213 ζ1 i ¡ ¢ 0 0224 1224 + s1;44 − s1;42 S1214 + s0;42 S1422 2ϑ¯ .
We may rewrite (58) as 6 X
0 00 00 0 cij e¯j = Fi3 ζ¯3 − Fi2 ζ¯2 − Fi1 ζ¯1 + Fi4 ϑ¯
(59)
j=1
for i = 3, 4, where 0223 1223 Fi1 = (x1 cˆi3 − x1 cˆi2 S1213 + cˆi2 S2213 ) ¢ ¡ 1223 0223 1214 0214 + s0,42 S2213 ) s1,43 − s1,42 S1213 + cˆi2 S0422 (ˆ ci4 − x1 cˆi2 S0412 , − 1214 0214 + s s0,44 − s1,42 S0412 0,42 S0422
Fi2 = x2 Fi3 , 0213 1213 Fi3 = (ˆ ci3 − x1 cˆi2 S0312 + cˆi2 S0322 )
−
1213 0213 1214 0214 ) + s0,42 S0322 )(s0,43 − s1,42 S0312 + cˆi2 S0422 (ˆ ci4 − x1 cˆi2 S0412 , 0214 1214 s0,44 − s1,42 S0412 + s0,42 S0422
0224 1224 Fi4 = 2(x1 cˆi4 − x1 cˆi2 S1214 + cˆi2 S1422 )
−2
1224 0224 1214 0214 ) + s0,42 S1422 )(s1,44 − s1,42 S1214 + cˆi2 S0422 (ˆ ci4 − x1 cˆi2 S0412 . 0214 1214 s0,44 − s1,42 S0412 + s0,42 S0422
Let
Z Aij (x3 ) :=
Fij (·, ·, x3 ) dx1 dx2 Ω12
and
Z
Kij (x3 ) :=
Z x1 Fij (·, ·, x3 ) dx1 dx2 , Lij (x3 ) :=
Ω12
x2 Fij (·, ·, x3 ) dx1 dx2 . Ω12
Then, from (43), (44), and (59) we finally deduce the following system of equilibrium differential equations: 0 00 00 0 0 (A33 ζ¯3 − A31 ζ¯1 − A32 ζ¯2 + A34 ϑ¯ ) = −p3 , (K33 ζ¯30 − K31 ζ¯100 − K32 ζ¯200 + K34 ϑ¯0 )00 = −p1 , (60) 0 00 00 0 00 (L33 ζ¯3 − L31 ζ¯1 − L32 ζ¯2 + L34 ϑ¯ ) = −p2 , 0 00 00 0 0 2(K43 ζ¯ − K41 ζ¯ − K42 ζ¯ + K44 ϑ¯ ) = −m, 3
1
2
ANISOTROPIC INHOMOGENEOUS THIN-WALLED BEAMS where
µZ
¶0
p1 = Ω12
µZ p2 =
Z +
Ω12
p3 =
b1 dx1 dx2 , Ω12
¶0 x2 b3 dx1 dx2
Z
Z +
x1 b3 dx1 dx2
31
b2 dx1 dx2 , Ω12
b3 dx1 dx2 . Ω12
The system (60) should then be completed with the suitable boundary conditions.
7.1
The homogeneous beam
In the general inhomogeneous and anisotropic case, the torsional, flexional, and extensional problems are all coupled together in the equilibrium differential system (60). A partial decoupling occurs already in the homogeneous fully anisotropic case. Indeed, in this case, from (56), we have s0;ij = cˆij ,
s1;ij = 0,
s2;ij =
a31 cˆij , 12
and therefore 0213 1223 0214 1224 S0312 = S2213 = S0412 = S1422 = 0,
0223 S1213 =
cˆ23 , cˆ22
1213 S0322 =−
cˆ23 , cˆ22
1214 S1422 =
which causes many of the coefficients of the system Aij , Kij , and Lij to be zero. In this case, the system (60) simply rewrites as 00 A33 ζ¯3 = −p3 , (−K31 ζ¯00 + K34 ϑ¯0 )00 = −p1 , 1 (61) −L32 ζ¯(iv) = −p2 , 2 00 0 0 2(−K41 ζ¯1 + K44 ϑ¯ ) = −m, where
· A33 = a1 a2 K31 = K41 = L32
¸ cˆ22 cˆ33 − cˆ223 (ˆ c22 cˆ34 − cˆ23 cˆ24 )2 , + cˆ22 cˆ22 (ˆ c22 cˆ44 − cˆ224 )
a31 a2 cˆ22 cˆ33 − cˆ223 a3 a2 cˆ22 cˆ34 − cˆ23 cˆ24 , K34 = 1 , 12 cˆ22 6 cˆ22 a31 a2 cˆ22 cˆ34 − cˆ23 cˆ24 a3 a2 cˆ22 cˆ44 − cˆ224 , K44 = 1 , 12 cˆ22 6 cˆ22 · ¸ a1 a32 cˆ22 cˆ33 − cˆ223 (ˆ c22 cˆ34 − cˆ23 cˆ24 )2 = + . 12 cˆ22 cˆ22 (ˆ c22 cˆ44 − cˆ224 )
Thus for a fully anisotropic but homogeneous beam there is only coupling between twisting and bending in direction 1.
cˆ24 , cˆ22
ANISOTROPIC INHOMOGENEOUS THIN-WALLED BEAMS
7.2
32
The homogeneous orthotropic/isotropic beam
When the material is orthotropic a complete decoupling occurs. Indeed, for orthotropic material we have that cki = 0 for k = 1, 2, 3 and i = 4, 5, 6, and c45 = c46 = c56 = 0. It then follows that K34 = K41 = 0 and the system (61) reduces to 00 A33 ζ¯3 = −p3 , −K ζ¯(iv) = −p , 31 1 1 (62) (iv) ¯ −L32 ζ2 = −p2 , 00 2K44 ϑ¯ = −m. Finally, if the material is isotropic, that is, if it is orthotropic and c12 = c23 = c13 =: λ, c44 = c55 = c66 =: µ, and c11 = c22 = c33 = λ + 2µ, where λ and µ are the Lam´e moduli, then we have that A33 = a1 a2 E,
K31 =
a31 a2 E, 12
2K44 =
a31 a2 µ, 3
L32 =
a1 a32 E, 12
where E := µ 3λ+2µ λ+µ is the Young modulus of the material. Hence, in the isotropic and homogeneous case we recover the usual form of the differential system of equilibrium equations.
References [1] B. A. Auld, Acoustic Fields and Waves in Solids, Vol. 1, John Wiley and Sons, New York, 1973. [2] R. Chandra, A. D. Stemple, and I. Chopra, Thin-walled composite beams under bending, torsional, and extensional effects, J. Aircraft, 27 (1990), pp. 619–626. [3] L. Della Longa and A. Londero, Thin Walled Beams with Residual Stress, submitted. [4] L. Freddi, A. Morassi, and R. Paroni, Thin-walled beams: The case of the rectangular cross-section, J. Elasticity, 76 (2004), pp. 45–66. [5] H. Le Dret, Probl`emes Variationnels dans les Multi-Domaines, Recherches en Math´ematiques Appliqu´ees 19, Masson, Paris, 1991. [6] H. Le Dret, Convergence of displacements and stresses in linearly elastic slender rods as the thickness goes to zero, Asymptot. Anal., 10 (1995), pp. 367–402. Monneau, F. Murat, and A. Sili, A [7] R. Partial Korn’s Inequality and Error Estimates for the 3d-1d Dimension Reduction in Anisotropic Heterogeneous Linearized Elasticity, in preparation.
ANISOTROPIC INHOMOGENEOUS THIN-WALLED BEAMS
33
[8] F. Murat and A. Sili, Anisotropic, Heterogeneous, Linearized Elasticity Problems in Thin Cylinders, in preparation. [9] L. Trabucho and J. M. Viano, Mathematical modelling of rods, in Handbook of Numerical Analysis, Vol. 4, North–Holland, Amsterdam, 1996, pp. 487–974. [10] F. Treves, Topological Vector Spaces, Distributions and Kernels, Academic Press, New York, London, 1967. [11] V. V. Volovoi and D. H. Hodges, Theory of anisotropic thin-walled beams, J. Appl. Mech., 67 (2000), pp. 453–459. [12] V. V. Volovoi, D. H. Hodges, C. E. S. Cesnik, and B. Popescu, Assessment of beam modeling methods for rotor blade applications, Math. Comput. Modelling, 33 (2001), pp. 1099–1112. [13] R. L. Wheeden and A. Zygmund, Measure and Integral. An Introduction to Real Analysis, Monogr. Textbooks Pure Appl. Math. 43, Marcel Dekker, New York, Basel, 1977.
∗
Franc ¸ ois Murat†
Roberto Paroni‡
Abstract This paper is devoted to the asymptotic analysis of the problem of linear elasticity for an anisotropic and inhomogeneous body occupying, in its reference configuration, a cylindrical domain with a rectangular cross section with sides proportional to ε and ε2 and clamped on one of its bases. The sequence of solutions uε of the equilibrium problem is shown to converge in an appropriate topology, as ε goes to zero, to the solution of a problem for a beam in which the extensional, flexural, and torsional effects are all coupled together. Keywords: asymptotic analysis, calculus of variations, thin-walled beams, dimension reduction, variational convergence, linear elasticity
1
Introduction
Geometrically, a thin-walled beam is a slender structural element whose length is much larger than the diameter of the cross section which, on its hand, is larger than the thickness of the thin wall. These kinds of beams have been used for a long time in civil and mechanical engineering and, most of all, in flight vehicle structures because of their high ratio between maximum strength and weight. More recently, their importance has increased because of the introduction of fiber-reinforced composite materials in structural components. These materials are finding more and more applications for their high resistance to corrosion and high strength. Composite beams are usually made up by fiber-reinforced laminates and, hence, are anisotropic and inhomogeneous, even in cross-section planes. These peculiarities make classical thin-walled beam theories not applicable. The problem though has attracted the interest of several researchers ∗ Dipartimento di Matematica e Informatica, Universit` a di Udine, via delle Scienze 206, 33100 Udine, Italy, email: [email protected] † Laboratoire Jacques-Louis Lions, Universit´ e Paris VI, Boˆıte courrier 187, 75252 Paris Cedex 05, France, email: [email protected] ‡ Dipartimento di Architettura e Pianificazione, Universit` a degli Studi di Sassari, Palazzo del Pou Salit, Piazza Duomo, 07041 Alghero, Italy, email: [email protected]
1
ANISOTROPIC INHOMOGENEOUS THIN-WALLED BEAMS
2
and by now a huge number of articles can be found on the subject; see, for instance, [11] and the references therein. This is strongly remarked also in the first sentences of the abstract of [12]: “There is no lack of composite beam theories. Quite to the contrary, there might be too many of them. Different approaches, notations, etc., are used by the authors of those theories, so it is not always straightforward to compare the assumptions made and to assess the quantitative consequences of those assumptions.” The problem under study has a huge technological interest. One very suggestive, mentioned in [2], concerns the rotor blades of helicopters. The blades are composite beams and, hence, anisotropic and inhomogeneous. The anisotropy and the inhomogeneity introduce, as we shall also deduce, structural couplings between bending, extension, and twisting behaviors. It has been observed experimentally that these couplings have a powerful influence on blade dynamics including vibrations and the aeroelastic stability; see [2]. If a model of composite beams that accurately describes the structural couplings was at our disposal, then we could try to vary the anisotropy and inhomogeneity so as to minimize undesired effects like, for instance, vibrations. Through the control of lamination parameters (ply orientation and stacking sequence), it would then be possible for industry to minimize the undesired effects. Our aim here is to deduce a “rigorous” model for a composite thin-walled beam, that is, a inhomogeneous and anisotropic beam. We shall achieve our goal by means of well-established asymptotic methods starting from the threedimensional linear theory of elasticity. This paper is devoted to the asymptotic analysis of the linearized system of equilibrium equations of a body which occupies, in its reference configuration, a cylindrical domain with a rectangular cross section with sides proportional to ε and ε2 and clamped on one of its two bases. In particular, we study the compactness properties of the sequence of solutions uε of the equilibrium problems and, letting ε go to zero, we are concerned with the identification of the limit problem. The same problem has been studied from the point of view of Γconvergence: in [4] in the simpler setting of homogeneous and isotropic material and in [3] in the case of an anisotropic material which is inhomogeneous only along the longitudinal axis and subject to residual stress. Trabucho and Viano [9] also studied the same problem by superimposing two asymptotic analyses where the lengths of the two sides of the cross section go to zero independently. Besides the material properties of the body, our treatment differs from the preceding works also in the topology used in the passage from the threedimensional problem to the one-dimensional: the one used in the present paper delivers much more information on the deformation of the beam. The approach is close to the one developed in a recent paper of Murat and Sili [8] for a thin cylinder of radius ε. The lack of isotropy or homogeneity assumptions leads to a limit problem where the extensional, flexural, and torsional effects are coupled together. In fact, we prove that the limit problem can be written as a system of five equations in a 5-tuple of unknowns (u, v, w, p, q) (see Theorem 6.1) and that uε − (u + εv + ε2 w + ε3 p + ε4 q) converges strongly to zero in H 1 (Ω), under some regularity assumptions on v, w, p, and q (see Corollary 6.1). We also
ANISOTROPIC INHOMOGENEOUS THIN-WALLED BEAMS
3
derive the set of Euler equations of the variational limit problem, that is, the system of equilibrium differential equations, in the fully general case. Then we show that a strong simplification and a partial decoupling occurs when the material is homogeneous, and a complete decoupling is obtained for a homogeneous orthotropic material. Notation. Throughout this paper Ω1 , Ω2 , and Ω3 will denote the following three intervals: Ωα := (−aα /2, +aα /2) for α = 1, 2 and Ω3 := (0, `), where a1 , a2 , and ` are three positive real numbers. Also, for i, j = 1, 2, 3 we set Ωij := Ωi × Ωj and Ω := Ω1 × Ω2 × Ω3 . Unless otherwise specified, we use the Einstein summation convention. Moreover, we use the following convention for indexing vector and tensor components: Greek indices α and β take their values in the set {1, 2} and Latin indices i, j, and k in the set {1, 2, 3}. With a little abuse of notation, and because this is a common practice and does not give rise to any mistakes, we call “sequences” even those families indicized by a continuous parameter ε ∈ (0, 1). The component k of a vector v will be denoted either with (v)k or vk , and an analogous notation will be used to denote tensor components. Eαβ denotes the Ricci’s symbol, that is, E11 = E22 = 0, E12 = 1, and E21 = −1. Since usually x = (x1 , x2 , x3 ), we shall then denote by x0 := (x1 , x2 ). A wide use will be made of vector valued distributions and Sobolev spaces; for a brief account of which and for the current notation we refer the reader to the book of Le Dret [5]. Throughout this paper C will denote a constant which may change line by line.
2
The three-dimensional problem
We consider a body which occupies, in its reference configuration, the region Ωε := ε2 Ω1 × εΩ2 × Ω3 ⊂ R3 . We denote by E(u) the strain of the displacement u, whose components are µ ¶ 1 ∂ui ∂uj Eij (u) = + . 2 ∂xj ∂xi The elasticity tensor, with respect to the reference configuration Ωε , of the material will be denoted by Cε . We assume it to be essentially bounded, Cε ∈ L∞ (Ωε ; R3×3×3×3 ); to have the minor symmetries, Cεijkl = Cεjikl = Cεijlk ;
ANISOTROPIC INHOMOGENEOUS THIN-WALLED BEAMS
4
and to be positive definite. That is, there exists a constant c > 0 such that Cε A · A ≥ c|A|2 ,
(1)
for all three by three symmetric matrices A and for all ε. We consider the body clamped on Γεb := ∂Ωε ∩ {x3 = 0} and we denote by © ª 1 Hdn (Ωε ; R3 ) := ϕ ∈ H 1 (Ωε ; R3 ) : ϕ = 0 on Γεb . The weak form of the equilibrium problem can be written as ε 1 ˜ ∈ Hdn (Ωε ; R3 ), u Z Z ε ε 1 C E(˜ u ) · E(ϕ) dv = F˜ ε · E(ϕ) dv ∀ϕ ∈ Hdn (Ωε ; R3 ), Ωε
(2)
Ωε
where the matrix field F˜ ε , which takes into account the presence of external forces, is assumed to be an element of L2 (Ωε ; R3×3 sym ). If F˜ ε is not just in L2 (Ωε ; R3×3 ) but in sym 2 3 H(div, Ωε ) := {T ∈ L2 (Ωε ; R3×3 sym ) : div T ∈ L (Ωε ; R )},
then the previous problem can be seen as the weak form of the following problem: ε uε ) + ˜bε = 0 in Ωε , div C E(˜ (3) Cε E(˜ uε )nε = c˜ε on Γεc , ε ε u ˜ =0 on Γb , where Γεc := ∂Ωε \Γεb , and nε denotes the unit outward normal vector to Ωε , while the body loads ˜bε and the contact loads c˜ε are simply given by ˜bε := −divF˜ ε
in Ωε ,
c˜ε := F˜ ε nε
in Γεc .
(4)
Note that given ˜bε ∈ L2 (Ωε ; R3 ) and c˜ε ∈ H −1/2 (∂Ωε ; R3 ) it is always possible to find an F˜ ε ∈ H(div, Ωε ) which satisfies (4).
3
The rescaled problem
It is convenient to work with the domain Ω instead of the domain Ωε . We therefore rescale the problem by means of the scaling map sε : Ω → Ωε , sε (x1 , x2 , x3 ) = (ε2 x1 , εx2 , x3 ). Let u ˜ε be the solution of (2); then we define the “rescaled solution” uε by uε1 := ε2 u ˜ε1 ◦ sε ,
uε2 := ε˜ uε2 ◦ sε ,
uε3 := u ˜ε3 ◦ sε .
ANISOTROPIC INHOMOGENEOUS THIN-WALLED BEAMS Let E ε be the “rescaled strain” defined by 1 1 ε4 E11 (ϕ) ε3 E12 (ϕ) E ε (ϕ) := ε13 E21 (ϕ) ε12 E22 (ϕ) 1 ε2 E31 (ϕ)
1 ε E32 (ϕ)
1 ε2 E13 (ϕ) 1 ε E23 (ϕ)
5
.
(5)
E33 (ϕ)
It follows that E ε (uε ) = E(˜ uε ) ◦ sε . We further assume that there exists a C ∈ L∞ (Ω; R3×3×3×3 ) such that C ε = C ◦ sε , ε 2 3×3 and we denote with F ε = F˜ ε ◦ s−1 ε ∈ L (Ω; Rsym ). With this notation u turns out to be the unique solution of ε 1 3 u ∈ Hdn (Ω; R ), Z Z (6) ε ε ε 1 CE F ε · E ε (ϕ) dx ∀ϕ ∈ Hdn (Ω; R3 ), (u ) · E (ϕ) dx = Ω
Ω
where Γb := ∂Ω ∩ {x3 = 0} and n o 1 Hdn (Ω; R3 ) := ϕ ∈ H 1 (Ω; R3 ) : ϕ = 0 on Γb . By taking ϕ = uε and using (1), we find ckE ε (uε )kL2 (Ω) ≤ kF ε kL2 (Ω) .
(7)
Thus a uniform bound on kF ε kL2 (Ω) would lead to rescaled strains uniformly bounded in ε. We augment this requirement by assuming Fε → F
in L2 (Ω; R3×3 sym ).
(8)
Remark 3.1 Let S ε := diag(1/ε2 , 1/ε, 1); then E ε (ϕ) = S ε E(ϕ)S ε . If we assume F ε ∈ H(div, Ω), then we can write Z Z ε ε F · E (ϕ) dx = S ε F ε S ε · E(ϕ) dx Ω
Ω
Z div(S ε F ε S ε ) · ϕ dx + hS ε F ε S ε n, ϕi∂Ω ,
=− Ω
1 (Ω; R3 ), and conclude that instead of considering F ε we could for all ϕ ∈ Hdn have been using the following body and contact forces
bε := −div(S ε F ε S ε ) in Ω −1
−1
and cε := S ε F ε S ε n
in Γc .
Note that if F ε = S ε F (0) S ε for some F (0) ∈ H(div, Ω) the body and −1 the contact forces would be independent of ε. Since S ε = diag(ε2 , ε, 1), −1 −1 the sequence {S ε F (0) S ε }, with F (0) ∈ H(div, Ω), strongly converges in
ANISOTROPIC INHOMOGENEOUS THIN-WALLED BEAMS
6
L2 (Ω; R3×3 sym ) and hence satisfies assumption (8). Assumption (8) though allows −1 −1 us to consider “stronger” forces than F ε = S ε F (0) S ε , like F ε = F (0) or, more generally, F ε = F (0) + εF (1) + ε2 F (2) + ε3 F (3) + ε4 F (4) , (4)
with F (i) ∈ H(div, Ω) for i = 0, 1, . . . , 4, where, for instance, the term ε4 F11 would lead to the definition of body and contact forces independent of ε.
4
Partial Korn’s inequalities
In this section we state and prove several Korn’s inequalities. The proofs of Theorems 4.2 and 4.3 follow some of the lines of that of Theorem 4.4, which is due to Monneau, Murat, and Sili [7]. Theorem 4.1 There exists a constant C such that Z ° ° ° ∂u ° ° ° ° 1° ≤C° °u1 − − u1 dx1 ° 2 ° ∂x1 L2 (Ω) L (Ω) Ω1 for every u1 ∈ H 1 (Ω1 ; L2 (Ω23 )). 1 ¯ Proof. By density we may restrict ourselves to considering u1 ∈ C R (Ω). For every x2 and x3 there exists a ξ = ξ(x2 , x3 ) such that u1 (ξ, x2 , x3 ) = −Ω1 u1 (s, x2 , x3 ) ds and Z x1 ∂u1 u1 (x1 , ·, ·) − u1 (ξ, ·, ·) = (s, ·, ·) ds. ∂x1 ξ
Taking squares and applying Jensen’s inequality we conclude the proof.
2
Theorem 4.2 There exists a constant C such that ° µZ ¶° Z ° ° °u2 − − u2 dx1 − x1 ∂ − u1 dx1 ° ° ° −1 ∂x2 Ω1 Ω1 H (Ω2 ;L2 (Ω13 )) ° ° ° ¡° ¢ ≤ C °E11 (u)°L2 (Ω) + °E12 (u)°L2 (Ω) for every u ∈ H 1 (Ω; R2 ). Proof. Let Z u ¯1 := u1 − − u1 dx1 , Ω1
Z Z ∂ u ¯2 := u2 − − u2 dx1 + x1 − u1 dx1 , ∂x2 Ω1 Ω1
ANISOTROPIC INHOMOGENEOUS THIN-WALLED BEAMS
7
R and note that −Ω1 u ¯2 dx1 = 0 and u ¯2 ∈ H 1 (Ω1 ; L2 (Ω23 )). Let ψ ∈ H01 (Ω2 ); then ¶ Z ∂u2 ∂ ψu ¯2 dx2 = ψ + − u1 dx1 dx2 ∂x1 ∂x2 Ω1 Ω2 Ω2 µ ¶ Z ∂u ¯1 = dx2 ψ 2E12 (u) − ∂x2 Ω2 Z dψ = 2ψE12 (u) + u ¯1 dx2 . dx 2 Ω2 R R R Since Ω2 ψ¯ u2 dx2 ∈ H 1 (Ω1 ; L2 (Ω23 )) and Ω1 Ω2 ψ¯ u2 dx2 dx1 = 0, by Theorem 4.1 and the above equation we deduce ° °Z ° ° Z ° ° ° ∂ ° ° ° ° ≤C° ψ¯ u2 dx2 ° ψ¯ u2 dx2 ° ° ° 2 ∂x1 Ω2 Ω2 L2 (Ω) L (Ω) ∂ ∂x1
Z
Z
µ
≤ C kψkH 1 (Ω2 ) (kE12 (u)kL2 (Ω) + k¯ u1 kL2 (Ω) ) ≤ C kψkH 1 (Ω2 ) (kE12 (u)kL2 (Ω) + kE11 (u)kL2 (Ω) ). Let ϕ ∈ L2 (Ω13 ); then ¯Z ¯ °Z ¯ ¯ ° ¯ ¯ 2 ϕψ u ¯2 dx¯ ≤ kϕkL (Ω13 ) ° ¯ ° Ω
Ω2
° ° ψu ¯2 dx2 ° °
L2 (Ω13 )
≤ C kϕkL2 (Ω13 ) kψkH 1 (Ω2 ) (kE11 (u)kL2 (Ω13 ) + kE12 (u)kL2 (Ω) ). A density argument concludes the proof. Indeed, let {ϕn } be an orthonormal basis of L2 (Ω13 ) and for any v ∈ H01 (Ω2 ; L2 (Ω13 )), let Z ψn (x2 ) := ϕn (x1 , x3 )v(x) dx1 dx3 ∈ H01 (Ω2 ). Ω13
PN
Then for vN := n=1 ψn ϕn we have ¯Z ¯ ¯ ¯ ¯ vN u ¯2 dx¯¯ ≤ kvN kH01 (Ω2 ;L2 (Ω13 )) (kE11 (u)kL2 (Ω) + kE12 (u)kL2 (Ω) ), ¯ Ω
and letting N go to infinity we conclude the proof.
2
Remark 4.1 In spite of the fact that the left-hand side belongs to L2 (Ω), the inequality of Theorem 4.2 does not hold true if one replaces the norm H −1 with the norm of L2 , because of the following counterexample, which is inspired by an example contained in [7] in a quite similar framework. ¯ α ) with ϕ1 satisfying Consider two scalar smooth functions ϕα ∈ C ∞ (Ω Z Z ∂ϕ1 ϕ1 dx1 = dx1 = 0. Ω1 Ω1 ∂x1
ANISOTROPIC INHOMOGENEOUS THIN-WALLED BEAMS Define u1 := −ϕ2
∂ϕ1 , ∂x1
u2 := ϕ1
∂ϕ2 , ∂x2
8
u3 := 0.
Then u ∈ H 1 (Ω; R3 ), and the inequality of Theorem 4.2 reduces to ° ° ° ° ° ∂ϕ2 ° ° ∂ 2 ϕ1 ° °ϕ1 ° ° ° , ≤ C °ϕ2 ° ∂x2 ° −1 ∂x21 °L2 (Ω) H (Ω2 ;L2 (Ω1 )) which cannot be true if we replace H −1 by L2 because in such a case, taking kϕ1 kL2 (Ω1 ) = k∂ 2 ϕ1 /∂x21 kL2 (Ω1 ) would imply that ° ° ° ∂ϕ2 ° ° ° ≤ Ckϕ2 kL2 (Ω) ° ∂x2 ° 2 L (Ω) ¯ 2 ), which is clearly impossible. for any ϕ2 ∈ C ∞ (Ω Define (using the summation convention) rd2 = {r ∈ L2 (Ω12 ; R2 ) : ∃ c ∈ R, d ∈ R2 such that rα (y) = Eβα xβ d + cα }, where E denotes the Ricci’s symbol. The elements of rd2 are the infinitesimal rigid displacements on Ω12 . It is easy to see that rd2 ⊂ H 1 (Ω12 ; R2 ); moreover, being finite-dimensional, it is closed in L2 (Ω12 ; R2 ). Thus, the orthogonal projection operator of L2 (Ω12 ; R2 ) on rd2 , which will be denoted by ℘, is well defined. Given a vector function v ∈ L2 (Ω12 , Rm ) with m ≥ 2, we define Z Z 1 0 (9) ϑ(v) := (x1 v2 − x2 v1 ) dx , where IO = (x21 + x22 ) dx0 . IO Ω12 Ω12 If v ∈ L2 (Ω12 ; R2 ), then the components of ℘ turn out to be Z ℘α (v) = Eβα xβ ϑ(v) + − vα dx0 .
(10)
Ω12
Furthermore, the two-dimensional Korn’s inequality can be written as X kEαβ (v)kL2 (Ω12 ;R2×2 ) kv − ℘(v)kH 1 (Ω12 ;R2 ) ≤ C
(11)
α,β
for all v ∈ H 1 (Ω12 ; R2 ) with a constant C which is independent of v. Similarly, given a vector function u ∈ L2 (Ω3 ; L2 (Ω12 , Rm )) with m ≥ 2, we analogously define ϑ(u) ∈ L2 (Ω3 ) and ℘α (u). The operator ℘ associates to any u ∈ L2 (Ω3 ; L2 (Ω12 , Rm )) a function ℘(u) ∈ L2 (Ω3 ; L2 (Ω12 , R2 )) which is an infinitesimal rigid displacement on Ω12 for almost every x3 ∈ Ω3 . Let us observe that the orthogonal complement, with respect to the L2 (Ω12 ; R2 ) inner product, of rd2 in H 1 (Ω12 ; R2 ) can be then characterized as 1 2 1 2 rd⊥ 2 = {v ∈ H (Ω12 ; R ) : ℘(v) = 0} = {v ∈ Hm (Ω12 ; R ) : ϑ(v) = 0}.
ANISOTROPIC INHOMOGENEOUS THIN-WALLED BEAMS
9
Moreover, we denote by 1 RD2⊥ (Ω) = {v ∈ L2 (Ω3 ; Hm (Ω12 ; R2 )) : ϑ(v) = 0 a.e. x3 ∈ Ω3 }.
(12)
Hereafter, for any u ∈ L2 (Ω3 ; H 1 (Ω12 ; Rm )), m ≥ 2, we set u ˜α := uα − ℘α (u). (13) R Of course u ˜ ∈ L2 (Ω3 ; H 1 (Ω12 ; R2 )) and −Ω12 u ˜ dx1 dx2 = 0 and ϑ(˜ u) = 0 a.e. in Ω3 , where the latter follows from the linearity of ϑ and the fact that ϑ(u) = ϑ(℘(u)). Thus u ˜ ∈ RD2⊥ (Ω). Lemma 4.1 There exists a constant C such that X k˜ ukL2 (Ω3 ;H 1 (Ω12 ;R2 )) ≤ C kEαβ (u)kL2 (Ω3 ;L2 (Ω12 )) α,β
for every u ∈ L2 (Ω3 ; H 1 (Ω12 ; Rm )), m ≥ 2. Proof. Since uα (·, ·, x3 ) ∈ L2 (Ω12 ) for almost every x3 ∈ Ω3 , from (10) and (11) we have that the relations Z 1 ℘α (u) = Eβα xβ ϑ(u) + uα dx0 , (14) |Ω12 | Ω12 X k˜ ukH 1 (Ω12 ;R2 ) ≤ C kEαβ (u)kL2 (Ω12 ) (15) α,β
hold for almost every x3 in Ω3 , and the claimed inequality follows by integration. 2 A different proof of the lemma above can be found in Le Dret [6]. Proposition 4.1 RD2⊥ (Ω) is a Hilbert space with the norm kvkRD2⊥ (Ω) :=
µX
¶1/2 kEαβ (v)k2L2 (Ω)
.
α,β
Proof. We have only to prove that kvkRD2⊥ (Ω) is equivalent to the norm induced on RD2⊥ (Ω) by L2 (Ω3 ; H 1 (Ω12 ; R2 )) since the former space is a closed subspace of the latter. For any v ∈ RD2⊥ (Ω), recalling that ℘(v) = 0, so that v = v˜, and using Lemma 4.1, we then have kvkL2 (Ω3 ;H 1 (Ω12 ;R2 )) = k˜ v kL2 (Ω3 ;H 1 (Ω12 ;R2 )) ≤ CkvkRD2⊥ (Ω) while the opposite inequality is trivially satisfied.
2
ANISOTROPIC INHOMOGENEOUS THIN-WALLED BEAMS
10
Theorem 4.3 There exists a constant C such that ° µ ¶° Z Z ° ° °u3 − x1 x2 dϑ(u) − x1 d − u1 dx0 + − u3 dx1 ° ° ° −1 dx3 dx3 Ω12 Ω1 H (Ω3 ;L2 (Ω12 )) µX ¶ ≤C kEαβ (u)kL2 (Ω) + kE13 (u)kL2 (Ω) α,β
for every u ∈ H 1 (Ω; R3 ). Proof. Let u ˜α := uα − ℘α (u) as in (13) and µ ¶ Z Z dϑ(u) d u ˜3 := u3 − x1 x2 − x1 − u1 dx0 + − u3 dx1 . dx3 dx3 Ω12 Ω1 Since E13 (u) = E13 (˜ u),
∂u ˜3 ∂u ˜1 = 2E13 (u) − . ∂x1 ∂x3
Let ψ ∈ H01 (Ω3 ); then Z ∂ ψ˜ u3 dx3 ∂x1 Ω3
Z =
³ ∂u ˜1 ´ ψ 2E13 (u) − dx3 ∂x3 Ω3
Z
³ dψ ´ 2 ψE13 (u) + u ˜1 dx3 . dx3 Ω3 R R R Since Ω3 ψ˜ u3 dx3 ∈ H 1 (Ω) and Ω1 Ω3 ψ u ˜3 dx3 dx1 = 0, by Theorem 4.1, Lemma 4.1, and the previous equality we deduce °Z ° ° ° Z ` ° ° ° ° ∂ ° ° ° ° ψ˜ u dx ψ u ˜ dx ≤ C 3 3° 3 3° ° ° ∂x1 0 Ω3 L2 (Ω) L2 (Ω) =
≤ ≤ Let ϕ ∈ L2 (Ω12 ); ¯Z ¯ ¯ ¯ ¯ ϕψ u ¯ = ˜ dx 3 ¯ ¯ Ω
≤
then ¯Z ¯ ¯ ¯
Ω12
CkψkH 1 (Ω3 ) (kE13 (u)kL2 (Ω) + k˜ u1 kL2 (Ω) ) P CkψkH 1 (Ω3 ) (kE13 (u)kL2 (Ω) + α,β kEαβ (u)kL2 (Ω) ). Z
ϕ Ω3
¯ °Z ¯ ° ψu ˜3 dx3 dx0 ¯¯ ≤ kϕkL2 (Ω12 ) ° °
Ω3
° ° ψ˜ u3 dx3 ° °
L2 (Ω12 )
¶ µ X kEαβ (u)kL2 (Ω) . CkϕkL2 (Ω12 ) kψkH 1 (Ω3 ) kE13 (u)kL2 (Ω) + α,β
Arguing as in the proof of Theorem 4.2, we conclude the proof.
2
ANISOTROPIC INHOMOGENEOUS THIN-WALLED BEAMS
11
Remark 4.2 As in Remark 4.1, in spite of the fact that the left-hand side belongs to L2 (Ω), the inequality of Theorem 4.3 does not hold true if one replaces the norm H −1 with the norm of L2 , because of the following counterexample. ¯ i ) with ϕα satisfying Consider three scalar smooth functions ϕi ∈ C ∞ (Ω Z Z ∂ϕα ϕα dxα = dxα = 0. Ωα Ωα ∂xα Define u1 := −ϕ2
∂ϕ1 ϕ3 , ∂x1
∂ϕ2 ϕ3 , ∂x2
u2 := −ϕ1
u3 := +ϕ1 ϕ2
∂ϕ3 . ∂x3
Then u ∈ H 1 (Ω; R3 ), and the inequality of Theorem 4.3 reduces to ° ° ° µ 2 ¶° 2 ° ° ° ° °ϕ3 ∂ ϕ1 ϕ2 + ϕ1 ∂ ϕ2 ° °ϕ1 ϕ2 ∂ϕ3 ° ≤ C , 2 2 ° ° ° ∂x3 H −1 (Ω2 ;L2 (Ω1 )) ∂x1 ∂x2 °L2 (Ω) which cannot be true if we replace H −1 by L2 , because in such a case taking kϕα kL2 (Ωα ) = k∂ 2 ϕα /∂xα 1 kL2 (Ωα ) would imply that ° ° ° ∂ϕ3 ° ° ° ≤ Ckϕ3 kL2 (Ω) ° ∂x3 ° 2 L (Ω) ¯ 3 ), which is clearly impossible. for any ϕ3 ∈ C ∞ (Ω The next partial Korn’s inequality is proved in Monneau, Murat, and Sili [7]. Theorem 4.4 There exists a constant C such that ° ¶° µZ Z ° ° °u3 − − u3 dx0 − xα d − uα dx0 ° ° ° −1 dx3 Ω12 Ω12 H (Ω3 ;L2 (Ω12 )) µX ¶ X kEαβ (u)kL2 (Ω) + ≤C kEα3 (u)kL2 (Ω) α
αβ 1 for every u ∈ Hdn (Ω; R3 ).
5
Limit strain characterization
Let 1 Hm (Ω12 ) := {z ∈ H 1 (Ω12 ) :
R
z dx0 = 0}, R 1 Hm (Ω1 ; L2 (Ω23 )) := {z ∈ H 1 (Ω1 ; L2 (Ω23 )) : Ω1 z dx1 = 0 a.e. in Ω23 }, 1 Ω12
−1 Hm (Ω3 ; L2 (Ω12 )) := {v ∈ H −1 (Ω3 ; L2 (Ω12 )) : hz3 , ϕi = 0 1
∀ ϕ ∈ H01 (Ω3 ; L2 (Ω2 ))},
ANISOTROPIC INHOMOGENEOUS THIN-WALLED BEAMS
12
where the bracket in the last definition has to be understood in the sense of the duality H −1 (Ω3 ; L2 (Ω12 )) × H01 (Ω3 ; L2 (Ω12 )). Let (uε ) be the sequence of solutions to problems (6). From (7) and assumption (8) it follows that sup kE ε (uε )kL2 (Ω) < +∞.
(16)
ε
Hence, possibly passing to a subsequence, we have that E ε (uε ) * E
in L2 (Ω; R3×3 sym )
for some E ∈ L2 (Ω; R3×3 sym ). In this section we characterize the limit strain E. For clarity we state several lemmas. Lemma 5.1 (component 33) There exists a function u ¯ in the set of the socalled Bernoulli–Navier displacements 1 U := {u ∈ Hdn (Ω; R3 ) : Eαi (u) = 0}
such that
∂u ¯3 = E33 (¯ u). ∂x3 Proof. From (16), the structure of E ε , and Korn’s inequality, we have that E33 =
C ≥ kE ε (uε )kL2 (Ω) ≥ kE(uε )kL2 (Ω) ≥ CK kuε kH 1 (Ω) , where C is a constant independent of ε and CK is Korn’s constant. Hence, up 1 to a subsequence uε * u ¯ in H 1 (Ω; R3 ), for some u ¯ ∈ Hdn (Ω; R3 ). The claim ε ε ε 2 follows by noticing that kEαi (u )kL (Ω) ≤ Cε, and E33 (u ) = ∂uε3 /∂x3 . In fact 1 it follows that u ¯ ∈ {u ∈ Hdn (Ω; R3 ) : Eαi (u) = 0}. 2 Remark 5.1 (representation of the space U ) It is well known (see, for instance, Le Dret [5]) that the space of Bernoulli–Navier displacements admits the following representation: ½ 2 1 2 1 U := u ∈ Hdn (Ω)2 × Hdn (Ω) : exists ζ ∈ Hdn (Ω3 )2 × Hdn (Ω3 ) such that u1 = ζ1 , u2 = ζ2 , u3 = ζ3 − x1 Moreover, U is a Hilbert space with the norm kukU := kE33 (u)kL2 (Ω) , 1 (Ω; R3 ) (see [8]). which is equivalent to that induced by Hdn
dζ1 dζ2 − x2 dx3 dx3
¾ .
ANISOTROPIC INHOMOGENEOUS THIN-WALLED BEAMS
13
Lemma 5.2 (component 23) There exists a function v¯ in the space ½ 1 1 (Ω)2 × L2 (Ω3 ; Hm (Ω12 )) : exist ϑ ∈ H 1 (Ω3 ) such that ϑ(0) = 0 V := v ∈ Hdn and % ∈ L2 (Ω3 ; H 1 (Ω2 )) such that v1 (x) = −x2 ϑ(x3 ), v2 (x) = x1 ϑ(x3 ), ¾ dϑ (x3 ) + %(x2 , x3 ) v3 (x) = x1 x2 dx3 such that
E23 = E23 (¯ v ). Proof. Let viε := 1ε uεi and ϑε := ϑ(v ε ) (see (9)). First of all, by adapting an argument of [4] and Lemmas 4.4 and 4.5, we prove that there exists ϑ ∈ H 1 (Ω3 ) with ϑ(0) = 0, such that, up to subsequences, ϑε → ϑ in L2 (Ω3 ). Applying Lemma 4.1, there exists a constant C such that X k˜ v ε kL2 (Ω3 ;H 1 (Ω12 ;R2 )) ≤ C kEαβ (v ε )kL2 (Ω3 ;L2 (Ω12 )) .
(17)
(18)
α,β
Since, furthermore, ε ε ε (uε ), (uε ), E22 (v ε ) = εE22 (uε ), E12 (v ε ) = ε2 E12 E11 (v ε )11 = ε3 E11
and using (16) we have ε kEαβ (v ε )kL2 (Ω) ≤ εkEαβ (uε )kL2 (Ω) ≤ C ε.
From (18) we get hence Now let η ∈
Cc∞ (Ω12 )
(19)
k˜ v ε kL2 (Ω3 ;H 1 (Ω12 ;R2 )) ≤ Cε;
(20)
v˜ε → 0 in L2 (Ω3 ; H 1 (Ω12 ; R2 )).
(21)
be such that Z η dx0 = − Ω12
IO . 2
ANISOTROPIC INHOMOGENEOUS THIN-WALLED BEAMS Then, taking into account (14), we have Z Z IO ϑε = −2ϑε η dx0 = −ϑε Ω12
14
ηDα xα dx0
Ω12
Z
Z
= ϑε
Dα η xα dx0 = ϑε
Eαγ Eβγ Dα η xβ dx0
Ω12
Ω12
Z Eαγ Dα ηEβγ xβ ϑε dx0
= Ω12
µ Eαγ Dα η ℘γ (v ε ) −
Z = Ω12
1 |Ω12 |
Z Ω12
¶ vγε dx0 dx0
Z Eαγ Dα η ℘γ (v ε ) dx0
= Ω12
Z
Z
= Ω12
Eαγ Dα η vγε dx0 −
Hence, denoting by
1 ϑ˜ε = IO
¡ ¢ Eαγ Dα η v ε − ℘γ (v ε ) dx0 .
Ω12
Z Ω12
Eαγ Dα η wγε dx0
and recalling (21), we find ϑε − ϑ˜ε → 0 in L2 (Ω3 ).
(22)
We now show that D3 ϑ˜ε is bounded in L2 . Since Eαγ Dα Dγ η = 0 everywhere in Ω3 and Dα η = 0 on ∂Ω3 , we have Z Z IO D3 ϑ˜ε = Eαγ Dα η D3 vγε dx0 = 2 Eαγ Dα η Eγ3 (v ε ) dx0 , Ω12
Ω12
ε ε (uε ) and E23 (v ε ) = E23 (uε ), and therefore D3 ϑ˜ε is bounded but E13 (v ε ) = εE13 2 ε ε ˜ ˜ in L (Ω3 ). Since ϑ (0) = 0, ϑ is then bounded in H 1 (Ω3 ) so that there exists ϑ ∈ H 1 (Ω3 ) with ϑ(0) = 0, such that, up to subsequences,
ϑ˜ε * ϑ in H 1 (Ω3 ). Thus, from (22) we obtain (17). Let us now set Z v¯1ε := v1ε − − v1ε dx0 ,
Z v¯2ε := v2ε − −
Ω12
and v¯3ε
:=
v3ε
µZ − − Ω12
v3ε
Ω12
v2ε dx0 ,
¶ Z d ε 0 − v dx . dx − xα dx3 Ω12 α 0
ANISOTROPIC INHOMOGENEOUS THIN-WALLED BEAMS
15
Observing that, by the definitions, v¯1ε = v˜1ε − x2 ϑε ,
v¯2ε = v˜2ε + x1 ϑε ,
from (17) and (21) we have that v¯1ε → −x2 ϑ,
v¯2ε → +x1 ϑ
in L2 (Ω3 ; H 1 (Ω12 )).
By Theorem 4.4, we have that k¯ v3ε kH −1 (Ω3 ;L2 (Ω12 )) ≤ C, and hence there exists v¯3 ∈ H −1 (Ω3 ; L2 (Ω12 )) such that, up to subsequences, v¯3ε * : v¯3 in H −1 (Ω3 ; L2 (Ω12 )).
(23)
Moreover, let us set v¯1 = −x2 ϑ,
v¯2 = +x1 ϑ,
and check that the vector field v¯ so defined satisfies the properties claimed in the statement of Lemma 4.1. A simple computation shows that E13 (¯ v ε ) = E13 (v ε ) and
E23 (¯ v ε ) = E23 (v ε ).
ε (uε ) → 0 in L2 (Ω) and E13 (¯ v ε ) → E13 (¯ v ) in Noticing that E13 (v ε ) = εE13 0 D (Ω), we obtain that E13 (¯ v ) = 0. Hence,
∂¯ v3 ∂¯ v1 dϑ =− = x2 ∂x1 ∂x3 dx3 and, integrating with respect to x1 , v¯3 = x1 x2
dϑ + %(x2 , x3 ) dx3
for some function % ∈ L2 (Ω3 ; H 1 (Ω2 )). Moreover, v¯3 ∈ L2 (Ω3 ; H 1 (Ω R 12 )) and, from (23) and the fact that v¯3 , v¯3ε ∈ L2 (Ω), we then obtain easily that −Ω12 v¯3 dx0 = 0, which concludes the proof. 2 Lemma 5.3 (characterization of the space V ) The space V admits the following characterization: 1 1 V = {v ∈ Hdn (Ω)2 × L2 (Ω3 ; Hm (Ω12 )) : Eαβ (v) = 0, E13 (v) = 0, R E23 (v) ∈ L2 (Ω) and Ω12 vα dx0 = 0 a.e.}.
Moreover, it is a Hilbert space with the norm kvkV := kW13 (v)kL2 (Ω) + kE23 (v)kL2 (Ω) , where W13 (v) =
µ ¶ ∂v3 1 ∂v1 − . 2 ∂x3 ∂x1
(24)
ANISOTROPIC INHOMOGENEOUS THIN-WALLED BEAMS
16
Proof. Let us call V the space at the right-hand side of equality (24) and let V be as in the statement of Lemma 5.2. It is trivial to check that V ⊆ V . Let 1 us prove the opposite inclusion. Let v ∈ V . Since vα ∈ Hdn (Ω) and Eαβ (v) = 0, 2 by integration there exists ϑ ∈ L (Ω3 ) such that v1 = −x2 ϑ(x3 ) + a1 (x3 ),
v2 = x1 ϑ(x3 ) + a2 (x3 ),
R
and since Ω12 vα dx0 = 0, we have a1 = a2 = 0 a.e.. From the resulting 1 expression of vα it follows that ϑ ∈ Hdn (Ω) and ϑ(0) = 0. Since E13 (v) = 0 we obtain that ∂v3 /∂x1 = x2 ϑ0 (x3 ). Then there exists % ∈ L2 (Ω23 ) such that v3 = x1 x2 ϑ0 (x3 ) + %(x2 , x3 ), 1 from which it follows also that % ∈ L2 (Ω3 ; Hm (Ω2 )) and that E23 (v) ∈ L2 (Ω). 1 The last part of the claim follows from the fact that L2 (Ω3 ; Hm (Ω12 ) is a Hilbert space with the scalar product Z 1 (Ω )) = 1 (Ω ) dx3 , hu3 , v3 iL2 (Ω3 ;Hm hu3 (x3 ), v3 (x3 )iHm 12 12 Ω3
1 1 and V is a closed subspace of Hdn (Ω)2 × L2 (Ω3 ; Hm (Ω12 )) which is Hilbert with the product norm 1 (Ω) + kv2 kH 1 (Ω) + kv3 kL2 (Ω ;H 1 (Ω )) kv1 kHdn 3 12 m dn
1 1 induced by Hdn (Ω)2 × L2 (Ω3 ; Hm (Ω12 )). The proof that this norm is equivalent to kvkV is an easy consequence of the Poincar´e inequality, the representation lemma, Lemma 5.2, and the characterization of the space V proved above. 2
Lemma 5.4 (components 22 and 13) There exists a function w ¯ in the space −1 W = {w ∈ RD2⊥ (Ω) × Hm (Ω3 ; L2 (Ω12 )) : 1
E11 (w) = E12 (w) = 0, E13 (w) ∈ L2 (Ω)}
such that E22 =
∂w ¯2 = E22 (w), ¯ ∂x2
E13 =
1 2
µ
∂w ¯3 ∂w ¯1 + ∂x3 ∂x1
¶ = E13 (w). ¯
Moreover, W is a Hilbert space with the norm −1 kwkW := kE22 (w)kL2 (Ω) + kE13 (w)kL2 (Ω) + kw3 kHm (Ω3 ;L2 (Ω12 )) . 1 Proof. Let 1 wε := 2 uε ; ε
ANISOTROPIC INHOMOGENEOUS THIN-WALLED BEAMS then
kE11 (wε )kL2 (Ω) ≤ Cε2 ,
kE12 (wε )kL2 (Ω) ≤ Cε,
kE22 (wε )kL2 (Ω) ≤ C,
kE13 (wε )kL2 (Ω) ≤ C.
17
(25)
Let us recall that w ˜αε := wαε − ℘α (wε ). By Lemma 4.1 and using (25), we have X kw ˜ ε kL2 (Ω3 ;H 1 (Ω12 ;R2 )) ≤ C kEαβ (wε )kL2 (Ω) ≤ C, α,β
R −Ω12 w ˜ ε dx0 = 0, and ϑ(w ˜ ε ) = 0. Thus, up to a subsequence, w ˜αε * w ¯α moreover, a.e.,
in L2 (Ω3 ; H 1 (Ω12 )) for α = 1, 2;
Z −
w ¯α dx0 = 0, ϑ(w) ¯ = 0.
(26)
(27)
Ω12
Hence (w ¯1 , w ¯2 ) ∈ RD2⊥ (Ω) (see (12)). Using (25) we also obtain E11 (w) ¯ = E12 (w) ¯ = 0. Moreover, ε E22 (uε ) = E22 (wε ) = E22 (w ˜ ε ) * E22 (w) ¯ in L2 (Ω).
Let w ˜3ε
:=
w3ε
µ ¶ Z Z dϑ(wε ) d ε 0 ε − x1 x2 − x1 − w dx + − w3 dx1 . dx3 dx3 Ω12 1 Ω1
By Theorem 4.3 we have that X kEαβ (wε )kL2 (Ω) + kE13 (wε )kL2 (Ω) ≤ C. kw ˜3ε kH −1 (Ω3 ;L2 (Ω12 )) ≤ C α,β
Hence, up to a subsequence, w ˜3ε * w3
in H −1 (Ω3 ; L2 (Ω12 )).
Taking into account that (see (16)) Z ℘1 (wε ) = − w1ε dx0 − x2 ϑ(wε ), Ω12
Z ℘2 (wε ) = − Ω12
w2ε dx0 + x1 ϑ(wε ),
ε we easily deduce that E13 (uε ) = E13 (wε ) = E13 (w ˜ ε ) and hence that µ ¶ 1 ∂w ¯1 ∂w ¯3 E13 = + . 2 ∂x3 ∂x1 R Finally, since −Ω1 w ˜3ε dx1 = 0, we have that
hw ¯3 , ϕiH −1 (Ω3 ;L2 (Ω12 ))×H01 (Ω3 ;L2 (Ω12 )) = 0
ANISOTROPIC INHOMOGENEOUS THIN-WALLED BEAMS
18
for all ϕ ∈ H01 (Ω3 ; L2 (Ω2 )). The last part of the claim follows from the fact that W is a closed subspace of −1 {z ∈ RD2⊥ (Ω) × Hm (Ω3 ; L2 (Ω12 )) : E13 (z) ∈ L2 (Ω)} 1 which, in turn, is a Hilbert space under the scalar product Z −1 hz, ζi := hz, ζiRD⊥ (Ω)×Hm + E13 (z)E13 (ζ) dx (Ω3 ;L2 (Ω12 )) 2
1
Ω
and from an application of Proposition 4.1.
2
Lemma 5.5 (representation of the space W ) The space W admits the following representation: ½ −1 W = w ∈ L2 (Ω3 ; H 1 (Ω12 ))2 × Hm (Ω3 ; L2 (Ω12 )) : E13 (w) ∈ L2 (Ω), 1 there exists η1 ∈ L2 (Ω3 ; H 2 (Ω2 )) and η2 ∈ L2 (Ω3 ; H 1 (Ω2 )) such that ∂η1 (x2 , x3 ) + η2 (x2 , x3 ), ∂x2 ¶ ¾ Z µ 2 a1 ∂η1 + x2 η1 dx2 = 0 . 12 ∂x2 Ω2
w1 (x) = η1 (x2 , x3 ), Z ηα dx2 = 0, Ω2
w2 (x) = −x1
(28) Proof. Let us call W the set on the right-hand side of equality (28), and let W be as in the statement of Lemma 5.4. Then it is trivial to check that W ⊆ W . Let us prove the converse inequality. Let w ∈ W as in Lemma 5.4. Since E11 (w) = E12 (w) = 0, by integration we deduce that there exist η1 ∈ L2 (Ω3 ; H 2 (Ω2 )) and η2 ∈ L2 (Ω3 ; H 1 (Ω2 )) such that w1 (x) = η1 (x2 , x3 ), Since
Z −
w2 (x) = −x1
wα dx0 = 0,
∂η1 (x2 , x3 ) + η2 (x2 , x3 ). ∂x2
ϑ(w) = 0,
(29)
Ω12
we have that
Z ηα dx2 = 0
a.e. for α = 1, 2,
Ω2
and
Z Ω2
a.e..
µ
¶ a21 ∂η1 + x2 η1 dx2 = 0, 12 ∂x2 2
ANISOTROPIC INHOMOGENEOUS THIN-WALLED BEAMS
19
Lemma 5.6 (component 12) There exists a vector function p¯ in the set 1 P = {0} × Hm (Ω1 ; L2 (Ω23 )) × {0} 1
such that E12 =
1 ∂ p¯2 = E12 (¯ p). 2 ∂x1
(30)
Moreover, P is a Hilbert space with the norm kpkP := kE12 (p)k. Proof. Let pεα :=
1 ε u ε3 α
for α = 1, 2,
and p¯ε1
:=
pε1
Z − − pε1 dx1 ,
p¯ε2
Ω1
:=
Z Z ∂ ε − − p2 dx1 + x1 − pε dx1 . ∂x2 Ω1 1 Ω1
pε2
ε (uε ); hence, by Theorem 4.1 we have Then E11 (¯ pε ) = E11 (pε ) = εE11
p¯ε1 → 0,
∂ p¯ε1 →0 ∂x1
in L2 (Ω).
ε (uε ), by Theorem 4.2 we also have that, up to Since E12 (¯ pε ) = E12 (pε ) = E12 subsequences, p¯ε2 * p¯2 in H −1 (Ω2 ; L2 (Ω13 )).
Setting p¯ := (0, p¯2 , 0), we have E12 = E12 (¯ p) =
1 ∂ p¯2 , 2 ∂x1
R that is, (30). It remains then to prove that p¯2 ∈ H 1 (Ω1 ; L2 (Ω23 )) and that −Ω1 p¯2 dx1 = 0. Since p¯2 ∈ H −1 (Ω2 ; L2 (Ω13 )), for any ϕ ∈ H01 (Ω2 ) and any ψ ∈ L2 (Ω13 ) the product ϕ ⊗ ψ belongs to H01 (Ω2 ; L2 (Ω13 )), and the linear map Pϕ : L2 (Ω13 ) → R,
hPϕ , ψi := h¯ p2 , ϕ ⊗ ψi
satisfies the estimate |hPϕ , ψi| ≤ k¯ p2 kH −1 kϕkH01 kψkL2 . Thus Pϕ ∈ L2 (Ω13 ). Moreover, from the definition of Pϕ and the fact that ∂Pϕ ∂ p¯2 2 2 ∂x1 ∈ L (Ω), we obtain that ∂x1 ∈ L (Ω13 ) and also that ∂Pϕ = ∂x1
Z Ω2
∂ p¯2 ϕ dx2 . ∂x1
(31)
ANISOTROPIC INHOMOGENEOUS THIN-WALLED BEAMS
20
R Since −Ω1 p¯ε2 dx1 = 0 it follows from the definitions that Z − Pϕ (s, x3 ) ds = h¯ p2 , ϕi = 0
(32)
Ω1
for almost every x3 ∈ Ω3 ; using this fact, the following Poincar´e inequality holds: ° ° ° ∂Pϕ ° ° , kPϕ (·, x3 )kL2 (Ω1 ) ≤ C ° (·, x ) 3 ° ° ∂x1 L2 (Ω1 ) where the constant C depends only on the domain Ω1 and is therefore independent of x3 . By substituting (31) inside the Poincar´e inequality, we obtain that ° ° ° ∂ p¯2 ° ° kPϕ kL2 (Ω13 ) ≤ C ° (33) ° ∂x1 ° 2 kϕkL2 (Ω2 ) . L (Ω) Using the density of Cc∞ (Ω2 ) ⊗ Cc∞ (Ω13 ) in L2 (Ω) (see, for instance, Treves [10, Theorem 39.2 and subsequent Corollary 3]), the fact that Pϕ ∈ L2 (Ω13 ), and inequality (33), we have that k¯ p2 kL2 (Ω)
=
|h¯ p2 , ϕψi| ψ∈Cc∞ (Ω13 ) kϕkL2 (Ω2 ) kψkL2 (Ω13 )
sup ϕ∈Cc∞ (Ω2 ),
|hPϕ , ψi| ϕ∈Cc∞ (Ω2 ), ψ∈Cc∞ (Ω13 ) kϕkL2 (Ω2 ) kψkL2 (Ω13 ) ° ° ° ∂ p¯2 ° kPϕ kL2 (Ω13 ) ° ° ≤ sup ≤ C° ; ∂x1 °L2 (Ω) ϕ∈Cc∞ (Ω2 ) kϕkL2 (Ω2 ) =
sup
R hence p¯2 ∈ L2 (Ω). Thus p¯2 ∈ H 1 (Ω1 ; L2 (Ω23 )), and (32) implies −Ω1 p¯2 dx1 = 0. 1 The last part of the claim follows from the fact that Hm (Ω1 ; L2 (Ω23 )) is a 1 Hilbert space with the norm ° ° ° ∂p2 ° ° ° 1 (Ω ;L2 (Ω )) := kpkHm 1 23 ° ∂x1 ° 2 , 1 L (Ω) which, by Theorem 4.1, turns out to be equivalent to the canonical one. Lemma 5.7 (component 11) There exists a function q¯ in the space 1 Q := Hm (Ω1 ; L2 (Ω23 )) × {0}2 1
such that E11 =
∂ q¯1 = E11 (¯ q ). ∂x1
Moreover, Q is a Hilbert space with the norm kqkQ := kE11 (q)k.
2
ANISOTROPIC INHOMOGENEOUS THIN-WALLED BEAMS Proof. Let q1ε := then
21
µ ¶ Z 1 ε ε ; u − − u dx 1 1 ε4 1 Ω1
° ε° ° ∂q1 ° ε ° sup ° = sup kE11 (uε )kL2 (Ω) ≤ C, ° ∂x1 ° 2 ε ε L (Ω)
and by Theorem 4.1 we have supε kq1ε kL2 (Ω) ≤ C. Then, up to a subsequence, ε (uε ) = ∂ q¯1ε /∂x1 * ∂ q¯1 /∂x1 in L2 (Ω) for some q1ε * q¯1 in L2 (Ω), and E11 2 q¯1 ∈ L (Ω). 1 The last part of the claim follows from the fact that Hm (Ω1 ; L2 (Ω23 )) is a 1 Hilbert space with the norm 1 kqkHm
1
(Ω1 ;L2 (Ω23 ))
:= kq1,1 kL2 (Ω) ,
which, by Theorem 4.1, turns out to be equivalent to the canonical one.
6
2
The limit problem
Let us consider the space A := U × V × W × P × Q. According to the notation and the results proved in the previous section, A is a Hilbert space when endowed with the product norm k(u, v, w, p, q)kA := kukU + kvkV + kwkW + kpkP + kqkQ . Given a 5-tuple of vector valued distributions (u, v, w, p, q) ∈ D0 (Ω; R3 )5 , let us define E11 (q) E12 (p) E13 (w) E22 (w) E23 (v) . E(u, v, w, p, q) := (34) Sym. E33 (u) We are now in a position to state the main result of this paper. Theorem 6.1 Let C be a positive definite fourth order tensor field on Ω with the minor symmetries, i.e., Cijkl = Cjikl = Cijlk . Let F ε be a second order symmetric tensor field which belongs to L2 (Ω; R3×3 ). Then problem (6), that is, ε 1 3 u ∈ Hdn (Ω; R ), Z Z (35) 1 CE ε (uε ) · E ε (ϕ) dx = F ε · E ε (ϕ) dx ∀ϕ ∈ Hdn (Ω; R3 ), Ω
Ω
admits a unique solution uε . Moreover, if F ε → F in L2 (Ω; R3×3 ), then we have the following:
ANISOTROPIC INHOMOGENEOUS THIN-WALLED BEAMS
22
1. the problem Z Z CE(¯ u, v¯, w, ¯ p¯, q¯) · E(u, v, w, p, q) dx = F · E(u, v, w, p, q) dx Ω
Ω
∀ (u, v, w, p, q) ∈ A (36) admits a unique solution (¯ u, v¯, w, ¯ p¯, q¯) ∈ A ; 2. uε * u ¯ in H 1 (Ω; R3 ); 3. E ε (uε ) → E(¯ u, v¯, w, ¯ p¯, q¯) in L2 (Ω; R3×3 ). The following corollary can be seen as a corrector result. Corollary 6.1 If the solution (¯ u, v¯, w, ¯ p¯, q¯) of problem (36) is such that ∂¯ v3 ∂w ¯3 ∂ q¯1 ∂ q¯1 ∂ p¯2 ∂ p¯2 , E23 (w), ¯ , , , , ∈ L2 (Ω), ∂x3 ∂x3 ∂x2 ∂x3 ∂x2 ∂x3 then where Proof. Since
(37)
kE ε (uε ) − E ε (¯ uε )kL2 (Ω;R3×3 ) → 0, u ¯ε := u ¯ + ε¯ v + ε2 w ¯ + ε3 p¯ + ε4 q¯.
E12 (¯ q) 0 0 0 E13 (¯ q) E22 (¯ p) E23 (w) ¯ +ε2 0 E23 (¯ p) E ε (¯ uε ) = E(¯ u, v¯, w, ¯ p¯, q¯)+ε Sym. E33 (¯ v) Sym. E33 (w) ¯ 0
the additional regularity assumptions imply that E ε (¯ uε ) ∈ L2 (Ω; R3×3 ) and kE ε (¯ uε ) − E(¯ u, v¯, w, ¯ p¯, q¯)kL2 (Ω;R3×3 ) → 0. Then the claim follows from step 3 of Theorem 6.1.
2
1 In order to prove Theorem 6.1, we introduce the subspaces of Hdn (Ω; R3 ), 1 Uˆ := {u ∈ Hdn (Ω; R3 ) : Eαβ (u) = Eα3 (u) = 0}, 1 Vˆ := {v ∈ Hdn (Ω; R3 ) : Eαβ (v) = E13 (v) = 0}, 1 Wˆ := {w ∈ Hdn (Ω; R3 ) : E1β (w) = 0}, 1 Pˆ := {p ∈ Hdn (Ω; R3 ) : E11 (p) = 0}, 1 Qˆ := Hdn (Ω; R3 ),
ˆ and define Aˆ := Uˆ × Vˆ × Wˆ × Pˆ × Q. ˆ Let us note that U = U , but similar equalities are not true for the spaces V , W , P, and Q. Nevertheless, for such spaces we can prove the following approximation lemma.
ANISOTROPIC INHOMOGENEOUS THIN-WALLED BEAMS
23
Lemma 6.1 For every v ∈ V , w ∈ W , p ∈ P, and q ∈ Q there exist sequences ˆ and (ˆ (ˆ v n ) in Vˆ, (w ˆ n ) in Wˆ , (ˆ pn ) in P, q n ) in Qˆ such that the following 2 convergences hold in the norm of L (Ω): (i) E23 (ˆ v n ) → E23 (v), (ii) E13 (w ˆ n ) → E13 (w) and E22 (w ˆ n ) → E22 (w), pn ) → E12 (p), (iii) E12 (ˆ (iv) E11 (ˆ q n ) → E11 (q). Proof. Let us prove (i). Since v ∈ V (see Lemma 5.2), there exist ϑ ∈ H 1 (Ω3 ) with ϑ(0) = 0 and % ∈ L2 (Ω3 ; H 1 (Ω2 )) such that v1 (x) = −x2 ϑ(x3 ), v3 (x) = x1 x2
v2 (x) = x1 ϑ(x3 ),
dϑ (x3 ) + %(x2 , x3 ). dx3
2 1 Then there exist sequences ϑn ∈ Hdn (Ω3 ) and %n ∈ Hdn (Ω23 ) such that
ϑn → ϑ in H 1 (Ω3 ), Setting
%n → % in L2 (Ω3 ; H 1 (Ω2 )).
vˆ1n (x) = −x2 ϑn (x3 ), vˆ3n (x) = x1 x2
vˆ2n (x) = x1 ϑn (x3 ),
dϑn (x3 ) + %n (x2 , x3 ), dx3
we obtain the claim. Let us prove (ii). As w ∈ W (see Lemma 5.4), there exist η1 ∈ L2 (Ω3 ; H 2 (Ω2 )) and η2 ∈ L2 (Ω3 ; H 1 (Ω2 )) such that w1 (x) = η1 (x2 , x3 ),
w2 (x) = −x1
∂η1 (x2 , x3 ) + η2 (x2 , x3 ). ∂x2
2 (Ω) with ∂η1n /∂x1 = 0 such that Then, there exist a sequence η1n ∈ Hdn
η1n → η1
in L2 (Ω3 ; H 2 (Ω2 ))
1 (Ω) with ∂η2n /∂x1 = 0 such that and a sequence η2n ∈ Hdn
η2n → η2
in L2 (Ω3 ; H 1 (Ω2 )).
Since ∂w3 /∂x1 = 2E13 (w)−∂η1 /∂x3 and E13 (w) ∈ L2 (Ω), by integration, there exists G13 ∈ H 1 (Ω1 ; L2 (Ω23 )) such that ∂G13 = E13 (w) ∂x1
and
w3 = 2G13 − x1
∂η1 , ∂x3
where we have also used the fact that η1 does not depend on x1 .
ANISOTROPIC INHOMOGENEOUS THIN-WALLED BEAMS
24
1 We may also find a sequence Gn13 ∈ Hdn (Ω) such that
Gn13 → G13
in H 1 (Ω1 ; L2 (Ω23 ))
and define w ˆ1n := η1n ,
w ˆ2n := −x1
∂η1n + η2n , ∂x2
w ˆ3n := Gn13 − x1
∂η1n . ∂x3
1 Then w ˆ n ∈ Hdn (Ω), E11 (w ˆ n ) = ∂η1n /∂x1 = 0, and E12 (w ˆ n ) = 0 so that w ˆn ∈ Wˆ . Moreover,
E13 (w ˆn) =
∂Gn13 ∂G13 → = E13 (w) ∂x1 ∂x1
in L2 (Ω),
and E22 (w ˆn) =
∂η n ∂w ˆ2n ∂ 2 η1n ∂w ˆ2 + 2 → = −x1 = E22 (w) ˆ 2 ∂x2 ∂x2 ∂x2 ∂x2
in L2 (Ω).
To prove (iii) it is enough to consider, for a given p = (0, p2 , 0) ∈ P, 1 (Ω), which converges to p2 in the norm of H 1 (Ω1 ; L2 (Ω23 )), a sequence pˆn2 ∈ Hdn and set pˆn 1 n = (0, pˆ2 , 0). Finally, claim (iv) simply follows from the density of Hdn (Ω) 1 in Hm (Ω1 ; L2 (Ω23 )). 2 1 Proof of Theorem 6.1. The existence and uniqueness of the solution of problem (36) follows from an application of the Lax–Milgram lemma to the symmetric bilinear form defined on A by Z a[(u, v, w, p, q), (ˆ u, vˆ, w, ˆ pˆ, qˆ)] := CE(u, v, w, p, q) · E(ˆ u, vˆ, w, ˆ pˆ, qˆ) dx, Ω
which is continuous and coercive with respect to the Hilbertian norm on A defined at the beginning of this section. Part 2 of the statement of Theorem 6.1 is actually a consequence of step 3. Let us now prove part 3. According to the results proved in the previous section, we have E ε (uε ) * E in L2 (Ω; R3×3 ), (38) and there exists a (¯ u, v¯, w, ¯ p¯, q¯) ∈ A such that E11 (¯ q ) E12 (¯ p) E13 (w) ¯ E22 (w) ¯ E23 (¯ v ) = E(¯ E= u, v¯, w, ¯ p¯, q¯). Sym. E33 (¯ u) The result will be achieved in two steps: (i) we prove that (¯ u, v¯, w, ¯ p¯, q¯) satisfies equality (36) and therefore coincides with the unique solution of the variational problem, and (ii) we show that the convergence in (38) is indeed strong.
ANISOTROPIC INHOMOGENEOUS THIN-WALLED BEAMS
25
Let (ˆ u, vˆ, w, ˆ pˆ, qˆ) ∈ Aˆ and set ϕˆε := u ˆ + εˆ v + ε2 w ˆ + ε3 pˆ + ε4 qˆ; 1 then ϕˆε ∈ Hdn (Ω; R3 ) and an easy computation shows that, as ε → 0, E11 (ˆ q ) E12 (ˆ p) E13 (w) ˆ E22 (w) ˆ E23 (ˆ v ) = E(ˆ E ε (ϕˆε ) → u, vˆ, w, ˆ pˆ, qˆ) Sym. E33 (ˆ u)
(39)
in the norm convergence of L2 (Ω; R3×3 ). Taking ϕ = ϕˆε in (35) and passing to the limit we find Z Z CE(¯ u, v¯, w, ¯ p¯, q¯) · E(ˆ u, vˆ, w, ˆ pˆ, qˆ) dx = F · E(ˆ u, vˆ, w, ˆ pˆ, qˆ) dx, Ω
Ω
for every (ˆ u, vˆ, w, ˆ pˆ, qˆ) ∈ Aˆ. This equality holds in fact for any (u, v, w, p, q) ∈ A in place of (ˆ u, vˆ, w, ˆ pˆ, qˆ) ∈ Aˆ because of the approximation Lemma 6.1 which ensures that there exists a sequence (u, vˆn , w ˆ n , pˆn , qˆn ) ∈ Aˆ such that k(u, vˆn , w ˆ n , pˆn , qˆn ) − (u, v, w, p, q)kA → 0. To show that the convergence in (38) is indeed strong, it suffices to prove that limε→0 kE ε (uε )kL2 (Ω;R3×3 ) = kE(¯ u, v¯, w, ¯ p¯, q¯)kL2 (Ω;R3×3 ) or, equivalently, that Z Z lim CE ε (uε ) · E ε (uε ) dx = lim F ε · E ε (uε ) dx ε→0
ε→0
Ω
Ω
Z =
F · E(¯ u, v¯, w, ¯ p¯, q¯) dx Ω
Z =
CE(¯ u, v¯, w, ¯ p¯, q¯) · E(¯ u, v¯, w, ¯ p¯, q¯) dx, Ω
where we passed to the limit thanks to the strong convergence of F ε .
7
2
Equilibrium differential equations
In this last section we derive the differential formulation of the limit problem. For simplicity we assume here that the elasticity tensor also satisfies the major symmetries; that is, Cijkl = Cklij for any i, j, k, l. Nevertheless, the same computation can be performed also in the general case. To make (36) more explicit and to keep the notation compact, in writing the elasticity tensor components Cijkl we associate to a pair of components ij a single component s following the rule 11 7→ 1, 22 7→ 2, 33 7→ 3, 23 7→ 4, 13 7→ 5, 12 7→ 6, and we write, for instance, c14 for C1123 ; see Auld [1] for more details
ANISOTROPIC INHOMOGENEOUS THIN-WALLED BEAMS
26
on the notation used. Clearly cij = cji . Still, for brevity, define e¯1 = E11 (¯ q ), e¯2 = E22 (w), ¯ e¯3 = E33 (¯ u), e¯4 = 2E23 (¯ v ), e¯5 = 2E13 (w), ¯ e¯6 = 2E12 (¯ p). Letting Z L (u, v, w, p, q) := F · E(u, v, w, p, q) dx, Ω
we can then rewrite (36) as Z X 6 £ c1j e¯j E11 (q) + c2j e¯j E22 (w) + c3j e¯j E33 (u) + 2c4j e¯j E23 (v) Ω j=1
¤ + 2c5j e¯j E13 (w) + 2c6j e¯j E12 (p) dx = L (u, v, w, p, q),
for every (u, v, w, p, q) ∈ A . Thus Z X 6
c1j e¯j E11 (q) dx = L (0, 0, 0, 0, q),
(40)
2c6j e¯j E12 (p) dx = L (0, 0, 0, p, 0),
(41)
Ω j=1
Z X 6 Ω j=1
Z X 6 £ ¤ c2j e¯j E22 (w) + 2c5j e¯j E13 (w) dx = L (0, 0, w, 0, 0),
(42)
Ω j=1
Z X 6
2c4j e¯j E23 (v) dx = L (0, v, 0, 0, 0),
(43)
c3j e¯j E33 (u) dx = L (u, 0, 0, 0, 0).
(44)
Ω j=1
Z X 6 Ω j=1
In this section, for simplicity, we assume Z L (0, 0, ·, ·, ·) = 0,
`
L (0, v, 0, 0, 0) = Z
L (u, 0, 0, 0, 0) =
m(x3 )ϑ(x3 ) dx3 ,
Z
0
b · u dx + Ω
(45) 2
s · u dH , Γ`
where Γ` = ∂Ω ∩ {x3 = `}; u ∈ U ; b ∈ L2 (Ω); s ∈ L2 (Γ` ); v ∈ V ; ϑ ∈ H 1 (Ω3 ), ϑ(0) = 0, is related to v as in Lemma 5.2; and m ∈ L2 (Ω3 ). Such assumptions are quite often satisfied in engineering applications. We now derive the equilibrium equations in differential form. Let ψ ∈ L2 (Ω) and define Z x1 Z Z x1 q1 := ψ(s, ·, ·) ds − − ψ(s, ·, ·) ds dx1 . −a1 /2
Ω1
−a1 /2
ANISOTROPIC INHOMOGENEOUS THIN-WALLED BEAMS
27
Then q := (q1 , 0, 0) ∈ Q and E11 (q) = ψ; hence, from (40) and (45) we deduce 6 X
c1j e¯j = 0 a.e..
(46)
j=1
With the same argument it follows from (41) that 6 X
c6j e¯j = 0 a.e..
(47)
j=1
From (46) and (47) we deduce, since c11 c66 − c216 > 0, that e¯1 = E11 (¯ q) = −
5 X c66 c1j − c16 c6j e¯j c11 c66 − c216 j=2
e¯6 = 2E12 (¯ p) = −
5 X c11 c6j − c16 c1j e¯j c11 c66 − c216 j=2
a.e.,
(48)
a.e..
(49)
Using (45), (48), and (49) we can rewrite (42), after setting c˜ij = cij − ci1
c11 c6j − c16 c1j c66 c1j − c16 c6j − ci6 , c11 c66 − c216 c11 c66 − c216
for i, j = 2, . . . , 5, as Z X 5 £ ¤ c˜2j e¯j E22 (w) + 2˜ c5j e¯j E13 (w) dx = L (0, 0, w, 0, 0) = 0.
(50)
Ω j=2
Since w ∈ W , it then admits the representation given in Lemma 5.5 in terms of functions η1 and η2 . Choosing η1 = η2 = 0, so that E22 (w) = 0, and w3 like it has been chosen q1 previously, we find from (50) that 5 X
c˜5j e¯j = 0 a.e..
(51)
j=2
Let ψ ∈ L2 (Ω23 ). Taking η1 = w3 = 0 and Z x2 Z Z η2 := ψ(s, ·) ds − − −a2 /2
Ω2
x2
ψ(s, ·) ds dx2
−a2 /2
so that E22 (w) = ψ, we find from (50) that 5 Z X j=2
Ω1
c˜2j e¯j dx1 = 0
a.e..
(52)
ANISOTROPIC INHOMOGENEOUS THIN-WALLED BEAMS Taking instead η2 = w3 = 0 and Z x2 Z t η1 := −a2 /2
28
ψ(s, ·) ds dt − K1 x2 − K2 ,
−a2 /2
where the constants K1 and K2 are chosen in order to satisfy the mean integral conditions required on η1 by (28), we have E22 (w) = −x1 ψ, and hence, from (50) we deduce 5 Z X x1 c˜2j e¯j dx1 = 0 a.e.. (53) Ω1
j=2
From (51), and observing that the positive definiteness of the elastic tensor implies c˜55 > 0, we find e¯5 = 2E13 (w) ¯ =−
4 X c˜5j e¯j c˜ j=2 55
a.e..
(54)
To solve (52) and (53) we need to write explicitly e¯2 = E22 (w). ¯ Since w ¯ ∈W, by Lemma 5.5, we can write w ¯1 (x) = η¯1 (x2 , x3 ),
w ¯2 (x) = −x1
∂ η¯1 (x2 , x3 ) + η¯2 (x2 , x3 ), ∂x2
where η¯1 and η¯2 belong to the appropriate spaces. Since e¯2 = −x1
∂ η¯2 ∂ 2 η¯1 + 2 ∂x2 ∂x2
(55)
and using (54), we can rewrite (52) and (53) as ∂ η¯2 ∂ 2 η¯1 − s0;22 s1;22 2 ∂x2 ∂x2 s2;22
∂ η¯2 ∂ 2 η¯1 − s1;22 ∂x22 ∂x2
where we have set cˆij :=
c˜55 c˜ij − c˜i5 c˜j5 , c˜55
4 Z X = − cˆ2j e¯j dx1 , j=3 Ω1
=
4 Z X − x1 cˆ2j e¯j dx1 , j=3 Ω1
Z sk;ij := − xk1 cˆij dx1 ,
(56)
Ω1
for i, j = 2, 3, 4 and k = 0, 1, 2. From these equations we find µ ¶ 4 Z 4 Z X X ∂ 2 η¯1 1 = s − x c ˆ e ¯ dx − s − c ˆ e ¯ dx , 0;22 1 2j j 1 1;22 2j j 1 ∂x22 s0;22 s2;22 − s21;22 j=3 Ω1 j=3 Ω1 µ ¶ 4 Z 4 Z X X 1 ∂ η¯2 = s − x c ˆ e ¯ dx − s − c ˆ e ¯ dx 1;22 1 2j j 1 2;22 2j j 1 , ∂x2 s0;22 s2;22 − s21;22 j=3 Ω1 j=3 Ω1
ANISOTROPIC INHOMOGENEOUS THIN-WALLED BEAMS
29
and then by integration η¯1 and η¯2 (the fact that s0;22 s2;22 − s21;22 > 0 can be checked, for instance, by using H¨older’s inequality; see Wheeden and Zygmund [13, Chapter 8, Exercise 4]). According to Remark 5.1 and Lemma 5.2, we let 0
0
00
00
e¯4 = 2x1 ϑ¯ +
e¯3 = ζ¯3 − x1 ζ¯1 − x2 ζ¯2 , Setting ijkl := Smpqr
∂ %¯ . ∂x2
(57)
si;2j sk;2l − sm;2p sq;2r , s0;22 s2;22 − s21;22
we then have ∂ 2 η¯1 ∂x22
¡ 00 ¢ ¯ 0213 ¯0 0223 ¯00 0224 ¯0 0214 ∂ % = S0312 ζ3 − x2 ζ¯2 − S1213 ζ1 + 2S1214 ϑ + S0412 , ∂x2
∂ η¯2 ∂x2
¡ 00 ¢ ¯ 1213 ¯0 1223 ¯00 1224 ¯0 1214 ∂ % = S0322 ζ3 − x2 ζ¯2 − S2213 ζ1 + 2S1422 ϑ + S0422 , ∂x2
and taking into account the relations (48), (49), (54), (55), and (57), we find 6 X
¢¡ 0 ¡ 00 ¢ 1213 0213 + cˆi2 S0322 ζ¯3 − x2 ζ¯2 cij e¯j = cˆi3 − x1 cˆi2 S0312
j=1
¡ ¢ 0223 1223 ¯00 − x1 cˆi3 − x1 cˆi2 S1213 + cˆi2 S2213 ζ1 ¡ ¢ 0224 1224 ¯0 +2 x1 cˆi4 − x1 cˆi2 S1214 + cˆi2 S1422 ϑ
(58)
¢ ¯ ¡ 1214 ∂ % 0214 + cˆi2 S0422 + cˆi4 − x1 cˆi2 S0412 , ∂x2 for i = 3, 4. Now let v ∈ V and ϑ and % be as in Lemma 5.2. With ψ ∈ L2 (Ω23 ) and ϑ = 0 and Z x2 Z Z x2 % := ψ(s, ·) ds − − ψ(s, ·) ds dx2 , −a2 /2
we find from (43) that
Ω2
−a2 /2
Z X 6 − c4j e¯j dx1 = 0. Ω1 j=1
It then follows that ¡ ¢¡ 0 00 ¢ 0213 1213 0 = s0;43 − s1;42 S0312 + s0;42 S0322 ζ¯3 − x2 ζ¯2 ¡ ¢ 0223 1223 ¯00 − s1;43 − s1;42 S1213 + s0;42 S2213 ζ1 ¡ ¢ 0224 1224 ¯0 +2 s1;44 − s1;42 S1214 + s0;42 S1422 ϑ ¡ ¢ ¯ 0214 1214 ∂ % , + s0;44 − s1;42 S0412 + s0;42 S0422 ∂x2
ANISOTROPIC INHOMOGENEOUS THIN-WALLED BEAMS
30
0214 1214 and provided that the last coefficient s0;44 − s1;42 S0412 + s0;42 S0422 6= 0, one finds ∂ %¯ −1 = 0214 1214 ∂x2 s0;44 − s1;42 S0412 + s0;42 S0422 h¡ ¢¡ 0 00 ¢ 0213 1213 · s0;43 − s1;42 S0312 + s0;42 S0322 ζ¯3 − x2 ζ¯2 ¡ ¢ 0223 1223 ¯00 − s1;43 − s1;42 S1213 + s0;42 S2213 ζ1 i ¡ ¢ 0 0224 1224 + s1;44 − s1;42 S1214 + s0;42 S1422 2ϑ¯ .
We may rewrite (58) as 6 X
0 00 00 0 cij e¯j = Fi3 ζ¯3 − Fi2 ζ¯2 − Fi1 ζ¯1 + Fi4 ϑ¯
(59)
j=1
for i = 3, 4, where 0223 1223 Fi1 = (x1 cˆi3 − x1 cˆi2 S1213 + cˆi2 S2213 ) ¢ ¡ 1223 0223 1214 0214 + s0,42 S2213 ) s1,43 − s1,42 S1213 + cˆi2 S0422 (ˆ ci4 − x1 cˆi2 S0412 , − 1214 0214 + s s0,44 − s1,42 S0412 0,42 S0422
Fi2 = x2 Fi3 , 0213 1213 Fi3 = (ˆ ci3 − x1 cˆi2 S0312 + cˆi2 S0322 )
−
1213 0213 1214 0214 ) + s0,42 S0322 )(s0,43 − s1,42 S0312 + cˆi2 S0422 (ˆ ci4 − x1 cˆi2 S0412 , 0214 1214 s0,44 − s1,42 S0412 + s0,42 S0422
0224 1224 Fi4 = 2(x1 cˆi4 − x1 cˆi2 S1214 + cˆi2 S1422 )
−2
1224 0224 1214 0214 ) + s0,42 S1422 )(s1,44 − s1,42 S1214 + cˆi2 S0422 (ˆ ci4 − x1 cˆi2 S0412 . 0214 1214 s0,44 − s1,42 S0412 + s0,42 S0422
Let
Z Aij (x3 ) :=
Fij (·, ·, x3 ) dx1 dx2 Ω12
and
Z
Kij (x3 ) :=
Z x1 Fij (·, ·, x3 ) dx1 dx2 , Lij (x3 ) :=
Ω12
x2 Fij (·, ·, x3 ) dx1 dx2 . Ω12
Then, from (43), (44), and (59) we finally deduce the following system of equilibrium differential equations: 0 00 00 0 0 (A33 ζ¯3 − A31 ζ¯1 − A32 ζ¯2 + A34 ϑ¯ ) = −p3 , (K33 ζ¯30 − K31 ζ¯100 − K32 ζ¯200 + K34 ϑ¯0 )00 = −p1 , (60) 0 00 00 0 00 (L33 ζ¯3 − L31 ζ¯1 − L32 ζ¯2 + L34 ϑ¯ ) = −p2 , 0 00 00 0 0 2(K43 ζ¯ − K41 ζ¯ − K42 ζ¯ + K44 ϑ¯ ) = −m, 3
1
2
ANISOTROPIC INHOMOGENEOUS THIN-WALLED BEAMS where
µZ
¶0
p1 = Ω12
µZ p2 =
Z +
Ω12
p3 =
b1 dx1 dx2 , Ω12
¶0 x2 b3 dx1 dx2
Z
Z +
x1 b3 dx1 dx2
31
b2 dx1 dx2 , Ω12
b3 dx1 dx2 . Ω12
The system (60) should then be completed with the suitable boundary conditions.
7.1
The homogeneous beam
In the general inhomogeneous and anisotropic case, the torsional, flexional, and extensional problems are all coupled together in the equilibrium differential system (60). A partial decoupling occurs already in the homogeneous fully anisotropic case. Indeed, in this case, from (56), we have s0;ij = cˆij ,
s1;ij = 0,
s2;ij =
a31 cˆij , 12
and therefore 0213 1223 0214 1224 S0312 = S2213 = S0412 = S1422 = 0,
0223 S1213 =
cˆ23 , cˆ22
1213 S0322 =−
cˆ23 , cˆ22
1214 S1422 =
which causes many of the coefficients of the system Aij , Kij , and Lij to be zero. In this case, the system (60) simply rewrites as 00 A33 ζ¯3 = −p3 , (−K31 ζ¯00 + K34 ϑ¯0 )00 = −p1 , 1 (61) −L32 ζ¯(iv) = −p2 , 2 00 0 0 2(−K41 ζ¯1 + K44 ϑ¯ ) = −m, where
· A33 = a1 a2 K31 = K41 = L32
¸ cˆ22 cˆ33 − cˆ223 (ˆ c22 cˆ34 − cˆ23 cˆ24 )2 , + cˆ22 cˆ22 (ˆ c22 cˆ44 − cˆ224 )
a31 a2 cˆ22 cˆ33 − cˆ223 a3 a2 cˆ22 cˆ34 − cˆ23 cˆ24 , K34 = 1 , 12 cˆ22 6 cˆ22 a31 a2 cˆ22 cˆ34 − cˆ23 cˆ24 a3 a2 cˆ22 cˆ44 − cˆ224 , K44 = 1 , 12 cˆ22 6 cˆ22 · ¸ a1 a32 cˆ22 cˆ33 − cˆ223 (ˆ c22 cˆ34 − cˆ23 cˆ24 )2 = + . 12 cˆ22 cˆ22 (ˆ c22 cˆ44 − cˆ224 )
Thus for a fully anisotropic but homogeneous beam there is only coupling between twisting and bending in direction 1.
cˆ24 , cˆ22
ANISOTROPIC INHOMOGENEOUS THIN-WALLED BEAMS
7.2
32
The homogeneous orthotropic/isotropic beam
When the material is orthotropic a complete decoupling occurs. Indeed, for orthotropic material we have that cki = 0 for k = 1, 2, 3 and i = 4, 5, 6, and c45 = c46 = c56 = 0. It then follows that K34 = K41 = 0 and the system (61) reduces to 00 A33 ζ¯3 = −p3 , −K ζ¯(iv) = −p , 31 1 1 (62) (iv) ¯ −L32 ζ2 = −p2 , 00 2K44 ϑ¯ = −m. Finally, if the material is isotropic, that is, if it is orthotropic and c12 = c23 = c13 =: λ, c44 = c55 = c66 =: µ, and c11 = c22 = c33 = λ + 2µ, where λ and µ are the Lam´e moduli, then we have that A33 = a1 a2 E,
K31 =
a31 a2 E, 12
2K44 =
a31 a2 µ, 3
L32 =
a1 a32 E, 12
where E := µ 3λ+2µ λ+µ is the Young modulus of the material. Hence, in the isotropic and homogeneous case we recover the usual form of the differential system of equilibrium equations.
References [1] B. A. Auld, Acoustic Fields and Waves in Solids, Vol. 1, John Wiley and Sons, New York, 1973. [2] R. Chandra, A. D. Stemple, and I. Chopra, Thin-walled composite beams under bending, torsional, and extensional effects, J. Aircraft, 27 (1990), pp. 619–626. [3] L. Della Longa and A. Londero, Thin Walled Beams with Residual Stress, submitted. [4] L. Freddi, A. Morassi, and R. Paroni, Thin-walled beams: The case of the rectangular cross-section, J. Elasticity, 76 (2004), pp. 45–66. [5] H. Le Dret, Probl`emes Variationnels dans les Multi-Domaines, Recherches en Math´ematiques Appliqu´ees 19, Masson, Paris, 1991. [6] H. Le Dret, Convergence of displacements and stresses in linearly elastic slender rods as the thickness goes to zero, Asymptot. Anal., 10 (1995), pp. 367–402. Monneau, F. Murat, and A. Sili, A [7] R. Partial Korn’s Inequality and Error Estimates for the 3d-1d Dimension Reduction in Anisotropic Heterogeneous Linearized Elasticity, in preparation.
ANISOTROPIC INHOMOGENEOUS THIN-WALLED BEAMS
33
[8] F. Murat and A. Sili, Anisotropic, Heterogeneous, Linearized Elasticity Problems in Thin Cylinders, in preparation. [9] L. Trabucho and J. M. Viano, Mathematical modelling of rods, in Handbook of Numerical Analysis, Vol. 4, North–Holland, Amsterdam, 1996, pp. 487–974. [10] F. Treves, Topological Vector Spaces, Distributions and Kernels, Academic Press, New York, London, 1967. [11] V. V. Volovoi and D. H. Hodges, Theory of anisotropic thin-walled beams, J. Appl. Mech., 67 (2000), pp. 453–459. [12] V. V. Volovoi, D. H. Hodges, C. E. S. Cesnik, and B. Popescu, Assessment of beam modeling methods for rotor blade applications, Math. Comput. Modelling, 33 (2001), pp. 1099–1112. [13] R. L. Wheeden and A. Zygmund, Measure and Integral. An Introduction to Real Analysis, Monogr. Textbooks Pure Appl. Math. 43, Marcel Dekker, New York, Basel, 1977.
