Asymptotic Behavior of 3D Unstable Structures Made of Beams

In our previous papers (Griso et al. in J. Elast. 141:181–225, 2020; J. Elast., 2021, https:// doi.org/10.1007/s10659-021-09816-w), we considered thick periodic structures (first paper) and thin stable periodic structures (second paper) made of small cylinders (length of order ε and cross-sections of radius r). In the first paper r = κε with κ a fixed constant, ε → 0, while in the second ε → 0 and r/ε → 0. In this paper, our aim is to give the asymptotic behavior of thin periodic unstable structures, when ε → 0, r/ε → 0 and ε2/r → 0. Our analysis is again based on decompositions of displacements. As for stable periodic structures, Korn type inequalities are proved. Several classes of unstable and auxetic structures are introduced. The unfolding and limit homogenized problems are really different of those obtained for the thin stable periodic structures. The limit homogenized operators are anisotropic, the spaces containing the macroscopic limit displacements depend on the periodicity cells. It was not the case in the two previous studies. Some examples are given.

circular cross-section whose radius is r, the periodicity parameter is ε, we assume that r/ε and ε 2 /r tend to 0.
There are many types of unstable structures or unstable states in structures in all or in some specific directions. The instabilities can be wished if well understood and modeled, they can also be used to better design materials or develop new auxetic structures. It is well known to engineers that for stable structures (wire trusses, lattices) made of very thin beams, bending dominates the stretching-compression. A contrario, if the same structures are made of thick beams the stretching-compression dominates. If structures are unstable, they work on rotation around nodes mostly.
This paper is the continuation of [23] which dealt with the 3D-stable periodic structures. Here, we investigate the unstable and auxetic 3D-periodic structures made of thin beams. The first difference between 3D-stable (see [23,Definition 5]) or -quasi stable periodic structures (see Definition 14) and those 3D-unstable lies in the Korn inequalities. For 3Dstable and -quasi-stable periodic structures we have (see [23,Proposition 2]) u L 2 (Sε,r ) ≤ C 1 + ε 2 r e(u) L 2 (Sε,r ) , ∇u L 2 (Sε,r ) ≤ C ε r e(u) L 2 (Sε,r ) , while for 3D-unstable periodic structures, one has (see Proposition 1) where S ε,r is the structure made of beams. That is why for 3D-periodic structures made of "thick" rods (the cross sections being of the same order as the period r ∼ ε), distinguishing stable structures from unstable ones is not really useful (see [19]).
Our analysis of the thin structures provides more than these above inequalities, it gives estimates of the centerline displacements and also of the small rotations of the cross-sections (see [23,Proposition 2] and Proposition 2 in Sect. 2.3).
The second and most important difference between 3D-stable and unstable periodic structures appears in the local behavior of cells. In the stable case we have found the relation where S is the running point in S, S is the 3D-periodic cell made of segments, Ω the macroscopic domain, U stands for the local displacement of the centerlines of the beams, R for the rotations of the cross-sections, t 1 (S) being the direction of a beam-centerline belonging to S. Both fields U and R are periodic with respect to the second variable belonging to S. The above relation means that the local displacements are of Bernoulli-Navier type. Then, the displacement of the nodes is given by the macroscopic displacement.
In the unstable case we have the relation (see Lemma 15) ∇U(x) t 1 (S) + ∂ U ∂S (x, S) = R (x, S) ∧ t 1 (S), (x, S) ∈ Ω × S where U is the macroscopic displacement, U stands for the local displacement of the centerlines, R for the rotations of the cross-sections, t 1 (S) being the direction of a beam-centerline (see (2.1)). Here also, both fields U and R are periodic with respect to the second variable. The above relation means that the local displacements are not of Bernoulli-Navier type. The macroscopic displacements are subject to the conditions of existence of solutions for the above equation (see Sect. 3). By way of example, for some auxetic structures we obtain that the macroscopic displacements satisfy some a priori conditions, e.g., where κ i1 > 0, i ∈ {2, 3}, are constant coefficients (see Sect. 14.2).
In [23], we have shown that the asymptotic behavior of a 3D-periodic stable structure is given by a classical elasticity problem, the stress tensor is given via the strain tensor and a 6 × 6 matrix whose coefficients depend on the geometry of the 3D cell. The obtained model is of extensional type, the macroscopic limit displacement is the limit of the extensional displacements of the set of centerlines S ε (it only depends on the stretching-compression of the small beams). Here, for a 3D-periodic unstable structure, we show that the macroscopic limit displacement is of inextensional type. It never depends on the stretching-compression of the small beams. The limit model is not a classical elasticity problem.
Our analysis relies on decompositions of displacements, as in our previous papers [19,23], first for a single beam (see [13][14][15]) and then for the macroscopic structure. According to these studies, a beam displacement is the sum of an elementary displacement and a warping. An elementary displacement has two components. The first one is the displacement of the beam centerline while the second stands for the small rotation of the beam cross-sections (see [13,15]). The warping takes into account the deformations of the cross sections. This decomposition has been extended for structures made of a large number of beams in [14] (see [4] for beam structures in the framework of nonlinear elasticity). Here, similar displacement decompositions are obtained.
To study the asymptotic behavior of periodic unstable structures and derive the limit problems we use the periodic unfolding method introduced in [9] and then developed in [10,11]. This method has been applied to a large number of different types of problems. We mention only a few of them which deal with periodic structures in the framework of the linear elasticity (see [5,16,[18][19][20][21][22]31]). As general references on the theory of beams or structures made of beams, we refer to [2,7,27,28,34,35].
The paper is organized as follows. Section 2 introduces structures made of segments (examples of 3D cell S). We recall known results concerning the decomposition of a beam displacement. This section also gives estimates of the terms appearing in the decomposition with respect to the L 2 -norm of the strain tensor. Then, we extend these results to structures made of beams. Complete estimates of our decomposition terms and Korn-type inequalities are obtained for general unstable 3D-periodic structures.
In Sect. 3, we solve the o.d.e. (see (3.1)-(3.2)) posed on the periodic cell S. It plays a fundamental role for unstable periodic structures. This o.d.e. admits solutions under some conditions. We will show in the following section that these conditions allow to define the space of macroscopic admissible displacements. In Sect. 4, several examples of 3D-periodic unstable structures are presented. Section 5 is dedicated to some properties of the various unstable structures introduced in Sect. 4. The statement of the elasticity system is given in Sect. 6. The scalings of the applied forces are given with respect to ε and r. That leads to an upper bound for the L 2 -norm of the strain tensor of the solution to the elasticity problem. Section 7 deals with the unfolding operators (see also [23]).
In Sect. 8, we give the asymptotic behavior of a sequence of displacements and their strain tensors. Then, in Sect. 9, in order to obtain the limit unfolded problem we split it into three problems: the first involving the limit warpings (these fields are concentrated in the cross-sections, this step corresponds to the process of dimension reduction), the second involving the microscopic inextensional limit displacements posed on the periodic cell S and the third the macroscopic limit problem involving the macroscopic displacements posed in the whole domain Ω.
Section 12 leads to the complete unfolding problem for all types of 3D-periodic unstable structures. To do that, different correctors are introduced, they allow to write the limit homogenized problem. We obtain a linear elasticity problem with constant coefficients calculated using the correctors. In Sect. 13 we apply the previously obtained results in the case when a periodic 3D beam structure is made of an isotropic and homogeneous material. In Sect. 14.2, we detail the spaces containing the macroscopic limit displacements for some structures presented in Sect. 4 (see also Fig. 1).
In the Appendix, some technical results are shown (proof of some lemmas, the way to build test functions and a new lemma of the periodic unfolding method).
Finally, we give mechanical engineers a translation in their terminology, and explain the obtained result, i.e. the limit problem in terms of known models for constitutive laws.
We restrict solution φ of (6.4) to the mean lines of the rods, i.e. the skeleton of the structure, S ε . Then, we approximate this restricted to the skeleton or graph S ε solution by a piece-wise affine (linear) approximation U ∈ U(S ε ), (2.2). This space is further decomposed on the static elastic vector field, V ∈ D E (S ε ), satisfying, e.g., (5.1), and its orthogonal complement, kinematic field, U − V ∈ D I (S ε ), see (2.4). In the case, when S ε is a stable structures, this complement is just rigid displacement. (5.1) is the strain equilibrium problem for a truss-system on S and describes the equilibrium of all axial (tensile) strains (forces normalized by the Young's modulus of fibers) in rods, acting on each node of the graph, see e.g. chapter about trusses in [29]. And after fixing of 3 scalar non-collinear displacements on one or different nodes, (5.1), will be uniquely solvable on the graph S for almost all x.
In terminology of physicist and dynamical systems, the elasto-static field V satisfies a Hamiltonian, while the kinematic, U − V , a Lagrangian (see [26, pages 33-34]). We will call the kinematic field rotations.
Our structure and its skeleton are periodic. In Sect. 3, matrices M denote unit perturbations from 6 standard experiments on the unit periodicity cell of the structure, 3 axial tensions and 3 shear experiments. System of equations (3.1) is equivalent to the tensile force balance on a rod-(truss-) system, S, normalized by the elastic property, Young's modulus, of rods, for each of such experiments. And (3.2) is equivalent to the moment balance equation on the same rod-system, also normalized by the tensile elastic property of rods. B V (M) denotes the mean or averaged rotation of each rod (segment), while B(M) is the equivalent reformulation for the rotation field for a frame of beams, restricted to an edge or beam. In the frame of beams the angles between beams are fixed, therefor this field vanishes closed to the nodes (see Chap. about FEM (finite element method) for frames in [29]).
In the limit (cell problems (12.3)) we have on segments, or beams, or elements just four scalar degrees of freedom (variables), the axial tension, torsion and two bending rotations. They correspond to the finite element (FE)-interpolation of the frame of beams from [29]. The tensor decomposition for 1D-system on a frame of graph is given by (12.4) and the 1D bilinear form for microscopic fields, U , R is given as a sum of 4 terms, the beam axial tension, torsion and 2 bending terms (energies). The same 1D bilinear form can be found in (6.5) of [31], where authors did not pass to the limit with the beam thickness and just approximated the cell solution, solving it by FEM for frames. Actual paper justifies this step in [31] mathematically.
While for the stable structures (see [19]), the homogenized macroscopic problem was pure elastic, corresponding to the first tensile energy, for the unstable case, it is rotation dominated, see (12.9). It can be interpreted as micro-polar elasticity, [3], [25] and it was used in our work [17].

Geometric Setting
In this paper we consider structures made of a large number of segments.
γ be a set of segments and K the set of the extremities of these segments.
S is called structure if -S is a connected set, -S is not included in a plane, 1 -for any segment γ = [A , B ] ∈ S, one has γ \ {A , B } ∩ K = ∅, -for any point of K belonging to only two segments, the directions of these segments are noncollinear.
Hereinafter, S is called a 3D-structure. The segment γ = [A , B ] ∈ S of length l is parameterized by S 1 ∈ [0, l ] and its direction is given by the unit vector S is the running point of S. On S we define a space of continuous fields U(S) with values in R 3 as follows: The space of rigid displacements is denoted by R: On U(S) we consider the semi-norm 2 3) 1 Here we only want to consider 3D cells, we can easily transpose the results of this paper for planar cells.
2 dU dS is equal to dU dS 1 |γ on every segment γ of S.
For p ∈ [1, +∞], we denote 3 This paper is dedicated to unstable structures, examples of which are given in Fig. 1. Stable structures have been considered in [23].
on every segment γ of S.

Notations
Denote -e 1 , e 2 , e 3 the usual basis of R 3 , -Y = (0, 1) 3 the open parallelotope associated with this basis, 4 -S a 3D-structure, in the sense of Definition 1, included in Y .

Definition 3
A structure S is a 3D-periodic structure if for every i ∈ {1, 2, 3} S ∪ S + e i is a structure in the sense of Definition 1.
From now on, S is a 3D-periodic structure.
Let Ω be a bounded domain in R 3 with a Lipschitz boundary and Γ be a subset of ∂Ω with non null measure. We assume that there exists an open set Ω with a Lipschitz boundary such that Ω ⊂ Ω and Ω ∩ ∂Ω = Γ . Denote One has The open sets Ω ε , Ω ε , Ω ε , Ω int ε and Ω int ε are connected, and satisfy The running point of S ε is denoted s. S ε,r is the structure made of beams. The cross-sections of the beams are discs of radius r and the centerlines of the beams are the segments of S ε , it also contains the balls of radius r centered on the points of K ε . The general beam P ξ ε, ,r is referred to an orthonormal frame εξ The set of junctions is denoted by J r . There exists c 0 which only depends on S such that The set J r is defined in such a way that S ε,r \ J r only consists of distinct straight beams. 3 such that u |Sε,r = u and u = 0 in S ε,r , \S ε,r , It means that the displacements belonging to V ε,r "vanish" on a part Γ ε,r included in ∂S ε,r ∩ ∂Ω.

Displacements Decomposition
In [14] it is shown that every displacement u of a beam structure can be decomposed as where U e is an elementary beam-structure displacement and u is a warping. For the beamstructure S ε,r we remind some definition and results.
Definition 4 (see [14]) An elementary beam-structure displacement is a displacement belonging to H 1 (S ε,r ) 3 whose restriction to each beam is an elementary displacement and whose restriction to each junction is a rigid displacement: U e is the elementary beam-structure displacement and u the warping, they belong to H 1 (S ε,r ) 3 . Here, the pair (U e , u) is not uniquely determined. The warping satisfies (see [14,15]) the following conditions "outside" the domain J r : For every displacement u ∈ H 1 (S ε,r ) 3 , we denote by e the strain tensor (or symmetric gradient) e(u) .
Here, as like as [23], we split the field U into the sum of two fields U h and U , where U h coincides with U in the nodes of S ε and is affine between two contiguous nodes, U is the residual part. In the same way, the fields R h and R are introduced. It is obvious, but important to note that U h describes the displacement of the nodes, i.e., the macroscopic behavior of the structure, whereas U stands for the local displacement of the beams.

Lemma 2 For every
(2.10) The constants do not depend on ε and r.
Observe that since the displacements in V ε,r are the restrictions of displacements belonging to H 1 (S ε,r ) 3 , all the estimates of the above Lemma 2 are valid replacing S ε by S ε . By construction, the fields U h , R h are affine on every segment of the structure S ε (resp. S ε ) and they vanish on the segments belonging to S ε \ S ε . Let u be in H 1 (S ε,r ) 3 . Applying the Poincaré-Wirtinger inequality in εξ + εS and using (2.10) 8 give a piecewise constant function b ∈ L ∞ (Ω ε ) 3 (constant in the cell εξ + εY ) such that Again the Poincaré-Wirtinger inequality in εξ + εS and the above estimate give another piecewise constant function a ∈ L ∞ (Ω ε ) 3 (constant in the cell εξ + εY ) such that where r is a rigid displacement in every cell εξ + εY . Then, as above, applying the Poincaré-Wirtinger twice (in εξ + εS ∪ ε S + e i and εξ + ε S + e i ) lead to (see also [23,Sect. 5 3 ) in the cell ε(ξ + Y ), ξ ∈ Ξ ε , as the Q 1 interpolate of its values on the vertices of this parallelotope.

A Preliminary Result
Denote We endow H 1 per,0 (S) 3 with the scalar product We define D E,per (S) as the orthogonal subspace of D I,per (S) in U per (S) for the above scalar product. Observe that since S is a 3D-periodic structure, one has D I,per (S) ∩ R = {0}. Set As for [23], we equip D I,per (S), with the semi-norm Since S is 3D-periodic structure, this semi-norm is a norm equivalent to the usual norm of the product space H 1 per,0 (S) 3 × H 1 per (S) 3 . The elements of D I,per (S) (resp. the first terms of the pairs in D I,per (S)) are the inextensional displacements.
Let M be a 3 × 3 constant matrix, equation a.e. in S, (3.1) admits at most one solution. Indeed, if we have two solutions then the difference belongs to D I,per (S). Denote M s (S) the subspace of the 3 × 3 symmetric matrices such that equation (3.1) admits a solution.
For every M ∈ M s (S). We denote V (M) the unique solution to (3.1). Now, consider the following equation: 2) It will play an important role in this study (see Sect. 8 and the following). Now, let M be in M s (S), one has Hence, there exists a field B V (M) defined on S, constant on every segment of S, satisfying Remind the following result: the function φ a , a > 0, defined by   Hence, there exits a field A V (M) ∈ H 1 per (S) 3 such that 2). Note that in the neighborhood of every node A ∈ K, one has (M ∈ M s (S)) (3.7) We have

Some Types of Unstable Structures (see
Definition 6 (Structure of type S 1 ) A 3D-periodic structure S ⊂ S is of type S 1 , if at least one segment in every line of S (1) is removed in such a way that the remaining segments form a 3D-periodic structure (see Fig. 1(d), (e), (f)).
obtained from a structure of type S 1 where at least one segment in every "zig-zag" line of S (2) is removed in such a way that the remaining segments form a 3D-periodic structure.
obtained from a structure of type S 2 where at least one segment in every "zig-zag" line of S (3) is removed in such a way that the remaining segments form a 3D-periodic structure.
. Definition 12 (Structure of type S 5 ) A 3D-periodic structure S is of type S 5 if it is obtained from a 3D-periodic structure of type S j , j ∈ {0, 1, 2, 3}, where we replace every node by a not necessarily regular octahedron 7 (see Fig. 3).
The structures of type S 3 or S 4 are of type S 6 (see . A structure of type S 5 which derives from a structure of type S 3 is of type S 6 (see Corollary 2). Definition 14 (Quasi-stable structure) A 3D-periodic structure S is quasi-stable, if it contains a substructure S which is a stable 3D-periodic structure (see [23,Definition 5]) such that

Structures of Type S 0
For type S 0 structures, we make the following additional assumptions: , meets Γ at most one point and L ∩ Ω is a connected set, -Assumption A Z : all the couples of contiguous lines parallel to Re 1 ⊕ Re i and belonging to S (1) are connected by a segment in S (i) whose direction is not collinear to e i , i ∈ {2, 3}.
If S contains only straight lines then the solution to (5.1) 1 satisfies The constant does not depend on ε.
Proof From equality (5.1) 1 , we first get Consider a line in S (1) ε , if one extremity of this line belongs to Ω \ Ω, we choose V 1 = 0 on this extremity then we solve the above equation. If both extremities are not in Ω \ Ω, we choose the solution to the above equation, the mean value of which on this line vanishes.
Since Ω is bounded, the Poincaré and Poincaré-Wirtinger inequalities give The constant is independent of ε. Since the values of V 1 are defined for every node of K ε , one extends this function in an element affine on every small segment of Hence (5.1) 2 . Now, consider a zig-zag line in S (2) ε , on this line, equation (5.1) 1 becomes Hence, one has to solve Again as for V 1 , if one extremity of the zig-zag line belongs to Ω \ Ω, we choose V 2 = 0 on this extremity then we determine V 2 using the above equality. If both extremities are not in Ω \ Ω, we choose the solution whose mean value on this line vanishes. Then, one extends this function in an element affine on every small segment of S 1) ε ∪ S (3) ε still denoted V 2 . Again, the Poincaré and Poincaré-Wirtinger inequalities and the above estimate lead to the L 2 norm of V 2 .
From the above equality (5.3) and the estimate (5.1) 2 , we get Proceeding in the same way gives V 3 and then its estimates. .

(5.4)
The constants do not depend on ε.
Moreover, one has Proof The results of this proposition are the immediate consequences of Lemma 4.

Remark 2
In the above lemma, since U − V ∈ D I (S ε ), one has Hence .
(5.5) If S contains only straight lines then we obtain . (5.6)

Structures of Type
The constant does not depend on ε.
Proof The "short" straight lines of S (i) ε , i ∈ {1, 2, 3}, have a length of order ε. We solve dV 1 ds = E on every "short" line of S (1) ε choosing the solution whose mean value is equal to 0 on every "short" line (possibly we set V 1 = 0 if an extremity of the "short" line belongs to Ω ∩ Ω). Hence, we get Then, we proceed as in the proof of Lemma 4.
Proof We proceed as in the proof of Lemma 5.
Proof It is an immediate consequence of Lemma 6 since for every 3 × 3 symmetric matrix problem (3.1) admits a unique solution.
The constant does not depend on ε.

Structures of Type
There is at least a solution (only one if we choose b 1 and b 1 ∈ Ra 1 ⊕ Ra 1 ). Second, now consider three segments [A, with two by two non-collinear directions. On these three segments we define a by There is at least a solution (only one if a 1 , a 1 , a 1 are independent). Observe that in these two situations above, one has W (A) = W (B) = 0. Now, consider n+1 segments [A, A 1 ], . . ., [A n , B] (n ≥ 1) with two by two non-collinear directions. Combining the two cases above, we can build a field W satisfying where t 1 stands for a unit vector in the direction of the segments.
Proof It is an immediate consequence of Lemma 8 since for every 3 × 3 symmetric matrix problem (3.1) admits a unique solution. Proof The proof of this lemma is a direct consequence of Lemmas 8 and 4.
Proof This lemma is a direct consequence of Lemma 9.

Structures of Type S 5
Lemma 11 Let S be a structure of type S 5 deriving from a structure S of type 14) The estimates of V depends on the type of the structure S . One has -the estimates of V are the same as those in Proof For simplicity, we assume E constant on every segment of S ε . The lines Aa, Bb, Cc, Dd, Ee and Ff intersect at the point O.
Let A be a node of S . Consider the octahedron εξ There exists a unique field One has Observe that the vectors Now, we proceed as to prove the Lemma 4. One first determine the component V 1 of the solution to (5.14).
Consider the segment [A , B ] ∈ S (if it exists) whose direction is collinear to e 1 . If In such a way that If the segment [A , B ] ∈ S (always whose direction is collinear to e 1 ) does not belong and V 1 (εξ + εA + εc) are known. We extend V 1 as an affine function in the segments joining two contiguous octahedra. The estimates of V 1 are similar to those obtained in the Lemma 4.
We determine V 3 in the same way.

Corollary 2
If S is a 3D-periodic structure of type S 5 and deriving from a structure S of type S 3 then for all E ∈ L 2 (S) (constant on every segment) there exists a unique field Proof The fact that equation (5.15) admits a unique solution is an immediate consequence of Lemma 11. The second statement is an immediate consequence of the first of this lemma.
Proposition 5 Let S be a structure of type S 5 deriving from a structure S of type S 0 or S 3 .
The estimates of V depends on the type of the substructure S (see the corresponding cases in Propositions 3 or 4).

Structures of Type
Proof This lemma is an immediate consequence of the definition of the structures of type S 6 .

Quasi-Stable Structures
Remind that M is a 3 × 3 symmetric matrix, thus M = 0 and b = 0. If S is a 3D-periodic quasi-stable structure then it contains a 3D-periodic stable structure. Applying above gives the result.
The constant C is independent of ε.
Proof First observe that due to the definition of quasi-stable structures, the set D I (S) of inextensional displacements is where Every element of D I,0S (S) is extended by 0 outside S. As a consequence 3 . Since S ε is a stable 3D-periodic stable structure, we know (see [23,Proposition 1]) that .
The constant does not depend on ε.
Then, the above two estimates lead to ξ ∈Ξε .
A straightforward calculation gives ξ ∈Ξε and finally By construction V belongs to H 1 Γ (S ε ) 3 . The constant does not depend on ε.
6 Statement of the Problem

Elasticity Problem
Let a ε ij kl ∈ L ∞ (S ε,r ), (i,j,k,l) ∈ {1, 2, 3} 4 , be the components of the elasticity tensor, these functions satisfy the usual symmetry and positivity conditions a ε ij kl = a ε,r jikl = a ε,r klij a.e. in S ε,r ; -for any τ ∈ M 3 s , where M 3 s is the space of 3 × 3 symmetric matrices, there exists C 0 > 0 (independent of ε and r) such that a ε ij kl τ ij τ kl ≥ C 0 τ ij τ ij a.e. in S ε,r . (6.1) The constitutive law for the material occupying the domain S ε,r is given by the relation between the linearized strain tensor and the stress tensor We assume that every beam is made of an orthotropic material, in the reference frame of the beams one has ⎛ The coefficients a ε ij kl of the above 6 × 6 matrix are functions in L ∞ (S ε ) The unknown displacement 9 u ε : S ε,r → R 3 is the solution to the linearized elasticity system: where ν ε is the outward normal vector to ∂S ε,r \ Γ ε,r , f ε is the density of volume forces. The variational formulation of problem (6.3) is

Force Assumptions and Apriori Estimates of the Solution to (6.4)
As in [23], we distinguish two types of applied forces, the first ones are applied between the junctions and the second ones in the junctions. Let (f, F, G) be in C(Ω) 9 and u ∈ V ε,r .
The applied forces f ε ∈ L ∞ (S ε,r ) 3 are where 1 B (A,r) is the characteristic function of the ball B (A, r).
The last term f |Sε stands for the applied forces in the set of beams Proceeding as in [23] and using the estimates of Proposition 2 give Sε,r (6.6) The constant does not depend on ε and r.

Lemma 14
The solution u ε of problem (6.4) satisfies Proof In order to obtain a priori estimate of u ε , we test (6.4) with v = u ε . From (6.6), one obtains which leads to (6.7).

The Unfolding Operators
The classical unfolding operator T ε was developed in [10,11]. As in [23], in this work we use unfolding operators for structures made of thin beams. One for the centerlines and another for the cross-sections of the beams.
Let us recall their definitions, for their properties we refer the reader to [23,Sect. 6].
In the definitions below (see Definitions 15,16) For simplicity we will still refer to them as S. The set of nodes is always denoted K, the number of beams of S will be still denoted m.

Definition 15 (Centerlines unfolding)
For φ measurable function on S ε , the unfolding operator T S ε is defined as follows: Definition 16 (Beams unfolding) For u measurable function on S ε,r , the unfolding operator is defined as follows: Let φ be measurable on S ε , if S belongs to the segment γ then we have Below we recall two of the main properties of these operators. For every φ ∈ L 2 (S ε ) (resp. ψ in L 2 (S ε,r )) one has for all ∈ {1, . . . , m}).

Asymptotic Behaviors
From now on, we assume that If the structure is of type S j , j ∈ {0, 1, 2}, we also assume that

Asymptotic Behavior of a Sequence of Displacements
In this section we consider a sequence {u ε } ε of displacements belonging to V ε,r and satisfying The fields U , U and R satisfy If the structure is of type S 0 one has
In order to show convergence (8.9) 2 , note that from (2.9) 2 and (8.1) it follows that Therefore, convergence (8.9) 2 follows. Denote This space is a closed subspace of H 1 Γ (Ω) 3 . Note that if S is of type S 0 , it is an immediate consequence of this definition to get U 1 = 0 a.e. in Ω (1)  Proof This result is an immediate consequence of (8.5), Lemma 3 and the equality

Remark 3
Since dU ε ds ·t 1 is smaller than U ε , it should be noted that the limit macroscopic field U does not depend on the limit of dU ε ds · t 1 . This last term takes into account the stretchingcompression of the small beams. 10 If S is of type S 0 and contains only straight lines then 3 | e(V)(x) ∈ M s (S) for a.e. x ∈ Ω and U i = 0 a.e. in Ω (i) , i ∈ {1, 2, 3} .

Proposition 7
Under the assumptions of Lemma 15, the following convergence holds: (8.14) Moreover ε r

From Definition 4 we have
The convergences (8.4) yield Hence, convergence (8.14) holds. Now we consider the asymptotic behavior of the strain tensors T b, ε (e s (u ε )) .

Remark 4 Due to (2.6), the warping u satisfies
For the sake of simplicity, if v belongs to L 2 (Ω × S; H 1 (D) 3 ) and is such that then we will write that v belongs to L 2 (Ω × S; D w ).

First Steps to the Limit Unfolded Problem
In this section we assume that S is a 3D-periodic structure neither quasi-stable nor stable (see Definitions 3,14

and 2 or [23, Definitions 2 and 5]).
To obtain the limit of the rescale LHS of (6.4), we only want to compute the unfolded limit of this term. To do so, we will choose test displacements v ε in V ε,r whose contribution in the junction domain J r goes to 0. Using (6.7), since we have  3 . Since V belongs to D(γ ) and r/ε tends to 0, the support of the above test-displacement is only included in the beams whose centerlines are εξ + εγ , ξ ∈ Ξ ε . By construction, this displacement vanishes in the junction domain J r .
Choosing v ε as a test function in (6.4), and then proceeding as in [23], we obtain 3 we obtain (9.3). Step 1. Preliminary results.

Lemma 17 One has
In this segment one has
We define v ε in the beam whose centerline is εξ for a.e.
By construction v ε belongs to V ε since for every Hence e(v ε ) = 0 a.e. in J r . This test displacement satisfies the condition (9.2). In the beam whose center line is εξ + εγ , one has Then, the above convergence and those in (9.6) lead to the following strong convergence in Step 3. Contribution to the unfolded limit problem.
Choosing v ε as a test function in (6.4), then unfolding the LHS of (6.4) and passing to the limit gives lim (ε,r)→(0,0) Proceeding as in [23], we obtain

The Limit Unfolded Problem Involving the Macroscopic Displacements
Step 1. The test displacement.
This test displacement satisfies (see estimates (A.12)) where O ε 3 r 2 stands for terms whose L ∞ -norm is bounded by a constant (independent of ε and r) multiply by ε 3 r 2 . Therefore Remind that from Sect. 3 and (A.12) one has Hence Then, going to the limit in the strain tensor gives Step 3. Contribution to the unfolded limit problem.
Choosing v ε as a test function in (6.4), then unfolding the LHS of (6.4) and passing to the limit, we get Now, we consider the RHS of (6.4) with v = v ε . As in [23], we easily prove that Hence, (9.9) is proved.

Expression of Z U
For every structure S of type S 0 , we set (i ∈ {1, 2, 3})   (1) ), Φ = 0 a.e. on Γ , and For the structures of type S 0 or S 6 we set

is an affine function on every segment of S.
We endow One has One has 11 Due to Assumption A Z , this space is in fact in Ω \ Ω and -D E,per (S) with the semi-norm which is a norm equivalent to the usual norm of the space H 1 per,0 (S) 3 .

Remark 5
• Given K 2 K 3 functions φ k , k ∈ K 1 belonging to L 2 Γ (Ω, ∂ 1 ), we can easily build and el- • Let S be a 3D-periodic structure of type S 0 . Observe that every function φ in H 1 per (S (i) ) can be extended in a function belonging to H 1 per (S), still denoted φ, affine on the segments belonging to S \ S (i) , i ∈ {1, 2, 3} and one has (1) )) ( V 1 being the restriction of an element belonging to L 2 (Ω; H 1 per (S)) also denoted V 1 ) such that and we have (1) ) .

Remark 6 Note that if S contains only straight lines then
Proof Equality (10.7) is the immediate consequence of Lemma 19. Now, from Lemma 3, there Observe that by construction, V ε,1 = U h ε,1 on every straight line of S (1) ε which meets Γ . Then, since V ε is an affine function on every segment of S ε , we have Besides, (2.10) gives (we recall that U ε vanishes on every node) ≤ C ε r e(u ε ) L 2 (Sε,r ) ≤ C r ε . (10.10) Then, up to a subsequence, Lemma 29 in the Appendix gives As a consequence of the above convergences, one gets ε r T S ε dU ε,1 ds (1) ).

Lemma 20
Let S be a 3D-periodic structure S is of type S 6 . There exists U ∈ L 2 (Ω; D E,per (S)) such that in Ω × S. (10.12) Proof We decompose Z U in the following way: where Z U (x, ·) is constant on every segment of S for a.e. x ∈ Ω and where the mean value of Z U (x, ·) is equal to zero on every segment of S for a.e. x ∈ Ω. Set U ∈ L 2 (Ω; H 1 0,K (S)) as the solution to The field U = U + U t 1 belongs to L 2 (Ω; D E,per (S)) and satisfies (10.12).

Remark 7
Let S be a 3D-periodic structure and E a field in L 2 (S ε ) such that εξ +εγ There exists φ, a function belonging to H 1 (S ε ) satisfying dφ ds = E, a.e. in S ε , φ = 0 on every node of S ε .
One has The constant does not depend on ε.
in other cases.

(10.13)
Proof Let φ be in D(Ω) and V ∈ H 1 per (S) 3 . We assume V constant in the neighborhood of every node of S.
Consider the field It belongs to H 1 Γ (S ε ) 3 . One has In the beam whose center line is εξ (10.14) By construction v ε belongs to V ε since for every x in B(εξ + εA, c 0 r) In the beam whose center line is εξ + εγ , one has ∂v ε ∂s Hence, passing to the limit in the rescaled stain tensor gives Now, unfolding the LHS of (6.4), passing to the limit and taking into account the above convergence together with (9.3)-(9.4) give Then, we obtain Finally, a density argument followed by a projection on L 2 (Ω; D E,per (S)) allows to replace φ V by any function V ∈ L 2 (Ω; D E,per (S)).

Lemma 22
Suppose the structure of type S 0 . One has Step 1. Preliminary considerations. In One has The function S → (e 1 · S) belongs to H 1 per (S (2) ∪ S (3) ) We extend it as an affine function on every segment of S (1) belonging to H 1 per (S). Denote e 1 this function. Set It belongs to L 2 (Ω; H 1 per (S)) 3 . Hence Consider the field (see Sect. A.5 in the Appendix) V ε . = V [2] ε , it belongs to H 1 Γ (S ε ) 3 . One has In the beam whose center line is εξ In the beam whose center line is εξ + εγ , one has ∂v ε ∂s Hence, passing to the limit in the rescaled stain tensor gives Thus, due to (10.16) Now, unfolding the LHS of (6.4), passing to the limit and taking into account the above convergence together with (9.3) give Then, we obtain Eventually, a density argument ends the proof. We remind (see [23,Lemma 25]) that for every v ∈ D w ⊂ H 1 (D) 3 and every ζ ∈ R 4 , there exists a strictly positive constant C such that

The Limit Unfolded Problem
where

Lemma 23
There exists a strictly positive constant C such that Thus, taking into account the fact that A(M) + A ∈ H 1 per,0 (S) 3 , there exists another constant vector C ∈ R 3 such that a.e. in S.
Since A(M) + A is a periodic function, this leads to Since M is a symmetric matrix, this implies that M = 0 and B = 0. As a consequence we get A(M) = B(M) = 0. Hence B = 0 and then A = 0 since A ∈ H 1 per,0 (S) 3 . The semi-norm is a norm.
By contradiction, as in [23,Lemma 16] we easily show that this norm is equivalent to the following: The space of 3 × 3 matrices is equipped with the Froebinius norm.
Since x ∈ Ω is a parameter, we get Finally, inequality (11.2) holds true thanks to the Korn inequality.
Theorem 1 Let u ε be the solution to (6.4). The fields and functions introduced in Lemma 15 and its corollary satisfy u is the unique solution to the following unfolded problem: (11.5) Furthermore, for all ∈ {1, . . . , m} one has 3×3 . (11.6) Proof This theorem summarizes the results of 22 and 21. We prove the coercivity of problem (11.5). From (11.1) and (11.2), one has The inequality above ensures the coercivity of problem (11.5). Then, since this problem admits a unique solution, the whole sequences in Lemma 15 and Proposition 7 (with u ε the solution to problem (6.4)) converge to their limits. Now, we prove the strong convergence (11.6). First, due to the inclusion of J r in From (8.15) and the fact that r/ε goes to 0 the following convergence holds ( ∈ {1, . . . , m}): Hence, choosing u ε as a test function in (6.4) and using the weak lower semi-continuity of convex functionals, one obtains Thus, all inequalities above are equalities and which in turn leads to the strong convergence (11.6).

Expression of the Warping u
As in [23,Sect. 9.1], we introduce the four warping-correctors (see Sect. A.1 in the Appendix). They belong to L ∞ (S; D w ). We have

Expression of the Microscopic Fields U , R
In this subsection we give the expression of the microscopic fields U , R in terms of U . To this end, we use the formulation (11.4).
Taking v = 0 in (11.4) and then replacing u by its expression (12.2), we obtain the following problem: Here, (12.4) One has As in [23], the symmetric matrix A belongs to L ∞ (S) 4×4 and it satisfies (12.6) Now, we introduce the correctors to solve the problem (12.6) 2 . They are the solutions to the following variational problems: where e 1 = 1 0 0 T , e 2 = 0 1 0 T and e 3 = 0 0 1 T . Hence, a.e. in Ω × S, (12.8) where

Now, Let's Go to the Homogenized Problem
First, observe that from (12.7) we have Hence, in problems (12.3), we replace ( U , R) by (12.8) and we choose ( A, B) = P q=1 e q (V) χ q . Taking into account the above equality, we obtain The above equality leads to the homogenized problem (12.9) where e(V) stands for the column e 1 (V) . . . e P (V) T and where B hom is a symmetric (12.11) In the RHS of (12.9) C hom is a P × 3 matrix, with entries c hom pi :

Lemma 24
The bilinear forms B hom satisfies the following properties: -symmetry, -coercivity, namely there exists C * 0 > 0 such that for every ζ ∈ R P , one has The symmetry of B hom is the consequence of the symmetry of the matrix A. Now we prove (12.13). From equality (12.10) and (12.5) we have Now, we claim that the map is a norm. Indeed, first it is a semi-norm. Now, if |ζ | = 0 then P p=1 ζ p χ p + B(M p ) = C ∈ R 3 . Then, proceeding as in the proof of Lemma 23 we obtain ζ p = 0, p ∈ {1, . . . , P }. The semi-norm is a norm. As a consequence, there exists C * 0 > 0 such that The coercivity follows. Theorem 2 (The homogenized limit problem) If S is a 3D-periodic unstable structure then, the limit field U ∈ V Γ (Ω, S) is the unique solution to the homogenized problem 14) where B hom is given by (12.10) and C hom by (12.12).

Determination of Z U in the Case S of Type S 6
Lemma 25 Let S be a structure of type S 6 then Z U = 0.

The Case of an Isotropic and Homogeneous Material
In the case of an isotropic and homogeneous material the stress tensor is given by where I 3 is the unit 3 × 3 matrix. λ and μ are the material Lamé constants. The correctors χ q ∈ L ∞ (S; D w ), q ∈ {1, 2, 3, 4} are those obtained in [23] (see also [13]).
Hence, we have is the Poisson coefficient.

Determination of U 1 in Ω (1) in the Case S 0
First, remind that U 1 = 0 in Ω (1) . This is why we need to determine the component in the direction e 1 of the limit displacement in Ω (1) .
In this paper we have not introduced applied forces which act with the extensional macroscopic displacements. In fact, this type of displacements is not really important for unstable structures because this only happens for structures of type S 0 with the component of direction e 1 in the open set Ω (1) . 12 Based on Remark 2, for structures of type S 0 we can add to the applied forces given by (6.5) the following: without changing the estimate (6.7). Below we revisit (12.6) 1 .
We assume the structure made of an isotropic and homogeneous material. 13 (see Sect. 13).
That gives: (1) , ∂ 1 ). 12 If S contains only straight lines, we can also consider such forces acting in the whole domains Ω (i) ∩ S ε , i ∈ {1, 2, 3}. We leave this case to the reader. 13 We can proceed in a similar way if A E is constant on every line of S (1) .

Approximate Solution to Problem (6.4)
For our ε-periodic r-thin unstable structure, the solution to the linearized elasticity problem can be reconstructed in the following form: The first term in the above writing gives the macroscopic displacement of the structure. The third term represents the small rotations of the cross-sections while the fourth and last term O r 2 ε stands for the deformations of the cross-sections. Now, we pay attention to the second term, it represents the main part of the local displacement of the centerlines of the beams. Consider a cell εξ + εS; we focus on the points of this cell. In the unfolding transformation, forgetting the macroscopic displacement, a point S of this cell is transformed to give The couple ( U, R) is given by (12.8). It belongs to L 2 (Ω; D I,per (S)). The map S −→ S + ε U εξ, S is of inextensional type, it means that under this transformation the lengths of the centerlines are not modified (neither stretching or compression). Near a node A, we get It means that near a node, this transformation is approximatively a rotation. As a result, the angles between the centerlines are preserved. Now, let's take a look at the transformation For simplicity, we replace the symmetric matrix e(U)(εξ ) by M. In Sect. 3 we have shown that where C(M) ∈ R 3 . Let γ be a segment of S. The components of the restriction to this segment of the associated displacement are Near a node A, we get As a consequence, the angle between two contiguous segments is generally not preserved. The local behavior of the structure is mainly determined by the knowledge of the matrices M (thus of the space M s (S)) and the corresponding solution V (M) to equation (3.1). But the character of the structure (auxetic or not) cannot be deduced from the local behavior since the map S −→ A e(U)(εξ ) (S) + U εξ, S is periodic. In our work, we have considered several basic types of unstable structures, some of these structures are auxetic. Remember, that auxetics are structures or materials that have a negative Poisson's ratio. When stretched, they become thicker perpendicular to the applied forces. This occurs due to their peculiar internal structure and how it deforms when the sample is uniaxially loaded, e.g., if we have simultaneously both inequalities in Ω: In our different types of unstable structures, we distinguish two main kinds, the first which may or may not be "a priori" auxetic: some among those of types S i , i ∈ {0, 1, 2} (see Fig. 1(a), (b), (c) (non-auxetic), Fig. 1(e) (auxetic) and Fig. 1(f) (partially auxetic)). By "a priori" it is meant that the auxetic character of these structures only depend on the space V Γ (Ω, S) of the macroscopic displacements. The second are or are not auxetic "a posteri" (some of types S i , i ∈ {3, 4, 5, 6}); it depends on the applied forces since the space V Γ (Ω, S) of the macroscopic displacements is H 1 Γ (Ω) 3 .

Examples of Cells and Spaces
where κ 12 < 0 and κ 13 < 0. These coefficients depend on the slopes of the oblique segments and their signs mean that the Poisson's ratios in planes parallel to Re 1 ⊕ Re 2 and Re 1 ⊕ Re 3 are positive. We also get This relation means that the structure is auxetic in planes parallel to Re 2 ⊕ Re 3 . The matrix B hom is of size 4 × 4, -cell S of Fig. 1 where κ 12 > 0. This coefficient depends on the slopes of the oblique segments, the matrix B hom is of size 5 × 5.

Concluding Remark Concerning the Mechanical Impact
We recall the solution procedure, to compute the cell problems and find the homogenized elastic coefficients. First, as explained in Sect. 3, we determine the conditions for which equation ( We also want to draw the attention on the mechanical impact of the paper. Since we want to stay in the linear elasticity regime, we need to choose forces, such that the right-hand side functional is bounded in the same order, as the elastic energy. In [19], we gave the order of each single loading component, i.e., the externally applied nodal forces F (A), the moments G(A), and the constant in the cross-section axial expansion forces, f |Sε , in the set of beams without changing the estimate below ((6.7) in this paper), e(u ε ) L 2 (Sε,r ) ≤ C r 2 ε 2 f L ∞ (Ω) + F L ∞ (Ω) + G L ∞ (Ω) . (14.5) This criteria on the applied forces, can be used to design structures, i.e., to find a correct proportion between r and ε, in order to stay in the linear elastic regime under certain required loading.