Thin-walled beams: a derivation ofVlassov theory via Γ-convergence

Lorenzo Freddi ∗ Antonino Morassi†

Roberto Paroni‡

Abstract

This paper deals with the asymptotic analysis of the three-dimen-sional problem for a linearly elastic cantilever having an open cross-section which is the union of rectangles with sides of order ε and ε2,as ε goes to zero. Under suitable assumptions on the given loadsand for homogeneous and isotropic material, we show that the three-dimensional problem Γ-converges to the classical one-dimensional Vlas-sov model for thin-walled beams.

2001 AMS Mathematics Classification Numbers: 74K20, 74B10, 49J45

Keywords: thin-walled cross-section beams, linear elasticity, Γ-convergence, di-mension reduction

1 Introduction

In this paper we continue a line of research initiated in [4] which aims at a rigorousvariational deduction of the one-dimensional theory for thin-walled beams from thethree-dimensional linear elasticity.

In [4] we considered a thin-walled cantilever Ωε = ωε× (0, `) of length `, madeof homogeneous linear isotropic material, with a rectangular cross-section ωε of

∗Dipartimento di Matematica e Informatica, via delle Scienze 206, 33100 Udine, Italy,email: freddi@dimi.uniud.it

†Dipartimento di Georisorse e Territorio, via Cotonificio 114, 33100 Udine, Italy, email:antonino.morassi@uniud.it

‡Dipartimento di Architettura e Pianificazione, Universita degli Studi di Sassari,Palazzo del Pou Salit, Piazza Duomo, 07041 Alghero, Italy, email: paroni@uniss.it

1

sides ε and ε2. By working in the framework of Γ-convergence, see, for example,[3] and [2], we proved that the three-dimensional elasticity problem converges in asuitable variational sense to a one-dimensional problem as ε goes to zero. The limitproblem is defined by a functional which includes the extensional, the flexural andthe torsional strain energies of the classical thin-walled model of beam, as theycan be deduced from De Saint-Venant’s theory. In particular, the strain energydensity of the limit model can be written as a diagonal homogeneous quadraticform of the longitudinal strain, the curvatures of the beam axis evaluated withrespect to the principal planes of bending, and the first derivative of the torsionaltwist. Therefore, the equations of equilibrium show a full decoupling betweenextensional, flexural and torsional effects.

The present paper extends the results obtained in [4] to the case of thin-walled beams with open (i.e., simply connected) multi-rectangular cross-section.More precisely, we consider a three-dimensional cylinder Ωε =

⋃3i=1 Ω(i)

ε , whereΩ(i)

ε = ω(i)ε × (0, `) and ω

(i)ε is a rectangle having sides of order ε and ε2. The

rectangles ω(i)ε partially overlap and are joined so as to achieve, for instance, cross-

sections having a >-like or @-like shape.Under the assumption that a end cross-section of the three-dimensional body is

fixed and the material is homogeneous and isotropic, the theory of Γ-convergenceis used to study the asymptotic behavior of the energy functional as ε goes to zero.

The limit functional strongly depends on the geometry of the cross-section. Inparticular, it is a homogeneous quadratic form of the longitudinal strain, of thetwo bending curvatures of the beam axis, of the first derivative of the twist andalso of the square of the second derivative of the twist, in the case of a sectioncomposed by at least three not aligned rectangles. This last term is concerned withthe so-called nonuniform torsional effects, which are responsible for the presenceof normal stresses induced by torsional deformations and, as a consequence, forthe coupling between extension, flexure and torsion. It should be remarked thatthe above mentioned effects are not included in the classical De Saint-Venant’stheory and have proved to be important in several engineering fields, especially inaeronautical applications where the presence of open cross-sections formed by partswith dimension of different order of magnitude requires more refined mechanicalbeam models.

A direct analysis of the limit energy functional shows that the full decouplingbetween extensional, flexural and torsional problems can be obtained by choosingthe axes of the reference system as the principal axes of the cross-section centeredin his center of mass, and by determining the sector coordinate function - whichis, roughly speaking, the limit of the warping function of the cross-section - withrespect to the shear center.

2

The nonuniform torsion theory is well known in the engineering literature ofthin-walled beams since the old paper by Timoshenko [11] and the fundamentalcontribution by Vlassov [12]. These approximate theories are usually based onsome a-priori assumptions on the deformation of the body and on the inducedstress field. Little attention has been given to the mathematical justification ofVlassov’s theory; Rodriguez and Viano, for example, presented in [10] an extensionof Vlassov’s theory for thin-walled beams as an asymptotic approximation of thethree-dimensional model as the area and the thickness of the cross-section areassumed to tend to zero independently in a suitable order.

The paper is organized as follows. In Section 2 we introduce the three-di-mensional problem and in Section 3 we rewrite it in a variational form on a fixeddomain. The proof of some compactness results for a scaled displacement field ispresented in Section 4. Section 5 is devoted to the establishment of the junctionconditions and to use them to characterize the essential kinematic fields of thecross-section. Γ-convergence results are presented in Sections 6 and 7. The limitenergy functional is investigated in Section 8 and some examples are discussed inSection 9. The contribution of the external loads and the strong convergence ofminimizers are studied in Section 10.Notation. Throughout this article, and unless otherwise stated, we use the Ein-stein summation convention and we index vector and tensor components as follows:Greek indices α, β and γ take values in the set 1, 2 and Latin indices i, j, h inthe set 1, 2, 3. Accordingly, α(i) will denote the parity of the index i, that isthe function which takes the value 1 if i is odd and the value 2 if i is even. Thecomponent k of a vector v will be denoted either with (v)k or vk and an analogousnotation will be used to denote tensor components. Eαβ denotes the Ricci’s sym-bol, that is E11 = E22 = 0, E12 = 1 and E21 = −1. L2(A; B) and Hs(A; B) are thestandard Lebesgue and Sobolev spaces of functions defined on the domain A andtaking values in B, with the usual norms ‖ · ‖L2(A;B) and ‖ · ‖Hs(A;B), respectively.When B = R or when the right set B is clear from the context, we will simply writeL2(A) or Hs(A), sometimes even in the notation used for norms. Convergence inthe norm, that is the so called strong convergence, will be denoted by → whileweak convergence is denoted with .

With a little abuse of notation, and because this is a common practice anddoes not give rise to any mistake, we use to call “sequences” even those familiesindicized by a continuous parameter ε which, throughout the whole paper, will beassumed to belong to the interval (0, 1].

3

2 The 3-dimensional problem

We consider a three-dimensional body which is at rest in the placement Ωε ⊂ R3,where Ωε := ωε × (0, `), ωε := ∪3

i=1ω(i)ε , and

ω(i)ε := (εq1 + εd(i) − εb(i), εq1 + εd(i))× (εq(i)

2 − ε2 s(i)

2, εq

(i)2 + ε2 s(i)

2), i = 1, 3

ω(2)ε := (εq1 − ε2 s(2)

2, εq1 + ε2 s(2)

2)× (εq(1)

2 , εq(3)2 ),

are three non-empty rectangles. Moreover we set h := q(3)2 − q

(1)2 .

s h

d

b

s

ε

ε

ε

ε

ε

qε

x1

x2

ε

qε

ωε(1)

ωε(2)

ωε(3)

(1)

(1)

2 (1)

2

(1)

2

sε 2 (3)

(2)

q2

(3)

bε(3)

dε (3)1

For later convenience we also set

Ω(i)ε := ω(i)

ε × (0, `), i = 1, 2, 3,

and we note that Ωε = Ω(1)ε ∪ Ω(2)

ε ∪ Ω(3)ε and that they are not pairwise disjoint.

Remark 2.1 Within our framework we can cover also cross-sections having a >-like shape by simply setting, for instance, s(3) = b(3) = 0 (of course, in this casethe rectangle ω

(3)ε is empty). A section having a @-like shape can be achieved, for

instance, by replacing the quantities d(1) and d(3) in the figure above with other twoparameters d

(1)ε and d

(3)ε satisfying the properties d

(1)ε /ε → d(1) and d

(3)ε /ε → d(3),

respectively, as ε goes to 0. The @-like shape is reached then by making the choice

4

d(1) = d(3) = s(2)/2 (and, for instance, d(1)ε = d

(3)ε = εs(2)/2). Similarly one can

consider also sections with a or -like shape. Our analysis covers also thesemore general settings but, for simplicity, we will concentrate ourselves to the casedisplayed in the figure above.

In what follows we consider an homogeneous isotropic material, so that theelasticity tensor C writes as

CA = 2µA + λ(trA) I

for every symmetric matrix A. Above, I denotes the 3 × 3 identity matrix. Weassume µ > 0 and λ ≥ 0 so to have, for every symmetric tensor A,

CA ·A ≥ µ|A|2, (1)

where · denotes the scalar product. We shall consider the spaces

H1#(Ωε;R3) :=

u ∈ H1(Ωε;R3) : u = 0 on ωε × 0

and H1#(Ω(i)

ε ;R3) defined in a similar way. We further denote by

Eu(x) := sym(Du(x)) :=Du(x) + DuT (x)

2,

Wu(x) := skw(Du(x)) :=Du(x)−DuT (x)

2,

(2)

the strain of u : Ωε → R3 and the skew symmetric part of the gradient Du.We consider the following total energy functionals

Fε(u) := Jε(u)−∫

Ωε

bε · u dx, (3)

whereJε(u) :=

12

∫

Ωε

CEu ·Eu dx (4)

are the bulk energies and bε ∈ L2(Ωε;R3) are the body forces.Due to the coerciveness inequality (1) and the strict convexity of the integrand,

the total energy functionals Fε admit for every ε a unique minimizer among allcompeting displacements u ∈ H1

#(Ωε;R3). As already explained in the introduc-tion our aim is to study the asymptotic behavior of such minimizers as ε goes to0, through the theory of Γ-convergence, for an account of it we refer to the booksof Braides [2] and Dal Maso [3].

5

3 The rescaled problem

To state our results it is convenient to stretch the domains Ω(i)ε along the transverse

directions x1 and x2 in suitable ways so that the transformed domains do notdepend on ε. Hereafter we denote by ω(i) := ω

(i)1 and let

p(i)ε : Ω(i) → Ω(i)

ε ,

be the (unique) affine invertible transformation between the sets Ω(i) and Ω(i)ε . It

turns out to be defined by

p(i)ε (y1, y2, y3) =

(εα(i)y1 + (ε− εα(i))q1, ε

α(i+1)y2 + (ε− εα(i+1))q(i)2 , y3

)

where α(i) is the parity of i, that is the function which takes the value 1 if i is oddand the value 2 if i is even.

s h

d

b

d

b

s

s

2ε

2ε

ε

ε

ε

ε

ε

2ε

q2

x1

x 2

q2

ε

qε 1

h

q2

y1

y 2

q2

d

b

s

s

s

d

b

q1

pε

pε

pε

ω

ω

ω

ε

(1)

(1)

(1)

(1)

(2)

(3)

(3)

(3)

(3)

(3)

(3)

(2)

(1)

(1)

(1)

(1) (1)

(1)

(3)

(2)

(3)

(2)

(3)

(3)

Given u ∈ H1#(Ωε;R3) we define three functions u(i) ∈ H1

#(Ω(i);R3) by

u(i) := u p(i)ε .

Of course, in the regions where the domains overlap the following two “junctionconditions” must be satisfied

u(i) p(i)ε

−1= u(2) p(2)

ε

−1in Ω(i)

ε ∩ Ω(2)ε , i = 1, 3. (5)

Let us consider the following 3× 3 matrix valued differential operators

H(i)ε u := Du Dp(i)

ε

−1=

(D1uεα(i)

,D2u

εα(i+1), D3u

)

6

where Diu denotes the column vector of the partial derivatives of u with respectto yi. We also set

E(i)ε u := sym(H(i)

ε u), W(i)ε u := skw(H(i)

ε u). (6)

Let us split the bulk energy Jε, defined in (4), into the sum of three energies eachone defined on a rectangular component Ω(i); more precisely

Jε(u) =3∑

i=1

J (i)ε (u), where J (i)

ε (u) :=12

∫

Ω(i)ε

χε(x)CEu ·Eu dx,

and χε : Ωε → 1/2, 1 is defined by

χε(x) :=

1/2 if x ∈ (Ω(1)

ε ∪ Ω(3)ε ) ∩ Ω(2)

ε

1 otherwise.

Let Aε :=(u(1),u(2),u(3)) ∈ ×3

i=1H1#(Ω(i);R3) : conditions (5) are satisfied

;

then we can consider the rescaled bulk energies Iε : Aε → [0, +∞) obtained byrescaling each term I

(i)ε of the sum on the corresponding domain Ω(i) with the

suitable change of variable, that is

Iε(u(1),u(2),u(3)) :=3∑

i=1

I(i)ε (u(i)) (7)

where

I(i)ε (u(i)) :=

1ε3

J (i)ε (u(i) p(i)

ε

−1) =

12

∫

Ω(i)

χ(i)ε (y)CE(i)

ε u(i) ·E(i)ε u(i) dy, (8)

andχ(i)

ε := χε p(i)ε .

Note thatIε(u(1),u(2),u(3)) =

1ε3

Jε(u).

4 Compactness lemmata

In this section we establish the compactness of appropriately rescaled sequences ofdisplacements and prove that the limit functions are displacements of Bernoulli-Navier type. The proof of the next lemma follows immediately from (1).

7

Lemma 4.1 Let (u(1)ε ,u(2)

ε ,u(3)ε ) be a sequence in Aε. If

supε∈(0,1]

1ε4

Iε(u(1)ε ,u(2)

ε ,u(3)ε ) < +∞, (9)

then there exists a constant C > 0 such that

3∑

i=1

‖E(i)ε u(i)

ε ‖L2(Ω(i);R3×3) ≤ Cε2 (10)

for every ε ∈ (0, 1].

To prove the compactness of the displacements we need the following scaled Korninequalities (already obtained in [4] in the particular case i = 2).

Theorem 4.2 Let i ∈ 1, 2, 3. There exists a constant C > 0 such that∫

Ω(i)

(|( u1

εα(i+1)−1,

u2

εα(i)−1,u3

ε2)|2 + |H(i)

ε u|2)

dy ≤ C

ε4

∫

Ω(i)

|E(i)ε u|2 dy (11)

for every u ∈ H1#(Ω(i);R3) and every ε ∈ (0, 1].

Proof. Let us divide the section ω(i)ε in squares of size ε2 and apply Korn’s

inequality (the one obtained by Anzellotti, Baldo and Percivale in [1]; see also [9]and Kondrat’ev and Oleinik [6], Theorem 2) to each beam of length ` and withsection a square with side proportional to ε2. Then, summing over all the obtainedinequalities we have that there exists a constant C > 0 such that

∫

Ω(i)ε

(|u|2 + |Du|2) dx ≤ C

ε4

∫

Ω(i)ε

|Eu|2 dx (12)

for every u ∈ H1#(Ω(i)

ε ;R3).

The inequality∫Ω(i) |H(i)

ε u|2 dy ≤ Cε4

∫Ω(i) |E(i)

ε u|2 dy is simply obtained byrescaling inequality (12). To show that

∫

Ω(i)

|( u1

εα(i+1)−1,

u2

εα(i)−1,u3

ε2)|2 dy ≤ C

ε4

∫

Ω(i)

|E(i)ε u|2 dy,

it suffices to set vε :=( u1

εα(i+1)−1,

u2

εα(i)−1,u3

ε2

), notice that |Eεu| ≥ ε2|Evε| and

apply the standard Korn inequality to vε on the domain Ω (see for instance [8],Theorem 2.7). 2

8

Let us consider the usual space of Bernoulli-Navier displacements on Ω(i)

HBN (Ω(i);R3) :=v ∈ H1

#(Ω(i);R3) : (Ev)jα = 0, j = 1, 2, 3, α = 1, 2

(13)

and setHBN := ×3

i=1HBN (Ω(i);R3). (14)

Analogously we denote by

L2 := ×3i=1L

2(Ω(i)) and H1 := ×3i=1H

1(Ω(i);R3).

Inequality (11) motivates the introduction of the following scaling operators

S(i)ε u :=

( u1

εα(i+1)−1,

u2

εα(i)−1,u3

ε2

). (15)

Lemma 4.3 Let (u(1)ε ,u(2)

ε ,u(3)ε ) be a sequence in Aε which satisfies (10). Then,

for any sequence of positive numbers εn converging to 0 there exist a subsequence(not relabeled) and functions (v(1),v(2),v(3)) ∈ HBN and (ϑ(1), ϑ(2), ϑ(3)) ∈ L2

such that, as n goes to ∞,

S(i)εn

u(i)εn

v(i) in H1(Ω(i);R3), (16)

(W(i)εn

u(i)εn

)12 −ϑ(i) in L2(Ω(i)). (17)

Proof. It is convenient to set v(i)ε := S(i)

ε u(i)ε . It is easily checked that, for ε ≤ 1,

|E(i)ε u(i)

ε | ≥ ε2|Ev(i)ε |, hence, by (10), Ev(i)

ε is uniformly bounded in L2(Ω(i);R3×3),and by Korn’s inequality v(i)

ε is uniformly bounded in H1(Ω(i);R3). It then exists av(i) ∈ H1

#(Ω(i);R3), and a subsequence (not relabeled) of εn such that v(i)εn v(i)

in H1(Ω(i);R3). Again, it is easy to check that |(E(i)ε u(i)

ε )jα| ≥ ε|(Ev(i)ε )jα| for

j = 1, 2, 3 and α = 1, 2, thus, using (10), we deduce that Cε ≥ ‖(Ev(i)ε )jα‖L2(Ω(i))

and consequently (Ev(i))jα = 0. Hence v ∈ HBN (Ω(i);R3).Using assumption (10) together with Theorem 4.2 we obtain that the sequence

H(i)εnu(i)

εn is bounded in L2 so that, up to subsequences, it weakly converges inL2(Ω(i);R3×3) to some H(i) ∈ L2(Ω(i);R3×3). Since, from (10), E(i)

εnu(i)εn → 0 in

L2(Ω(i);R3×3) we have W(i)εnu(i)

εn → H(i) weakly in L2(Ω(i);R3×3). In particular,H(i) is, almost everywhere, a skew-symmetric matrix. Denoting (H(i))12 = −ϑ(i)

we obtain (17). 2

9

Remark 4.4 Using the same notation as in the proof of Lemma 4.3, and since(H(i)

ε u(i)ε )13 = D3u

(i)ε1 = εα(i+1)−1D3v

(i)ε1 , and (H(i)

ε u(i)ε )23 = D3u

(i)ε2 = εα(i)−1D3v

(i)ε2

we note that we have also

(H(i))13 =(α(i)− 1

)D3v

(i)1 and (H(i))23 =

(α(i + 1)− 1

)D3v

(i)2 .

By Lemma 4.3, the displacements v(i) are of Bernoulli-Navier type and there-fore they can be written as

v(i)α = ξ(i)

α (y3), α = 1, 2, v(i)3 = ξ

(i)3 (y3)− yαξ(i)

α

′(y3), (18)

whereξ(i)α ∈ H2

#(0, `) := ξ ∈ H2(0, `) : ξ(0) = ξ′(0) = 0and

ξ(i)3 ∈ H1

#(0, `).

The notation established in this section, in particular in Lemma 4.3 and in (18),will be used throughout the rest of the paper.

5 Junction conditions

The present section is devoted to establish the relationship existing between thelimit fields (v(1),v(2),v(3)) and (ϑ(1), ϑ(2), ϑ(3)) introduced in Lemma 4.3. Hence,during the whole section, we assume that (u(1)

ε ,u(2)ε ,u(3)

ε ) be a sequence in Aε

which satisfies (10), that εn be a sequence of positive numbers converging to 0 andthat a subsequence (not relabeled) and triples of functions (v(1),v(2),v(3)) ∈ HBN

and (ϑ(1), ϑ(2), ϑ(3)) ∈ L2 have been chosen in order to satisfy (16) and (17). Letus moreover use notation (18).

To find relations between the limit fields we study the junction conditions, thatis the system (5), by adapting some inspiring ideas of Le Dret [7] and [5]. Since

p(i)ε

−1(x1, x2, x3) =

(x1 + (εα(i) − ε)q1

εα(i),x2 + (εα(i+1) − ε)q(i)

2

εα(i+1), x3

),

the two junction conditions can be equivalently written as

u(i)(ε(z1 − q1) + q1, z2, z3

)= u(2)

(z1, ε(z2 − q

(i)2 ) + q

(i)2 , z3

),

z ∈ Ω(i) ∩ Ω(2), i = 1, 3,(19)

10

where

Ω(1) ∩ Ω(2) = (q1 − s(2)

2, q1 +

s(2)

2)× (q(1)

2 , q(1)2 +

s(1)

2)× (0, `),

Ω(3) ∩ Ω(2) = (q1 − s(2)

2, q1 +

s(2)

2)× (q(3)

2 − s(3)

2, q

(3)2 )× (0, `).

It is worth notice that in this way also the two junction regions, which originallydepend on ε, have been trasformed into the fixed domains Ω(i) ∩ Ω(2), i = 1, 3.

The following lemma is stated for Ω(1) ∩ Ω(2) but, with straightforward adap-tations, it holds also for Ω(3) ∩ Ω(2).

Lemma 5.1 Let w ∈ H1(Ω(1)∩Ω(2)) and wε ∈ H1(Ω(1)∩Ω(2)) be a sequence suchthat

wε w in H1(Ω(1) ∩ Ω(2)).

Then the sequence of functions

(z2, z3) 7→ −∫ q1+s(2)/2

q1−s(2)/2wε

(ε(z1 − q1) + q1, z2, z3

)dz1

converges in the norm of L2((q(1)2 , q

(1)2 +s(1)/2)× (0, `)) to the trace of the function

w on q1 × (q(1)2 , q

(1)2 + s(1)/2) × (0, `). We will denote such a trace simply by

w(q1, z2, z3).

Proof. As

∫ `

0

∫ q(1)2 +s(1)/2

q(1)2

∣∣∣−∫ q1+s(2)/2

q1−s(2)/2wε

(ε(z1 − q1) + q1, z2, z3

)dz1 − wε(q1, z2, z3)

∣∣∣2dz2dz3 =

=∫ `

0

∫ q(1)2 +s(1)/2

q(1)2

∣∣∣−∫ q1+εs(2)/2

q1−εs(2)/2wε(z1, z2, z3)− wε(q1, z2, z3) dz1

∣∣∣2dz2dz3

=∫ `

0

∫ q(1)2 +s(1)/2

q(1)2

∣∣∣−∫ q1+εs(2)/2

q1−εs(2)/2

∫ z1

q1

D1wε(t, z2, z3) dt dz1

∣∣∣2dz2dz3

≤ s(2)ε

∫ `

0

∫ q(1)2 +s(1)/2

q(1)2

∫ q1+εs(2)/2

q1−εs(2)/2|D1wε(t, z2, z3)|2 dtdz2dz3

≤ s(2)ε‖D1wε‖2L2(Ω(1)∩Ω(2))

≤ Cε,

the claim follows by continuity of the trace. 2

11

Lemma 5.2 The following equalities hold for almost every y3 ∈ (0, `).

1. ξ(i)2 (y3) = 0, i = 1, 3;

2. ξ(2)1 (y3) = 0;

3. ξ(i)3 (y3)− q1ξ

(i)1

′(y3) = ξ

(2)3 (y3)− q

(i)2 ξ

(2)2

′(y3), i = 1, 3.

Proof. As the pair (u(1)εn ,u(2)

εn ) satisfies (19) with i = 1, averaging the secondcomponents with respect to z1 we have

−∫ q1+s(2)/2

q1−s(2)/2u

(1)εn2

(εn(z1 − q1) + q1, z2, z3

)dz1 =

= −∫ q1+s(2)/2

q1−s(2)/2u

(2)εn2

(z1, εn(z2 − q

(1)2 ) + q

(1)2 , z3

)dz1.

Applying Lemma 5.1 to the sequence u(1)εn2 and using (16), we deduce that the

left hand side of the equality above converges to (z2, z3) 7→ v(1)2 (q1, z2, z3) in

L2((q(1)2 , q

(1)2 + s(1)/2) × (0, `)). On the other hand the right hand side converges

to zero in the same space, indeed

∫ `

0

∫ q(1)2 +s(1)/2

q(1)2

∣∣∣−∫ q1+s(2)/2

q1−s(2)/2u

(2)εn2

(z1, εn(z2 − q

(1)2 ) + q

(1)2 , z3

)dz1

∣∣∣2dz2dz3 ≤

≤∫ `

0−∫ q1+s(2)/2

q1−s(2)/2

∫ q(1)2 +s(1)/2

q(1)2

|u(2)εn2

(z1, εn(z2 − q

(1)2 ) + q

(1)2 , z3

)|2 dz2dz1dz3

≤∫ `

0−∫ q1+s(2)/2

q1−s(2)/2

1εn

∫ q(1)2 +εns(1)/2

q(1)2

|u(2)εn2(z1, z2, z3)|2 dz2dz1dz3

≤ εn

s(2)

∫

Ω(2)

∣∣∣u(2)εn2

εn

∣∣∣2dz

and the last term of the chain tends to zero due to (16). Taking into account (18),this proves 1 for i = 1. The case i = 3, follows by the same argument applied tothe other junction condition. The item 2 of the statement, concerning ξ

(2)1 , can be

proved similarly by considering the first component of (19) with i = 1.In order to prove 3 we consider the following scaled average of the third com-

12

ponent of (19) with i = 1

−∫ q

(1)2 +s(1)/2

q(1)2

−∫ q1+s(2)/2

q1−s(2)/2

u(1)εn3

(εn(z1 − q1) + q1, z2, z3

)

ε2n

dz1dz2 =

= −∫ q1+s(2)/2

q1−s(2)/2−∫ q

(1)2 +s(1)/2

q(1)2

u(2)εn3

(z1, εn(z2 − q

(1)2 ) + q

(1)2 , z3

)

ε2n

dz2dz1

and by applying twice Lemma 5.1 and (16) we deduce

−∫ q

(1)2 +s(1)/2

q(1)2

v(1)3 (q1, z2, z3) dz2 = −

∫ q1+s(2)/2

q1−s(2)/2v

(2)3 (z1, q

(1)2 , z3) dz1.

Taking into account (18) and the statements 1 and 2 of the present lemma wededuce item 3 for i = 1. The case i = 3 follows similarly by considering the secondjunction condition. 2

To deduce further “limit” junction conditions we need the following two-dimen-sional Korn’s inequality (see [8], Theorem 2.5)

‖w − ℘w‖H1(ω;R2) ≤ C ‖Ew‖L2(ω;R2×2), (20)

which holds for all w ∈ H1(ω;R2) and where ω is any Lipschitz bounded subset ofR2. If we denote with (y1(G), y2(G)) the center of mass of ω, we have that the α-component of the projection of w on the space of “two-dimensional” infinitesimalrigid displacements, see [4], is (with the summation convention)

℘wα = tα(w) + Eβα (yβ − yβ(G))ϑ(w) (21)

where

ϑ(w) =1

IG(ω)

∫

ωEγδ (yγ − yγ(G))wδ dy1 dy2,

tα(w) =1|ω|

∫

ωwα dy1 dy2.

Above IG(ω) denotes the polar moment of inertia of the section ω with respect tothe center of mass,

IG(ω) :=∫

ω(y1 − y1(G))2 + (y2 − y2(G))2 dy1 dy2.

Analogously, IG(ω(i)) will denote the polar moment of inertia with respect to thecenter of mass (y1(G(i)), y2(G(i)) of the section ω(i). For later convenience we set

w(i)ε :=

S(i)ε u(i)

ε

ε(22)

13

andϑ(i)

ε :=1

IG(ω(i))

∫

ω(i)

Eγδ (yγ − yγ(G(i)))w(i)εδ dy1 dy2,

so that(℘w(i)

ε )α = tα(w(i)ε ) + Eβα (yβ − yβ(G(i)))ϑ(i)

ε . (23)

Lemma 5.3 There exists a constant C > 0 such that

‖w(i)ε − ℘w(i)

ε ‖L2(0,`;H1(ω(i);R2)) ≤ Cε,

for every ε ∈ (0, 1].

Proof. Since (Ew(i)ε )11 = ε(−1)i

(E(i)ε u(i)

ε )11, (Ew(i)ε )22 = ε(−1)i+1

(E(i)ε u(i)

ε )22, and(Ew(i)

ε )12 = (E(i)ε u(i)

ε )12, we have

‖(Ew(i)ε )αβ‖L2(Ω(i)) ≤

1ε‖(E(i)

ε u(i)ε )αβ‖L2(Ω(i)). (24)

Hence, taking into account (20), (24) and (10), we have

∫ `

0‖w(i)

ε − ℘w(i)ε ‖2

H1(ω(i);R2)dy3 ≤ C

∫ `

0

∑

αβ

‖(Ew(i)ε )αβ‖2

L2(ω(i))dy3

≤ C

ε2

∑

αβ

‖(E(i)ε u(i)

ε )αβ‖2L2(Ω(i))

≤ Cε2,

which concludes the proof. 2

Lemma 5.4 We have

1. ϑ(i)εn

ϑ(i) in L2(Ω(i)). Therefore, ϑ(i) does not depend on y1 and y2;

2. ϑ(i) ∈ H1#(0, `);

3. ϑ(1)(y3) = ϑ(2)(y3) = ϑ(3)(y3) =: ϑ(y3) for a.e. y3 in (0, `).

Proof. Let’s prove 1. From Lemma 5.3 we have that

‖Dα(w(i)ε − ℘w(i)

ε )‖L2(Ω(i)) ≤ Cε. (25)

Since (W℘w(i)ε )12 = −ϑ

(i)ε and (Ww(i)

ε )12 = (W(i)ε u(i)

ε )12, from the identity

ϑ(i)ε = −(W℘w(i)

ε )12 = −(W(i)ε u(i)

ε )12 + (W(w(i)ε − ℘w(i)

ε ))12, (26)

14

and using (25), we get the following estimate

‖ϑ(i)ε + (W(i)

ε u(i)ε )12‖L2(Ω(i)) = ‖(W(w(i)

ε − ℘w(i)ε ))12‖L2(Ω(i)) ≤ Cε. (27)

The claim 1 follows then by taking into account (17) and from the fact that, bydefinition, the functions ϑ

(i)εn do not depend on y1 and y2.

Part 2 of the statement will be proven by following the same argument ofLemma 4.6 of [4]. Let ξ ∈ C∞

0 (ω(i)) be such that

∫

ω(i)

ξ dy1 dy2 = −IG(ω(i))2

.

Then, taking into account (23), we have

IG(ω(i))ϑ(i)ε = −2ϑ(i)

ε

∫

ω(i)

ξ dy1 dy2 = −ϑ(i)ε

∫

ω(i)

ξDαyα dy1 dy2

= ϑ(i)ε

∫

ω(i)

yα Dαξ dy1 dy2 = ϑ(i)ε

∫

ω(i)

EαγEβγyβ Dαξ dy1 dy2

=∫

ω(i)

Eαγ(Eβγyβϑ(i)ε )Dαξ dy1 dy2

=∫

ω(i)

Eαγ

((℘w(i)

ε )γ − 1|ω(i)|

∫

ω(i)

w(i)εγ dy1 dy2

)Dαξ dy1 dy2

=∫

ω(i)

Eαγ(℘w(i)ε )γDαξ dy1 dy2

=∫

ω(i)

Eαγw(i)εγ Dαξ dy1 dy2 −

∫

ω(i)

Eαγ(w(i)ε − ℘w(i)

ε )γDαξ dy1 dy2.

Hence, denoting by

ϑ(i)ε :=

1IG(ω(i))

∫

ω(i)

Eαγw(i)εγ Dαξ dy1 dy2,

and recalling (25), we find

ϑ(i)ε − ϑ(i)

ε → 0 in L2(Ω(i)). (28)

We now show that D3ϑ(i)ε is bounded in L2. Since EαγDαDγξ = 0 in ω(i) and

15

Dαξ = 0 on ∂ω(i), we have

IG(ω(i))D3ϑ(i)ε =

∫

ω(i)

EαγDαξ D3w(i)εγ dy1dy2

= 2∫

ω(i)

EαγDαξ (Ewε)γ3 dy1dy2 −∫

ω(i)

EαγDαξ Dγw(i)ε3 dy1dy2

= 2∫

ω(i)

EαγDαξ (Ew(i)ε )γ3 dy1dy2−

∫

ω(i)

Dγ(EαγDαξ w(i)ε3 ) dy1dy2

+∫

ω(i)

EαγDαDγξw(i)ε3 dy1dy2

= 2∫

ω(i)

EαγDαξ (Ew(i)ε )γ3 dy1dy2,

but (Ewε)13 = (Eεuε)13/εα(i+1) and (Ewε)23 = (Eεuε)23/εα(i), and thereforeD3ϑ

(i)ε is bounded in L2(0, `). Thus, from (28) and Lemma 5.4 we conclude that

ϑ(i)ε ϑ(i) in H1(Ω(i)).

Therefore, since ϑ(i)ε (0) = 0, we conclude that ϑ(i) ∈ H1

#(0, `), that is 2.In order to prove 3, let us prove the equality ϑ(i)(y3) = ϑ(2)(y3) for i = 1, 3.

By differentiating (5) we find

εnD1u(i)εn p

(i)εn

−1= D1u

(2)εn p

(2)εn

−1

D2u(i)εn p

(i)εn

−1= εnD2u

(2)εn p

(2)εn

−1in Ω(i)

εn∩ Ω(2)

εn

which immediately lead to

(W(i)

εnu(i)

εn

)12 p(i)

εn

−1=

(W(2)

εnu(2)

εn

)12 p(2)

εn

−1in Ω(i)

εn∩ Ω(2)

εn.

Then, from the equality

ϑ(2)εn

(z3)− ϑ(i)εn

(z3) = −∫

ω(i)εn∩ω

(2)εn

[(W(2)

εnu(2)

εn

)12 p(2)

εn

−1+ ϑ(2)

εn(z3)

]dx1dx2+

−−∫

ω(i)εn∩ω

(2)εn

[(W(i)

εnu(i)

εn

)12 p(i)

εn

−1+ ϑ(i)

εn(z3)

]dx1dx2

=ε3n

|ω(i)εn ∩ ω

(2)εn |

∫

p(2)εn

−1(ω

(i)εn )∩ω(2)

[(W(2)

εnu(2)

εn

)12

+ ϑ(2)εn

(z3)]dy1dy2+

− ε3n

|ω(i)εn ∩ ω

(2)εn |

∫

ω(i)∩p(i)εn

−1(ω

(2)εn )

[(W(i)

εnu(i)

εn

)12

+ ϑ(i)εn

(z3)]dy1dy2,

16

where, with a small abuse of notation, p(i)εn

−1(ω(2)

εn ) and p(2)εn

−1(ω(i)

εn ) denote theinverse of the restriction to ω(i) and ω(2), respectively, of the projection on thefirst two factors of p

(i)εn and p

(2)εn .

As |ω(i)εn ∩ ω

(2)εn | = s(i)s(2)ε4

n, and using Holder’s inequality, we have

∫ `

0|ϑ(i)

εn− ϑ(2)

εn| dy3 ≤

≤ 1s(i)s(2)εn

∫ `

0

∫

p(2)εn

−1(ω

(i)εn )∩ω(2)

∣∣∣(W(2)

εnu(2)

εn

)12

+ ϑ(2)εn

(y3)∣∣∣ dy1dy2dy3+

+1

s(i)s(2)εn

∫ `

0

∫

ω(i)∩p(i)εn

−1(ω

(2)εn )

∣∣∣(W(i)

εnu(i)

εn

)12

+ ϑ(i)εn

(y3)∣∣∣ dy1dy2dy3

≤ Cε1/2n

εn

(‖(W(2)

εnu(2)

εn

)12

+ ϑ(2)εn‖L2(Ω(2)) + ‖(W(i)

εnu(i)

εn

)12

+ ϑ(i)ε ‖L2(Ω(i))

)

≤ Cε1/2n ,

where the estimates (27) have been used to conclude the computation. Hence theright hand side goes to zero as εn → 0 while the liminf of the left hand side isgreater then the L1-norm of ϑ(i) − ϑ(2). Thus, ϑ(i) = ϑ(2). 2

5.1 The case without any junction

It is the simplest case of the rectangular cross-section which has been consideredin detail in [4]. It can be obtained as a particular case of the present setting bytaking b(i) = s(i) = d(i) = 0 for i = 1, 3.

5.2 The case with only one junction

This case arises, for instance, when b(3) = s(3) = d(3) = 0 and turns out to be sim-pler than the general case. Indeed the displacement fields v(1) and v(2), by (18),can be written in terms of six fields ξ

(i)j , i = 1, 2, j = 1, 2, 3, which depend only

on the coordinate y3. These six fields can be reduced, by using Lemma 5.2 toonly three fields, which together with the rotation angle ϑ will fully describe thekinematics of the beam.

Lemma 5.5 We have that

ϑ = ϑ(1) = ϑ(2) ∈ H1#(0, `),

17

and there exist η1, η2 ∈ H2#(0, `) and η3 ∈ H1

#(0, `) such that

v(1)1 = η1(y3), v

(1)2 = 0, v

(1)3 = η3(y3)− y1η

′1(y3)− q

(1)2 η′2(y3),

v(2)1 = 0, v

(2)2 = η2(y3), v

(2)3 = η3(y3)− q1η

′1(y3)− y2η

′2(y3).

(29)

for almost every y3 ∈ (0, `).

Proof. The first part of the statement has been already proven in Lemma 5.4. Itremains to show that equalities (29) hold. That v

(2)1 = v

(1)2 = 0 follows from (18)

and Lemma 5.2, and that v(1)1 = η1(y3) and v

(2)2 = η2(y3) follows after setting

η1 = ξ(1)1 and η2 = ξ

(2)2 . From 3 of Lemma 5.2 follows that

ξ(1)3 − q1η

′1 = ξ

(2)3 − q

(1)2 η′2,

thus settingη3 := ξ

(2)3 + q1η

′1 = ξ

(1)3 + q

(2)2 η′2,

we deduce the desired expressions of v(1)3 and v

(2)3 . 2

5.3 The case with two junctions

In the case with a beam section composed by three rectangles the displacement ofthe beam is described by the rotation angle ϑ and by nine fields ξ

(i)j , i, j = 1, 2, 3,

all depending only on the coordinate y3. Lemma 5.2 gives us five conditions towhich these fields have to satisfy. Thus, using these five conditions, we can reducethe ten original fields to only five fields and not to four as in the case of only onejunction. This fact points out that we are still missing a junction condition.

The next lemma holds only if there are two distinct junctions.

Lemma 5.6 We have that

ξ(1)1 (y3)− ξ

(3)1 (y3) = hϑ(y3)

for almost every y3 ∈ (0, `). Moreover ϑ ∈ H2#(0, `).

Proof. From the first component of (19), rescaling, integrating and making achange of variables, we obtain for a.e. z3 ∈ (0, `)

−∫ q1+εns(2)/2

q1−εns(2)/2−∫ q

(1)2 +s(1)/2

q(1)2

u(1)εn1

εndz2dz1 = −

∫ q1+s(2)/2

q1−s(2)/2−∫ q

(1)2 +εns(1)/2

q(1)2

u(2)εn1

εndz2dz1,

−∫ q1+εns(2)/2

q1−εns(2)/2−∫ q

(3)2

q(3)2 −s(3)/2

u(3)εn1

εndz2dz1 = −

∫ q1+s(2)/2

q1−s(2)/2−∫ q

(3)2

q(3)2 −εns(3)/2

u(2)εn1

εndz2dz1.

(30)

18

Recalling (22), the difference of the right hand sides of the two equations abovecan be rewritten as

−∫ q1+s(2)/2

q1−s(2)/2

(−∫ q

(1)2 +εns(1)/2

q(1)2

w(2)εn1 dz2 −−

∫ q(3)2

q(3)2 −εns(3)/2

w(2)εn1 dz2

)dz1 =

= −∫ q1+s(2)/2

q1−s(2)/2

(−∫ q

(1)2 +εns(1)/2

q(1)2

w(2)εn1 − (℘w(2)

εn)1 dz2+

−−∫ q

(3)2

q(3)2 −εns(3)/2

w(2)εn1 − (℘w(2)

εn)1 dz2

)dz1+

+−∫ q1+s(2)/2

q1−s(2)/2

(−∫ q

(1)2 +εns(1)/2

q(1)2

(℘w(2)εn

)1 dz2 −−∫ q

(3)2

q(3)2 −εns(3)/2

(℘w(2)εn

)1 dz2

)dz1,

but (℘w(2)εn

)1(z) =(q

(1)2 + q

(3)2

2− z2

)ϑ(2)

εn+−

∫

ω(2)

w(2)εn1 dz1dz2 and therefore

−∫ q1+s(2)/2

q1−s(2)/2

(−∫ q

(1)2 +εns(1)/2

q(1)2

(℘w(2)εn

)1 dz2 −−∫ q

(3)2

q(3)2 −εns(3)/2

(℘w(2)εn

)1 dz2

)dz1 =

= hϑ(2)εn− εn

s(1) + s(3)

2ϑ(2)

εn.

Taking into account the identities above and using Lemma 5.3, from the differenceof (30) we deduce

limn→∞

∫ `

0

∣∣∣−∫ q1+εns(2)/2

q1−εns(2)/2

(−∫ q

(1)2 +s(1)/2

q(1)2

u(1)εn1

εndz2 −−

∫ q(3)2

q(3)2 −s(3)/2

u(3)εn1

εndz2

)dz1+

−hϑ(2)εn

∣∣∣ dz3 = 0

and, applying Lemma 5.1, we get

∫ `

0

∣∣∣−∫ q

(1)2 +s(1)/2

q(1)2

v(1)1 (q1, z2, z3) dz2 −−

∫ q(3)2

q(3)2 −s/2

v(3)1 (q1, z2, z3) dz2 − hϑ

∣∣∣ dy3 = 0.

The statement of the lemma follows from (18). 2

The next lemma, which summarizes the results for a beam with a sectionhaving two junctions, states that the transverse displacement of the cross sectionis described by a rigid translation, of components η1 and η2, and a rotation ϑaround a point c.

19

Lemma 5.7 For every c = (c1, c2) ∈ R2 there exist η1, η2 ∈ H2#(0, `) and η3 ∈

H1#(0, `) such that

v(i)1 = η1(y3)− (q(i)

2 − c2)ϑ(y3), v(i)2 = 0, i = 1, 3,

v(2)1 = 0, v

(2)2 = η2(y3) + (q1 − c1)ϑ(y3),

v(i)3 = η3(y3)− y1η

′1(y3)− q

(i)2 η′2(y3) + ψ(i)(y1)ϑ′(y3), i = 1, 3,

v(2)3 = η3(y3)− q1η

′1(y3)− y2η

′2(y3) + ψ(2)(y2)ϑ′(y3),

(31)where the so-called “sector coordinates”

ψ(i)(y1) := y1(q(i)2 − c2)− q1(q

(i)2 − c2)− (q1 − c1)q

(i)2 + K, i = 1, 3,

ψ(2)(y2) := −(q1 − c1)y2 + K,

were y1 ∈ (q1+d(i)−b(i), q1+d(i)) and y2 ∈ (q(1)2 , q

(3)2 ), are defined up to an additive

constant K.

Proof. The equalities v(2)1 = v

(1)2 = v

(3)2 = 0 follow from (18) and Lemma

5.2. From (18) and Lemma 5.6, setting η1 := ξ(1)1 + (q(1)

2 − c2)ϑ, we deducev

(1)1 = η1 − (q(1)

2 − c2)ϑ and ξ(3)1 = η1 − (q(1)

2 + h− c2)ϑ = η1 − (q(3)2 − c2)ϑ. Thus

v(3)1 = η1−(q(3)

2 −c2)ϑ. Setting η2 := ξ(2)2 −(q1−c1)ϑ, we have v

(2)2 = η2+(q1−c1)ϑ.

Moreover, by definition, η1, η2 ∈ H2#(0, `).

Let K be any constant. Setting

η3 := ξ(2)3 + q1η

′1 −Kϑ′,

from (18) we immediately obtain the desired expression for v(2)3 and η3 ∈ H1

#(0, `).From statement 3 of Lemma 5.2 we deduce

ξ(i)3 = η3 − q

(i)2 η′2 +

(− q1(q(i)2 − c2)− (q1 − c1)q

(i)2 + K

)ϑ′,

from which follow the expressions for v(i)3 , i = 1, 3. 2

We finally note that the sector coordinates depend on the coordinates of thepoint c and an additive constant K. For simplicity of notation, in the sequel wedo not explicitly stress this dependence.

20

Remark 5.8 The main difference between Lemma 5.5 and 5.7, which respectivelyhold for one and two junction conditions, is that in the former ϑ is only oncedifferentiable while in the latter it is twice differentiable. This higher regularitywill be used to construct the recovery sequence in Section 7. On the other hand,the same procedure can be applied also in the case of only one junction by simplyassuming ϑ smooth at the starting step (38).

We also note that formally we can deduce the displacements for a beam withonly one junction from Lemma 5.7 by setting c1 = q1, c2 = q

(1)2 and K = 0.

Remark 5.9 The same technique can be used to treat hollow cross-sections. Inthis case it can be proved that the twist angle ϑ vanishes, which suggests thata different scaling of the rotation of the section, at level ε, is needed. This willrequire a further study.

6 A liminf inequality

In this section we deduce a lower bound for the limiting energy. We start bydeducing the convergence of some of the components of the rescaled strain. Theresults of this section hold, with straightforward adaptations, also in the case ofonly one junction.

Lemma 6.1 We have, up to subsequences,

(E(i)εnu(i)

εn )33

ε2n

D3v(i)3 = η′3 − y1η

′′1 − q

(i)2 η′′2 + ψ(i)(y1)ϑ′′ in L2(Ω(i)), i = 1, 3,

(E(2)εn u(2)

εn )33

ε2n

D3v(2)3 = η′3 − q1η

′′1 − y2η

′′2 + ψ(2)(y2)ϑ′′ in L2(Ω(2)),

(E(i)εnu(i)

εn )13

ε2n

−(y2 − q(i)2 )ϑ′ + η(i) in L2(Ω(i)), i = 1, 3,

(E(2)εn u(2)

εn )23

ε2n

+(y1 − q1)ϑ′ + η(2) in L2(Ω(2)),

(32)where η(i) ∈ L2(Ω(i)), i = 1, 3, are independent of y2, while η(2) ∈ L2(Ω(2)) isindependent of y1.

Proof. To prove the first and the second of the (32) it suffices to notice that(E(i)

ε u(i)ε )33 = D3u

(i)ε3 , divide by ε2 and apply (16).

The statements concerning the quantities defined on Ω(2) are proven in [4],Lemma 4.7. The similar statements on the quantities defined on Ω(i), i = 1, 3, canbe obtained by carefully exchanging the exponents of ε.

21

From (10) we deduce that, up to subsequences,(E(i)

ε u(i)ε )α3

ε2 E

(i)α3 in L2(Ω(i)).

To characterize E(i)α3 ∈ L2(Ω(i)), note that

2D3(W(i)ε u(i)

ε )12 = D3

(D2u

(i)ε1

εα(i+1) − D1u(i)ε2

εα(i)

)

= D2

(D3u

(i)ε1

εα(i+1)+

D1u(i)ε3

ε3

)−D1

(D2u

(i)ε3

ε3+

D3u(i)ε2

εα(i)

)

= 2D2(E(i)

ε u(i)ε )13

εα(i+1)− 2D1

(E(i)ε u(i)

ε )23

εα(i),

in the sense of distributions. Hence for ψ ∈ C∞0 (Ω(i)) we have

∫

Ω(i)

(W(i)ε u(i)

ε )12D3ψ dy =∫

Ω(i)

(E(i)ε u(i)

ε )13

εα(i+1)D2ψ dy −

∫

Ω(i)

(E(i)ε u(i)

ε )23

εα(i)D1ψ dy.

On the other hand, using (10) we have that

(E(i)ε u(i)

ε )13

εα(i+1) 0 in L2(Ω(i)), if i = 2

(E(i)ε u(i)

ε )23

εα(i) 0 in L2(Ω(i)), if i = 1, 3.

Hence, passing to the limit in the previous equality we find∫

Ω(i)

−ϑD3ψ dy = −∫

Ω(i)

E(i)23 D1ψ dy, if i = 2,

∫

Ω(i)

−ϑD3ψ dy =∫

Ω(i)

E(i)13 D2ψ dy, if i = 1, 3.

ThusD1E

(i)23 = D3ϑ, if i = 2,

D2E(i)13 = −D3ϑ, if i = 1, 3.

in the sense of distributions, and therefore, taking into account that ϑ does notdepend on y1 and y2 we have that

E(i)23 = y1D3ϑ + γ(i), if i = 2,

E(i)13 = −y2D3ϑ + γ(i), if i = 1, 3.

with γ(i) independent of y2 if i = 1, 3 and of y1 if i = 2. The conclusion is obtainedby setting η(i) := γ(i) − q2D3ϑ if i = 1, 3 and η(2) := γ(2) + q1D3ϑ. 2

22

Remark 6.2 In order to shorten notation, let us observe that the statement ofLemma 6.1 can be summarized as follows

(E(i)εnu(i)

εn )α(i)3

ε2n

Eβα(i)(yβ − q(i)β )ϑ′ + η(i) in L2(Ω(i)), i = 1, 2, 3,

where we agree that q(i)1 := q1 for any i, and α(i) is the parity of the index i.

As previously said, we consider homogeneous isotropic materials and denoteby

f(A) :=12CA ·A = µ|A|2 +

λ

2|trA|2, (33)

the stored energy density. Let

f0(α, β) : = minf(A) : A ∈ Sym, A23 = α, A33 = β= minf(A) : A ∈ Sym, A13 = α, A33 = β

Let us remark that, by isotropy, in the definition of f0, A23 can be replaced withA13. A simple computation shows that

f0(α, β) = 2µα2 +12Eβ2 (34)

where E is the Young modulus, which is given by

E =µ(2µ + 3λ)

µ + λ.

Theorem 6.3 For every sequence of positive numbers εn converging to 0 and forevery sequence (u(1)

εn ,u(2)εn ,u(3)

εn ) ∈ Aεn which satisfies (32), we have

lim infn→+∞

1ε4n

Iεn(u(1)εn

,u(2)εn

,u(3)εn

) ≥3∑

i=1

∫

Ω(i)

f0

(Eβα(i)(yβ − q(i)β )ϑ′, D3v

(i)3

)dy,

where q(i)1 := q1.

Proof. Taking into account the decomposition given in (7), it suffices to showthat

lim infn→+∞

1ε4n

I(i)εn

(u(i)εn

) ≥∫

Ω(i)

f0

(Eβα(i)(yβ − q(i)β )ϑ′, D3v

(i)3

)dy.

Looking at the expression (8) of the functional I(i)ε and observing that, by the

definitions of f and f0 given in (33) and (34),

12CA ·A ≥ f0(Aα(i)3, A33),

23

we have

1ε4n

I(i)εn

(u(i)εn

) ≥∫

Ω(i)

χ(i)εn

f0((E(i)

εnu(i)εn )α(i)3

ε2n

,(E(3)

εn u(3)εn )33

ε2n

) dy.

Using the convexity of f0,

∫

Ω(i)

χ(i)εn

f0((E(3)

εn u(i)εn )α(i)3

ε2n

,(E(i)

εnu(i)εn )33

ε2n

) dy ≥

≥∫

Ω(i)

χ(i)εn

4µ[(E(i)

εnu(i)εn )α(i)3

ε2n

− Eβα(i)(yβ − q(i)β )ϑ′ − η(i)

]·

·[Eβα(i)(yβ − q(i)β )ϑ′ + η(i)

]dy

+∫

Ω(i)

χ(i)εn

E[(E(i)

εnu(i)εn )33

ε2n

−D3v(i)3

]D3v

(i)3 dy

+∫

Ω(i)

χ(i)εn

f0

(Eβα(i)(yβ − q(i)β )ϑ′ + η(i), D3v

(i)3

)dy

and by Lemma 6.1 we find

lim infk→+∞

1ε4n

I(i)εn

(u(i)εn

) ≥∫

Ω(i)

f0

(Eβα(i)(yβ − q(i)β )ϑ′ + η(i), D3v

(i)3

)dy.

On the other hand, by (34),∫

Ω(i)

f0

(Eβα(i)(yβ − q(i)β )ϑ′ + η(i), D3v

(i)3

)dy =

=∫

Ω(i)

f0(Eβα(i)(yβ − q(i)β )ϑ′, D3v

(i)3 )+

+4µ

∫

Ω(i)

Eβα(i)(yβ − q(i)β )ϑ′η(i) dy + 2µ

∫

Ω(i)

η(i)2 dy.

and the proof is concluded by observing that∫

Ω(i)

Eβα(i)(yβ − q(i)β )ϑ′η(i) dy = 0,

because η(i) is independent of y3−α(i), by Lemma 6.1. 2

24

7 The limit energy

The content of this section refers to the case of two junctions, but it can be adaptedwith straightforward modifications to the case of a single junction.

Let

A :=

(v(1),v(2),v(3), ϑ) ∈ HBN ×H2#(0, `) :

: ∃ η1, η2 ∈ H2#(0, `), and η3 ∈ H1

#(0, `) satisfying (31)

.(35)

Theorem 7.1 Let I : H1 ×H2(0, `) → [0, +∞) be defined by

I(v(1),v(2),v(3), ϑ) :=

3∑

i=1

I(i)(v(i), ϑ) if (v(1),v(2),v(3), ϑ) ∈ A,

+∞ otherwise,

(36)

whereI(i)(v(i), ϑ) =

∫

Ω(i)

f0

(Eβα(i)(yβ − q(i)β )ϑ′, D3v

(i)3

)dy.

As ε → 0+, the sequence of functionals1ε4

Iε Γ-converges to the functional I, inthe following sense:

1. [liminf inequality] for every sequence of positive numbers εn converging to 0and for every sequence (u(1)

εn ,u(2)εn ,u(3)

εn ) ∈ Aεn such that

S(i)εn

u(i)εn

v(i) in H1(Ω(i);R3),

(W(i)εn

u(i)εn

)12 −ϑ in L2(Ω(i)),

we havelim infn→+∞

1ε4n

Iεn(u(1)εn

,u(2)εn

,u(3)εn

) ≥ I(v(1),v(2),v(3), ϑ);

2. [recovery sequence] for every sequence of positive numbers εn converging to 0and for every (v(1),v(2),v(3), ϑ) ∈ A there exists a sequence (u(1)

n ,u(2)n ,u(3)

n ) ∈Aεn such that

S(i)εn

u(i)n v(i) in H1(Ω(i);R3),

(W(i)εn

u(i)n )12 −ϑ in L2(Ω(i)),

andlim supn→+∞

1ε4n

Iεn(u(1)n ,u(2)

n ,u(3)n ) ≤ I(v(1),v(2),v(3), ϑ).

25

Proof. We start by proving the liminf inequality. Without loss of generality wemay suppose that

lim infn→+∞

1ε4n

Iεn(u(1)εn

,u(2)εn

,u(3)εn

) < +∞,

and therefore thatsup

n

1ε4n

Iεn(u(1)εn

,u(2)εn

,u(3)εn

) < +∞.

Thus the assumptions of Lemma 4.1 are satisfied and therefore Lemma 4.3, Lemma5.7 and Lemma 6.1 hold. The liminf inequality then follows from Theorem 6.3.

We now find a recovery sequence. Let us first note that

f0(α, β) = f(Λ(i)(α, β)),

where Λ(i) are symmetric matrices with

Λ(i)11 (α, β) = Λ(i)

22 (α, β) = −νβ, Λ(i)12 (α, β) = 0, Λ(i)

33 (α, β) = β, i = 1, 2, 3,

Λ(i)23 (α, β) = Λ(2)

13 (α, β) = 0, Λ(i)13 (α, β) = Λ(2)

23 (α, β) = α, i = 1, 3,(37)

and ν denotes the Poisson’s coefficient

ν =λ

2(λ + µ).

Let us assume that I(v(1),v(2),v(3), ϑ) < +∞, otherwise there is nothing to prove.Then (v(1),v(2),v(3), ϑ) ∈ A and therefore there exist η1, η2 ∈ H2

#(0, `), and η3 ∈H1

#(0, `) such that (31) hold.To start, we further assume η1, η2, η3 and ϑ to be equal to zero in a neigh-

borhood of y3 = 0. Let u0,ε : Ωε → R3, be the sequence of functions definedby

u0,ε := uf,ε + ut,ε (38)

with(uf,ε)1 := εη1 − ν

(ε2x1η

′3 +

ε

2(x2

2 − x21)η

′′1 − εx1x2η

′′2

),

(uf,ε)2 := εη2 − ν(ε2x2η

′3 − εx1x2η

′′1 +

ε

2(x2

1 − x22)η

′′2

),

(uf,ε)3 := ε2η3 − εx1η′1 − εx2η

′2,

(39)

and(ut,ε)1 := −(x2 − εc2)ϑ− νrε

1ϑ′′,

(ut,ε)2 := (x1 − εc1)ϑ− νrε2ϑ′′,

(ut,ε)3 := ψεϑ′,

(40)

26

where

rε1(x1, x2) :=

ε3(∫ y1

q1

ψ(1)(s) ds κ(1)ε ) p(1)

ε

−1in ω

(1)ε ,

ε4(ψ(2)(y2)y1κ(2)ε κ(2)

ε ) p(2)ε

−1in ω

(2)ε ,

ε3(∫ y1

q1

ψ(3)(s) ds κ(3)ε ) p(3)

ε

−1in ω

(3)ε ,

rε2(x1, x2) :=

ε4(ψ(1)(y1)y2 κ(1)ε ) p(1)

ε

−1in ω

(1)ε ,

ε3( ∫ y2

q(1)2

ψ(2)(s) dsκ

(2)ε + κ

(2)ε

2) p(2)

ε

−1in ω

(2)ε ,

ε4(ψ(3)(y1)y2 κ(3)ε ) p(3)

ε

−1in ω

(3)ε ,

and

ψε(x1, x2) :=

ε2ψ(1)ε p

(1)ε

−1in ω

(1)ε ,

ε2ψ(2)ε p

(2)ε

−1in ω

(2)ε ,

ε2ψ(3)ε p

(3)ε

−1in ω

(3)ε ,

where

ψ(1)ε := −ε(y1 − q1)(y2 − q

(1)2 )κ(1)

ε + ψ(1) − ε(q1 − c1)(y2 − q(1)2 ),

ψ(2)ε := κ

(2)ε κ

(2)ε (ψ(2) + ε(y1 − q1)(y2 − c2))+

+(1− κ(2)ε )(ψ(2) + ε(y1 − q1)(q

(3)2 − c2))+

+ (1− κ(2)ε )(ψ(2) + ε(y1 − q1)(q

(1)2 − c2)),

ψ(3)ε := −ε(y1 − q1)(y2 − q

(3)2 )κ(3)

ε + ψ(3) − ε(q1 − c1)(y2 − q(3)2 ).

In the definitions above the cut-off functions κ(1)ε : ω(1) → [0, 1], κ

(3)ε : ω(3) → [0, 1],

and κ(2)ε , κ

(2)ε : ω(2) → [0, 1], have the following properties

D2κ(1)ε = D2κ

(3)ε = D1κ

(2)ε = D1κ

(2)ε = 0,

|D1κ(1)ε |, |D1κ

(3)ε |, |D2κ

(2)ε |, |D2κ

(2)ε | ≤ 2

ε,

κ(3)ε (y1, y2) = κ(1)

ε (y1, y2) = 0 for q1 − εs(2)

2≤ y1 ≤ q1 + ε

s(2)

2,

κ(3)ε (y1, y2) = κ(1)

ε (y1, y2) = 1 for y1 ≥ q1 + εs(2)

2+ ε and y1 ≤ q1 − ε

s(2)

2− ε,

27

κ(2)ε (y1, y2) = 0 for y2 ≤ q

(1)2 + ε

s(1)

2, κ(2)

ε (y1, y2) = 1 for y2 ≥ q(1)2 + ε

s(1)

2+ ε,

κ(2)ε (y1, y2) = 0 for y2 ≥ q

(3)2 − ε

s(3)

2, κ(2)

ε (y1, y2) = 1 for y2 ≤ q(3)2 − ε

s(3)

2− ε.

It can be easily checked that ψε is well defined.From the definitions above one can easily compute u(i)

0,ε := u(i)0,ε p

(i)ε and verify

that u(i)0,ε satisfies the following estimates

‖E(i)ε u(i)

0,ε

ε2−Λ(i)(Eβα(i)(yβ − q

(i)β )ϑ′, D3v

(i)3 )‖L2(Ω(i)) ≤ εC(v(i), ϑ),

‖(W(i)ε u(i)

0,ε)12 + ϑ‖L2(Ω(i)) ≤ εC(v(i), ϑ),

‖S(i)ε u(i)

0,ε − v(i)‖H1(Ω(i)) ≤ εC(v(i), ϑ),

(41)

where C(v(i), ϑ) depends only on v(i) and ϑ. Hence in this case (u0,εn) is a recoverysequence.

In the general case, i.e., (v(1),v(2),v(3), ϑ) ∈ A, we can find, by convolution,functions (v(1)

k ,v(2)k ,v(3)

k , ϑk) ∈ A which are smooth, equal to zero near by y3 = 0and such that

‖Λ(i)(Eβα(i)(yβ − q(i)β )ϑ′k, D3v

(i)3k )−Λ(i)(Eβα(i)(yβ − q

(i)β )ϑ′, D3v

(i)3 )‖L2(Ω(i)) ≤ 1/k,

‖ϑk − ϑ‖L2(Ω(i)) ≤ 1/k,

‖v(i)k − v(i)‖H1(Ω(i)) ≤ 1/k,

for every k. Denoting by uk,ε the sequence defined as u0,ε in (38) but with η(i)’sand ϑ replaced by the η

(i)k ’s and ϑk used in the definition of (v(1)

k ,v(3)k ,v(2)

k , ϑk) ∈A, given a sequence εn converging to zero, we can find an increasing sequenceof integers (kn) and therefore a diagonal u(i)

n := u(i)kn,εn

such that the sequence

un = (u(1)n ,u(2)

n ,u(3)n ) satisfies the required recovery sequence conditions. 2

8 The elastic energy and the shear center

The limit energy functional I(v(1),v(2),v(3), ϑ) can be written in a more explicitform by using Lemma 5.7 and the fact that η1, η2, η3 and ϑ depend only on y3.

28

Indeed, the limit strain energy rewrites as

I(η, ϑ) =∫ `

0

µ

2Jtϑ

′2 +E

2

A −S2 −S1 Sψ

I22 I12 −Iψ2

I11 −Iψ1

sym Iψψ

η′3η′′1η′′2ϑ′′

·

η′3η′′1η′′2ϑ′′

dy3

(42)where (denoting with da = dy1dy2)

A =∫

ω(1)

da +∫

ω(3)

da +∫

ω(2)

da = b(1)s(1) + b(3)s(3) + hs(2)

is the total area,

S1 =∫

ω(1)

q(1)2 da +

∫

ω(3)

q(3)2 da +

∫

ω(2)

y2 da,

S2 =∫

ω(1)

y1 da +∫

ω(3)

y1 da +∫

ω(2)

q1 da

are the static moments,

Jt = 4(∫

ω(1)

(y2 − q(1)2 )2 da +

∫

ω(3)

(y2 − q(3)2 )2 da +

∫

ω(2)

(y1 − q1)2 da)

=13(b(1)s(1)3 + b(3)s(3)3 + hs(2)3)

is the “torsional” moment of inertia,

I11 =∫

ω(1)

q(1)2

2da +

∫

ω(3)

q(3)2

2da +

∫

ω(2)

y22 da,

I22 =∫

ω(1)

y21 da +

∫

ω(3)

y21 da +

∫

ω(2)

q21 da,

I12 =∫

ω(1)

y1q(1)2 da +

∫

ω(3)

y1q(3)2 da +

∫

ω(2)

q1y2 da,

are the moments of inertia, and

Sψ =∫

ω(1)

ψ(1)(y1) da +∫

ω(3)

ψ(3)(y1) da +∫

ω(2)

ψ(2)(y2) da,

Iψ1 =∫

ω(1)

q(1)2 ψ(1)(y1) da +

∫

ω(3)

q(3)2 ψ(3)(y1)da +

∫

ω(2)

y2ψ(2)(y2) da,

Iψ2 =∫

ω(1)

y1ψ(1)(y1) da +

∫

ω(3)

y1ψ(3)(y1) da +

∫

ω(2)

q1ψ(2)(y2) da,

Iψψ =∫

ω(1)

ψ(1)(y1)2 da +∫

ω(3)

ψ(3)(y1)2 da +∫

ω(2)

ψ(2)(y2)2 da,

29

are the sectorial statical moment, the sectorial products of inertia and the sectorialmoment of inertia, respectively.

We note that, as it stands, the extensional, flexural and torsional problems areall coupled together. If the axes are centered in the center of mass, i.e. S1 = S2 = 0,and are principal axes, i.e. I12 = 0, then there is coupling only between extensionand torsion through the coupling term Sψη′3ϑ

′′, and flexure and torsion throughthe coupling terms Iψαη′′αϑ′′. We will have full decoupling only in the case in whichalso Sψ = Iψ1 = Iψ2 = 0. We recall (see Lemma 5.7) that the sector coordinatesdepend on three parameters, the coordinates of the point c = (c1, c2) and theadditive constant K. Hence we can solve the system

Sψ = 0Iψ1 = 0Iψ2 = 0

(43)

of three equations in the three unknowns c1, c2 and K. Roughly speaking we cansay that we choose the constant K so that the sector coordinate will have nullmean, i.e., Sψ = 0, and then the point c so to make the sectorial products ofinertia Iψ1, Iψ2 equal to zero. This latter point is usually called, in the engineeringliterature, the shear center and the resulting Iψψ is the warping stiffness.

Remark 8.1 When only one junction is present, which happens for instance ifb(3) = s(3) = d(3) = 0 (see Section 5.2) then, from Remark 5.8 follows that thesector coordinates can be chosen equal to zero and the limit strain energy functionaltakes the form

I(η, ϑ) =∫ `

0

µ

2Jtϑ

′2 +E

2

A −S2 −S1

I22 I12

sym I11

η′3η′′1η′′2

·

η′3η′′1η′′2

dy3.

It is worth notice that this De Saint-Venant limit energy is formally the samewhich holds in the case when there are no junctions (see [4], Section 7).

9 Some examples

9.1 Single-junction sections

By Lemma 5.5, Lemma 5.7 and Remark 5.8 (see also Remark 8.1) follows thatwhen the cross-section consists of straight segments connected by a unique commonjunction, that is when the one dimensional limit structure is star-shaped, then thewarping stiffness is zero, the shear center is the common point of intersection. Thefigure below shows three examples of this kind of sections.

30

9.2 Channel-section

Consider the so-called Channel-section shown in the figure below.

s

b

h=2b

x =y

x

1

2

G

1

y2

c

The thickness s is much smaller then the height h. Let us follow the scheme ofSection 8. The area is

A = 4bs.

The point G with coordinates x1(G) = −b/4 and x2(G) = 0 is the center of massof the cross-section. Choosing a reference system Gy1y2 with origin in G we havethat q1 = b/4, q

(1)2 = b and q

(1)2 = −b. Hence, from the integration formulae of

Section 8, we easily get

S1 = 0, S2 = 0, Jt =43bs3, I11 =

83b3s, I22 =

512

b3s, I12 = 0,

Sψ = 4Kbs, Iψ1 = −sb3

3(5b− 8c1), Iψ2 = − 5

12sb3c2.

Solving the system (43) we obtain that K = 0 and the shear center c has coordi-nates c1 = 5

8b and c2 = 0, in the reference Gy1y2, as shown in the figure above.The resulting warping stiffness is Iψψ = 7

24b5s.

31

9.3 Symmetric double-T section

Consider the double-T section shown in the figure below.

s

b

h/2

x

x =y

1

2

G

2

c

b

s

s

y1

h/2

(1)

(1)

(2)

(3)

(3)

The thicknesses s(3), s(2) and s(1) are much smaller then the height h. Let usfollow the scheme of Section 8. It is useful to set A(3) := b(3)s(3), A(1) := b(1)s(1)

and A(2) := hs(2). Then the total area is

A = A(1) + A(2) + A(3).

The center of mass G of the cross-section has the following coordinates

x1(G) = 0, x2(G) =h

2(A(3) −A(1))

A.

Choosing a reference system Gy1y2 we have that

q1 = 0, q(1)2 =

h

2(1− A(3) −A(1)

A), q

(1)2 = −h

2(1 +

A(3) −A(1)

A).

Hence, from the integration formulae of Section 8, and setting also

I(1)2 :=

A(1)b(1)2

12, I

(3)2 :=

A(3)b(3)2

12,

which are the axial moments of inertia of ω(1) and ω(3) with respect to the axis y2,after some simple calculations we get

S1 = 0, S2 = 0, I12 = 0, Jt =13(b(1)s(1)3 + b(3)s(3)3 + hs(2)3),

I11 =h2

4[A(3) −A(1)

A(A(1) −A(3) + A(2)) + A(1) + A(3) +

A(2)

3], I22 = I

(1)2 + I

(3)2 ,

32

Sψ = AK, Iψ1 = c1I11, Iψ2 =h

2(I(3)

2 − I(1)2 )−

(c2 +

h

2A(3) −A(1)

A

)(I(3)

2 + I(1)2 ).

Solving the system (43) we obtain that K = 0 and the shear center c has coordi-nates

c1 = 0, c2 =h

2

(I(3)2 − I

(1)2

I(3)2 + I

(1)2

− A(3) −A(1)

A

),

in the reference Gy1y2, as shown in the figure above. The resulting warping stiffnessis

Iψψ = h2 I(3)2 I

(1)2

I(3)2 + I

(1)2

.

Let us remark that from the expressions above one can deduce, in particular, thatif A(3) = A(1) and I

(3)2 = I

(1)2 , then c = (0, 0), that is c ≡ G.

10 The loaded beam

As stated in (3), the total energy Fε is equal to the elastic energy Jε minus thework done by the external loads:

Fε(u) = Jε(u)−∫

Ωε

bε · u dx.

Let us denote by

Lε(u) :=∫

Ωε

bε · u dx =3∑

i=1

∫

Ω(i)ε

χ(i)ε bε · u dx

=: L(1)ε (u) + L(2)

ε (u) + L(3)ε (u)

where χ(i)ε is defined in Section 3. Then we can write

Fε(u) = F (1)ε (u) + F (3)

ε (u) + F (2)ε (u)

:= J(1)ε (u)− L(1)

ε (u) + J(3)ε (u)− L(3)

ε (u) + J(2)ε (u)− L(2)

ε (u).

With change of variables, as done in section 3, we deduce and set

F (i)ε (u(i)) : =

1ε3F (i)

ε (u(i) p(i)ε

−1)

= I(i)ε (u(i))−

∫

Ω(i)

b(i)ε · u(i) dy =: I(i)

ε (u(i))− L(i)ε (u(i)),

(44)

33

whereb(i)

ε = χ(i)ε bε p(i)

ε .

Let us assume that

b(i)ε1 = εα(i)

(ε2b

(i)1 (y)− ε

m(i)(y3)IG(ω(i))

(y2 − y2(G(i)))),

b(i)ε2 = εα(i+1)

(ε2b

(i)2 (y) + ε

m(i)(y3)IG(ω(i))

(y1 − y1(G(i)))),

b(i)ε3 = ε2b

(i)3 (y),

(45)

with b(i) = (b(i)1 , b

(i)2 , b

(i)3 ) ∈ L2(Ω(i);R3) and m(i) ∈ L2(0, `), while IG(ω(i)) is

always the moment of inertia with respect to the center of mass (y1(G(i)), y2(G(i))of the section ω(i). With this choice, the work of the external loads can be rewrittenas

Lε(u(1),u(2),u(3)) =3∑

i=1

L(i)ε (u(i)) (46)

where

L(i)ε (u(i)) = ε4

∫

Ω(i)

b(i) · S(i)ε u(i) dy + ε4

∫

Ω(i)

m(i) ϑ(i)ε (u(i)) dy.

The sequence of functionals Lε/ε4 is continuously convergent with respect to theconvergence used in Theorem 7.1, that is, for any sequence (u(1)

ε ,u(2)ε ,u(3)

ε ) ∈ Aε

such that

S(i)ε u(i)

ε v(i) in H1(Ω(i);R3),

(W(i)ε u(i))12 −ϑ in L2(Ω(i)),

we have

1ε4

L(i)ε (u(i)

ε ) → L(i)(v(i), θ) :=∫

Ω(i)

b(i) · v(i) dy +∫ `

0m(i) ϑdy3,

so that1ε4

Lε(u(1)ε ,u(2)

ε ,u(3)ε ) → L(v(1),v(2),v(3), θ) (47)

where

L(v(1),v(2),v(3), θ) :=3∑

i=1

L(i)(v(i), θ) =3∑

i=1

(∫

Ω(i)

b(i) · v(i) dy +∫ `

0m(i) ϑ dy3

).

34

In terms of the variables (η, ϑ) the limit load can be written in the form

L(η, ϑ) =∫ `

0l1η1 + l2η2 + l3η3 + mϑ−mαη′α + bϑ′ dy3 (48)

where

l1 = b(1)1 + b

(3)1 , l2 = b

(2)2 , l3 = b

(1)3 + b

(2)3 + b

(3)3 ,

m =3∑

i=1

m(i) −∫

ω(1)

(q(1)2 − c2)b

(1)1 da−

∫

ω(3)

(q(3)2 − c2)b

(3)1 da+

+∫

ω(2)

(q1 − c1)b(2)2 da,

m1 =∫

ω(1)

y1b(1)3 da +

∫

ω(3)

y1b(3)3 da +

∫

ω(2)

q1b(2)3 da,

m2 =∫

ω(1)

q(1)2 b

(1)3 da +

∫

ω(3)

q(3)2 b

(3)3 da +

∫

ω(2)

y2b(2)3 da,

b =∫

ω(1)

b(1)3 ψ(1) da +

∫

ω(3)

b(3)3 ψ(3) da +

∫

ω(2)

b(2)3 ψ(2) da,

and da = dy1dy2.We note that the contribution to the moments m of the loads b(i) are moments

evaluated with respect to the shear center and not with respect the center of massof the section.

The theorem below follows from the stability of Γ-convergence with respectto continuously convergent, real valued, perturbations (see Dal Maso [3], Proposi-tion 6.20).

Theorem 10.1 Let the rescaled total energy Fε be defined as

Fε(u(1),u(2),u(3)) =3∑

i=1

F (i)ε (u(i)). (49)

As ε → 0+, the sequence of functionals1ε4

Fε Γ-converges to the limit functionalF := I − L in the sense specified in Theorem 7.1.

For every ε ∈ (0, 1] let us denote by uε the solution of the following minimiza-tion problem

minFε(u) : u ∈ H1(Ωε;R3), u = 0 on Sε(0). (50)

The existence of a solution can be proved by the direct method of the Calculusof Variations and the uniqueness follows by the strict convexity of the functionalsFε. The next theorem describes the behaviour of the sequence of minimizers uε.

35

Theorem 10.2 The following minimization problem for the Γ-limit functionalF = I − L

minF (η, ϑ) : η1, η2 ∈ H2#(0, `), η3 ∈ H1

#(0, `) and ϑ ∈ H2#(0, `) (51)

admits a unique solution (η, ϑ). Moreover, denoting with (v(1), v(2), v(3)) a fieldrelated to (η, ϑ) throughout equations (31) we have that, as ε → 0+,

1. S(i)ε u(i)

ε → v(i) strongly in H i(Ω(i);R3),

2. (W(i)ε u(i)

ε )12 → −ϑ strongly in L2(Ω(i)),

3.1ε4

Fε(u(1)ε , u(2)

ε , u(3)ε ) converges to F (η, ϑ).

Proof. From the Γ-convergence Theorem 10.1 and from well known propertiesof Γ-limits and in particular by putting together Propositions 6.8 and 8.16 (lowersemicontinuity of sequential Γ-limits), Theorem 7.8 (coercivity of the Γ-limit) andCorollary 7.24 (convergence of minima and minimizers) of Dal Maso [3], followsthat

1 ′. S(i)ε u(i)

ε v(i) in H i(Ω(i);R3),

2 ′. (W(i)ε u(i)

ε )12 − ϑ in L2(Ω(i)),

3 ′.1ε4

Fε(u(1)ε , u(2)

ε , u(3)ε ) converges to F (η, ϑ).

It remains to prove that in 1 ′ and 2 ′ the convergence is, in fact, strong.Let us denote by aε the approximate minimizers defined as the sequence u0,ε

in (38), (39), (40) with (η, ϑ) replaced by (η, ϑ). Even if we cannot say thatit is a recovery sequence, for it does not satisfy the boundary conditions, theestimates (41) with (v, ϑ) replaced by (v, ϑ) hold and therefore

limε→0+

1ε4

Fε(a(1)ε , a(2)

ε ,a(3)ε ) = F (η, ϑ),

limε→0+

1ε4

Lε(a(1)ε , a(2)

ε ,a(3)ε ) = L(η, ϑ).

(52)

In particular we have that

limε→0+

F(i)ε (u(i)

ε )− F(i)ε (a(i)

ε )ε4

= 0, limε→0+

L(i)ε (u(i)

ε − a(i)ε )

ε4= 0. (53)

36

A key point in the proof of strong convergence of 1 ′ consists in showing that

limε→0+

∥∥∥E(i)ε (u(i)

ε − aε)ε2

∥∥∥L2(Ω(i))

= 0. (54)

To start we observe that the quadratic form f(A) defined in (33) satisfies theidentity

f(U) = f(A) + CA · (U−A) + f(U−A)

for every pair of 3× 3 matrices A and U. By (1) we thus obtain the inequality

f(U) ≥ f(A) + CA · (U−A) + µ|U−A|2,which can be used in the expression of the integral functional F

(i)ε to obtain that

F (i)ε (u(i)

ε ) ≥ F (i)ε (a(i)

ε ) +∫

Ω(i)

χ(i)ε CE(i)

ε a(i)ε ·E(i)

ε (u(i)ε − a(i)

ε ) dy +

+µ

∫

Ω(i)

χ(i)ε |E(i)

ε u(i)ε −E(i)

ε a(i)ε |2 dy − L(i)

ε (u(i)ε − a(i)

ε ).

Hence, taking also into account that χ(i)ε ≥ 1/2, we have

F(i)ε (u(i)

ε )− F(i)ε (a(i)

ε )ε4

≥ 12

∫

Ω(i)

CE(i)ε a

(i)ε ·E(i)

ε (u(i)ε − a

(i)ε )

ε4dy +

+µ

2

∫

Ω(i)

|E(i)ε u(i)

ε −E(i)ε a

(i)ε |2

ε4dy − L(i)

ε (u(i)ε − a(i)

ε ).

(55)

Let us then prove that

limε→0+

∫

Ω(i)

CE(i)ε a

(i)ε ·E(i)

ε (u(i)ε − a

(i)ε )

ε4dy = 0 (56)

and (54) is obtained by passing to the upper limit as ε → 0+ in (55). The proofof (56) proceeds along the same lines of Theorem 8.1 of [4] but we give here fulldetails for convenience of the reader.

In order to prove (56) we observe that for every pair of matrices A and B

CA ·B = 2µA ·B + λ tr(A) tr(B).

Then∫

Ω(i)

CE(i)ε a

(i)ε ·E(i)

ε (u(i)ε − a

(i)ε )

ε4dy = 2µ

∫

Ω(i)

E(i)ε a

(i)ε ·E(i)

ε (u(i)ε − a

(i)ε )

ε4dy +

+λ

∫

Ω(i)

tr(E(i)ε a

(i)ε ) tr(E(i)

ε (u(i)ε − a

(i)ε ))

ε4dy.

(57)

37

In order to perform the computation, it is convenient to shorten the notation bysetting

Aεhj :=

(E(i)ε a

(i)ε )hj

ε2, U ε

hj :=(E(i)

ε u(i)ε )hj

ε2,

∆εhj := U ε

hj −Aεhj =

(E(i)ε (u(i)

ε − a(i)ε ))hj

ε2,

so that the integrand in (57) takes the following expression

2µ(Aεhj∆

εhj) + λAε

hh∆εjj .

By (41) we have

Aε → Λ(i)(Eβα(i)(yβ − q(i)β )ϑ′, D3v

(i)3 ) in L2(Ω(i)), (58)

where Λ(i) is defined in (37), and by the equation above and Lemma 4.1 it followsthat ∆ε is bounded in L2(Ω(i)). Thus, from (58) we immediately deduce that

limε→0+

∫

Ω(i)

Aε12∆

ε12 dy = 0,

andlim

ε→0+

∫

Ω(i)

Aεα(i+1)3∆

εα(i+1)3 dy = 0. (59)

From (58) and 1 ′ follows that ∆ε33 0 in L2(Ω(i)), and therefore

limε→0+

∫

Ω(i)

Aεhj∆

ε33 dy = 0, h, j = 1, 2, 3.

From 1 ′, 2 ′, 3 ′ and Lemma 6.1, follows that, up to subsequences, U εα(i)3 weakly

converges in L2(Ω(i)) to Eβα(i)(yβ − q(i)β )ϑ′ + η(i), for some η(i) as specified in the

statement of Lemma 6.1. By (58) we have that Aεα(i)3 → Eβα(i)(yβ−q

(i)β )ϑ′ strongly

in L2(Ω(i)) and hence, up to subsequences, ∆εα(i)3 → η(i) weakly in L2(Ω(i)). Thus

limε→0+

∫

Ω(i)

Aεα(i)3∆

εα(i)3 dy =

∫

Ω(i)

Eβα(i)(yβ − q(i)β )ϑ′η(i) dy = 0.

Putting together with (59) we have that

limε→0+

∫

Ω(i)

Aεh3∆

εh3 dy = 0, h, i = 1, 2, 3.

38

Let ∆11 and ∆22 be, up to subsequences, the weak limits in L2(Ω(i)) of ∆ε11 and

∆ε22, respectively. Summarizing and taking the limit as ε → 0+ in (57) we obtain

(even for the whole sequence)

limε→0+

∫

Ω(i)

CE(i)ε a

(i)ε ·E(i)

ε (u(i)ε − a

(i)ε )

ε4dy =

= limε→0+

∫

Ω(i)

2µ(Aε11∆

ε11 + Aε

22∆ε22) + λAε

hh∆εαα dy

=∫

Ω(i)

D3v(i)3 (∆11 + ∆22)[−2ν(µ + λ) + λ] dy = 0

because −2ν(µ + λ) + λ = −λ + λ = 0, and (56) and hence (54) are proven.Setting z(i)

ε = S(i)ε (u(i)

ε − a(i)ε ), we have

‖Ez(i)ε ‖L2(Ω(i)) ≤ ‖E

(i)ε (u(i)

ε − a(i)ε )

ε2‖L2(Ω(i))

and by the application of the standard Korn inequality to z(i)ε , and using (54), we

obtain that‖z(i)

ε ‖H1(Ω(i)) → 0.

From this fact and the third equation of (41) applied to the sequence aε we gainthe strong convergence in 1 ′. This proves 1.

On the other hand, by (54) and inequality (11), we have that

limε→0+

∫

Ω(i)

|H(i)ε (u(i)

ε − a(i)ε )|2 dy = 0

and since the definition of H(i)ε we deduce from here that

D1(u(i)ε2 − a

(i)ε2 )

εα(i)→ 0,

D2(u(i)ε1 − a

(i)ε1 )

εα(i+1)→ 0, in L2(Ω(i)).

Thus, from the second equation of (41) applied to the sequence aε follows that

(W(i)ε u(i)

ε )12 =D2(u

(i)ε1 − a

(i)ε1 )

2εα(i+1)− D1(u

(i)ε2 − a

(i)ε2 )

2εα(i)+ (W(i)

ε a(i)ε )12 → −ϑ in L2(Ω(i)),

which proves 2 . 2

AcknowledgementsThe work of L.F. and R.P. has been partially supported by Progetto Cofinan-

ziato 2005 “Modellazione matematica di strutture e materiali complessi”, whileA.M. has been partially supported by MURST, grant no. 2003082352.

39

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