continuous modeling of lattice structures by homogenization

Continuous modeling of lattice structuresby homogenization

H. Tollenaere & D. Caillerie*Laboratoire Sols Solides Structures, BP 53X, 38041 Grenoble, France

(Received 1 December 1995; accepted 1 November 1997)

We present a method of continuous modeling of quasi-repetitive lattice structurescalled discrete homogenization and derived from the periodic continuous mediumhomogenization. The structure studied here is a two-dimensional quasi-periodic latticetruss with pin-jointed nodes in the case of static deformation. A one-dimensionalsimple illustrative example is given.q 1998 Elsevier Science Limited and Civil-Comp Limited. All rights reserved.

1 INTRODUCTION

Taking advantage of the smallness of each elememtarybeam compared with the size of the whole structure, differ-ent attempts have been made to model repetitive trussescomposed of a large number of beams by continuous equa-tions.1–9So, a crane jib may be modeled as a beam, a roof ora space antenna as a plate or a shell. The aim of this model-ing is to simplify the analysis or the optimization of thestructure.

The method developed in this work is an adaptation of thetwo-scale asymptotic expansions used in homogenization ofperiodic media;10,11it differs partly from the other methodsproposed in the literature and does not suffer some of thedefects of them. Though related to the studies of refs12,13,our model is different in the sense that we consider thebeams as one-dimensional structures. The method hadbeen developed for beam-like structures14–17 and is nowsettled for beam-like, plate-like or three-dimensional repe-titive trusses of very general design, even for structures withmechanism and for nearly repetitive structures (here it isdeveloped for two-dimensional nearly repetitive structures).The same method has been applied to structures in largedisplacement framework,18 thanks to its characteristic fea-tures (asymptotic expansion of displacements and tensions),the method could be extended to such non-linear problemsas plasticity.

The structures under consideration are very large pin-jointed trusses, so the unknowns are the displacements ofnodes and the tensions in the beams; the loads are applied on

the nodes. The structures are assumed to be large, whichmeans that they consist of a large number of elementarybeams, the lengths of which are small compared with thesize of the structures; the ratio of beam lengths over struc-ture size is the small parametere of the problem. Themethod of discrete homogenization consists in assumingasymptotic series expansions in powers ofe for the nodedisplacements and the tensions in beams. The balance equa-tions of nodes and force displacement relations of beams arethen developed by carrying these series expansions in themand by using Taylor’s expansion of finite differences. Lasty,an identification of the terms of same power ofe yieldspartial differential equations satisfied by the first terms ofthe series expansions of the displacements and tensions.Following the type of the structure under considerationthese partial differential equations are those of three- ortwo-dimensional elasticity (the case of this study) or thoseof a beam or a plate, so they define the continuous model ofthe truss.

The continuous model is valid for the study of eitherstatic deformations or vibrations; only the case of staticdeformations is presented here. The calculations havebeen completed for a quite general truss and the resultsgive general expressions of the elastic moduli of the equiva-lent continuous medium. The reckoning of these modulineeds only the solving the self-balance equations of oneelementary cell, which can be performed analytically forsimple cells and can be easily programmed on a computerfor more complicated cells.

A two-dimensional example would be too long to bedeveloped here and only a one-dimensional one is pre-sented. It is however illustrative, indeed it shows that the

Advances in Engineering SoftwareVol. 29, No. 7–9, pp. 699–705, 1998q 1998 Elsevier Science Ltd and Civil-Comp Ltd

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ADES 376

699

*Author to whom all correspondence should be addressed.

method enables the expected well-known result to be found,and it is sufficient to point out the differences of resultsbetween this method and one used in litterature.

The Einstein convention of repeated index is usedthroughout this paper.

2 NUMBERING OF NODES AND BEAMS

The truss under consideration is made of beams only linkedby their ends, so the geometry of the truss is completelydefined by the positions of the nodes and the connectivitybetween them, that is which of them are linked by a beam.

The lattice truss is assumed to be quasi-periodic, whichmeans that it is the repetition of nearly identical cells. Weuse a system of numbering of nodes and beams that reflectsthe regularity of the structure and leads to the curvilinearcoordinates of the equivalent continuum.

The truss is two-dimensional, so the cells composing itare numbered by a system of two integers (v1, v2). The nodesare then numbered by a tripletn¼ (n, v1, v2), which meansthat the noden is the noden in the cell (v1, v2). The samesystem is used for beams which are numberedb¼ (b, v1, v2).

Each beam links two nodes and is oriented, which meansthat one the two nodes is the origin of the oriented beam,denotedO(b). The other node is the end and is denotedE(b).O(b) andE(b) are then mappings from the set of beamsBinto the set of nodesN andO¹ 1(n) (respectivelyE¹ 1(n))denotes the set of beams the origin (respectively the end) ofwhich is n.

In n¼ (n, v1, v2) andb¼ (b, v1, v2), the indicesn andbnumber the nodes and beams of a reference cell the sets ofwhich are denotedNR andBR.

The beams link nodes of the same cell or of neighboringcells, as the orientation of a beam is arbitrary, it is chosen sothat the originO(b) of the beamb¼ (b, v1, v2) is in the cell(v1, v2), that is, a node numbered (n, v1, v2). The endE(b) ofthe beamb¼ (b, v1, v2) is in the cell (v1, v2) or in a neigh-boring one, anyway in a cell numbered (v1 þ d1, v2 þ d2)with d ie{-1, 0, 1}, andE(b) is a node numbered (m, v1 þ d1,v2 þ d2).

As the truss is repetitive,n, m, d1, d2 depend only onb,they are denoted respectivelyOR(b), ER(b), d1b, d2b.

3 GEOMETRY OF A QUASI-PERIODIC TRUSS

The lattices under consideration are structures with a largenumber of cells, which is equivalent to saying thate ¼ 1=(Nc)1=2 is a small parameter (Nc being the number ofcells in the structure).

We use an asymptotic expansion method which assumesthate tends to zero. Actually, that means that we consider asequence of structures with more and more cells which aresmaller and smaller. The sequence of structures has to be

completely defined, that is to say that for each value ofe it isnecessary to know where the nodes of the structure arelocated in the space. Moreover, we intend to considertrusses which are not exactly repetitive but quasi-periodicwhich means that their geometry can vary slightly.

Then, we assume that the positions of the nodes are givenby an expansion of the form:

;n¼ (n,v1, v2), r (n) ¼ r0(l1e,l2e) þ ern1(l1e,l2e)

þ e2rn2(l1e,l2e) þ … ð1Þ

wherel ie ¼ evi and wherer0(l1e,l2e), rn1(l1e,l2e), … aregiven functions of (l1,l2).

At first order in e, all the nodes of a cell have approxi-mately the same location given byr0(l1,l2), their positionswith respect to the pointr 0(l1,l2) are of the order ofe andgiven byern1(l1e,l2e) þ ….

In this way, an exactly periodic lattice of periodseL1 andeL2 is defined by:

r (n) ¼ l1eL1 þ l2eL2 þ ern

which fits eqn (1) withr0 ¼ l1L1 þ l2L2 and withrn1 ¼ rn

independent of (l1, l2).Fig. 1 displays an example of a quasi-periodic structure

which is described by:

r ((1,v1, v2)) ¼ R(l1e) cos l2e p

2

� �i þ sin l2e p

2

� �j

h ir ((2,v1, v2)) ¼ R(l1e)

hcos l2e p

2þ e

p

5

� �i

þ sin l2e p

2þ e

p

5

� �ji

with R(l1) ¼ Ri þ l1(Re ¹ Ri), l1e ¼ ev1 andl2e ¼ ev2.

Fig. 1. Example of a quasi-periodic structure in the cell (4,4).

700 H. Tollenaere, D. Caillerie

For this example, ther (n) can be developed in:

r ((1,v1, v2)) ¼ R(l1e) cos l2e p

2


2

� �j

h ir ((2,v1, v2)) ¼ R(l1e) cos l2e p

2


2

� �j

h iþ e

p

5R(l1e) ¹ sin l2e p

2

� �i þ cos l2e p

2

� �j

h iþ e2[…]

Eqn (1) enables the geometrical characteristics of thebeams to be expanded. The length 1b ¼ kr (E(b)) ¹

r (O(b))k and the unitary vector of the beameb ¼ [r (E(b)) ¹ r (O(b))]=1b can be expanded in:

1b ¼ eLb0(l1e,l2e) þ e2Lb1(l1e,l2e) þ …

eb ¼ eb0(l1e,l2e) þ eeb1(l1e,l2e) þ …

The expressions of theLbk andebk can be found in terms ofthe fieldsrnk(l1,l2) by using Taylor expansions:

eb0 ¼1

Lb0 rER(b) ¹ rOR(b) þ]r0

]lidib

� �(2)

It is assumed that, after deformation, the geometry ofthe structure remains quasi-periodic, which means thatthe positions of the nodes are given by expansions ofthe form eqn (1). The difference between the positionsof a node before and after deformation is its displace-ment u(n), which is then given by an expansion of theform:

u(n) ¼ u0(l1e,l2e) þ eun1(l1e,l2e)

þ e2un2(l1e,l2e) þ …, lie ¼ evi ð3Þ

Eqn (3) shows that at first order ine, the displacements ofthe nodes of the structures are approximately given by thedisplacement fieldu0(l1,l2). Consequently, the assump-tion that the structure is quasi-periodic (as described byeqn (1)) and remains quasi-periodic after deformationyields that the displacement of nodes becomes, ase

decreases, a function of a domain.Very often in literature4–6 a displacement field is defined

on the whole space domain occupied by the structure. Thisfield is an extension of the displacements of the nodes,recalling the interpolations of finite elements. This kind ofextension would probably be necessary if a convergenceproof was considered; it is not for the use of the asymptoticexpansions developed here.

The following Sections 4–8 are devoted to the derivationof the equations governing the fieldu0(l1,l2) and givingthe equivalent continuous model.

4 EXPANSION OF BEAM TENSIONS

Consistently with the assumption about geometry, weassume that the Young modulusEb and the sectionAb ofa beam are given by series expansion in power ofe. Thisimplies that the stiffnesskb ¼ (EbAb)=1b of the beam can be

expanded in:

kb ¼1ekb0(l1e,l2e) þ kb1(l1e,l2e) þ ekb2(l1e,l2e) þ …

(4)

In the tension displacement relations for beams:

Nb ¼ kb[u(E(b)) ¹ u(O(b))]·eb (5)

the differences such as u0(l1e þ ed1,l2e þ ed2) ¹

u0(l1e,l2e) can be expanded using a Taylor expansion(ed1 is small). Consequently the beam tensionsNb can beexpanded in:

Nb ¼ NbO(l1e,l2e) þ eNb1(l1e,l2e) þ e2Nb2(l1e,l2e) þ …

(6)

and by identification of terms of same power ine, theexpansion of the tension displacement relation yields:

Nb0 ¼ kb0 uER(b)1 ¹ uOR(b)1 þ]u0

]li dib

� �·eb0 (7)

Nb1 ¼ …, …

5 EXPANSION OF BALANCE EQUATIONS—BALANCE EQUATION OF THE EQUIVALENTCONTINUUM

To find the equivalent continuum of the truss, it remains todevelop the balance equations of the nodes, which are:

;n,∑

beO¹ 1(n)Nbeb ¹

∑beE¹ 1(n)

Nbeb þ fe=n ¼ 0 (8)

wheref e=n is the external force applied on the noden. Thisforce is assumed to be:

fe=n ¼ ef e=n(l1e,l2e)

This shows thatf e=n is assumed to be of the order ofe, thisassumption will be proved consistent with the continuummodeling and the order of magnitude ofkb, eqn (4).

The expansion of the balance eqn (8) is much easier whenthe equations are written with a virtual power formulationwhich reads:

;v(n),∑beB

Nbeb·[v(O(b)) ¹ v(E(b))] þ∑neN

fe=n·v(n) ¼ 0

(9)

Sums such as∑

beB can be split in∑

v1, v2

∑beBR

and, in theprocess of expansion, which implies thate is very small, thesumse2

∑v1,v2 are approximated by integrals

�q dl1 dl2.

We get then the virtual power formulation of the equili-brium of the equivalent continuum by taking in eqn (9):

;n¼ (n,v1, v2), v(n) ¼ v0(l1e,l2e),lie ¼ evi

where v0(l1,l2) is a smooth given virtual macroscopicvelocity field.

Continuous modeling of lattice structures by homogenization 701

We get then:

;v0, ¹

∫qSi0·

]v0

]li dl1 dl2 þ

∫qf ·v0 dl1 dl2 ¼ 0 (10)

with:

Si0 ¼∑beBR

Nb0dibeb0 (11)

and

f ¼∑neNR

fe=n (12)

The previous process explains why the external forcesf e=n

are assumed to be of the order ofe.The vectorsSi0 describe the internal efforts of the equi-

valent continuum in the curvilinear coordinates system (l1,l2). They are related to the stress tensorjO in Section 9.Eqn (10) is the virtual power formulation of the balanceequation of the continuum equivalent to the structure.

6 SELF-EQUILIBRIUM OF THE REFERENCECELL

To establish the constitutive relation of the equivalent con-tinuum we need to use the self balance equations of the cellwhich is obtained from eqn (9) by taking:

;n¼ (n,v1, v2), v(n) ¼ ev(l1e,l2e)vn(l1e,l2e), lie ¼ evi

Using the same process as in Section 5, we get:

;v,vn

∫q

∑beBR

NbOeb0·[vER(b) ¹ vOR(b)]

" #v dl1 dl2 ¼ 0 (13)

which yields:

;vn,∑beBR

Nb0eb0·[vER(b) ¹ vOR(b)] ¼ 0 (14)

This relation is the virtual formulation of the self-equilibrium of the cell.

7 CONSTITUTIVE RELATIONS OF THEEQUIVALENT CONTINUUM

In order to get the description of the continuum equivalentto the structure, its balance eqn (10) has to be completed bya constitutive relation linkingSi0 to ]u0=]li . The constitu-tive relation is in fact given by eqns (7), (14) and (11).Indeeed,]u0=]li being considered as data, eqns (7) and(14) are equations the unknowns of which areNb0 andun1. Solving these equations givesNb0 as a function of]u0=]li , then eqn (11) yields the wanted constitutiverelation.

As the eqn (14) is discrete, the procedure described abovecan be carried out in two steps which makes the solving of

eqns (7) and (14) easier. First a baseJbk of solutions of eqn(14) is determined, then this base is orthogonolized in such away to eliminate theun1 and to expressNb0 in terms of]u0=]li .

Let Jbk, the indexk numbering the elements of the base,denote a base of solutions of eqn (14). The tensionsNb0

which are also solutions of eqn (14) are then a linear com-bination of theJbks:

Nb0 ¼∑

k

akJbk

Carrying this relation in the tension displacement eqn (7)yields:∑

k

akJbk ¼ kb0 uER(b) ¹ uOR(b) þ]u0

]li dib

� �·eb0

Orthogonalizing theJbks by:∑beBR

JbkJbh

kb0 ¼ dkh

enables theun1 to be eliminated and theak to becalculated:

ak ¼ fik·]u0

]li

where

fih ¼∑beBR

Jbhdibeb0

Then, we can reckonNb0 andSi0 in terms of the]u0=]li :

Nb0 ¼∑

k

Jbkfik

" #·]u0

]li (15)

Si0 ¼∑

k

fik fjk·]u0

]li

� �This last relation is the constitutive relation of the equiva-lent continuum, it reads:

Si0 ¼∑

k

fik # fjk

!@

]u0

]li (16)

where fik # fjk is the linear operator called tensorial ordiadic product of the two vectorsfik andfjk.

A@y denotes the image of the vectory by the operatorA.WhenA is the diadic producta # b, A@y, is such that:

a # b@y ¼ (b·y)a

8 CONTINUOUS MODEL OF THE TRUSS INCURVILINEAR COORDINATES

The continuous model of the truss in curvilinear coordinatesis given then by the constitutive eqn (16) and the balance


eqn (10) which yields:

]Si0

]li þ f ¼ 0 (17)

Some boundary conditions have to be added to this prob-lem. They can be deduced by expanding the boundary con-ditions of the truss.

The well-posedness of boundary value problems for theequivalent continuum comes down to the positive definite-ness of the following quadratic form:

12

∑k

fik # fjk

!@

]u0

]lj

" #·]u0

]li

It can be studied and deduced from general assumptions onthe truss.

Remark: carryingu0, solution of eqns (16) and (17) com-plete with boundary conditions, into eqn (15) gives the firsttermNb0 of the expansion, eqn (6) ofNb, that is an approx-imation of the tensions in the beams of the truss.

9 TWO-DIMENSIONAL ELASTIC MODEL INSPACE COORDINATES

The equivalent continuum constitutive relation, eqn (16), islinear, which could be expected since the considered truss islinear. Moreover the continuous model, eqns (16) and (17),should be a two-dimensional elastic model. This does notappear obviously in eqns (16) and (17) written in curvilinearcoordinates. The best way to prove that the continuousmodel is actually a two-dimensional elastic model is towrite the balance and constitutive equations in usual spacecoordinates.

The cells of the lattice truss are located by the two inte-gers (v1, v2) or in an equivalent manner by the two realnumbers (l1e, l2e) with l ie ¼ evi, which take discrete values.In the homogenization process, the discrete variablesbecome the continuous variables (l1, l2), which meansthat (l1, l2) locate the material points of the equivalentcontinuum. From the eqn (1), we see that the position ofthe material point located by (l1, l2) is r0(l1,l2), then thecontinuum model in space coordinates is given by thechange of variables:

(l1,l2) → r0(l1,l2): (18)

The easiest way to find the relation between the vectorsSi0

and the stress tensorj0 in the continuum is to make thechange of variables, eqn (18), in the virtual power formula-tion eqn (9).

LetLxv be the gradient tensor of the virtual velocity fieldv. The change of variables, eqn (18), yields:

]v]li ¼Lxv@

]r0

]li (19)

The change of variables, eqn (18), in eqn (10) yields:

;v, ¹

∫q

Si0 #]r0

]li

� �: Lxv

1g

dx1 dx2

þ

∫qf ·v

1g

dx1 dx2 ¼ 0

whereg is the determinant of the matrix of the componentsof the two vectors]r0=]l1 and]r0=]l2 and where : denotesthe scalar (contracted) product of two tensors.

The stress tensorj0 is then:

j0 ¼1gSi0 #

]r0

]li (20)

which verifies the balance equation:

divxj0 þ

1gf ¼ 0 (21)

To derive the constitutive relation betweenj0 and the straintensore(u0), it is useful to obtain the expression ofj0 interms ofNb0, Lb0 andeb0. This expression can be deducedfrom eqns (20) and (11), but it is easiest to derive it from asuitable choice of virtual velocities in the virtual powerformulation of the self-equilibrium of the cell, eqn (14).

Let E be any tensor and letvn ¼ E@rn, then eqn (2)yields:

vER(b) ¹ vOR(b) ¼ E@ Lb0eb0 ¹]r0

]lidib

� �and eqn (14) becomes:

∑beBR

Nb0eb0 # Lb0eb0 ¹]r0

]lidib

� �" #: E¼ 0

asE is any tensor, the previous relation added to eqns (11)and (20) yield:

j0 ¼1g

∑beBR

Nb0Lb0eb0 # eb0 (22)

This shows that, as expected for an elastic model, the stresstensor is symmetrical.

To get the constitutive relation in space coordinates, wethen have to carry the expression of]u0=]li (see eqn (19)) ineqn (15), itself carried in eqn (22). The result is then:

j0 ¼1g

∑k

zk[zk : e(u0)] (23)

where zk ¼ (]r0=]li) # fik and e(u0) is the symmetricalpart ofLxu

0.The continuous model of the truss in space coordinates is

then given by the balance eqn (21) and the constitutiverelation, eqn (23). This model is clearly a two-dimensionalelastic one, its elastic energy density being:

12g

∑k

[zk : e(u0)]2


10 A SIMPLE ONE-DIMENSIONAL EXAMPLE

We study a one-dimensional straight structure built by theperiodic repetition of two beams of stiffnesseskc/e andkd/eand lengthsl c ¼ eLc and l d ¼ eLd submitted to externalforces f e=n ¼ ef e=ne parallel toe, the unit vector parallel tothe structure, so the problem is entirely one-dimensional.

Only one indexv is needed to number the cells. There aretwo nodes 1 and 2 and two beamsc andd in the referencecell.

The beamc is oriented from node 1 to node 2, the beamd isoriented from node 2 to node 1, then:

OR(c) ¼ 1, ER(c) þ 2, dc ¼ 0

OR(d) ¼ 2, ER(d) ¼ 1, dd ¼ 0

(there is only oned per beam).The problem is one-dimensional, so there is only one

coordinatel and only one vectorS0 ¼ S0e proportional toe. The balance equation of the equivalent continuum readsin this example:

]S0

]lþ f ¼ 0

The self equilibrium of the reference cell reads:

Nc0 ¹ Nd0 ¼ 0

Nd0 ¼ Nc0 ¼ 0

Then there is only oneJ and only onef which are:

J ¼

��Jc

Jd¼

kckd

kc þ kd

!1=2��11

f ¼ (dcJc þ ddJd)e¼kckd

kc þ kd

!1=2

e

u0 ¼ u0e is parallel toe, it depends onl and the constitutiverelation is the well-known relation:

S0 ¼kckd

kc þ kd

]u0

]l

The structure is exactly periodic then:

r0(l) ¼l(Lc þ Ld)e

Setting x ¼ l(Lc þ Ld) gives the constitutive relation inspace variablex:

j0 ¼ (Lc þ Ld)kckd

kc þ kd

]u0

]x

In the following, the previous result is compared to that ofthe ‘static condensation’ method.

The method used in4,5 consists in looking for the dis-placementu(n) of any noden as the value of a functionU(x)at the locationx(n) of the node. The differencesu(m) ¹ u(n)in the tension displacement relations of the beams areexpanded using Taylor expansions to the second order, anorigin being chosen in the cell.

Choosing the node 2 as origin yields for example:

u((1,vþ 1)) ¹ u((2, v)) ¼ 1dU9 þ(1d)2

2U0

u((1,v)) ¹ u((2, v)) ¼ ¹ 1cU9 þ(1c)2

2U0

The contribution of the beamsc andd of the cellv to theelastic energy of the structure is:

Ev ¼12

"kc

e¹ 1cU9 þ

(1c)2

2U0

� �2

þkd

e1dU9 þ

(1d)2

2U0

!2#The method, called ‘static condensation’, used in4,5 con-sists then of setting]Ev=]U" equal to zero, this enables toexpressU0 andEv in term of U9:

Ev ¼12

kckd(lc)2(ld)2(lc þ ld)e[kc(lc)4 þ kd(ld)4]

(U9)2(lc þ ld)

The length (l c þ l d) of the cell is in Ev the element ofintegration dx in the passage of the energy of the discretestructure to that of the equivalent continuous bar. Thus theelastic stiffness of the equivalent continuous bar, given bythis method, is:

kckd(Lc)2(Ld)2(Lc þ Ld)kc(Lc)4 þ kd(Ld)4

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continuous modeling of lattice structures by homogenization

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