fracture modeling using a micro-structural mechanics approach––i. theory and formulation

Fracture modeling using a micro-structuralmechanics approach––I. Theory and formulation

C.S. Chang a,*, T.K. Wang b, L.J. Sluys c, J.G.M. van Mier c

a Department of Civil Engineering, University of Massachusetts, 253 Marston Hall, Box 35205, Amherst, MA 01003-5205, USAb Department of Structural and Foundation Engineering, Polytechnic School of the University of Sao Paulo, SP, Brazil

c Faculty of Civil Engineering and Geosciences, Delft University of Technology, Delft, The Netherlands

Received 22 March 2002; accepted 16 April 2002

Abstract

Stress–strain relationships used for modeling the mechanical behavior of materials have been traditionally derived

following a phenomenological approach, without explicit considerations of the micro-structures of material. Here, we

adopt a micro-structural mechanics approach to model the development of fracture in concrete. The continuum is

assumed to have an underlying micro-structure of lattice type, which has been demonstrated as a useful description for

concrete fracture. A finite element formulation is also described that incorporates the developed stress–strain rela-

tionship. In an accompanying paper, we will show the results of finite element analyses, discuss its characteristics with

respect to mesh size independency, and evaluate the applicability of this method.

� 2002 Elsevier Science Ltd. All rights reserved.

Keywords: Fracture; Damage mechanics; Concrete; Micro-structural model; Stress–strain relationship

1. Introduction

Stress–strain relationships used for modeling the mechanical behavior of materials have been tradi-tionally derived following a phenomenological approach, without explicit considerations of the micro-structures of material. Here we adopted a micro-structural mechanics approach to model the developmentof fractures in concrete, assuming the material has an underlying micro-structure of lattice type. Althoughthe lattice does not directly reflect the micro-structure of concrete, it has been demonstrated as a usefulmodel for the description of concrete fracture [1–3], in particular when a micro-structure is projected on alattice and corresponding properties are assigned to relevant elements in the lattice. The lattice merelyserves as a convenient computational backbone.

In the micro-structural mechanics approach, the stress–strain relationship of the concrete is derivedbased on the geometric and material properties of the underlying lattice structure. Since a lattice network of

Engineering Fracture Mechanics 69 (2002) 1941–1958

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beams can be represented by a lattice network of particles, we adopt the approach used in granular me-chanics [4–9] to derive the stress–strain relationship for concrete.

The contribution consists of two parts. The first part focuses on the theoretical aspects on the derivationof stress–strain relationships of a continuum material with an underlying lattice structure. In the secondpart the computational issues and model performance based on the results of finite element analysis thatincorporates the developed stress–strain relationships are discussed.

In the current paper, we first briefly show the relationship between a lattice network of beams and alattice network of particles. On this basis, the stress–stain relationship of micro-polar type is derived for acontinuum with underlying lattice structure. Then, the finite element formulation that incorporates thisstress–strain relationship is also discussed.

2. Lattice network

A set of randomly located points can be represented as a lattice network by connecting these points withlines. If the lines are regarded as beams, the lattice network becomes a frame structure which has beendemonstrated as a useful model for the description of concrete fracture [2,3], in particular when a micro-structure is projected on a lattice and corresponding properties are assigned to relevant elements in thelattice.

Alternatively the points can be considered as particles, which are connected by springs. In general, threetypes of spring are used to represent the stiffness between two contact particles, namely, a normal spring, ashear spring and a rotational spring (see Fig. 1(c)). For convenience, the symbol of a cross in a circle asshown in Fig. 1(c) will be used in the future figures to represent the three springs. The lattice network ofparticles is used extensively in granular mechanics to represent a granular medium [10–13]. In principle, therepresentation of a lattice network of beams is equivalent to a lattice network of particles as schematicallyshown in Fig. 1(a) and (b). In what follows, we first describe the kinematics of two interacting particles, and

Fig. 1. (a) Lattice network of beams; (b) equivalent lattice network of particles, and (c) three types of springs in between particles.

1942 C.S. Chang et al. / Engineering Fracture Mechanics 69 (2002) 1941–1958

derive the force–displacement relationship for the two particles system. Then we will demonstrate that thedeformation behavior of a lattice beam can be represented by a two-particle system connected by springs.

In a lattice network of particles, the movement of the nth particle can be defined by means of thetranslation uni and the rotation xn

i at the particle centroid. For two contact particles connected by springs,the relative displacement dnmi and the relative rotation hnmi between two particles can be obtained from themotion of particles, given as follows:

dnmi ¼ umi � uni þ eijkðrmk xmj � rnkx

nj Þ ð1Þ

hnmi ¼ xmi � xn

i ð2Þ

where the quantity eijk is the permutation symbol used in tensor representation for the cross product ofvectors, and rnci , rmci are the position vectors of contact point ‘c’ relative to the centroid of particle ‘n’ and‘m’, respectively.

The particle interaction can be modeled by two types of spring stiffness: displacement stiffness, androtational stiffness. The displacement stiffness includes the normal stiffness knmn and the shear stiffness knms ,and represents the resistance to compression and sliding of two particles. The rotational stiffness includesthe twisting stiffness gnmn and the rolling stiffness gnms , which corresponds to the resistance to twisting androlling of two particles. The different types of contact stiffness are schematically shown as springs betweenparticles in Fig. 1(c). The relative displacement dnmi and the relative rotation hnmi are related to the contactforce f nmi and the contact moment mnm

i at the inter-particle contact. In a general form

f nmi ¼ Knmij dnmj ð3Þ

mnmi ¼ Gnm

ij hnmj ð4Þ

in which the contact stiffness tensor Knmij and rotation stiffness tensor Gnm

ij are given in terms of contactstiffness

Knmij ¼ knmn nnmi n

nmj þ knms ðsnmi snmj þ tnmi t

nmj Þ ð5Þ

Gnmij ¼ gnmn nnmi n

nmj þ gnms ðsnmi snmj þ tnmi t

nmj Þ ð6Þ

where ni, si and ti are the basic unit vectors of the local coordinate system constructed at each inter-particlecontact. The vector ni is the outward normal to the contact plane. The other two orthogonal vectors si and tiare on the contact plane. In a spherical coordinate system, a vector can be defined by two angles c and b asshown in Fig. 2, e.g.,

~nn ¼ sin c cos b~iiþ sin c sin b~jjþ cos c~kk

~ss ¼ cos c cos b~iiþ cos c sin b~jj� sin c~kk

~tt ¼ � sin b~iiþ cos b~jj

ð7Þ

For a two dimensional case (gn ¼ 0) and for two equal-size particles in contact (i.e., r ¼ l=2 where l is thelength between two particle centroids, which is also the length of lattice beams, see Fig. 3), the interaction oftwo particles can be given as follows in a matrix form (see Appendix A):

C.S. Chang et al. / Engineering Fracture Mechanics 69 (2002) 1941–1958 1943

f 1xf 1y

m1

f 2xf 2y

m2

8>>>>>><>>>>>>:

9>>>>>>=>>>>>>;

¼

kn 0 0 �kn 0 00 ks ksl

20 �ks ksl

2

0 ksl2

ksl2

4þ gs 0 � ksl

2ksl2

4� gs

�kn 0 0 kn 0 00 �ks � ksl

20 ks � ksl

2

0 ksl2

ksl2

4� gs 0 � ksl

2ksl2

4þ gs

266666664

377777775

u1xu1yx1

u2xu2yx2

8>>>>>><>>>>>>:

9>>>>>>=>>>>>>;

ð8Þ

where ux, uy , fx, fy represent respectively the displacements and forces in x- and y-directions at the centroidof a particle. The value x and m are respectively the rotation and the moment for the particle. The su-perscripts refer to particle number 1 or 2.

For a beam, the stiffness matrix is

f 1xf 1y

m1

f 2xf 2y

m2

8>>>>>><>>>>>>:

9>>>>>>=>>>>>>;

¼

EbAb

l 0 0 � EbAb

l 0 0

0 12EbIl3

6EbIl2 0 � 12EbI

l36EbIl2

0 6EbIl2

4EbIl 0 � 6EbI

l22EbIl

� EbAb

l 0 0 EbAb

l 0 0

0 � 12EbIl3 � 6EbI

l2 0 12EbIl3 � 6EbI

l2

0 6EbIl2

2EbIl 0 � 6EbI

l24EbIl

266666664

377777775

u1xu1yx1

u2xu2yx2

8>>>>>><>>>>>>:

9>>>>>>=>>>>>>;

ð9Þ

Fig. 2. Local coordinate system.

Fig. 3. A beam represented by a two-particle system connected by springs.


where Eb, l, Ab are the Young modulus, length and cross section area of the beam respectively. The areaAb ¼ bh (see Fig. 3) with h as the depth of the beam. For the two-dimensional case considered here, thewidth of the beam b is taken equal to one. The values of ux, uy , fx, fy represent the displacements and forcesin x- and y-directions, and the value of x and m are the rotation and moment. The superscripts now refer tothe two ends of the beam.

It is noted that Eqs. (8) and (9) are identical if the following relationships hold true between theproperties of beam and inter-particle springs:

kn ¼EbAb

l

ks ¼12EbIl3

ð10Þ

gs ¼EbIl

ð11Þ

By selecting appropriate spring constants, the behavior of a beam can be identically represented by a two-particle system as schematically shown in Fig. 3. Thus, the lattice network of beams can be replaced by anequivalent lattice network of particles.

3. Equivalent continuum

A lattice network of beams and a lattice network of particles are discrete systems as shown in Fig. 1.We now treat the lattice network of beams as an ‘equivalent’ continuum and derive the stress–strainrelationship of the continuum based on the geometric and material properties of the lattice structure.Since we have shown that a lattice network of beams can be represented by a lattice network of parti-cles, the approach used in granular mechanics is adopted to proceed with the process of homogeniza-tion.

We select one regular triangle as a unit cell, which can be conceived as three particles connected bysprings. Over the cell, we construct a linear field of displacement and a linear field of rotation.

ui ¼ uoi þ uoi;jxj; xj ¼ xoj þ xo

j;pxp ð12Þ

where the superscript ‘o’ refers to the selected reference point (e.g. the centroid) of the cell (see Fig. 4). Wealso define two length measures for the micro-structure of the cell, namely, gi which represents the lengthvector from centroid to the ith side, and li which represents the length vector of the ith side.

Substituting Eq. (12) into the kinematic equations of two contact particles (Eqs. (1) and (2)), it follows:

dci ¼ ðuoi;k � eijkxoj Þlck � xo

j;peijklckg

cp ð13Þ

hci ¼ xoi;jl

cj ð14Þ

where the superscript ‘c’ refers to the contact of two particles. Based on Eqs. (13) and (14), two continuumvariables are identified for the cell: the rotation gradient xo

i;j and the ‘micro-polar strain’ given by

eoki ¼ uoi;k � eijkxoj ð15Þ

which has the same form as those used in the micro-polar model [14] and Cosserat model [15]. The energycounterpart of eoij is the usual Cauchy stress rij, which can be characterized as forces on a plane of unit area.


The energy counterpart of xoj;i is couple stress lij, which can be characterized as force couples on a plane of

unit area. Furthermore, using the principle of energy balanceZrije

oij

�þ lijx

oj;i

�dV ¼

X3

c¼1

f ci dci

�þ mc

i hci

�ð16Þ

Substituting dci , hci in Eqs. (13) and (14) into Eq. (16), the Cauchy stresses and couple stress can be derived as

follows:

rki ¼1

V

Xc

f ci lck

lpj ¼1

V

Xc

½�eijkf ci lckgcp þ mcjlcp�

ð17Þ

3.1. Stress–strain relationship

Substituting f ci , mci in Eqs. (3) and (4) into Eq. (17), the stress–strain relationship becomes

rmq ¼ Cmqkieoki þ Dmqpjx

oj;p

lpj ¼ Dpjmqeomq þ Fpjrtxo

t;r

ð18Þ

where

Cmqki ¼1

V

Xc

Kcqil

cklcm

Dmqpj ¼ � 1

V

Xc

eijkKcqil

cml

ckg

cp

Dpjmq ¼ � 1

V

Xc

eijkKciql

cklcmgcp

Fpjrt ¼1

V

Xc

½eijkeqtmKciql

cklcmgcpg

cr þ Gc

jtlcrlcp�

ð19Þ

Fig. 4. The regular triangular unit cell of a lattice network.


In Eq. (18), eoij and xoi;j represent the strain measures of a triangular cell. For convenience, the superscript ‘o’

of the two variables will be eliminated in the subsequent text.For a two-dimensional case of a hexagonal lattice network based on regular triangular cells, there are

two basic configurations of regular triangles as marked ‘1’ and ‘2’ in Fig. 5, which are in opposite di-rection. For the triangle cell 2, the stress–strain relationship is the same as Eq. (19), except that the valuesof Dmqpi and Dpimq have negative signs. If we select the representative volume for the stress–strain rela-tionship to be the region of the hexagonal lattice that covers six triangles, the average values of Dmqpj andDpjmq in Eq. (19) are zero. Thus the stress–strain relationship according to Eq. (19) is given in a matrixform by

r11

r22

r12

r21

l13

l23

8>>>>>><>>>>>>:

9>>>>>>=>>>>>>;

¼ffiffiffi3

p

4

3kn þ ks kn � ks 0 0 0 0kn � ks 3kn þ ks 0 0 0 0

0 0 kn þ 3ks kn � ks 0 00 0 kn � ks kn þ 3ks 0 00 0 0 0 4gs þ ksl2

30

0 0 0 0 0 4gs þ ksl2

3

26666664

37777775

e11e22e12e21x3;1

x3;2

8>>>>>><>>>>>>:

9>>>>>>=>>>>>>;

ð20Þ

Note that

e12 ¼ u2;1 � x; e21 ¼ u1;2 þ x ð21Þ

In order to compare with the usual stress–strain relationship, we define the symmetric and anti-sym-metric terms of stress

s ¼ r12 þ r21

2; T ¼ r12 � r21

2ð22Þ

and the symmetric and anti-symmetric terms of strain

c ¼ e12 þ e21 ¼ u2;1 þ u1;2; w ¼ u2;1 � u1;2;C ¼ ðu2;1 � u1;2Þ � 2x ¼ e12 � e21

ð23Þ

where C is the relative rotation between rigid body rotation w and particle rotation 2x. Note that the workdone by sc þ TC is the same as that by r12e12 þ r21e21. Using the new variables, Eq. (20) becomes

Fig. 5. A hexagonal lattice.


r11

r22

s

Tl13

l23

8>>>>>>><>>>>>>>:

9>>>>>>>=>>>>>>>;

¼ffiffiffi3

p

4

3kn þ ks kn � ks 0 0 0 0

kn � ks 3kn þ ks 0 0 0 0

0 0 kn þ ks 0 0 0

0 0 0 2ks 0 0

0 0 0 0 4gs þ ks3l2 0

0 0 0 0 0 4gs þ ks3l2

2666666664

3777777775

e11e22c

C

x3;1

x3;2

8>>>>>>><>>>>>>>:

9>>>>>>>=>>>>>>>;

ð24Þ

Compare Eq. (24) with the plane stress condition of Cosserat model (see for example Chang and Ma [6];Sluys [16])

r11

r22

s

Tl13

l23

8>>>>>>><>>>>>>>:

9>>>>>>>=>>>>>>>;

¼

2G1�t

2Gt1�t 0 0 0 0

2Gt1�t

2G1�t 0 0 0 0

0 0 G 0 0 0

0 0 0 W 0 0

0 0 0 0 j 0

0 0 0 0 0 j

266666664

377777775

e11e22c

C

x3;1

x3;2

8>>>>>>><>>>>>>>:

9>>>>>>>=>>>>>>>;

ð25Þ

where G is the shear modulus and t is the Poisson’s ratio. W is the ‘spin’ modulus and j is the ‘bending’modulus. It is noted that the elastic behavior is isotropic for a hexagonal lattice network. By comparing thetwo Eqs. (24) and (25), the constants for the equivalent continuum can be related to the spring constants ina lattice spring system or to the beam properties in a lattice beam system (see Tables 1–3). In Table 1, theexpressions of Poisson’s ratio and bulk modulus in terms of properties of lattice beams are identical to thatgiven by Schlangen and Garboczi [17]. It is noted that when ks and gs are zero, the lattice spring system isreduced to a central force spring network. In that case, the Poisson’s ratio is 1/3. When ks is very largecompared to kn, the Poisson’s ratio is close to �1.

Table 1

Continuum properties related to lattice spring and lattice beam properties

Continuum Lattice spring Lattice beam

tkn � ks3kn þ ks

1� hl

� �2

3þ hl

� �2

G

ffiffiffi3

p

4kn þ ksð Þ

ffiffiffi3

p

4

EbAb

lþ 12EbIb

l3

� �

E ¼ 2Gð1þ tÞ 2ffiffiffi3

pknðkn þ ksÞ3kn þ ks

2ffiffiffi3

p 1þ hl

� �2

3þ hl

� �2

EbAb

l

B ¼ E2ð1� tÞ

ffiffiffi3

p

2kn

ffiffiffi3

pEbAb

2l

W

ffiffiffi3

p

2ks

6ffiffiffi3

pEbIbl3

jffiffiffi3

pgs þ

ffiffiffi3

p

12ksl2

2ffiffiffi3

pEbIbl


3.2. Links between properties of lattice beams and the equivalent continuum

A network of lattice beams can be defined by three material constants, Eb, h and l, as described in Eq. (9).A network of lattice springs can be defined by three material constants, kn, ks and gs, as described in Eq. (8).The four continuum material constants shown in the stress–strain relationship are G, t, W, j. There are onlythree independent continuum material constants out of the four constants. The bending modulus j can beobtained from G and t

W ¼ 1� 3t1� t

G ð26Þ

The continuum material constants as a function of lattice spring properties and lattice beam propertiesare given in Table 1, where the cross-sectional area of the lattice beam Ab ¼ bh and for the two-dimensionalcondition b ¼ 1. The moment of inertia Ib, about the axis passing through centroid of the cross-sectionalarea, is bh3=12. For convenience, the lattice spring properties as a function of continuum material constantsand lattice beam properties are given in Table 2. The lattice beam properties as a function of continuummaterial constants and lattice spring properties are given in Table 3.

For a central force spring network, ks ¼ gs ¼ 0, the lattice network of beams becomes a lattice networkof bars. The members cannot carry moment and shear force. The central force spring network contains onlyone spring constant, ks. The corresponding bar property is given by, EbAb=l. For this special case, we shoulduse the following relationship from Table 1:

G ¼ffiffiffi3

p

4kn ð27Þ

Then use Table 2, the corresponding property for the bars can be obtained by

EbAb

l¼ 4Gffiffiffi

3p ð28Þ

Table 3

Lattice beam properties related to continuum and lattice spring properties

Lattice beam Continuum Lattice spring

hl

ffiffiffiffiffiffiffiffiffiffiffiffiffi1� 3t1þ t

r ffiffiffiffiffikskn

s

Eb Effiffiffi3

pð1� tÞ

ffiffiffiffiffiffiffi1�3t1þt

qkn

ffiffiffiffiffiknks

s

Table 2

Lattice spring related to continuum and lattice beam properties

Lattice spring Continuum Lattice beam

knEffiffiffi

3p

ð1� tÞEbAb

l

ks2Gð1� 3tÞffiffiffi

3p

ð1� tÞ12EbIbl3

gs1ffiffiffi3

p j � Gð1� 3tÞ6ð1� tÞ l2

� �EbIbl


3.3. Material length and the micro-structure

In a micro-polar stress–strain material, the ratio of bending modulus j and Young’s modulus E has aunit of length square. This length is commonly termed as internal length of the material. In the presentmodel, the ratio of bending modulus and Young’s modulus can be evaluated using the relationships givenin Table 1. The ratio is found to be directly related to the size of lattice beam, given by

jE¼ 1

12

3þ h=lð Þ2

l=hð Þ2 þ 1

!l2 ð29Þ

Thus based on the current material model, the internal length of the material stems from the size of itsunderlying lattice structure. The internal length plays an important role in regularization for achievingmesh size independency in a finite element method [16].

3.4. Fracture criterion for media

The damage in this continuum is dictated by the damage of its underlying lattice structure. A fracturelaw for lattice beams is established as follows:

F =Ab

ftþ jV j=Ab

fsh> 1 ð30Þ

where F is the axial tensile force and V is the shear force applied to the lattice beam underlying eachcontinuum element, ft is tensile strength and fsh is shear strength of the underlying lattice beams and Ab isthe cross-section area of beams. As soon as a beam reached its allowable strength, it is broken and com-pletely loses its load bearing capacity. No softening law is imposed. The failure criterion is similar to thatused by van Mier [3].

For the case that considers only tensile failure mode, the fracture law is simplified to be

F =Ab

ft> 1 ð31Þ

and for the case that considers only shear failure mode, the fracture law becomes

jV j=Ab

fsh> 1 ð32Þ

The axial force and shear force of a lattice beam can be obtained from Eq. (9), given by

F ¼ EbAb

l

� �dn

V ¼ 12EbIl3

� �ds þ

6EbIl2

� �x

ð33Þ

where Eb is the Young’s modulus of beams, dn is the elongation, ds is the relative shear displacement, and xis the averaged angular rotation at the two ends of a lattice beam. The value of di in a Cartesian coordinatesystem is defined in Eq. (2) and can be calculated from Eq. (13) using the displacement and rotation fieldcomputed from the finite element results. The values of dn and ds can be obtained from di using a trans-formation process from the x–y coordinate to the n–s coordinate system.

Since the underlying lattice network has three directions of lattice beams, the fracture criterion has to bechecked in all three directions. It is noted that the strength of concrete is influenced by the alignment of itsunderlying lattice structure. Therefore, the material strength is anisotropic. At the initial fracture, only a


portion of the lattice beams in the material is broken. If the load continues to increase, the intact beams inother directions would fracture thus causing additional damage of the material. The damage degradation ina continuum is a very difficult process to model, because it is a complicated phenomenon affected by thematerial axis that changes during the damage progress. The underlying structure provides a useful schemefor the description of degradation of the material.

3.5. Stress–strain relationship for a damaged medium

Corresponding to Eq. (18), the stress–strain relationship for the equivalent continuum of the system oflattice beams after the occurrence of damage can be expressed as

rmq ¼ Cdmqkieki þ Dd

mqpjxj;p

lpj ¼ Ddpjmqemq þ F d

pjrtxt;r

ð34Þ

where Cdmqki, D

dmqpj, and F d

pjrt with superscript ‘d’ represent the stiffness tensors of the damaged material.Degradation of the stiffness tensor is caused by progressive micro-cracking in the bonds of grains, which isnow represented by the failure of lattice beams. Since the stiffness tensors of a representative volume de-rived in Eq. (19) are evaluated from the stiffness of each individual lattice beam, the failure of beams can beeasily incorporated by slightly modifying Eq. (19) into the following set of equations:

Cdmqki ¼

1

V

Xc

dcKcqil

cklcm

Ddmqpi ¼ � 1

V

Xc

dceijkKcqil

cml

ckg

cp

Ddpjmq ¼ � 1

V

Xc

dceijkKciql

cklcmgcp

F dpjrt ¼

1

V

Xc

dc½eijkeqtmKciql

cklcmgcpg

cr þ Gc

jtlcrlcp�

ð35Þ

where the damage scalar dc for each beam is either zero or one (i.e., dc ¼ 1 for undamaged beam and 0 fordamaged beam). For the case of dc ¼ 1 for all beams, Eq. (35) is reduced to the same form as Eq. (19). Notethat the stress–strain relationship for a damaged medium is highly non-linear, since the damage scalar dc isdetermined by the force in each beam, which is in turn a function of the strain of the representative element.

If the fractured beams are known, Eq. (35) can be integrated to obtain a closed form expression. This isshown as an example for the lattice structure in Fig. 6(a). After one set of the beams (see the gray lines inFig. 6(b)) has reached its allowable strength, these beams lost their load bearing capacity. Among the three

Fig. 6. Original and damaged lattice structures of a representative volume.


sets of load-bearing members, only two sets are now active. Using Eq. (35) and taking into account thedamaged beams, we have the following closed-form expressions of the stress–strain matrix,

r11

r22

sTl13

l23

8>>>>>><>>>>>>:

9>>>>>>=>>>>>>;

¼ffiffiffi3

p

4

17knþ3ks6

kn�ks2

ffiffi3

pðkn�ksÞ6

ffiffi3

pðknþ3ksÞ

6� ksl

2ffiffi3

p ksl6

kn�ks2

3knþks2

ffiffi3

pð3knþksÞ

6

ffiffi3

pðkn�ksÞ2

ksl2ffiffi3

p � ksl6ffiffi

3p

ðkn�ksÞ6

ffiffi3

pð3knþksÞ

63knþ5ks

6ks3

ksl6

7ksl6ffiffi3

pffiffi3

pðknþ3ksÞ

6

ffiffi3

pðkn�ksÞ2

ks3

4ks3

� ksl2

ksl2ffiffi3

p

� ksl2ffiffi3

p ksl2ffiffi3

p ksl6

� ksl2

103gs þ ksl2

6� ksl2

6ffiffi3

p

ksl6

� ksl6

7ksl6ffiffi3

p ksl2ffiffi3

p � ksl2

6ffiffi3

p 2gs þ 5ksl2

18

266666666664

377777777775

e11e22cC

x3;1

x3;2

8>>>>>><>>>>>>:

9>>>>>>=>>>>>>;

ð36Þ

The stiffness matrix indicates that the damaged material becomes anisotropic and its material axis hasrotated. The damaged material still possess a load bearing capacity even along the direction of the fracturedbeams. The strongest material axis for this damaged material is not perpendicular to the direction ofdamage (i.e, direction of the broken beam). This is a feature that differs from standard continuum damagemodels.

In the same manner, after the failure of two sets of beams (see Fig. 6(c)), the three sets of load bearingmembers are now reduced to one. The stress–strain matrix is further reduced to

r11

r22

sTl13

l23

8>>>>>><>>>>>>:

9>>>>>>=>>>>>>;

¼ffiffiffi3

p

4

83kn 0 0 0 0 00 0 0 0 0 00 0 2

3ks 2

3ks 0 4ksl

3ffiffi3

p

0 0 23ks 2

3ks 0 0

0 0 0 0 83gs 0

0 0 4ksl3ffiffi3

p 0 0 2ksl2

9

266666664

377777775

e11e22cC

x3;1

x3;2

8>>>>>><>>>>>>:

9>>>>>>=>>>>>>;

ð37Þ

This stiffness matrix is singular. The material has now completely lost its load bearing capacity in thedirection perpendicular to the intact (non-broken) beams, although it can still carry load along the directionof intact beams.

3.6. A simplified non-polar model

By neglecting all variables associated with rotation (i.e., the stress components fT ; l13; l23g and straincomponents fC;x13;x23g), the micro-polar continuum is reduced to the classical continuum. It is desirableto study the non-polar model because of three reasons:

(1) The non-polar model does not have a bending modulus in the stress–strain relationship. Therefore itdoes not have an explicit internal length in the stress–strain relationship. On the other hand, thenon-polar stress–strain relationship is determined from the underlying structure. Thus it implicitly con-tains an internal length. It is interesting to investigate whether the non-polar model still has the abilityof regularization for mesh size independency, as found in models which contain an explicit internallength [16].

(2) Many of the damage/fracture models presented in literature have been based on the classical contin-uum. The simplified non-polar stress–strain model can be directly compared with these damage models.

(3) It is also interesting to know how much difference there is between the polar and non-polar models.This will provide an understanding on the effect of moment transfer in the system. For these purposes,we examine the damaged stress–strain relationship here taking the same example of the lattice structureas shown in Fig. 6(a). The issues described above will be discussed in detail in the accompanying paper.


Here, after one set of the beams (see Fig. 6(b)) has reached its allowable strength, using Eq. (35) or (36)and taking into account the damaged beams, we have the following closed-form expression of the stress–strain matrix,

r11

r22

s

8<:

9=; ¼

ffiffiffi3

p

24

17kn þ 3ks 3ðkn � ksÞffiffiffi3

pðkn � ksÞ

3ð3kn þ ksÞffiffiffi3

pð3kn þ ksÞ

3kn þ 17ks

24

35 e11

e22c

8<:

9=; ð38Þ

The stiffness matrix also indicates that the damaged material becomes anisotropic with a rotated materialaxis. The damaged material still possess a load bearing capacity even along the direction of the fracturedbeams. The strongest material axis for this damaged material is not perpendicular to the direction ofdamage (i.e, direction of the broken beam). This is a feature that differs from the usual damage models. Inthe same manner, after the failure of two sets of beams (see Fig. 6(c)), the three sets of load bearingmembers are now reduced to one. The stress–strain matrix is further reduced to

r11

r22

s

8<:

9=; ¼ 2

ffiffiffi3

p

3

kn 0 00 0

ks

24

35 e11

e22c

8<:

9=; ð39Þ

This matrix is singular and clearly shows that the material completely lost its load bearing capacity in thedirection 2.

4. Finite element formulation

The non-polar stress–strain relationship is written in a conventional form, and can, therefore, be easilyincorporated in a finite element method for the analysis of fracture. For the micro-polar stress–strain re-lationship, the finite element formulation can also be found in many works e.g., [5,16] etc. In this paper, wediscretize the continuum into constant strain triangular elements. For completeness, the strain–displace-ment relationship and shape functions used in the FEM formulation are given in Appendix B.

4.1. Finite element mesh with an underlying lattice structure

The finite element model can be used to model aggregates and fine particles in the real material ofconcrete, similar to the discrete lattice simulation used by Van Mier [3] or Schlangen and Garboczi [17].However, this requires the use of very small elements, and for this reason a considerable computationaleffort. In this paper, the underlying lattice structure is not intended to be used as a real structure in Ref. [1].It is rather used here for modeling and computational purposes. As described previously, the underlyingstructure provides a reasonable model for characterizing the damage of materials. Thus it not only helps toformulate a more realistic stress–strain relationship but it also makes the stress–strain modeling an easiertask.

For the present finite element model, at each integration point of the element, the geometry of theunderlying lattice structure is hexagonal containing uniform cell sizes. For various integration points, thesize and alignment of lattice may be different. By assigning randomly the alignment of the lattice for eachelement, the model would represent a heterogeneous material with random properties. It is noted that,before fracture occurs, the elastic properties are isotropic (i.e., independent of lattice alignment). However,the alignment has an effect on the strength of the material under a specific axis of applied stress. The tworeference alignments are vertical and horizontal alignments as shown in Fig. 7(a) and (b). It takes 30�rotation from vertical alignment to horizontal alignment. All possible configurations are between these twocases. Rotation with an angle of more than 30�, the configuration repeats. Therefore, the geometrical


property of hexagonal lattice structure is close to isotropic. The strength behavior is bound by the behaviorof the two reference alignments: vertical and horizontal alignments.

The size of the lattice structure underlying the element can be either smaller or larger than the actual sizeof the element. We want the element size to be small enough to model the crack zone. If the element size islarger than the crack zone, it will not be able to capture the detailed behavior and thus a discretization errorwill appear. If the purpose of the finite element analysis is to model the behavior within the crack zone, theelement size should be smaller than the crack zone. The subsequent section will discuss the choice of theactual size of lattice beams to be used in the analysis.

4.2. Material constants for the underlying lattice structure

Five constants are needed to specify a lattice structure: two of the constants are the dimension of thelattice beam h and l (b ¼ 1), and the other three are the tensile strength ft, the shear strength fsh andmodulus Eb for lattice beams. Given a Poisson’s ratio of the concrete material, the ratio of h=l for a latticebeam can be obtained from (see Table 1 with b ¼ 1)

t ¼1� h

l

� �23þ h

l

� �2 ð40Þ

which is similar to the method of determining beam sizes in a discrete lattice structure used by Schlangenand Van Mier [1].

The value of the stiffness modulus for the beam, Eb, can be obtained from (see Table 1 with Ab ¼ bh andb ¼ 1)

E ¼ 2ffiffiffi3

p 1þ hl

� �23þ h

l

� �2 Ebhl

ð41Þ

where E is the Young’s modulus of the concrete material.The size of latttice beams l can be determined from the ratio of the bending modulus and the Young’s

modulus as shown in Eq. (29), given by

l2 ¼ 12l=hð Þ2 þ 1

3þ h=lð Þ2

!jE

ð42Þ

The tensile strength ft and the shear strength fsh for lattice beams can be estimated by a combinedexperimental–computational study.

Fig. 7. The vertical and horizontal lattice alignments.


4.3. Regularization

For fracture analysis using the finite element method, a major problem is the occurrence of mesh sizedependency. The problem of mesh size dependency can be overcome if a stress–strain law explicitly containsa length parameter (termed as the internal length), such as that in micro-polar continua, non-local or high-order gradient continua [16,18–23]. The internal length in the stress–strain law plays a role that regularizesthe results of different mesh sizes. The present stress–strain model is of micro-polar type, which explicitlycontains an internal length (see Eqs. (24), (36) and (37) that is related to the dimension of the underlyinglattice structure. Therefore, it is expected that if the couple stresses are activated the results are independentof the mesh size, which will be verified in part II of this paper.

On the other hand, as a simplified case of the micro-polar model, the non-polar stress–strain model (seeEqs. (38) and (39)) takes the form of a classical stress–strain relationship, which does not explicitly containan internal length. Therefore, it is expected that the non-polar stress–strain law, similar to other classicalmodels, would display problems of mesh size dependency. However, the present non-polar stress–strain lawis micro-structurally based. Although the stress–strain relationship does not explicitly contain an internallength, the material parameters in the stress–strain relationship are dependent on the dimensions of theunderlying lattice structure, and thus implicitly the stress–strain law contains a length scale parameter. Thequestion is: can the non-polar model also lead to mesh size independent results? Or does the regularizationeffect only comes from the explicit length parameter in the polar version of the model. These issues will bediscussed in part II of this paper.

5. Conclusion

In this paper, we adopt a micro-structural approach to formulate the stress–strain behavior of concretematerial, which is treated as a continuum but has an underlying micro-structure of lattice type. As a resultof the moment transferring capability of lattice beams, the derived stress–strain relationship for the con-crete material reveals to be of micro-polar type. The internal length is dependent on the dimension of theunderlying lattice structure. The properties of the continuum can be linked to the properties of the un-derlying lattice beams.

Damage evolution of the material is dictated by fracture of its underlying lattice structure. Therefore themodel does not involve assumptions for the damage law of the material. After damage, the stiffness tensorshows a rotation of material axis and becomes anisotropic. As the material continue to degrade, the ma-terial axis continue to rotate. These features make the model more realistic and show significant conse-quences with crack development. In order to study the performance of this model, a finite elementformulation is described that incorporates the developed stress–strain relationship. In an accompanyingpaper the results of finite element analyses will be discussed to evaluate the performance of this micro-structurally based model.

Appendix A. Two dimensional interaction of a two-particle system

The corresponding particle force and particle moment acting on the particle centroid, f ni and mni are

related to the inter-particle force f nmi and the inter-particle force couple mnmi . For a packing of particles, one

particle has more than one neighbors, the equilibrium equations can be given by

f ni þXm

f nmi ¼ 0 ðA:1Þ


mni þ

Xm

mnmi

�þ eijkrnmk f

nmj

�¼ 0 ðA:2Þ

For a two-particle system in a two-dimensional configuration, Eqs. (A.1) and (A.2) can be written as

f mx ¼ �f nx ¼ f nmx ðA:3Þ

f my ¼ �f ny ¼ f nmy ðA:4Þ

mmz ¼ �mn

z ¼ mnmz � rnmx f

nmy þ rnmy f

nmx ðA:5Þ

and the kinematics of two particles (see Eqs. (1) and (2)) can be reduced to

dnmx ¼ umx � unx � rmy xmz þ rnyx

nz ðA:6Þ

dnmy ¼ umy � uny þ rmx xmz � rnxx

nz ðA:7Þ

hnmz ¼ xmz � xn

z ðA:8ÞFor a lattice-beam parallel to the x-axis (rmnx ¼ �rnmx ¼ l=2; rmny ¼ rnmy ¼ 0), the stiffness tensor for the

contact as defined in Eqs. (5) and (6) becomes

Kxx ¼ kn; Kxy ¼ Kyx ¼ 0; Kyy ¼ ks ðA:9Þ

Gzz ¼ gs ðA:10ÞBased on Eqs. (3), (4), (A.9) and (A.10), the inter-particle contact forces can be expressed as follows:

f nmx ¼ kndnmx ¼ knumx � knunx ðA:11Þ

f nmy ¼ ksdnmy ¼ ksumy � ksuny þ

ksl2

xmz þ ksl

2xnz ðA:12Þ

mnmz ¼ gsh

nmz ¼ gsxm

z � gsxnz ðA:13Þ

Substituting Eqs. (A.12) and (A.13) into Eqs. (A.5), the moment acting at the particle centroid can beobtained according to

mmz ¼ �mn

z ¼ksl2

umy�

� uny�þ ksl2

4

�þ gs

�xmz þ ksl2

4

�� gs

�xnz ðA:14Þ

Similarly the forces acting on the particle centroid can be obtained from Eqs. (A.11), (A.12), (A.3) and(A.4). The final form is given in Eq. (8).

Appendix B. The strain–displacement relationship and shape functions used in the FEM formulation

A constant-strain triangular element is considered here for the finite element formulation. The shapefunction is defined to relate any variable /ðx; yÞ in the element to the three nodal values /i, given by

/ðx; yÞ ¼X3

i¼1

Niðx; yÞ/i ðB:1Þ

The shape function Niðx; yÞ is given by

Niðx; yÞ ¼1

2Dai½ þ bixþ ciy� ðB:2Þ


where ai, bi and ci can be determined from the coordinates of the three nodal points:

ai ¼ xjym � xmyjbi ¼ yj � ymci ¼ xm � xj

ðB:3Þ

The displacement vector consists of the following variables on the three nodes, given by uT ¼½u1x ; u1y ;x1; u2x ; u

2y ;x

2; u3x ; u3y ;x

3�. The displacement vector is related to the strains of the element by

ef g ¼ B½ � uf g ðB:4Þ

where the strain vector is defined as eT ¼ e11; e22; c;C;x3;1;x3;2½ �. The components of this strain vector isdefined in Eq. (23). The matrix [B] can thus be derived from the shape functions, given by

B½ � ¼

oN1

ox 0 0 oN2

ox 0 0 oN3

ox 0 0

0 oN1

oy 0 0 oN2

oy 0 0 oN3

oy 0oN1

oyoN1

ox 0 oN2

oyoN2

ox 0 oN3

oyoN3

ox 0

� oN1

oyoN1

ox � 23

� oN2

oyoN2

ox � 23

� oN3

oyoN3

ox � 23

0 0 oN1

ox 0 0 oN2

ox 0 0 oN3

ox

0 0 oN1

oy 0 0 oN2

oy 0 0 oN3

oy

26666666664

37777777775

ðB:5Þ

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fracture modeling using a micro-structural mechanics approach––i. theory and formulation

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