MICROMECHANICAL ANALYSIS OFCOMPOSITES

Michal ŠejnohaJan Zeman

Faculty of Civil EngineeringCzech Technical University in Prague

SESSION CONTENT

• Review of basic continuum mechanics principles for deriving effective thermoe-lastic material properties of macroscopically homogeneous composite materials,introduction to the ellipsoidal inclusion problem and related averaging schemessuch as the dilute approximation, the self-consistent method and the Mori-Tanakamethod for evaluation of effective elastic moduli of statistically isotropic and er-godic composite

• Introduction to basic principles of homogenization of composite materials withperiodic microstructures, concept of the unit cell, periodic fields, formulation ofthe thermoelastic problem, formulation of the homogenization problem for perfo-rated plates, formulation of the homogenization problem for masonry, resolutionof the problem by the Finite Element Method, implementation of the periodicboundary conditions into the existing commercial codes

• Extension of basic mechanics principles for deriving elastic and inelastic macro-scopic constitutive equations of random composite materials, general aspects,statistically equivalent unit cell, extension of the Hashin-Shtrikman variationalprinciples to treat statistically homogeneous and ergodic random media witheigenstrains, application to unidirectional composites with random distribution offibers within the transverse plane section bonded by nonlinear viscoelastic polymermatrix, resolution of the problem by the Fast Fourier Transform

MICROMECHANICAL ANALYSIS OF COMPOSITES

SESSION I

• Governing equations of elasticity

• Volume averages and minimum energy principles

• Overall moduli - elementary bounds on effective properties

ELASTIC AND INELASTIC ANALYSIS OF HETEROGENEOUS MATERIALS

heterogeneous material: material having different material properties from pointto point (concrete, soils, composites, jointed rocks, etc.)

anisotropic material: material having different material properties in different directions,no planes of material symmetry - 81 independentmaterial constants

orthotropic material: three orthogonal planes of material symmetry - 9 independentmaterial constants

transversally isotropic material: one plane of symmetry (plane of isotropy) witha perpendicular axis of symmetry (elastic moduli are invariant withrespect to rotation about this axis) - 5 independent constants

isotropic material: three axis of rotational symmetry - 2 independent constants

micromechanics: mechanics of composites at the constituents levelmacromechanics: mechanics of composite structures such as laminates

micromechanical analysis:

1. mechanics of materials (simplifying assumptions unnecessary to specify the detailsof the stress and strain distribution at the micromechanical level, geometry of themicrostructure is arbitrary)

2. elasticity theory (elasticity theory models involves the solution of actual localstress and strain fields and actual geometry of the microstructure is taken intoaccount)

3. empirical solutions (curve fitting to elasticity solutions or experiments)

EVALUATION OF EFFECTIVE PROPERTIES OF COMPOSITE MATERIALS - 1

Eshelby, Hill, Walpole, Laws, Dvorak, Hutchinson, Pagano, Christensen, Mura

......... the concept of an effective modulus is essential to the development of practicalengineering stress-strain relationships for composite materials.

GOVERNING EQUATIONS OF LINEAR ELASTICITY

In this study we shall consider only statics in the absence of the body forces. Individualparticles of the body will be identified by their coordinates xi (i = 1, 2, 3) in theundeformed configuration.

Displacement fieldui = ui(x1, x2, x3) = ui(xj).

Strain field

�ij =1

2(ui,j + uj,i) . (1)

Equations of equilibrium in the absence of body forces

σij,j = 0 (2)

σij = σji.

Surface tractionspi = σijnj. (3)

Constitutive equations

σij = Lijkl(�kl − �0kl) (4)Lijkl = Ljikl = Lijlk.

The fourth-order tensor Lijkl is known as the stiffness tensor. Suppose that a strainenergy density function U(�ij) per unit volume volume exists such that

σij =∂U

∂�ij. (5)

Eqs. (2) and (4) readily provideLijkl = Lklij

and that

U =1

2Lijkl�ij�kl. (6)

Providing the strain energy U has a minimum in the stress-free state then Lijkl ispositive definite

Lijkl�ij�kl > 0 (7)

for all non-zero symmetric tensors �ij. It is now possible to invert Eq. (4) to get

�ij = Mijklσkl + �0ij (8)

Mijkl is known as the compliance tensor. Note that

Mijkl = Mjikl = Mijlk = Mklij

and also

LijrsMrskl =1

2(δikδjl + δilδjk) = Iijkl, (9)

where δij is the Kronecker delta and Iijkl represents the fourth-order identity tensor.Note: we used standard Cartesian tensor notation in which repeated suffixes are summedover the range 1, 2, 3.

AVERAGES

[Laws, 1980]

In preparation for evaluation of the overall moduli we first review some basic formulaefor the determination of average stresses and strains. To that end, we assume that thedisplacements fields are continuous, and the strain fields are compatible; also, the stressfields and tractions are continuous and in equilibrium

Consider first an arbitrary homogeneous medium of volume V with the boundary S. Ingeneral, the volume average of a quantity is just the ordinary volume average given by

〈f〉 = 1V

∫V

fdV. (10)

Let ε(x) and σ(x) be certain fields in V . Their volume averages are defined as

〈ε〉 = 1V

∫V

ε(x) dV 〈σ〉 = 1V

∫V

σ(x) dV (11)

After applying the divergence theorem we arrive at

〈εij〉 =1

2V

∫S

(uinj + ujni)dS (12)

〈σij〉 =1

2V

∫S

(pixj + pjxi)dS (13)

Next, consider a heterogeneous elastic medium which consists of a homogeneous matrix

V2 and homogeneous inclusion V1. Evaluation of the above volume averages requires anapplication of a generalized (but still standard) divergence theorem. Let f be continuousin V and continuously differentiable in the interior of V1 and V2. We may now applythe divergence theorem separately to V1 and V2 to conclude that∫

V

∂f

∂xidV +

∫Σ

[f ]mjdS =

∫S

fnidS (14)

where [f ] denotes the jump in the value of f as we travel across Σ from V1 to V2. Now,assume that perfect bonding exists. When setting f = ui in Eq. (14) we immediatelyrecover Eq. (12). Since tractions are continuous across Σ

[σij]mj = 0

setting f = σikxj yields Eq. (13). We may now conclude that Eqs. (12) and (13)apply to any heterogeneous material, generally anisotropic, consisting of a homogeneousmatrix and an arbitrary number of homogeneous inclusions.

EXAMPLES 1.1

Consider an arbitrary composite material with outer boundary S

1. Suppose that the composite is loaded by displacements ui on S, which are com-patible with the uniform strain Eij, i.e. ui = Eijxj (affine displacements). Showthat

< εij >= Eij.

2. Suppose that the composite is loaded by prescribed tractions pi on S, which arecompatible with the uniform stress Σij, i.e. pi = Σijnj. Show that

< σij >= Σij.

3. Let σij be a self-equilibrated stress field (σij,j=0) and ui is a displacement fieldassociated with strain εij =

12(ui,j +uj,i). Show that if either ui = 0 or σijnj = 0

on the boundary then ∫V

σijεij dV = 0.

4. For the boundary conditions of Exs. (1) and (2) show that (Hill’s lemma)

< σijεij >=< σij >< εij > and < U >=1

2< σij >< εij > .

MINIMUM ENERGY PRINCIPLES

We now give a brief review of the classical energy principles as they have been extensivelyused in assessing the bounds on the overall elastic properties of composites.

First, consider an arbitrary anisotropic elastic medium with prescribed displacements uialong its boundary. Let εij, σij, U be the associated strain, stress, and strain energydensity, respectively. The purpose of this investigation is to show that the energy densityU∗ associated with any kinematically admissible displacement field u∗i is greater thanthe energy function U associated with the true solution. Let

ε∗ij =1

2(u∗i,j + u

∗j,i) σ

∗ij = Lijklε

∗kl

U∗ =1

2σ∗ijε

∗ij.

In the next step, calculate the energy of the difference state with displacements (u∗i−ui),which is positive

1

2(σ∗ij − σij)(ε∗ij − εij) ≥ 0

Therefore,

1

2

∫V

(σ∗ijε∗ij − σijεij) dV ≥

1

2

∫V

(σ∗ijεij + σijε∗ij − 2σijεij) dV

Applying Betti’s theorem σ∗ijεij = σijε∗ij yields

1

2

∫V

(σ∗ijε∗ij − σijεij) dV ≥

∫V

σij(ε∗ij − εij) dV = 0

and finally ∫V

U∗ dV ≥∫V

U dV (15)

we recover a special case of the theorem of minimum potential energy.

Next, consider the second boundary value problem with prescribed tractions along theboundary of the anisotropic solid. Once again, let ui be the required solution andεij, σij,W be the corresponding strain, stress, and stress (complementary) energy den-sity function, respectively. Suppose that τij is any statically admissible stress field anddefine the associated field ηij

ηij = Mijklτkl.

Again, using the trick of computing the positive energy associated with difference state

1

2(τij − σij)(ηij − εij) ≥ 0

yields

1

2

∫V

(τijηij − σijεij) dV ≥1

2

∫V

(τijεij + σijηij − 2σijεij) dV∫V

(τij − σij)εij dV = 0.

It now follows that ∫V

Mijklτijτkl dV ≥∫V

Mijklσijσkl dV, (16)

which is the special case of the theorem of minimum complementary energy.

EXAMPLES 1.2

Consider an arbitrary heterogeneous body with outer boundary S.

1. Suppose that the medium is loaded by prescribed displacements ui on S. Then, ifthe material is stiffened in any way (keeping the boundary fixed) show that strainenergy increases (Hill’s stiffening theorem).

2. Show that if the stiffness tensor Lijkl is increased by a positive amount, then thecorresponding compliance tensor Mijkl decreases by the positive amount.

OVERALL MODULI

Even with a powerful hardware at hand the detailed evaluation of local fields on themicroscale may prove to be prohibitively expensive and in most practical applications isout of questions. We must therefore settle for a somewhat limited objective.

......... Since the dimension of the reinforcement (the microscale) are often of the orderof 1µm and the dimension of laboratory samples (the macroscale) are of the order of1 cm, one hopes that, for certain types of loading, knowledge of the actual stress fieldis not necessary. Rather, the overall, or macroscopic response is the important physicalvariable. Thus we are led to study the restricted problem of determining overall elasticmoduli of composites...... (Norman Laws, 1980).

Assumptions:

1. A heterogeneous medium made of bonded phases or constituents separated by welldefined interfaces of zero thickness (interfaces or coatings are separate phases)

2. Matrix-based composites: inclusions or inhomogeneities embedded in a commonmatrix (Fibers, particles, whiskers, etc.), two-phase, three-phase, multiphase.

3. Loading conditions are limited to macroscopically uniform fields (recall examples1.1.1 and 1.1.2)

4. Each phase is elastically homogeneous and presumably anisotropic. The com-posite aggregate is assumed to be macroscopically homogeneous continuum withcertain effective (overall, macroscopic) properties.

representative volume (RVE) - must be geometrically typical of the composite

1. It contains a sufficient number of inclusions so that under macroscopically uniformtractions or displacements, its properties are independent of its size. In general,a representative volume element is much larger than typical inclusion size or fiberdiameter (1-150 µm).

2. In some specific models the RVE may have a well defined geometry and boundaryconditions (periodic arrays, unit cells).

RVE

Objectives:

Evaluate the effective elastic material properties of the composite system in terms ofphase elastic moduli, phase volume fractions, and geometry of the microstructure.

ELEMENTARY THEORY OF OVERALL MODULI

Consider a representative volume V of a heterogeneous medium with phases r =1, 2, . . . N under uniform overall strain E or stress Σ. Introducing certain mechani-cal influence functions, we wish to find the local fields in the phases as

εr(x) = Ar(x)E, σr(x) = Br(x)Σ. (17)

With the help of Eq. (11) the phase volume averages are provided by

εr =1

Vr

∫Vr

Ar(x)E dV = ArE, (18)

σr =1

Vr

∫Vr

Br(x)Σ dV = BrΣ.

Ar and Br are mechanical strain and stress concentration factors.

The concentration factors posses only the minor symmetry

Aijkl = Ajikl = Aijlk AT 6= A

Bijkl = Bjikl = Bijlk BT 6= B

In any elastically homogeneous phase r the phase constitutive relations read

εr(x) = Mrσr(x) σr(x) = Lrεr(x). (19)

For the representative volume of a statistically isotropic and ergodic composite mediumsubjected to uniform overall strain and stress

< ε >= E < σ >= Σ,

the overall constitutive relations are given by

< ε >= MΣ < σ >= LE with M = L−1. (20)

Recall Eq. (18) to show that

crεr =1

V

∫Vr

Ar(x)E dV and hence (21)

N∑r=1

crεr =1

V

N∑r=1

∫Vr

Ar(x)E dV = < ε >,

where cr = Vr/V is the volume fraction of the phase r. Thus

< ε >=N∑r=1

crεr and similarly < σ >=N∑r=1

crσr (22)

Eqs. (19), (20) and (22) then readily provide the overall constitutive relations in termsof phase moduli and phase volume fractions as

M < σ >=N∑r=1

crMrσr L < ε >=N∑r=1

crLrεr. (23)

Recall that for the present boundary conditions, the following relation holds

< σijεij >=< σij >< εij > and < U >=1

2< σij >< εij > .

Then, invoking the Reciprocal theorem of elasticity (Betti’s theorem) for any two setsof elastic fields (< σij >,< εij >) and (< σ

∗ij >,< ε

∗ij >) in volume V

< σij >< ε∗ij >=< σ

∗ij >< εij > Lijkl < εkl >< ε

∗ij >= Lijkl < ε

∗kl >< εij >,

suggests overall symmetry relations for any actual heterogeneous medium.

L = LT and M = MT.

Introducing phase concentration factors yields

M < σ > = c1M1B1 < σ > +c2M2B2 < σ > + . . .+ cNMNBN < σ >

L < ε > = c1L1A1 < ε > +c2L2A2 < ε > + . . .+ cNLNAN < ε > . (24)

Eqs. (24) must hold for any < σ >,< ε >. Hence, the overall elastic stiffness andcompliance matrices are

L =N∑r=1

crLrAr M =N∑r=1

crMrBr. (25)

Although we proved the diagonal symmetry of L and M, it is advisable to proceed withcaution when evaluating the strain and stress concentration factors only approximatelyusing various averaging methods. Then the diagonal symmetry must be examined foreach case.

EXAMPLES 1.3

Consider a two-phase or binary system (r = 1, 2).

1. Show thatc1A1 + c2A2 = c1B1 + c2B2 = I

2. Using the above relations, show that

L = L2 + c1(L1 − L2)A1 M = M2 + c1(M1 −M2)B1

3. If the overall elastic properties are known, show that

c1A1 = (L1 − L2)−1(L− L2) c1B1 = (M1 −M2)−1(M−M2)

c2A2 = −(L1 − L2)−1(L− L1) c2B2 = −(M1 −M2)−1(M−M1)

ELEMENTARY ESTIMATES OF OVERALL MODULI OFUNIDIRECTIONAL COMPOSITES

The objective is to present the elementary mechanics of materials for predicting four in-dependent effective moduli of an orthotropic continuous fiber-reinforced lamina. Recallthat in the elementary mechanics of materials approach to micromechanical modeling,the geometry of the microstructure is not specified. Hence, the RVE may be a genericblock consisting of fiber material bonded to matrix material. The only information re-quired are elastic moduli of individual phases and the fiber volume fraction. In addition,the matrix is assumed to be isotropic whereas the fiber is orthotropic.

A 1

A f

A m

c1σ

2

1

Matrix

FiberLf

Matrix

c2σ c12σ

Lm

2

Lm

2

2L

LONGITUDINAL MODULUS

Suppose that the macroscopic longitudinal normal stress σc11 is prescribed (for the timebeing, we drop < > and consider average quantities only). Then the force equilibriumcondition requires

σc11 = cfσf11 + cmσ

m11.

Neglecting the Poisson strains, a one-dimensional Hook’s law read

σc11 = Ec11ε

c11, σ

f11 = E

f11ε

f11, σ

m11 = E

m11ε

m11.

Therefore

Ec11εc11 = cfE

f11ε

f11 + cmE

m11ε

m11.

Assuming that average strains in the composite, fiber and matrix along the fiber direc-

tion are equal (εc11 = εf11 = ε

m11) provides the rule of mixtures

Ec11 = cfEf11 + cmE

m11.

The same results can be obtained using the strain energy approach, which yet confirmsthe validity of the assumption of equal strains.

TRANSVERSE MODULUS

Suppose that the macroscopic transverse normal stress σc22 is prescribed. Geometricalcompatibility requires that the total transverse composite displacement, uc2, must beequal to the sum of the transverse displacement in the fiber, uf2 , and in the matrix, u

m2

uc2 = uf2 + u

m2

Using the elementary definition of normal strain we get

uc2 = εc22L2, u

f2 = ε

f22Lf , u

m2 = ε

m22Lm; ε

c22L2 = ε

f22Lf + ε

m22Lm.

Combining a one-dimensional Hook’s law with compatibility of strains

σc22 = Ec22ε

c22, σ

f22 = E

f22ε

f22, σ

m22 = E

m22ε

m22; ε

c22 = cfε

f22 + cmε

m22,

and assuming that the stresses in the composite, fiber, and matrix are all equal, we getthe inverse rule of mixture

1

Ec22=

cf

Ef22+

cmEm22

.

Note: The simplified geometry of the RVE suggests that the assumption of equalstresses is valid (analogy with the series arrangement of springs). In reality, however, theactual fiber-packing is far from the idealized one. Employing the strain energy approachwe show that indeed the assumption of equal stresses is not generally justified. The

total energy stored in the composite Uc is represented as the sum of the strain energyin the fibers, Uf , and the strain energy in the matrix Um

Uc = Uf + Um.

To proceed, recall Eq. (15) and write the fiber and matrix strains in terms of thecomposite strain as

εf22 = afεc22, ε

m22 = amε

c22

Then, substitution of the above conditions into expression for the composite energy Uc(neglecting the strain energy due to the Poisson strain mismatch), leads to

Ec22 = cf (af )2Ef22 + cm(am)

2Em22.

Example:

Consider an E-glass/epoxy composite with the following material properties

Ec11 = 34.82 GPa Ec22 = 10.55 GPa

Ef11 = Ef22 = 72.4 GPa cf = 0.45

Em11 = Em22 = 3.79 GPa cm = 0.55

Solving the following system of equations

cfaf + cmam = 1

Ec22 = cf (af )2Ef22 + cm(am)

2Em22,

we get

af = 0.432, am = 1.465,afam

=εf22εm22

= 0.259,σf22σm22

= 5.63

Thus the assumption of equal stresses in fibers and matrix is not justified for this ma-

terial and is not valid for most other composites as well.

LONGITUDINAL SHEAR MODULUS AND POISSON’S RATIO

Similar considerations, which led to rule of mixtures for the longitudinal and transverseelastic moduli, provide another rules of mixture for the longitudinal shear moduli andthe Poisson ratio

1

Gc12=

cf

Gf12+

cmGm12

νc12 = cfνf12 + cmν

m12.

As for the longitudinal modulus, the above estimate of the overall Poisson ratio isgenerally accepted, while the estimate of overall shear modulus is rather poor becausethe shear stresses are not equal as assumed.

EXAMPLE 1.4

Using the strain energy approach, show that the geometrical assumption of equal axialstrains is valid. Use the material properties for E-glass/epoxy composite system.

ELEMENTARY BOUNDS ON THE OVERALL MODULI

Here we derive elementary bounds on the overall elastic moduli for a two-phase com-posite medium using the classical extremum principles.

Suppose that the RVE is subjected to surface displacements corresponding to macro-scopic uniform strain E. The actual strain energy per unit volume is given by

U =1

2ETLE. (26)

To invoke the principle of minimum potential energy, Eq. (15), consider a trial strain ε∗

uniform throughout the RVE such that ε∗ = E. Note that such a trial strain complieswith the prescribed boundary conditions, but leaves an artificial unequilibrated stressfield. When introducing ε∗ into Eq. (15) we immediately discover that

1

2ETLE ≤ c1

1

2ETL1E + c2

1

2ETL2E. (27)

Since Eq. (27) holds for any E, we get the upper bound on the overall elastic stiffnesstensor (constant strain bound) in the form

L ≤ c1L1 + c2L2. (28)

Next, consider the loading conditions due to prescribed tractions associated with a uni-form stress Σ. In view of the previous steps, assume a trial stress σ∗ uniform throughoutthe RVE such that σ∗ = Σ. Note that such a trial stress is self-equilibrated, but thenecessary bonding between the different phases is not maintained. Next, introduc-ing σ∗ into the principle of minimum complementary energy, Eq. (16), renders thethe upper bound on the overall elastic compliance tensor(constant stress bound) in the form

M ≤ c1M1 + c2M2. (29)

Example:

We now apply the above procedure to the simplest composite medium consisting ofisotropic phases and which has overall isotropy. For such a medium, the overall stiffnesstensor can be written in terms of the overall bulk modulus κ and the overall shearmodulus µ as

Lijkl = κδijδkl + µ(δikδjl + δilδjk −2

3δijδkl). (30)

Substitution of Eq. (30) into Eq. (28) leads to

κ ≤ c1κ1 + c2κ2 = κV (31)µ ≤ c1µ1 + c2µ2 = µV ,

where subscript V stands for [Voight, 1887], who was the first to propose these boundsfor a polycrystalline aggregate.

Dual assumption about uniform trial stress, put forward by [Reuss, 1929], gives

1

κ≤ c1

κ1+c2κ2

=1

κR(32)

1

µ≤ c1

µ1+c2µ2

=1

µR.

Eqs. (31) and (32) provide upper and lower bounds on the overall moduli

κR ≤ κ ≤ κV µR ≤ µ ≤ µV . (33)

Unfortunately these bounds are too far apart to be of any practical significance.

EXAMPLES 1.5

1. Show that Hill’s stiffening theorem implies the following. For a given compositesuppose that we replace some of the material by a stiffer material. Then theoverall stiffness of the new material is greater then the overall stiffness of theoriginal material.

2. Since 3E

= 1µ

+ 13κ

, show that

EV = c1E1 + c2E2 if ν1 = ν2.

EXACT SOLUTION FOR TWO-PHASE COMPOSITES WITH ISOTROPICPHASES AND EQUAL SHEAR MODULI

Consider a two-phase composite medium with phases of equal shear moduli. We wishto determine the overall bulk modulus in terms of local stress and strain fields. Thecomplete solution was given by [Hill, 1964] for a two phase composite solid of arbitrarygeometry, but macroscopically homogeneous. Extension to N -phase composite mediumcan be found in [Walpole, 1981].

Following Hill, displacement fields in each phase assume the form

uri = φr,i r = 1, 2

so that the associated strains are given by

εrij = φr,ij.

The corresponding stresses are

σrij = 2µφr,ij +

(κr −

2

3µ

)∇2φrδij.

It follows from the above equation that the stresses are self-equilibrated provided theLaplacian is piecewise constant

∇2φr = θr.

It is now evident that θ1 and θ2 represent the phase dilatation. For the tractions to becontinuous across the interface, [σij]nj = 0, we require

(κ1 +4

3µ)θ1 = (κ2 +

4

3µ)θ2.

The overall dilatation < εkk > is given by

< εkk >= θ = c1θ1 + c2θ2 =⇒θ1

κ2 +43µ

=θ2

κ1 +43µ

=θ

c1κ2 + c2κ1 +43µ.

The macroscopic constitutive equations are written as

< σij >= 2µ < εij > +(κ−2

3µ) < εkk > δij, (34)

with the overall bulk modulus κ given by

κ =c1κ1(κ2 +

43µ) + c2κ2(κ1 +

43µ)

c1(κ2 +43µ) + c2(κ1 +

43µ)

(35)

Note: The overall stress-strain relationship is isotropic despite the fact that the phasegeometry has not been specified. Also, the overall bulk modulus depends only onthe phase moduli and phase volume fractions. The above solution corresponds to amacroscopically uniform pure dilatation of a representative volume.

IMPROVED BOUNDS

Recall Hill’s stiffening theorem∫V

U(εij) dV ≤∫V

U∗(ε∗ij) dV, (36)

where the (∗) system is associated with the stiffened material and V is volume of theRVE. Consider now a two-phase isotropic composite medium with elastic moduli (κ1, µ1)and (κ2, µ2), respectively, with volume fractions c1 and c2. In addition, consider twogeometrically similar composites with phases of equal shear moduli and κ1, κ2 fixed.The first one is associated with the shear modulus µL while the the shear modulus ofthe second one is µU . Suppose that

µL = min (µ1, µ2) µU = max (µ1, µ2).

Eqs. (34) and (36) provide the best possible bounds on κ in terms of phase moduli andphase concentrations alone

κ(l.b.) = κ1 +c2 (κ2 − κ1)

1 + c1 (κ2 − κ1) /(κ1 + 43µL)(37)

κ(u.b.) = κ1 +c2 (κ2 − κ1)

1 + c1 (κ2 − κ1) /(κ1 + 43µU).

MICROMECHANICAL ANALYSIS OF COMPOSITES

SESSION II

• Micromechanical analysis of random composites

• Equivalent unit cell for composites with aligned fibers - fiber tow, unidirectionalfibrous plies and laminates

• Equivalent unit cell for textile composites

MICROMECHANICAL MODELING OF RANDOM COMPOSITESPart I - Quantification of random microstructure

Objectives

Micromechanical analysis of composite materials with disordered microstructure

Tools

Microscopic color or binary images of real microstructures

Graphite-fiber tow (12000 fibers) Random cut (300 fibers)

CONCEPT OF AN ENSEMBLE

To reflect the random character of the media it is convenient to introduce the conceptof an ensemble – the collection of a large number of systems (micrographs taken fromdifferent samples of the material) which are different in their microscopical details butthey are identical in their macroscopic details.

Then, effective or expected value of some quantity corresponds to the process of itsaveraging through all systems, forming the ensemble.

Consider a sample space S with individual members denoted as α. Define p(α) as theprobability density of α in S. Then an ensemble average of function F (x, α) at pointx is provided by

F (x) =

∫SF (x, α)p(α) dα. (38)

Following the above definition would require experimental determination of the ensem-ble average of function F (x, α) for a given point x through the cumbersome procedureof manufacturing a large number of samples (which form the ensemble space S), mea-suring F (x, α) for every sample and then its averaging for all samples. Therefore, itappears meaningful to introduce certain hypotheses regarding the ensemble average,which substantially simplify this task.

STATISTICAL HYPOTHESIS

Ergodic hypothesisThis hypothesis demands all states available to an ensemble of the systems to beavailable to every member of the system in the ensemble as well [Beran, 1968]. Oncethis hypothesis is adopted, spatial or volume average of function F (x, α) given by

〈F (x, α)〉 = limV→∞

1

V

∫V

F (x + y, α) dy (39)

is independent of α and identical to the ensemble average

〈F (x)〉 = F (x). (40)

Statistical homogeneitySuppose that function F depends on n vectors x1,x2, . . .xn. The ensemble average ofF is invariant with respect to translation

F (x1, . . . ,xn) = F (0,x2 − x1, . . . ,xn − x1) = F (x12, . . . ,x1n), (41)

where xij = xj − xi.

Statistical isotropyThe ensemble average of F is invariant with respect to both translation and rotation

F (x12, . . . ,x1n) = F (rij) (42)

where rij = ‖xij‖, i = 1, . . . , n, j = 1, . . . , n.

MICROSTRUCTURE DESCRIPTION

Indicator functionTo describe a random microstructure we introduce a characteristic or indicator functionχr(x, α), which is equal to one when point x lies in phase r in the sample α and equalto zero otherwise

χr(x, α) =

{1 x ∈ Dr(α)0 otherwise.

(43)

The symbol Dr(α) denotes here the domain occupied by r-th phase. For a two-phasefibrous composite, r = f,m, characteristic functions χf (x, α) and χm(x, α) are relatedby

χm(x, α) + χf (x, α) = 1. (44)

General n-point probability functionWith the aid of function χ, the general n-point probability function Sr1,...,rn is given by[Beran, 1968, Torquato and Stell, 1982]

Sr1,...,rn(x1, . . . ,xn) = χr1(x1, α) · · ·χrn(xn, α). (45)

Thus, Sr1,...,rn gives the probability of finding n points x1, . . . ,xn randomly thrown intomedia located in the phases r1, . . . , rn. We limit our attention to functions of the orderof one and two.

FUNCTIONS OF THE FIRST AND SECOND ORDER

One-point probability function Sr(x)Gives the probability of finding phase r at x

Sr(x) = χr(x, α), (46)

Two-point probability function Srs(x,x′)Gives the probability of finding simultaneously phase r at x and phase s at x′.

Srs(x,x′) = χr(x, α)χs(x′, α), (47)

Statistical hypothesisHomogeneity:

Sr(x) = Sr Srs(x,x′) = Srs(x− x′)

Isotropy:Srs(x− x′) = Srs(‖x− x′‖)

Ergodicity:Sr = cr

Limiting valuesfor x→ x′ : Srs(x,x′) = δrsSr(x)

for ‖x− x′‖ → ∞ : lim‖x−x′‖→∞

Srs(x,x′) = Sr(x)Ss(x

′)

Homogeneous Isotropic Ergodic

Sr(x) Sr Sr cr

Srs(x1,x2) Srs(x12) Srs(r12) Srs(r12)

NUMERICAL EVALUATION OF STATISTICAL DESCRIPTORS

One-point matrix probability function SmRecall that the one–point matrix probability function Sm gives the chance of findinga randomly placed point located in the matrix phase. To determine this quantity, asimple Monte-Carlo like simulation can be utilized – we throw randomly point into themicrostructure and count successful “hits“ into the matrix phase. Then, the value offunction Sm can be estimated as

Sm ≈n′

n

Two-point matrix probability function Smm(x)a) Using sampling templateInstead of tossing a line corresponding to x into a medium, sampling template is formed.The center of such a sampling template is randomly thrown into a medium and corre-sponding successful hits are counted. Furthermore, if the medium under considerationis statistically isotropic, values found for points located on the same circumference canbe averaged as well, which allows large number of tests to be performed within oneplacement of the template. Although simple, such simulations are computationally veryintensive.

b) Exploiting binary imagesSuch a digitized micrograph can be imagined as a discretization of the characteristicfunction χr(x, α), usually presented in terms of a M ×N bitmap. Denote the value ofχr for the pixel located in the i

th row and jth column as a χr(i, j)Ergodic and statistically homogeneous medium

Srs(m,n) =1

(im − iM + 1)(jn − jN + 1)

iM∑i=im

jN∑j=jn

χr(i, j)χs(i+m, j + n) (48)

where im = max(1 − m, 1), iM = min(M,M − m) and jn = max(1 − n, 1), jN =min(N,N − n). Observe that O((MN)2) operations are needed for function Srs.Periodic ergodic mediumFunction Srs can be written as a correlation of functions χr and χs

Srs(x) =1

Ω

∫Ω

χr(y)χs(x + y) dy (49)

FOURIER TRANSFORM

The d-dimensional Fourier Transform of function f(x)

F (f(x)) = f̃(ξ) =∫

Ω

f(x)eiξ·x dx, (50)

where i is the imaginary unit. The operator F is called the Fourier transform operator.

Inverse Fourier Transform

F−1(f̃(ξ)

)= f(x) = (2π)−d

∫Ω

f(ξ)e−iξ·x dξ. (51)

Simple algebra shows that the operator F satisfies the following equation

F−1 (F (f(x))) = f(x). (52)

Fourier Transform of derivativeProvided that function f(x) decays “sufficiently rapidly” to 0 for |x| → ∞ we have

˜( ∂f∂xi

)(ξ) =

∫Ω

∂f

∂xieiξ·x dx = −iξi

∫Ω

∂f

∂xieiξ·x dx = −iξif̃(ξ). (53)

DISCRETE FOURIER TRANSFORM

The Discrete Fourier Transform (DFT) often replaces its continuous counterpart whenanalyzing discrete systems such as digitized images of real microstructures. The com-plexity of their geometries usually calls for sampling of large micrographs. The actualmicrostructure is then approximated by the measured segment periodically extendedoutside the measured region → allows application of DFT .Properties of DFT

1. The discrete Fourier representation results in periodic representation in real space

2. The spectrum of the discrete real space is also periodic

One-dimensional DFTConsider a discrete set of N points defined on the interval 0 ≤ n ≤ N − 1. Applyinga discrete version of the Fourier series this set is given by

x(n) =1

N

N−1∑k=0

ξ(k)ei(2π/N)kn, (54)

where the coefficients ξ(k) are provided by the Discrete Fourier Transform of x(n)

ξ(k) =1

N

N−1∑n=0

x(n)e−i(2π/N)kn. (55)

CONVOLUTION THEOREM

Convolution of two functions f & g∫Ω

f(x− x′)g(x′) dx′.

Fourier Transform of Convolution

F(∫

Ω

f(x− x′)g(x′) dx′)

= F (f(x))F (g(x)) . (56)

Inverse Fourier Transform (Recall Eq. (51) to get)

F−1(∫

Ω

f̃(ξ − ξ′)g̃(ξ′) dξ′)

= (2π)df(x)g(x), (57)

which implies that

F (f(x)g(x)) = (2π)−d∫

Ω

f̃(ξ − ξ′)g̃(ξ′) dξ′. (58)

Correlation of two functions f & g (g(x) is a real function)∫Ω

f(x + x′)g(x′) dx′,

Fourier Transform of Correlation

F(∫

Ω

f(x + x′)g(x′) dx′)

= F (f(x))F (g(x)), (59)

where ·means complex conjugate and should not be mistaken with the ensemble averageused before.

EVALUATION OF FUNCTION Srs FROM BINARY IMAGES

Fourier Transform of SrsRecall Eq. (49) and relation (59) to get

S̃rs(ξ) =1

Ωχ̃r(ξ)χ̃s(ξ), (60)

Taking advantage of the periodicity of function χr one may implement the DiscreteFourier Transform (DFT) to evaluate Eq. (60).

Discrete version of of Eq. (49)

Srs(m,n) =1

MN

M∑i=1

N∑j=1

χr(i, j)χs((i+m)%(M), (j + n)%N), (61)

where symbol “%” stands for modulo. The above equation, usually termed the cycliccorrelation, readily implies periodicity of function Srs. Note that the correlation propertyof DFT holds for cyclic correlation. Referring to Eq. (60) it is given by the followingrelation

DFT{Srs(m,n)} = DFT{χr(m,n)}DFT{χs(m,n)}. (62)The inverse DFT denoted as IDFT then serves to derive function Srs at the final set ofdiscrete points as

Srs(m,n) = IDFT{DFT{χr(m,n)}DFT{χs(m,n)}}. (63)

Usually, the Fast Fourier Transform, which needs only O(MN log(MN)) operations, iscalled to carry out the numerical computation.

EFFECT OF RESOLUTION OF DIGITIZED MEDIA

Resolution 488x358 pixels Resolution 244x179 pixels

CPU time in seconds required to evaluate function Smm

Method Bitmap resolution

976 × 716 488 × 358 244 × 179Direct with periodicity Eq. (61) × 9095.27 446.03

Direct without periodicity Eq. (48) × 5200.07 238.62FFT based Eq. (63) 6.24 1.54 0.37

TESTING ERGODIC ASSUMPTION

To test the ergodic hypotheses, we shall require

cr =1

n

n∑i=1

Sir, r = f,m , (64)

where n is the number of members in the ensemble. Functions Sir can be derived byrandomly placing a point in the member i in a certain number of times while countingthe number of hits in the phase r and then dividing by the total number of throws.When setting the number of throws equal to 500 we found Sf = 0.42, which agreeswell with the average fiber volume fraction cf = 0.435.

TESTING STATISTICAL ISOTROPY

To that end, we examine the distribution of the two-point matrix probability functionSmm for a statistically uniform medium as a function of the absolute difference of pointsx and x′ and orientation φ. Should the material be statistically isotropic (independentof orientation) the variation coefficient v(φ) of Smm(φ)|r/R for a given ratio r/R wouldbe equal to zero.

0.0 2.0 4.0 6.0 8.0 10.0

r/R

(b)

0.0

1.0

2.0

3.0

4.0

5.0

6.0

v[%

]

USEFUL RELATIONS BETWEEN TWO-POINT PROBABILITYFUNCTIONS

0.0 2.0 4.0 6.0 8.0 10.0

r/R

(a)

0.20

0.25

0.30

0.35

0.40

0.45

0.50

0.55

0.60

Sm

m

n = 500n = 1000n = 5000n = 10000n = 50000

0.0 2.0 4.0 6.0 8.0 10.0

r/R

(b)

−0.10

0.00

0.10

0.20

0.30

0.40

0.50

0.60

S2

SmmSfmSffSmm + SfmSfm + Sff1 − (Smm + 2Sfm + Sff)

Known function

Smm(r) Smf (r) Sff (r)

Smm(r) Smm(r) cm − Smf (r) cm − cf + Sff (r)Smf (r) cm − Smm(r) Smf (r) cf − Sff (r)Sff (r) cf − cm + Smm(r) cf − Smf (r) Sff (r)

STATISTICALLY EQUIVALENT RVE

We wish to replace the real microstructure by a material representative volume element(RVE), represented here by a periodic unit cell (PUC) consisting of a small number ofparticles, and yet statistically resembles the actual composite. We argue, that if thePUC has a statistically similar spatial distribution of fibers as the real microstructure,it will also posses similar mechanical properties.

CONSTRUCTION OF PERIODIC UNIT CELLS

Such a PUC can be constructed by matching the two-point probability functions of thereal microstructure and the unit cell.

F (xN , H1, H2) =Nm∑k=1

(Smm(rk)− Smm(rk)

)2

where xN = {x1, y1, . . . , xN , yN} stores the positions of particle centers of the periodicunit cell, xi and yi correspond to the x and y coordinates of the i-th particle, H1 andH2 are the dimensions of the unit cell, Smm(rk) is the value of S corresponding to theoriginal media evaluated at the distance rk and Nm is the number of points, in whichboth functions are evaluated. Statistical isotropy is assumed.

OPTIMIZATION PROCESS

Optimal fiber configuration

For a given number of fibers N , dimensions of a unit cell H1 and H2 and valuesof the original function Smm(r) evaluated at points ri, i = 1, . . . , Nm find theconfiguration of particle centers xN(H1, H2) such that:

xN(H1, H2) = arg minxN∈S

F (xN , H1, H2),

where S denotes a set of admissible vectors xN .

Optimal ratio H1/H2

For known values of xN(η) and for the fixed volume fraction of phases, find theratio ηN such that:

ηN = arg minη∈〈ηa;ηb〉

F (xN(η)),

where values ηa and ηb should be chosen to cover all the reasonable dimensions ofthe unit cell.

EXAMPLE OF AN ADMISSIBLE UNIT CELL

0 5 10 15 20 250

5

10

15

20

25

The objective function F is multi-dimensional with large number of plateaus and localminima −→ efficient optimization methods such as Genetic algorithms or Simulatedannealing methods, (see [Matouš et al., 2000, Lepš and Šejnoha, 2000, Sejnoha, 2000]and [Zeman and Šejnoha, 2001] for details on their implementation) are needed.

EXAMPLES OF PERIODIC UNIT CELLS FOR FIBER TOW

MICROMECHANICAL ANALYSIS OF COMPOSITES

SESSION III

• Ellipsoidal inclusion problem - the Eshelby problem

• Overall moduli - Inhomogeneity problem, the Mori-Tanaka and Self consistentmethods

THE ELLIPSOIDAL INCLUSION PROBLEM

[Eshelby, 1957], [Hill, 1965], [Laws, 1980]

Most of the useful theories of composite materials depend one way or the other on thesolution of the ellipsoidal inclusion problem. The problem was addressed by Eshelby in1957. He proved that the local stress and strain fields in an ellipsoidal inhomogeneityLr, embedded into an infinite homogeneous elastic material L 6= Lr, that is loaded byremotely applied uniform stress or strain fields, respectively, are also uniform. This resulthas then provided a stepping stone for evaluation of effective properties in variety ofcomposite systems. The Eshelby problem thus deserves a special attention, particularlyin view of forthcoming discussions, and will be described herein.

Note that ellipsoidal shapes include:spheres, spheroids, circular and elliptical cylinders, and elliptical or circular (penny-shaped) discs and platelets.

THE ESHELBY PROBLEM

Solution of the Eshelby problem relies on a certain transformation problem. In particular,consider a homogeneous infinite anisotropic material L, which contains an ellipsoidalinclusion of the same material, in a certain volume Vr. The surrounding material isneither constrained nor loaded.

Suppose that the inclusion is now loaded by a certain transformation uniform strain µ.If free, it would undergo a homogeneous deformation µ, but due to constraint of thematrix it attains in situ deformation �r = Sµ, where S is the Eshelby tensor. Thusknowing the Eshelby tensor S we may pose the following problem:

Find a transformation strain µ, which must be applied in the homogeneous inclusion Lin order to create the same stress and strain fields, which would be developed in theinhomogeneity Lr of the same shape and size bonded to a large volume of unconstrainedmaterial L, that is subjected to uniform overall stress or strain fields.

E S+ε = µr εr

L r

LL

E E

L

Homogeneous inclusion Inhomogeneity

The above problem can be rephrased as follows:

For a given Lrfind the required transformation strain µ:

�r = E + Sµ

σr = L(�r − µ) = Lr�r =⇒ µ = L−1(L− Lr)�r,

where the stain �r is

�r = E + Sµ = E + SL−1(L− Lr)�r

(65)

�r =[I− SL−1(L− Lr)

]−1E = ArE.

Recall that Ar represents the strain mechanical concentration factor relating the overallstrain E to the local strain �r. Hence, the Eshelby tensor is sufficient to determine thelocal strains (stresses) for a given Lr. The above approach is known as the equivalentinclusion method.

EVALUATION OF THE ESHELBY PROBLEM

Solution of the Eshelby problem proceeds as follows ([Eshelby, 1957]):

I. Remove the inclusion from the matrix. Then, allow the inclusion to undergo astress-free deformation µ. Let

λij = Lijklµkl,

be the stress derived from µij by Hook’s law. Both the inclusion and matrixmedium are stress-free.

II. Apply surface tractions −λijnj on the inclusion boundary Sr to restore its originalshape and size. The stress in the inclusion is now −λij, while the matrix is stillstress free. Put it back in the matrix and reweld across Sr. The surface tractionscan be interpreted as a layer of body forces spread over Sr.

III. Remove this unwanted layer of body forces by applying an equal and oppositesurface tractions λijnj over Sr. The body is now free of external forces but inthe state of self-stress owing to the presence of transformation strain µ in theinclusion.

Then, the displacement at point x′ due to this layer of body forces is equal to

up(x′) =

∫Sr

u∗pi(x− x′)λijnj(x) dx. (66)

In Eq. (66), Sr is the surface separating matrix and inclusion Vr and u∗pi(x − x′) is

fundamental solution, which corresponds to the displacement at point x in the directionof i due to the concentrated unit force applied at point x′ in the direction of p to theinfinite medium. Since

∂u∗pi(x− x′)∂x

′j

= −∂u∗pi(x− x′)

∂xj,

the divergence theorem provides

up(x′) = −1

2

∫Vr

(∂u∗pi(x− x′)

∂x′j

λij +∂u∗pj(x− x′)

∂x′i

λij

)dx.

The resulting strain field than can be found in the form

�pq(x′) = Ppqij(x

′)λij = Ppqij(x′)Lijklµkl, (67)

where P is given by

Ppqij(x′) =

1

4

∫Vr

[∂2u∗pi(x− x′)

∂x′j∂x′q

+∂2u∗pj(x− x′)

∂x′i∂x′q

+∂2u∗qi(x− x′)∂x′j∂x

′p

+∂2u∗qj(x− x′)

∂x′i∂x′p

]dx′ =

∫Vr

�∗pqij(x− x′) dx (68)

Once the value of P is known, the Eshelby tensor S follows directly from Eq. (67)

S = PL. (69)

The following relations will be useful when deriving the tensor P.

FOURIER TRANSFORM

The d-dimensional Fourier transform of function f(x) is provided by

f̃(ξ) =

∫Rdf(x)eiξ·x dx, (70)

where i is the imaginary unit. Once f̃(ξ) is known, the inverse transform is given by

f(x) =1

(2π)d

∫Rdf̃(ξ)e−iξ·x dξ (71)

Provided that function f(x) decays “sufficiently rapidly” to 0 for |x| → ∞ we have

˜( ∂f∂xi

)(ξ) =

∫Rd

∂f

∂xieiξ·x dx = −iξi

∫Rdfeiξ·x dx = −iξif̃(ξ), (72)

THE DIRAC DELTA FUNCTION

Recall the following property of Dirac’s functions δ(x− x′)

1. δ(x− x′) = 0 for x 6= x′

2. δ(x− x′) =∞ for x = x′

3.

∫Ω

δ(x− x′)g(x) dx = g(x′)

The third property of the Dirac delta function suggests that it’s 1D Fourier transformyields ∫ ∞

−∞δ(x)eiξ·x dx = 1. (73)

Hence, the inverse relation gives

1

2π

∫ ∞−∞

1 · e−iξ·x dξ = δ(x). (74)

FUNDAMENTAL SOLUTION

Fundamental solution u∗pi complies with the solution of the following equation

Lijkl∂u∗pk(x)

∂xl∂xj+ δpiδ(x) = 0. (75)

For general anisotropic medium the closed form solution of Eq. (75) does not exist.However, the knowledge of fundamental solution u∗pi is not necessary for evaluation ofEq. (68). Instead, its Fourier representation will be used to deliver the result. Hence,taking Fourier’s transform of Eq. (75) and using Eqs. (72) and (75), we arrive at

− Lijklξlξjũ∗pk(ξ) + δpi = 0 (76)

This yields the desired relation

ũ∗ik(ξ) = (Lijklξlξj)−1. (77)

It is worth to mention the following property of ũ∗ik. When ξ = rω, then

ũ∗(rω) =1

r2ũ∗(ω). (78)

After reviewing all necessary relations, we return to the original task and continue withdetermination of P tensor. First, note that

u∗ik(x) =1

8π3

∫R3ũ∗ik(ξ)e

−iξ·x dξ (79)

When introducing the following substitution ξ = rω, where r is the magnitude of vectorξ and ω = ξ/r and noticing that dξ = r2 dr dω, we arrive at

u∗ik(x) =1

8π3

∫S

∫ ∞0

ũ∗ik(rω)e−iξ·xr2 dr dω, (80)

where S corresponds to the surface of the unit sphere with center located at the originof the coordinate system. Since the above expression is even in r it can be recast as

u∗ik(x) =1

16π3

∫S

∫ ∞−∞

ũ∗ik(rω)e−iξ·xr2 dr dω. (81)

Substitution of Eq. (78) into the above expression leads to

u∗ik(x) =1

16π3

∫S

ũ∗ik(ω)

[∫ ∞−∞

e−irω·x dr

]dω. (82)

In view of Eq. (74) it becomes clear that the bracketed term is equal to 2πδ(ω · x).Eq. (82) then becomes

u∗ik(x) =1

8π2

∫S

ũ∗ik(ω)δ(ω · x) dω. (83)

However, to obtain tensor P we need to evaluate terms like∫Vr

∂2u∗pi(x− x′)∂x′j∂x

′q

dx =∂

∂x′j∂x′q

∫S

[∫Vr

ũ∗pi(ω)δ(ω · (x− x′)) dx]

dω. (84)

To that end, we introduce a function ψ(x′,ω)

ψ(x′,ω) =

∫Vr

δ(ω · (x− x′)) dx. (85)

Hereafter, we limit our attention to an ellipsoidal inclusion only. It can be shown thatif point x′ is contained by the inclusion then, for an ellipsoid given by αijxixj = 1,t2 = α−1ij ωiωj and a = detαij ([Laws, 1980], [Mura, 1982]), the function ψ(x

′,ω)reads

ψ(x′,ω) =π

t3√a

(t2 + (x′ · ω)2

), (86)

Note that ψ(x′,ω) is the only term in Eq. (84), which depends on x′. Hence, we needto evaluate

∂2ψ(x′,ω)

∂x′j∂x′q

=2πωjωqt3√a. (87)

It should be emphasized that the term ∂2ψ(x′,ω)/∂x′q∂x′j and consequently the tensor

Ppqij do not depend on the value of x′ so the strain inside the ellipsoidal inclusion is

constant. After combining Eqs. (87) and (68) we finally arrive at

Ppqij =1

16π√a

∫S

1

t3(ũ∗piωjωq + ũ

∗pjωiωq + ũ

∗qiωjωp + ũ

∗qjωiωp

)dω (88)

EXAMPLE 1.6

As an example, we evaluate the P tensor for a spherical inclusion in an isotropicmaterial. Stiffness tensor of such a medium can be written in the form

Lijkl = κδijδkl + µ(δikδjk + δilδjk −2

3δijδkl).

After some algebra it becomes evident that

ũ∗pi(ω) =1

µδpi −

3κ+ µ

µ(3κ+ 4µ)ωpωi.

Next, verify that Lijklωlωjũ∗pk(ξ) = δpi.

Examining Eq. (88) suggests that the following integrals must be evaluated∫S

ωjωq dω

∫S

ωpωqωiωj dω. (89)

Introducing the polar coordinate system readily provides

ω1 = cos θ cosφ, ω2 = cos θ sinφ, ω3 = sin θ,

θ ∈ 〈−π/2, π/2〉, φ ∈ 〈0, 2π〉, dω = cos θ dθ dφ. (90)

Substitution from (90) into (89) gives∫ 2π0

cos2 φ dφ =

∫ 2π0

sin2 φ dφ =[x

2

]2π0

= π∫ 2π0

sinφ dφ =

∫ 2π0

cosφ dφ =

∫ 2π0

sinφ cosφ dφ = 0∫ π/2−π/2

cos3(θ) dθ =

[sin θ − 1

3cos θ

]π/2−π/2

=4

3∫ π/2−π/2

sin2(θ) cos(θ) dθ =

[1

3sin3(θ)

]π/2−π/2

=2

3,

so that ∫S

ω1ω1 dω =

∫S

ω2ω2 dω =

∫S

ω3ω3 dω =4π

3otherwise 0.

Therefore ∫S

ωpωi dω =4π

3δpi.

Similarly we get ∫S

ωpωqωiωj dω =4π

15(δpjδqi + δqjδpi + δpqδij).

The above relations provide the desired result in the form.

Ppqij =1

3(3κ+ 4µ)δpqδij +

3(κ+ 2µ)

10µ(3κ+ 4µ)(δpiδqj + δpjδqi −

2

3δpqδij). (91)

In general, the P tensor depends on the aspect ratio of the ellipsoid and the coefficientsof L. Some useful forms follow ([Walpole, 1969]).

A SPHERICAL INCLUSION IN ISOTROPIC MEDIUM

P =

13( σ

3κ+ η

µ) 1

3( σ

3κ− η

2µ) 1

3( σ

3κ− η

2µ) 0 0 0

13( σ

3κ+ η

µ) 1

3( σ

3κ− η

2µ) 0 0 0

13( σ

3κ+ η

µ) 0 0 0

ηµ

0 0

ηµ

0

ηµ

where

σ =1 + ν

3(1− ν), η =

2(4− 5ν)15(1− ν)

and κ, µ, ν are the bulk and shear moduli and the Poisson ratio of the isotropic medium.References: [Eshelby, 1957], [Walpole, 1969]

A CIRCULAR CYLINDER

(prolate spheroid) in a transversely isotropic medium (xA = x1).

P =

0 0 0 0 0 0

k+4m8m(k+m)

−k8m(k+m)

0 0 0

k+4m8m(k+m)

0 0 0

k+2m2m(k+m)

0 0

12p

0

12p

where k, m, p are Hill’s moduli; [Walpole, 1969].

A PENNY-SHAPED INCLUSION

(circular disc) in a transversely isotropic matrix (xA − x1)

P =

1n

0 0 0 0 0

0 0 0 0 0

0 0 0 0

0 0 0

1p

0

1p

where p, n, are Hill’s moduli of the medium [Walpole, 1969].

APPROXIMATE EVALUATION OF OVERALL MODULI - 2

Dvorak (lecture notes), [Šejnoha, 1999]

Here we address the problem of evaluating overall elastic moduli of a composite mediumwith many inclusions embedded in a matrix material. We have already shown that thisproblem is solved once the mechanical stress or strain concentration tensors are known.The problem is rather difficult and an exact solution in general does not exist, mainly be-cause the geometry is not well defined. Although a valuable piece of information can beobtained from high contrast micrographs, we usually content ourself with the knowledgeof the volume fractions of the inclusions and their orientation and shape. It becomesclear that certain assumptions are needed to resolve this matter. Three approximatemethods are now available in the literature, which are worthwhile mentioning:

1. The dilute approximation2. The self-consistent method3. The Mori-Tanaka method

Before we contemplate with individual methods we recall the basic problem associatedwith a single inclusion (Lr) in an infinite matrix (L) subjected to uniform remote strainsor stresses. For such a problem the exact solution for the concentration tensor Ar existsand can be written in terms of the Eshelby tensor as

Ar =[I− SL−1 (L− Lr)

]−1.

Also, it is perhaps appropriate to recall the basic results already derived in precedingsections

L =N∑r=1

crLrAr M =N∑r=1

crMrBr

I =N∑r=1

crAr I =N∑r=1

crBr,

so we can express the overall stiffness and compliance matrices in the forms

L = L1 +N∑r=2

cr(Lr − L1)Ar

M = M1 +N∑r=2

cr(Mr −M1)Br.

Hereafter, we use suffix r = 1 to refer to the matrix and suffix r = 2, . . . , N to refer tothe inclusions. If a phase has more than one shape (as specified by the P tensor), thenan index r must be allocated to each such shape.

SCHEMATIC REPRESENTATION OF INDIVIDUAL METHODS

Σ

L

L

Σ

LL L

L

DILUTE APPROXIMATION

SELF - CONSITENT METHOD

MORI - TANAKA METHOD

INHOMOGENEITY PROBLEM

[Hill, 1965] used the principle feature of the Eshelby solution, that an ellipsoidal inho-mogeneity is strained uniformly, to pose an inhomogeneity problem. He introduced anoverall constraint tensor L∗ for material L, with inverse M∗, associated with the uniformcavity strain �∗ caused by distribution of tractions over the inhomogeneity surface thatare compatible with a uniform field of stress σ∗ such that

σ∗ = −L∗�∗, �∗ = −M∗σ∗. (92)

σ = Σ + σr *

ΣΣ

=+

+ +

LL

L

ΣσLL

Lσ r

rr

r

σ*

1

2

3

A schematic representation of the inhomogeneity problem suggests that σr = Σ +σ∗.

Thenσr −Σ = L∗ (E − �r) , �r −E = M∗ (Σ− σr) , (93)

and so

(L∗ + Lr) �r = (L∗ + L)E, (M∗ + Mr)σr = (M

∗ + M) Σ. (94)

Introducing the mechanical strain and stress concentration factors Ar and Br we find

�r = ArE, σr = BrΣ,

whereAr = (L

∗ + Lr)−1 (L∗ + L) Br = (M

∗ + Mr)−1 (M∗ + M) . (95)

After setting Σ = 0 it becomes clear that

�r = �∗ = Sµ, σr = σ

∗ = L (�∗ − µ) .Then, using Eq. (92), we discover

σ∗ = −L∗Sµ = L(S− I)µ

L(I− S) = L∗S

P = SM = (I− S)M∗

(M + M∗)−1 = L∗S = −L(S− I)

L∗ = −L(I− S−1)

S = (L∗ + L)−1L

P = (L∗ + L)−1 = PT

Introducing another tensor Q = (L∗+L)−1 = PT yields the concentration factor tensors

A−1r = P(L∗ + Lr) = I− P(L− Lr)

B−1r = Q(M∗ + Mr) = I−Q(M−Mr)

. (96)

EXAMPLE 2.1

1. Verify that PL + MQ = I.

2. For a spherical inclusion in an isotropic material, let

L∗ = (3κ∗, 2µ∗)

Show that

κ∗ =4

3µ

µ∗ =µ(9κ+ 8µ)

6(κ+ 2µ)

THE DILUTE APPROXIMATION

Consider a matrix-based composite consisting of N phases. When using the diluteapproximation we assume that there is no interaction between inclusions r = 2, . . . , N .This means that the concentration factor of each phase can be evaluated as if a singleinclusion of each phase was embedded in an infinite volume of matrix. When solvingan inclusion problem for each phase we find

A−1r = P(r)(L∗ + Lr) = I− P(r)(L1 − Lr)

B−1r = Q(r)(M∗ + Mr) = I−Q(r)(M1 −Mr).

(97)

Note that P(r), Q(r) and L∗, M∗ depend on the inclusion geometry and elastic moduliof the matrix L1. The overall stiffness and compliance matrices thus attain the formgiven by Eq. (92)

L = L1 +N∑r=2

cr(Lr − L1)Ar

M = M1 +N∑r=2

cr(Mr −M1)Br.

THE SELF-CONSISTENT METHOD

The self-consistent method of [Hershey, 1954] and [Kroner, 1958] was originally pro-posed for aggregates of crystals and then reviewed and elaborated by [Hill, 1965] inconnection with composites.

The method draws on the familiar solution to an auxiliary inclusion problem. In particu-lar, it assumes that interaction of phases is accounted for by imagining each phase to bean inclusion embedded in a homogeneous medium that has the overall properties (L,M)of the composite. To proceed, denote here the inclusion and the matrix by subscripts1 and 2, respectively. Recall Eq. (22) and write the elementary relations between thephase and overall averages of stress field

c1(σ1 −Σ) + c2(σ2 −Σ) = 0. (98)

The basic postulate of the self-consistent method suggests that

σ1 −Σ = L∗(E − �1), (99)

and from Eq. (98)σ2 −Σ = L∗(E − �2). (100)

Evidently, both phases are regarded on the same footing (concentration factors for boththe inclusion and the matrix are derived from the same L∗). It means that the sameoverall moduli are predicted for another composite in which the roles of the phases arereversed. Recall the following connections

P(r) = (L∗r − L)−1 Q(r) = (M∗r −M)−1,

to findA−1r = I + P

(r)(Lr − L) = P(r)(L∗ + Lr)

B−1r = I + Q(r)(Mr −M) = Q(r)(M∗ + Mr),

where L∗r, M∗r depend only on Lij, Mij and shape of the inclusion. Consider now aligned

inclusions of similar shapes (P(r) = P,Q(r) = Q) to get

L =

[∑r

cr(L∗ + Lr)

−1

]−1− L∗ (101)

M =

[∑r

cr(M∗ + Mr)

−1

]−1−M∗.

To derive the above form of the overall stiffness matrix L we exploited the relations∑crAr = I and Ar = (L

∗ + Lr)−1(L∗ + L). Changing A → B and L → M leads to

the overall compliance matrix M. Eq. (101) represents an implicit set of 21 algebraicequations. Thus a suitable iteration scheme must be introduce to solve for unknownstiffnesses or compliances.

OVERALL MODULI OF UNIDIRECTIONAL FIBER COMPOSITES - SCM

Here we present approximate values of elastic overall moduli of fibrous materials withtransversely isotropic phases provided by the self-consistent method (see [Hill, 1965] forreference). For a transversely isotropic solid, with x1 as the axis of rotational symmetry,the stress-strain relation

σ = L�,

is usually written in terms of Hill’s moduli

σ1

σ2

σ3

σ4

σ5

σ6

=

n l l 0 0 0

l (k + m) (k−m) 0 0 0l (k−m) (k + m) 0 0 00 0 0 m 0 0

0 0 0 0 p 0

0 0 0 0 0 p

�1

�2

�3

�4

�5

�6

, (102)

where

k = −[1/G23 − 4/E22 + 4ν212/E11]−1 l = 2kν12n = E11 + 4kν

212 = E11 + l

2/k m = G23, p = G12.

1. Transverse shear modulus m is found first, as a solution of the cubic equation:

c1k1k1 + m

+c2k2

k2 + m= 2

(c1m2

m2 −m+

c2m1m1 −m

).

2. Longitudinal shear modulus p is found as a solution of a quadratic equation:

c1p− p2

+c2

p− p1=

1

2p.

3. Moduli k, l, and n, in terms of m, are found from:

1

k + m=

c1k1 + m

+c2

k2 + m

l

k + m=

c1l1k1 + m

+c2l2

k2 + m

n− c1n1 − c2n2l − c1l1 − c2l2

=l1 − l2k1 − k2

.

Word of caution: The estimates become unreliable when the fiber concentration is highand when the phase moduli differ considerably.

EXAMPLE 2.2

Estimate the effective elastic properties for the composite made from isotropic dispersionof spheres

THE MORI-TANAKA METHOD

Consider a matrix-based composite consisting of N phases. The method approximatesthe effect of phase interaction on the local stresses by assuming that the stress ineach phase is equal to that of a single inclusion embedded into an unbounded matrixsubjected to as yet unknown average matrix strain �1 = A1E or stress σ1 = B1Σ.

In Benveniste’s reformulation of the Mori-Tanaka method ([Benveniste, 1987]), thesolution of a single inclusion in a large volume of matrix loaded by �1 or σ1 assumesthe form

�r = Tr�1 , σr = Wrσ1 (103)

and since E =N∑r

cr�r, Σ =N∑r

crσr, one can find that the average matrix strain

and stress vectors are

�1 =

[c1I +

N∑n=2

crTr

]−1E σ1 =

[c1I +

N∑n=2

crWr

]−1Σ. (104)

Eqs. (104) also suggest that the concentration factors As and Bs are

Ar = Tr

[c1I +

N∑n=2

crTr

]−1, Br = Wr

[c1I +

N∑n=2

crWr

]−1, (105)

where, recall Eq. (96)

T−1r = P(r)(L∗ + Lr) = I− P(r)(L1 − Lr)

W−1r = Q(r)(M∗ + Mr) = I−Q(r)(M1 −Mr)

(106)

Tensors P(r), Q(r), L∗ and M∗ are now functions of the moduli in L1. It becomesclear that the Mori-Tanaka method provides an explicit relations for the mechanicalconcentration factors. Substitution of Eqs. (105) into Eq. (92) leads to

L =

[N∑r=1

crLrTr

][N∑r=1

crTr

]−1, M =

[N∑r=1

crMrWr

][N∑r=1

crWr

]−1. (107)

OVERALL MODULI OF UNIDIRECTIONAL FIBER COMPOSITES -MTCM

Mori-Tanaka estimates of elastic overall moduli of fibrous materials with transverselyisotropic phases written in terms of phase moduli and phase volume fractions are

k =k2k1 + m1 (c2k2 + c1k1)

c2k1 + c1k2 + m1

l =c2l2 (k1 + m1) + c1l1 (k2 + m1)

c2 (k1 + m1) + c1 (k2 + m1)

n = c2n2 + c1n1 + (l− c2l2 − c1l1)l2 − l1k2 − k1

m =m2m1 (k1 + 2m1) + k1m1 (c2m2 + c1m1)

k1m1 + (k1 + 2m1) (c2m1 + c1m2)

p =2c2p2p1 + c1 (p2p1 + p

21)

2c2p1 + c1 (p2 + p1).

OVERALL ELASTIC SYMMETRY OF SCM AND MT METHODS

First, assume that all phases s = 2, . . . N are of the same shape and alignment, sothat there is only one pair of P and L∗ tensors, L∗ = P−1 − L.

L = L in the self-consistent methodL = L1 in the Mori-Tanaka method.

This formal similarity between the SCM and M-T method allows to write the overallstiffness matrix in the form

L =

[∑r

cr(L∗ + Lr)

−1

]−1− L∗.

This shows that both the self-consistent and Mori-Tanaka methods give L = LT formultiphase systems reinforced by inclusions of the same shape and alignment. Thesame is not true, however, when multiphase materials with inclusions of different shapeare considered. Numerical examples show that in such a case L 6= LT. Note that thedilute approximation preserves the diagonal symmetry in both cases.

NUMERICAL EXAMPLE - MT METHOD

Consider a multiphase system consisting of 3 isotropic phases:

1. Ti3Al matrix2. Carbon circular discs with mid-plane normal to the x3-axis3. SiC fibers parallel to the x3-axis

E1=96.5 GPa, G1=37.1 GPa, c1=0.55E2=34.4 GPa, G2=14.3 GPa, c2=0.25E3=431.0 GPa, G3=172.0 GPa, c3=0.20

L =

123.62 43.84 35.89 0 0 0

43.84 123.62 35.89 0 0 0

21.89 21.89 124.05 0 0 0

0 0 0 33.04 0 0

0 0 0 0 33.04 0

0 0 0 0 0 39.89

GPa

L 6= LT

Example 2 : Elastic porous media

Consider an elastic isotropic body weakened by spherical pores. Determine the influenceof the porosity (≡ volume fraction of the pores) on the overall elastic moduli of thebody using the dilute approximation and the Mori-Tanaka method.

Solution: We will model a porous medium as a two-phase material (N = 2) composedof an elastic matrix characterized by the material stiffness tensor L1 and spherical pores(L2 = 0) with the volume fraction c2. The relation for the effective stiffness matrix ofthe material has the following form, recall Eq. (92):

L = L1 + c2(L2 − L1)A2 = L1 (I− c2A2) ,

Dilute approximation

Recall that for the dilute approximation, the concentration tensor can be obtained as,see Eq. (97),

A−12 = I− P(2)(L1 − L2) = I− P(2)L1. (108)

For the matrix phase, the stiffness tensor has the following form:

(L1)ijkl = κ1δijδkl + µ1(δikδjk + δilδjk −2

3δijδkl).

To simplify the algebra, we introduce the fourth-order projection tensors Λh and Λs [Milton and Kohn, 1988]

(Λh)ijkl =1

3δijδkl, (Λs)ijkl =

1

2[δikδjl + δilδjk]−

1

3δijδkl,

which posses the following properties:

ΛhΛh = Λh, ΛsΛs = Λs, ΛhΛs = 0,

ΛsΛh = 0, Λs + Λh = I, (αΛh + βΛs)−1 =

1

αΛh +

1

βΛs

This allows us to express the tensor L1 as

L1 = 3κ1Λh + 2µ1Λs.

Similarly, the P(2) for the spherical pore can be expressed as, compare with Eq. (91),

P(2)ijkl =

1

3(3κ1 + 4µ1)δijδkl +

3(κ1 + 2µ1)

10µ1(3κ1 + 4µ1)(δikδjl + δilδjk −

2

3δijδkl)

P(2) =1

3κ1 + 4µ1Λh +

3(κ1 + 2µ1)

5µ1(3κ1 + 4µ1)Λs

Hence,

A−12 = I− P(2)L1

= Λh + Λs −(

1

3κ1 + 4µ1Λh +

3(κ1 + 2µ1)

5µ1(3κ1 + 4µ1)Λs

)(3κ1Λh + 2µ1Λs)

=4µ1

3κ1 + 4µ1Λh +

9κ1 + 8µ15(3κ1 + 4µ1)

Λs

A2 =3κ1 + 4µ1

4µ1Λh +

5(3κ1 + 4µ1)

9κ1 + 8µ1Λs = αhΛh + αsΛs

Finally,

L = L1 (I− c2A2)= (3κ1Λh + 2µ1Λs) (Λh + Λs − c2 (αhΛh + αsΛs))= 3κ1 (1− c2αh) Λh + 2µ1 (1− c2αs) Λs

Effective moduli can be expressed as

κdil

κ1= 1− c2αh

µdil

µ1= 1− c2αs

Mori-Tanaka method

The concentration tensors for the Mori-Tanaka method follow from the identity, comparewith Eq. (105)

A2 = T2 [c1I + c2T2]−1 ,

where the partial concentration factor yields is defined as, see Eq. (106),

T−12 = I− P(2)L1,

i.e. it coincides with the concentration factor for the dilute approximation, see Eq. (108).This allows us to introduce the concentration tensor for the Mori-Tanaka method in theform

A2 = T2 [I− c2 (I− T2)]−1

= (αhΛh + αsΛs) [Λh + Λs − c2 (Λh + Λs − αhΛh − αsΛs)]−1

=αh

1− c2 + c2αhΛh +

αs1− c2 + c2αs

Λs

Similarly to the dilute approximation case, we finally obtain

κMT

κ1= 1− c2αh

1− c2 + c2αhµMT

µ1= 1− c2αs

1− c2 + c2αs

Comparison of the dilute approximation and the Mori-Tanaka scheme

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1c2

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

κ/κ 1

ν1=0.10ν1=0.25ν1=0.40

Dilute

MT

0 0.2 0.4 0.6 0.8 1c2

0

0.2

0.4

0.6

0.8

1

µ/µ 1

ν1=0.10ν1=0.25ν1=0.40

Dilute

MT

Materials with elastic moduli E = 1 Pa, ν varies

Engineering moduli predictions

Note that the engineering moduli are related to material parameters κ and µ by relations

E =9κµ

3κ+ µ, ν =

3κ− 2µ6κ+ 2µ

.

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1c2

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

E/E 1

ν1=0.10ν1=0.20ν1=0.30ν1=0.40

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1c2

0.7

0.8

0.9

1

1.1

1.2

1.3

1.4

1.5

1.6

ν/ν 1

ν1=0.10ν1=0.20ν1=0.30ν1=0.40

Materials with elastic moduli E = 1 Pa, ν varies

MICROMECHANICAL ANALYSIS OF COMPOSITES

SESSION IV

• Overall moduli - homogenization of periodic structures

• Homogenization of masonry structures - use of available commercial software

• Homogenization of perforated plates

EVALUATION OF EFFECTIVE PROPERTIES OF COMPOSITEMATERIALS - PERIODIC UNIT CELL APPROACH

Suquet, Moulinec, Fish, Oden, Dvorak

Here, we are concerned with the RVE having well defined geometry and boundaryconditions. In particular, we consider a periodic representative volume defined in termsof a unit cell (UC) with prescribed periodic boundary conditions.

⇓ ⇓

EVALUATION OF EFFECTIVE PROPERTIESUSING UNIT CELL MODELS

Consider a composite material characterized by a periodic unit cell (PUC). The overallelastic behavior of such a composite is then governed by its microstructure and by thebehavior of individual phases conveniently described by displacement, strain and stressfields in the form

∂σ(x) = −b(x), (109)ε(x) = ∂Tu(x), (110)

σ(x) = L(x)ε(x) + λ(x), ε(x) = M(x)σ(x) + µ(x), (111)

where u is the displacement field; ε and σ are the strain and stress fields, respectively; Lis the stiffness tensor and M is the compliance tensor such that L−1 = M; µ representscertain stress free strains and λ = −Lµ; b is the vector of body forces and ∂ is thedifferential operator matrix [Bittnar and Šejnoha, 1996, p. 9]

The following discussion is limited to two-phase fibrous composites with fibers alignedalong the x3 axis. We further assume that each phase is transversally isotropic with x3being the axis of rotational symmetry. Additional simplification arises when neglectingthe out-of-plane shear response. The generalized plane strain is then a natural assump-tion. In such a state, the only non-zero components of the strain and stress tensors areε11, ε12, ε22, ε33 and σ11, σ12, σ22, σ33, respectively. Note that due to perfect bondingbetween individual phases the components ε33 and σ33 attain constant values.

PERIODIC STRAIN FIELDS

Suppose that the UC is subjected to boundary displacements u resulting in a uniformstrain E throughout the UC.

Let u∗(x) be a periodic displacement field in the UC. Then, the strain and displacementfields in the UC admit the following decomposition

u(x) = E · x + u∗(x) ε(x) = E + ε∗(x). (112)

The periodicity of u∗ implies that the average of ε∗ on the unit cell vanishes. Hence

〈ε(x)〉 = E + 〈ε∗(x)〉 , 〈ε∗(x)〉 = 1Ω

∫Ω

ε∗(x) dx = 0 (113)

Proof: Using Green’s formula gives

Ω〈ε∗ij(x)

〉=

1

2

∫Ω

(u∗i,j + u

∗j,i

)dΩ =

1

2

∫∂Ω

(u∗inj + u

∗jni)

dS = 0

Ω 〈ε∗(x)〉 =∫∂Ω

u∗ ⊗ nΩ dS,

since u∗i (x) is identical on opposite sides and nj is opposite on opposite sides of theunit cell.

GENERALIZATION FOR A MEDIUM WITH PORES P (voids and cracks)

Let P and M are the subdomains occupied by the pores and by other material. Then,the following generalization is valid∫

Mε∗(x) dx +

∫∂Pu∗ ⊗ nP dS = 0 (114)

Proof:

∫M

1

2

(u∗i,j + u

∗j,i

)dΩ =

1

2

=0 due to periodicity︷ ︸︸ ︷∫∂Ω

(u∗inj + u

∗jni)

dS−12

∫∂P

(u∗injP + u

∗jniP

)dS

nM

Ω int

ΓΩ

n

M

Γ

P

The following generalization also holds

ΩE =

∫Mε dx +

∫∂Pu⊗ nP dS (115)

ΩEij =

∫Mεij dΩ +

∫∂P

1

2(uinPj + ujnPi) dS

Proof:

ΩE =

=

∫Mε(x) dΩ︷ ︸︸ ︷

ME +∫Mε∗ dx +

=

∫∂Pu⊗ nP dS︷ ︸︸ ︷∫

∂P(E · x)⊗ nP dS +

∫∂Pu∗ ⊗ nP dS (116)

Substitution of Eq. (114) into Eq. (116) leads to

ΩE =ME +∫∂P

(E · x)⊗ nP dS

Denoting Ω−M = P yields

PE =∫∂P

(E · x)⊗ nP dS

EXAMPLES 3.1

Consider two elementary macroscopic strain fields E11 and E12 in 2D. Using the diver-gence theorem verify that

PE =∫∂P

(E · x)⊗ nP dS.

OVERALL MODULI - PURE MECHANICAL RESPONSE

Consider a two phase fibrous composite with transversally isotropic phases under gen-eralized plane strain and write the the 4×4 stiffness matrix L of each phase as

L =

(k + m) (k−m) 0 l(k−m) (k + m) 0 l

0 0 m 0

l l 0 n

, (117)

k = −[1/GT − 4/ET + 4ν2A/EA]−1 l = 2kνAn = EA + 4kν

2A = EA + l

2/k m = GT.

Phase constitutive equations - assume constant phase moduli

σr(x) = Lrεr(x), εr(x) = Mrσr(x), r = f,m. (118)

Mechanical loading problem

u0(x) = E · x x ∈ S (119)p0(x) = Σ · n(x) x ∈ S (120)

u0 and p0 are the displacement and traction vectors at the external boundary S of arepresentative volume element Ω of the composite; n is the outer unit normal to S; Eand Σ are the applied macroscopic uniform strain and stress fields, respectively.

Macroscopic constitutive relations

〈σ(x)〉 = 〈L(x)ε(x)〉 =2∑r=1

crLr 〈εr(x)〉 = LE (121)

〈ε(x)〉 = 〈M(x)σ(x)〉 =2∑r=1

crMr 〈σr(x)〉 = M Σ (122)

〈·〉 stands for the spatial average of a given field, cr is the volume fraction of therth phase, and L and M are the effective stiffness and compliance matrices of theheterogenous material, respectively. Eqs. (121) and (122) follow directly from Hill’slemma [Hill, 1963].

Hill’s lemmaFor compatible strain and equilibrated stress fields the following relation holds〈

ε(x)Tσ(x)〉

= 〈ε(x)〉 T 〈σ(x)〉 , (123)

and consequently

ET LE =〈ε(x)TL(x)ε(x)

〉, (124)

ΣT M Σ =〈σ(x)TM(x)σ(x)

〉. (125)

Eq. (123) states in fact that the average of “microscopic” internal work is equal to themacroscopic work done by internal forces.

OVERALL MODULI - continue

Michel, Moulinec, Suquet (1998)

Here we present two specific procedures for evaluation of overall moduli of periodicmicrostructures. The stepping stone in the derivation is the Hill lemma.

Assume a virtual displacement δu = δE · x + δu∗, with δu∗ being periodic. Then

〈σ : δε〉 = Σ : δE Σ = 〈σ〉 . (126)

Proof: (1

Ω

∫Ω

σij dΩ

): δEij +

=0︷ ︸︸ ︷∫∂Ω

σijnjδu∗i dS = Σij : δEij.

An alternate formula for overall average stress Σ can be deduced for the compositescontaining rigid inclusions where σ is not defined. Assuming a self-equilibrated stressfield (σij,j = 0) we write

σij = (σikxj),k.

Then writing the volume average of σij yields

ΩΣij =

∫Ω

σij dΩ =

∫∂Ω

σikxjnk dS. (127)

NUMERICAL RESOLUTION OF THE PROBLEM BY THE FINITEELEMENT METHOD

Here we present a numerical solution of certain boundary value problem using the FiniteElement Method (FEM). The UC is discretized into finite elements and subjected toeither uniform macroscopic states of strain, E, or stress, Σ together with prescribedperiodic boundary conditions.

PRESCRIBED OVERALL STRAIN

Substitute the constitutive law

σ(x) = L(x)ε(x) = L(x) (E + ε∗(x)) (128)

into Eq. (126) with δε∗ (≡ periodic) instead of δε. Eq. (126) then assumes this form

〈σδε〉 = Σ 〈δε∗〉 = 0. (129)

HenceΩ 〈L(x) (E + ε∗(x)) δε∗(x)〉 = 0. (130)

Solving the above relation calls for a suitable numerical technique such as the FiniteElement Method (FEM), [Bittnar and Šejnoha, 1996].

Displacement field in Eq. (112)

u(x) = E · x + N(x)r. (131)

N(x) represent shape functions of a given element and r is the vector of unknowndegrees of freedom.

Strain field

ε(x) = E + B(x)r. (132)

Introducing Eq. (132) into Eq. (130) gives

δrT(∫

Ω

BT(x)L(x)B(x) dΩ

)r + δrT

(∫Ω

BT(x)L(x) dΩ

)E = 0. (133)

Linear system of governing equationsFor any kinematically admissible displacements δr and for the prescribed strain EEq. (133) yields

Kr = f , (134)

where

K =∑e

Ke where Ke =1

Ω

∫Ae

BTLeB dAe

f =∑e

f e where f e = − 1Ω

∫Ae

BTLeE dAe. (135)

K is the stiffness matrix of the system and f is the vector of global nodal forces resultingfrom the loading by E; e stands for the number of elements, Ae is the area of elemente and Ω is the area of the PUC.

In Eq. (134), K represents the problem stiffness matrix and f is the vector of nodalforces due to material heterogeneity. Note that the work done by external forces isapriory zero.

System (134) can be used to provide the coefficients of the effective stiffness matrixL as volume averages of the local fields derived from the solution of four successiveelasticity problems. To that end, the periodic unit cell is loaded, in turn, by each ofthe four components of E, while the other three components vanish. The volumestress averages normalized with respect to E then furnish individual columns of L. Therequired periodicity conditions (same displacements u∗ on opposite sides of the unitcell) is accounted for through multi-point constraints.

PRESCRIBED OVERALL STRESS

Suppose that the surface tractions compatible with the macroscopic uniform state ofstress Σ are prescribed. Such a loading conditions leaves us with unknown overall strainE and periodic displacement field u∗ to be determined. Substituting the microscopicconstitutive equation

σ(x) = L(x)ε (u(x)) (136)

into Eq. (126) results in the following expression

〈σε(δu)〉 = 〈ε(x)Lε(δu)〉 = ΣδE. (137)

Substituting Eq. (112) into Eq. (137) yields

δET 〈L(x) (E + ε∗(x))〉+〈δε∗(x)TL(x)E

〉+〈δε∗(x)TL(x)ε∗(x)

〉= δETΣ. (138)

Since δE and δε∗(x) are independent, the preceding equation can be split into twoequalities

δETΣ = δET [〈L(x)〉E + 〈L(x)ε∗(x)〉] (139)0 =

〈δε∗(x)TL(x)

〉E +

〈δε∗(x)TL(x)ε∗(x)

〉In the FE approach the matrix B, relating strains and displacements in the form ε∗ = Brand consequently, δε∗ = Bδr is applied to Eq. (139) to get

Linear system of governing equations1

Ω

∫Ω

L dΩ1

Ω

∫Ω

LB dΩ

1

Ω

∫Ω

BTL dΩ1

Ω

∫Ω

BTLB dΩ

{ Er

}=

{Σ

0

}. (140)

The above system of equations serves to derive the coefficients of the effective compli-ance matrix M. In analogy with the strain approach, the periodic unit cell is loaded,in turn, by each of the six components of Σ, while the other three components vanish.The volume strain averages normalized with respect to Σ then supply individual entriesof M.

GENERALIZED PLANE STRAIN

Consider a unidirectional fiber reinforced composite medium. Such a medium is invari-ant under translation along the axial direction (invariance of geometrical and materialproperties). The there-dimensional UC is represented by align cylinders. Their length,although arbitrary, is usually taken such that the overall volume of the UC is equal tounity. If we admit the axial direction as a direction of orthotropy (all phases are assumedto orthotropic) and if no out-of-plane shear stresses or strains are prescribed, then the3D UC can be replaced by the 2D UC of the same cross-section.

Suppose that the overall uniform stress vector, given in contracted notation, reads

Σ = {Σ11,Σ22,Σ12,Σ33}T.

Then, because of translational invariance in the fiber direction, the displacement fieldsolution of the three-dimensional local problem corresponds to a generalized plane state

uα = uα(x1, x2) α = 1, 2 u3 = E33x3.

The corresponding macroscopic strain vector is given by

E = {E11, E22, 2E12, E33}T.

For a given prescribed boundary conditions, the solution of the local problem then boils

down to finding the following unknown local fields

ε∗ =

ε∗11ε∗222ε∗12

0

, ε =

ε11

ε22

2ε12

E33

, σ =

σ11

σ22

σ12

σ33

.

PERIODIC BOUNDARY CONDITIONS

Elimination

Here we present a general way of applying the periodic boundary conditions for dis-placement field u∗. You can imagine the periodicity conditions as constraints of thetype

Pr = 0, P = [0, I,−I], r = {r1, r2, r3}T, (141)

where r1 denotes the unknowns corresponding to interior nodes, whereas r2 and r3denote unknowns corresponding to nodes located on the boundary of the unit cell. Thediscrete formulation of Eq. (134) can be expanded into:

{δr1, δr2, δr3}

K11 K12 K13KT12 K22 K23KT13 K

T23 K33

r1

r2

r3

= {δr1, δr2, δr3}f 1f 2f 3

for every δr such that δr2 = δr3. Since r2 = r3 we get[

K11 K12 + K13

KT12 + KT13 K22 + K23 + K

T23 + K33

]{r1

r2

}=

{f 1

f 2 + f 3

}(142)

Note that this procedure is equivalent to assigning the same code numbers to corre-sponding degrees of freedom on opposite sides of the UC.

NUMERICAL RESULTS

Material properties of T30/Epoxy system

phase EA ET GT νA

[GPa] [GPa] [GPa]

fiber 386 7.6 2.6 0.41

matrix 5.5 5.5 1.96 0.40

Components of the effective stiffness matrix [GPa]

Unit cell L11 L22 L33 L44 cf

Original 10.76 10.73 2.215 177.2 0.44

2 fibers PUC 10.78 10.75 2.202 177.2 0.44

5 fibers PUC 10.76 10.73 2.215 177.2 0.44

10 fibers PUC 10.76 10.73 2.215 177.2 0.44

Hexagonal array 10.74 10.74 2.213 177.3 0.44

NUMERICAL RESULTS

Variation of effective stiffnesses for five ten-particle optimal PUC

Modulus Mean value Standard deviation Variation coefficient

[GPa] [GPa] [%]

L11 10.76 0.013 0.12

L22 10.73 0.013 0.12

L33 2.215 0.003 0.13

Variation of effective stiffnesses for five randomly picked ten-particle PUC

Modulus Mean value Standard deviation Variation coefficient

[GPa] [GPa] [%]

L11 10.73 0.32 2.97

L22 10.71 0.38 3.54

L33 2.210 0.07 3.48

THERMOMECHANICAL PROBLEM

Suppose that in addition to the mechanical loading the representative volume is sub-jected to a uniform temperature change ∆θ.

Local constitutive equations - recall Eq. (111)

σ(x) = L(x)ε(x) + λ(x), ε(x) = M(x)σ(x) + µ(x), r = f,m. (143)

Phase constitutive equations

σr(x) = Lrεr(x) + λr, εr(x) = Mrσr(x) + µr, r = f,m. (144)

Local phase eigenstrains µr = mr∆θ and eigenstresses λr

µr = −Mrλr, λr = −Lrµr, r = f,m. (145)

The thermal strain vector mr lists the coefficients of thermal expansion of the phase r.

Hill’s lemma - recall Eq. (126)〈δε(x)Tσ(x)

〉=〈δε(x)TL(x) (ε(x)− µ(x))

〉= δETΣ. (146)

Linear system of governing equations - recall Eq. (140)

1

Ω

∫Ω

L dΩ1

Ω

∫Ω

LB dΩ

1

Ω

∫Ω

BTL dΩ1

Ω

∫Ω

BTLB dΩ

{ Er

}=

Σ +

1

Ω

∫Ω

Lµ dΩ

1

Ω

∫Ω

BTLµ dΩ

. (147)

When excluding the thermal effects the above equation reduces to Eq. (140). However,when the loading conditions are limited to the uniform temperature change equal tounity, the components of the overall average strain comply with the effective coefficientsof thermal expansion m. Note that the present formulation is not applicable with thestrain control conditions when admitting the thermal loading. Clearly, the overall strainE is then not known and cannot be prescribed.

NUMERICAL RESULTS

Material properties of T30/Epoxy system

phase EA ET GT νA αA αT

[GPa] [GPa] [GPa] [K−1] [K−1]

fiber 386 7.6 2.6 0.41 −1.2× 10−6 7× 10−6

matrix 5.5 5.5 1.96 0.40 2.4× 10−5 2.4× 10−5

Components of the effective thermal expansion coefficients [K−1]

Unit cell αx αy αA cf

Original 2.269× 10−5 2.248× 10−5 −7.463× 10−7 0.4362 fibers PUC 2.273× 10−5 2.244× 10−5 −7.463× 10−7 0.4365 fibers PUC 2.269× 10−5 2.248× 10−5 −7.462× 10−7 0.43610 fibers PUC 2.269× 10−5 2.249× 10−5 −7.462× 10−7 0.436Hexagonal array 2.259× 10−5 2.259× 10−5 −7.462× 10−7 0.436Mori-Tanaka 2.250× 10−5 2.250× 10−5 −7.464× 10−7 0.436

EXAMPLE - H