MICROMECHANICAL ANALYSIS OF COMPOSITES

Transcription

1 MICROMECHANICAL ANALYSIS OF COMPOSITES Michal Šejnoha Jan Zeman Faculty of Civil Engineering Czech Technical University in Prague

2 SESSION CONTENT Review of basic continuum mechanics principles for deriving effective thermoelastic material properties of macroscopically homogeneous composite materials, introduction to the ellipsoidal inclusion problem and related averaging schemes such as the dilute approximation, the self-consistent method and the Mori-Tanaka method for evaluation of effective elastic moduli of statistically isotropic and ergodic composite Introduction to basic principles of homogenization of composite materials with periodic microstructures, concept of the unit cell, periodic fields, formulation of the thermoelastic problem, formulation of the homogenization problem for perforated plates, formulation of the homogenization problem for masonry, resolution of the problem by the Finite Element Method, implementation of the periodic boundary conditions into the existing commercial codes Extension of basic mechanics principles for deriving elastic and inelastic macroscopic constitutive equations of random composite materials, general aspects, statistically equivalent unit cell, extension of the Hashin-Shtrikman variational principles to treat statistically homogeneous and ergodic random media with eigenstrains, application to unidirectional composites with random distribution of fibers within the transverse plane section bonded by nonlinear viscoelastic polymer matrix, resolution of the problem by the Fast Fourier Transform

3 MICROMECHANICAL ANALYSIS OF COMPOSITES SESSION I Governing equations of elasticity Volume averages and minimum energy principles Overall moduli - elementary bounds on effective properties

4 ELASTIC AND INELASTIC ANALYSIS OF HETEROGENEOUS MATERIALS heterogeneous material: material having different material properties from point to point (concrete, soils, composites, jointed rocks, etc.) anisotropic material: material having different material properties in different directions, no planes of material symmetry - 81 independent material constants orthotropic material: three orthogonal planes of material symmetry - 9 independent material constants transversally isotropic material: one plane of symmetry (plane of isotropy) with a perpendicular axis of symmetry (elastic moduli are invariant with respect to rotation about this axis) - 5 independent constants isotropic material: three axis of rotational symmetry - 2 independent constants

5 micromechanics: mechanics of composites at the constituents level macromechanics: mechanics of composite structures such as laminates micromechanical analysis: 1. mechanics of materials (simplifying assumptions unnecessary to specify the details of the stress and strain distribution at the micromechanical level, geometry of the microstructure is arbitrary) 2. elasticity theory (elasticity theory models involves the solution of actual local stress and strain fields and actual geometry of the microstructure is taken into account) 3. empirical solutions (curve fitting to elasticity solutions or experiments) EVALUATION OF EFFECTIVE PROPERTIES OF COMPOSITE MATERIALS Eshelby, Hill, Walpole, Laws, Dvorak, Hutchinson, Pagano, Christensen, Mura... the concept of an effective modulus is essential to the development of practical engineering stress-strain relationships for composite materials.

6 GOVERNING EQUATIONS OF LINEAR ELASTICITY In this study we shall consider only statics in the absence of the body forces. Individual particles of the body will be identified by their coordinates x i (i = 1, 2, 3) in the undeformed configuration. Displacement field Strain field u i = u i (x 1, x 2, x 3 ) = u i (x j ). ɛ ij = 1 2 (u i,j + u j,i ). (1) Equations of equilibrium in the absence of body forces σ ij,j = 0 (2) σ ij = σ ji. Surface tractions Constitutive equations p i = σ ij n j. (3) σ ij = L ijkl (ɛ kl ɛ 0 kl) (4) L ijkl = L jikl = L ijlk.

7 The fourth-order tensor L ijkl is known as the stiffness tensor. Suppose that a strain energy density function U(ɛ ij ) per unit volume volume exists such that σ ij = U ɛ ij. (5) Eqs. (2) and (4) readily provide and that L ijkl = L klij U = 1 2 L ijklɛ ij ɛ kl. (6) Providing the strain energy U has a minimum in the stress-free state then L ijkl is positive definite L ijkl ɛ ij ɛ kl > 0 (7) for all non-zero symmetric tensors ɛ ij. It is now possible to invert Eq. (4) to get ɛ ij = M ijkl σ kl + ɛ 0 ij (8) M ijkl is known as the compliance tensor. Note that M ijkl = M jikl = M ijlk = M klij and also L ijrs M rskl = 1 2 (δ ikδ jl + δ il δ jk ) = I ijkl, (9)

8 where δ ij is the Kronecker delta and I ijkl represents the fourth-order identity tensor. Note: we used standard Cartesian tensor notation in which repeated suffixes are summed over the range 1, 2, 3. [Laws, 1980] AVERAGES In preparation for evaluation of the overall moduli we first review some basic formulae for the determination of average stresses and strains. To that end, we assume that the displacements fields are continuous, and the strain fields are compatible; also, the stress fields and tractions are continuous and in equilibrium Consider first an arbitrary homogeneous medium of volume V with the boundary S. In general, the volume average of a quantity is just the ordinary volume average given by f = 1 fdv. (10) V Let ε(x) and σ(x) be certain fields in V. Their volume averages are defined as ε = 1 ε(x) dv σ = 1 σ(x) dv (11) V V V V V

9 After applying the divergence theorem we arrive at ε ij = 1 (u i n j + u j n i )ds (12) 2V S σ ij = 1 (p i x j + p j x i )ds (13) 2V S Next, consider a heterogeneous elastic medium which consists of a homogeneous matrix V 2 and homogeneous inclusion V 1. Evaluation of the above volume averages requires an application of a generalized (but still standard) divergence theorem. Let f be continuous in V and continuously differentiable in the interior of V 1 and V 2. We may now apply the divergence theorem separately to V 1 and V 2 to conclude that V f x i dv + Σ [f]m j ds = S fn i ds (14) where [f] denotes the jump in the value of f as we travel across Σ from V 1 to V 2. Now, assume that perfect bonding exists. When setting f = u i in Eq. (14) we immediately recover Eq. (12). Since tractions are continuous across Σ [σ ij ]m j = 0 setting f = σ ik x j yields Eq. (13). We may now conclude that Eqs. (12) and (13) apply to any heterogeneous material, generally anisotropic, consisting of a homogeneous matrix and an arbitrary number of homogeneous inclusions.

10 EXAMPLES 1.1 Consider an arbitrary composite material with outer boundary S 1. Suppose that the composite is loaded by displacements u i on S, which are compatible with the uniform strain E ij, i.e. u i = E ij x j (affine displacements). Show that < ε ij >= E ij. 2. Suppose that the composite is loaded by prescribed tractions p i on S, which are compatible with the uniform stress Σ ij, i.e. p i = Σ ij n j. Show that < σ ij >= Σ ij. 3. Let σ ij be a self-equilibrated stress field (σ ij,j =0) and u i is a displacement field associated with strain ε ij = 1(u 2 i,j +u j,i ). Show that if either u i = 0 or σ ij n j = 0 on the boundary then σ ij ε ij dv = 0. V 4. For the boundary conditions of Exs. (1) and (2) show that (Hill s lemma) < σ ij ε ij >=< σ ij >< ε ij > and < U >= 1 2 < σ ij >< ε ij >.

11 MINIMUM ENERGY PRINCIPLES We now give a brief review of the classical energy principles as they have been extensively used in assessing the bounds on the overall elastic properties of composites. First, consider an arbitrary anisotropic elastic medium with prescribed displacements u i along its boundary. Let ε ij, σ ij, U be the associated strain, stress, and strain energy density, respectively. The purpose of this investigation is to show that the energy density U associated with any kinematically admissible displacement field u i is greater than the energy function U associated with the true solution. Let ε ij = 1 2 (u i,j + u j,i) σ ij = L ijkl ε kl U = 1 2 σ ijε ij. In the next step, calculate the energy of the difference state with displacements (u i u i ), which is positive 1 2 (σ ij σ ij )(ε ij ε ij ) 0 Therefore, 1 (σ 2 ijε ij σ ij ε ij ) dv 1 (σ V 2 ijε ij + σ ij ε ij 2σ ij ε ij ) dv V Applying Betti s theorem σijε ij = σ ij ε ij yields 1 (σ 2 ijε ij σ ij ε ij ) dv V V σ ij (ε ij ε ij ) dv = 0

12 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 the boundary of the anisotropic solid. Once again, let u i be the required solution and ε ij, σ ij, W be the corresponding strain, stress, and stress (complementary) energy density function, respectively. Suppose that τ ij is any statically admissible stress field and define the associated field η ij η ij = M ijklτ kl. Again, using the trick of computing the positive energy associated with difference state yields 1 2 V It now follows that (τ ij η ij σ ij ε ij ) dv 1 2 V 1 2 (τ ij σ ij )(η ij ε ij ) 0 V V M ijkl τ ij τ kl dv (τ ij ε ij + σ ij η ij 2σ ij ε ij ) dv (τ ij σ ij )ε ij dv = 0. V M ijkl σ ij σ kl dv, (16) which is the special case of the theorem of minimum complementary energy.

13 EXAMPLES 1.2 Consider an arbitrary heterogeneous body with outer boundary S. 1. Suppose that the medium is loaded by prescribed displacements u i on S. Then, if the material is stiffened in any way (keeping the boundary fixed) show that strain energy increases (Hill s stiffening theorem). 2. Show that if the stiffness tensor L ijkl is increased by a positive amount, then the corresponding compliance tensor M ijkl decreases by the positive amount.

14 OVERALL MODULI Even with a powerful hardware at hand the detailed evaluation of local fields on the microscale may prove to be prohibitively expensive and in most practical applications is out of questions. We must therefore settle for a somewhat limited objective.... Since the dimension of the reinforcement (the microscale) are often of the order of 1µm and the dimension of laboratory samples (the macroscale) are of the order of 1 cm, one hopes that, for certain types of loading, knowledge of the actual stress field is not necessary. Rather, the overall, or macroscopic response is the important physical variable. Thus we are led to study the restricted problem of determining overall elastic moduli of composites... (Norman Laws, 1980). Assumptions: 1. A heterogeneous medium made of bonded phases or constituents separated by well defined interfaces of zero thickness (interfaces or coatings are separate phases) 2. Matrix-based composites: inclusions or inhomogeneities embedded in a common matrix (Fibers, particles, whiskers, etc.), two-phase, three-phase, multiphase. 3. Loading conditions are limited to macroscopically uniform fields (recall examples and 1.1.2) 4. Each phase is elastically homogeneous and presumably anisotropic. The composite aggregate is assumed to be macroscopically homogeneous continuum with certain effective (overall, macroscopic) properties.

15 representative volume (RVE) - must be geometrically typical of the composite 1. It contains a sufficient number of inclusions so that under macroscopically uniform tractions or displacements, its properties are independent of its size. In general, a representative volume element is much larger than typical inclusion size or fiber diameter (1-150 µm). 2. In some specific models the RVE may have a well defined geometry and boundary conditions (periodic arrays, unit cells). RVE Objectives: Evaluate the effective elastic material properties of the composite system in terms of phase elastic moduli, phase volume fractions, and geometry of the microstructure.

16 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 mechanical influence functions, we wish to find the local fields in the phases as ε r (x) = A r (x)e, σ r (x) = B r (x)σ. (17) With the help of Eq. (11) the phase volume averages are provided by ε r = 1 V r V r A r (x)e dv = A r E, (18) σ r = 1 V r V r B r (x)σ dv = B r Σ. A r and B r are mechanical strain and stress concentration factors. The concentration factors posses only the minor symmetry A ijkl = A jikl = A ijlk B ijkl = B jikl = B ijlk A T A B T B

17 In any elastically homogeneous phase r the phase constitutive relations read ε r (x) = M r σ r (x) σ r (x) = L r ε r (x). (19) For the representative volume of a statistically isotropic and ergodic composite medium subjected 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 c r ε r = 1 V N c r ε r = 1 V r=1 A r (x)e dv and hence (21) V r N A r (x)e dv = < ε >, V r r=1 where c r = V r /V is the volume fraction of the phase r. Thus < ε >= N c r ε r and similarly < σ >= r=1 N c r σ r (22) r=1

18 Eqs. (19), (20) and (22) then readily provide the overall constitutive relations in terms of phase moduli and phase volume fractions as M < σ >= N c r M r σ r L < ε >= r=1 N c r L r ε r. (23) r=1 Recall that for the present boundary conditions, the following relation holds < σ ij ε ij >=< σ ij >< ε ij > and < U >= 1 2 < σ ij >< ε ij >. Then, invoking the Reciprocal theorem of elasticity (Betti s theorem) for any two sets of elastic fields (< σ ij >, < ε ij >) and (< σ ij >, < ε ij >) in volume V < σ ij >< ε ij >=< σ ij >< ε ij > L ijkl < ε kl >< ε ij >= L ijkl < ε kl >< ε ij >, suggests overall symmetry relations for any actual heterogeneous medium. L = L T and M = M T. Introducing phase concentration factors yields M < σ > = c 1 M 1 B 1 < σ > +c 2 M 2 B 2 < σ > c N M N B N < σ > L < ε > = c 1 L 1 A 1 < ε > +c 2 L 2 A 2 < ε > c N L N A N < ε >. (24)

19 Eqs. (24) must hold for any < σ >, < ε >. Hence, the overall elastic stiffness and compliance matrices are L = N c r L r A r M = r=1 N c r M r B r. (25) Although we proved the diagonal symmetry of L and M, it is advisable to proceed with caution when evaluating the strain and stress concentration factors only approximately using various averaging methods. Then the diagonal symmetry must be examined for each case. r=1 EXAMPLES 1.3 Consider a two-phase or binary system (r = 1, 2). 1. Show that c 1 A 1 + c 2 A 2 = c 1 B 1 + c 2 B 2 = I 2. Using the above relations, show that L = L 2 + c 1 (L 1 L 2 )A 1 M = M 2 + c 1 (M 1 M 2 )B 1 3. If the overall elastic properties are known, show that c 1 A 1 = (L 1 L 2 ) 1 (L L 2 ) c 1 B 1 = (M 1 M 2 ) 1 (M M 2 ) c 2 A 2 = (L 1 L 2 ) 1 (L L 1 ) c 2 B 2 = (M 1 M 2 ) 1 (M M 1 )

20 ELEMENTARY ESTIMATES OF OVERALL MODULI OF UNIDIRECTIONAL COMPOSITES The objective is to present the elementary mechanics of materials for predicting four independent effective moduli of an orthotropic continuous fiber-reinforced lamina. Recall that in the elementary mechanics of materials approach to micromechanical modeling, the geometry of the microstructure is not specified. Hence, the RVE may be a generic block consisting of fiber material bonded to matrix material. The only information required 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. Lm 2 Matrix 2 A 1 L 2 L f Fiber 1 Lm 2 Matrix A f A m σ c2 σc12 σ c1

21 LONGITUDINAL MODULUS Suppose that the macroscopic longitudinal normal stress σ c 11 is prescribed (for the time being, we drop < > and consider average quantities only). Then the force equilibrium condition requires σ c 11 = c f σ f 11 + c m σ m 11. Neglecting the Poisson strains, a one-dimensional Hook s law read σ11 c = E11ε c c 11, σ f 11 = E11ε f f 11, σ11 m = E11ε m m 11. Therefore E11ε c c 11 = c f E11ε f f 11 + c m E11ε m m 11. Assuming that average strains in the composite, fiber and matrix along the fiber direction are equal (ε c 11 = ε f 11 = ε m 11) provides the rule of mixtures E11 c = c f E f 11 + c m E11. m The same results can be obtained using the strain energy approach, which yet confirms the validity of the assumption of equal strains.

22 TRANSVERSE MODULUS Suppose that the macroscopic transverse normal stress σ c 22 is prescribed. Geometrical compatibility requires that the total transverse composite displacement, u c 2, must be equal to the sum of the transverse displacement in the fiber, u f 2, and in the matrix, u m 2 u c 2 = u f 2 + u m 2 Using the elementary definition of normal strain we get u c 2 = ε c 22L 2, u f 2 = ε f 22L f, u m 2 = ε m 22L m ; ε c 22L 2 = ε f 22L f + ε m 22L m. Combining a one-dimensional Hook s law with compatibility of strains σ c 22 = E c 22ε c 22, σ f 22 = E f 22ε f 22, σ m 22 = E m 22ε m 22; ε c 22 = c f ε f 22 + c m ε m 22, and assuming that the stresses in the composite, fiber, and matrix are all equal, we get the inverse rule of mixture 1 = c f + c m. E22 c E f E m Note: The simplified geometry of the RVE suggests that the assumption of equal stresses is valid (analogy with the series arrangement of springs). In reality, however, the actual fiber-packing is far from the idealized one. Employing the strain energy approach we show that indeed the assumption of equal stresses is not generally justified. The

23 total energy stored in the composite U c is represented as the sum of the strain energy in the fibers, U f, and the strain energy in the matrix U m U c = U f + U m. To proceed, recall Eq. (15) and write the fiber and matrix strains in terms of the composite strain as ε f 22 = a f ε c 22, ε m 22 = a m ε c 22 Then, substitution of the above conditions into expression for the composite energy U c (neglecting the strain energy due to the Poisson strain mismatch), leads to E c 22 = c f (a f ) 2 E f 22 + c m (a m ) 2 E m 22.

24 Example: Consider an E-glass/epoxy composite with the following material properties Solving the following system of equations E c 11 = GPa E c 22 = GPa E f 11 = E f 22 = 72.4 GPa c f = 0.45 E m 11 = E m 22 = 3.79 GPa c m = 0.55 c f a f + c m a m = 1 E c 22 = c f (a f ) 2 E f 22 + c m (a m ) 2 E m 22, we get a f a f = 0.432, a m = 1.465, = εf 22 σ f 22 = 0.259, = 5.63 a m ε m 22 σ22 m Thus the assumption of equal stresses in fibers and matrix is not justified for this material and is not valid for most other composites as well.

25 LONGITUDINAL SHEAR MODULUS AND POISSON S RATIO Similar considerations, which led to rule of mixtures for the longitudinal and transverse elastic moduli, provide another rules of mixture for the longitudinal shear moduli and the Poisson ratio 1 = c f + c m G f G m ν12 c = c f ν f 12 + c m ν12. m G c 12 As for the longitudinal modulus, the above estimate of the overall Poisson ratio is generally accepted, while the estimate of overall shear modulus is rather poor because the shear stresses are not equal as assumed. EXAMPLE 1.4 Using the strain energy approach, show that the geometrical assumption of equal axial strains is valid. Use the material properties for E-glass/epoxy composite system.

26 ELEMENTARY BOUNDS ON THE OVERALL MODULI Here we derive elementary bounds on the overall elastic moduli for a two-phase composite medium using the classical extremum principles. Suppose that the RVE is subjected to surface displacements corresponding to macroscopic uniform strain E. The actual strain energy per unit volume is given by U = 1 2 ET LE. (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 complies with the prescribed boundary conditions, but leaves an artificial unequilibrated stress field. When introducing ε into Eq. (15) we immediately discover that 1 2 ET LE c ET L 1 E + c ET L 2 E. (27) Since Eq. (27) holds for any E, we get the upper bound on the overall elastic stiffness tensor (constant strain bound) in the form L c 1 L 1 + c 2 L 2. (28)

27 Next, consider the loading conditions due to prescribed tractions associated with a uniform stress Σ. In view of the previous steps, assume a trial stress σ uniform throughout the RVE such that σ = Σ. Note that such a trial stress is self-equilibrated, but the necessary bonding between the different phases is not maintained. Next, introducing σ into the principle of minimum complementary energy, Eq. (16), renders the the upper bound on the overall elastic compliance tensor (constant stress bound) in the form Example: M c 1 M 1 + c 2 M 2. (29) We now apply the above procedure to the simplest composite medium consisting of isotropic phases and which has overall isotropy. For such a medium, the overall stiffness tensor can be written in terms of the overall bulk modulus κ and the overall shear modulus µ as L ijkl = κδ ij δ kl + µ(δ ik δ jl + δ il δ jk 2 3 δ ijδ kl ). (30) Substitution of Eq. (30) into Eq. (28) leads to κ c 1 κ 1 + c 2 κ 2 = κ V (31) µ c 1 µ 1 + c 2 µ 2 = µ V, where subscript V stands for [Voight, 1887], who was the first to propose these bounds for a polycrystalline aggregate.

28 Dual assumption about uniform trial stress, put forward by [Reuss, 1929], gives 1 κ c 1 + c 2 = 1 (32) κ 1 κ 2 κ R 1 µ c 1 + c 2 = 1. µ 1 µ 2 µ R Eqs. (31) and (32) provide upper and lower bounds on the overall moduli κ R κ κ V µ R µ µ V. (33) Unfortunately these bounds are too far apart to be of any practical significance. EXAMPLES Show that Hill s stiffening theorem implies the following. For a given composite suppose that we replace some of the material by a stiffer material. Then the overall stiffness of the new material is greater then the overall stiffness of the original material. 2. Since 3 = 1 + 1, show that E µ 3κ E V = c 1 E 1 + c 2 E 2 if ν 1 = ν 2.

29 EXACT SOLUTION FOR TWO-PHASE COMPOSITES WITH ISOTROPIC PHASES AND EQUAL SHEAR MODULI Consider a two-phase composite medium with phases of equal shear moduli. We wish to determine the overall bulk modulus in terms of local stress and strain fields. The complete solution was given by [Hill, 1964] for a two phase composite solid of arbitrary geometry, but macroscopically homogeneous. Extension to N-phase composite medium can be found in [Walpole, 1981]. Following Hill, displacement fields in each phase assume the form so that the associated strains are given by The corresponding stresses are u r i = φ r,i r = 1, 2 σ r ij = 2µφ r,ij + ε r ij = φ r,ij. (κ r 23 µ ) 2 φ r δ ij. It follows from the above equation that the stresses are self-equilibrated provided the Laplacian is piecewise constant 2 φ r = θ r.

30 It is now evident that θ 1 and θ 2 represent the phase dilatation. For the tractions to be continuous across the interface, [σ ij ]n j = 0, we require The overall dilatation < ε kk > is given by (κ µ)θ 1 = (κ µ)θ 2. < ε kk >= θ = c 1 θ 1 + c 2 θ 2 = θ 1 κ µ = θ 2 κ µ = The macroscopic constitutive equations are written as θ c 1 κ 2 + c 2 κ µ. < σ ij >= 2µ < ε ij > +(κ 2 3 µ) < ε kk > δ ij, (34) with the overall bulk modulus κ given by κ = c 1κ 1 (κ µ) + c 2κ 2 (κ µ) c 1 (κ µ) + c 2(κ µ) (35) Note: The overall stress-strain relationship is isotropic despite the fact that the phase geometry has not been specified. Also, the overall bulk modulus depends only on the phase moduli and phase volume fractions. The above solution corresponds to a macroscopically uniform pure dilatation of a representative volume.

31 Recall Hill s stiffening theorem V IMPROVED BOUNDS U(ε ij ) dv V U (ε ij) dv, (36) where the ( ) system is associated with the stiffened material and V is volume of the RVE. Consider now a two-phase isotropic composite medium with elastic moduli (κ 1, µ 1 ) and (κ 2, µ 2 ), respectively, with volume fractions c 1 and c 2. In addition, consider two geometrically similar composites with phases of equal shear moduli and κ 1, κ 2 fixed. The first one is associated with the shear modulus µ L while the the shear modulus of the second one is µ U. Suppose that µ L = min (µ 1, µ 2 ) µ U = max (µ 1, µ 2 ). Eqs. (34) and (36) provide the best possible bounds on κ in terms of phase moduli and phase concentrations alone κ(l.b.) = c 2 (κ 2 κ 1 ) κ c 1 (κ 2 κ 1 ) /(κ 1 + 4µ 3 L) κ(u.b.) = c 2 (κ 2 κ 1 ) κ c 1 (κ 2 κ 1 ) /(κ 1 + 4µ 3 U). (37)

32 MICROMECHANICAL ANALYSIS OF COMPOSITES SESSION II Micromechanical analysis of random composites Equivalent unit cell for composites with aligned fibers - fiber tow, unidirectional fibrous plies and laminates Equivalent unit cell for textile composites

33 MICROMECHANICAL MODELING OF RANDOM COMPOSITES Part 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)

34 CONCEPT OF AN ENSEMBLE To reflect the random character of the media it is convenient to introduce the concept of an ensemble the collection of a large number of systems (micrographs taken from different samples of the material) which are different in their microscopical details but they are identical in their macroscopic details. Then, effective or expected value of some quantity corresponds to the process of its averaging through all systems, forming the ensemble. Consider a sample space S with individual members denoted as α. Define p(α) as the probability density of α in S. Then an ensemble average of function F (x, α) at point x is provided by F (x) = F (x, α)p(α) dα. (38) S Following the above definition would require experimental determination of the ensemble average of function F (x, α) for a given point x through the cumbersome procedure of manufacturing a large number of samples (which form the ensemble space S), measuring F (x, α) for every sample and then its averaging for all samples. Therefore, it appears meaningful to introduce certain hypotheses regarding the ensemble average, which substantially simplify this task.

35 STATISTICAL HYPOTHESIS Ergodic hypothesis This hypothesis demands all states available to an ensemble of the systems to be available to every member of the system in the ensemble as well [Beran, 1968]. Once this hypothesis is adopted, spatial or volume average of function F (x, α) given by 1 F (x, α) = lim V V is independent of α and identical to the ensemble average V F (x + y, α) dy (39) F (x) = F (x). (40) Statistical homogeneity Suppose that function F depends on n vectors x 1, x 2,... x n. The ensemble average of F is invariant with respect to translation where x ij = x j x i. F (x 1,..., x n ) = F (0, x 2 x 1,..., x n x 1 ) = F (x 12,..., x 1n ), (41) Statistical isotropy The ensemble average of F is invariant with respect to both translation and rotation where r ij = x ij, i = 1,..., n, j = 1,..., n. F (x 12,..., x 1n ) = F (r ij ) (42)

36 MICROSTRUCTURE DESCRIPTION Indicator function To describe a random microstructure we introduce a characteristic or indicator function χ r (x, α), which is equal to one when point x lies in phase r in the sample α and equal to zero otherwise { 1 x D r (α) χ r (x, α) = (43) 0 otherwise. The symbol D r (α) denotes here the domain occupied by r-th phase. For a two-phase fibrous composite, r = f, m, characteristic functions χ f (x, α) and χ m (x, α) are related by χ m (x, α) + χ f (x, α) = 1. (44) General n-point probability function With the aid of function χ, the general n-point probability function S r1,...,r n is given by [Beran, 1968, Torquato and Stell, 1982] S r1,...,r n (x 1,..., x n ) = χ r1 (x 1, α) χ rn (x n, α). (45) Thus, S r1,...,r n gives the probability of finding n points x 1,..., x n randomly thrown into media located in the phases r 1,..., r n. We limit our attention to functions of the order of one and two.

37 FUNCTIONS OF THE FIRST AND SECOND ORDER One-point probability function S r (x) Gives the probability of finding phase r at x S r (x) = χ r (x, α), (46) Two-point probability function S rs (x, x ) Gives the probability of finding simultaneously phase r at x and phase s at x. S rs (x, x ) = χ r (x, α)χ s (x, α), (47) Statistical hypothesis Homogeneity: S r (x) = S r S rs (x, x ) = S rs (x x ) Isotropy: Ergodicity: Limiting values for x x : S rs (x x ) = S rs ( x x ) S r = c r for x x : S rs (x, x ) = δ rs S r (x) lim S rs(x, x ) = S r (x)s s (x ) x x

38 Homogeneous Isotropic Ergodic S r (x) S r S r c r S rs (x 1, x 2 ) S rs (x 12 ) S rs (r 12 ) S rs (r 12 )

39 NUMERICAL EVALUATION OF STATISTICAL DESCRIPTORS One-point matrix probability function S m Recall that the one point matrix probability function S m gives the chance of finding a randomly placed point located in the matrix phase. To determine this quantity, a simple Monte-Carlo like simulation can be utilized we throw randomly point into the microstructure and count successful hits into the matrix phase. Then, the value of function S m can be estimated as S m n n

40 Two-point matrix probability function S mm (x) a) Using sampling template Instead 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 corresponding successful hits are counted. Furthermore, if the medium under consideration is statistically isotropic, values found for points located on the same circumference can be averaged as well, which allows large number of tests to be performed within one placement of the template. Although simple, such simulations are computationally very intensive. b) Exploiting binary images Such a digitized micrograph can be imagined as a discretization of the characteristic function χ r (x, α), usually presented in terms of a M N bitmap. Denote the value of χ r for the pixel located in the i th row and j th column as a χ r (i, j) Ergodic and statistically homogeneous medium S rs (m, n) = 1 (i m i M + 1)(j n j N + 1) i M i=i m j N j=j n χ r (i, j)χ s (i + m, j + n) (48) where i m = max(1 m, 1), i M = min(m, M m) and j n = max(1 n, 1), j N = min(n, N n). Observe that O((MN) 2 ) operations are needed for function S rs. Periodic ergodic medium Function S rs can be written as a correlation of functions χ r and χ s S rs (x) = 1 χ r (y)χ s (x + y) dy (49)

41 FOURIER TRANSFORM The d-dimensional Fourier Transform of function f(x) F (f(x)) = f(ξ) = f(x)e iξ x dx, (50) where i is the imaginary unit. The operator F is called the Fourier transform operator. Inverse Fourier Transform F 1 ( f(ξ) ) = f(x) = (2π) d 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 derivative Provided that function f(x) decays sufficiently rapidly to 0 for x we have ( f x i ) (ξ) = f e iξ x dx = iξ i x i f x i e iξ x dx = iξ i f(ξ). (53)

42 DISCRETE FOURIER TRANSFORM The Discrete Fourier Transform (DFT) often replaces its continuous counterpart when analyzing discrete systems such as digitized images of real microstructures. The complexity of their geometries usually calls for sampling of large micrographs. The actual microstructure is then approximated by the measured segment periodically extended outside 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 DFT Consider a discrete set of N points defined on the interval 0 n N 1. Applying a discrete version of the Fourier series this set is given by x(n) = 1 N N 1 k=0 ξ(k)e i(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)

43 CONVOLUTION THEOREM Convolution of two functions f & g f(x x )g(x ) dx. Fourier Transform of Convolution ( ) F f(x x )g(x ) dx = F (f(x)) F (g(x)). (56) Inverse Fourier Transform (Recall Eq. (51) to get) ( ) F 1 f(ξ ξ ) g(ξ ) dξ = (2π) d f(x)g(x), (57) which implies that F (f(x)g(x)) = (2π) d Correlation of two functions f & g (g(x) is a real function) f(x + x )g(x ) dx, f(ξ ξ ) g(ξ ) dξ. (58) Fourier Transform of Correlation ( ) F 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 average used before.

44 EVALUATION OF FUNCTION S rs FROM BINARY IMAGES Fourier Transform of S rs Recall Eq. (49) and relation (59) to get S rs (ξ) = 1 χ r(ξ) χ s (ξ), (60) Taking advantage of the periodicity of function χ r one may implement the Discrete Fourier Transform (DFT) to evaluate Eq. (60). Discrete version of of Eq. (49) S rs (m, n) = 1 MN M i=1 N χ r (i, j)χ s ((i + m)%(m), (j + n)%n), (61) j=1 where symbol % stands for modulo. The above equation, usually termed the cyclic correlation, readily implies periodicity of function S rs. Note that the correlation property of DFT holds for cyclic correlation. Referring to Eq. (60) it is given by the following relation DFT{S rs (m, n)} = DFT{χ r (m, n)}dft{χ s (m, n)}. (62) The inverse DFT denoted as IDFT then serves to derive function S rs at the final set of discrete points as S rs (m, n) = IDFT{DFT{χ r (m, n)}dft{χ s (m, n)}}. (63) Usually, the Fast Fourier Transform, which needs only O(M N log(m N)) operations, is called to carry out the numerical computation.

45 EFFECT OF RESOLUTION OF DIGITIZED MEDIA Resolution 488x358 pixels Resolution 244x179 pixels CPU time in seconds required to evaluate function S mm Method Bitmap resolution Direct with periodicity Eq. (61) Direct without periodicity Eq. (48) FFT based Eq. (63)

46 TESTING ERGODIC ASSUMPTION To test the ergodic hypotheses, we shall require c r = 1 n n Sr, i r = f, m, (64) i=1 where n is the number of members in the ensemble. Functions S i r can be derived by randomly placing a point in the member i in a certain number of times while counting the 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 S f = 0.42, which agrees well with the average fiber volume fraction c f =

47 TESTING STATISTICAL ISOTROPY To that end, we examine the distribution of the two-point matrix probability function S mm for a statistically uniform medium as a function of the absolute difference of points x and x and orientation φ. Should the material be statistically isotropic (independent of orientation) the variation coefficient v(φ) of S mm (φ) r/r for a given ratio r/r would be equal to zero v[%] r/r (b)

48 USEFUL RELATIONS BETWEEN TWO-POINT PROBABILITY FUNCTIONS S mm r/r (a) n = 500 n = 1000 n = 5000 n = n = S r/r (b) S mm S fm S ff S mm + S fm S fm + S ff 1 (S mm + 2S fm + S ff ) Known function S mm (r) S mf (r) S ff (r) S mm (r) S mm (r) c m S mf (r) c m c f + S ff (r) S mf (r) c m S mm (r) S mf (r) c f S ff (r) S ff (r) c f c m + S mm (r) c f S mf (r) S ff (r)

49 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 of particles, and yet statistically resembles the actual composite. We argue, that if the PUC has a statistically similar spatial distribution of fibers as the real microstructure, it will also posses similar mechanical properties.

50 CONSTRUCTION OF PERIODIC UNIT CELLS Such a PUC can be constructed by matching the two-point probability functions of the real microstructure and the unit cell. F (x N, H 1, H 2 ) = N m k=1 ( Smm (r k ) S mm (r k ) ) 2 where x N = {x 1, y 1,..., x N, y N } stores the positions of particle centers of the periodic unit cell, x i and y i correspond to the x and y coordinates of the i-th particle, H 1 and H 2 are the dimensions of the unit cell, S mm (r k ) is the value of S corresponding to the original media evaluated at the distance r k and N m is the number of points, in which both functions are evaluated. Statistical isotropy is assumed.

51 OPTIMIZATION PROCESS Optimal fiber configuration For a given number of fibers N, dimensions of a unit cell H 1 and H 2 and values of the original function S mm (r) evaluated at points r i, i = 1,..., N m find the configuration of particle centers x N (H 1, H 2 ) such that: x N (H 1, H 2 ) = arg min F (x N, H 1, H 2 ), x N S where S denotes a set of admissible vectors x N. Optimal ratio H 1 /H 2 For known values of x N (η) and for the fixed volume fraction of phases, find the ratio η N such that: η N = arg min F η η a;η b (xn (η)), where values η a and η b should be chosen to cover all the reasonable dimensions of the unit cell.

52 EXAMPLE OF AN ADMISSIBLE UNIT CELL The objective function F is multi-dimensional with large number of plateaus and local minima efficient optimization methods such as Genetic algorithms or Simulated annealing methods, (see [Matouš et al., 2000, Lepš and Šejnoha, 2000, Sejnoha, 2000] and [Zeman and Šejnoha, 2001] for details on their implementation) are needed.

53 EXAMPLES OF PERIODIC UNIT CELLS FOR FIBER TOW

54 MICROMECHANICAL ANALYSIS OF COMPOSITES SESSION III Ellipsoidal inclusion problem - the Eshelby problem Overall moduli - Inhomogeneity problem, the Mori-Tanaka and Self consistent methods

55 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 the solution of the ellipsoidal inclusion problem. The problem was addressed by Eshelby in He proved that the local stress and strain fields in an ellipsoidal inhomogeneity L r, embedded into an infinite homogeneous elastic material L L r, that is loaded by remotely applied uniform stress or strain fields, respectively, are also uniform. This result has then provided a stepping stone for evaluation of effective properties in variety of composite systems. The Eshelby problem thus deserves a special attention, particularly in view of forthcoming discussions, and will be described herein. Note that ellipsoidal shapes include: spheres, spheroids, circular and elliptical cylinders, and elliptical or circular (pennyshaped) discs and platelets.

56 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 ellipsoidal inclusion of the same material, in a certain volume V r. The surrounding material is neither 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 the matrix it attains in situ deformation ɛ r = Sµ, where S is the Eshelby tensor. Thus knowing the Eshelby tensor S we may pose the following problem: Find a transformation strain µ, which must be applied in the homogeneous inclusion L in order to create the same stress and strain fields, which would be developed in the inhomogeneity L r of the same shape and size bonded to a large volume of unconstrained material L, that is subjected to uniform overall stress or strain fields. Homogeneous inclusion Inhomogeneity E E L L L r L ε = E + Sµ r ε r

57 The above problem can be rephrased as follows: For a given L r find the required transformation strain µ: ɛ r = E + Sµ σ r = L(ɛ r µ) = L r ɛ r = µ = L 1 (L L r )ɛ r, where the stain ɛ r is ɛ r = E + Sµ = E + SL 1 (L L r )ɛ r (65) ɛ r = [ I SL 1 (L L r ) ] 1 E = Ar E. Recall that A r represents the strain mechanical concentration factor relating the overall strain E to the local strain ɛ r. Hence, the Eshelby tensor is sufficient to determine the local strains (stresses) for a given L r. The above approach is known as the equivalent inclusion method.

58 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 a stress-free deformation µ. Let λ ij = L ijkl µ kl, be the stress derived from µ ij by Hook s law. Both the inclusion and matrix medium are stress-free. II. Apply surface tractions λ ij n j on the inclusion boundary S r to restore its original shape and size. The stress in the inclusion is now λ ij, while the matrix is still stress free. Put it back in the matrix and reweld across S r. The surface tractions can be interpreted as a layer of body forces spread over S r. III. Remove this unwanted layer of body forces by applying an equal and opposite surface tractions λ ij n j over S r. The body is now free of external forces but in the state of self-stress owing to the presence of transformation strain µ in the inclusion. Then, the displacement at point x due to this layer of body forces is equal to u p (x ) = u pi(x x )λ ij n j (x) dx. (66) S r

59 In Eq. (66), S r is the surface separating matrix and inclusion V r and u pi(x x ) is fundamental solution, which corresponds to the displacement at point x in the direction of i due to the concentrated unit force applied at point x in the direction of p to the infinite medium. Since u pi(x x ) x j = u pi(x x ) x j, the divergence theorem provides u p (x ) = 1 ( ) u pi (x x ) λ 2 V r x ij + u pj(x x ) λ j x ij i The resulting strain field than can be found in the form dx. ɛ pq (x ) = P pqij (x )λ ij = P pqij (x )L ijkl µ kl, (67) where P is given by P pqij (x ) = 1 [ 2 u pi(x x ) + 2 u pj(x x ) + 2 u qi(x x ) 4 V r x j x q x i x q x j x p + 2 u qj(x x ] ) dx = ɛ x pqij(x x ) dx (68) i x p V r Once the value of P is known, the Eshelby tensor S follows directly from Eq. (67) S = PL. (69)

60 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 f(ξ) = f(x)e iξ x dx, (70) R d where i is the imaginary unit. Once f(ξ) is known, the inverse transform is given by f(x) = 1 f(ξ)e iξ x dξ (71) (2π) d R d Provided that function f(x) decays sufficiently rapidly to 0 for x we have ( f x i ) (ξ) = R d f x i e iξ x dx = iξ i R d fe iξ x dx = iξ i f(ξ), (72)

61 THE DIRAC DELTA FUNCTION Recall the following property of Dirac s functions δ(x x ) 1. δ(x x ) = 0 for x x 2. δ(x x ) = for x = x 3. δ(x x )g(x) dx = g(x ) The third property of the Dirac delta function suggests that it s 1D Fourier transform yields Hence, the inverse relation gives 1 2π δ(x)e iξ x dx = 1. (73) 1 e iξ x dξ = δ(x). (74)

62 FUNDAMENTAL SOLUTION Fundamental solution u pi complies with the solution of the following equation L ijkl u pk (x) x l x j + δ pi δ(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 of Eq. (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 This yields the desired relation L ijkl ξ l ξ j ũ pk(ξ) + δ pi = 0 (76) ũ ik(ξ) = (L ijkl ξ l ξ j ) 1. (77) It is worth to mention the following property of ũ ik. When ξ = rω, then ũ (rω) = 1 r 2 ũ (ω). (78) After reviewing all necessary relations, we return to the original task and continue with determination of P tensor. First, note that u ik(x) = 1 ũ 8π ik(ξ)e iξ x dξ (79) 3 R 3

63 When introducing the following substitution ξ = rω, where r is the magnitude of vector ξ and ω = ξ/r and noticing that dξ = r 2 dr dω, we arrive at u ik(x) = 1 ũ 8π ik(rω)e iξ x r 2 dr dω, (80) 3 S 0 where S corresponds to the surface of the unit sphere with center located at the origin of the coordinate system. Since the above expression is even in r it can be recast as u ik(x) = 1 ũ 16π ik(rω)e iξ x r 2 dr dω. (81) 3 Substitution of Eq. (78) into the above expression leads to u ik(x) = 1 [ ] ũ 16π ik(ω) e irω x dr 3 S S 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 ũ 8π ik(ω)δ(ω x) dω. (83) 2 S However, to obtain tensor P we need to evaluate terms like Vr 2 u pi(x x [ ] ) dx = ũ x j x q x pi(ω)δ(ω (x x )) dx dω. (84) j x q S V r

64 To that end, we introduce a function ψ(x, ω) ψ(x, ω) = δ(ω (x x )) dx. V r (85) Hereafter, we limit our attention to an ellipsoidal inclusion only. It can be shown that if point x is contained by the inclusion then, for an ellipsoid given by α ij x i x j = 1, t 2 = α 1 ij ω iω j and a = det α ij ([Laws, 1980], [Mura, 1982]), the function ψ(x, ω) reads ψ(x, ω) = π ( t 3 t 2 + (x ω) 2), (86) a Note that ψ(x, ω) is the only term in Eq. (84), which depends on x. Hence, we need to evaluate 2 ψ(x, ω) x j x q = 2πω jω q t 3 a. (87) It should be emphasized that the term 2 ψ(x, ω)/ x q x j and consequently the tensor P pqij 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 P pqij = 1 16π a S 1 t 3 (ũ pi ω j ω q + ũ pjω i ω q + ũ qiω j ω p + ũ qjω i ω p ) dω (88)

65 EXAMPLE 1.6 As an example, we evaluate the P tensor for a spherical inclusion in an isotropic material. Stiffness tensor of such a medium can be written in the form L ijkl = κδ ij δ kl + µ(δ ik δ jk + δ il δ jk 2 3 δ ijδ kl ). After some algebra it becomes evident that ũ pi(ω) = 1 µ δ pi 3κ + µ µ(3κ + 4µ) ω pω i. Next, verify that L ijkl ω l ω j ũ pk (ξ) = δ pi. Examining Eq. (88) suggests that the following integrals must be evaluated ω j ω q dω ω p ω q ω i ω j dω. (89) S Introducing the polar coordinate system readily provides ω 1 = cos θ cos φ, ω 2 = cos θ sin φ, ω 3 = sin θ, S θ π/2, π/2, φ 0, 2π, dω = cos θ dθ dφ. (90)

66 Substitution from (90) into (89) gives 2π 0 2π 0 sin φ dφ = π/2 π/2 π/2 π/2 cos 2 φ dφ = 2π 0 2π 0 cos φ dφ = [ x ] 2π sin 2 φ dφ = = π 2 2π 0 sin φ cos φ dφ = 0 cos 3 (θ) dθ = [sin θ 13 ] π/2 cos θ = 4 π/2 3 [ ] π/2 1 sin 2 (θ) cos(θ) dθ = 3 sin3 (θ) = 2 3, 0 π/2 so that Therefore ω 1 ω 1 dω = ω 2 ω 2 dω = ω 3 ω 3 dω = 4π S S S 3 ω p ω i dω = 4π 3 δ pi. S otherwise 0. Similarly we get S ω p ω q ω i ω j dω = 4π 15 (δ pjδ qi + δ qj δ pi + δ pq δ ij ).

67 The above relations provide the desired result in the form. P pqij = 1 3(3κ + 4µ) δ 3(κ + 2µ) pqδ ij + 10µ(3κ + 4µ) (δ piδ qj + δ pj δ qi 2 3 δ pqδ ij ). (91) In general, the P tensor depends on the aspect ratio of the ellipsoid and the coefficients of L. Some useful forms follow ([Walpole, 1969]).

68 A SPHERICAL INCLUSION IN ISOTROPIC MEDIUM P = 1 ( σ + η ) 1 ( σ η ) 1 ( σ η ) κ µ 3 3κ 2µ 3 3κ 2µ 1 ( σ + η ) 1 ( σ η ) κ µ 3 3κ 2µ 1 ( σ + η ) κ µ η µ 0 0 η µ 0 η µ 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]

69 A CIRCULAR CYLINDER (prolate spheroid) in a transversely isotropic medium (x A = x 1 ). P = k+4m 8m(k+m) k m(k+m) k+4m m(k+m) k+2m 0 0 2m(k+m) 1 0 2p 1 2p where k, m, p are Hill s moduli; [Walpole, 1969].

70 A PENNY-SHAPED INCLUSION (circular disc) in a transversely isotropic matrix (x A x 1 ) P = 1 n p 0 1 p where p, n, are Hill s moduli of the medium [Walpole, 1969].

71 APPROXIMATE EVALUATION OF OVERALL MODULI - 2 Dvorak (lecture notes), [Šejnoha, 1999] Here we address the problem of evaluating overall elastic moduli of a composite medium with many inclusions embedded in a matrix material. We have already shown that this problem 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 because the geometry is not well defined. Although a valuable piece of information can be obtained from high contrast micrographs, we usually content ourself with the knowledge of the volume fractions of the inclusions and their orientation and shape. It becomes clear that certain assumptions are needed to resolve this matter. Three approximate methods are now available in the literature, which are worthwhile mentioning: 1. The dilute approximation 2. The self-consistent method 3. The Mori-Tanaka method Before we contemplate with individual methods we recall the basic problem associated with a single inclusion (L r ) in an infinite matrix (L) subjected to uniform remote strains or stresses. For such a problem the exact solution for the concentration tensor A r exists and can be written in terms of the Eshelby tensor as A r = [ I SL 1 (L L r ) ] 1.

72 Also, it is perhaps appropriate to recall the basic results already derived in preceding sections N N L = c r L r A r M = c r M r B r r=1 r=1 I = N c r A r I = r=1 N c r B r, r=1 so we can express the overall stiffness and compliance matrices in the forms N L = L 1 + c r (L r L 1 )A r M = M 1 + r=2 N c r (M r M 1 )B r. r=2 Hereafter, we use suffix r = 1 to refer to the matrix and suffix r = 2,..., N to refer to the inclusions. If a phase has more than one shape (as specified by the P tensor), then an index r must be allocated to each such shape.

73 SCHEMATIC REPRESENTATION OF INDIVIDUAL METHODS DILUTE APPROXIMATION Σ Σ SELF - CONSITENT METHOD L L L L MORI - TANAKA METHOD L L <σ >

74 INHOMOGENEITY PROBLEM [Hill, 1965] used the principle feature of the Eshelby solution, that an ellipsoidal inhomogeneity is strained uniformly, to pose an inhomogeneity problem. He introduced an overall constraint tensor L for material L, with inverse M, associated with the uniform cavity strain ɛ caused by distribution of tractions over the inhomogeneity surface that are compatible with a uniform field of stress σ such that σ = L ɛ, ɛ = M σ. (92) Σ Σ 1 σr Lr L = Σ L L + 2 L σ* σ r = Σ + σ* σr L r A schematic representation of the inhomogeneity problem suggests that σ r = Σ + σ. Then σ r Σ = L (E ɛ r ), ɛ r E = M (Σ σ r ), (93) and so (L + L r ) ɛ r = (L + L) E, (M + M r ) σ r = (M + M) Σ. (94)

75 Introducing the mechanical strain and stress concentration factors A r and B r we find ɛ r = A r E, σ r = B r Σ, where A r = (L + L r ) 1 (L + L) B r = (M + M r ) 1 (M + M). (95) After setting Σ = 0 it becomes clear that ɛ r = ɛ = Sµ, σ r = σ = L (ɛ 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) 1 L P = (L + L) 1 = P T

76 Introducing another tensor Q = (L +L) 1 = P T yields the concentration factor tensors A 1 r = P(L + L r ) = I P(L L r ) B 1 r = Q(M + M r ) = I Q(M M r ). (96) EXAMPLE Verify that PL + MQ = I. 2. For a spherical inclusion in an isotropic material, let L = (3κ, 2µ ) Show that κ = 4 3 µ µ = µ(9κ + 8µ) 6(κ + 2µ)

77 THE DILUTE APPROXIMATION Consider a matrix-based composite consisting of N phases. When using the dilute approximation 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 single inclusion of each phase was embedded in an infinite volume of matrix. When solving an inclusion problem for each phase we find A 1 r = P (r) (L + L r ) = I P (r) (L 1 L r ) B 1 r = Q (r) (M + M r ) = I Q (r) (M 1 M r ). (97) Note that P (r), Q (r) and L, M depend on the inclusion geometry and elastic moduli of the matrix L 1. The overall stiffness and compliance matrices thus attain the form given by Eq. (92) N L = L 1 + c r (L r L 1 )A r r=2 N M = M 1 + c r (M r M 1 )B r. r=2

78 THE SELF-CONSISTENT METHOD The self-consistent method of [Hershey, 1954] and [Kroner, 1958] was originally proposed for aggregates of crystals and then reviewed and elaborated by [Hill, 1965] in connection with composites. The method draws on the familiar solution to an auxiliary inclusion problem. In particular, it assumes that interaction of phases is accounted for by imagining each phase to be an 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 subscripts 1 and 2, respectively. Recall Eq. (22) and write the elementary relations between the phase and overall averages of stress field c 1 (σ 1 Σ) + c 2 (σ 2 Σ) = 0. (98) The basic postulate of the self-consistent method suggests that and from Eq. (98) σ 1 Σ = L (E ɛ 1 ), (99) σ 2 Σ = L (E ɛ 2 ). (100) Evidently, both phases are regarded on the same footing (concentration factors for both the inclusion and the matrix are derived from the same L ). It means that the same overall moduli are predicted for another composite in which the roles of the phases are reversed. Recall the following connections

79 P (r) = (L r L) 1 Q (r) = (M r M) 1, to find A 1 r = I + P (r) (L r L) = P (r) (L + L r ) B 1 r = I + Q (r) (M r M) = Q (r) (M + M r ), where L r, M r depend only on L ij, M ij and shape of the inclusion. Consider now aligned inclusions of similar shapes (P (r) = P, Q (r) = Q) to get L = M = [ ] 1 c r (L + L r ) 1 L (101) r [ ] 1 c r (M + M r ) 1 M. r To derive the above form of the overall stiffness matrix L we exploited the relations cr A r = I and A r = (L + L r ) 1 (L + L). Changing A B and L M leads to the overall compliance matrix M. Eq. (101) represents an implicit set of 21 algebraic equations. Thus a suitable iteration scheme must be introduce to solve for unknown stiffnesses or compliances.

80 OVERALL MODULI OF UNIDIRECTIONAL FIBER COMPOSITES - SCM Here we present approximate values of elastic overall moduli of fibrous materials with transversely isotropic phases provided by the self-consistent method (see [Hill, 1965] for reference). For a transversely isotropic solid, with x 1 as the axis of rotational symmetry, the stress-strain relation σ = Lɛ, is usually written in terms of Hill s moduli where σ 1 σ 2 σ 3 σ 4 σ 5 σ 6 n l l l (k + m) (k m) = l (k m) (k + m) m p p ɛ 1 ɛ 2 ɛ 3 ɛ 4 ɛ 5 ɛ 6, (102) k = [1/G 23 4/E ν 2 12/E 11 ] 1 l = 2kν 12 n = E kν 2 12 = E 11 + l 2 /k m = G 23, p = G Transverse shear modulus m is found first, as a solution of the cubic equation: c 1 k 1 k 1 + m + c ( 2k 2 k 2 + m = 2 c1 m 2 m 2 m + c ) 2m 1. m 1 m

81 2. Longitudinal shear modulus p is found as a solution of a quadratic equation: c 1 p p 2 + c 2 p p 1 = 1 2p. 3. Moduli k, l, and n, in terms of m, are found from: 1 k + m = c 1 k 1 + m + c 2 k 2 + m l k + m = c 1l 1 k 1 + m + c 2l 2 k 2 + m n c 1 n 1 c 2 n 2 l c 1 l 1 c 2 l 2 = l 1 l 2 k 1 k 2. Word of caution: The estimates become unreliable when the fiber concentration is high and when the phase moduli differ considerably. EXAMPLE 2.2 Estimate the effective elastic properties for the composite made from isotropic dispersion of spheres

82 THE MORI-TANAKA METHOD Consider a matrix-based composite consisting of N phases. The method approximates the effect of phase interaction on the local stresses by assuming that the stress in each phase is equal to that of a single inclusion embedded into an unbounded matrix subjected to as yet unknown average matrix strain ɛ 1 = A 1 E or stress σ 1 = B 1 Σ. In Benveniste s reformulation of the Mori-Tanaka method ([Benveniste, 1987]), the solution of a single inclusion in a large volume of matrix loaded by ɛ 1 or σ 1 assumes the form ɛ r = T r ɛ 1, σ r = W r σ 1 (103) N and since E = c r ɛ r, Σ = and stress vectors are r N c r σ r, one can find that the average matrix strain r [ ] 1 [ ] 1 N N ɛ 1 = c 1 I + c r T r E σ 1 = c 1 I + c r W r Σ. (104) n=2 n=2

83 Eqs. (104) also suggest that the concentration factors A s and B s are A r = T r [c 1 I + where, recall Eq. (96) ] 1 ] 1 N N c r T r, B r = W r [c 1 I + c r W r, (105) n=2 n=2 T 1 r = P (r) (L + L r ) = I P (r) (L 1 L r ) W 1 r = Q (r) (M + M r ) = I Q (r) (M 1 M r ) (106) Tensors P (r), Q (r), L and M are now functions of the moduli in L 1. It becomes clear that the Mori-Tanaka method provides an explicit relations for the mechanical concentration factors. Substitution of Eqs. (105) into Eq. (92) leads to [ N ] [ N ] 1 [ N ] [ N ] 1 L = c r L r T r c r T r, M = c r M r W r c r W r. (107) r=1 r=1 r=1 r=1

84 OVERALL MODULI OF UNIDIRECTIONAL FIBER COMPOSITES - MTCM Mori-Tanaka estimates of elastic overall moduli of fibrous materials with transversely isotropic phases written in terms of phase moduli and phase volume fractions are k = k 2k 1 + m 1 (c 2 k 2 + c 1 k 1 ) c 2 k 1 + c 1 k 2 + m 1 l = c 2l 2 (k 1 + m 1 ) + c 1 l 1 (k 2 + m 1 ) c 2 (k 1 + m 1 ) + c 1 (k 2 + m 1 ) n = c 2 n 2 + c 1 n 1 + (l c 2 l 2 c 1 l 1 ) l 2 l 1 k 2 k 1 m = m 2m 1 (k 1 + 2m 1 ) + k 1 m 1 (c 2 m 2 + c 1 m 1 ) k 1 m 1 + (k 1 + 2m 1 ) (c 2 m 1 + c 1 m 2 ) p = 2c 2p 2 p 1 + c 1 (p 2 p 1 + p 2 1). 2c 2 p 1 + c 1 (p 2 + p 1 )

85 OVERALL ELASTIC SYMMETRY OF SCM AND MT METHODS First, assume that all phases s = 2,... N are of the same shape and alignment, so that there is only one pair of P and L tensors, L = P 1 L. L = L in the self-consistent method L = L 1 in the Mori-Tanaka method. This formal similarity between the SCM and M-T method allows to write the overall stiffness matrix in the form L = [ ] 1 c r (L + L r ) 1 L. r This shows that both the self-consistent and Mori-Tanaka methods give L = L T for multiphase systems reinforced by inclusions of the same shape and alignment. The same is not true, however, when multiphase materials with inclusions of different shape are considered. Numerical examples show that in such a case L L T. Note that the dilute approximation preserves the diagonal symmetry in both cases.

86 NUMERICAL EXAMPLE - MT METHOD Consider a multiphase system consisting of 3 isotropic phases: 1. Ti 3 Al matrix 2. Carbon circular discs with mid-plane normal to the x 3 -axis 3. SiC fibers parallel to the x 3 -axis E 1 =96.5 GPa, G 1 =37.1 GPa, c 1 =0.55 E 2 =34.4 GPa, G 2 =14.3 GPa, c 2 =0.25 E 3 =431.0 GPa, G 3 =172.0 GPa, c 3 = L = GPa Example 2 : Elastic porous media L L T Consider an elastic isotropic body weakened by spherical pores. Determine the influence of the porosity ( volume fraction of the pores) on the overall elastic moduli of the body using the dilute approximation and the Mori-Tanaka method.

87 Solution: We will model a porous medium as a two-phase material (N = 2) composed of an elastic matrix characterized by the material stiffness tensor L 1 and spherical pores (L 2 = 0) with the volume fraction c 2. The relation for the effective stiffness matrix of the material has the following form, recall Eq. (92): L = L 1 + c 2 (L 2 L 1 )A 2 = L 1 (I c 2 A 2 ), Dilute approximation Recall that for the dilute approximation, the concentration tensor can be obtained as, see Eq. (97), A 1 2 = I P (2) (L 1 L 2 ) = I P (2) L 1. (108) For the matrix phase, the stiffness tensor has the following form: (L 1 ) ijkl = κ 1 δ ij δ kl + µ 1 (δ ik δ jk + δ il δ jk 2 3 δ ijδ kl ). To simplify the algebra, we introduce the fourth-order projection tensors Λ h and Λ s [Milton (Λ h ) ijkl = 1 3 δ ijδ kl, (Λ s ) ijkl = 1 2 [δ ikδ jl + δ il δ jk ] 1 3 δ ijδ kl, which posses the following properties: Λ h Λ h = Λ h, Λ s Λ s = Λ s, Λ h Λ s = 0, Λ s Λ h = 0, Λ s + Λ h = I, (αλ h + βλ s ) 1 = 1 α Λ h + 1 β Λ s

88 This allows us to express the tensor L 1 as L 1 = 3κ 1 Λ h + 2µ 1 Λ s. Similarly, the P (2) for the spherical pore can be expressed as, compare with Eq. (91), P (2) ijkl = P (2) = 1 3(3κ 1 + 4µ 1 ) δ ijδ kl + 3(κ 1 + 2µ 1 ) 10µ 1 (3κ 1 + 4µ 1 ) (δ ikδ jl + δ il δ jk 2 3 δ ijδ kl ) 1 Λ h + 3(κ 1 + 2µ 1 ) 3κ 1 + 4µ 1 5µ 1 (3κ 1 + 4µ 1 ) Λ s Hence, Finally, A 1 2 = I P (2) L 1 ( = Λ h + Λ s = 1 3κ 1 + 4µ 1 Λ h + 3(κ 1 + 2µ 1 ) 5µ 1 (3κ 1 + 4µ 1 ) Λ s 4µ 1 3κ 1 + 4µ 1 Λ h + 9κ 1 + 8µ 1 5(3κ 1 + 4µ 1 ) Λ s A 2 = 3κ 1 + 4µ 1 Λ h + 5(3κ 1 + 4µ 1 ) Λ s = α h Λ h + α s Λ s 4µ 1 9κ 1 + 8µ 1 L = L 1 (I c 2 A 2 ) = (3κ 1 Λ h + 2µ 1 Λ s ) (Λ h + Λ s c 2 (α h Λ h + α s Λ s )) = 3κ 1 (1 c 2 α h ) Λ h + 2µ 1 (1 c 2 α s ) Λ s ) (3κ 1 Λ h + 2µ 1 Λ s )

89 Effective moduli can be expressed as Mori-Tanaka method κ dil κ 1 = 1 c 2 α h µ dil µ 1 = 1 c 2 α s The concentration tensors for the Mori-Tanaka method follow from the identity, compare with Eq. (105) A 2 = T 2 [c 1 I + c 2 T 2 ] 1, where the partial concentration factor yields is defined as, see Eq. (106), T 1 2 = I P (2) L 1, 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 the form A 2 = T 2 [I c 2 (I T 2 )] 1 = (α h Λ h + α s Λ s ) [Λ h + Λ s c 2 (Λ h + Λ s α h Λ h α s Λ s )] 1 α h α s = Λ h + Λ s 1 c 2 + c 2 α h 1 c 2 + c 2 α s

90 Similarly to the dilute approximation case, we finally obtain κ MT c 2 α h = 1 κ 1 1 c 2 + c 2 α h µ MT c 2 α s = 1 µ 1 1 c 2 + c 2 α s Comparison of the dilute approximation and the Mori-Tanaka scheme κ/κ Dilute MT ν 1 =0.10 ν 1 =0.25 ν 1 = c 2 Engineering moduli predictions µ/µ Dilute MT ν 1 =0.10 ν 1 =0.25 ν 1 = c 2 Materials with elastic moduli E = 1 Pa, ν varies Note that the engineering moduli are related to material parameters κ and µ by relations E = 9κµ 3κ 2µ, ν = 3κ + µ 6κ + 2µ.

91 E/E ν 1 =0.10 ν 1 =0.20 ν 1 =0.30 ν 1 = c 2 ν/ν ν 1 =0.10 ν 1 =0.20 ν 1 =0.30 ν 1 = c 2 Materials with elastic moduli E = 1 Pa, ν varies

92 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

93 EVALUATION OF EFFECTIVE PROPERTIES OF COMPOSITE MATERIALS - PERIODIC UNIT CELL APPROACH Suquet, Moulinec, Fish, Oden, Dvorak Here, we are concerned with the RVE having well defined geometry and boundary conditions. In particular, we consider a periodic representative volume defined in terms of a unit cell (UC) with prescribed periodic boundary conditions.

94 EVALUATION OF EFFECTIVE PROPERTIES USING UNIT CELL MODELS Consider a composite material characterized by a periodic unit cell (PUC). The overall elastic behavior of such a composite is then governed by its microstructure and by the behavior of individual phases conveniently described by displacement, strain and stress fields in the form σ(x) = b(x), ε(x) = T u(x), (109) (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; L is the stiffness tensor and M is the compliance tensor such that L 1 = M; µ represents certain stress free strains and λ = Lµ; b is the vector of body forces and is the differential operator matrix [Bittnar and Šejnoha, 1996, p. 9] The following discussion is limited to two-phase fibrous composites with fibers aligned along the x 3 axis. We further assume that each phase is transversally isotropic with x 3 being the axis of rotational symmetry. Additional simplification arises when neglecting the out-of-plane shear response. The generalized plane strain is then a natural assumption. In such a state, the only non-zero components of the strain and stress tensors are ε 11, ε 12, ε 22, ε 33 and σ 11, σ 12, σ 22, σ 33, respectively. Note that due to perfect bonding between individual phases the components ε 33 and σ 33 attain constant values.

95 PERIODIC STRAIN FIELDS Suppose that the UC is subjected to boundary displacements u resulting in a uniform strain E throughout the UC. Let u (x) be a periodic displacement field in the UC. Then, the strain and displacement fields 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) = 1 ε (x) dx = 0 (113) Proof: Using Green s formula gives ε ij(x) = 1 ( u 2 i,j + uj,i) 1 d = 2 ε (x) = u n ds, ( u i n j + u jn i ) ds = 0 since u i (x) is identical on opposite sides and n j is opposite on opposite sides of the unit cell.

96 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 ε (x) dx + u n P ds = 0 (114) Proof: M M P =0 due to periodicity {}}{ 1 ( u 2 i,j + uj,i) 1 ( ) d = u 2 i n j + u 1 jn i ds 2 P ( u i n jp + u jn ip ) ds Γ n P Γ M n int M The following generalization also holds E = ε dx + u n P ds (115) E ij = M P ε ij d (u in Pj + u j n Pi ) ds M P

97 Proof: = ε(x) d = u n P ds M P {}} {{ }} { E = ME + ε dx + (E x) n P ds + u n P ds (116) M P P Substitution of Eq. (114) into Eq. (116) leads to E = ME + (E x) n P ds Denoting M = P yields PE = (E x) n P ds P P EXAMPLES 3.1 Consider two elementary macroscopic strain fields E 11 and E 12 in 2D. Using the divergence theorem verify that PE = (E x) n P ds. P

98 OVERALL MODULI - PURE MECHANICAL RESPONSE Consider a two phase fibrous composite with transversally isotropic phases under generalized plane strain and write the the 4 4 stiffness matrix L of each phase as (k + m) (k m) 0 l (k m) (k + m) 0 l L = 0 0 m 0, (117) l l 0 n k = [1/G T 4/E T + 4ν 2 A/E A ] 1 l = 2kν A n = E A + 4kν 2 A = E A + l 2 /k m = G T. Phase constitutive equations - assume constant phase moduli σ r (x) = L r ε r (x), ε r (x) = M r σ r (x), r = f, m. (118) Mechanical loading problem u 0 (x) = E x x S (119) p 0 (x) = Σ n(x) x S (120) u 0 and p 0 are the displacement and traction vectors at the external boundary S of a representative volume element of the composite; n is the outer unit normal to S; E and Σ are the applied macroscopic uniform strain and stress fields, respectively.

99 Macroscopic constitutive relations σ(x) = L(x)ε(x) = ε(x) = M(x)σ(x) = 2 c r L r ε r (x) = L E (121) r=1 2 c r M r σ r (x) = M Σ (122) stands for the spatial average of a given field, c r is the volume fraction of the r th phase, and L and M are the effective stiffness and compliance matrices of the heterogenous material, respectively. Eqs. (121) and (122) follow directly from Hill s lemma [Hill, 1963]. Hill s lemma For compatible strain and equilibrated stress fields the following relation holds ε(x) T σ(x) = ε(x) T σ(x), (123) r=1 and consequently E T L E = ε(x) T L(x)ε(x), (124) Σ T M Σ = σ(x) T M(x)σ(x). (125) Eq. (123) states in fact that the average of microscopic internal work is equal to the macroscopic work done by internal forces.

100 Michel, Moulinec, Suquet (1998) OVERALL MODULI - continue Here we present two specific procedures for evaluation of overall moduli of periodic microstructures. 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: =0 ( ) {}}{ 1 σ ij d : δe ij + σ ij n j δu i ds = Σ ij : δe ij. An alternate formula for overall average stress Σ can be deduced for the composites containing rigid inclusions where σ is not defined. Assuming a self-equilibrated stress field (σ ij,j = 0) we write σ ij = (σ ik x j ),k. Then writing the volume average of σ ij yields Σ ij = σ ij d = σ ik x j n k ds. (127)

101 NUMERICAL RESOLUTION OF THE PROBLEM BY THE FINITE ELEMENT METHOD Here we present a numerical solution of certain boundary value problem using the Finite Element Method (FEM). The UC is discretized into finite elements and subjected to either uniform macroscopic states of strain, E, or stress, Σ together with prescribed periodic boundary conditions. Substitute the constitutive law PRESCRIBED OVERALL STRAIN 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 Hence σδε = Σ δε = 0. (129) L(x) (E + ε (x)) δε (x) = 0. (130) Solving the above relation calls for a suitable numerical technique such as the Finite Element 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 unknown degrees of freedom.

102 Strain field ε(x) = E + B(x)r. (132) Introducing Eq. (132) into Eq. (130) gives ( ) ( ) δr T B T (x)l(x)b(x) d r + δr T B T (x)l(x) d E = 0. (133) Linear system of governing equations For any kinematically admissible displacements δr and for the prescribed strain E Eq. (133) yields Kr = f, (134) where K = e f = e K e where K e = 1 f e where f e = 1 B T L e B da e A e A e B T L e E da e. (135) K is the stiffness matrix of the system and f is the vector of global nodal forces resulting from the loading by E; e stands for the number of elements, A e is the area of element e and is the area of the PUC. In Eq. (134), K represents the problem stiffness matrix and f is the vector of nodal forces due to material heterogeneity. Note that the work done by external forces is apriory zero.

103 System (134) can be used to provide the coefficients of the effective stiffness matrix L as volume averages of the local fields derived from the solution of four successive elasticity problems. To that end, the periodic unit cell is loaded, in turn, by each of the four components of E, while the other three components vanish. The volume stress averages normalized with respect to E then furnish individual columns of L. The required periodicity conditions (same displacements u on opposite sides of the unit cell) is accounted for through multi-point constraints. PRESCRIBED OVERALL STRESS Suppose that the surface tractions compatible with the macroscopic uniform state of stress Σ are prescribed. Such a loading conditions leaves us with unknown overall strain E and periodic displacement field u to be determined. Substituting the microscopic constitutive equation into Eq. (126) results in the following expression σ(x) = L(x)ε (u(x)) (136) σε(δu) = ε(x)lε(δu) = ΣδE. (137)

104 Substituting Eq. (112) into Eq. (137) yields δe T L(x) (E + ε (x)) + δε (x) T L(x)E + δε (x) T L(x)ε (x) = δe T Σ. (138) Since δe and δε (x) are independent, the preceding equation can be split into two equalities δe T Σ = δe T [ L(x) E + L(x)ε (x) ] (139) 0 = δε (x) T L(x) E + δε (x) T L(x)ε (x) In the FE approach the matrix B, relating strains and displacements in the form ε = Br and consequently, δε = Bδr is applied to Eq. (139) to get Linear system of governing equations 1 1 { L d LB d 1 B T L d 1 B T LB d E r } = { Σ 0 }. (140) The above system of equations serves to derive the coefficients of the effective compliance 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 entries of M.

105 GENERALIZED PLANE STRAIN Consider a unidirectional fiber reinforced composite medium. Such a medium is invariant under translation along the axial direction (invariance of geometrical and material properties). 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 to unity. If we admit the axial direction as a direction of orthotropy (all phases are assumed to orthotropic) and if no out-of-plane shear stresses or strains are prescribed, then the 3D 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 = {Σ 11, Σ 22, Σ 12, Σ 33 } T. Then, because of translational invariance in the fiber direction, the displacement field solution of the three-dimensional local problem corresponds to a generalized plane state u α = u α (x 1, x 2 ) α = 1, 2 u 3 = E 33 x 3. The corresponding macroscopic strain vector is given by E = {E 11, E 22, 2E 12, E 33 } T. For a given prescribed boundary conditions, the solution of the local problem then boils

106 down to finding the following unknown local fields ε = ε 11 ε 22 2ε 12 0, ε = ε 11 ε 22 2ε 12 E 33, σ = σ 11 σ 22 σ 12 σ 33.

107 Elimination PERIODIC BOUNDARY CONDITIONS Here we present a general way of applying the periodic boundary conditions for displacement field u. You can imagine the periodicity conditions as constraints of the type Pr = 0, P = [0, I, I], r = {r 1, r 2, r 3 } T, (141) where r 1 denotes the unknowns corresponding to interior nodes, whereas r 2 and r 3 denote unknowns corresponding to nodes located on the boundary of the unit cell. The discrete formulation of Eq. (134) can be expanded into: K 11 K 12 K 13 {δr 1, δr 2, δr 3 } K T 12 K 22 K 23 K T 13 K T 23 K 33 r 1 r 2 r 3 = {δr 1, δr 2, δr 3 } f 1 f 2 f 3 for every δr such that δr 2 = δr 3. Since r 2 = r 3 we get [ ] { } { } K 11 K 12 + K 13 K T 12 + K T 13 K 22 + K 23 + K T 23 + K 33 r 1 r 2 = f 1 f 2 + f 3 (142) Note that this procedure is equivalent to assigning the same code numbers to corresponding degrees of freedom on opposite sides of the UC.

108 NUMERICAL RESULTS Material properties of T30/Epoxy system phase E A E T G T ν A [GPa] [GPa] [GPa] fiber matrix Components of the effective stiffness matrix [GPa] Unit cell L 11 L 22 L 33 L 44 c f Original fibers PUC fibers PUC fibers PUC Hexagonal array

109 NUMERICAL RESULTS Variation of effective stiffnesses for five ten-particle optimal PUC Modulus Mean value Standard deviation Variation coefficient [GPa] [GPa] [%] L L L Variation of effective stiffnesses for five randomly picked ten-particle PUC Modulus Mean value Standard deviation Variation coefficient [GPa] [GPa] [%] L L L

110 THERMOMECHANICAL PROBLEM Suppose that in addition to the mechanical loading the representative volume is subjected 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 σ r (x) = L r ε r (x) + λ r, ε r (x) = M r σ r (x) + µ r, r = f, m. (144) Local phase eigenstrains µ r = m r θ and eigenstresses λ r µ r = M r λ r, λ r = L r µ r, r = f, m. (145) The thermal strain vector m r lists the coefficients of thermal expansion of the phase r. Hill s lemma - recall Eq. (126) δε(x) T σ(x) = δε(x) T L(x) (ε(x) µ(x)) = δe T Σ. (146)

111 Linear system of governing equations - recall Eq. (140) 1 1 L d B T L d 1 1 { LB d B T LB d E r } Σ + 1 Lµ d = 1 B T Lµ d. (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 to unity, the components of the overall average strain comply with the effective coefficients of thermal expansion m. Note that the present formulation is not applicable with the strain control conditions when admitting the thermal loading. Clearly, the overall strain E is then not known and cannot be prescribed.

112 NUMERICAL RESULTS Material properties of T30/Epoxy system phase E A E T G T ν A α A α T [GPa] [GPa] [GPa] [K 1 ] [K 1 ] fiber matrix Components of the effective thermal expansion coefficients [K 1 ] Unit cell α x α y α A c f Original fibers PUC fibers PUC fibers PUC Hexagonal array Mori-Tanaka

113 EXAMPLE - HOMOGENIZATION OF MASONRY STRUCTURES To introduce the subject we consider a view of a typical bridge arc showing a regular arrangement of stones in the masonry together with an existing crack running both across a stone and along a head joint. The main objectives are: determination of the homogenized thermoelastic material properties, fracture energies and construction of failure envelopes as a design aid for a quick assessment of the load-bearing capacity of homogenized medium using a suitable commercial software such as ATENA implementation of periodic boundary condition and introduction of uniform macroscopic strain and stress fields via nodal displacements and nodal forces, respectively

114 RVE - Periodic fields Consider a representative volume element (RVE) constructed by taking the smallest segment cut from a masonry that could be periodically repeated to cover the whole structure (periodic unit cell (PUC)), and subjected to an overall strain E. In particular, the geometry of PUC is fully described by specifying the width and height of basic blocks, the thickness of head and bead joints and the type of a bond of a given structure. Such information can be easily obtained, e.g., by in-situ measurements or analysis of digital photographs of a structure. Displacement and strain field - review u(x) = E x + u (x) (148) ε(x) = E + ε(u (x)) = E + ε (x) (149) where u is a Y -periodic part of the displacement field due to microstructure periodicity