Lecture on Application to Composite Fracture Mechanics

<strong>Lecture</strong> onApplication to Composite Fracture MechanicsProf. F. G. YuanMars Mission Research CenterDepartment of Mechanical and Aerospace EngineeringNorth Carolina State UniversityRaleigh, NC 27695July 28-31, 1998Mechanics of Materials BranchNASA Langley Research CenterHampton, VA 23681

Anisotropic Elasticity: Application to Composite Fracture MechanicsTable of ContentsChapter1. Constitutive Law and Transformation of Axes2. The Lekhnitskii Formalism3. The Stroh Formalism4. Determination of Stress Coefficient Terms in Cracked Solids forMonoclinic Materials with Plane Symmetry at x 3 = 0N.C. State Univ., Raleigh, NC F. G. Yuan, July 28-31, 1998

Anisotropic Elasticity: Application to Composite Fracture MechanicsChapter 1 Constitutive Law and Transformation of Axes1.1 Constitutive law in terms of stress and strain tensors1.2 Contracted matrix notation of constitutive law and stress and engineering strain components1.3 Some properties of Q1.4 Laws of transformation for C and s1.5 Rotation between the x 3 - axis and invariants1.6 Elastic symmetry1.7 Properties of reduced elastic and compliance constants s’1.8 Problems1.9 ReferencesAppendix A: Matrices1.1 Constitutive law in terms of stress and strain tensorsReferring to a fixed rectangular coordinate system x i , the stress tensor, σij, and straintensor, eij, for a homogeneous linearly anisotropic elastic material are related through stressstrainlaw byσ = C e , (1.1)ijijklklwhere i, j, k, l = 1, 2, or 3. The repeated indices imply summation andstiffness tensors.Cijklare the elasticThe inverse of (1.1) is written ase= σ , (1.2)ijs ijklklwhere s ijklare the elastic compliance tensors.Eq. (1.1) is the general form of the linear constitutive relation for an elastic solid and is knownas the generalized Hooke’s law. As the equations (1.1) and (1.2) indicate, these are 3 4 = 81stiffnesses and compliances. Based on the symmetry of stress and strain tensors, the elasticstiffness tensors have the following symmetry propertiesC = C = C(1.3)ijkljiklThe equalities on the first two and the last two suffixes reduce the independent stiffnesses andcompliances to 36.If a strain energy density function W exists which is defined byWand it is independent of the path e ij , then=epq∫ ijdeij=0∫epqijlkσ C e de(1.4)0ijklklijN.C. State Univ., Raleigh, NC 1-1 F. G. Yuan, July 28-31, 1998

Anisotropic Elasticity: Application to Composite Fracture Mechanics∂WdW = deij= C∂eijijklekldeij∂W∂eij= Cijklekl=σij∂ 2 W = Cijkl(1.5)∂e∂eThe differentiations on the right sides of (1.4) are interchangeable. ThusijklCijkl= C klij(1.6)Same relations hold for s ijkl . The relation in (1.6) further reduces the number to 21 in the mostgeneral case.Suppose there is a point P in a Cartesian coordinate system Ox1, Ox2, Ox3, thecoordinates of P with respect to the system being x1, x2, x3. The numbers x1, x2, x3thatrepresent the coordinates of the point P in Fig. 1.1 are also the components of the vector OP .Then on transforming to a second Cartesian system O x1 , Ox2, Ox3shown in Fig. 1,x 3_x3 P_x 2ox 2_x 1x 1Fig. 1.1 Rotation between two coordinate systemsthe coordinates of P in the new coordinate system xi, related to x i byxi= Ωijxjor x = Ω x(1.7)where Ωij= cos( xiOxj) , i.e., Ωijare the direction cosines of the axes. Ω is an orthogonalmatrix.Ω Ω = Ω Ω = δ(1.8)where δ ik is the Kronecker delta,or in matrix notation:ijikjikijkN.C. State Univ., Raleigh, NC 1-2 F. G. Yuan, July 28-31, 1998

Anisotropic Elasticity: Application to Composite Fracture MechanicsTTΩ Ω = Ω Ω = I ,−1whence Ω T = Ω .The transformed stresses, strains and stiffnesses following tensor transformation are given byσ = Ω Ω σ(1.9)An equation similar to (1.11) can be written for s ijkl.Cijkleijijipipjqjqpq= Ω Ω e(1.10)ipjqkrpq= Ω Ω Ω Ω C(1.11)1.2 Contracted matrix notation of constitutive law and stress and engineering straincomponentsIt is obvious from the fact that (1.1) or (1.2) is a set of linear homogeneous equationsrelating the six independent stress components to the six independent strain components. Thestress and strain components are frequently written in contracted notation of σ and ε in acolumn formΤ Τσ = σ , σ , σ , σ , σ , ) , ε = e , e , e , 2e, 2e, 2 ) (1.12)(11 22 33 23 13σ12lspqrs(11 22 33 23 13e12The superscript T stands for the transpose. Note that the occurrence of the factor 2 in theabove strain expression defines ε i as engineering strains. It is also worth noting that theengineering strains do not form the components of a tensor as eij. Eqs (1.1) and (1.2) can beexpressed in matrix notation asσp= Cpqεqor σ = C ε(1.13)ε = or ε = sσ(1.14)pspqσqwhere p, q = 1, 2, 3,···, 6 and Cs = sC = I. Note that, by comparing eqs (1.13) and (1.14) with(1.1) and (1.2), the strains in the contracted notation are engineering strains. In the s ijkland theC ijkl the first two suffixes are abbreviated into a single one running from 1 to 6, and the lasttwo are abbreviated in the same way, according to the scheme:tensor notationmatrix notation11122233323,32431,13512,216Besides the brevity in replacing two suffices (i,j) and (k,l) by single suffices p and q, thisexpression has an advantage that it enables the contracted variables, C pq or s pq , to beconsidered as the elements of a matrix, but has the disadvantage that the contracted variablesare not tensors so cannot be manipulated by the rule of tensor transformation. In the sequel,either four suffix notation C ijkl or s ijkl or two suffix notation C pq or s pq will be used according towhich is the more convenient in each case.In the following, the relations between the C ijkl, s ijkl and C pq , s pq are derived. If weexpand (1.1) for i = j = 1, we getN.C. State Univ., Raleigh, NC 1-3 F. G. Yuan, July 28-31, 1998

Anisotropic Elasticity: Application to Composite Fracture Mechanicsσ = C1+ C= C111111231111eee112311+ C+ C11121131ee11121231+ C12+ C11131132ee13321113+ C+ C131121113321331122221122+ 2Ce + 2Ce + C e + 2Ce + C eee+ C+e221123+23113333(1.15)Introducing the contracted stress and strain notation in (1.12), we have1 11σ1= C1111ε1+ 2C1112ε6+ 2C1113ε5+ C1122ε2+ 2C1123ε4+ C1113ε3(1.16)2 22Comparing with the expanded form of (1.13)σ = (1.17)1C11ε1+ C12ε2+ C13ε3+ C14ε4+ C15ε5+ C16ε6we find C1111= C11, C1122= C12, , C1112= C16, (1.18)and by similar expansion and comparison for other values of i, j, it can be shown thatCijkl= C pq(1.19)On the other hand, expanding (1.2) for i = j = 1 and contracting by (1.12), we getε = s + 2s+ s + s + s σ + s σ(1.20)1 1111σ1 1112σ621113σ5 1122σ22Comparing this with the expanding form of (1.14), we find thats1111= s 11,1122s12112341133s = , , 2s 1112= s16(1.21)By similar expansion and comparison for other values i, j, it can be shown that3sijkl2s4sijklijkl= spq= s= spqpqif both p and q ≤ 3if either p or q ≤ 3if both p and q > 3(1.22)From (1.19) and (1.22), it is clearly seen that C and s are symmetric matrices.The strain energy density function W in (1.4) can be rewritten asW1212= C pqεpεqor W = s pqσpσqThe positive definiteness of W impliesW11C pqεpεq= spqσpσ > 0(1.23)22=qprovided not all theε pare zero. This indicates that matrices C pq and s pq are both positivedefinite. From (1.23), the determinant Cpq, spq, and their principal minors of all orders arepositive.N.C. State Univ., Raleigh, NC 1-4 F. G. Yuan, July 28-31, 1998

Anisotropic Elasticity: Application to Composite Fracture MechanicsIn the new coordinate system xi, we haveσ = C ε(1.24)iiijijjjε = s σ(1.25)Using the contracted notation of (1.12a), (1.9) can be written in matrix formσ = Qσ(1.26)where Q is the 6 × 6 transformation matrix (not symmetric) which can be subdivided into four3 × 3 matrices as⎡K12K2⎤Q = ⎢ ⎥(1.27)⎣K3K4 ⎦The expressions of the four sub-matrices are2 2 2⎡Ω11Ω12Ω13⎤⎢ 2 2 2 ⎥K1= ⎢Ω21Ω22Ω23⎥(1.28a)2 2 2⎢⎥⎣Ω31Ω32Ω33⎦⎡Ω12Ω13Ω13Ω11Ω11Ω12⎤K⎢Ω Ω Ω Ω Ω Ω⎥2=⎢22 23 23 21 21 22(1.28b)⎥⎢⎣Ω32Ω33Ω33Ω31Ω31Ω32⎥⎦⎡Ω21Ω31Ω22Ω32Ω23Ω33⎤K⎢Ω Ω Ω Ω Ω Ω⎥3=⎢31 11 32 12 33 13(1.28c)⎥⎢⎣Ω11Ω21Ω12Ω22Ω13Ω23⎥⎦⎡Ω22Ω33+ Ω23Ω32Ω23Ω31+ Ω21Ω33Ω21Ω32+ Ω22Ω31⎤K⎢Ω Ω + Ω Ω Ω Ω + Ω Ω Ω Ω + Ω Ω⎥4=⎢32 13 33 12 33 11 31 13 31 12 32 11(1.28d)⎥⎢⎣Ω12Ω23+ Ω13Ω22Ω13Ω21+ Ω11Ω23Ω11Ω22+ Ω12Ω21⎥⎦Notice that the elements of K 1 , K 2 , K 3 , K 4 can be written in compact form as( ) 2K1 ij= Ωij( K2) = Ω Ω , i not summed, j ≠ k ≠ p ,ijijikrjip( K 3) = Ω Ω , j not summed, i ≠ r ≠ s , (1.29)ijrksj( K 4) = Ω Ω + Ω Ω , j ≠ k ≠ p , i ≠ r ≠s .spSince Ω ij is an orthogonal matrix and hence, by (1.8),rpskN.C. State Univ., Raleigh, NC 1-5 F. G. Yuan, July 28-31, 1998

Anisotropic Elasticity: Application to Composite Fracture Mechanics3∑i=1( K32)ij0 = ∑j=1= ( K3)ij3∑i=131)ij1 = ∑j=1( K = ( K )(1.30)For the transformation of ε, (1.26) cannot be used simply because the definition of ε isdifferent from that of σ. However, if we define1ije T = e , e , e , e , e , )(1.31)(11 22 33 23 31e12we then have the same stress transformation matrix for ee = Q e(1.32)We generalize the 3 × 3 matrix introduced by Reuter to the following 6 × 6 matrix⎡I0 ⎤R = ⎢ ⎥(1.33)⎣02I⎦in which I is the 3 × 3 unit matrix. We haveε = R e , ε = R e(1.34)and (1.32) becomes−1ε = R Q R ε(1.35)A more compact form of the transformation for strain components (eq. (1.35)) will be shownin eq. (1.42). It should be pointed out that the dissimilar definitions of σ and ε given by (1.12)have one advantage. With the symmetry properties of C ijkl given by (1.3), the 6 × 6 matrices Cand s are symmetric. If we had used the same definition for the column matrices σ and ε, thenC and s would not be symmetric.1.3 Some properties of QThe inverse of (1.26) is−1σ = Q σ(1.36)−1PhysicallyQalso characterizes a negative rotation of axes. Instead of actually performing theinversion of Q given by (1.27), we write the inverse of the tensor transformation (1.9) asσ = Ω Ω σ(1.37)and rewrite this in the form of (1.36) using (1.12). In this way we obtainijN.C. State Univ., Raleigh, NC 1-6 F. G. Yuan, July 28-31, 1998piT T-1= ⎡K12K3⎤Q ⎢ T ⎥(1.38)⎣KT2K4 ⎦It can be seen that Q -1 is the transpose of Q defined in (1.27) except that the factor 2 remainsin the same place.qjpq

Anisotropic Elasticity: Application to Composite Fracture Mechanicsor crack axes not coinciding with principal material axes, material properties with respect tox 1 , x 2 are required for modeling the crack behavior. Let θ be the angle of rotation about thex 3 - axis shown in Fig. 2,_x 2x 2_x 3 (x 3 )θx 1_x 1then (1.7) givesFig. 1.2 Rotation counterclockwise by θ angle with respect to x 3 - axiswhere m = cos θ, n = sin θ.The transformation matrix Q given by (1.28) becomesand from (1.38),⎡ m n 0⎤Ω⎢ ⎥ij= − n m 0(1.48)⎢ ⎥⎢⎣0 0 1⎥⎦2 2⎡ m n 0 0 0 2mn⎤⎢ 2 2⎥⎢n m 0 0 0 − 2mn⎥⎢ 0 0 1 0 0 0 ⎥Q ( θ ) = ⎢⎥(1.49)⎢ 0 0 0 m − n 0 ⎥⎢ 0 0 0 n m 0 ⎥⎢2 2⎥⎢⎣− mn mn 0 0 0 m − n ⎥⎦2 2⎡mn 0 0 0 − 2mn⎤⎢ 2 2⎥⎢ n m 0 0 0 2mn⎥⎢⎥-10 0 1 0 0 0Q ( θ ) = Q(−θ) = ⎢⎥ (1.50)⎢ 0 0 0 m n 0 ⎥⎢ 0 0 0 − n m 0 ⎥⎢⎥2 2⎢⎣mn − mn 0 0 0 m − n ⎥⎦From (1.44) and (1.45) or (1.47), the transformed elastic and compliance constants can beexpressed in the following tablesN.C. State Univ., Raleigh, NC 1-8 F. G. Yuan, July 28-31, 1998

Anisotropic Elasticity: Application to Composite Fracture Mechanicss11 ss1216s22s26 s66( C 11) ( C 12) ( 2C16) ( C22) ( 2C26) ( 4C66)s 11( C 11) m 4 2m 2 n 2 2m 3 n n 4 2mn 3 m 2 n 2s 12( C 12) m 2 n 2 m 4 +n 4 mn 3 -m 3 n m 2 n 2 m 3 n-mn 3 -m 2 n 2s 2 ) -2m 3 n 2(m 3 n-mn 3 ) m 4 -3m 2 n 2 2mn 3 3m 2 n 2 -n 4 m 3 n-mn 316( C16s (22) n 4 2m 2 n 2 -2mn 3 m 4 -2m 3 n m 2 n 222C26( C26s 2 ) -2mn 3 2(mn 3 -m 3 n) 3m 2 n 2 -n 4 2m 3 n m 4 -3m 2 n 2 mn 3 -m 3 ns 4 ) 4m 2 n 2 -8m 2 n 2 4(mn 3 -m 3 n) 4m 2 n 2 4(m 3 n-mn 3 ) (m 2 -n 2 ) 266( C66s13( C13)s23( C23)s2C36(36s 13( C 13) m 2 n 2 mns23( C23)n 2 m 2 -mns 2 ) -2mn 2mn m 2 -n 236( C36)s44( C44)s45( C45)s55( C55)s44( C44 ) m 2 -2mn n 2s45( C45)mn m 2 -n 2 -mns55( C55)n 2 2mn m 2s34( C34)s35( C35)s34( C34 ) m -ns35( C35)n ms14( C14)s15( C15)s24( C24)s25( C25)s46(2C 46)s56( 2C56s ( 14) m 3 -m 2 n mn 2 -n 3 m 2 n -mn 214Cs 15( C 15) m 2 n m 3 n 3 mn 2 mn 2 m 2 ns24( C24 ) mn 2 -n 3 m 3 -m 2 n -m 2 n mn 2s25( C25)n 3 mn 2 m 2 n m 3 -mn 2 -m 2 ns 2 ) -2m 2 n 2mn 2 2m 2 n -2mn 2 m 3 -mn 2 n 3 -m 2 n46( C46s 2 ) -2mn 2 -2m 2 n 2mn 2 2m 2 n m 2 n-n 3 m 3 -mn 256( C56)N.C. State Univ., Raleigh, NC 1-9 F. G. Yuan, July 28-31, 1998

Anisotropic Elasticity: Application to Composite Fracture Mechanicss33( C33)= s33(C33)With the above equations, a number of invariants of the elastic constants can beconstructed. For example,IIII1234= s= s11= s= s664413+ s22− 4s+ s+ s5523+ 2s12= s= s441312= s66= s552311− 4s+ s+ s+ s1222+ 2s12(1.51)2 2 2 2I5= s34+ s35= s34+ s35are all invariant with respect to rotation about the x 3- axis. Corresponding invariants exist forrotation about the other axis, for example,I ′1= s11+ s33+ 2s13= s11+ s33+ 2s13(1.52)I1′′= s22+ s33+ 2s23= s22+ s33+ 2s23are invariant for rotation about the x 2and x 1axes respectively. The cubic compressibilityβ = s11+ s22+ s33+ 2(s12+ s13+ s23)(1.53)is invariant for all rotations.In the next chapters, several important anisotropic parameters are related between therotated and original coordinates:Eigenvalue:µ =kµkcosθ− sinθµ sinθ+ cosθkand furthermore, the Stroh eigenvectors A and B transform like a vector, namelyAij= ΩikAkj, Bij= ΩikBkj1.6 Elastic symmetryA body that permits certain transformations of reference axes without change inproperties along these axes is said to possess certain types of symmetry. If the body allowsreflection in a plane, this plane is called a plane of symmetry. If its properties are independentof rotation about an axis, this axis is an axis of symmetry. Further, when symmetries are usedin structural mechanics, three factors have to be taken into consideration: geometry, loading,and material. In terms of material symmetry in linear materials, the symmetry is generallycalled elastic symmetry.If the x i axes are referred to Cartesian coordinates (x, y, z) or ( x1, x2, x3), the materialanisotropy is called rectilinear anisotropy, Similarly, the axes are expressed by cylindrical (r, θ,z) or spherical coordinates (ρ, θ, ϕ), the material possesses cylindrical anisotropy or sphericalanisotropy respectively.N.C. State Univ., Raleigh, NC 1-10 F. G. Yuan, July 28-31, 1998

Anisotropic Elasticity: Application to Composite Fracture Mechanics⎡s11⎢s⎢12⎢s13⎢⎢0⎢ 0⎢⎣ 0sss122223000sss132333000s0004400s00005500 ⎤0⎥⎥0 ⎥⎥0⎥0 ⎥⎥s66⎦(1.62)In general, the engineering constants are defined based on the above equation in theorthogonal coordinate. The coordinate which exhibits three planes of elastic symmetry is oftencalled material principal axes. If the components of stress and strain are defined in terms ofcoordinate axes (say, structural axes) other than material principal axes, the transformationequations corresponding to (1.62) will change the form back to that of (1.61), indicating thata shear stress will produce a normal component of strain, and vice versa.are:The compliance constants in terms of engineering constants for orthotropic material(1) Cartesian coordinate:⎡ 1⎢ E1⎢⎢ ν−⎢ E⎢ ν⎢−⎢ E⎢⎢0⎢⎢ 0⎢⎢⎢ 0⎣121131ν−E1E2ν−E000212232ν−Eν−E1E32300031330001G230000001G130000001G12⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦(1.63)(2) Cylindrical coordinate:N.C. State Univ., Raleigh, NC 1-13 F. G. Yuan, July 28-31, 1998

Anisotropic Elasticity: Application to Composite Fracture Mechanics⎡ 1 νθrνzr⎤⎢− − 0 0 0E⎥rEθEz⎢⎥⎢ νrθ1 νzθ−− 0 0 0 ⎥⎢ ErEθEz⎥⎢ ν ν⎥rz θz1⎢−−0 0 0 ⎥⎢ ErEθEz⎥(1.64)⎢1⎥⎢0 0 0 0 0G⎥θz⎢⎥⎢10 0 0 0 0 ⎥⎢Grz⎥⎢1 ⎥⎢ 0 0 0 0 0 ⎥⎣Grθ ⎦(2) Spherical coordinate:⎡ 1 νθρνϕρ⎤⎢ − − 0 0 0E⎥⎢ρEθEϕ⎥⎢ νρθ 1 νϕθ⎥⎢−− 0 0 0E⎥ρEθEϕ⎢⎥⎢ νρϕνθϕ 1− −0 0 0⎥⎢ E⎥ρEθEϕ⎢⎥(1.65)⎢10 0 00 0 ⎥⎢Gθϕ⎥⎢1 ⎥⎢ 0 0 0 0 0 ⎥⎢Gρϕ⎥⎢1 ⎥⎢ 0 0 0 0 0 ⎥⎣Gρθ⎦νijνjiFrom symmetry of the matrix, = . The material independent constants reduce to 9.EiEj(c) A Plane of Isotropy (Transversely isotropic material)If the x 1 x 2- plane is isotropic, then⎡s11⎢s⎢12⎢s13⎢⎢0⎢ 0⎢⎣ 0The independent material constants reduce to 5.(d) Isotropysss121113000sss131333000s0004400s00005502( s110 ⎤0⎥⎥0 ⎥⎥0⎥0 ⎥⎥− s12) ⎦(1.66)N.C. State Univ., Raleigh, NC 1-14 F. G. Yuan, July 28-31, 1998

Anisotropic Elasticity: Application to Composite Fracture MechanicsThus1 0′)− .( s = CSince the strain energy density is1 T 1 ~ T ~ 1 ~ TW = σ ε = σ ε = σ s′~ σ > 0 ,2 2 2s′ is positive definite. In the above equation, the two-dimensional deofmation, ε 3 = 0 hasbeen assumed. From (1.73), C 0 is also positive definite.For monoclinic material with plane of elastic symmetry x 3 = 0, the C 0 and s′ can berearranged in the two separate matrix forms:C0⎡C⎢C⎢= ⎢C⎢⎢0⎢⎣0111216CCC12222600CCC16266600000CC44450 ⎤0⎥⎥ ⎡C10 ⎥ = ⎢⎥ ⎣ 0C45⎥C ⎥55⎦0 ⎤C⎥2 ⎦(1.74)⎡s′′ ′11s12s160 0 ⎤⎢s′′ ′ ⎥⎢12s22s260 0⎥ ⎡s1′0 ⎤s′= ⎢s′′ ′16s26s660 0 ⎥ = ⎢ ⎥⎢⎥ ⎣0s2′⎦⎢0 0 0 s44′ s45′⎥⎢⎣0 0 0 s′′ ⎥45s55⎦It can be easily proved that the subsets of elastic stiffnesses and subsets of reducedcompliances are the inverses of each otherC ′ ′1s1= s1C1= IC s′= s′C = I2222(1.75)(1.76)The subscripts 1 and 2 represent the in-plane and antiplane deformation respectively.For in-plane deformation, let C ) 1be the adjoint of C 1 so thatC 1C ) 1= J I ,It follows from (1.76) 1 that)C1= J s 1′~From (1.77) and (1.78) the components of C1areJ = detC= s′(1.77)−11(det1)(1.78)N.C. State Univ., Raleigh, NC 1-16 F. G. Yuan, July 28-31, 1998

Anisotropic Elasticity: Application to Composite Fracture Mechanics)C)C)C)C)C)C112266121626= C= C= C= C= C= C221111161212CCCCCC666622262616− C− C− C− C− C− C)Similarly, if s′1 is the adjoint of s′1, we haves ′ s) ′ = JComparing with (1.76) 2 ,orCCCCCC11226612162611)′ J226216212121611= J s′CCC−166222611= J s′22= J s′I−1s 1= C1= ( s′22s′= ( s′s′11= ( s′s′11= ( s′s′16= ( s′s′12= ( s′s′12666622262616− s′226− s′216− s′212) J) J) J− s′s′12− s′s′16− s′s′1166= J s′12= J s′16= J s′662226) J) J) J26(1.79)(1.80)(1.81)Eq. (1.78) provides s′ 1in terms of the components of C 1 while (1.80) gives C 1 in terms ofcomponents of s′1. Using (1.79) the expression γ(µ) in terms of C’s can be transformed intoreduced compliance constantsγ ( µ ) = ( C16C= J ( s′1226− C) − ( C2− s′µ + s′µ )1612C661112C26− C16C22) µ + ( C22C66− C2262) µ(1.82)From the identity in the Lekhnitskii formulation432s ′ µ − 2s′16µ+ (2s′′12+ s66)µ − 2s′′26µ+ s2211=shown in the next chapter and (1.82), it can be readily proven that the relation betweenreduced compliances and elastic stiffnesses0( s′122− s′µ + s′µ ) γ ( µ ) = J ( s′16Similarly,1112= J[(s′= −(C2− s′µ + s′µ )2166616− s′s′11+ 2C2666112) µ + 2( s′s′µ + C222112µ )26− s′s′−1′22− s′26+ s′2µ) γ ( µ ) = C16+ ( C12+ C66)1216) µ + ( s′2( s µ µ + C µ21226− s′s′1122)](1.83)(1.84)N.C. State Univ., Raleigh, NC 1-17 F. G. Yuan, July 28-31, 1998

Anisotropic Elasticity: Application to Composite Fracture MechanicsFrom Lekhnitskii’s formulation, (1.83) and (1.84) are written asorand2p = −(C + 2Cµ + C µ ) / γ ( µ )(1.85)k66 26 k 22 k k22pk = −(C + 2Cµk+ C µk) / J6626222q = [ C + ( C + C ) µ + C µ ]/ γ ( µ )(1.86)k16 12 66 k 26 k kwhere p k and q k are the displacement coefficients in the Lekhnitskii formulation.1.8 Problems1. For a thick composite coupon shown below, derive the matrices Cijand sijin terms of x 1and x 2 , angle α, and engineering material constants._x (x 3 3)_x 2x 1x 2Fiber Orientationα_x 1Fig. 3 A composite coupon with off-axis angle αAns:a. Determine orthotropic compliance constants sijfrom engineering constants w.r.t.material principal axes xi;b. Using tables to determine transformed compliance constants s ij with rotation angle –α;c. For thick composite specimens, calculate the reduced compliance constants s′ij.−12. Prove Q using linear algebra.3. Are the following terms are invariant ?2 2 1 2s13+ s23+ s3622 2 2 2 2 1 2s11+ s22+ 2s12+ s16+ s26+ s6644. If the body has elastic symmetry for θ = 60 o , which of compliance terms are zero?N.C. State Univ., Raleigh, NC 1-18 F. G. Yuan, July 28-31, 1998

1.9 ReferencesAnisotropic Elasticity: Application to Composite Fracture Mechanics1. R. F. S. Hearmon, An Introduction to Applied Anisotropic Elasticity, Oxford UniversityPress, Great Britain, 1961.2. J. F. Nye, F. R. S., Physical Properties of Crystals, Clarendon Press, Oxford, 1985.3. S. G. Lekhnitskii, Theory of Elasticity of an Anisotropic Body, Mir Publishers, Moscow,1981.4. R. C. Reuter, "Concise Property Transformation Relations for an Anisotropic Lamina",Journal of Composite Materials, Vol. 5, pp. 270-271, 1971.5. E. Prince, Mathematical Techniques in Crystallography and Material Science, Springer-Verlag New York, Inc., 1982.5. A. Kelly and N. H. Macmillan, Strong Solids, Oxford Science Publications, 1986.6. T. C. T. Ting, “Effects of Change of Reference Coordinates on the Stress Analyses ofAnisotropic Elastic Materials”, International Journal of Solids and Structures, Vol. 18,No. 2, pp. 139-152, 1982.7. T. C. T. Ting, “Invariants of Anisotropic Elastic Constants”, Quarterly Journal ofMechanics and Applied Mathematics, Vol. 40, Pt. 3, pp. 431-448. 1987.Appendix A: MatricesA number of matrices have their special characteristics. Some definitions are givenbelow:Matrix A is calledifrealA = AsymmetricA T = AantisymmetricHermitianA T= − AA T = AorthogonalT -1A = Aunitary-1 TA = Adiagonal A = 0 for i ≠ jidempotentA 2 = AijFor a nonsingular matrix A, its inverse is expressed by)A T−1A =ATwhere A ) is the adjoint matrix.A few relations involving products of square matrices are listed :N.C. State Univ., Raleigh, NC 1-19 F. G. Yuan, July 28-31, 1998

Anisotropic Elasticity: Application to Composite Fracture Mechanics( ABC )− 1= −1−1A −1C BT T T T( ABC ) = C B ATr ( AB)= Tr(BA)Tr A = ∑iA iidet( AB ) = (det A)(detB)= det( BA)conjudgate ( AB ) =AB11Re( A )Re( B)= Re[( A + A)B]= Re[ A(B + B)]22N.C. State Univ., Raleigh, NC 1-20 F. G. Yuan, July 28-31, 1998

Anisotropic Elasticity: Application to Composite Fracture MechanicsA composite with material properties:E L = 11.773 Msi, E T = 5.162 Msi, E Z = 1.53 MsiG LT = 2.479 Msi, G LZ = 0.64 Msi, G TZ = 0.57 Msiν LT = 0.401, ν LZ = 0.22, ν TZ = 0.29For θ = -α = -45 o ,sij=⎡1⎢⎢4⎢1⎢⎢4⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣⎡1−2νLT1 1 ⎤⎢ + + ⎥⎣ ELETGLT⎦⎡1−2νLT1 1 ⎤⎢ + − ⎥⎣ ELETGLT⎦1 ⎡νLTνTZ⎤− ⎢ +2⎥⎣ ELET⎦001 ⎡ 1 1 ⎤⎢ − ⎥2 ⎣ ELET⎦−1′s = C⎡ 0.151⎢− 0.05⎢= ⎢ 0⎢⎢0⎢⎣− 0.0521 ⎡1−2νLT1 1 ⎤⎢ + −4⎥⎣ ELETGLT⎦1 ⎡1−2νLT1 1 ⎤⎢ + +4⎥⎣ ELETGLT⎦1 ⎡νLTνTZ⎤− ⎢ +2⎥⎣ ELET⎦⎡8.352⎢3.395⎢C = ⎢ 0⎢⎢0⎢⎣1.781001 ⎡ 1 1 ⎤⎢ − ⎥2 ⎣ ELET⎦− 0.050.15100− 0.0523.3948.352001.7811 ⎡νLTνTZ⎤− ⎢ +2⎥⎣ ELET⎦1 ⎡νLTνTZ⎤− ⎢ +2⎥⎣ ELET⎦1Eν−ELZL00Z001.658ν+E− 0.0960000.6050.0350TZT0001 ⎡ 1 1 ⎤⎢ +2⎥⎣GTZGLZ⎦1 ⎡ 1 1 ⎤− ⎢ − ⎥2 ⎣GTZGLZ⎦000− 0.096000.0350.60501.65800001 ⎡ 1 1 ⎤− ⎢ −2⎥⎣GTZGLZ⎦1 ⎡ 1 1 ⎤⎢ + ⎥2 ⎣GTZGLZ⎦− 0.052⎤− 0.052⎥⎥0 ⎥ (1/Msi)⎥0⎥0.345 ⎥⎦1.781⎤1.781⎥⎥0 ⎥ Msi⎥0⎥3.441⎥⎦01 ⎡ 1⎢ −2 ⎣ EL1 ⎡ 1⎢ −2 ⎣ ELνLZ− +E1+2νELL00LT1 ⎤ ⎤⎥E⎥T ⎦ ⎥1 ⎤ ⎥⎥E⎥T ⎦ ⎥ν ⎥TZ⎥ET⎥⎥⎥⎥⎥⎥⎥1 ⎥+ ⎥ET⎦N.C. State Univ., Raleigh, NC 1-21 F. G. Yuan, July 28-31, 1998

Anisotropic Elasticity: Application to Composite Fracture Mechanics2.1 Basic equations2.2 General solutions for the in-plane deformation fieldChapter 2 The Lekhnitskii Formalism2.1 Basic equationsA two-dimensional deformation means that the displacements or stresses are functionsof x 1 and x 2 only. Beginning with stress and assuming that they depend on x 1 and x 2 only,Lekhnitskii was the first to formulate the anisotropic problem with stress functions similar tothe Airy’s stress function approach for isotropic solids. Under the framework of the stressfunction approach, the strain compatibility equations need to be satisfied. Therefore theLekhnitskii formalism is in terms of reduced compliance constants, while the Strohformulation (to be discussed in Ch. 3) is formulated in terms of elastic constants. The mainobjective of this chapter is to briefly derive the general solutions using stress functions F andΨ and to provide a link with Stroh formalsim.Since the stresses are independent of x 3 - axis, the equilibrium equations in therectangular Cartesian coordinates are∂σ11∂σ12∂σ12∂σ22∂σ13∂σ23+ = 0 , + = 0 , + = 0(2.1)∂x∂x∂x∂x∂x∂xor in a simple index form121∂σ i∂σ i∂x1 2+ =1∂x2Introducing potential functions φi22120 , (i = 1, 2, 3) (2.2)∂φi∂iσi1= − , σi2= φ (i = 1, 2, 3) (2.3)∂x∂xThe eq. (2.2) are automatically satisfied. Since σ 21 = σ 12 ,1∂φ∂φ+∂x∂x1 2=120(2.4)In order to satisfy eq. (2.4), an Airy stress function F is introduced∂Fφ1= − ,∂x2∂Fφ2=(2.5)∂x1Combining (2.3) and (2.5) and setting φ3= −Ψ, we obtainN.C. State Univ., Raleigh, NC 2-1 F. G. Yuan, July 28-31, 1998

Anisotropic Elasticity: Application to Composite Fracture Mechanicsσ112∂ F= ,∂xσ2213σ22∂Ψ= ,∂x22∂ F= ,∂xσ2123σ12∂Ψ= −∂x12∂ F= −∂x∂x12,(2.6)The two stress functions F and Ψ are the foundation for the Lekhnitskii formulation.Assuming u i are independent of x 3 , the strain components are∂uε1=∂x11ε,4∂uε2=∂x∂u=∂x32,22,∂uε5=∂x∂uε6=∂x3112∂u+∂x21(2.7)The strain compatibility equations for (2.7) are∂ ε ∂ ε+∂ ∂212x2222x12∂ ε6−∂x∂x12= 0,∂ε4∂x1∂ε5−∂x2= 0(2.8)For monoclinic materials with a plane of material symmetry at x 3 = 0, the constitutive lawreduces toε = s′σ (i, j = 1, 2, 4, 5, 6) (2.9)iijjwhere s′ ij= sij− s3 is3j/ s33(i, j = 1, 2, 4, 5, 6) are reduced elastic compliances and0 0s′ C = C s′= I from eq. (1.73). σ3is determined by σ3= −s3jσj/ s33(j = 1, 2, 4, 5, 6). Therelation (2.9) can be written in a more explicit form:⎧ε1⎫ ⎡s′⎪ε⎪ ⎢ ′⎪2s⎪ ⎢⎨ε4 ⎬ = ⎢ 0⎪ ⎪ ⎢ε5⎪ ⎪ ⎢0⎪⎩ε ⎪⎭⎢⎣′6s111216s′12s′2200s′2600s′44s′45000s′45s′550s16′ ⎤ ⎧σ1 ⎫s⎥ ⎪ ⎪26′ σ⎥ ⎪2⎪0 ⎥ ⎨σ4 ⎬⎥0 ⎪ ⎪⎥σ5⎪ ⎪s′⎥66 ⎦ ⎪⎩σ6⎪⎭(2.10)It is worth noting that for plane stress problems, σ 0 , reduced elastic constants0 0C′ = C − C (i, j = 1, 2, 4, 5, 6) and C′ s = s C′= I should be used.ijij3 iC3j/ C33Substituting (2.6) into (2.10), the compatibility equations (2.8) in terms of stress functions Fand Ψ lead to3=whereL F = and L Ψ 0(2.11)402=L∂∂4444= s22′ − 2s26′ + (2s4312′ + s66′ ) − 2s2 2 16′∂x1∂x1∂x2∂x1∂x2∂4∂∂x∂x132+ s′11∂∂x442N.C. State Univ., Raleigh, NC 2-2 F. G. Yuan, July 28-31, 1998

Anisotropic Elasticity: Application to Composite Fracture MechanicsL∂22= s44′ − 22∂x1s′452∂∂x∂x12+ s′552∂∂x22From eq. (2.11), it is clear the in-plane deformation and antiplane deformation aredecoupled.2.2 General solutions for the in-plane deformation fieldIn the following, only in-plane deformation case is considered. The constitutive law canbe written in a more compact form:⎧ε⎡ ′ ′ ′1 ⎫ s11s12s16⎤ ⎡σ1 ⎤⎪ ⎪ ⎢ ′ ′ ′ ⎥ ⎢ ⎥⎨ε2 ⎬ = s⎢12s22s26σ⎥ ⎢2(2.12)⎥⎪ ⎪⎩ε⎭ ⎢⎣′ ′ ′6s16s26s66⎥⎦⎢⎣σ12⎥⎦The general solution of F is expressed asF x , x ) = F()(2.13)(1 2zwhere z = x 1+ µ x2Substituting (2.13) into (2.11) 1 , eigenvalues µ can be determined by432s ′ µ − s′µ + (2s′+ s′) µ − 2s′µ + s′0(2.14)1121612 6626 22=Clearly, the four roots are dependent on the material properties only, independent of geometryof the solids and boundary conditions. The roots associated with (2.14) are arranged so thatµ 1 and µ 2 have positive imaginary parts and µ3= µ1and µ4= µ2.Similarly for anti-plane deformation, the eigenvalue can be obtained from L Ψ 0,s ′ ′ µ s′The function of F can be written as:255µ − 2s45+44=021, x2) = 2∑ReF kz k)k=1where zk = x 1+ µkx2. The stress components are[ ( ]2=F ( x(2.15)222[ µkFk′′( zk)],σ2= 2∑Re[ Fk′′( zk)],σ6= 2∑Re[ ( −µk) Fk′′( zk]21= 2∑Re)k=1k=1k=1σ (2.16)and the displacements are221= 2∑Re)k=1k=1[ p ′ ] = ∑ [ ′kFk( zk) , u22 Re qkFk( zk]u (2.17)2where pk = s′11µ k− s′16µk+ s′12, qk= s′12µk− s′26+ s′22/ µkIf the two functions φ1and φ2, instead of F, are used, from (2.5) we obtainN.C. State Univ., Raleigh, NC 2-3 F. G. Yuan, July 28-31, 1998

Anisotropic Elasticity: Application to Composite Fracture Mechanics221= 2∑Re ()k = 1k=1[ −µk) Fk′( zk)],φ2= 2∑Re[ Fk′( zk]φ (2.18)Assuming for each k, F′ k( z k) has the same function form, we letF ′k( zk) ≡ f ( zk) qk(2.19)Then the displacements, (2.17), and potential function φ , (2.18), can be expressed in a matrixformu = 2 Re[ A f q](2.20)φ = 2 Re[ B f q]wheref is a diagonal matrix defined as f = diag f ( z ), f ( )][1z2⎡ p1p2⎤A = [ a1 , a2] = ⎢ ⎥ , (2.21)⎣q1q2⎦⎡−µ1− µ2 ⎤B = [ b1, b2] = ⎢ ⎥ , (2.22)⎣ 1 1 ⎦B −11 ⎡−1− µ2⎤=µ −⎢ ⎥1µ2 ⎣ 1 µ(2.23)1 ⎦The simple form of the complex matrices A and B will be used to derive some importanttensors in the next chapter based on Stroh formulation.For a given problem, the unknowns in (2.20) remain to be solved are:(1) determine the complex-valued function form f(z) from loading (point force),dislocation, geometry (wedge, crack);(2) determine the complex vector q from the boundary conditions.N.C. State Univ., Raleigh, NC 2-4 F. G. Yuan, July 28-31, 1998

Anisotropic Elasticity: Application to Composite Fracture MechanicsChapter 3 The Stroh Formalism3.1 Stroh’s formalsim3.2 Eigenvalues µ3.3 The Sextic formalism of Stroh3.4 Normalization of A and B3.5 Explicit expressions of S, H, L3.6 Tractions, resultant forces and moments in terms of stress functions3.7 Example 1: Uniform stress solution3.8 Example 2: Green’s function of a point force and a dislocation3.9 Example 3: Green’s function of a point force applied at the crack tip3.10 Example 4: Energy release rate expression3.11 Example 5: A crack subjected to a uniform load at infinity3.12 Example 6: An elliptical hole subjected to uniform load at infinity3.13 Example 7: An elliptical hole subjected to bending load at infinity3.14 Problems3.15 ReferencesAppendix A: J-integralAppendix B: Energy release rate gIn the previous chapter, Lekhnitskii formulation can directly derive the eigenvalues µ k ,and A, B matrices in terms of compliance constants and present in a simpler expression.However, Stroh formulation for treating two-dimensional anisotropic elasticity problems forgeneral anisotropic solids in a rectangular coordinate system possesses several unique featuresover Lekhnitskii formulation:(1) The formulation starts with displacement formulation. Thus there is no need toconsider strain compatibility conditions;(2) The formulation utilizes the stress functions φ which have the same order as thedisplacement expression u;(3) Using matrix notation, the coefficients A and B associated with u and φ have richorthogonality and closure relations which can simplify the calculation significantlyand make results in a very compact form;(4) If the physical parameters can be related to Barnett-Lothe tensors, S, H, L, therelations can directly apply to degenerate cases such as isotropic solids;(5) The formulation is mathematically elegant and has been proved a very powerfultechnique in the two-dimensional deformation of anisotropic elasticity problems;(6) The formulation has potential in solving wave equations, bending in anisotropicplate, three-dimensional anisotropic elasticity problems, etc.3.1 Stroh’s FormalismIn a fixed Cartesian coordinate system x i, (i = 1, 2, 3), consider a two-dimensionaldeformation of an anisotropic linear elastic body in which the deformation field is independentof the x 3coordinate. This class of the problems was first introduced by Eshelby et al., (1953).The stresses σijare related to displacements u i by the generalized Hooke’s lawij= Cijkluk, lσ (3.1)N.C. State Univ., Raleigh, NC 3-1 F. G. Yuan, July 28-31, 1998

Anisotropic Elasticity: Application to Composite Fracture Mechanicswhere C ijkl are the elastic stiffness tensor. The equilibrium equations in the absence of bodyforces expressed in terms of displacements areC u ijkl k , lj= 0(3.2)Let the displacement be assumed to be in the following form:whereui = aif (z) or u = a f (z)(3.3)z = x 1+ µ x 2(3.4)In the above f is an arbitrary function of z, and µ and ai are unknown constants to bedetermined. Usingu = ( δ + µδ2) a f ( )(3.5)and substituting (3.5) into (3.2) gives′k , l l1l kzorC+ µδ ) ( δ + ) a f ′′ijkl( δj1 j2l1µδl2k=2[ C ( C + C ) µ + C µ ] a 0(3.6)i1 k1+i1k2 i2k1i2k2 k=0(3.6) can be written in matrix notation asT2[ Q + ( R + R ) µ + Tµ ] a = 0(3.7)where Q, R, and T are 3 × 3 matrices whose components areQ = C=(3.8)iki1 k1, Rik= Ci1k 2, TikCi2k2For a nontrivial solution of a, we haveT2Q + ( R + R ) µ + Tµ= 0(3.9)The determinant gives six roots for the eigenvalue µ. The associated eigenvector a isdetermined from (3.7). The stresses σijare obtained by inserting (3.5) into (3.1) and areexpressed by two vectorsτ1= [ σ11, σ12, σ13](3.10)τ2= [ σ12, σ22, σ23]andσi1= ( Qik+ Rikµ) a f ′k( z)(3.11)σ = ( R + T µ ) a f ′(z)i2kiikIt can be proved that µ of (3.9) cannot be real for the strain energy density being positivedefinite. Since the coefficients of the sextic equation for µ arising from (3.9) are real there arekN.C. State Univ., Raleigh, NC 3-2 F. G. Yuan, July 28-31, 1998

Anisotropic Elasticity: Application to Composite Fracture Mechanicsthree pairs of complex conjugates for µ. If µ k, ak(k = 1, 2, …, 6) are the eigenvalues and theassociated eigenvectors, we can rearrange the first three eigenvalues have positive imaginaryparts andµ µ a = a ( 1, 2 , 3)(3.12)k+ 3=k,k+3 kk =where the overbar denotes the complex conjugate. Assuming that µ are distinct, the generalsolution obtained by superposing six solutions of the form (3.3) is3[kfk k k k+3k=1u = ∑ a ( z ) + a f ( z )](3.13)where f k (k = 1, 2, …, 6) are arbitrary functions of their arguments andzk= x 1+kx 2(3.14)kµSimilarly, the general solution for the stresses obtained from (3.10) can be written asτ =τ12=3∑k=13∑k=1[( Q + µ R)a[( RTk+ µ T)akkf ′(zkkf ′(zkk) + ( Q + µ R)ak) + ( RTk+ µ T)akkf ′k+3k( zf ′kk+3)]( zk)](3.15)The stress component σ3is determined from σ3is determined by σ3= −s3jσj/ s33(j = 1, 2,4, 5, 6), which is the plane strain condition, ε3= 0.3.2 Eigenvalues µFor monoclinic materials with the symmetry plane at x 3= 0, employing the contractednotation for C ijkl introduced in Section 1.2 the three matrices Q, R, and T defined in (3.8)have the expressions⎡C11C160 ⎤ ⎡C16C120 ⎤ ⎡C66C260 ⎤Q =⎢C⎥⎢16C660 , R =⎢C⎥⎥ ⎢66C260 , T =⎢C⎥⎥ ⎢26C220 (3.16)⎥⎢⎣0 0 C55⎥⎦⎢⎣0 0 C45⎥⎦⎢⎣0 0 C44⎥⎦Eq. (3.9) reduces to⎡⎢⎢C⎢⎣16C11+ ( C+ 2C1216+ Cµ + C066662µ2) µ + C µ26C16+ ( CC6612+ C+ 2C260662) µ + C µµ + C µ22262C55+ 2C0045µ + C44⎤⎥⎥a=02µ ⎥⎦(3.17)µ 1 and µ 2 are the roots ofN.C. State Univ., Raleigh, NC 3-3 F. G. Yuan, July 28-31, 1998

Anisotropic Elasticity: Application to Composite Fracture Mechanics222 2( C + 2Cµ + C µ ) ( C + 2Cµ + C µ ) − [ C + ( C + C ) µ + C µ ] 0 (3.18)11 16 66 66 26 2216 12 66 26=µ 3 is the root of a quadratic equation2C55+ 2C45µ + C44µ= 0Recall from the Lekhnitskii formulation (2.14), a more compact form can be written in term ofreduced compliance constantsand432s ′ µ − s′µ + (2s′+ s′) µ − 2s′µ + s′0(3.19)1121612 6626 22=s ′ ′ µ s′255µ − 2s45+44=03.3 The Sextic formalism of StrohIntroducing a new vector b1b = ( R T + Tµ ) a = − ( Q + Rµ) a(3.20)µthe stresses in (3.11) can be rewritten asσ = −µf ( ), σ = f ( )(3.21)i 1b i′ zi 2b i′ zThe second equality in (3.20) follows from (3.7). Defining the stress functionφ i= b if (z) or φ = b f (z)(3.22)(3.21) are equivalent toσ = −φσ = φ(3.23)i1 i,2,i2i,1Therefore, it is sufficient to consider the stress function φ because the stresses can beobtained from (3.23) through differentiation.From (3.23), σ21= σ12impliesand, by (3.22),φ + φ 0(3.24)1 ,1 2,2=(1 2=b ) + µ ( b)0(3.25)where () k defines as the kth component of the associated vector. The general solution for thestress function φ is obtained by superposing six solutions of the form (3.22) associated withsix eigenvalues µ k. Together with (3.13) we haveN.C. State Univ., Raleigh, NC 3-4 F. G. Yuan, July 28-31, 1998

Anisotropic Elasticity: Application to Composite Fracture Mechanicsu =φ =3∑k=13∑k=1[ ak[ bkf ( zkkkk) + af ( z ) + b fkkfk+3k+3( z( zkk)])](3.26)In the above b kis related to a kthrough (3.20) andb = ( 1, 2 , 3)(3.27)k+ 3b kk =The vectors a k , b k are the Stroh eigenvectors. The displacements are given by (3.26) 1 whilethe stresses are obtained from (3.26) 2 and (3.23) by differentiation.In most applications the arbitrary functions f k appearing in (3.26) have the samefunctions form for each k. We may therefore letf ( z ) = f ( z ) q , f+ 3( z ) = f ( z ) q(3.28)kkkkkkkk(k = 1, 2, 3) where q kare arbitrary complex constants. The second equation is useful forobtaining real form solutions for u and φ. Equations (3.26) can then be written asu = 2 Re[ Aφ = 2 Re[ Bffq]q](3.29)where A and B are complex matrices,z[ ]f () z = diag f ( z1), f ( z2), f ( z3 ).= x 1+kx 2, Im[µ k ] > 0 , k = 1, 2, 3.kµA = [ a , a a ] , [ b , b b ]1 2,3B = (3.30)1 2,3In the above, f(z) is an arbitrary function, q is a unknown complex constant vector to bedetermined. µ k , a k , and b k are the Stroh eigenvalues and eigenvectors which depend on theelastic constants only. Hence, regardless of the problems to be solved, Stroh eigenvalues andeigenvectors µ k , a k , and b k form the basic elements in the Stroh formulation. Although thesolutions to an anisotropic elasticity problem is, in general, expressed in terms of the Stroheigenvalues and eigenvectors are complex, there are identities which express certaincombination of the eigenvalues and eigenvectors in terms of real matrices, such as N i andBarnett-Lothe tensors, S, H, and L.When q is replaced by− i q , (3.29) is expressed in an alternative formu = 2 Im[ Aφ = 2 Im[ Bffq]q](3.31)N.C. State Univ., Raleigh, NC 3-5 F. G. Yuan, July 28-31, 1998

Anisotropic Elasticity: Application to Composite Fracture MechanicsIt has been shown by Stroh (1962) and Ingebrigtsen and Tonning (1969), that the entireproblem can be recast into a six-dimensional framework and the unknowns µ k, , a k , and b k maybe simultaneously determined from the six-dimensional eignevalue problem from (3.20)N ξ = µ ξ(3.32)whereN NN = ⎡ ⎣ ⎢ 1 2 ⎤ ⎡a⎤TN N⎥ , ξ = ⎢ ⎥ (3.33)3 1 ⎦ ⎣ b⎦ −1T−1−1TN =− T R , N = T , N = RT R −Q(3.34)1ξ spans a six-dimensional space. It has been shown that the vectors2⎡a⎤⎡b⎤ξ = ⎢ ⎥ and η =⎣ b⎢ ⎥⎦ ⎣ a⎦ associated with same eigenvalue µ are the right and left eigenvectors of the unsymmetricmatrix N, respectively. The orthogonality between the right and left eigenvectors leads toi.e., in the matrix form:η T = 0 , α ≠ β , α, β = 1, 2, ⋅⋅⋅, 6αξ β3⎡b⎢⎣aTα ⎤α⎡a⎥ ⎢⎦ ⎣bββ⎤⎥ = [ b⎦Tα, aTα⎡a] ⎢⎣bββ⎤⎥ = 0 ,⎦µ ≠ µαβ(3.35)T⎡b1⎢ T⎢b2T⎢b3⎢ T⎢b1⎢Tb2⎢T⎢⎣b3aaaaT1T2T3T1T2T3aa⎤⎥⎥⎥ ⎡a⎥ ⎢⎥ ⎣b⎥⎥⎥⎦11ab22ab33ab11ab22⎡*⎢0⎢a3⎤⎢0⎥ = ⎢b3⎦ ⎢0⎢0⎢⎣00*000000*000000*000000*00⎤0⎥⎥0⎥⎥0⎥0⎥⎥* ⎦or⎡B⎢⎣BTTAATT⎤ ⎡A⎥ ⎢⎦ ⎣B⎡*⎢0⎢A⎤⎢0⎥ = ⎢B⎦⎢0⎢0⎢⎣00*000000*000000*000000*00⎤0⎥⎥0⎥⎥0⎥0⎥⎥* ⎦N.C. State Univ., Raleigh, NC 3-6 F. G. Yuan, July 28-31, 1998

Anisotropic Elasticity: Application to Composite Fracture Mechanicswhere the * symbol denotes a possible nonzero element.Since the eigenvectors a α , b α are unique up to a multiplicative constant, the eigenvectors a α ,b α can be normalized under the assumption of µ α being distinct such thatThusTTb a + a b = δ(3.36)α β α β αβ⎡B⎢⎣BTTAATT⎤ A⎥ ⎡ ⎦ ⎣ ⎢ BA⎤⎥ = ⎡ I 0B⎦⎣ ⎢⎤0 I⎥⎦(3.37)orT TT TB A + A B = I = B A + A B(3.38)T TT TB A + A B = 0 = B A + A B(3.38) are called orthogonality relations. Since the two matrices are inverse to each other,then⎡A⎢⎣BA⎤⎡B⎥ ⎢B⎦⎣BTTAATT⎤ ⎡⎥ = I⎢⎦ ⎣00⎤I⎥⎦(3.39)orT TT TAB + AB = I = BA + BA(3.40)T TT TAA + AA = 0 = BB + BB(3.40) are called closure relations. (3.40) imply Re[AB T ] = I/2, AA T and BB T are pureimaginary. Hence the Barnett-Lothe tensors S, H, and L defined byS 2TTT= i( 2AB− I ) , H = 2iAA , L = − i BB(3.41)are real. It can be shown that they are positive definite and tensors of rank two. The matrixproducts, AB T , AA T , and BB T , appeared in the final solutions of the two-dimensionalanisotropic elasticity problems can be often replaced by the real tensors, S, H, and L.Moreover, S, H, L are related byHL − SS = I(3.42)Another two tensors, impedance tensor M and its inverse M -1 defined byM−1−1−1−1−1−1= −iB A , M = ( −i)( BA ) = i A BM and M -1 Tare positive definite, Hermitian ( M = M ), and tenor of second order. Barnett(1985) has shown that M is convenient for use in problems involving dislocations and cracksalong the interface. In general, the Stroh formulation can not be directly applied to degeneratematerials where µ have repeated roots (3.18) or (3.19). However, if the final solution requiresonly S, H, L, then the solution also holds for degenerate materials (Barnett and Lothe, 1973).N.C. State Univ., Raleigh, NC 3-7 F. G. Yuan, July 28-31, 1998

Anisotropic Elasticity: Application to Composite Fracture MechanicsThe eigenrelation (3.32) can be combined into a matrix form⎡AN ⎢⎣BA⎤⎡A⎥ = ⎢B⎦⎣BA⎤⎡µ⎥B⎢⎦ ⎣00 ⎤µ⎥⎦where µ = µ = diag µ , µ , ] .[1 2µ3By (3.39), we obtain3.4 Normalization of A and BT T⎡AA⎤⎡µ0 ⎤ ⎡BA ⎤N = ⎢ ⎥ ⎢ ⎥ ⎢ T T ⎥(3.43)⎣BB⎦⎣0µ ⎦ ⎣BA ⎦In order to satisfy the normalization condition in (3.36), normalized factors k 1 , k 2 , andk 3 for A and B need to be quantified. In this procedure, we determine B first and A. Formonoclinic materials, it is easily seen that both A and B have the structure:⎡*⎢*⎢⎢⎣0Therefore, the determination of k 1 and k 2 can be separated from that of k 3 . The in-planedeformation, B from (2.22) or from (3.25) can be written with normalization factors as**0⎡−k1µ1− k2µ2 ⎤B = ⎢⎥(3.44)⎣ k1k2⎦From (3.17), a 1 and a 2 can be written as0⎤0⎥⎥* ⎥⎦2k ⎡ − ( C66+ 2C26µ+ C22µ) ⎤a = ⎢2 ⎥(3.45)γ ( µ ) ⎣C16+ ( C12+ C66)µ + C26µ⎦where γ(µ) is a function of µ to be determined. Substituting into (3.20), we obtain b 1 and b 2shown in (3.44) ifγ ( µ ) = ( C16C= J ( s′1226− C) − ( C2− s′µ + s′µ )1612C661112C26− C16C22) µ + ( CThe above equality hae been shown in (1.82). From (1.85) and (1.86) with (3.45) or directlyfrom (2.21) without going these intermediate steps,⎡k1p1k2p2⎤A = ⎢ ⎥(3.46)⎣k1q1k2q2⎦The normalization factors which satisfy (3.36) are obtained from the following relations22C66− C2262) µN.C. State Univ., Raleigh, NC 3-8 F. G. Yuan, July 28-31, 1998

Anisotropic Elasticity: Application to Composite Fracture Mechanics2k2k2122( q1( q2− µ p ) = 1121− µ p2) = 1(3.47)3.5 Explicit expressions of S, H, LFrom the inverse of B -1 in (3.44),we have−1−1B −1 1 ⎡−k1− µ2k1⎤= ⎢ −1−1⎥µ1− µ2 ⎣ k2µ1k, (3.48)2 ⎦−1 1 ⎡ p1− p2p1µ2− p2µ1⎤− AB =−⎢⎥(3.49)µ1µ2 ⎣ q1− q2q1µ2− q2µ1 ⎦Clearly the normalization factors are cancelled out for the multiplication. Using2p = s′µ − s′µ + and qk= s′12µk− s′26+ s′22/ µkfrom Lekhntskii formulation11 16s′k k k 12⎡ ′′ ′ ′−1s11(µ1+ µ2) − s16s11µ1µ2− s12⎤− AB = ⎢⎥ (3.50)⎣ s12′ − s22′ / µ1µ2s26′ − s22′ ( µ1+ µ2) / µ1µ2 ⎦Further, employing the relations between µiand the coefficients of the quartic equation in(3.19), we can obtain the following form through some manipulations⎡ ′′ ′−1is11Im( µ1+ µ2) s11µ1µ2− s12⎤- AB = ⎢⎥(3.51)⎣ s12′ − s11′ µ1µ2is11′ Im[ µ1µ2( µ1+ µ2)] ⎦(3.51) can be used to calculate the inverse impedance tensor M -1 1 −1( M − = i A B ) and can beapplied to degenerated materials. Sinceusing (3.41), we have−−1T T −1AB = −(AB )( BB )(3.52)−−1 −1−1AB = SL + i L(3.53)−1Therefore the real and imaginary parts of - AB are SL -1 and L -1 , respectively. With SL -1 andL -1 determined, L and S are then obtained. Finally H is determined from (3.42)The compact form of S, H, and L matrices are1 −1H = L− + S(SL )(3.54)s′s′− ⎡ − ⎤=)2( / d b11c[( s′′ − ⎡bd12 12/ s11)c]⎤S2be − d⎢ ⎥ , H = s11′ ( 1 −)⎣e− d⎢ ⎥ (3.55)2⎦be − d ⎣de ⎦N.C. State Univ., Raleigh, NC 3-9 F. G. Yuan, July 28-31, 1998

Anisotropic Elasticity: Application to Composite Fracture Mechanics−1⎡bd⎤1 ⎡ e − d ⎤L = s11′ ⎢ ⎥ , L =′ −⎢ ⎥(3.56)2⎣de⎦s11(be d ) ⎣−d b ⎦µ + µ = a + ib,µ µ = c + id12e = ad − bc = Im[ µ µ ( µ + µ )]In the isotropic solids under in-plane deformation, the Barnett-Lothe tensors are112212L =12s′I,(3 − 4ν)(1 + ν )H =I,2E(1−ν)1 − 2ν⎡0S =2(1 − )⎢⎣111ν−1⎤0⎥⎦(3.57)2where s′ = ( 1−ν )/ E .113.6 Tractions, resultant forces and moments in terms of stress functionsThe traction vector t at a point on a curved surface Γ with unit outward normal n = [n 1 ,n 2 , 0] T is given byt = σ n(3.58)i ij jLet s be an arc length measured counter-clockwise along Γ as shown in Fig. 3.1,Using (3.23) and (3.59),dx1= ,dsn2dx= , n3= 0(3.59)ds1n2 −d dx1dx2φi= φi,1+ φi,2ds ds ds= −σn − σ n = −σni11i22ijjHence, t i can be expressed asddti= − φior t = − φ(3.60)ds dsThus if the boundary Γ is traction-free surface, thenφ = 0 on Γ (3.61)N.C. State Univ., Raleigh, NC 3-10 F. G. Yuan, July 28-31, 1998

Anisotropic Elasticity: Application to Composite Fracture Mechanicsx 2tnΓsFig. 3.1 The surface traction t on a curved boundary Γ with a unit outward normal vector n.x 1areThe resultant force due to the surface traction t acting on Γ between s 1 and s 2 (s 2 > s 1)s2∫ t ( s)ds = φ ( s1) − φ(s2) = −∆φ(3.62)s1If Γ encloses a region and there exists a concentrated force f inside the region, theequilibrium of the solid gives∫ Γt(s) ds= − f∆φ = f(3.63)This implies that if we traverse Γ counter-clockwise one complete circle, φ increases by theamount f. Therefore φ must be a multiple-valued function. A Riemann surface with a propercut has to be introduced to maintain a unique solution for φ. If there is no concentrated forceinside the enclosed region Γ, φ is a single-valued function.The moment about the x 3-axis due to the surface traction t acting on Γ between s 1 ands 2 (s 2 > s 1) arewheres2s2∫ ( x1t2− x2t1) ds = −[x1φ2− x2φ1− χ ](3.64)ss11∫χ z)= z φ ( λ)dλ(2As illustrated before, using Stroh formulation, the two-dimensional anisotropic problemsrequire f(z) and q (complex constant) to be solved. The following example demonstrates howf(z) is selected for a specific problem and q is then determined from boundary conditions.3.7 Example 1: Uniform stress solutionN.C. State Univ., Raleigh, NC 3-11 F. G. Yuan, July 28-31, 1998

Anisotropic Elasticity: Application to Composite Fracture MechanicsThe displacement u and the stress function φ for a uniform stress solution can bewritten asu = x ε∞1 1φ = x τ∞1 2+ x ε2− x τ∞2∞2 1(3.65)where τ ∞ 1= [ σ ∞ ∞ ∞11, σ 12, σ 13] T ∞ ∞ ∞ ∞, τ 2= [ σ 21, σ 22, σ 23]ε ∞ 1ε∞ 110 2ε∞13= [ , , ] T, ε ∞ 2= [ 2ε ∞ ∞ 12, ε 22, 2ε∞23]σ ij∞are the stresses at infinity. The equivalent strainsε ij∞applied at infinity can be obtained fromthe stress-strain law.The uniform stress solution can also be expressed byTTwhereu = 2 Re[ Aφ = 2 Re[ Bzzq]q](3.66)T Tq = A g + B h(3.67)g ,∞= τ 2h . (3.68)∞= ε 1The proof of equivalence between (3.65) and (3.66) with (3.67) and (3.68) is givenbelow:byFirst, let g and h be real constant vectors to be determined, (3.66) can be representedu = x [( AB12φ = x [( BB12T+ x [( AµBT+ ABT+ x [( BµB+ AµB+ BBTTT+ BµB) h + ( AAT) h + ( AµA) h + ( BATTT+ AA+ BA) h + ( BµATTTT) g]+ AµA) g]+ BµATT) g]) g](3.69)Comparing (3.69) with (3.65) yields⎡h⎤⎡AB⎢ ⎥ = ⎢⎣g⎦⎣BBTT+ AB+ BBTTAABATT+ AA+ BATT⎤⎥⎦−1∞⎡ε1⎤⎢ ∞ ⎥⎣τ2 ⎦(3.70)∞⎡ ε2⎤ ⎡A⎢ ⎥ = ⎢⎣−τ∞1 ⎦ ⎣BA⎤⎡µ⎥B⎢⎦ ⎣00 ⎤ ⎡Bµ⎥ ⎢⎦ ⎣BTTAATT⎤ ⎡h⎤⎥ ⎢ ⎥⎦ ⎣g⎦orN.C. State Univ., Raleigh, NC 3-12 F. G. Yuan, July 28-31, 1998

Anisotropic Elasticity: Application to Composite Fracture Mechanics∞⎡ ε ⎤ ⎡h2 ⎤⎢ ⎥ = N ⎢ ⎥⎣−τ∞⎦ ⎣g1 ⎦where (3.43) has been used in deriving (3.71). By virtue of (3.40),(3.71)ABT+ ABT= I = BAT+ BATAAT+ AAT= 0 = BBT+ BBT(3.70) becomes∞⎡h⎤⎡ε1⎤⎢ ⎥ = ⎢ ∞ ⎥⎣g⎦⎣τ2 ⎦Thus, the uniform stress solution expression, (3.66) - (3.68), is confirmed.∞ε2andτ∞ 1are determined by (3.71). i.e.,∞∞⎡ ε2⎤ ⎡ε1⎤⎢ ⎥ = N∞ ⎢ ∞ ⎥⎣−τ1 ⎦ ⎣τ2 ⎦3.8 Example 2: Green’s function of a point force and a dislocationFor an infinite plane subjected to a line force f and a line dislocation with Burgersvector b both located at x 1 = x 2 = 0, the solution can be written as1u = Im[ A ln zπq1φ = Im[ B ln zπq00]](3.72)where q 0 is a complex constant to be determined. Since lnz is a multi-valued function, weintroduce a branch cut along the negative x 1 -axis. In the polar coordinate system x 1 = rcosθ,x 2 = rsinθ, the above solution applies toTherefore− π ≤ θ ≤ π , r > 0± iπln z = ln( x + µαx = re = r ± iπθ = ± π 1 2) ln( ) ln at θ = ± π , for α = 1, 2, 3θ =± πEqs. (3.72) must satisfy the conditions(ln z)θ = π − (ln z)θ = −π= 2πiu ( r,π ) − u(r,−π) = b , φ ( r,π ) − φ(r,−π) = f(3.73)The first equation satisfies the displacement discontinuity due to dislocation; while thesecond equation arises from (3.63).N.C. State Univ., Raleigh, NC 3-13 F. G. Yuan, July 28-31, 1998

Anisotropic Elasticity: Application to Composite Fracture Mechanics(3.74) can be written as20Re( Aq0) = b , 2 Re( Bq ) = f(3.74)⎡A⎢⎣BFrom orthogonality relations (3.37),A⎤⎡q⎥B⎢⎦ ⎣q00⎤ ⎡b⎤⎥ = ⎢ ⎥⎦ ⎣ f ⎦(3.75)Hence⎡q⎢⎣q00⎤ ⎡B⎥ = ⎢⎦ ⎣BTTAATT⎤ ⎡b⎤⎥ ⎢ ⎥⎦ ⎣ f ⎦(3.76)T Tq0 = A f + B b ←(3.77)3.9 Example 3: Green’s function of a point force applied at the crack tipThe above solution form can be utilized to construct the solution of a wedge subjectedto a point force at the wedge apex. If the wedge angle equals 2π, the solution for a cracksubjected to a line force at the tip can be obtained.The infinite plane subjected to a line force f and a line dislocation with Burgers vector bboth located at x 1 = x 2 = 0, the solution can be converted into a real formwhereln r2u= − h − S(θ ) h − H(θ ) gπln rT2φ= − g + L(θ ) h − S ( θ ) gπ(3.78)h = Sb + Hf ,g = STf− LbTTS = i( 2AB− I ), H = 2iAA , L = −22TS(θ ) = Re[ A ln(cosθ+ µ sinθ) B ]π2TH(θ ) = Re[ A ln(cosθ+ µ sinθ) A ]π2TL(θ ) = − Re[ B ln(cosθ+ µ sinθ) B ]πS, H, L, S(θ), H(θ), and L(θ) are real matrices.i BBIn a cylindrical coordinate system (r, θ, x 3 ), let t r and t θ be the traction vectors on acylindrical surface r = constant and on a radial plane θ = constant, thenTN.C. State Univ., Raleigh, NC 3-14 F. G. Yuan, July 28-31, 1998

Anisotropic Elasticity: Application to Composite Fracture Mechanics1φ,θtr= − , tθ= φ , rrTT Tσrr= n tr, σr θm tr= n tθTTwhere n = [cosθ, sinθ] and m = [ − sinθ, cosθ]For the real form solution (3.78),= , σθθ= m T tθtθ1= −2πrgThe equation indicates that t θ is independent of θ. Hence if t θ vanishes at a particular θ, itvanishes for all θ. The solution for a crack subjected to a line force f at the crack tip may beobtained from the real form solution (3.78) by setting g = 0, i.e.,⎡lnr ⎤2u= −⎢I + S(θ ) h⎣ π ⎥⎦2φ= L(θ ) h(3.79)The concentrated force f requires thatφ ( π ) − φ(−π) =forh = 2[ L(π ) − L(−π)]−1fSinceL( π ) = L,L(−π) = −L−1h = L f(3.80)The following example demonstrates that using one of the Barnett-Lothe tensors, L,the energy release rate can be written in a compact form and can be applied to degeneratematerials.3.10 Example 4: Energy release rate expressionAs shown in the Appendices, the energy release rate for a crack in an anisotropic solidcan be expressed by1 −T 1G = k L k(3.81)2where k = [k II , k I , k III ] T . For isotropic solids, using augmented form (3.57) for 3 × 3 matrix L -1 ,N.C. State Univ., Raleigh, NC 3-15 F. G. Yuan, July 28-31, 1998

Anisotropic Elasticity: Application to Composite Fracture MechanicsL−1⎡⎢122(1 −ν)= ⎢0E ⎢⎢0⎣0100⎤⎥0 ⎥1 ⎥1 + ν⎥⎦we obtain21 −ν2 2 1 2G = ( kI+ kII) + kIIIfor plane strain (3.82)E2GE(1+ 2ν)For plane stress, E and ν are replaced by and2(1 + ν )νfor Mode I and II respectively.1+ ν1 2 2 1 2G = ( kI+ kII) + kIIIfor plane stress (3.83)E2GIn the following three examples, two real matrices G1(θ ) and G3(θ ) are introduced indescribing stress terms along the hole or crack boundary.From (3.32),N ξ =where µ = µ = diag µ , µ , ] .µ ξ , the equation can be written in an explicit matrix form[1 2µ3Post-multipling [B T , A T ], we have⎡A⎤⎡Aµ⎤N ⎢ ⎥ = ⎢ ⎥(3.84)⎣B⎦⎣Bµ⎦T T⎡N1N2 ⎤ ⎡ABAA ⎤ ⎡Aµ B⎢ ⎥ ⎢ ⎥ =T T T ⎢⎣N3N1⎦ ⎣BBBA ⎦ ⎣Bµ BUsing definition of Barnett-Lothe tensors,TTTA µ A ⎤T ⎥B µ A ⎦(3.85)⎡ S⎢⎣−Leq. (3.85) can be expressed byT TH ⎤ ⎡ABAA ⎤⎥ = 2 i ⎢− i IT T ⎥(3.86)TS ⎦ ⎣BBBA ⎦⎡Aµ B2⎢⎣Bµ BTTTA µ A ⎤ ⎡N⎥ =TB µ A⎢⎦ ⎣N13N2 ⎤ ⎡N1⎥ − iTN⎢1 ⎦ ⎣N3N2 ⎤ ⎡ STN⎥ ⎢1 ⎦ ⎣−LH ⎤TS⎥⎦(3.87)One of these eqs. in (3.87) is expressed byTTB µ B = N − i(N S − N )(3.88)23 3 1LUsingB−1T T −1T −1T −= B ( BB ) = −2i B L leads to B = i B1 L / 2. (3.88) becomesN.C. State Univ., Raleigh, NC 3-16 F. G. Yuan, July 28-31, 1998

Anisotropic Elasticity: Application to Composite Fracture MechanicsB−1T-1−1µ B = ( N1- N3SL) − iN3L(3.89)The real and imaginary parts of (3.89) is defined as G 1 (0) and G 3 (0)−1B µ B = G1( 0) + iG3(0)(3.90)whereG 3SL-11( 0)= N T1- N ,−1G3 (0)= −N 3L.Referring to a new coordinate system obtained by rotating x i about the x 3 - axis an angleθ, the dependence of µ (θ ) on θ isµ cosθ− sinθµ ( θ ) = (3.91)µ sinθ+ cosθWe can definewhereG−1B µ ( θ ) B ≡ G1(θ ) + i G3(θ )µ ( θ ) = diag[µ1(θ ), µ2(θ ), µ3(θ )]−11( θ ) Re[ B µ ( θ ) B ] = N T 1( θ ) − N3(θ )1= SL−1(3.92)G−13( ) Im[ B µ ( θ ) B ] = −3(θ )θ = N L−It can be shown that G 1 (θ) and G 3 (θ) for specific angles have the following structures⎡ * * * ⎤ ⎡**G =⎢−⎥1(0) 1 0 0 , G⎢⎢ ⎥ 3(0) = 0 0⎢⎢⎣* * * ⎥⎦⎢⎣* *G i( π / 2) = G i( −π/ 2) , i = 1, 3⎡01 0⎤⎡0G =⎢ ⎥1(π / 2) * * * , G⎢⎢ ⎥3(π / 2) = *⎢⎢⎣* * * ⎥⎦⎢⎣** ⎤0⎥⎥* ⎥⎦0 0⎤* *⎥⎥* * ⎥⎦For monoclinic materials,−1B µ ( θ ) B =⎡(µ1µ2−1)mn + ( µ1+ µ2) m1 ⎢⎢− 1ς1ς2⎢⎣0where m = cosθ, n = sinθ, ςThe real G 1 and G 3 matrices areα2= cos θ + µ sinθ.αµ1µ2( µ µ − 1) mn − ( µ + µ ) n1201220 ⎤⎥0⎥µ ⎥3(θ ) ς1ς2 ⎦(3.93)N.C. State Univ., Raleigh, NC 3-17 F. G. Yuan, July 28-31, 1998

Anisotropic Elasticity: Application to Composite Fracture Mechanics⎡ a c 0 ⎤G⎢ ⎥1(0)= G1= −10 0 ,⎢ ⎥⎢⎣0 0 µ ′3⎥⎦and µ3= µ ′3+ i µ ′3For isotropic materials,⎡bd 0 ⎤G⎢ ⎥3(0)= G3= 0 0 0(3.94)⎢ ⎥⎢⎣0 0 µ ′′3⎥⎦⎡ sin 2θ− cos2θ0⎤⎡1+ cos2θsin 2θ0⎤G ( ) =⎢− cos2 − sin 2 0⎥1θ θ θ , G⎢⎥⎢⎥3(θ ) = sin 2θ1−cos2θ0 (3.95)⎢⎥⎢⎣0 0 0⎥⎦⎢⎣0 0 1⎥⎦3.11 Example 5: A crack subjected to a uniform load at infinityσ 22∞σ 23∞σ 12∞x 2σ 13∞σ 13∞σ 11∞2ax 1σ 11∞σ 12∞σ 23∞σ 22∞Fig. 3.2 A crack subjected to uniform remote load at infinityReferring to a fixed coordinate system $x i, consider a crack of length 2a centrallylocated at $x 2= 0 in a homogeneous anisotropic material. A uniform stress σ ij∞is applied atinfinity. The crack surfaces are traction-free. The solution consists of a uniform stress solutiondue to the uniform loading σ ij∞and a disturbed solution due to the presence of the crack. It iseasily shown that the solution is given byu = Re[ Aφ = Re[ Bf ( zˆ)f ( zˆ)BB−1−1] τ] τ∞2∞2+ xˆε∞1 1+ xˆτ∞1 2+ xˆε2− xˆτ∞2∞2 1(3.96)wheref ( z$) = diag[ f ($ z1), f ($ z2), f ($ z3 )], zˆα = xˆ1 + µ α xˆ2f ( z$) = z$ 2 −a 2 −z$(3.97)N.C. State Univ., Raleigh, NC 3-18 F. G. Yuan, July 28-31, 1998

Anisotropic Elasticity: Application to Composite Fracture MechanicsUsing τ1= −φ , 2, τ2= φ , 1, andτ ∞ 1σ ∞ 11σ ∞ ∞12σ 13= [ , , ] T ∞ ∞ ∞ ∞, τ 2= [ σ 21, σ 22, σ 23]ε ∞ 1ε∞ 110 2ε∞13= [ , , ] T, ε ∞ 2= [ 2ε ∞ ∞ 12, ε 22, 2ε∞23]TTz2− a2⎪⎧2 2± x1− a ,= ⎨2 2⎪⎩ ± i a − x1,at xat x22= 0, ± x= ± 0, x11> a< ait can be shown that⎫⎪2 2 −1∞ ∞u = ± [ xˆ1− xˆ1− a ] SL τ2+ xˆ1ε1⎪xˆ⎪1 ∞τ2= τ ⎬2 22xˆ1− a⎪xˆ⎪1∞ ∞ ∞τ1= − G1τ2+ G1τ2+ τ1 ⎪2 2xˆ1− a⎪⎭for xˆ2= 0, ± xˆ1> a(3.98)⎫−12 2 −1∞ ∞u = [ xˆa x x⎪1SL± − ˆ1L ] τ2+ ˆ1ε1⎪τ2= 0⎬ forxˆ⎪1∞ ∞ ∞τ1= m G3τ2+ G1τ2+ τ1 ⎪2 2a − xˆ1⎪⎭xˆ2= ± 0,xˆ1< a(3.99)Note that τ2is independent of material property. The crack opening displacement ∆uand thestress intensity factor k arewhere k = [ k , k , k ]III+−2 2 −1∞∆u( ˆ1)= u(xˆ1,0) − u(xˆ1,0) = 2 a − xˆ1L τ2IIITx (3.100)k ˆ(3.101)∞= lim 2π( x1− a ) τ2= π a τ2xˆ1 →aIntroducing a new coordinate system (x1, x2) at the right crack tip such thatf ( z$ ) can be expanded asαx$ = x + a, x$= x1 1 2 2(3.102)where3/ 2 5/ 2⎡ zαzαzα⎤f ( zˆα) = f ( zα+ a)= −a+ 2a⎢ zα− + − + O(z22a4a32a⎥⎣⎦zα= x + x = (cosθ+ µ sinθ),1µα 2rα− π ≤ θ ≤ π7 / 2α)N.C. State Univ., Raleigh, NC 3-19 F. G. Yuan, July 28-31, 1998

Anisotropic Elasticity: Application to Composite Fracture MechanicsTherefore the asymptotic crack tip stress fields are expressed asτ1= −φ,2= −1 ⎡ µ 3µ5µ3/ 2 −1⎤∞ ∞ 5/ 2Re⎢B+ z − z B +1 2+1+ O ( r )22 z 4a32a⎥ k G τ τπ ⎣⎦orτ1= −3/ 21 ⎡ a 3µz 5µ⎛ z ⎞ ⎤−1∞ ∞ 5/ 2Re⎢Bµ + − ⎜ ⎟ B ⎥ k + G1τ2+ τ1+ O ( r )2πa⎢ z 4 a 32 a⎣⎝ ⎠ ⎥⎦(3.103)1 ⎡ 1 3 5 3/ 2 −1⎤5/ 2τ2= φ,1= Re⎢B+ z − z B + O ( r )22 z 4a32a⎥ kπ ⎣⎦(3.104)wheref ( z;µ ) = diag[f ( z1 , µ1) , f ( z2, µ2) , f ( z3, µ3)]and f(z; µ) is a given function with µ being a parameter. From the above expressions, thestress distribution does not have z n terms, n = 1, 2, 3, ⋅⋅⋅, for the infinite plane under remoteuniform loading.It is clear that the second-order terms, T-stress terms, areFor monoclinic materials(2) ∞τ1= G1τ2+ τ(2)τ = 02∞1(3.105)(2)∞ ∞ ∞⎧σ11⎫ ⎧σ12s16′ + σ22c+ σ11⎫(2) ⎪ (2) ⎪ ⎪⎪τ1= ⎨σ12 ⎬ = ⎨ 0 ⎬(3.106)⎪ (2) ⎪ ⎪ ∞∞ ⎪⎩σ⎭ ⎩′ ′13σ23s45/ s55+ σ13 ⎭where c = Re( µ 1µ2) .For orthotropic materials, T-stress terms reduce toτ⎧σ⎪= ⎨σ⎪⎩σ∞⎫ ⎧ − σ22s22′ / s11′ + σ⎪ ⎪⎬ = ⎨⎪ ⎪∞⎭ ⎩σ13(2)11(2) (2)1 120(2)13∞11⎫⎪⎬⎪⎭3.12 Example 6: An elliptical hole subjected to uniform load at infinityN.C. State Univ., Raleigh, NC 3-20 F. G. Yuan, July 28-31, 1998

Anisotropic Elasticity: Application to Composite Fracture Mechanicsx 2t mmt n= 0 t Γnθ ΓBσ nnaAσ nnbΓx 1Fig. 3.3 Tractions and stresses along the boundary of the elliptical holeTo consider this problem with elliptical hole, it is convenient to express all the functionsin terms of transformed variables ζ α instead of z α . The solution form is suggested as:u = Re[ A ζ−1B-1−1-1φ = Re[ B ζ B q]q]+ u+ φ∞∞(3.107)whereandu x x x x , (3.108)∞ ∞ ∞ ∞ ∞ ∞=1ε1+2ε2, φ =1τ2−2τ1ζα=zα+z2α− ( a2a − i µ bα2 2+ µ b )αζz−z− ( a2+ µ b2 22− 1 α αα)α=a + i µαbIn the above, the infinite plane with an ellipse opening has been transformed to the ζ-planeiψwith an opening of a unit circle, ζ = e . The mapping function isczcd−1α=αζα+αζα11= ( a − iµ αb),dα= ( a + i22α µ αwhere 2a and 2b are the major and minor axes of the ellipse. This mapping function maps theouter ellipse onto the exterior of a unit circle.The ellipse boundary Γ can be parameterized byThe infinitesimal arc length of the ellipse isb)x1= a cosψ , x2= bsinψ(3.109)ds = ρ dψN.C. State Univ., Raleigh, NC 3-21 F. G. Yuan, July 28-31, 1998

Anisotropic Elasticity: Application to Composite Fracture Mechanics2 2 2 2ρ = a sin ψ + b cos ψ(3.110)The parameter ψ can be related to θ Γ , which is the angle directed from the x 1 -axis to thetangent n, byρ cosθΓ= a sinψ, ρ sinθΓ= −bcosψ(3.111)The unit vectors tangential and normal to the ellipse as shown in the Fig. 3.3 areTTn ( θΓ) = [cosθΓ, sinθΓ, 0] , m ( θΓ) = [ −sinθΓ, cosθΓ, 0](3.112)The stress function φ at the elliptic boundary Γ , x 1 = acosψ and x 2 = bsinψ, isφΓ− ∞ ∞= Re[ e iψ ( aτ2− ibτ1+ q)](3.113)For the traction-free on the surface of the elliptic hole, setting φ = 0 , we obtain∞ ∞= −aτ 2+ ibτ1Γq (3.114)If t n is the traction on a surface whose unit normal vector n(θ) and unit tangentialm(θ) areTn ( θ ) = [cosθ, sinθ, 0] ,Tm ( θ ) = [ −sinθ, cosθ, 0], (3.115)then from (3.60), we havet= φ φ m( θ θ φ θ φn−, m= −∇ ⋅ ) = sin,1− cos,2(3.116)The subscripts m followed by a comma denote differentiation in direction of m. The normalstress σ nn and the shear σ nm and σ n3 on the surface are determined fromσ = n θ t σ m θ t σ = t(3.117)Notice that on the boundary, it can be shown thatnnTT( )n,nm= ( )n,n3 (n) 3dζζ = (3.118)iψαζ α = e , α , mzα , mdzαiψdζ= i e dψ, dz = −ρ(cosθµ sinθ) dψ(3.119)ααΓ+αz = ( θ )(cosθµ sinθ)(3.120)α , mµ α Γ Γ+ α ΓUtilizing (3.118)-(3.120), the traction t n acting on the surface normal to the ellipse boundary,which is called hoop stress vector, can be derived from (3.107) 2 and (3.116)1 −iψ− ∞tn = Im[ e B µ ( θ1 Γ) B q]− φ,m(3.121)ρand the hoop stress σ nn is determined from σnn= n T ( θ Γ) tn.Since−1B µ θ ) B = G ( θ ) i G ( θ )(3.122)(Γ 1 Γ+3ΓΓN.C. State Univ., Raleigh, NC 3-22 F. G. Yuan, July 28-31, 1998

Anisotropic Elasticity: Application to Composite Fracture Mechanicst n and σ nn can be expressed as, after substituting (3.122), (3.114), and (3.108) into (3.121)tn= cosθ[( τΓ+ sinθ[( τσnnΓ= nT− mT∞1∞2+ G ( θ ) τ1− G ( θ ) τ1ΓΓ(0)[ G ( θ ) τ1(0)[ G ( θ ) τ1ΓΓ∞2∞1∞2∞1+ bG3+ a G+ bG3− a G( θ ) τ3Γ( θ ) τΓ( θ ) τ3Γ( θ ) τΓ∞1∞2∞1∞2/ a)]/ a]/ b)]/ b](3.123)(3.124)Consider a special case of uniform tension σ applied at infinity, we have∞∞ ∞τ1= 0,τ2= σ22m(0)andσσ∞22= [ G ( θ )] + a[G ( θ )] b(3.125)∞nn/22 1 Γ 12 3 Γ 22/In the above, [ ] ij denotes the (ij) elements of the matrix.3.13 Example 7: An elliptical hole subjected to bending load at infinityx*2x 2 x1 *Mbaαx 1MFig. 3.4 An elliptical hole under bendingConsider the body subjected to pure bending M in a direction at an angle α with thepositive x 1 – axis shown in Fig. 3.4. The solution isu = Re[ A ζ−2B-1−2-1φ = Re[ B ζ B q]q]+ u∞∞+ φ(3.126)where n(α) = [cosα, sinα, 0] T ,∞ M2φ = ( −x1 sin α + x2cos α)n(α),2IN.C. State Univ., Raleigh, NC 3-23 F. G. Yuan, July 28-31, 1998

Anisotropic Elasticity: Application to Composite Fracture Mechanicsu ∞ can be determined from φ ∞ through constitutive relation and (3.23). q is a constant vectorto be determined, I is the moment inertia of the cross section normal to the loading axis.The stress function φ at the elliptical boundary Γ is−2ψ M2 M 2 2 2 2φ Re{ e i Γ= [ q + ( a sin α − ib cos α ) n(α )]} + ( a sin α + b cos α)n(α ) (3.127)4I4ITraction-free condition on the surface of the elliptical hole φΓ= constant leads toM2q = − ( a sin α − ib cosα) n(α)(3.128)4IFollowing the same procedure as the case of uniform loading, the hoop stress vector t nand hoop stress, σ nn , at the elliptical opening can be written ast2ρ−i2ψ−1∞n= Im[ e B µ ( θΓ)B q]− φ,m(3.129)σ = θ t(3.130)nnTn ( Γ)The hoop stress along the opening boundary can also be expressed by− σwheressccsnn*0**22c =*** 2 ac Tas T/( Ma / I ) = s0 cos ( θΓ− α)− n ( θΓ) G3(θΓ)n(α)+ n ( θΓ) G1(θΓ) n(α)2ρ2ρ(3.131)= sin( ψ − α)− (1 − c)cosαsinψ= −2(c sin 2ψ+ s cos2ψ)= −2(s sin 2ψ+ c cos2ψ)1 2= [1 − c − (1 + c4c= − sin 2α2ba22222) cos2α]This case can be reduced to a crack when b = 0. Then we have2 2222 ˆ 2 ˆ ˆ−za− a − zαzaα=2N.C. State Univ., Raleigh, NC 3-24 F. G. Yuan, July 28-31, 1998n2− aξ (3.132)aM 2 2q = − a sin α n(α)4IIntroducing a coordinate system at the right crack tip such that

Anisotropic Elasticity: Application to Composite Fracture Mechanicsand expandingφ near the crack tip1= xˆ1+ a , x2xˆ2x =M ⎡a⎤φ ⎢ˆ ˆ ˆ ⎥ )4I⎢⎣2⎥⎦222−1∞= − sin α Re B − (2a)3/ z + 4az+ 5 z3/ + ⋅⋅⋅ B n(α + φ(3.133)and T has the structure1 ⎡ 1 −1⎤τ2= φ,x ˆ = Re+ O(zˆ)1 ⎢BB2 zˆ⎥ k(3.134)π ⎣ ⎦1 ⎡−1⎤τ1= −φ, ˆ = − Re+ O(zˆ)x 2 ⎢B µ B ⎥ k T2πzˆ+(3.135)⎣ ⎦k = limTr→0π aa M 22πrτ 2= sin α n(α)(3.136)θ = 0 2 IM= a sinα(sinαG1 + cosαI)n(α )Iz ˆ = xˆµ+2 21+ xˆ2, r = xˆ1xˆ2⎡*⎤T =⎢0⎥⎢ ⎥⎢⎣* ⎥⎦(3.137)3.14 Problems1. For monoclinic materials with plane of symmetry at x 2 = 0, the matrix in determining µ kand a is22⎡ C11+ C66µ( C12+ C66)µ C15+ C46µ⎤⎢2⎥⎢( C12+ C66)µ C66+ C22µ( C25+ C46)µ ⎥a = 022⎢15 46(25 46)⎥⎣C + C µ C + C µ C55+ C44µ⎦The in-plane and antiplane deformation are coupled.2. For a thick composite specimen made of AS4 carbon warp-knit fabric under tension, thematerial properties are:E1= 10.4 Msi,E2= 5.22 Msi,ν12= 0.403, GE = 1.45Msi, ν = ν = 0.493132312= 2.54 MsiIf the crack is parallel to x 1 – axis or oriented 45 0 from the x1-axis and the stiffer direction isalong the x2-axis, determineN.C. State Univ., Raleigh, NC 3-25 F. G. Yuan, July 28-31, 1998

(a)Anisotropic Elasticity: Application to Composite Fracture Mechanicsµk(k = 1, 2)(b) A and B(c) k 1 and k 2 (normalized factors)(d) Barnett-Lothe tensors, S, H, Lx _2 x 2x 145 οx 2x 1_x 1Fig. 3.2 A crack with inclined angle 0 0 and 45 0 in a thick composite couponTTT3. Prove HL − SS = I [ S = i( 2AB− I), H = 2iAA , L = −2iBB]= 2 i − i =TTHL AA ( 2 BB ) 4AATBBTTTT TSS = [ i(2AB− I)][i(2AB− I)]= −4ABAB + 4ABT− ITHL - SS = 4AABB= 4A(AT= 4ABTTB + BT− 4ABT+ 4ABABTA)BT+ I = IT− 4AB− 4ABTT+ I+ I3.15 References1. D. M. Barnett and R. J. Asaro, “The Fracture Mechanics of Slit-like Cracks in AnisotropicElastic Media”, Journal of Mechanics Physics and Solids, Vol. 20, pp. 353-366, 1972.2. D. M. Barnett, “” Structural and Deformation of Boundaries, Eds. K. A. Subramanian andM. A. Imam, pp. 31, AIME, 1985.3. J. D. Eshelby, W. T. Read, and W. Shockley, “Anisotropic Elasticity with Applications toDislocation Theory”, Acta Metall., Vol. 1, pp. 251, 1953.N.C. State Univ., Raleigh, NC 3-26 F. G. Yuan, July 28-31, 1998

Anisotropic Elasticity: Application to Composite Fracture Mechanics4. G. R. Irwin, “Analysis of Stresses and Strains near the End of Crack Traversing a Plate”,ASME, Journal of Applied Mechanics, Vol. 24, pp. 361-364, 1957.5. A. N. Stroh, “Dislocations and Cracks in Anisotropic Elasticity”, Phil. Mag. Vol. 3, pp.625-646, 1958.6. A. N. Stroh, “Steady State problems in Anisotropic Elasticity”, J. Math. Phys., Vol. 41,pp. 77-103, 962.7. K. A. Ingerbrigtsen and A. Tonning, “Elastic Surface Waves in Crystal”, Physics Review,Vol. 184, pp. 942-, 1969.Appendix AJ-integralnx 2r 2r 1x 1crackΓThe J-integral can be written asFig. 3.5 A cracked body and a contour around a crackJ=∫Γ( W n − t u11 dui= (itiui) ds2∫ σ2−,1(A1)Γ ds1 T T= (2du− t u,1) ds2∫ τΓSubstituting the expressions for τ 2 , u and t and using the orthogonality relationsii,1) dsBTTT TA + A B = I , B A + A B = 0(A2)J can be further expressed byJ1 δ⎡Re4⎢⎣It can be readily shown thatT −T= ∑∑gmB∫mnδ δzα m + n∫ dzα= ln( r2 / r1) + i 2Γδ⎤m + n −1( δm+ 1) ( δn+ 1) z dz B gnΓ⎥ (A3)⎦π1δ m= δ n= −(A4)2N.C. State Univ., Raleigh, NC 3-27 F. G. Yuan, July 28-31, 1998

Anisotropic Elasticity: Application to Composite Fracture MechanicsThe term contributing to J is the first term andT −T−1{[ln(r / r ) + i 2 g B B g }1J = Re2 1π ]1116(A5)By the use of the identity−T−1−1B B = −2iL(A6)eq. (A5) becomes1T −11 T −1J = Im{ [ln( r2/ r1) + i 2π ] g1L g1} = g1L g184(A7)The above derivation has used the real values of g 1 and L -1 . Sincewe have2g1 = k(A8)π1 −T 1J = k L k(A9)2Appendix BEnergy Release Rate Gx 2u(r,+π )x 1∆arx 2∆a-rx 1τ (∆a-r,0)2Fig. 3.6 Crack opening displacements for the extended crack and stress distribution ahead ofthe crack tip prior to crack extension.N.C. State Univ., Raleigh, NC 3-28 F. G. Yuan, July 28-31, 1998

Anisotropic Elasticity: Application to Composite Fracture MechanicsIrwin (1957) realized that if a crack extends by a small amount ∆ a , the energyabsorbed in the process is equal to the work required to close the crack to its original length.Using a polar coordinate system with the origin at the extended crack tip, the energy releasedfor a unit area to extend is written as1 aG = lim σi2(∆a− r,0)[ui( r,π ) − ui( r,−π)] dra 0 2∆a∫ ∆(B1)∆ → 0where Gis the energy release rate; ∆ais the crack extension at the crack tip;σi2(∆a− r,0)are the stresses prior to the crack extension; u i( r,± π ) are the displacementsdue to extension. (B1) can be rewritten by1 aTG = lim2( ∆a− r,0)[( r,π ) − ( r,−π )] dra 0 2∆a∫ ∆τu u(B2)∆ → 0By Stroh formulation, the stress vector τ 2 is written asδm−1τ2( r.θ ) = φ,1= Re ∑ B ( δm+ 1)z B gmAlong the crack plane, θ = 0,τ ( ∆a− r,0)=2==∑mm=1,3,5, ⋅⋅⋅∑j=0,1,2, ⋅⋅⋅m=1,2,3, ⋅⋅⋅mm=1,2,3, ⋅⋅⋅( δ∑( j +( δ+ 1)( ∆a− r)1)(2+ 1)( ∆a− r)∆a− r)N.C. State Univ., Raleigh, NC 3-29 F. G. Yuan, July 28-31, 1998δmgmj−1/2δgmRe[ gNote that g m is pure imaginary for m = 2, 4, 6, ⋅⋅⋅. For the displacement vector u,At the crack flanks, θ = ± π ,Thus= 2= 2= 2u = Re[z = r )= Re{= Re{n=1,3,5, ⋅⋅⋅n=1,3,5, ⋅⋅⋅l=0,1,2, ⋅⋅⋅∑An=1,2,3, ⋅⋅⋅± iπcos( ± π = re ,∆u= u(r,π ) − u(r,−π)n=1,2,3, ⋅⋅⋅n=1,2,3, ⋅⋅⋅∑∑∑∑∑rδ n + 1( −1)[ A rrδn+ 1( −1)( n−1) / 2l( −1)Lδn+ 1 i(δ n + 1) π−1e( n−1)/2−1l+1/ 22l+1zδ n + 1zRe[ LB−1rgn2 j+1]emδ n + 1 δ n + 1 ± i(δn+ 1)π=− r−1δn+ 1 −i(δ n + 1) π2isin[( δ + 1) π ] ABgLgrnrδn+ 1ne−1− iSL−1 −1−1where the identity − AB = SL + iLhas been used in deriving eq. (B4).−1gn] g}nB−1g]n]}(B3)(B4)

Anisotropic Elasticity: Application to Composite Fracture MechanicsSubstituting (B3), (B4) into (B2)11−∑∑∫ ∆ al T 1lj rG = lim 2( j + ) ( −1)g2j+1L g2l+1r ( ∆a− r)dr (B5)∆a→02∆a 0j l 2∆a− rIntroducing a nondimensional variable, x = r / ∆a111l T −1j+l+1 l j xG = lim ∑∑2(j + ) ( −1)g2j+1L g2l+1(∆a)∫ x (1 − x)dx∆a→0 2∆a0j l 21 − xIn the integrand, the linear term of ∆a corresponds to the first term or j = l = 01 T −1πG = lim [ g1L g1∆a+ o(∆a)]∆a→02∆a2(B6)1 T −1= k L k2For the monoclinic solid with plane of symmetry at x 3 = 0, L -1 under in-plane deformation canbe expressed by−1⎡bd⎤L = s 11′ ⎢ ⎥(B7)⎣de⎦G =12s′111= s′11[ek21= s11′2[ k , k ]212µ + µ = a + ib,µ µ = c + id12e = ad − bc = Im[ µ µ ( µ + µ )]112{ Im[ µ µ ( µ + µ )] k + 2 Im( µ µ ) k k + Im( µ + µ ) k }11+ 2dkk2⎡b⎢⎣d121d⎤⎡k2⎤e⎥ ⎢k⎥⎦ ⎣ 1 ⎦+ bk222]1This expression can be reduced to orthotropic solids without cross term of k 1 and k 2 in eq.(B8). This special case has been formulated from Lekhnitskii formulation and the results are ineq. (26) of a paper entitled “On Cracks in Rectlinearly Anisotropic Bodies” by G. C. Sih, P.C. Paris, and G. R. Irwin, International Journal of Fracture, Vol. 1, pp. 189-203, 1965.22112122122 2(B8)N.C. State Univ., Raleigh, NC 3-30 F. G. Yuan, July 28-31, 1998

Anisotropic Elasticity: Application to Composite Fracture MechanicsChapter 4 Determination of Stress Coefficient Terms in Cracked Solidsfor Monoclinic Materials with Plane Symmetry at x 3 = 04.1 Abstract4.2 Introduction4.3 Mathematical formulation4.4 Crack-tip fields4.5 Auxiliary fields associated with the crack tip fields4.6 J-integral4.7 Betti’s reciprocal theorem4.8 ReferencesAppendix A: Determining the unknown coefficients in the crack field using the J-integralAppendix B: Determining the unknown coefficients in the crack field using Betti’s theoremAppendix C: Deformation field for degenerate materials4.1 AbstractDetermination of all the coefficients in the crack tip field expansion for monoclinicmaterials under two-dimensional deformation is presented in this paper. For monoclinic materialswith a plane of material symmetry at x 3 = 0, the in-plane deformation is decoupled from the antiplanedeformation. In the case of in-plane deformation, utilizing conservation laws of elasticityand Betti’s reciprocal theorem, together with selected auxiliary fields, T-stress and third-orderstress coefficients near the crack tip are evaluated first from path-independent line integrals. Todetermine the T-stress terms using the J-integral and Betti’s reciprocal work theorem, auxiliaryfields under a concentrated force and moment acting at the crack tip are used respectively.Through the use of the Stroh formalism in anisotropic elasticity, analytical expressions for all thecoefficients including the stress intensity factors are derived in a compact form that hassurprisingly simple structure in terms of one of the Barnett-Lothe tensors, L. The solution formsfor degenerated materials, orthotropic, and isotropic materials are also presented.4.2 IntroductionThe use of fracture mechanics to assess the failure behavior in a flawed structure requiresthe identification of critical parameters which govern the severity of stress and deformation fieldin the vicinity of the flaw, and which can be evaluated using information obtained from the flawgeometry, loading, and material properties. In the linear elastic solids, stress intensity factors, k i (i= I, II, III), represent the leading singular terms in the Williams eigenfunction expansion seriesnear a crack tip. k i are often assumed to be unique parameters associated with crack extension.The physical implications of the higher-order non-singular terms have been noted by Cotterell(1966). Especially, the so-called T-stress, second term of the crack tip stress field whichrepresents the constant normal stress parallel to the crack surfaces, has been found as anadditional parameter in characterizing the behavior of a crack (Larsson and Carlsson, 1973; Rice,1974). Cotterell and Rice (1980) showed that T-stress substantially influences the fracture pathstability of a mode-I crack. The stress biaxiality parameter (Leevers and Radon, 1982; Sham,1989 and 1991) has been tabulated as a function of relative crack lengths and overall geometry inmany fracture test specimens for the isotropic solid using computational techniques (e.g., Kfouri,1986 and Sham, 1991). Kardomateas et al. (1993) examined the third-term of the Williamssolution and concluded its significance in the center-cracked and single-edge specimens with shortcrack lengths.N.C. State Univ., Raleigh, NC 4-1 F. G. Yuan, July 28-31, 1998

Anisotropic Elasticity: Application to Composite Fracture MechanicsIn anisotropic linear elastic solids, Gao and Chiu (1992) examined the T-stress term of acrack in infinite orthotropic solids under mode-I loading. Because of the material anisotropyinvolved, the T-stress term is affected by the material properties. It is also expected that, ingeneral, mixed-mode crack behavior and the biaxiality parameter are also dependent on thematerial anisotropy. Thus, it is essential to develop efficient computational techniques todetermine T-stress term coefficients including the stress intensity factors in anisotropic crackedmaterials with finite geometry. In this paper, two methods based on the J-integral and Betti’sreciprocal theorem are proposed to obtain compact forms in calculating all the stress coefficientterms in the crack tip field expansion for monoclinic materials with a plane of material symmetryat x 3 = 0. To determine T-stress term using the J-integral, the method by Kfouri (1986) isextended to anisotropic solids. The closed form solution of the auxiliary field, a point force actingat the crack tip, is derived for this purpose. A path-independent integral based on the Betti’sreciprocal work concept has been used for determining the stress intensity factors by Stern,Becker, and Dunham (1976), Hong and Stern (1978), Sinclair, Okajima, and Griffin (1984) forisotropic materials; Soni and Stern (1976) for orthotropic materials; and An (1987) forrectlinearly anisotropic materials. This path-independent line integral is also extended to determineall the stress coefficient terms with auxiliary fields.4.3 Mathematical FormulationIn a fixed Cartesian coordinate system x i , (i = 1, 2, 3), consider a two-dimensionaldeformation of an anisotropic elastic body in which the deformation field is independent of the x 3coordinate. In this paper, attention focuses on the monoclinic material having three mutuallyperpendicular symmetry planes and one of the planes coinciding with the coordinate plane x 3 = 0.In this case, the in-plane and out-of-plane deformations are uncoupled. For in-plane deformationthe strain and stress relations can be written asTwhere ε = [ ε1 , ε2, γ12] , σ = [ σ11, σ22, σ12]orε = s ′ σ(4.1)Tε = s′ σ , i, j = 1, 2, 6i ij jwhere s′= s′are reduced compliance coefficients defined by s ′= s − s s / s .ij jiij ij i3 j3 33Throughout this paper, all indices range from 1 to 2 and the summation convention isapplied to repeated Latin index unless otherwise noted. The bold-face letters are used to representmatrices or vectors. A comma stands for differentiation; overbar denotes complex conjugate. Asymbol Re stands for real part; Im for imaginary part.In the absence of body forces, general solutions of the displacement vector u, the stressfunction φ, and stresses σ , for in-plane deformation, according to Stroh formalism (Ting, 1996),can be represented byN.C. State Univ., Raleigh, NC 4-2 F. G. Yuan, July 28-31, 1998

Anisotropic Elasticity: Application to Composite Fracture Mechanicsoru = Re[φ = Re[2∑α=12∑u = Reφ = Reabααα = 1ddααf ( zf ( zαα[ A f ( z)d][ B f ( z)d])])](4.2)(4.3)σ =− φ , σ = φ(4.4)i1 i, 2 i2 i,1wheref ( z)= diag[f ( z1),f ( z2)]zα= x1 + µ αx, Im[µ 2 α ] > 0f(z) is an arbitrary function, d is a unknown complex constant vector to be determined. µ α ,a α , and b α are the Stroh eigenvalues and corresponding eigenvectors determined by elasticconstants only. For in-plane deformation, µ α are given by the roots of the characteristics equation:( )432s′ µ − 2s′ µ + 2s′ + s′ µ − 2s′ µ + s′ = 0(4.5)111612 6626 22with positive imaginary parts. From energy consideration, Lekhnitskii (1963) showed that theroots are either complex or purely imaginary and cannot be real. A and B are Stroh matrices givenbyA= [ a a ] = ⎡ p⎣ ⎢ 1p2⎤1 2q⎥1q2⎦(4.6)⎡−µ1− µ2 ⎤ −11 ⎡−1 − µ2 ⎤B = [ b1,b2] = ⎢ ⎥ , B =−⎢ ⎥ (4.7)⎣ 1 1 ⎦ µ1µ2 ⎣ 1 µ1 ⎦p = ′ ′ ′ ′ ′ ′ . (4.8)2αs11µ α− s16µα+ s12, qα= s12µα− s26+ s22/ µαThe eigenvectors a α and b α are unique to an arbitrary multiplier. Introducing normalizationfactors k α , we haveThe values of k α satisfy the conditions⎡k1p1k2p2⎤ ⎡−k1µ1− k2µ2 ⎤A = ⎢ ⎥ , B =⎣k1q1k2q⎢⎥ (4.9)2 ⎦ ⎣ k1k2⎦2a Tαbβ= δαβN.C. State Univ., Raleigh, NC 4-3 F. G. Yuan, July 28-31, 1998

Anisotropic Elasticity: Application to Composite Fracture MechanicsBased on the Stroh formalism, the matrices A and B defined in eq. (4.9) satisfyorthogonality relations. For in-plane deformation, these relations can be expressed byBTTA + A B = I(4.10)whereIt can be proved thatT TB A + A B = 0 (4.11)222 k ( q1− p1µ 1)= 1, 2 k2( q2− p2µ2)1=1B− TB− 112i−= − L, L −1 = − Im[ AB−1 ](4.12)where L is a real, symmetric, and positive definite matrix which will be used frequently in thesequel.Note that the normalization factors cancel each other for the term AB -1 in the secondequation of (4.12). Therefore there is no need to introduce the normalization factors in computingL -1 . From eq. (4.12) 2 , it is easy to getL−1= ′ ⎡ b dL⎣ ⎢ ⎤1 ⎡ e − d⎤s11⎥ , =⎦ ′ −2d e s⎢⎣−⎥(4.13)11( be d ) d b ⎦µ + µ = a+ ib, µ µ = c+id,1 2 1 2e = ad − bc = Im[ µ1µ 2( µ1+ µ2)](4.14)For crack problems, it may be convenient to introduce a complex potential function Φ(Guo, 1991) such thatΦ = B-1f () z B g(4.15)where g = Bd, then the displacement expression in eq. (4.3) and stresses in eq. (4.4) can berewritten in terms of the potential function Φσu= Re[ AB -1 Φ ](4.16)=− Re[ Φ ], σ = Re[ Φ ](4.17)i1 i, 2 i2 i,1The traction vector t at a point on a curve Γ with unit outward normal n is given by⎡dΦi⎤ ⎡dΦ⎤ti=−Re ⎢⎣ ds ⎥ , t =−Re⎦ ⎣⎢ ds ⎦⎥(4.18)where s is an arc length measured along Γ as shown in Fig. 4.1. Thus, without loss of generality,the traction free boundary conditions on a boundary may be written asRe[ Φ ] = 0 on Γ (4.19)N.C. State Univ., Raleigh, NC 4-4 F. G. Yuan, July 28-31, 1998

Anisotropic Elasticity: Application to Composite Fracture Mechanicsx 2tnΓsFig. 4.1 The surface traction t on a curved boundary Γ with a unit outward normal vector n.The resultant force and moment about the x 3 axis due to the surface traction t acting on Γbetween s 1 and s 2 (s 2 > s 1 ) arex 1s2∫ t( sds ) = Re[ Φ( s1) −Φ( s2)]s1(4.20)s2s2∫ ( xt1 2− xt2 1) ds= −Re[ x1Φ2 −x2Φ1−χ ](4.21)ss11wherezχ() z = ∫ Φ 2( λ)dλIf Γ encloses a region and there are concentrated force f and moment M inside the region, then theequilibrium of the body demands that− ∫ t()sds=Γf(4.22)−∫ ( xt − )1 2xt2 1ds=MΓ(4.23)4.4 Crack-tip fieldsConsider a crack in the anisotropic body. Let a coordinate system be attached to the cracktip and crack plane lies on the x 1 - x 3 coordinate plane. The configuration is shown in Fig. 4.2. Thecrack faces are assumed to be traction-free. Note that the crack plane may not coincide with thesymmetry plane of the material.N.C. State Univ., Raleigh, NC 4-5 F. G. Yuan, July 28-31, 1998

Anisotropic Elasticity: Application to Composite Fracture Mechanicsnx 2r 2r 1x 1crackΓFig. 4. 2 A cracked body and a contour around a crackTo find the solution for the crack tip field, employing eq. (4.15) and (4.16) withf ( zδ) = z, we take the solutions u and Φ in the formα α α+ -1u=Re[ A z δ 1 B g] (4.24)Φ = B+ -1z δ 1 B g(4.25)where the complex variable z α is defined byzα= r(cosθ + µ sin θ) , −π ≤θ ≤παg is a complex constant vector and δ is the complex constant. We seek admissible values of δsubjected to a restriction in which the strain energy is bounded as r → 0, that is− 1

Anisotropic Elasticity: Application to Composite Fracture MechanicsorWith eq. (4.30), from eq. (4.26) or (4.27), we haveSince g is a constant, from eq. (4.31), (4.30),Re( δ ) = ( n −2) / 2, n = 1, 2, 3, ⋅⋅⋅ (4.30)i 2 Im( δ )g = −r cos[2πRe( δ )] g(4.31)Im( δ ) = 0 , δ = ( n −2)/2 (4.32)⎪⎧g,g = ⎨⎪⎩ − g,1 1 3δ = − , , , ⋅⋅⋅2 2 2δ = 0 , 1, 2 , ⋅⋅that is, δ is real, the vector g associated with δ is real if δ = -1/2, ½, 3/2, …; and g is pureimaginary if δ = 0, 1, 2, …. It is clear from the determinant given in eq. (4.29) that each of theeigenvalues δ is a root of multiplicity two. Since g has two arbitrary components, we have twoindependent eigenfunctions associated with the double eigenvalues δ. Therefore the assumedforms of u and Φ given by eq. (4.24) and (4.25) are justified; no logarithmic type of solution formexists.Superimposing all the solutions with different orders of r, the crack tip field can beconstructed asu =Φ =∑n=1∑n=1Re[ A zBzδn+ 1δn+ 1B−1Bg−1ngn](4.33)where δ n= ( n −2)/2. gnis real for n = 1, 3, 5, ⋅⋅⋅, gnis pure imaginary for n = 2, 4, 6,⋅⋅⋅, and gnare dependent on the geometry of the cracked body, material properties, and loading conditions.Similarly, there is no need to introduce k α in calculating σ ij and u i from eq. (4.33).Performing algebraic calculation and defining gcomponents can be written asσσσ112212===n=T[ gn1 , gn2], the stress and displacementδ ⎧ 1n2 δ n 2 δ nδ n δn∑ ( δn+ 1) r Re⎨[ gn1( µ2ς2− µ1ς1) + gn2µ1µ2( µ2ς2− µ1ς1)]n=1⎩µ1− µ2δ ⎧ 1nδnδ nδ n δ n∑ ( δn+ 1) r Re⎨[ gn1( ς2− ς1) + gn2( µ1ς2− µ2ς1)]n=1⎩µ1− µ2δ ⎧ 1nδ n δ nδnδ n∑ ( δn+ 1) r Re⎨[ gn1( µ1ς1− µ2ς2) + gn2µ1µ2( ς1− ς2)]n=1⎩µ1− µ2⎫⎬⎭⎫⎬⎭⎫⎬⎭(4.34)N.C. State Univ., Raleigh, NC 4-7 F. G. Yuan, July 28-31, 1998

Anisotropic Elasticity: Application to Composite Fracture Mechanicsuu21==∑n=1∑n=1rr⎧Re⎨⎩µ1− µ2δ + 1 1n⎧Re⎨⎩µ1− µ2δ + 1 1nδ n + 1 δ n + 1δ n + 1δn+ 1[ g ( p ς − p ς ) + g ( µ p ς − µ p ς )]n12211δ n + 1 δ n + 1δ n + 1δn+ 1[ g ( q ς − q ς ) + g ( µ q ς − µ q ς )]n12211n2n2112222221111⎫⎬⎭⎫⎬⎭(4.35)where ςα= cos θ + µ sinθ.αFor the leading-order term, δ 1 = -1/2, letting g1 = 2 / π k , the singular solution and T-stress term for the monoclinic solids can be written in the form:⎡1⎤(1) (1)1/( k σ + k ) + T⎢0⎥O()1 21 I+ rσ =IIσII2πr⎢ ⎥⎢⎣0⎥⎦111 21 I 12+ r(1) (1)1/( k ε + k ) + T⎢s′⎥O()ε =IIεII2πr⎡s′⎤⎢ ⎥⎢⎣s′16⎥⎦(4.36)(4.37)ε = s′ ,(1) (1)Iσ Iε = s′(1) (1)IIσ II2r(1)(1 ⎡s11′ cos s16′)θ + sinθ⎤ ⎡− sinθ⎤ 3 / 2u = ( kIuI+ kIIuII) + T r ⎢+ r + O(r )s12sin⎥ ω ⎢cos⎥ (4.38)π⎣ ′ θ ⎦ ⎣ θ ⎦( n)( n)where σ , u , ⋅⋅⋅,are the n-th order terms of the crack tip field; the subscripts I or II indicatesthe distribution associated with mode-I or mode-II. k = [ k k ]T ; k I and k II are stress intensityfactors for mode-I and mode-II respectively, ω is a constant representing rigid body rotation, andII ,I( σ )( σ1122( σ )12)III⎡ µ ⎛1µ2= Re⎢⎜⎢⎣µ1− µ2 ⎝⎡ 1 ⎛= Re⎢⎜⎢⎣µ1− µ2 ⎝⎡ µ ⎛1µ2= Re⎢⎜⎢⎣µ1− µ2 ⎝µ2ς2−µ1ς2−1−ς1µ ⎞⎤1 ⎟⎥ς1 ⎠⎥⎦µ ⎞⎤2 ⎟⎥,ς1 ⎠⎥⎦1 ⎞⎤⎟⎥ς2 ⎠⎥⎦( σ )( σ11 II22( σ )12)IIII⎡ 1 ⎛= Re⎢⎜⎢⎣µ1− µ2 ⎝⎡ 1 ⎛= Re⎢⎜⎢⎣µ1− µ2 ⎝⎡ 1 ⎛= Re⎢⎜⎢⎣µ1− µ2 ⎝2µ2ς2−1ς2−µ1−ς12µ ⎞⎤1 ⎟⎥ς1 ⎠⎥⎦1 ⎞⎤⎟⎥ς1 ⎠⎥⎦µ ⎞⎤2 ⎟⎥ς2 ⎠⎥⎦N.C. State Univ., Raleigh, NC 4-8 F. G. Yuan, July 28-31, 1998

Anisotropic Elasticity: Application to Composite Fracture Mechanics( u )1( u2)II⎡ 1= Re⎢⎣µ1− µ2⎡ 1= Re⎢⎣µ1− µ2( µ p ς − µ p ς )1⎤( µ q ς − µ q ς ) ⎥⎦12222221111⎤⎥⎦,( u )1( u2)IIII⎡ 1= Re⎢⎣µ1− µ2⎡ 1= Re⎢⎣µ1− µ2( p ς − p ς )2⎤⎥⎦⎤( q ς − q ς ) ⎥⎦2221111For the second-order term, δ 2 = 0,⎡ ′ ′(2)s11cosθ+ s16sin θ ⎤ ⎡− sin θ ⎤u = T r ⎢⎥ + ωr⎢ ⎥(4.39)⎣ s12′ sin θ ⎦ ⎣ cos θ ⎦(2) (2)σ = T , σ = σ 0(4.40)(2)11where ω represents the rigid body rotation, and22 12=T = −ig b + g d), ω = −is′( g d + g )(4.41)(21 2211 21 22eClearly the T-stress term is dependent on the material properties in anisotropic solids.For orthotropic materials where x 1 and x 2 coincide with the material symmetry axes, the T-terms become⎡ ′(2)s11cosθ⎤ ⎡− sinθ⎤ (2)u = Tr ⎢ ⎥ + ωr⎢ ⎥ , σ11= T⎣s12′ sinθ⎦ ⎣ cosθ⎦In isotropic solids, the second-order terms are(2) T r ⎡ cosθ⎤ ⎡− sinθ⎤ (2)u = ⎢ ⎥ + ωr⎢ ⎥ , σ11= T for plane stressE ⎣−νsinθ⎦ ⎣ cosθ⎦( 1 −ν)(2) T (1 + ν ) r ⎡ cosθ⎤ ⎡− sinθ⎤ (2)u = ⎢ ⎥ + ωr⎢ ⎥ , σ11= T for plane strainE ⎣ −νsinθ⎦ ⎣ cosθ⎦For the degenerate materials, the crack tip fields are derived in Appendix C.4.5 Auxiliary Fields associated with the Crack Tip FieldsSome auxiliary fields with higher-order singularities are needed in order to determine thecoefficients in the expansion of the crack tip field by the use of conservation laws of elasticity andBetti’s reciprocal theorem. Since the negative integers are also the eigenvalues of the crackproblem which satisfy zero traction on the crack surfaces and satisfy the field governing equationsof the anisotropic solids, the associated eigenfunctions can be conveniently used as auxiliary(pseduo) fields. Note that each eigenfunction with higher-order singularity has unbounded energynear the crack tip and thus corresponds to some concentrated source at the tip. This eigenfunctioncan be imagined as a self-equilibrated solution to the crack problem under some specified loads.These auxiliary fields may be obtained by choosing the values of n in eq. (4.33) as negativeintegers, that is,N.C. State Univ., Raleigh, NC 4-9 F. G. Yuan, July 28-31, 1998

Anisotropic Elasticity: Application to Composite Fracture Mechanicsmu a ∆ + 1= Re[ A za ∆m+ 1 −1Φ= BzBBh−1mhm](4.42)aσa=− Re[ Φ ], σ = Re[ Φ ](4.43)i1 i, 2 i2 i,1∆ m=− m /2, m = 1, 3, 4, ⋅⋅⋅, (4.44)hm=[m1, hm2h ]Twhere h m are arbitrary constant vectors. h mis real for m = 1, 3, 5, ⋅⋅⋅; h mis pure imaginary for m =4, 6, 8,⋅⋅⋅, Utilizing eq. (4.22-4.23), the auxiliary fields, except m = 4, defined by eq. (4.42) yieldzero resultant force on any contour Γ which encloses the crack tip shown in Fig. 4.2. Thecorresponding resultant moment about the x 3-axis, produced by the tractions acting on thecontour Γ is also zero for the auxiliary fields in eq. (4.42). The special case of ∆ m = -2 or m = 4Tcan be also directly explained from eq. (4.23). In this case, the function associated with [ 0, h42]corresponds to the particular solution for a crack under a concentrated moment about x 3 -Taxis, ( −2π ih42), applied at the crack tip; the function associated with [ h 41, 0]corresponds tothe homogeneous solution which satisfies zero concentrated force and moment at the crack tip.From eq. (4.42)-(4.44), the stress and displacement components of the auxiliary fields areσσσ( a )11( a )22( a )12uu= ( ∆= ( ∆= ( ∆mmm+ 1) r+ 1) r+ 1) r∆m∆m∆m⎧ 1Re⎨⎩µ1− µ2⎧ 1Re⎨⎩µ1− µ2⎧ 1Re⎨⎩µ1− µ2⎧Re⎨⎩µ1− µ22 ∆m2 ∆m∆m∆m[ h ( µ ς − µ ς ) + h µ µ ( µ ς − µ ς )]m1∆m∆m∆m∆m[ h ( ς − ς ) + h ( µ ς − µ ς )]m1∆m∆m∆m∆m[ h ( µ ς − µ ς ) + h µ µ ( ς − ς )]m12121212121m21m22m2∆m+ 1 ∆m+ 1∆m+ 1∆m+[ h ( p ς − p ς ) + h ( µ p ς − µ p )]a ∆ + 1 1m11= rm12 2 1 1m21 2 2 2 1ς1⎧Re⎨⎩µ1− µ2∆m+ 1 ∆m+ 1∆m+ 1∆m+[ h ( q ς − q ς ) + h ( µ q ς − µ q )]a ∆ + 1 1m12= rm12 2 1 1m21 2 2 2 1ς1112221122⎫⎬⎭21⎫⎬⎭1⎫⎬⎭⎫⎬⎭⎫⎬⎭(4.45)(4.46)In the following two sections, stress intensity factors, T-stress term, and coefficients ofhigher-order terms are determined using the J-integral and the Betti’s reciprocal theoremincluding the use of above auxiliary fields.4.6 J-Integral(a) T-stress TermN.C. State Univ., Raleigh, NC 4-10 F. G. Yuan, July 28-31, 1998

Anisotropic Elasticity: Application to Composite Fracture MechanicsJ i conservation laws (Knowles and Sternberg, 1972) for a plane anisotropic elasticityproblem may be written as( )Jk = ∫ Wnk − tiui, kds = 0 , k = 1, 2 (4.47)Cfor an arbitrary closed contour C that encloses no defects, cracks, or material inhomogeneities. Inthe above equations, W is the strain energy density, W = σijε ij/2, where σ ij and ε ij are the stressesand strains respectively; t i are the traction components defined along the contour, t i = σ ij n j; n k arethe unit outward vector normal to the contour path. Letting k = 1, the conservation law isreduced to the Rice’s path-independent J-integral or the rate of energy release rate per unit ofcrack extension along the x 1 - axis, which is given by( )TT/ 2 t u,J = ∫ σ εn 1− 1dsΓ(4.48)where Γ is an arbitrary path which starts on the straight lower face of the crack, encloses thecrack tip and ends on the upper straight face with the positive direction in a counterclockwisedirection shown in Fig. 4.2. Here, the crack surfaces are assumed to be traction-free.Consider a cracked body under the two-dimensional deformation. The stress, strain, anddisplacement fields are represented by σ ij , ε ij , and u i , respectively. As r → 0, the asymptotic fieldsincluding the constant T-stress terms are given before. Now the coefficients of the T-stress termsand third-order terms are derived using conservation laws.In general, for the purpose of determining the coefficients g n of the term r δ n ( δ n≥−1/ 2 ) inthe actual crack tip stress field, we may employ the J-integral method and follow the followingprocedure:(i) find an auxiliary (pseudo) field that has singularity σ a −δn−ij~ r 1 as r → 0. It is convenient toselect auxiliary stress field which gives zero traction on the crack surfaces and contains only−the stress singular term rn −δ 1 ;(ii) superimpose the actual field (the mixed-mode boundary value problem, in general) on theauxiliary field and represent the J-integral for the superimposed state aswhereJs = J + Ja + JM(4.49)a T aa TaJs= ∫ σ + σ ) ( ε + ε ) n / 2 − ( t + t ) ( u + )] dsΓ[(1 ,1u,1TTJa= ∫ [( σ a ) ε an / − ( a) a]12 t u ,1dsΓandJM= Js − J − JaT a a T T a a T= ∫ {[ σ ε + ( σ ) ε] n1 / 2 −t u,1−( t ) u,1}dsΓN.C. State Univ., Raleigh, NC 4-11 F. G. Yuan, July 28-31, 1998

Anisotropic Elasticity: Application to Composite Fracture Mechanics= ∫ [ σ T εa − T a− ( a ) Tn]1t u,1t u,1dsΓ(4.50)= ∫ ( aa aσ − − )iju i , jn1 tiui, 1tiui,1dsΓwhere the superscript or subscript “a” denote quantities referred to the auxiliary field; Js is the J-integral for the superimposed state; J for the actual state; and J a for the auxiliary field and J M is theinteraction integral. In the sequel, we assume that the J-integral is path-independent for both theactual field and the selected auxiliary fields, denoted by J and J a . Then the integral J s for thesuperimposed state, thus J M , is also path-independent. If the auxiliary fields given by eq. (4.42)are used, it is readily proved thatJJ aa≠ 0, for ∆ = −1/2= 0 m ; (4.51), for ∆ < −1/2m(iii) evaluate J M as Γ → 0. For simplicity, Γ may be taken as a circle with radius r, as r → 0, theonly terms in the integrand that contribute to J M are the cross terms between r δnin the actual−δ stress field and the auxiliary stress term with order rn − 1 ;(iv) carry out the routine manipulation, the exact expression for J M can be obtained asJ = J ( g ) when r → 0;M M n(v) evaluate J M for a finite contour Γ using the computed actual field and the exact auxiliarysolution; and determine the coefficients g n from the value of J M and the expression ofJ = J ( g ) as r → 0.M M nIn extracting the T-stress in eq. (33) or g 2 , we make use of another auxiliary field, that is thea −1solution to a point force f (per unit thickness) applied at the crack tip. In this case, σij∝ r . Notethat the point force f must be resisted by traction t applied to some boundary C in achievingequilibrium. In the Stroh formalism (Ting, 1996), the real form solutions due to the point-forceapplication can be written asa ⎡lnr⎤2u= −⎢+ S(θ )⎥h⎣ π I⎦(4.52)a2φ = L ( θ )h(4.53)where−1h = L f ,f=[1 2f , f ]T2TS(θ ) = Re[ A ln(cosθ+ µ sinθ) B ]π2TL(θ ) = − Re[ B ln(cosθ+ µ sinθ) B ]πx 1 = rcosθ, x 2 = rsinθ.It assumes the valuesN.C. State Univ., Raleigh, NC 4-12 F. G. Yuan, July 28-31, 1998

Anisotropic Elasticity: Application to Composite Fracture Mechanics⎧ 0 θ = 0ln(cos θ + µ α sinθ) = ⎨⎩±iπθ = ± πIn a cylindrical coordinate system (r, θ, x 3 ), let t r and t θ be the traction vectors on acylindrical surface r = constant and on a radial plane θ = constant, then1 atr= − φ, θ ,ratθ(4.54)= φ ,ra T a T Tσr= n tr, σr θm tr= n tθTTwhere n = [cosθ, sinθ] and m = [ − sinθ, cosθ].It follows from eq. (4.52) and (4.53) that= , σa θ= m T tθatθ= φ, r= 0 , σθσθ= 0σ φ ,θ= a ra T ar= −n /roraσr[( µ µ −1) sinθ+ ( µ + µ ) cosθ]2f ⎧⎫1 1 21 2cos θ − sinθ= − Im⎨⎬2πr ⎩ς1ς2⎭f ⎧⎪− Im⎨2πr⎩⎪[( 1) ( ) ]2µµ − cosθ − µ + µ sinθ sin θ + µµ ⎫1 2cosθ⎪⎬ςς1 2⎭⎪2 1 2 1 2(4.55)andauf ⎪⎧⎡ b1= − ⎨ s11′ ln r2π⎢⎪⎩ ⎣ df ⎪⎧2⎡ d− ⎨ s ′11ln r2π⎢⎪⎩ ⎣ e⎤⎥⎦⎤⎥⎦−−ImIm⎡⎢⎣⎡⎢⎣1µ − µ111µ − µ22⎡⎢⎣pq22⎡ µ1⎢⎣ µ1ln ςln ςpq2222ln ςln ς−−22p ⎪⎫1ln ς1 ⎤ ⎤⎥ ⎥ ⎬q1ln ς1 ⎦ ⎦ ⎪⎭− µ2p− µ q211ln ς ⎪⎫1 ⎤ ⎤⎥ ⎥ ⎬ln ς1 ⎦ ⎦ ⎪⎭(4.56)For isotropic materials,a = aa 1σrθσθ= 0, σr= − ( f1 cosθ+ f2sinθ)(4.57)π r2⎪⎧⎪⎫a 1 ⎡(κ + 1) ln r − ( κ −1)θ ⎤ ⎡− 2sin θ sin 2θ⎤ ⎡ f1⎤u = − ⎨⎢⎥ − ⎢2 ⎥⎬⎢⎥ (4.58)8πG ⎪⎩ ⎣ ( κ −1)θ ( κ + 1)ln r⎦⎣ sin 2θ2sin θ ⎦⎪⎭⎣ f2⎦κ = 3 − 4νN.C. State Univ., Raleigh, NC 4-13 F. G. Yuan, July 28-31, 1998

Anisotropic Elasticity: Application to Composite Fracture MechanicsBy superimposing the actual field on the auxiliary field, and using the path-independent J-integral for the monoclinic elastic cracked body, it can be proved thatJs = J + Ts′ f + ω f11 1 2JM= Ts11 ′ f1 + ω f2(4.59)Note that because σ ija∝r u r−1 a −1,i,j∝a a( i i )a,Ja= ∫ W n1 − t u1ds = 0ΓKfouri (1986) used the method to calculate the T-term for isotropic materials. Wang et al.(1980) and Wu (1989) applied the J-integral to determine the stress intensity factors forrectilinear anisotropic solids and general anisotropic materials respectively. In this paper, themethod is extended to determine all the coefficients in the crack tip field expansion for monoclinicmaterials. From eq. (4.59), it follows thatT =JMs′f 2 = 0f11 1and ω =JM f1 = 0f2Detailed proof of eq. (4.59) is given below:For general anisotropic linear elastic solids, we have the following relationswhere c ijkl = c klij. From eq. (4.50),aa aaσε = c εε = σε and σ ε σ uijijijkl kl ijijijijija=ij i , jaaa aduiaJM= ∫ ( σij ui, jn −tiui, − tiui, ) ds= ∫ ( σi−tiui,) dsΓ1 1 1Γ2 1(4.60)dswhere dua / ids is the tangential derivative of u a . As r → 0, we evaluate J and note that thei Monly terms that contribute to J M are the cross terms between T and f. After substituting these fieldsinto the integral and performing the routine algebra, the integral J M may be evaluated asJMaduia= limi−tiuids→∫ ( σ,)Γ Γ2 10 dsa( 2)duia ( 2)= ∫ ( σi−tiui,) dsΓ21ds( 2)a=− u t ds=−∫∫ΓΓi,1( 2)=−( u )( 2)= ( u ), 1TTi( 2)( u ), 1, 1f∫TΓttaadsds(4.61)In the above derivation, eq. (4.22) has been used. From eq. (4.33) and (4.39) via (4.12) 2 ,N.C. State Univ., Raleigh, NC 4-14 F. G. Yuan, July 28-31, 1998

Anisotropic Elasticity: Application to Composite Fracture Mechanics⎡ ′(2) −1Ts11⎤u,1= −iL g2= ⎢ ⎥(4.62)⎣ ω ⎦Insertion eq. (4.62) into (4.61) leads to−1T −1J = −iL f = −if L g(4.63)Mg T 2and J M= T s′11f1+ ω f2(4.64)Eq. (4.63) and (4.64) can be used to calculate g 2 and T. As f is arbitrary, it is convenient tochoose f to be the following valuesT[ 1, 0] ≡ eT1, [ 0 , 1] ≡ e2respectively. Note that e k has a dimension force/length. Correspondingly, eq. (4.63) yields twolinear equations and they are, in matrix notation,~-1JM=−i L g2where~ () 1 ( 2)TJ M= [ J M, J M]2( k )and J Mis the value of J M when f = e k . Therefore,~g 2= i LJ M(4.65)Using eq. (4.64), The T-stress and ω can be obtained as(1)T J / s′M 11= ,(2)ω = J M(4.66)However, the choice of auxiliary field is not unique. For instance, we may choose theauxiliary fields as the sum of the field ( σ ′, u ′)for the point force f and a field ( σ ′′, u′′) which is aknown solution for the same cracked body under some loads on the outer boundary, thenaσ = σ′ + σ′′au = u′+ u ′′,Superimposing the actual field on the new auxiliary field, we haveFrom the definition of J M ,aaaJ [ σ + σ ] = J[ σ] + J [ σ ] + J [ σ, σ ](4.67)saaJ [σ, σ ] = J [ σσ , ′+ σ′′ ] = J [ σσ , ′] + J [ σσ , ′′](4.68)MM M MMJ [ σ , σ ′′] = J [ σ + σ ′′] − J[ σ ] − J[ σ ′′](4.69)MsIn elastic materials, J is equal to the energy release rate G in the absence of body forces anddislocations and is related to the stress intensity factors throughN.C. State Univ., Raleigh, NC 4-15 F. G. Yuan, July 28-31, 1998

Anisotropic Elasticity: Application to Composite Fracture MechanicsTherefore, eq. (4.69) leads toG = k21 T −1LkJ M[ , ]T −σσ′′ = k L k′′1 (4.70)where k is the stress intensity factor for the state ( σ ,u), and k ′′ for ( σ ′′ ,u′′).Using eq. (4.70) and (4.64), eq. (4.68) can be expressed byJMaT −1[ σ , σ ] = T s′f + ω f + k L k ′′(4.71)Inserting eq. (4.71) into (4.67) and utilizing eq. (4.13) yield1112aaJ [ σ + σ ] = J[ σ] + J [ σ ] + Ts′ f + ω fsFor isotropic case,a11 1 2[ µ µII IIµµII I II Iµµ µ µI I ]+ s′ Im ( + ) k k′′ + ( k k′′+ k′′ k ) + ( + ) k k′′11 1 2 1 2 1 2 1 2(4.72)aaJ [ σ + σ ] = J[ σ] + J [ σ ] + Ts′ f + ω f ++ 2s′ ( k k′′ + k k′′) (4.73)sa11 1 2 11II II I IFor plane strain under mode-I loading, if f = (f 1 , 0), then eq. (4.64) and (4.73) reduce to the2results given by Kfouri (1986) and s ′ = (1 − ) / E(b) The third term11νThe third-term coefficient in the asymptotic solution can be also obtained from the J-a2integral method. An auxiliary field with singularity σ ~ O(r−3/ij) can be introduced by selecting m= 3 in eq. (4.42). By superimposing the actual field (the mixed mode boundary value problem) onthe auxiliary field, the interaction integral J M may be evaluated asFollowing in a similar manner,3= −2hLgT 1J −Mπ3 3(4.74)g32 ~= − L J M(4.75)3π~ )(1) (2 Twhere JM= [ JM, JM] andof force/(length) 1/2 .J is the value of J M when h = , ( k 1, 2)(k )M3e k=. ekhas a dimensionIn general, superimposing of an auxiliary field with σ a ij∝ r ∆ n on the actual field andapplying the J-integral to this combined state with derivations proved in Appendix A for n ≠ 2, wecan get the interaction integral J M denoted by J Mn, that isThen=− 2πδ( δ + 1)h L g , n = 1, 3, 4, 5, ⋅⋅⋅. (4.76)T −1J Mn n n n nN.C. State Univ., Raleigh, NC 4-16 F. G. Yuan, July 28-31, 1998

Anisotropic Elasticity: Application to Composite Fracture Mechanicsgn⎧~LJMn−n = ⋅⋅⋅⎪n( n+ ) , 1 , 3 , 52πδδ 1,= ⎨ ~⎪i LJMnn = ⋅⋅⋅⎩⎪ 2n( n+ 1) , 4 , 6 , 8πδ δ,(4.77)where~ (1) (2JM n= [M n, J)M nJ ]T( k )and J Mnis the value of J Mn whenhn⎧ e= ⎨⎩iekk, n = 1, 3, 5,⋅⋅⋅, n = 4, 6, 8,⋅⋅⋅Here, e k possesses dimension force /( length) 1−δ n. For the first singular term, introducing stressTintensity factors, k = [ kII, kI] = π / 2 g 1for the actual field and k a = a a T[ k , k ] =II Iπ / 2h 1forthe auxiliary field, J M1 and k can be rewritten from eq. (4.76), (4.77) asa T −1J M 1= ( k ) L kk= LJ$ M1(4.78)whereJ ˆ = (1) (2)TM1 [ JM1, JM1] and J ( k )M1is the value of J M1whenak = ek.4.7 Betti’s Reciprocal Theorem(a) T-stress TermFor a linear elastic plane problem, Betti’s reciprocal theorem can be stated as∫Ca( t u t a u)⋅ − ⋅ ds = 0 (4.79)where C is an any closed contour enclosing a simple connected region in the elastic body; u is thedisplacement vector and t the traction on C corresponding to the solution of any particular elasticboundary value problem for the elastic body; u a and t a are corresponding quantities of the solutionof any other problem for the body. Considering a crack in an anisotropic linear elastic material,and suppose the crack surfaces are free of tractions for both elastic states. If the closed contour Cencloses the crack tip and extends along the crack surfaces, then it can be shown that the integrala a( )I = ∫ t⋅u −t ⋅udsΓ(4.80)is path independent where Γ is an any path which starts from the lower crack face and ends on theupper. Let (t, u) be an actual state for the crack under consideration, then eq. (4.80) providessufficient information for determining the amplitude for each term in the asymptotic crack-tipfields if proper auxiliary solutions (t a , u a ) are provided. In this section the Betti’s reciprocal workcontour integral is used for computing stress intensity factors, T-stress and other higher-orderN.C. State Univ., Raleigh, NC 4-17 F. G. Yuan, July 28-31, 1998

Anisotropic Elasticity: Application to Composite Fracture Mechanicscoefficients for monoclinic materials. The procedure can be evaluated from the analysis asfollows.δFor determining the coefficients g n of the term r n ( δn≥−1/ 2)in the actual crack tip stressfield, an auxiliary (pseudo) field with σ a ij∝ − δrn −2anor u ∝ r−δ−1can be chosen. As r → 0, take aΓ as a circle around the crack tip and evaluate integral I. When r → 0, the only product betweengn and the auxiliary terms in the integrand given above can contribute to the integral I. Therefore,the expression for I = I(g n) can be obtained as r → 0. The value of I for a finite contour Γ shownin Fig. 4.2 is available from the numerical solutions for t and u of the boundary value problemsand the exact auxiliary solution. The g n can be computed from the expression for I = I(g n) and thevalue of I.To determine the T-stress or g 2 for the crack-tip field from eq. (4.34), the auxiliary elastica −2field with stress singularity σij∝ r as r → 0 is used and can be obtained from the eq. (4.42) bychoosing m = 4, that is, in Stroh formalism,iuaΦ= Re[ A za= B z−1−1B−1Bh−14h ]4(4.81)The moment about x 3-axis applied at the crack tip, using eq. (4.23) and (4.81), is given byM=−2π i h42When Γ shrinks to the crack tip, it is clear that only those parts of the integrand in eq. (4.80)which behaves like O(1/r) as r → 0 can contribute this portion of the integral. Substituting thesefields of the two states into eq. (4.80), performing the integration for the circle surrounding thecrack tip and evaluating the results in the limit of vanishing radius, the results may be derived, andr→0a aT −1( t ⋅ u − t ⋅ u) ds = −2h4L g2I = lim ∫π (4.82)Γg2i ~= L I(4.83)2π~ )(1) (2where I = [ I , I ] and I ( k ) is the value of I when h = i 4e , (k = 1, 2). (Dimension of e k k isforce).From eq. (4.13), (4.41), and (4.83),(1)IT = ,2 π s′11(2)Iω = (4.84)2 πFor isotropic materials, the auxiliary displacement vector and stress functions can be modified assinθ⎡1+ cos2θ=r⎢⎣ sin 2θsin 2θ⎤1 − cos2θ⎥⎦[ h ]aΦ Im4(4.85)N.C. State Univ., Raleigh, NC 4-18 F. G. Yuan, July 28-31, 1998

Anisotropic Elasticity: Application to Composite Fracture Mechanicsa⎡σ11⎤⎢ a ⎥ 1⎢σ22 ⎥ =2ar⎢ ⎥⎣σ12 ⎦⎡−2 cos3θcosθ⎢− 2sin 3θsinθ⎢⎢⎣− sin 4θ− sin 4θ⎤2(cos2θ−1)sin 2θ⎥Im⎥− 2sin 3θsinθ⎥⎦[ h ]4(4.86)⎡u⎢⎣ua1a2⎤ 1 ⎡−( κ + 1) cosθ+ 2sin 2θsinθ⎥ = −4⎢⎦ Gr ⎣ ( κ −1)sinθ− cos2θsinθ− ( κ − 1) sinϑ− cos2θsinθ⎤Im− ( κ + 1) cosθ− 2sin 2θsinθ⎥⎦[ h ]4(4.87)Then the path-independent integral I has the same form as eq. (4.82). Eq. (4.83) and (4.84) arestill valid.(b) The Third TermThe coefficients of the third term in the eigenfunction expansion of the stress field can alsobe obtained using Betti’s theorem. Selecting m = 5 in eq. (4.42), an auxiliary field with stress2singularity σ ~ r−5/ desired for this purpose can be obtained. Applying the Betti theorem ofaijreciprocity to the actual field and the auxiliary field and evaluating the integral I as Γ → 0 near thecrack tip, we obtainI = − π (4.88)T 13 h5L− g3Eq. (4.88) will be used to calculate g 3 for mixed-mode problem when the two proper auxiliaryfield solutions are provided. g 3 can be expressed in the form1 ~g3= − L I(4.89)3πApplying Betti’s reciprocal theorem to the actual fields and auxiliary fields withσ a ∆ij∝ + 2, the path independent I denoted by I n+2 can be evaluated byr nIt follows from eq. (4.90)whereT −1I n + 2=− 2π ( δ n+ 1) h n + 2L g n(4.90)~⎧ LIn+2⎪−, n = 1, 3 , 5 , ⋅⋅⋅2π( δn+ 1)gn= ⎨ ~(4.91)⎪iLIn+2, n = 2 , 4 , 6 , ⋅⋅⋅⎪⎩2π( δn+ 1)= [ I , I ]I ( 1) ( 2)Tj j j( k )and I jis the value of I j whenhn⎧ e= ⎨⎩iekk, n = 1, 3, 5,⋅⋅⋅, n = 2, 4, 6,⋅⋅⋅N.C. State Univ., Raleigh, NC 4-19 F. G. Yuan, July 28-31, 1998

Anisotropic Elasticity: Application to Composite Fracture MechanicsHere, e k possesses dimension force × ( length) δ n. The detailed proof is shown in Appendix B.For the first singular term from eq. (4.90) and (4.91), I3 and k can be written asπT −1T −1I 3=− h 3L g1 =− 2 h3L kπ(4.92)4.8 Referencesk =− 1 ~L I2π3(4.93)1. D. An, “Weight Function Theory for a Rectilinear Anisotropic Body”, International Journalof Fracture, Vol. 34, pp. 85-109, 1987.2. B. Cotterell, “Notes on the Paths and Stability of Cracks”, International Journal of Fracture,Vol. 2, pp. 526-533, 1966.3. B. Cotterell and J. R. Rice, “Slightly Curved or Kinked Cracks”, International Journal ofFracture, Vol. 16, No. 2, pp. 155-169, 1980.4. H. Gao and C. H. Chiu, “Slightly Curved or Kinked Cracks in Anisotropic Elastic Solids”,International Journal of Solids and Structures, Vol. 29, No. 8, pp. 947-972, 1992.5. C. C. Hong, and M. Stern, “The Computation of Stress Intensity Factors in DissimilarMaterials”, Journal of Elasticity, Vol. 8, No. 1, pp. 21-34, 1978.6. G. A. Kardomateas, R. L. Carlson, A. H. Soediono and D. P. Schrage, “Near Tip Stress andStrain Fields for Short Elastic Cracks”, International Journal of Fracture, Vol. 62, pp. 219-232, 1993.7. A. P. Kfouri, “Some Evaluations of the Elastic T-term using Eshelby’s Method”,International Journal of Fracture, Vol. 30, pp. 301-315, 1986.8. J. K. Knowles and E. Sternberg, “On a Class of Conservation Laws in Linearized and FiniteElastostatics”, Archive for Rational Mechanics and Analysis, Vol. 44, pp. 187-211, 1972.9. S. G. Larsson and A. J. Carlsson, “Influence of Non-Singular Stress Terms and SpecimenGeometry on Small-Scale Yielding at Crack Tips in Elastic-Plastic Materials”, Journal ofMechanics Physics and Solids, Vol. 21, pp. 263-277, 1973.10. P. S. Leevers and J. C. Radon, “Inherent Stress Biaxiality in Various Fracture SpecimenGeometries”, International Journal of Fracture, Vol. 19, pp. 311-325, 1982.11. S. G. Lekhnitskii, Theory of an Anisotropic Elastic Body, Holden-Day, San Francisco, 1963.12. J. R. Rice, “Limitations to the Small-Scale Yielding Approximation for Crack-Tip Plasticity”,”, Journal of Mechanics Physics and Solids, Vol. 22, pp. 17-26, 1974.13. T. L. Sham, “The Theory of Higher Order Weight Functions for Linear Elastic PlaneProblems”, International Journal of Solids and Structures, Vol. 25, No. 4, pp. 357-380,1989.N.C. State Univ., Raleigh, NC 4-20 F. G. Yuan, July 28-31, 1998

Anisotropic Elasticity: Application to Composite Fracture Mechanics14. T. L. Sham, “The Determination of the Elastic T-term using Higher Order Weight Functions”,International Journal of Fracture, Vol. 48, pp. 81-102, 1991.15. G. B. Sinclair, M. Okajima, and J. H. Griffin, “Path-Independent Integrals for ComputingStress Intensity Factors at Sharp Notches in Elastic Plates”, International Journal forNumerical Methods in Engineering, Vol. 20, pp. 999-1008, 1984.16. M. Soni and M. Stern, “On the Computation of Stress Intensity Factors in Fiber CompositeMedia Using a Contour Integral Method, International Journal of Fracture, Vol. 12, No. 3,pp. 331-344, 1976.17. M. Stern, E. B. Becker, and R. S. Dunham, “A Contour Integral Computation of Mixed-Mode Stress Intensity Factors”, International Journal of Fracture, Vol. 12, No. 3, pp. 359-368, 1976.18. T. C. T, Ting, Anisotropic Elasticity: Theory and Applications, Oxford University Press,Oxford, 1996.19. S. S. Wang, J. F. Yau, and H. T. Corten, “A Mixed-Mode Crack Analysis of RectilinearAnisotropic Solids using Conservation Laws of Elasticity”, International Journal ofFracture, Vol. 16, No. 3, pp. 247-259, 1980.20. K. C. Wu, “Representtions of Stress Inetnsity Factors by Path-Independent Integrals”, ASME,Journal of Applied Mechanics, Vol. 56, pp. 780-785, 1989.Appendix ADetermining the Unknown Coefficients in the Crack Field Using the J-integralThe path-independent integral J M from eq. (4.60) isLet the actual and auxiliary crack fieldsJMadui= ∫ ( σi−t aiuidsds)Γ 2 , 1(A1)Φ =∑Β f n() zn−B g n1 (A2)Φ a=fa −Β () z B h1 (A3)whereδn+f ()= z z1 , δ n= ( n −2)/2 , n = 1, 2, 3, ⋅⋅⋅.na∆ m +f ()= z z1 , ∆ m=− m / 2 , m = 1, 3, 4, ⋅⋅⋅.From the identity11Re( C) Re( D) = Re[( C + C) D] = Re[ C( D+D)]22where C and D are complex matrices, we have(A4)N.C. State Univ., Raleigh, NC 4-21 F. G. Yuan, July 28-31, 1998

Anisotropic Elasticity: Application to Composite Fracture Mechanicsanda aa Tσ i 2du i= duiΦi, 1= d u Φ,1−1a T= d [Re( AB Φ ) ]Re[ Φ,1]1Re[( AB ) ( cc . .)]2−1a T= dΦ Φ,1+Re[ ] ( ) Re[ ](A5)aadΦitiui, 1ds=−Re( ) ui,1dsdsa T−1=−Re( dΦ) Re[ AB Φ,1]1a T −1=− Re[( dΦ) (AB Φ,1+ c. c.)]2where c.c. denotes the complex conjugate of the preceding term, i.e.,(A6)ThereforeF + c. c.= F + F1a T a TJM=⎧−1−1Re d + c c + d + c c⎫⎨⎩∫[( AB Φ ) ( Φ,1. .) ( Φ ) ( AB Φ . .)] ⎬2 Γ,1⎭1 ⎡T −T a T−1 T −T a T−1= Re ⎢ ∑∫ ( h B df A B fn′ B gn+ h B df A B fn′B gn2Γ⎣ nT −T a T−1 T −T a T−1+ h B df B A fn′ B gn+ h B df B A fn′B gn)]1 ⎡T −T a T T−1= Re ( h B A B+ B A B g2⎢ ∑∫ df ( ) fn′n⎣ ΓnT −T a T T−1+ h B df ( A B + B A)f ′ B g )Using the orthogonality relations in eq. (4.10) and (4.11),nn]J M can be further rewritten asT TA B + B A =T TI and A B + B A= 0, (A7)Defining1 ⎡T −T a−1⎤JM= Re⎢∑∫h B df fn′B gn2Γ⎥⎣ n⎦1 ⎡T −Ta −1⎤= Re ⎢∑h B ∫ fn′() z df () z B gn2Γ⎥⎣ n⎦aδn+∆mR ≡ f ′( z ) df ( z ) = ( δ + 1)( ∆ + 1 ) z dzmn∫Γnα α n mΓα∫α(A8)it is readily shown that(a) n = m or δ n+ ∆ m+ 1=0,N.C. State Univ., Raleigh, NC 4-22 F. G. Yuan, July 28-31, 1998

Anisotropic Elasticity: Application to Composite Fracture MechanicsR(b) n ≠ m orδ n+ ∆ m+ 1≠0,i.e.RmnRmndz= ( δ + 1)( ∆ + 1) ∫α = − δ ( δ + 1)[ln( r2 / r1) + i2π];Γ zmn n m m mα( δn + 1)( 2−m) n m n m n m n m={[ r ( − )/ 2−r ( − )/ 2 ]cos(− −n− m ) / + i [ r ( )/ 2+ r ( )/ 221π 221 ]sin( n−m) π / 2 }n−m⎧ j( δn + 1)( 2−m) n−m n−m⎪( − )[( )/ 2 ( )/ 21r2−r1]for n− m=2 j= ⎨n−mj( δn + 1)( 2−m) n−m n−m⎪ i ( − )[( )/ 2 ( )/ 21r2+ r1]for n− m= 2j+ 1⎩n−mUsing R mn and the identitydenoting J M as J Mm, h as h m , eq. (A8) becomesT[ m ]TB B =−2i L− − 1 − 11 ⎡T −T−1⎤J Mm= Re⎢∑RmnhmB B gn2⎥⎣ n⎦1 ⎡T −1 ⎤= ⎢n⎥ n⎣⎦= ⎡T −1⎤Re ∑Rmn( -2i)hmL g Im⎢∑RmnhmL g2⎥n⎣ n⎦= Im R h L g + Im( R ) h L g + Im( R h L g )∑−1 T −1 T −1mm m mn m n mn m nnnn=m+2j,n≠mn=m+2j+1The last two terms of the above equation are zero. Thus the term contributing to the J Mm is theterm between g m and h m only, andorT −1[ RmmhmL gm]JMm= ImT −1= hmL gmIm{ − δm( δm+ 1)[ln( r2 / r1) + i2π]}, n ≠ 0T −1=−2πδ ( δ + 1) h L gm m mm∑J Mn−1=−2πδ( δ + 1) h L g(A9)Tn n nnAs h n is arbitrary, we choosehn⎧ e= ⎨⎩iekk, n = 1, 3, 5,⋅⋅⋅, n = 4, 6, 8,⋅⋅⋅Here, e k (k = 1, 2) possesses dimension force() 1two different values of J Mndenoted by J Mneq. (A9) leads to/( length) 1−δ n. Therefore, for a given n, there will be(and J 2 )Mncorrespondingly. For the two choices of h n ,N.C. State Univ., Raleigh, NC 4-23 F. G. Yuan, July 28-31, 1998

Anisotropic Elasticity: Application to Composite Fracture Mechanicswheregn⎧~LJMn−n = ⋅⋅⋅⎪n( n+ ) , 1352πδδ 1, , ,= ⎨ ~⎪i LJMnn = ⋅⋅⋅⎩⎪ 2n( n+ 1) , 4 , 6 , 8πδ δ,~ (1) (2JM n= [M n, J)M nJ ]T(A10)Appendix BDetermining the Unknown Coefficients in the Crack Field Using Betti’s TheoremFollowing arguments similar to those presented for the J M integral in Appendix A, one canget the expression for the integral I. The path-independent integral from Betti’s theorem isIT= ∫ [( u a ) t − ( t a )T u]dsΓ(B1)Using the complex potential functions, Φ and Φ a and identity eq. (A4),a−1a T( u ) t ds =−Re( AB Φ ) ]Re[ dΦ]1−1aT=− Re[( AB dΦ + cc . .) dΦ]2a Ta T−1( t ) u ds =−Re( dΦ) Re[ AB Φ]1aT −1=− Re{[ d( Φ + c. c.) ] AB Φ}21a T −1 a T −1=− Re{ d[( Φ + c. c.) AB Φ] − ( Φ + c. c.) AB dΦ}2(B2)(B3)∫Γ( r , π )2a 1( t ) Ta TaTu ds=− + cc AB − 1 −1Re{[( Φ . .) Φ] − ∫ ( Φ + cc . .) AB d Φ}(B4)2Γ( r1, −π)According to eq. (4.19), the traction free conditions on the crack faces for the auxiliary field maybe written asRe[ Φ a ( z )] = 0 or Φ a + c. c. = 0 at θ =± π .Therefore the first term on the right hand side of (B4) vanishes and∫Γ1( t a ) T u dsa c cT −1= Re[ ∫ ( Φ + . .) AB d Φ](B5)2 ΓSubstituting (B2) and (B5) into (B1) and using the expressions for Φ and Φ a (eq. (A2) and (A3))yieldN.C. State Univ., Raleigh, NC 4-24 F. G. Yuan, July 28-31, 1998

Anisotropic Elasticity: Application to Composite Fracture Mechanics1a T a TI =−⎧ −1 −1Re + c c d + + c c d⎫⎨⎩∫( AB Φ . .) Φ ( Φ . .) AB Φ⎬2 Γ⎭1 ⎡T −T a T−1 T −T a T−1=− Re⎢ ∑∫ h B f A B dfnB gn+ h B f A B dfnB g2Γ⎣ nT −T a T−1 T −T a T−1+ h B f B A df B g + h B f B A df B gn1 ⎡T −T a T T−1I =− Re ⎢∑∫h B f ( A B+ B A)dfnB gn+2Γ⎣ nUsing the orthogonality relations, eq. (A7),nT −T a T T−1[ h B f ( A B+ B A)dfnB gn]]1 ⎡T −Ta−1⎤I =− Re⎢∑∫h B f dfnB gn2Γ⎥⎣ n⎦nn]nwhereDefining1 ⎡T −T a−1⎤=− Re ⎢∑h B2∫ f () z dfn()z B gΓn⎥⎣ n⎦a∆ m +f ()= z z1 , ∆ m=− m /2, m = 1, 3, 4, ⋅⋅⋅.aδn+ ∆m+1Q ≡ f ( z ) df ( z ) = ( δ + 1)z dzmn∫Γα n α nΓα∫α(B6)it is readily shown that(a) n = m - 2 or δ n+ ∆ m+ 2=0Q(b) n ≠ m -2 or δ n+ ∆ m+ 2 ≠ 0dz, −2 = ( δ−2 + 1) ∫α = ( δ−2 + 1)(ln r2 / r1+ i2π);Γ zm m m mαQmnδn + 1n− m+ 2 2 n− m+ 2 2n− m+ 2 2 n− m+2 2=−2[( r −r n− m 2 + i r + r ( n−m)21) cos[( ) π / ] (21) sin[ π / 2]}n− m+2( )/ ( )/ ( )/ ( )/ ,⎧ j+ 1( δn + 1) n− m+ n− m+⎪( − )(( 2)/ 2 ( 2)/21 2 r2−r1) for n− m=2 ji.e. Q n mmn= ⎨− + 2j+ 1( δn + 1) n− m+ n− m+⎪ i ( − )(( 2)/ 2 ( 2)/22 1r2+ r1) for n− m= 2 j+1⎩ n− m+2Using Q mn and expression for L -1 , writing I as I m , and h as h m , the I -integral from eq. (B6), for agiven ∆m, becomesN.C. State Univ., Raleigh, NC 4-25 F. G. Yuan, July 28-31, 1998

Anisotropic Elasticity: Application to Composite Fracture Mechanics1T −T−1Im=− Re{ ∑QmnhmB B gn}2 n1T −1 T −1=− Re{ ∑Qmn( − 2i) hmL gn} =−Im[ ∑QmnhmL gn]2 nnT −1T −1 T −1=−Im{ Q h L g } −Im { Q h L g } −Im { Q h L g }∑m,m−2m m−2mn m nnnn= m+ 2 j , n≠m− 2 m= m+ 2 j+1Since the last two terms of the above equation are equal to zero, it can be clearly seen that theterm contributing to the I is the cross terms between g m-2 and h m only.∑mnmnThusT −1[ m m−2]T −1I =− Im Q m,m-2h L g =− h L g Im{( δ + 1)[ln( r / r ) + i2π]}m=−2π δ 1)T −1(m-2+ hmL gm−2mm−2 m−2 2 1T −1or I n=−2π( δ + 1) h L g(B7)+ 2n n+2From eq. (B7), following a similar procedure as before, g n can be expressed bywhere( k )I n +2gis the value of I n+2 whenn⎧~LIn+2−n = ⋅⋅⋅⎪ (n+ ) , 1 , 3 , 52π δ 1,= ⎨ ~⎪i LIn+2n = ⋅⋅⋅⎩⎪2 (n+ 1) , 2 , 4 , 6π δ,~ (1) (2In[n, I)+ 2=+ 2 n+2I ]Tn(B8)hn⎧ e= ⎨⎩iekk, n = 1, 3, 5,⋅⋅⋅, n = 2, 4, 6,⋅⋅⋅Here, e k possesses dimension force × ( length) δ n .Appendix CDeformation Field for Degenerate MaterialsThe solutions for the nongenerate materials can be modified so that they can be applied fordegenerate materials. We write the general solutions for nongenerate materials aswhereu = ReΦ = X(z)g-1[ AB X(z)g]X() z = B f () z B−1N.C. State Univ., Raleigh, NC 4-26 F. G. Yuan, July 28-31, 1998

Anisotropic Elasticity: Application to Composite Fracture MechanicsIn the limit µ1= µ2= µ , it can be proved that the matrix reduces towhere z = x 1 + µ x 2 ;X() z = f () z I + x f ′()z V2⎡ µ µ ⎤V = ⎢ ⎥⎣ − 1 − µ ⎦Hence, for isotropic materials, we can obtain1u= Re () E− ′()V g2Gwhere{[ f z ix2f z ] }[ f () z I x2f () z V]gΦ = + ′21 ⎡ − i2s11 ′ s11 ′ + s12′ ⎤ ⎡ iE =s ′ − ′⎢⎣− ′ + ′ − 2 ′⎥ , V =11s12( s11 s12)i s⎢11 ⎦ ⎣−1− 1⎤−i⎥⎦For a crack in isotropic materials, choosing f(z) = z δ+1 and performing routine manipulations, thecrack tip fields can be represented as⎡σ11 ⎤⎢σ⎥=⎢22⎥⎢⎣σ12⎥⎦∑n( δ + 1) rnδni⎧⎡2e⎪⎢Re⎨⎢⎪ i⎢⎩⎣e( δ nθ−π/2)δ θn− δ en+ δ eni(δn−1)θiδnθi[ π / 2+( δn−1)θ ]+ δ+⎤ ⎫nesinθe δnesinθi(δ −1)ii[ /2+(−1)] ⎥ ⎡gn1⎤⎪ n θδ nθπ δnθsinθe − δnesinθ⎥ ⎢ ⎥ ⎬i[ π / 2+( δ −1)θ ] i(δ −1)θgn2⎪⎭nsinθ− δ ennsinθ⎥ ⎣⎦⎦⎡u1⎤⎢ ⎥ =⎣u2 ⎦∑nr−i[⎪⎧⎡ ( κ + 1) eRe⎨⎢i(⎪⎩ ⎣−( κ − 1) e+ 2( δ + 1) esinθ( κ − 1) e+ 2( δ + 1) esinθ⎤ ⎡g⎥sinθ⎢⎦ ⎣gδ n + 1π / 2−(δ n + 1) θ ]iδnθi(δ n + 1) θi(π / 2+δ nθ)nnn1δ n + 1) θi(π / 2+δ nθ)−i[π / 2−(δ n + 1) θ ]iδnθ4G+ 2( δ + 1) e sinθ( κ + 1) e − 2( δ + 1) ennn 2where κ = 3 −4 ν for plane strain; κ = ( 3− ν)/( 1+ν)for plane stress.⎤⎪⎫⎥⎬⎦⎪⎭N.C. State Univ., Raleigh, NC 4-27 F. G. Yuan, July 28-31, 1998

Anisotropic Elasticity: Application to Composite Fracture MechanicsReferences on Anisotropic Bodies1. S. A. Ambartsumyan, Theory of Anisotropic Plates, Technomic Publishing Co., Inc., 1970.2. S. A. Ambartsumian, Fragments of theTheory of Anisotropic Shells, World ScientificPublishing Co., Singapore, 1991.3. R. F. S. Hearmon, Applied Anisotropic Elasticity, Oxford University Press, 1961.4. S. G. Lekhnitskii, Theory of Elastcity of an Anisotropic Elastic Body, Holden-Day, SanFrancisco, California, 1963, (translation of 1950 Russian edition).5. S. G. Lekhnitskii, Anisotropic Plates, Gordon and Breach Science Publishers, New York,1968, (translation of 1957 Russian edition).6. G. N. Savin, Stress Concentration Around Holes, Pergamon Press, 1961, (translcation ofamended 1951 Russian edition).7. T. C. T. Ting, Anisotropic Elasticity, Theory and Applications, Oxford University Press,Oxford, 1996.8. J. M. Whitney, Structural Analysis of Laminated Anisotropic Plates, Technomic PublishingCo., Lancaster, Pennsylania, 1987.N.C. State Univ., Raleigh, NC 4-28 F. G. Yuan, July 28-31, 1998