Compositional elastic field in three dimensions for anisotropic materials

Transcription

1 Chapter 5 Compositional elastic field in three dimensions for anisotropic materials 5. Introduction In this chapter, an analytical solution for the three-dimensional compositional elastic field in a film on a compliant substrate will be developed, both the film and substrate are elastically anisotropic. The methodologies used to develop analytical solutions for three-dimensional elasticity problems are based primarily on the modern theory of generalized solutions of differential equations, or on the theory of multidimensional singular potentials and singular integral equations 69]. However, the construction of anisotropic solutions are often limited to transverse isotropy and mainly deal with isothermal and infinite or semi-infinite geometries, or bodies with axial symmetry. A general three-dimensional anisotropic elasticity solution still remains to be developed. Anisotropic elasticity theories in two dimensions, namely the Stroh formalism 37, 38, 43, 44, 45] and the Lekhnitskii formalism 39], are hard to extend into three dimensions. However, some progress has been made 44, 7]. In 44], Ting proposed extensions and applications of Stroh s two-dimensional formalism to certain three-dimensional deformations of anisotropic elastic solids. In 7], Vel and Batra used the generalized Eshelby-Stroh formalism to analyze the three-dimensional deformations of a multilayered, linear elastic, anisotropic rectangular plate subjected to arbitrary boundary conditions at its edges. Their solution is presented in terms of infinite series. However, the extensions of the two-dimensional formalisms generally depend on the assumed function form of the displacement in terms of the coordinate variables. In this chapter, general displacement potentials are used to solve the three-dimensional elasticity problem. This is inspired by some related three-dimensional composites mechanics problems 7, 5, 5], it is observed that the formulation of the governing equation and the choice of basic variables greatly alter the algebraic work involved in the problem solving. We have shown that in the two-dimensional case in Chapters 3 and 4, an Airy stress function can be adopted to solve both isotropic and anisotropic problems. In contrast, the three-dimensional Airy stress function, which was introduced by Maxwell and Morera, and generalized later by Beltrami 7], involves six unknown components and

2 5. Governing equations 6 is technically hard to implement 73]. Hence, a displacement potential method is sought instead. Despite the many displacement potentials available in two dimensions for various situations 74], and for isotropic three-dimensional cases (such as Papkovich-Neuber potentials, there are very few suitable for three-dimensional anisotropic problems. In 75], a modified Helmholtz decomposition of a vector field was introduced to resolve the issue of anisotropy, but the high-order PDEs obtained are still difficult to solve. In this work, displacement potentials introduced by Carrier 76] are adopted. A multidimensional Laplace transform 77, 78] is used as a tool to facilitate the process. We assume periodic deformation in the in-plane directions, to avoid problems with singularities. Note that this is different from the solutions presented in 7, 5], it is assumed that the displacement field is periodic. This chapter is arranged as follows. In Section 5., the governing equations are derived. The displacement potentials are presented in Section 5.3, and the strain/stress fields in the film and substrate are derived in Sections 5.4 and 5.5. In Section 5.6, the relevant unknowns are determined by traction continuity conditions and matching of applied loads and moments. Section 5.7 verifies the solution by a comparison with FEM. Finally, the extension to materials with orthotropic symmetry is given in Section Governing equations 5.. Governing equation and boundary conditions Consider a film occupying the region, L ], L ] H f, H f ], and a substrate occupying the region, L ], L ] H s, H s ], as shown in Figure 5.. As in Section 3.5, local coordinate systems for the film and substrate are adopted. The film and substrate both are made of materials with cubic symmetry. The elastic constants are C, C, C 44 and C, C, C 44 in the film and substrate respectively. The composition in the film is assumed to be a function periodic in x and y. The substrate has a uniform composition. The interface between the film and substrate is taken to be coherent, so displacements and tractions match at the interface. The constitutive equation for the film is given in Eqn. (.. Again, similar to Eqn. (3.6, the compositional misfit strain is taken to obey Vegard s law, i.e., α in the film is α = ɛ T + η(c C. (5. In Eqn. (5., ɛ T is the epitaxial misfit strain between the substrate and the film at its average composition, η is a constant, and C is the average composition of the film. Also,

3 5. Governing equations :<; *,+.-/3/465 '( $&%! "# Figure 5.: The film-substrate system and their local coordinate systems are shown. The film and substrate both are made of materials with cubic symmetry. The elastic constants are C, C, C 44 and C, C, C 44 in the film and substrate respectively. The composition in the film is C(x, y, z and uniform in the substrate. The interface between the film and the substrate is assumed to be coherent. for a cubic material, the elastic moduli tensor C ijkl has the form given by Eqn. (.8. In 944, Carrier 76] introduced displacement potentials φ i (i =,, 3 as: u = u x, u y, u z ] T u x = φ,x, u y = φ,y, u z = φ 3,z. (5. By substituting displacement components in Eqn. (5. into Navier s equations for the general orthotropic materials (see Eqn. (.39, for cubic materials, we have: (C φ,xx + C 44 (φ,yy + φ,zz + (C + C 44 (φ,yy + φ 3,zz ],x = (C + C α,x, (C φ,yy + C 44 (φ,xx + φ,zz + (C + C 44 (φ,xx + φ 3,zz ],y = (C + C α,y, (C φ 3,zz + C 44 (φ 3,xx + φ 3,yy + (C + C 44 (φ,xx + φ,yy ],z = (C + C α,z. (5.3a (5.3b (5.3c Integrating Eqn. (5.3a, we have (C φ,xx +C 44 (φ,yy +φ,zz +(C +C 44 (φ,yy +φ 3,zz = (C +C α +ψ (y, z. (5.4 Let φ (x, y, z = φ o (x, y, z + φ (y, z. By substituting into Eqn. (5.4, and choosing φ (y, z such that yz φ = ψ (y, z/c 44, we find (C φ o,xx + C 44 (φ o,yy + φ o,zz + (C + C 44 (φ,yy + φ 3,zz = (C + C α. (5.5

4 5. Governing equations 64 Note the term φ (y, z has no effect on the displacement u x. Hence the integration function ψ (y, z has no effect on the solution of displacement field and can be set to zero. By applying the same procedure to the remaining equations in (5.3, we obtain the following governing equations: (C φ,xx + C 44 (φ,yy + φ,zz + (C + C 44 (φ,yy + φ 3,zz = (C + C α, (5.6a (C φ,yy + C 44 (φ,xx + φ,zz + (C + C 44 (φ,xx + φ 3,zz = (C + C α, (5.6b (C φ 3,zz + C 44 (φ 3,xx + φ 3,yy + (C + C 44 (φ,xx + φ,yy = (C + C α. (5.6c The traction free boundary conditions on the top surface of the film z = H f, when expressed in terms of φ i, are σ zz = C (φ,xx + φ,yy + C φ 3,zz (C + C α =, σ yz = C 44 (φ + φ 3,yz =, σ xz = C 44 (φ + φ 3,xz =. (5.7a (5.7b (5.7c Because the film-substrate interface is coherent, the displacements on the film-substrate interface, as well as the normal and shearing tractions must be continuous across the interface, i.e., φ,x, φ,y, φ 3,z, σ zz, σ zx, σ zy are matched on the interface (5.8 The adoption of Carrier s displacement potentials (in Eqn. (5. gives a form of the Navier s equations that has only three potentials and a highest order derivative of two. Other types of displacement potentials, such as the classical Helmholtz decomposition of a vector field, introduce four potentials and equations with three derivatives in their forms of Navier s equations. Our strategy for solving the film-substrate problem is based on first finding separate solutions for each, and coupling these solutions through the boundary conditions at the film-substrate interface. The idea is based on the following observation. If the film is detached from the substrate and the displacements on the interface are assumed to be given as (see Figure 5.: φ,x = u x, φ,y = u y, φ 3,z = u z on the interface, (5.9 then the elastic problems in the film and the substrate become similar. Both problems are well-posed and can be solved independently. Moreover, the solution for the substrate can

5 5. Governing equations 65 be readily obtained from that for the film because the substrate can be considered as a special case of the film with uniform composition. Note with this strategy, the displacements at the interface automatically match; to make the solution physical, however, the tractions σ zz, σ xz and σ yz also need to be matched. Given this strategy, we solve the problem as follows. First, we take the Laplace transform of the governing equation (5.6 in both x and y; this gives a transformed potential φ. Second, φ is solved using the boundary conditions in Eqns. (5.7 and (5.9. The transformed strains are then found from φ. All the inverse transform operations are performed on the transformed strains. A periodic deformation state is assumed, so that the inverse transform can be constructed from the residues of the periodically distributed poles. Finally, the coefficients introduced in the above process are determined by matching the tractions at the interface as well as the resultant forces and moments on the system. 5.. Double Laplace transform The composition field C(x, y, z in the film is assumed to be periodic in the x and y directions. Following the preliminaries in Section 3., we decompose C(x, y, z as C(x, y, z = b(x, yz + C (x, y, z, (5.a b(x, y = C(x, y, Hf C(x, y, H f H f = m,n= mπ i( x+ b mn e nπ y L L, (5.b and C (x, y, z = m,n,k= mπ i( x+ C mnk e nπ y+ kπ L L H f z. (5.c Note that by separating the linear piece in z, we have C (x, y, H f = C (x, y, H f. Then C (x, y, z is periodic in all directions. This will assure the convergence of its Fourier series representation in Eqn. (5.c. Double Laplace transforms were introduced by Van der Pol and have been applied to problems involving PDEs, see 77, 78] for example. The transform pair is given by f(p, p = e p y f(x, ye p x dxdy, (5.a

6 5. Governing equations 66 f(x, y = e p y f(p, p e px dp dp. (5.b πi πi Here we list some basic properties of the double Laplace transform: Br. Br. f,x = p f f(, y, f,y = p f f(x,, f,xx = p f p f(, y f,x (, y, f,yy = p f p f(x, f,y (x,, f,xy = p p f p f(x, p f(, y + f(,. (5.a (5.b (5.c (5.d (5.e Note the full D transform of f(x, y is denoted by the notation f; while the notation f and f are used to denote the Laplace transform of f on a single variable x and y respectively. Applying the two-dimensional Laplace transform to Eqn. (5.6, we have: φ,zz + φ = b, (5.3 φ = φ φ, = C 44 (C + C 44 C 44 (C + C 44, (5.4a = φ 3 C = p + p, (5.4b C C 44 (C + C 44 (C + C 44 C 44, = C. (5.4c (C + C 44 C 44 Also, the right hand side of Eqn. (5.3 is b = (C + C α (C + C 44 C 44 + f, (5.5a

7 5. Governing equations 67 f = C 44 p φ (x,, z + φ,y (x,, z p φ (, y, z + φ,x (, y, z (p φ3 (, y, z + φ 3,x (, y, z + (p φ3 (x,, z + C (C + C 44 + (C + C 44 C (C + C 44 (C + C 44 φ 3,y (x,, z p φ (, y, z + φ,x (, y, z p φ (x,, z + φ,y (x,, z. (5.5b The matrices in Eqn. (5.4a, and, in Eqn. (5.4c are related by =, =, = (5.6 = (5.7 is the ] three-fold rotation matrix for cubic symmetry. The boundary conditions (5.7 are transformed to φ,zz + φ,z + 3 φ = g at z = H f, (5.8a =, = p p p p, 3 = sp sp, s = C C. (5.8b The displacement boundary conditions in (5.9 are transformed to φ,z + φ = w at z = H f, (5.9a =, = p p. (5.9b

8 5.3 Displacement potential φ 68 The term g in Eqns. (5.8a is ( + s α(z = H f + g s g = ( φ(+3,z(x,, H f ( φ(+3,z(, y, H f, (5.a ( g s = s p φ (, y, H f + φ,x (, y, H f + p φ (x,, H f + φ,y (x,, H f. (5.b Finally, w in Eqn. (5.9a is w = û x + φ (, y, H f û y + φ (x,, H f û z. (5. Note that w contains information on the assumed displacement at the film-substrate interface. 5.3 Displacement potential φ We now solve Eqn. (5.3. Multiply both sides of Eqn. (5.3 by to obtain φ,zz + φ = b, (5. = p A + p B, A =, B =. (5.3 If p, p are not zero simultaneously, the general solution of Eqn. (5. is: φ g (z = e Dz, ] e Dz c, (5.4 c c, c, c 3, c 4, c 5, c 6 ] T is an arbitrary vector of coefficients, and is a matrix whose columns are the eigenvectors of, with corresponding eigenvalues λ k = d k (k =

9 5.3 Displacement potential φ 69,, 3. Hence = D, D = diag(d, d, d 3. (5.5 In Eqn. (5.4, the following notation is used: Since we have included both ±D in Eqn. (5.4, we take without loss of generality. When p = p =, e ±Dz = diag(e ±d z, e ±d z, e ±d 3z. (5.6 d k = i λ k (5.7 is indefinite and so the solution above cannot be used. In this case, we use a direct method to generate the solution, which corresponds to a one-dimensional problem in the thickness (z direction. This is presented in Section 5.4. In order to find a particular solution for Eqn. (5., we need an explicit expression for b in terms of z. By Eqns. (5. and (5., α can be expressed in terms of e i kπ H f z (k Z and z. By applying the same decomposition as in Eqn. (5. to f in Eqn. (5.5a, and using Eqn. (5.5b, we express b in terms of e i kπ H f z (k Z and z. Since (in Eqn. (5.4a is a constant matrix, b can be written as b = k= e i kπ H f z b k + dz, (5.8 b k and d are straightforward to obtain and are not presented here. From (5.8, one can verify that a particular solution of Eqn. (5. is φ p (z = k= e i kπ H f z b k + dz, (5.9 = ( kπ H f. (5.3 By substituting Eqns. (5.4 and (5.9 into boundary conditions (5.8a and (5.9a, we solve the coefficient vector c: c = v, (5.3

10 5.3 Displacement potential φ 7 = = = = e DHf e DHf e DHf e DHf D + D D + D + D ], (5.3a, (5.3b, (5.3c, (5.3d = D +, (5.3e g v = φ p,zz φ p,z 3 φ p ]. (5.3f w φ p,z φ p The inverse of the block matrix is = e DHf e DHf e DHf e DHf ] S S ], (5.33a S and S are the Schür complements: S = e DHf e DHf ( e DHf e DHf, (5.33b S = e DHf e DHf ( e DHf e DHf. (5.33c By substituting Eqn. (5.33 into Eqn. (5.3, with Eqns. (5.4 and (5.9, we have = = = φ = φ p +, ]v ( ( = e i kπ H f z kπ + + k= z H f ] H f e D(z+Hf e D(z+Hf ( e D(z Hf e D(z Hf ( 3 +, = ( kπ i H f ( k ] b k d + g + w, (5.34 ] e DHf ( ] e DHf ( 3, = e DHf ( e DHf ( + ] ], (5.35a, (5.35b. (5.35c

11 5.4 Strain and stress fields in the film Strain and stress fields in the film 5.4. Strains in terms of φ The strain components can be calculated from φ. Define normal strain and shear strain vectors by ɛ N (ɛ xx, ɛ yy, ɛ zz T and ɛ S (ɛ xy, ɛ yz, ɛ xz T. (5.36 Then, by the transform properties given in Eqns. (5., the transformed normal strains ɛ N = u x,x u y,y = φ,xx φ,yy u z,z φ 3,zz = diag(,, φ,zz + diag(p, p, φ p φ (, y, z + φ,x (, y, z p φ (x,, z + φ,y (x,, z. (5.37 By Eqn. (5., φ,zz can be expressed in terms of φ. Hence Eqn. (5.37 can be further simplified to ɛ N = N φ + v N, (5.38 N = p p µ C p µ C p C 44 C (p + p, (5.39a µ = C + C 44, (5.39b (p φ (, y, z + φ,x (, y, z v N = (p φ (x,, z + φ,y (x,, z. ( + s α + C f(3 (5.39c Similarly, we can write the transformed shear strain as ɛ S = ( S φ + S φ,z v S, (5.4

12 5.4 Strain and stress fields in the film 7 S = p p p p, S = p p p p, p φ(+ (x,, z + p φ(+ (, y, z φ (+ (,, z v S = φ (+3,z (x,, z φ (+3,z (, y, z (5.4a. (5.4b As stated earlier, we consider solutions periodic in x and y, so there are no singularities at the film-substrate interface. The assumption of periodicity also simplifies the inverse transform of (5.38 and (5.4, since a straightforward pole analysis immediately shows that the poles are i mπ L (m Z for p, and i nπ L (n Z for p, see Section 3.3. for more details. Based on the pole analysis, we have the following possibilities: ( p = and p =, which corresponds to a D problem in the z direction; ( p = i mπ L and p =, which corresponds to a D plain strain problem in the xz plane; (3 p = and p = i nπ L, which corresponds to a D plain strain problem in the yz plane; and (4 p = i mπ L p = i nπ L, which corresponds to a periodic 3D problem. Note for cases (, (3 and (4, m, n Z, mn. In the following text, we will call case ( as mode M d, cases ( and (3 as mode M d, and case (4 as mode M 3d. From the residue theorem, a successive inverse transform of ɛ about p and p on the poles gives the general structure of the strain components: ( ɛ = Res p = p = + n= n ɛ e (p x+p y + Res p = p =i nπ L =R f (zɛ] + m= m m= m Res p =i mπ L p = ( ɛ e (p x+p y + R f mπ x (zɛ]ei L + ( ɛ e (p x+p y m,n= mn n= n Res p =i mπ L p =i nπ L ( ɛ e (p x+p y R f nπ y (zɛ]ei L + R f 3 m,n= mn and (zɛ]ei( mπ L x+ nπ L y, (5.4 the notation R f i (zɛ] (i =..3 is used to distinguish whether ɛ is ɛ N or ɛ S. Note that R f (z, Rf (z, Rf (z and Rf 3 (z correspond to cases (, (, (3 and (4, respectively.

13 5.4 Strain and stress fields in the film 73 Finally, we note that this solution requires distinct eigenvalues for. It can be shown that for any anisotropy ratio A = C 44 C C, this requirement is fulfilled. The isotropic case A = has an eigenvalue of triple multiplicity and can be readily solved. This case will not be covered in this thesis. For the cases only one p is, the eigenvalue problem for is important for later inverse transform operation. For such cases, we note the following decomposition for the matrices A and B in from Eqn. (5.3: A = A D A A, B = BD B B, (5.43a B = xy A, D A = D B = diag(, e/a, f/a, (5.43b (e b (f b µ µ (e b+µ (f b+µ A = (a b+µ (a b+µ, xy =, (5.43c a = C, b = C 44, µ = C + C 44, (5.43d d = (a + b + µ(a + b µ(a b µ(a b + µ, (5.43e e = b (b µ + a + d, f = b (b µ + a d. (5.43f Note the matrix xy is the ( mirroring operation. For anisotropy ratio A = C 44 C C, if A >, then d < and D A contains a complex conjugate pair; while for A <, we have d > and D A are all real numbers. In both cases, we have e f and e a f a =, hence the three eigenvalues, e/a and f/a are distinct. For A =, we have d =, then the eigenvalue is with triple multiplicity Inverse transform preliminaries Because of the periodic strain/stress assumption, p = i mπ L (m Z and p = i nπ L (n Z are all simple poles. Hence we only need to evaluate the limits of the R f i (z for different combinations of p and p. As stated earlier, the combinations having p = or p = simplify the structure of in Eqn. (5.3, and hence the relevant eigenfunctions,,, and in the final solution of φ in Eqn. (5.34. However,, etc. in Eqn. (5.35 contain p in both their denominator and numerator. Thus the limit p must be carried out symbolically. Here, we use ɛ N as an example to show the process. The residues associated with the shear strain ɛ S can be obtained in the same spirit. Also, the properties of some related

14 5.4 Strain and stress fields in the film 74 matrices needed to obtain R f i (z (i =, via the inverse transform of ɛ N and ɛ S when taking p or p are given in Appendix F. Note that, when taking p, we have N = p N ; similarly, when taking p, we have N = p N. The matrices N and N are given as N =, N =, a = C, b = C 44. (5.44 µ a b a µ a b a Then we have N = N = N (z, p N (z, p z, p, p ] = N H f ] N z (z, p H f (z, p B, p ] N z (z, p H f (z, p A, p N (z, p, p N = N (z, p, p N (z, p p, p N (z, p p, p (5.45a (5.45b (5.45c (5.45d Another important issue is the constraint from the film-substrate interface. Recall that from our strategy, the displacements on the interface are taken as known (see Eqn. (5.9, and enter the boundary condition (5.9a through w in Eqn. (5.. The interface contributes to the solution in the film through terms such as N w in Eqn. (5.38 for ɛ N, and ( S w + S,zw in Eqn. (5.4 for ɛ S. These terms appear in the inverse transform operation through R f i (zɛ N] and R f i (zɛ S] in Eqn. (5.4. However, by using the properties of in Eqns. (F- and (5.45d, along with the properties of the double Laplace transform in (5., it can be shown that for mode M d, the R f i (z (i =, depend on the residues of û x,x, û x,y, û y,x, û y,y, û z,xx and û z,yy for both ɛ N and ɛ S. Hence, for mode M d, we introduce the following definitions: ( T ] T ξm, n ξm, s ξm] b Res û x,x, û y,x, û p =i mπ z,xx L p = (m Z, m, (5.46a

15 5.4 Strain and stress fields in the film 75 T ξn, n ξn, s ξn] b Res p = p =i nπ L ( ] T û x,y, û y,y, û z,yy (n Z, n, (5.46b In contrast, for mode M 3d, R f 3 (zɛ N] and R f 3 (zɛ S] depend on the residues of û x,y,z]. Thus we introduce ( ] T ξmn, x ξmn, y ξmn] z T Res û x, û y, û p =i mπ z L p =i nπ L (m, n Z, mn. (5.46c The ξ n,s,b] m,n] and ξx,y,z] mn carry all the information of the constraint from the film-substrate interface. They enter the solution for the elastic fields and will be determined later using the traction continuity conditions on the film-substrate interface. In fact, the ξ n,s,b] m,n] and ξ x,y,z] mn are the Fourier coefficients of the functions on the right hand side of Eqns. (5.46a (5.46c. That is, ξ n m (m Z, m are the Fourier components in x of u x,x, etc Evaluation of R f i (z (i =..3 We now evaluate the R f i (z (i =..3 for ɛ N and ɛ S in Eqn. (5.4 in terms of ξ (see Eqn. (5.46 and the eigenfunctions,. As mentioned previously, if p = p =, composition is a function of z only, with α = α (z = ɛ T + η b z + is indeterminate. However, in this case, the k= k C k e i kπ H f z. (5.47 Thus all the strain components are independent of x and y. This allows the elasticity problem to be solved directly using the compatibility and equilibrium equations and boundary conditions. The six compatibility equations reduce to three nontrivial ones This tells that ɛ xx,zz =, ɛ yy,zz =, ɛ xy,zz =. (5.48 ɛ xx = a z + b, ɛ yy = a z + b, ɛ xy = a 3 z + b 3, (5.49 a i, b i are constants. By using the shear free traction boundary condition, we have

16 5.4 Strain and stress fields in the film 76 a 3 = b 3 =, hence σ xy =. The equilibrium equations are: σ xz,z =, σ yz,z =, σ zz,z =. (5.5 With the traction free boundary conditions at z = H f, we derive that σ xz = σ yz = σ zz = throughout the film. This results a biaxial stress state. The ɛ zz can be solved from σ zz =. After some algebra, we have R f (zɛ N] = a z + b a z + b a i, b i (i =, are constant coefficients. C C ((a + a z + (b + b + ( + C C α (z, (5.5 By using Eqns. (5.38, (5.4, (5.45 and (5.46, the remaining three R f i (zɛ N] (i =,, 3 can be written as: R f (zɛ N] = Θ f N + N (z, p diag(,, /p ξ n m, ξ s m, ξ b m] T, (5.5a R f (zɛ N] = Θ f N + N (z, p diag(,, /p ξ s n, ξ n n, ξ b n] T, (5.5b R f 3 (zɛ N] = Θ f 3N + N (zξ x mn, ξ y mn, ξ z mn] T. (5.5c Note each R f i (zɛ N] is split into two parts. The first part, Θ f in, is the contribution from the composition. These Θ f in (i =,, 3 are given by with Ω (z = ( k= p e i kπ H f z + Θ f N = N Ω (z + ( + sα m (z,, ] T, Θ f N = N Ω (z + ( + sα n (z,, ] T, (5.53a (5.53b Θ f 3N = N Ω 3 (z + ( + sα mn (z,, ] T, (5.53c ( ( kπ H f (z, p ( kπ ] p (z, p i H f p (z, p ] ( kπ ] ( (p k A H f C mk + z (z, p H f (z, p A b m η(c + C + ( + sα m (H f (z, p, (5.54a

17 5.5 Strain and stress fields in the substrate 77 ( ( ( Ω (z = p e i kπ H f z kπ ( kπ + H f (z, p ] p (z, p i H f p (z, p k= ] ( kπ ] ( (p k B H f C nk + z (z, p H f (z, p B b n η(c + C + ( + sα n (H f (z, p, (5.54b Ω 3 (z = ( k= + z e i kπ H f z + H f ( ( kπ H f ] b mn ( ] kπ i H f ( k C mnk η(c + C + ( + sα mn (H f, (5.54c α m α n α mn = η k= C mk C nk C mnk e i kπ H f z + b m b n b mn z. (5.55 The second part of the R f i (z in Eqn. (5.5 is the contribution from the interface. This contribution depends on the ξ. The unknowns ξ n,s,b] m, ξ n,s,b] n, and ξ mn x,y,z] will be determined later from the traction continuity conditions on the interface (see Section 5.6. Results parallel to Eqns. (5.5 for R f i (zɛ S] (i =..3 are given in Appendix H. Finally, once ɛ N and ɛ S are known, the stress field in the film can be computed as σ xx σ yy σ zz σ xy σ yz σ xz = C C C C C C ɛ N (C + C (ɛ T + η(c C C C C = C 44 ɛ S., (5.56a (5.56b 5.5 Strain and stress fields in the substrate The strain and stress fields in the substrate can be constructed from the solution for the film. In order to do this, first, change the y and z axis in the substrate to their opposite

18 5.5 Strain and stress fields in the substrate 78 directions and label the new coordinate system {xoy}. Note in {xoy}, for the substrate, u x retains the same value, while u y and u z change their signs. Hence, we have ξ n mn = ξm( n n and s,b] ξ mn = ξ s,b] m( n for (m, n (,. In addition to these changes, we change the notation z z, y y and reverse the sign of σ xy and σ xz (due to the sign convention in the {xoy} coordinate system. This gives the solution of the substrate in its original coordinate system {xoy} (see Figure 5.. Finally, we obtain the strain field in the substrate by setting the misfit strain α = and eliminating all composition related terms. Define χ ij ] = ( z, p and ωij ] = ( z, p. Then after some algebra, we find the strain field in the substrate (in {xoy} : ɛ =R s (zɛ]+ m= m ɛ can be ɛ N or ɛ S and R s (zɛ]e i mπ L x + a z + b n= n R s nπ i (zɛ]e, y L + R s (zɛ N ] = a z + b C C ( (a + a z + (b + b R s (zɛ N ] = N ( z, p diag(,, /p ξm, n ξm, s ξm] b T, m,n= mn R s (zɛ N ] = N ( z, p diag(,, /p ξ( n s, ξn ( n, ξb ( n ]T, R s 3(zɛ N ] = R s mπ i( x L 3(zɛ]e nπ y L, (5.57 (5.58a (5.58b (5.58c N ( zξ x m( n, ξy m( n, ξz m( n ]T, (5.58d R s (zɛ S ] =,, ] T, R s (zɛ S ] = χ (χ + χ 3 (χ 3 + χ 33 /p R s (zɛ S ] = ω + ω 3 (ω3 + ω 33 /p ω ξ n m ξ s m ξ b m ξ s ( n ξ n ( n ξ b ( n + + (5.58e ( z, p, ]ξm s, (5.58f ( z, p, ]ξ( n s, (5.58g R s 3(zɛ S ] = diag(,, ( S ( z + S,z( z ξ x m( n, ξy m( n, ξz m( n ]T. (5.58h Note functions superscripted with an asterisk are identical to their counterparts in the film, except the elastic constants are taken as those of the substrate. The stress field in the

19 5.6 Determination of the unknown coefficients 79 substrate can be calculated using its constitutive relations. 5.6 Determination of the unknown coefficients With the strain/stress fields in the substrate solved, we next determine the unknown Fourier coefficients. These coefficients are ξ n,s,b] m,n] and ξx,y,z] mn in Eqn. (5.46 for modes M d and M 3d, and a i, b i and a i, b i (i =, in Eqns. (5.5 and (5.58a for mode Md. Note the full elastic solution is a superposition of the above modes, therefore one can match each Fourier mode independently through the coherency conditions ( Modes M d and M 3d The ξ n,s,b] mn and ξ x,y,z] mn (m, n Z and (m, n (, can be solved by the stress continuity conditions, i.e. σxz f = σxz, s σyz f = σyz s and σzz f = σzz s on the film-substrate interface. For p = i mπ L, p = i nπ L, we have: ] C 44 diag(,, R f (m, Rf (n, Rf 3 (m, n ɛ S ] z= H f = C44diag(,, R s (m, R s ( n, R s 3(m, n] ɛ S ] z=h s (5.59a ] C, C, C ] R f (m, Rf (n, Rf 3 (m, n ɛ N ] z= H f (C + C α m, α n, α mn ] z= H f = C, C, C ] R s (m, R s ( n, R s 3(m, n] ɛ N ] z=h s (5.59b For each pair of (m, n, when m = or n =, each matching condition in Eqn. (5.59 provides an equation for ξ mn n,s,b] ; similarly, when mn, the matching conditions provide equations for ξ mn x,y,z]. For example, using the continuity of shearing tractions of σ yz and σ xz in R f,s], one obtain the following equations for ξm s and ξs n : C 44 χ ( H f ξ s m = C 44χ (H s ξ s m, C 44 ω ( H f ξ s n = C 44ω (H s, nξ s n. (5.6a (5.6b Since Eqn. (5.6 becomes χ (z = sin(p (z H f cos(p H f, ω (z = sin(p (z H f cos(p H f, (5.6 ( C 44 tan(p H f + C44 tan(p H s ξm s =, (5.6a

20 5.6 Determination of the unknown coefficients 8 ( C 44 tan(p H f + C44 tan(p H s ξn s =. (5.6b Hence ξm s = ξs n =. Equations (H-a and (H- immediately confirm that in the xz plane strain case, σ yz = while in the yz plane strain case, σ xz =. ξ x,y,z] mn The remaining unknowns ξm n and ξb m (m Z, m, ξn n and ξb n (n Z, n, and (m, n Z, mn can be determined in a similar way. The results are presented in Appendix G Mode M d first order bending analysis To determine the unknowns a i, b i and a i, b i (i =, from mode Md, the displacement continuity conditions will be used since the traction continuity conditions are satisfied trivially. It can be shown that except for a rigid translation, the displacement matching conditions on the interface are equivalent to matching the normal strains and bending curvatures in the x and y directions. Therefore, ] u x,x = ɛ xx = a H f + b = a H s + b, (5.63a ] u y,y = ɛ yy = a H f + b = a H s + b, (5.63b u z,xx = ɛ xz,x ɛ xx,z = a = a ], u z,yy = ɛ yz,y ɛ yy,z = a = a ]. (5.63c (5.63d The relations within the square brackets in Eqn. (5.63 give four conditions for the above unknowns. The other four conditions to evaluate a i, b i and a i, b i (i =, come from matching resultant forces and moments on the system. Specifically, we match the resultant forces F x and F y and the out-of-plane bending moments M x and M y from our solution to those from the external loading F e x, F e y, M e x and M e y. This is because on the lateral boundaries x = and y =, only mode M d contributes to these resultants. The twisting moments and the in-plane bending moments will be discussed later. We also note that in order to properly evaluate F x, F y, M x and M y, we must include contributions from modes M d and M 3d as well as M d. A common Cartesian coordinate system {xoy} I is set up on the interface for this purpose. Contribution of normal stress components The contribution of each normal stress component to F x, F y, M x and M y on the lateral boundaries x = and y = can be identified by evaluating the contribution of R f,s] i ɛ N ]

21 5.6 Determination of the unknown coefficients 8 (i =..3 respectively. We first evaluate the contribution of R f,s] i ɛ N ] (i =,, 3 to F x and M x. In ɛ f,s] N, since Rf,s] i (i =, 3 are associated with e i nπ y L, their contributions to σxx on F x and M x are all zero. Then we only need to focus on R f,s] (from xz plane strain. For each Fourier mode e i mπ x L f,s] in the series associated with R, we have σ xx,x = p σ xx, p = i mπ L. From the equilibrium equation σ xx,x + σ xz,z =, we obtain σ xx = p σ xz,z. (5.64 Hence, evaluating in their local coordinates, we have F f x R ] = M f x R ] = F s xr ] = M s xr ] = H f σ xx dz = σ xz Hf H f p = σ xz ( H f, (5.65a H f p σ xx zdz = ] H (zσ xz Hf f H f p σ xz dz, (5.65b H f H f σ xx dz = σ xz Hs H s p = σ xz (H s, (5.65c H s p σ xx zdz = H (zσ xz Hs s ] H s p σ xz dz. (5.65d H s H s H f H s H s Collecting the results in the {xoy} I system, we have F x R f,s] (,,3 ] = F f x R ] + F s xr ] = p σ xz ( H f p σ xz (H s =, (5.66a M x R f,s] (,,3 ] = M x f R ] + MxR s ] + H f Fx f R ] H s FxR s ] ( = H f H s σ xz R f,s] ɛ S ]]dz =. (5.66b p H f + H s Note in Eqn. (5.66b, we have used the result that the resultant shear force from σ xz vanishes. This is presented later in Eqn. (5.7. We reach a similar conclusion when we consider the lateral boundary y = : F y R f,s] (,,3 ] =, M yr f,s] (,,3] =. (5.67 Then the resultants F x, F y, M x and M y are all from R f,s], the contribution of mode

22 5.6 Determination of the unknown coefficients 8 M d. Thus, matching forces and moments requires (a H f (a H f b = b ] F e x/(l H f C γɛ T M e x/(l H f C γ ( ηµ C + ɛ T F e y /(L H f C γɛ T M e y /(L H f C γ ( ηµ C + ɛ T, (5.68 = = ] ] r( + r r, = 3 3 (3 + 4rr r, (5.69a ( s ] ] ( s ( s C C s C C s ( s, =, (5.69b r = Hs H f, s = C, s = C, (5.69c C C γ = ( C ( C µ C = H f H f H f + C C α ɛ T zdz = η, (5.69d H f L L L L H f Czdxdydz, H f (5.69e Note in Eqn. (5.68, F e x, M e x, F e y and M e y are the applied loads and moments on the two lateral surfaces yz and xz. Finally, the a i, b i can be solved by Eqn. (5.63. Contribution of shear stress components In the above calculation, we only considered the contribution from normal stresses. To show that the resultant shear forces and relevant twisting moments from our solution are correct in the Saint Venant sense, we evaluate them for each shear stress component on the lateral boundaries x = and y =. We consider σ xy first. For both the film and the substrate, the in-plane shear stress σ xy has the form of R f,s] mπ i( x+ 3 (ze nπ y L L (note that shear stresses only appear in modes M d and M 3d. Thus, for fixed z, on either lateral boundary x = or y =, an integration shows that the resultant force vanishes, as does the twisting moment about z in that plane. Hence σ xy has no contribution to the resultant external loading. We then consider σ xz. Note σ xz are generated by R f,s] ɛ S ] and R f,s] 3 ɛ S ]. For the case of R f,s] ɛ S ], we have σ xz,x = p σ xz. From the equilibrium equation σ xz,x +

23 5.6 Determination of the unknown coefficients 83 σ zz,z =, we obtain σ xz = p σ zz,z. (5.7 Hence integrating σ xz over the system along thickness direction, with Eqn. (5.7, we get ( H f H s + H f H s σ xz R f,s] ɛ S ]]dz = p ( σ zz H f + σ H f zz Hs H s R ɛ N ]] =. (5.7 Thus σ xz R f,s] ɛ S ]] has no effect on the external loading. For the case of R f,s] 3 ɛ S ], since σ xz R f,s] mπ i( x+ 3 ɛ S ]] is in form of e nπ y L L, its resultant force on yz plane vanish. Hence σ xz make no contribution to the resultant shear force. However, σ xz may contribute to the twisting moment in the yz plane. This can be seen as follows. By σ xz,x = p σ xz and σ yz,y = p σ yz as well as the equilibrium equation σ xz,x + σ yz,y + σ zz,z =, we have p = i mπ L ( H f H s + H f H s and p = i nπ L. Then σ xz = p (σ zz,z + p σ yz, (5.7 σ xz R f,s] 3 ɛ S ]]dz = p p ( H f H s + H f H s σ yz R f,s] 3 ɛ S ]]dz. (5.73 The right hand side of Eqn. (5.73 in general is not zero. This may lead to a twisting moment in the yz plane. Finally, we consider σ yz. As the case of σ xz, the only possible effect of σ yz is a twisting moment in the xz plane Remark on in-plane and twisting moments As shown above, under the assumption of periodic strain/stress in x and y, the resultants F x, M x, F y and M y are determined solely from mode M d. All the unknown Fourier coefficients involved in modes M d and M 3d are determined by the stress continuity conditions on the film-substrate interface. This suggests that the in-plane moments yz yσ xxdydz and xz xσ yydxdz, as well as the twisting moments from the transverse shearing stresses yz yσ xzdydz and xz xσ yzdxdz cannot to be imposed as boundary conditions on the periodic problem. We argue that if the thickness of the film-substrate system is small compared to its extension in xy plane, the twisting moments can be ignored due to the free traction on top and bottom surfaces. This is a common assumption in classic plate theory.

24 5.7 Numerical verification with FEM Numerical verification with FEM We verify our results using the finite element method Ansys R. We rely on the analogy between compositional misfit strain and thermal loading. Specifically, we suppose the temperature field T in Ansys and the composition field C in our analytical solution satisfy the following relation: ɛ T + η(c C = α T (T T, (5.74 α T is the thermal expansion coefficient and T is the reference temperature. For example, if take η = α T, C = T, then T = C ɛ T /η. We use the following nondimensional parameters in the model problem verification: L = 8, L = 6, H f = H s =, C = 4., C =.5, C 44 =.7, C =., C =.9, C 44 =, and ɛt =., η =.. Figure 5.: The composition/temperature profile used to compare the analytical and FEM (Ansys R solution is shown. The value (C or T decreases gradually from the center of the spheres to the outer region with a uniform reference value. The temperature/composition field is shown in Figure 5.. We use dimensionless meshsize /8 in the analytical solution to evaluate the discrete composition field. In the FEM model, we use meshsize /8 for the non-periodic single film-substrate configuration, as well as /4 and /5 for the 3 3 block matrix configuration (to simulate the periodic case, results are taken from the middle block. The comparison of stress components on three axes (y = 3, z =, (x = 4, z = and (x = 4, y = 3 passing the centroid of the

25 5.7 Numerical verification with FEM σ xx (x,3,.5..5 FEM, Periodic, h=/4 FEM, Periodic, h=/5 Analytical, h=/8 FEM, Non periodic, h=/8 σ xy (x,3, x. σ yy (x,3,.5. σ yz (x,3,...3 x σ zz (x,3, x σ xz (x,3,... x x x Figure 5.3: The stress components for the temperature/composition field along axis (y = 3, z = are shown for the temperature/composition field in Figure. 5.. The dotted and dash-dot lines are for periodic FEM results with meshsize of h = /4 and h = /5 respectively; the dash line represents the non-periodic (single block FEM result. The analytical result is shown by a solid line. The periodic case in FEM are simulated by taking 9 non-periodic blocks of the same film-substrate configurations together as a 3 3 matrix alignment. The results from the central block are used to make the comparison. For the analytical solution, 4, and 7 terms are used to sum m, n, k respectively. Three FEM results are presented to show the meshsize effect (convergence and the boundary effect. rectangular body are shown in Figures 5.3, 5.4 and 5.5 respectively. The results from the analytical method and the FEM model show a good agreement except in a region close to the interface (see Figure 5.5. We believe the FEM results are suspect in this region. This is because in the FEM model, we assign different reference temperatures to the film and the substrate to simulate the misfit. However, the discretization leads to a layer of shared grids on the film-substrate interface. This introduces a transition

26 5.8 Extension to orthotropic materials x 3 σ xx (4,y,.. σ xy (4,y, 4 σ yy (4,y, FEM, Periodic, h=/4 FEM, Periodic, h=/5 Analytical, h=/8 FEM, Non periodic, h=/ σ yz (4,y, x 3. 3 σ zz (4,y,...3 σ xz (4,y, y y Figure 5.4: The stress components along axis (x = 4, z = are shown for the temperature/composition field in Figure. 5.. zone with finite width that depends on the discretization. It can be seen that when the meshsize changes from /4 to /5, the width of the zone decreases. On the other hand, the analytical solution is continuous in the film and transits exactly on the interface. Also, the periodic FEM results show convergence to the analytical results with decreased meshsize. 5.8 Extension to orthotropic materials From the previous sections, we see that the periodic 3D solution with p and p (mode M 3d is simpler than the two D problem either p or p equals zero (mode M d. The primary reason for this is that extensive algebra is required for mode M d to take the p limit before performing the inverse transform. This algebra can only be carried out symbolically. On the other hand, for mode M 3d, the eigenvalue problem can be

27 5.8 Extension to orthotropic materials x 3 σ xx (4,3,z. FEM, Periodic, h=/4 FEM, Periodic, h=/5. Analytical, h=/8 FEM, Non periodic, h=/ σ xy (4,3,z σ yy (4,3,z σ yz (4,3,z x 3 σ zz (4,3,z σ xz (4,3,z z z Figure 5.5: The stress components along axis (x = 4, y = 3 are shown for the temperature/composition field in Figure. 5.. The analytical results show very good agreement of boundary conditions on the top traction free surface. Note the FEM results near the interface are suspect owing to a transition layer caused by the discretization. evaluated numerically. For a general orthotropic material with 9 elastic constants C, C, C 33, C, C 3, C 3, C 44, C 55, C 66 ], and misfit strains ɛ T i and η i (i =,, 3, the matrices involved in mode M d become more complicated because there is less symmetry. Therefore the eigenvalue problem for the p = or p = cases is also more complicated. However, by using the same strategy for cubic symmetry, our method can be extended to general orthotropic materials. Some matrices need to be altered to reflect the new set of elastic constants. With these changes, one can show that the procedure and the form of the solution remain essentially the same. Some important changes are listed in the following text.

28 5.8 Extension to orthotropic materials 88 In Eqn. (5.4a, changes to = C 55 (C 3 + C 44 C 44 (C 3 + C 44. (5.75 C 33 Also, the matrices, in Eqn. (5.4c change to = C (C + C 66 C 66 (C 3 + C 55 C 55, = Among the matrices for boundary conditions in Eqns. changed: 3 = s p s p C 66 (C + C 66 C (C 3 + C 44 C 44. (5.76 (5.8b and (5.9b, only 3 is, s = C 3 C 33, s = C 3 C 33. (5.77 There are minor changes to the right hand side vectors in the relevant equations, but the most important parts are in b, g and v N : Eqn. (5.5a, in b : (C + C α C C C 3 α C C C 3 α C 3 C 3 C 33 α 3, (5.78a Eqn. (5.a, in g : ( + s α(z = H f s α + s α + α 3 ](z = H f, (5.78b Eqn. (5.39c, in v N : ( + s α s α + s α + α 3 ], (5.78c α i (i =,, 3 are the transformed misfit strains for material with orthogonal symmetry: Also, in the transformed strains, the matrix N = α i = ɛ T i + η i (C C (i =,, 3. (5.79 N in Eqn. (5.39a changes to p p C 3+C 55 C 33 p C 3+C 44 C 33 p ( C 55 C 33 p + C 44 C 33 p As stated before, the relevant eigenvalue problems for. (5.8 are very important. For materials with less symmetry, these eigenvalue problems become more complicated. Using

29 5.8 Extension to orthotropic materials 89 the same notation as in Section 5.3, we find the eigenvalue problem for A (Eqns. (5.3 and (5.43a as follows: A = bc+e(h a b(h d G f b bc+e(g a b(g d H f b, D A = diag(d, H, G, (5.8 a = C (C 3 + C 55, (5.8a C 55 C 33 C 55 H = (a + f + (a f 4b, (5.8b G = (a + f (a f 4b ; (5.8c and for B (Eqns. (5.3 and (5.43a, B = e(h d ec+b(h a (G f(h d ec+b(h a e(g d ec+b(g a (H f(g d ec+b(g a, D B = C 44 C 55 diag(d, H, G, (5.83 ( C C55 a = (C 3 + C 44, (5.84a C 44 C 33 C 44 C 44 H = (a + f + (a f 4e, (5.84b G = (a + f (a f 4e. (5.84c Note in Eqns. (5.8 and (5.83, b, c, d, e, f are common and are given as follows: b = C 3 + C 55 C 33, c = C + C 66 C 44 (C 3 + C 44 (C 3 + C 55 C 33 C 44, (5.85a d = C 66 C 44, e = (C 3 + C 44 C 55 C 33 C 44, f = C 55 C 33. To save space the full solution will not be presented here. (5.85b

30 5.9 Summary and discussion Summary and discussion In this chapter, we use a displacement potential to solve for the elastic fields in a 3D filmsubstrate system with anisotropic symmetry. A double Laplace transform technique is adopted to solve the resultant PDEs. Under the assumption of periodic deformation gradient, the analytical elastic solution can be obtained via an inverse transform. This is a natural extension to the D results given in Chapters 3 and 4. The method demonstrates the usefulness of the displacement potential and the integral transform. Our analytical solution will be used in the simulation of composition evolution. In this case, its main advantage will be in computational cost. All eigenfunctions in our solution need to be evaluated only once. Also, by Eqn. (5.68 and Appendix G, the determination of all the unknowns introduced by the film-substrate coupling takes the same effort independent of the thickness of the system. However, with a fully numerical method, the size of the resulting linear system scales linearly with the total system thickness, hence the computational work scales greater than linearly depending on the solver.