on of zirconia Proof. If a rotation Q satisfies(2.7)for a given positive definite symmetric 33 matrix U. then it leaves the eigenspace of matrix U invariant. i.e Uu= nueU(Qu)=n(Qu) Part(n). Let (n, e),(nz, G2),(n3.G3)be eigenvalue eigenvector pairs of matrix U Noting that Qe=-1+2e@e, we obtain QFU(Q2)=U Let e,=e, Ae.Since Qae=,Q=一e1,Q}t=±e, and Qe e=一, we require that the eigenvectors of U not lie along &, e Suppose Qi,=ou, and Qu,=wu, with ∈{1、-1;andQ∈{Q,Qa}, then the necessary and sufficient conditions for QE. Qa to satisfy (2.7)is n,= n so we set n2* n3 Part(In. Let (n, G,),(n2, u2),(3, e,) be the three eigenvalue-eigenvector pairs of matrix the们=mS0 we set nr≠ Let e be a20 Id axis of p,then QE U(Q)'=U Necessary and sufficient conditions for the 4-fold otations to sati ee,=0, and suppose that e,=e, Ae. Since Qie=e. Qe=-e, and Qi e3 e3, we require that the eigenvectors of U not lie along a 2-fold axis Theorem I is another case of Theorem 2. 10 of Ball and James(1992). Hertog(1987) has obtained the Bain strain when the tetragonal symmetry group is 1, Q(z/2.e3) Q(3T/2, e3), Q(z, e3) and the monoclinic symmetry group is 1, Q(. e3)). Sets Wi -m (2. 8)and whm(2.9)characterise all possible Bain strains that accompany a tetragonal to monoclinic transformation. Examples of Bain strains belonging to set wi mare cases(i)and(ii)in Fig. 5 and to set wii m are cases(iii)and(iv). Lattice parameters of the tetragonal and monoclinic phases determine the values of (n 2. 3)(see Section 3). Since vector e can lie along any of the four 2-fold axes of j, five Bain strains are possible for a tetragonal to monoclinic transformation, and we will use experimental observations to decide upon the bain strains that operate in zirconia 2.2. Variants and energy wells I he free energy is constructed such that at higher temperatures austenite has lower energy, whereas at lower temperatures martensite has lower energy; the energy of both phases is assumed to be equal at a critical transformation temperature e It is natural to identify the austenite at the critical temperature as the reference ,日)=φ(Un,B), where Ua is a bain strain accompanying an austenite to martensite transformation at temperature 0. Due to material symmetry (2.5)we see that the energy o has the same value at matrice U,=Q..Qi, Q Distinct matrices u.i=1.2.,. v are called variants of martensite and the number of variants is given by
Transformation of zirconia 267 Proqf: If a rotation Q satisfies (2.7) for a given positive definite symmetric 3 x 3 matrix U. then it leaves the eigenspace of matrix U invariant, i.e. Uu = Y/U- U(Qu) = Y](Qu). Parr (I). Let (r/, C), (q2, a,), (vi, a,) be eigenvalueeeigenvector pairs of matrix U. Noting that Qz = -I+26 @ @, we obtain QzU(Qz)’ = U. Let &I = Gj A 6. Since Q;;‘$ = &, Q:r”G = --gI, Q$,ei,i? = +6,~, and Q;,i$ = -Cl. we require that the eigenvectors of U not lie along Cl, &. Suppose QO, = rut% and Qti, = ~6, with CL) E (1, - 1) and Q E {Q&, Qa,}, then the necessary and sufficient conditions for QZ,, QZ, to satisfy (2.7) is q2 = Y/~, so we set q2 # yli. Part (II). Let (vi, ai), (yap, a,), (q3, e,) be the three eigenvalueeeigenvector pairs of matrix U. then Qi,U(Qi,)T = U. Necessary and sufficient conditions for the 4-fold rotations to satisfy (2.7) is PI, = q2. so we set YI, # ylz. Let 6 be a 2-fold axis of Y,, then 6.6, = 0, and suppose that &I = 8, A 6. Since QzC = 6, Qt GI = -Cl, and Qz 8, = -&, we require that the eigenvectors of U not lie along a 2-fold axis. Theorem 1 is another case of Theorem 2. IO of Ball and James (1992). Hertog ( 1987) has obtained the Bain strain when the tetragonal symmetry group is {I, Q(rc/2,6,), Q(3rc/2, e,), Q(rc, iZ,)] and the monoclinic symmetry group is [I, Q(x, 6,)). Sets 9&i *“’ (2.8) and ~$,ul;“’ (2.9) characterise all possible Bain strains that accompany a tetragonal to monoclinic transformation. Examples of Bain strains belonging to set +Y)+“’ are cases (i) and (ii) in Fig. 5 and to set &I;+” are cases (iii) and (iv). Lattice parameters of the tetragonal and monoclinic phases determine the values of (q,, q2, qj) (see Section 3). Since vector i? can lie along any of the four 2-fold axes of .P,, five Bain strains are possible for a tetragonal to monoclinic transformation, and we will use experimental observations to decide upon the Bain strains that operate in zirconia. 2.2. Variants and energy wells The free energy 4 is constructed such that at higher temperatures austenite has lower energy, whereas at lower temperatures martensite has lower energy ; the energy of both phases is assumed to be equal at a critical transformation temperature 0,. It is natural to identify the austenite at the critical temperature as the reference configuration, then $(I,@,) = &U”, O,), (2.10) where U,) is a Bain strain accompanying an austenite to martensite transformation at temperature 8,. Due to material symmetry (2.5) we see that the energy 4 has the same value at matrices U, = Q,U,Q’, Q;E:Y,,. (2.1 I) Distinct matrices U,, i = 1,2, . , v are called variants of martensite, and the number of variants is given by
N. K SIMHA 1- QRU ig. 2. The tetragonal (bold)and four monoclinic wells two twin-related variants. the bold dashed line denotes habit microstructures with tetragonal on one side and a monoclinic laminate on the other Order of austenite point group gpa Order of martensite point group m see Van Tandeloo and Amelinckx (1974). For a tetragonal to monoclinic trans- formation(2. 12)gives v= 4 Frame indifference (2. 1 )implies that the free energy has the same value at matrices u and ru where r is a rotation the sct =RI: Ris a rotation, is called the austenite well. and the union of martensite wells M=Ur-1;: R is a rotation and U, a martensite variant). Following Ball and James(1992), we schematically depict wells as circles in the space of deformation gradients. Figure 2 shows the austenite and four martensite wells for a tetragonal to monoclinic transformation. The free energy o is lower on the austenite well at temperatures 0>0, whereas at temperatures 0 B it is lower on the mar tensite wells, and at the critical temperature 0 it is the same on both austenite and martensite wells(2.10) 2.3 Twins a deformation that results in a twin is continuous(coherent) and has different constant deformation gradients on either side of the twin plane Suppose deformation y: Q2+9 has constant gradients F+ and F- on either side of a plane with normal
268 N. K. SIMHA Fig. 2. The tetragonal (bold) and four monoclinic wells are depicted schematically by circles. Each light dashed line indicates a twin, and points on it correspond to average deformation gradients of laminates of two twin-related variants. The bold dashed line denotes habit plane microstructures with tetragonal phase on one side and a monoclinic laminate on the other. Order of austenite point group .P’, ’ = Order of martensite point group P, ’ (2.12) see Van Tandeloo and Amelinckx (1974). For a tetragonal to monoclinic transformation (2.12) gives v = 4. Frame indifference (2.1) implies that the free energy 4 has the same value at matrices U and RU where R is a rotation. The set zz! = {RI : R is a rotation) is called the austenite well, and the union of martensite wells is (2.13) A= (J:'=,{Ru,:R IS a rotation and U, a martensite variant}. (2.14) Following Ball and James (1992), we schematically depict wells as circles in the space of deformation gradients. Figure 2 shows the austenite and four martensite wells for a tetragonal to monoclinic transformation. The free energy 4 is lower on the austenite well at temperatures 8 > O,, whereas at temperatures fl < 19, it is lower on the martensite wells, and at the critical temperature 8, it is the same on both austenite and martensite wells (2. IO). 2.3. Twins A deformation that results in a twin is continuous (coherent) and has different constant deformation gradients on either side of the twin plane. Suppose deformation y : R + @ has constant gradients F, and F_ on either side of a plane with normal ij
Transformation of zirconia y=y(x) y=RU X RU reference deformed Fig. 3. A twinning deformation in the reference configuration, then the Hadamard compatibility condition says that deformation y is continuous on Q2 if and only if F+=F+p⑧q for some pes. Let F, and F, belong to different martensite wells and let R be a rotation. Suppose F; and R, are constant deformation gradients on either side of a win plane with normal iEg' in the reference configuration, then by using the Hadamard jump condition (2. 15)we get F,=F,+ag where ae is the twinning shear. Let F=R U; and F;=RU; where U; and U,are two martensite variants(2. 11), then(2.16)becomes RRRU=U+Ra⑧i R=RRR a=Ra andn=n (2.17) to obtain the twinning equation RU =U+an Equation (2. 18)can be interpreted as the requirement for a coherent planc interfacc, having normal f in the reference configuration, between regions with constant defor- mation gradients U; and RU, (see Fig 3)and with a as the shear vector. The twin plane normal in the deformed configuration is given by i=U7 f/UF Al, and by taking determinants on either side of (2. 18), it follows that a i=0(Ball and James 1987) thus the twin normal ii and shear a correspond to crystallographic twinning elements K and n, respectively
Transformation of zirconia reference deformed Fig. 3. A twinning deformation. in the reference configuration, then the Hadamard compatibility condition says that deformation y is continuous on CI if and only if F, =F_+p@tj (2.15) for some PE @. Let Fi and F, belong to different martensite wells and let R be a rotation. Suppose F, and RF, are constant deformation gradients on either side of a twin plane with normal HEW’ in the reference configuration, then by using the Hadamard jump condition (2.15) we get RF, = F, + I @ ;. (2.16) where 5 E g3 is the twinning shear. Let F, = R,U, and F, = R,U, where U, and U, are two martensite variants (2.1 l), then (2.16) becomes R;RR;U, = U, + Rf 10 r? Set R=RTRR,, a=RT1, andfi=h to obtain the twinning equation (2.17) RU,= U,+a@ii. (2.18) Equation (2.18) can be interpreted as the requirement for a coherent plane interface, having normal ii in the reference configuration, between regions with constant deformation gradients U, and RU, (see Fig. 3) and with a as the shear vector. The twin plane normal in the deformed configuration is given by ii = U,-’ fi/lU,’ Al, and by taking determinants on either side of (2.18), it follows that a * ii = 0 (Ball and James. 1987) ; thus the twin normal ii and shear a correspond to crystallographic twinning elements K, and y,, respectively
N. K. SIMHA Given a pair of martensite variants (U, U) we now find a rotation R, shear vector and twin plane normal f that satisfy the twinning equation(2. 18). Define C:=UFIUZUI then C=(I+U; 8a)(+aUF n), and the six parameters that determine shear a and normal n can be obtained from proposition 4 of Ball and James(1987)[also see Ericksen(1980)and Gurtin(1983). Suppose A1< A2<h, are the eigenvalues of the matrix CL, solutions for twinning shear a and normal f exist, if and only if λ1≥0andλ2=1 (2.20) When Ct+ l, shear a and normal i are given by -4,e+yA2 (2.21) where x=+l, p*0 is a constant chosen such that a=I and e, e3 are normalised eigenvectors of C, corresponding to ii, a,, respectively. Rotation R is found using R=(U+a⑧n)U (2.23) For given U. U (Ui+U, there are either two solutions(R, a, n) of the twinning equation(2. 18)(corresponding to x=+I)or none. Twin related variants are atically depicted on the well diagram(Fig. 2)by means of a light dashed line connecting variant U, with deformation gradicnt RU Thus there are cither nonc or ' o dashed lines between any pair of martensite wells. Suppose there are v martensite variants, then there are v(v-1)/2 pairs of martensite variants, and if all pairs are twin elated, then the martensite can form v(v-I)twins. Suppose we take(2. 16)to be the ic twinning equation with a solution(R, a, h); then it is not possible to say how many such twins the martensite can form since each variant U: corresponds to infinitely many deformation gradients F,=R U, where R, is a rotation. Moreover, (2. 18)involves only variants U, and U Thus it is advantageous to consider(2. 18)as he basic twinning cquation To classify twins, we rewrite the twinning equation(2. 18)as RU;=[+a nju where i=U-i is the twin plane normal in the deformed configuration; variant RU can be obtained by shearing variant U:. One class of solutions(R, a, n) has the property that the two twin related variants are related also by a 2-fold rotation(Zanzotto 1988),i RU,=[+a⑧司U;=Ql (2.24) where He sa, the austenite point group, and the rolation Q=-+2n g ninl, for type I twins, (2.25 for type II tw
270 N. K. SIMHA Given a pair of martensite variants (U, Uj) we now find a rotation R, shear vector a, and twin plane normal ti that satisfy the twinning equation (2.18). Define c, : = U,:’ u; u,-‘, (2.19) then C, = (I + U; ’ ii 0 a)(1 + a @ U,: ’ a), and the six parameters that determine shear a and normal ti can be obtained from proposition 4 of Ball and James (1987) [also see Ericksen (1980) and Gurtin (1983)]. Suppose i, < A2 d A3 are the eigenvalues of the matrix C, ; solutions for twinning shear a and normal fi exist, if and only if I, >,Oand;l, = I. (2.20) When C, # I, shear a and normal fi are given by (2.21) where x = f 1, p # 0 is a constant chosen such that ]Ei] = 1 and i?,, & are normalised eigenvectors of C, corresponding to &, &, respectively. Rotation R is found using R =(Ui+a@n)U,-‘. (2.23) For given Uj, U, (Ui # U,) there are either two solutions (R, a, n) of the twinning equation (2.18) (corresponding to x = + 1) or none. Twin related variants are schematically depicted on the well diagram (Fig. 2) by means of a light dashed line connecting variant U, with deformation gradient RU,. Thus there are either none or two dashed lines between any pair of martensite wells. Suppose there are v martensite variants, then there are v(v - 1)/2 pairs of martensite variants, and if all pairs are twin related, then the martensite can form v(v - 1) twins, Suppose we take (2.16) to be the basic twinning equation with a solution (8, H, ?I) ; then it is not possible to say how many such twins the martensite can form, since each variant Ui corresponds to infinitely many deformation gradients F, = RiUi, where R, is a rotation. Moreover, (2.18) involves only variants Uj and U,. Thus it is advantageous to consider (2.18) as the basic twinning equation. To classify twins, we rewrite the twinning equation (2.18) as RLJ, = [I+ a 0 ii]U, where ii = U; ’ ii is the twin plane normal in the deformed configuration ; variant RU, can be obtained by shearing variant Ui. One class of solutions (R, a, fi) has the property that the two twin related variants are related also by a 2-fold rotation (Zanzotto, 1988), i.e. RU, = [I+ a @ ii]U, = QU,H, where HE P’,, the austenite point group, and the rotation (2.24) Q= i -1+2ii @ ii/liil*, for type I twins, -1+2a@ ailal*, for type II twins ; (2.25)
Transformation of zirconia martensite austenite 围园昌喜 Fig, 4 Schematic representation of Fig. I. The transition zone(shaded )corresponds to the region where he martensite bands cease to he parallel and split in Fig. I rotation Q is of r radians about the twin plane normal in the deformed configuration for type I twins and about the shear vector for type II twins. If a twin is both type I and type Il, then it is called a compound twin In summary, then, the point groups and lattice parameters of the austenite and martensite determine the bain strains for a martensitic transformation For each bai strain, the symmetry of the free energy delivers a finite set of martensite variants The twinning equation(2.18)and its solution (2.21),(2.22)show that twins are ompletely determined by these variants. Consequently one can find all the possible wins between the martensite variants 2. 4. The habit plane microstructure In Fig. 4. we schematically represent the habit plane microstructure shown in Fig 1. Austenite, with the deformation gradient L, is separated frill Martensite by tlie habit plane with the normal he, and the martensite comprises of a periodic arrangement of parallel bands with deformation gradients QU, and QRU: Q and R are rotations(2. 2), and the martensite variants UU, satisfy the twinning equation (2.18), i.e. Q(RU -U,=a@n). Let the volume fraction of the variant with the deformation gradient QU, be !E(0, 1), then the average deformation gradient in the martensite region Is F=nQU, +(1-HQRU (2.26) Austenite, with the deformation gradient identity l, is separated by the habit plane viewed tant deformation gradient equal the average deformation gradient Fu. Then the Hadamard condition(2. 15)gives
Transformation of zirconia Fig. 4. Schematic representation of Fig. 1. The transition zone (shaded) corresponds to the region where the martensite bands cease to be parallel and split in Fig. I. rotation Q is of 71 radians about the twin plane normal in the deformed configuration for type I twins and about the shear vector for type II twins. If a twin is both type I and type II, then it is called a compound twin. In summary, then, the point groups and lattice parameters of the austenite and martensite determine the Bain strains for a martensitic transformation. For each Bain strain, the symmetry of the free energy 4 delivers a finite set of martensite variants. The twinning equation (2.18) and its solution (2.21), (2.22) show that twins are completely determined by these variants. Consequently one can find all the possible twins between the martensite variants. 2.4. The habit plane microstructure In Fig. 4, we schematically represent the habit plane microstructure shown in Fig. 1. Austenite, with the deformation gradient I, is separated from martensite by the habit plane with the normal ii~.G%~, and the martensite comprises of a periodic arrangement of parallel bands with deformation gradients QU, and QRU, : Q and R are rotations (2.2), and the martensite variants U,. U, satisfy the twinning equation (2.18), i.e. Q(RU,-U, = a @ ii). Let the volume fraction of the variant with the deformation gradient QU, be ~E(O, l), then the average deformation gradient in the martensite region is F,, = PQU,+(~-P)QRU,. (2.26) Austenite, with the deformation gradient identity I, is separated by the habit plane from martensite, viewed as a region with a constant deformation gradient equal to the average deformation gradient F,,. Then the Hadamard condition (2.15) gives