CONSTITUTIVE MODELING AND SIMULATION OF THE SUPERELASTIC EFFECT IN - PowerPoint PPT Presentation

Experiments 500 450 400 350 300 σ eq (MPa) 250 200 150 100 Torsion 2.0%, Tension 0.7% 50 Torsion 2.0%, Tension 1.05% Torsion 2.0%, Tension 3.0% Torsion 2.0%, Tension 5.8% 0 0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 ε eq Equivalent stress-strain plot of torsion followed by tension ( T YPE II) SMA – p.29/108

Experiments 500 450 400 350 300 σ eq (MPa) 250 200 150 100 Tension 0.7%, Torsion 2.0% 50 Tension 1.5%, Torsion 2.0% Tension 3.0%, Torsion 2.0% Tension 6.0%, Torsion 2.0% 0 0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 ε eq Equivalent stress-strain plot of simultaneous tension-torsion ( T YPE III) SMA – p.30/108

Closure In superelastic materials, austenite-to-martensite phase transformation occurs by nucleation of martensite twins, followed by detwinning. The stress-strain response is rich and complex, especially under non-proportional loading. The response to torsion defies the classical metal plasticity paradigm. The modeling of the mechanical response of superelastic materials leads to a multiscale problem. Martensite laminates, crystal grains, continuum SMA – p.31/108

References 1. J.M. Ball and R.D. James, Arch. Rat. Mech. Anal. , 100 , pp. 13-52, (1987) [Non-linearly elastic theory of martensitic transformation]. 2. K.F . Hane and T.W. Shield, Acta Mater. , 47 , pp. 2603-17, (1999) [Exhaustive description of the microstructure in NiTi phase transition]. 3. K. Gall and H. Sehitoglu, Int. J. Plast. , 15 , pp. 69-92, (1999) [Role of texture in NiTi]. 4. T.W. Duerig, A. Pelton and D. Stöckel, Mater. Sci. Engrg. , A273-275 , pp. 149-160, (1999) [A review of biomedical applications of Nitinol]. 5. J.M. McNaney, V. Imbeni, Y. Jung, P . Papadopoulos and R.O. Ritchie, Mech. Mat. , 35 , pp. 969-986, (2003) [Multiaxial experiments on polycrystal NiTi]. SMA – p.32/108

Contents-II C ONTENTS OF P ART II Continuum modeling at the crystal grain level Loading and transformation criteria Connection between continuum and microstructural modeling A constrained optimization problem Twinned and detwinned martensite SMA – p.33/108

Nomenclature Nomenclature: : deformation gradient F : Lagrangian strain tensor E : Right stretch tensor U : Infinitesimal strain tensor ǫ : 1st (unsymmetric) Piola-Kirchhoff stress tensor P : 2nd (symmetric) Piola-Kirchhoff stress tensor S : Infinitesimal stress tensor σ : general rotation tensor Q : second-order identity tensor I SMA – p.34/108

Modeling Consider constitutive modeling at the crystal grain level. The total deformation induced by the martensitic transformation can be expressed in the form: F = F 1 F 2 R 3 , where : lattice deformation F 1 : twinning deformation F 2 : rigid rotation R 3 Of the three deformation components, F 1 is dominant. SMA – p.35/108

Modeling There are three basic modeling choices at the crystal grain level: Model the transformation of austenite to detwinned martensite. austenite austenite 1 austenite SMA – p.36/108

Modeling There are three basic modeling choices at the crystal grain level: Model the transformation of austenite to twinned martensite. austenite 1 2 12 1 1 2 2 austenite austenite SMA – p.36/108

Modeling There are three basic modeling choices at the crystal grain level: Model the transformation of austenite first to twinned martensite, then to detwinned martensite. austenite 1 2 12 1 1 2 2 austenite 1 austenite SMA – p.36/108

Modeling Consider first modeling the transformation of austenite to detwinned martensite. In the case of Ni-Ti (cubic to monoclinic transformation), there are a total of 12 martensite “lattice correspondence variants”, i.e., F 1 ,i = R ( Q i UQ T i ) , i = 1 , . . . , 12 . In the above polar decompositions, R is the rotation tensor, U is the unique stretch tensor, and Q i are the 12 rotations corresponding to the 12 ways in which a cube maps onto itself. The tensor U can be deduced from the crystallography of the austenite-martensite transformation. SMA – p.37/108

Modeling Consider next modeling the transformation of austenite to twinned martensite. In the case of Ni-Ti (cubic to monoclinic transformation), one may expect up to 66x8 = 528 martensite “habit plane variants”. Compatibility requirements reduce this number to 192 feasible habit plane variants. These variants can be obtained by an energy minimization method. Of the 192 feasible habit plane variants, only 24 are experimentally observed to be dominant. SMA – p.38/108

Modeling In the case of a typical habit plane variant α , the microscopic deformation gradient takes the form F 12 ,α = I + g m α ⊗ n α , where g is the trans- g m α formation displacement, twinned martensite m α is the unit vector in the transformation direction, and n α is the outward unit normal to the habit plane. n α habit plane SMA – p.39/108

Modeling Proceed with modeling by analogy to the theory of elastic-plastic materials. A plastic (resp. phase-tranforming) material can be views as an elastic material whose response is parametrized by the plastic (resp. transformation) variables. Two key differences: Martensitic transformation is fully reversible. Loading conditions in multi-surface plasticity are determined per surface, while in martensitic transformation they are cumulative. SMA – p.40/108

Modeling Micromechanically-motivated constitutive assumptions: 1. Existence of a Lagrangian transformation strain E t depending on the martensitic volume fractions nv � � � { ξ α } = ( ξ 1 , ξ 2 , · · · , ξ nv ) | ξ β ≤ 1 , ξ β ≥ 0 , β =1 such that E t = ˆ E t ( { ξ α } ) , where nv stands for the number of potentially present variants. Homogeneity condition: ˆ E t ( { 0 } ) = 0 . Note the analogy (and contrast) with plasticity theory. SMA – p.41/108

Modeling 2. Admittance of a Helmholtz free energy Ψ = ρ 0 ( ǫ − ηθ ) , where Ψ = ˆ Ψ( E , { ξ α } , θ ) . Here, ρ 0 is the mass density, ǫ is the internal energy per unit mass, η is the entropy, and θ the absolute temperature. Recall the local form of the energy equation ǫ = ρ 0 r − Div q 0 + S · ˙ ρ 0 ˙ E , where r is the heat supply per unit mass and q 0 the referential heat flux vector. Also recall the Clausius-Duhem inequality ηθ ≥ ρ 0 r − Div q 0 + q 0 · Grad θ ρ 0 ˙ . θ SMA – p.42/108

Modeling For any homothermal superelastic process, � � nv S − ∂ ˆ ∂ ˆ Ψ Ψ � · ˙ · ˙ E − ξ α ≥ 0 . ∂ E ∂ξ α α =1 Since E and { ξ α } can be varied independently, a standard process leads to S = ∂ ˆ Ψ ∂ E , while the Clausius-Duhem inequality further implies that nv ( − ∂ ˆ Ψ � ˙ ) ˙ D = ξ α ≥ 0 . ∂ξ α α =1 SMA – p.43/108

Modeling 3. Existence of functions ˆ α and ˆ Y f Y r α in the form α = ˆ α = ˆ Y f Y f Y r Y r α ( E , { ξ β } , θ ) α ( E , { ξ β } , θ ) , associated with the forward and reverse transformation of variant α , respectively, where ˆ α < ˆ Y f Y r α . Y f ⇔ α = 0 variant α active in forward transformation Y r ⇔ α = 0 variant α active in reverse transformation Forward and reverse transformation active sets: J f ( E , θ ) = { α | ˆ Y f α ( E , { ξ β } , θ ) = 0 , ξ α > 0 } , J r ( E , θ ) = { α | ˆ Y r α ( E , { ξ β } , θ ) = 0 , ξ α > 0 } . SMA – p.44/108

✐ ❢ s q ♣ r q ❞ ❡ ❣ ❳ ❤ ♣ ❥ ❤ ❦ ❧ ♠ ❨ t ♦ ④ ⑩ ⑦ ⑨ ⑧ ⑦ ⑥ ⑤ ③ ✉ ① ✇ ② ① ✇ ✈ ◗ ♥ Modeling Yield surfaces in stress space (b) (a) (e) (c) (d) 0 (c) ▼❖◆P ❯❖❱❲ (c) (b) (a) (b) (a) (d) (d) (e) (e) ❘✞❙❚ ❩❭❬ ❪✞❫ ❴❛❵ ❜✼❝ SMA – p.45/108

Modeling Persistency of a forward-active variant α , α = ∂ ˆ Y f � Q αβ ˙ ˙ ∂ E · ˙ α Y f E − ξ β = 0 , β ∈J f αβ = − ∂ ˆ Y f where Q f α are the components of the forward ∂ξ β coupling matrix (assumed invertible) which quantifies the coupling between variants during forward transformation. It follows that during forward transformation of variant α , ∂ ˆ Y f � ˙ ∂ E · ˙ β Q f − 1 ξ α = E . βα β ∈J f This is a rate-type, rate-independent equation (note connection with rate-independent plasticity). SMA – p.46/108

Modeling 4. Stipulate of forward and reverse transformation criteria. From a state of a forward transformation ( J f � = ∅ ):  ⇔ > 0 forward transformation   ∂ ˆ Y f  � ∂ E · ˙ α W f . E = 0 ⇔ neutral forward transformation α   α ∈J f  ⇔ < 0 elastic unloading From a state of a reverse transformation ( J f = ∅ , J r � = ∅ ):  > 0 ⇔ elastic reloading   ∂ ˆ Y r  � ∂ E · ˙ W r α . E ⇔ = 0 neutral reverse transformation α   α ∈J r  < 0 ⇔ reverse transformation SMA – p.47/108

Modeling To determine the weight function W f α , note that during forward transformation α − ∂ ˆ Ψ � ) ˙ ( − Y f ξ α ≥ 0 . ∂ξ α α ∈J f Recalling the equation for ˙ ξ α , it follows that β − ∂ ˆ ) ∂ ˆ Y f Ψ � � ∂ E · ˙ Q f − 1 αβ ( − Y f α E ≥ 0 , ∂ξ β α ∈J f β ∈J f which implies that β − ∂ ˆ Ψ � Q f − 1 αβ ( − Y f W f α = ) . ∂ξ β β ∈J f A similar analysis applies to W r α . SMA – p.48/108

Modeling Consider an alternative option: postulate a single forward transformation function of the form Y f = � − ∂ ˆ Ψ � q − F c , ∂ξ α where � ( · ) � q denotes the standard vector q -norm. q = 1 q = 2 q = ∞ Note analogy to yield functions in plasticity theory. Why use non-smooth transformation criteria? SMA – p.49/108

Modeling How to incorporate microstructural information into the crystal grain constitutive model? Homogenize the kinematics and kinetics of the microstructure over a representative referential volume element (RVE) V that characterizes the physics of the martensitic transformation. austenite 1 2 12 1 1 2 2 austenite austenite SMA – p.50/108

Modeling “Descend” into the limiting case of infinitesimal deformations and let the mechanical part of the Helmholtz free energy be of the form nv +1 1 � 2( ǫ − ǫ t α ) · C ( ǫ − ǫ t ψ ( ǫ , { ξ α } ) = + ψ m ( { ξ α } ) ξ α α ) , � �� � α = mixing � �� � single phases where C is the elasticity tensor (assumed phase-invariant). Subsequently, write the mechanical part of the Gibbs free energy of the mixture as nv nv 1 � � 2 σ · C − 1 σ + ξ α σ · ǫ t γ ( σ , { ξ α } ) = α + . . . α =1 α =1 SMA – p.51/108

Modeling Assume a spatially homogeneous stress σ 0 over the ensemble and employ a Legendre transformation to write � � − γ ( σ 0 , { ξ α } ) + σ 0 · ǫ � � sup dV = ψ R ( ǫ , { ξ α } ) dV , σ 0 V V where ψ R is the Reuss (lower) bound to the mechanical part of the free energy. It follows from the above that the Reuss bound is nv nv ǫ , { ξ α } ) = 1 � � ξ α ǫ t ξ α ǫ t ψ R (¯ 2(¯ ǫ − α ) · C (¯ ǫ − α ) α =1 α =1 � 1 in terms of the volume-averaged strain ¯ ǫ = ǫ dV . vol ( V ) V SMA – p.52/108

Modeling Alternatively, assume a spatially homogeneous strain ǫ 0 over the ensemble and again homogenize the elastic part of the Helmholtz free energy to find that nv nv nv ǫ , { ξ α } ) = 1 α ) − 1 � � � ξ α ǫ t ξ α ǫ t σ · ξ α ǫ t ψ V (¯ 2(¯ ǫ − α ) · C (¯ ǫ − ˜ α , 2 α =1 α =1 α =1 where ˜ σ is the self-equilibrated zero-mean fluctuation stress. One may estimate the contribution of the fluctuation stress, although the existing estimates are generally not very accurate. SMA – p.53/108

Modeling Motivated by the micromechanical model, let the Helmholtz free energy be written as Ψ = ˆ Ψ( E , { ξ α } , θ ) = nv 1 � � E − E t � � E − E t � · C + B ( θ − θ 0 ) ξ α . 2 α =1 � �� � � �� � mechanical chemical where E t is the cumulative transformation strain, B is a chemical energy constant (latent heat of transformation), and θ 0 is the equilibrium temperature. Note that the strains are moderate (typically, less than 6-10%). SMA – p.54/108

Modeling The cummulative transformation strain is now defined as nv � E t = ξ α E t α , α =1 where the variant transformation strain E t α is given by α = 1 E t 2 g ( m α ⊗ n α + n α ⊗ m α + g n α ⊗ n α ) , in terms of the microscopic gm twinned martensite deformation gradient F α = I + g m α ⊗ n α . n habit plane SMA – p.55/108

Modeling Further, admit critical thermodynamic force-based transformation functions α = − ∂ ˆ α = − ∂ ˆ Ψ Ψ + F c , Y f − F c Y r , ∂ξ α ∂ξ α to characterize forward and reverse transformation, respectively. Given the particular form of the Helmholtz free energy, nv � � � � B ( θ − θ 0 ) + F c � α = ˆ Y f Y f ξ β E t · E t α ( E , { ξ β } , θ ) = C E − α − , β β =1 nv � � � � B ( θ − θ 0 ) − F c � α = ˆ Y r Y r ξ β E t · E t α ( E , { ξ β } , θ ) = C E − α − . β β =1 SMA – p.56/108

Modeling During elastic loading or forward transformation, the loading conditions simplify to � ˙ ξ α ≥ 0 , α ∈J f namely, the total production of martensite is non-negative. Recalling further the form of the forward transformation criterion, it follows that during elastic loading or forward transformation − ∂ ˆ − F c � ˙ Ψ � � ξ α = 0 . ∂ξ α α ∈J f An analogous result applies to reverse transformation. SMA – p.57/108

Modeling Note that here ∂ 2 ˆ Ψ = Q αβ = E t α · C E t β , ∂ξ α ∂ξ β where the nv × nv coupling matrix [ Q α β ] is symmetric and of rank at most 6. This implies that for nv > 6 , the matrix [ Q αβ ] is necessarily positive semi-definite. Recalling again that − ∂ ˆ − F c � ˙ Ψ � � ξ α = 0 , ∂ξ α α ∈J f it follows that the identification of active variants and volume fractions can be cast as a constrained optimization problem at fixed strain and temperature. SMA – p.58/108

Modeling Specifically, in the case of forward transformation, the martensitic content { ξ α } satisfies { ξ γ }∈J f Φ f , { ξ α } = argmin subject to inequality constraints nv � − ξ α ≤ 0 ξ α ≤ 1 . , α =1 where the functional Φ f is defined as � Φ f = Ψ + F c ξ α , α ∈J f and may attain multiple minima. A similar conclusion applies to reverse transformation for the functional Φ r = Ψ − � α ∈J r F c ξ α . SMA – p.59/108

Modeling How to model detwinning? The twinned and detwinned variants may coexist, so that Ψ = ˆ Ψ( E , { ξ hab α , ξ lat β } , θ ) = nv 1 � � E − E t � � E − E t � · C + B ( θ − θ 0 ) ξ α , 2 α =1 where now nvh nvl � � E t = ξ hab α E t ξ lat β E t α + β . α =1 β =1 In the above, nvh and and nvl denote the total number of habit and lattice correspondence variants, respectively. SMA – p.60/108

Closure Plasticity theory can be used as a guide toward the development of superelastic constitutive models. It is possible to motivate continuum formulations of superelasticity using micromechanics. It is sensible to consolidate the loading criteria, while keeping the transformation criteria separate. Under certain assumptions, superelasticity can be cast as a constraint minimization problem. The role of interaction energy between variants remains an open question. SMA – p.61/108

References 1. J.G. Boyd and D.C. Lagoudas, Int. J. Plast. , 12 , pp. 805-842, (1996) [Phenomenological modeling of phase transformation]. 2. S. Leclercq and C. Lexcellent, Int. J. Mech Phys Sol. , 44 , pp. 953-980, (1996) [Phenomenological modeling of phase transformation including detwinning]. 3. M. Huang and L.C. Brinson, J. Mech. Phys. Sol. , 46 , pp. 1379-1409, (1998) [Micromechanically motivated model for single crystal] 4. N. Siredey, E. Patoor, M. Berveiller and A. Eberhardt, Int. J. Sol. Struct. , 36 , pp. 4289-4315, (1999) [Micromechanically motivated multivariant modeling] SMA – p.62/108

Contents-III C ONTENTS OF P ART III Operator-split algorithms at the crystal grain level Optimization-based algorithms at the crystal grain level Modeling of textured polycrystals Experiments and simulation An application to the design of biomedical devices SMA – p.63/108

Algorithmic treatment Within the finite element paradigm, the algorithmic problem for a single crystal is set up as follows: “Given the state at time t n , and the total displacement u n +1 at time t n +1 , determine the state at time t n +1 .” The preceding problem is iteratively nested inside the global momentum balance problem. material global u ( i ) (Gauss pt.) n +1 momentum balance i ← i + 1 i−th iterate local state i−th iterate global state assemble SMA – p.64/108

Algorithmic treatment Two potential approaches: Unified approach Determine the phase state by solving a constrained optimization problem. Unified treatment of transforming and non-transforming states Operator-split approach Analogy with computational plasticity for rate-independent materials Different treatment of transforming and non-transforming states SMA – p.65/108

Unified approach Treat elastic loading/forward transformation in a unified manner. σ eq ¯ ξ u = 1 ξ l = � ξ α ¯ 0 ǫ eq Express the constraint conditions as nv � ¯ ¯ − ξ α ≤ 0 ξ l − ξ α ≤ 0 ξ u ≤ 1 , , , α =1 where ¯ ξ l and ¯ ξ u are lower and upper values of the total martensitic volume fraction. SMA – p.66/108

Unified approach Treat elastic unloading/reverse transformation in a unified manner. σ eq ξ u = � ξ α ¯ ¯ ξ l = 0 0 ǫ eq Express the constraint conditions as nv � ξ α − ¯ 0 ≤ ¯ − ξ α ≤ 0 ξ u ≤ 0 , , ξ l , α =1 where ¯ ξ l and ¯ ξ u are lower and upper values of the total martensitic volume fraction. SMA – p.66/108

Unified approach Forward transformation Reverse transformation ❽❸❾ ❻❸❼ ❹❸❺ ➁❸➂ ❶❸❷ ❿❸➀ The polytope constraints for nv = 3 SMA – p.67/108

Unified approach Recall that the forward transformation problem can be cast in the form { ξ γ }∈J f Φ f , { ξ α } = argmin subject to the preceding polytope constraints. Introduce Lagrange multipliers { λ α , α = 1 , . . . , nv } , λ l , and λ u , and write ∂ Φ f + λ α + λ l − λ u = 0 , ∂ξ α where nv nv nv � � � λ α ( − ξ α ) + λ l (¯ ξ α − ¯ ξ l − ξ α ) + λ u ( ξ u ) = 0 . α =1 α =1 α =1 SMA – p.68/108

Unified approach Three key algorithmic tasks: Identify the state as “loading” (forward transforming) or “unloading” (reverse transforming). Identify the set C of potentially active variants. Select up to 6 such variants from the set of all nv habit plane variants. Identify the variants from the set C that are active. All three tasks are deformation-dependent (albeit history-independent) and generally need to be performed in each loading step! SMA – p.69/108

Unified approach Recall that the functional Φ f is defined as nv Φ f = 1 � � � E − E t � � E − E t � F c ξ α , · C B ( θ − θ 0 ) ξ α + 2 α =1 α ∈J f ∂ 2 Φ f hence is quadratic in { ξ α } , and the 24 × 24 Hessian ∂ξ α ∂ξ β is at most of rank 6. This leads to a semi-definite quadratic programming problem. To “relax” this problem, identify a set C of 6 candidate active variants at each step. Subsequently, a standard active set strategy of positive-definite quadratic programming is applied to determine the active variants and volume fractions. SMA – p.70/108

Unified approach There are several ways to identify a “reasonable” set C of 6 candidate active variants. One possibility is to first determine the extrema of Φ f with respect to each variant separately, i.e., solve α ( E , ξ α , θ ) = ∂ ˆ Φ f ˆ Y f = 0 , ∂ξ α for each ξ α , α = 1 , . . . , 24 . Subsequently, select the 6 variants that correspond to the lowest values of Φ f = ˆ Φ f ( E , ξ α , θ ) . Alternatively, one may perform a pair-wise minimization to determine 3 pairs of potentially active variants, as motivated by experiments. The set of active variants depends crucially on the load. SMA – p.71/108

Unified approach Given the set C of candidate variants, express the discrete counterpart of the first-order conditions ∂ Φ f + λ α + λ l − λ u = 0 , ∂ξ α nv nv nv � � � λ α ( − ξ α ) + λ l (¯ ξ α − ¯ ξ l − ξ α ) + λ u ( ξ u ) = 0 α =1 α =1 α =1 in matrix form as [ Q ][ ξ n +1 ] + [Π n +1 ] T [ λ n +1 ] = [ c n +1 ] , [Π n +1 ][ ξ n +1 ] = [ h n +1 ] . This system possesses a unique solution, as [ Q ] is positive-definite and [Π n +1 ] has full row-rank. SMA – p.72/108

Unified approach Use an active set strategy starting from any feasible initial guess of volume fractions and active variants (working set). 1. Solve the first-order equations and determine new values of [ ξ n +1 ] and [ λ n +1 ] . 2. If one or more of the inactive constraints are violated, then add the most offending constraint to the working set and go to step 1. 3. If one or more of the Lagrange multipliers become negative, then drop from the working set the constraint that corresponds to the lowest value over all multipliers and go to step 1. Slow, but reliable! SMA – p.73/108

Unified approach The discrete counterparts of the loading criteria from a forward transformation state at t = t n +1 are  > 0 forward transformation ⇔     ( ξ f � α,n +1 − ξ α,n ) , = 0 and λ l = 0 neutral forward transformation ⇔  α ∈C  = 0 and λ l > 0 elastic unloading  ⇔  where ξ f α,n +1 is calculated by constrained minimization of Φ f . Note that the latter two conditions are modified in order to account for the explicit enforcement of the constraint ξ l − � ¯ α ∈C ξ α ≤ 0 . SMA – p.74/108

Unified approach The discrete counterparts of the loading criteria from a reverse transformation state at t = t n +1 are  < 0 reverse transformation ⇔     � ( ξ r α,n +1 − ξ α,n ) , = 0 and λ u = 0 neutral reverse transformation ⇔  α ∈C  = 0 and λ u > 0 elastic reloading  ⇔  where ξ r α,n +1 is calculated by constrained minimization of Φ r . Note that the first two conditions are modified in order to account for the explicit enforcement of the constraint � α ∈C ξ α − ¯ ξ u ≤ 0 . SMA – p.75/108

Unified approach In the unified approach, there is no distinction in the treatment of elastic or transforming states. This implies that one only needs to check whether the state is “forward” or “reverse”. Use a flag to handle the characterization (set initially to “forward”). and resolve all possible combinations of forward/reverse loading: ❍❍❍❍❍❍ ∆ ξ f n + 0 ∆ ξ r n ❍ !flag ‘reverse’ − 0 ‘forward’ flag SMA – p.76/108

Unified approach The unified approach is also applicable for problems that lead to non-linear programming. For example, assume nv 1 � � E − E t � � E − E t � · ¯ + B ( θ − θ 0 ) Ψ = ξ α , C 2 α =1 where, using the rule of mixtures, � � nv nv � � ¯ 1 − C = ξ α C a + C m . α =1 α =1 Use non-linear programming with active set strategy or reduce the problem to one of quadratic programming by using time-lagging estimates of ¯ C . SMA – p.77/108

Operator-split approach Examine the operator-split approach in connection with the preceding model. Given the state at t n (i.e., { ξ β,n } ), the temperature θ n +1 , and the i − th iterate of the displacement u ( i ) n +1 : 1. Assume that the total strain E ( i ) n +1 is elastic, i.e., there is no phase forward transformation. 2a. If ˆ α ( E ( i ) Y f n +1 , { ξ β,n } , θ n +1 ) < 0 for all variants α , then the process in ( t n , t n +1 ] is elastic. 2b. If ˆ α ( E ( i ) Y f n +1 , { ξ β,n } , θ n +1 ) ≥ 0 for some variant(s) α , then there is some phase transformation. SMA – p.78/108

Operator-split approach As in the unified approach, one again needs to determine: The set C of potentially active variants. The variants from the set C that are active. The operator-split approach is popular in computational plasticity because of the simplicity of the elastic predictor and the physical interpretation of the plastic corrector. Are these attractive features preserved in the case of micromechanically motivated phase transition models? SMA – p.79/108

Operator-split approach The operator-split approach has two complications: 1. It is not readily obvious how to effect the “plastic” (i.e., transformation) correction. It is sometimes assumed that ξ α = κ ∂Y f ˙ α , ∂ξ α which corresponds to an “associated” flow rule. Y f 1 = 0 Y 2 f = 0 There is no constitutive justification for this assumption! SMA – p.80/108

Operator-split approach 2. The evolution of the variants is subject to the polytope constraints. ξ < 0 Y f > 0 Y f = 0 ξ = 0 One needs to perform constrained projections in multi-dimensional space. In some cases, the projections indicate unsuitable selection of potentially active variants. SMA – p.81/108

Polycrystal modeling Texture can potentially play a pivotal role in the mechanical response of polycrystalline superelastic alloys. Two ways to model the textured polycrystal structure: Direct simulation Resolve the polycrystal structure using individual finite elements. This is a simple, but potentially expensive approach. Two-scale analysis Solve fine-scale boundary-value problem on a representative domain and extract volume-averaged stress to be used in a continuum-level simulation. This approach can radically reduce the computational cost. SMA – p.82/108

➋ ➌ ➣ → ➔ ➑ ➐ ➏ ➃ ➎ ➍ ➉ ➊ Polycrystal modeling To conduct either direct simulation or two-scale analysis, one needs to utilize the texture information. As an example, consider the Nitinol tubes described earlier. Here, the manufacturing process induces primarily � 111 �{ 110 } -type sheet texture “wrapped” around the cylindrical surface, such that the � 111 � austenite lattice direction is aligned with the longitudinal axis of the tube. ➇➆➈ ➑➓➒ ➑➓➒ ➄➆➅ SMA – p.83/108

Polycrystal modeling Pole figures for Nitinol tubes (R.D. horizontal, T.D. vertical) SMA – p.84/108

Polycrystal modeling Generally, pole figure data are directly processed by the software that handles crystallographic data acquisition. Relevant software: BEARTEX, LaboTex, popLA, etc. Typical software output: ω (1) ω (1) ω (1) q (1) 1 2 3 ω (2) ω (2) ω (2) q (2) , 1 2 3 · · · · · · · · · · · · where ω ( i ) I , I = 1 , 2 , 3 , are orientation-defining angles for the i -th bin and q ( i ) is the corresponding diffraction intensity. SMA – p.85/108

PSfr Polycrystal modeling q ′ = q ′′ q x p ′ r ′′ = z ω 2 ω 1 p ω 3 r = r ′ q ′′′ = y Matthies-Roe angle convention ( p , q , r ) → ( x, y, z ) SMA – p.86/108

Polycrystal modeling Also, recall that the austenite is wrapped around the tube. q p ω 0 r This induces an initial rotation ω 0 of the cubic crystals relative to the r axis. SMA – p.87/108

Polycrystal modeling In summary, the total rotation Q that characterizes the texture can be expressed as Q = Q 3 Q 2 Q 1 Q 0 , where Q 0 : rotation by ω 0 relative to r , Q 1 : rotation by ω 1 relative to r , Q 2 : rotation by ω 2 relative to q ′ , Q 3 : rotation by ω 2 relative to r ′′ . To obtain a matrix representation of the rotation, recall the Rodrigues formula for rotation in { p , q , r } by ω with respect to p : Q r = p ⊗ p + cos ω ( q ⊗ q + r ⊗ r ) − sin θ ( q ⊗ r − r ⊗ q ) . SMA – p.88/108

Polycrystal modeling For any given grain, the angles ω I , I = 1 , 2 , 3 , are picked from the pole figure data using a random processes weighted by the diffraction intensities. The angle ω 0 is computed directly as a function of the location of the grain on the tube. Once the cumulative rotation Q is known, the habit plane transformation strains { E t α } of a typical crystal grain are expressed as α Q T , E t a,tex = QE t where E t α are the transformation strains relative to the austenite (cubic) frame. SMA – p.89/108

Polycrystal modeling The direct simulation approach is straightforward. The two-scale approach requires the solution of a boundary-value problem at each Gauss point: Use the macro-scale deformation as input for the fine-scale problem, determine the mean stress, and use it to define the constitutive behavior in the macro-scale problem. SMA – p.90/108

Polycrystal modeling Use Taylor assumption, i.e., impose constant deformation gradient field ¯ F = F in the fine-scale problem. Recalling the Hill-Mandel averaging theorem, compute and output the mean 1st Piola-Kirchhoff stress ¯ P . Interesting issues: Size of the fine-scale problem (convergence of the microstructure) Implicit global solution options (Newton’s method and its variants) Other consistent averaging options SMA – p.91/108

Numerical simulations Material parameters: E = 38 . 0 GPa Austenite Young’s modulus : Martensite Young’s modulus : E = 10 . 0 GPa ν = 0 . 3 Poisson’s ratio : B = 0 . 607 MPa / ◦ C Chemical energy constant : θ − θ 0 = 22 . 3 ◦ C Temperature : F c = 7 . 5 MPa. Critical force : The crystallographic vectors are taken to have components [ n ] = [ − 0 . 88888 , 0 . 21523 , 0 . 40443] T [ m ] = [0 . 43448 , 0 . 75743 , 0 . 48737] T , relative to the austenite lattice, while g = 0 . 13078 . SMA – p.92/108