Algorithmic aspects and finite element solutions for advanced phase field approach to martensitic phase transformation under large strains

Abstract

A new problem formulation and numerical algorithm for an advanced phase-field approach (PFA) to martensitic phase transformation (PT) are presented. Finite elastic and transformational strains are considered using a fully geometrically-nonlinear formulation, which includes different anisotropic elastic properties of phases. The requirements for the thermodynamic potentials and transformation deformation gradient tensor are advanced to reproduce crystal lattice instability conditions under a general stress tensor obtained by molecular dynamics (MD) simulations. The PFA parameters are calibrated, in particular, based on the results of MD simulations for PTs between semiconducting Si I and metallic Si II phases under complex action of all six components of the stress tensor (Levitas et al. in Phys Rev Lett 118:025701, 2017a; Phys Rev B 96:054118, 2017b). The independence of the PFA instability conditions of the prescribed stress measure is demonstrated numerically for the initiation of the PT. However, it is observed that the PT cannot be completed unless the stress exceeds the stress peak points that depend on which stress measure is prescribed. Various 3D problems on lattice instability and following nanostructure evolution in single-crystal Si are solved. The effect of stress hysteresis on the nanostructure evolution is studied through analysis of the local driving force and stress fields. It is demonstrated that variation of internal stress fields due to differing boundary conditions may lead to completely different PT mechanisms.

Appendix

1.1 A: Derivation of the weak form of the mechanical equilibrium equation and linearization

In Sect. 3, we outline the weak forms for the mechanical equilibrium equations without details. We present the detailed derivation in this appendix. As described in Sect. 3, we have used a non-monolithic scheme to solve the governing equations, i.e. for solving for the displacements, we assume that the order parameter is constant in all iterations.

The weak form of the equilibrium equation (Eq. 13) is given by Eq. (39). Integrating Eq. (39) by parts and using the Gauss divergence theorem, the weak form is rewritten and can be presented as

$$\begin{aligned} R({\pmb {u}},\delta {\pmb {u}})=\int _{\Omega _0} {\pmb {P}}^T\varvec{:}{\varvec{\nabla }}_0\delta {\pmb {u}} dV_0 - \int _{S_{0T}} \bar{{\pmb {p}}}{{\cdot }}\delta {{\pmb {u}}}\,dS_0=0 \end{aligned}$$

(74)

where we have used the identity ${\varvec{\nabla }}_0\cdot ({\pmb {P}}^T\cdot \delta {\pmb {u}})= ({\varvec{\nabla }}_0\cdot {\pmb {P}})\cdot \delta {\pmb {u}}+{\pmb {P}}^T:{\varvec{\nabla }}_0\delta {\pmb {u}}$ and recall that $\bar{{\pmb {p}}}$ is the specified traction on the traction boundary $S_{0T}$. Noticing that

$$\begin{aligned}&{\pmb {E}}=\frac{1}{2}({\pmb {F}}^T{\pmb {.}}{\pmb {F}}-{\varvec{I}}),\quad \delta {\pmb {F}}={\varvec{{\nabla }}}_0\delta {\pmb {u}}\quad \text {and}\quad \nonumber \\&{\varvec{{\nabla }}}_0(.)={\varvec{\nabla }}(.){\pmb {.}}{\pmb {F}} \end{aligned}$$

(75)

the variation of the Lagrangian strain is expressed as

$$\begin{aligned} \delta {\pmb {E}}= & {} \frac{1}{2}({\varvec{{\nabla }}}_0\delta {\pmb {u}}^T{{\cdot }} {\pmb {F}}+{\pmb {F}}^T{{\cdot }}{\varvec{{\nabla }}}_0\delta {\pmb {u}})\nonumber \\= & {} \frac{1}{2}{\pmb {F}}^T{{\cdot }}({\varvec{\nabla }}\delta {\pmb {u}}^T+{\varvec{\nabla }}\delta {\pmb {u}}){{\cdot }}{\pmb {F}} ={\pmb {F}}^T{{\cdot }}\delta {\varvec{\varepsilon }}{{\cdot }}{\pmb {F}} \end{aligned}$$

(76)

with $\delta {\varvec{\varepsilon }}$ given by Eq. (41). Utilizing the relations between the Piola–Kirchhoff and Cauchy stresses given by Eqs. (10) and (11), as well as using Eq. (76), the first integrand in Eq. (74) can be rewritten as

$$\begin{aligned} {\pmb {P}}\varvec{:}{\varvec{\nabla }}_0(\delta {\pmb {u}})= {\pmb {S}}\varvec{:}{\pmb {F}}^T{{\cdot }}{{\varvec{\nabla }}}(\delta {\pmb {u}}){{\cdot }}{\pmb {F}}= {\pmb {S}}\varvec{:}\delta {\pmb {E}}={\varvec{{\tau }}}\varvec{:}\delta {\varvec{\varepsilon }}, \end{aligned}$$

(77)

where ${\varvec{{\tau }}}=J {\varvec{\sigma }}$ is the Kirchhoff stress. Therefore, the weak form of the equilibrium equation Eq. (74) can be written in the form given by Eq. (40).

Because we will use the Newton’s iteration for computing the displacements, we must linearize the weak form given by Eq. (40). In doing so we expand the weak form in a Taylor series about ${\pmb {u}}$

$$\begin{aligned} R({\pmb {u}}+ \Delta {\pmb {u}}, \delta {\pmb {u}})= & {} R({\pmb {u}}, \delta {\pmb {u}})\nonumber \\&+\,\Delta R({\pmb {u}}, \Delta {\pmb {u}}, \delta {\pmb {u}})+o(\Delta {\pmb {u}}) = 0, \end{aligned}$$

(78)

where $\Delta {\pmb {u}}$ is an increment of the displacement vector, $\delta {\pmb {u}}$ has been kept fixed, $o(\Delta {\pmb {u}})$ consists of the higher order terms in $\Delta {\pmb {u}}$ such that $\lim _{\Delta {\pmb {u}}\rightarrow {\pmb {0}}} o(\Delta {\pmb {u}})/|\Delta {\pmb {u}}|= 0$, and $\Delta R({\pmb {u}}, \Delta {\pmb {u}}, \delta {\pmb {u}})$ is the directional derivative of R defined as [48]

$$\begin{aligned} \Delta { F}({\pmb {u}}, \Delta {\pmb {u}}, \delta {\pmb {u}})= & {} D F({\pmb {u}},\delta {\pmb {u}})\cdot \Delta {\pmb {u}} \nonumber \\= & {} \left. \frac{d}{d\epsilon } F({\pmb {u}}+\epsilon \Delta {\pmb {u}},\delta {\pmb {u}})\right| _{\epsilon =0} \end{aligned}$$

(79)

for any differentiable functional or function F.

The linearized form of residual R in Eq. (40) can be expressed as

$$\begin{aligned} D R\cdot \Delta {\pmb {u}}=\int _{\Omega _0}\Delta {\pmb {S}}\varvec{:}\delta {\pmb {E}} dV_0+\int _{\Omega _0}{\pmb {S}}\varvec{:}\Delta (\delta {\pmb {E}})dV_0. \end{aligned}$$

(80)

We will now derive an amenable form of the integrands in Eq. (80).

Noticing that ${\pmb {U}}_t$ is independent of ${\pmb {u}}$, we derive the expression for the increment of ${\pmb {S}}$ using Eq. (11):

$$\begin{aligned} \Delta {\pmb {S}} = J_t{\pmb {U}}_t^{-1} \cdot \Delta \hat{{\pmb {S}}}\cdot {\pmb {U}}_t^{-1} = J_t{\pmb {U}}_t^{-1} \cdot ({{\pmb {C}}}:\Delta {{\varvec{E}}}_e) \cdot {\pmb {U}}_t^{-1}, \nonumber \\ \end{aligned}$$

(81)

where ${{\pmb { C}}}$ is the fourth order elastic modulus tensor with respect to $\Omega _t$ and is given by Eqs. (27) and (44). Using Eq. (3)$_1$, we show that the increments $\Delta {\pmb {E}}_e$ and $\Delta {\pmb {E}}$ are related by

$$\begin{aligned} \Delta {\pmb {E}}_e = {\pmb {U}}_t^{-1}\cdot \Delta {\pmb {E}}\cdot {\pmb {U}}_t^{-1}, \end{aligned}$$

(82)

which we use to rewrite Eq. (81) as

$$\begin{aligned} \Delta {\pmb {S}}= & {} J_t {\pmb {U}}_t^{-1} \cdot ({{\pmb { C}}}:{\pmb {U}}_t^{-1}\cdot \Delta {{\varvec{E}}}\cdot {\pmb {U}}_t^{-1})\cdot {\pmb {U}}_t^{-1} \nonumber \\= & {} {\varvec{{{\mathcal {C}}}}}: \Delta {\pmb {E}}, \end{aligned}$$

(83)

where ${\varvec{{{\mathcal {C}}}}}$ is the fourth order elasticity tensor defined in the reference configuration $\Omega _0$, which is related to ${{\pmb { C}}}$ by

$$\begin{aligned} {{\mathcal {C}}}^{IJKL}=J_t (F_t^{-1})^{Ii}(F_t^{-1})^{Jj}(F_t^{-1})^{Kk}(F_t^{-1})^{Ll}{ C}^{ijkl}. \end{aligned}$$

(84)

Note that the indices in upper case, i.e. I, J,  etc. are for $\Omega _0$ and the indices in lower case, i.e. i, j,  etc., are for $\Omega _t$. Using Eqs. (76) and (83), we rewrite the first integrand of Eq. (80) as

$$\begin{aligned} \Delta {\pmb {S}}:\delta {\pmb {E}}= & {} \delta {\pmb {E}}: {\varvec{{{\mathcal {C}}}}}: \Delta {\pmb {E}} \nonumber \\= & {} {\pmb {F}}^T\cdot \delta {\varvec{\varepsilon }}\cdot {\pmb {F}}:({\varvec{{{\mathcal {C}}}}}: {\pmb {F}}^T\cdot \Delta {\varvec{\varepsilon }}\cdot {\pmb {F}} )\nonumber \\= & {} \delta {\varvec{\varepsilon }}:J{{\pmb {\textsf {C}}}}:\Delta {\varvec{\varepsilon }}, \end{aligned}$$

(85)

where we have used (see Chapter 10 of [51])

$$\begin{aligned} \Delta {\pmb {E}} ={\pmb {F}}^T\cdot \Delta {\varvec{\varepsilon }}\cdot {\pmb {F}}, \end{aligned}$$

(86)

with $\Delta {\varvec{\varepsilon }}$ given by Eq. (45) and ${{\pmb {\textsf {C}}}}$ as the fourth order elasticity tensor defined in $\Omega$, which is given by Eq. (43).

Next, let us simplify the second integrand in Eq. (80). It can be obtained from Eq. (76) that

$$\begin{aligned} \Delta (\delta {\pmb {E}})= & {} \frac{1}{2}({\varvec{\nabla }}_0 \Delta {\pmb {u}}^T{{\cdot }}{\varvec{\nabla }}_0\delta {\pmb {u}}+{\varvec{\nabla }}_0 \delta {\pmb {u}}^T{{\cdot }}{\varvec{\nabla }}_0 \Delta {\pmb {u}})\nonumber \\= & {} \frac{1}{2}{\pmb {F}}^T{{\cdot }}({\varvec{\nabla }} \Delta {\pmb {u}}^T{{\cdot }}{\varvec{\nabla }}\delta {\pmb {u}}+{\varvec{\nabla }} \delta {\pmb {u}}^T{{\cdot }}{\varvec{\nabla }} \Delta {\pmb {u}}){{\cdot }}{\pmb {F}}. \end{aligned}$$

(87)

Thus, noticing that ${\pmb {S}}={\pmb {F}}^{-1}{{\cdot }}{\varvec{{\tau }}}{{\cdot }}{\pmb {F}}^{-T}$, the second integrand of Eq. (80) is expressed as

$$\begin{aligned} {\pmb {S}}\varvec{:}\Delta (\delta {\pmb {E}})={\varvec{\nabla }}\delta {\pmb {u}}\varvec{:}{\varvec{{\tau }}}{{\cdot }}{\varvec{\nabla }}\Delta {\pmb {u}}^T. \end{aligned}$$

(88)

Therefore, Eq. (80) simplifies to the form given by Eq. (42).