A phase-field model for fracture in piezoelectric ceramics

Abstract

A phase-field model is presented for modeling the fracture of piezoelectric ceramics. The implementation of several different crack face boundary conditions, including conducting, permeable, and insulating or impermeable, as well as energetically consistent is described. The approach to the latter involves a finite deformation framework for piezoelectricity. In addition, a new function that governs material degradation is proposed to eliminate the presence of high phase-field values in the vicinity of large electric fields. The new function is found to lead to improved brittle material behavior as well. Results are presented that demonstrate the capability of the model to capture complicated phenemona that arise in piezoelectric fracture, including crack retardation, acceleration, and turning.

Appendices

Appendix A

Here we demonstrate the transformations between energy functions. Consider a Helmholtz free energy for conducting crack-face boundary conditions of the form,

$$\begin{aligned}&\psi = {\bar{\psi }} (\underbrace{f{\varepsilon _{ij}}}_{{{\bar{\varepsilon }}_{ij}}},\underbrace{f{D_i}}_{{{\bar{D}}_i}}) + {\psi ^\mu }(\mu ,{\mu _{,i}})\\&{\sigma _{ij}} = f\frac{{\partial {\bar{\psi }} }}{{\partial (f{\varepsilon _{ij}})}}, \quad {E_i} = f\frac{{\partial {\bar{\psi }} }}{{\partial (f{D_i})}}, \nonumber \\&\eta =f'{\varepsilon _{ij}}\frac{{\partial {\bar{\psi }} }}{{\partial (f{\varepsilon _{ij}})}} + f'{D_i}\frac{{\partial {\bar{\psi }} }}{{\partial (f{D_i})}}+ \frac{{\partial {\psi ^\mu }}}{{\partial \mu }},\nonumber \\&{\xi _i} = \frac{{\partial {\psi ^\mu }}}{{\partial {\mu _{,i}}}}\nonumber \end{aligned}$$

(6.1)

Next, consider a full Legendre transformation on the electrical and mechanical terms to define the Gibbs free energy as,

$$\begin{aligned} g = \psi - {\sigma _{ij}}{\varepsilon _{ij}} - {E_i}{D_i} \end{aligned}$$

(6.2)

The variation of the Gibbs free energy is,

$$\begin{aligned} \delta g&= \delta \psi - {\sigma _{ij}}\delta {\varepsilon _{ij}} - {\varepsilon _{ij}}\delta {\sigma _{ij}} - {E_i}\delta {D_i} - {D_i}\delta {E_i}\nonumber \\&= \frac{{\partial {\bar{\psi }} }}{{\partial (f{\varepsilon _{ij}})}} \left( { f'{\varepsilon _{ij}}\delta \mu + {f\delta {\varepsilon _{ij}}} }\right) \nonumber \\&+\, \frac{{\partial {\bar{\psi }} }}{{\partial (f{D_i})}} \left( { f'{D_i}\delta \mu + {f\delta {D_i}} }\right) + \frac{{\partial {\psi ^\mu }}}{{\partial \mu }}\delta \mu \nonumber \\&+\,\frac{{\partial {\psi ^\mu }}}{{\partial {\mu _{,i}}}}\delta {\mu _{,i}}- {{\sigma _{ij}}\delta {\varepsilon _{ij}}}\nonumber \\&-\, {\varepsilon _{ij}}\delta {\sigma _{ij}} - {{E_i}\delta {D_i}} - {D_i}\delta {E_i}\nonumber \\&= - {\varepsilon _{ij}}\delta {\sigma _{ij}} - {D_i}\delta {E_i} + \frac{{\partial {\psi ^\mu }}}{{\partial {\mu _{,i}}}}\delta {\mu _{,i}} \nonumber \\&+\, \left[ {f'{\varepsilon _{ij}}\frac{{\partial {\bar{\psi }} }}{{\partial (f{\varepsilon _{ij}})}} + f'{D_i}\frac{{\partial {\bar{\psi }} }}{{\partial (f{D_i})}} + \frac{{\partial {\psi ^\mu }}}{{\partial \mu }}} \right] \delta \mu \nonumber \\&= - {\varepsilon _{ij}}\delta {\sigma _{ij}} - {D_i}\delta {E_i} + \frac{{\partial {\psi ^\mu }}}{{\partial {\mu _{,i}}}}\delta {\mu _{,i}} \nonumber \\&+ \left[ {\frac{{f'}}{f}{\varepsilon _{ij}}{\sigma _{ij}} + \frac{{f'}}{f}{D_i}{E_i} + \frac{{\partial {\psi ^\mu }}}{{\partial \mu }}} \right] \delta \mu \end{aligned}$$

(6.3)

Next, assume the Gibbs free energy can be written in the form,

$$\begin{aligned} g = \bar{g}(\underbrace{{\sigma _{ij}}/f}_{{{\bar{\sigma }}_{ij}}},\underbrace{{E_i}/f}_{{{\bar{E}}_i}}) + {g^\mu }(\mu ,{\mu _{,i}}) \end{aligned}$$

(6.4)

Now, the variation of the Gibbs free energy is written as,

$$\begin{aligned} \delta g&= \frac{{\partial \bar{g}}}{{\partial ({\sigma _{ij}}/f)}}(\delta {\sigma _{ij}}/f - \delta \mu f'{\sigma _{ij}}/{f^2}) \nonumber \\&+\, \frac{{\partial \bar{g}}}{{\partial ({E_i}/f)}}(\delta {E_i}/f - \delta \mu f'{E_i}/{f^2}) \nonumber \\&+\, \frac{{\partial {g^\mu }}}{{\partial \mu }}\delta \mu + \frac{{\partial {g^\mu }}}{{\partial {\mu _{,i}}}}\delta {\mu _{,i}} \nonumber \\&= \frac{1}{f}\frac{{\partial \bar{g}}}{{\partial ({\sigma _{ij}}/f)}}\delta {\sigma _{ij}} + \frac{1}{f}\frac{{\partial \bar{g}}}{{\partial ({E_i}/f)}}\delta {E_i} + \frac{{\partial {g^\mu }}}{{\partial {\mu _{,i}}}}\delta {\mu _{,i}} \nonumber \\&-\, \left[ {\frac{{f'}}{{{f^2}}}\frac{{\partial \bar{g}}}{{\partial ({\sigma _{ij}}/f)}}{\sigma _{ij}} + \frac{{f'}}{{{f^2}}}\frac{{\partial \bar{g}}}{{\partial ({E_i}/f)}}{E_i} - \frac{{\partial {g^\mu }}}{{\partial \mu }}} \right] \delta \mu \nonumber \\ \end{aligned}$$

(6.5)

However, enforcing the equality of the two forms of $\delta g$ for arbitrary variations of the stress and electric field yields,

$$\begin{aligned} \delta {\sigma _{ij}},\delta {E_i}{\text { arbitrary}} \rightarrow {\varepsilon _{ij}}&= - \frac{1}{f}\frac{{\partial \bar{g}}}{{\partial ({\sigma _{ij}}/f)}} \nonumber \\ {D_i}&= - \frac{1}{f}\frac{{\partial \bar{g}}}{{\partial ({E_i}/f)}} \end{aligned}$$

(6.6)

This allows us to write a reduced form for $\delta g$ as,

$$\begin{aligned} \delta g&= { - {\varepsilon _{ij}}\delta {\sigma _{ij}}} { - {D_i}\delta {E_i}} \nonumber \\&+ \left[ { \frac{{f'}}{f}{\varepsilon _{ij}}{\sigma _{ij}} + {\frac{{f'}}{f}{D_i}{E_i}} + \frac{{\partial {\psi ^\mu }}}{{\partial \mu }}} \right] \delta \mu \nonumber \\&+\, \frac{{\partial {\psi ^\mu }}}{{\partial {\mu _{,i}}}}\delta {\mu _{,i}}\nonumber \\&= \frac{{\partial {g^\mu }}}{{\partial \mu }} \delta \mu + \frac{{\partial {g^\mu }}}{{\partial {\mu _{,i}}}}\delta {\mu _{,i}} \end{aligned}$$

(6.7)

Finally, this reduced equality must hold for arbitrary variations of $\delta \mu$ and $\delta \mu _{,i}$ yielding,

$$\begin{aligned} \delta \mu ,\delta {\mu _{,i}}{\text { arbitrary}} \rightarrow \frac{{\partial {\psi ^\mu }}}{{\partial \mu }} = \frac{{\partial {g^\mu }}}{{\partial \mu }},\quad \frac{{\partial {\psi ^\mu }}}{{\partial {\mu _{,i}}}} = \frac{{\partial {g^\mu }}}{{\partial {\mu _{,i}}}}\nonumber \\ \end{aligned}$$

(6.8)

The final set of relationships can be collected in the following forms.

$$\begin{aligned} \frac{{\partial {\psi ^\mu }}}{{\partial \mu }}&= \frac{{\partial {g^\mu }}}{{\partial \mu }},\quad \frac{{\partial {\psi ^\mu }}}{{\partial {\mu _{,i}}}} = \frac{{\partial {g^\mu }}}{{\partial {\mu _{,i}}}} \nonumber \\ {\sigma _{ij}}/f&= \frac{{\partial {\bar{\psi }} }}{{\partial (f{\varepsilon _{ij}})}} = {\bar{\sigma }_{ij}}({\bar{\varepsilon }_{ij}},{\bar{D}_i})\nonumber \\ {E_i}/f&= \frac{{\partial {\bar{\psi }} }}{{\partial (f{D_i})}} = {\bar{E}_i}({\bar{\varepsilon }_{ij}},{\bar{D}_i}) \nonumber \\ f{\varepsilon _{ij}}&= - \frac{{\partial \bar{g}}}{{\partial ({\sigma _{ij}}/f)}} = {\bar{\varepsilon }_{ij}}({\bar{\sigma }_{ij}},{\bar{E}_i})\nonumber \\ f{D_i}&= - \frac{{\partial \bar{g}}}{{\partial ({E_i}/f)}} = {\bar{D}_{ij}}({\bar{\sigma }_{ij}},{\bar{E}_i}) \end{aligned}$$

(6.9)

The first set of equations implies that, without any effect on the derived micro-force quantities, we can take $\psi ^{\mu }=g^{\mu }$. The next sets of equations imply that the ${\bar{\varepsilon }_{ij}}({\bar{\sigma }_{ij}},{\bar{E}_i})$ and ${\bar{D}_{ij}}({\bar{\sigma }_{ij}},{\bar{E}_i})$ constitutive relationships are the inverse of the ${\bar{\sigma }_{ij}}({\bar{\varepsilon }_{ij}},{\bar{D}_i})$ and ${\bar{E}_i}({\bar{\varepsilon }_{ij}},{\bar{D}_i})$ form. For a general nonlinear constitutive relationship, such an inverse may not be possible to obtain. However, for linear piezoelectric materials the inversion is well-known. Hence, the Helmholtz and Gibbs from energies for a material with conducting cracks are written as,

$$\begin{aligned} {\bar{\psi }}&= {\textstyle {1 \over 2}}{f^2}c_{ijkl}^D{\varepsilon _{ij}}{\varepsilon _{kl}} - {f^2}{h_{kij}}{D_k}{\varepsilon _{ij}} + {\textstyle {1 \over 2}}{f^2}\beta _{ij}^\varepsilon {D_i}{D_j} \nonumber \\ \bar{g}&= - {\textstyle {1 \over 2}}s_{ijkl}^E{\sigma _{ij}}{\sigma _{kl}}/{f^2} - {d_{kij}}{E_k}{\sigma _{ij}}/{f^2} - {\textstyle {1 \over 2}}\kappa _{ij}^\sigma {E_i}{E_j}/{f^2}\nonumber \\ \end{aligned}$$

(6.10)

The electrical enthalpy, is then just a partial Legendre transformation on the electrical terms, and a partial inversion of the constitutive relationships, and is written as,

$$\begin{aligned} \bar{h}&= \bar{h}(f{\varepsilon _{ij}},{E_i}/f) \nonumber \\&= {\textstyle {1 \over 2}}{f^2}c_{ijkl}^E{\varepsilon _{ij}}{\varepsilon _{kl}} - {e_{kij}}{E_k}{\varepsilon _{ij}} - {\textstyle {1 \over 2}}\kappa _{ij}^\varepsilon {E_i}{E_j}/{f^2}\nonumber \\ \end{aligned}$$

(6.11)

Note that the relationships between the various forms of the elastic, piezoelectric, and dielectric coefficients are well-known, and once one set has been determined the others can be computed via simple linear algebraic manipulations.

Appendix B

Here we compare our approach for the construction of the phase-field electrical enthalpy for permeable crack face boundary conditions to another appearing in the literature. To greatly simplify the demonstration we resort to a simple one-dimensional model with homogeneous fields. For this case, the electrical enthalpy, $h$, of a linear piezoelectric material is given as,

$$\begin{aligned} h = {\textstyle {1 \over 2}}{c^E}{\varepsilon ^2} - e\varepsilon E - {\textstyle {1 \over 2}}{\kappa ^\varepsilon }{E^2} \end{aligned}$$

(7.1)

Here $c^E$ is the elastic stiffness at constant electric field, $E$, $e$ is the piezoelectric coefficient, and $\kappa ^{\varepsilon }$ is the dielectric permittivity at constant strain, $\varepsilon$. For the construction of the phase-field free energy we will use the degradation function $f(\mu )=\mu$, which leads to the phase-field free energy,

$$\begin{aligned} h = {\textstyle {1 \over 2}}{\mu ^2}{c^E}{\varepsilon ^2} - \mu e\varepsilon E - {\textstyle {1 \over 2}}{\kappa ^\varepsilon }{E^2} + \frac{{{G_c}}}{{4{l_0}}}{(\mu - 1)^2}\nonumber \\ \end{aligned}$$

(7.2)

This is in contrast to that used by Abdollahi and Arias (2012) (AA) who proposed to degrade all energy terms including strain in with the same functional form,

$$\begin{aligned} {h^{AA}} = {\textstyle {1 \over 2}}{\mu ^2}{c^E}{\varepsilon ^2} - {\mu ^2}e\varepsilon E - {\textstyle {1 \over 2}}{\kappa ^\varepsilon }{E^2} + \frac{{{G_c}}}{{4{l_0}}}{(\mu - 1)^2}\nonumber \\ \end{aligned}$$

(7.3)

The solutions to the phase-field order parameter for each of these cases is,

$$\begin{aligned} \begin{aligned}&\frac{{\partial h}}{{\partial \mu }} = \mu {c^E}{\varepsilon ^2} - e\varepsilon E + \frac{{{G_c}}}{{2{l_0}}}(\mu - 1) = 0 \\&\quad \rightarrow \,\mu = \frac{{{{{G_c}} / {2{l_0}}} + e\varepsilon E}}{{{{{G_c}} / {2{l_0} + {c^E}{\varepsilon ^2}}}}} \end{aligned} \end{aligned}$$

(7.4)

and

$$\begin{aligned} \begin{aligned}&\frac{{\partial {h^{AA}}}}{{\partial \mu }} = \mu {c^E}{\varepsilon ^2} - \mu e\varepsilon E + \frac{{{G_c}}}{{2{l_0}}}(\mu - 1) = 0\\&\quad \rightarrow \,{\mu ^{AA}} = \frac{{{{{G_c}} /{2{l_0}}}}}{{{{{G_c}}/{2{l_0} + {c^E}{\varepsilon ^2} - e\varepsilon E}}}} \end{aligned} \end{aligned}$$

(7.5)

Within the regions identified as cracks the strain becomes singular and the limiting forms for the order parameters become,

$$\begin{aligned} \mu \approx \frac{{eE}}{{{c^E}\varepsilon }} \quad {\text {and}} \quad {\mu ^{AA}} \approx \frac{1}{{{c^E}{\varepsilon ^2}}} \quad \text {for large } \varepsilon \end{aligned}$$

(7.6)

These results can now be applied within the stress and electric displacement relationships as,

$$\begin{aligned} \left. \begin{aligned}&\sigma = \frac{{\partial h}}{{\partial \varepsilon }} = {\mu ^2}{c^E}\varepsilon - \mu eE = 0\\&D = - \frac{{\partial h}}{{\partial \varepsilon }} = \mu e\varepsilon + {\kappa ^\varepsilon }E = \left( {{\kappa ^\varepsilon } + \frac{{{e^2}}}{{{c^E}}}} \right) E \end{aligned} \right\} {\text { as }}\varepsilon \rightarrow \infty \nonumber \\ \end{aligned}$$

(7.7)

Notice that the electric displacement is now related to the electric field by the dielectric permittivity at constant stress $\sigma$, ${\kappa ^\sigma } = {\kappa ^\varepsilon } + {e^2}/{c^E}$, which is exactly the behavior that should be recovered near a permeable crack. In contrast, the stress and electric displacement from the enthalpy form used by AA are,

$$\begin{aligned} \left. \begin{aligned}&\sigma = \frac{{\partial h}}{{\partial \varepsilon }} = {\mu ^2}{c^E}\varepsilon - {\mu ^2}eE = 0\\&D = - \frac{{\partial h}}{{\partial \varepsilon }} = {\mu ^2}e\varepsilon + {\kappa ^\varepsilon }E = {\kappa ^\varepsilon }E \end{aligned} \right\} {\text { as }}\varepsilon \rightarrow \infty \nonumber \\ \end{aligned}$$

(7.8)

which does not recover the desired behavior near permeable cracks. This may be viewed as a small correction, but the question persists for nonlinear material models. Our approach offers a concrete procedure for transforming any electromechanical energy functional into one suitable for applications within the phase-field fracture framework.

Appendix C

Here we present the finite element calculation for the potential drop across two electrodes for two-dimensional geometries using the vector potential formulation for conducting cracks. The unit vector s is chosen such that $\mathbf{{n}} \times \mathbf{{s}} = {\mathbf{{e}}_z}$, where ${\mathbf{{e}}_z}$ is the unit vector normal to the two-dimensional plane containing the structure. The electrical virtual work statement can be written as,

$$\begin{aligned} \sum \limits _I {\delta {\varphi ^I}} {F^I}&= \int \limits _s {\delta \omega \phi \;} ds\\&= \int \limits _{{s_\omega }} {\frac{{\partial \delta \varphi }}{{\partial s}}\phi \;} ds + \int \limits _{{s_\phi }} {\frac{{\partial \delta \varphi }}{{\partial s}}\phi \;} ds\\&= \int \limits _{{s_\omega }} {\frac{\partial }{{\partial s}}(\delta \varphi \phi ) - \delta \varphi \frac{{\partial \phi }}{{\partial s}}\;} ds + \int \limits _{{s_\phi }} {\frac{{\partial \delta \varphi }}{{\partial s}}\phi \;} ds\\&= \delta {\varphi ^A}{\phi ^{AB}} - \delta {\varphi ^C}{\phi ^{CD}} - \int \limits _C^A {\delta \varphi \frac{{\partial \phi }}{{\partial s}}\;} ds\\&+\, (\delta {\varphi ^B} - \delta {\varphi ^A}){\phi ^{AB}} + \delta {\varphi ^D}{\phi ^{CD}}\\&- \,\delta {\varphi ^B}{\phi ^{AB}}- \int \limits _B^D {\delta \varphi \frac{{\partial \phi }}{{\partial s}}\;} ds \\&\quad + (\delta {\varphi ^C} - \delta {\varphi ^D}){\phi ^{CD}}\\&= - \int \limits _C^A {\delta \varphi \frac{{\partial \phi }}{{\partial s}}\;} ds - \int \limits _B^D {\delta \varphi \frac{{\partial \phi }}{{\partial s}}\;} ds \end{aligned}$$

Here, $\delta \varphi ^I$ represent arbitrary variations of the nodal vector potential at the $I^{\text {th}}$ node, and $F^I$ is the associated work-conjugate nodal electrical “force” quantity. This formula reveals two features of the nodal electrical forces for the vector potential formulation. First, if we consider any node $I$ on an electrode that is not on any termination point of that electrode, and taking $\delta \varphi ^I = 1$ and all other $\delta {\varphi ^{J \ne I}} = 0$, the formula indicates that $F^I=0$. So, on a bounding surface, the force-free condition corresponds to a constant electric potential, i.e. an electroded surface. Then, if we take $\delta \varphi = 1$ on the surface between points $A$ and $C$ and zero elsewhere, the formula shows that ${\phi ^{CD}} - {\phi ^{AB}} = \sum \nolimits _{A \rightarrow C} {{F^I}}$, where the summation is on the nodes that lie between points $A$ and $C$ on the surface. The same formula can be obtained by taking $\delta \varphi = 1$ on the surface between points $B$ and $D$ and zero elsewhere, in which case the right-hand-side changes to summing the forces between nodes $B$ and $D$. In either case, this formula demonstrates that the potential drop between two electrodes can be computed by summing the forces along the boundary connecting two adjacent termination points. Finally the total charge residing between any two points, $A$ and $B$, on an electrode can also be computed in a straightforward manner using,

$$\begin{aligned} \int \limits _s {\omega \;} ds = - \int \limits _s {{D_i}{n_i}\;} ds = \int \limits _A^B {\frac{{\partial \varphi }}{{\partial s}}\,} ds = {\varphi ^B} - {\varphi ^A}.\nonumber \\ \end{aligned}$$

(8.1)