Simulation of finite-strain inelastic phenomena governed by creep and plasticity

Abstract

Inelastic mechanical behavior plays an important role in many applications in science and engineering. Phenomenologically, this behavior is often modeled as plasticity or creep. Plasticity is used to represent the rate-independent component of inelastic deformation and creep is used to represent the rate-dependent component. In several applications, especially those at elevated temperatures and stresses, these processes occur simultaneously. In order to model these process, we develop a rate-objective, finite-deformation constitutive model for plasticity and creep. The plastic component of this model is based on rate-independent $J_2$ plasticity, and the creep component is based on a thermally activated Norton model. We describe the implementation of this model within a finite element formulation, and present a radial return mapping algorithm for it. This approach reduces the additional complexity of modeling plasticity and creep, over thermoelasticity, to just solving one nonlinear scalar equation at each quadrature point. We implement this algorithm within a multiphysics finite element code and evaluate the consistent tangent through automatic differentiation. We verify and validate the implementation, apply it to modeling the evolution of stresses in the flip chip manufacturing process, and test its parallel strong-scaling performance.

Appendix A

1.1 Objectivity of discrete Kirchhoff stress tensor ${{\varvec{\tau }}}$

The expression for the updated Kirchhoff stress is given by (94). Since the Kirchhoff stress is a measure of stress in the current configuration, in order ensure material frame indifference it must transform as,

$$\begin{aligned} {{\varvec{\tau }}}_{n+1}^* = {\mathbf {Q}}{{\varvec{\tau }}}_{n+1}{\mathbf {Q}}^T, \end{aligned}$$

(107)

when the incremental deformation gradient is superposed with a rotation ${\mathbf {Q}}$,

$$\begin{aligned} \mathbf {f}_{n+1}^* = {\mathbf {Q}}\mathbf {f}_{n+1}. \end{aligned}$$

(108)

In the development below we prove that this is indeed the case for our algorithm.

Proof

Since: $\bar{\mathbf{f}}_{n+1} = \left[ \det \mathbf {f}_{n+1}\right] ^{-\frac{1}{3}}\mathbf {f}_{n+1}$, so if $\mathbf {f}_{n+1}^* = {\mathbf {Q}}\mathbf {f}_{n+1}$, from (108) we have

$$\begin{aligned} \bar{\mathbf{f}}_{n+1}^*= & {} \left[ \det \mathbf {f}_{n+1}^*\right] ^{-\frac{1}{3}}\mathbf {f}_{n+1}^* \nonumber \\= & {} \left[ \det {\mathbf {Q}}\mathbf {f}_{n+1} \right] ^{-\frac{1}{3}} {\mathbf {Q}}\mathbf {f}_{n+1} \nonumber \\= & {} \left[ \det \mathbf {f}_{n+1} \right] ^{-\frac{1}{3}} {\mathbf {Q}}\mathbf {f}_{n+1} \nonumber \\= & {} {\mathbf {Q}} \left[ \det \mathbf {f}_{n+1} \right] ^{-\frac{1}{3}} \mathbf {f}_{n+1} \nonumber \\= & {} {\mathbf {Q}} \bar{\mathbf{f}}_{n+1}. \end{aligned}$$

(109)

Further, from (68) we have

$$\begin{aligned} \bar{\mathbf{b}}_{n+1}^\mathrm{e\;trial *}= & {} \bar{\mathbf{f}}_{n+1}^*\bar{\mathbf{b}}_{n}^\mathrm{e}\bar{\mathbf{f}}_{n+1}^{T*} \nonumber \\= & {} {\mathbf {Q}} \bar{\mathbf{f}}_{n+1}\bar{\mathbf{b}}_{n}^\mathrm{e}[{\mathbf {Q}} \bar{\mathbf{f}}_{n+1}]^T \nonumber \\= & {} {\mathbf {Q}} \bar{\mathbf{f}}_{n+1}\bar{\mathbf{b}}_{n}^\mathrm{e} \bar{\mathbf{f}}_{n+1}^T {\mathbf {Q}}^T \nonumber \\= & {} {\mathbf {Q}} \bar{\mathbf{b}}_{n+1}^\mathrm{e\;trial} {\mathbf {Q}}^T. \end{aligned}$$

(110)

From this we also conclude that

$$\begin{aligned} \mathrm{tr}\,[\bar{\mathbf{b}}_{n+1}^\mathrm{e\;trial*}]= & {} \hbox {tr}\,[{\mathbf {Q}}\bar{\mathbf{b}}_{n+1}^\mathrm{e\;trial*} {\mathbf {Q}}^T] \nonumber \\= & {} \mathrm{tr}\,[\bar{\mathbf{b}}_{n+1}^\mathrm{e\;trial}], \mathrm{since } \, \mathrm{tr}[\cdot ]\,\, \hbox {is invariant under rotation}. \nonumber \\ \end{aligned}$$

(111)

From (70) we have

$$\begin{aligned} \mathbf {s}_{n+1}^\mathrm{trial*}= & {} \mu \, \hbox {dev} \, [\bar{\mathbf{b}}_{n+1}^\mathrm{e\;trial *}] \nonumber \\= & {} \mu \, \big ( \bar{\mathbf{b}}_{n+1}^\mathrm{e\;trial *} - \frac{1}{3} \hbox {tr}\,[\bar{\mathbf{b}}_{n+1}^\mathrm{e\;trial *}]{\mathbf {Q}}\mathbf {1}{\mathbf {Q}}^\mathrm{T} \big ) \nonumber \\= & {} \mu \, {\mathbf {Q}} \big ( \bar{\mathbf{b}}_{n+1}^\mathrm{e\;trial} - \frac{1}{3} \hbox {tr}\,[\bar{\mathbf{b}}_{n+1}^\mathrm{e\;trial}]\mathbf {1}\big ){\mathbf {Q}}^\mathrm{T} \nonumber \\= & {} {\mathbf {Q}} \mathbf {s}_{n+1}^\mathrm{trial} {\mathbf {Q}}^T. \end{aligned}$$

(112)

Since the superposed deformation is a rigid body rotation, we have $J_{n+1}^* = J_{n+1}$, and from (71)

$$\begin{aligned} p_{n+1}^*= & {} \frac{\kappa }{2} (J_{n+1}^{*2} - 1) /J_{n+1}^{*} \nonumber \\= & {} \frac{\kappa }{2} (J_{n+1}^2 - 1) /J_{n+1} \nonumber \\= & {} p_{n+1}. \end{aligned}$$

(113)

Further, we have

$$\begin{aligned} \alpha _{n+1}^\mathrm{trial*} = \alpha _n = \alpha _{n+1}^\mathrm{trial}. \end{aligned}$$

(114)

Thus we conclude that the trial Kirchhoff stress ${{\varvec{\tau }}}$ transforms objectively, since from (94), we have

$$\begin{aligned} {{\varvec{\tau }}}_{n+1}^\mathrm{trial*}= & {} p_{n+1}^* J_{n+1}^*\mathbf {1}^* + \mathbf {s}_{n+1}^\mathrm{trial*} - \beta ( T ( \mathbf {X}, t_{n+1})-T_0 )\mathbf {1}^* J_{n+1}^* \nonumber \\= & {} p_{n+1} J_{n+1} {\mathbf {Q}} \mathbf {1} {\mathbf {Q}}^T + {\mathbf {Q}} \mathbf {s}_{n+1}^\mathrm{trial} {\mathbf {Q}}^T \nonumber \\&- \beta ( T ( \mathbf {X}, t_{n+1})-T_0 ){\mathbf {Q}} \mathbf {1} {\mathbf {Q}}^T J_{n+1} \nonumber \\= & {} {\mathbf {Q}} \big ( p_{n+1} J_{n+1}\mathbf {1} + \mathbf {s}_{n+1}^\mathrm{trial} \nonumber \\&- \beta ( T ( \mathbf {X}, t_{n+1})-T_0 )\mathbf {1} J_{n+1} \big ) {\mathbf {Q}}^T \nonumber \\= & {} {\mathbf {Q}} {{\varvec{\tau }}}_{n+1}^\mathrm{trial} {\mathbf {Q}}^T. \end{aligned}$$

(115)

From (83), the “direction” of the updated Kirchhoff stress is given by,

$$\begin{aligned} \mathbf {n}_{n+1}^\mathrm{trial*}= & {} \frac{\mathbf {s}_{n+1}^\mathrm{trial*}}{\Vert \mathbf {s}_{n+1}^\mathrm{trial*}\Vert } \nonumber \\= & {} \frac{{\mathbf {Q}} \mathbf {s}_{n+1}^\mathrm{trial} {\mathbf {Q}}^T}{\Vert \mathbf {s}_{n+1}^\mathrm{trial} \Vert } \nonumber \\= & {} {\mathbf {Q}} \mathbf {n}_{n+1}^\mathrm{trial} {\mathbf {Q}}^T\nonumber \\= & {} {\mathbf {Q}} \mathbf {n}_{n+1} {\mathbf {Q}}^T. \end{aligned}$$

(116)

Now we consider the increment of inelastic strain.

For pure creep, from (88), the increment of the creep strain transforms like

$$\begin{aligned} \varDelta {\gamma ^\mathrm{cr}}^*= & {} \varDelta t \gamma _{n+1}^\mathrm{cr*} \nonumber \nonumber \\= & {} \varDelta t B_{n+1} \left[ \Vert \mathbf {s}_{n+1}^\mathrm{trial*}\Vert - \frac{2}{3} \mu \varDelta {\gamma ^\mathrm{cr*}} \text{ tr }\left[ \bar{\mathbf{b}}_{n+1}^\mathrm{e\,trial *}\right] \right] ^{c_1} \nonumber \\= & {} \varDelta t B_{n+1} \left[ \Vert \mathbf {s}_{n+1}^\mathrm{trial}\Vert - \frac{2}{3} \mu \varDelta {\gamma ^\mathrm{cr*}} \text{ tr }\left[ \bar{\mathbf{b}}_{n+1}^\mathrm{e\,trial}\right] \right] ^{c_1}.\nonumber \\ \end{aligned}$$

(117)

We note the equation for $\varDelta \gamma ^\mathrm{cr*}$ above, is the same as the equation for $\varDelta \gamma ^\mathrm{cr}$ (88). Consequently, $\varDelta {\gamma }^\mathrm{cr*} = \varDelta {\gamma }^\mathrm{cr}$.

Next we consider plasticity and creep. From (90), we have

$$\begin{aligned} \bar{\mu }^* = \frac{1}{3}\mu \, \text{ tr }\left[ \bar{\mathbf{b}}_{n+1}^\mathrm{e\;trial *}\right] = \frac{1}{3}\mu \, \text{ tr }\left[ \bar{\mathbf{b}}_{n+1}^\mathrm{e\;trial}\right] . \end{aligned}$$

(118)

From (73), we have

$$\begin{aligned} {f}^\mathrm{trial*}_{n+1}= & {} {f}({{\varvec{\tau }}}^\mathrm{trial*}_{n+1}, \alpha _{n+1}^*) \nonumber \\= & {} \Vert \mathbf {s}^\mathrm{trial*}_{n+1}\Vert - \sqrt{\frac{2}{3}}(\sigma _{Y} +K \alpha _{n+1}^\mathrm{trial*}) \nonumber \\= & {} \Vert \mathbf {s}^\mathrm{trial}_{n+1}\Vert - \sqrt{\frac{2}{3}}(\sigma _{Y} +K \alpha _{n+1}^\mathrm{trial}) \nonumber \\= & {} {f}^\mathrm{trial}_{n+1}. \end{aligned}$$

(119)

Form (93), the equation for $\varDelta \gamma ^*$ is given by,

$$\begin{aligned}&f_{n+1}^\mathrm{trial*} - 2\bar{\mu }^*\left( 1+\frac{K}{3\bar{\mu }^*}\right) \varDelta \gamma ^* \nonumber \\&\quad = 2\bar{\mu }^*\varDelta t B \left( \Vert \mathbf {s}_{n+1}^\mathrm{trial*} \Vert - f_{n+1}^\mathrm{trial*} + \frac{2}{3}K\varDelta \gamma ^* \right) ^{c_1},\nonumber \\&\Rightarrow \,\,\, f_{n+1}^\mathrm{trial} - 2\bar{\mu }\left( 1+\frac{K}{3\bar{\mu }}\right) \varDelta \gamma ^* \nonumber \\&\quad = 2\bar{\mu }\varDelta t B \left( \Vert \mathbf {s}_{n+1}^\mathrm{trial} \Vert - f_{n+1}^\mathrm{trial} + \frac{2}{3}K\varDelta \gamma ^* \right) ^{c_1}. \end{aligned}$$

(120)

By comparing (120) with (93), we conclude that the equations for $\varDelta \gamma ^*$ and $\varDelta \gamma$ are the same, and therefore, $\varDelta \gamma ^* = \varDelta \gamma$.

From (83), for the deviatoric part of Kirchhoff stress, we have

$$\begin{aligned} \mathbf {s}_{n+1}^*= & {} \Vert \mathbf {s}_{n+1}^* \Vert \mathbf {n}_{n+1}^\mathrm{trial*} \nonumber \\= & {} \bigg [ \Vert \mathbf {s}_{n+1}^\mathrm{trial*}\Vert - \frac{2}{3} \mu \left( \varDelta \gamma ^* + \varDelta {\gamma ^\mathrm{cr*}} \right) \text{ tr }\left[ \bar{\mathbf{b}}_{n+1}^\mathrm{e\,trial*}\right] \bigg ] \mathbf {n}_{n+1}^\mathrm{trial*} \nonumber \\= & {} \bigg [ \Vert \mathbf {s}_{n+1}^\mathrm{trial}\Vert - \frac{2}{3} \mu \left( \varDelta \gamma + \varDelta {\gamma ^\mathrm{cr}} \right) \text{ tr }\left[ \bar{\mathbf{b}}_{n+1}^\mathrm{e\,trial}\right] \bigg ] {\mathbf {Q}} \mathbf {n}_{n+1}^\mathrm{trial} {\mathbf {Q}}^T \nonumber \\= & {} {\mathbf {Q}} \mathbf {s}_{n+1} {\mathbf {Q}}^T. \end{aligned}$$

(121)

In the equation above, we have made use of several results derived in this section.

Finally, form the definition of Kirchhoff stress (94) we have,

$$\begin{aligned} {{\varvec{\tau }}}_{n+1}^*= & {} p_{n+1}^* J_{n+1}^*\mathbf {1} + \mathbf {s}_{n+1}^* - \beta ( T^* ( \mathbf {X}, t_{n+1}) - T_0 ) \mathbf {1} J_{n+1}^* \nonumber \\= & {} p_{n+1} J_{n+1} \mathbf {1} + {\mathbf {Q}} \mathbf {s}_{n+1} {\mathbf {Q}}^T \nonumber \\&- \beta ( T ( \mathbf {X}, t_{n+1}) - T_0 ) \mathbf {1} J_{n+1} \nonumber \\= & {} {\mathbf {Q}} \bigg ( p_{n+1} J_{n+1}\mathbf {1} + \mathbf {s}_{n+1} \nonumber \\&- \beta ( T ( \mathbf {X}, t_{n+1}) - T_0 ) \mathbf {1} J_{n+1} \bigg ) {\mathbf {Q}}^T \nonumber \\= & {} {\mathbf {Q}} {{\varvec{\tau }}}_{n+1} {\mathbf {Q}}^T. \end{aligned}$$

(122)

This concludes our proof. $\square$