4. Presentation of the behavior model#

4.1. Kinematic aspect#

This model assumes, just as in small deformations, the existence of a relaxed configuration \({\Omega }^{r}\), that is to say locally free of stress, which then makes it possible to decompose the total deformation into elastic and plastic parts, this decomposition being multiplicative.

Subsequently, we will note by \(F\) the gradient tensor which makes it go from the initial configuration \({O}_{0}\) to the current configuration \(\Omega (t)\), by \({F}^{p}\) the gradient tensor which makes it go from the configuration \({O}_{0}\) to the relaxed configuration \({\Omega }^{r}\), and \({F}^{e}\) from the configuration \({O}^{r}\) to \(\Omega (t)\). The index \(p\) refers to the plastic part, the index \(e\) to the elastic part.

_images/10000FCA00001CD6000013A398ECE64086423371.svg

Figure 4.1-a: Breakdown of the gradient tensor \(F\) into an elastic part \({F}^{e}\) and plastic \({F}^{p}\)

By composition of the movements, the following multiplicative decomposition is obtained:

\(F={F}^{e}{F}^{p}\)

The elastic deformations are measured in the current configuration with the left Cauchy-Green Eulerian tensor \({b}^{e}\) and the plastic deformations in the initial configuration by the \({G}^{p}\) tensor (Lagrangian description). These two tensors are defined by:

\({b}^{e}={F}^{e}{F}^{\text{eT}}\), \({G}^{p}=({F}^{\text{pT}}{F}^{p}{)}^{-1}\) from where \({b}^{e}={\text{FG}}^{p}{F}^{T}\)

The model presented is written in such a way as to distinguish isochoric terms from volume change terms. For this purpose, the following two tensors are introduced:

\(\stackrel{ˉ}{F}={J}^{-1/3}F\) and \({\stackrel{ˉ}{b}}^{e}={J}^{-2/3}{b}^{e}\) with \(J=\text{det}F\)

By definition, we have: \(\text{det}\stackrel{ˉ}{F}=1\) and \(\text{det}\stackrel{ˉ}{{b}^{e}}=1\).

4.2. Behavioral relationships#

The law presented is a thermoelasto (visco) plastic model with isotropic work hardening which, under the hypothesis of small deformations, tends towards the [R5.03.02] model with Von Mises criterion (this is the plastic model). The plastic deformations take place at a constant volume so that:

\({J}^{p}=\text{det}{F}^{p}=1\) from where \(J={J}^{e}=\text{det}{F}^{e}\)

Behavioral relationships are given by:

  • the definition of hyperelastic free energy:

\(\psi ={\psi }_{\mathrm{ther}}(T,J)+{\psi }_{\mathrm{elas}}(J,{\stackrel{ˉ}{b}}^{e})+K(\alpha )\),

with in particular \({\psi }_{\mathrm{elas}}(J,{\stackrel{ˉ}{b}}^{e})=\frac{1}{2}\frac{E}{3(1-2\nu )}\left[\frac{1}{2}({J}^{2}-1)-\mathrm{ln}J\right]+\frac{\mu }{2}(\mathrm{tr}{\stackrel{ˉ}{b}}^{e}-3)\)

and \({\psi }_{\mathrm{ther}}(T,J)=-3\alpha \Delta T(J-\frac{1}{J})\)

  • stress-thermoelastic deformation relationship:

\(\tilde{\tau }=\mu {\stackrel{ˉ}{\tilde{b}}}^{e}\)

\(\text{tr}\tau =\frac{\mathrm{3K}}{2}({J}^{2}-1)-\frac{\mathrm{9K}}{2}\alpha (T-{T}_{\text{ref}})(J+\frac{1}{J})\)

  • plasticity threshold (it is assumed that it is expressed with Kirchhoff constraints):

\(f={\tau }_{\text{eq}}-R(p)-{\sigma }_{y}\)

where \(R\) is the isotropic work hardening variable, a function of the cumulative plastic deformation*p*.

  • flow laws:

\(\stackrel{ˉ}{F}{\dot{G}}^{p}{\stackrel{ˉ}{F}}^{T}=-\dot{\lambda }\frac{3}{{\tau }_{\text{eq}}}\tilde{\tau }{\stackrel{ˉ}{b}}^{e}=-3\dot{\lambda }(\frac{1}{3}\text{tr}\stackrel{ˉ}{{b}^{e}}+\frac{\tilde{\tau }}{\mu })\frac{\tilde{\tau }}{{\tau }_{\text{eq}}}\)

\(\dot{p}=\dot{\lambda }\)

For the plasticity model, the plastic multiplier is obtained by writing the coherence condition \(\dot{f}=0\) and we have:

\(\dot{p}\ge \mathrm{0,}f\le 0\text{et}\dot{p}f=0\)

In the viscous case, we take \(\dot{p}\) equal to:

\(\dot{p}={\dot{\varepsilon }}_{0}{\left[\text{sh}(\frac{\langle f\rangle }{{\sigma }_{0}})\right]}^{m}\)

where \({\dot{\varepsilon }}_{0}\), \({\sigma }_{0}\), and \(m\) are the viscosity coefficients. Note that this law is reduced to a Norton law when the 2 material parameters \({\dot{\varepsilon }}_{0}\) and \({\sigma }_{0}\) are very large.

It is recalled that:

\({\stackrel{ˉ}{b}}^{e}={J}^{-2/3}{b}^{e}\)

\(\stackrel{ˉ}{F}={J}^{-1/3}F\)

and that the partition of the deformations is written as:

\({\stackrel{ˉ}{b}}^{e}=\stackrel{ˉ}{F}{G}^{p}{\stackrel{ˉ}{F}}^{T}\)

For metallic materials where the ratio \({\tau }_{\text{eq}}/\mu\) is small compared to 1, the expression of the flow law can be approximated by:

\(\stackrel{ˉ}{F}{\dot{G}}^{p}{\stackrel{ˉ}{F}}^{T}=-\dot{\lambda }\text{tr}\stackrel{ˉ}{{b}^{e}}\frac{\tilde{\tau }}{{\tau }_{\text{eq}}}+O(\frac{{\tau }_{\text{eq}}}{\mu })\) eq.4.2-1

where \(O(\frac{{\tau }_{\text{eq}}}{\mu })\) is negligible against the first term.

It is this last expression that is implemented in the*Code_Aster*.

Note:

If the deformations are small, we have:

\(\begin{array}{}J\simeq 1+\text{tr}\varepsilon \\ {b}^{e}\simeq \text{Id}+2{\varepsilon }^{e}\\ {G}^{p}\simeq \text{Id}-2{\varepsilon }^{p}\end{array}\)

where \(\varepsilon\) is the total deformation, \({\varepsilon }^{e}\) the elastic deformation, and \({\varepsilon }^{p}\) the plastic deformation in small deformations.

By replacing these three expressions in the equations of the law of behavior presented here, we find the classical thermo-elasto-plastic model with isotropic work hardening and Von Mises criterion.

4.3. Correction of elastic energy in the presence of thermal#

Expressing hyperelastic energy \({\psi }_{\mathrm{elas}}\) poses some difficulties. In fact, it depends on \(J\); in the presence of thermal deformation, \(J\) includes a thermal component that disrupts the expression and that should be corrected.

The approach for correcting Simo-Miehe energy in the presence of thermal is as follows:

  1. cancellation of pure thermal energy when the stress is zero;

  2. resolution of the equation in \(J\) obtained, for which \({J}_{0}\) is called the solution.

  3. final calculation of Simo-Miehe elastic energy in the presence of thermal:

\({\psi }_{\mathrm{elas}}^{\mathrm{corrigee}}={\psi }_{\mathrm{elas}}(J,{\stackrel{ˉ}{b}}^{e})+{\psi }_{\mathrm{ther}}(T,J)+{\psi }_{\mathrm{elas}}(J,{\stackrel{ˉ}{b}}^{e})-{\psi }_{\mathrm{elas}}({J}_{\mathrm{0,}}{\stackrel{ˉ}{b}}^{e}={I}_{d})-{\psi }_{\mathrm{ther}}(T,{J}_{0})\)

The cancellation of the stress in pure thermal leads to the equation:

\(\text{tr}\tau =0\iff ({J}^{2}-1)-3\alpha (T-{T}_{\text{ref}})(J+\frac{1}{J})=0\Leftarrow \{\begin{array}{}{J}^{3}-J-3\alpha \Delta T({J}^{2}-1)\\ J\ne 0\end{array}\)

This equation, under hypothesis \(3\alpha \Delta T\ll 1\), has solutions close to -1, 0, and 1. Only the largest is physically acceptable. We therefore pose:

\({J}_{0}=\mathrm{MAX}\{J\mathrm{tq}{J}^{3}-J-3\alpha \Delta T({J}^{2}-1)=0\}\)

The corrected energy in the presence of thermal is finally written:

\({\psi }_{\mathrm{elas}}^{\mathrm{corrigee}}={\psi }_{\mathrm{elas}}(J,{\stackrel{ˉ}{b}}^{e})+{\psi }_{\mathrm{ther}}(T,J)+{\psi }_{\mathrm{elas}}(J,{\stackrel{ˉ}{b}}^{e})-{\psi }_{\mathrm{elas}}({J}_{\mathrm{0,}}{\stackrel{ˉ}{b}}^{e}={I}_{d})-{\psi }_{\mathrm{ther}}(T,{J}_{0})\)

4.4. Choice of the work hardening function#

This behavior relationship is available in operator STAT_NON_LINE, under the keyword factor COMPORTEMENT and the argument “SIMO_MIEHE” of the keyword factor DEFORMATION. For the work hardening function, linear work hardening can be chosen or a traction curve can be provided. Five relationships can be used.

RELATION =/'VMIS_ISOT_TRAC'

/”VMIS_ISOT_PUIS” /”VMIS_ISOT_LINE” /”VISC_ISOT_TRAC” /”VISC_ISOT_LINE”

For purely thermoelastic behavior, the user chooses the argument 'VMIS_ISOT_LINE' for example, with SY very large (the behavior is then hyperelastic); for isotropic work hardening given by a traction curve, the user chooses the argument 'VMIS_ISOT_TRAC' in the plastic case or 'VISC_ISOT_TRAC' in the plastic case or '' in the viscous case and for linear isotropic work hardening, the argument 'VMIS_ISOT_LINE' in the plastic case or ' VISC_ISOT_LINE 'in the viscous case. For elastoplastic behavior whose work hardening curve (rational traction curve) is given by a power law, of the form


:math:`R(p)={\sigma }_{y}+{\sigma }_{y}{(\frac{E}{a{\sigma }_{y}}p)}^{\frac{1}{n}}`, the user chooses the argument 'VMIS_ISOT_PUIS'.

The various characteristics of the material are entered in the operator DEFI_MATERIAU ([U4.23.01]) under the keywords:

  • ELAS regardless of the law (we give the Young’s modulus, the Poisson’s ratio and possibly the thermal expansion coefficient),

  • TRACTION for “VMIS_ISOT_TRAC” and “VISC_ISOT_TRAC” (we give the rational traction curve),

  • ECRO_PUIS for “VMIS_ISOT_PUIS” (we give the parameters of the power law),

  • ECRO_LINE for “VMIS_ISOT_LINE” and “VISC_ISOT_LINE” (we give the elastic limit and the work-hardening slope),

  • VISC_SINH for “VISC_ISOT_TRAC” and “VISC_ISOT_LINE” (the three viscosity coefficients are given).

Note:

The user must ensure that the « experimental » tensile curve used, either directly or to derive/to establish the work-hardening slope, is given in the rational stress plane \(\sigma =F/S\) - logarithmic deformation \(\text{ln}(1+\Delta l/{l}_{0})\) where where where \({l}_{0}\) is the initial length of the useful part of the specimen, \(\Delta l\) the length variation after deformation, \(F\) the force applied and \(S\) the current area. Note that \(\sigma =F/S=\frac{F}{{S}_{0}}\frac{{l}_{0}}{l}\frac{1}{J}\) whereuse \(\tau =J\sigma =\frac{F}{{S}_{0}}\frac{{l}_{0}}{l}\). In general, it is the quantity \(\frac{F}{{S}_{0}}\frac{{l}_{0}}{l}\) that is measured by the experimenters and this directly gives the Kirchhoff constraint used in the Simo and Miehe model.

4.5. Constraints and internal variables#

The constraints are Cauchy \(\) constraints, calculated therefore on the current configuration (six components in 3D, four in 2D).

The internal variables produced in the*Code_Aster* are:

  • \(\mathit{V1}\), the cumulative plastic deformation \(p\),

  • \(\mathit{V2}\) the plasticity indicator (0 if the last calculated increment is elastic, 1 otherwise).

  • \(\mathit{V3}\) to \(\mathit{V8}\): the opposite of elastic deformation \({\stackrel{ˉ}{b}}^{e}\)

Note:

If the user wants to possibly recover deformations by post-processing his calculation, he must plot the Green-Lagrange \(E\) deformations, which represents a measure of deformations into large deformations (options EPSG_ELGA or EPSG_ELNO ). Classical linearized \(\varepsilon\) deformations measure deformations under the assumption of small deformations and do not make sense in large deformations.

4.6. Field of use#

The choice of kinematics DEFORMATION: “PETIT_REAC” also makes it possible to treat a law of thermo-elastoplastic behavior with isotropic work hardening and Von Mises criterion in large deformations. The law is written in small deformations and large deformations are taken into account by updating the geometry.

Between the law presented here (SIMO_MIEHE) and PETIT_REAC,

  • there is no difference if the deformations are small

  • there is no difference if the deformations are large but the rotations are small

  • there are differences if the rotations are important.

In particular, the solution obtained with kinematics PETIT_REAC may differ significantly from the exact solution in the presence of large rotations, regardless of the size of the time steps chosen by the user, unlike kinematics SIMO_MIEHE.

4.7. Integrating the law of behavior#

In the case of incremental behavior, keyword factor COMPORTEMENT, knowing the stress \({\sigma }^{-}\), knowing the stress, the cumulative plastic deformation \({p}^{-}\), the trace divided by three of the elastic deformation tensor \(\frac{1}{3}\text{tr}{\stackrel{ˉ}{b}}^{e-}\), the displacements \({u}^{-}\) and \(\Delta u\) and the temperatures and and the temperatures \({T}^{-}\) and \(T\), we want to determine \((\sigma ,p,\frac{1}{3}\text{tr}{\stackrel{ˉ}{b}}^{e})\).

Since the displacements are known, the gradients of the transformation from \({\Omega }_{0}\) to \({\Omega }^{-}\), noted \({F}^{-}\), and from \({\Omega }^{-}\) to \(\Omega\), noted \(\Delta F\), are known.

The implicit discretization of the law gives:

\(F=\Delta F{F}^{-}\)

\(J=\text{det}F\)

\(\stackrel{ˉ}{F}={J}^{-1/3}F\)

\({\stackrel{ˉ}{b}}^{e}=\stackrel{ˉ}{F}{G}^{p}{\stackrel{ˉ}{F}}^{T}\)

\(J\sigma =\tau\)

\(\tilde{\tau }=\mu {\stackrel{ˉ}{\tilde{b}}}^{e}\)

\(\frac{1}{3}\text{tr}\tau =\frac{1}{2}K({J}^{2}-1)-\frac{3}{2}K\alpha (T-{T}_{\text{ref}})(J+\frac{1}{J})\)

\(f={\tau }_{\text{eq}}-R({p}^{-}+\Delta p)-{\sigma }_{y}\)

\(\stackrel{ˉ}{F}({G}^{p}-{G}^{p-}){\stackrel{ˉ}{F}}^{T}=-\text{tr}{\stackrel{ˉ}{b}}^{e}\frac{\tilde{\tau }}{{\tau }_{\text{eq}}}\Delta p\) from where \({\stackrel{ˉ}{b}}^{e}=\stackrel{ˉ}{F}{G}^{p-}{\stackrel{ˉ}{F}}^{T}-\text{tr}{\stackrel{ˉ}{b}}^{e}\frac{\tilde{\tau }}{{\tau }_{\text{eq}}}\Delta p\)

In the plastic case: \(\Delta p\ge \mathrm{0,}f\le 0\text{et}f\Delta p=0\)

In the viscous case: \(\langle {\tau }_{\text{eq}}-R({p}^{-}+\Delta p)-{\sigma }_{y}\rangle -{\sigma }_{0}{\text{sh}}^{-1}\left[{(\frac{\Delta p}{{\dot{\varepsilon }}_{0}\Delta t})}^{\frac{1}{m}}\right]=0\)

Note:

This formulation is incrementally objective because the only incremental tensor quantity that is involved in discretization is \({\dot{G}}^{p}\). As \({G}^{p}\) and \({G}^{p-}\) are measured on the same configuration, i.e. the initial configuration, the discretization of \({\dot{G}}^{p}\) , i.e. \(\Delta {G}^{p}={G}^{p}-{G}^{p-}\) , is incrementally objective.

We introduce \({\tau }^{\text{Tr}}\), the Kirchhoff tensor which results from an elastic prediction (Tr: trial, in English test):

\({\tilde{\tau }}^{\text{Tr}}=\mu {\tilde{\stackrel{ˉ}{b}}}^{\text{eTr}}\)

where

\({\stackrel{ˉ}{b}}^{\text{eTr}}=\stackrel{ˉ}{F}{G}^{p-}{\stackrel{ˉ}{F}}^{T}=\Delta \stackrel{ˉ}{F}{\stackrel{ˉ}{b}}^{e-}\Delta {\stackrel{ˉ}{F}}^{T}\), \(\Delta \stackrel{ˉ}{F}=(\Delta J{)}^{-1/3}\Delta F\), and \(\Delta J=\text{det}(\Delta F)\)

We obtain \({\stackrel{ˉ}{b}}^{e-}\) from the stresses \({\tau }^{-}\) by the thermoelastic stress-deformation relationship and from the trace of the elastic deformation tensor.

\({\stackrel{ˉ}{b}}^{e-}=\frac{{\tilde{\tau }}^{-}}{{\mu }^{-}}+\frac{1}{3}\text{tr}{\stackrel{ˉ}{b}}^{e-}\)

Note:

The advantage of this formulation is that it is not necessary to calculate the plastic deformation \({G}^{p-}\) which would require us to reverse the transformation gradient \(\stackrel{ˉ}{F}\). You only need to know. \(\stackrel{ˉ}{F}{G}^{p-}{\stackrel{ˉ}{F}}^{T}\) .

If \({\tau }_{\text{eq}}^{\text{Tr}}<R({p}^{-})+{\sigma }_{y}\), you stay elastic. In this case, we have:

\(p={p}^{-}\), \(\tau ={\tilde{\tau }}^{\text{Tr}}+\frac{1}{3}\text{tr}{\tau }^{\text{Tr}}\text{Id}\), and \(\frac{1}{3}\text{tr}{\stackrel{ˉ}{b}}^{e}=\frac{1}{3}\text{tr}{\stackrel{ˉ}{b}}^{\text{eTr}}\)

if not, we get:

\(\text{tr}{\stackrel{ˉ}{b}}^{e}=\text{tr}{\stackrel{ˉ}{b}}^{\text{eTr}}\), thanks to the simplification of the flow law: \({\stackrel{ˉ}{b}}^{e}={\stackrel{ˉ}{b}}^{{e}^{\text{Tr}}}-\text{tr}{\stackrel{ˉ}{b}}^{e}\frac{\tilde{\tau }}{{\tau }_{\text{eq}}}\Delta p\)

Taking the deviatory parts of this equation, and multiplying them by \(\mu\), we get: \({\tilde{\tau }}^{\text{Tr}}=\tilde{\tau }(1+\frac{\mu \text{tr}{\stackrel{ˉ}{b}}^{\text{eTr}}\Delta p}{{\tau }_{\text{eq}}})\)

By calculating the equivalent stress, we come back to a non-linear scalar equation in \(\Delta p\):

\({\tau }_{\text{eq}}^{\text{Tr}}-{\tau }_{\text{eq}}-\mu \text{tr}{\stackrel{ˉ}{b}}^{\text{eTr}}\Delta p=0\)

In the plastic case: \({\tau }_{\text{eq}}={\sigma }_{y}+R({p}^{-}+\Delta p)\), which leads to \(\Delta p\) solution of the equation:

\({\tau }_{\text{eq}}^{\text{Tr}}-{\sigma }_{y}-R({p}^{-}+\Delta p)-\mu \text{tr}{\stackrel{ˉ}{b}}^{\text{eTr}}\Delta p=0\)

In the viscoplastic case: \({\tau }_{\text{eq}}={\sigma }_{y}+R({p}^{-}+\Delta p)+{\sigma }_{0}{\text{sh}}^{-1}\left[{(\frac{\Delta p}{{\dot{\varepsilon }}_{0}\Delta t})}^{\frac{1}{m}}\right]\), which leads to \(\Delta p\) solution of the equation:

\({\tau }_{\text{eq}}^{\text{Tr}}-{\sigma }_{y}-R({p}^{-}+\Delta p)-{\sigma }_{0}{\text{sh}}^{-1}\left[{(\frac{\Delta p}{{\dot{\varepsilon }}_{0}\Delta t})}^{\frac{1}{m}}\right]-\mu \text{tr}{\stackrel{ˉ}{b}}^{\text{eTr}}\Delta p=0\)

In the case where the work hardening is linear, or given by a traction curve defined point by point, and therefore refined by pieces, the equation to be solved is linear. The solution is obtained directly. \(\Delta p\) In other cases, the resolution is performed in Code_Aster by a secant method with a search interval (cf. [R5.03.05]). The integration can be controlled by parameters RESI_INTE_RELA and ITER_INTE_MAXI.

Once \(\Delta p\) is known, we can then to derive/to establish the Kirchhoff tensor, i.e.:

\(\tau =\frac{{\tau }_{\text{eq}}}{{\tau }_{\text{eq}}^{\text{Tr}}}{\tilde{\tau }}^{\text{Tr}}+\left[\frac{K}{2}({J}^{2}-1)-\frac{\mathrm{3K}}{2}(T-{T}_{\text{ref}})(J+\frac{1}{J})\right]\text{Id}\)

Once the cumulative plastic deformation, the stress tensor and the tangent matrix have been calculated, a correction is made on the trace of the elastic deformation tensor \({\stackrel{ˉ}{b}}^{e}\) to take into account the plastic incompressibility, which is not maintained with the simplification made on the flow law [éq 4.2.1]. This correction is carried out by using a relationship between the invariants of \({\stackrel{ˉ}{b}}^{e}\) and \({\tilde{\stackrel{ˉ}{b}}}^{e}\) and by exploiting the plastic incompressibility condition \({J}^{p}=1\) (or equivalently \(\text{det}{\stackrel{ˉ}{b}}^{e}=1\)). This relationship is written as:

\({x}^{3}-{\stackrel{ˉ}{J}}_{2}^{e}x-(1-{\stackrel{ˉ}{J}}_{3}^{e})=0\)

with \({\stackrel{ˉ}{J}}_{2}^{e}=\frac{1}{2}{\stackrel{ˉ}{b}}_{\text{eq}}^{{e}^{2}}=\frac{{\tau }_{\text{eq}}^{2}}{2{\mu }^{2}}\), \({\stackrel{ˉ}{J}}_{3}^{e}=\text{det}{\tilde{\stackrel{ˉ}{b}}}^{e}=\text{det}\frac{\tilde{\tau }}{\mu }\), and \(x=\frac{1}{3}\text{tr}{\stackrel{ˉ}{b}}^{e}\)

Solving this third-degree equation makes it possible to obtain \(\text{tr}{\stackrel{ˉ}{b}}^{e}\) and therefore the thermoelastic deformation \({\stackrel{ˉ}{b}}^{e-}\) at the following time step. In the case where this equation admits several solutions, we take the solution that is closest to the solution of the previous time step. This is also why we store in an internal variable \(\frac{1}{3}\text{tr}{\stackrel{ˉ}{b}}^{e}\).