5. Tangent operator

The tangent operator called by option RIGI_MECA_TANG corresponds to the tangent operator deduced from the speed problem and calculated from the results known at time t.

The tangent operator called by option FULL_MECA should correspond to the tangent operator for the discretized problem implicitly. In reality, we did not perform this calculation. When option FULL_MECA is selected, we then take the tangent operator deduced from the speed problem and calculated from the results known at the time t+dt.

Below we detail the tangent operator deduced from the speed problem according to the mechanism (s) involved.

5.1. Tangent operator of the nonlinear elastic mechanism

We simply have the following nonlinear elastic relationship:

\({\dot{\mathrm{\sigma }}}_{\text{ij}}={D}_{\text{ijkl}}(\mathrm{\sigma }){\dot{\mathrm{\varepsilon }}}_{\text{kl}}={(\frac{{I}_{1}+{Q}_{\text{init}}}{3{P}_{a}})}^{n}{D}_{\text{ijkl}}^{\text{lineaire}}{\dot{\mathrm{\varepsilon }}}_{\text{kl}}\)

Hence immediately the tangent operator:

\({H}_{\text{ijkl}}^{\text{elas}\text{.}\text{nl}}={(\frac{{I}_{1}+{Q}_{\text{init}}}{3{P}_{a}})}^{n}{D}_{\text{ijkl}}^{\text{lineaire}}\)

5.2. Tangent operator of isotropic elastic and plastic mechanisms

In this case, we have the following relationship:

\({\dot{\mathrm{\sigma }}}_{\text{ij}}={D}_{\text{ijkl}}(\mathrm{\sigma })({\dot{\mathrm{\varepsilon }}}_{\text{kl}}-{\dot{\mathrm{\varepsilon }}}_{\text{kl}}^{\text{ip}})={D}_{\text{ijkl}}(\mathrm{\sigma })({\dot{\mathrm{\varepsilon }}}_{\text{kl}}+\frac{1}{3}{\dot{\mathrm{\lambda }}}^{i}{\mathrm{\delta }}_{\text{kl}})\)

He is coming: \({\dot{I}}_{1}=3K({\dot{\mathrm{\varepsilon }}}_{v}+{\dot{\mathrm{\lambda }}}^{i})\)

Taking into account this relationship and the work-hardening law of \({Q}_{\text{iso}}\), condition \({\dot{f}}^{i}=0\) becomes:

\({\dot{f}}^{i}=-\frac{{\dot{I}}_{1}}{3}+{\dot{Q}}_{\text{iso}}=-K({\dot{\mathrm{\varepsilon }}}_{v}+{\dot{\mathrm{\lambda }}}^{i})-{K}^{p}{\dot{\mathrm{\lambda }}}^{i}=0\)

That is: \({\dot{\mathrm{\lambda }}}^{i}=-\frac{K}{K+{K}^{p}}{\dot{\mathrm{\varepsilon }}}_{v}\)

Reporting this result into the expression for \({\dot{\mathrm{\sigma }}}_{\text{ij}}\), we get:

\({\dot{\mathrm{\sigma }}}_{\text{ij}}={D}_{\text{ijkl}}({\dot{\mathrm{\varepsilon }}}_{\text{kl}}-\frac{1}{3}\frac{K}{K+{K}^{p}}{\dot{\mathrm{\varepsilon }}}_{\text{mm}}{\mathrm{\delta }}_{\text{kl}})=({D}_{\text{ijkl}}-\frac{1}{3}\frac{K}{K+{K}^{p}}{D}_{\text{ijmn}}{\mathrm{\delta }}_{\text{mn}}{\mathrm{\delta }}_{\text{kl}}){\dot{\mathrm{\varepsilon }}}_{\text{kl}}\)

Hence the tangent operator:

\({H}_{\text{ijkl}}^{\text{ip}}={D}_{\text{ijkl}}-\frac{1}{3}\frac{K}{K+{K}^{p}}{D}_{\text{ijmn}}{\mathrm{\delta }}_{\text{mn}}{\mathrm{\delta }}_{\text{kl}}\)

We can also write in matrix form:

\({H}^{\text{ip}}={(\frac{{I}_{1}}{3{P}_{a}})}^{n}\left[\begin{array}{cccccc}\mathrm{\lambda }-\mathrm{\khi }+2\mathrm{\mu }& \mathrm{\lambda }-\mathrm{\khi }& \mathrm{\lambda }-\mathrm{\khi }& 0& 0& 0\\ \mathrm{\lambda }-\mathrm{\khi }& \mathrm{\lambda }-\mathrm{\khi }+2\mathrm{\mu }& \mathrm{\lambda }-\mathrm{\khi }& 0& 0& 0\\ \mathrm{\lambda }-\mathrm{\khi }& \mathrm{\lambda }-\mathrm{\khi }& \mathrm{\lambda }-\mathrm{\khi }+2\mathrm{\mu }& 0& 0& 0\\ 0& 0& 0& 2\mathrm{\mu }& 0& 0\\ 0& 0& 0& 0& 2\mathrm{\mu }& 0\\ 0& 0& 0& 0& 0& 2\mathrm{\mu }\end{array}\right]\)

where for this formula only \(\mathrm{\lambda }\) and \(\mathrm{\mu }\) are the Lamé and \(\mathrm{\khi }=\frac{{K}_{o}^{{e}^{2}}}{{K}_{o}^{e}+{K}_{o}^{p}}\) coefficients.

5.3. Tangent operator of elastic and plastic deviatory mechanisms

Condition \({\dot{f}}^{d}=0\) is written as:

\({\dot{f}}^{d}=\frac{\partial {f}^{d}}{\partial {\mathrm{\sigma }}_{\text{ij}}}{\dot{\mathrm{\sigma }}}_{\text{ij}}+\frac{\partial {f}^{d}}{\partial R}\dot{R}+\frac{\partial {f}^{d}}{\partial {X}_{\text{ij}}}{\dot{X}}_{\text{ij}}=\frac{\partial {f}^{d}}{\partial {\mathrm{\sigma }}_{\text{ij}}}{\dot{\mathrm{\sigma }}}_{\text{ij}}+\frac{\partial {f}^{d}}{\partial R}{\dot{\mathrm{\lambda }}}^{d}{G}^{R}+\frac{\partial {f}^{d}}{\partial {X}_{\text{ij}}}{\dot{\mathrm{\lambda }}}^{d}{G}_{\text{ij}}^{X}=0\)

Since the tensor \({G}^{X}\) is purely deviatory, the product \(\frac{\partial {f}^{d}}{\partial {X}_{\text{ij}}}{G}_{\text{ij}}^{X}\) is reduced to:

\(\frac{\partial {f}^{d}}{\partial {X}_{\text{ij}}}{G}_{\text{ij}}^{X}=\text{dev}(\frac{\partial {f}^{d}}{\partial {X}_{\text{ij}}}){G}_{\text{ij}}^{X}=-{I}_{1}{Q}_{\text{ij}}{G}_{\text{ij}}^{X}\)

The plastic multiplier can therefore be in the form of:

\({\dot{\mathrm{\lambda }}}^{d}=\frac{1}{{H}^{\text{dev}}}\frac{\partial {f}^{d}}{\partial {\mathrm{\sigma }}_{\text{ij}}}{\dot{\mathrm{\sigma }}}_{\text{ij}}\)

by showing the plastic module \({H}^{\text{dev}}\), given by:

\({H}^{\text{dev}}={I}_{1}^{2}{(\frac{{I}_{1}+{Q}_{\text{init}}}{3{P}_{a}})}^{-\mathrm{1,5}}\left[A{(1-\frac{R}{{R}_{m}})}^{2}+\frac{1}{b}{Q}_{\text{ij}}({Q}_{\text{ij}}+\mathrm{\varphi }{X}_{\text{ij}})\right]\)

The constraints-deformations relationship then makes it possible to write:

\(\frac{\partial {f}^{d}}{\partial {\mathrm{\sigma }}_{\text{ij}}}{\dot{\mathrm{\sigma }}}_{\text{ij}}=\frac{\partial {f}^{d}}{\partial {\mathrm{\sigma }}_{\text{ij}}}{D}_{\text{ijkl}}({\dot{\mathrm{\varepsilon }}}_{\text{kl}}-{\dot{\mathrm{\lambda }}}^{d}{G}_{\text{kl}}^{d})=\frac{\partial {f}^{d}}{\partial {\mathrm{\sigma }}_{\text{ij}}}{D}_{\text{ijkl}}{\dot{\mathrm{\varepsilon }}}_{\text{kl}}-{\dot{\mathrm{\lambda }}}^{d}\frac{\partial {f}^{d}}{\partial {\mathrm{\sigma }}_{\text{ij}}}{D}_{\text{ijkl}}{G}_{\text{kl}}^{d}\)

This finally gives for the plastic multiplier:

\({\dot{\mathrm{\lambda }}}^{d}=\frac{\frac{\partial {f}^{d}}{\partial {\mathrm{\sigma }}_{\text{ij}}}{D}_{\text{ijkl}}{\dot{\mathrm{\varepsilon }}}_{\text{kl}}}{{H}^{\text{dev}}+\frac{\partial {f}^{d}}{\partial {\mathrm{\sigma }}_{\text{ij}}}{D}_{\text{ijkl}}{G}_{\text{kl}}^{d}}\)

Reporting this result into the expression for \({\dot{\mathrm{\sigma }}}_{\text{ij}}\), we get:

\({\dot{\mathrm{\sigma }}}_{\text{ij}}={D}_{\text{ijkl}}({\dot{\mathrm{\varepsilon }}}_{\text{kl}}-\frac{\frac{\partial {f}^{d}}{\partial {\mathrm{\sigma }}_{\text{pq}}}{D}_{\text{pqmn}}{\dot{\mathrm{\varepsilon }}}_{\text{mn}}}{{H}^{\text{dev}}+\frac{\partial {f}^{d}}{\partial {\mathrm{\sigma }}_{\text{rs}}}{D}_{\text{rstu}}{G}_{\text{tu}}^{d}}{G}_{\text{kl}}^{d})\)

Hence the tangent operator:

\({H}_{\text{ijkl}}^{\text{dp}}={D}_{\text{ijkl}}-{D}_{\text{ijmn}}{G}_{\text{mn}}^{d}\frac{\frac{\partial {f}^{d}}{\partial {\mathrm{\sigma }}_{\text{pq}}}{D}_{\text{pqkl}}}{{H}^{\text{dev}}+\frac{\partial {f}^{d}}{\partial {\mathrm{\sigma }}_{\text{rs}}}{D}_{\text{rstu}}{G}_{\text{tu}}^{d}}\)

The tangent operator thus obtained is not symmetric. However, for the moment, law CJS is based on finite elements that require a symmetric operator. In the end, we remember not \({H}_{\text{ijkl}}^{\text{dp}}\) but \({\tilde{H}}_{\text{ijkl}}^{\text{dp}}\) which is given by:

\({\tilde{H}}_{\text{ijkl}}^{\text{dp}}=\frac{{H}_{\text{ijkl}}^{\text{dp}}+{H}_{\text{klij}}^{\text{dp}}}{2}\) with \(\text{ij}\) and \(\text{kl}\) caught in \((\text{11},\text{22},\text{33},\text{12},\text{13},\text{23})\)

5.4. Tangent operator of elastic, plastic, isotropic and deviatory mechanisms

The following two conditions must be met: \({\dot{f}}^{i}=0\) and \({\dot{f}}^{d}=0\). Taking into account the constraints-deformations relationship, which is written as:

\({\dot{\mathrm{\sigma }}}_{\text{ij}}={D}_{\text{ijkl}}({\dot{\mathrm{\varepsilon }}}_{\text{kl}}+\frac{1}{3}{\dot{\mathrm{\lambda }}}^{i}{\mathrm{\delta }}_{\text{kl}}-{\dot{\mathrm{\lambda }}}^{d}{G}_{\text{kl}}^{d})\)

the first condition gives:

\({\dot{f}}^{i}=-K({\dot{\mathrm{\varepsilon }}}_{v}+{\dot{\mathrm{\lambda }}}^{i}-{\dot{\mathrm{\lambda }}}^{d}{G}_{v}^{d})-{K}^{p}{\dot{\mathrm{\lambda }}}^{i}=0\)

where we put \({G}_{v}^{d}={G}_{\text{kk}}^{d}=\text{tr}({G}^{d})\).

The second condition results in:

\(\frac{\partial {f}^{d}}{\partial {\mathrm{\sigma }}_{\text{ij}}}{D}_{\text{ijkl}}{\dot{\mathrm{\varepsilon }}}_{\text{kl}}+\frac{1}{3}{\dot{\mathrm{\lambda }}}^{i}\frac{\partial {f}^{d}}{\partial {\mathrm{\sigma }}_{\text{ij}}}{D}_{\text{ijkl}}{\mathrm{\delta }}_{\text{kl}}-{\dot{\mathrm{\lambda }}}^{d}\frac{\partial {f}^{d}}{\partial {\mathrm{\sigma }}_{\text{ij}}}{D}_{\text{ijkl}}{G}_{\text{kl}}^{d}-{H}^{\text{dev}}{\dot{\mathrm{\lambda }}}^{d}=0\)

Thus, the plastic multipliers \({\dot{\mathrm{\lambda }}}^{i}\) and \({\dot{\mathrm{\lambda }}}^{d}\) are obtained by solving the system:

\(\{\begin{array}{}-(K+{K}^{p}){\dot{\mathrm{\lambda }}}^{i}+K{G}_{v}^{d}{\dot{\mathrm{\lambda }}}^{d}=K{\dot{\mathrm{\varepsilon }}}_{v}\\ -\frac{1}{3}\frac{\partial {f}^{d}}{\partial {\mathrm{\sigma }}_{\text{ij}}}{D}_{\text{ijkl}}{\mathrm{\delta }}_{\text{kl}}{\dot{\mathrm{\lambda }}}^{i}+(\frac{\partial {f}^{d}}{\partial {\mathrm{\sigma }}_{\text{ij}}}{D}_{\text{ijkl}}{G}_{\text{kl}}^{d}+{H}^{\text{dev}}){\dot{\mathrm{\lambda }}}^{d}=\frac{\partial {f}^{d}}{\partial {\mathrm{\sigma }}_{\text{ij}}}{D}_{\text{ijkl}}{\dot{\mathrm{\varepsilon }}}_{\text{kl}}\end{array}\)

either:

\({\dot{\mathrm{\lambda }}}^{i}=\frac{K{G}_{v}^{d}\frac{\partial {f}^{d}}{\partial {\mathrm{\sigma }}_{\text{ij}}}{D}_{\text{ijkl}}{\dot{\mathrm{\varepsilon }}}_{\text{kl}}-(\frac{\partial {f}^{d}}{\partial {\mathrm{\sigma }}_{\text{mn}}}{D}_{\text{mnpq}}{G}_{\text{pq}}^{d}+{H}^{\text{dev}})K{\dot{\mathrm{\varepsilon }}}_{v}}{(K+{K}^{p})(\frac{\partial {f}^{d}}{\partial {\mathrm{\sigma }}_{\text{rs}}}{D}_{\text{rstu}}{G}_{\text{tu}}^{d}+{H}^{\text{dev}})-\frac{1}{3}K{G}_{v}^{d}\frac{\partial {f}^{d}}{\partial {\mathrm{\sigma }}_{\text{vw}}}{D}_{\text{wvxy}}{\mathrm{\delta }}_{\text{xy}}}\)

\({\dot{\mathrm{\lambda }}}^{d}=\frac{(K+{K}^{p})\frac{\partial {f}^{d}}{\partial {\mathrm{\sigma }}_{\text{ij}}}{D}_{\text{ijkl}}{\dot{\mathrm{\varepsilon }}}_{\text{kl}}-\frac{1}{3}K\frac{\partial {f}^{d}}{\partial {\mathrm{\sigma }}_{\text{mn}}}{D}_{\text{mnpq}}{\mathrm{\delta }}_{\text{pq}}{\dot{\mathrm{\varepsilon }}}_{v}}{(K+{K}^{p})(\frac{\partial {f}^{d}}{\partial {\mathrm{\sigma }}_{\text{rs}}}{D}_{\text{rstu}}{G}_{\text{tu}}^{d}+{H}^{\text{dev}})-\frac{1}{3}K{G}_{v}^{d}\frac{\partial {f}^{d}}{\partial {\mathrm{\sigma }}_{\text{vw}}}{D}_{\text{vwxy}}{\mathrm{\delta }}_{\text{xy}}}\)

These expressions are still written:

\({\dot{\mathrm{\lambda }}}^{i}={T}_{{1}_{\text{kl}}}{\dot{\mathrm{\varepsilon }}}_{\text{kl}}\) and \({\dot{\mathrm{\lambda }}}^{d}={T}_{{2}_{\text{kl}}}{\dot{\mathrm{\varepsilon }}}_{\text{kl}}\)

where tensors \({T}_{1}\) and \({T}_{2}\) are given by:

\({T}_{{1}_{\text{kl}}}=\frac{K{G}_{v}^{d}\frac{\partial {f}^{d}}{\partial {\mathrm{\sigma }}_{\text{ij}}}{D}_{\text{ijkl}}-(\frac{\partial {f}^{d}}{\partial {\mathrm{\sigma }}_{\text{mn}}}{D}_{\text{mnpq}}{G}_{\text{pq}}^{d}+{H}^{\text{dev}})K{\mathrm{\delta }}_{\text{kl}}}{(K+{K}^{p})(\frac{\partial {f}^{d}}{\partial {\mathrm{\sigma }}_{\text{rs}}}{D}_{\text{rstu}}{G}_{\text{tu}}^{d}+{H}^{\text{dev}})-\frac{1}{3}K{G}_{v}^{d}\frac{\partial {f}^{d}}{\partial {\mathrm{\sigma }}_{\text{vw}}}{D}_{\text{vwxy}}{\mathrm{\delta }}_{\text{xy}}}\)

\({T}_{{2}_{\text{kl}}}=\frac{(K+{K}^{p})\frac{\partial {f}^{d}}{\partial {\mathrm{\sigma }}_{\text{ij}}}{D}_{\text{ijkl}}-\frac{1}{3}K\frac{\partial {f}^{d}}{\partial {\mathrm{\sigma }}_{\text{mn}}}{D}_{\text{mnpq}}{\mathrm{\delta }}_{\text{pq}}{\mathrm{\delta }}_{\text{kl}}}{(K+{K}^{p})(\frac{\partial {f}^{d}}{\partial {\mathrm{\sigma }}_{\text{rs}}}{D}_{\text{rstu}}{G}_{\text{tu}}^{d}+{H}^{\text{dev}})-\frac{1}{3}K{G}_{v}^{d}\frac{\partial {f}^{d}}{\partial {\mathrm{\sigma }}_{\text{vw}}}{D}_{\text{vwxy}}{\mathrm{\delta }}_{\text{xy}}}\)

By putting the expressions \({\dot{\mathrm{\lambda }}}^{i}\) and \({\dot{\mathrm{\lambda }}}^{d}\) from into the formula for \({\dot{\mathrm{\sigma }}}_{\text{ij}}\), we find:

\({\dot{\mathrm{\sigma }}}_{\text{ij}}={D}_{\text{ijkl}}({\dot{\mathrm{\varepsilon }}}_{\text{kl}}+\frac{1}{3}{T}_{{1}_{\text{nm}}}{\dot{\mathrm{\varepsilon }}}_{\text{nm}}{\mathrm{\delta }}_{\text{kl}}-{T}_{{2}_{\text{pq}}}{\dot{\mathrm{\varepsilon }}}_{\text{pq}}{G}_{\text{kl}}^{d})\)

Hence the tangent operator:

\({H}_{\text{ijkl}}^{\text{idp}}={D}_{\text{ijkl}}+\frac{1}{3}{D}_{\text{ijmn}}{\mathrm{\delta }}_{\text{mn}}{T}_{{1}_{\text{kl}}}-{D}_{\text{ijpq}}{G}_{\text{pq}}^{d}{T}_{{2}_{\text{kl}}}\)

As this tangent operator is not symmetric, we use not \({H}_{\text{ijkl}}^{\text{idp}}\) but \({\tilde{H}}_{\text{ijkl}}^{\text{idp}}\) which is given by:

\({\tilde{H}}_{\text{ijkl}}^{\text{idp}}=\frac{{H}_{\text{ijkl}}^{\text{idp}}+{H}_{\text{klij}}^{\text{idp}}}{2}\) with \(\text{ij}\) and \(\text{kl}\) caught in \((\text{11},\text{22},\text{33},\text{12},\text{13},\text{23})\)