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. Tangent operator of the nonlinear elastic mechanism ----------------------------------------------------- We simply have the following nonlinear elastic relationship: :math:`{\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: :math:`{H}_{\text{ijkl}}^{\text{elas}\text{.}\text{nl}}={(\frac{{I}_{1}+{Q}_{\text{init}}}{3{P}_{a}})}^{n}{D}_{\text{ijkl}}^{\text{lineaire}}` Tangent operator of isotropic elastic and plastic mechanisms ----------------------------------------------------------------- In this case, we have the following relationship: :math:`{\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: :math:`{\dot{I}}_{1}=3K({\dot{\mathrm{\varepsilon }}}_{v}+{\dot{\mathrm{\lambda }}}^{i})` Taking into account this relationship and the work-hardening law of :math:`{Q}_{\text{iso}}`, condition :math:`{\dot{f}}^{i}=0` becomes: :math:`{\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: :math:`{\dot{\mathrm{\lambda }}}^{i}=-\frac{K}{K+{K}^{p}}{\dot{\mathrm{\varepsilon }}}_{v}` Reporting this result into the expression for :math:`{\dot{\mathrm{\sigma }}}_{\text{ij}}`, we get: :math:`{\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: :math:`{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: :math:`{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 :math:`\mathrm{\lambda }` and :math:`\mathrm{\mu }` are the Lamé and :math:`\mathrm{\khi }=\frac{{K}_{o}^{{e}^{2}}}{{K}_{o}^{e}+{K}_{o}^{p}}` coefficients. Tangent operator of elastic and plastic deviatory mechanisms ------------------------------------------------------------------ Condition :math:`{\dot{f}}^{d}=0` is written as: :math:`{\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 :math:`{G}^{X}` is purely deviatory, the product :math:`\frac{\partial {f}^{d}}{\partial {X}_{\text{ij}}}{G}_{\text{ij}}^{X}` is reduced to: :math:`\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: :math:`{\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 :math:`{H}^{\text{dev}}`, given by: :math:`{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: :math:`\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: :math:`{\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 :math:`{\dot{\mathrm{\sigma }}}_{\text{ij}}`, we get: :math:`{\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: :math:`{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 :math:`{H}_{\text{ijkl}}^{\text{dp}}` but :math:`{\tilde{H}}_{\text{ijkl}}^{\text{dp}}` which is given by: :math:`{\tilde{H}}_{\text{ijkl}}^{\text{dp}}=\frac{{H}_{\text{ijkl}}^{\text{dp}}+{H}_{\text{klij}}^{\text{dp}}}{2}` with :math:`\text{ij}` and :math:`\text{kl}` caught in :math:`(\text{11},\text{22},\text{33},\text{12},\text{13},\text{23})` Tangent operator of elastic, plastic, isotropic and deviatory mechanisms ----------------------------------------------------------------------------- The following two conditions must be met: :math:`{\dot{f}}^{i}=0` and :math:`{\dot{f}}^{d}=0`. Taking into account the constraints-deformations relationship, which is written as: :math:`{\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: :math:`{\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 :math:`{G}_{v}^{d}={G}_{\text{kk}}^{d}=\text{tr}({G}^{d})`. The second condition results in: :math:`\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 :math:`{\dot{\mathrm{\lambda }}}^{i}` and :math:`{\dot{\mathrm{\lambda }}}^{d}` are obtained by solving the system: :math:`\{\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: :math:`{\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}}}` :math:`{\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: :math:`{\dot{\mathrm{\lambda }}}^{i}={T}_{{1}_{\text{kl}}}{\dot{\mathrm{\varepsilon }}}_{\text{kl}}` and :math:`{\dot{\mathrm{\lambda }}}^{d}={T}_{{2}_{\text{kl}}}{\dot{\mathrm{\varepsilon }}}_{\text{kl}}` where tensors :math:`{T}_{1}` and :math:`{T}_{2}` are given by: :math:`{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}}}` :math:`{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 :math:`{\dot{\mathrm{\lambda }}}^{i}` and :math:`{\dot{\mathrm{\lambda }}}^{d}` from into the formula for :math:`{\dot{\mathrm{\sigma }}}_{\text{ij}}`, we find: :math:`{\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: :math:`{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 :math:`{H}_{\text{ijkl}}^{\text{idp}}` but :math:`{\tilde{H}}_{\text{ijkl}}^{\text{idp}}` which is given by: :math:`{\tilde{H}}_{\text{ijkl}}^{\text{idp}}=\frac{{H}_{\text{ijkl}}^{\text{idp}}+{H}_{\text{klij}}^{\text{idp}}}{2}` with :math:`\text{ij}` and :math:`\text{kl}` caught in :math:`(\text{11},\text{22},\text{33},\text{12},\text{13},\text{23})`