4. Incremental shape

We are placing ourselves here in the context of finite increases. The index — refers to a component at the beginning of a loading step and the absence of an index a component at the end of a loading step. The \(\Delta\) operator refers to the increase of a component.

In pure mechanics or when the law is used in modeling THM in effective constraints, the equations translating elastic behavior are written:

\(\begin{array}{}s\text{=}{s}^{\text{-}}\text{+}2\mu (\Delta e\text{-}\Delta {e}^{p})\text{=}{s}^{e}\text{-}2\mu \Delta {e}^{p}\text{}\begin{array}{cc}\text{où}& {s}^{e}\text{=}{s}^{\text{-}}\text{+}2\mu \Delta e\end{array}\\ {I}_{1}\text{=}{I}_{{1}^{\text{-}}}\text{+}\mathrm{3K}(\Delta {\varepsilon }_{\nu }\text{-}\Delta {\varepsilon }_{\nu }^{p})\text{=}{I}_{1}^{e}\text{-}\mathrm{3K}\Delta {\varepsilon }_{\nu }^{p}\text{}\begin{array}{cc}\text{où}& {I}_{1}^{e}\text{=}{I}_{1}^{\text{-}}+\mathrm{3K}\Delta {\varepsilon }_{\nu }\end{array}\end{array}\)

When the law is used in total stress modeling THM, the stress tensor and the equations translating elastic behavior are written as follows:

\(\begin{array}{}\sigma \text{=}H(\varepsilon \text{-}{\varepsilon }^{p})\text{+}{\sigma }_{p}I\\ {\sigma }^{e}\text{=}{\sigma }^{\text{-}}\text{+}H\Delta \varepsilon +\Delta {\sigma }_{p}I\text{=}\sigma {\text{'}}^{\text{-}}\text{+}H\Delta \varepsilon \text{+}{\sigma }_{p}I\\ s\text{=}{s}^{e}\text{-}2\mu \Delta {e}^{p}\text{où}{s}^{e}\text{=}{s}^{\text{-}}\text{+}2\mu \Delta e\\ {I}_{1}\text{=}{I}_{1}^{e}\text{-}\mathrm{3K}\Delta {\varepsilon }_{v}^{p}\text{où}{I}_{1}^{e}\text{=}{I}_{1}^{\text{-}}\text{+}\mathrm{3K}\Delta {\varepsilon }_{v}\text{+}3\Delta {\sigma }_{p}\end{array}\)

\(\Delta {\sigma }_{p}\text{=}\text{-}b(S\Delta {p}_{\text{lq}}\text{+}(1\text{-}S)\Delta {p}_{\text{gz}})\text{=}b(S\Delta {p}_{c}\text{-}\Delta {p}_{\text{gz}})\)

In addition, the flow rule is written as:

\(d{\varepsilon }_{\text{ij}}^{p}\text{=}d\lambda \text{.}\frac{\partial G}{\partial {\sigma }_{\text{ij}}}(\sigma ,\gamma )\text{=}d\lambda (\eta (\gamma ){\delta }_{\text{ij}}\text{+}\sqrt{\frac{3}{2}}\frac{{s}_{\text{ij}}}{{s}_{\text{II}}})\)

Like \({e}^{p}\text{=}{\varepsilon }^{p}\text{-}\frac{\text{tr}({\varepsilon }^{p})}{3}I\), \(\Delta {\varepsilon }^{p}\text{=}\Delta {e}^{p}\text{+}\frac{\Delta {\varepsilon }_{\nu }^{p}}{3}I\). From this we deduce that \(\{\begin{array}{c}\Delta {e}^{p}\text{=}\frac{3}{2}\frac{s}{{\sigma }_{\text{eq}}}\Delta \lambda \\ \Delta {\varepsilon }_{\nu }^{p}\text{=}3\eta (\gamma )\Delta \lambda \end{array}\), and therefore, like \({\sigma }_{\text{eq}}^{e}s\text{=}{\sigma }_{\text{eq}}{s}^{e}\), the following relationships are identical for both aspects of the law:

\(\{\begin{array}{c}s\text{=}{s}^{e}\text{-}3\mu \frac{s}{{\sigma }_{\text{eq}}}\Delta \lambda \text{=}{s}^{e}(1\text{-}3\mu \frac{\Delta \lambda }{{\sigma }_{\text{eq}}^{e}})\\ \begin{array}{}{I}_{1}\text{=}{I}_{1}^{e}\text{-}\mathrm{9K}\eta (\gamma )\Delta \lambda \text{}\\ {\sigma }_{\text{eq}}\text{=}{\sigma }_{\text{eq}}^{e}\text{-}3\mu \Delta \lambda \text{}\end{array}\end{array}\)

Finally, \(\Delta {\gamma }^{p}\text{=}\sqrt{\frac{2}{3}\Delta {e}^{p}:\Delta {e}^{p}}\text{=}\sqrt{\frac{2}{3}\Delta {\lambda }^{2}(\frac{3}{2}\frac{s}{{\sigma }_{\text{eq}}}):(\frac{3}{2}\frac{s}{{\sigma }_{\text{eq}}})}\text{=}\Delta \lambda\). So, taking up the result obtained in paragraph 3.2.3, we see that \(\Delta \gamma \text{=}\Delta \lambda [:ref:\)eta (gamma )pm 1 <eta (gamma )pm 1>`] and like :math:mathrm{Delta gamma },mathrm{Delta lambda }ge 0`, we necessarily have \(\Delta \gamma \text{=}\Delta \lambda [:ref:\)eta (gamma )+1 <eta (gamma )+1>`]ge 0text { . } `

Note : The calculation of \(\mathrm{\Delta \lambda }\) is the same in all formulations.

4.1. Calculation of \(\Delta \lambda\)

We denote \(P\) the transition matrix such as:

\(\tilde{P}\text{.}{s}^{e}\text{.}P\text{=}{\stackrel{ˉ}{s}}^{e}\text{avec}{\stackrel{ˉ}{s}}^{e}\text{=}\text{diag}({s}_{1}^{e},{s}_{2}^{e},{s}_{3}^{e})\), or

\(\tilde{P}\text{.}{\sigma }^{e}\text{.}P\text{=}{\stackrel{ˉ}{\sigma }}^{e}\text{avec}{\stackrel{ˉ}{\sigma }}^{e}=\text{diag}({\sigma }_{1}^{e},{\sigma }_{2}^{e},{\sigma }_{3}^{e})\text{et}{\sigma }_{i}^{e}={s}_{i}^{e}+\frac{{I}_{1}^{e}}{3}\)

So:

\(\begin{array}{}\tilde{P}\text{.}\sigma \text{.}P\text{=}\tilde{P}\text{.}(s\text{+}\frac{{I}_{1}}{3}I)\text{.}P\text{=}\tilde{P}\text{.}({s}^{e}(1\text{-}3\mu \frac{\Delta \lambda }{{\sigma }_{\text{eq}}^{e}})\text{+}\frac{1}{3}({I}_{1}^{e}\text{-}\mathrm{9K}\eta (\gamma )\Delta \lambda )I)\text{.}P\\ \text{}\text{=}(1\text{-}3\mu \frac{\Delta \lambda }{{\sigma }_{\text{eq}}^{e}}){\stackrel{ˉ}{s}}^{e}\text{+}\frac{1}{3}({I}_{1}^{e}\text{-}\mathrm{9K}\eta (\gamma )\Delta \lambda )I\\ \text{}=\text{diag}({\sigma }_{1},{\sigma }_{2},{\sigma }_{3})\\ \text{avec}{\sigma }_{i}\text{=}(1\text{-}3\mu \frac{\Delta \lambda }{{\sigma }_{\text{eq}}^{e}}){s}_{i}^{e}\text{+}\frac{1}{3}({I}_{1}^{e}\text{-}\mathrm{9K}\eta (\gamma )\Delta \lambda )\text{=}{\sigma }_{i}^{e}\text{-}3\Delta \lambda (\frac{\mu {s}_{i}^{e}}{{\sigma }_{\text{eq}}^{e}}\text{+}K\eta (\gamma ))\end{array}\)

So, if \((1\text{-}3\mu \frac{\Delta \lambda }{{\sigma }_{\text{eq}}^{e}})>0\), that is \({s}_{\text{II}}>0\), the \({\sigma }_{i}\) are ordered like the \({s}_{i}^{e}\). If \((1\text{-}3\mu \frac{\Delta \lambda }{{\sigma }_{\text{eq}}^{e}})\text{=}0\), that is \({s}_{\text{II}}\text{=}0\), the \({\sigma }_{i}\) are all equal to \(\frac{1}{3}({I}_{1}^{e}\text{-}\mathrm{9K}\eta \Delta \lambda )\text{=}\frac{1}{3}({I}_{1}^{e}\text{-}\mathrm{3K}\eta \frac{{\sigma }_{\text{eq}}^{e}}{\mu })\). We can thus write the criterion \(F\) as a function of \(\Delta \lambda \text{ou}\Delta \gamma\) only, the \({s}_{i}^{e}\) being given and ordered, by taking:

\(\begin{array}{}{\sigma }_{3}\text{-}{\sigma }_{1}\text{=}(1\text{-}3\mu \frac{\Delta \lambda }{{\sigma }_{\text{eq}}^{e}})({s}_{3}^{e}\text{-}{s}_{1}^{e})\\ {\sigma }_{3}\text{=}{s}_{3}^{e}(1\text{-}3\mu \frac{\Delta \lambda }{{\sigma }_{\text{eq}}^{e}})\text{+}\frac{1}{3}({I}_{1}^{e}\text{-}\mathrm{9K}\eta (\gamma )\Delta \lambda )\end{array}\)

We then obtain a non-linear function \(F(\Delta \lambda )\) solved by a Newton algorithm, given by:

\(\begin{array}{}F(\Delta \gamma )\text{=}0\text{=}({s}_{3}^{e}\text{-}{s}_{1}^{e})\left[1\text{-}\frac{3\mu }{{\sigma }_{\text{eq}}^{e}}h(\Delta \gamma )\Delta \gamma \right]\text{-}b(\Delta \gamma )\left[1\text{-}\frac{1}{{\sigma }_{3}^{b\text{-}d}}({s}_{3}^{e}\text{+}\frac{{I}_{1}^{e}}{3}\text{-}g(\Delta \gamma )\Delta \gamma )\right]\\ \text{-}{(S(\Delta \gamma ){\sigma }_{c}^{2}(\Delta \gamma )\text{-}m(\Delta \gamma ){\sigma }_{c}(\Delta \gamma )\left[{s}_{3}^{e}\text{+}\frac{{I}_{1}^{e}}{3}\text{-}g(\Delta \gamma )\Delta \gamma \right])}^{\frac{1}{2}}\end{array}\) Eq 3.1

where we noted:

\(\begin{array}{}\\ \begin{array}{cc}h(\Delta \gamma )\text{=}\frac{1}{\eta +1}\text{=}\frac{3\text{+}\text{sin}\phi ({\gamma }^{\text{-}}\text{+}\Delta \gamma )}{3(1\text{+}\text{sin}\phi ({\gamma }^{\text{-}}\text{+}\Delta \gamma ))}& \text{et}g(\Delta \gamma )\text{=}\frac{3\text{+}\text{sin}\phi ({\gamma }^{\text{-}}\text{+}\Delta \gamma )}{3(1\text{+}\text{sin}\phi ({\gamma }^{\text{-}}+\Delta \gamma ))}(\frac{\mathrm{6K}\text{sin}\phi ({\gamma }^{\text{-}}+\Delta \gamma )}{3\text{+}\text{sin}\phi ({\gamma }^{\text{-}}\text{+}\Delta \gamma )}\text{+}\frac{3\mu {s}_{3}^{e}}{{\sigma }_{\text{eq}}^{e}})\end{array}\end{array}\).

In practice, we will apply Newton’s algorithm to the function

\(\begin{array}{}\stackrel{ˉ}{F}(\Delta \gamma )\text{=}{\left[({\sigma }_{3}\text{-}{\sigma }_{1})\text{-}b(1\text{-}\frac{{\sigma }_{3}}{{\sigma }_{3}^{b\text{-}d}})\right]}^{2}\text{-}(S{\sigma }_{c}^{2}\text{-}m{\sigma }_{c}{\sigma }_{3})\\ \text{=}{\left[({s}_{3}^{e}\text{-}{s}_{1}^{e})\left[1\text{-}\frac{3\mu }{{\sigma }_{\text{eq}}^{e}}h(\Delta \gamma )\Delta \gamma \right]\text{-}b(\Delta \gamma )\left[1\text{-}\frac{1}{{\sigma }_{3}^{b\text{-}d}}({s}_{3}^{e}\text{+}\frac{{I}_{1}^{e}}{3}\text{-}g(\Delta \gamma )\Delta \gamma )\right]\right]}^{2}\\ \text{-}S(\Delta \gamma ){\sigma }_{c}^{2}(\Delta \gamma )\text{+}m(\Delta \gamma ){\sigma }_{c}(\Delta \gamma )\left[{s}_{3}^{e}\text{+}\frac{{I}_{1}^{e}}{3}\text{-}g(\Delta \gamma )\Delta \gamma \right]\end{array}\)

in order to overcome the difficulties associated with the sign of the element under the root during iterations. The derivative of \(\stackrel{ˉ}{F}\) with respect to \(\Delta \gamma\) is given in Appendix 2.

4.2. Existence of the solution

The principle of analytical resolution consists in determining point \(({I}_{1},s)\) as the projection of point \(({I}_{1}^{e},{s}^{e})\) on the charge surface with respect to the plastic flow potential:

\({s}_{\text{II}}\ge 0\)

\(\mathrm{\Delta \lambda }\le \frac{{\mathrm{\sigma }}_{\text{eq}}^{e}}{\mathrm{3\mu }}\) \(\mathrm{\Delta \lambda }>\frac{{\mathrm{\sigma }}_{\text{eq}}^{e}}{\mathrm{3\mu }}\) \(\frac{\partial F}{\partial \mathrm{\sigma }}=\frac{\partial F}{\partial \stackrel{ˉ}{\mathrm{\sigma }}}\frac{\partial \stackrel{ˉ}{\mathrm{\sigma }}}{\partial \mathrm{\sigma }}\)

\(\stackrel{ˉ}{\mathrm{\sigma }}\)

No solution

\(\stackrel{ˉ}{\mathrm{\sigma }}\)

However, the solution must respect condition \({s}_{\text{II}}>0\), that is to say \(F(\stackrel{ˉ}{\sigma },\gamma )=({\stackrel{ˉ}{\sigma }}_{3}-{\stackrel{ˉ}{\sigma }}_{1})-\sqrt{-{\stackrel{ˉ}{\sigma }}_{3}\text{.}m{\sigma }_{c}+S{\sigma }_{c}^{2}}-b\text{.}(1-\frac{{\stackrel{ˉ}{\sigma }}_{3}}{{\sigma }_{3}^{b-d}})\): we can therefore see that there is an area in which the problem does not admit a solution, which corresponds to \(\frac{\partial F}{\partial {\stackrel{ˉ}{\mathrm{\sigma }}}_{i}}={\mathrm{\delta }}_{\mathrm{i3}}-{\mathrm{\delta }}_{\mathrm{i1}}+\frac{1}{2}{\mathrm{\delta }}_{\mathrm{i3}}{\mathrm{\sigma }}_{c}m{\left[-{\stackrel{ˉ}{\mathrm{\sigma }}}_{3}\text{.}{\mathrm{m\sigma }}_{c}+{\mathrm{S\sigma }}_{c}^{2}\right]}^{-\frac{1}{2}}+b\frac{{\mathrm{\delta }}_{\mathrm{i3}}}{{\mathrm{\sigma }}_{3}^{b-d}}\).