5. Tangent operator in speed

The condition

\(\dot{f}=0\) eq 5-1

is written:

\(\dot{f}=\frac{\partial f}{\partial {\sigma }_{\text{ij}}}\dot{{\sigma }_{\text{ij}}}+\frac{\partial f}{\partial {\gamma }^{p}}\dot{{\gamma }^{p}}=0\)

From the expression of the cumulative plastic deviatoric deformation and \({\gamma }^{p}=\sqrt{\frac{2}{3}{e}_{\text{ij}}^{p}{e}_{\text{ij}}^{p}}\) from the \(\dot{{e}^{p}}=\dot{\lambda }\tilde{G}\) relationship, we then find the condition:

\(\dot{f}=\frac{\partial f}{\partial {\sigma }_{\text{ij}}}\dot{{\sigma }_{\text{ij}}}+\frac{\partial f}{\partial {\gamma }^{p}}\dot{\sqrt{\frac{2}{3}}}\dot{\lambda }\tilde{{G}_{\text{II}}}=0\)

This gives us for the plastic multiplier:

\(\dot{\lambda }=\frac{-\frac{\partial f}{\partial {\sigma }_{\text{ij}}}\dot{{\sigma }_{\text{ij}}}}{\sqrt{\frac{2}{3}}\frac{\partial f}{\partial {\gamma }^{p}}\tilde{{G}_{\text{II}}}}\)

Then considering the constraints/deformations relationship:

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

and by including it in the expression for \(\dot{\lambda }\) we can write:

\(\dot{\lambda }=-\frac{\frac{\partial f}{\partial {\sigma }_{\text{ij}}}{D}_{\text{ijkl}}\dot{{\varepsilon }_{\text{kl}}}-\dot{\lambda }\frac{\partial f}{\partial {\sigma }_{\text{ij}}}{D}_{\text{ijkl}}{G}_{\text{kl}}}{\sqrt{\frac{2}{3}}\frac{\partial f}{\partial {\gamma }^{p}}\tilde{{G}_{\text{II}}}}\)

Either:

\(\dot{\lambda }=-\frac{\frac{\partial f}{\partial {\sigma }_{\text{ij}}}{D}_{\text{ijkl}}\dot{{\varepsilon }_{\text{kl}}}}{\sqrt{\frac{2}{3}}\frac{\partial f}{\partial {\gamma }^{p}}\tilde{{G}_{\text{II}}}-\frac{\partial f}{\partial {\sigma }_{\text{ij}}}{D}_{\text{ijkl}}{G}_{\text{kl}}}\) eq 5-2

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

\(\dot{{\sigma }_{\text{ab}}}={D}_{\text{abcd}}(\dot{{\varepsilon }_{\text{cd}}}+\frac{\frac{\partial f}{\partial {\sigma }_{\text{ij}}}{D}_{\text{ijkl}}\dot{{\varepsilon }_{\text{kl}}}}{\sqrt{\frac{2}{3}}\frac{\partial f}{\partial {\gamma }^{p}}\tilde{{G}_{\text{II}}}-\frac{\partial f}{\partial {\sigma }_{\text{ij}}}{D}_{\text{ijkl}}{G}_{\text{kl}}}{G}_{\text{cd}})\) eq 5-3