4. Calculus of derivatives#

4.1. Derivative of the criterion#

4.1.1. Derivative with respect to constraints#

4.1.1.1. Intermediate derivative with respect to the deviator#

We start from: \(\frac{\partial g}{\partial {s}_{\text{ij}}}=h(\theta )\frac{\partial {s}_{\text{II}}}{\partial {s}_{\text{ij}}}+{s}_{\text{II}}\frac{\partial h(\theta )}{\partial {s}_{\text{ij}}}\)

where \(\frac{\partial {s}_{\text{II}}}{\partial {s}_{\text{ij}}}\) and \(\frac{\partial h(\theta )}{{s}_{\text{ij}}}\) are respectively given by:

\(\frac{{s}_{\text{II}}}{{s}_{\text{ij}}}=\frac{{s}_{\text{ij}}}{{s}_{\text{II}}}\)

\(\begin{array}{}\frac{\partial h(\theta )}{\partial {s}_{\text{ij}}}=\frac{1}{\mathrm{6h}{(\theta )}_{5}}\frac{\partial }{\partial {s}_{\text{ij}}}(1+{\gamma }_{\text{cjs}}\sqrt{54}\frac{\mathrm{det}(\underline{\underline{s}})}{{s}_{\text{II}}^{3}})\\ \text{}=\frac{-{\gamma }_{\text{cjs}}\mathrm{cos}(3\theta )}{\mathrm{2h}{(\theta )}^{5}{s}_{\text{II}}^{2}}{s}_{\text{ij}}+\frac{{\gamma }_{\text{cjs}}\sqrt{54}}{\mathrm{6h}{(\theta )}^{5}{s}_{\text{II}}^{3}}(\frac{\partial \mathrm{det}(\underline{\underline{s}})}{\partial {s}_{\text{ij}}})\end{array}\)

Finally:

\(\frac{\partial g}{\partial {s}_{\text{ij}}}=\frac{1}{h{(\theta )}^{5}}\left[(1+\frac{{\gamma }_{\text{cjs}}}{2}\mathrm{cos}(3\theta ))\frac{{s}_{\text{ij}}}{{s}_{\text{II}}}+\frac{{\gamma }_{\text{cjs}}\sqrt{54}}{{\mathrm{6s}}_{\text{II}}^{2}}(\frac{\partial \mathrm{det}(\underline{\underline{s}})}{\partial {s}_{\text{ij}}})\right]\)

And as a result:

\(\frac{\partial g}{\partial {s}_{\text{ij}}}=\frac{1}{h{(\theta )}^{5}}\left[(1+\frac{{\gamma }_{\text{cjs}}}{2}\mathrm{cos}(3\theta ))\frac{s}{{s}_{\text{II}}}+\frac{{\gamma }_{\text{cjs}}\sqrt{54}}{{\mathrm{6s}}_{\text{II}}^{2}}(\frac{\partial \mathrm{det}(\underline{\underline{s}})}{\partial {s}_{\text{ij}}})\right]\) eq 4.1.1.1-1

4.1.1.2. Intermediate derivative with respect to constraints#

We pose by definition: \({Q}_{\text{ij}}=\mathrm{dev}(\frac{\partial g}{\partial {s}_{\text{ij}}})\)

\(\frac{\partial g}{\partial {\sigma }_{\text{ij}}}=\frac{\partial g}{\partial {s}_{\text{kl}}}\frac{\partial {s}_{\text{kl}}}{\partial {\sigma }_{\text{ij}}}=\left[\mathrm{dev}(\frac{\partial g}{\partial {s}_{\text{kl}}})+\frac{1}{3}\frac{\partial g}{\partial {s}_{\text{mm}}}{\delta }_{\text{kl}}\right]\left[{\delta }_{\text{ik}}{\delta }_{\text{jl}}-\frac{1}{3}{\delta }_{\text{ij}}{\delta }_{\text{kl}}\right]\)

\(\frac{\partial g}{\partial {\sigma }_{\text{ij}}}={Q}_{\text{kl}}{\delta }_{\text{ik}}{\delta }_{\text{jl}}-\frac{1}{3}{\delta }_{\text{ij}}{Q}_{\text{kl}}{\delta }_{\text{kl}}+\frac{1}{3}\frac{\partial g}{\partial {q}_{\text{mm}}}\left[{\delta }_{\text{ik}}{\delta }_{\text{jl}}{\delta }_{\text{kl}}-\frac{1}{3}{\delta }_{\text{ij}}{\delta }_{\text{kl}}{\delta }_{\text{kl}}\right]\)

\(\frac{\partial g}{\partial {\sigma }_{\text{ij}}}={Q}_{\text{ij}}\)

All you have to do is take the deviatory part of \(\frac{\partial g}{\partial {s}_{\text{ij}}}\) to get:

\(\frac{\partial g}{\partial {\sigma }_{\text{ij}}}={Q}_{\text{ij}}=\mathrm{dev}(\frac{\partial g}{\partial {s}_{\text{ij}}})=\frac{1}{h{(\theta )}^{5}}\left[(1+\frac{{\gamma }_{\text{cjs}}}{2}\mathrm{cos}(3\theta ))\frac{{s}_{\text{ij}}}{{s}_{\text{II}}}\text{+}\frac{{\gamma }_{\text{cjs}}\sqrt{54}}{6{s}_{\text{II}}^{2}}\mathrm{dev}(\frac{\partial \mathrm{det}(\underline{\underline{s}})}{\partial {s}_{\text{ij}}})\right]\)

And as a result:

\(Q=\frac{\partial g}{\partial \sigma }=\frac{1}{h{(\theta )}^{5}}\left[(1+\frac{{\gamma }_{\text{cjs}}}{2}\mathrm{cos}(3\theta ))\frac{s}{{s}_{\text{II}}}\text{+}\frac{{\gamma }_{\text{cjs}}\sqrt{54}}{6{s}_{\text{II}}^{2}}\mathrm{dev}(\frac{\partial \mathrm{det}(\underline{\underline{s}})}{\partial s})\right]\) eq 4.1.1.2-1

4.1.1.3. Final expression of the derivative of the criterion with respect to the constraints#

The derivative of the criterion with respect to the constraints is then:

\(\frac{\partial f}{\partial \sigma }=\frac{1}{a({\gamma }^{p})}{\frac{1}{{\sigma }_{c}{h}_{c}^{0}}}^{\frac{1}{a({\gamma }^{p})}}(g){r}^{\frac{1-a({\gamma }^{p})}{a({\gamma }^{p})}}Q-\frac{\partial u}{\partial \sigma }\) eq 4.1.1.3-1

with

\(\frac{\partial u}{\partial \sigma }=-\frac{m({\gamma }^{p})k({\gamma }^{p})}{{\sigma }_{c}}(\frac{1}{\sqrt{6}{h}_{c}^{0}}Q+\frac{1}{3}I)\) eq 4.1.1.3-2

4.1.2. Derivative with respect to the work hardening variable#

\(\frac{\partial f}{\partial {\gamma }^{p}}=-{(\frac{1}{a({\gamma }^{p})})}^{2}{(\frac{g(s)}{{\sigma }_{c}{h}_{c}^{0}})}^{\frac{1}{a({\gamma }^{p})}}\mathrm{log}(\frac{g(s)}{{\sigma }_{c}{h}_{c}^{0}})\mathrm{.}\frac{\partial a}{\partial {\gamma }^{p}}-\frac{\partial u}{\partial {\gamma }^{p}}\) eq 4.1.2-1

with

\(\frac{\partial u}{\partial {\gamma }^{p}}=-\frac{1}{\sqrt{6}{\sigma }_{c}}\frac{\partial (\mathrm{km})}{\partial {\gamma }^{p}}({\gamma }^{p})\frac{g}{{h}_{c}^{0}}-\frac{1}{3{\sigma }_{c}}\frac{\partial (\mathrm{km})}{\partial {\gamma }^{p}}({\gamma }^{p}){I}_{1}+\frac{\partial (\mathrm{ks})}{\partial {\gamma }^{p}}({\gamma }^{p})\) eq 4.1.2-2

4.2. Total derivative of the criterion with respect to the plastic multiplier#

Let’s consider the function:

\({f}^{}(\Delta \lambda )=f({s}^{e}-2\mu \Delta \lambda \tilde{G},{I}_{1}^{e}-\mathrm{3K}\Delta \lambdag ,{\gamma }^{{p}^{\text{-}}}+\Delta \lambda \sqrt{\frac{2}{3}}\tilde{{G}_{\text{II}}})\) eq 4.2-1

Where

_images/Object_139.svg

is a fixed tensor independent of. \(\Delta \lambda\) It is from this function that we are looking for the zero to find the stress state:

\(\frac{\partial f\text{*}}{\partial \Delta \lambda }=-\frac{\partial f}{\partial \sigma }\mathrm{.}(2\mu \tilde{G}+\mathrm{KG}I)+\frac{\partial f}{\partial {\gamma }^{p}}\sqrt{\frac{2}{3}}\tilde{{G}_{\text{II}}}\) eq 4.2-2

4.3. Derivatives of the parameters with respect to the work hardening variable#

\(\{\begin{array}{}\frac{\partial s}{\partial {\gamma }^{p}}=-\frac{1}{{\gamma }_{e}}\text{}\mathrm{si}{\gamma }^{p}<{\gamma }_{e}\\ \frac{\partial s}{\partial {\gamma }^{p}}=\text{0}\mathrm{si}{\gamma }^{p}\ge {\gamma }_{e}\end{array}\) eq 4.3-1

\(\{\begin{array}{}\frac{\partial m}{\partial s}=-\frac{{\sigma }_{c}}{{\sigma }_{\text{p1}}}\text{}\mathrm{si}{\gamma }^{p}<{\gamma }_{e}\\ \frac{\partial m}{\partial s}=\text{0}\mathrm{si}{\gamma }^{p}\ge {\gamma }_{e}\end{array}\) eq 4.3-2

\(\{\begin{array}{}\frac{\partial m}{\partial a}=-\frac{{\sigma }_{c}}{{\sigma }_{\text{p1}}}\mathrm{log}({m}_{\text{pic}}\frac{{\sigma }_{\text{p1}}}{{\sigma }_{c}}+1)\frac{{a}_{\text{pic}}}{{a}^{2}}{({m}_{\text{pic}}\frac{{\sigma }_{\text{p1}}}{{\sigma }_{c}}+1)}^{\frac{{a}_{\text{pic}}}{a}}\text{si}{\gamma }^{p}<{\gamma }_{e}\\ \frac{\partial m}{\partial a}=-\frac{{\sigma }_{c}}{{\sigma }_{\text{p2}}}\mathrm{log}({m}_{e}\frac{{\sigma }_{\text{p2}}}{{\sigma }_{c}})\frac{{a}_{e}}{{a}^{2}}{({m}_{e}\frac{{\sigma }_{\text{p2}}}{{\sigma }_{c}})}^{\frac{{a}_{\text{pic}}}{a}}\text{si}{\gamma }^{p}<{\gamma }_{e}\end{array}\) eq 4.3-3

\(\frac{\partial \Omega }{\partial {\gamma }^{p}}=\frac{({\gamma }_{\text{ult}}-{\gamma }_{e})}{{({\gamma }_{e})}^{\eta }}\frac{{a}_{e}-{a}_{\text{pic}}}{1-{a}_{e}}(\frac{\eta }{{\gamma }_{\text{ult}}-{\gamma }^{p}}{({\gamma }^{p})}^{\eta -1}+{({\gamma }^{p})}^{\eta }\frac{1}{{({\gamma }_{\text{ult}}-{\gamma }^{p})}^{2}})\) eq 4.3-4

\(\frac{\partial a}{\partial \Omega }=\frac{1-{a}_{\text{pic}}}{{(1+\Omega )}^{2}}\) eq 4.3-5

\(\begin{array}{cc}\frac{\partial m}{\partial {\gamma }^{p}}=\frac{\partial m}{\partial a}\frac{\partial a}{\partial {\gamma }^{p}}+\frac{\partial m}{\partial s}\frac{\partial s}{\partial {\gamma }^{p}}& \text{si}{\gamma }^{p}<{\gamma }_{e}\\ \frac{\partial m}{\partial {\gamma }^{p}}=\frac{\partial m}{\partial a}\frac{\partial a}{\partial {\gamma }^{p}}& \text{si}{\gamma }_{\text{ult}}(1-{10}^{\text{-3}})>{\gamma }^{p}\ge {\gamma }_{e}\\ \frac{\partial m}{\partial {\gamma }^{p}}=0& \text{si}{\gamma }_{\text{ult}}(1-{10}^{\text{-3}})<{\gamma }^{p}\end{array}\) eq 4.3-6

\(\begin{array}{cc}\frac{\partial k}{\partial {\gamma }^{p}}=-{(\frac{2}{3})}^{\frac{1}{\mathrm{2a}}}\mathrm{log}(\frac{2}{3})\frac{1}{{\mathrm{2a}}^{2}}\frac{\partial a}{\partial {\gamma }^{p}}& {\gamma }_{\text{ult}}(1-{10}^{\text{-3}})>{\gamma }^{p}\\ \frac{\partial k}{\partial {\gamma }^{p}}=0& \text{sinon}\end{array}\) eq 4.3-7