11. Description of document versions#

Version Aster

Author (s) Organization (s)

Description of changes

8.5

C.Chavant, J.El-Gharib EDF -R&D/ AMA V.Gervais, CS

Initial text

Derivation of main stress

Let \(\sigma\) be a symmetric tensor and \({\sigma }_{d}\) this tensor in the base that diagonalizes it. Let’s denote by \(P(\sigma )\) the passage matrix that diagonalizes the \(\mathrm{\sigma }\): \(\sigma \text{:}\sigma \text{=}P(\sigma )\text{.}{\sigma }_{d}\text{.}\tilde{P}(\sigma )\) tensor. In index writing, we adopt the following convention for raster entries:

_images/10000234000069D500002972DB170B0A78C3E88D.svg

so that the matrix product is written: \({(A\text{.}B)}_{j}^{i}\text{=}{A}_{m}^{i}{B}_{j}^{m}\) with the rule of summation of repeated indices.

So we have the relationship:

\(\frac{\partial {\sigma }_{d}}{\partial {\sigma }_{j}^{i}}\text{=}\tilde{P}(\sigma )\text{.}\frac{\partial \sigma }{\partial {\sigma }_{j}^{i}}\text{.}P(\sigma )\)

\(\frac{\partial {\sigma }_{{d}_{k}}}{\partial {\sigma }_{j}^{i}}={P}_{k}^{i}{P}_{k}^{j}\) or in index form*without index indication* \(k\)

Demonstration:

In what follows, we’ll note \(\mathrm{\sigma }\) any component of the \(\sigma\) tensor without specifying the indices when they don’t play a role.

We have \(\sigma \text{=}P(\sigma )\text{.}{\sigma }_{d}\text{.}\tilde{P}(\sigma )\), and as a result:

  1. \(\frac{\partial {\sigma }_{d}}{\partial \sigma }\text{=}\tilde{P}(\sigma )\text{.}\frac{\partial \sigma }{\partial \sigma }\text{.}P(\sigma )+\frac{\partial \tilde{P}(\sigma )}{\partial \sigma }\text{.}\sigma \text{.}P(\sigma )+\tilde{P}(\sigma )\text{.}\sigma \text{.}\frac{P(\sigma )}{\partial \sigma }\)

2..

By reporting the \(\sigma \text{=}P(\sigma )\text{.}{\sigma }_{d}\text{.}\tilde{P}(\sigma )\) equality in the last two terms, we get:

  1. \(\frac{\partial {\sigma }_{d}}{\partial \sigma }\text{=}\tilde{P}(\sigma )\text{.}\frac{\partial \sigma }{\partial \sigma }\text{.}P(\sigma )+\frac{\partial P(\sigma )}{\partial \sigma }\text{.}\tilde{P}(\sigma )\text{.}{\sigma }_{d}\text{.}P(\sigma )\text{.}\tilde{P}(\sigma )+\tilde{P}(\sigma )\text{.}P(\sigma )\text{.}{\sigma }_{d}\text{.}\tilde{P}(\sigma )\text{.}\frac{\partial P(\sigma )}{\partial \sigma }\)

that is to say:

  1. \(\frac{\partial {\sigma }_{d}}{\partial \sigma }=\tilde{P}(\sigma )\text{.}\frac{\partial \sigma }{\partial \sigma }\text{.}P(\sigma )+\frac{\partial \tilde{P}(\sigma )}{\partial \sigma }\text{.}P(\sigma )\text{.}{\sigma }_{d}+{\sigma }_{d}\text{.}\tilde{P}(\sigma )\text{.}\frac{\partial P(\sigma )}{\partial \sigma }\)

In matrix writing, this is written:

\(\frac{\partial {\sigma }_{{}_{d}}^{i}}{\partial \sigma }\text{=}{(\tilde{P}(\sigma )\text{.}\frac{\partial \sigma }{\partial \sigma }\text{.}P(\sigma )+\frac{\partial \tilde{P}(\sigma )}{\partial \sigma }\text{.}P(\sigma )\text{.}{\sigma }_{d}+{\sigma }_{d}\text{.}\tilde{P}(\sigma )\text{.}\frac{\partial P(\sigma )}{\partial \sigma })}_{i}^{i}\)

Let’s show that the sum of the last two terms in this expression is zero:

  • \({(\frac{\partial \tilde{P}(\sigma )}{\partial \sigma }\text{.}P(\sigma )\text{.}{\sigma }_{d}+{\sigma }_{d}\text{.}\tilde{P}(\sigma )\text{.}\frac{\partial P(\sigma )}{\partial \sigma })}_{j}^{i}=\partial {\tilde{P}}_{m}^{i}\text{.}{P}_{l}^{m}\text{.}{\sigma }_{{d}_{j}^{l}}+{\sigma }_{{d}_{m}^{i}}\text{.}{\tilde{P}}_{l}^{m}\text{.}\partial {P}_{j}^{l}\)

where \(\partial a\) refers to \(\frac{\partial a}{\partial \mathrm{\sigma }}\) in order to simplify the writing.

We then write that \(i=j\) and that only the terms \({\sigma }_{{d}_{p}}^{p}\) are non-zero.

We get:

\(\begin{array}{}{(\frac{\partial \tilde{P}(\sigma )}{\partial \sigma }\text{.}P(\sigma )\text{.}{\sigma }_{d}+{\sigma }_{d}\text{.}\tilde{P}(\sigma )\text{.}\frac{\partial P(\sigma )}{\partial \sigma })}_{i}^{i}\text{=}\partial {\tilde{P}}_{m}^{i}\text{.}{P}_{i}^{m}\text{.}{\sigma }_{{d}_{i}^{i}}+{\sigma }_{{d}_{i}^{i}}\text{.}{\tilde{P}}_{l}^{i}\text{.}\partial {P}_{i}^{l}\\ \text{=}(\partial {\tilde{P}}_{m}^{i}\text{.}{P}_{i}^{m}+{\tilde{P}}_{m}^{i}\text{.}\partial {P}_{i}^{m}){\sigma }_{{d}_{i}^{i}}\underline{\text{sans}\text{sommation}\text{sur}\text{l'indice}i}\end{array}\)

which is clearly rubbish since \(\tilde{P}\text{.}P\text{=}I\). Hence finally \(\frac{\partial {\mathrm{\sigma }}_{{d}_{k}}}{\partial {\mathrm{\sigma }}_{j}^{i}}=\frac{\partial {\mathrm{\sigma }}_{{d}_{k}^{k}}}{\partial {\mathrm{\sigma }}_{j}^{i}}={P}_{k}^{i}{P}_{k}^{j}\).

Derived from the function \(\stackrel{ˉ}{F}(\mathrm{\Delta \lambda })\)

\(\begin{array}{}\frac{\partial h}{\partial \Delta \gamma }\text{=}\frac{\partial }{\partial \Delta \gamma }(\frac{3\text{+}\text{sin}\phi ({\gamma }^{\text{-}}\text{+}\Delta \gamma )}{3(1\text{+}\text{sin}\phi ({\gamma }^{\text{-}}\text{+}\Delta \gamma ))})\text{=}\{\begin{array}{cc}\text{-}\frac{2\frac{\partial \phi }{\partial \gamma }\text{cos}\phi (\gamma )}{3(1\text{+}\text{sin}\phi (\gamma ){)}^{2}}& \text{si}\gamma <{\gamma }^{\text{res}}\\ 0& \text{sinon}\end{array}\text{}\\ \frac{\partial g}{\partial \Delta \gamma }\text{=}\frac{\partial }{\partial \Delta \gamma }\left[\frac{3\text{+}\text{sin}\phi ({\gamma }^{\text{-}}\text{+}\Delta \gamma )}{3(1\text{+}\text{sin}\phi ({\gamma }^{\text{-}}\text{+}\Delta \gamma ))}(\frac{\mathrm{6K}\text{sin}\phi ({\gamma }^{\text{-}}\text{+}\Delta \gamma )}{3\text{+}\text{sin}\phi ({\gamma }^{\text{-}}\text{+}\Delta \gamma )}\text{+}\frac{3{m}_{3}^{e}}{{\sigma }_{\text{eq}}^{e}})\right]\\ \text{}\text{=}\frac{\mathrm{6K}\frac{\partial \phi }{\partial \gamma }\text{cos}\phi (\gamma )}{(3\text{+}\text{sin}\phi (\gamma ))(1\text{+}\text{sin}\phi (\gamma ))}\text{-}\frac{2\frac{\partial \phi }{\partial \gamma }\text{cos}\phi (\gamma )}{3(1\text{+}\text{sin}\phi (\gamma ){)}^{2}}(\frac{\mathrm{6K}\text{sin}\phi (\gamma )}{3\text{+}\text{sin}\phi (\gamma )}\text{+}\frac{3{m}_{3}^{e}}{{\sigma }_{\text{eq}}^{e}})\end{array}\)

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

Calculating derivatives \(\frac{\partial {\sigma }_{p}}{\partial {p}_{g}}\) and \(\frac{\partial {\sigma }_{p}}{\partial {p}_{c}}\)

  • LIQU_SATU (PRE1 = \({p}_{\mathrm{lq}}\)): \(\frac{\partial {\sigma }_{p}}{\partial {p}_{c}}\text{=}\text{-}\frac{\partial {\sigma }_{p}}{\partial {p}_{\text{lq}}}\text{=}\mathrm{bS}\)

  • LIQU_GAZ_ATM (PRE1 =- \({p}_{\mathrm{lq}}\)): \(\frac{\partial {\sigma }_{p}}{\partial {p}_{c}}\text{=}\text{-}\frac{\partial {\sigma }_{p}}{\partial {p}_{\text{lq}}}\text{=}\mathrm{bS}\)

  • GAZ (PRE1 = \({p}_{g}\)): \(\frac{\partial {\sigma }_{p}}{\partial {p}_{g}}\text{=}\text{-}b(1\text{-}S)\)

  • LIQU_VAPE_GAZ (PRE1 = \({p}_{c}\), PRE2 = \({p}_{g}\)): \(\frac{\partial {\sigma }_{p}}{\partial {p}_{g}}\text{=}\text{-}b\), \(\frac{\partial {\sigma }_{p}}{\partial {p}_{c}}\text{=}\mathrm{bS}\)

  • LIQU_GAZ (PRE1 = \({p}_{c}\), PRE2 = \({p}_{g}\)): \(\frac{\partial {\sigma }_{p}}{\partial {p}_{g}}\text{=}\text{-}b\), \(\frac{\partial {\sigma }_{p}}{\partial {p}_{c}}\text{=}\mathrm{bS}\)

  • LIQU_VAPE (PRE1 = \({p}_{\mathrm{lq}}\)): \(\frac{\partial {\sigma }_{p}}{\partial {p}_{c}}\text{=}\text{-}\frac{\partial {\sigma }_{p}}{\partial {p}_{\text{lq}}}\text{=}\mathrm{bS}\)

  • LIQU_AD_GAZ_VAPE (PRE1 =

    _images/Object_314.svg

, PRE2 =

_images/Object_315.svg

): \(\frac{\partial {\sigma }_{p}}{\partial {p}_{g}}\text{=}\text{-}b\), \(\frac{\partial {\sigma }_{p}}{\partial {p}_{c}}\text{=}\mathrm{bS}\)