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## [Getfem-commits] (no subject)

 From: Konstantinos Poulios Subject: [Getfem-commits] (no subject) Date: Sun, 27 May 2018 09:28:14 -0400 (EDT)

branch: devel-logari81-plasticity
commit ca00b69be06ee2e49357aeae3df0ffd7af650666
Author: Konstantinos Poulios <address@hidden>
Date:   Sun May 27 15:27:00 2018 +0200

Fix kinematic hardening modulus definition in small strain plasticity
documentation
---
.../userdoc/model_plasticity_small_strain.rst      | 24 +++++++++++-----------
1 file changed, 12 insertions(+), 12 deletions(-)

diff --git a/doc/sphinx/source/userdoc/model_plasticity_small_strain.rst
b/doc/sphinx/source/userdoc/model_plasticity_small_strain.rst
index 634a2d5..b872c59 100644
--- a/doc/sphinx/source/userdoc/model_plasticity_small_strain.rst
+++ b/doc/sphinx/source/userdoc/model_plasticity_small_strain.rst
@@ -374,26 +374,26 @@ i.e. :math:A = H_i\alpha and a uniaxial yield stress
defined by

for :math:\sigma_{y0} the initial uniaxial yield stress. The yield function
(and plastic potential since this is an associated plastic model) can be
defined by

-.. math:: \Psi(\sigma, A) = f(\sigma, A) = \|\mbox{Dev}(\sigma -
H_k\varepsilon^p)\| - \sqrt{\frac{2}{3}}(\sigma_{y0} + A),
+.. math:: \Psi(\sigma, A) = f(\sigma, A) = \|\mbox{Dev}(\sigma -
\frac{2}{3}H_k\varepsilon^p)\| - \sqrt{\frac{2}{3}}(\sigma_{y0} + A),

where :math:H_k is the kinematic hardening modulus. The same computation as
in the previous section leads to

-.. math:: \tilde{\mathscr E}^p(u_{n+1}, \theta\Delta t \xi_{n+1}, \zeta_n) =
\zeta_n + \Frac{1}{2\mu+H_k}\left(1 - \Frac{1}{1+(2\mu+H_k)\theta\Delta
t\xi_{n+1}}\right)(2\mu\mbox{Dev}(\varepsilon(u_{n+1}))-(2\mu+H_k)\zeta_n)
+.. math:: \tilde{\mathscr E}^p(u_{n+1}, \theta\Delta t \xi_{n+1}, \zeta_n) =
\zeta_n + \Frac{1}{2(\mu+H_k/3)}\left(1 - \Frac{1}{1+2(\mu+H_k/3)\theta\Delta
t\xi_{n+1}}\right)(2\mu\mbox{Dev}(\varepsilon(u_{n+1}))-2(\mu+H_k/3)\zeta_n)

-.. math:: \begin{array}{rcl} \tilde{\mathscr A}(u_{n+1}, \theta \Delta t
\xi_{n+1}, \zeta_{n}, \eta_n) &=& \eta_n + \sqrt{\Frac{2}{3}} \theta \Delta t
\xi_{n+1}\|\mbox{Dev}(\sigma_{n+1} - H_k\varepsilon^p_{n+1})\| \\ &=& \eta_n +
\sqrt{\Frac{2}{3}} \theta \Delta t
\xi_{n+1}\|2\mu\mbox{Dev}(\varepsilon(u_{n+1})) -
(2\mu+H_k)\varepsilon^p_{n+1}\| \\ &=&  \eta_n + \sqrt{\Frac{2}{3}}
\Frac{\theta \Delta t \xi_{n+1}}{1+(2\mu+H_k)\theta\Delta
t\xi_{n+1}}\|2\mu\mbox{Dev}(\varepsilon(u_{n+1})) - [...]
+.. math:: \begin{array}{rcl} \tilde{\mathscr A}(u_{n+1}, \theta \Delta t
\xi_{n+1}, \zeta_{n}, \eta_n) &=& \eta_n + \sqrt{\Frac{2}{3}} \theta \Delta t
\xi_{n+1}\|\mbox{Dev}(\sigma_{n+1} - \frac{2}{3}H_k\varepsilon^p_{n+1})\| \\
&=& \eta_n + \sqrt{\Frac{2}{3}} \theta \Delta t
\xi_{n+1}\|2\mu\mbox{Dev}(\varepsilon(u_{n+1})) -
2(\mu+H_k/3)\varepsilon^p_{n+1}\| \\ &=&  \eta_n + \sqrt{\Frac{2}{3}}
\Frac{\theta \Delta t \xi_{n+1}}{1+2(\mu+H_k/3)\theta\Delta
t\xi_{n+1}}\|2\mu\mbox{Dev}(\varepsi [...]

where :math:\zeta_n and :math:\eta_n are defined by

-.. math:: \zeta_n = \varepsilon^p_n+(1-\theta)\Delta t \xi_n
(\mbox{Dev}(\sigma_n)-H_k\varepsilon^n_p) = \varepsilon^p_n+(1-\theta)\Delta t
\xi_n \left(2\mu\mbox{Dev}(\varepsilon(u_{n}))-(2\mu+H_k)\varepsilon^n_p\right),
+.. math:: \zeta_n = \varepsilon^p_n+(1-\theta)\Delta t \xi_n
(\mbox{Dev}(\sigma_n)-\frac{2}{3}H_k\varepsilon^n_p) =
\varepsilon^p_n+(1-\theta)\Delta t \xi_n
\left(2\mu\mbox{Dev}(\varepsilon(u_{n}))-2(\mu+H_k/3)\varepsilon^n_p\right),

-.. math:: \eta_n  = \alpha_n+(1-\theta)\sqrt{\Frac{2}{3}}\Delta t \xi_n
\|\mbox{Dev}(\sigma_n)-H_k\varepsilon^n_p\| =
\alpha_n+(1-\theta)\sqrt{\Frac{2}{3}}\Delta t \xi_n
\|2\mu\mbox{Dev}(\varepsilon(u_{n}))-(2\mu+H_k)\varepsilon^n_p\|.
+.. math:: \eta_n  = \alpha_n+(1-\theta)\sqrt{\Frac{2}{3}}\Delta t \xi_n
\|\mbox{Dev}(\sigma_n)-\frac{2}{3}H_k\varepsilon^n_p\| =
\alpha_n+(1-\theta)\sqrt{\Frac{2}{3}}\Delta t \xi_n
\|2\mu\mbox{Dev}(\varepsilon(u_{n}))-2(\mu+H_k/3)\varepsilon^n_p\|.

Note that the isotropic hardening modulus do not intervene in
:math:\tilde{\mathscr E}^p(u_{n+1}, \theta \Delta \xi, \varepsilon^p_{n}) but
only in :math:f(\sigma, A).

**Elimination of the multiplier (for the return mapping approach)**

-Denoting :math:\delta = \Frac{1}{1+(2\mu+H_k)\theta\Delta t\xi_{n+1}},
:math:\beta = \Frac{1-\delta}{2\mu+H_k} and :math:B =
2\mu\mbox{Dev}(\varepsilon(u_{n+1}))-(2\mu+H_k)\zeta_n the expression for
:math:\varepsilon^p_{n+1} and :math:\alpha_{n+1} becomes
+Denoting :math:\delta = \Frac{1}{1+2(\mu+H_k/3)\theta\Delta t\xi_{n+1}},
:math:\beta = \Frac{1-\delta}{2(\mu+H_k/3)} and :math:B =
2\mu\mbox{Dev}(\varepsilon(u_{n+1}))-2(\mu+H_k/3)\zeta_n the expression for
:math:\varepsilon^p_{n+1} and :math:\alpha_{n+1} becomes

.. math:: \varepsilon^p_{n+1} = \zeta_n+\beta B, ~~~ \alpha_{n+1} = \eta_n +
\sqrt{\Frac{2}{3}}\beta \|B\|,
:label: hardeningepsalp
@@ -408,24 +408,24 @@ Thus, either we are in the elastic case, i.e.
:math:\xi_{n+1} = 0, \delta = 1

or we are in the plastic case and :math:\xi_{n+1} > 0, \delta < 1,
:math:\delta \|B\| = \sqrt{\Frac{2}{3}}(\sigma_{y0}+H_i \alpha_{n+1}) and
:math:(1-\delta) solves the equation

-.. math:: \|B\| - (1-\delta)\|B\| = \sqrt{\Frac{2}{3}}\left(\sigma_{y0}+H_i
\eta_n + \sqrt{\Frac{2}{3}} \Frac{H_i}{2\mu+H_k}(1-\delta)\|B\|\right),
+.. math:: \|B\| - (1-\delta)\|B\| = \sqrt{\Frac{2}{3}}\left(\sigma_{y0}+H_i
\eta_n + \sqrt{\Frac{2}{3}} \Frac{H_i}{2(\mu+H_k/3)}(1-\delta)\|B\|\right),

-.. math:: 1-\delta =
\Frac{2\mu+H_k}{\|B\|(2\mu+H_k+\frac{2}{3}H_i)}\left(\|B\|-\sqrt{\Frac{2}{3}}(\sigma_{y0}+H_i
\eta_n) \right)
+.. math:: 1-\delta =
\Frac{2(\mu+H_k/3)}{\|B\|(2\mu+\frac{2}{3}(H_k+H_i))}\left(\|B\|-\sqrt{\Frac{2}{3}}(\sigma_{y0}+H_i
\eta_n) \right)

The two cases can be summarized by

-.. math:: \beta =
\Frac{1}{\|B\|(2\mu+H_k+\frac{2}{3}H_i)}\left(\|B\|-\sqrt{\Frac{2}{3}}(\sigma_{y0}+H_i
\eta_n) \right)_+
+.. math:: \beta =
\Frac{1}{\|B\|(2\mu+\frac{2}{3}(H_k+H_i))}\left(\|B\|-\sqrt{\Frac{2}{3}}(\sigma_{y0}+H_i
\eta_n) \right)_+

which directly gives :math:{\mathscr E}^p(u_{n+1}, \zeta_n, \eta_n) and
:math:{\mathscr A}(u_{n+1}, \zeta_n, \eta_n) thanks to :eq:hardeningepsalp.
The multiplier :math:\xi_{n+1} being given by

-.. math:: \xi_{n+1} = \Frac{1}{(2\mu+H_k)\theta\Delta t}(\Frac{1}{\delta}-1) =
\Frac{1}{\theta\Delta t}~\Frac{\beta}{1-(2\mu+H_k)\beta}.
+.. math:: \xi_{n+1} = \Frac{1}{(2(\mu+H_k/3))\theta\Delta
t}(\Frac{1}{\delta}-1) = \Frac{1}{\theta\Delta
t}~\Frac{\beta}{1-2(\mu+H_k/3)\beta}.

**Plane strain approximation**

-Still denoting  :math:\delta = \Frac{1}{1+(2\mu+H_k)\theta\Delta
t\xi_{n+1}}, :math:\beta = \Frac{1-\delta}{2\mu+H_k}, :math:B =
2\mu\mbox{Dev}(\varepsilon(u_{n+1}))-(2\mu+H_k)\zeta_n and :math:\overline{B}
= 2\mu\overline{Dev}(\bar{\varepsilon}(u_{n+1}))-(2\mu+H_k)\bar{\zeta}_n its
in-plane part, one has
+Still denoting  :math:\delta = \Frac{1}{1+2(\mu+H_k/3)\theta\Delta
t\xi_{n+1}}, :math:\beta = \Frac{1-\delta}{2(\mu+H_k/3)}, :math:B =
2\mu\mbox{Dev}(\varepsilon(u_{n+1}))-2(\mu+H_k/3)\zeta_n and
:math:\overline{B} =
2\mu\overline{Dev}(\bar{\varepsilon}(u_{n+1}))-2(\mu+H_k/3)\bar{\zeta}_n its
in-plane part, one has

.. math:: \bar{\tilde{\mathscr E}}^p(u_{n+1}, \theta\Delta t \xi_{n+1},
\bar{\zeta}_n) = \bar{\zeta}_n + \beta \overline{B},

@@ -434,7 +434,7 @@ Still denoting  :math:\delta =
\Frac{1}{1+(2\mu+H_k)\theta\Delta t\xi_{n+1}},

with

-.. math:: \|B\|^2 = \|2\mu\overline{\mbox{Dev}}(\bar{\varepsilon}(u_{n+1})) -
(2\mu+H_k)\bar{\zeta}_n\|^2 +
\left(2\mu\Frac{\mbox{tr}(\bar{\varepsilon}(u_{n+1}))}{3}
-(2\mu+H_k)\mbox{tr}(\bar{\zeta}_n) \right)^2.
+.. math:: \|B\|^2 = \|2\mu\overline{\mbox{Dev}}(\bar{\varepsilon}(u_{n+1})) -
2(\mu+H_k/3)\bar{\zeta}_n\|^2 +
\left(2\mu\Frac{\mbox{tr}(\bar{\varepsilon}(u_{n+1}))}{3}
-2(\mu+H_k/3)\mbox{tr}(\bar{\zeta}_n) \right)^2.

The yield condition still reads



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