[Getfem-commits] (no subject)

[Konstantinos Poulios]

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),
 
 which leads to
 
-.. 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