From: frohne Date: Wed, 30 Jan 2013 18:04:36 +0000 (+0000) Subject: extension of section 7 X-Git-Url: https://gitweb.dealii.org/cgi-bin/gitweb.cgi?a=commitdiff_plain;h=7a8b9dc224d7edf8625fb89a3bd026027974e2f3;p=dealii-svn.git extension of section 7 git-svn-id: https://svn.dealii.org/trunk@28198 0785d39b-7218-0410-832d-ea1e28bc413d --- diff --git a/deal.II/examples/step-42/doc/intro-step-42.tex b/deal.II/examples/step-42/doc/intro-step-42.tex index b8af117870..854946cd40 100644 --- a/deal.II/examples/step-42/doc/intro-step-42.tex +++ b/deal.II/examples/step-42/doc/intro-step-42.tex @@ -52,7 +52,7 @@ $$\mathcal{F}(\tau) = \vert\tau^D\vert - \sigma_0$$ with $\sigma_0$ as yield stress. If there is no plastic deformation - that is $\lambda=0$ - this yields $\vert\sigma^D\vert < \sigma_0$ and otherwise if $\lambda > 0$ it follows that $\vert\sigma^D\vert = \sigma_0$. That means if the stress is smaller as the yield stress there are only elastic deformations. Therein the Index $D$ denotes the deviator part of the stress $\sigma$ which -is dedined as +is defined as $$\sigma^D = \sigma - \dfrac{1}{3}tr(\sigma).$$ It describes the hydrostatic part of the stress tensor in contrast to the volumetric part. For metal the hydrostatic stress composes the main indicator for the plastic deformation.\\ @@ -196,8 +196,8 @@ as the additional, positive force $\sigma(u)n$ that the obstacle exerts on the boundary $\Gamma_C$ of the body.\\ \noindent -The existence and uniqueness of $(u,\lambda)\in V\times K$ of this saddle point -problem has been stated in Glowinski, Lions and Tr\'{e}moli\`{e}res: Numerical +The existence and uniqueness of the analytical solution $(u,\lambda)\in V\times +K$ of this saddle point problem has been stated in Glowinski, Lions and Tr\'{e}moli\`{e}res: Numerical Analysis of Variational Inequalities, North-Holland, 1981.\\ \noindent @@ -219,7 +219,7 @@ Newton method} Now we describe an algorithm that combines the Newton-method, which we use for the nonlinear constitutive law, with the semismooth Newton method for the contact. It -works as follows: +sums up the results of the sections before and works as follows: \begin{itemize} \item[(0)] Initialize $\mathcal{A}_k$ and $\mathcal{F}_k$, such that $\mathcal{S} = \mathcal{A}_k \cup \mathcal{F}_k$ and $\mathcal{A}_k \cap \mathcal{F}_k = \emptyset$ and set $k = 1$. \item[(1)] Assemble the Newton matrix $a'(U^k;\varphi_i,\varphi_j)$ and the right-hand-side $F(U^k)$. @@ -239,6 +239,12 @@ works as follows: F\left(U^{k+1}\right) \vert < \delta$ then stop, else set $k=k+1$ and go to step (1). \end{itemize} +\noindent +Compare to step-41 step (1) is added but it should be clear from the sections +above that we only linearize the problem. In step (2) we have to solve a linear +system of equations again. And now the solution has to fulfill two stopping criterias. $\mathcal{A}_{k+1} = \mathcal{A}_k$ makes sure that the contact zones +are iterated and the second ensures an accurate enough residual which means that +the plastic zones are also iterated. \section{Implementation}