From 9e74a51e4a5919a2f2ade4958166b9f3f43b4255 Mon Sep 17 00:00:00 2001 From: frohne Date: Thu, 21 Feb 2013 16:32:58 +0000 Subject: [PATCH] commit again because of an wrong revert git-svn-id: https://svn.dealii.org/trunk@28512 0785d39b-7218-0410-832d-ea1e28bc413d --- deal.II/examples/step-42/Makefile | 2 +- .../examples/step-42/doc/intro-step-42.tex | 48 +++++++++++++------ 2 files changed, 34 insertions(+), 16 deletions(-) diff --git a/deal.II/examples/step-42/Makefile b/deal.II/examples/step-42/Makefile index 1c6563053e..4a005b2d7f 100644 --- a/deal.II/examples/step-42/Makefile +++ b/deal.II/examples/step-42/Makefile @@ -14,7 +14,7 @@ target = step-42 # run-time checking of parameters and internal states is performed, so # you should set this value to `on' while you develop your program, # and to `off' when running production computations. -debug-mode = off +debug-mode = on # As third field, we need to give the path to the top-level deal.II 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 980227e73e..b92a1d6b61 100644 --- a/deal.II/examples/step-42/doc/intro-step-42.tex +++ b/deal.II/examples/step-42/doc/intro-step-42.tex @@ -80,7 +80,8 @@ $$ \sigma_n = \sigma n\cdot n$$ is zero with the outward normal $n$. If there is contact ($u_n = g$) the tangential stress $\sigma_t = \sigma\cdot n - \sigma_n n$ vanishes, because we consider a frictionless situation and the normal stress is negative. The gap $g$ comes with the start configuration of the obstacle and the -deformable body. +deformable body. We refer that you have to ensure that the obstacle does not hit +the boundary of $\Gamma_C$ \section{Derivation of the variational inequality} @@ -115,14 +116,14 @@ Therein $\varepsilon$ denotes the linearised deformation tensor with $\varepsilo Most materials - especially metals - have the property that they show some hardening effects during the forming process. There are different constitutive laws to describe those material behaviors. The simplest one is called linear isotropic hardening described by the flow function -$\mathcal{F}(\tau,\eta) = \vert\tau^D\vert - (\sigma_0 + \gamma\eta)$ where +$\mathcal{F}(\tau,\eta) = \vert\tau^D\vert - (\sigma_0 + \gamma^{\text{iso}}\eta)$ where $\eta$ is the norm of the plastic strain $\eta = \vert \varepsilon - A\sigma\vert$. It can be considered by establishing an additional term in our primal-mixed formulation:\\ Find a pair $\lbrace(\sigma,\xi),u\rbrace\in \Pi (W\times L^2(\Omega,\mathbb{R}))\times V^+$ with -$$\left(\sigma,\tau - \sigma\right) - \left(C\varepsilon(u), \tau - \sigma\right) + \gamma\left( \xi, \eta - \xi\right) \geq 0,\quad \forall (\tau,\eta)\in \Pi (W,L^2(\Omega,\mathbb{R}))$$ +$$\left(\sigma,\tau - \sigma\right) - \left(C\varepsilon(u), \tau - \sigma\right) + \gamma^{\text{iso}}\left( \xi, \eta - \xi\right) \geq 0,\quad \forall (\tau,\eta)\in \Pi (W,L^2(\Omega,\mathbb{R}))$$ $$\left(\sigma,\varepsilon(\varphi) - \varepsilon(u)\right) \geq 0,\quad \forall \varphi\in V^+,$$ -with the hardening parameter $\gamma > 0$.\\ +with the hardening parameter $\gamma^{\text{iso}} > 0$.\\ Now we want to derive a primal problem which only depends on the displacement $u$. For that purpose we set $\eta = \xi$ and eliminate the stress $\sigma$ by applying the projection theorem (see Grossmann, Roos: Numerical Treatment of Partial Differential @@ -135,21 +136,34 @@ Find the displacement $u\in V^+$ with $$\left(P_{\Pi}(C\varepsilon(u)),\varepsilon(\varphi) - \varepsilon(u)\right) \geq 0,\quad \forall \varphi\in V^+,$$ with the projection: $$P_{\Pi}(\tau):=\begin{cases} - \tau, & \text{if }\vert\tau^D\vert \leq \sigma_0 + \gamma\xi,\\ - \hat\alpha\dfrac{\tau^D}{\vert\tau^D\vert} + \dfrac{1}{3}tr(\tau), & \text{if }\vert\tau^D\vert > \sigma_0 + \gamma\xi, + \tau, & \text{if }\vert\tau^D\vert \leq \sigma_0 + \gamma^{\text{iso}}\xi,\\ + \hat\alpha\dfrac{\tau^D}{\vert\tau^D\vert} + \dfrac{1}{3}tr(\tau), & \text{if }\vert\tau^D\vert > \sigma_0 + \gamma^{\text{iso}}\xi, \end{cases}$$ with the radius -$$\hat\alpha := \sigma_0 + \gamma\xi .$$ +$$\hat\alpha := \sigma_0 + \gamma^{\text{iso}}\xi .$$ With the relation $\xi = \vert\varepsilon(u) - A\sigma\vert$ it is possible to eliminate $\xi$ inside the projection $P_{\Pi}$:\\ $$P_{\Pi}(\tau):=\begin{cases} \tau, & \text{if }\vert\tau^D\vert \leq \sigma_0,\\ \alpha\dfrac{\tau^D}{\vert\tau^D\vert} + \dfrac{1}{3}tr(\tau), & \text{if }\vert\tau^D\vert > \sigma_0, \end{cases}$$ -$$\alpha := \sigma_0 + \dfrac{\gamma}{2\mu+\gamma}\left(\vert\tau^D\vert - \sigma_0\right) ,$$ +$$\alpha := \sigma_0 + \dfrac{\gamma^{\text{iso}}}{2\mu+\gamma^{\text{iso}}}\left(\vert\tau^D\vert - \sigma_0\right) ,$$ with a further material parameter $\mu>0$ called shear modulus. We refer that this only possible for isotropic plasticity.\\ -So what we do is to calculate the stresses by using Hooke's law for linear elastic, isotropic materials -$$\sigma = C \varepsilon(u) = 2\mu \varepsilon^D(u) + \kappa tr(\varepsilon(u))I = \left[2\mu\left(\mathbb{I} -\dfrac{1}{3} I\otimes I\right) + \kappa I\otimes I\right]\varepsilon(u)$$ +To make things a bit easier from now on we denote +$$\gamma := \dfrac{\gamma^{\text{iso}}}{2\mu + +\gamma^{\text{iso}}}\in[0,1)\text{ with }\gamma^{\text{iso}}\in[0,\infty),$$ +$$\beta :=\dfrac{\sigma_0}{\vert\tau^D\vert}.$$ If $\gamma^{\text{iso}}$ tends to zero $\gamma$ tends also to zero. And if $\gamma^{\text{iso}}$ tends to +infinity $\gamma$ tends to one. This allows us to reformulate our problem as +follows $$P_{\Pi}(\tau):=\begin{cases} + \tau, & \text{if }\vert\tau^D\vert \leq \sigma_0,\\ + \gamma\tau^D + (1-\gamma)\beta\tau^D + + \dfrac{1}{3}tr(\tau), & \text{if }\vert\tau^D\vert > + \sigma_0, \end{cases}.$$ +For further details see Suttmeier: On Plasticity with Hardening: +An Adaptive Finite Element Discretisation, International Mathematical Forum, 5, +2010, no. 52, 2591-2601.\\ +So what we do is to calculate the stresses +by using Hooke's law for linear elastic, isotropic materials $$\sigma = C \varepsilon(u) = 2\mu \varepsilon^D(u) + \kappa tr(\varepsilon(u))I = \left[2\mu\left(\mathbb{I} -\dfrac{1}{3} I\otimes I\right) + \kappa I\otimes I\right]\varepsilon(u)$$ with the material parameter $\kappa>0$ (bulk modulus). The variables $I$ and $\mathbb{I}$ denote the identity tensors of second and forth order. In that notation $2\mu \varepsilon^D(u)$ is the deviatoric part and $\kappa @@ -188,15 +202,19 @@ semi-linearform $a(.;.)$ at the point $u^i$ is $$a'(u^i;\psi,\varphi) = (I(x)\varepsilon(\psi),\varepsilon(\varphi)),\quad x\in\Omega,$$ $$ I(x) := \begin{cases} -2\mu\left(\mathbb{I} - \dfrac{1}{3} I\otimes I\right) + \kappa I\otimes I, & +C_{\mu} + C_{\kappa}, & \quad \vert \tau^D \vert \leq \sigma_0\\ -\dfrac{\alpha}{\vert\tau^D\vert}2\mu\left(\mathbb{I} - \dfrac{1}{3} I\otimes I -- \dfrac{\tau^D\otimes\tau^D}{\vert\tau^D\vert}\right) + \kappa I\otimes I, -&\quad \vert \tau^D \vert > \sigma_0 +\gamma C_{\mu} + (1-\gamma)\beta\left(C_{\mu} - +2\mu\dfrac{\tau^D\otimes\tau^D}{\vert\tau^D\vert^2}\right) + C_{\kappa}, &\quad +\vert \tau^D \vert > \sigma_0 \end{cases} $$ with -$$\tau^D := C\varepsilon^D(u^i).$$ +$$C_{\mu} := 2\mu\left(\mathbb{I} - \dfrac{1}{3} I\otimes +I\right)\quad\text{(shear part of the stress strain tensor)},$$ +$$C_{\kappa} := \kappa I\otimes I\quad\text{(bulk part of the stress strain +tensor)},$$ +$$\tau^D := C\varepsilon^D(u^i).$$ Remark that $a(.;.)$ is not differentiable in the common sense but it is slantly differentiable like the function for the contact problem and again we refer to Hintermueller, Ito, Kunisch: The primal-dual active set strategy as a semismooth newton method, SIAM J. OPTIM., 2003, Vol. 13, No. 3, pp. 865-888. -- 2.39.5