Since you can very easily reach a few million degrees of freedom in three
dimensions, even with adaptive mesh refinement, we decided to use Trilinos and
p4est to run our code in parallel, building on the framework of step-40 for
-the parallelization.
+the parallelization.\\
\begin{huge}
{distributed}
\sigma_0$ and otherwise if $\lambda > 0$ it follows that $\vert\sigma^D\vert = \sigma_0$.
That means if the stress is smaller than the yield stress there are only elastic
deformations in that point.\\
-Or to consider it the other way around. If the deviator stress $\sigma^D$ is in
-a norm bigger than the yield stress then $\sigma^D$ has to be projected back to
-the yield surface and there are plastic deformations which means $\lambda$
+To consider it the other way around if the deviator stress $\sigma^D$ is in a
+norm bigger than the yield stress then $\sigma^D$ has to be projected back to the yield surface and there are plastic deformations which means $\lambda$
would be positiv for that particular point. We refer that the stresses are
computed by Hooke's law for isotorpic materials. You can find the description at the end of section 3. Else if the norm of the deviator stress tensor is smaller or equal the yield stress then $\lambda$ is zero and there are no plastic deformations in
that point.\\
$$\left(\sigma,\varepsilon(\varphi) - \varepsilon(u)\right) \geq 0,\quad \forall \varphi\in V^+.$$
Therein $\varepsilon$ denotes the linearised deformation tensor with $\varepsilon(u) := \dfrac{1}{2}\left(\nabla u + \nabla u^T\right)$ for small deformations.\\
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 behavior. The most simple one is called linear isotropic hardening
-with the flow function $\mathcal{F}(\tau,\eta) = \vert\tau^D\vert - (\sigma_0 + \gamma\eta)$.
+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
+$\eta$ 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
$$\left(\sigma,\tau - \sigma\right) - \left(C\varepsilon(u), \tau - \sigma\right) + \gamma\left( \xi, \eta - \xi\right) \geq 0,\quad \forall (\tau,\eta)\in \Pi (W,L^2(\Omega,\mathbb{R}))$$
$$\left(\sigma,\varepsilon(\varphi) - \varepsilon(u)\right) \geq 0,\quad \forall \varphi\in V^+,$$
with the hardening parameter $\gamma > 0$.\\
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 on\\
+set $\eta = \xi$ and eliminate the stress $\sigma$ by applying the projection
+theorem () on\\
$$\left(\sigma - C\varepsilon(u), \tau - \sigma\right) \geq 0,\quad \forall \tau\in \Pi W,$$
which yields with the second inequality:\\
Find the displacement $u\in V^+$ with
$\mathbb{I}$ denote the identity tensors of second and forth order. In that
notation $2\mu \varepsilon^D(u)$ is the deviatoric part and $\kappa
tr(\varepsilon(u))$ the volumetric part of the stress tensor.\\
-In the next step we test in a pointwise sense where the deviator part of the stress in a norm is bigger as the yield stress.
-If there are such points we project the deviator stress in those points back to the yield surface. Methods of this kind
-are called projections algorithm or radial-return-algorithm.\\
+In the next step we test in a pointwise sense where the deviator part of the
+stress in a norm is bigger than the yield stress. If there are such points we
+project the deviator stress in those points back to the yield surface. Methods of this kind are called projections algorithm or radial-return-algorithm.\\
Now we have a primal formulation of our elasto-plastic contact problem which only depends on the displacement $u$.
-It consists of a nonlinear variational inequality and has a unique solution as it shows the theorem of Lions and Stampaccia
-(A proof can be found in Rodrigues: Obstacle Problems in Mathematical Physics, North-Holland, Amsterdam, 1987).\\
+It consists of a nonlinear variational inequality and has a unique solution as
+it satisfies the theorem of Lions and Stampaccia. A proof can be found in
+Rodrigues: Obstacle Problems in Mathematical Physics, North-Holland, Amsterdam,
+1987).\\
To handle the nonlinearity of the constitutive law we use a Newton method and to deal with the contact we apply an
active set method like in step-41. To be more concrete we combine both methods to an inexact semi smooth Newton
method - inexact since we use an iterative solver for the linearised problems in each Newton step.
For the Newton method we have to linearise the following semi-linearform
$$a(\psi;\varphi) := \left(P_{\Pi}(C\varepsilon(\varphi)),\varepsilon(\varphi)\right).$$
-Becaus we have to find the solution $u$ in the convex set $V^+$, we have to apply an SQP-method (SQP: sequential quadratic
-programming). That means we have to solve a minimisation problem for a known $u^i$ in every SQP-step of the form
+Because we have to find the solution $u$ in the convex set $V^+$, we have to
+apply an SQP-method (SQP: sequential quadratic programming). That means we have to solve a minimisation problem for a known $u^i$ in every SQP-step of the form
\begin{eqnarray*}
& & a(u^{i};u^{i+1} - u^i) + \dfrac{1}{2}a'(u^i;u^{i+1} - u^i,u^{i+1} - u^i)\\
&=& a(u^i;u^{i+1}) - a(u^i;u^i) +\\
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