From 87b6ab0a319d6b3725bf27916cb78302e501008b Mon Sep 17 00:00:00 2001 From: wolf Date: Wed, 25 May 2005 17:59:22 +0000 Subject: [PATCH] Minor last fixes. git-svn-id: https://svn.dealii.org/trunk@10744 0785d39b-7218-0410-832d-ea1e28bc413d --- .../step-18.data/intro.tex | 24 +++++++++++-------- 1 file changed, 14 insertions(+), 10 deletions(-) diff --git a/deal.II/doc/tutorial/chapter-2.step-by-step/step-18.data/intro.tex b/deal.II/doc/tutorial/chapter-2.step-by-step/step-18.data/intro.tex index e88fcad5f6..3fddef76cf 100644 --- a/deal.II/doc/tutorial/chapter-2.step-by-step/step-18.data/intro.tex +++ b/deal.II/doc/tutorial/chapter-2.step-by-step/step-18.data/intro.tex @@ -129,11 +129,13 @@ strain $\varepsilon(\vec u)$ are symmetric tensors of rank 2. Numerically, this system is solved as follows: first, we discretize the time component using a backward Euler scheme. This leads to a discrete equilibrium of force at time step $n$: -\begin{align} - -\div \sigma^n &= f^n, -\intertext{where} - \sigma^n &= \sigma^{n-1} + C \varepsilon (\Delta \vec u^n), -\end{align} +\begin{gather} + -\div \sigma^n = f^n, +\end{gather} +where +\begin{gather} + \sigma^n = \sigma^{n-1} + C \varepsilon (\Delta \vec u^n), +\end{gather} and $\Delta \vec u^n$ the incremental displacement for time step $n$. This way, if we want to solve for the displacement increment, we have to solve the following system: @@ -153,7 +155,8 @@ The weak form of this set of equations, which as usual is the basis for the finite element formulation, reads as follows: find $\Delta \vec u^n \in \{v\in H^1(\Omega(t_{n-1}))^d: v|_{\Gamma_D}=\vec d(\cdot,t_n) - \vec d(\cdot,t_{n-1})\}$ such that -\begin{multline} +\begin{gather} + \begin{split} \label{eq:linear-system} (C \varepsilon(\Delta\vec u^n), \varepsilon(\varphi) )_{\Omega(t_{n-1})} = @@ -163,8 +166,9 @@ such that +(\vec b(\vec x,t_n)-\vec b(\vec x,t_{n-1}), \varphi)_{\Gamma_N} \\ \forall \varphi \in \{\vec v\in H^1(\Omega(t_{n-1}))^d: \vec - v|_{\Gamma_D}=0\}. -\end{multline} + v|_{\Gamma_D}=0\}. + \end{split} +\end{gather} We note that, for simplicity, in the program we will always assume that there are no boundary forces, i.e.~$\vec b = 0$, and that the deformation of the body is driven by body forces $\vec f$ and prescribed boundary displacements @@ -375,8 +379,8 @@ sequence of operations on the present mesh: In practice, $A^K$ is computed using numerical quadrature according to the formula \begin{gather} - A^K_{ij} = \sum_q w_q \varepsilon(\varphi_i(\vec x_q)) : C : - \varepsilon(\varphi_j(\vec x_q)), + A^K_{ij} = \sum_q w_q [\varepsilon(\varphi_i(\vec x_q)) : C : + \varepsilon(\varphi_j(\vec x_q))], \end{gather} with quadrature points $\vec x_q$ and weights $w_q$. We have built these contributions before, in step-8 and step-17, but in both of these cases we -- 2.39.5