From: wolf Date: Thu, 20 Oct 2005 22:04:32 +0000 (+0000) Subject: Use a number of suggestions by Francesco Freddi. X-Git-Url: https://gitweb.dealii.org/cgi-bin/gitweb.cgi?a=commitdiff_plain;h=839865028486276b5672eaea1bef50d71174fbe0;p=dealii-svn.git Use a number of suggestions by Francesco Freddi. git-svn-id: https://svn.dealii.org/trunk@11640 0785d39b-7218-0410-832d-ea1e28bc413d --- diff --git a/deal.II/doc/tutorial/chapter-2.step-by-step/step-18.data/intro.html b/deal.II/doc/tutorial/chapter-2.step-by-step/step-18.data/intro.html index 5b8eeb9c1a..22a9f287ba 100644 --- a/deal.II/doc/tutorial/chapter-2.step-by-step/step-18.data/intro.html +++ b/deal.II/doc/tutorial/chapter-2.step-by-step/step-18.data/intro.html @@ -6,6 +6,7 @@ file by clicking here]

+

This tutorial program is another one in the series on the elasticity problem that we have already started with step-8 and step-17. It extends it into two @@ -294,23 +295,31 @@ possibly time-varying force function $ \Gamma_D\subset\partial\Omega(t)$ $ \Gamma_N=\partial\Omega(t)\backslash\Gamma_D$

Note that these equations are posed on a domain $ \Omega(t)$ that changes with time, with the boundary moving according to the @@ -386,17 +395,17 @@ displacements $ \vec u(\vec x,t)$ of the points on the boundary. To -complete this system, we have to specify the relationship between the -stress and the strain, as follows: +complete this system, we have to specify the incremental relationship between +the stress and the strain, as follows:

@@ -404,7 +413,7 @@ stress and the strain, as follows:

where a dot indicates a time derivative. Both the stress $ \sigma$ and the strain $ \varepsilon(\vec u)$ are symmetric tensors of rank 2. @@ -427,7 +436,7 @@ 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$ :

@@ -435,7 +444,7 @@ discrete equilibrium of force at time step @@ -447,7 +456,7 @@ where @@ -458,25 +467,33 @@ and $ \Delta \vec u^n$ the incremental displacement for time step $ n$ -. This way, if we want to solve for the displacement increment, we +. In addition, we have to specify initial data +$ \vec u(\cdot,0)=\vec u_0$ +. +This way, if we want to solve for the displacement increment, we have to solve the following system:

$\displaystyle \dot\sigma = C \varepsilon (\dot{\vec u}),$ (11)
$\displaystyle -\div\sigma^n = f^n,$ (12)
$\displaystyle \sigma^n = \sigma^{n-1} + C \varepsilon (\Delta \vec u^n),$ (13)
@@ -694,7 +711,7 @@ when computing time step $ \sigma^n$ : even if we compute the incremental updates $ \Delta \vec u^n$ using lowest-order finite elements, then its symmetric gradient $ \varepsilon(\Delta\vec u^n)$ is in general still a function that is not easy to describe. In particular, it is @@ -723,7 +740,7 @@ stress-strain tensors $ \sigma^n$ in a computer program. @@ -735,7 +752,7 @@ require the stress is in the term --> $ (\sigma^{n-1},\varepsilon(\varphi))_{\Omega(t_{n-1})}$ . In practice, we of course replace this term by numerical quadrature: @@ -744,7 +761,7 @@ course replace this term by numerical quadrature: @@ -1008,18 +1026,19 @@ sequence of operations on the present mesh:
  • assemble_system () [via solve_timestep ()]: This first function is also the most interesting one. It assembles the linear system corresponding to the discretized version of equation - (17). This leads to a system matrix $ A_{ij} = \sum_K
 A^K_{ij}$ built up of local contributions on each cell $ K$ with entries

    @@ -1027,7 +1046,7 @@ A^K_{ij}$">
  • @@ -1035,7 +1054,7 @@ A^K_{ij}$">

    In practice, $ A^K$ is computed using numerical quadrature according to the formula @@ -1044,7 +1063,7 @@ In practice, @@ -1052,11 +1071,11 @@ In practice,

    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 @@ -1071,7 +1090,7 @@ with quadrature points $ \varepsilon(\varphi_i),\varepsilon(\varphi_j)$ . This is not really convenient, in particular if we want to consider more complicated elasticity @@ -1086,7 +1105,7 @@ with quadrature points $ C_{ijkl} = \lambda \delta_{ij} \delta_{kl} + \mu (\delta_{ik} \delta_{jl}
 + \delta_{il} \delta_{jk})$ . While we in fact do not use a more complicated @@ -1111,23 +1130,23 @@ What we then need is two things: a way to create the stress-strain rank-4 as well as to create a symmetric tensor of rank 2 (the strain tensor) from the gradients of a shape function $ \varphi_i$ at a quadrature point $ \vec x_q$ on a given cell. At the top of the implementation of this example program, you will find such functions. The first one, get_stress_strain_tensor, takes two arguments corresponding to the Lamé constants $ \lambda$ and $ \mu$ and returns the stress-strain tensor for the isotropic case corresponding to these constants (in the program, we @@ -1137,11 +1156,11 @@ What we then need is two things: a way to create the stress-strain rank-4 get_strain takes an object of type FEValues and indices $ i$ and $ q$ and returns the symmetric gradient, i.e. the strain, corresponding to shape function $ \varphi_i(\vec x_q)$ , evaluated on the cell on which the FEValues object was last reinitialized. @@ -1193,17 +1212,18 @@ Assembling the local contributions
    $\displaystyle - \div C \varepsilon(\Delta\vec u^n)$ $\displaystyle = \vec f + \div\sigma^{n-1}$   in $ \Omega(t_{n-1})$ $\displaystyle \Delta \vec u^n(\vec x,t)$ $\displaystyle = \vec d(\vec x,t_n) - \vec d(\vec x,t_{n-1}) \qquad$   on $ \Gamma_D\subset\partial\Omega(t_{n-1})$ $\displaystyle \vec n \ C \varepsilon(\Delta \vec u^n(\vec x,t))$ $\displaystyle = \vec b(\vec x,t_n)-\vec b(\vec x,t_{n-1}) \qquad$   on $ \Gamma_N=\partial\Omega(t_{n-1})\backslash\Gamma_D$ $ \Delta \vec u^n \in
 \{v\in H^1(\Omega(t_{n-1}))^d: v\vert _{\Gamma_D}=\vec d(\cdot,t_n) - \vec d(\cdot,t_{n-1})\}$ @@ -557,7 +574,7 @@ such that
    \begin{gather*}\begin{split}(C \varepsilon(\Delta\vec u^n), \varepsilon(\varphi)...
 ...in H^1(\Omega(t_{n-1}))^d: \vec v\vert _{\Gamma_D}=0\}. \end{split}\end{gather*} @@ -570,7 +587,7 @@ are no boundary forces, i.e.  $ \vec b = 0$ , and that the deformation of the body is driven by body forces $ \vec d$ alone. It is also worth noting that when integrating by parts, we would get terms of the form $ (C \varepsilon(\Delta\vec u^n), \nabla \varphi
 )_{\Omega(t_{n-1})}$ , but that we replace it with the term involving the @@ -598,14 +615,14 @@ symmetric gradient $ \varepsilon(\varphi)$ instead of $ \nabla\varphi$ . Due to the symmetry of The system at time step $ n$ , to be solved on the old domain $ \Omega(t_{n-1})$ , has exactly the form of a stationary elastic problem, and is therefore similar to what we have already implemented @@ -639,7 +656,7 @@ continuous finite elements. There are differences, however:
      -
    1. We have to move the mesh after each time step, in order to be +
    2. We have to move (update) the mesh after each time step, in order to be able to solve the next time step on a new domain;

      @@ -649,7 +666,7 @@ There are differences, however: --> $ \sigma^{n-1}$ to compute the next incremental displacement, i.e. we need to compute it at the end of the time step @@ -672,12 +689,12 @@ Updating the stress variable

      As indicated above, we need to have the stress variable $ \sigma^n$ available when computing time step $ n+1$ , and we can compute it using

      @@ -685,7 +702,7 @@ when computing time step
    $\displaystyle \sigma^n = \sigma^{n-1} + C \varepsilon (\Delta \vec u^n).$ (18)
    $\displaystyle (\sigma^{n-1},\varepsilon(\varphi))_{\Omega(t_{n-1})} = \sum_{K\s...
 ...athbb{T}}} \sum_q w_q \ \sigma^{n-1}(\vec x_q) : \varepsilon(\varphi(\vec x_q),$ @@ -753,16 +770,16 @@ course replace this term by numerical quadrature:

    where $ w_q$ are the quadrature weights and $ \vec x_q$ the quadrature points on cell $ K$ . This should make clear that what we really need is not the stress $ \sigma^{n-1}$ in itself, but only the values of the stress in the quadrature points on all cells. This, however, is a simpler task: we only have to provide @@ -782,7 +799,7 @@ end of each time step we then only have to evaluate $ \varepsilon(\Delta \vec u^n(\vec x_q))$ , multiply it by the stress-strain tensor $ \sigma^n(\vec x_q)$ at quadrature point $ q$ . @@ -809,14 +826,14 @@ due to the fact that we compute $ \Delta u^n$ on the domain $ \Omega(t_{n-1})$ , and then use this displacement increment to both update the stress as well as @@ -825,7 +842,7 @@ move the mesh nodes around to get to $ \Omega(t_n)$ on which the next increment is computed. What we have to make sure, in this context, is that moving the @@ -838,10 +855,11 @@ can be understood as follows: locally, the incremental deformation $ \Delta\vec u$ can be decomposed into three parts, a linear translation (the constant part -of the displacement field in the neighborhood of a point), a dilational +of the displacement increment field in the neighborhood of a point), a +dilational component (that part of the gradient of the displacement field that has a nonzero divergence), and a rotation. A linear translation of the material does not affect the stresses that are frozen into it - the stress values are @@ -853,14 +871,14 @@ situation where $ \Delta\vec u=(y, -x)^T$ , with which $ \varepsilon(\Delta \vec u)=0$ ). Nevertheless, if the the material was pre-stressed in a certain direction, then this direction will be rotated along with the material. To @@ -869,20 +887,20 @@ this end, we have to define a rotation matrix $ R(\Delta \vec u^n)$ that describes, in each point the rotation due to the displacement increments. It is not hard to see that the actual dependence of $ R$ on $ \Delta \vec u^n$ can only be through the curl of the displacement, rather than the displacement @@ -890,7 +908,7 @@ itself or its full gradient (as mentioned above, the constant components of the increment describe translations, its divergence the dilational modes, and the curl the rotational modes). Since the exact form of $ R$ is cumbersome, we only state it in the program code, and note that the correct updating formula @@ -900,7 +918,7 @@ for the stress variable is then
    $\displaystyle \sigma^n = R(\Delta \vec u^n)^T [\sigma^{n-1} + C \varepsilon (\Delta \vec u^n)] R(\Delta \vec u^n).$ (20)
    $\displaystyle A^K_{ij} = (C \varepsilon(\varphi_j), \varepsilon(\varphi_i))_K;$ (21)
    $\displaystyle A^K_{ij} = \sum_q w_q [\varepsilon(\varphi_i(\vec x_q)) : C : \varepsilon(\varphi_j(\vec x_q))],$ (22)
    \begin{gather*}\begin{split}f^K_i &= (\vec f, \varphi_i)_K -(\sigma^{n-1},\varep...
 ...gma^{n-1}_q : \varepsilon(\varphi_i(\vec x_q)) \right\} \end{split}\end{gather*} (23)

    -to the right hand side of (17) is equally +to the right hand side of ([*]) is equally straightforward (note that we do not consider any boundary tractions $ \vec b$ here). Remember that we only had to store the old stress in the quadrature points of cells. In the program, we will provide a variable @@ -1213,7 +1233,7 @@ to the right hand side of (17) is equally --> $ \sigma^{n-1}_q$ in each quadrature point. With this the code for the right hand side looks as this, again rather elegant: @@ -1243,12 +1263,12 @@ for (unsigned int i=0; i<dofs_per_cell; ++i) --> $ \vec f(\vec x_q) \cdot \varphi_i(\vec x_q)$ , we have made use of the fact that for the chosen finite element, only one vector component (namely component_i) of $ \varphi_i$ is nonzero, and that we therefore also have to consider only one component of @@ -1257,7 +1277,7 @@ for (unsigned int i=0; i<dofs_per_cell; ++i) --> $ \vec f(\vec x_q)$ . @@ -1285,10 +1305,12 @@ This essentially concludes the new material we present in this function. It --> $ \Delta \vec u^n$ computed before, we update the stress values in all quadrature points - according to (18) and (20), + according to ([*]) and ([*]), including the rotation of the coordinate system.

    @@ -1346,7 +1368,8 @@ Plasticity models. The most obvious extension is to use a more realistic material model for large-scale quasistatic deformation. The natural choice for this would be plasticity, in which a nonlinear relationship between -stress and strain replaces equation (11). Plasticity +stress and strain replaces equation ([*]). Plasticity models are usually rather complicated to program since the stress-strain dependence is generally non-smooth. The material can be thought of being able to withstand only a maximal stress (the yield stress) after which it starts to @@ -1358,7 +1381,7 @@ elastic energy $\displaystyle E(\vec u) = (\varepsilon(\vec u), C\varepsilon(\vec u))_{\Omega} - (\vec f, \vec u)_{\Omega} - (\vec b, \vec u)_{\Gamma_N},$ (24) @@ -1370,7 +1393,7 @@ subject to the constraint $\displaystyle f(\sigma(\vec u)) \le 0$ (25) @@ -1413,7 +1436,7 @@ Incompressibility is characterized by Poisson's ratio $\displaystyle \nu = \frac{\lambda}{2(\lambda+\mu)},$     @@ -1424,7 +1447,7 @@ where $ \lambda,\mu$ are the Lamé constants of the material. Physical constraints indicate that $ -1\le \nu\le \tfrac 12$ -. If $ \nu$ approaches $ \tfrac 12$ , then the material becomes incompressible. 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