From 09c11d29c6341afab6300b2d9267ee6f098bf84b Mon Sep 17 00:00:00 2001 From: wolf Date: Tue, 3 May 2005 14:51:36 +0000 Subject: [PATCH] Fix one TODO. git-svn-id: https://svn.dealii.org/trunk@10618 0785d39b-7218-0410-832d-ea1e28bc413d --- deal.II/examples/step-18/step-18.cc | 96 +++++++++++++++++------------ 1 file changed, 58 insertions(+), 38 deletions(-) diff --git a/deal.II/examples/step-18/step-18.cc b/deal.II/examples/step-18/step-18.cc index 9854802bd5..b679fed048 100644 --- a/deal.II/examples/step-18/step-18.cc +++ b/deal.II/examples/step-18/step-18.cc @@ -2337,47 +2337,67 @@ namespace QuasiStaticElasticity // @sect4{TopLevel::update_quadrature_point_history} - // At the end of each time step, we should - // have computed an incremental - // displacement update so that the material - // in its new configuration accomodates for - // the difference between the external body - // and boundary forces applied during this - // time step minus the forces exerted + // At the end of each time step, we + // should have computed an + // incremental displacement update + // so that the material in its new + // configuration accomodates for + // the difference between the + // external body and boundary + // forces applied during this time + // step minus the forces exerted // through pre-existing internal // stresses. In order to have the - // pre-existing stresses available at the - // next time step, we therefore have to - // update the pre-existing stresses with - // the stresses due to the incremental - // displacement computed during the present - // time step. Ideally, the resulting sum of - // internal stresses would exactly counter - // all external forces. Indeed, a simple - // experiment can make sure that this is - // so: if we choose boundary conditions and - // body forces to be time independent, then - // the forcing terms (the sum of external - // forces and internal stresses) should be - // exactly zero. If you make this - // experiment, you will realize from the - // output of the norm of the right hand - // side in each time step that this is - // almost the case: it is not exactly zero, + // pre-existing stresses available + // at the next time step, we + // therefore have to update the + // pre-existing stresses with the + // stresses due to the incremental + // displacement computed during the + // present time step. Ideally, the + // resulting sum of internal + // stresses would exactly counter + // all external forces. Indeed, a + // simple experiment can make sure + // that this is so: if we choose + // boundary conditions and body + // forces to be time independent, + // then the forcing terms (the sum + // of external forces and internal + // stresses) should be exactly + // zero. If you make this + // experiment, you will realize + // from the output of the norm of + // the right hand side in each time + // step that this is almost the + // case: it is not exactly zero, // since in the first time step the - // incremental displacement and stress - // updates were computed relative to the - // undeformed mesh, which was then - // deformed. In the second time step, we - // again compute displacement and stress - // updates, but this time in the deformed - // mesh -- there, the resulting updates are - // very small but not quite zero. This can - // be iterated, and in each such iteration - // the residual, i.e. the norm of the right - // hand side vector, is reduced by - -//TODO: compute amount + // incremental displacement and + // stress updates were computed + // relative to the undeformed mesh, + // which was then deformed. In the + // second time step, we again + // compute displacement and stress + // updates, but this time in the + // deformed mesh -- there, the + // resulting updates are very small + // but not quite zero. This can be + // iterated, and in each such + // iteration the residual, i.e. the + // norm of the right hand side + // vector, is reduced; if one makes + // this little experiment, one + // realizes that the norm of this + // residual decays exponentially + // with the number of iterations, + // and after an initial very rapid + // decline is reduced by roughly a + // factor of about 3.5 in each + // iteration (for one testcase I + // looked at, other testcases, and + // other numbers of unknowns change + // the factor, but not the + // exponential decay). // In a sense, this can then be considered // as a quasi-timestepping scheme to -- 2.39.5