| Author | Message | ||||
     Joe Pierce (jpierce) New member Username: jpierce Post Number: 1 Registered: 04-2008 |
I have a stress problem that has a symmetry plane that is a 45deg angle to the x axis. I think the boundary condition should be value(Up)=Vp and value(Vp)=Up. This does not work with FlexPDE. How do you create a BC where Up=Vp in FlexPDE? Thanks
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| Robert G. Nelson (rgnelson) Moderator Username: rgnelson Post Number: 1099 Registered: 06-2003 |
value(Up)=Vp and value(Vp)=Up do not impose two boundary conditions. It is simply one condition stated twice. This creates two identical rows in the matrix and makes the system singular. I do not know what the proper combination of boundary conditions is on the diagonal boundary. I tried using Natural(Up)=normal(Sx,Txy), but this is a tautological equation that adds no new information to the system. The solution is similar to the Quarter_Sym problem, but not exactly the same. You need some condition that actually imposes the same kind of constraint that the upper leg frozen at the left imposes. I assume this will be conditions on the normal and tangential stresses. The Quarter_Sym system solves very quickly, so I'm not sure what the motivation is for imposing the diagonal symmetry plane. | ||||
| Joe Pierce (jpierce) New member Username: jpierce Post Number: 2 Registered: 04-2008 |
Thanks for the fast response. The models I posted are very simple examples of the diagonal symmetry plane. The real problem is much more complicated and has small relevant features that slow the solution down quite a bit. Will the addition of a 3rd variable like w=Up-Vp with value(w)=0 on the diagonal sym plane? I tried this and the solution diverged. I may have applied it wrong. |
Quarter Symmetry Works