Discussion Closed This discussion was created more than 6 months ago and has been closed. To start a new discussion with a link back to this one, click here.
Block Verification
Posted 24 apr 2012, 16:05 GMT-4 Structural Mechanics 2 Replies
Please login with a confirmed email address before reporting spam
Why must you have all 3 roller boundary conditions (RBC) to get a correct solution? If you just keep the RBC opposite the prescribed displacement boundary, and turn the others to free, you don't get a correct result (forget the boundary loads and plasticity, just a simple linear block). Doesn't make sense, with all 4 sides free the block should't rotate or asymmetrically move under a uniaxial and uniform displacement, but it does.
What am I missing?
~C
Hello Chris Layman
Your Discussion has gone 30 days without a reply. If you still need help with COMSOL and have an on-subscription license, please visit our Support Center for help.
If you do not hold an on-subscription license, you may find an answer in another Discussion or in the Knowledge Base.
Please login with a confirmed email address before reporting spam
In a stationary structures problem you need to have enough boundary conditions to remove all rigid body modes even if your loading does not trigger these rigid body modes. Without proper constraints the stiffness matrix is singular and there is no unique solution since you can add a rigid body displacement and stiff satisfy equilibrium. If you pose this as a transient problem you should get a unique solution.
Nagi Elabbasi
Veryst Engineering
Please login with a confirmed email address before reporting spam