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total DOF > 3*(# total nodes) for a linear tetrahedral mesh
Posted 24.10.2011, 08:21 GMT-4 Mesh, Modeling Tools & Definitions, Parameters, Variables, & Functions, Structural Mechanics Version 4.2 4 Replies
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Hi,
I find it strange and puzzling! I have a simple unit cube meshed with tetrahedral elements. I chose "Linear" for the interpolation order, linear elastic material and static study. Used displacement BCs, no other type of constraints. See the attached model file.
When I export mesh, I see 33 linear tets and total number of nodes is 15. Since, each node has 3 degrees of freedom (viz. {u v w}, I checked in the equation view as well for the field variables in the solver node), I must have 3*15 = 45 DOF.
However, while solving in the log file it reports 222 DOF! What are these extra DOF, which are nearly 5 times the "actual" DOF?
In a 3D model with 20k nodes, this difference in DOF is a big blow on the computation time and memory for me. So I really need to understand what is this and how to get rid of these excess DOF.
Best regards,
Kodanda
I find it strange and puzzling! I have a simple unit cube meshed with tetrahedral elements. I chose "Linear" for the interpolation order, linear elastic material and static study. Used displacement BCs, no other type of constraints. See the attached model file.
When I export mesh, I see 33 linear tets and total number of nodes is 15. Since, each node has 3 degrees of freedom (viz. {u v w}, I checked in the equation view as well for the field variables in the solver node), I must have 3*15 = 45 DOF.
However, while solving in the log file it reports 222 DOF! What are these extra DOF, which are nearly 5 times the "actual" DOF?
In a 3D model with 20k nodes, this difference in DOF is a big blow on the computation time and memory for me. So I really need to understand what is this and how to get rid of these excess DOF.
Best regards,
Kodanda
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4 Replies Last Post 25.10.2011, 07:17 GMT-4