Degrees of Freedom 4

[OptiEng]

I recently read a paper mentioning that and FE analysis was conducted a mesh density of approx 10,000 degrees of freedom, can anyone explain this?

Thanks.

 

[rstupplebeen]

DOFs are used as a measure of problem size instead of saying how many elements or nodes were used. It is a more accurate measure of size because nodes can have a variable number of DOFs and elements can have a variable number of nodes. For nonlinear problems none of these measures are that good for total solution time though. I hope this helps.


Rob Stupplebeen

 

[corus]

A mesh density of so many degrees of freedom means nothing much, but it does help to hype up the size of the model. If there were 3 dof per node then there would be only 3000 nodes, and obviously a much smaller number of elements. And that's not very impressive at all. A politician probably prepared the model.

corus

 

[rb1957]

assuming a -ve outlook ... bullsh!t trying to baffle brains ... # of dofs is an easy number meant to show that the FEM is complex (not quite the same as detailled) enough to do the job. much more meaningful would be if they said they'd run convergence studies on the model and settled on a mesh that had < 1% change.

assuming a +ve outlook, the guy just grabbed a number (or was told to grab a number) to quantify the statement "the structure was modelled with a FEM".

 

[OptiEng]

Ok thanks, I am coming across this more and more, just seen an abaqus technology brief TB-03-HTB-1 (Jan 2004), where they stated the no. elements = 828,853, no. nodes = 1,380,834 and No. dof = 4,142,502. Would this number be picked up from the abaqus output file or calculated?

Thanks

 

[drawoh]

OptiEng,

Finite Element Method involves breaking a structure down into elements with nodes, writing stiffness equations for each degree of freedom of each node, and then solving them simultaneously. 10,000 DsOF means that they solved a 10k[&times;]10k matrix. Be glad you did not have to do this by hand.

JHG

 

[rb1957]

it (obviously) easy to calculate the number of nodes and elements ... count them or the diagnostic messages should tell you. # of dofs depends on the code (are the dofs assigned to the nodes or to the elements, eg NASTRAN gives each node 6 dofs and neutralises the ones that elements don't pick up). i suspect that (as drawoh above points out) since the # of dofs is the number of equations solved, then this too should be in the diagnostics.

but i wouldn't bother checking it (if that's what you're intending on doing). it may be the same as the report, or the report could have rouned some. it'd be more relevent to check the results (compare the report data with the FE output). if you're really suspicious, rerun the model.

 

[GregLocock]

Actually I have solved plenty of real life problems with fewer than 10000 dofs. In fact a significant part of my career has been based on 2 dof (not FEA though).

Numerical complexity is often a substitute for understanding.

Cheers

Greg Locock

I rarely exceed 1.79 x 10^12 furlongs per fortnight