Learning Forum

[John Hicks]

Is there a general number of eqillibrium iterations a model runs, in a substep, before it's considered "unefficient"? I understand that it'll depend based on how a model is setup, along with the contacts and the forces/displacements/etc. Personally, I've been told 3-8 iterations per substep is good. But, if one of my substeps is taking 50 iterations to solve, i'd like to think that it's not solving efficiently and taking too much time. Therefor I need to go back into my model's setup and change some things. 

Thanks

[peteroznewman]

The 3-8 iterations per substep is a reasonable target, but there is nothing wrong with the solver taking many more substeps when it encounters a difficult part of the solution. Look at the Newton Raphson Force Convergence Plot. If you see several iterations resulting in a Bisection or two, you can prevent that by adding a Step to the solution with the End Time equal to the last converged substep prior to all those wasted iterations that ended in a bisection, and starting the next Step with more substeps in the Initial and Minimum Substeps prior to the solver starting the difficult part of the solution.

[John Hicks]

Thanks Peter ????

So (for example), if I see that my run is bisecting at substep 6, in my next run, I can "add" a Step at the end of substep 5 that will give substep 6 more substeps? Aka it will increment in a smaller time step than what the program automatically chooses. 

Is that a control in workbench mechanical or does it need to be a command block?

[peteroznewman]

In Mechanical, under the Static Structural branch is the Analysis Settings. That is where you can change the number of Steps, turn on Auto Time Stepping, set the Initial, Minimum and Maximum Substeps and turn on Large Deflection.

Let's say you have a 1-step solution with Initial and Minimum Substeps of 10. If your model converged nicely at substep 5 then started bisections, note the time at substep 5, say it was 0.5 s and in the next run, change the 1-step solution into a 2-step solution.

Step 1 has an End Time that you wrote down from substep 5 say 0.5 s but you would change the Initial Substeps to 5 instead of 10 because it is only half as long. Step 2 has the original end time, likely 1.0 but use an Inital Substeps of 10 (which is half the time increment of the original run) if the orginal run had a single bisection then converged.  If the original run had 2 bisections then converged, use Initial Substeps of 20 in Step 2.  If the original run had 3 bisections, use Initial Substeps of 40 in Step 2.