Dear STKOteam,STKO Team wrote: ↑Tue Jun 14, 2022 2:08 pmKn and Kt are penalty coefficients, so there is the typical rule:
It should be a number large enough to enforce the constraint (Kn for non-penetration, Kt for stick condition), but not too large to avoid numerical problems.
As usual, estimate the typical stiffness (Ktyp) of your problem (E*L for solid elements, EA/L for trusses, and so on...).
Then evaluate the order or magnitude: OOM = round(log10(Ktyp))
Now the penalty coefficient should be P = 10^X
where is X > OOM.
2 or 3 orders of magnitude is enough (so X = OOM + 2). At most use X = OOM+8. Why 8? because double-floating-points numbers have 16 significant digits, so OOM+8 means something in between accuracy of the constraint and numerical stability
Regarding mu, it is the friction coefficient (i.e. the tangent of the friction angle). You can find online typical friction coefficients based on the two materials in contact.
As you mentioned above, Kn stands for non-penetrating, so it may be larger than the typical stiffness. But does this work for kt? As far as I know, kt is subject to Mohr-Coulomb's theory. Is it just defined by the experiment?
Looking forward to your reply!