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[Václav Šmilauer <eudoxos@xxxxxxxx>]

> > Fs_tot=ks₁us=ks₁∫dus+ks₁Δus
> > Fs_incr     =ks₀∫dus+ks₁Δus
> >
> >   
> Heh... There is no friction in those equations, but I still see an effect!!
> If I understand correctly now, u_s is total displacement 
> (elastic+plastic, from contact creation to now), a "candidate" force is 
> computed from it, and corrected according to the plastic rule. But if I 
> change the plasticity criterion, the plastic part is computed 
> differently, and the elastic part is apparently changing. Correct?

I don't think so; I am not sure if I understand, though, but let me try:

1. The total formulation tracks displacement. Therefore, shearForce is
always computed from current displacement, using current value of ks
(note ks₁ in the above eqs) (Plastic displacement _is_ forgotten).

2. The incremental formulation tracks force directly. (Plastic part of
the force is forgotten here)

> Initially, I thought the plastic part was really removed forever, 
> meaning u_s from the previous step would have been only the elastic 
> displacement. In that case, changing friction should not affect current 
> forces.

But you are not changing friction (frictionAngle), but stiffness (ks)?
If you change frictionAngle, it will not affect current forces (with
total formulation), except for possible additional plastic slip where
you would be beyond plastic surface.

v