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["Sergei D." <dorofeenko@xxxxxxxxx>]

But you speak about elastoplastic law with different stiffness for 
loading/unloading.
I speak about viscoelastic law (with constant stiffness):
F = k x + c dx/dt
So, if dx/dt>0 (loading) then F > F_elastic=k x,
and if dx/dt<0 (unloading) then F<F_elastic
ok?


> I think I do not have my mind in the viscous framework.
>
> But, in a general case, some of your phrases do not correspond to my 
> current point of view:
>
> > If more stiffness then more force for given overlap.
> In my former simulations, I used this "NormalInelasticityLaw" which 
> describes (Normal Force; overlap) states such as illustrated by 
> attached figure. So, in this case, you can observe that the highest 
> (tangent) stifness corresponds to lowest values of force : in 
> unloading the stifness I used is higher than the one of loading; but 
> unloading values of force are nevertheless lower than loading (initial 
> loading) values of force, for given overlap.
>
> Then, when you say:
> > I consider only a loading because a unloading is less the loading, so 
> > if loading is stable (not overestimate) then unloading is stable also.
> Using this contact law I speak about, I had to decrease the time step. 
> Let "K" be the ratio "unloading stifness" / "loading stifness" (= the 
> ratio of the two slopes of the two type of lines you can see on the 
> figure). The time step I could use with K=1 (loading = unloading), 
> could not be used anymore if K>1 (the case of the figure).
>
> I hope I could explain what I think ?
>     It's completely ok for me when you say "So, for stability you need 
> smaller time step for more stiffness". But I tend currently to 
> disagree if you speak only of forces.
>
> Jerome
>
>
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--
Best regards,
Sergei D.