yade-dev team mailing list archive

[Bruno Chareyre <bruno.chareyre@xxxxxxxxxxx>]

>     I have one big guess. How can this formulation being incremental
>     using trialForce? Let's take a simple case: normal force is set to
>     a constant value. Then say that trialForce will increase. In this
>     case shearForce is the same since normal force is as well (like
>     maintaining constant the maximum elastic shear displacement). If
>     any time you get the work done as above, I would say that
>     1/ks*(trialForce-shearForce)
>     is the total plastic displacement, is it?
No. It is the increment of displacement.

Example :
1) time t : us_total=1, us_elastic=0.1, us_plastic=0.9, fs=0.1*ks
2) assume an increment dus between t and t+dt
3) trialFs = previousFs + dus*ks = us_elastic*ks + dus*ks = 
ks*(us_elastic+dus)
4) 1/ks*(trialFs-previousFs) = ? .... dus

us_total is never used and remains unknown.

Cheers.

Bruno