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
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