New question #700748 on Yade:
https://answers.launchpad.net/yade/+question/700748
Hi,
I want to calculate the plastic shear displacement of any specific contact
during every time step. I'm using Law2_ScGeom6D_CohFrictPhys_CohesionMoment[1].
In the source code, the plastic displacement is calculated by the following
code (L140-L159):
Vector3r& shearForce = geom->rotate(phys->shearForce);
const Vector3r& dus = geom->shearIncrement();
//Linear elasticity giving "trial" shear force
shearForce -= phys->ks * dus;
Real Fs = phys->shearForce.norm();
Real maxFs = phys->shearAdhesion;
if (!phys->cohesionDisablesFriction || maxFs == 0) maxFs += Fn
* phys->tangensOfFrictionAngle;
maxFs = math::max((Real)0, maxFs);
if (Fs > maxFs) { //Plasticity condition on shear force
if (phys->fragile && !phys->cohesionBroken) {
phys->SetBreakingState();
maxFs = max((Real)0, Fn *
phys->tangensOfFrictionAngle);
}
maxFs = maxFs / Fs;
Vector3r trialForce = shearForce;
shearForce *= maxFs;
if (scene->trackEnergy || traceEnergy) {
Real sheardissip = ((1 / phys->ks) *
(trialForce - shearForce)) /*plastic disp*/.dot(shearForce) /*active force*/;
Briefly, the plastic shear displacement is calculated by 1/ks*(trialForce -
shearForce). However, it seems trialForce is not available during the
simulation process. Though shearInc is available, which is dus in the source
code, it is not exactly the same as 1/ks*(trialForce - shearForce). Therefore I
want to ask if there is a simple way to get the plastic shear displacement
instead of modify the source code?
Thank you!
[1]
https://gitlab.com/yade-dev/trunk/-/blob/master/pkg/dem/CohesiveFrictionalContactLaw.cpp
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