yade-dev team mailing list archive

[chiara modenese <c.modenese@xxxxxxxxx>]

Thanks for the example! Now I see that why is right that the shear force
change sign. It's global but its application is related to the ids of the
bodies.
cheers, chia

2010/3/31 Václav Šmilauer <eudoxos@xxxxxxxx>

>
> > Hi V., if I correctly understood, the shear force (the incremental
> > one) behaves as you describe, but the TOTAL shear force (the one that
> > we store in the simulation) is really expressed in the global system
> > (right?).
>
> (let's suppose that https://bugs.launchpad.net/yade/+bug/493102 is
> solved for the rest of this discussion)
>
> shearForce (the total, current one) is defined for interaction (not for
> particle), in global coordinates. Now, let's say shearForce is
> (-1,-1,-1), and that we have 2 bodies, A and B, which are assigned to
> id1 and id2. It means that you apply (-1,-1,-1) to A (id1) and
> -(-1,-1,-1)=(1,1,1) to B (id2).
>
> If you swap the particles (A=id2, B=id1), then shearForce will be
> computed in the inverse sense:
>
> See e.g. ScGeom::updateShear (or updateShearForce), where it is clear:
> normal is already inverse (always from id1 to id2), then everything is
> symmetric for id1 and id2 (therefore the same if you swap id1/id2);
> relativeVelocity will be inverted, so will be shearVelocity,
> shearDisplacement and shearIncrement / shearForce increment.
>
> (This will happen similarly in Dem3DofGeom, but it is less obvious to
> follow in the code)
>
> With swapped particled (A=id2, B=id1), therefore, shearForce will be
> (1,1,1). But we apply now (1,1,1) to B (which is now id1) and (-1,-1,-1)
> to A (which is now id2).
>
> So you see real forces on bodies are the same.
>
> > Moreover if my external force on the particle is in one direction, I
> > would always expect an opposite sign for the total shear force (we
> > oscillates around the same value but the sign should always be
> > opposite to it).
> Sorry, I am not clear as to what you mean here.
>
> HTH, Vaclav
>
>
>
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