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Re: [Question #691584]: About periodic simple shear

 

Question #691584 on Yade changed:
https://answers.launchpad.net/yade/+question/691584

Jan Stránský posted a new comment:
Hello,

> why does the code first simulate a constant isotropic compression of O.cell.velGrad=Matrix3(-.1,0,0, 0,-.1,0, 0,0,-.1)?
> Is isotropic compression necessary prior to the simple shear of O.cell.velGrad=Matrix3(0,0,.1, 0,0,0, 0,0,0)?

Because the author decided it :-)
Probably because it corresponds to simulated real situation.
Definitely it is not necessary.

> Does this step ultimately cause the periodic boundary condition to be
flush with the particle regularHexa pack? Because without it, the
periodic boundary condition seems to simulate a regularHexa pack within
the representative volume element but it is suspended inside of it at
iteration 0.

Sorry, here I am lost...
Could you please reword the question and idea behind?

> exz ... Would this definition be analogous to the cell strain or
deformation at a given time step?

In this very specific case yes, exz=O.cell.trsf[0,2] is shear strain (in small strain sense)
BUT, in general, be careful with mapping O.cell.trsf to strain (!)

> I am fairly lost on what the data column 'i' could represent

do you mean i=O.iter?
If yes, then "i" is iteration number

> Finally, I am curious about tanPhi.
> Why is a change in angle being outputted when the rotational DOFs are being blocked? Does tanPhi represent something else here?

Even you have rotational DOFs of particle blocked, the cell as a whole
deforms (after all you apply some shear).

Anyway, here,
tanPhi=(stress[0,2]/stress[2,2]) = shear stress / normal stress
so it is not a geometric angle, but some stress measure.
To me it seems like reformulated cohesion-less (c=0) Mohr-Coulomb condition
shear stress = normal stress * tan(phi)
tan(phi) = shear stress / normal stress
so you basically output tangent of friction angle.

cheers
Jan

[8] https://en.wikipedia.org/wiki/Lode_coordinates

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