Implementation of Normal and Shear Cohesion

Asked by Kneib François

Hi all,

I'm just wondering about the Physics unity of the normalCohesion and shearCohesion parameters of a cohesive frictional material.

Here is the part of the Ip2_CohFrictMat_CohFrictMat_CohFrictPhys :

contactPhysics->normalAdhesion = std::min(sdec1->normalCohesion,sdec2->normalCohesion)*pow(std::min(geom->radius2, geom->radius1),2);
contactPhysics->shearAdhesion = std::min(sdec1->shearCohesion,sdec2->shearCohesion)*pow(std::min(geom->radius2, geom->radius1),2);

So is it some Pascals ? Why using the area of a square of length=radius ?

Tx

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Bruno Chareyre
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Revision history for this message
Chiara Modenese (chiara-modenese) said :
#1

Hi Kneib,

I am not the author of the concerned code but, from what I can see, it
computes a normal and shear adhesive forces so you are right that the unit
of measure of normalCohesion is Pa. Generally, the squared of particle
radius is regarded as some crude measure of the contact area between two
contact spheres (I have seen using it in a few papers).

HTH, Chiara

On 30 August 2011 13:50, Kneib François <
<email address hidden>> wrote:

> New question #169595 on Yade:
> https://answers.launchpad.net/yade/+question/169595
>
> Hi all,
>
> I'm just wondering about the Physics unity of the normalCohesion and
> shearCohesion parameters of a cohesive frictional material.
>
> Here is the part of the Ip2_CohFrictMat_CohFrictMat_CohFrictPhys :
>
> contactPhysics->normalAdhesion =
> std::min(sdec1->normalCohesion,sdec2->normalCohesion)*pow(std::min(geom->radius2,
> geom->radius1),2);
> contactPhysics->shearAdhesion =
> std::min(sdec1->shearCohesion,sdec2->shearCohesion)*pow(std::min(geom->radius2,
> geom->radius1),2);
>
> So is it some Pascals ? Why using the area of a square of length=radius ?
>
> Tx
>
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Revision history for this message
Best Bruno Chareyre (bruno-chareyre) said :
#2

That is right, it is in Pa.
Multiplication by r^2 is for dimensional consistency. With these
equations the strength of the packing is not a function of particles sizes.
If you want the area of a disk instead of a square, I guess you will
multiply cohesion by pi/2 or something like that. ;)

Revision history for this message
Kneib François (francois-kneib) said :
#3

Thanks Chareyre, that solved my question.