rolling friction model

Submitted by JoshuaP on Wed, 11/26/2014 - 17:12

Hey ,

I tested a bit the rolling friction model epsd2 and read also the paper from Ai, Chen, Rotter and Ooi.
Where the limiting spring torque is calculated by:
nu_rollfrict*R*F_n

If you consider now that the maximum torque from rolling in one contact that can occur is:
nu_friction*R*F_n

So if I choose my friction coefficient less than my rolling friction coefficient I could never reach the limiting spring torque.
That means I would stay all the time in the spring model and cant roll, in case of free rolling from one particle under inclined gravity.
Whereas the angle of the gravity doesnt mind.

The question now is: should the rolling friction coefficient in every case be lower than the friction angle? Otherwise a single particle could never roll on an inclined plane.

I would be very grateful for every answer

best regards

Forums:

zamir | Wed, 11/26/2014 - 22:01

Hi Joshua,

Let me know if I am wrong here, because I am also very interested in particle rotations. As I understand it, the moment due to rolling friction is only limited by nu_rollfrict*R*F_n. The actual instantaneous torque due to rolling friction is given by kt*(R^2)*dthetaR where kt=tangential stiffness and dthetaR = relative rotation (ignoring velocity, of course).

This means that torque due to rolling friction is very much dependent on tangential deformation, whereas torque due to sliding friction is due to normal deformation. In a hertzian contact model where the normal and tangential stiffnesses are coupled, its not very likely that mu_R and mu_S will equally contribute to torque. However, in a hookean model, you could set kn and kt such that mu_R and mu_S have equal contributions to torque in certain conditions.

Of course, I am still learning every day, and I cannot be certain of my logic, so to test rotational problems, I wrote an integrator that tracks particle rotations. I posted it in the developers forum. Look for fix_nve/sphere/orientation. There is an example of how to use it posted there as well.

JoshuaP | Thu, 11/27/2014 - 09:38

hey Zamir,

the equation
kt*R^2*dThetaR
is just for the case that you are below the full mobilisation rolling angle (so no real rolling just wiggle). Besides I'm just observing the rolling resistance torque without viscous damping.
So if you want a real rolling of particles the torque to the particle must be higher than the limiting spring torque which is:
nu_roll*R_r*F_n
where nu_roll is rolling friction coef.

and if you regard a single particle rolling down an inclined plane you have the downhill force =F_g* sin(beta)
where beta is the incline angle and F_g the massforce. But this force is in the rotational centre so it creates no torque. The only force that creates a torque is the the friction force between particle and plane.
This force is F_r=F_n*nu_r
with nu_r being the friction coefficient (not the rolling frict coef !!)
The torque created by this is F_r*R_r

so summarized The created torque by rolling down a plane is F_n*nu_r*R_r and must be taller than the limiting spring torque nu_roll*R_r*F_n.

That means nu_r must be taller than nu_roll if you want real rolling and not just wiggle.

zamir | Mon, 12/01/2014 - 18:00

Sounds right to me. Test it out Joshua, I think it would be really great to see some empirical data to verify this!!

aaigner's picture

aaigner | Wed, 12/03/2014 - 22:30

Hi JoshuaP!

You are correct.. thus a particle start to roll down an inclined plate, the torque due to gravity and friction has to exceed the rolling resistance torque.
Otherwise the particle won't start to roll and not move at all. In case you have a very high rolling friction coefficient and if you increase further the pitch, the particle will slide down the wall (not roll!).

Bests
Andreas

JoshuaP | Tue, 06/16/2015 - 13:03

Hey,
I would like to discuss the rolling model a bit more. Maybe there are a few of you who are interested also in this. My interpretation of the EPSD is:
The EPSD model represents the behaviour of an ellipse up to a certain point. The resistance against rotation increases while the ellipse is rotating till a certain point. At this point the EPSD rolling model adds a constant resistance moment, which does not agree with the behaviour of the ellipse anymore.
The rolling resistance of a cube would be different. The highest resistance moment is at the beginning of the rotation, whereas with increasing rotation, the resistance moment will decrease.

The problem I get with the EPSD and EPSD2 is that after loading and unloading i get to much swelling. This has been tested and is mainly caused by the spring part of the rolling model. With the cdt model it is better, but there occurs a creeping behaviour caused by the residual kinetic energy of the constant torque on particles. This behaviour is already identified by J. AI "Assessment of rolling resistance models in discrete element simulations" (Its written in the last few sentences).

I'm actually checking out some changes in the EPSD2 model to get a better behaviour in the uniaxial compression test. Maybe some of you have also changed already a bit the rolling models? I would be very interested in your opinions.

Kind regards,
Joshua

JoshuaP | Fri, 06/19/2015 - 10:51

The EPSD calculates a rolling stiffness dependent on k_n and EPSD2 on k_t. But when I'm using the hertzian contact model, my k_n and k_t are dependent on my overlap. So is my rolling stiffness then also dependent on the overlap?

Thanks
Joshua

aaigner's picture

aaigner | Mon, 06/22/2015 - 10:27

That's the reason for the non-linear behaviour of the EPSD/EPSD2 model.

JoshuaP | Mon, 06/22/2015 - 11:52

But what happens then is that if you compress a specimen the particles have high overlap and a stiff rolling behaviour. If you unload now, the overlap decreases and you get a soft rolling behaviour that results in a higher back rotation than before. There will be a high swelling behaviour at unloading which is unphysical.