# Changing "breakN", even drastically, produces very little alteration in total force

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Hi Vaclav,

This question will be slightly general/broad, but if you wish, I can post a sample script that will illustrate some of my problems.

As you may be aware, I have been running a set of simulations in which I am investigating the total global force exerted by a bonded body, made of an IceMat, against a surface. In an attempt to vary the total exerted force, we have been altering the breakN (bond strength) values for the IceMat. It seems that this should be one of the most important variables in determining the magnitude of the resultant force.

However, in our experiments, changing the breakN of the impinging body, even by a drastic amount, produces very little total effect on the force exerted on the surface. We reduced breakN by 90% (going from 0.025 to 0.0025, for instance) and it only lowered the exerted force by about 5%; we turned cohesion off entirely and it dropped by only another 5%.

Might there be bugs in the IcePhys bonding, or are there other factors I am overlooking here? Once again, I can post a simplified version of my script if this would be beneficial to you.

Cheers,

Christopher Stanbridge

answered Jul 29, 2015 by (46,530 points)

Hi, I am just guessing without seeing your actual script, but likely you changed IceMat.breakN but it is only used when computing IcePhys.brkNT for the first time (when the contact is created), so those few newly created contacts make the different you see. If you want to update already existing contacts, you have to set ContactLoop.updatePhys to True, then run one step, then set it back to False (you can keep it at True, but it will be recomputing things uselessly all the time).

m.breakN=1e4
S.lab.contactLoop.updatePhys=True
S.one()
S.lab.contactLoop.updatePhys=False


Which gives me the idea that having something like "ContactLoop.updatePhysOnce" which one would set when material is changed, and everything would be recomputed at the next step only. How does it sound? I actually like the idea, thanks! It would also allow to make the change inside PyRunner entirely (the snippet above will not work that way, as obviously calling S.one() from within an engine will lead to deadlock). I will follow-up on this in a while.

Cheers, Václav

commented Jul 29, 2015 by (46,530 points)

The latest revision 8bdbf3c allows one to say S.lab.contactLoop.updatePhys='once' which will update contacts in the next step and then switch itself back to 'never' (default).

commented Jul 29, 2015 by (1,030 points)

Hi Vaclav,

Thanks for the additional info (the ability to change updatePhys will benefit my simulations), but I'm not sure the problem is with updatePhys. I am not talking about a single simulation in which I vary breakN; rather, I am referring to three separate trials I have run with different breakN values (0, 0.001, and 0.01).

If you'd like, I can set up a simplified version of the simulation and send you the code for it.

Thanks again, CS

commented Jul 29, 2015 by (46,530 points)

Hi Christoph, are you sure the sample is loaded in a way which makes breakN relevant? Let me see, send the script. You can attach/paste it here or use PM. Cheers, v.

commented Jul 30, 2015 by (1,030 points)

Hi Vaclav, I have emailed you a script at your woodem.org email address.

commented Jul 31, 2015 by (46,530 points)

Hi Christoph, I need some more information about the script. Do I understand correctly that it does essentially gravity deposition, followed by setting the upper wall to motion, eventually loading the sample in compression, and looking at vertical force value (which the "exerted force" you refer to the in the origina post?)? v.

commented Jul 31, 2015 by (1,030 points)

Yes, that is the basic gist of what our script does. The exerted force is the total force on the stationary wall.

commented Jul 31, 2015 by (46,530 points)

As you are doing confined compresison, cohesion (tensile breakage) will have very little influence. I am actually surprised it is as big as 5% then. EDIT: sorry, not confined, you are not in periodic space. Still, there is friction which will carry most of the load, and lateral expansion will be small due to that.