cracks number for different discretization

Hi,

I think your goal is completely impossible, and not meaningful.

If 10 people get the flu this winter in Nancy while 1000 people get it in Beijing, it does not necessarily mean the flu outbreak is worse in Beijing than in Nancy.. I guess we should have to divide by the population to have a meaningful quantity and to compare flu outbreak..

In your case, crack number alone is also a meaningless quantity, I think. I would rather work with crack relative numbers, e.g. crack number / initial number of cohesive interactions. This would be a better measurement of how cracked your medium is.

Jerome

Hello Jerome

Thank you for your answer.
I ask this question because i try to simulate the same test performed by Wassermann (phD thesis 2006) with the grey iron on spherical sample with diameter 70 mm and height 140 mm.
In this context, i try to reproduce the same number of cracks and the dissipated energy.
I have a second problem in this test, in the experimental test the dissipated energy is equal to 550 Joules but in the simulated test i obtain 37 Joules with the numerical sample calibrated on the mechanical response of grey iron.
My goal is to obtain the dissipated energy by the numerical sample equal to the dissipated energy by the real sample.
I try to modify the calibration with increasing the value of tensile strength and cohesion between particles but i don't get what i want.
Also i remarked that the energy dissipated is dependent to the volume. In fact, when i increase the dimension of the numerical sample, the energy dissipated become important.
My question is : there is a method that enables to incearse the dissipated energy with the same sample ?

I wish you Merry Christmas.
Best regards.
Jabrane.

Regarding energy, continuum mechanics theory imposes that energy quantities (in J) obviously increase with sample volume for a given stress-strain behavior. Which is why energy quantities in J are again not so meaningful in my opinion, compared with energy densities in J/m3

Keeping the same sample volume and the same stress-strain behavior (I guess you would also like to match this one to the experiments), you thus will have no possibility to change the energy input to the sample...

You may just possibly play with micro-parameters to maybe change the relative weights of distinct energy features in your simulation (energy dissipated through plastic sliding i.e. friction, vs elastic energy losses when cohesive interactions break)

Jerome