Stress-strain curve for the direct shear test

Hi Thao,

I sort of did what you describe in my paper here: https://www.sciencedirect.com/science/article/abs/pii/S0266352X18302477. You can email me if you want a copy: fegarcia@umich.edu.

The only issue is I did my post-processing after the fact using Eqn. 2 in the paper, which is a classic stress homogenization equation that utilizes contact force vectors and branch vectors. I output the contacts data at sequential stages of the direct shear test and then processed it in Matlab. I did not actually do these calculations directly within LIGGGHTS. If you want to modify the LIGGGHTS source code, you could probably output the product of branch vector and contact force vector and then run a compute reduce in a specific region.

Sincerely,
Estefan Garcia

Hi Thao,

I'm honored you're using my paper :-). You can email me anytime.

As for your two methods, the first one will likely not work on its own. It is a variation of the same equation I used, but it uses the volume of the entire domain. Assuming your simulation contains direct shear test walls within the model domain, the volume of the particle assemblage is less than the volume used by the compute pressure command. So your stress calculation value will be less than the actual value. The numerator of the equation will actually be correct for the entire particle assemblage, so you can adjust your results by the total volume of the direct shear test box (including any displacements of the loading platen). However, you will still underestimate your stress as measured at the wall boundaries because those shear stresses are concentrated within the shear band along the mid-plane of the direct shear box.

The equation in the second approach does look very similar (although not exactly the same) to that for the first approach. They both utilize the products of branch vector and force vector. If you just sum up all of the per-atom stress tensors, it looks like you'll double count the forces between each grain, so you'll need to factor in a 1/2. It's also not dividing by the volume, so you'll need to do that too. I'm curious how your first approach will compare with your second approach if you divide your answer in the second approach by 2*(Domain Volume). Keep in mind, this will still be an underestimation of the shear stresses measured at the wall boundaries for the reasons mentioned earlier.