Hi Jan,
My two cents about symmetry of stress :
f*branch.transpose()
R*T*n[i]*t[j]
This definition of stress is assuming static equilibrium. In this
situation, there is a mathematical proof that the result will be
symmetric, even if it there is no obvious symmetry in the formula.
IIRC, the proof is based on (1) the decomposition of branch into
O1C-O2C, and (2) the fact that the total torque of all forces on one
grain is 0.
If we are not stricly at static equilibrium, this definition does not
apply. DEM is never stricly at equlibrium, so this stress is never
exactly symmetric, but it is usually very close.
If you do this :
.5*(f*branch.transpose() + branch*f.transpose())
R*T*.5*(n[i]*t[j]+n[j]*t[i])
which is standard way of tensors symetrization.
you get something symmetric. Ok, it looks better, but you are in fact
only hidding the initial mistake : abusing static equlibrium
assumption, and the result in itself is not better.
I'd keep the non-symmetric definition, and the non symmetry of the
result would just remind people to be carefull with that.
For dynamic cases, there are other definitions of stress, considering
kinetic energy of grains, probably not what you want at the moment.
Cheers.
Bruno
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