[lammps-users] how does LAMMPS shear a box ?

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1

Hi,

In “stress/compute” command manual, it’s said that LAMMPS returns six components of symmetric stress tensor which are xx,yy,zz,xy, xz and yz when a box is deformed. My question is that when I am deforming a crystal in shear, does lammps do it by developing both XY and YX stress (to keep the stress tensor symmetric) or only by XY stress element in the stress tensor (YX being zero in that case ) ?

In other words, when a box is being sheared, is it being applied upon two stress elements (XY and YX in two opposite sense ). In solid mechanics, shearing of a box can be done either by application of only XY stress on one face defined as ‘simple shear’ or by applying both XY and YX at the same time defined as ‘pure phear’. Could anyone please point out which one of these LAMMPS is doing when deforming a box in shear ? If it is doing it by pure shear , how am I supposed to apply simple shear then and vice versa ?

I am using July 2010 version of LAMMPS.

For clarity, I have attached a schematic to help you visualize my problem. Any help would be greatly appreciated.

Thanks,

  • Piyas

Schematic.ppt (230 KB)

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Compute pressure and stress/atom calculate the stress state
of the system whether the box is deformed or not. They
are simply diagnositic measures. The fix deform command
works by changing the shape/size of the box to set
a strain, not a stress. Whatever stress results from
that is an a consequence, not a cause. The shape of
the box is described in the fix deform doc page in terms
of "tilt factors" which are explained in a geometric sense on the region prism
doc page.

Steve

3

Thanks a lot, Steve, for the comment.

  • Piyas