I'm running SW Premium 2017 SP2.0, so it's linear static simulations only. (This post has now been edited multiple times as I learn more.) I guess there are really three questions:
1) Engineering: Am I getting wrong answers by using linear static on this problem?
2) Solidworks: Is there a way to get the Large-Displacement option to work with this model, as SW Help implies that it should?
2a) Combined: If not, can anyone think of a work-around to get the same boundary conditions without using a shell?
The problem is shearing a right-circular cylinder between rigid, parallel planes. The bottom plane is represented by a fixture. The top plane is displaced parallel to itself (X direction in image below) and must be allowed to move perpendicular to itself (Y direction) but not to tilt. The only way I found to do this was to use a very stiff shell for the top plane and to constrain the edges perpendicular to that displacement so that they could not rotate around themselves (Z direction) like this:
(The dimensions of the cylinder are 3/8" X 1/8".) Linear static simulations with small X displacement give near-zero Y displacement of the shell. Large-displacement solutions give essentially the same until I make the prescribed X displacement 8X larger than the .001" shown here. Then they fail during "Equilibrium iterations" within a single load step. I've tried many combinations of settings in Study Properties to no avail. If I make the X displacement significantly larger (say .02"), LD will set up more than one load step, but it still fails the same way. Converting the problem to applied force instead of displacement doesn't appear to help either, although I can get it to go much further into the first load step before reporting, "Equilibrium satisfaction is not achieved."
I expect near-zero perpendicular displacements for very small parallel displacements (cosines of small angles). I believe the planes must be pulled closer together, however, as the shear becomes large because the material begins to be stretched in tension (the volume is not changed, of course, under "simple shear," I think it's called), but I'm guessing it takes fairly large shear for this to show up. Is my intuition wrong, or is my linear static solution just letting me down? -- John Willett
Donut with No Hard Point 2.zip 288.2 KB