From my understanding, option 1 will fix the thing in space, and anything that touches it will thus be fixed to that contact point. Option 2 will make the connecting faces rigid with respect to each other, but they could still move in space if there is some other load (gravity, a side force, moment, etc.).
So if you want some big lug to weigh on your machine frame and cause some deflection, use option 2. The contacting faces will maintain the same inital orientation wrt each other, and they will displace in the same manner.
If your part connects to a bracket that is essentially bolted to the ground, you could just select the bracket and say "make rigid". You care about the other stuff, not the bracket. I don't think this will drastically increase simulation time, because the displacement equations for those nodes will all be equal to zero (it's fixed, right?) and therefore easy to eliminate from the solution matrix. If it's rigid, the displacement is zero, the strain is zero, and the stress is zero.
(I just read your other post, so I guess everything after the first paragraph is not what you need, but maybe somebody else would like to know.)
I hope that helps you decide which way to go.
Thanks for the input, David.
Regarding option 1---I wonder how long it is before I run into a meshing problem with a "rigid" part. I just don't see the point in meshing it.
Perhaps it is because of the other case---where you're attaching a rigid part to a non-rigid one, which is grounded somewhere. Imagine you have a cantilever beam with a big weight fixed at the free end. You can imagine how you can use the three options I've listed to simulate that weight, since we don't care about the deformation of the weight itself. I guess in this respect, the mesh is "necessary" for the rigid weight just so you can calculate the displacements (essentially an extension of the free end of the beam). So the strain is zero in this case, but not the displacement. Could this be why the part is meshed?
I'm convinced the results would be the same for the cantilever beam case with all three options. When I have some time, I'll make the model and post results.
I was actually a bit confused up there...
"Fix" is the FEA option that makes the thing rigid AND fixed in space. "Make rigid" just means that nothing in the body can deform. It could all translate if it was connected to something deformable.
I see what you mean, "I wonder how long it is before I run into a meshing problem with a "rigid" part. I just don't see the point in meshing it." Also, from a small test I just did, it actually takes a bit longer to run a study with a rigid (or "fixed") component in it.
Good luck! I guess one does what one must!
Attached is a sample case of a cantilever beam with a weight at the end with a load of 1 g. The beam is 1x1x24 inches, 6061, and the weight is 4x4x4, steel. The mass of the weight is considerable. The results are interesting.... For those who choose to play with it, I put a sensor at the end of the beam on the "top" face, so you can quickly probe for calculated defleciton differences. The mesh is identical (meaning it was generated with identical parameters) in all cases.
Here are the calculated deflections (mm) at the end of the beam:
1. With no weight at all (self-weight deflection of the beam) - 0.1227.
2. With the weight treated as rigid: simulation tree->parts->weight->make rigid. - 0.3681.
3. With everything meshed* - 2.927.
4. Remote mass: simulation tree->parts->weight->treat as remote mass**; default "global bonded" contact set - 2.927.
5. Remote mass; manual bonded contact - 2.927.
6. Remote mass, no connections*** - 2.927.
7. Manual remote mass---exclude the weight part, then: simulation tree->external loads->remote load/mass->load/mass (rigid connection)**** - 2.926.
8. Remote load: simulation tree->external loads->remote load/mass->load (direct transfer)***** - 2.927.
9. Remote load: simulation tree->external loads->remote load/mass->load/mass (rigid connection) - 2.926.
10. "Hand calc" (using Engineering Power Tools) - 2.499 (if you add case 1, you get 2.622).
* When both parts are in the analysis, there is a manually - created contact set (bonded) between the weight and the end face of the beam.
** Remote loads are applied to the [free] end face of the beam.
*** If you have no connections defined, the mesher will warn you that the weight part is unconstrained and ask you if you want to remove it from the analysis. I said yes, which, apparently, did not suppress its treatment as a remote mass. Evidently this is a bug; I shouldn't have to define a contact set for a part that I've replaced with a remote mass.
**** I used my CG macro to put a coordinate system at the CG of the weight first (see https://forum.solidworks.com/thread/27603?tstart=0) .
***** Only add a force equivalent to the mass times the acceleration due to gravity, and make it in the same direction as the overall gravity force. According to the help file, Solidworks will calculate the force and moment to apply to the face you select.
It is a mystery to me what Solidworks is doing in case 2. Seems like total garbage. Moreover, it is the one case that is substantially slower than the others. There is a step called "forming surface to surface bonding" which makes this case take about twice as long as the others (they are all very fast in this case).
cantilever_weight.zip 159.6 KB
Good analysis, Emilio!
I think I know why case 2 is so different...
If you make the weight "rigid", then it helps to stiffen the beam (a lot, apparently!) because the end face of the beam has to remain in perfect contact with the rigid face of the weight. Normally, a beam with a cantilever load like that would have its bottom edge a bit further out than its top end edge after deformation, right? But since you told it to be rigid, then the end face of the beam can't deform into a curved surface - it's been stiffened by the rigid weight.
So we should use "make rigid" carefully in our studies - it may have more effect than we expect.
David, I think you are implying making a component rigid also constrains it somehow. This doesn't seem to be the case. Moreover, at the interface, you have 1 square inch of aluminum trying to deform a steel body. I think it's safe to say at that point that this face is undeformed. Only when shear becomes important ("fat" beams) does the shape of the face matter, and I made this beam in a 24:1 ratio to try to avoid that.
Attached are some screenshots of the deformations. You'll see that the rigid mass can indeed translate, and also rotate---but it is not nearly the same amount of displacement. The coordinate system you see is the original location of the CG of the weight. The case with no mass at the end is representative of all cases where the weight part is excluded and replaced with a remote mass or load, either "manually" or "automatically". The deformation scale in the plots below is equal (but the colors do not correspond to the same number).
First, the weight treated as rigid.
Now, the weight meshed along with the beam, and "bonded" at the end face.
The weight replaced with a load, and the end face presumably made "rigid" so that it must remain planar (and a 1-inch square).
The issue with the "rigid" body not behaving correctly is a problem in the current version (SP4.1) of simulation. It is now SPR 522706.
The fix is implemented in 2010 SP2