9 Replies Latest reply on Oct 8, 2016 4:22 PM by John Willett

    "Free Body Force" Definition in SolidWorks

    John Willett

      What exactly is a "free body force" as reported by SW under Simulation/Result Tools/List Free Body Force?

       

      SW KB entry S-035204 says in part:  "In SolidWorks Simulation, ...and free body forces are used in accordance with their respective conceptual definitions in basic physics...  Free body forces... show the equivalent forces that are acting upon a body or portion of a body, including contact forces, external loads, restraints, and connectors. If you look at the free body force for an entire component, you should see that all the forces sum to zero if the body does not have any rigid body motion."  [Emphasis mine.]

       

      This "definition" is not completely correct since there appears to be no way to select "an entire component;" and in any case it leaves the question open, especially with regard to gravity, "Inertial Relief," and other body forces.  Body and surface forces would normally appear together in the kind of "Free Body Diagram" found in elementary physics, often all acting at the center of mass of the body to determine its linear acceleration.  (Obviously I'm ignoring torques here for simplicity.)  As far as I can find with Google (which of course knows all), there are no "conceptual definitions in basic physics" for the term, "free body forces."

       

      In this forum the best I can find is the thread at Free Body vs Reaction Force , in which Mauricio Martinez-Saez provides an beautiful discussion of reaction forces but skirts the question of free body forces.

       

      Can anyone further clarify the meaning of the term?  My guess is that the "free body forces" reported when a particular surface is selected include contact forces and torques (normal and friction) + all external loads + reaction forces and torques + the small fraction of body forces that is associated with the nodes actually on that surface.  This could be pretty misleading when body forces are present.  (Interestingly, free body forces can be measured across bonded contacts as long as they are created as separate contact sets.  Global bonding is apparently not distinguishable from a single part.)

       

      SW's use of the related terms "contact force" and "reaction force" appears pretty straightforward (except with regard to torques, which are apparently not included even when present) in that the former applies to other bodies (though only through contact conditions other than bonded), the latter to restraints, and neither measures external loads nor (I think!) fractional body forces.

       

      Do I have it about right, or am I still muddled? -- John Willett

        • Re: "Free Body Force" Definition in SolidWorks
          John Willett

          A bit more on the above:  The following simple assembly has two square columns in contact, with no-penetration conditions between the contacting faces (there are no split lines):

          Two Forces.png

          The upper column is "pulled" down on the lower with a uniformly distributed external load (including the contacting region).  Simulation/Result Tools/Contact Force shows 10 lbf equal and opposite between the two faces as expected.  Simulation/Result Tools/List Free Body Force shows the same -10 lbf on the upper face of the lower block (contact force), but 0 lbf on the lower face of the upper block (contact force cancels external load).  Conclusion:  Free Body Force equals the sum of Contact Force and External Load.  (I haven't figured out a way to get reaction forces into the summation because they apparently cannot exist on the same surface as either of the other two.)  Comments solicited... -- John Willett

            • Re: "Free Body Force" Definition in SolidWorks
              Janko Stellaard

              Hi John,

               

              It is indeed the sum (or overview) of ALL forces acting on all the faces of a component.

               

              In my opinion it is just a tool, that makes postprocessing a bit quicker. You don't have to switch between contact/friction force and reaction force. In one view you can see all the forces acting on a body.

               

              Janko Stellaard

              Cadmes BV

              The Netherlands

                • Re: "Free Body Force" Definition in SolidWorks
                  John Willett

                  >>[Free body force] is indeed the sum (or overview) of ALL forces acting on all the faces of a component.<<

                   

                  Janko et al. -- That's what I had thought, but look at this example.  Here is a simple model of a glass bar supported by two near-weightless metal plates.  Gravity acts on the bar in the -X direction, with the foot of the "upper" (right-hand) plate fixed and bearing all the weight while the foot of the "lower" (left-hand) plate is sliding so cannot bear any weight:

                  Full Model.png

                  Looking at the sliding foot of the left plate, we see that "reaction force" gives only a +Y-directed force supporting the plate, as expected because of the torque generated around the "upper" foot by the weight of the plate:

                  Reaction Force Sliding Foot.png

                  "Free body force" gives essentially zero Y-directed force (which would be expected, since the fixture counteracts the -Y-directed force exerted by that foot); BUT it also gives an unbalanced force in the -X (sliding) direction, which makes no sense to me at all:

                  Free Body Force Sliding Foot.png

                  (I guess it doesn't bother me too much that all the moments are near zero, even though the bar does seem to be experiencing some bending...)

                   

                  Is it coincidental that this -X-directed free body force is identical in magnitude to the +Y-directed reaction force found in the previous illustration?  Does anyone have an explanation for this paradoxical result? -- John Willett

                    • Re: "Free Body Force" Definition in SolidWorks
                      Bill McEachern

                      It would not surprise me if the there is an error in reporting the free body forces and they have the Y component reporting in the X component spot and vise versa. Try another example with non trivial loads and see how it looks. This type of error is not that uncommon in SWX Sim.

                        • Re: "Free Body Force" Definition in SolidWorks
                          John Willett

                          Thanks, Bill -- My reflex is to trust its answers more than I know I should and then to get upset when I find a contradiction.  How many of these little details are worth reporting through the VAR to SW? -- John Willett

                            • Re: "Free Body Force" Definition in SolidWorks
                              Bill McEachern

                              Hi John,

                              they are all worth reporting but you will get fatigued doing it If you reportEd everything. The Swx support guys if you get to them just want to get you off the phone and typically think you don't know a thing. The VARs try, well some of them, and they typically aren't up to the task having sufficient convection on a given issue, particularity given how unreceptive Swx support typically is, well at least that is my experience over the decades. They are a CAD company, analysis just doesn't get the love it deserves given its criticality. I would be a fan of charging a lot more and having a far more bullet proof program but that's way above my level of influence.

                            • Re: "Free Body Force" Definition in SolidWorks
                              John Willett

                              Previously I wrote about the sliding foot of the left plate:

                               

                              "'Free body force' gives essentially zero Y-directed force (which would be expected, since the fixture counteracts the -Y-directed force exerted by that foot); BUT it also gives an unbalanced force in the -X (sliding) direction, which makes no sense to me at all...

                              (I guess it doesn't bother me too much that all the moments are near zero, even though the bar does seem to be experiencing some bending...)"

                               

                              The above result came from the Iterative solver, but I just repeated the exercise with the Direct Sparse solver (had to decrease mesh resolution slightly so as avoid the "running out of integer limit" error) and got much more credible results:

                               

                              "Free body force" now gives a non-zero Y-directed force (neglecting the reaction from the fixture for a much more useful result), zero force in the -X (sliding) direction as expected, and a non-zero Z-directed torque that is consistent with the torque on the other end of the same bar.  Much better!

                               

                              I guess you are correct that it's just another silly SW error. -- John Willett

                        • Re: "Free Body Force" Definition in SolidWorks
                          James Riddell

                          I have a suspicion that this might have to do with the common edge between the applied force 'downward' and the contact force upward.  Have you tried a spit surface for the applied force a short distance away from the supporting column? 

                          I, too, would have expected to see a -10 force on the upper contact surface.  Seems like it would be a logical improvement to SW to have the option of not/including adjacent edge/surface forces.  It's always a good sanity-check when the FBDs work out the way theory suggests they should.

                            • Re: "Free Body Force" Definition in SolidWorks
                              John Willett

                              James -- I tried your suggestion of eliminating the common edges by making the lower block narrower.  Same result.  (Note the sign correction to the force on the upper face of the lower block in my 2nd post, above -- should have read -10 lbf.)

                               

                              Of course this example is somewhat contrived; you wouldn't normally have an external load acting on a face that's also in contact with another body.

                               

                              I also tried applying the force only outside of split lines that exclude the contact area.  In this case the Free Body Force on the section of the lower face of the upper block that includes the contact area goes back to equal the contact force of +10 lbf.

                               

                              I think Janko Stellard is right and zero is the correct answer for the original Free Body Force on the bottom face of the upper block (contact force of +10 plus external force of -10 = 0).