The problem of support reaction forces

[a126z]

1. In ABAQUS, for a three-dimensional cantilevered cylinder structure, there is only a force in the Y direction at the top of the free end, so why is there a branch reaction force in the X and Z directions at the solid end (with all six degrees of freedom constrained)?
2. The support reaction force is in equilibrium with the external force. But the external force is only Fy, so the support reaction force should only be Fy and Mx. In fact, the values of the support reaction forces Fx,Fz,My,Mz are not 0, why is this? I want a scientific and reasonable explanation.
3. I have turned off geometric nonlinearity and the material is linear. I have only defined the modulus of elasticity and Poisson's ratio. In addition, the structure is subjected to a force of 1E4 kN, even if I change it to 100 kN, the phenomenon is still the same.
4. The following figure shows the resultant bearing reaction force calculated at 1E4kN with the model.

 

[rb1957]

1) round off error ? ... 13 orders of magnitude = pretty small error



"Hoffen wir mal, dass alles gut geht !"
General Paulus, Nov 1942, outside Stalingrad after the launch of Operation Uranus.

 

[SWComposites]

Yeah, who cares? Those Y and Z values are tiny.

FEM = an approximate numerical solution of an approximate representation of reality.

 

[a126z]

Thanks for the answer. Because I tried to load the cantilever beam with the same equivalent force for the combined forces, I found that these tiny amounts are not equal at the supports. Although this is a small order of magnitude, I wanted to explore what this small amount is due to. According to the balance of internal and external forces, these values would be equal to 0. But in reality they are not equal, so I would like to inquire what might be the cause?

 

[rb1957]

I'm not sure what you mean. I'm not sure what this table is.

Is it the table of applied loads ? (then it "must" be your applied load, with moments about the origin)

Is it the table of reactions ? then it should equal your applied load. these round-off errors (small though they are) are quite large.

I would try a simpler model and see if you get the same.

I would ask Abaqus helpdesk.

I would try some Abaqus example problems.

Have you turned off "double precision" ?

"Hoffen wir mal, dass alles gut geht !"
General Paulus, Nov 1942, outside Stalingrad after the launch of Operation Uranus.

 

[a126z]

No, it is not the apllied load. The applied load is on the concentrate force which is 1e7 N in the top of the free section.

Yes, the values in the table refer to the reactions of the support.

Even though I tried a simpler 3D beam model, the model still has a small reaction force in the direction of the unloaded external force at the solid end, why is this?

As shown in the figure, the numerical magnitude of the shear force in the section cut out is not equal to 1e7 using a single-precision static universal analysis in the above case of a force of only 1e4 kN (i.e., 1e7 N)?

 

[BrianE22]

I'd guess the digital nature of the calculations. The computer is solving thousands of digital conversions of a continuous load and deflection problem. There's rounding errors of the continuous load due to the digitization. They add up throughout the part.

Try a coarser or finer resolution. I'm guessing the error will be more or less for different resolutions.

 

[SWComposites]

1) plot the deformed geometry (this should ALWAYS be the first thing plotted); this should give some insight
2) plot the X, Y, Z reaction forces at the fixed end, separately.
3) what you are seeing is likely a result of the left end fixed from displacement in all directions

 

[rb1957]

"single-precision" ... yep, round-off. The round off is 13 orders of magnitude less than the the applied force. If you question this, ask the Abaqus helpdesk.

"Hoffen wir mal, dass alles gut geht !"
General Paulus, Nov 1942, outside Stalingrad after the launch of Operation Uranus.

 

[SWComposites]

sigh ... when I was in engineering school long ago, just after the days of slide rules and the last class to use punch cards, it was beat into our thick heads to pay attention to significant digits in calculations, and if one put too many digits in the calcs for homework or exams the score would be 0, no partial credit; most of us learned quickly.

what part of "finite element analysis is an APPROXIMATE numerical solution" is so hard to understand?