Radial stiffness of PreCalibrated Bushings ( Wöhler - Curve; W-N curve )

AlexJagow

Hello "Mecway Wold", at first I´m a newbie in using Mecway for FEM. After searching at the 35 Forum pages for combining loads and looking into the manuels for preloads, I would like to ask you directly for any kind of support/ help. I´m looking for a possible calculation method for rotationally symmetric rubber bushing(Outer diameter~Ø60mm). The rubber bushing has an outer and inner sleeve out of metal. It should be calibrated by radial force(or pressure) at first( for approximately 3mm of diameter) and than I would like to add some additional radial force onto it to calculate the radial stiffness. Inside the manual of Mecway I found the pre-tension method only. Do you see any possibility to conduct such analysis? Like, conducting the calibrating at first and save the results to use it for further analysis, or combining the calibration and radial force in one analysis depending on time steps? Looking forward hearing from you Alex

disla

Hi AlexJagow,

¿What do you mean by preloading the bushing?.
I have done an example like loading in one direction first and then radially on a rubber bushing. ¿Is it like what you are looking for?
It needs ccx to run.

File edited

Victor

Just concurring with how disla's example does it. You don't need anything complicated like multiple steps or restarting the solver to apply different loads in sequence. Just define their time dependence with a table or formula. Do that with Nonlinear Static 3D analysis type and Quasi-static turned on.

Sergio

Hello @Alex, for rubber bushes is very important to take in count the thermal restrain prior to the assembly operation and then the mechanical load, due to the high stresses related to high thermal restrain of rubber compared to metal or plastics armatures. I did a lot of that kind of simulations using Abaqus and few using Mecway/CalculiX. The easier way to me was define one step for thermal restrain, second for assembly and the last for main loading.

Sergio

Some pictures and animations of a very simple bush:

https://drive.google.com/drive/folders/1BNMXJYl-5QSG4qEw49QTRvr0yvvjuDiV?usp=sharing

Be carefull in case you include the outer/iner rings, as CCX will promediate the stress between rubber and metal at the interfaces if you use coincident nodes.

JohnM

A trick I have used for a quick "press fit" check is to apply a frictionless boundary condition on the radial surface, but use an actual value to displace/crush. This works well, especially if the parent material is relatively stiff.

I will also recommend getting to know the hyper elastic material models, they can be useful.

Sergio

If is only for compute the stiffness or stress in the part, for a cylindrical bush the inner/outer sleeves are not needed for the simulation (unles they get in contact with the rubber or have some special shape). The press fit can be modeled using a cylindrical coordinate system (TRANSFORM card) asociated to the outer set of nodes of the part and applying a radial displacement (1st degree of freedom in a cylindrical CS).

AlexJagow

Hello MECWAY Users & creators, wow I´m really impressed about your feedback. I have to do some homework now, as it seems. :-) Thank you so far and may the finite element be with you. With best regards Alex

disla

I have edited my previous file to add some thermal restrain as Sergio suggest and contacts between rubber and metal. Still below 1000 nodes.

Sergio

Woooow, nice example @disla, but there is a mistake. The order of the loads must be:

1) Cool down the part from mold to room temperature (so the temperature and restrain is a lot more). The rubber will get streched in this kind of assembly as in your model
2) Mechanical swage or assembly operation, so normally the outer sleve is pressed fit radially, in part to reduce the stresses due to 1), and to keep it in place. The rubber will get without stress or better, with a little of compresive stress
3) Now that the part is at room temperature and at assembly condition, you can apply the main loads (radial, axial, torsional, conical... or any combination). Sometimes you have a preload step with the gravity loads of what you are keeping with the bush, and then yes, the main load acting in the part.

disla

Thanks Sergio,

You are right. My simulation could be a bracelet on a cold day. .
I hope AlexJagow didn’t take it too much verbatim.

I couldn’t figure what was the real application for that. I couldn’t access to your example as I was linked to google drive and I’m not registered. I will try to adjust numbers to make everything meaningful now you have explain the procedure and everything has more sense. This could be some kind of silent block isn’t it ?¿.
I think most of the steps ,boundary conditions and external actions are still ok to make it work. I can only foresee one delicate step ,..the rubber assembly to recover some of its flat shape after shrinking. I have an idea on how to solve it that could be interesting if it works.

Thanks again for your comments.

Sergio

@disla, the procedure for simulation is the normal for a silent block or any part that is molded with plastic or metal sleeves or inserts, due to the big difference in thermal restriction of rubber and those other materials. About the press fit, the radial displacement using a cilindrical coordinate system is the easier way, tested several times. You just define a cilindrical CS with the TRANSFORM card aligned with the bush, and then affect the exterior nodes of the part. Then the first degree of freedom is a radial displacement for those nodes, and is easy to apply a press fit directly without any kind of contacts. I have attached the files here directly, is an old example that I did for a training

disla

Thank you very much Sergio,

The transform Card is new for me and it works perfectly. I wasn’t able until today to apply a radial displacement properly.
How I feared the most tricky part is that adjustment to set the Rubber in place again and out of stress.
I found that the price to pay is that the external sleeve ends up in a large compressive stress due to the displacement but I can imagine the Rubber is the main object of interest.
I will do a second try to see if there is some workaround.

@John Thanks also for the comment. I still do not understand your trick for a quick "press fit".

“apply a frictionless boundary condition on the radial surface, but use an actual value to displace/crush”

¿Does it avoid the Compressive stress remaining on the Outer Sleeve.?

Sergio

Very very nice model... a last trick. In order to better/quick convergence all the loads are applied as displacements and not forces, and, a RIGID BODY is created between the movil side (inner sleeve faces in this case) and a node in the center of the model. Then the displacements are applied to this pilot node, so you can apply directly a rotation for the torsional or conical cases. Also this would allow you to ask for the reactions on this node to create the plot of load/deflection, that is one of the main results requested for this kind of analysis.

disla

Yes, you read my mind.
I think the "REF" node is still there but rigid body and shell elements ...mmhh, not yet...and if I convert them to solids the model had more than 1.000 nodes.
AlexJagow could give it a try for sure.

Sergio

For analizing this simple bush the outer/inner sleeves are not needed.

AlexJagow

Dear Sergio and disla, really cool. I start understanding your contacts and contrains. Without your examples I wouldn`t catch that. I will try to conduct the analysis with solids. Thumbs up, you made my day. Alex

disla

Updated for v17. Now it works again.

From Victor solution ( https://mecway.com/forum/discussion/1180/two-concentrated-loads-fluxes-different-amplitudes)

"Version 16/17 applies shell edge forces in a more accurate way that includes moments that account for the geometry change due to the curvature. It looks like CCX has trouble with these moments in this case.
A workaround is to change the two forces on the shell edges to be on the nodes instead. It should be pretty close to the same thing since these are linear elements. Then it does solve."