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Polyurethane De-Lamination Formula

Polyurethane De-Lamination Formula

Polyurethane De-Lamination Formula

(OP)
I'm a design engineer for a well established a bearing manufacturer. We have recently in the last few years added products to our catalog that include a Cam follower and Yoke type bearing with a Polyurethane tire molded to the outer ring by means of a bonding agent. We did some test studies to determine maximum load and de-lamination. We discovered that The 1/8" thick tire would de-laminate if we mounted the bearing so it had a radial load and had more than 1/32" of polyurethane compression. The polyurethane we tested is our standard that has a durometer of Shore A 80. I was wondering if there is a standard formula or calculation to determine this point of de-lamination with other thicknesses and durometers? I'm sure there are lots of variables to consider, but I'm just looking for a basic formula so I can code a calculator for our engineering department which would save us some time. I haven't been able to get any information on this anywhere. Does a formula for this exist and if so could someone point me in the right direction?

RE: Polyurethane De-Lamination Formula

1/32" compression on a 1/8" thickness is 25%. You probably have even higher peak shear deformations. I don't think any elastomer will last long cycling that much. This is something that requires a great deal of specific relevant experience in elastomer formulation and application. Adding some fiber or wire mesh reinforcement to the rubber tire would help. Good luck.

RE: Polyurethane De-Lamination Formula

The main mistake people make approaching elastomers is they think the elastomers are compressible. If they were then 25% would be not a big deal. In fact, their compressibility is near enough to zero to not matter.

The upshot of this is that the shear loads that are generated by attempting to compress them will easily exceed the material or any adhesive bond. I've seen this play out several times; the successful resolution was to stop trying to restrain the elastomer, instead just capturing it to keep it from leaving. In your case, just limiting the allowable load is an option.

Best example I saw was a thick wall tube and matching pin (demo from a honing company) that had a fit close enough that it would hold moderate air pressure. It also included a rubber cylinder. One could put that rubber cylinder into the tube, set it onto a table top, and put the pin in after. If the pin was pushed to contact the constrained rubber and then the pin hit with a metal hammer it made the solid metallic ping as if the pin was on an anvil. No noticeable compression in the rubber.

What makes it worse is that the edge of the adhesive contact on that tire has zero beyond it to resist the shear - guaranteeing that the failure will start there and progress inward until complete failure. And that is in a static condition. Make it dynamic and the mating surface cannot be counted on to supply friction to help resist the shear load.

Best of luck, but low strength, high Poisson's ratio is a problem for such applications.

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