I’ve been trying to help some other teams figure out why their wheels are coming apart, and I’d like to have the CD engineering-types check my math on this. If you have the time and inclination and expertise in the area, I’d very much appreciate a look-see. (I’m a physics teacher, not an engineer, so I think this is right, but I’m not positive that I’ve got all the details where they need to be.)
Assuming this is right (and/or once corrected), I might take the time to throw it into a spreadsheet so people could plug in their own numbers for their particular setup and see what kind of expansion they might expect (and thus determine if they need to use banding wire, or a different wheel, or whatnot).
These numbers are taken from the 80A Andymark kitbot traction wheels, estimating that the 8 oz of wheel has about half of that in rubber weight on the outer rim.
Four ounces (0.12 kg) of rubber on a 6" (0.15 m) wheel spinning at 7000 rpm has a surface speed of (7000/60) x pi x 0.15 = 55 m/s, and requires a force of Fc = m x v^2 / r = 0.12 x 55^2 / 0.075 = 4840 N to keep it that way.
Assuming that’s spread uniformly over the surface of a 1" wide wheel (which would have a total surface area of 1" x 2 x pi x 0.075 = 0.012 m^2), it will experience a pressure of P = F / A = 4840 / 0.012 = 403 kPa. The extent to which your rubber is heavier or wheel is skinnier will make this pressure go up, and the extent to which your wheel is smaller will make this pressure go down.
Young’s Modulus for Neoprene rubber is 2.5 MPa.
P = gamma x %deltaL, so %deltaL = P / gamma = 403 kPa / 2.5 MPa = 0.1612 = 16.12%.
The diameter of the rubber on your wheel is going to go from 6" to 6.97", which is more than enough to pull it right off the hub.
This is where it gets less than exact, though at first glance I think the method is right there. First, AndyMark wheels are all polyurethane rubber, not neoprene. Second, the Young’s Modulus varies dramatically with durometer.
The other thing worth mentioning, which is far more minor, is that your model assumes the polycarbonate does not itself expand, which it does somewhat.
There’s a wrench in the gears, thanks! I’m looking around online for the Young’s Modulus of 80A polyurethane, thus far to no avail. Will adjust when found.
Good call on the polycarbonate. I was ignoring it because the Young’s Modulus of neoprene is so much lower, but with the right values that might become a nontrivial contribution.
Cool. That’s still a 0.4" increase in diameter, not counting for the polycarbonate stretching inside that to hold things in, at 7,000 rpm – and 0.8" at 10,000 rpm. Enough, apparently, to cause them issues.
I found an article about converting from durometer to Young’s Modulus:
Pretty much every wheel in common use in FRC is measured using Shore A (thus the “A” in 80A). Thus, for a given durometer D, the Young’s Modulus (YM) in PSI is:
YM = 145 * e^(D * 0.0235 - 0.6403)
Multiply by 6.89476 to convert to Kpa. I can’t vouch for the accuracy of the article, but the results at least fall into a reasonable range!
For an 80A wheel, that gives you a Young’s Modulus of 3.45 KPa. For a 60A wheel, it would be 2.19 KPa.
Right, but that cross section would have proportionally less mass as well. Wouldn’t those two factors cancel each other out, because they’re both directly proportional?
Is this assuming that the rubber is an infinitely thin layer located exactly at 3" from the center? Wouldn’t it be more correct to say that the mass of the rubber is nearer to 2" or so since there is more mass near the outer edge of the wheel?
Yes. Once I’m happy with the math, you should be able to (mostly) account for that by putting in an average diameter based on your wheel construction. For the kitbot wheel, the average diameter is closer to 5.5" than 6". For a solid rubber wheel, it would be closer to 3".
I’d be interested to see the results for a smaller, denser wheel - something like a 4" colson or the 4" solid rubber wheel that WCP is selling. I’m less worried about a HiGrip style wheel - the solid rubber wheels have more mass, and more of that mass is from the thick layer of rubber.
So… I see what you are doing, considering the radial (centripetal) load needed to keep a lump of material from flying off into space.
But this is not really the way a disk of material behaves. The hoop stress is typically a larger component of holding the disk together than the radial stress is.
The challenge you have with this problem is that these wheels are not homogeneous. The tread on the wheel in your example is a different material from the wheel itself. I think you are attempting to simplify the problem by saying that the “stronger” part of the wheel does not grow significantly and the “rubber” part of the wheel wants to grow so the adhesion of the rubber material to the rim of the wheel is the weak link where the failure will occur.
I can’t find a simple calculator that can solve that problem. But if you run the calculator for the “stronger” part and then run the calculator a second time for the ring of the rim as a separate part, you could calculate how much difference in radial growth the two materials would have in their independent states. You may find that the rim itself is not strong enough to support these hoop stresses on its own and it is relying on the “stronger” material of the wheel to hold it together. There are a couple of ways that I can think of to try to calculate the stress at the joint between the two materials, but I have no reference to support these methods. This would be something that I would typically rely on FEA analysis to solve for me. But , suffice it to say that if the rim material is not strong enough to hold itself together and it is fairly thick such that the inner portions of that material cannot carry the stress of the outer portions of that material, then chunks are likely to be liberated.
TL:DR: Run the calculator for the hoop stress of the rubber tread and see if it predicts failure. If it does, then you are relying on the adhesion of the tread to the wheel to prevent the failure.
I’ll look at @wgorgen’s links for a more detailed analysis and get back to this later today if I can. (Oddly, they expect me to teach classes and stuff while I’m at school rather than chase math down engineering robo-rabbit-holes. Bummer.)
Meantime, here’s the spreadsheet thus far. Make a copy and play around if you want to put in your own data/assumptions and see what happens with the simple analysis I’ve got so far.
I ran this calculator for the 4" diameter solid rubber wheel (with a 1.25" diameter bore) at 7000 RPM.
Here are the other values I used:
rho = 1230 kg/m^3
poisson’s ratio = 0.50
Y.M. = 2.19 kpa (0.00219 Mpa)
The results show that the rim is going to try to grow by 0.33 inches (and the ID will try to grow by 0.43 inches). The tangential (hoop) stress at the rim is 0.359 Mpa and the hoop stress in the bore is 1.513 Mpa (which seems within the tensile strength limits of Neoprene rubber but not a huge safety factor to account for molding imperfections, etc.).
This result is not completely realistic however because the steel tube that runs through the bore is actually helping to hold things together. That steel tube is not accounted for and will actually carry a significant fraction of the stress and reduce the rim stresses significantly as long as the molded rubber adheres well to the tube.
In practice, when we spun our fairlane wheel up to ~7000 rpm, we only saw maybe 1/8" of rim growth. We were estimating this based on a high speed video on a phone so it was not an accurate measurement, but it seemed reasonably small (certainly much smaller than the 1/3" predicted by this calculation).
EDIT: updated the Poisson’s ratio to be more reflective of rubber and updated the results of the calculation.
Playing around with the calculator, it looks like Poisson’s ratio has a significant effect on the result, and 0.30 is typical for most metals. I found various other types of rubber (but not specifically 80A polyurethane) in the 0.50 range.
Rubber is essentially an incompressible substance that deflects by changing shape
rather than changing volume. It has a Poisson’s ratio of approximately 0.5.
So, I think you are on the right track by using 0.5 in your calculations.