



joed
Mar 30, 2010, 12:56 PM
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Right, I'm a bit confused. While an engineer at heart, I've not the education nor experience of one, so I'm hoping someone can point out any mistakes or otherwise shed some light on the situation. Am considering axle material and strength. See attached images (yes, I got a little carried away with CAD). I measured the axle diameter of a size 2 friend at 1/4” = 6.35mm. Noting that 1 MPa = 1 N/mm^2, gonna stick with mm. Assumed a fall load of 10KN, giving 2.5KN vertical force per lobe, and with a cam angle of 13.75*, get tan(13.75) ~ 0.25, giving 10KN per lobe normal force. Forces shown are those exerted by lobes on axle. Internet research tells me that bending stress is given by bending moment / section modulus. Section modulus for a circular shaft, given by pi x d^3 / 32. So for this shaft, I get section modulus = 25.14. Now, first considering the forces aligned with the stem, the 2.5KNs. Looking at the bending moment about the outer edge of the stem head. Moment = 2500 x (14 + 3) = 42500 Giving stress = 42500 / 25.14 = 1690 N / mm^2 And next, the 10KN forces. Firstly, moment caused by the outer lobe about the outer edge of the inner lobe (the 8mm distance drawn). Moment = 10000 x 8 = 80000 Stress = 3182 N / mm^2 Now, both 10KN forces about the outer edge of the stem head. Moment = 10000 x (14 – 3) = 110000 Stress = 4375 N / mm^2 So, here's my issue. What on earth is the axle made of??! Wild Country website say Chrome Molybdenum. Various material specifications from the web indicate that yield stress is unlikely to exceed 1000 N / mm^2, and ultimate tensile strength may exceed this only by 10% or so. Any thoughts? What have I forgotten to consider?

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dimensions.png
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forces.png
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soNVclimbing
Mar 30, 2010, 1:07 PM
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I am not sure but it sounds like chrome molybdenim is chromoly as we refer to it and is probably 4130. I weld on it all day long, we use it for race car chassis and components. Use ER70s2 weld rod.





Rudmin
Mar 30, 2010, 1:27 PM
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joed wrote: Now, both 10KN forces about the outer edge of the stem head. Moment = 10000 x (14 – 3) = 110000 Stress = 4375 N / mm^2 Not sure about the other calculations, but this one is not correct. The horizontal moment force is only between opposing lobes, not between the lobe and the stem. Also, consider that your actual maximum stress will probably be the tensile stress caused by bending. You can't just add the maximum stress from different calculations together. Your bending moments are being applied at 90 degrees to each other and in different locations, so the maximum stress is not simply the addition of the maximum stress caused by each bending moment because those maximum stresses will be in different locations. I would double check your answers against a beam stress calculator.





adatesman
Mar 30, 2010, 1:29 PM
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joed
Mar 30, 2010, 2:04 PM
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Thanks, I'll give that calculator some thought.
Rudmin wrote: Not sure about the other calculations, but this one is not correct. The horizontal moment force is only between opposing lobes, not between the lobe and the stem. . . Your bending moments are being applied at 90 degrees to each other and in different locations, so the maximum stress is not simply the addition of the maximum stress caused by each bending moment because those maximum stresses will be in different locations. Are you sure about the horizontal moments?? Why would they not contribute a moment about the stem head (or any other point for that matter)? They cancel out when resolving forces horizontally, but that doesn't exclude them from contributing a moment. Take them individually:  the outer lobe produces a moment about the stem head.  the inner lobe equally produces a moment, but of opposite direction.  therefore net moment is one minus the other? Place a weight on the end of a ruler, and a support in the middle. Applying a torque about the other end (as opposed to second weight) will put it in equilibrium  the two forces (weight and support) do contribute a moment about the other end. And yes, I hadn't been intending to add the horizontal and vertical moments directly. I doubt that stresses add like this, but my first guess trying to add them would be as per pythagoras.





adatesman
Mar 30, 2010, 2:20 PM
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(This post was edited by adatesman on Aug 12, 2010, 1:02 PM)





jeremy11
Apr 2, 2010, 4:03 PM
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Great discussion, great link. This would have been potentially helpful back when I spec'd the Al axle for my monster cam. I overbuilt it significantly since I could calculate the forces at the axle but didn't know how to figure out what diameter of what alloy I would need to resist those forces. Looks like this is the answer! The formulas are also surprisingly simple.





