justin_le wrote: ↑
Apr 05, 2018 3:48 pm
Even if the machining (not lasering, fwiw) was off and one tooth initially had most of the contact, that contact patch would almost immediately deform until other teeth or splines are making contact and things get distributed. And it does this well before
actual shear failure of any tooth.
Okay. I did this calculation last night, but it was nearly 3.00am my time, which is never a good time for math and given my previous attempts, I decided to leave till today and check it.
I have, I believe it is correct, but check it, and do with it what you will. Below is the inevitable image showing most of the salient data. (I've always drawn sketches for this stuff.)
What I set out to do was see how far the face of one of the Al splines would compress before the widening area of that face -- compress a curve, the contact area widens as it compresses -- would be sufficient for the shear strength of 6061-T6 to arrest the compression.
The formula I started with is that for the Shear modulus
: G = Fl / AΔx
G = Shear modulus (26Gpa for 6061-T6)
F = Force acting (100Nm @8.5mm = 11764N (2.644ksi))
l = Initial length (perpendicular to force for shear) = 8.5mm
A = Cross-sectional area. (I start with the contact area taken from the image: 0.34mm * 5mm(~3/16") = 1.7e-6m^2)
Δx = the compression distance I sought.
Plugging those numbers into the formula I get Δx = 0.00226230m. 2.26mm give or take.
The model of the fluted shaft and torque plate is drawn from one of your close-up photos above, and should be more or less correct. The arc of short wiggly lines through the section of the TP flute are tangents to the force arc every 0.5° (where the shear planes go as the axle rotates), and the arc length is 3.73410mm.
That means the axle would compress the Al by 60.5% of the section of the tooth before stopping!.
Ah,but the area would increase as it progresses, and as it compresses, other teeth will come into contact and the load will start to be distributed.
So then I started looking at how the width of the section of a 2mm radius tooth increases as the axle cut across its diameter; and trying to make some assumptions for the depth the first tooth would need to penetrate before the second tooth came into contact, assuming some reasonable manufacturing tolerance.
I started with 50microns (a gnat's under 2thou") which is as good as most people can achieve (on steel) with a surface grinder and way better than I've ever seen achieved with a wiggle (rotary) broach; and they tend to 'wander' in aluminium.
The best case scenario I calculated was with ±0.000 tolerance on both parts -- impossible -- and I got a little under 0.5mm of compression before things equalised and the axle stopped compressing the TP teeth. 0.5/3.7341*100 = 13.4%
The problem is, the 'Elongation at break' figure given for 6061-T6 is 10%. It would never stop cutting!
Which is pretty much exactly what your tests showed.
You could try using 7075-T6, but it's shear modulus is exactly the same at 26GPa, and it elongation at break is only 7.9%. It's ultimate, tensile and fatigue strength are all about double that of 6061, but it achieves it by being over 50% harder (150 brinell vs 96 for 6061).
Harder means less ductile, hence less elongation before failure.And that's before you start considering the bogeymen of cyclic fatigue, shock loads, et. al.
Double the thickness of the TPs and use very expensive Al, or move to steel (flutes work for gearbox shafts at the clutch, and sliding joints in drive shafts), but I think persuading manufacturers to start cutting flutes rather than flats on their axles -- given the price pressure -- is asking a lot.
Once again. Good luck.