Engineering calculation, fastening 2 gears

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I hear a collective sigh of relief. 2 pins should adequately transmit the torque, so the glue is only holding it all together. My biggest concern with the use of adhesive on the faces to transmit torque is simply that ANY axial misalignment (a fraction of a thou even?) could cause some tightness at some point in the gear meshing, that would transmit to oscillating high shear forces on the adhesive. This would be bound to cause premature failure of the assembly. As a starter gear-box you may get away with it, but with 2 pins taking the shear forces I am sure you now have a much stronger and more durable assembly. But you need to try and get perfect concentricity of one gear on the other. (As a designer I should arrange the small gear to have an extended cylinder upon which the larger gear if fitted, as concentric cylindrical surfaces can easily be made when making the gear blanks. But you are past the design stage!).
Well done on the machining!
Not quite sure I follow some of your comments but maybe this will help.

Re concentricity the gears were purchased but required re-machining the widths & bores. The large gear was dialed in on the original bore. The small gear was fixtured to a spigot turned in-situ. Its ~ 0.001" clearance fit diameter basis, so 0.0005" annular gap. I could have made it interference fit & still used Loctite, but that's water under the bridge. The reason is because the gears still need to be timed & I wasn't quite sure which would be the best way. I now have clearer line of sight (or maybe a defined plan to paint myself into a corner LOL). I will permanently attach crankshaft gear & this gear cluster, leaving the ring gear to cam cup the one to time relative to cam events. Anyway, with the gears dry assembled as present I have put a dial on both gear OD's relative to axle shaft & they are within 0.001" runout of each other. The adhesive wont vary the fit so that will be as good as it gets.

When I was talking about Loctite strength, I just meant the intersecting (circular green area) on the 10T hub which intersects the 15T ID hole, not the face of the gears. Any Loctite that gets squeezed in there is bonus points but not expecting it to contribute anything. I intend to do that assembly as the first step like my mockup. Then the key pins get drilled & those are also retained by Loctite.


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OK, I guess I'm going to have to just volunteer my ignorance with my attempted back of envelope calculation. Which is what prompted the post to begin with.
Hopefully the spreadsheet shows what I was trying to evaluate - the resultant force of shearing a steel pin key at the hub line vs shearing the Loctite adhesive applied to the hub. So is resultant 'force' at the hub line radius an apples to apples comparison, all other things equal? I'm not saying this is the failure mode & maybe my assumptions are all messed up.


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Thanks Petertha. I have probably been talking mostly nonsense, as I initially thought you had been describing 2 gears on a single shaft - stuck together by the side faces, and onto the shaft, then you explained they rotated on the layshaft, so thought you were sticking the 2 side faces together, and their individual bores were the bearing surfaces on the layshaft. Now looking at the Snag thumbnails in the 2 previous posts, (that somehow are all "Blue" and indistinct when I try to open them) I can see what looks like a boss on the smaller gear upon which is mounted the larger gear. That is excellent for concentricity - as you have measured - and would prevent any shear forces within loctite between the faces if there were any eccentricity.
Anyway, your post 23 is exactly the right calculation to compare Glue versus steel in shear. The "2-pins" design is certainly the way to go. 2 x 468lb = 936lbs beats 657lb in my books!
Do you know the max torque possible from the motor (stall torque) - which can be multiplied by the gear ratios to get the max torque across this joint? (Then the shear force).
Your calculations are essentially the same as used to design shear pin applications, where should something dramatically fail, the shear pins actually shear and let everything spin freely, rather than break a gear tooth or worse.
Thanks for sharing your design work with us all.
Thanks, I'm feeling more confident now.

No, I haven't figured out how to back calculate torque & see how this relates to equivalent force at the hub shear line. I can't think it could be much because its ultimately cam plates (essentially attached to the ring gear) which are lifting against valve springs through the gear train described in post#5.
Hi Petertha: The key to the design is the maximum torque that can be applied - So you can avoid failure.
When you decide on the starter motor (you mention a 390? - I should look that up?) you need to understand the stall torque. Electric motors (generally) cannot exceed the stall torque. Like steam engines, the max torque is developed at zero revs. (until you get into the more complex steam engines - like turbines - and more complex motors - like Tesla cars use?). So knowing the stall torque of an electric motor is a good maximum torque to use. While this torque may never be realised in practice (It relies on the engine being driven being seized!) it is good to know that the Driving motor cannot strip teeth from gears, or otherwise destroy the gearbox. Also, as a rule of thumb, I should increase the max torque if the maximum oscillation of torque is more that 20% of the max torque of the driving motor. (This oscillating force of a tooth of a gear can fatigue the root of the gear). Actually, even that may not be safe enough for industrial lifetimes, but for a model should be OK.
I have attached a "quick guide" pdf - but it is probably teaching you stuff you know better than I do?
Now how to determine the torque when the motor is driving the engine? You are right to consider spring pressures for valves, cam effects to give the mechanical advantage and drive the valve from the camshaft - your post #5. So having determined the max spring load when appropriate valves are depressed, then worked backwards to the cam follower, via all the changes of velocity ratio (if any?) by rockers, you can estimate the torque on the camshaft by the radius of the cam peak times the force from the valve. In reality, the torque is less than this, due to the profile of the cam doing the work at a smaller radius while the spring is not fully compressed, but unless you research a "cam-profile calculator" against your profile, I can't suggest an "easy" calculation to give you the reaction torque from a cam. BUT you could use a torque wrench (with a max reading gauge) to measure it... either from the end of the crank, or camshaft? As the starter is only cranking the engine at low speed, the dynamic loading on the valve train (necessary calculations during original engine design) can be ignored. (At the Car design office I worked in, the engineers doing valve train design had numerical models that looked at the forces reacting on cams from accelerating the masses of the valve train to operate valves at high speed. Much higher than spring forces alone!).
I hope this helps? And I apologise in advance if there are errors in my stuff - the brain is getting addled with age..


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Steamchick, I rather think you need to read this whole thread again. Petertha's gear cluster is part of his timing gear train. That is my starter gearbox, which I posted merely to illustrate gears being Loctited.
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