2 small CIMs, with a 12 tooth gear on each that mate with a single 45 tooth gear on a separate shaft. On the same shaft as the 45 tooth gear is a 13 tooth gear. These gears are 20 pitch, 14.5 degree pressure angle, and 3/8" face width. For now, let’s say the 13 tooth gear is made out of common cold rolled steel.
Assuming a worst case scenario (motors reach stall torque/current), will the 13 tooth gear be able to withstand the force?
What calculations are used for this?
Stall Torque on Two Motors Combined = 39.3 in-lbs. Edit: Oops… gotta be careful here. They are applying stress on different teeth. Use 19.65 in-lbs)
12:45 Ratio, so 19.65 * (45 / 12) = 73.7 in-lbs.
This 13 tooth gear has a radius of (13/20) / 2, or 0.325 inches.
Hence, the applied force is 73.7 in lbs / 0.325 lbs or 226.8 lbs.
A quick Lewis Bending Equation I’m showing says this is not fairing too well. This admittedly conservative equation is showing 0.68 factor of safety, assuming 40 kpsi yield stress on your ‘common cold rolled steel’.
(Estimated Stress: 58 kpsi for 20 degree Pressure Angle. 14.5 will be roughly 20-30% weaker)
A few things to ask:
I used the stall torque of the CIMs. Is it possible for this geartrain to actually ‘Lock Up’? Are you doing anything fancy that could cause locking or braking down the line in your drivetrain? If not, and your gear the drivetrain properly to ensure you don’t trip breakers while pushing up against a wall, then the true peak torque on these gears will be less than what I used. (Perhaps even only 40% of the torque I used.)
A few things you can do to reduce stress on the gear:
Also note, a quality gear bought from some place like Stock Drive Products may not be your average strength steel. 40kpsi is pretty conservative for a heat-treated gear, though I’ve seen some of the Martin stuff out of Mcmaster run at 30kspi out of the box.
Calcs attached (EDIT: for 20 degrees, its what I had handy) for reference.
Bottom Line:
I won’t make any promises, but assuming a standard drive train setup with slipping tires right around stall current, I’d say this gear, as is, is on on the edge. I’d sleep better with 1/2" face width.
Matt
Gear Tooth Stress Calculator.xls (17 KB)
Gear Tooth Stress Calculator.xls (17 KB)
Dave, since we plan to make that gear for you guys anyway, why not just make it out of 4340, or even something like 1018 heat treated? You shouldn’t ever have to worry about it then!
Travis, I wasn’t sure what your gear guy would want to cut. Does 4340 machine well?
I have not gone through the calculations above, but if the Lewis formula give gives tells you your gear tooth is 70% of the strength you need, I am pretty confident that your gear is plenty strong enough for a FIRST applicition.
I have had good success using just a straight beam calculation as an estimate of the strength of the gear teeth. Here is a relatively easy web application from efunda that can do the calculations for you.
Joe J.
We have never broken a gear on a drivetrain. We have blown apart some gearboxes on our arms in years past by either falling on them or having someone else fall on them.
I confess that my ‘edit’ about the two CIMs working on different teeth had poor logic - they’re running on different teeth on the 45 tooth gear, but presumably the 13 tooth gear will have one contact point.
Since Dr. Joe seems to have a bit more confidence, from experience in this working well than I do, I want to take another swing just to see what the numbers show.
First step I think is to be a bit more conservative with the torque these gears should actually see. I’ll assume that you’ll design the max torque to be right around 40 Amps, which I calced out to be about 14 in-lbs from a pair of CIMs. After a 45:12 ratio, this becomes about 53.25 in-lbs.
Using this number, and the standard AGMA gear equation
Lewis Form Factor (Y) for a 14.5 degree PA gear = 0.223
Radius = 13/20 / 2 inches = 0.325 inches
Force on Tooth (Wt) = 53.25 / .325 = 163.85 lbs.
Face Width (F) = .375
Pitch § = 20
Stress = (Wt * P) / (F * Y)
I have this showing 38,266 PSI, or a factor of safety of 1.04 (at a megar 40,000 PSI)
I think you have a few factors to consider:
Positive Effects:
Negative Effects:
Another Note:
A roughly equivalent pinion in the FIRST kit drive train has a face with of 0.809 inches. This has been designed to take a single CIM motor’s peak torque output with an overall reduction of (50/14 * 50/14 * 50/14) or a totally reduction of 45.5 times.
However, since you have 2 times as many motors, your face width is 2.16 times smaller, and your overall reduction is 3.75, this gear will see (14/12 = 1.16) 1.16 times more torque due to their smaller tooth count, and the kit material (assuming 100ksi yield) is 2.5 stronger than plain steel.
This means the strength of this gear, relative to the one in the drive train, based on application, is (2 * 2.16 * 3.75 * 1.16 * 2.5 = 46.98 ) about (45.5/46.98) 0.96 times as robust. However, there’s presumably some factor of safety built in to the kit, so being withing 4% is probably well within their factor safety.
Hence, with average steel, I’m still going to call this border lined, but probably more above the border than below. It sounds like Travis is proposing some stronger material, which is probably the easiest fix.
Good luck,
Matt
I had a situation like this two years ago. Lets just say, the calculations all worked out to about a 1.7 safety margin, with a 12 tooth 20 pitch hardened 3/8" face gear. Yet, somehow, the teeth still magically disappeared. Later, we found out that the gears were running misaligned. If properly machined and screwed down, your drivetrain should be fine. However, if I were you I would still make a few spares…
The Lewis form factors are very very conservative for an intermittent duty or short life application (such as a FIRST robot). I have designed many production actuators for automotive applications (power liftgate actuators, power sliding door actuators, power lock/unlock actuators, power cinching latches, power cinching strikers, power unlatch actuators, power window actuators, etc.) I have never had a case where the Lewis forumulas gave me useful data.
I have had good success approximating a gear tooth as a uniform beam with its cross section equal to the cross section of the base of the tooth and its length equal to the height of the tooth with a load applied at the end of the beam.
Pcircular = 2 Pi/P = .31"
Circular Tooth Thickness = Pcircular/2 = .16
Tooth Height = 2.25/P = .11
This can be approximated as a beam
L = .11 in
H = Circ Tooth Thickness = .16 in (this is conservative but a first order approx)
T = .375 in
F = 164 lbsf
I = (1/12) T * H^3 = (1/12) * (.375) * (.16)^3 = .000128 in^4
c = H/2 = .08
Going to the efunda.com site I linked to above, I get the
Stress = 11,000 psi
Unless your gear is plastic, I don’t think you have to worry about 11,000psi stress levels.
Joe J.
Hmph. Sometimes I lothe what academia has (not) taught me. :]
Thank you, Joe!
Matt
I confess I double checked your circular tooth thickness. Perhaps we’re just using different terminology, but if you mean the thickness of the gear tooth on the circular pitch, I think the formulas are different.
I have:
pCircular * Pdiametral = pi
and
Circular tooth thickness:
t = pi / (2 * P) or 1.5708 / P
So I get a tooth height (thickness) of (1.57 / 20) = 0.07854 in. or exactly half of what you found.
Calcs:
I = (1/12) T * H^3 = (1/12) * (.375) * (.07854)^3 = 0.0000151 in^4
c = H/2 = .0392
Hence, the stresses are substantialy higher.
I know that there’s lot of terminology floating around, are we just using a differnent lingo?
Matt
Matt,
Good, thanks for checking the calcs.
Nope, no new lingo, I just put an extra factor of 2 in my formula.
The correct formula is Pcirc = pi / Pdiametral not “2 pi” as I said in my note.
Recalculating we get the stress as:
45,000psi – which is to be expected because the stress goes up like the square of the beam thickness so we would expect 4X the stress for halving the thickness.
45Kpsi is highish but not too bad. While it is higher than a typical Low Carbon steel, it is sort of middle of the road for a good steel.
If all you can use in low carbon steel, I think it would be fine if you could case harden the pinion (any reasonable depth of case hardening will effectively harden the high stressed areas of the tooth – the edged of a beam take the highest stress – if you can get the upper 1/3 and lower 1/3 hardened the fact that the middle is soft will not matter very much because the stress level in the middle 1/3 is so much less than the outer 1/3’s).
Also, note that there is conciderable safety factor even in this calculation since the geartooth is not a beam but closer to a parabola and the force is not all on one tooth AND at the tip. So… …you’ve got some margin even in this calculation, though I would not depend on it too much. My experience is that if the material can take the stress you calculate in this manner you will be fine and not have too much safety margin in the design.
Joe J.