Will the gear tooth break?

[Matt Adams]

Originally Posted by Joe Johnson

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

Joe J.

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

[Joe Johnson]

Originally Posted by Matt Adams

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.

Hence:
I = (1/12) T * H^3 = (1/12) * (.375) * (.07854)^3 = 0.0000151 in^4
c = H/2 = .0392

and well... to put it lightly, the stresses are substantial and it doesn't work out so well.

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.