A little help please

I looked through the books I have, and apparently the load on a square shaft and journal isn't something "common", at least it wasn't easy to find, or I'm asking the wrong question. I ran the numbers again with a slightly more accurate model, using all 4 sides, with an accurate length measurement. I did assume the load was even across the contact area, which isn't really the case, but I'm hoping it's close enough. With that, I get about 9ksi, which gives you a safety factor on the journal of just over 6 assuming annealed 4140. You would be doing better if you had normalized 4140, but if you don't know what you have, assume the worst case. In any case, if the shaft holds, the journal should be fine. I think Ray is working on calcs for the shaft.

I've also realized I've forgotten most of what I learned in school, and remembering it hurts :headache:
 
For torsional calculation, Diameter of shaft = SQRT [ (60 x HP)/RPM] provided no stress area exceeds about 4000-6000psi.

So far you're at SQRT [ (60x20)/10,000] = 0.346 in. You're good to go there with a tiny bit of headroom provided we go back and do the other calculations to see how many PSI are being placed at the 4 sides.

I'm going to do a quick integral (calculus) equation to calculate it as when I did a mean calculation based on half the distance up from the .134 side, I came up with 9700psi. Can guarantee that the integral equation will improve the number in your favor.

Still though, untreated 4140 is good for 40,000+ and 316 SS is 30,000 psi. No matter how you slice it, you're less than 1/3 the ripping point of the steel.

It's past midnight an I don't do calculus past dinner time :). See you in the morning and I'll show you the detailed calculations.

Ray
 
BTW: I don't do stuff like this too often. I'm a physicist by training, not an mechanical engineer so, I have to start with the basic principals and formulate the answer. M-E guys probably have this stuff memorized because they do it all the time.
 
It's reassuring that Ray is coming up with similar numbers to me. Syaminab has a good point about fatigue, though I was taught to use the full yield strength of the material, and apply a factor of safety after the fact. The results should be more or less the same though. You definitely don't want to be right up against the yield strength for the material in any case.
 
OK, I plugged the integral into mathcad and the difference between that value and using the mean distance was about 1%. -Insignificant!

Here's the calculations:
Assuming:
HP=20
RPM=9500

Torque = 5252 x HP x RPM ftlbs (this is a fact of geometry and rotational kinematics. multiply x 12 for inlbs).
T = 5252 x 20 x 9500 = 11.06 ftlbs = 132.72 inlbs

Force = Torque/Distance (where distance will be the mean radius of the contact area).

Distance of the center of contact area from the geometry of the circle circumscribing the square is:

(0.375 / 2) - (0.134/ 2) = 0.1205 in

so...

Force = 132.72 / 0.1205 = 1101 lbs

Pressure = Force / Area (fact of math and geometry).

Total area from the geometry is the 4 sides coming into contact between the square shaft and it's receptacle.
Total Area = 4 x Length x Width
Total Area = 4 x 0.25 x 0.134 = 0.134 insquared.

Therefore:
Pressure = 1101 / 0.134 = 8216 psi

I took the liberty of performing the same calculations assuming there was no pilot hole. If you trust my math, the pressure was 7552 psi.

The pilot hole creates an 8.08% weakening of the fixture.

Metal Strength....

The weakest link is the 316 Stainless which has a yield strength of 30,000 PSI (conservatively. I actually found two different numbers in various sources and 30,000 was the lowest).

If you multiply yield by 0.4 (as suggested), you get: 12,000

8216 is a heck of a lot less than 12,000!

If you look a couple posts down that mentioned Torsional Calculation there was a formula that calculated optimal diameter based on 4000-6000 psi. It came-up with an optimal diameter value of 0.346.

We've now checked against 2 methods and it seems OK.

Two is never enough...

There's another method called "Resistance to Twisting" which is based on shearing force calculations (see that link). D = 4 x SQRT (HP/RPM).

This formula produces an even smaller recommeded diameter. I personally would dismiss this formula as an incorrect application for the model it was based. I'm guessing it was intended for larger diameter applications.

Question: Does the shaft fit tightly in the receptacle? How many cylinders is this engine? Does this fixture receive the "pulsing" from each cylinder fired? Are there any simultaneous opposing forces counteracting a piston firing (I assume yes and if so, this works in your favor).

The upshot is that there's "Impulse" issues to consider but... It's not easy to calculate and if other people have already made this thing and it works then, go with it. If the parts fit very tightly, it will minimize Impulse Deformation.


Ray

EDITED: See the "Question"s above.
 
Thank you all. Some real good info there. Gonna go with the Standard Square and pin it. ------ "Billy G"
 
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