Instantaneous torque in a shaft

[KMBatchelor]

I just broke a shaft and it makes no sense to me. The driving motor cannot break the shaft if the shaft is held still at start-up but it snapped like a twig when it seized.

If one is driving a shaft with a motor of W rotational inertia, with a capacity of X in-lbs of torque, at Y rotational velocity, and the shaft seizes to a stop in Z seconds, how does one calculate any torque spikes the shaft might see?

Thanks,

[PinkertonD]

Have you first examined the break in the shaft? Could it be metal failure? Fatigue etc?

As to the second part, try calculating for there being no time for the seizing, instantaneous, and see if it is still within the strength parameters of the shaft. If so, then you need to look at the shaft and not the motive source.

[KMBatchelor]

Thank you.

[zeke]

Startup torque is maybe 5* running torque and is a fraction of the torque you will get at seizure.

If the motor seizes ,i.e. quick stoppage at the input and an inertia ,I at a distance L from the point of seizure, it is very possiblr that the resulting induced torque, T to the shaft, which acts like a torsion spring, will break. Depends on the speed of the stoppage. Worst case instaneous stoppage

You do the problem by equating the inertila energy,.5Iw^2 to the tosional windup energy stored in the shaft

.5*I*w^2=.5*T^2/GJ*L

I moment of inertia of rotating mass

J polar moment of inertia of shaft
G shear modulus of elasticity
w angular speed
L distance of shaft between stop point and mass inerta
T peak torque in shaft
If seizure occurs somewhere in the motor train , then problem is done similarly.


THere is usuall no point in solving the equation, since one doesn't design fot this.

[KMBatchelor]

Thank you. I was just curious and had never tried to do it before. My boss was gobsmacked that the shaft broke. He'll enjoy knowing a bit more.

[RWOLFEJR]

Couple quick thoughts on this...

There are instances where the design of a shaft should handle lock-up at full design rpm. With a large safety factor on top of that. Example being forging equipment or punch presses have some degree of risk of lock-up by design. Also... until there's some magic motor out there that can reach full speed instantly... lock-up will always be a greater load on the shaft than start-up. Depending on the application there might be times where it'd be surest to just design for lock-up.

There are certainly times when design doesn't need to be nearly that stout. When all possible modes of failure are considered, and the resulting condition of things afterward... a design could very well be reduced to minimum required torque carrying capacity. Stuff like the drive shafts in a watch or maybe instances where full lock-up would be impossible somehow by design. I suppose that so long as the power transmitting gadgetry can't destroy things if it should fail, then leaning out the components could very well be the ideal.

KMBatchelor... I don't know exactly what your looking at... can't see your design. One thing you can be certain of is that the shaft in the condition it was in wasn't able to carry the load. (Duh... obviously...) So now you need to figure out why.

First thing I'd want to know is why this... stuff, locked up. And what was it exactly that did lock up? Then you need to look at the broken pieces and determine if the break was due to a flaw in the part or if it was under designed for the application.

When I'm looking for max potential load on a shaft I use the flywheel lock-up formula in Machinery's Handbook. It's the easiest thing out there to fudge out a quick number that'll give you a good idea where you're at. Just call your connected or keyed rotating parts the weight and ballpark an average radius of the weight and you'll get a good idea of the potential torque load your shaft could see. It can get you close enough and if you have any doubts error to the safe. Then run the shafting formulas on twisting etc. and see what you get.

Good Luck,
Bob

[KMBatchelor]

Thanks again. This was a custom designed walkway bridge. The lifting mechanism drove the bridge up, pivoting it at the end. The drive screw, a 1 1/2" Acme screw, appears to have frozen in the nut and sheared. It was an exciting failure.