Calculate hold down force for rotating shaft 1

[localhero]

I have a .850" diameter steel shaft that will have a torque of 125 ft-lbs put on it. I need to calculate how much down force I need to stop the shaft from rotating. I suppose we can assume that it will be steel on steel.

Going with a coefficient of friction of about .35 I tried to calculate with the typical CoF problem of a box and an incline, but I don't know how steep to make the incline.

How can I figure out how much force it will take to stop a .850" diameter shaft from rotating under a torque of 125 ft-lbs?
Thanks

 

[MintJulep]

Assuming that you can neglect shaft deflection. Which you can't. About 10,000 pounds

 

[desertfox]

Hi

Took a look at the attachment

“Do not worry about your problems with mathematics, I assure you mine are far greater.” Albert Einstein

 

[localhero]

Thanks.

 

[CostasV]

If you have 2, or better 3, contact points, the force is getting 2, or 3 times less, and you can eliminate shaft deflection issues.

 

[chicopee]

Desertfox, the direction of P1 that you scratched out was the correct direction, not that of P2.

 

[chicopee]

Localhero, can you explain that portion of your OP "... but I don't know how steep to make the incline."

 

[gruntguru]

A pipe wrench will amplify the force for you.

je suis charlie

 

[3DDave]

How am I missing this?

(125 ft-lbf *12 inches/ft) /.85 inches = 1764 lbf tangentially at the surface of the rod across the diameter of the rod.

Since the rod is slippery at .35 CoF the Normal force required is 1764 lbf/.35 = ~ 5000 lbf.

It doesn't get multiplied by two because the moment applied at each contact spot relative to that spot is zero, with the torque resisted by the force *CoF * diameter at the location opposite each contact spot.

To be sure this is a simplification as it assumes the load is actually a couple, not a torque.

 

[MintJulep]

Since it's a poorly posed question different responders have different interpretations of the problem.

Regardless, the 0.85 inch dimension was stated as a diameter, and MUST be divided by 2 to get a radius.

If your interpretation of the problem involves the applied down force and an opposing reaction then go right ahead and multiply by 2.

 

[dvd]

Is it a holding force only, or is it also necessary to absorb all of the kinetic energy of the rotating system? Braking on a small shaft diameter may be problematic.

 

[3DDave]

If the bar was diamond (square on edge) and the distance was .85 across the corners, would that distance be divided by 2? Free body diagram shows two forces as a couple resolved into separate forces applied at opposite corners, each with a magnitude of torque required divided by distance of separation.

I'm going with this as the original question is related to down force, not uniform radial pressure.

 

[gruntguru]

3DDave, rotating shafts are usually supported by bearings. The OP speaks of "stopping" the shaft with a "down force".

je suis charlie

 

[3DDave]

The OP did not supply information to support either calculation. 10,000 pounds is a lot for a bearing on an .875 shaft to take, so there should be a back-up block on the opposite side of the load to prevent overloading the bearing or bending the shaft, which reduces the load to 5000 pounds and does not load the bearings (as much, depending on construction.)

 

[MikeHalloran]

Localhero, read up on a classical device called a 'prony brake'.
You will probably find useful equations in any text that includes an analysis.


Mike Halloran
Pembroke Pines, FL, USA

 

[gruntguru]

"The OP did not supply information to support either calculation."

In that case lets go straight to the pipe wrench. Self weight plus Torque/Length - say 50 lbs total?

je suis charlie