Torque converted to load 1

[jbknudsen]

I am trying to calculate the load required to remove a circular shell from a circular urethane mold. The process currently employes a 3/4" dia 6 pitch acme rod x 24" long and a pneumatic impact wrench in conjucnction with a spredder bar. I am wanting to replace this operation with pneumatic cylinders (safety and speed). I am having a hard time finding a calculation that would allow one to take the known torque figure and convert that into load. Someone suggested to look at Shgleys book "Mechanical Engineering Design" I am looking to buy the book used, however, i need the formula asap. Can anyone give any more details.

PS: I have suggested fastening the core to the shop floor and placing a scale on the crane/forklift and lifting the shell and noting the reading on the scale at the point the shell moves. Managment has shot down that idea.

Thanks, Jay

 

[sbi]

Look up teh torque ratting of the air wrench to see what its maximun is, most 1\2" are around 240 lbs/torque

next get a large manual torque wrench and start at say 200 lb and increase the setting in a 25 lb rate till you get high enough that you are turning the acme and removing the mold with the aplied force. from there you can get a good idea of the force required to remove the mold, after you hvae done a dozen or so to get a average, you should have a good place to start on the change over design to true pnumatic cylindars

 

[jbknudsen]

The rating on the pneumatic impact is ft/lbs of torque. Is there a direct relation between the ft/lbs of torque to load. For example, if 236 ft/lbs of torque is the average, could I then design the cylinders for 250 lbs of force and achieve the same result (lifting the shell over the core)

Thanks for you help

Jay

 

[strokersix]

Jay,

There is a direct relation between axial force and torque on screw threads. Unfortunately for typical 60 degree screw threads only about 10 percent of the torque actually gets transferred to screw tension. The other 90 percent goes into friction of the threads and friction under the screw head or nut. In your case an acme screw improves this percentage. If you have a thrust bearing or at least good lubrication under the nut that helps too.

Here is a formula: Torque = Force [(P/2*pi)+(Ut*Rt/cosB)+(Un*Rn)] where P = thread pitch, Ut = friction coeff in thread, Rt = thread contact radius, B = half angle of thread, Un = friction coeff under nut, and Rn = nut contact radius. This formula requires assumptions for friction coefficients, therein is the problem.

Too bad management shot down your idea for a direct measurement of the required force. That would eliminate most of the uncertainty in a calculation.

Mike

 

[jbknudsen]

Thanks Mike, I'll give it a shot. It would still be nice to put a scale on the shell and pull it off, that would help confirm or decline the accuracy of the coeficient of friction.

Thanks,
Jay

 

[chicopee]

Remember what the pneumatic impact wrench and spredder bar are for and that is to make an initial separation between the molds pieces. Without this contraption it would be extremely hard to make the separation.

Torque can be calculated by two fundamental equations. T=F*X and T=63030*H/N. These equations can be modified for threaded rods, rotating and pivoting elements and any other design situation that you will be involved. Once you have developed some numbers for torque be sure to stress evaluation for your linkage.

 

[LesJr]

I've tried the calculation above and had some "funny" numbers come from it. Please help clarify some of the symbols and how they are to be used. I have a 1.4375-20 thread, a bolt head of 1.570in. How do plug those number into the above equations. I believe my units are off?

 

[sancat]

Hi Les,
English units are so complicated, that I prefer to translate to metric, and back.
236lb.ft are 320 N.m
20 pitch is 1.27mm per thread your screw diameter seems 36.5mm (20 seems a lot of tpi for an acme 36.5mm diameter!!)
Your nut diameter is 39.9mm, and therefore I would asume
Rt is around (36.5-1.27)mm, or 0.03523m, and Rn is around (36.5+39.9)/2= 0.0382m
Ut and Un, well, there lies the problem. I would tend to think that the screw is oiled, and 0.3 seems a good guess, and the nut is not oiled, so, 0.5 should do it. Said that, your design's safe side is when the applied force is the highest (and therefore friction is the lowest). Yes you could end with a very big piston, but it wont fail (remember that no one forgets the person who underestimates a force, but when you over estimate you pass unnoticed). A better method (and a cheapper selection) would be done with your method of clamping the piece to the ground and lifting it with a crane. In your case I would assume much lower friction coefficients to be on the safe side.
CosB, well the helix angle tangent is 1.27/(36.5xpi),so b is 0.6348 degrees.
Apply all that in the formula, and the force should come up in Newtons. (9.8 newts is 1 kgf, 2.2lbf is 1 kgf)
hope i helped
sancat

 

[LesJr]

sancat,

thanks for the clarification. I'm not using an acme thread. I was unclear by not stating that the thread for my issue is a UNC thread. Does your solution still apply, in particular the half angle calculation?
les

 

[sancat]

Les,
Sorry for the delay, but, I wasnt around. If you are using a standard bolt (60 degrees) and, strokersix is right (I hope he is, since I used his formulas myself), I dont recomend you use that method to estimate the force.
a 10% efficiency in the torque/force conversion, could give you an error of 100% in your calculation with a slightly variation in efficiency from 10 to 20%.
Your approach: to clamp the thing to the floor, or to something, and use a dynamometer, is much much better.
The theoretical method above will be "usable" for acme threads, I wont reccomend it for unc threadding.
sancat