Torque required for rotation 1

[jreamer]

Hi,

I am trying to figure out how much torque we need to spin a compressed object. I am applying 6000 pounds of compression force to an object between 2 plattens. In between the plattens we have 2 round plates supported by 8 crowned cam follower bearings per side. The manufacturer of the bearings say that the coefficient of friction for the bearings is .0025. The plates are 24" diameter. We want to apply a torque to the bottom plate to spin the entire object. I am looking for an equation to figure this out. I looked one up in my old text book and applied that, but we were greatly undersized. The equation I used was (2/3)*(coefficient of friction)*(radius)*(force)= Torque Required. I might be applying the forces wrong or the coefficient of friction wrong, or even using the wrong equation, I am not sure. Any insight would be greatly appreciated.

Thanks.

 

[whammett]

118.8 lbs? How undersized were you?

"I came, I saw, I made it better."
-Ode to Industrial Engineers
Will ChevronTexaco Corp.

 

[imcjoek]

The equation you used makes sense for an equilibrium case (not accelerating), except that I'm not sure where the 2/3 comes from. You will need more torque (substantially more, depending on inertia, desired velocity, and desired time frame) to accelerate your system.

Also the hardness of the material that bears on the cam followers will have an effect on their actual effective friction from the rotating plates perspective. Picture it with a rubber film between your plate and cam followers and the reason should become apparent. A good hardened and ground/polished race would work best.

 

[desertfox]

Hi jreamer

Can't quite visualise your set up any chance of uploading a sketch.

desertfox

 

[MintJulep]

I come up with 118.9 lbs. Whammet, did you forget to multiply by the factor of near unity?

 

[jreamer]

I have a sketch attached. We are bringing the top plate down to apply the compressive force. Whammet/Mintjulep, that is right around what I got and we undersized. We are going through our setup and making sure we are not binding anywhere but I just want to check my numbers. Desertfox, any help would be appreciated.

Thanks again.

 

[whammett]

Of coure not MintJulep. I blame it on a typo.
Jreamer, like imcjoek stated, that is a good equation to get you in the ball park, but the actual initial force will have to be greater to get it started.

"I came, I saw, I made it better."
-Ode to Industrial Engineers
Will ChevronTexaco Corp.

 

[MintJulep]

I don't understand the formula presented.

Assuming that the rollers are all at the perimeter of the plates, and are equally loaded (unlikely):

2 plates x (6000 pounds/8 rollers) x 8 rollers x .0025 x 1 foot = 30 lb-ft.

How stiff are the plates?

I suspect they are deflecting, causing things to bind.

Are you within the load rating of the bearings?

 

[jreamer]

MintJulep, the plates are standard A-36 hot roll steel (probably not hard enough to withstand the deflection caused by the bearings). We are also within the load rating of the bearings so that is not a concern.

 

[desertfox]

Hi jreamer

Slightly confused still, you want torque all the answers I see are only in lbs ie units of force,
Using your formula I get 120lbs-in.
Looking also at the sketch am I right in thinking that you want both top and bottom plate to spin as one, after its crushed the object in the middle?
If that is the case your relying on the friction between the crushed object and top plate to act as a driver, if you only apply torque to the bottom plate.
Finally when you say the torque was under sized what happened ie:- bottom plate didn't turn, or bottom plate did turn, but top plate was stationary.
Sorry I just want to get clear in my head what exactly is happening.
I assume the 0.0025 is the coefficient for rolling friction.

desertfox

 

[jreamer]

Desertfox,

Yes I am looking for the torque required to turn the object along with both top and bottom plates as one while the force is being applied. You are also correct in saying that I want the friction between the object and the top plate to turn the top plate. When I said undersized, I ment that when the force was applied, the motor driving the bottom plate (which I am using as the drive plate) would not turn, it was stalled. When the force was removed, the motor would turn the bottom plate and the object. We then decreased the force applied from the top and were able to get rotation. The most force we were able to get was around 1500 pounds before the motor stalled. I am trying to figure out where I went wrong and how much we actually need. And the .0025 is the coefficient of rolling friction. I know that the coefficient of static friction will be more so I will add more torque to the final answer to get us moving since the manufacturer did not know the coefficient of static friction for these bearings.

Thanks

 

[dvd]

1. Your cylinders aren't guided, there is no reason that the plates will stay concentric with respect to each other. When you try to spin the compressed object the material being squeezed has to deform to allow top and bottom plates to spin on misaligned axes.

2. The plates deflect elastically which means your yoke rollers are always running uphill out of the wave of elastically deformed material.

3. Page 70 of the loadrunner catalog (

http://www.loadrunners.com/lit/LR120.pdf)

discusses track capacity factor. Material hardness affects the load bearing capacity of the rail (platens in your case) but not necessarily the frictional load. That will be more of a function of the surface finish and Young's modulus for the platen since deflection is a function of Young's modulus not material hardness.

 

[desertfox]

Hi jreamer

I might be off course here but wouldn't the torque be double the 120lbs-in because you have to overcome the friction over both sets of cams and you have 6000lbs force equal and opposite on both sets of cams.Just as an additional point the coefficient for rolling friction 0.0025
would nomally be multiplied by the normal load on each cam wheel.
So if you take 6000lb and assume each cam takes an equal load then for one plate:-

force horizontal to rolling resistance =(6000/8 )* 0.0025*8
obviously the 8 cancels out

therefore force to overcome friction= 15lb on one plate

therefore 30lb on 2 plates, if the the cam rollers are on a 24" diameter then the torque required would be

30*24"/2 = 360lb-in if you multiply that by your 2/3 it equals 240lb-in.
I am not sure where your 2/3 factor came from so I can't comment.
Whats the torque rating of you motor?

desertfox

 

[desertfox]

hi jreamer

I should of added that I think your cams are possibly digging into your plate's as you have vertually a point or line contact with the cams.

desertfox

 

[btrueblood]

What keeps the rollers in line with their direction of travel, i.e. is there a slight caster set on each roller? If not, then the wheels may be skidding sideways. Even with crowned rollers, some skid has to occur for the rotary motion to occur, and the amount of skid increases with load.

In any case, who cares what the theoretical equation is, you have data. Take 6000/1500 * .0025 to find your new coefficient of friction, or 6000/1500 * 120 in-lb. to find your new required torque.

 

[zekeman]

I also think at a minimum,for design purposes you should analyze the deformation of your plate under a center load of 6000 lb simply supported at the periphery and your cam follower bearings should each be designed to take a load of 6000/2 lb which assumes that under deformation, a worst case of only two rollers on each plate will be in contact. Also a contact stress on the plate must be performed to assure the integrity of the plate,using Hertz formula
S= F{1/rho)/[w{(1-mu1^2)/E1^2+(1-mu2^2/E2^2}}]
F force on cam follower roller
w width of cam follower roller
mu1, mu2 Poisson's ratio for follower, plate
E1,E2 Modulus of elasticity for follower, plate

 

[zekeman]

correction, the Hertz formula should be

S=0.57*sqrt( F*{1/rho)/[w{(1-mu1^2)/E1^2+(1-mu2^2)/E2^2}])

 

[IRstuff]

Since, the two sets of bearings are in series, one would think that the force would be doubled. However, this should still be only the static breakaway torque, i.e., enough to keeping the object spinning, if it were already spinning, and you just wanted to overcome the frictional losses.

From a dead stop, then, you would essentially take forever to spin up to any speed, since you have no excess torque for accelerating the structure.

Your equation looks funny. The 2/3 factor is something I use to approximate the rotational inertia of a flat plate, so you see to be mixing concepts.

Your equation should have a torque term equal to the breakaway torque, plus an I*alpha term that represents the
acceleration you are trying to achieve. Obviously, if you allow alpha, the angular acceleration to be zero, you get basically the equation you're using. The higher the acceleration you want, the more torque you need to apply. Once you get up to speed, the torque can be reduced to the first term that counters the friction.


TTFN

FAQ731-376

 

[dhengr]

jreamer:
It seems like btrueblood 8JAN 17:51, gives you some pretty good advice about running torque and friction coef. on the system (structure) you have, you actually measure it (approximated it, at that time). With crowned rollers I don’t know how much I would worry about caster, they move (roll) perpendicular to their radius through the center of rotation.

Even with the sketch it is very difficult to fathom what you are really trying to do or exactly how you are doing it. There is no info. on the stiffness or dimensions of the various system parts. I assume this system is built and being tested, per your 8JAN 16:43 post, and am not sure what you are willing to change to make it work better, whatever work better means. Obviously, one solution is a bigger motor.

You have deflection and Hertz stress/deformation problems all over the place. Are there only two hydraulic cylinders, if so, what prevents the upper head frame (platten) from trying to rotate wrt the bed frame? If this happens and the cylinders try to extend to allow this, do they impart a force greater than 6k? What are the mechanical properties of the material being compresses, how is it contained laterally? Obviously it must have enough shear strength to couple the t & b round plates. But, how does it distribute the 6k force to the t & b round plates, uniformly? What are its elastic properties as it deflects and loads the plates? This loading on the cam rollers is probably pretty uniform to all 8 rollers. But, the rollers are loaded on their inner edges due to the round plate deflection. However, what’s the stiffness of the head and bed frames, they will tend to load the cam rollers (roller reaction points) nearest the jacks and unload the cam rollers at 90° away (in plan) from the jacks. That’s true even without considering the cross stiffness of the head and bed frames. You have a significant material compatibility problem between the A36 pl. (whatever its thickness) and the cam rollers, for this type of application.

You have designed a number of non-symmetrical bending and deflection axis into this system, vs. your round load plates and cam rollers, which just seem to complicate your problem and system. Once you have addressed these stiffness and deflection problems, I think you will have gone a long way toward resolving your cam roller and torque problems. I would go to one pushing cylinder, on center, add some structure to maintain alignment, and possibly a new bearing arrangement. You don’t explain how you impart the motor torque into the bottom round plate, obviously that has to fit into the new structure. How is this system loaded and unloaded. What does it do, other than spin a “compressed object,” without giving away any state secrets, of course.