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Force generated by rotating mass

Force generated by rotating mass

Force generated by rotating mass

(OP)
Hello

Could someone be kind enough to confirm or correct my calculation of the force generated by a mass of 80 pounds rotating at a radius of 7.5 inches at 2200 rpm. My calculation is 81,666 pounds of force... am I close?

Thank you for your time  

RE: Force generated by rotating mass

Ralph2,

   That is a weight of 80lb!

   I am getting 82,500lb.  Don't do it while I am in the area, okay? smile

               JHG

RE: Force generated by rotating mass

IRstuff,

   That sounds about right.  I rounded mine off.

               JHG

RE: Force generated by rotating mass

(OP)
Thanks guys.. close enough. Before I go to my customer I wanted to be sure of my numbers.
This is from a machine used to clear brush and small trees, the environmentally correct way to clear seismic survey trails, among other uses. A 8 foot drum attached to the front of an all terrain vehicle. There are various designs but this one has 6 rows equally spaced with 8 carbide tipped teeth on each row. Each tooth is about 10 pounds.
The issue is that the customer wants to balance the rotors without the teeth, his argument is that each row has 8 teeth and is directly opposed by another row.
My argument is that because the teeth are offset there will be considerably dynamic unbalance generated once the teeth are installed.
I wanted to try and quantify just how much of a force is generated at each bearing. I have not yet completed my calculations, perhaps... it will be low enough to be acceptable and his request will work... but I doubt it

Ralph
  

RE: Force generated by rotating mass

OK, that's a totally different problem, then.  Now, you're looking strictly at an imbalance.  The odds are that they're randomly unbalanced, and very unlikely that they'll all line up.  

So, if you assume a worst case of a 2lb imbalance per row, AND a worst case of all 6 rows having the same imbalance in the same orientation, you'd have something like 12,372 lbf.  Still, a rather large number, but clearly, the most likely imbalance contributor is going to be the teeth, particularly since they're located at the maximum radius.

TTFN

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RE: Force generated by rotating mass

IRstuff, I think the worst case would be three adjacent rows of imbalance.  If all the rows have the same imbalance will there not be some resulting balancing of opposed forces?

Ted

RE: Force generated by rotating mass

(OP)
The question... how much force is generated at each bearing by adding teeth to a previously balanced rotor without teeth. The numbers I have come up with are scary to the point I wonder about my logic.
This rotor has 6 rows with 6 evenly spaced teeth. there is a combined force due to adding the teeth of 15312 pounds acting at the center of each row of teeth. All references are from the right hand side bearing in inches.
The center of row 1 (where the cumulative force of the teeth react) is 36.375. The center of the row opposite is at 38.25 (1.875 inch offset).
Row 2 is at 34.5, and opposite at 40 (5.5 inch offset)
Row 3 is at 32.625 and opposite at 41.75 (9.125 offset)
From here I calculate what the force at the bearing is, eg. row 3 15312*9.125/32.625=4282. If I do this for the other 2 sets and combine the values (@ 120 degree intervals 789; 2441; 4282) I get 3,026 pounds.
This seems high, to the point I feel I am overlooking some basic math.
One of the issues is that the center of the force of each set is not directly in the center of the rotor. Set 3 for example is offset by 0.437 inches, how to include this is giving me a headache.
The rotor is 73.5 inches on bearing centers. Picture of rotor included (I hope)
Any advice / thoughts / lessons in math appreciated
Ralph

RE: Force generated by rotating mass

Oh, that's why a picture is always preferred.  I think you are grossly overestimating the imbalance forces.  Each row of teeth is matched by a diametrically placed set of teeth, and even though they are laterally displaced, there is no imbalance from the offset, per se, since the system as a whole is rigid.  The only imbalance comes from material distribution uniformity, i.e., if, say, row 3 were missing 2 teeth, while its diametrically opposed row wasn't missing any teeth.  

The whole thing weighs, what, several hundred pounds?  If the shaft were light enough to be affected by the moment created by the tooth offset, it would be too light to maintain its rotational momentum when the teeth actually bite into something.  The tearing force generated by this structure has to be high, and the rotational inertia has to be even higher, for this system to work correctly in the first place.  Therefore, it cannot be affected lateral displacement of teeth.

TTFN

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RE: Force generated by rotating mass

Ralph2, I'd have to agree with your customer.  Get a  reasonable balance on the rotor.  Adding the teeth will tend to be self-balancing.  The remaining imbalance will not be significant in this application as IRstuff points out.  Compared to the cutting force reaction, imbalance due to the tooth distribution will probably not be noticed.

Worry more about the design of tooth attachment and getting good welds so that teeth don't fly off the rotor.

Ted

RE: Force generated by rotating mass

(OP)
Why would a lateral displacement not create some dynamic unbalance. In the worst case scenario some 9.125 inches. Just how much is what I was trying to quantify while balancing this rotor.(I have a fair amount of "think-time" while waiting for welds to cool)

This rotor sans teeth weighs 900 ish pounds and rotates at 2200 rpm. We are guessing that each tooth weighs 2.5 pounds with a center of gravity at about 7.5 inches.

Unfortunately, at the moment I have no feedback on how this is going to work out. Ideally I would hope the customer mounts the rotor and runs it with no teeth, makes a judgment on the level of vibration. Then install the teeth and makes a new assessment.

We will see, more work for us if it comes back. These are a bugger to balance, as one welds on weights the rotor distorts, creating runout on the bearing stubs. This then offsets the mass and requires yet more weight. To keep it "straight" I weld short passes on the opposite side. All very time consuming it takes at least an 8 hour day to do one

RE: Force generated by rotating mass

Ralph2,

   I have worked on some reciprocating devices, where I thought the back and forwards forces were rather high.  Vibration turned out to not be a problem, because the device was rigidly coupled to a relatively heavy structure.  The forces, however high, were exerted for a short period of time, on a substantial mass, and the resulting acceleration was not that high.  The results were not as noticeable as I had anticipated.

   The forces I was looking at were nowhere near your ballpark, but the principal works.  Your 80klb imbalance would be manageable if the system were attached rigidly to a D9 bulldozer or an Abrams tank.  Do you have a heavy vehicle to attach this thing to?

   If you balance thing perfectly, you still are going to have your cutting forces to contend with.

               JHG

RE: Force generated by rotating mass

(OP)
We will have to wait and see if the customer is happy with this. Personally I have not seen one of these in operation, the guy from our shop who tries to do the field balancing tells me that they do run "rough". Operator comfort is the bottom line criteria, if his coffee thermos stays in the holder it seems to be good enough.

RE: Force generated by rotating mass

Ralph2,
your calculations for bearing reations to each resolved couple are incorrect.  You should divide each resolved couple by the bearing to bearing distance, 73.500 inches.  The couples act anywhere along the length of the shaft, not at any particular point of application.  The bearing force reacting to each couple is acting through the length of the shaft, sum moments around each bearing.  Reaction force at a bearing multiplied by the bearing separation equals the couple due to imbalance in the plane of the couple.  Then the couple divided by the length of bearing separation equals the resultant bearing reaction in the plane of the couple.
The bearing forces are much lower than what you calculated.

Do a search on two plane balancing for a methodology on balancing multiple imbalances on a stiff rotor.

Ted

RE: Force generated by rotating mass

(OP)
Thanks hydtools (Ted)
So in my worst case set I have a (15312*9.125/32.625)=4282 pounds at the bearing. You are now saying this number should be divided by the distance between bearings? = 58.25 pounds. This is a much more realistic value.
I appreciate your analysis / help
Thank you

RE: Force generated by rotating mass

(OP)
I have used hydtools advice and have come up with a combined force of 41.24 pounds on one bearing and 42.28 pounds on the other. Values that are reasonable and possibly acceptable to the customer. These numbers work out to a balance spec ISO 1940 G4 (roughly)

Thanks all

RE: Force generated by rotating mass

Ralph2,
No, I'm saying 15312*9.125/32.625 = 4282lbs. should be 15312*9.125/73.5 = 1900lbs.

15132*9.125 in-lb is the resolved couple in the row 3-6 plane and 73.5 in. is the bearing spread.  1900lbs is the bearing reaction in that plane.  Resolve this reaction with the other two reactions to the other two couples to find the resultant bearing reaction.

Ted

RE: Force generated by rotating mass

(OP)
That changes things a bit.
I now have bearing reactions in rows 1/4 =390.62; rows 2/5 =1145.83; rows 3/6 =1901.047. Adding these up give a total of 1308.06 pounds of force at each bearing.. ouch!
 

RE: Force generated by rotating mass

Either balance the rotor or make sure the bearings can handle the load.
What will the cutting load be on the bearings?
Vibratory shakers apply 22,000 lbs to bearings.

Ted

RE: Force generated by rotating mass

How big is this ATV? How is this device mounted to the ATV? On 2 pivots like a typical snow plow?   

RE: Force generated by rotating mass

(OP)
There are quite a few styles out there. I have never seen the machine that holds the rotor in question but here is a link to something similar http://www.edgeequipment.com/equipment/mulcher_b500b-3.html A search for brush mulcher will bring up quite a few samples.

The original question / discussion arose.. Because the customer wanted it balanced SANS teeth and... while in the process of balancing this rotor I wondered what the actual value (force) of the unbalance would be at the bearings.

 

RE: Force generated by rotating mass

You may add counterbalance weights to reduce the imbalance of the staggered teeth rows.  Add the weights after the teeth are installed if you need to balance the rotor before adding the teeth.  It would be a calculated balance method rather than a measure and balance method.  But close enough to reduce the imbalance and feedback to the operator.

Ted

RE: Force generated by rotating mass

(OP)
Okay.. I think I am happy with my latest rational. I have two equal opposing forces acting on a rotor at (set 3/6) 32.625 and 41.75 inches from one end of my rotor which is 73.5 inches long.
Around one end I will have F(15321.54)@ D(73.5-32.625)/73.5=8515.648 Around the other end I will have 15312.54*41.75/73.5=8697.941
I do this for all 3 sets and add the results at each end giving me a net value of 675.59 and 631.48. A realistic value and is roughly equivalent to an ISO 1940 value of 60.

Have I gone wrong?? my calculations (spreadsheet) are attached

RE: Force generated by rotating mass

I will make you less happy.

For your row set 3/6 example you did not take into account the moment caused by row 3 about the rh bearing, 15312*32.625/73.5 = 6797, in order to determine the lh bearing reaction 8698 - 6797 = 1901.  Calculate 15312*(41.75-32.625)/73.5 for the bearing reaction of 1901 lbs.  There are two moments around each bearing.  One caused by each row of teeth.  Each row generates a moment countered unequally by its opposing row resulting in a twisting couple which is the product of the force 15312 and the row center offset from each other.  Divide this couple by the bearing spacing 73.5 to get the bearing reaction to that row's imbalance couple.

Combine each bearing reaction to arrive at the resultant reaction.  Keep in mind the direction of all row set reactions when you calculate the resulting bearing reaction.

Ted

RE: Force generated by rotating mass

the bearing reaction will depend upon the stiffness of the supports. To develop any where near the calculated reaction force, the supports would have to be very stiff indeed. With the vehicle on its suspension the 2200 rpm may be well above some rigid body mode, and the vehicle (or rotor frame on flexible arms) motion will be mass controlled, not stiffness controlled. If so, the rotor will be rotating about its CG, and the bearing supports will be moving an amount related to the offset or eccentricity of the CG to the shaft mechanical center.  

G60 would be on the order of 0.01 inch eccentricity, or 0.02 inch peak-to-peak or TIR.
See page 3 or 4 here -
http://www.irdbalancing.com/downloads/TechPaper1BalQualityReqmts.pdf

RE: Force generated by rotating mass

(OP)
Thanks hydtools... I thought of that this AM while having my first cup of coffee. Back to my spread sheet <grrr>

RE: Force generated by rotating mass

(OP)
Okay.. re done and I hope for the last time. Am I right this time????

Set 1/4, the right side bearing would see a force of (37.125*15312.54=35.25*15312.54+73.5*X) =390.626 pounds. The left side bearing would see the same force (36.375*15312.54=38.25*15312.54+73.5*X) but as a negative value of -390.626
 
Set 2/5, RS (39*15312.54=33.5*15312.54+73.5*X) =1145.836  LS (34.5*15312.54=40*15312.54+73.5*X)

Set 3/6 RS, (40.875*15312.54=31.75*15312.54+73.5*X) = 1308.064 LS (32.625*15312.54=41.75*15312.54+73.5*X)

When these forces are combined the net result is a pull of 1308.064 pounds on one bearing with an equal pull in the opposite direction on the other bearing.

This works out roughly to an ISO value of G120.. I am pretty sure I will see this rotor back soon to balance with the teeth installed.
 

RE: Force generated by rotating mass

That's what I got.

Ted

RE: Force generated by rotating mass

(OP)
Thanks for hanging in and giving me the help and direction I needed.

RE: Force generated by rotating mass

Sorry I answered without looking at my work.  The set calculations are the same except check your 3/6 set calculation.  I think it should be 1900 lbs.

My resultant force at the right hand bearing is 2640 lbs. acting nearly 90 degrees clockwise from row 1.  I assumed rows numbered 1 thru 6 in a counter-clockwise direction looking end on at the left hand bearing.

There are some strange keystrokes in your equations.  Typos?

Ted

RE: Force generated by rotating mass

(OP)
Thanks Ted
You are quite right, I copied and pasted something that is different than the spreadsheet I was using. I do have 1901.047 as a result of set 3/6. From this computer I can not add the resulting forces so will take your word for the resulting 2640 pounds.  

RE: Force generated by rotating mass

(OP)
Ted, your resulting force (2640) is almost exactly double (1308.66) what a little DOS program (from IRD)I have used for years to combine weights. I used a drafting program (Solidworks) to calculate the results and agree with your figure (2638.710)...
I may have to retire my DOS program or at least be skeptical of the results.. Thanks for your help

RE: Force generated by rotating mass

You are welcome.

Ted

RE: Force generated by rotating mass

All - I'm new to the forum, just ran across this topic on a google search today. I'm trying to recreate the original ~82,500lb force from the 80lb weight at r=7.5". The only way I can do so is to divide the 80lb weight by 9.8 m/s^2 - was there a unit error in the calculation? I believe the actual value is approx. 2090 lbf.

F = mrw^2

m = 80lb/32.17 ft/s^2 = 2.49 slug
r = 7.5 in = 0.625 feet
w = 2200 RPM/60 = 36.67 RPS

F = 2092 lbf

I may be way off base.

RE: Force generated by rotating mass

The units for 'w' are rad/sec, not rev/sec.

Ted

RE: Force generated by rotating mass

(OP)
The formula I use is somewhat bastardized to give the results in pounds of force while dealing with a weight in grams. So the 80 pounds ends up being 80*16*28.349=36286.72 gram. Then my formula .062*(N/1000)^2*radius*weight in grams .062*4.84*7.5*36286.72=81666.892 pounds of force.
How accurate the number is..is what led to the original question. I don't recall from where I got this (perhaps some balancing course). I believe if it gives an approximate value but only over a certain speed range... For example if one uses the speed of 10 RPM the value is only 3 pounds.. obviously not true  

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