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Force generated by rotating mass 1
- Thread starter Ralph2
- Start date
![[smile]](/data/assets/smilies/smile.gif "[smile] [smile]")
I get 82483 lbf, not sure why our answers are that different.
F = m*r*[ω]2
TTFN
FAQ731-376
F = m*r*[ω]2
TTFN
FAQ731-376
- Thread starter
- #5
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
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
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
FAQ731-376
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
FAQ731-376
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1
- #7
- Thread starter
- #8
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
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
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
FAQ731-376
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
FAQ731-376
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
Worry more about the design of tooth attachment and getting good welds so that teeth don't fly off the rotor.
Ted
- Thread starter
- #11
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
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
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
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.
- Thread starter
- #13
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.
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
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
- Thread starter
- #15
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
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
- Thread starter
- #16
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
Thanks all
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
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
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- #18
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