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Counterweight for crankshaft

  1. Nov 17, 2015 #1
    I have a crank shaft that lifts a flatplatform up 0.5 inches and down 0.5 inches. The load here is about 250 lbs. So just one load on this 3" long steel shaft with 1 inch thickness. Is it necessary to have a counter weight for this? If so, how much should the counterweight weight?

    The shaft will rotate at 650 RPM.

    Thank you,
     
  2. jcsd
  3. Nov 17, 2015 #2

    berkeman

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    Staff: Mentor

    Can you post a picture of the setup? Or a drawing? :smile:
     
  4. Nov 17, 2015 #3
    Attached is just the shaft component. So on the left of the flywheel will be the load (250lbs). On the right is the counter weight that I modelled...I do not think its neccessary. Flywheel is 6 inches in diameter.

    The reason why the counterweights are so offset from the flywheel is because there will be a mount inbetween the flywheel and the counterweight which keeps this on the ground.
     

    Attached Files:

  5. Nov 17, 2015 #4

    JBA

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    How is the crank motion being applied to the platform?
     
  6. Nov 18, 2015 #5
    The shaft on the left of the flywheel is offset by 0.25 Inches from the center of the flywheel. So it will cause the table to move up and down vertically.
     
  7. Nov 18, 2015 #6

    JBA

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    OK, I now see that offset, I missed that when I first viewed the drawing.

    I think that the addition of a counterbalance to offset the weight of the offset shaft and its attached load would definitely help reduce the lateral vibrations and eccentric loading on the main bearing, and increase the operating life of the bearing, similar to the counterbalances on an automobile crankshaft; but, to be effective the counterbalance needs to be directly adjacent to the eccentric shaft.

    Unfortunately, I have had only limited experience, many years ago, on this issue as it applies to positive displacement pump crankshaft design; so, I can't give you a strictly professional response to how important it is on your particular application. At the same time, the below site includes the formula for determining the amount of counterbalance for an engine crank journal; so, you might run a quick calculation for the counterbalance weight for your application and then based on the amount of weight specified decide how important its effect will be for your application.

    http://www.modelaengine.com/13-crankshaft-design-a-6303.html
     
  8. Nov 18, 2015 #7
    So what your saying is the counterweight should be as close to the flywheel as possible? Or even closer I guess if it wasn't for the flywheel blocking it? Right now thats the closest it can be because of space constraint. The counterweight(s) will be made of mild steel...so approximately ~10lbs
     
  9. Nov 18, 2015 #8

    JBA

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    If you place the counterweight at its present location and orientation, it will make the eccentric loading on the crankshaft main bearing worse (think of the main bearing located between the crank and c'wt as the center pivot between two loads at opposite ends of a beam); but, if your rotate the c'wt 180° to eliminate that eccentric loading then it increases lateral vibrations on the whole drive assembly. As a result, the current c'wt location is a no win situation. Assuming that the small shaft diameter next to the c'wt is the shaft bearing location, is there enough clearance between that bearing mount and the back of the flywheel that could allow a semicircular c'wt to be bolted on that back face?
     
  10. Nov 18, 2015 #9
    Yes the main bearing is a journal bearing that will be located between the flywheel and the c'wt. it will pretty much take up all the space inbetween the two parts with maybe 0.1-0.2 inch clearance.

    Check picture below for full assembly.

    Thank you
     

    Attached Files:

  11. Nov 18, 2015 #10

    JBA

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    Ignoring the current location of the c'wt for the present, in your original picture, there appears to be a clearance space between the main bearing journal and the back of the flywheel. How wide is that clearance?
     
  12. Nov 18, 2015 #11
    0.8inches approx
     
  13. Nov 18, 2015 #12

    JBA

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    OK, stay with me for a bit. Now, what is the outside diameter of your flywheel and the diameter of the hub section that separates the flywheel from the main bearing journal?
     
  14. Nov 18, 2015 #13
    flywheel outside diameter is 6 inches
    hub diameter is 1.47 in
     
  15. Nov 18, 2015 #14

    JBA

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    OK, what I wanted to determine was the weight of a steel 6" OD x 1.47 ID x .8 thick semicircular disc segment to see if something like that could be bolted on to the back face of the flywheel between the flywheel and main journal bearing. Unfortunately, that weight is only 3 lbs; which definitely does not get us anywhere the 10 lb requirement you calculated from my reference site.

    Well, at this point, I have to admit that I am running out of suggestions to potentially help solve your crankshaft imbalance issue; but, I am a still concerned about the eccentric loading and vibrations that will exist without a proper counterbalance.
     
  16. Nov 18, 2015 #15
    Well the load on the left of the flywheel will be 250 lbs (vertical load). As suggested from my original post, it may not be significant to require a counterWeight. The 10 lbs I got was through solidworks material properties of steel for the given geometry of the cw't.... I do not know how much the weight should actually be for this if any weight is necessary at all.

    Edit: realistically more like 185-210 lbs
     
  17. Nov 18, 2015 #16

    JBA

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    If you have not done so, I strongly suggest you analyze the bending stress at the junction of the flywheel hub and the main bearing journal using the dynamic loading of vertically moving that 200 lb load over a vertical distance of 1 inch in just under .05 sec. and I will leave it to you to calculate the resulting acceleration and dynamic loading; but, I feel you are seriously under estimating both the cantilever beam stresses, fatigue stress limits and the potentially the motor horsepower required for your planned design. This is not simply a rotating mass problem.

    In conclusion, I can only caution you that, unless you have an extensive background in designing this type of application, you potentially not taking its analysis as seriously as is required.
     
  18. Nov 18, 2015 #17
    Just to provide a little more background,
    This is a vibration platform that I will be using to vibrate/"shake" packaged boxes for package integrity testing...
    Could you explain the 1 inch per .05 sec?
    650 RPM = 650 Inchs/Min (in y axis)
    So in 1 second it travels 10.83 inches (total distance)
    I will be using a 2HP motor which will will drive the flywheel via belt.

    Also wouldn't the fact that the cw'ts are off to the right help create a larger moment despite its relatively lower mass with respect to the crank load?
     
    Last edited: Nov 18, 2015
  19. Nov 18, 2015 #18

    JBA

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    I wondered if that was the application.

    The key thing to remember here is that while the crank motion is continuously rotational the 200 lb table motion strictly in the Y direction and as a result that in each 1/2 of a revolution, the 200 lbs is being accelerated from a 0 vertical speed to a maximum velocity in the first quarter of that 1/2 revolution and then decelerated back to a 0 vertical speed in the second 1/4 of that 1/2 revolution and then this is repeated again in the second 1/2 revolution of the crank.


    As an experiment, I have performed the following calculations that need to be confirmed by you before you accept them:
    While the time for one revolution is .093 sec, the table will travel 1 inch vertically every 1/2 revolution so the time for it to travel either up or down 1 inch is 1/2 x .093 sec = .046 sec per stroke; and this means the table wt will travel vertically at an average Y velocity of 1/.046 = 21.66 in/sec = 108 ft/min on both the up and down stroke.

    As I see it, the full force of and maximum crank stresses from the combination of table's weight and its KE should be at the bottom of the down stroke, with the KE being offset by the table weight at the top of the stroke.



    The biggest remaining issue, and one I am not knowledgeable enough to address, is the actual torque load on the drive system by the crank. Obviously, for the down stroke the table weight will handle the problem. The difficulty here is that at the bottom dead center position the crank leverage to lift the table weight is essentially zero; and, I guess that is where the stored flywheel and table inertia from the down stroke comes into effect but I know nothing about type of calculation.


    With regard to the position of the c'wt's, with the c'wts separated across the main bearing and placed as shown, when you spin the crank the crank load and the c'wt are going to act in concert to place a combined eccentric load on the main bearing shaft and the bearing. Just take a look at your drawing and visualize how centrifugal force is going to act on both of them. They are going to try to wobble the whole assembly around the center point of the main bearing.

    I hope this is of some help in your project efforts.

    PS The only shake table I have seen was driven by hydraulic pistons.
     
    Last edited: Nov 18, 2015
  20. Nov 18, 2015 #19

    JBA

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    Just as an added note, there is something else I remember about that shaker table and that is that it was setting on springs to balance the dead weight of the table (and maybe part of the load, I do't know about that part). That way the only energy input required was to cycle the table with assistance from the springs. Of course, in that case there has to be some careful design work done on the spring selection as to its springs' rate and natural frequency to make sure it does hit a resonance point at the desired operating cycling speed(s) of the table.
     
  21. Nov 19, 2015 #20
    Thanks JBA, the calcs make sense. Yes I am using bushings, primarily used to keep the platform horizontal but it will be absorbing/dissipating energy which will help the table from stopping at the dead bottom point
     
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