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Lever and fulcrum calculation

  1. Oct 22, 2016 #1

    as B moves to over A, ( A is a fixed point ) C slides up its guide slot and is stopped by a non compressible object and no amount of force can shift it further up.

    measurements in units.

    At this point >> how do I know how much pressure C excerts upwards. psi? Mpa?

    I hope I'm making myself sufficiently clear. I wouldn't be surprised if there are simpler ways to look at this.

    I've no doubt I'm using wrong words. (how to frame Q correctly?)

    Is this a fairly standard configuration? if so what's it called?

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  3. Oct 22, 2016 #2


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    This is a bell crank / connecting rod / slide arrangement . It does not have any general name .

    You need to work with forces . Newtons or Lbf .

    A force applied to the the upper arm on the bell crank will be balanced by a reaction force from the stop fitted to the slide .

    The relationship between the applied and reaction forces can easily be calculated . To do an actual calculation you need to assign an effective length to the upper arm of the bell crank .
  4. Oct 22, 2016 #3
    Thank you. Much appreciated.

    bellcrank/connecting rod/slider helped a lot. I found images of examples (and lots of formulas). So I've taken steps towards understanding.


    This image explains a bit more what I'm trying to do.

    There's a box as a piston inside another (open ended) box.

    Fixed pivot point A is on a lid.

    Between the lid and piston is earth mix.

    When the lever pulls up the piston the compressed earth becomes very strong and can be used as building block.

    It is recommended to apply a compression of 20Mpa.

    I'm building the 'knuckle' at the moment out of 3/4 inch steel.

    I envisage the lever on this to be about 6 foot.

    I'm working with inches as units (for scaling) and see myself pulling on the lever at about 5 foot or 60 units.

    If I know how hard I have to pull I think can see if I'm building the parts strong enough to not deform.
  5. Oct 22, 2016 #4


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    Force acting on piston is pressure x area of piston .

    Recommended pressure is given as 20MPa . So need to know the dimensions of the piston .

    It is important to use consistent units in any calculations .
  6. Oct 22, 2016 #5
    OK. I'll give all dimensions as inches. In all of the above 1 unit = 1 inch.

    I was hoping to work out the dimensions of the piston/box from this.

    So, let's say the area of the piston is 6 inches by 12 inches (to produce a block 4 inches thick) or 72 square inches.

    edit add: It takes me a while to absorb what you're getting at.

    Using an online converter 20Mpa is 2900psi.

    Does this then mean the piston needs to apply 2900 lb x 72 = 208,800 lb of force to achieve the required compression on the earth block?
    Last edited: Oct 22, 2016
  7. Oct 22, 2016 #6


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    20 MPa is about 200 atmospheres or 2940 psi . That is a very high pressure . Are you sure that this is the correct figure ?
  8. Oct 22, 2016 #7


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    Yes it does .
  9. Oct 22, 2016 #8
    Good morning.

    Yes, it's recommended in a paper by an Indian scientist who is not the Indian engineer who designed the first of these types of earth compressors. I don't know what pressures the engineer worked with.

    The scientist compared 5 and 20 Mpa in the paper and concluded 20 is better. I suppose that doesn't mean that 5 is not sufficient.

    Anyway, I accept 20Mpa is a lot.

    If I assume that is what I need or want. How do I work out how much pressure to excert on the lever?

    While I can build the mechanism I don't really understand the overall transfer of forces. The lever/knuckle pushes down on the lid while pulling up on the piston. In between the lid is a mass of soil that when squeezed between the lid and piston quickly becomes incompressible. The more that mass is compressed the stronger the resultant block becomes. How much pressure can a mechanism like this excert.

    (AFK for another 8 hours.) Thank you Nidum for the responses.
  10. Oct 23, 2016 #9
    ok. I find a formula F1L1 =F2L2 for a right angle bellcrank. ie. F2 = F1L1 / L2 or using above figures F2 = F1 x 60/3 = 20F1

    That's all very well.

    What's got me stumped is as the bellcrank short lever swings around A the lifting shaft B to C is less and less right angle to it. The angle becomes smaller.

    While the bellcrank continues to move the lift shaft moves the piston less and less upwards. It's at this time maximum pressure is applied to the block. That's the pressure I'm after.

    Is my thinking all wrong. What am I missing.?.
  11. Oct 23, 2016 #10


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    See message .
  12. Oct 25, 2016 #11

    this is from a site demonstrating how to use it.

    It's at this point I want to know the pressure.

    The guy is pushing down.

    The short bellcrank lever is moving towards him.

    The 'piston puller shaft' is still moving up as maximum compression is achieved.

    What's that pressure in relation to the force exerted on the long lever of the bell crank?
  13. Oct 26, 2016 #12


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    Can't tell for sure but I think that the lever mechanism uses the toggle clamp principle to get a very high final force level on the piston puller .

    See if you can draw the mechanism . Put in some estimated dimensions .
    Last edited: Oct 26, 2016
  14. Oct 26, 2016 #13
    Sorry, I don't see how I can make the already posted images clearer.

    The dimensions are as already given.

    Perhaps. The bellcrank pivot point A rests on two 'cradle' plates welded to the lid on either side of it and turns at that point as the long lever is moved.

    The short arm of the bellcrank shares a pivot point B with the assembly that pulls up the piston.

    That assembly is a square bar with the two long plates or shafts welded to its ends that pull up the piston from pivot point C.

    I don't see a toggle mechanism. ?
  15. Oct 26, 2016 #14


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    The toggle effect comes into play as the handle is moves down toward a horizontal position and perpendicular arm on the handle rotates to a vertical position inline with the links connected to the bottom piston.
  16. Oct 26, 2016 #15
    I think I'm beginning to understand.

    If right it reminds me of the cam lock mechanism on sliding tripod legs. They always seem to grip particularly hard (and prone to break) when fully applied.

    So then (awaiting input) toggle is not necessarily a particular mechanism but an effect. If so then perhaps I've experienced this in other ways and instinctively felt there is something particular about this bellcrank configuration that does apply very high force at optimum use.

    So how to calculate this.?

    I don't want to get ahead of myself at this point and would still like to understand fully this toggle effect.
  17. Oct 27, 2016 #16


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    To start, I need to correct my above terminology error(s).

    A "bellcrank" is actually the name for a simple lever with one arm at an angle or perpendicular to the other and a pivot at the joint of those two arms as described above by john101. In this case the bellcrank is the handle with the short arm acting as one of the links in the toggle mechanism.
    A "toggle mechanism", in a general mechanical application sense, is a mechanism that has an unstable point at which a small lateral force will cause it to snap to either one side or the other (simple examples are a standard electric wall switch and what are sold as "toggle switches" for use in electrical circuits); and in your application that is achieved by two links joined by a center pivot that can be straightened to apply an extreme force or as an "over center" locking device like you have seen on a tripod.

    Now, back to your specific application issues. First, I want to discuss some issues related to how this mechanism is used. One advantage of it is that it can provide a method that as a lateral force is applied at the two links center connection can as they approach an inline orientation can act as an extreme force multiplier. The downside of this assembly is that at the same time at that point the amount of lateral motion and lengthening of the linkage are very small, while the increase in possible toggle force rapidly increases. As a result, the design angle between the links at which the maximum force is to be applied is must very carefully controlled to prevent either excessive or insufficient loading when the linkage simply snaps through its alignment point without reaching the desired application load. At same time, the opposite is also true; in that, the amount of toggle force relative to the amount of lateral force applied decreases rapidly as the distance between the toggle links connection and their inline alignment increases.

    Mathematically (Using the simplest form being a straight lateral force against the center of the toggle assembly): F toggle = F lateral x cos Θ , where Θ is the angle between a line drawn between the top and bottom pivots of the toggle and a line drawn between the two pivot points of one of the toggle arms.
    (For your actual case, where the lateral force is being applied by the handle torque applied to the top link of the toggle the mathematics is more complicated but the above gives a representative example of the force ratios and how the required operating handle force will be effected.)

    For this reason, in many applications the toggle is applied in series with another force creating element like a spring. As an illustration of this arrangement using the above press, it could (or, since we cannot see the details of the bottom portion of the press) may actually be a loading spring, or set of springs, between the bottom plate of the press and the cross bar between the two links being used to lift the bottom plate that provide the necessary compression loading when the toggle is fully extended.

    I am focusing on this issue because there some specific design factors on this type of application that can effect the actual point at which a desired load will be achieved at a reasonable press handle load as the toggle approaches its full extension on the press.
    1. The accuracy of the amount of soil that is loaded into the press each time.
    2. The consistency of the density and compressibility of the soil that is being loaded into the press each time.
    3 The desire or requirement for each finished brick to be of a specified thickness and the allowed variance of that dimension.

    OK, At this point, I am going to stop to give you some time to review all of the above and post any comments and questions you may have before addressing any further issues.
  18. Oct 27, 2016 #17
    OK, Thank you.

    I find the idea of using a spring (if I understand correctly, to ameliorate the toggle force in extremis very interesting.) I'm not sure I understand how.

    I will be incorporating stops to prevent "the flip".

    I have a few different car coils.

    As yet the press doesn't include any springs.

    Another part of the 'machine' is when one pivots the long handle back the other way the two linkages rest on rods sticking out of the case and when pressed down the finished brick rises up ( after lifting the lid/cap ) out of the case to be removed, the piston is then retracted, soil reloaded, lid replaced, repressed, etc producing a smooth workflow.

    the two linkages simply pull up a box piston inside a case.

    The soil will be consistent, sifted, clayey soil on site with roughly 10% lime and 10% water with little organic material loaded from a measured container.

    I'll have to work out exactly how much later through trial and error. I assume if soil mix is measured and prepared consistently all bricks/blocks will be sufficiently similar.

    (I suspect I'll have to rebuild the bellcrank/'knuckle' as I may have underestimated just how much force is involved.)
  19. Oct 27, 2016 #18
    edit add: to post 17. The long handle is removed from the 'knuckle' before moving the linkages from the lid and raising the brick/block.

    (I used inkscape and youtube tutorial on using it to make this image.)


    Based on this image.

    What is the relationship (graph) between F1 and F2 as angle ABC* approaches 0 as the soil between lid and piston becomes incompressible.

    edit correction: based on notation by TomG, I had angle BA BC meaning the angle ABC. F1AB is 90 degrees (I'm having a bit of trouble understanding the formula. Not so much calculating with cos, tan perhaps, using online calculator, but if I get it F2 is 0 when ABC is 0 and BAC is 180.? And the distance to F1 doesn't matter?)
    Last edited: Oct 28, 2016
  20. Oct 28, 2016 #19


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    • the angle F1,A,B is 90o
    • when B, A, C are in a straight line with each other, the angle is 180o.
    Angle 'A' = the included angle BAC
    Angle 'B = the included angle ABC

    F2 = F1 x cos(B) x tan(90- (A/2))

    p.s. A graph is left as an exercise for the reader. :smile:
  21. Oct 28, 2016 #20


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    As a suggestion for the "spring" I mentioned in my last post; obviously, you are going to want something that is as simple and compact as possible and one way to achieve this may be by placing a sheet or sheets of high hardness elastomer cut to the dimensions of the bottom of your mold cavity on top of the bottom plate and then placing a similar size metal plate on top of that which will serve as the actual bottom of the mold cavity.

    Because of the high compressive pressure you want, a good standard sheet material might be 90 durometer nitrile (rubber) and this material is available in several thicknesses up to 1/4" thicknesses; but, you may want to stack multiple sheets so you can adjust the amount of compression to obtain the toggle load position you want.. A sheet of that material or something similar may be available from a local gasket material supplier or can be found online from someone like:


    Just a suggestion.

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