How to Calculate Pressure Exerted by C in a Lever and Fulcrum Configuration?

[john101]

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?

 

[Nidum]

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 .

 

[john101]

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.

 

[Nidum]

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 .

 

[john101]

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?

 

[Nidum]

20 MPa is about 200 atmospheres or 2940 psi . That is a very high pressure . Are you sure that this is the correct figure ?

 

[Nidum]

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?

Yes it does .

 

[john101]

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 exert 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.

 

[john101]

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.?.

 

[Nidum]

See message .

 

[john101]

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?

 

[Nidum]

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 .

 

[john101]

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. ?

 

[JBA]

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.

 

[john101]

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.

 

[JBA]

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.

 

[john101]

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.)

 

[john101]

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?)

 

[Tom.G]

Based on this image. What is the relationship (graph) between F1 and F2 as angle BA BC approaches 0

Given:

• the angle F1,A,B is 90^o
• when B, A, C are in a straight line with each other, the angle is 180^o.

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.

 

[JBA]

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:

https://www.atlanticgasket.com/gasket-manufacturing/types-of-gaskets/nitrile-gaskets.html

Just a suggestion..

 

[Tom.G]

Correction to post #19:

Based on this image. What is the relationship (graph) between F1 and F2 as angle BA BC approaches 0
Given:

• the angle F1,A,B is 90o
• when B, A, C are in a straight line with each other, the angle is 180^o.

Angle 'A' = the included angle BAC
Angle 'B = the included angle ABC
L_H = length of handle, F1 to A
L_AB = length of arm A to B


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

p.s. A graph is left as an exercise for the reader.

(That's what happens when trying to think around midnite!)

 

[john101]

That seems like a brilliant suggestion, and I've got a nice thick sheet of rubber (off a truck) to experiment with. Thank you for that.

I explain how I understand it. As the bellcrank long handle moves towards horizontal and the brick becomes incompressible,(and the fact that no two soil mixes will be exactly the same so the distance C will have to move is never exactly the same.) if the rubber is there it allows the bellcrank to continue to move and apply the force achieved at the optimum toggle point. Otherwise ... what?

As you can see I'm still struggling to fully understand. So far I've found various resources that describe the toggle mechanism. None of them explain it to me. It's as if they are written for someone who already understand.

Further, I don't understand Tom.G's formula. Is it correct. Can a better explanation of it be given or a resource that does so be pointed at?(edit add : OK Tom has corrected formula. :) cool )

Thank you.

 

[JBA]

You understand what I was describing perfectly.
Without the rubber, if there is too little sand or the fill material density is a bit lower than needed, then the desired brick molding compression load will not be applied at the toggle's full stroke; and, alternatively, if there is too much sand or there is a higher density material, then rotation of the toggle may be stopped at an angle that does not provide the force multiplication required due to the limit of the force the human operator can apply to the handle.
This is because, as I described earlier, achieving the maximum amount of force multiplication by the toggle is dependent upon the toggle achieving its optimum stroke, or at very close to it.

 

[john101]

OK, thank you for that. I feel more confident about my thinking now.

That seems to me a brilliant innovation or addition to the press. I thank you and possibly many other people trying to build or has built this type of press will be grateful too. Now thinking about it I'm surprised no-one has thought about it before. Simple, easily done. Excellent. Now to try it.

Now I'll work on Tom's formula. I'll see if I can make a graph.

 

[Tom.G]

Now I'll work on Tom's formula. I'll see if I can make a graph.

If you come up with a graph, please post it. I'd like to see it... to see if my mental picture of it matches.

 

[john101]

I'll use distance from A to F1 as X, AB as Y, angle BAC as a (alpha), angle ABC as b (beta)

So, F2 = F1*(X/Y)*cos(b)*tan(90-(a/2))

If that's ok, image and formula, I'll be AFK for a few hours and hope to have something to post by tomorrow.

 

[Tom.G]

Looks good.
(I see you cleaned up my typo of an extra "(" too!)

 

[john101]

my brain hurts.

 

[JBA]

If that is the result of the equation then it appears there is a problem with the equation; and, a complicating factor is that the amount of F1 force that the operator can apply will not be constant due to the fact that his weight contribution vector is always vertically downward. From a practical application standpoint, it might be better to start at the point that the handle is at a 45° angle to the ground with the handle F1 force vector vertically down and calculate the F2 toggle force based upon the operator's weight in a range from 45° to 0° for the handle angle. Unfortunately this adds another angle factor into the equation.

Edit: A simpler way might be to keep the same calculation range and assume a good estimation of the operator's weight force will always be tangent to the handle and as a result the torque force applied to point B will be constant as it rotates through its corresponding 45° angle.

 

[john101]

I was quite careful in setting out the formula in LibreOffice Calc and making the graph and used an online triangle solver to get angle alpha. (from this I also got length AC which will be useful later.). I'll recheck later. I have to be afk for the next 8 hours or so.

I used 100 pounds of force at 60". I figured this to be a reasonable starting point. As you say it will vary depending on the angle.

Meanwhile I've improved the knuckle, increasing the pin sizes and general thickness of material.

Later, and thank you.

edit add: I think I got the problem : I need to convert degrees to radians before calc calculates cos and tan. Later.

 

[john101]

that's a bit better... something is missing though.

while trying to understand the toggle effect I found this :

helpful?

Later

 

[Tom.G]

Nice graph.

that's a bit better... something is missing though.

Sure is. I expected F2 to rise steeply as angle BAC approached 180^o.

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

Ahh! Think I found it. Get rid of the "90-" in the formula so it becomes:

F2 = F1 x (L_H / L_AB) x cos(B) x tan(A/2)
(that makes the argument to tan() cover the range of 45^o to 90^o, rather than 45^o to 0^o )

You should see F2 approach infinity as angle BAC approaches 180^o.

 

[john101]

OK, Now we're getting somewhere...

Given :

Formula is correct, both as given and as used in spreadsheet.

All prior reasoning is correct and understood and applied correctly.

then

at 179 degrees the force on a 72 square inch piston is around what corresponds to the looked for 20Mpa.

(I think I can see clearly the value of the rubber.)

I think from this it is possible by adjusting variables to finish a design and know what size piston, arm lengths et.c. will be best.

It now occurs to me that :

While the triangle is measured from the centers of the pivot points is that actually where these forces are? Is it rather the surfaces of the shafts and the distances between them that need to be considered.

edit add : and what about material for shafts etc . size, shear, wear, deformation, fatigue... ? I don't really have a good concept of the magnitude of forces involved. Examples ?

 

[JBA]

Well, we are all finally converging, after trying to calculate the press force due to handle load, I decided to take the reverse approach and calculate the handle load based upon a given required compression pressure (I have inserted arbitrary brick length and width values) as you will see in the attached Excel calculation workbook. I decided to attach the workbook rather than spend time trying to describe everything. In the end, once the relationship of the angle between the bottom link and the vertical in terms of the handle angle was found the rest turned out to very simple. A key to applying this to the design is to design the initial contact pressure point at the minimum handle angle from the horizontal that will still guarantee the required compression pressure at full travel. A big help in this respect would be to calculate the bottom plate travel vs handle angle as well; but, I haven't added that yet. (Due to time restraints). See the below graph for an assumed 6" x 12" brick face area with 20 MPa pressure loading for an initial view of the worksheet results.

 

[Tom.G]

While the triangle is measured from the centers of the pivot points is that actually where these forces are? Is it rather the surfaces of the shafts

For calculating leverage or mechanical advantage yes, that is the effective length of a lever.

edit add : and what about material for shafts etc . size, shear, wear, deformation, fatigue... ? I don't really have a good concept of the magnitude of forces involved. Examples ?

The supporting shaft must be sized for the total force. If I did the math correctly (no guarantee), the force on the bearing surfaces, tensile force on BC and compressive force on AB is 208,800lbf, or 104 Tons. WOW!

Time for the Mechanical Guru's to chip in here.

...calculate the handle load based upon a given required compression pressure...

Interesting information to have. Thanks.