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.

 

[john101]

Wow indeed. That is a lot...
____________

I use LibreOffice Calc. I've had trouble saving and reopening worksheets in any format in it,. Your excel file opened fine. It'll take me some time to digest it. It's been more than 30 years since I did this sort of maths. Bit slow in my dotage. I don't get a couple of things re forces. If I can't figure it it out I'll ask.

Meanwhile here's a triangle solver. I can only upload images atm. I've tried to standardise the naming of angles and lengths.

The main measurements I'm sticking with are AB 3" and AF1 60". Here, for example, I'm making a, or BC, 16.5 and get a stroke of 2.725 inches.

 

[john101]

At this point I'm not overly worried about the forces as I have some very heavy machinery parts to scavenge. It would be nice to have some input about shaft shear. Is it right to think that the 100 tons would be shared by the two rests that the shaft sits on ie. 50 tons each?

ATM I'm looking at a 1 1/4 inch polished round bar off a railcar wheel assembly and the 1 1/4 inch x 2 1/2 inch section swivel arm it is mounted on which is pre-drilled for the axle. Likewise I can cut one of those rectangular arms in half and get the half round rests to put on the lid.

This is a major upgrade from what I've been building so far but looks like it's worth it. (I haven't seen a version of this press this massive before. I have seen bent ones.)

Another thing to consider is the thickness of the walls of the box the piston slides up and down in.

Off hand I haven't thought it's much to worry about as I don't expect much lateral force on it from the compressed soil. There will be ample clearances for water to pass through.

 

[JBA]

At this point I'm not overly worried about the forces as I have some very heavy machinery parts to scavenge. It would be nice to have some input about shaft shear. Is it right to think that the 100 tons would be shared by the two rests that the shaft sits on ie. 50 tons each?

The force will be 50 tons each.

The principal loads will be carried by the linkage pivot shafts, the AB arm and the two BC links. Particular attention needs to be given to the tension stress on the BC link areas on each side of its holes and shearing stress between those holes and end of the link. At the highest load point, the short AB arm will be principally in compression so the bearing stress of the shafts on its holes will be the major concern for that arm. As you have seen, there can be very high bearing stresses between the shafts and their hole surfaces so they will be of the most concern as points of wear and possible galling if the shafts are not surface hardened; and, hardened inserts for the linkage holes would be also be a good idea for reducing hole wear and friction and extending the operating life of those critical areas.

Edit; Ref letters correction also error on excel calculation sheet: I just discovered I used the wrong ref cell for getting the pressure load.
I used the area x 20 MPa when I should have used the area x 2900 psi (dumb error) Sorry.
See corrected Excel program and graph on below Post.

 

[JBA]

Corrected Excel Program and Sample Graph.

 

[john101]

_____________

Here is a

"MSI-70 70 Ton Bar Shear( https://www.msi-tx.com/hydraulic-bar-shears )
The MSI-70 is a Semi-Automatic 70 ton shear for cutting round bars up to 1-1/4” diameter..."

so perhaps the 1 1/4 inch round I have will do, perhaps with a significant safetymargin.

I'm thinking using cotter pins to hold them in place so they can regularly be removed for inspection, and also to fit a grease nipple to apply Extreme Pressure grease. Given the bar is the axle for a traincar wheel, polished surface as well as the rectangular bars' hole it passes through is a snug fit, that might do it. I'll start on the bigger knuckle while continuing to built a relatively lightweight press with the one I've almost finished.

 

[JBA]

A polished axle bar should be a good choice, but, as a side note, I am having a bit of a problem understanding how a 1 1/4" diameter bar is an axle for a traincar. What type of train is this and what is the load capacity of the cars for that train? The railway car axles I have seen while I was involved in engineering one type of equipment for railway freight cars are considerably larger than 1 1/4" diameter.

 

[john101]

I should say one of those rail-trucks that service rails. They have a hydraulic arrangement front and back that raises and lowers a rig with the wheels that lifts the truck up and allows it to run on rails. Its axle is about 10 inches long and machined to various widths with a section at the back I can cut off that is 1 1/4 inches thick. I realized I should have made that clearer. Stupid of me. You're right of course. My apologies.

I don't know how heavy. Perhaps 10 tons. I'll ask next time one pulls into be serviced next door and post then.

I'm still trying to understand what the excel sheet tells about forces. While I can go from an object, real or visualised, to the maths (is careful steps) I have trouble going from the maths to the real world tangibles. Can you explain it in a bit more detail, please?

 

[JBA]

I need a little clearer description of what you want me to clarify.

One thing that you need for design that I have not provided to you so far is a graph of the increase of forces on the machine as the handle rotates; but, the minimum force on the all of the pivot pins and on their mountings to achieve your desired 20 MPa pressure on a 6" x 12" brick is equal to the Fc = 208,800 lbs shown upon the sheet.

Unfortunately, that 208,800 lb force is not the maximum force the that the linkage can apply when fully aligned, that force is only limited by yielding at the highest stress point(s) in the machine assembly and and this where the rubber pad becomes important, by including that pad (if sufficiently thick), as I think you already understand, it can act as that weakest yielding member rather than one of your machines parts.
The thing that makes a toggle assembly so dangerous to the machine is the fact that as the toggle begins to reach it s full alignment, the amount of force that the operator has to apply to the handle reduces to essentially zero (as shown on my graph) so the operator doesn't have any real sense that the machine load is increasing to the point that something in the machine has to yield.

Just as a note at this point, I don't know how much education or experience you have with stress analysis calculations but as things move along I ( and I believe other forum members) will be glad to give you some assistance in that process.

 

[john101]

Thank you, I think you've answered it. I just hadn't got it before. Now I realize that as I use the tripod clamp, say on a camera tripod, as it grips it suddenly snaps into position. I presume this is at the point where the grip is greatest but it doesn't take much to get it there when close. Now the graph begins to make sense to me.

I did view the rubber as a kind of failsafe that allows the toggle point to be reached without the rigidity of the mechanism being a hindrance, or become a problem.

I wonder if a stop with a rubber pad like the rubber stopper on a rear suspension on a car like this :

placed so that when the long lever approaches horizontal it comes up against that so to push the last few degrees, or fractions of degrees, there is a sense of resistance could be helpful. Another thought I have is a simple mechanical dial that gives degrees and/or resistance readout to the operator a bit like that on a mechanical scale.

re: "stress analysis calculations". Offhand I know nothing about this. I look forward to learning.edit add : also, one would not want to go past the toggle point. (?) If it flips past that it would be very difficult to get back if there is any elasticity in the whole thing. My meaning is that if it's all rigid including the brick and it has got past the toggle point there would be nothing left to compress so getting it back would be no problem. (?) Seems bit like a damned if you do damned if you don't kind of thing.

 

[john101]

I just found I can upload zip files so I compressed the LibreOffice Calc file. Then I downloaded and opened it. OK. Don't know if excel will, but there it is.

 

[JBA]

I have not yet tried to download your calculation but I want to respond to another item in your prior post. A critical thing to understand is that a mechanism constructed of metal is not actually a rigid because all metals have an elastic modulus that causes any component act essentially as a very stiff spring when subjected to load.

For all materials this is known as Young's Modulus E = stress (psi) / % strain (inches of extension per inch of the region being stressed) and for steel E = 29 x10^6.
As a result, whenever you apply a load, and particularly in your case a very high load that can result in a high stress in the components of a mechanism there will be always be a spring back effect when the load is reduced and/or released.

Now for a bit of a lesson in stress analysis:
To give you an example, I will use one of your lower toggle links. If you use a A36 alloy mild steel for the link, this material has a yield stress S yield = 32,000 psi.
(What this means is that the link is in its "elastic strain" region and will act like a spring up to the point that its tension stress reaches 35,000 psi. So now we can calculate how much one of your 19" center to center links will stretch if loaded to that maximum stress point.
The strain = stress / E = 32,000 / 29x10^6 = 0.0011 in / in of length so the link's stretch = 19" x .0011 in/in = 0.021 inches or 0.53 mm.
Now, in fact, you would not design the link to this high of a stress so if you use a safety factor of 2 in your design then the maximum stress would be only 16,000 psi and the amount of stretch would only be 0.0105 inches.

The whole point of this is that each of your assembly should , in addition to being designed for an acceptable maximum safe stress, also be analyzed for its amount of deflection at full load design load and all of those values summed for the complete assembly because that will determine how much the handle of your unit will "kick back" when the operator releases the load on the finished brick. Fortunately, since the finished brick can actually be considered "rigid" then the resulting total handle kick back will be small; and this is very critical for your unit with the toggle because as you can see by reading my graph backwards the force on the handle increases rapidly as the handle rises from the horizontal.

Because of that factor in a machine of your type using a toggle, if there were any real amount of elasticity in your compressed material then the unit would be so potentially dangerous to operate that I would have warned you not to proceed with its construction and refused to assist in this project.

And, as a bit of a caution, for the same reason it is important for any rubber sheet that you use as I suggested to help protect the machine to be relatively thin and hard.

Now that I have submerged you in all of that, I want to say that you are correct that the toggle be blocked from going over center because this would result in a downward force and the operator would have to pull up on the handle to release the load; whereupon crossing the toggle center force the force on the handle will suddenly reverse and fly upward under his combined pulling and the toggle's effect. additionally, I think your rubber pad at the end of the handle is also a good idea because combined with the toggle stop it will guarantee there is always a bit of upward force on the handle at the end of the stroke.

I am going to stop at this and let you review and think through all of the above.

 

[john101]

Thank you. I will need some time to think through all that.

As a note: Clayey soil swells and shrinks when wet or dry. ie. I wonder if squeezing the soil at high pressure 'dries' it. It would make sense to me if it does. so that might be a mitigating factor in reducing any 'snap back' of full toggle effect.

In searching for any information about that I came across this :

http://strawbale.pbworks.com/w/page/18605635/Compressed Earth Blocks

"When a block machine compresses a block, it reduces the volume by 30%. It does this by mechanically aligning the moist clay particles, removing the air pockets and sticking the clay to the sand. If too much water is in the mix, there will be more air space between the particles when the brick dries."

...ie. if one wants a 12 x 6 x 6 brick 432 cubic inches, one needs to start with 617 cu", or a space above the piston of 12 x 6 x 8.6 which means a stroke of 2.6 inches which fits nicely with the calculated stroke of 2.73 in post #37. Therefore c or AB needs to be not less than 3".

 

[john101]

While I don't know if there is anything meaningful in this I found it interesting that while a and c are fixed, A and B graphs a line, b and C graphs are curved. I'm sure it's all very logical but given the curves in things maybe there's some direct relationship between b and or C and the forces calculated. ?

Here I input (in the triangle solver) a as 8, 16.5 and 22 inches

edit add:

...ie. if one wants a 12 x 6 x 6 brick 432 cubic inches, one needs to start with 617 cu", or a space above the piston of 12 x 6 x 8.6 which means a stroke of 2.6 inches which fits nicely with the calculated stroke of 2.73 in post #37. Therefore c or AB needs to be not less than 3".

On the other hand when inputting different lengths of a I find that a must be more than about 12".

 

[JBA]

Sorry, but I am having a problem understanding what the above curves represent or how they apply; but on a much more important note, I am curious about where the 30% compression value comes from in your above reference, because the in the article by Wayne Nelson he states:

"The other test is to build a form 2' long and 1.5" wide x 1.5" tall that is then filled with moist soil. Wait a week, letting it dry in the shade. Once the soil sample is dry, you will notice a gap between the end of the clay body and the end of the box. This gap shows how much the clay shrinks. Acceptable shrinkage is less than two inches and preferably 1/4" or less. The less shrinkage the better the soil. If you have a lot of shrinkage, indicating a high clay soil (over 25% clay), you might add sand to make a better mix. A soil with a very expansive clay may be rejected after this test, but soils in close proximity or lower in the ground may have less expansive clay. When you find a good soil, it is important to keep testing the soil even after you start to make bricks to insure quality."

So, without knowing the percent of clay in the 30% reduction statement only, a test of the actual soil you intend to use can determine the amount of compression travel you will require; and this critical factor in the design of your unit because of its low force multiplication during the majority of its early handle travel arc. For example, by adding more degree increments at the right end of my calculation it shows that with the handle only 1" above the end of its travel the operator force required to attain your desired 20 KPa brick pressure will still be 182 lbs.

The displacement of the water and air is more a factor of the time of force application to allow them to percolate out of the soil, so the amount of force required for 10+% of the the compression travel depends principally on how quickly the operator wants to squeeze out the water and air based upon the porosity of the soil; and, there is no reason to expect any significant spring back from this part of the pressing process. I would think the amount of handle spring back travel should be very low and limited to some residual pressed clay expansion combined with the total strain in all of the parts of your machine.

As to your question about the pressing "drying the clay" it is doing that in the since that the removal of water from a substance is the description of drying. At the same time there will most likely be some residual moisture in the mix that will just be eliminated by evaporation and a hot dry climate can accelerate that portion of the process.

 

[JBA]

Have you seen the below video of a Chinese brick making machine. It has no compression loading advantages over your planned machine because it uses essentially another configuration of the same linkage you are using. I am just sending it as a general interest.item.