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

[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.

 

[john101]

The curves : b is the distance between points A and C, C is the angle ACB. The curves are plots of how they change in moving the long handle from 90 degrees to horizontal. If it is useful or not I don't know. I find it interesting that while the other values change linearly these don't.

That paragraph you quoted refers to shrinkage due to drying. The one I quoted, that mentions 30 %, I'm pretty certain refers to the compression of a prepared mix of soil (that's loose) in the machine. I'm assuming he got that figure from experience. There needs to be a consistency of preparation of the mix and some experimentation.

This press is a variant of the Cinva-Ram invented by a Colombian (not Indian as I previousle said) inventor in the 50's. As I played around with the numbers it struck me that the dimensions of the press as I've found on the net are pretty optimal and great variations from it is not really of value. I found the 30% calculation interesting re stroke, again seemingly confirming he got it right back then.

That Chinese variant is pretty interesting in its 'auto loader' and the way it flips the lid back before elevating the brick. I'll use those ideas. Thank you.

As can be seen in the video the production process can be quick. I think I'll be taking it more slow.

_______

Another thing that I mentioned in early post is I'll be making stabilised compressed blocks by mixing in lime. Roughly 80 % soil, 12 % lime, with soil and water the remaining 8 % depending on experimentation.

The blocks need to cure slowly starting with keeping them moist for a few days then slowly drying over the next three weeks. ( plain soil without stabiliser produces very good blocks but they need to be maintained as over time they can erode, particularly if exposed to rain. )

I've just finished digging a 60 foot diameter bowl 3 feet deep in the middle. In this I plan to set up the press and a curing area. About a foot and a half down the soil becomes damp and consistent. Some say use damp soil but as I'll be stabilising it I'll have to experiment adding water.

 

[john101]

a screencap from the posted vid appears to confirm about 30% compression.

Comparing loaded soil height to compressed brick :

 

[john101]

here's a good one...

 

[JBA]

Based upon the above, I think you are going to need a longer two man handle; or, maybe a T handle would actually be a good idea.

 

[JBA]

Here is a video that shows the potential operating method that I am concerned you may run into with the linkages low force multiplication for the majority of your compressing handle stroke.

 

[john101]

Great. Nice to see the process through.

I'm thinking about extending the length of the lever.

I'm getting close to finishing the 'knuckle'. Before welding it all up I'll post a pic with measurements and description with the hope of getting comments.

 

[john101]

Not welded yet:

AB = 3", for scaling: caliper set to 1", hammer 3lb, glove covers ugly grind job, the long lever is inserted in the red rectangular box, the short angle iron represents crosspiece.

turns out the axle 'guides' have hardened inserts. I cut one guide in half and use it as rests. Everything below that is just a 'sketch' for now.

 

[JBA]

I'm a bit concerned about the combined smaller apparent 1 1/4" diameter shaft and narrower side plates for the B shaft because that pivot has to carry the same 200,000 lbs maximum load as the handle's A shaft with a larger shaft diameter and wider hardened half bushings.

 

[john101]

Ok, the reason for the larger shaft and plate is that I have a stack of these 'railcar' hydraulic machines but only one of the wider pair found, unfortunate, I understand the weak link here is the 1 1/4" shaft and 1" plate.

How concerned are you? (afk for 8hrs)

 

[JBA]

One concern is that the shearing load the shaft is 210,000 lbs and that does not even include any safety factor.
As an comparison, the allowable double shear load rating of a high strength 1 1/4" SAE Grade 8 A490 steel bolt is 221,000 lbs; so, to be safe your shaft material should have a tensile strength equal to or greater than that of quenched and tempered A490 Alloy steel.
Since you are using an axle shaft, I would expect it to be a heat treated alloy bar, so if you are unable to determine the material of the axle shaft, you may just have to take the "try it and see" route. The risk is that, because it is an axle shaft it may hardened to the point of being a brittle material that will suddenly break rather than just deform as a more malleable quenched and tempered alloy would do.

Apart from the above, with respect to the small shaft and holes, due to their small contact areas, I am mainly concerned about possible lubricant displacement and galling in the segment of the shaft and hole that will be repeatedly exposed to a very high load in the region of the maximum load at the end of each pressing stroke. Unfortunately, I cannot find any reliable reference as to what an acceptable contact pressure should be.

Note: Edited for more clarity