40Ton Hydraulic Press Brake

[farmersammm]

So, I guess you figured your force applied to the right side "slave" by figuring out how much force would be needed to do the work the same as a wheelbarrow. Then you figured out how far the left side "slave" pivot had to be from the fulcrum to do the same amount of work you're applying to the right side "slave".

[farmersammm]

And, yeah..........you're reducing the mechanical advantage of your lever. It's "lifting" the wheelbarrow load, and still has to "lift" the load on the left side of the press. But I guess your software, or whatever, has this all figured out.

[BrooklynBravest]

I was just thinking(with no sleep) about the lever.

Attachment 1735203 The work is done by the lever on the left side of your press. Yes, it's inefficient, but it's the nature of this kind of fulcrum with the load in the middle, not at the side opposite the fulcrum(although at least half the load is). You're limited by the amount of work it can do when exerting force on the link. That link is the actual place to look at effective force, because it applies force to the right "slave". The force on the left "slave" is reduced by the force bled off by the right "slave". This I guess is where the equalization occurs.

Your bottom pivot on the main arm bears all of the applied force. It's carrying the load of both slaves. As a fulcrum would.

I think I got it right this time. Takes me a while to figure this stuff out.

Right, but the concern is where do we calculate mechanical advantage from. From where you labelled force, or load (the linkage attachment point)

[BrooklynBravest]

Bit better drawing

The level here is is the 14" span to the 6" span. Which would be 2.33:1

But is that truly the advantage in the system, when the linkage is at 7.5". Which 7.5" to 6" is barely an advantage at all.

I'm inclined to believe what JDM said, especially because it takes 10" of movement in the cylinder to produce 4" of movement in the press. Which is a bit more than 2:1.

[farmersammm]

The first thing I'd figure, is how long the lever at the circled point has to be in order to apply 20 tons(half your target force) down force to the die carriage. Then I'd work back from there towards the cylinder.

[farmersammm]

For simplicity, let's forget about the link that goes to the die carriage (it's inefficient to a degree). You're looking at a simple lever that has to take, let's say 10,000 pounds of force, and transmit 20,000 of force to the carriage. I know this isn't your target figure, but it's simple.

[BrooklynBravest]

The first thing I'd figure, is how long the lever at the circled point has to be in order to apply 20 tons(half your target force) down force to the die carriage. Then I'd work back from there towards the cylinder.

I haven't a clue. But that lever is greater than 1:1

[farmersammm]

Now we have to figure out how to get that required force to the previous lever. This is akin to a wheelbarrow type lever system. We need 10,000 of force transmitted thru the adjustable link to the first lever we looked at.

[farmersammm]

If you stay at 1:1 on the slave fulcrum, just substitute the required amount of force necessary to apply the desired force (20 tons).

[farmersammm]

Then we have to figure out how to get the required force from the cylinder to the linkage

My arrow on the linkage is pointing in the wrong direction........sorry. It should point to the left

[BrooklynBravest]

Then we have to figure out how to get the required force from the cylinder to the linkage

My arrow on the linkage is pointing in the wrong direction........sorry. It should point to the left

well that linkage is seeing 20t from the ram. And its pushing on greater than a 1:1 so...

[farmersammm]

Now we can find out how much of a cylinder we need. You have the "wheelbarrow" load + the simple lever load. Start with the "wheelbarrow" load, then figure out how long the leg has to be to the left "slave" in order to apply the residual force left over from the first load(adjustable link).

[farmersammm]

I think a simple hydraulic system would be easier, unless you're either good at math, or have the correct software. (I'd have to have the software)

The only real hydraulic synchronizing issues would be present in "free" bends. A full bend to the depth of the bottom die simply requires the cylinders to bottom out, which they will if all the lines are the same length. It's no different than a front end loader I'd imagine. I suppose one side could lag the other, but I'd bet it wouldn't be that much of an issue.

[farmersammm]

Think of a wheelbarrow to determine how much distance should be from the linkage point to either end of the lever.

[farmersammm]

My pic editing software sometimes gets the better of me I moved the text to the proper place on the lower portion

[whtbaron]

I think you're letting the tie rod linkage complicate things. All it does is split the load from one end of the brake to the other. The downforce at the #5 pins will be equal because they are tied together. Do all your calculations on the tall fulcrum and that will give you the total force down on the brake, which is what you need to know. The extra lever at the end of your tie rod doesn't multiply your force, it splits it.

[farmersammm]

I probably go at this different from other folks. It's just the way my mind works.

Either way, it's just a system of relatively simple levers. Engineers make the big bucks because they know how to actually generate the numbers with a formula.

I suppose you could break it down to simple straight lines, and just use measurements of length in proportion to the force you need go transfer/generate. Hell, it's just all a bunch of proportions anyways.

I promise to lay off now. I hope I haven't set you on the wrong path, or confused the crap out of you.

[farmersammm]

i think you're letting the tie rod linkage complicate things. All it does is split the load from one end of the brake to the other. The downforce at the #5 pins will be equal because they are tied together. Do all your calculations on the tall fulcrum and that will give you the total force down on the brake, which is what you need to know. The extra lever at the end of your tie rod doesn't multiply your force, it splits it.

what pin numbers?????????????????????????? I didn't see any pin numbers

[whtbaron]

The 2 pins on your #5 distance...

[farmersammm]

The 2 pins on your #5 distance...

Oh absolutely. I never considered those. They just add some inefficiency because of their angles to the lever.

[farmersammm]

The left side lever is a class 2 lever with a twist. You have to generate 20 tons above the fulcrum point, and generate 20 tons below the fulcrum point.

[BrooklynBravest]

I think you're letting the tie rod linkage complicate things. All it does is split the load from one end of the brake to the other. The downforce at the #5 pins will be equal because they are tied together. Do all your calculations on the tall fulcrum and that will give you the total force down on the brake, which is what you need to know. The extra lever at the end of your tie rod doesn't multiply your force, it splits it.

I am strictly just trying to understand why it works for educational purposes.

The way I am viewing it, The main level is producing a 2.3:1 of 40T. But the second lever attached to it is stealing half of that, and both are pressing the same blade down.

I didnt pick the height of the linkage for any specific reason though. I am mostly concerned if it is in a good place. The 18" link.

[farmersammm]

I am strictly just trying to understand why it works for educational purposes.

The way I am viewing it, The main level is producing a 2.3:1 of 40T. But the second lever attached to it is stealing half of that, and both are pressing the same blade down.

I didnt pick the height of the linkage for any specific reason though. I am mostly concerned if it is in a good place. The 18" link.

So............you know that 40 tons of force is being generated at point B given the distance between fulcrum, and both points A and B

This is your simple class 1 lever in terms of calculating the force generated.

[farmersammm]

And yes..........the adjustable linkage, and the other lever, are stealing force from your main lever. It's as good a way as any other to think about it if it makes sense.

[whtbaron]

The way I am viewing it, The main level is producing a 2.3:1 of 40T. But the second lever attached to it is stealing half of that, and both are pressing the same blade down.

Getting back to Sam's analogy with the front end loader, adding more arms to the loader doesn't increase it's lift capacity. Lift capacity is determined by the a) output of the cylinders and b) leverage of the arm length. Adding more arms wouldn't add more leverage, it just adds more attachment points.

In your case, the "loader" only has one arm and it's pushing down. TOTAL downward force will be determined by a) your cylinders output and b) the leverage it gains on that right cam or lever. In reality, depending on the length of your brake, you could add one , two or three extra cams to even out the downward forces on the brake, but the total downward force won't increase, it will merely be divided equally along the top of the brake by the number of attachment points (less a small amount for friction, yada,yada). Determining the split is only complicating your math, it should be split evenly at measurement points B. Nothing gets "stolen", only split. Total downward force remains the same if you calculate it for the right cam.