I am building a grape press for harvesting the juice. I figure with only 10# pressure on my crank I am generating about 3,000# force. Will that force be distributed between the top and bottom evenly or will I have 3000# force on both ends. I am worried about my upper beam not standing up to the load. I used a 2X4 and an rethinking it now.
Something like this? PRESS
If so, though i do not know how to figure force distribution exactly, i am pretty sure the force on the bottom will be spread out if it is like the picture, but on the top will be concentrated at 3 points.
Where the screw crank is, and at the fixing bolts on the ends of the cross beam.
Seems like 4x4 would be a better way to go?
Very similar to that, I have finer threads so more leverage. I have 2 each 2X4 's across the top side by side. My question really had more to do with top and bottom total force and not so much the points where it would be applied. I know when I test the breaking strength of string each loop has 1/2 the total tension when pulling. I am thinking it would be the same when pushing against two points but not positive.
Your descriptions are very confusing. At least to me.
How do the two “2x4s across the top” share the load? How are they arranged?
As a *general *rule forces will apportion themselves symmetrically around the geometry. But, and it’s a biggie, …
Depending on tolerances and such you may end up with the vast majority of the force going down one path at first until the whole device bends into a force-symmetric shape. Thereafter the forces pass symmetrically, but meanwhile you’ve applied a massive bending force to the device. Which may break it promptly or start pulling out fasteners, etc.
Until we can see a drawing or a picture we’re guessing about what you’re describing.
Think of 2 upright 2X4’s about 40" long, the narrow side is facing you, these will be feeling tension stresses. Now at the top and bottom of these two upright I have screwed in a cross 2X4 joining them at about 16" apart, I have one going across the face and once going across the back at the very top and bottom of the uprights, So a total of 4 cross 2X4s. Think of a jack in the middle of this frame pushing the top and bottom apart.
The more I think about it I am pretty sure both top and bottom would feel the full force of the load and even though they are equal one would certainly fail before the other.
Your 3000 lb spreads out over the cross sectional area “A”, being the area of the moving plate of the press.
So P , pressure = 3000/A , (units of lb per square inch ).
The grapes being mostly fluid, the pressure is then applies to the top plate, the sides and the bottom, equally , everywhere. Except gravity is adding a little pressure with depth, but thats a tiny amount of force , if its only a foot deep of grapes and grape juice, so lets ignore that.
So… you can calculate the force on each piece of wall and each piece of floor by pressure * area too… Its the same pressure produced by the moving plate…
Since you’re making a press it’s not quite as you describe with a jack between the top and bottom cross beams. In a press the jackscrew will have be imparting force through the threads in the upper cross beam while imparting force to the bottom cross beam through a flat plate spread across some portion of that beam, but it’s not fixed to the lower beam so the beam can distort more, but in a press that plate will actually be quite large. Which end has the greatest contact area depends on the size of the threads but in a press the bottom plate probably has much more surface area and the bottom of the press is not just a crossbeam but a large platform to hold the bucket. If you look at most presses like the link in post #2 the cross beam at the top will use some form of arch to strengthen it and probably be may of metal instead of wood while the much larger bottom has a lot area that the force is spread over.
I figured it like this. I have a 12" bar cranking a 8 threads per inch all thread. So I have about a 300 to 1 advantage. So 10# force on the bar would give me 3000# distributed over a 95 sq in plate for about 31# per sq in pressure on the grapes. I didn't calculate out the forces on the frame until after I built it. Not a big deal if I have to rebuild it. I was just a little surprised at how much force would be generated with such little effort. I may end of using a courser thread on the screw press.
I tried to look up ideal pressures for crushing grapes but couldn’t find anything. I did find that too much pressing reduces the quality of the juice and flavor. Something I had not considered.
The full force isn’t applied to the cross beams until the grapes are well crushed so you may never reach that maximum if you avoid overcrushing. Just to cut down on the number of turns you need an Acme screw may be what you want. A little more costly for the screw and nut though.
Actually my example was my wine press because that is what made me think of the question but the question really has nothing to do with the wine press. It is a load question. I will rephrase it.
Think of a metal square welded together that has zero chance of failing or even bending. I put a jack inside that square and jack it up to full pressure, say 1 ton. If I were to put a scale on top or below that jack it would read 1 ton. But is the force 1/2 ton on the top and 1/2 ton on the bottom or 1 ton in both places?
I have no idea but it’s an interesting question. If the structure is suspended in the air, and both ends of the jack contact the cross beams with the same surface area, and contacting in the same manner like flat plates fixed or free, ignoring gravity, the pressure on both ends should be equal. In reality the bottom is usually sitting on or fixed to the ground so there will be some inequality.