Physics Help Forum How To Calculate Forces Around This HInge?

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 Sep 14th 2018, 02:13 AM #1 Junior Member   Join Date: Jul 2016 Posts: 6 How To Calculate Forces Around This HInge? I am not a young student. I'm an old man with rotten physics and maths. My plan is to build a sheet metal folder. This kind of thing: https://www.kincrome.com.au/metal-bender-730mm Just like a hinge you lay the sheet metal on, across the hinge bit, and then close the hinge up which forces the metal to bend. I need to know how strong to make the various parts. There's not many, a top clamp and the bending leaf and the hinges at each side is about it. Those are the places that will fail if I get it wrong. So I've learned that mild steel has a Yield Tension strength of about 250MPa So I guess that force must be applied by the bending leaf to the job at the hinging point. And one should be able to work out what force must be used to lift the folding leaf given its size and what the forces will be on the hinges and therefore how strong they must be and what force on the clamping bar... BUT: I don't know how to do it. To show I'm trying I'll set out what I've figured so far, such as it is: mild steel sheet metal Yield Tension Strength 250MPa. Use 1mm thick sheet metal for convenience (pretty real, too). Use 1 metre wide sheet of metal. Use flap of metal on the folding side of 1m long. So we've got a moving flap of metal 1m x 1m. So the cross sectional area of the bend is 1mm x 1m = 0.001mē So 256MPa = 256M Newton/mē So I need 0.001 x 256M Newtons = 256K Newtons. I think that's the force that will need to be exerted at the bend and that force will bear back on the hinges of which there's only two. So the hinges will bear 128K Newtons each. Now 1 kg weight = 10 Newtons. So this is like 12.8 K kilograms each. That's fantastic. That's like 12.8 tonnes on each hinge. So that's me stymied. I reckon that's got to be wrong but I don't know how to make it right. And when I consider the whole hinge thing I can't understand it as a lever. Where is the fulcrum? At the bend it must be? Or where? And so on. I would really like some help. There's plenty of help, tables and formula on the web for press brakes - that is bending machines that bend sheet metal by hitting it with a die. But I can't find anything about this. Yet this seems the most basic and primitive. A real starting point.
 Sep 16th 2018, 09:45 AM #2 Physics Team     Join Date: Jun 2010 Location: Morristown, NJ USA Posts: 2,324 Where you're going wrong is not using proper calculations of stress in the metal sheet that's being worked, and the force that must be applied to cause enough stress to cause the sheet to deform plastically. You can model the sheet in its initial flat condition as being supported by the two supports on either side of the V gap. Then you apply a load between those two supports. The stress induced in the sheet can be modeled as stress = Mc/I, where M is the moment applied by the brake (the part that moves down when you operate the lever), c is the the half height of the sheet, and I is the moment of inertia of the sheet. If F is the force applied to the sheet by your mechanism, for this case you have: M = F x d/2, where d = distance between supports (i.e at the top of the "V") c = 1/2 x 1 mm = 0.5 mm I = bh^3/12, where b = width of the sheet (1 m), h = thickness (1 mm). Be sure to convert all dimensions to common units - I suggest consistently use meters. Set Mc/I = yield stress of the aluminum, and solve for F. Hope this helps! topsquark likes this.
 Sep 16th 2018, 03:27 PM #3 Junior Member   Join Date: Jul 2016 Posts: 6 I posted a reply and it didn't turn up. Essentially I said this is not a press brake problem but a folding machine problem, they're different. So I made a drawing. https://imgur.com/a/XljBilZ out of time now. hope this helps. regards... p.s. just added another drawing - the first one you'll see, the tiny one. It's what I think is the essence of the thing. Just a simple question about bending around a corner. What force must be applied at C? I'm late...... Last edited by abrogard; Sep 16th 2018 at 03:44 PM.
 Sep 16th 2018, 06:12 PM #4 Physics Team     Join Date: Jun 2010 Location: Morristown, NJ USA Posts: 2,324 The drawing helps. You ask what force must be applied at C - well, if that's the only point of contact between the bending arm and the steel sheet then you could use: yield stress = Mc/I = Fdc/I where 'F' is the force and 'd' is the distance from where the you want the sheet to bend (i.e. at the clamp edge) to point C. But if point C is the only point of contact you won't get a nice sharp 90 degree bend like you've shown it, so there must be contact between the full length of the swing arm and the sheet, not just at C. This makes it a more difficult calculation. You can estimate the worst case by changing the value of 'd' to be the distance from the bend to point A (i.e. the closest point of contact of the swing arm to the bend). This will make for a much higher force having to be applied.
 Sep 16th 2018, 07:26 PM #5 Senior Member   Join Date: Apr 2017 Posts: 428 This is not the sort of problem that is best cracked by calculations ... From your drawing the bed and bending arm are 16mm thick , more than adequate . It follows the pin for the hinge should be about between 4 and 8 mm dia. The 'teeth' on the hinge are not critical , perhaps anything between 10 to 100 mm long whichever is easy to machine .... Your design looks way over engineered ( plenty strong enough).
Sep 16th 2018, 08:54 PM   #6
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 Originally Posted by ChipB The drawing helps. You ask what force must be applied at C - well, if that's the only point of contact between the bending arm and the steel sheet then you could use: yield stress = Mc/I = Fdc/I where 'F' is the force and 'd' is the distance from where the you want the sheet to bend (i.e. at the clamp edge) to point C. But if point C is the only point of contact you won't get a nice sharp 90 degree bend like you've shown it, so there must be contact between the full length of the swing arm and the sheet, not just at C. This makes it a more difficult calculation. You can estimate the worst case by changing the value of 'd' to be the distance from the bend to point A (i.e. the closest point of contact of the swing arm to the bend). This will make for a much higher force having to be applied.
Thankyou for the input.

I don't understand your yield stress equation I'm afraid. Sorry. What is 'I' ?

And I'm thinking I don't need to calculate any Yield Stress figure. The tabled Yield Tension for Mild Steel is what I've in mind when I say it. Variously quoted such as here 370MPa https://www.azom.com/article.aspx?ArticleID=6115

I don't much care the actual true value. I'm trying to ascertain if that's a value that I should use in this calculation. I'm even hypothesizing that's the actual answer to my question; it's not just a value I should use but it the answer, the Force required.

My errors again are misleading. I only put the 'point C' in order to try simplify, get my question clearer.

I was thinking that forces applied to the bending arm, picking it up, operating the bender, would be effective at that point and therefore only that point need be considered.

Like they do all that kinda reductionist clever stuff in maths and physics.

But I'm wrong again. Due to my poor understanding. Sorry.

Yes, the whole arm presses against the job. So if it makes a difference then okay it must be taken into consideration.

And 16mm thick bending arm etc,... Again just my off the cuff attempts to put in values that make my question clearer.

The bender I propose to make currently would have material thickness of bed and bending arm of only 10mm, that's all I've got.

The question is intended to ascertain what sort of stuff I'm going to need - will my 10mm stuff be good enough to build a bender that will bend 1mm sheet metal, a 1 metre wide bend? That's the guts of it. What I'm on about.

 Sep 17th 2018, 01:58 PM #7 Junior Member   Join Date: Jul 2016 Posts: 6 I posted a reply but it hasn't appeared. I more or less said: . I don't know what 'I' is in your equation. Nor 'M', come to that. You could say I can't understand equation essentially, sorry. . The force will be applied over the whole bending arm length, not only at C. I'm not really asking what force will be applied at C. My question is what is the max force going to be experienced by the parts of the construction. It is a practical question. In order to facilitate building the thing. I simply guess, in my ignorance, that C is where the max force will operate. . The dimensions shown are similarly only for illustration. There is no 16mm bending arm, no 8mm hinge. The actual material I have is more like 3" x 1/4" angle. Knowing the max force should enable me with a bit of luck to calculate the dimensions required to handle it. And I think I added: I'm guessing the Tensile Yield Strength of Mild Steel will turn out to be the max force (or just a little more than that). I already surmised that and attempted to translate that figure - say 250MPa - into a force required to lift the bending arm and a force experienced at the hinges. I think I failed. I would much like help in doing that little calculation, too. That questions goes like this, I guess: If the force required to make a 90 degree bend 1m long (with a inside radius say 1mm because I know no better) is 250MPa then what Force in Newtons will be experienced on the bending bar, the clamp, the hinges and what force will be required for the operator to lift the bending arm?
 Sep 20th 2018, 07:05 AM #8 Junior Member   Join Date: Sep 2016 Location: London Posts: 21 Physics is a good subject and used in the new invention. Physics teachers learn a lot from this forum.
 Sep 20th 2018, 03:10 PM #9 Junior Member   Join Date: Jul 2016 Posts: 6 The system seems to be preventing me pointing out the question is not answered yet. I've tried posting twice and off they go 'for moderation' and never turn up - nor does a moderator's advice that they won't be. I'll try state it again and much more concisely than I originally did: How to calculate the forces experienced within a sheet metal bender bending a 1mm thick sheet of mild steel and making a 1 metre long bend? Well that's good: this posted immediately. p.s. And I had a sort of second question about my maths. Is my figuring correct? If the answer for force on the bed was simply the Tensile Yield Strength for mild steel and we take that as, say, 250MPa Then: force in the machine = 250MPa x area of bend = 250 x 0.001 = 250KP =250K Newtons =25K kilograms. = 25 tonnes = 12.5 tonnes on each hinge. right? p.p.s. Now my earlier posts have turned up. So perhaps I seem to be stating the same thing over and over... sorry... just a little glitch. Where we're at I think is all right, interested parties could start right here. Last edited by abrogard; Sep 20th 2018 at 04:32 PM.
Sep 20th 2018, 04:24 PM   #10

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 Originally Posted by abrogard The system seems to be preventing me pointing out the question is not answered yet. I've tried posting twice and off they go 'for moderation' and never turn up - nor does a moderator's advice that they won't be.
Some members have a worse problem than others. My account can't do anything with the spam filter. I have forwarded the problem to mash but I haven't heard from him yet. (That was a while ago.) Until then if a post doesn't show up let me know and I'll approve it. That's the best I can do.

-Dan
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