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B How/Why Is So Much Force Generated In A Bender?

  1. Sep 9, 2018 #1
    I am talking about a sheet metal bender. Sometimes called a 'brake'.

    Bending metal of 1mm thickness or less.

    For those who don't know that's merely a long horizontal hinged metal bed upon which another sheet of metal is laid and clamped down just short of the hinge.

    Then closing the hinge forces the metal to bend at the clamp.

    If I take a piece of such metal say 1" wide, 6" long (and 1mm thick) and put one end in a vice leaving say 4" ( 100mm for convenience in measuring) sticking out I can attach a spring scale to the free end and pull on it while watching the scale.

    I note that the metal bends at 2kg of force applied.

    Continue that same 2kg and I can bend the metal down 90 degrees.

    So I am thinking that if I had a piece of that same metal 4' long then the force required to bend it would be 48 x 2kg = 96kg.

    And this is not an extraordinary force. And at the clamp there's no more than 2kg applied at each inch.

    It all sounds very simple and ordinary.

    But I am told that constructing such a device would require very strong construction indeed and that the force at the clamp would easily distort the clamp and cause a poor bend.

    That trying to bend heavier metal - say 2mm - would be sufficient to break the device, even the great heavy ones such as that found in my local hardware store for instance.

    Hinges snap off. The clamp breaks.

    So my understanding of the forces is obviously totally wrong.

    How are they to be understood?
     
  2. jcsd
  3. Sep 9, 2018 #2

    Simon Bridge

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    The key is in material properties ... like rigidity and strength.
    You just need a better model.

    Note: in a bender you are pretty much applying forces at a particular place. The action has to be transmitted to where you want. You are basically trying to make a lever that will minimise the effort needed to bend a bit of metal, out of metal, that you do not want to bend.
     
  4. Sep 9, 2018 #3

    sophiecentaur

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    Whatever Force is being applied to the actual workpiece will also be applied to the tool (Newton's Third Law), locally where it is pressing on the workpiece. The tool has to be much stronger than the workpiece or it will also distort. The tool has to be made of a hard material and the 'business end' needs to be as thick as possible, consistent with fitting into the space. That allows it to bend you metal sheet without bending itself.
     
  5. Sep 9, 2018 #4
    Well, I don’t think your understanding is too bad, but there are some other factors to consider.

    I think the primary thing you are leaving out is the lever arm. The brake is used to put a very sharp radius of curvature in the bend, and more importantly to NOT bend the sheet anywhere else. When you pull down on the end of a 4” strip protruding from a clamp you don’t tend to get a sharp crease with flat wings. Perhaps the metal yields more near the vice, so it sort of creases, but it also yields further out resulting in something more curved and messy. To insure a sharp crease, a brake applies the force very close to the ledge against which the metal is being bent. (Let’s call that the fulcrum). Take your same strip and instead of hanging the weight at 4” from the vice, try 1”. Now it takes 4X more force to bend the sheet. Now try 1/2”: 8X or 1/4”: 16X. The tighter you want the bend, the closer you have to apply the force to avoid bending any other part of the sheet. Now try to bend the strip the same way the brake does. Place a thin flat tool directly up against the vice and see if you can push down hard enough to make a really nice sharp crease. Much harder, right?
     
  6. Sep 9, 2018 #5
    There's the hub of my misunderstanding. I was naively thinking that the extra force that you indicate is required gets provided by the leverage - by the fact that I'm a applying a force out at 100mm instead of 1mm.

    I'm still not clear. Far from it. But I think you're saying that there's a distinct 'local' force required in order to produce the sharp bend.

    The leverage I'm thinking of will not produce that sharp bend. No way.

    The leverage I'm thinking of is all wrong in my head because of the lack of stiffness, right? IF the metal were rigid, totally stiff, then the force would be directed right down to the clamping point - and if it were that rigid then it would snap rather than bend?

    So how would I calculate what forces would be required and therefore the strengths of material required there at the bending point?

    I'd guess it'd have to do with some highly technical material properties with names like modulus of elasticity and that kind of thing? And it would have its roots in the intermolecular forces of the material in question ( mild steel ) ?
     
  7. Sep 9, 2018 #6
    That is what I am saying.
     
  8. Sep 9, 2018 #7
    Have you ever tried to bend a bar or piece of pipe over your knee? If you have a short piece you can’t bend it. If you have a long piece you can. Why? Leverage. The distance from the fulcrum multiplies the force giving a larger bending moment.
     
  9. Sep 9, 2018 #8

    Stephen Tashi

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    It isn't clear what you mean. Two distinct interpretations:

    1) The heavy duty machines at your hardware store cannot bend a sheet of metal 4 ft wide and 2mm thick.

    versus

    2) The way you are calculating the required force to bend a 4 foot sheet of metal 2mm thick, the heavy duty machines should not be able to bend a 2mm thick sheet of metal 4 ft long, but the machines specifications say that they actually can.

    By the way, what kind of metal are you talking about?
     
  10. Sep 9, 2018 #9
    Yes I know. I tried to indicate I've got the message about this. I always knew in fact. We all do, don't we. Though we may fail to take that knowledge into account sometimes.

    I have tried to say here that my way of 'taking it into account' was thinking that the sheet metal itself provided the lever - i.e. I apply the force at 100mm and there's my leverage: 100mm, multiplying my force until it reaches the required amount there at the fulcrum point, the jaws of the clamped piece.

    I said this:


    mild steel sheet metal at say 1mm for the sake of calculations. Given a 4' sheet. So we want a tight 90° bend 4' long on such.

    How to calculate the forces required to be exerted at some point on the bending platform of the 'hinge', the bender ?

    How to calculate the forces required at the bending point to bend the steel to 90°?
     
  11. Sep 9, 2018 #10
    You first calculate the bending moment required. One of the first hits using search term mechanics of sheet metal forming is a book titled Mechanics of Sheet Metal Forming, by Marciniak, Duncan, and Hu. Chapter six tells you how to calculate the bending moment. From that, the force is calculated. Spend some time reading that book, and you will find out why it's so difficult to tell somebody how to do it in an internet forum. BTW, the material in this book is normally taught as a senior or graduate level course in mechanical engineering.

    On the other hand, you can do like I did. I made a sheet metal brake from some angle iron, C-clamps, and hinges. It did a very nice job of bending the thin sheet metal for the fan coil unit (photo below), but bent the brake when I formed the 16 gauge mounting brackets.
    P9090013.jpg
     
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