What Force Is Required To Bend Mild Steel?

[abrogard]

In a Sheet Metal Bender or Folder (different times/places get called different names) what maximum force is experienced and where?

Not a 'press brake' that uses a die punching into the metal.

But an old fashioned manually operated bender that is like a hinge laying flat.  The job gets laid on top.  Job gets clamped just before the intended bend line and then the 'hinge' is closed and this forces the job to bend.

We are talking bending to 90 degrees.

And generally making as tight a bend as possible, smallest internal radius.

The purpose of the query is to enable choosing of materials to construct a viable machine.

Proposed material:  mild steel, 1mm thick,  1metre wide.    If it makes any difference we could say to make a bend 100mm (4") from the end.

The internal radius of the bend could be, for the sake of the exercise, say,  2mm.  

I have a couple of rough drawings on the web attempting to show clearly what I mean.  There's a point 'C' marked which indicates where I expect the greatest force to have to be applied for it is the area of the actual small radius bend.

But I don't know.

I'd love to know.

Here's the drawings:

https://imgur.com/a/XljBilZ

 

 

 

 

 

[Enthalpy]

Hi abrogard, there is no simple answer.

It depends on how your material hardens during the deformation. Usual stainless steel needs more force than mild steel. Knowing the yield strength prior to deformation does not suffice here.

A first approach would compute a bending moment from the 'deformed!) yield stress and the thickness, then compare with a lever length deduced from the bending radius, and compute a force, not forgetting the symmetries and factor-of-two. Alas, this simple model is known to fail, by too much for practical purposes.

So you have to use empirical knowledge.

For 1m width it takes a big force, usually by hydraulic cylinders. This force suffices to deform the bending tools (matrix etc) and make the unusable when the item to be bent is narrow, so your machine needs a force limiter AND a smart operator.

[Phi for All]

The sheet will stretch, so you need to calculate the bend allowance.

[abrogard]

8 minutes ago, Phi for All said:

The sheet will stretch, so you need to calculate the bend allowance

I don't understand.  Thank you for the link, a good one.  I'll download the table and keep the link.

But I don't see what I care about the material stretching.  I just want to know what force I must apply to bend and what force that will generate in the machine and therefore how strongly it must be constructed.

26 minutes ago, Enthalpy said:

For 1m width it takes a big force, usually by hydraulic cylinders. This force suffices to deform the bending tools (matrix etc) and make the unusable when the item to be bent is narrow, so your machine needs a force limiter AND a smart operator

 

I think we must be confusing different machines or something.   Im of 1mm sheet metal is easily bent by a hand manually operated  (no hydraulics) bender in my local hardware store.  As a lad I used to bend such material with such a bender in sheet metal shops I worked in.

Unless my memory is at fault in both instances.  Possible, I admit.  But I think unlikely.  I can check the hardware store.

29 minutes ago, Enthalpy said:

It depends on how your material hardens during the deformation. Usual stainless steel needs more force than mild steel. Knowing the yield strength prior to deformation does not suffice here.

This is a new one.  I can see how significant this factor would be.  All bets are off if the material effectively changes its nature during the process.

But we can find calculations and even calculators, even tables, for press brake operations that are ultra simple, something like a constant x cross sectional area x Tensile Yield strength.   I think that's about it.  And I'd assume they cover the 'margin of error' implied by the 'hardening' factor somewhere within their constant.

A similar formula with the same technique is all that I'd require here.

Can that  be found?

If

35 minutes ago, Enthalpy said:

So you have to use empirical knowledge.

 

Relating to that: how about if I take a piece of 1mm steel 25mm wide x 150mm and fasten it in a  hinge with 100mm protruding - like a miniature bender and bend it using a spring scale hooked on the top end 100mm away from the bending point and observe the maximum force reached? 

This result could be multiplied to reach a figure valid for a 1m wide bend?  i.e. result x 40 ?

Would give me a  ballpark figure perhaps or a completely invalid figure for some reason?