manage different k bending and rolling with autodesk inventor

[S.O.1212]

Well I think the k of the rolling is different from the k of the bending so they are normal to have the possibility to use 2 k different because they are actually different.

 

[Catafratto]

Why don't you like the solution proposed by us?

 

[PIETRO2002]

If development calculation is a matter of life or death, I would proceed like this:

calculation of the development of the fallen part, with its coefficient k.

calculation of the development of the two folds from 30 ( doing one l) with its k.

I make the sum of these developments and I find total development, which will certainly be

different, from the piece I will get in practice!

Ergo, if you want the tenth, do a little bit of proof and rule you accordingly!

 

[PIETRO2002]

another possible method is to find a medium k coefficient, of course calculating the percentage in a weighted way, i.e. how much they affect the lengths of the calandated part compared to the length of the two folded lembi.

 

[Cast16]

you can set a different k-factor to fold. you have to use a different "unfold rule" for the fold that should not follow the k-factor of the part. see video in this link: https://www.screenpresso.com/=vzrde

 

[eugenesis]

However, boys, maybe you lose yourself in a glass of water, at each processing you can give a different factor k desired --> explanation options --> explanation rules .. etc.

 

[PIETRO2002]

However, boys, maybe you lose yourself in a glass of water, at each processing you can give a different factor k desired --> explanation options --> explanation rules .. etc.

then everything varies according to the material, to the radius of the fold blade, to the width of the matrix, to the bending system if in air, in support or in coniatura, to the direction of lamination of the sheet from which the piece is obtained.
the various coefficients k according to the various working conditions.

 

[MassiVonWeizen]

I'm sorry if I interfere, but @s. 12.12 did not ask what must be and how to use factor k in reality, not even alternative solutions to calculate development; asked how to manage multiple values of factor k in inventor.
so you seem to me to be a request linked purely to the software settings.

everything written remains useful to the forum, but the main request that I imagine - I do not know inventor - has been recovered from @eugenesis.
It's fair to say that having initially posted the wrong section may have deceived the various users.

 

[Marco F inox]

hello excuse the question but why use 2k different? the value k on inventor is managed according to the thickness of the sheet that checks, having a unique thickness the k will be unique.
maybe you can get something going on set sheet values and varying values

when you make such a narrow fold on a thickness of 3mm (half), there are many more fibers ironing than those that you buy. the neutra line moves towards the fold center and then k is small (0,33). vice versa when you calandra, also this thickness, with a wide radius, you can consider the compressed fibers, equal to those that iron, from here the k close to 0.5.
therefore the factor k does not depend only on the thickness and, in this case they need absolutely two, different from each other.
I also remember that among the data to be inserted, there are also the shape of the punch and the quarry, you saw that changing the inner radius, also changes the development.

 

[Marco F inox]

another possible method is to find a medium k coefficient, of course calculating the percentage in a weighted way, i.e. how much they affect the lengths of the calandated part compared to the length of the two folded lembi.

no, you can't, it's incorrect, it doesn't create carding and it just creates confusion.

 

[PIETRO2002]

no, you can't, it's incorrect, it doesn't create carding and it just creates confusion.

as it has already been shown, for each fold you can attribute the k value you want!

 

[PIETRO2002]

no, you can't, it's incorrect, it doesn't create carding and it just creates confusion.

But on the card you also put:
type of blade.
type of matrix.
state of supply of sheet metal.
sense of lamination of the piece compared to the matrix.
working conditions ( bending in air - bending in support - coniatura )
and on the design indicate to each fold the coefficient k you attributed.

 

[PIETRO2002]

I'm sorry if I interfere, but @s. 12.12 did not ask what must be and how to use factor k in reality, not even alternative solutions to calculate development; asked how to manage multiple values of factor k in inventor.
so you seem to me to be a request linked purely to the software settings.

everything written remains useful to the forum, but the main request that I imagine - I do not know inventor - has been recovered from @eugenesis.
It's fair to say that having initially posted the wrong section may have deceived the various users.

He's already given the answer to his question!

 

[Marco F inox]

But on the card you also put:
type of blade.
type of matrix.
state of supply of sheet metal.
sense of lamination of the piece compared to the matrix.
working conditions ( bending in air - bending in support - coniatura )
and on the design indicate to each fold the coefficient k you attributed.

certainly pietro, but on the card or cartiglio, I also put the result of a practical test, which is and remains the best and most accurate solution to get what the designer wants.

 

[Tarkus]

Good morning to all, I hope to write in the right section.
I am struggling with a dilemma, I should make a sheet metal component, where there is both a downside and a fold as a result I would need to handle two different k things I have previously set up in the appropriate section.
only that since the surface and therefore the corner where I should bend is not parallel and the software does not bend me, and if I do it with the surfaces or with the contour flange does not bend me (bend). place an image here.

Has anyone found himself with my own problem yet?
Can you give me advice on how to handle this situation?
thanks in advance

I made a rule with different k factors depending on the radius.
that is the small rays that are made in bending machine will have a factor k, the larger rays, "from calandra", will have the factor k 0.5.
I'm fine.
place image

 

[S.O.1212]

hello excuse the question but why use 2k different? the value k on inventor is managed according to the thickness of the sheet that checks, having a unique thickness the k will be unique.
maybe you can get something going on set sheet values and varying values

actually the k of the bend is different from the k of the rolling

 

[batleo]

actually the k of the bend is different from the k of the rolling

Thanks I am a little acerbo with the plates

 

[Marco F inox]

Thanks I am a little acerbo with the plates

the factor k does not depend on the thickness of the sheet, but on the relationship between the inner radius of fold and the thickness, Ir (internal connection) / mt (oftensor), something quite different.
the greater this relationship, the greater the k, whose value goes from 0.3 to 0.5.
this is why of the remarkable difference between bending and rolling.
inventor should take note of this technology and adapt its data.