manage different k bending and rolling with autodesk inventor

[S.O.1212]

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

 

[PIETRO2002]

Did you check if the flange's inner fold radius is equal to the thickness?

 

[Marco F inox]

Did you check if the flange's inner fold radius is equal to the thickness?

from the drawing it is evident that the fold would not be feasible if not by force, as there is collision with the table of the prism. you could make the fold first, then turn it off, or turn it down only the ends, fold, then finish the displacement, as long as the calandra allows it. for the displacement you can use a k factor of 0.5 while for the fold a k of 0.3 can go well.

 

[PIETRO2002]

from the drawing it is evident that the fold would not be feasible if not by force, as there is collision with the table of the prism. you could make the fold first, then turn it off, or turn it down only the ends, fold, then finish the displacement, as long as the calandra allows it. for the displacement you can use a k factor of 0.5 while for the fold a k of 0.3 can go well.

the measurements are not known, if small it could be obtained by moulding, if large, the curved part obtained by rolling and the two saucers welded perhaps to tig, always that the artifact allows this operation.

 

[PIETRO2002]

the measurements are not known, if small it could be obtained by moulding, if large, the curved part obtained by rolling and the two saucers welded perhaps to tig, always that the artifact allows this operation.

I'm running. It's a big piece, building a mold of that size would be honorable, unless you have so many pieces to do.

 

[MassiVonWeizen]

look that the user is asking for information on how to do it in inventor... .

 

[S.O.1212]

I thank all for the answers, as Massivonweizen suggests, the sheet should create it with inventor so that we can then make the development so that we can bend it and turn it. the measures are indicative. I also confirm that the thicknesses are equal. I could try to force it, but this involves the fact that I can't develop. I really don't know how to develop!

 

[MassiVonWeizen]

opening the discussion in the inventor section would return more visibility inherent to the request.@ Stefanobruno can you please move the discussion? Thank you.

 

[PIETRO2002]

I attach the pdf with the design made in inventor.

procedure:

Make a 2d sketch and extrude it as a superfice.

make a thick surface thickening you need.

become the sheet metal piece.

revenues flat development.

Note: Put the radius of the same thickness of the sheet!

 

[S.O.1212]

ok but in this way I have a unique k, I would need a different k since one and the other is bending

 

[Peppe]

give an eye here.

 

[Marco F inox]

i thank all for the answers, as massivonweizen suggests, the sheet should create it with inventor so that we can then make the development so that we can bend it and turn it. the measures are indicative. i also confirm that the thicknesses are equal. i could try to force it, but this involves the fact that i can't develop. i really don't know how to develop!

as you cannot with inventor, i can tell you how to calculate development with formulas, so you have an alternative solution, of course with all the numerical data.

 

[Marco F inox]

did you check if the flange's inner fold radius is equal to the thickness?

serves little this comparison, as the inner bending radius depends mainly on the width of the used quarry, the punch and the material. for the calculation of development, it is necessary to measure the actual internal radius and to use a k of approximately 0.3.

 

[Marco F inox]

serves little this comparison, as the inner bending radius depends mainly on the width of the used quarry, the punch and the material. for the calculation of development, it is necessary to measure the actual internal radius and to use a k of approximately 0.3.

i saw the design and i have to say that to achieve an internal radius of 2mm. as thick, you need a very narrow quarry (9-12mm) and a lot of strength.

 

[S.O.1212]

I repeat, the measures are purely indicative, it only serves to make it clear that I have a dizzying (which should have its k) and the fold (which should have its k). since inventor does not give the possibility to put different k if you do not have different solids, but doing so does not make you do the development, and however does not allow you to do the function (fold band) to do the radiation.

 

[PIETRO2002]

I apologize for my curiosity, but what error do you think you have by attributing only the coefficent k you use for the displacement? It's a piece of carpentry. If you want tenths of precision, you have to do it by funnel mold. . . like the ones used for the automotive.
more than the cofficient k, I would say to be attentive to the state of supply of the material, to the sense of rolling of the sheet, to the type of matrix and punch that you use to make the fold of the two legs. .

 

[S.O.1212]

Perhaps in this case as you say the error is negligible but in other similar processes I assure you that the development error is relevant. in any case they asked me to hold against both k factors for this type of processing.

 

[Marco F inox]

Perhaps in this case as you say the error is negligible but in other similar processes I assure you that the development error is relevant. in any case they asked me to hold against both k factors for this type of processing.

It is not understandable that a program, applied to the same piece, does not allow you to use two different ks, when there are different folds. I offer you a comparison solution. make a real example with all odds, I calculate the development, then you compare it to yours, using the k you prefer.

 

[S.O.1212]

then we take a step back, I have to realize this sheet, using 2 k different development is very different, so I would need a way to manage the k of the fold and that of the rolling

 

[batleo]

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