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Custom sheet metal rules

12 REPLIES 12
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Message 1 of 13
klaas.oudejans
833 Views, 12 Replies

Custom sheet metal rules

Dear readers, 

 

I am going nuts. To the point that I am convinced that Inventor has a bug! Please convince me otherwise & guide me in the right direction! 

 

I want to use the custom unfolding rules using Bend Compensation with an angular reference to the bending angle. 

 

I have an Excel sheet where I have determined the bend allowance, the set back and from that, the bend compensation. When I compare these values to the values I get when using a K-factor linear unfolding method in Inventor, I get almost exactly the same results. 

 

Tells me the formulas are working. Yeah! 

 

So I implement the formulas in the custom equations. Exactly the same as in the Excel sheet. 10 degrees angle, great results! Yeah! 45 degrees, great results! More yeah! 90 Degrees, still going strong, I think I might make it! 

 

100 degrees.... Fail! Every time an angle goes over 90 degrees, the result is crap.

 

I have been struggling with this all day long!! So in the end I completely erased the equations from the Inventor custom equations.I just say: Compensation is 1 mm. For all angles. This should no room for interpretation. 

Again I try. I have a sheet metal part with two legs. A 100mm leg and a 50mm leg. Flat pattern should always be 151mm. Period. 

Again, under 90 degrees and on 90 degrees, I get nice flat patterns. Above 90 degrees, it fails again. 

 

Am I completely overlooking something obvious?? Or is Inventor really bugging on this? 

 

I am working on 2012 SP 2

 

I hope somebody can enlighten me on this matter, because it is very urgent and I need to get it fixed somehow. 

 

Thanks in advance!

12 REPLIES 12
Message 2 of 13
mrattray
in reply to: klaas.oudejans

Bend compensation formulae are a bit more complicated then I think you think they are. Post more details about exactly what you're trying to achieve and we'll help you out. The Excel file of formulas would be helpful. I went through this myself a couple of years ago, so I emphasize with you here.
Mike (not Matt) Rattray

Message 3 of 13
mcgyvr
in reply to: mrattray

did you try it with kfactors instead of a custom equation?

I just use kfactors and it works perfectly everytime..

 



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Message 4 of 13

K-factors is what we normally use, but in this case, I want to create flat patterns as if the sheet metals had a different bend radius than they actually have. 

 

So I want to use a custom rule to "fake" a bend radius that the part doesn't actually have to get to the rigth flat pattern. Makes sense? Probably not, but to get into the full details of why this is, I would have to write a book about it. So I'll skip that ok? 🙂

 

I have been testing with theoretic results according to the formulas. If I replicate all those scenario's and Inventor's linear unfold with k-factor, I get equal results to the results from the Excel. 

 

Then I recreate the exact some formulas as in the Excel sheet, but the results do not match. 

 

As soon as I let the custom rule drive the flat pattern lengths, the results are off for every angle above 90 degrees. 90 and under are fine and consistent.  

 

So instead of keying in the equations in Inventor, I just type in fixed values and check if I get the expected results. If you look at the screenshot, it should start to make sense right? 

 

Thanks for helping! 

 

/Klaas

 

 

Message 5 of 13
mrattray
in reply to: klaas.oudejans

What exactly are you plugging into Inventor for the formulae?
What are you doing with the K-factor? What K value are you using for comparisons?
Mike (not Matt) Rattray

Message 6 of 13
mrattray
in reply to: mrattray

Never mind on the K-factor, I found it in the Excel sheet.
So, your goal is to replicate the bend compensation of a .28 K-factor but with a correction factor for a fictional bend radius?
Can I get a screen shot of your Inventor unfold style screen with your formulae shown? I strongly suspect a simple error.
Mike (not Matt) Rattray

Message 7 of 13

I have looked at the formula for hours, had some others sit at my desk to check with me, then I created the unfold rule that does no calculations at all. Just adds 10 mm to the flat pattern. 

And it doesn't work for angles over 90 degrees. 

 

I hope you can find an error in there 🙂 

 

Thanks, 

 

Klaas

Message 8 of 13
klaas.oudejans
in reply to: mrattray

Just realized that you probably don\t get all the styles in that part, so an additional screenshot supplied: 

 

 

Message 9 of 13
klaas.oudejans
in reply to: mrattray

Forgot to add: Yes, that's exactly what I want to do!
Message 10 of 13

Nobody has seen / solved this issue or knows a solution? 

 

/Klaas

Message 11 of 13
Hochenauer
in reply to: klaas.oudejans

Here is a brief observation I got from looking at your screen shot:

 

You are using similar formulas for the angle < 90 degrees and bigger 90 degrees. That does not make sense.

 

The tan() expression only makes sense for the case < 90 degrees. it correlates to a piece of a triangle in the bend region to define L1.

 

For angles > 90 degrees L1 goes up to the quadrant of the bend arc, it is Inner Radius + Thickness instead.

 

Unfolded Length = L1 + L2 - Correction.

 

The terms L1 and L2 are constant for angles > 90 degrees. You need to set up your correction term to be a function of Thickness, angle, radius,... to match your manufacturing environment.

 

 

I hope this helps at least a little.

 

regards,

Gerald

 



Gerald Hochenauer
Senior Principal Engineer, Inventor
Autodesk, Inc.

Message 12 of 13
jaybear
in reply to: Hochenauer

I had this problem trying to develop a bend deduction formula.  It worked perfectly for  < 90 deg and blew up on > 90 deg.  I finally switched to a bend allowance and it works through all angles.  Don't laugh; but our company uses this formula to calculate blanks: inside dim of flange + inside dim of flange = blank, regardless of bend radius or angle.  The formula shown in the attached picture works perfectly for us; but the the flange has to be created with the options shown in the picture, with the flange specified by the INSIDE dim of the flange.

 

As an example, If i create a 1" x 6" 12 ga. sheet metal part and add a 1" inside flange along the 6" edge, I get a formed piece with 1" inside legs, 1.105" outside legs and it will create a 2" x 6" blank at any angle of bend with any bend radius.

 

 

I know this formula won't produce the blank you need; but, I think you can use it as a base to get you started.

Message 13 of 13
jaybear
in reply to: klaas.oudejans

I've attached an example ipt file that was bent with my formula.  Hope it helps.

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