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wolterh6
Posts: 38
Registered: ‎11-02-2011
Message 1 of 5 (180 Views)

Folding length cost

180 Views, 4 Replies
11-19-2011 04:41 PM

I would like to know what calculation on the length of a folding feature is performed by Inventor when one folds it. Read my example to better understand my question.

 

I am trying to fold a part of my sheet 90 degrees, I am going to call the folding part sheet B and the rest of the sheet, sheet A. Now, understand that when I fold sheet B 90 degrees, its face will no longer be parallel to sheet A, but the face of its thickness will. Consider this, now: when I fold the sheet B, there will be an inner part of the folding (which will have the faces of both sheet sections 90 degrees from each other) and the outer part (which will have an angle of 270 degrees between them).

 

I want the distance between the thickness face and the outer face of sheet A to be X in length. Which is the intial length of sheet B for this to happen? Consider using the BendRadius, Thickness, and other Inventor Sheet Metal parameters that will participate in the equation.

 

The question surged because I wanted such distance to be 5 in a part with a Thickness setting of 0.5 (as well as BendRadius), but because of this folding length calculation, I had to make sheet B 5.631 in length.

*Expert Elite*
Posts: 873
Registered: ‎02-16-2006
Message 2 of 5 (172 Views)

Re: Folding length cost

11-20-2011 12:05 AM in reply to: wolterh6

Make sure you are in sheet metal environment.

The k value regulates the amount of material added to a bend.

Go to sheet metal styles and edit your k value to get the flat length you require.

*Expert Elite*
Posts: 25,338
Registered: ‎04-20-2006
Message 3 of 5 (170 Views)

Re: Folding length cost

11-20-2011 04:49 AM in reply to: wolterh6

Attach your file here.
Are you aware of Bend Allowance calculations (material compresses on inside of bend and streches on outside of bend - see Machinerys' Handbook).

 

Why are you using Fold rather than modeling as folded (using Fold is almost never the best way to model a sheet metal part - attach your part here).

Please mark this response as "Accept as Solution" if it answers your question.
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Active Contributor
wolterh6
Posts: 38
Registered: ‎11-02-2011
Message 4 of 5 (159 Views)

Re: Folding length cost

11-20-2011 06:27 PM in reply to: JDMather

Well, honestly my drawing professor didn't even teach my class how to use Sheet Metal, so I'm here on my own. I wasn't aware of either of the things you said hehe. But yes, the term you specified sounded just right; I do not own or know that Handbook, will give it a search though. The part I am working with its attached.

*Expert Elite*
Posts: 25,338
Registered: ‎04-20-2006
Message 5 of 5 (147 Views)

Re: Folding length cost

11-21-2011 05:51 AM in reply to: wolterh6

Notice that by using Countoured Flange you don't have to do any calculations - Inventor will calculate your flat pattern taking into account the Bend Allowance.

Also, only takes two lines in the sketch.

 

Please mark this response as "Accept as Solution" if it answers your question.
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