I would dimension the holes from center-to-center in the folded part rather than from the end.
This is the critical dimension for parts to go together.
Then once the correct k-factor or bend table is calculated you can dimension the hole location in the flat pattern from the end.
Also I would put the hole in from the other side so that it goes all the way through the bend (do you really want it to be in the bent section - usually there is some sort of fastener that would clamp better to the planar face rather than hitting the fillet).
The Machinerys' Handbook has k-factors for various materials, but you can simply take whatever dimensions Inventor gives you for the flat (forget the AutoCAD flat dimensions - no bend allowance was taken into account) and make a part from those dimensions. Compare the finished size to design intent. Adjust the k-factor (or bend table) to increase or decrease overall length of flat. Test again. Should be able to narrow down quickly.
Now that you know the number for your material and processes you can use that k-factor (or bend table) for future designs. (my understanding is that you could have 5 brand new machines set up "identically" to make the same part and the bend allowance would be slightly different for each machine - such is the nature of bend stretching. fortunately sheet metal parts usually have generous tolerances so slight difference don't matter. If tolerences are tight then operator must keep a bend table up to date as machine/tooling wears and you must feed that info into Inventor).
BTW - good job in modeling the part. I see so much stuff here that is unconstrained sketches.
"Adjust the k-factor (or bend table) to increase or decrease overall length of flat." I didn't know this was possible..I guess if you can edit the factor then that should work...just never thought of it. Knowing this will save me lots of time and your typing fingers...LOL
Nope, I don't want the hole in the radius, but currently that is the way they are making them. I will get the ok to change that once I get things fixed with the K Factor.
All the machines we have here are pretty wore out even if we wanted a certain tolerance we couldn't hold it well. We use a sigle ram press to make the basic shape then brake the sharp angle and basicly WYSIWUG. So if we get brakes to vary 2 deg in a batch were doing great..LOL. The guys do a reall good job of getting things as close as they can, but they aren't GOD, so they can only do so much.
One thing I did learn early on CONSTRAIN EVERYTHING then dimension. I also don't leave pieces grounded in assemblies, I remove the grounding and use constraints to lock down parts. Had to many early assemblies and iAssemblies(especially) blow on me.
When I create new drawing in ACAD I try to make sure that loss of material is built in. But some of the drawings that are much older (ones that hadn't been changed since they were brought over from P&P) and once done by someone else I am not sure of, so that is why I am OCD about my stuff..Sometimes to much...LOL
"Adjust the k-factor (or bend table) to increase or decrease overall length of flat." things as close as they can, but they aren't GOD, so they can only do so much.
I forgot to mention that different radii bends can each have a different k-factor (which you can assign by bend). Especially bend radii that are less than the thickness of the material. I haven't experimented - but I suspect the other end of the spectrum is also true, that is, the very large radius bend would have a different k-factor than "normal" bends where normal bends radii = material thickness.
Fortunately in sheet metal work the tolarances are usually liberal enough that getting exact isn't critical.
Do an experiment - simply change that angle a couple of degrees and measure the length of the flat in comparison.
Inventor is making an allowance for stretching of the flat in the bending zone.
Attached is a PDF that might help with calculating bend deductions that will match what your shop floor produces. This is more accurate than the K-factor method, although K-factors are often plenty accurate for many shops.
This file was originally posted at this link http://inthemachine-autodesk.typepad.com/blog/2008
I'm posting it here in hopes that others might find it if they search in the future (I just searched high and low on the internet and had to resort to digging through my personal stash of reference material that I had squirreled away).
Once the test bend information is gathered you can add this information to a bend table in Inventor.
I hope this helps.
Best of luck to you in all of your Inventor pursuits,
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Well I was able to get a hold of a 20th edition of the Machinery's handbook. I found the section on Punches and Dies and from what I can tell it seems I need to use the formula L= (.71*T) + (1.57+R). I use this formula x2 for each side of the 5" arch and the formula L = L* angle of bend / 90 for the bend over 90deg
I then used the formula L= (.71*T) + (1.57+R) * 180/90 to figure out the length arch is supposed to be, but I came up with a number much larger ( 1+ inches lager) than the over all sheared lenght of the original part.
Could use some help figuring this out. If you need more info just ask and it will be yours.
My numbers used in calculation.
Material is mild steel sheet
material thickness = .1196
Lg Inner Radius = 5 in
Md Inner radius = .75
Sm Inner Radius = .0625
The flat areas are
.5625 -> End with double brake and this is the flat with the hole in it.
1.0625 -> End with just one hole and a single brake.
.625 -> This flat for short 110 deg brake.
This is what I came up with
L = .076544 + 1.1755 = 1.254044 * 2 = 2.508088 (*2 since there is 2 3/4 in brakes)
L = .076544 + .098125 * 110/90 = .213458433
L = .076544 + 7.85 * 2 = 15.85088
Total length should be 2.508088 + .213458433 + 15.85088 = 18.57242643
I know that going out the number of places I did isn't realistic tolerances but that is what the calculator gave me..LOL
It seems I missed a number when typing...the amount 1.1755 should be 1.17655
I almost forgot...In order to get the right length I changed the KFactor to .3826 ul. Does this seem out of line?
I attached an updated file.
I am not sure...but should I have add the flat lengths into the total length as well? If so my total lenght of part is 20.543972.....
Looking good so far.
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