![]() ![]() Most plasma cutters are off a bit anyway so you end up with stuff that does not fit from a quarter million dollar machine.Home / math / volume calculator Volume Calculator So although these might seem like horrible answers that is how it is done in the real world. If you want it exact or loose, you probably want it a little large because the weld will shrink it. Most guys that are doing these projects have one piece of metal, so they get one shot.Įspecially if you have a piece or object to fit the band to, and you want a tight fit you may want a close fit before you weld because the weld will shrink it substantially. So if your piece is a quarter of an inch too long, there is no problem at all, you just trim that one piece down to an exact fit. However that center line may not be exactly in the center by actual testing and building. What many people are saying is that there is a center line. And then I would start shearing up whole sheets of metal to make pipe.īut if you are asking these questions you do not know all the little nuances, that creep into it. And it would come out perfect almost every time. And I can assure you that if I had to make a lot of the same sized pipe, I would create a test piece, a test strip, using the method I suggested. I think we should be versed in both.I used to make a lot of round pipe. Granted, theory is different than practice. I have to REALLY sit down and do the calculations. But he was asking for the mathematical calculation to use. Yeah, we can all cut long, roll and then trim to fit. I just want to further my understanding of fabrication 02-27-2017 at 10:49 PM.įor instance say taking a 3/8" rod and bending it into a 1" ring.Thank you i appreciate this and everybody else who has answered! this is not a real job for me i just made it up so that i could get exactly this type of feedback. ![]() If the material thickness in relation to the bend would have been greater, then the formula above would have to be adjusted because the neutral axis would not be in the center anymore.įor instance say taking a 3/8" rod and bending it into a 1" ring. Length of the bar is then 63.6" (length x 3.14). That means the center of the circle is 20"+1/4" = 20.25". Length of the bar is then 62" (length x 3.14). That means the center of the circle is 20"-1/4" = 19.75". Since it's a large bend, radius 10" while the thickness of the bar is only 1/4" then neutral line is in the middle. I found this as a guideline - Basic Diemaking Eugene Ostergaard R4T C=.5T R=Inside Bend Radius, T=Thickness of material, C=Location of neutral axis.Īnyway, back to the question - 1"X1/4" flat bar that has been rolled and has a diameter of 20". On tight bends it is closer to the inside of the bend. On a large bend radius in relation to the material thickness the neutral line is in the middle. So the thickness of the material has to be taken into account. So the length of the original bar is somewhere in the middle, called the neutral axis. When you bend something the outside will stretch and the inside will compress. I'll leave my somewhat redundant post below since it has a nice picture I stole on the web. ![]() ![]() Reading closely again I see that people have commented above on the same thing. But that difference of pi would only amount to 1/32" difference in the length of the metal to make a 20 inch circle. If the metal is one eighth thick, then you should add one eighth to the diameter before you multiply it by pi. My thought is if you roll it and it comes out almost an eighth inch in diameter too large, you can just grind a quarter inch off one end and your are good to go. The inside up against your die, is going to compress, the outside is going to expand, the actual size is somewhere in between. The reason is the thickness of the metal. You tend to need more length than just diameter times 3.14 Does this mean that in order to replicate that circle i would need to cut my flatbar to 62.8"? would that at least be very close? Does this math make sense? That math gives me a circumference of 62.8". Is this correct? I take the diameter of the existing rolled flatbar (20") and X that by 3.14 (pie). Lets say i have a magical break press that will only bend the straight flat bar once it senses the required flat length of the existing rolled flat bar. I am instructed to build one more of these starting from a straight piece of 1"X1/4" flat bar. So visually, this flat bar is in the shape of a circle now. Lets say there is a 1"X1/4" flat bar that has been rolled and has a diameter of 20". ![]()
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