petef Posted May 16, 2010 Share Posted May 16, 2010 Hello All, I am new to posting in this forum but I think this is the right one. Can anyone tell me a good method for calculating the amount of shrinkage in length when one takes a .5" square cross section of stock and you twist it? I would like to be able to control the overall length of a piece and know what it should end up when the twisting is done. I am assuming there is a relationship between the number of turns and the cross sectional size of the stock. As an example: lets say I have a square .5" stock 10" long. I am going to put a twist of 360 degrees along the lengthwise axis. What will the length be after the twist is complete? I am looking for a book reference, formula, rule of thumb or whatever other means there might be of calculating this. I appreciate any inputs. Pete Quote Link to comment Share on other sites More sharing options...
John B Posted May 16, 2010 Share Posted May 16, 2010 Same length as what you started with, you are twisting around a central axis. Quote Link to comment Share on other sites More sharing options...
petef Posted May 16, 2010 Author Share Posted May 16, 2010 John B, Thanks for putting my head on straight. Now that I think about it I see how that makes sense. Pete Quote Link to comment Share on other sites More sharing options...
John B Posted May 16, 2010 Share Posted May 16, 2010 You are welcome Pete, its only cagetwists and the like that will alter lengths. Welcome to the site and your journey into the world of hot metal. Quote Link to comment Share on other sites More sharing options...
petef Posted May 18, 2010 Author Share Posted May 18, 2010 In this same vein I would like to find a source of formulas or rules of thumb that apply to calculating lengths when bending stock. Do you or anyone know of a compilation like this? Pete Quote Link to comment Share on other sites More sharing options...
TASMITH Posted May 18, 2010 Share Posted May 18, 2010 I can give you one "quick and dirty" formula for stock required when making a ring Pete. example: you want to bend a ring with a six inch ID and are using 1/2 inch round stock Multiply the required diameter (inside) by three and the diameter of the stock by three and add them together (3X6)+(3x1/2)=18 + 1 1/2= 19 1/2 inches of material to make a six inch Id ring If you calculate the same thing using Pi x D 3.142 x 6 =18.875 inches But this gives you the diameter as measured in the center line of the stock and your inside diameter would end up just under six inches so you would have to add about a 1/2 inch to that length to achieve a six in inside diameter making it 19.375 inches or almost perfectly matching the "quick and dirty" calculation of 19.5 inches. Terry Quote Link to comment Share on other sites More sharing options...
petef Posted May 19, 2010 Author Share Posted May 19, 2010 Great just the kind of thing I was looking for. Pete Quote Link to comment Share on other sites More sharing options...
Richard Furrer Posted May 21, 2010 Share Posted May 21, 2010 Same length as what you started with, you are twisting around a central axis. John, An odd thing..I agree with you...no shrinking, but I did a very tight twist (think threading) on a 3/4" square bar that was 66" long. It lost four inches. I never quite figured out what was occurring....assuming I measured correctly before and after. Ric Quote Link to comment Share on other sites More sharing options...
pkrankow Posted May 21, 2010 Share Posted May 21, 2010 The corners would put some tension on the center, so for enough twists I would expect a reduction in length as the center upsets slightly, but that would be a lot of twisting! I say this because rubberbands shrink as they twist, and we have all played with wind up rubber band powered airplanes at some point in our lives! Phil Quote Link to comment Share on other sites More sharing options...
Shakeypm Posted May 21, 2010 Share Posted May 21, 2010 Hi Everyone, I hope this helps, sorry it is so boring! I have learned that as you twist a bar of steel, you increase the outside diameter. especially when you do many twists back and forth in a short length, so essentially you are tuning a square bar into a round bar with ridges. so the width of the bar is changing from the original width to the width from corner to corner. using the Pythagorean theory of "A squared + B squared = C squared" you can have an increase in cross section up to about 44 %. which means a decrease in length of up to about 44% in the section you are twisting. All that xxxx being said I would generally do a test piece to measure my finished pieces length versus my starting length (there will be slight variances between two twists because of temperature of metal when twisting). or just test the shortening of the twisted sections length for the twist pattern you are doing. this is especially important if you are fitting the piece between two fixed points. remember that if the piece is a little short between two pieces you can do a very slight "drawing" of the piece to make it fit. conversely you can "upset" it slightly to shorten it. Steve Long Beach, CA Quote Link to comment Share on other sites More sharing options...
Richard Furrer Posted May 21, 2010 Share Posted May 21, 2010 I don't follow you Steve...if a bar were 1" square and 12" long...how long do you think it would be after twisting? I would say under all but extreme cases of near 90 degree twists angles the bar will remain 12" long. Ric Quote Link to comment Share on other sites More sharing options...
pkrankow Posted May 21, 2010 Share Posted May 21, 2010 I think someone with a power twister needs to do some empirical testing on some bars of know length. How many turns per foot are needed to see a measurable length change? What size(s) of stock, and hot vs cold can be further parameters. I am going to suggest that a one full turn (4 flats) per foot a ruler will not measure a change in length. Phil Quote Link to comment Share on other sites More sharing options...
Shakeypm Posted May 21, 2010 Share Posted May 21, 2010 Phil, I don't know how to link photos from the gallery but here is a link for a pretty simple twist www.iforgeiron.com/gallery/image/32465-a-series-of-reverse-twists The reason I wanted to include the link is so we have common ground to work from. In this twist you can see that the length of the corners as measured along the corner in the twist is much longer than the length of the bar from the start of the twist to the end measured in a straight line. this is the amount that the bars width or mass along the bar is increased. in other words if the bar was 2" square and spun into a very tight twist the overall diameter of the nearly round rod would be close to 2.8 inchs, where do we get this increase of diameter? we get it from the length of the bar. so the bars length decreases up to 44% in the length of the twist. Clear as mud! If you hold the bar in a rigid clamp not allowing it to shorten you will diminish the width of the bar proportionately to the amount of twists. because you are trying to make the corners longer without adding mass, thus the dimensions of the bar decrease to give the corners the length you are forcing upon them. Does this help? or am I not explaining myself correctly? Steve P.S. anyone going to the ABANA conference in Memphis next month? Quote Link to comment Share on other sites More sharing options...
pkrankow Posted May 22, 2010 Share Posted May 22, 2010 You explain well, now...is there a rate if the end is unsupported that the bar shortens? How many twists to get an appreciable, measurable change in length? I have a ruler ticked in 1/16, so that can be called a measurable change, but only have manual twisting options. Is there an estimation of the rate of length change for twisting per number of turns taken? Phil Quote Link to comment Share on other sites More sharing options...
Shakeypm Posted May 22, 2010 Share Posted May 22, 2010 Phil, So I had an extra piece of 5/8" square bar and here it is: 5/8" square bar 11.5 inches long. Subtract 1.5 inch and have a twist length of 10 inch. 10" bar single twist length change following corner edge total 10.25" overall length change Negligible. 10" bar 2 full twists length change following corner edge total 10.5" overall length change Negligible. width between edges from 5/8" to 9/16" 10" bar 4 full twists length change following corner edge total 14.25 overall length change 9.75". width between edges from 5/8" to 7/16" 10" bar 8 full twists length change following corner edge total a pain to find overall length change 9 11/16". over all diameter from 27/32" to 3/4 " All dimensions gave up something in the twist. at about 6 twists in 10" almost all measurable length was given up converting to decorative round rod. until an equilibrium was met between a square and round shape. so 10" of 5/8 rod twisted 4 times gave up a 1/4" or 2.5 % TaDa, Steve P.S. I also accomplished some iron work today Quote Link to comment Share on other sites More sharing options...
petef Posted May 22, 2010 Author Share Posted May 22, 2010 Steve and Phil, Thanks for both of your inputs. It looks like I will need to do some test pieces as I will need to make several that all need to be the same length. I will repeat your experiment and see what I get with 1/2" square. Pete Quote Link to comment Share on other sites More sharing options...
petef Posted May 23, 2010 Author Share Posted May 23, 2010 OK I fired up the forge this morning and performed the experiment. I took a 1/2" square stock 10" long and made file marks at the 1" and 9" positions. This would mark the edge of my vice jaws and twisting wrench. I heated it up and put a 180 degree twist in it, quenched it and put it up against the tape measure. There was no discernible change in the length measurements. I then heated it again and continued the twist to a full 720 degrees which is equal to two full twists. Again, after quenching, I put it up to the tape measure and I could still see no change in the length measurements. I had also used the calipers to measure the cross sectional distance from corner to corner and again after the full twist. The before number was 11/16" and the after was 21/32". So there was a shrinkage in the cross section width from corner to corner. Below are the pictures showing the various measurements. This experiment answered the questions I had so I can move ahead with planning my project. Quote Link to comment Share on other sites More sharing options...
Shakeypm Posted May 25, 2010 Share Posted May 25, 2010 well done petef Steve Quote Link to comment Share on other sites More sharing options...
JDB Posted October 26, 2010 Share Posted October 26, 2010 If you want to figure the bend allowance the easy way go to a website called: engineersedge.com/bend_allowance and look at the formula there. You punch in your stock thickness and the radius you are bending and it will give you the extra stock allowance you need. Quote Link to comment Share on other sites More sharing options...
ThomasPowers Posted October 27, 2010 Share Posted October 27, 2010 JDB; does that site make allowances for bending hot rather than cold? Quote Link to comment Share on other sites More sharing options...
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