Spring winding advice



When I make springs, either carbon or stainless, I set the springs in my oven. I was looking for the temperatures when I ran into this.
http://www.rockfordspring.com/relaxationofsprings.asp
Not what I was looking for, sounds like it is geared to making sure a spring keeps its characteristics.
http://home.earthlink.net/~bazillion/finish.html#stress Is more what I was looking for.
Whenever I make springs, I ask uncle for the stress relieving temps since I never can remember them. Uncle ran coil winders for a number of years before trying selling Orchids.
Wes
-- "Additionally as a security officer, I carry a gun to protect government officials but my life isn't worth protecting at home in their eyes." Dick Anthony Heller
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I'm not too worried about relaxation. These springs will not be under load except while actually being used.
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On Thu, 22 Jul 2010 13:32:31 -0500, "Lloyd E. Sponenburgh"

Could you turn an arbitrary shallow taper, wind a spring, mark the spring before releasing it (maybe a paint strip to tell you where each turn ends up) and then find the mandrel size directly?
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OOOO! Yer a smart guy! I have a whole junkbox full of "arbritrary tapers"!
I do like that one.
Thanks, LLoyd
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On Thu, 22 Jul 2010 11:18:43 -0700, snipped-for-privacy@krl.org wrote:

I realize you (LS) have already made the springs but am posting anyway about a possible method. AFAICT, springback in degrees for springs like you are bending will be about e = a*d*k, where a = total angle bent; d = diameter of bend, at wire midline; e = springback angle; and k = arbitrary constant that wraps up materials properties (yield strength, Young's modulus), units (mm, degrees, etc), and wire thickness (which has an inverse effect on k).
To determine k, wind a test spring on a test mandrel, measure springback, and compute it. For example, if A" = 1800 degrees = 5 turns and mandrel diameter is 0.5 (so D' = 0.5+0.013) and springback is e60 degrees, compute k = e/(A"*D") = 360/(1800*0.513) = 0.390.
In following, let A = target angle and D = target diameter (0.625-0.013 0.612"). Let A', D' be the winding angle and diameter that springs back to the target angle and diameter, via A = A'-e' = A' - k*A'*D' equation.
Springback doesn't change wire length, so A*D = A'*D' or A' = A*D/D'. Thus A = A*D/D' - k*A*D*D'/D' or 1 = D/D' - k*D, whence D' = D/(1+k*D). Using the made-up example numbers gives D' = 0.612/(1+.239) = 0.49
Although the e = a*d*k model that I mentioned above doesn't show any "knee in overcoming the elastic limit" behavior as you talked about in a later post (at 2010.07.22 23:22:57), it still would be interesting to apply the D' = D/(1+k*D) equation to your springs and check its accuracy.
--
jiw

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All good info. Thanks. I'm compiling a mini "library" of spring-winding stuff.
As time permits, I will challenge the math you've presented. It looks rational and reasonable.
I'm still confused about the "knee", though, because it's _really_ there. Perhaps it's explained by the math...
LLoyd
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On Sat, 24 Jul 2010 18:51:00 -0500, Lloyd E. Sponenburgh wrote:

The thing to challenge is the model, e = k*A*D (e=springback angle, A=total angle wound, D=diameter wound, k=material and units constant) which is linear in A and D. That agrees with some of the online references I looked at, but not all. If I could have read the exponents in the four equations just before halfway thru <http://www.thefabricator.com/article/tc/roll-forming-high-strength-materials I would have tried to use a non-linear model.
--
jiw

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James Waldby wrote:

I'm pretty sure it's a "3". Here's the best that I could do, but it was only a 3kB image:
http://home.comcast.net/~bobengelhardt/formula1.jpg
Bob
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On Sat, 24 Jul 2010 18:51:00 -0500 "Lloyd E. Sponenburgh" <lloydspinsidemindspring.com> wrote:
<snip>

There is a book in the Workshop Practice series, "Spring Design and Manufacture" by Tubal Cain. It does have at least one chapter on winding springs. I gave it a cursory glance, some of the same problems discussed here were covered. See:
(Amazon.com product link shortened) />/
There was a crappy pdf version at www.freebookspot.com a while back. It is readable, may give you some ideas (shrug).
--
Leon Fisk
Grand Rapids MI/Zone 5b
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Oh, MAN! That's a good one! I got it off the freebookspot site.
THANKS!
LLoyd
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Lloyd E. Sponenburgh wrote:

Might find some info here. I believe he has a formula to determine the arbor size you need.
http://home.earthlink.net/~bazillion/intro.html
--
Steve W.

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That was the first search I hit, too.
There's no formula, just "pick an arbor just a little smaller than the desired size of the spring."
It's also a very non-technical approach to the subject, with a lot of "this'n'that will vary, but there's no way to know how..." stuff.
I was kind of hoping someone had a table with different wire metals and rules, plus maybe a nomograph plotting diameter vs. wiresize vs. abor size.
(Hopin' too much, huh?)
LLoyd
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"Lloyd E. Sponenburgh" <lloydspinsidemindspring.com> wrote in message

I have an entire book (a little old MAP book, or similar) on spring-winding and other non-cutting jobs on small lathes. It's in a box, in my storage area, three miles away. Otherwise I'd pull it out for you. <g>
But I'd be skeptical of any specific formula. You might get a ball-park number out of it. But there's nothing you won't learn by wrapping the wire, hard, around a few test "arbors," like pieces of pipe or bar stock.
--
Ed Huntress



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"Lloyd E. Sponenburgh" <lloydspinsidemindspring.com> fired this volley in

Well, I learned a lot in an hour of shop time.
I ended up using drills as arbors, as someone suggested here.
The first thing I discovered is that the underside of the wire guide must not "back bend" the wire -- at all, if the wire is straight to begin with. This wire was so close to straight as not to be an issue -- it comes off the roll with a bend radius of about 40".
The second thing I discovered confirmed my impression that the knee in the "overcoming the elastic limit" curve would be sharp. It was almost a right angle <G>.
I started with a 3/8" arbor tonight. I got a spring about 1-1/4" in diameter.
I did it with a 5/16" arbor, and got a 3/4" spring.
I did it with a 1/4" arbor, and got a 5/16" spring ---BOING! Hit the knee.
Tension wasn't as much of an issue as everyone including me suspected. Everything from "nearly breaking" to "just keeping it straight over the arbor" turned out stuff varying only about 25% in diameter. So it's an issue, but not a big one. Consistent tension is the magic number.
I fiddled with 9/32" and tension, and got two springs made (extension wound) that would fit, then drew them out to the correct length to form compression springs.
They work fine as-is. A 1-hour soak at 430F should take the strains out, to make them stay in the shape they're in now.
LLoyd
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"Lloyd E. Sponenburgh" <lloydspinsidemindspring.com> wrote in message

Another RCM success story...now we can move on to solving the Gulf oil spill..... <g>
--
Ed Huntress



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Ed Huntress wrote:

You are out of fate, Ed. That sucker, er.. gusher, is capped.
And a tropical storm is heading that way to "test" it!
--

Richard Lamb



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Lloyd E. Sponenburgh wrote:

ISTR Machinery's Handbook having a section on that, but it's been awhile since I browsed through a copy.
Jon
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There was an article published in the Home Shop Machinist many years ago with nomograms. I saw it on the web awhile ago. Email me. I think I saved it somewhere.
RWL
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On 2010-07-22, Lloyd E. Sponenburgh <lloydspinsidemindspring.com> wrote:

    Oh -- only 5/8". The way you were going on, I was picturing something like 5" to 10" diameter.

    Why not try just a few turns on 1/2" and 3/8" and measure what you get. That will probably let you calculate something quite close to your needs. Maybe it will show you that you need a third try at 1/4", but I doubt it.
    For the really large ones that I was picturing, I would have suggested three staggered rollers to put a pre-bend into the wire separate from the arbor.
    Good Luck,         DoN.
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On Jul 22, 12:57pm, "Lloyd E. Sponenburgh" <lloydspinsidemindspring.com> wrote:

can you do this quick&dirty by having a roller (narrow guided with the wire or wide, length of the arbor) press the wire against the arbor right where the wire contacts the arbor? (arbor and roller make a pinch roller)
dave
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