Spring winding advice

"Lloyd E. Sponenburgh" fired this volley in news:Xns9DBDA1E4D7B12lloydspmindspringcom@216.168.3.70:

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 .

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

Another RCM success story...now we can move on to solving the Gulf oil spill.....

Years ago a place I worked made some springs for some project or another. The Shop Foreman, who seemed to know what he was doing, had us build a gizmo, that fitted in the tool post and tensioned the wire as it wound onto the arbor. It has been a long time but I seem to remember that the arbor was basically the approximate desired inside diameter of the spring and the turns per inch was governed by the carriage, driven by the lead screw.

I also seem to remember that tension on the wire was the critical point in the whole operation.

Cheers,

John D. Slocomb (jdslocombatgmail)

You are out of fate, Ed. That sucker, er.. gusher, is capped.

And a tropical storm is heading that way to "test" it!

J. D. Slocomb fired this volley in news: snipped-for-privacy@4ax.com:

Think relative diameter of the wire and the arbor.

I'm working with 13-thou wire to finish out to a 5/8" o.d. spring.

The rules for "thick wire" don't hold.

But the problem's solved.

LLoyd

Depending on where you are, I might be able to help....I have a spring winder sitting on a bench at work....if you have the wire and can get to southern wisconsin, I suspect we could get you a spring wound.

Mike

snipped-for-privacy@iname.com fired this volley in news: snipped-for-privacy@4ax.com:

Thanks... got it wound.

I cobbled up a winder very similar to the old manual Porter bench mounted ones. The problem was in the determination of arbor size. That fine wire didn't take a set unless it was wound on an arbor MUCH smaller than the finished size.

LLoyd

ISTR Machinery's Handbook having a section on that, but it's been awhile since I browsed through a copy.

Jon

LLoyd,

I've not had a look at any for some years but the dimensions you give remind me of the piston spring in SU carbs. Maybe you want to give these guys an email and see if they can provide any information

.

What kind of diameters are we talking about? Could you do some test bends around, say, drill bits?

Good Luck! Rich

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

you know, don't you, that the 'real' spring winding machines extrude the wire against an angled anvil that is tilted in X,Y (presuming the wire is fed in Z) so that it causes the wire to bend to the desired spiral - if you build that, then it's simple enough to adjust to get the diameter and spacing you want, no adjustable mandrils required

The first time uncle explained a spring winder it took me a while to realize he was describing what you wrote above.

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

"Bill Noble" fired this volley in news:i2dmta$kte$ snipped-for-privacy@news.eternal-september.org:

Yep; well, _fed_, not extruded, in most cases. I get to see one in animation every time an episode of "How it's Made" comes on.

But Bill, I need two springs. I have a spring winder.

And FWIW, I've already made them successfully.

LLoyd

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 e=360 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.

James Waldby fired this volley in news:i2f376$p9c$1 @news.eternal-september.org:

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

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

I would have tried to use a non-linear model.

I'm pretty sure it's a "3". Here's the best that I could do, but it was only a 3kB image:

Bob

I wonder if an equivalent could be made from a scissors knurl with grooved rollers. Tilt the near end down to grab the wire between the lower roller and the shaft in the spindle, and adjust the upper roller or the crossfeed screw to change the wire loop diameter.

There is an adjustable fine wire feeder in your MIG welder.

jsw