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Spring Theory
8

Spring Theory

Spring Theory

(OP)
Howdie.
Would anyone happen to know of a good book that explains the dynamics of springs?  I have searched and searched on the web forever for the equations governing the dynamics of all types of springs and come up zero, and can't seem to find anything on something that MUST be covered in detail SOMEWHERE.  You know it's getting bad when you reach somebody's New Age page about "Spring" theory and how it will change your life....Jeez!

Thanks,
treddie

RE: Spring Theory

2
Hi treddie

Try getting a copy of "spring design" by W.R.Berry its out of print but you might find a copy in your local libary.
Here is a site which gives info about spring design:-
http://www.mech.uwa.edu.au/DANotes/springs/home.html


Regards

desertfox

RE: Spring Theory

Pick up Schaum's Outline Series
Theory and Problems of Machine Design.
Most University Bookstores have good
Machine Design Books if you cannot find
it at your library.
Machinery's Handbook might suffice.
How little do you know?
What kind of spring?

RE: Spring Theory

What do you mean by spring dynamics? Are you referring to the dynamics of masses loaded by springs? If yes, then any university school book such as Dynamics by Meriams's or the books mentioned by dimjim.

However, if you you refer to the the dynamic shock wave in a spring which is loaded by an impact force or fatigue life of springs, then you should geT:

Spring Design and application by Chironis
Mechanical springs by Wahl

or

Spring Design by W.R.Berry as desertfox suggested.

http://israelkk.googlepages.com/home

RE: Spring Theory

(OP)
Thanks everyone for the replies.  I will check the references you have suggested.

Israelkk>
The problem is one of CAD modelling the shape of a flat spiral spring (clock spring) accurately, while taking into consideration that the last winding is pushing against the spring and causing the coils to compress inward on that side (therefore, outward on the other side). The compression dampens out as you get closer to the center of the spring, I guess, due to the fact that the closer you get to center, the more opposition there is to the side force.

Also, am trying to accurately model conventional coil spring behaviour as it rests inside of a curved space with a ball at one end.

treddie

RE: Spring Theory

treddie

When you say CAD modeling do you mean FEA analysis of the clock/brush spring?

If so, I do not exactly understand what are you looking for. Are you referring to the phenomena that the coils are moving radially when the spring is deflected by some angle of rotation?

Anyway, the number of turns that a clock/brush spring can have is up to three full turns (1180 degrees), beyond this the spring will buckle such the coils will move in the direction of the spring axis of rotation forming a cone like shape of spring.

Due to the large deflections (three turns max) of such a spring a linear FEA analysis will not fit.

More than that, the formulations for such springs are well established by theory and practice for many years. Therefore, I do not see any value in using a cannon to kill an ant.

One more problem to consider is that all formulations are based on experience and you can not manufacture any shape you like using a coiled strip. The formulations takes this into account when it put limits to the strip length to strip thickness ratio selection.

As to accurately model conventional coil spring behaviour as it rests inside of a curved space with a ball at one end, friction will play a major role. Therefore, not all the coils will participate all the time. Therefore I expect a very erratic and inconsistent behavior of such a spring arrangement, not to mention wear of the coils and very limited life cycle.

In a coiled compression spring the diameter of the coils varies from coil to coil and they are not completely centered to each other.

RE: Spring Theory

It's not clear to me that the clock spring problem is actually a dynamic problem at all.  But sounds like you need information on FEA, not springs- unless some has exhaustively approached that very problem.

RE: Spring Theory

(OP)
Looks like someone has directed me to a forum on Springs.  I will check there, too.

But to answer your question, israelkk>
Visualize a flat spiral spring of, say 20 turns.  It has an axle, ofcourse, perpendicular to the direction of the spring's turns.  Put one end of the spring axle in a hole in a tabltop so that the axle is sticking straight up out of the tabletop.  The plane of the spring turns is now parallel to the tabletop.  Now, push on the spring itself, pushing parallel to the tabletop, using your finger.  The windings compress on that side of the spring, and expand on the other.

This is all for a visual I need to build; it's not for building an actual spring, although I will be building it in ProE, nonetheless. That is, once I can get a fairly accurate plan view of the spring windings, by building a VB program to generate it.  I'll then save out as a DXF to import into ProE for a trajectory, along which the spring will follow.

treddie

RE: Spring Theory

It's an interesting probem, and I am sure it has been solved empirically, or even analytically before. However, in my experience spring designers often try and do things that result in non-analytical designs.  Your coil spring in a tube example is a good one.

So, you could use a non linear (in geometry) FEA package. You could build it in ADAMS, which would cope with all the non linearities, but it would be a very slow running model, and would be very 'fragile'.





Cheers

Greg Locock

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.

RE: Spring Theory

2
My attempt to the solution of this problem is as follows.
Let's assume O is the center of the spiral and B is the free end, where a force P is acting in the direction BO.
One can observe that each half winding of the spiral may be approximated as a semicircular arc with radius r equal to half the distance of the endpoints.
Now the problem is that of the deflection of a semicircular arc with two fixed ends at A and at B, the ends being free to slide along the diameter. The arc is loaded at A and B by two forces P directed along the diameter.
This problem is solved in textbooks treating curved beams: my own is a monumental work by Belluzzi (only in italian), but I guess it can be found elsewhere.
The solution to this problem is as follows (hope with no mistake on my part, as some simplifications were necessary):
ξ=(Pr3/16EJ)(2θ-π+2sinθcosθ+8(θsinθ+cosθ)/π-4sinθ)
η=(Pr3/16EJ)(4cosθ-2-2cos2θ+8(sinθ-θcosθ)/π)
where
ξ is the displacement in the direction AB
η is the displacement in the direction orthogonal to AB
θ is an angle measured from A (where it is zero) to B (where it equals π)
and the other symbols are easy to figure out.
Note also that the above equations take the center O as not moving (because of symmetry): this should be accounted for in the following.
Now it shouldn't be too hard to do the following: take first the innermost half winding and calculate its deflection, considering that one end in this case doesn't move, not the center as in the equations above; then take the following one that will start its deflection from the former and so on.
Of course all this assumes that each half winding is a true circular arc and that the deflections are small. However as the solution is obtained from the equivalent of the well known equation y'=-M/EJ, and this one is known to work well with relatively large deformations, I guess that the results could be acceptable even for a spring, where the deflection may be many times the beam depth (or thickness). Anyway don't think that more could be done by analytical means.

Good luck! (you'll need much of itsmile)

prex

http://www.xcalcs.com
Online tools for structural design

RE: Spring Theory

prex

In such a spring the force P is NOT directed to the center "O", it is acting perpendicular to OB (tangent to the outside diameter).

Usually the B point is held in position and the spring center tab is rotating with respect to the spring axis. Unless the B point is clamped the spring end at the B point can rotate with respect to the B point.

RE: Spring Theory

(OP)
Thanks everyone for your excellent comments.
prex > I will give your solution a try.  Looks very straight forward, though in just quickly sketching it out, It looks like it's behaving more like a squashed onion (with all its layers) than a spring.  But I'll run the equations to find out.
Anyway, I'm going to see if I can find a used copy of Spring Design on Amazon, plus the other books, but I'll check out the University library, first.

Thanks again, everbody!
Time to hit the books!
treddie

RE: Spring Theory

Ooops!
The factor Pr3/16EJ above is incorrect: should be Pr3/4EJ.

prex

http://www.xcalcs.com
Online tools for structural design

RE: Spring Theory

(OP)
Cool. Thanks.

RE: Spring Theory

treddie,
just to complete:
1)of course the above formulae become invalid if there is contact between the turns; in that case the problem becomes hyperstatic and I guess there is no simple solution
2)the outermost half turn, to be precise, is supported, not fixed, at one end. If you succeed in using the above formulae, drop me a message here or in the contact section of the site below: in my reference I have also the formulae for the fixed-supported boundary condition.

prex

http://www.xcalcs.com
Online tools for structural design

RE: Spring Theory

(OP)
prex:
I am assuming that in your equations, the following are:
    M = Bending moment
    E = Modulus of Elasticity
    J = Polar Moment of Inertia

If so, I have one question for you.
Since my spring has a rectangular crossection, would I not replace J with I (Moment of Inertia)?

treddie

RE: Spring Theory

treddie,
J is the moment of inertia about the neutral axis of bending. If t is the thickness of your spring and b the width, J=bt3/12

prex

http://www.xcalcs.com
Online tools for structural design

RE: Spring Theory

(OP)
prex:
Gotcha. Thanks.
I have to get out some jobs for clients, so I won't be able to run a test program just yet.  I'm really curious to set this thing up, though, so hopefully I'll be able to get on it shortly.  I'll let you know of my results.
treddie

RE: Spring Theory

(OP)
prex:
Have had the chance to set up the "guts" of the program.  Getting ready to set up procedure for the "re-joining" of the beams, and then debug.  Hopefully soon.
treddie

RE: Spring Theory

(OP)
prex:
I'm assuming I have your math heiarchy correct here:

 Let F = (Pr^3) / (4EJ)

Then:

?=F * ((2?) - (?) + (2sin?cos?) + (8(?sin?+cos?)/?) - (4sin?))

?=F * ( (4cos?) - (2) - (2(cos?)^2) + (8*(sin?-?cos?) / ?) )

Right now, I'm getting strange behaviour and want to check the
equations first before I proceed to debug further.

Incidentally, I dug out my old Strength of Materials book and found the discussion on the First Moment-Area Theorem, which you allude to above.  Just out of curiosity, I’m also going to play  with the idea of the spiral as a straightened out piece of metal laid horizontally, with a sinusoidally changing applied load from above, representing the continuously changing load with THETA.
Then working backwards to “assemble” those loads onto the wrapped-up spiral.  Kind of like reducing the problem to one of a single long beam with vertical loads and typical shear and moment diagrams.

RE: Spring Theory

Everything is OK for me, of course replacing the ? with the corresponding symbols (look in the Process TGML link at bottom of the Reply window, then 'Special characters', for how to use symbols on this site).
Look here to see what I got with those formulae using Excel: in this exercise the factor Pr3/EJ was 0.04 (at the inner end of the spiral) and the radius was increasing by 20% at each half turn (we are dealing with archimedean spirals, aren't we?).
To obtain these result you only need to be careful about signs and origin of displacements: the above formulae are valid only between 0 and π.
You'll see that, for a spiral with many turns, the turns will come into contact for a very small deformation.
Is this what you were looking for?

prex

http://www.xcalcs.com
Online tools for structural design

RE: Spring Theory

(OP)
Howdie prex.
Finally got a chance to get back to this thing.
Have been running some test values, and one thing has me stumped.  The 2 equations for, ξ and η, have only 2 variables, (P) and (θ); (E and J are constants for a given spring).  How are ξ and η applied to the x and y outputs of the circle?  My tests simply scale the size but not the shape of the half-segments, because simply changing for instance, load (P), does nothing more than scale both outputs, which is obvious from the equations.  Therefore, doing any of the following really doesn't make sense (where xc and yc are outputs from a simple circular arc from 0 to π):
    As offsets:   
       x = xc + ξ
       y = yc + η

    As coefficients:
       x = xc *ξ
       y = yc * η

    As "self-contained" coords:
       x = ξ
       y = η

(E), modulus of elasticity and (J), moment of inertia are constants, for a given spring.  This leaves (P) to do nothing more than scale the results.  What's going on here that I'm missing?

It's difficult without seeing the actual discussion by Belluzzi.

treddie
p.s.  Thanks for the tip on special characters.

RE: Spring Theory

ξ and η are displacements, so that your first equation (as offsets) is the correct one.
And OK the independent variable of the equations for ξ and η is θ: can't see what worries you. P is more a parameter than a variable (just like EJ).

prex

http://www.xcalcs.com
Online tools for structural design

RE: Spring Theory

(OP)
I agree (P) being more a parameter.
Let me go back to my code and see if I can find anything stupid.  Right now, I get a sort of elliptical shape for one 1/2 turn (to be expected), but changing the load does not change the proportions of the curve, just its size.  Unless my understanding of the behaviour of the equations is wrong to begin with; that the curve proportions WOULD stay the same regardless of load, only the DISPLACEMENT AND SCALE of the curves should change.  This is easy to test...I've only been looking at 1 half-segment up to now to test the basic code.  Time to connect them all and look at the overall shape.

treddie

RE: Spring Theory

(OP)
My last post doesn't make sense either.  In looking at your image again, it's clear that each of your segments is definitely changing its proportions.  This leaves only one other possibility I can think of:
  x = Σ(xc + ξc)
  y = Σ(yc + ηc)

  where:
    xc = rcosθ
    yc = rsinθ

accumulating the displacements after each step θ.

treddie

RE: Spring Theory

(OP)
God! I’m losing it!
What I MEANT to say, was…
  x = xc +  Σ( ξT)
  y = yc + Σ( ηT)

  where:
    xc = rcosθ
    yc = rsinθ

    ξT = Total accumulated ξ up to the present θ. ( ξT not  
            reset to zero after each half-segment)

    ηT = Total accumulated η up to the present θ. ( ηT not
            reset to zero after each half-segment)

treddie

RE: Spring Theory

Yes, you are lost!smile
Try to follow the following steps (or ask for help to a mathematically minded colleague):
1- Rewrite my equation for ξ so that ξ=0 for θ=0, you should obtain:
ξ=(Pr3/4EJ)(2θ+2sinθcosθ+8(θsinθ+cosθ-1)/π-4sinθ)
2-For the first semicircle
x=rcosθ+ξ(θ)
y=rsinθ+η(θ)
3-Now call xe,ye the endpoint of the first semicircle, namely
xe=-r+ξ(π)
ye=0
4-The next semicircle must start from this endpoint end has a new radius R=rr
Using again θ going from 0 to π its undeformed equation is
x=xe+R(1-cosθ)
y=-Rsinθ
(hope with no error on my part)
5-Add to the above ξ and η calculated with the new R
6-Loop from point 3 for subsequent semicircles.

prex

http://www.xcalcs.com
Online tools for structural design

RE: Spring Theory

Follow up: step 5 above should read:
5-Add to the above ξ and η calculated with the new R and with the sign of P reversed

prex

http://www.xcalcs.com
Online tools for structural design

RE: Spring Theory

(OP)
Well, it likes like I DID understand you correctly in the beginning, because that is exactly the approach I took, except that I let the centerpoint of all the semicircles line up; I was going to deal with that problem after all of the semicircles were completed and in an array.
I also think that one of the problems I had was that I was putting in unrealistic values for the load (P), resulting in wacked out results.  But to know for sure could you check these three equations for correctness, if you can one last time, because I STILL am getting “puckered” like deformations (I would upload the image to you, but didn’t want you to have to go through the web server hassle to retrieve it):

For distribution relative to spiral center:

            Pr3                                                   8(θsinθ + cosθ)
 ξ  =  ______  *   ( 2θ  -  π  +  2sinθcosθ  +  ______________  -  4sinθ )

           4EJ                                                             π



For distribution relative to spiral end:

            Pr3                                           8(θsinθ + cosθ - 1)
 ξ  =  ______  *   ( 2θ  +  2sinθcosθ  +  ________________  -  4sinθ )

           4EJ                                                       π


“Vertical” displacement:

            Pr3                                                   8(sinθ  -  θ cosθ)
 η  =  ______  *   ( 4cosθ  -  2  -  2cos 2 θ  +  ______________  )

           4EJ                                                              π

RE: Spring Theory

(OP)
Got it!
As I thought, a stupid high-school error in my code.  Shapes are correct, now I'll link up all the half-segments.
treddie

RE: Spring Theory

(OP)
It's working.  FINALLY!  
Pretty much gives me what I need.  I think I'll toy with it and get it to do applied loads at other than along the x-axis, so I can make it more versatile.

Anyway, prex, thank you very much for all of your help and patience.  I really appreciate your time.

treddie

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