Technical Questions on Spring Design

KiwiBird

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Jan 30, 2000
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Talking to Jer last night I asked him a question that he suggested I post here.

It started out simply when I asked him what difference it made if I used a long or short spring (same rate) on my shock.

I thought you went for a short spring to save weight but Jer said there is a lot more to it than that.

What do you guys have to say?
 

burtlamborn

Member
Dec 21, 2000
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I'm no suspension expert, but no one else has posted yet so I'll mention a few things (that you've probably already thought of anyway.) And all of this is just my opinion, for what it's worth.

The length of the spring has to be chosen so it works with the amount of stroke the shock has, taking into account the amount of preload that will likely be seen by the large variety of riders using the machine. You want the spring to have enough travel so that when it's preloaded heavily, it still has enough travel that it doesn't bottom coil-to-coil before the other suspension components can do their jobs at the end of the shock stroke. (Rising-rate linkage, bottoming cone, PDS damping, etc.)

I assume you meant purchasing a shorter spring of the same rate, in which case you already know that if you modify a spring by shortening it, you change its spring rate.

I'm sure there's a million other things to think about. Anybody else?

Burt

2.
 

RM_guy

Moderator
Damn Yankees
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Nov 21, 2000
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I hope I’m not getting in over my head but here it goes. I understand spring design but my understanding of how it works as a suspension component on a dirt dike may be a bit shaky.

Spring rate is roughly based on the flex modulus (springiness) and diameter of the wire. The units for spring rate are pounds/inch (or kg/mm). This means it takes ‘x’ pounds to compress ‘y’ inches. The length of the spring determines how much total load the spring can take for any particular spring rate. A spring with a rate of 10 lb/in would need 100 lbs to compress it 10 inches. To change the compressing force to 200 lbs at 10 inches you would have to increase the spring rate (via the wire diameter) to 20 lb/in.

For a particular shock, any optional springs would have to be the same length in order to allow the shock to fully stroke before the spring bottoms out (coils come in contact with each other-which should not be allowed to happen in an optimum design). As far as I know the only optional springs would be variations in the spring weight (wire diameter), not length.

It might be confusing when sag and spring preload is involved. In my example above of a 10lb/in spring, at free length it would take 10 lbs to compress it 1 inch. If it was preloaded by 1 inch to begin with (via the shock spring adjusters) it would now take 20 lbs to compress it an additional inch. This makes the spring stiffer and allows it to support a heavier rider and reduce sag. It doesn’t change the spring rate because that is controlled by the wire diameter. When the spring is preloaded it looks shorter but it is due to the extra compression, it is in an area the has more force. Remember, the further you compress the spring, the more force you need.

As far as the weight of the spring, it is a balance between reliability, as the springs is compressed a zillion times, and weight. It is important that the fatigue stress of the wire material is not exceeded. In general less material makes it less reliable. I would just as soon accept the extra weight and keep it reliable.

I hope I didn’t make this more confusing than it already is. I did skim over some of the more technical details. Feel free to blast me
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I don’t have a riding problem...I ride, I fall down...NO PROBLEM!!
 

James Dean

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May 17, 2000
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Many dimensions go into the design of a spring. The spring "rate" is very sensitive to wire size and overall diameter.

Here is the equation:

Spring rate K= Gd**4/8nD**3

K is the stiffness (lb/in)

G is the material elastic modulus

d is the wire size

n is the number of active coils

D is the mean coil diameter

--------------------------

The wire size (d) and coil diameter (D) are by far more sensitive than the number of coils. By several orders of magnitude.

The reliability of the spring not to sack out over time or break in two may be the tradeoff here.

This equation can be used for the fork, shock, or clutch springs when making changes.

-------------------------

For the lightest possible bike we really should be looking for the fewest number of coils to reduce weight. The Engineers have left an extra pound or two on your shock spring that could be trimmed off just for the sake of reliability.

I used this equation once to drop a pound off my KX by using an RM spring. Fewer coils and a smaller wire size offset each other to make a lighter spring with the same stiffness. It fit perfectly and never had a problem.
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James Dean
 

yzguy15

Sprayin tha game
N. Texas SP
Oct 27, 2000
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James, that's a fascinating equation. It's giving me a little bit of trouble though. Do you think you could give us a little demonstration, put this equation to work? What is the "material elastic modulus"? How do you measure the wire size? In that equation what's the 4/8 mean? Thats not a fraction is it? Thanks a lot.

Also, on a side note, how did you figure this stuff out?

[This message has been edited by yzguy15 (edited 01-03-2001).]
 

KiwiBird

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Jan 30, 2000
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Most of the answers here have talked about static measurement of the spring, how does the spring length/form affect the dynamic characteristics?

What is the idea of a barrel shaped spring?

Why do some springs have one end narrower than the other?

Jer talked about (I'm about to murder his explanation here) how the spring is compressed and the load is transferred down through the coils - like a wave, imagine whacking the spring with a hammer. If you have different lengths/same rate does the harmonic resonance (don't know if that's the term) the same?
 

James Dean

Member
May 17, 2000
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Writing that equation so more people understand:

K= (G x d x d x d x d) / (8 x n x D x D x D)

The / means divide. This equation can be found in the Machinery Handbook along with examples. I find mostly theoretical equations in engineering books and many useful things like this in that fat little handbook.

The dynamic (resonance) frequencies in these springs are very high because they are relatively stiff and will be damped out by larger masses. If you are referring to the spring/bike mass natural frequency then it will be lower. Proportional to (K/M)**.5 . (The square root of K/M where M is mass.) Shock waves can travel down the spring if a sharp enough impact occurs. The frequencies will differ based on mass and stiffness. Whether it is noticeable is uncertain.

Springs can taper for a number of reasons. The springs on the WP compression valve check can be flattened to a pancake and spring back. Shock springs may be tapered for convenience of a larger shock body using an existing clevice design. (not sure) The KTM PDS springs are dual rate so part of the coils bottom on each other which reduces active coils (n) in the equation to make the higher stiffness (K).

James
 

burtlamborn

Member
Dec 21, 2000
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YZguy15,

One reason to have a spring with different diameter coils is to have a progressive rate spring. The larger diameter coils will deflect first, at a 'softer' rate. The smaller diameter coils will have a higher rate.

The 'G' term is a property of the material. It's actually the modulus of rigidity, also called the shear modulus. Not to be confused with the modulus of elasticity, E, which is a more commonly used property for engineering calculations.

E, the modulus of elasticity, also called Young's modulus, is the slope of the straight line portion of the stress/strain diagram, derived by pulling on a sample of the material.

G, the modulus of rigidity or shear modulus, is similar, but for shearing stress and strain, derived by putting the material in shear, not tension. For most materials they are related by Poisson's ratio, nu, something like

G=E/2(1+nu)

For example, for common 300-series stainless steel, E is about 29 million psi, while G is about 11 million psi.

References:

MIL-HDBK-5F - "Metallic materials and elements for aerospace vehicle structures"

"Mechanics of Materials", Higdon, et. al., 4th edition

"Mechanical Engineering Design", Shigley, 5th edition.

Of course, this is all just engineering garbage. To get my suspension working it's best, I ask my friend the fast desert racer to give me advice. There's no substitute for experience...

Burt
 

MACE

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Nov 13, 1999
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Just to stir the pot....

And being torsion dependent, we carrying a lot of dead weight not winding these springs out of hollow tubing.
 

Jeremy Wilkey

Owner, MX-Tech
Jan 28, 2000
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I love this one.. I have some things I dieing to talk about here.. I've done lots of work with this and have a question I can't find an asnswer to based on obserable documneted testing.. I'm working on a manipulation of this for a Road Race aplication but.. I don't have a solid answer about why... I'm going to let this go for a while longer then post what I find interesting and then ask my question to the mighty Mace, and JD... For insight.. I know of only 2 people in suspensnion who have ever acknolowegded what Kiwi is talking about but, I suspect that many others are silent on this.. More to come I'm sure..

Jer
 

James Dean

Member
May 17, 2000
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Burt is right, G is modulus of rigidity not the "Elastic Modulus - E".

yzguy15,

Here is an example of a shock spring calculation:
------------------
G = 11400000 psi
d = .451 in
n = 7 1/8 coils (9 1/8 coils but the ends are ground flat making them rigid 9 1/8 - 2 = 7 1/8)
D = 2.94 in

K = 11400000 x (.451)**4 / (8 x 7.125 x (2.94)**3)
K = 325 lb/in

-------------------

Fork spring:

d = .189 in
n = 46 active coils (48 - 2 = 46)
D = 1.26 in

K = 20 lb/in

----------------
( These examples were both mid-eighties Honda springs from past bikes. )

James Dean
 

KiwiBird

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Mace - hollow Titanium tubing...oooohhhh!!!
 

c3hammer

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May 20, 2000
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Getting back to the original question....

Springs will come up against stress/strain limits very quickly when the length, pitch (n) and wire (d) are reduced. All of these can be reduced for any coil spring to achieve the same rate. But.....

The strain to failure comes into play very quickly. Effectively what happens is that when the pitch / angle (n) of the coils gets steeper the spring gets stiffer, but the load is being transfered as a combination of compression and shear (twist) rather than flexure.

Most materials have lower compression / shear strength than flexural strength and thus the spring breaks too easily / soon.

I've worked with carbon fiber springs for McLaren F1. They could never achieve high enough strain to failure numbers. I just built tooling, so I'm not that good with the math though. It was pretty cool to see how they were braiding and winding the carbon over a thermoplastic rod at very specific angles so that when coiled into a spring and cured all the loads would be transfered more in either compression or flexure and not as much in shear.

The only carbon spring I've ever seen work was a torsion spring for aircraft doors. They're coiled up to preload the door so that someone could lift it open with one hand. The cross section was about .25" x .45". It was amazing how a 3.4 lb. spring could allow a huge door to be opened with one hand. I think the steel springs that are used in most of these applications weight almost 5 times as much. Carbon is really good stuff if you just flex it!

Pete
 

James Dean

Member
May 17, 2000
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How often do shock springs break on dirt bikes??

Does it happen mostly when the preload is set too firm, with frequent hard landings, or just random flaws?

If you could reduce the wire diameter by 5% the number of active coils will be reduced by 23% to achieve the same stiffness. The spring weight would drop about 20-25%. This isn't as exotic as carbon fiber, titanium, or hollow springs but every pound counts. Doesn't get much cheaper than this.
 

MACE

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Nov 13, 1999
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Air Springs.

JD, you're right. I think if you look at some of the factory SX bikes you'll see some special shock springs. Of course, they don't worry about durability beyond a few hours.

Carbon Hammer, interesting story....
 

KiwiBird

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Jan 30, 2000
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So Jer, you going to spill what ya know?
 

DENNY

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Nov 24, 1999
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I tried to stay out of this after our previous spring rate discussions but it's like loosing a tooth. Your tongue just keeps going back.

Short and simple example for the original question.

You have a 1-lb./inch spring 12 inches long and cut it in half. Now you have 2 1-lb./inch springs 6 inches long. The travel is one half and the load bearing capabilities is one half, 12 lbs. to 6 lbs.
 

marcusgunby

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Jan 9, 2000
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I seem to remember from my college days that if you cut a spring in half it increases the rate by 50%.Its load bearing would not of changed would it?
 

James Dean

Member
May 17, 2000
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When a spring is shortened to have fewer coils it gets stiffer. 1/2 as many coils is twice as stiff. 1/3 as many coils is 3 times as stiff. 2/3 as many coils is 1.5 times as stiff.

-----------
Using springs in PARALLEL like your forks allow you to just add the 2 stiffnesses for twice the stiffness.

K = k1 + k2

So you can get a rate half way between by using different spring rates in your fork.

------

Using different rate springs in SERIES like a PDS spring, follows a much different equation.

K = (k1 + k2)/(k1 x k2)

The resulting spring is softer putting them end to end. The progressive wound coils in the PDS and other dual rate springs need to bottom against each other to get to the higher second spring rate. (K = k2) This is why the coils are close together at one end.
 

DENNY

Member
Nov 24, 1999
218
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Sorry to bring this post back but I had a chance to experiment and I thought I would share my results. I put a valve spring into a spring tester and fully compressed it and the tester read 120 lbs./ half inch. Then I compressed the spring a quarter of an inch and it read 60 lbs. Then I compressed another like spring that I cut in half. At the .25 inch mark on the tester it read 53 lbs. Remember that I used 2 different springs from the same engine so the rates were a little different. I hope this clears up the controversy.
 

mxneagle

Member
Jan 7, 2001
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If you reduce the number of active coils and maintain the same spring rate you must also increase the stress in the spring wire (based on the previously noted equations) You would really need to calculate the wire stress required to achieve the desired rate and make sure that wire materials fatigue life is acceptable. I don't have my engineering books handy but all that is out there for the finding.

My guess is that for OEM springs a slightly shorter spring would not be an issue. A spring half as long would probably be a disaster.

Plus with the increased wire stress the spring would "sack out" faster.
 

osheen

Member
Feb 27, 2000
202
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Why is there not a factor in the rate equation for distance between coils? If a spring has less distance between the coils it will be softer than one with a larger space between coils as in progressive wound springs.

Another thing that I'm puzzled about is, I have two shock springs that are 5.3 Kg/mm. One is from RT and the other an OEM Yamaha (KYB). I trust that RT is acurate with their rating. The Yam spring was supposed to be a 5.2 Kg/mm. I had it tested by a reliable source and it came out to 5.3. The Yam spring was about 20mm shorter and I believe a tad smaller in dia. (the spring itself, not the wire dia.)

Here's the dilema. They were each installed on identical 99 KX250s. The one with the RT spring was definitely softer. The one with the Yam spring felt way stiff and had too much static sag (without rider) to accomplish correct sag with rider aboard.It actually only had a couple mm of spring preload. Hence the spring being to stiff. I installed a 20mm spacer on the spring to make up for it being shorter, just to make sure I didn't run out of threads. But the spacer shouldn't matter in the equation anyway, because the spring was not shortened. When I had the spring on my YZ250 it felt just slightly stiffer than the 5.0 I took off, just as it should have.

So what gives? Since the RT spring had more active coils for the same given rate would that change the characteristics dynamically?

I just love these spring questions. It sounds like we have some intellegent guys among us too.

------------------
osheen

BRC, AMA life member

'01 YZ250F '77 KD100m
'95 KDX200 '75 Road Toad
'79 KX250 '74 Z50A
 

PedroMx

Member
Jan 12, 2001
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Hi:
Since so much (and good)have been written in this post I would like to share some information that must be of help.
In order to measure the rate of a spring in a test machine it must be noted that springs usually come with ground and closed ends to seat squarely in the shock absorber, this takes out some of the linearity of the spring (if it is linear-rate) and confuses things a lot if it is not linear, therefore the spring must be compressed a certain amount (30 mm being OK for suspension springs) then taken the readout as starting point and compressed further to take the real rate of the spring.
One thing regarding progressive springs by winding them at different pitching angles is that the progressivness is achieved by collapsing the tighter wound coils (as stated), but this makes for a bumpy spring rate, as the coils contact each other. The best way to go is achieving progressivness by tapering the wire from wich the spring is made, this was the way Porsche went in its LeMans cars in the late '60s (in addition of making them of titanium wire).
Last it must be noted that a coil spring is nothing but a coiled torsion bar and thus any reasoning valid for them regarding lenght/rate/diameter is equal (and easier to visualize). F1, Indy and other types of auto racing have been using tubular torsion bars for decades now and I think that there is a matter of time since somebody makes a hollow spring, assuming they gain something doing it so.

Hope I helps and that I made myself clear.
Excuse my english.
smile.gif
 

osheen

Member
Feb 27, 2000
202
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Excuse your English? Heck, it's better than most everyone else.......

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osheen

BRC, AMA life member

'01 YZ250F '77 KD100m
'95 KDX200 '75 Road Toad
'79 KX250 '74 Z50A
 

semi old guy

Member
Dec 9, 1999
53
0
I like pedro's description. he point's out the rate change due to the end effect. as far as lighter (shorter smaller wire diameter) springs I suspect that increased loads would cause yielding or failer. Spring steel is real hard and stuff. in fact you can probably recall a few years back when springs sacked out much sooner than they do today. I've also seen them shatter into many pieces in a fork.

the shock wave suggestion is interesting
 
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