Spring stiffness

[jtbo]

Okay, one guy from live for speed forum made little formula and I made little software, basicly it is how to calculate spring stiffness from few known variables.

I need little help now to check out how this thing goes, is results even near?

Units used are metric, but I'm sure there is some conversion calculators available from google.

I have used Carbon steel (78600) as Modulus Of Rigidity (N/mm^2), don't know if there is better value for spring steel?

I got 15.3 N/mm (same as kN/m) for GLT front springs, now do anyone have reliable data what is GLT stock spring stiffness? Imperial or metric is ok for me, just would need little info to verify my calculations.

Software is available from here:
http://www.janiervast.com/tiedostot/Spr ... _beta1.zip

Stock GLT spring as I measured is:
diameter of wire (mm) = 12
Major external diamter of spring (mm) = 127
Number of coils = 6,5

Thanks from your help

[Ronnie]

that's going to be a difficult thing to measure unless someone's got a balance big enough and strong enough to drive a wheel onto. Also it needs to be level with the other three wheels.

Given the above set up the load on one given wheel can be measured.

Then the length of the spring has to be measured with the car jacked up (without the wheel on would be better to minimise negative forces on the spring), and under the known load.

Simple maths will convert weight into force under gravity. Force = measured weight x 9.8 (acceleration of gravity)

This force will deform the spring the distance measured unloaded MINUS the distance measured loaded .

Can anyone come up with a better way to do this?

I'd give it a got, jtbo - but I don't have access to a balance!

[jtbo]

This is data from my front springs, GLT version:
Length 312mm
Coil outer diameter 127mm
Coil inner diameter 104mm
Wire width 12mm, there was black plastic surface that I guessed to be 0,2mm actual measure was 12,2mm
length from Dead coil to active coil top 53mm bottom 52mm Measured from end of wire.
Active coil to active coil 61mm
Total coils 6,5 I did measure to wire tip to wire tip (end point of wire, is that tip?)

So we don't need to calculate how much spring is compressed, idea is to have a program that would calculate stiffness of spring when you have spring on table so no need to guess how stiff that spring is.
Makes things bit more difficult?

It looks like that formula is not yet complete, there is some variables that need to be added to calculation and I'm not math genius, Tristan (guy from lfs forum) is much better with math and I do leave all math to him

If you like to know how much force there is affecting one spring when car is standing still on smooth surface, that is quite easy to calculate.

We need to know car weight, also weight distribution, then let's say we would like to know front springs, that is weight of car - % of weigh distribution to rear / 2 so there is how many kg is one front spring currently carrying.

If car weights 1100kg and weight distribution is 53% fron 47% rear, then it is (1100-47%)/2=291,5kg

One teacher told me that math is only writing things differently, I still don't quite understand it but I would like to be better with math, it makes many things so much easier to solve

[jtbo]

Got that one cracked, GLT front spring stiffness is around 30kN/m (+/- 2kN/m I think).

My calculator does calculation properly, thing is that when you count number of coils you need to substract 2 one from top and one from bottom, because those are not affecting spring stiffness. Then coil/spring diameter has to be measured from middle of wire to middle of wire and lastly spring steel is 80000 kN/mm^2

So when you go to buy Mk1 or Mk2 escort springs from junkyard, you can now measure and calculate their stiffness and compare to original GLT springs. Oh, those springs will fit, as will Mercedes 190 springs fit too if you cut them bit shorter and when you cut springs they will come stiffer too.

I have learned lot today

[SteveP]

when you cut springs they will come stiffer too.

I don't think thats quite correct, from what I've heard/read cutting springs results in a shorter spring and the same stiffness (giving problems!).

[jtbo]

At least what those math geniouses talked it would be stiffer, but I think that it is not enough stiffer so they bottom out easier.

One thing that I'm still bit worried about is that spring length is not taken account in formula, but I'm not math genious so I can't think nothing of it

This is what they told me:

A coil spring is bascially a wound up torsion bar. Removing coils is the same as shorting a torsion bar. Take a plastic rules and grip the ends. Twist -> easy to twist. Now hold one end and the middle so the effective length of the ruler is halved, and it will be harder to twist, hence the spring rate increases.

However I think that this length variable must be taken account here, so short and long springs are different stiffness if only variable changing is length. Must still ask from them.

Here is also mentioned how increasing number of coils will make spring less stiff:
http://home.earthlink.net/~bazillion/finish.html

[jtbo]

I was thinking more of this.

If we have two springs, one is 400mm long and other 800mm, both have other values same, 4 coils. Now If I cut longer one so that it is as long as shorter one I would have one spring with 4 coils and one spring with 2 coils. Because less coils is stiffer that means that cutted spring is stiffer, but I don't fully understand how this works as if spring is longer it should be less stiff with same amount of coils as shorter one if it works like torsion bar.

I'm confused

I think that thing that makes spring stiffer is that wire length is smaller and that is how torsion bar effect comes in?
Why everything that has round shapes is so hard to understand?

[Ronnie]

Imagine each coil of the spring deforms by a set amount for a set load - for example 1 cm deflection for each Vicky you pile on the bonnet

So a spring with 6 coils will give 6cm deflection if you have a nearly naked Vicky on the bonnet

A spring with 3 coils will give a 3cm deflection with the same girl

It will feel stiffer! (oops, I'm going to get into trouble now), because there is less travel even if it isn't any stronger.

Does this make sense or am I talking nonsense?

[jtbo]

Less coils is stiffer spring, that seem to be true. I don't understand that, but I pretend that I understand it

[classicswede]

Cutting coil does NOT make the spring stiffer.

Less coils for the same length and dia wire does make a spring a little stiffer but not a great deal.

[jtbo]

Cutting coil does NOT make the spring stiffer.

Less coils for the same length and dia wire does make a spring a little stiffer but not a great deal.

Many ppl seem to think otherwise, yet no one has proved evidence to direction or other. Would be nice to know this for sure.

[340GLT]

Not being funny but the ruler trick you mentioned is irrelavent to springs as it is a shear and torsion whereas springs are a compression force.
Also if you consider the complete CSA of a spring multiplyed by its length to give its overall volume. If you do it with two springs and multiply by rate then surely the rate of the cut one will be less. As stiffer springs may be shorter but the coil thickness is greater to keep the volume the same. if your with me?
Cheers Adam

[Ronnie]

Indeed - researching the internet I have found repeated claims that shorter springs are inherently stiffer. The closest thing I can find to evidence is from an empirical research study by a dentist! Regardless, a spring is a spring:

http://www.angle.org/anglonline/?reques ... &page=0027

jtbo, watch out for progressive springs - that is springs with varying spacing between the coils. This will throw your calculations out of the window!

Cheers folks

[Ronnie]

Adam - springs don't really have to deal with compression forces. The material itself is more than capable of resisting attempts to squash it. Imagine all the forces operating on a spring, now uncoil that spring into a straight bar while retaining all the force vectors, the directions which the forces are acting in... Does that make sense? When you compress a spring you are primarily trying to bend the lumen. jtbo's friend was correct in his analogy - a spring is a wound up torsion bar.

This is similar to the concept of a screw thread being an inclined plane (and thus giving mechanical advantage) - if you unravel the thread while maintaining it's shape on the Y axis, you get a simple gradient

[jtbo]

Little miniature spring, but hey, it is a spring and not even one of those evil progressive ones

1,6mm wirediameter
17mm outer coil diameter
15,4 inner coil diameter
That makes 16,2 average that is used in calculation
3mm coil to coil
3 coils

with 1kg load compression = 7,7mm

Sprint Rate (N/mm) = Applied Force(N)/Change in Length(mm)

Spring rate = 1,27 N/mm

Ok now important thing, I did cut one coil off and look:

with 1kg load compression = 4,4mm

Spring rate 2,23 N/mm !

One tiny little problem here, I don't exactly know what material this spring is made of, it won't rust and it is shiny, but I think that it is not stainless, but just with special coating or something.

As Material Modulus of Rigity = G (N/mm^2)
Carbon Steel 78,600
316 Stainless 68,900
Brass 34,500
Berylium copper 50,000

There is no match, also 19800 gives correct spring rate for case 1, but with only 2 coils it won't match anymore.
I must report this to Tristan, let's see what he thinks about it.