Dynamics of a series of springs 3

[garycvv]

Hello,
I'm wondering if someone can help me with a problem regarding the dynamics of a series of springs. If there are ~20 springs attached in series; the first spring is fixed at one end and a displacement is applied to the end spring, I would like to predict how much the first spring will extend before the next spring starts to extend. I've done some basics using first principles but I'm not sure I believe the results.
Can someone point me in the right direction?
Thanks

 

[rb1957]

springs in series ... won't they all stretch at the same time ? aren't they all subject to the same load ??

Quando Omni Flunkus Moritati

 

[cowski]

If all the springs are connected, they will all extend (or compress) at least a small amount when a load is applied.

A sketch or photo of your particular setup would really help as would knowing the spring rate(s) of the springs...

 

[Dougt115]

Provided that the springs are in series they will all move according to the force applied.

If they are identical they will act as one spring. With uniform changes in length of each spring. And the location change of each interface will be a simple matter of where it is in the chain. From the fixed end the fixed end does not move the second interface moves x, the third 2x, forth 3x and so on.

If they are different strengths, spring constants, they will each move proportional to their spring constant, the weakest spring moving the most. F=Kx Each spring absorbing 1/20 of the force applied.

If you have them in parallel they will all move the same amount just some will store more energy than the others. E=Kx^2/2 and the force in each spring is F=Kx with all experiencing the same displacement, x.

 

[drawoh]

garycvv,

If the force is applied slowly, I would expect that the springs all would compress together. You are talking about a fairly messy dynamics problem.

Have you looked at multiple-DOF vibrations?

--
JHG

 

[MintJulep]

All of the springs will extend simultaneously, with each spring extending inversely proportional to its "k"

The total displacement equals your applied displacement.

The force is the SAME on all springs.

D = F[(1/k1)+(1/k2)+(1/k3)...+(1/kn)]

 

[garycvv]

Hello,
Thanks for the responses. I've attached an image that shows the problem.
In the image I've only shown three springs.
All three springs are identical, a spring constant of 20N/mm and a mass of 10g. Spring 1 is at the left hand side and fized where shown. When the end of the third spring is displaced quickly I believe the first spring will extend before the second spring start to extend.
I think the amount of extension of spring 3 would be due to the spring constant of the spring and the mass of springs 1 and 2 together with the acceleration. (m1+m2). a = k3 . x3
So the extension of spring 3 would be the (mass of spring 1 + mass of spring 2) multimplplied by the acceleration / spring constant of spring 3.
I've created a simple FE model which gives me similar but not the same number as to what I would expect. It's out by ~10% which normally I'd be ok with but on such a simple model I'd like to get the two to agree. I've changing a few parameters such as time steps etc. which can lead to erroneous results when performing dynamic analysis but no joy.
Statically I agree when a force or a displacement is applied to spring 3 then the extension will be equal but I don't believe the extension would be equal in a dynamic case.
Thanks

 

[msquared48]

Dynamically speaking, they will not extend at the same rate with respect to time. This is getting into wave theory, for the force will propagate as a wave through the system.

Mike McCann
MMC Engineering

 

[prex]

The three springs being identical, your setup is equivalent to a single spring, and is also equivalent to the problem of axial vibrations in an elastic bar with distributed mass. A book on vibration theory will give you all the details.

prex
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[Denial]

Garycvv

You state at 15:24 that: "When the end of the third spring is displaced quickly I believe the first spring will extend before the second spring start to extend".[ ] Change that to "When the end of the third spring is displaced quickly I believe the THIRD spring will START TO extend before the second spring starts to extend".[ ] It's that tricky little "start to" that renders your approach overly simple.

Prex is on the money.[ ] This is wave-propagation problem, one that will be covered in a lot of text books. If it was my problem I would handle it using a standard FE program, proceeding as follows.
»[ ] Model it as a concatenated series of beam elements, each beam having the appropriate axial stiffness and mass per unit length. I would use an absolute minimum of two beams for each spring.
»[ ] Include in the model any additional masses that might be present.[ ] (For example, there will probably be some sort of mass at the loaded end.)
»[ ] Restrain all nodes against all displacements except axial movement.
»[ ] Perform a natural frequency analysis.
»[ ] Compare the period of the fundamental vibration mode with the length of time over which your load is applied.[ ] This will give you a feel for whether, from the structure's point of view, the load is applied quickly or slowly.
»[ ] If this suggests that the load is NOT applied slowly, then model the actual time-varying application of the load, using linear transient dynamic analysis.

 

[IRstuff]

The key difference will be determined by the period of the spring. If you do the math, the period of a single spring is 6.2 ms, so your 1 second duration motion is long compared to the spring time constant, which means the three springs will behave as a single spring. If the motion is, say, performed in 1 ms, then you would expect to see transient effects.

TTFN
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[boyze]

I've run across this dynamics problem a few times in my journeys. I've seen a preload spring go into resonance on a shaker table and using a strobe light to watch the motion. It does act as a wave. We were able to match the behavior very well be using the same modeling approach suggested by Denial. Interestingly (at least to us) the solution was to use a plastic sleeve as a friction damper on the outside of the coils.

Have Fun!

James A. Pike

http://www.xl4sim.com

http://www.erieztechnologies.com

 

[garycvv]

Hello,
Thanks for the tips. When I model the system of springs transiently in the FE programme I see that the 3rd spring extends a set length (but not the full length it could be), and then the 2nd spring extends to the same length and as the third spring, then the 1st spring extends to the same amount as the 2nd and 3rd springs.
I haven't checked fully but the velocity at which the extension passes from sping 3 to spring 2 appears to be consistant with the wave speed.
What I woudl like to clarify is why spring 3 extends a set amount not necessarily it's full allowable extension before spring 2 starts to extend. As I mentioned I thought it would be due to the spring rate of spring 3 and the mass of the remaining and at that point stationary springs. The FE programme gives me a close answer but not quite.
I'm wondering if it might unrelated to the mass of the stationary springs and might be due to the time taken for the wave to travel through spring 3.
I'll try a few more experiements and work out the fundamental frequency.
Thanks

 

[IRstuff]

I think what you're seeing is a quirk or bug in the FE program. Your time steps may be too coarse for the solvers to solve the dynamics correctly.

The speed of sound in steel is on the order of 6000 m/s, so any stress on the material should be propagating at that speed, which means that you should almost not see any propagation delay, unless you're rooting around in the microsecond time scale. Just pure inertia considerations should tell you that the displacement wave should propagate accordingly. I'm now fairly certain that what you are seeing is a consequence of your time steps being too coarse. I'm guessing that your time steps need to be on the order of 10 microseconds

TTFN
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[garycvv]

Hello,
Thank you for the replies. I believe that the speed of sound in a block of steel would be ~6000m/s but it would be a lot lower for a spring made of steel. I believe the wave speed is L x SQRT(k/m). This gives a velocity in the tens of m/s not ~6000m/s. I'm seeing the displacement of the springs move at tens of m/s. I've done a study of the time steps and I don't see any change when going to microsecond time steps.
Thanks

 

[GregLocock]

Spring surge calc

http://www.roymech.co.uk/Useful_Tables/Springs/Springs_Surge.html

Cheers

Greg Locock


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[IRstuff]

The natural frequency is not necessarily related to the speed of sound in the material. If you had a block of spring steel it would have same speed of sound, but its natural frequency would be considerably higher.

TTFN
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[Dougt115]

Natural frequency has nothing to do with the speed of sound for springs that I have ever heard of.

Your spring system will have several frequencies for you to watch out for. You will need to look out for second, third and maybe forth modes. Think of these as waves traveling down the springs much as when you play with a rope, fixed at the other end, by shaking it. The motion travels to the other end then comes back. Shake it faster two waves and so on. Time it right and the returning wave will have a max amplitude at the same time as the new wave is generated.

There are discrete frequencies at which the wave forms will add together. These are what you need to avoid. Oh, and your first mode may be less than a Hz and amplitude will not alter the resonate frequencies.

Modal analysis can find these for you.

 

[SomptingGuy]

Forget speeds of sounds in solids. The stress is mostly shear anyway (although a compression/rarefaction wave would travel around the length of the coils at the speed of sound if excited that way). Think instead "Slinky" for the longitudinal behaviour:

http://arxiv.org/pdf/1208.4629.pdf

Same physical problem, but with parameters that emphasize mass over stiffness.

- Steve

 

[tbuelna]

To make things easy, if we assume the spring(s) initial combined compressed distance is equal to the final spring(s) total travel (ie. 1000mm) after 1 second, since there are 20 identical compression springs working in series, each spring should have an initial compression of 50mm. And given a spring rate of 10N/mm, each spring would have an installed preload force of 500N. Since each spring also has a mass of just 10g (or .0066 cu.in. of steel), I would think this combination of forces, spring mass and displacement would be difficult, if not impossible, to achieve with conventional spring materials like CrSi alloy steel without yielding.