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How to calculate natural freq. with a spring/mass w/constant preload?Helpful Member!(8) 

stillfeelme (Mechanical) (OP)
15 Jul 11 16:21
Hello,

I hope I am asking this in the right location.  I have a fairly simple question that I have not been able to determine.  I want to know how to calculate the natural frequency of a spring mass system that undergoes a constant preload in the spring.  This would be for a simple compression spring.  I am trying to compare FEA results to hand calculations.  I have looked in a couple different textbooks and online and I can't find anything that has this. I did a search here and I couldn't find anything as well.
Helpful Member!(2)  desertfox (Mechanical)
15 Jul 11 16:27
stillfeelme (Mechanical) (OP)
15 Jul 11 16:51
desertfox,

Thanks I searched that link and I couldn't find anything with a constant preload.  It does have all the scenarios for single degree of freedom with damping etc.
Helpful Member!  israelkk (Aerospace)
15 Jul 11 17:01
To my best knowledge the natural frequency of mass spring system depends only on the mass and the spring rate and has nothing to do with the preload which doesn't change the spring rate.
desertfox (Mechanical)
15 Jul 11 17:21
Hi Again

The formula under "free natural vibrations" is what your looking for :-
Natural frequency"   n = ω n ( 2 π )= Sqrt (k /m )/ ( 2 π ).....(units = cycles/s)
 
forget the preload its just spring stiffness and mass


desertfox
Helpful Member!(2)  Tmoose (Mechanical)
15 Jul 11 17:44
In some situations preload can change stiffness/spring rate.  
Angular contact ball bearings come to mind.

http://img.photobucket.com/albums/0803/jstanley/pre02.jpg
http://www.docstoc.com/docs/74528752/PRELOADING-BEARINGS-Barden-Precision-Ball-Bearings
stillfeelme (Mechanical) (OP)
15 Jul 11 18:16
Desertfox,

I am aware of that equation you listed as I have been using it under non preloaded conditions to match pretty close to what the FEA model says.  What I experience is when I add the preload  the natural frequency goes up even though I didn't necessarily change the stiffness of the spring.  So I am trying to figure out what is the correlation under preload to the natural frequency so that I can do a hand calculation to predict or match analysis.  
desertfox (Mechanical)
15 Jul 11 18:32

hi stillfeelme

One possibility is that spring stiffness or rates for the first and last 15% of compression are not a straight lines,so the stiffness particularly towards the maximum compression could quite easily be an increase its stiffness which might account for the descrepancy.
Another source of error might be that the calculation your doing uses the theoretical spring stiffness as opposed the the actual spring stiffness unless you measured the spring stiffness first.

desertfox
IRstuff (Aerospace)
15 Jul 11 19:45
Natural frequency supposedy varies with the inverse square root of mass

TTFN

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GregLocock (Automotive)
15 Jul 11 21:00
is your fea model linear?

Cheers

Greg Locock


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zekeman (Mechanical)
16 Jul 11 12:11
"I have looked in a couple different textbooks and online and I can't find anything that has this. I did a search here and I couldn't find anything as well. "

And you are not likely, since preload never enters the differential equation.

It doesn't matter where the mass is located in the single degree system. What matters if you quasistatically  move the mass dx, then you get an opposing force dF and

the spring constant, k, is dF/dx remains independent of the preload and hence,

wn=sqrt(k/M)

T
Helpful Member!(2)  FeX32 (Mechanical)
16 Jul 11 14:14
In a linear system (one with linear stiffness), I agree with zekeman (and a few others above). No matter what, the square root of the systems eigenvalues only depend on the mass matrix and hence will not change with load, gravity (in any direction) or anything else for that matter.  
However, in a nonlinear system (which I think Greg is hinting at) a preloaded spring will exhibit different "fundamental frequencies" dependent on a few factors. A system with a nonlinear spring in some engineers eyes doesn't even have a fundamental frequency, just a frequency that, given certain load, deflection and mass conditions, it "mostly" vibrates at.
Many real life systems are inherently nonlinear. Thus, your answer in short to wether preload can effect a fundamental frequency is yes.  

peace
Fe

Tmoose (Mechanical)
17 Jul 11 11:25
"What I experience is when I add the preload  the natural frequency goes up even though I didn't necessarily change the stiffness of the spring."

How are you preloading the real life spring(s)?  And what kind of spring(s) are they ?
stillfeelme (Mechanical) (OP)
17 Jul 11 12:23
Hello,

All thanks for the feedback what I am  doing is forcing the compression by deflection in one direction.  So I displace the spring and the displacement is fixed, then calculate the natural frequency.  The only thing I can think of is maybe the deflection is causing it to behave non linear
desertfox (Mechanical)
17 Jul 11 13:59
Hi stillfeelme

How far are you compressing the spring, can you give us spring dimensions
 
desertfox (Mechanical)
17 Jul 11 14:35
Hi again

Take a look at page 58 of this ref it shows a typical load deflection curve for a compression spring :-

http://www.scribd.com/doc/29044778/Spring-Design-Handbook

notice the change in stiffness at each end of the curve

desertfox
drawoh (Mechanical)
18 Jul 11 13:03
stillfeelme,

   What do you mean by a constant preload?

   If the thing is vibrating along the axis of the preload spring, the preload is not constant, and this is (part of) your spring element in your vibration equations.  

   If the spring acts normal to the vibration, it is part of a friction damper, and the preload would be constant.   

               JHG

IRstuff (Aerospace)
18 Jul 11 13:18
Well, one possible analogy would be a car suspension, which has relatively low "sprung" mass, but is essentially "preloaded" by the weight of the vehicle.  And, static compression of the suspension springs do change the natural frequency of the suspension.

TTFN

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btrueblood (Mechanical)
18 Jul 11 13:29
"I want to know how to calculate the natural frequency of a spring mass system that undergoes a constant preload in the spring. "

"So I displace the spring and the displacement is fixed, then calculate the natural frequency.  "

Those appear to be mutually cancelling statements.

The first statement implies you have a spring (theoretically modelled as mass-less), with an attached mass, and want to know the fundamental frequency.  Posts have been given that give the correct answer, which is that the frequency is independent of any preload.

The second statment implies that you are fixing the spring ends, to achieve a preload, thus anything attached to the spring ends cannot vibrate.

Based on the second statement - are you trying to calculate the modes of vibration of the spring itself?  These can, and do vary depending on preloading, see a good spring design book.

The only other possibility I can see is that you are calculating/modelling transverse oscillations, not axial oscillations, of the spring/mass system?  Here again, the tranverse charactersitics of the spring can vary depending on its preloaded length.

If neither of the above assumptions by me are correct, then can you kindly explain what the heck you are talking about?  Perhaps a picture would help us understand your question.
ivymike (Mechanical)
18 Jul 11 21:50
sounds like he's calculating the surge frequency.
stillfeelme (Mechanical) (OP)
18 Jul 11 22:03
All,

The easiest way to explain it would be a spring that is precompressed to fit within a cavity.  I fix the spring by the the amount it is displaced and then I am trying to get the natural frequency of this system.  I was looking for a hand calculation that gives me the natural frequency something along the lines of a suspension system but I can't find an example that fits what I am trying to do.  Again thanks for the help but I am reaching the point where I don't know if a hand calculation exists for this.
FeX32 (Mechanical)
18 Jul 11 22:06
Make sure you are doing the FEA right.
How much different is the unpreloaded fn from the preloaded one?

peace
Fe

FeX32 (Mechanical)
18 Jul 11 22:08
Remember there is a tolerance in FEA for model convergence. And make sure the elements are linear.  

peace
Fe

stillfeelme (Mechanical) (OP)
18 Jul 11 22:10
FeX32,

The displacement is about 25% of the free length.
stillfeelme (Mechanical) (OP)
18 Jul 11 22:18
FeX32,

The difference between unloaded and loaded results in a factor close to 4.  Loaded giving me a natural frequency ~4 times higher.  Elements are linear.

ivymike,

I actually thought it would be close to the surge frequency but they were far apart.  I have looked into that equation already as well.
Tmoose (Mechanical)
18 Jul 11 22:19
So the mode would be axial surge of the spring at a fixed length?

Not the "preloaded" end bouncing axially like this?
http://comp.uark.edu/~jjrencis/femur/Learning-Modules/The-Basic-Finite-Element-Equation/sdof3-7.gif
stillfeelme (Mechanical) (OP)
18 Jul 11 22:30
Tmoose,

The mode is the preloaded end bouncing like that.  I will have to take a look tomorrow.  
FeX32 (Mechanical)
18 Jul 11 22:38
Alright.

From this:

Quote:


 I fix the spring by the the amount it is displaced and then I am trying to get the natural frequency of this system.

I can gather that you have effectively fixed the spring in a cavity. Thus resulting in a system in which this cavity (or whatever else is in your design) is now involved in the eigenvalue decomposition.
Basically I think you are calculating the fundamental freq. of you design with a spring fixed in a cavity (which means the spring now cony contributes to a small amount of the stiffness).  Where as before you had just the spring contributing to the stiffness.

btw, a factor of 4 is a very large difference. It is worth finding out the reasons for this, wether it is an modeling FEA error or the such.

 

peace
Fe

electricpete (Electrical)
19 Jul 11 0:05
I think it would help to provide a simple figure.

I'm going to assume a simple case that your "spring" has distributed mass and is captured at a fixed unmovable position on both ends. Then I think your result might be plausible (axial resonant frequency increases as the restrain points are pushed closer together).  This would be vaguely analogous to the familiar situation to a rubber band held a certain distance apart.  For the rubber band, we intuitively know that if we increase the preload (non-vibrating) strain  by making the distance between rubber band support points longer, we will increase the resonant frequency of rubber band.  Likewise I think that for the axial beam, increasing the preload (non-vibrating) strain by moving the support points away from their free=non-preloaded position (either longer or shorter) will  increase the resonant frequency of the axial-vibrating beam.  
 

=====================================
(2B)+(2B)'  ?

izax1 (Mechanical)
19 Jul 11 6:24
I'm not sure what your "spring" is like. but most mechanical springs are non-linear, which means a preload is changing the spring stiffness and thus the natural frequency. A natural frequency is by definition linear and will give a distinct numerical value dependent on the preload of a "normal" mechanical spring.

btw: Do you trust your FE results for the preloaded spring? Is the preload constant? Does the results give you a value for the  natural frequency?  
electricpete (Electrical)
19 Jul 11 8:49
Sorry, I think I was wrong in my comments.    There is a math analogy between string in tension vibrating transversely and a beam vibrating axially, however tension in string does not play similar role to tension/compression in beam.  My analogy was incorrect.  (Ref "Fundamentals of Vibration" by Meitrovic, Table 8.1)

=====================================
(2B)+(2B)'  ?

izax1 (Mechanical)
19 Jul 11 9:43
This has been a looong discussion, so I might have missed the point along the way. But it is fundamental if the problem is about calculating the natural frequency of a SPRING or calculating the natural frequency of a SPRING/MASS SYSTEM. The natural frequency in preloaded string (spring) is very much dependent on preload. Just look at the natual frequency variations in the transverse vibrations of a jet engine turbine. The higher the rpm, the higher the tansverse natural frequency. Each turbine blade is preloaded by the centrifugal forces as the turbine revs up. (You might even hear it when you are onboard the airplane and sitting in the "right" position relative to the engine.) When the engine revs up, the "singing" in the engine is changing with revs. (I am not talking about the aerodynamic sound for increased aerodynamic pressure)

To try to answer the original question: Find the stiffness of the preloaded spring, and use that for the natural frequency calculation. It should not be more complicated than that.
btrueblood (Mechanical)
19 Jul 11 12:35
Unless you know what the preload initial condition actually does in your FEA model, your output is worthless.  Suggest you look at the mode shapes between the unloaded and preloaded cases.  Further suggest that you post some imagery here, you've led a lot of smart people down the garden path and into a maze.  Without a lot more definition of what you are doing and how/why (like a PICTURE), it's going to be hard for a clear message of helpfulness to be conveyed...and that is annoying to people who are going out of their way to help you.  And probably annoying to you as well.
stillfeelme (Mechanical) (OP)
19 Jul 11 14:57
All,

I was looking for a hand calculation but is not simple to find.  In my head I thought maybe someone here may have done a calculation on a spring to calculate a natural frequency if the spring had been compressed.  I was just trying to figure out the relationship between preload amount and natural frequency.
electricpete (Electrical)
19 Jul 11 15:30
Just to clarify:
Are there any masses involved in your problem other than the spring?

What are the boundary conditions?....Both ends fixed at a certain distance (i.e. 75% of the uncompressed distance) ?
 

=====================================
(2B)+(2B)'  ?

stillfeelme (Mechanical) (OP)
19 Jul 11 16:21
Electricpete,

A mass is connected and the spring is compressed at the amount you said fixed at both ends.  If you know of a way to calculate the natural frequency I am interested.  
electricpete (Electrical)
19 Jul 11 16:47
So, in addition to the spring distributed mass, there is a lumped mass attached somewhere?  Where is this lumped mass attached.... in the middle of the spring?
Attachment at an end would not seem consistent with the comment that both coil ends are fixed boundary condition.

I'm not trying to be picky. But the problem description is still not understandable.
 

=====================================
(2B)+(2B)'  ?

drawoh (Mechanical)
19 Jul 11 17:06
stillfeelme,

   Does the mass move?  We are only interested in masses that move.

   It sounds like you are trying to find the resonant frequency of the compressed spring.  In that case, the only mass that does anything is that of the spring itself.

               JHG

hydtools (Mechanical)
19 Jul 11 17:17
stillfeelme,
a sketch would help.

Ted

stillfeelme (Mechanical) (OP)
19 Jul 11 17:40
Electricpete,

I tried to make the problem simple to see if anyone would respond with a hand calculation relating preload with natural frequency.  The design is a  mass connected to a spring.  The spring is compressed to an amount.  All I am trying to figure is what is the natural frequency in this example.  In my analysis I fixed the displacement and then calculated what the natural frequency is.  So I have one spring a mass  and preload disacement
FeX32 (Mechanical)
19 Jul 11 17:51

Quote:


All I am trying to figure is what is the natural frequency in this example.  In my analysis I fixed the displacement and then calculated what the natural frequency is.  So I have one spring a mass  and preload disacement

There doesn't exist a simple formula for 99% of engineering problems. Thus, don't be disappointed if the engineers here can't give you one. However, there is a theoretical solution (approximation or otherwise) for almost all problems. (I hope you understand what I mean here) I suspect that this problem of yours does have one. It is just hard to understand how you are setting up the system.
The above quote reinforces what I said before. Likely, you are fixing the displacement which must fix the added mass thus rendering it pointless.
A sketch would say a 1000 words smile

peace
Fe

stillfeelme (Mechanical) (OP)
19 Jul 11 20:12

All,

1. The spring is preloaded and has a mass attached
2. The spring is preloaded by about 25% and is always under preload.
3. The mass has more freedom to move in the positive direction but and a shorter stroke limit in the negative direction.

Hopefully the attachment helps but I tried to keep the problem simple asking on here what is the correlation between the preload deflection and the natural frequency?  In my case there is always preload.

 
electricpete (Electrical)
19 Jul 11 22:53

Quote:

3. The mass has more freedom to move in the positive direction but and a shorter stroke limit in the negative direction.
I don't know exactly what that means, but it's clearly non-linear.  I'm not sure how you decided to withhold that until now.  As stated above, there are some smart people willing to help you, but no mindreaders here.   

=====================================
(2B)+(2B)'  ?

stillfeelme (Mechanical) (OP)
19 Jul 11 23:54
Electricpete,

The spring has displacement as I said but I am interested in knowing the correlation between preload and natural frequency.  This is why I left out some of the details because I was trying to see what is the correlation between the preload and the natural frequency before I attempted to tackle anything else.  If I can't prove the effects of preload, I will have little chance proving anything else.  That is why I didn't ask anyone to solve the problem I just wanted to know is there a correlation between spring preload and natural frequency.

Quote:

I hope I am asking this in the right location.  I have a fairly simple question that I have not been able to determine.  I want to know how to calculate the natural frequency of a spring mass system that undergoes a constant preload in the spring.  This would be for a simple compression spring.  I am trying to compare FEA results to hand calculations.  I have looked in a couple different textbooks and online and I can't find anything that has this. I did a search here and I couldn't find anything as well.




 
drawoh (Mechanical)
20 Jul 11 1:54
stillfeelme,

   I still do not understand this fixed preload.  Is the diagram I have attached a better description of what you are doing?  

   Assume the spring is nominally compressed.

               JHG

stillfeelme (Mechanical) (OP)
20 Jul 11 7:04
Drawoh,

Yes that is it.  The spring is always under precompression and is never at a free length.  The spring load is suppose to keep the mass to the right just as your picture.
stillfeelme (Mechanical) (OP)
20 Jul 11 7:16
Desertfox,

I have looked at the surge  frequency but that is not the first mode.  The first mode is much lower than the surge frequency.  
Tmoose (Mechanical)
20 Jul 11 7:50
"The first mode is much lower than the surge frequency."

Maybe some lateral bending frequency? Like a buckling column?   
desertfox (Mechanical)
20 Jul 11 8:03
Hi stillfeelme

I'm having trouble understanding what the preload as to do with the natural frequency, on that site I gave you it states "Natural frequency of a loaded spring system" I assume the word loaded spring system means the spring is preload, so I cannot see any difference between that and your sketch.

desertfox


 


 
stillfeelme (Mechanical) (OP)
20 Jul 11 8:08
Tmoose,

First mode is not a buckling or lateral frequency.  It acts in the same direction as with no preload
izax1 (Mechanical)
20 Jul 11 9:30
desertfox

The preloaded spring will change the natural frequency only if the preload will shift the spring stiffness. And for most mechanical springs that is the case. f=k/m and if k changes, f will change.

The system illustrated by drawoh is a non-linear system (in the direction of the spring), and will not have a natural frequency. (Assuming the illustration is showing the initial position) The system need to be able to move in a sinusoidal pattern to have a natural frequency.
btrueblood (Mechanical)
20 Jul 11 11:32
A coil spring won't change its stiffness much for a 25% of free length deflection.  

The answer is that the frequency for the system that causes the mass to move now depends on the amplitude of the forcing function as a percentage of the spring preload.  The lowest frequency of vibration is still sqrt(k/m), with the caveat that the amplitude of the forced motion has to be high enough to overcome the preload and cause the mass to move.  The spring might move at higher frequencies, but lower amplitudes too.  The impact boundary condition imparts the significant nonlinearity that epete points out.  A part of the problem solution is to determine the coefficient of restitution for the impact that occurs when the mass impacts the fixed boundary, in order to find its rebound velocity.  I am not at all certain that an FEA will give you anything close to a correct eigenvalue solution for the problem you have described, and would ignore its output.

See how much a simple drawing can do?
drawoh (Mechanical)
20 Jul 11 13:38
stillfeelme,

   In the model I have shown, the mass remains forced against the side wall unless the inertia force exceeds the spring preload.  There is no mass spring interaction that would make vibration analysis relevant.

   The compressed spring will have internal resonant frequencies due to its own distributed mass.  

   If the maximum acceleration due to vibration exceeds the spring force, the mass will disengage the wall and move.  There will be a series of impacts.  I do not think an equation will help you here.  You will have to solve this numerically.

   This is how a lot of optical alignment fixtures work.  You apply a preload.  The preload and the resulting friction force grossly exceed any force your fixture is likely to see, and everything remains aligned without you having to lock it.

               JHG

stillfeelme (Mechanical) (OP)
20 Jul 11 16:07
All,

Thanks for the help.  How can this  be solved numerically? I guess my model  is useless back to the drawing board
drawoh (Mechanical)
20 Jul 11 17:38
stillfeelme,

   If you can write out the acceleration of your base as a function of time, you can work out the force on your mass.  The function probably is a sine wave of some kind.  

   You can then use numerical integration to work out how your mass moves with respect to your base.  Note how, much of the time, it won't.  

               JHG

PNachtwey (Electrical)
21 Jul 11 2:21
The formula for the calculating the natural frequency of a mass on a spring was given long ago.  However, there is a difference between the natural frequency and the damped frequency of oscillation.

I wonder if stillfeelme is asking the right question because I thought the original question was answered.  It isn't clear to me if there is more.

If stillfeelme knows the mass, damping, spring constant and forcing function then the solution should be easy.  I can do it.


 

Peter Nachtwey
Delta Computer Systems
http://www.deltamotion.com

 

electricpete (Electrical)
21 Jul 11 8:01
PNachtwey - you might want to read the recent posts starting 19 Jul 11 20:12 to understand the system being discussed.

=====================================
(2B)+(2B)'  ?

Helpful Member!  flash3780 (Mechanical)
21 Jul 11 14:15
In the situation that drawoh posted, the spring displacement is a step function, and hence non-linear. If the system is non-linear, the stiffness is a function of the displacement (f = k * dx ---> k(x) = df/dx).
The natural frequencies of the system becomes a bit more difficult to determine.

| k1 |-----| k2 |
|^^^^|  m  |^^^^|
|    |_____|    |
        |
        |->x(t), f(t)

You know that k1 is the spring stiffness, but k2 is the stiffness of the wall. It goes from zero to infinity at x=0.

Of course, the fundamental equation of motion still applies, but the math gets a lot uglier.
a*d2x/dt2 + c*dx/dt + k*x = f(t)
Or, in the undamped case:
a*d2x/dt2 + k*x = f(t)
k1(x)   = const
k2(x)   = inf (x> 0)
        = 0   (x<=0)
Hence,
ktot(x) = inf (x> 0)
        = k1  (x<=0)

The way to solve this problem is to solve the differential equation (unless you can find it worked out somewhere already). The tricky thing is, most nonlinear problems don't have closed-form solutions, so a numerical solution might be all that's available.

Perhaps a continuous approximation of the stiffness function might work (a heavyside step function or the like with a very high amplitude).

I'd look into Duffing's equation, which represents an un-damped, harmonically excited, single degree of freedom system with a nonlinear spring. It will require an iterative solution.
desertfox (Mechanical)
22 Jul 11 11:21
Hi stillfeelme

Attached are two links the second one is about the natural frequency of a preloaded spring in a mechanism.
I would like to ask what vibrations if any, the device you have will under go and the reason I ask is the first natural frequency is the one you have a formula for, so if your vibrations are lower than this value there shouldn't be a problem because if preloading the spring increases the natural frequency your moving it further away from danger.





http://www.efunda.com/designstandards/springs/spring_frequency.cfm


http://docs.google.com/viewer?a=v&;q=cache:S1XHFDFldDUJ:www.dtic.mil/cgi-bin/GetTRDoc%3FLocation%3DU2%26doc%3DGetTRDoc.pdf%26AD%3DADA800851+natural+frequency+of+a+preloaded+spring&hl=en&gl=uk&pid=bl&;srcid=ADGEEShwscX8My3iAA0EcaUQzPhbfh3NuZOlEL6-dl1GcmFQshydvR5NPZmf9zjcduxI5gI_Uf66cRc_xAHryzdrzNqG7X9PdRoR651uipXJo09CNBVc5KobqiWFaIv8jQ9WFleP7nVi&amp;sig=AHIEtbRxBYgsNNkL0YdseTJDf-AmzWidGA
stillfeelme (Mechanical) (OP)
22 Jul 11 16:21
Hi desertfox,

The first link yes I have seen that link before as it appears to describe the surge frequency of the spring.  The calculated surge frequency is 20 times higher than the natural frequency of the spring mass if they behave linearly SQRT(k/m).  We are trying to keep this device away from 50Hz and multiples of 50Hz and some other higher order natural frequencies.  I have seen evidence of a possible resonance so now I am trying to predict what I have seen in the field with calculations and FEA analysis to decide the next course of action.

The second link is very interesting and is exactly what I was asking for in my original post.  My spring is always under some sort of preload and is never in the free condition.  So I was trying to see what equation would let me hand calculate the frequency if it was under preload.  Now I am not sure that equation applies 100% but it does not match my model.  To be honest I don't know if my model is comparable at the moment with the boundary conditions applied.  However the hand calculation numbers I am getting are close to 50 Hz so it is making a case.

flash3780,

That looks like something a Fourier series can solve.
drawoh (Mechanical)
22 Jul 11 17:10
stillfeelme,

   I am still having problems with your preload.

   If a mass and spring are vibrating, the system starts out with the spring deflected some distance.  If there is not a lot of damping, the mass moves such that the spring passes through its free state, and winds up deflected opposite from where it started out.  

   If the spring is always preloaded, the mass and spring are not a vibrating system.  

               JHG

PNachtwey (Electrical)
22 Jul 11 22:50
electricpete, I am wondering WHY there is still a question about if the spring is linear or non-linear.  stillfeelme should know what he entered for the spring constant in the FEA. stillfeelme should know whether the FEA spring is linear or not.  It isn't up to us to guess.

The simulation using ODEs is trivial EVEN if the spring is non-linear.  I do simulations of hydraulic systems that are similar to the example flash3780 has shown but hydraulic systems ARE non-linear.  The natural frequency changes as a function of position.

electricpete, I don't like being told to read the posts. I can read.  I know what questions to ask. I can do the simulations using multiple non-linear ODEs with little effort.  The mechanical people are not scoring points, they aren't supplying a transfer function for the system. My experience with mechanical people is that most systems are kludged and not designed. The mechanical people can never tell the control guys the system gain, damping factor, and natural frequency.  Why don't we have it in this case?

Here is a trivial example mass on a spring simulation.
http://www.deltamotion.com/peter/Mathcad/Mathcad%20-%20T0C1%20MassOnASpring-PID.pdf
In this example the spring constant is a constant.  The motion of the spring is being controlled.  This worksheet was generated for a PLC forum to show the importance of using the derivative gain.  It was done over a year ago.

All I see is a lot of guessing or speculating. stillfeelme need to supply information.  




 







 

Peter Nachtwey
Delta Computer Systems
http://www.deltamotion.com

 

FeX32 (Mechanical)
22 Jul 11 23:31
PNachtwey,

With all due respect. Relax. I am a mechanical engineer and can tell you anything there is to know about control (and much more)....no bashing. Don't generalize!

The answer to this thread was given many many posts ago by about 5 intelligent people. However, some very helpful members have stuck around and tried to help the OP conceptually.

Quote:


The mechanical people are not scoring points, they aren't supplying a transfer function for the system.

This makes me laugh. Try thinking about an analytical solution to a nonlinear DE by Laplace. Then we can talk.

The rest of your post is nonsensical.

The problem with this tread was only one thing. The OP.
I mean no disrespect by this post.

cheers
 

 

peace
Fe

alister1370 (Industrial)
23 Jul 11 14:02
Hi,
The answer is in the second order motion equation of a simple spring-mas system.

wn=2*pi*fn
fn=1/tn
a2y^2+a1y+a0=0
r1,r2
  

stillfeelme (Mechanical) (OP)
23 Jul 11 17:16
All,

Thanks for the help.  I should have given out more information on the original post.  I would never thought it would cause a war of words between people on a forum.  Last post again thanks all

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