Dear All
In order to determine equivalent spring model of rubber i performed modal analysis of a sof system(mass on rubber).i measured accelerance FRF.Then
find receptance FRF by dividing accelerance by
-omega^2.Then i tried to find spring stiffness by the formula k=re(R)/(|R|^2*(1-r^2))
where k is spring stiffness,R is receptance(displacement/force),r=frequency/natural frequency.
The receptance values i calculated are positive at frequencies higher than natural frequency and negative at the frequencies below it.Therefore the calculated spring stiffness values are negative which seems to be impractical.But the natural frequency is found clearly.
Do u know the reason for such a problem?
Do u think i have problems with the measurements?
or is it just a phase shift and not a problem?
I will be thankful if u reply as soon as possible.

You said you are using a single degree of freedom model. That means one mass m, one spring k, one damping c, none of the above varies with frequency.

Your formula k=re(R)/(|R|^2*(1-r^2)) where you have frequency dependent parameters R and r and so resulting K would vary with frequency. Is that what you intended? If so you are talking about a frequency-dependent mass-spring model.

I'm going to make an assumption you only want non-frequency-dependent mass spring model. Under that assumption, you could find k is equal to the inverse of your receptance function at low frequencies. The FRF of displacement/force (you call it receptance) should be relatively constant far below resonant speed and it's value is the inverse of k

Also, considering only the regions far above and far below resonance where effects of damping are negligible, and the transfer function is almost completely real, we do expect a sign change in the transfer function between far above resonance and far below resonance. I would expect a positive transfer function between displacement and force (receptance) far below resonance where movement is in-phase with force, and a negative transfer function far above resonance where movement is 180-degrees out of phase with force. That seems the exact opposite of what you said.

This message has been edited. Last edited by: electricpete,

I'll try to develop an expression based on k varying with frequency w (this is NOT normal mass spring system but from your post I gather maybe this material acts in a way to suspect k a function of w?). Call it k(w) to emphasize the functonal dependence on frequency. I just take the simple approach of treating it like an m/k/c system whose paramaters do not change with frequency and solving for k(w). That is correct in the sense that for any single value of w, the force equation applies (using k(w) for k) and the frequency dependence is invisible. Assuming this system does not shift any frequencies, we can add the results together by superposition of linear systems. Even though the parameter varies with frequency, I believe mathematically this would be a linear system. If it were not a linear system, we would have right to characgterize it with an FTF function (superposition would not apply).

X(w) = displacement
F(w) = Force

R(w) = X(w)/F(w) = 1/[j*w*c + k(w) - m*w^2]


Solve for k(w)

R(w) = X(w)/F(w) = 1/[j*w*c + k(w) - m*w^2]

[j*w*c + k(w) - m*w^2] = 1/R(w)

k(w) = 1/R(w) +m*w^2 -j*w*C

Let R(w) = R*exp(j*theta) where R and theta still vary with w

k(w) = 1/[R*exp(j*theta)] +m*w^2 -j*w*C

k(w) = (1/R)*exp(-j*theta) +m*w^2 -j*w*C
k(w) = (1/R)*exp(-j*theta) +m*w^2 -j*w*C
k(w) = (1/R)(cos(theta)-j*sin(theta) + m*w^2 -j*w*C
k(w) = (1/R)cos(theta) +(-1/R)j*sin(theta) + m*w^2 -j*w*C
k(w) = [(1/R)cos(theta) +m*w^2] -j*[(1/R)sin(theta) + j*w*C]

Above we had the real part in left square brackets and imaginary part in right square brackets.
k(w) must be real, so we have:
k(w) = [(1/R)cos(theta) +m*w^2]

If we wanted to allow the possibility that k(w) may be complex (perhaps we have some frequency-dependent damping captured in k(w) as well, then we have the full expression above. The magnitude is the square root of the sum of the squares of the terms in square brackets and the angle is arctan(imaginary/real).


Symbols used:
w = radian frequency
F = force
X = displacement
R = F/X
k(w) = frequency dependent stiffness
C = damping
m = mass

This message has been edited. Last edited by: electricpete,

I should make one more comment to qualify my answers. I have never done an FRF test. My input on this question is limited to trying to provide a math analysis. (Always worthwhile to double-check my math).

I'm interested to hear others' comments on the original question.

Emre - can you clarify whether you wanted a single value k (traditional SDOF) or k as a function of w (implied by your post).

Going back to the traditional k independent of frequency (I think this is the more likely scenario of what you are after). In the very low frequency far below resonance, the system is "spring-controlled" in that spring forces are much larger than damping and mass-acceleration forces. That leads to the result I said above k~X/F = 1/R in the range far below resonance w<<sqrt(k/m).

The same result can also be proven starting with above expresion k(w):

quote:

k = 1/R(w) +m*w^2 -j*w*C

At low frequencies w<<sqrt(k/m), w approaches 0, the 2nd two terms disappear and we're left with
k ~ 1/R(w->0)
where R(w->0) is the low value that R far below resonance

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Dear All;
First of all Thank u very much for your attention.
Mr Kovac i could not understand the relation between the links you gave with the problem.
Dear electricpete my motivation is to determine the spring and dashpoth system equivalent od rubber.This is an approximation of a nonlinear system to a linear one.
I performed a similar experiment done by Lin described by "Evaluation of Frequency Dependent Rubber .."
The damping is hysteritic (structural) not viscous.The formula i wrote before is also derived from this assumption.
A constant Stiffness can be calculated by peak picking method.wn=sqrt(k/M).
But i also wanted to evaluate frequency dependent stiffness.
As you said the behaviour of the real part of the receptance is the inverse of the expected one.i could not find any sensible reason for that.
i want to know whether other guys experienced similar results from FRFs.
Thank You for your time and consideration.

I would say that if you set up your response accelerometer and your hammer sensor orientation randomly without regard for orientation or signal polarity, there is a 50% chance that the receptance ratio R will behave exactly as we expected based on paper model with force and displacement defined positive in same direction (Re{R}>0 below resonance and Re{R}<0 above resonance) and a 50% chance that the ratio will behave exactly opposite, as you measured (Re{R}<0 below resonance and Re{R}>0 above resonance).

Since yours measured opposite, I assume it has to do with a reversing of the polarity of the signal leads or reversing of the orientation of the sensors. Normally that aspect of polarity is not important in single-channel measurements and I'm not familiar with how easy it is to preserve that +/- orientation information in the signal output. Maybe others know better. I would think you should just assume there has been a reversal and multiply all your data by -1. Unless you have paid close attention to the polarities or would like someone to make suggestions about preserving polarities?

I looked in Harris' "Shock and Vibration Handbook", as you mentioend,they describe a frequency dependence of the stress/strain hysteresis curve of rubber. Then they suggest a simplified/idealized model they call "linear viscoelastic" ... the rubber is modeled as two parallel branches. One branch has spring k1, the other branch is series combintation of spring k2 and damper c1. There is an equation #35.2 governing the response of that model which I can give if you're interested but I suspect you understand far more about rubber characteristics than me.

This message has been edited. Last edited by: electricpete,

Pete thank u for your attention.
I will be thankful if u give the equation.
Can u suggest any online documents,books or journals regarding to your comments about polarity and orientation.

For the model discussed above, if excited by a force F0*sin(w*t), the response is given by

x = [F0/(k1+k2)]*
[<sin(w*t)> + k2*sin(w*t)/{k1*(1+w^2*u^2)} - k2*w*u*cos(w*t)/{k1*(1+w^2*u^2)}]

where u = c*(1/k1 + 1/k2)

It strikes me that this equation is not in the exact form needed to incorporate into the sdof model.

I think one could go back to the original equivalent (two parallel branches) and add the mass and solve the basic equation of motion to give R = X/F = function(m,k1,k2,c).

where m is known and k1, k2, c are unknown.

I can find that functional form if you want... It's probably very easy but I just don't have the time right now.

Once you have the functional form, it would not be too difficult to do a curve fit in excel using the "solver" add-in in excel. Here would be the procedure: Pull off a representative set of data points (w1,R1), (w2,R2)...(wn,Rn) from your measured data. Set up cells for the known mass and the unknown k1, k2, c. Then for your values w1, w2...w1 calculate the corresponding estimates Rhat1, Rhat2,Rhat3 using the cells with m, k1, k2, c. Calculate the square of the residuals as ri^2 = (Ri-Rhati)^2. Total up all the residuals. Then run solver and using the menu box tell it to minimize the cell containing the sum of the squares of the residuals by changing the cells containing m, k1, k2, c.

On second thought, the procedure above applies to fitting a scalar. I would do a set of residuals for the magnitudes and a set of residuals for the angles, and sum all of those square residuals into the objective cell.

I can reformulate the equation and provide more details on the excel curve fitting if you want. Just ask.

I don't have any knowledge at all about how to analyse and preserver the polarity of the measurement. Maybe someone else on the board can provide comments on that?

quote:

Originally posted by electricpete:
I would say that if you set up your response accelerometer and your hammer sensor orientation randomly without regard for orientation or signal polarity, there is a 50% chance that the receptance ratio R will behave exactly as we expected based on paper model with force and displacement defined positive in same direction (Re{R}>0 below resonance and Re{R}<0 above resonance) and a 50% chance that the ratio will behave exactly opposite, as you measured (Re{R}<0 below resonance and Re{R}>0 above resonance).

Since yours measured opposite, I assume it has to do with a reversing of the polarity of the signal leads or reversing of the orientation of the sensors. Normally that aspect of polarity is not important in single-channel measurements and I'm not familiar with how easy it is to preserve that +/- orientation information in the signal output. Maybe others know better. I would think you should just assume there has been a reversal and multiply all your data by -1. Unless you have paid close attention to the polarities or would like someone to make suggestions about preserving polarities?

I will write a report about this study.Therefore if there is any document regarding to your comments below,I may use them.

quote:

Originally posted by electricpete:
I would say that if you set up your response accelerometer and your hammer sensor orientation randomly without regard for orientation or signal polarity, there is a 50% chance that the receptance ratio R will behave exactly as we expected based on paper model with force and displacement defined positive in same direction (Re{R}>0 below resonance and Re{R}<0 above resonance) and a 50% chance that the ratio will behave exactly opposite, as you measured (Re{R}<0 below resonance and Re{R}>0 above resonance).

Since yours measured opposite, I assume it has to do with a reversing of the polarity of the signal leads or reversing of the orientation of the sensors. Normally that aspect of polarity is not important in single-channel measurements and I'm not familiar with how easy it is to preserve that +/- orientation information in the signal output. Maybe others know better. I would think you should just assume there has been a reversal and multiply all your data by -1. Unless you have paid close attention to the polarities or would like someone to make suggestions about preserving polarities?

[/QUOTE]

I thought that stress-strain hysteresis also depends on the amplitude of the strain cycles just like the magnetic B-H one. In that case, shooting for a linear model may not be the best idea after all. Am I wrong?
Incidentally this is a very interesting issue when dealing with the vibrations of endwindings in electrical machine like synchronous generators. These are supported by blocks made of polymers whose stiffness as well as hysteresis (≈damping) vary with temperature (around the glass transition temperature)and may cause resonances to dangerously approach frequencies close to 2XLF that get excited with electromagnetic forces caused by ac phase currents with very serious inconveniencies.
Any comment on this?

An interesting URL about hysteresis of polymers etc.


http://scholar.lib.vt.edu/theses/available/etd-16421214...unrestricted/etd.pdf

I agree a linear model cannot represent hysteresis. To what extent hysteresis is important, I don't know (one would suspect it depends on the type of rubber and the frequencies and magnitudes).

The one piece of information I have is Harris' shock and vib handbook has an entire chapter on rubber and spends a lot of time presenting linear models but barely mentions hysteresis. One excerpt:

"VISCOELASTICITY
Rubber has elastic properties similar to those of a metallic spring and has energyabsorbing
properties like those of a viscous liquid.23 These viscoelastic properties
allow rubber to maintain a constant shape after deformation, while simultaneously
absorbing mechanical energy.The viscosity (which varies with different elastomers)
increases with reduced temperature. The elasticity follows Hooke’s law and increases
with increased strain, while the viscosity follows Newton’s law and increases
with increased strain rate. Therefore, when applying a strain, the resultant stress will increase with increasing strain rate. Springs or dashpots are frequently used to make theoretical models which illustrate the interaction of the elastic and viscous components of rubber. The springs and dashpots can be combined in series or in parallel, representing the Maxwell or Voigt elements (see Table 36.2). Rubber actually consists of an infinite number of
such models with a wide spectrum of spring constants and viscosities."

It might be the case that this book is a relic from the old days before widely available computers, when linear models were a necessity for solutions and non-linearities were glossed over because they couldn't be incorporated into solutions. Now with computer simulations, a lot more details and non-linearities can be easily handled and therefore non-linearities are of more intense interest.

pete..... you have way too much time on your hands....

Rusty, I agree!!!
Used to do this all of the time. Measure accelerance with a two channel fft (acceleration/force). Make sure your calibrations are correct and in the proper units. Double integrate the FRF and take the reciprocal. You now have a plot where the cursor readout at any frequency is the stiffness at that frequency.