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Dunkerley for systems including distributed mass?
2

Dunkerley for systems including distributed mass?

Dunkerley for systems including distributed mass?

(OP)
Dunkerely's method tells us that if we have a system of discrete springs and masses, we can combine the resonant frequencies associated with each single mass alone attached to the system of springs using:
1/w^2 = 1/w1^2 + 1/w2^2 + ....

The only proof I have seen of Dunkerley's method uses discrete masses.  But it seems that Dunkerely gives a good estimate of the natural frequency for several systems that include distributed mass  (examples below).  Does Dunkerley apply for systems that include distributed masses?  Is the resulting calculated frequencies always a lower bound as it is when only discrete masses are present?  Has anyone seen any proof or reference to support using Dunkerley on systems that included distributed masses?

Here are two examples of using Dunkerly on systems that include both discrete and distributed masses that seem to work:

===================EXAMPLE 1 ===============
Let's say we have a  a system similar to the Jeffcott rotor:  a beam of length L simply supported on both ends with distributed mass (total distributed mass m) and concentrated mas M

From S&V Handbook Chapter 1 page 1.12, we get the answer by adding 50% of the beam mass to the concentrated mass. More specifically:
w^2= 48*E*I / (L^3*<M+0.5*m > )
where the constant 0.6 is not exact... more later.

We also know the solution to the simpler problems of distributed mass alone and lumped mass alone:
w1^2 =  Pi^4 * E*I / (m * L^3)
w2^2 = 48 * E*I / (M* L^3)

Applying Dunkerley's equation:
1/w^2 >=  1/w1^2 + 1/w2^2
1/w^2 >=  (m * L^3) /  (Pi^4 * E*I ) + (M * L^3) / (48 * E*I)

1/w^2 >=  ([0.493*m + M]* L^3) / (48 * E*I)
( where we have used 0.493 = 48/Pi^4)

w^2 <=  (48 * E*I) / ([0.492767*m+ M]* L^3)

This comes out pretty close to the S&V Handbook Chapter 1 value 0.5 except 0.49... instead of 0.5.   But one thing we know about the S&V chapter 1 value is that it is not exact.  A method to solve it "exactly" (within the simple Euler beam assumptions) is given in Shock and Vib Handbook chapter 7.  That gives the results are tabulated below.  The first column is the mass ratio m/M  and the second column is that number close to 0.5 (call it X) :

m/M       X
0.001 0.4857168063
0.01  0.4857480390
0.1   0.4860378548
0.2   0.4863333819
0.5   0.4870836626
1   0.4880107197
2   0.4891832972
5   0.4907094280
10   0.4915643189
100  0.4926253580
1000 0.4927527114

It looks to me like these stay below the 0.492767=48/Pi^4  for all values, and only approach this limiting value 0.492767 as m/M gets very high   (the latter part is expected since the beam approaches the continuous case when m/M gets very high).

So the approach above resulted in a number slightly too high in the denominator which corresponds to a frequency that was slightly too low, which is what we expect for the Dunkerley method.

============= EXAMPLE 2 ========================
Beam supported from fixed/cantilevered support at left side,  beam has total distributed mass m uniformly dstributed over length of the beam and concentrated mass M at the end of the beam.
From S&V Handbook Chapter 1 page 1.12, we get the answer by adding 50% of the beam mass to the concentrated mass. More specifically:
w^2= 3*E*I / (L^3*<M+0.23*m > )
I'm not sure how exact the 0.23 number is.

Let's solve using Dunkerley

w1 = resonant frequency of the distributed mass W1 per unit length.
w1^2 = 12.4*E*I/(m*L^3)

w2 = resonant frequency of the concentrated mass W2 at position a.
w2^2 = 3*E*I/(W2*a^3)


1/w^2 = 1/w1^2 + 1/w2^2
1/w^2 = (m*L^3) / (12.4*E*I) + (M*L^3) / (3*E*I)
1/w^2 = (0.24*m +M )*L^3 / ( 3*E*I) (note we have used 0.24 = 3/12.4)

w^2 = ( 3*E*I) / (L^3*<0.24*m +M> )
This is pretty close to what comes from the S&V handbook, just a 0.24 instead of 0.23.  Again a higher number in the denominator gives a lower resonant frequency which is what we expect from Dunkerley

===============================

Based on two these two data points (two solved examples above), we might conclude that the Dunkerley approach works to give a frequency slightly below the correct frequency, even when we include distributed mass.  Is this always the case? Any proof?

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RE: Dunkerley for systems including distributed mass?

(OP)
I forgot to say in my intro that for discrete systems, Dunkerley's method gives us an estimate of the frequency which is a lower bound.

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RE: Dunkerley for systems including distributed mass?

We use Dunkerley's equation for floor vibration calcs.  Calc the beam natural frequency, the girder natural frequency, and then combine them.  This has been done for many years, but has been shown lately to be less accurate than other methods for our systems.  Still, it does a fair job at estimating the system natural frequency.

RE: Dunkerley for systems including distributed mass?

You can assume that it approximates a continuous system given enough points, but as your examples suggest you have choose the mass distribution very carefully and then hope that the series is convergent.

The only formal derivation I've seen is for discrete masses.





RE: Dunkerley for systems including distributed mass?

I have a proof of sorts for Dunkerley's method, but it is too small to write in this box.


worst case (lowest frequency) would be the sum of the weights hanging off two springs in series

the two springs would have a rate of k1*k2/(k1+k2)

w^2=(k1*k2)/(k1+k2)/(m1+m2)

invert and expand to

1/w^2=m1/k2+1/w1^2+1/w2^2+m2/k1

both of the m/k arguments must be positive

so

m1/k2+1/w1^2+1/w2^2+m2/k1 > 1/w1^2+1/w2^2

so

1/w^2>1/w1^2+1/w2^2

Howzat? (yes, I'm bored, every cpu in the joint is busy!)



Cheers

Greg Locock

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RE: Dunkerley for systems including distributed mass?

(OP)
That's a little different than the original question since it doesn't address continuous. But I'm interested to understand it, if it provides an intuitive proof-by-example of the discrete Dunkerly.

What you're saying is that a general 2DOF discrete mass system
System A:   Ground -- K1 --- M1 --- K2 --- M2
will always have resonant frequencies higher than the lower bound given by the following system:

System B:   Ground -- K1 --- - K2 --- M1 -----M2
Note that system B can also be expressed as
System B:   Ground --- Kseries ---- Mseries   
(where Kseries is series combo of K1, K2 and Mseries is series combo of M1, M2)

If that were true, I agree it shows the Dunkerly bound.  But I don't immediately see how you came to that conclusion that System B always provides a bounding low frequency for system A.

Let
wA1, wA2 = resonant frequencies of the 2DOF system system A  (wA1<wA2)
wB = sqrt (Kseries/Mseries) = resonant frequency of system B
w1 = sqrt(k1/m1),  w2=sqrt(k2/m2)     

I can see that Kseries < K2, K2  and Mseries >  M1, M2
Therefore wB < w1,   wB < w2

But the relationship between w1, w2 frequencies of the 2DOF system (wA1, wA2) is not straightforward to me.  The 2DOF system can certainly have a resonant frequency wA1 lower either w1 or w2.

So what do we know?
wB < min(w1, w2)
wA1 can be < min(w1, w2)
the relationship between wB and wA1 still not proven ?
(we could prove it with Dunkerly, but that would be circular logic).

Is the conclusion wB < wA1 intuitively obvious?  

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RE: Dunkerley for systems including distributed mass?

Um,

"worst case (lowest frequency) would be the sum of the weights hanging off two springs in series"

is so obvious I didn't think I'd have to prove it. It IS obvious. I think...

Now, can I extend that to the multiple series of SDOF systems - yes, I think so. That 'proof' is robust, so long as we are talking about spring mass systems with one degree of freedom per mass.

Can I show that it works for the generalised case of two systems with distributed mass/stiffness equations? No.

I haven't actually seen D's method written down, but it seems to me you'd have to be very careful as to how you joined the systems.

Cheers

Greg Locock

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RE: Dunkerley for systems including distributed mass?

(OP)

Quote:

is so obvious I didn't think I'd have to prove it. It IS obvious. I think...
Obvious is in the eye of the beholder.  My gut reaction is that putting the masses together at the end of the springs (system B)  maximizes the movement of the masses compared to the stretching of the springs.   Maximizing the role of the masses and minimizing the role of the springs should drive toward lower frequency from a simple sqrt(k/m).  (but that leaves me a little short because sqrt(k/m) does not apply directly to system A's resonant frequencies).  Or talking energies we are maximizing ratio the ratio of KE / PE which drives towards a lower w in a handwaving Raleigh sort of way.   But for me it is still a little bit short of obvious.  I would be interested to hear any other explanations of why wB < wA1

Quote:

I haven't actually seen D's method written down, but it seems to me you'd have to be very careful as to how you joined the systems.
My understanding of Dunkerly is that the individual sub-systems include all the springs of the original system, but only one mass at a time.

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RE: Dunkerley for systems including distributed mass?

(OP)
OK, I think I have proven to myself why system B always has lower resonant frequency than system A.  It's probably more complicated than needs to be, but here it is:

Excite both systems (A and B) at the same frequency w (somewhere below the resonant frequency of both systems) with exactly enough force to reach the same displacement of mass 2 in both systems.  Call that displacement Y2

The max kinetic energy in system B is always higher than the max KE in system A because mass 1 moves full distance Y2 in system B but smaller distance in system A.

The PEmax in system B is always smaller than in system A .  That proof is a little tricker.  But consider the simple series combination of K1 and K2 and stretched statically to a total distance Y2.  Where does the midpoint between the springs lie?  Answer: It always lies at a position to minimize the total stored potential energy of those two springs. Any other distribution of stretching spring 1 and spring 2 to total distance Y2 will have a higher PE (therefore the midpoint is pulled toward that equilibrium position of lowest PE)..  System B will find that minimum PE configuration because it has no mass in between to change that optimal distribution.  But system A has a mass exerting force at the midpoint that will change the relative stretching of springs 1 and 2 to some ratio other than the optimal ratio that minimizes PE.

So, now we know that for any w, if we excite both systems to the same distance Y2, system B will always have a higher ratio KEmax/PEmax.

If we start with w=0, the ratio KEmax/PEmax is very small <<1 for both systems.  As we slowly increase w, the ratio will increase for both systems.   At any w, the ratio is always higher for system B than for system A, so system B will reach KEmax/PEmax first (at a lower w) than system A.

Or is there a simpler way (short of intuition)?

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RE: Dunkerley for systems including distributed mass?

(OP)
Correction in bold:
"If we start with w=0, the ratio KEmax/PEmax is very small <<1 for both systems.  As we slowly increase w, the ratio will increase for both systems.   At any w, the ratio is always higher for system B than for system A, so system B will reach KEmax/PEmax=1 (resonance) first (at a lower w) than system A."

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RE: Dunkerley for systems including distributed mass?

My hand waving explanation of the (not so) obvious:

in order to get a low resonant frequency we want to maximise the mass, and minimise the spring rate, of an SDOF system. Given the components available, the way to do that is to weld the two masses together and put the springs in series.

Well now you've set me a nice little problem - can I find a pair of systems for which D's method does not give the lower bound solution?

Incidentally, if my proof

m1/k2+1/w1^2+1/w2^2+m2/k1 > 1/w1^2+1/w2^2

 is the same as the standard one, note that the inequality is actually very unequal, we are ignoring terms of the same order as the RHS. This makes the curious agreement of your earlier results even stranger.



Cheers

Greg Locock

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RE: Dunkerley for systems including distributed mass?

(OP)
I can accept now that if we have a series collection of masses and springs, we can always create the lowest resonant frequency by lumping the masses together and springs together next to each other.  Seems obvious. Just took awhile.

The traditional proof of Dunkerley establishes that:

1/w1^2 + 1/w2^2 + 1/w3^2 +... = a11/m1 + a22/m2 + a33/m3 ...

where
* aii is stiffness of the total system to a push at position of mass mi.  
* w1, w2, w3 are the resonant frequencies

The approximation
1/w1^2 ~ a11/m1 + a22/m2 + a33/m3 ...
is a good approximation when the second resonant frequency is much higher than the first resonant frequency.

I have a hard time extending that condition for a good approximation to the distributed case.



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RE: Dunkerley for systems including distributed mass?

(OP)
Whoops
1/w1^2 + 1/w2^2 + 1/w3^2 +... = a11/m1 + a22/m2 + a33/m3 ...
should have been
1/w1^2 + 1/w2^2 + 1/w3^2 +... = m1/a11 + m2/a22 + m3/a33 ...

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RE: Dunkerley for systems including distributed mass?

(OP)
1/w1^2 ~ a11/m1 + a22/m2 + a33/m3 ...
should have been
1/w1^2 ~ m1/a11 + m2/a22 + m3/a33 ...

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RE: Dunkerley for systems including distributed mass?

k1=10^6
m1=1
w1=10^3
k2=10^10
m2=10^4
w2=10^3
Dunkerley: 1/w^2>=(10^-6 +10^-6)=2*10^-6
w=700

But by inspection

k~k1
m~m2

w=sqrt(k1/m2)=10

Dunkerley is non conservative, or I have made a mistake.


Cheers

Greg Locock

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RE: Dunkerley for systems including distributed mass?

(OP)
Dunkerly works as expected. The individual systems are the systems considering one mass at a time (but both springs present).  Assume as before
Ground = k1 == m1 ==k2 ==m2

 
> w1:=evalf(k1/m1);
                          w1 := .1000000 10^7

> Kseries:=evalf((k1*k2)/(k1+k2));
                        Kseries := 999900.0100

> w2:=evalf(Kseries/m2);
                          w2 := 99.99000100

> w:=sqrt(1/(1/w1^2+1/w2^2));
                           w := 99.99000050


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RE: Dunkerley for systems including distributed mass?

(OP)
Whoops. A few math errors of my own (forgot a few sqrt).  Here is a corrected repeat that matches your w=10
> w1:=evalf(sqrt(k1/m1));
                             w1 := 1000.

> Kseries:=evalf((k1*k2)/(k1+k2));
                        Kseries := 999900.0100

> w2:=evalf(sqrt(Kseries/m2));
                          w2 := 9.999500038

> w:=sqrt(1/(1/w1^2+1/w2^2));
                           w := 9.999000145

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RE: Dunkerley for systems including distributed mass?

(OP)
Since we are discussing Dunkerley and the 2DOF system (Ground==K1==M1==K2==M2), I will mention that I found a set of parameters where Dunkerley fails miserable.

Try the values
K1 = 100;  M1 = 100;  K2 = 1; M2 = 1
(picture a large single degree of freedom machine, on which is attached a small dynamic absorber tuned near the same resonant frequency).

You can do the calculation using my spreadsheet below which calculates the actual frequencies of the 2DOF system and the Dunkerley estimate:

http://home.comcast.net/~electricpete/eng-tips/2DOF_Calc1.xls

For these particular values of the parameters,  the exact frequencies are  w1 =0.95 and w2=1.05.
Dunkerley predicts w=0.71 which is 25% low !!!

The reason that Dunkerley breaks down for this particular problem can be seen by re-examining the proof of Dunkerley.   It can be shown that
     1/w1^2 + 1/w2^2 = k11/m1 + k22/m2
where
w1 is exact first resonant frequency, w2 is exact second resonant frequency of the system,
kii is the total system static stiffness at location of mass mi
  
The approximation is
1/w1^2 ~ k11/m1 + k22/m2
The error in the above approximation is 1/w2^2
If w2>>w1, then the error 1/w2^2 is small and the approximation is good.  
If w2 is not to much higher than w1, then the error 1/w2^2 is no longer small and the approximation is bad.

In the example above with w1 =0.95 and w2=1.05, we don't meet the assumption w2>>w1 and therefore the  error 1/w2^2 is not small.

For other sets of values where w1 and w2 are not very close, Dunkerley is usually within a few % of the lower resonant frequency.

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RE: Dunkerley for systems including distributed mass?

Good, I'm glad you've found one. I did find some abstracts of papers that seemed to assume that Dunkerley gave a reasonably accurate estimate all the time, which worried me.

Cheers

Greg Locock

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RE: Dunkerley for systems including distributed mass?

(OP)
Back to the original subject of Dunkeley for distributed systems

http://home.comcast.net/~electricpete/eng-tips/DunkerleyEnergyWork.pdf

At the link above, I have provided what appears to be a proof of Dunkerley's relationship based on energy considerations for the general case discrete OR distributed systems.

There is one teensy weensy (humongous) problem.  The conclusion at the end of my proof is that the natural frequency wn of the composite system is related to the natural frequency wni of the individual systems as follows:

1/wn^2  <   Sum{  (1/wni)^2 }

In the real Dunkerley approach, it is a > sign.

Either I have a gross conceptual error in the assumptions of my proof, or a minor dyslexic inversion buried in the algebra.   I have double-checked my proof and can't find either yet.

Any suggestions?

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RE: Dunkerley for systems including distributed mass?

(OP)
I figured it out.  

The result is correct.

My error was in not recognizing the right answer.  I was looking for a lower bound for frequency wn.

At first glance, 1/wn^2  <   Sum{  (1/wni)^2 } does not look like a lower bound for frequency wn, but it is.  The RHS is an upper bound for 1/wn^2.  The inverse is a lower bound for wn^2.

The proof is correct, just ignore that last paragraph I wrote.

Presumably the error is tied to how closely the individual mode shapes match the fundamental mode shape. I think in the two examples posted at the beginning, the invdividual mode shapes match the composite mode shape well.

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RE: Dunkerley for systems including distributed mass?

Well done.

The section in Rao that discusses D. is very careful to define HOW the systems are joined.

Cheers

Greg Locock

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RE: Dunkerley for systems including distributed mass?

(OP)
Rao 3rd ed p 463 starting at equation 7.7 says

Quote:

1/w1^2 ~ 1/w1n^2 + 1/w2n^2 + 1/w3n^2 +...+ 1/wnn^2
where win = ..sqrt(kii/mi) densotes the natural frequency of a SDOF system consisting of mass mi and spring stiffness kii, i=1,2...n.

To me it seems pretty straightforward.  The requirement for the individual systems is that they each include the entire set of springs and only a portion of the masses. When you add the individual systems all together, all the masses have to be accounted for exactly once.  From the energy proof, I don't see any reason we have to restrict the individual systems to be SDOF systems.

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RE: Dunkerley for systems including distributed mass?

(OP)
Just to expand a bit on my previous comments. The individual masses are placed onto the spring system in the same location as they were placed in the composite spring system.  So some springs may combine in parallel or series or be irrelevant... just have to look at how the individual mass would act if it were the only mass on that spring system.

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RE: Dunkerley for systems including distributed mass?

Quote from the "proof".
"
All of the ratio's [fki(s)/fpi(s)] are <1 since none of these systems are moving
at their own individual natural mode shape (s<>sni for any i). Therefore we
can remove the factors [fki(s)/fpi(s)] from the RHS which will increase the
RHS and change the equality to an inequality (<) as follows:
KEtot/PE < Sum{ (w/wni)^2 }
"

Maybe I have to read this again, but it is not clear how we got here.  The statement, "All of the ratio's [fki(s)/fpi(s)] are <1 ," basically says (?) that each (and hence the sum) natural frequency squared is greater than the ratio of KE/PE for a mode shape other than the natural frequency.  We know that the reverse is true.

The natural frequency relates to an eigenvalue (of these assumed positive [semi-]definite) matrices or operators.  The first natural frequency squared, which one trys to estimate is a lower bound for this ratio (ratio > 1st natural frequency).  Similarly the ratio will be an lower bound to the highest natural frequency squared, hence the sum of the frequencies squared.

So, the inequality (or whatever) that I see looks like

w1_hat^2> KE1/PE < KEtot/PE < sum(wi^2).

I don't see how we get anywhere with this.

Regards,

Bill

RE: Dunkerley for systems including distributed mass?

It is probably easier to say this in words then to use other's notation.

The Rayleigh quotient (KE/PE) produces something larger than the fist natual frequency squared when using something other than the 1st mode shape.  So the KE/PE >= w1^2 with and assumed modeshape.  

This was said very badly in the post above.

Regards,

Bill

RE: Dunkerley for systems including distributed mass?

(OP)
Thanks for comments Bill.  I value your opinion and your time on this as in most things.

Quote:

electricpete's proof:
For any s, the Raleigh estimate of w (call that estimate w_hat)
satisfies the relationship:
KE = PE
(w_hat/wn)^2 *fk(s) = fp(s)
fk(s)/fp(s) = (wn/w_hat)^2...
fk(s)/fp(s) = {1 if s=sn;   < 1 for s< > sn...
All of the ratio's [fki(s)/fpi(s)] are <1 since none of these systems are moving
at their own individual natural mode shape (s<>sni for any i)

Quote:

wcfoiles
The statement, "All of the ratio's [fki(s)/fpi(s)] are <1 ," basically says (?) that each (and hence the sum) natural frequency squared is greater than the ratio of KE/PE for a mode shape other than the natural frequency.  We know that the reverse is true.
It is proven above.  The Raleigh estimate of natural frequency (w_hat) is higher than the actual natural frequency (wn) when we don't choose the correct modeshape (s< > si).
w_hat>wn
(w_hat/wn )^2>1
substitute (w_hat/wn)^2 = fp(s)/fk(s)
fp(s)/fk(s)>1
fk(s)/fp(s)<1

I see you don't like the terms fk(s) and fp(s).  Just to remind you of the definitions:
fp(s) = PE     (not even a new variable)
fk(s) = (w/wn)^2 * KE

A small note - your statement appeared dimensionally incompatible which I'm sure is just an oversight: ("The Rayleigh quotient (KE/PE) produces something larger than the fist natual frequency squared when using something other than the 1st mode shape.  So the KE/PE >= w1^2 with and assumed modeshape.").  

There may be other ways to word your statement  that include ratio w/wn without introducing any new variable.  I got around it the issue by defining those new variables in the way that  made the most sense to me.  Also we can easily recognize the functional dependence (w or s) when we look at the variables involved in these ratios (functions of s are fp(s), fk(s), functions of w is w^2, constant is wn for a given system).  If you just write KE, it is not immediately obvious that KE is a function of both w and s.

As far as I can tell, I have provided a correct proof.  But I certainly may be missing something and definitely interested to hear any further comments.

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RE: Dunkerley for systems including distributed mass?

The basic relationships that exist are as follows (under manipulations similar to close to where you start)

1/wn^2 >= (KE/w_hat^2)/PE = 1/w_hat^2

With wn = system 1st natural frequency
KE = max. kinetic energy evaluated at shape s and frequency w_hat
PE = max. potential energy of shape s

KE/w_hat^2 should be free of a frequency component under assumptions.

I don’t see how an algebraic manipulation of KE =sum(Kei) can alter this relationship

It doesn’t look like PE has much to do with the manipulations.

KE =sum(KEi) = sum(wni^2/wni^2*KEi) – how do we get anywhere.  Even if we divid by w_hat^2, we have terms like wni^2/w_hat^2 – How do they compare?

I don’t know a relation amound the wni and the w_hat.  The wni are natural frequencies of sub-systems.  The energy inequality holds for the assembled system modes.
----------------------


1/w1^2 ~ 1/w1n^2 + 1/w2n^2 + 1/w3n^2 +...+ 1/wnn^2
where win = ..sqrt(kii/mi) densotes the natural frequency of a SDOF system consisting of mass mi and spring stiffness kii, i=1,2...n.

I am not sure that you will get good estimates of wni with sqrt(kii/mi).  This kii is acquired by applying (if test or this is the unit displacement method) a unit displacement to mi (ith degree of freedom where mi is) and holding all the other mj degree of freedoms to 0 displacement.  For example

g rd  -- k1 –m1 – k2 – m2 – k3 -- grnd
with k1 =1 = k3 and k2 =1000 – take m1 = m2 = 1

K= [ 1001 -1000
     -1000 1001]

M= [ 1  0
     0  1]   --- makes the following more simple.


k11 = 1001 , but when looking for wn1 we want to
sqrt(k11/m1) ~= 31.64 compared to a component mode frequency^2 of approximately sqrt(0.5).

If instead we look at the displacements after applying a unit force to this system, we have the IC’s as

IC = [0.5002499...      0.49975..
      0.49975…     0.5002499 …]

1/wn1^2 ~= 0.5 for the sub-system not the assembled system  

The first natural freq^2 w1^2 ~=1

Note wn1=wn2 ~= 1.0004998 and

1/w1^2 < 1/wn1^2 + 1/wn2^2 holds  

The estimate for wni is obtained by applying a force to the ith degree of freedom not by applying a unit displacement.  This corresponds to the difference between the IC and the stiffness matrix.



Regards,

Bill

RE: Dunkerley for systems including distributed mass?

(OP)
On the second part of your message, I agree. My notation and discussion involving kii was misguided.

On the first part of your message, regarding my proof, I still don't agree.  

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RE: Dunkerley for systems including distributed mass?

(OP)
Again following up on the second part of your message, I should correct my post from 15 Sep 07 20:05 to provide a more exact quote from Rao
1/w1^2 ~ 1/w1n^2 + 1/w2n^2 + 1/w3n^2 +...+ 1/wnn^2
where win = ..sqrt(1/<aii*mi>) densotes the natural frequency of a SDOF system consisting of mass mi and the entire spring system.  aii are the diagonal elements of the inverse of the stiffness matrix (i.e. the diagonal elements of the IC matrix.)

Thanks for straightening me out on that.

I think there interpretation is still correct that if I want to calculate the natural frequency win,  I take all the masses but one out of my system and calculate the natural frequency.  

======================

Returning to my proof.   First let me review the functional dependence of the variables one more time.
  • Assume a given m/k system.  
  • wn is a constant for that system.
  • PE is a function of s.  (PE = fp(s))
  • KE is a function of s and w (KE = fk(s)*w^2).
  • w_hat (estimate of wn) is a function of s

Quote (wcfoiles):


1/wn^2 >= (KE/w_hat^2)/PE = 1/w_hat^2
With wn = system 1st natural frequency
KE = max. kinetic energy evaluated at shape s and frequency w_hat
PE = max. potential energy of shape s
What you wrote is true, but you have defined KE in a manner different than I did in my proof.   

You have effectly defined KE as a function of w_hat.   I think it is a little bit of a circular defintion.  If you make the substitution w=w_hat early as you did, you end up with nowhere to go, as you described.   

I didn't define KE that way.  i have defined KE more generally as a function of w (and s of course).

At one location in my proof (bottom of page 1 and top of page 2), I have used temporarily substituted w=w_hat for purposes of finding the ratio fk(s)/fp(s).   Once that relationship is found, it does not restrict w to remain w_hat forever (the ratio depends on s and therefore we can use that relationship for any choice of w).  In the final part of the proof where we assemble the systems, we again have expressions for KE and PE in terms of w and s.  (not in terms of w_hat).



Quote (wcfoiles):

KE/w_hat^2 should be free of a frequency component under assumptions.
It is true for your definition of KE but not for mine.  My KE is a function of w and s, and my w_hat is a function of s only, so the ratio is a function of w.

Quote:

KE =sum(KEi) = sum(wni^2/wni^2*KEi) – how do we get anywhere.
I don't know.  That's not my equation.  Here is my equation that analyses what happens when we add the systems together:

KEtot = Sum { KEi }
KEtot = Sum{ PEi * <KEi /PEi> }
KEtot = Sum{ PEi * < (w/wni)^2 *[fki(s)/fpi(s)] > }

The remainder of the proof is there for you to see, but to summarize:
  • The fki(s)/fpi(s) we know are <= 1 from excercizing Raleigh on the invidual systems.
  • PEi = PEtot, so we can divide through to put in in denominator on LHS
  • Assume resonance condition (KEtot/PE = 1) and solve for w.
I think we are getting closer to reaching resolution.  Either you will agree with me or you will identify my error.  Either one would be good. Thanks for your help.

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RE: Dunkerley for systems including distributed mass?

(OP)
I will see if I can reformulate my proof to where you were heading... eliminate fk(s) and fp(s) and describe the same thing in terms of w_hat and wi_hat.  I think it can be done.  Give me a day or so.

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RE: Dunkerley for systems including distributed mass?

(OP)
After thinking about it a little while, I don't think I can reformulate it. I'll stick with what I linked above.

I think the big difference between the proof that I posted and the proof that you attempted is where we applied Raleigh.  You applied Raleigh to the composite system. I applied Raleigh to the individual systems.  Applying Raleigh to the individual systems is where the inequality comes from in my proof.

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RE: Dunkerley for systems including distributed mass?

(OP)
OK, I think I have a proof which doesn't use those distasteful (to some) fp(s) and fk(s).

Again we are looking to prove Dunkerley. Studying a composite system and individual systme who have the same springs as the original system but only portion of it's mass.

DECLARE  that we are only considering the system under conditions of the modeshape corresponding to the true mode shape of the composite system

Raleigh tells us that for the composite system, the estimate w_hat is exact (since the mode shape is correct)

w_hat^2 = PE / [KE/w^2] = wn^2     {equation 1}

(Note that quantity [KE/w^2] is not a function of w, since the KE is directly proportional to w^2)

Raleigh tells us that for the individual systems, the estimate wi_hat is high. (since the mode shape is incorrect)

wi_hat^2 =   PE / [KEi/w^2]  > wni   

For convenience, define unknown constant Ci > 1 to convert the above into an equality
PE / [KEi/w^2]  = Ci* wni  
[KEi/w^2]  = PE / (Ci*wni)  where Ci>1    {equation 2}

(note we don't need to distinguish PE from PEi, since they are all the same)

Now, we want to find the composite wn^2 from the individual wni.  Start with equation 1 and recognize the KE is the sum of the KEi
 wn^2 = PE / [KE/w^2]  = PE /  Sum[KEi / w^2]
 wn^2 = PE /  Sum[KEi / w^2]   {equation 3}

Substitute equation 2 into equation 3
 wn^2 = PE /  Sum {PE / (Ci*wni) }
 wn^2 = 1 /  Sum {1/(Ci*wni)}
1/ wn^2 =   Sum {1/(Ci*wni)}
Since Ci > 1
1/ wn^2 >   Sum {1/wni}

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RE: Dunkerley for systems including distributed mass?

(OP)
Whoops. I got dyslexic again with my < / >.  Should be
1/wn^2  <   Sum{  (1/wni)^2 }

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RE: Dunkerley for systems including distributed mass?

(OP)
In the midel of the proof, those wni should have obviously been wni^2.

Must be getting late.

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RE: Dunkerley for systems including distributed mass?

(OP)
Did I say midel? Really late.

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RE: Dunkerley for systems including distributed mass?

- [Statement from above] The fki(s)/fpi(s) we know are <= 1 from exercising Raleigh on the invidual systems

Much seems to rely upon the above statement.  Let's examine it.

This looks suspicious.  Take the example from

g rd  -- k1 –m1 – k2 – m2 – k3 -- grnd
with k1 =1 = k3 and k2 =1 – take m1 = m2 = 1E6


K= [ 2 -1
        -1 2]


M = 1E6 [1 0
                 0 1]
2(fk1(s)+fk2(s) = 2KE/w^2 = s^t M s and is a function of the masses.  So this can be made as large as one likes independent of PE.

 2fpi(s) = s^t K s

Let s = [1 1]^t


fk1(s) = fk2(s) by symmetry
fk1(s) = ½  m1 * 1^2 = 0.5 E6 or as large as one likes by changing the system

PE = 1 and

fki(s)/PE can be quite large = 0.5E6 >>1

Similarly, the ratio can be quite small.

Regards,

Bill

RE: Dunkerley for systems including distributed mass?


It is an interesting idea to break KE/w^2 into a sum.

    PE / [KEi/w^2]  = Ci* wni  
    [KEi/w^2]  = PE / (Ci*wni)  where Ci>1    {equation 2}

It is possible that some of the KEi/w^2 = 0 because of the mode shape for wn, i.e. nodes

In such cases where one needs to divide by KE, one divides by 0.  (If one allows dividing by zero one can create ‘proofs’ that 2=1.)

Even if KE is positive definite for the system, not all the KEi’s need to be >0 for all deflections – That is degrees of freedom don’t have to contribute to kinetic energy for each deflection, even though any permissible deflection (dynamic deflection, vibration) has positive kinetic energy.

Regards,

Bill

RE: Dunkerley for systems including distributed mass?

(OP)

Quote:

2(fk1(s)+fk2(s) = 2KE/w^2 = s^t M s and is a function of the masses.  So this can be made as large as one likes independent of PE.
You have left out a factor of wni^2 (i=1,2)  For each of the individual systems we have

KEi = (w/win)^2  *fki(s)
2*KE = (w/w1n)^2  *fk1(s) + (w/w2n)^2  *fk2(s)
2*KE/w^2 = (1/w1n)^2  *fk1(s) + (1/w2n)^2  *fk2(s)

We can increase KE without bound as we increase the masses, but this decreases the win and so (1/win)^2 also increases without bound to match.  We can't draw direct conclusion about fki(s), other than Raleigh tell us it is less than 1 if we don't have modeshpae s=sni.


Quote:

PE = 1 and...
I'm not sure exactly where you were headed.  But it is true we could force fp(s) lower without bounds  in a  variety of ways that suggest on the surface that the ratio goes above 1.  But if we decrease the displacements, we also decreate the KE (affects numerator of the ratio fk(s)/fp(s).).   Likewise, if we decrease PE by decreasing the spring stiffness for a given displacement, it has the effect of lowering wn which results in the same effect as increasing mass that we already discussed above.

** A very important point: ****
My statements about the ratio fk(s)fp(s) are equivalent to Raleigh. Here are the statements:
fki(s)/fpi(s) = 1 for s = sni
fki(s)/fpi(s) < 1 for s < >  sni

Proof:
For s=sn
w_hat^2 = PE / [KE/w^2] = wn^2   
PE / [KE/w^2] = wn^2   {equation 11}
Plug in definitions of fk(s) and fp(s) into equation 11:
fp(s)  / [{(w/wn)^2  *fk(s)}/w^2] = wn^2
Cancel the w on numerator and denominator LHS:
fp(s)  / [(1/wn)^2  *fk(s)] = wn^2
Cancel out wn^2 on LHS and RHS:
fp(s)  / fk(s) = 1
Invert:
fk(s)/fp(s) = 1


For s < > sn
wi_hat^2 =   PE / [KEi/w^2]  > wni  
PE / [KEi/w^2]  > wni  {equation 12}
Plug in definitions of fki(s) and fpi(s) into equation 12:
fpi(s)  / [{(w/wni)^2  *fki(s)}/w^2] > wni^2
Cancel the w on numerator and denominator LHS:
fpi(s)  / [(1/wni)^2  *fki(s)] > wni^2
Cancel out wni^2 on LHS and RHS:
fpi(s)  / fki(s) > 1
Invert
fki(s)/fpi(s) < 1
Do you agree that my statements about the ratio fki(s)/fpi(s) is equivalent to Raleigh?

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RE: Dunkerley for systems including distributed mass?

(OP)

Quote:

It is an interesting idea to break KE/w^2 into a sum.

    PE / [KEi/w^2]  = Ci* wni  
    [KEi/w^2]  = PE / (Ci*wni)  where Ci>1    {equation 2}

It is possible that some of the KEi/w^2 = 0 because of the mode shape for wn, i.e. nodes

In such cases where one needs to divide by KE, one divides by 0.  (If one allows dividing by zero one can create ‘proofs’ that 2=1.)

Regardless of what came before equation 2, equation 2 is used to support the final conclusion of the proof.  In the special case that you mention where the ith mass is a discrete mass at a node of the composite system, then KE=0 and equation 2 remains true (it corresponds to a choice  C=Infinity which does not violate our assumption Ci>1).  The inequality equation 2 remains true as stated and the conclusion remains true.

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RE: Dunkerley for systems including distributed mass?

** A very important point: ****
My statements about the ratio fk(s)fp(s) are equivalent to Raleigh. Here are the statements:
fki(s)/fpi(s) = 1 for s = sni
fki(s)/fpi(s) < 1 for s < >  sni
-------
Do I have your definitions wrong or is

fki(s)/fpi(s) = 1/wni^2 for s = sni  
and

fki(s)/fpi(s) =< 1/wni^2 for s < >  sni  = can’t do better than <= for this inequality (repeated roots are a possibility)
----

Later,

fp(s)  / [{(w/wn)^2  *fk(s)}/w^2] = wn^2  

Let’s go back to PE and KE by definitions

fk(s) =KE *w^2   fps =PE gives

PE/((w/wn)^2 KE) =  ?? wn^2  or since the KE is evaluated at sn and wn does it not =

                    = (wn/w)^2 because PE/KE = 1, evaluated at said points


I can’t follow this all the way from start to finish and get the desired inequality.  Since the inequality is true (Dunkerly), it is impossible to show that the result is incorrect.  I am still unconvinced that I have seen a valid proof of this from the KE and PE starting point.  

Regards,

Bill

RE: Dunkerley for systems including distributed mass?

(OP)

Quote:

Do I have your definitions wrong or is

fki(s)/fpi(s) = 1/wni^2 for s = sni  
and

fki(s)/fpi(s) =< 1/wni^2 for s < >  sni  = can’t do better than <= for this inequality (repeated roots are a possibility)
Yes, you have my definitions wrong.
See top of page 2:
fk(s)/fp(s) = 1 for s=sni
fk(s)/fp(s) < 1 for s< > sni

Actually those aren't definitions. The definitions inferred from page 1 are
fp(s) = PE
fk(s) = KE*(wn/w)^2


Quote:

Later,
fp(s)  / [{(w/wn)^2  *fk(s)}/w^2] = wn^2
fp(s)  / [{(w/wn)^2  *fk(s)}/w^2] = wn^2
divide each side by wn^2 and cancel w^2/w^2 on LHS:
fp(s)  / fk(s) = 1
Same as on the top of page 2

Quote:

Let’s go back to PE and KE by definitions

fk(s) =KE *w^2
That's not my definition.  My definition is
fk(s) = KE*(wn/w)^2

The proof seems fairly straightforward to me once the functional dependence is demonstrated and the newly defined variable definitions are noted.  In fact the proof proceeds in algebraic form.

Perhaps I will edit it again and make sure the variables are very clearly defined and the step-by-step algebraic manipulations are clear.

Quote:

Since the inequality is true (Dunkerly), it is impossible to show that the result is incorrect.
imo, if we ignore my proof, the conclusion that Dunkerley applies to distributed systems is not proven anywhere else.   At the beginning of the thread it was asked for proofs of Dunkerley for distributed masses, and no-one had any.  I have seen mentioned in a textbook where they said it only applies to discrete (I'll see if I can dig that up.

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RE: Dunkerley for systems including distributed mass?

(OP)
Below is an excerpt which leads me to believe Dunkerley might be restricted to discrete systems (along with the fact that all the proofs seem to only address discrete systems):

"Applied Structural and Mechanical Vibrations: Theory, methods and measuring instrumentation"
Paolo L.Gatti and Vittorio Ferrari
ISBN 0-203-01455-3; ISBN 0-203-13764-7;  ISBN 0-419-22710-5
Copyright 2003 Taylor and Francis

Chapter 9

Quote:


In the light of the fact that—unless the assumed shape coincides with the true eigenshape—the Rayleigh method always leads to an overestimate of the first eigenvalue, Section 9.2.2 considers Dunkerley’s formula which, in turn, always leads to an underestimate of the first eigenvalue. Although its use is generally limited to positive definite systems with lumped masses, Dunkerley’s formula can also be useful when we need to verify that the fundamental frequency of a given system is higher than a given prescribed value. The Rayleigh and Rayleigh-Ritz methods apply equally well to both discrete and continuous systems,[/b] and so does the assumed modes method, which is closely related to the Rayleigh-Ritz method but uses a set of time dependent generalized coordinates in conjunction with Lagrange equations....

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RE: Dunkerley for systems including distributed mass?

(OP)
But to reiterate, my proof is not restricted to discrete systems... can apply equally to continuous systems.

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RE: Dunkerley for systems including distributed mass?

OK - I see the light, maybe with a slight nuance.

1/wn^2>=[ KE/w^2]/PE = [SUM(KEi)/w^2]/PE <= SUM(1/wni^2) Holds.  Why?  PE may be too large for the individual sub-systems, because it involves other degrees of freedom, possibly (and K is positive definite or at lest non-negative for a mechanical system - this may not be completely trivial to show, but it seems intuitive.).  KEi is correct for a diagnonal mass matrix (lumped mass model).

The right side inequality holds for all deflections, in particular sn.  sn makes the equality on the left side, which leads to the simple inequality.  Even if KEi = 0 for some i, KEi/PE <= wni (PE may be the wrong PEi, but it errs on the large side, because PE has more allowable deflections than PEi, possibly.).

Regards,

Bill

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