Dunkerley for systems including distributed mass? 2

[electricpete]

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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[WCFoiles]

** 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

 

[electricpete]

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

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

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.

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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[electricpete]

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

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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[electricpete]

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

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[WCFoiles]

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