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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[electricpete]

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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[GregLocock]

Well done.

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

Cheers

Greg Locock

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[electricpete]

Rao 3rd ed p 463 starting at equation 7.7 says

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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[electricpete]

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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[WCFoiles]

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

 

[WCFoiles]

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

 

[electricpete]

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

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)

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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[WCFoiles]

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

 

[electricpete]

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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[electricpete]

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.
[ul]
[li]Assume a given m/k system. [/li]
[li]wn is a constant for that system.[/li]
[li]PE is a function of s. (PE = fp(s))[/li]
[li]KE is a function of s and w (KE = fk(s)*w^2).[/li]
[li]w_hat (estimate of wn) is a function of s[/li]
[/ul]

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).

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.

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:
[ul]
[li]The fki(s)/fpi(s) we know are <= 1 from excercizing Raleigh on the invidual systems.[/li]
[li]PEi = PEtot, so we can divide through to put in in denominator on LHS[/li]
[li]Assume resonance condition (KEtot/PE = 1) and solve for w.[/li]
[/ul]

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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[electricpete]

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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[electricpete]

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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[electricpete]

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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[electricpete]

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

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[electricpete]

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

Must be getting late.

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[electricpete]

Did I say midel? Really late.

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[WCFoiles]

- [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

 

[WCFoiles]

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

 

[electricpete]

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.

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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[electricpete]

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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