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?


=====================================
Eng-tips forums: The best place on the web for engineering discussions.

 

[electricpete]

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.

=====================================
Eng-tips forums: The best place on the web for engineering discussions.

 

[271828]

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.

 

[hacksaw]

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.

 

[GregLocock]

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

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

 

[electricpete]

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?


=====================================
Eng-tips forums: The best place on the web for engineering discussions.

 

[GregLocock]

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

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

 

[electricpete]

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

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.

=====================================
Eng-tips forums: The best place on the web for engineering discussions.

 

[electricpete]

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)?


=====================================
Eng-tips forums: The best place on the web for engineering discussions.

 

[electricpete]

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."


=====================================
Eng-tips forums: The best place on the web for engineering discussions.

 

[GregLocock]

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

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

 

[electricpete]

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.





=====================================
Eng-tips forums: The best place on the web for engineering discussions.

 

[electricpete]

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 ...

=====================================
Eng-tips forums: The best place on the web for engineering discussions.

 

[electricpete]

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

=====================================
Eng-tips forums: The best place on the web for engineering discussions.

 

[GregLocock]

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

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

 

[electricpete]

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




=====================================
Eng-tips forums: The best place on the web for engineering discussions.

 

[electricpete]

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

=====================================
Eng-tips forums: The best place on the web for engineering discussions.

 

[electricpete]

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.


=====================================
Eng-tips forums: The best place on the web for engineering discussions.

 

[GregLocock]

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

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

 

[electricpete]

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?

=====================================
Eng-tips forums: The best place on the web for engineering discussions.