M+0.5*m for adding shaft mass to Jeffcott rotor 4

[electricpete]

For a massless beam with simply-supported end conditions and a mass in the center, the Euler Beam model tells us we can easily find the stiffness (force divided by centerdisplacement) is
K = 48*E*I/L^3
and the first resonant frequency therefore
w1=sqrt(48*E*I/[M*L^3]))

Many references (Harris’ Shock and Vib Handbook, Mark’s Mechanical Engineering Handbook, Rao’s Mechanical Vibration, and Ehrlich’s Rotordynamics Handbook) suggest that you can enlarge the above formulation to provide for mass in the shaft as follows:
w1=sqrt(48*E*I/[(M + 0.5*m)*L^3]))
where M is center lumped mass and m is distributed mass of the beam (excluding center lumped mass).

Questions:
1 – Does anyone seen any proof or justification for M+0.5*m?

2 – What assumptions / approximations are made to arrive at this formulation (beyond Euler/Bernoulli beam assumptions)?

Note some more related discussion in thread384-156111


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[electricpete]

I realize Raleigh Ritz will be a lot more work for the problem that can be solved discretely. The main point for me was to confirm what the lumped-mass first mode would look like. That gives us a little idea what kind of mode-shapes might come close for Raleigh Ritz when distributed mass is present.

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[GregLocock]

I don't agree that the mode shape is necessarily the same as the summation of the modeshape for the individual masses. After all, the mode shape of a shaft with a distributed mass is not the sum of the mode shapes for each discretised mass element.

Hmm. Maybe it is. I don't /think/ it is.






Cheers

Greg Locock

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[electricpete]

Thanks for chiming in Greg. Sometimes what sounds right to me is wrong and it takes someone else to show me that. If I'm going down the wrong path I'd like to know it.

The way I look at it, the massless beam has to respond instantaneously to any applied force (unlike a distributed mass beam which cannot respond instantaneously). So look at the system vibrating at resonance take a snapshot in time. Each discrete mass is applying a force to the beam, and the beam is responding (instantly) to each of those forces with a deformation proportional to the associated static deflection pattern. So the mode shape has to be a linear combination of the static deflections associated with each of the masses individually. Or maybe that's faulty logic?


One example of this would be analysis of the simple massless-beam with concentrated rotor on the center. If we use discrete analysis, stiffness k=48*E*I/L^3, and radian frequency w=sqrt(k/m) = sqrt(48*E*I/[L^3*m])

If we use Raleigh assuming the static mode shape (derived in the file below as y(x)=-x^3/12+ L^2*x/16), we get exactly the same result for frequency w, which seems to approx that the static deflection is the correct mode-shape at least in this case. Raleigh analysis of this simple case using static mode shape is here:

http://home.houston.rr.com/electricpete/RaleighJeffcott.pdf

or here if you like pop-up screens:

http://electricpete1.tripod.com/LinkToRaleigh.htm

The idea of dividing the continuous beam into small pieces is interesting. It seems like we would have superposition of an infinite number of infinitessimal static deflections? I'm not sure I can see where that leads us.

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[electricpete]

Hmmm. One brain-lapse/typo changed my whole meaning. Here is correction:

"If we use Raleigh assuming the static mode shape (derived in the file below as y(x)=-x^3/12+ L^2*x/16), we get exactly the same result for frequency w, which seems to prove that the static deflection is the correct mode-shape at least in this case. Raleigh analysis of this simple case using static mode shape is here:"


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[electricpete]

Some of the stuff under "symbols" in the most recent Raleigh pdf file is left over from an earlier analysis (Raleigh Ritz). Please ignore the section entitled "symbols"

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[GregLocock]

"The idea of dividing the continuous beam into small pieces is interesting. It seems like we would have superposition of an infinite number of infinitessimal static deflections? I'm not sure I can see where that leads us. "

Well, Leibnitz and Newton found that a handy approach!

I'll see if it makes any sense tonight.





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Greg Locock

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[electricpete]

OK, I'll think about it as well.

In the meantime, I tried the excercize again. This time I used a massless beam with two attached concentrated masses. The masses are attached a distance "a" from each end. The beam is simply supported on both ends like the previous example.

First I calculated the static deflection. Then plugged that into Raleigh's to find the first critical frequency. The result was
w^2 := [6/a^2] * [I1*E] /[(3*L-4*a)*M]

This matches the frequency given in DeSilva's "Vibration and Shock Handbook" table 34.2. He references Blevins.

This would again seem to support (but doesn't prove) the idea that the first mode shape for a massless beam with discrete masses attached is a linear combination of static deflections from individual masses.

Here are the calculations. Sorry for the popups...I'm having trouble with my normal server at houston.rr.com

http://electricpete1.tripod.com/LinkTwoMassShaft.htm

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[electricpete]

Actually, more than a linear combination of the individual static deflections. In this case it was the static deflection associated with having all masses attached. In other words the weighting factor for the static deflections is the weight of the mass.... and more simply the mode shape is the static deflection with all masses attached. I'm not sure if this part will remain true in general.

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[electricpete]

The formula for a point load a distance a from LHS of simply supported beam is given on page 876 of Rao Mechanical Vibrations 3rd ed. It is a quadratic polynomial in a which is different for x<a and x>b.

We'd have to integrate that piecewise expressoin for a from 0 to L. The problem is that the dividing point between the pieces changes for every differential element (since a is the dividing point).

I can't see any way to do it.

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[electricpete]

I forgot to say that we hope the result of that integration would be the first mode shape that we know is sin(x*Pi/L). Hard to see how a quadratic is going to integrate to a sinusoid unless there is a funky infinite power series going on there. But then again it's not a straightforward integration as discussed above.

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[electricpete]

Sorry - cubic polynomial, not quadratic.

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[electricpete]

Last correction: "The formula for a point load a distance a from LHS of simply supported beam is given on page 876 of Rao Mechanical Vibrations 3rd ed. It is a quadratic polynomial in x which is different for x<a and x>a."


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[GregLocock]

I'll have to do the beam deflection shape from first principles, Rao is 50 miles away.

Here's my thoughts from the drive home:


(1)Static bending:

superposition applies (it's a passive linear system)

The deflected shape of a uniform beam under its own weight is therefore the sum of the deflections of a massless beam due to the weight of each beam element.

(2)Modal analysis:

superposition applies (it's a passive linear system)

The mode shape of a uniform beam should be the sum of the modeshapes of a massless beam with a notional point mass applied at even intervals along its length, independently.

Now, I am comfortable with (1)(but will check it), but very dubious of (2)(but will check it)- I don't know why I'm dubious.
--------------------------------------------------------
We also have (3) electricpete's hypothesis: the (first?)(flexural?) mode shape of a (uniform?) beam with distributed and point masses is the same as the static deflection shape under the weight of the same masses.

In itself it sounds reasonable. Can I be completely evil and point out that a uniform beam supported free-free has a first flexural mode, but has no static deflection shape? That's probably too much of a nit-pick.

Anyway, I'll wander off for an hour or two of calculus or worse.
--------------------------------------------------------
Before I do:

Two point masses, two equal springs in series, grounded at one end.

Statics: deflected shape is [2,3] (by inspection)

Dynamics: first mode shape is [2,1+5^.5] (Blevins)

Good. That encourages me. The mode shape is similar to, but not identical to, the static deflection shape.

Cheers

Greg Locock

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[GregLocock]

(2) is a bust, as presented it involves adding behaviours from different frequencies, that is not allowed.

Cheers

Greg Locock

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[GregLocock]

OK, I got waylaid.

I get that the deflected shape of a pin jointed uniform beam under its own weight q per unit length (usual approximations) is

EI/q*y=L/12*x^3-1/24*x^4-L^3/24*x

The first mode shape of the same beam is given by

sin (pi*x/L)

Hopefully if you drop the first expression into a Rayleigh equation you'll end up with a higher frequency than the second, which should give you pi/2/L^2*(EI/q)^.5



Cheers

Greg Locock

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[GregLocock]

... and therefore the static deflection shape is only a good approximation to the mode shape, and is not the correct (lowest frequency) solution.

Cheers

Greg Locock

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[electricpete]

I would revise (3) as follows:
(3a) = the continuous mode shapes of a beam with point masses only (no distributed masses) are a linear combination of the static deflection shapes under the weight of each of the same masses.

(3b?) = the first continuous mode shape of a beam with point masses only (no distributed masses) is the static deflection shapes under the weight of the same masses. (?)

Just to repeat the justification for 3a: A massless beam responds instantaneously to any applied forces (whether time-varying or static). If we take a snapshot in time of the forces applied to the beam by the accelerating masses, then we know enough to construct the deflection. It is the same as the static deflection under those same forces. It is also a linear combination of the static deflection pattern of each of those masses.

3b is still open to debate. I don’t have any proof other than in one case (the two-mass shaft above) it was true.

(2) was suggested as a possible way to check (3b) taking the limit of many small masses. I don’t think the integral is doable (not for me anyway). While we don't know the result of the integral, we do know it should be the the static deflection of the continous beam which by the 4-integration method is a fourth-order polynomial. So it would not match the first mode shape which is a sin. On the surface that disproves 3b, but I’m not sure whether 3b might somehow still apply for discrete masses.

The excercize of two masses I think proves 3b wrong (I’m still thinking about it). It does not prove 3a wrong. It would be impossible to prove 3a wrong using discrete analysis because of the “linear combination” part.

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[electricpete]

I wrote: “The excercize of two masses I think proves 3b wrong...”

To clarify which excercize I’m talking about, it should read: “Greg’s excercize of two masses I think proves 3b wrong..”


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[WCFoiles]

Let’s try some logic for the discrete system.

M – mass matrix - diagonal
K – stiffness matrix
w = weight vector – diagonal of M as a column time G
xs= static deflection = K^(-1)w
x1 = 1st modeshape, discrete
n1= first natural frequency

then

(n1)^2 M x1 = K x1 – eigenvector

Assume x1=xs then

(n1)^2 xs= M^(-1)K xs=M^(-1)K K^(-1)w
Or ,
(n1)^2 xs= M^(-1)w

Since M is diagonal and w=diag(M) G

M^(-1) w =G (1 1 … 1) ^t

Thus xs = G/(n1)^2 (1 1 … 1)^t

This would say that the static deflection is the same at all locations. It could happen for a particular choice of weights, but when there is more than weight, symmetry or not, this is unlikely.

Conclusion: xs is not x1 in general.

 

[electricpete]

That’s a great analysis and takes a lot of the mystery out of the questions.. Now we have three arrows to kill item 3b:
1 – If it were true, in the limit we would expect the static deflecton of a continuous beam (which we know to be a 4-th order polynomial) to be the same as first mode-shape (which we know to be a sinusoid). So it must not be true.
2 – Greg’s 2-DOF system had a fundamental modeshape which was not the static deflection shape.
3 – Bill’s analysis.

I should point out that 3b was just an off-hand comment that arose from analysing the beam with two masses. In that case not only was the continuous mode shape a linear combination of static deflections of individual masses (3A), but it was also THE static deflection shape from the combined weight loading (3B). That anomaly is explained in Bill’s writeup where it is shown this can occur when the static deflection involves the same movement of all masses.... which holds for the case of two masses equally distant from the two ends of the simply-supported beam.

I’m pretty confident that 3A remains true, although as Bill points out it doesnt’ have a lot of usefulness... other than as an aide to guessing approximate continuous shapes for Raleigh methods.

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