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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[svanels]

Pete in the cantilever beam, we calculate the stiffness by

k = 3*E*I/L³, E = tension type spring

For the torsional spring (normal coiled wire) we have

k = G*d⁴/(64*n*R³)

In this case:
n = number of coils
R = coil radius
d = wire diameter

About your calcualtions, I think I have to dust-off my mathcad copy, since it have been a while I put my teeth in some higher math.

 

[electricpete]

I'm thinking I messed up in assuming that the modeshape of the first mode (which is used in Raleigh's formula) will always be the same as the shape that the shape that the beam would assume under gravity (the static deflection shape)

The shape that the beam would assume under gravity can always be calculated by 4 integrations. It will be a polynomial of order at most 4 assuming concentrated masses and uniformly distributed masses.

But that is not the first shape of the pinned/pinned beam with uniformly distributed mass...it's mode shape I know from Den Hartog and others is a sinusoid.

I used the sinusoid for the above case, but the other cases I used static deflection (pinned/pinned beam with mass concentrated in the center, cantilever beam with uniform mass, cantilever beam with concentrated mass on the end).

It seems to me I saw in a book that I could use the static deflection shape but that is incorrect at least ine some cases. Are there circumstances under which the first mode shape is the same as the static deflection mode shape? Maybe that holds when there are lumped masses only?

(if this thread has become too muddled with my rambling, I may start a new thread with a more concise statement of my question).


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[WCFoiles]

Unless you use an exact mode shape there will be errors.

In the Thompson book mentioned, later he states (ex 10.5-2) that 0.5 M (if you use your example) would work. Still later Thompson, but only briefly, covers the Ritz method (Rayleigh-Ritz in the book). If you use this approach you can use another shape to approximate the mode.

Use assumed modes that meet the boundary conditions for best results. You can use two or more less accurate assumed mode shapes and arrive at a pretty good result.

 

[GregLocock]

What you could do is use Rayleigh Ritz and Fourier to build successive approximations to the mode shape, using a harmonic series of n+half-sine waves (for a symmetrical case).

As I mentioned above, using the static deflection curve as the mode shape gave much better results than a half sine wave only.



Cheers

Greg Locock

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

 

[WCFoiles]

The stiffness used, 48EI/L^3, corresponds to an assumed deflection shape equal to that when a force is applied to the center of the rotor. If one wants to use other shapes for the mass term (via Rayleigh) then one would have to re-compute the stiffness term as well through the potential energy term (virtual work, or other).

The 48EI/L^3 comes conveniently from beam equations, or one could look at the stored energy as 1/2 (48EI/L^3)*d^2 - with d as the deflection in the center. Other deflections besides the cubic that goes with a central force application may produce different potential energies.

 

[spciesla]

Pete,

"The shape that the beam would assume under gravity can always be calculated by 4 integrations. It will be a polynomial of order at most 4 assuming concentrated masses and uniformly distributed masses."

You get a 4th order polynomial because of the assumptions made in the derivation. Namely, dy/dx = tan (theta) = theta, for small values of theta.

It's like developing the differential equation for a pendulum. In order to solve it analytically, you need to substitute theta for sine(theta).

Steve

 

[electricpete]

Greg - you're right. I heard you mention mode shapes earlier and in my mind I was careful looking at the shape.... just the wrong shape. Sin(x) is right for that free/free uniform-mass beam but none of the others were right.

Steve - you're right that's an approximation inherent in many beam solutions. I don't lose much sleep over that one. Has got to be way out there in the umpteenth decimal place.

Bill. Your thought is important and it steers me towards a related question: how do we go from point A (A1,A2) to point B where:

A1 is w1=sqrt(48*E*I/[M*L^3])) for massless shaft with concentrated M in the center
A2 is w2=sqrt(48*E*I/[X*m*L^3])) where X~0.5 for beam with uniformly distributed mass.

B is w=sqrt(48*E*I/[(M + 0.5*m)*L^3])) for both distributed mass totaling m and concentrated mass M in the center.

I can think of two ways to combine A1, A2 to arrive at conclusion B and neither is exact (even if I knew some exact value of X calculated from beam alone):

The first way to combine A1 and A2 (which Bill correctly identified as my approach) is to combine them is to assume that the distributed mass beam acts like a SDOF mass system with same resonant frequency. The spring constant is assumed to be comparable to that taken from the other geoemtry (mass in middle), and calculate resulting effective mass X*m. Then if we believe the beam acts like spring with two masses on the end, we can just add the masses. Interesting Ehrich Handbook of Rotordynamics makes a similar related comparision on page 1.20. Here he compares the effect of adding series stifness to the uniform-mass beam and concludes it is the same as the effect upon frequency as adding series stiffness to the Jeffcot rotor (massless beam with concentrated mass in the middel) to within 2.5%. From my thinking, it is related in that in both cases the distributed mass beam is acting pretty much like a SDOF mass spring system.

The second way would be to apply Dunkerly's equation.
1/(w^2) <= 1/w1^2 + 1/w2^2
If we apply w1 and w2 as identified in A1 and A2, we get the correct w as identified in B.

Dunkerly's equation as far as I understand it comes close to being an equality if there are no other resonant frequencies which are close to the first (i.e. if the 2nd resonant frequency is 10x as high as the first the inequality is almost an equality).

That's the end of my discussion on the two ways to combine A1 and A2. Any comments on those or other ways to combine them?

One more thought: if I wanted to try to get a most-accurate estimate (still within the bounds of the Euler beam model), I guess the best approach (short of a numerical solution for a specific case), would be Raleigh Ritz of the entire system including both concentrated and distributed mass, rather than breaking it into two pieces. I'm sure that would not be practical by hand but maybe not too hard with Maple. Or are there other approaches... what do you guys think?

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[electricpete]

A quick qualitative look at the approximation involved in assuming sin(theta)=theta. Assume that the angle is 100 mils change over a one foot distance.

> theta:=0.1/12;
theta := 0.008333333333

> sin(theta);
0.008333236883

> (theta-sin(theta))/theta;
0.00001157400000

It seems negligible in most cases but still good to recognize what assumptions are built into our models.

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[electricpete]

Without realizing it, I used the approximation in my proof when I said theta = 0.1/12. I guess that should have been sin(theta). Then I could've used arcsin( ) to find theta. I don't think that affects the conclusion.

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[spciesla]

Actually, you should have:

tan (theta) = dy /dx = (0.1/12)

theta = arctan(0.1/12) = 0.00833314

(0.1 / 12) = 0.00833333

Like you said, not much of a difference!

 

[WCFoiles]

El Pete,

You’ve been talking about using Rayleigh’s principle. That’s all it takes to arrive at B.

Assume a shape, determine the beam potential energy to get the ‘k’. Determine the mass from the kinetic energy; this has two parts the continuous and the discrete.

Ke= 1/2*m/L (v^2)int(0 … L/2) shape dx + (1/2)v^2*M* shape(L/2)

shape(L/2) is the shape function evaluated at L/2; this gives the discrete kinetic energy due to the mass in the center, at L/2. Now, v – the velocity is a scaling factor, the time derivative of the displacement at the center (i.e. v=-omega*sin(omega t) *d).

When you scale the shape from the location of the discrete mass, you get the 1*M. The continuous part comes from the integral.

If you locate the mass elsewhere besides the middle, depending upon how you set up the shape and scaling, you can get different constants for M and m.

 

[WCFoiles]

The dy/dx in the beam equation is also an approximation resulting from the curvature formula. All-in-all it is not that big a deal.

 

[electricpete]

Going back to the subject of using "static deflection" pattern (deflection due to gravity) as a first mode shape.

I think the static deflection IS the correct first mode shape IF there is only a concentrated mass present. At any moment in time, pick a force applied to the beam by concentrated mass. The beam having no mass responds instantly to that force (using shape predicted by the static beam equations). The shape is the same as the static defleciton shape. Does anyone agree/disagree?

So I do think the deflection patterns that I used in my 21 Jun 06 23:20 post do correctly represent the deflection for the two extreme cases... one when M=0 (y=sin(pi*x/L) and one when m=0 (y=x^3/12-L^2*x/16).

Also note that using these two deflection patterns for their respective mass distributions with Raleigh's method does correctly predict the natural frequencies:
w=sqrt(48*E*I1/(mu*L^4)) for center mass only
w=sqrt(pi^4*E*I1/mu/L^4) for distributed mass only

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

Bill - I thought about it for awhile and I think I see what you are saying about Raleigh providing the means to get from A to B.

The Raleigh's formula is something like:

w^2 = [numerator]/[denominator] = [PE] / [KE/w^2]

Separate the problem into two different problems, one with the distributed mass m and one with concetnrated mass M.

Both have the same PE so both have the same numerator.

We have to add the KE's, so we "add" the denominators.

More precisely adding KE's amounts to adding the reciprocals of w^2 i.e.

Equation1: 1/w^2 = 1/w1^2 + 1/w2^2 where w1 and w2 come from analysis of the beam each with only one of the masses.

Now an interesting thing that confused me for awhile... this appears in conflict with Dunkerly's law that tells us that
Equation 2: 1/w^2 < 1/w1^2 + 1/w2^2 (different because here is an inequality vs equality above)

I think the resolution of that apparent conflict between Eq1 and Eq2 is that if we did a real solution of w1 and w2 for the different mass distributions per Dunkerly (Eq2), we would use a different deflection pattern y(x) for each one (depending on mode shape associated with that mass). But we used the same deflection pattern for both masses to come up with Equation 1. The difference between Equation 1 and 2 comes from this differnece in treatment of deflection pattern. Does anyone agree/disagree?

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Now I tried to use Raleigh Ritz to analyse the problem of shaft with distributed mass and center mass here:

http://home.houston.rr.com/electricpete/RaleighRitz2.pdf

By the way I forgot to define my symnbol mu. mu = m/L is mass per shaft lenght.

I used two different deflections: one corresponding to the uniform-density beam with no center mass, the other corresponding to the mass-less beam with center mass.

The result for the general case is obscenely complicated and doesn't give me any insight. I need to think about how to reformulate it to see if I can show something useful. (any suggestions are welcome?). At any rate it could be used to compute the frequency if exact values of M, mu, L and I are known.

At the end of the file, I did check the frequency when m=0 and when M=0 and results were as expected.

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[electricpete]

A small addition to the above Raleigh Ritz solution is here:

http://home.houston.rr.com/electricpete/RaleighRitz3a.pdf

I added a check that Raleigh Ritz selected the proper modeshape in the extreme conditions M=0 and mu=m=0. The results were as expected.

The general modeshape was:
y(x) = c1*y1(x) + c2*y2(x)
where
y1(x):=sin(Pi*x/L) is the known modeshape when M=0
y2(x):=-x^3/12+L^2*x/16 is the known modeshape when mu=0 (applies only over the range 0<x<L/2 which is the range used in these calcs).

The checks added at the end show that the solution forces c1 to 0 when mu goes to 0. This removes the center-concentrated-mass modeshape and leaves only the distributed-mass mode-shape as expected.

Also the equation for c1 blows up when M goes to 0 (can be seen by inspection of algebraic expression c1solution1 which has M in the denominator). This means we need to set C2 to 0 to keep C1 finite. C2 going to 0 removes the distributed mass mode-shape and leaves only the concentrated mass mode-shape as expected.

So in this case it seems like Raleigh Ritz did it's job of automatically selecting the right mode-shape.

Now I have one more question. If I have some general mu and M not equal to 0, Raleigh Ritz will select a mode shape. Is this guaranteed in this case to be the correct mode shape (within the usual assumptions of Euler beam model)?

It seems logical to me that we might apply some kind of superposition argument to say that any modeshape which arises from superposition of masses mu and M will be a superposition of the associated mode-shapes. Would this be a correct statements?

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[electricpete]

I had a dyslexic moment above. Please substitute correction as shown in bold:

"The checks added at the end show that the solution forces c1 to 0 when mu goes to 0. This removes the distributed-mass modeshape and leaves only the concentrated-mass mode-shape as expected.

Also the equation for c1 blows up when M goes to 0 (can be seen by inspection of algebraic expression c1solution1 which has M in the denominator). This means we need to set C2 to 0 to keep C1 finite. C2 going to 0 removes the concentrated mass mode-shape and leaves only the distributed mass mode-shape as expected."


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[electricpete]

Just to sort through the ramblings. I feel pretty good about all my conclusions above except one. My one remaining question is:

Is there any basis to conclude that the exact mode-shape for an arbitrary combination of concentrated and uniformlty-distributed mass will be a linear combinatino of the mode shapes associated with concentrated mass alone and distributed mass alone?

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[WCFoiles]

If you had a complete (in the mathematical sense) set of mode shapes then the exact mode shape would be a linear combination of these. The only problem is that this would be an infinte sum.

There is no reason to suspect that the actual mode shape if a finite combination of shapes or mode shapes from another system (i.e. wlithout the concentrated mass).

 

[electricpete]

Yes, that makes some sense.

Now another question. If my beam had only discrete masses attached (no distributed mass), then wouldn't the fundamental mode shape of the beam with all masses be a linear combination of each of the fundamental mode shapes for the beam with only one mass attached?

It seems the answer to this one should be yes. The fundamental mode shape for one mass attached is the static deflection resulting from force of the weight acting on the beam.

If we attach multiple discrete masses, we have at any given time multiple forces, each one causing an instantaneous reponse proportional to the static deflection. We can add the response to these forces by superposition. The response to the sum of the forces of all those masses should be a linear combination of the static deflections. Right?

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[WCFoiles]

This last point is somewhat philosophical. N discrete masses connected by a massless shaft. What is important about the beam in such a situation? Only the stiffness. Thus, one has only a discrete system, dynamically.

Since this is not a continuous system there would be only N modes. Thus the dimension of the solution space is N, and all deflections are linear combinations of the any N deflections, static deflections or not.

 

[electricpete]

Yes, you're right, this type of system with N discrete masses can be modeled and solved as discrete. So we could find fundamental frequency exactly by discrete Raleigh's quotient. If mode shapes are also wanted, we could find discrete mode shapes by an eigenvalue problem. Without a lot of work we could convert that discrete modeshape to the continuous modeshae which is at most a cubic in x in each segment.

Partly the purpose of my question is to confirm that a beam with one mass attached has fundamental modeshape equal to the static deflection pattern, and further if many masses attached the continuos fundamental modeshape is a linear combination of the deflection patterns of the individal masses.

I believe we could also solve the problem exactly by continuous Ritz-Raleigh. I also believe that if we chose the static deformations associated with each of the individual masses, then Ritz-Raleigh would give an exact solution. (but an arbitrary set of N linearly independent continuous mode shapes would of course not give an exact solution).

Might be a good excercize for me to try and compare both approaches someday.

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