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]

“other than as an aide to guessing approximate continuous shapes for Raleigh methods.”

To elaborate on the above statement a little further:
1 – If there is only one discrete mass (and no distributed mass), then we know the mode shape exactly and can calculate the frequency exactly using continuous Raleigh as was done above at

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

(although we could solve the problem easier using sqrt(k/m) with k from static beam theory).
2 – If there are multiple discrete masses (and no distributed mass), then we know the mode shape will exactly be some linear combination of the static mode shapes. Then we could use Raleigh Ritz to construct the exact continuous mode shape and find the exact frequency. (although we could solve the problem easier by discrete/matrix methods).
3 – If there is a combination of distributed and concentrated mass, the mode shape can’t be found from 3A. But the static deflection shape might be a fair starting point for an approximate solution from Raleigh or Raleigh Ritz.

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[electricpete]

I summarize periodically for my own benefit as much as anyone else's. If something doesn't sound right, let me know.

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[WCFoiles]

(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.

I have my doubts about this. Any mode is a linear combination of the static deflection shapes which are cubic polynomials. At best this forms a 4-dimensional vector space over the real numbers. Thus, linear combinations of this could only produce up to 4 independent and still forms a cubic polynomial.

With n masses, linear combinations of the n independent mode shapes can produce arbitrary displacements at the mass locations, an n-dimensional vector space. However, a cubic polynomica cannot produce n, for n>4, arbitrary configurations.

Thus, linear combinations of the static deflections, a cubic, can not produce all the possible deflections necessarily. Likewise, linear combinations of the static deflections can not produce every mode shape. Why should the first mode be special?

This may involve solving a system with > 4 masses to see an actual modeshape. This could be very difficult to solve analytically.


Regards,

Bill

 

[electricpete]

Assuming we have have n=200 discrete masses on the massless beam, the true mode-shape will be a different cubic polynomial between each mass. At each mass the cubic polynomials are tied together with a boundary condition that y, y’ and y’’ are continuous at the mass (y’’’ is discontinuous at the mass).
The building blocks for this set of piecewise polynomials comes from the set of static deformation polynomials. Each static deformation polynomial has a different cubic polynomial on each side of associated mass. This gives us all the degrees of freedom we need to construct the full mode-shape piecewise accross the whole beam.
The strongest argument is again the physical argument. The massless beam has no memory and deflection at any instant in time is determined soley by the appliled forces at that instant in time. We know how to find the beam deflecton shape for a beam with 200 point forces... by the 4-integration method which will give us the required 201 piecewise polynomials.

The above conclusion in my mind applies to all N modeshapes. The first modeshape doesn't have any special role in conclusion 3a. It had a special role in the now-deceased conclusion 3b.

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[electricpete]

“We know how to find the beam deflecton shape for a beam with 200 point forces... by the 4-integration method which will give us the required 201 piecewise polynomials.”

I’m sure it’s obvious to Bill and Greg and the others, but for the sake of completeness I’ll clarify that the 4-integration performed with 200 point forces is equivalent to the superposition of 200 static deformation solutions which was each obtained by 4-integrations of a single point force.


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[WCFoiles]

It should have been obvious that one can get n independent deflections from force distributions (linear combinations) at the mass locations. This beam doesn't allow moments at the mass locations. However, the stiffness matrix is invertible; so, forces at the masses span n-dimensions, and these must be all possible configurations.

Linear combinations of the individual gravity sags will produce any mode shape.

Regards,

Bill

 

[electricpete]

I feel like I understand the role of static deflection a lot better now thanks to comments from Greg and Bill.

Back to the subject of the beam with both distributed and concentrated mass. Assuming my goal were to accurately find the first resonant frequency:

What would be guidelines for choosing a set of shapes for Raleigh Ritz?
Should we have a set of shapes that approximate the first mode shape or should we try to have one shape for each of the first few mode shapes?
Greg – why did you suggest the set of sinusoids for the Raleigh Ritz approach?

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[WCFoiles]

What would be wrong with just using a transfer matrix solution using a field matrix with continuous mass, a point mass element, and another continuous mass field matrix? This should derive a solution equation.

Regards,

Bill

 

[electricpete]

I will look into that. It looks like it would be a straightforward modification to my scilab transfer matrix program, once I figure out what the matrices should look like. Most of the information I have on transfer matrices uses a massless field matrix with half of the mass lumped into the point matrix on each end. I’ll dig back through it some more and look for information on continuous mass (consistent mass?) field matrices.

In the meantime, I am still interested to know a little bit about selecting functions for Raleigh Ritz. Rao gives an example but doesn’t say much about selecting the functions.


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[electricpete]

Or maybe Maple would be better to see if it can find an algebraic solution.

Is the consistent mass method “exact” (with Euler assumptions), or just better than the lumped mass method?


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[WCFoiles]

The Transfer Matrix should take the begining state to the end state of the segment, both for the discrete portion and the continuous ones.

If you do it symbolicly you will have an equation that you have to solve numerically - at least for all but the simplist problems.

Regards,

Bill

 

[GregLocock]

I suggested sinusoids as a sort of faux Fourier series analysis. If you include enough of them then obviously you can match any mode shape, and analytically they are perhaps a little more tractable than polynomials. Incidentally did you try overplotting a half sine wave and the deflection shape I worked out above? Pretty close. Of course you'll need to include cos terms for non symmetric shapes, so a polynomial may be faster.







Cheers

Greg Locock

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[electricpete]

Yes I can see the logic of that choice now.

Returning to the problem from a different angle - I think maybe the distributed mass beam with concentrated mass can be solved exactly as follows.

Rao gives the solution for the general solution for uniform mass beam as
y(x)=Asin(beta*x) + Bcos(beta*x) + Csinh(beta*x) + Dcosh(beta(x)
where argument beta =sqrt(w) * (mu/E/I) ^(1/4)

Since y(x) is only determined to within an arbitary multiplicative constant, we can arbitrarily set A=1.

Then we have 4 unknowns (B,C,D,w).

We can solve with 4 boundary conditions. Several examples such as pinned/pinned beam are solved.

I think we might be able to use the same approach for the beam with distributed mass mu and concentrated mass M by examining the interval [0,L/2]

The 4 boundary conditions:
y(0)=0 (pinned boundary on the left)
y’’(0)=0 (pinned boundary on the left)
y’(L/2) = 0 (from symmetry)
y’’’’(L/2) = E*I*w^2 * M

The last boundary condition listed is the tricky one. IF we assume for the moment that this were a massless beam, we could apply the rules of static beam deformation to arrive at that 4th equation, using the force from the beam F=w^2 * M. But if we are considering a single point on the beam only (L/2), I don’t think that the distributed mass can possibly change this relationship as it applies at a single point.

What do you guys think?

I’ll try it on Maple this weekend and see if the solution looks anything close to the solution we expect.

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[electricpete]

Whoops, those units F=w^2*M aren't right.
Maybe F=w^2*M*Y(L/2)?

I'm not sure any more. What do you guys think... can the force from that concentrated mass be captured in a boundary condition on the shape of the beam?

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[electricpete]

F=w^2*M*Y(L/2) would give
y’’’’(L/2) = E*I*w^2*M*Y(L/2) which still provides a required equation for a solution. I'm just not sure if it's gonig to be the right equation/solution.

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[electricpete]

Two corrections are in order:
1 – One mass is shared among two half-sections of beam. Therefore should use M/2 in place of M.
y’’’’(L/2) = E*I*w^2*(M/2)M*Y(L/2)
2 – I have to look carefully at whether a minus sign will be required for y’’’’(L/2), based on the sign convention used in derivation of the general form of the equation listed above.

I’m pretty sure with these corrections, the 4th equation would be valid. The justifcation:
1 – We know that if there we no mass in the beam, we could determine shape from forces applied. Additionally I don’t see how the presence of distributed mass will change this conclusion when we are looking only at a single point boundary condition at L/2.
2 – The mode shape corresponds to a snapshot when the deflection is maximum. The force applied by the mass at L/2 will also be maximum at that point in time in a direction equal/opposite of y. Since the mass force is sinusoidal in time and we have matched the maximum force with the maximum deflection, I think the dynamic nature of the force is properly captured.


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[electricpete]

Hmmm. By my previous logic the presence of distributed mass doesn’t affect y’’’’(L/2). Therefore if I remove discrete mass M (leaving distributed mass mu) I should expect I could still predict y’’’’(L/2) using applied discrete force at that location which is 0.
But I know the solution of the case where M=0 and mu remains non-zero is y(x)=sin(x*Pi/L). Differentiating 4 times we have y’’’’(x)=sin(x*Pi/L). So Y’’’’(L/2) is not 0.
This seems to contradict the approach I outlined above.


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[WCFoiles]

y’’(0)=0 (pinned boundary on the left)


Shouldn't this be y'''(0)=0?

If you are using symmetry and a boundary conditon of y'(L/2)=0, then I think you have to use M/2.



Regards,

Bill

 

[electricpete]

The y’’(0)=0 on the left side correspons to the pinned (simply-supported) assumption on left end - no moment can be generated at this simple support (moment = E*I*y’’=0)

I agree with the M/2 instead of M.

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[electricpete]

I see now that Harris’ Shock and Vib Handbook 5th ed page 7.26 has the exact solution of this problem (simply-supported beam with distributed mass plus concentrated mass in center) using four boundary conditions somewhat similar to what is described above. If I get a chance I will summarize the results here.

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