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

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

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

(OP)
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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RE: M+0.5*m for adding shaft mass to Jeffcott rotor

2
Well, Blevins 8-8 says 0.49 rather than 0.5, he says it was derived using Rayleigh (good idea). One obvious comment I'd make is that the deflected shape of the beam will be distorted compared with the case when M=0.

of course when M=0 an exact solution is given,
f1=pi/(2*L^2)*sqrt(EIL/m)

hence w1 = pi^2/L^2*sqrt (EIL/m)
=sqrt(EI .5*pi^4 /(L^3*m*.5))

pi^4=97.4, so that's fairly close to 48*2







Cheers

Greg Locock

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

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

(OP)
If the solution works good where M=0 and the mode shape is sinusoidal (caused only by distributed mass), then one would think it would not be as accurate during a more typical case where the mode shape is most heavily influenced by the center mass (third-order polynomial).

With a change in ratio m/M we can surely cause a change in deflection shape and with varying m/M it can't be "exact" (within the bounds of the model) for all m/M.  Maybe the the deflection shapes are so close there's not more than 1 or 2% error regardless of m/M?

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

(OP)
I gotta getmea Blevins wonadeez days.

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

70 bucks from amazon. Used to be 200 secondhand.

In our one lecture on rayleigh the point that was most strongly emphasised was that best results relied on using a 'natural' deformed shape for the structure. I think off-hand we worked through the effects of assuming a half sine, and  a parabola, instead of the proper deflected beam shape, even for an unusual mass loading, the usual beam equation was best.



Cheers

Greg Locock

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

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

(OP)
$70 is still a little steep but I'll keep it on my wishlist at Alibris to see if a cheap one pops up (sometimes people do us a favor like rip a few pages which cuts the price in half).

I used Maple to look at the effect of different assumptions for deflection (distributed mass deflection pattern or concentrated mass deflection pattern) upon the distributed mass critical frequency. I found the associated coefficient  X~0.5 varies about 1% between these two extremes of deflection assumptions.

==============Symbols=============
w  = radial resonant frequency of distributed mass beam
eqw = equation to calculate w from Raleigh's law using symmetry about x=L/2.
wdist = calculated value of w assuming distributed mass defection pattern (y(x)=sin(Pi*x/L)

wconc = calculated value of w assuming concentrated mass (in center) defection pattern (y(x)=x^3/12-L^2*x/16) - derived from beam theory.

eqwdist = eqw with distributed-mass deflection pattern
eqconc = eqw with concentrated-mass deflection pattern

==============Calcuations=============
> eqw:=w=sqrt(E*I1/mu*int(diff(y(x),x,x)^2,x=0..L/2)/int(y(x)^2,x=0..L/2)):
>
> eqwdist:=subs(y(x)=sin(Pi*x/L),eqw):
> wdist:=simplify(rhs(eqwdist));
       wdist := Pi^2*(E*I1/mu/L^4)^(1/2)


> X:=evalf((sqrt(48*E*I1/(mu*L^4)))/wdist)^2;
                           X := .4927671479
         (Above is value of X assuming distributed deflection pattern)

> eqwconst:=subs(y(x)=x^3/12-L^2*x/16,eqw):
> wconst:=simplify(rhs(eqwconst));
      wconst := 4/17*105^(1/2)*17^(1/2)*(E*I1/mu/L^4)^(1/2)
      
> X:=evalf((sqrt(48*E*I1/(mu*L^4)))/wconst)^2;
                           X := .4857142865
        (Above is value of X assuming concentrated deflection pattern)
$70 is still a little steep but I'll keep it on my wishlist at Alibris to see if a cheap one pops up (sometimes people do us a favor like rip a few pages which cuts the price in half).

I used Maple to look at the effect of different deflections and found not much difference in the coefficient X~0.5

Symbols
w  = radial resonant frequency
eqw = equation to calculate w from Raleigh's law using symmetry about x=L/2.
wdist = calculated value of w assuming distributed mass defection patter (y(x)=sin(Pi*x/L)

wconc = calculated value of w assuming concentrated mass (in center) defection pattern (y(x)=x^3/12-L^2*x/16) - derived from beam theory.

> eqw:=w=sqrt(E*I1/mu*int(diff(y(x),x,x)^2,x=0..L/2)/int(y(x)^2,x=0..L/2)):
>
> eqwdist:=subs(y(x)=sin(Pi*x/L),eqw):
> wdist:=simplify(rhs(eqwdist));
       wdist := Pi^2*(E*I1/mu/L^4)^(1/2)


> X:=evalf((sqrt(48*E*I1/(mu*L^4)))/wdist)^2;
                           X := .4927671479
         (Above is value of X assuming distributed deflection pattern)

> eqwconst:=subs(y(x)=x^3/12-L^2*x/16,eqw):
> wconst:=simplify(rhs(eqwconst));
      wconst := 4/17*105^(1/2)*17^(1/2)*(E*I1/mu/L^4)^(1/2)
      
> X:=evalf((sqrt(48*E*I1/(mu*L^4)))/wconst)^2;
                           X := .4857142865
        (Above is value of X assuming concentrated deflection pattern)

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

(OP)
Hmm. I guess I wrote the same thing twice. You can skip the second one.

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

(OP)
I should have defined X.  X is the unknown to satisfy
w = sqrt(48*E*I/[X*m*L^3])

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

1*pi^4=2^5*3

I'm still admiring the neatness of that equation that uses all 5 of the first integers (cheating slightly) and ends up with pi.

So, the factor is 0.49 for all practical purposes. Excellent.

Cheers

Greg Locock

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

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

and yes I know it isn't a very good approximation!

Cheers

Greg Locock

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

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

(OP)
One double-check on the result.  Using the distributed-beam deflection pattern, we get the higher value of X which corresponds to the lower value of w.

If we were trying to find w for distributed mass beam by trying various deflections, we would expect the lowest w corresponds to the correct inertial-loading deflection pattern, which is what we found.

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

There is a similar equation for a mass attached to a spring, the effective mass me = m + 1/3*mspring

Theory of vibration with applications by William Thomson has an example applied to a cam with follower and valve as commonly found in engines.

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

Interesting, some philosophy must be used to find this magic number.  If one does have cubic deflections, one can use the cubic shape functions to find the consistent mass matrix for a FE solution.  The consistent mass matrix for a beam element would be (up to sign conventions) as follows.

mass matrix=  (m/420)  [156   22L   54   -13L
                        22L  4L^2   13L  -3L^2
                         54    13L   156  -22L
                        -13L  -3L^2 -22L  4L^2]


with the usual displacements at the nodes [x1 theta1 x2 theta2]^t .

Since one does not consider the kinetic energy (virtual work term in FE derivation) from other displacements, should you just set them to 0?  If so, this should reduce to (in the virtual work equation) m(156+54)/420 = m/2.  Really, the m would be for each half of the supporting beam - 2 elements, but still gives m/2.

Or, one could consider it as ElPete does.


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

(OP)
Hi Bill!  Welcome to eng-tips. Hope you will stick around awhile and share your wisdom like you do on www.maintenanceforums.com

I see your matrix matches equation 14.3 of "Vibration Simulation using Matlab and Ansys" by Hatch.  There is some discussion there about the consistent mass matrix being introduced to model distributed mass beam element.  It's a little bit above my head.  I'll study it a little more if I have a chance. (unless you want to explain it some more... )

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

My copy of Thompson's "Theory of Vibration With Applications" (4th Edition) develops this equation using Rayleigh's method (Example 2.4-2, page 25)

The kinetic energy is defined as 0.5 * Meff * (x')^2.  In this case, he uses y' instead of x' because of the vertical deflection.

To get y', he uses the deflection equation of a simply supported beam with a concentrated load at the middle of the span.  He provides an equation for the deflection of the first half of the span as y = ymax * f(x).  

He integrates over the length of the beam to get the maximum kinetic energy.  With a wave of the hands, a little smoke and mirrors, the effective mass is calculated to be M + .4857m.

Steve

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

Steve, I think I have a translation of the 1st edition, which don't have the beam example, but he uses a spring, integrating to calculate the maximum kinetic energy results in 1/3* mspring

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

Svanels,

My copy also has the spring example (Example 2.4-1, pages 24-25).  The one thing that I don't understand in the integral for kinetic energy is that the mass of the spring mspring is divided by its length, l.

Steve

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

Steve he assumes that the mass varies linear with the length of the spring, which for a coiled uniform wire would be reasonable assumption.

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

(OP)
The spring example sounds like longitudinal beam vibration.  That sounds a little trickier.

Back to the case of lateral vibration, if there is a cantilevered shaft with a lumped mass on the end, we express the natural frequency as
w = SQRT(3*E*I / ((Mlumped+0.23*Mshaft)*L^3)).

Solving for the coefficient X=0.23 using similar approach as above, I got
X = 0.236 applying Raleigh's method with deflection pattern corresopnding to concentrated mass only.
X = 0.241 applying Raleigh's method with deflection pattern corresopnding to distributed mass only.

Since the real deflection is influenced both by concentrated (lumped) mass at the end as well as distributed shaft mass, I assume actual deflection would be somewhere between the two, I assume X lies somewhere between those two numbers.

Here are the details of those calcualtions:
http://home.houston.rr.com/electricpete/CantileverShaftMassEffect.pdf
(please let me know if that links doesn't work)

Steve - did you notice M + .4857m is the same as I calculated using distributed mass defelction pattern in my post above (21 Jun 06 23:20)?  No smoke and mirrors there I hope.

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

(OP)
Correction,  X=0.4857 was calculated using the concentrated mass deflection pattern.

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

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.
thumbsup




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

(OP)
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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RE: M+0.5*m for adding shaft mass to Jeffcott rotor

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.

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

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.

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

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.

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

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
 

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

(OP)
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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RE: M+0.5*m for adding shaft mass to Jeffcott rotor

(OP)
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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RE: M+0.5*m for adding shaft mass to Jeffcott rotor

(OP)
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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RE: M+0.5*m for adding shaft mass to Jeffcott rotor

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!



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

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.

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

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.

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

(OP)
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?

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

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

(OP)
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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RE: M+0.5*m for adding shaft mass to Jeffcott rotor

(OP)
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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RE: M+0.5*m for adding shaft mass to Jeffcott rotor

(OP)
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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RE: M+0.5*m for adding shaft mass to Jeffcott rotor

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).

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

(OP)
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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RE: M+0.5*m for adding shaft mass to Jeffcott rotor

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.

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

(OP)
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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RE: M+0.5*m for adding shaft mass to Jeffcott rotor

(OP)
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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RE: M+0.5*m for adding shaft mass to Jeffcott rotor

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

(OP)
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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RE: M+0.5*m for adding shaft mass to Jeffcott rotor

(OP)
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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RE: M+0.5*m for adding shaft mass to Jeffcott rotor

(OP)
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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RE: M+0.5*m for adding shaft mass to Jeffcott rotor

"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.



Cheers

Greg Locock

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

(OP)
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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RE: M+0.5*m for adding shaft mass to Jeffcott rotor

(OP)
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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RE: M+0.5*m for adding shaft mass to Jeffcott rotor

(OP)
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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RE: M+0.5*m for adding shaft mass to Jeffcott rotor

(OP)
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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RE: M+0.5*m for adding shaft mass to Jeffcott rotor

(OP)
Sorry - cubic polynomial, not quadratic.

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

(OP)
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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RE: M+0.5*m for adding shaft mass to Jeffcott rotor

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

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

Cheers

Greg Locock

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

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

... 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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RE: M+0.5*m for adding shaft mass to Jeffcott rotor

(OP)
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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RE: M+0.5*m for adding shaft mass to Jeffcott rotor

(OP)
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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RE: M+0.5*m for adding shaft mass to Jeffcott rotor

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.

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

(OP)
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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RE: M+0.5*m for adding shaft mass to Jeffcott rotor

(OP)
“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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RE: M+0.5*m for adding shaft mass to Jeffcott rotor

(OP)
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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RE: M+0.5*m for adding shaft mass to Jeffcott rotor


(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

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

(OP)
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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RE: M+0.5*m for adding shaft mass to Jeffcott rotor

(OP)
“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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RE: M+0.5*m for adding shaft mass to Jeffcott rotor

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

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

(OP)
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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RE: M+0.5*m for adding shaft mass to Jeffcott rotor

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

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

(OP)
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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RE: M+0.5*m for adding shaft mass to Jeffcott rotor

(OP)
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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RE: M+0.5*m for adding shaft mass to Jeffcott rotor

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

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

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

(OP)
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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RE: M+0.5*m for adding shaft mass to Jeffcott rotor

(OP)
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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RE: M+0.5*m for adding shaft mass to Jeffcott rotor

(OP)
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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RE: M+0.5*m for adding shaft mass to Jeffcott rotor

(OP)
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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RE: M+0.5*m for adding shaft mass to Jeffcott rotor

(OP)
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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RE: M+0.5*m for adding shaft mass to Jeffcott rotor

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

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

(OP)
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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RE: M+0.5*m for adding shaft mass to Jeffcott rotor

(OP)
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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RE: M+0.5*m for adding shaft mass to Jeffcott rotor

(OP)
Attached is the exact solution using method described above (similar to S&V handbook).
http://home.houston.rr.com/electricpete/JeffcotWithDistribMassExact.pdf

The boundary condition that I wrote above as
y’’’’(L/2) = E*I*w^2*(M/2)*Y(L/2)
   should have been
y’’’(L/2) = -E*I*w^2*(M/2)*Y(L/2)

In the correction immediately above, there is a change to minus sign on right as expected.  Also the force from mass acceleration is related to the 3rd derivative (shear force), not the 4th derivative (force per length) as I had written before.  This can be verified by examining the units.

Results for X vary from 0.4857 for very small MassRatio (m<<M) to 0.4927 for very high MassRatio (m>>M) . These match the results which can be derived for the case of distributed mass alone or concentrated mass alone. A table of X vs MassRatio is at the end of the worksheet.

To two decimal places it is 0.49 no matter what, so the values really aren't needed for any practical calculations.  It was more of an excercize for me and I learned a lot by working through it and asking questions here.

I still have my assignment from Bill to solve using the transfer matrix which would be a useful excercize as well.  Sometime in the near future...

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

(OP)
On the off-change that someone may actually want these results at some time in the future (long after mywebsite is gone), I'll post the tabular results here;

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

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

(OP)
I tried to use a similar analytical approach to find the resonant frequency of a shaft on bearings... modeled by a beam with uniformly distributed mass and with springs on each end and moment=0 at each end.

The boundary conditions:
> # Equation:  y'(0)=0
> # Equation:   y'(L)=0
> # Equation 3: y''(0)=k*y
> # Equation 4: y''(L/2)=k*y

Maple couldn't find a solution. It looks like I have 4 equations in the 4 uknowns so I don't know why there would be a problem.  Maybe I'm overlooking something basic or did something stupid? Any ideas?

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

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

(OP)
Please disregard my last post.  I think I can work through it by checking for errors and trying a few more things. I don't want to waste your time with proofreading my code and using Maple to solve for unknowns.  I'll try to stick with vibration questions.

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