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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[GregLocock]

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

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[electricpete]

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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[electricpete]

I gotta getmea Blevins wonadeez days.

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[GregLocock]

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

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[electricpete]

$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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[electricpete]

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

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[electricpete]

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

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[GregLocock]

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.

 

[GregLocock]

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.

 

[electricpete]

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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[svanels]

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

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

 

[WCFoiles]

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.

 

[electricpete]

Hi Bill! Welcome to eng-tips. Hope you will stick around awhile and share your wisdom like you do on

http://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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[spciesla]

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

 

[svanels]

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

 

[spciesla]

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

 

[svanels]

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

 

[electricpete]

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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[electricpete]

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

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