effect of force on resonant frequency

[Tmoose]

The December issue of "Electrical Apparatus" has an article on motor ventilating fan design. One statement on page 24 is that the steady force from wind pressure (which might be "as great as the weight of the blade itself") has the effect of reducing the blade's resonant frequency.

I think that the force is being mistaken as interchangeable with blade mass. That is easy to do, since resonant frequency formulas and tables often are related to static deflection, which does depend on force.
I worked with a very useful shaft analysis program from a big bearing company in the 90s that allowed force inputs but carried the results calculation beyond deflection and to First bending mode resonant frequency. The instruction manual warned about the incorrect wn result, but I knew at least one user that was unaware of the distinction.

Dan T

 

[electricpete]

Sounds like a good catch. Linear system resonant frequency is not affected by any static force.

If non-linear, difficult to predict but I think most non-linearities such as clearance act in a direction where applied static force causes increase in effective stiffness and increase in resonant frequency.

Going off on a little tangent to things I don’t know much about.... “Rotordynamics Prediction in Engineering” does include an axial force in their lateral critical speed rotordynamic model. (equations 1.19 thru 1.25 and 4.25). I’m not sure I fully understand it. From a first scan, I see they’re saying axial force increases strain energy, which would be a stiffening effect and increase frequency. However I vaguely recall some discussion that shaft in axial compression can have lower resonant frequency due to something resembling buckling: If you compress unloaded shaft, there is no lateral deflection, but if you compress a shaft which already has center radial load applied, deflection in direction of applied load increases (right?).


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[Tmoose]

Yeah, that was how Cosmos/M FEA calculated eigenfrequencies.
Put enough axial compressive load on a shaft and the critical speed was zero.
Conversely, negative (axial) force raised Wn. I figured it was just as a tighter guitar string plays a higher note. And I think probably does whether oriented horizontally (gravity transverse), vertically (gravity axial), or in deep space (no gravity at all).

 

[electricpete]

Good points. I didn’t realize it until you mentioned it, but the axial-tension PE formula used by Rotordynamics Prediction in Engineering is exactly the same as the PE formula a used by Den Hartog when he analyses string in tension by Raleigh’s method.

Attached I tried to do an analysis to decide whether this effect would be significant by comparing PE of beam in bending to PE of string in tension, both assuming simple sinusoidal shape, using PE formula’s from Den Hartog.

> # For simply-supported shaft of length L, diameter D
> # mat properties rho, E
> # Applied tension is T
> # Assumed mode-shape is y(x)=Y0*sin(pi*x/L)
>
> # Symbols
> # x = axial coordinate
> # y(x) = transverse displacement
> # T = applied tension
> # Yp(x) = 1st deriv of y wrt x
> # Ypp(x)= 2nd derivative of y wrt x
> # (p for prime)
> # PEtension = PE due to tension
> # PEbending = PE due to bending

PEbending = (1/2)*E*I*Int{ [dy^2/dx^2]^2, x=0.L)
PEbending := E*Id*Pi^4*Y0^2/(4*L^3)

PEtension = (1/2)*E*I*Int{ [dy/dx]^2, x=0.L)
PEtension:= T*Pi^2*Y0^2/(4*L)

PEtension/PEbending = 64*T*L^2/(E*D^4*Pi^3)

To get a conservative estimate of the highest possible value of this ratio, we load our steel shaft to a stress level of 30ksi = 0.001 * E. Substituting T/(pi*D^2/4) = 0.001*E into above ratio, we get:
PEtension/PEbending ~ 0.5E-3*pi* (L/D)^2

That is a very small number. Even if we had ridiculously high L/D to make this ratio 10% (and that would only be a 1% increase in natural frequency.

That’s assuming I have no silly math errors or gross conceptual errors (open to comment).

This analysis does not lend itself to beam in compression (since we cannot create negative PE, so no way to lower natural frequency). I guess it must be a fundamentally different phenomenon for beam in compression than in tension.


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[electricpete]

Correction in bold:

"PEtension = (1/2)*E*I*Int{ [dy/dx]^2, x=0.L)
PEtension:= T*Pi^2*Y0^2/(4*L)"

should have obviously been

"PEtension = (1/2)*T*Int{ [dy/dx]^2, x=0.L)
PEtension:= T*Pi^2*Y0^2/(4*L)"

(The error is typo in that one equation, does not affect any later equations). The conclusion remains that the effect of axial tension on resonant frequency seems negligible.

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[Denial]

Many (and I mean many) years ago, unable to find it in any textbook, I derived a formula for the natural frequency of simply-supported bending beam when loaded with an axial force. The result was:

Omega = Sqrt[ (pi^2/(m*l^2))*(pi^2*E*I/l^2 + N) ]
where
Omega is in radians/sec
m is mass per unit length
l is length
E*I is beam's flexural stiffness
N is the axial force (+ve for an axial tension, -ve for compression)

As I thought when I derived it, it almost seems to simple to be true. Yet it ticks all the obvious check boxes. Unfortunately the sands of time and job-changes have buried my derivation: only the result remains in my files.

 

[electricpete]

That looks like what we'd expect from combining above expressions for PE with expression for KE and plugging into Raleigh's formula.

Examining your factor (pi^2*E*I/l^2 + N), we see the ratio of first term from PEbending to 2nd term from PEtension is
[pi^2*E*I/L^2]/[N] = (pi^2*E*I) / (N*L^2)

...which is identical to what you get from my ratio
PEbending/ PEtension= [E*Id*Pi^4*Y0^2/(4*L^3)]/[T*Pi^2*Y0^2/(4*L)]
PEbending/ PEtension= [E*Id*Pi^2/(L^2)] / [T]]
PEbending/ PEtension= (pi^2*E*Id) / (T*L^2)

I have wondered before why the tension expressions show up in the book "Rotordynamics Prediction for Engineering" but don't show up anywhere else. Now I'm pretty sure the reason is that the PE from tension is negligible compared to PE from bending for industrial machinery.






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[electricpete]

So, positive value of tension (T or P) clearly represents tension.

We could plug negative value into Denial's formula. Would it be a valid prediction of effect of compression? Or a non-physical number?

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[electricpete]

fyi - I have attached file is my previous file with something new added at the end.... I added derivation of an equation for natural frequency using Raleigh’s principle (equates KEmax to PEmax and solve for w) and the result is the same as Denial’s.

I’m still curious about how we would treat compression.


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[Denial]

ElectricPete. If you are wondering whether my formula is valid for compression, let me answer with a double negative: I can see no reason why it should not be equally valid for tension and compression. (Correct signs being used, of course.)

 

[electricpete]

I can see some reasons to question the applicability to compression:

1 - It is derived from Raleigh's principle using potential energy. Tension results in increased potential energy which results in increase in predicted resonant frequency:
what = sqrt(PEbending+PEaxial) / [KE/w]
[what goes up when we add PEaxial].

To give a decrease in resonant frequency would require PEaxial to be negative. It's not obvious to me what is the physical meaning of negative PE and how it would relate to Raleigh's formula.

2 - We can see physically there are differences between tension and compression for a long slender beam. In absence of radial force and in presence of symmetry, if we apply axial tension the system is in stable equilibrium at zero radial deflection. If we apply compressive force the system, it is in unstable equilibrium at zero deflection. Small radial perturbation can cause it to move radially. The phenomenon of buckling in compression has no counterpart in tension.

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[electricpete]

By the way, "what" should have been written as w_hat, which is the estimate of the unknown true w.

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[electricpete]

Maybe it is negative potential energy based on examination of energy expended by radial forces:

For normal positive potential energy from bearing stiffness or axial tension: to move rotor off-center against bearing stiffness or against axial tension, we have to put energy into the system (apply radial force in direction of radial motion), which increases the stored (positive) potential energy.

However to move rotor off-center in presence axial compression, we have to remove energy from the system (apply radial force in direction opposite radial motion), which would apparently be a decrease stored potential energy (or addition of negative potential energy)

It resembles the “negative stiffness” effect of magnetic field in a motor... as you move rotor off-center, the magnetic field acts in a direction to move it further offcenter. This also decreases resonant frequency. I have never thought of it as negative potential energy before, but to fit it into Raleigh’s model that’s how we’d have to view it.

I don't know much about buckling of static (non-vibrating) systems, but I know it's complicated and different than analysis of tension. I'm not sure whether or not any of those effects apply here....open to comment.

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[electricpete]

This is addressed by “Vibration Dynamics and Control” by Genta, chapter 12.
Denial’s equation does apply in tension and compression.

There is a critical value given by equation 12.112 which corresponds to buckling.

This point of buckling occurs when the predicted frequency is zero.

... which is the point where strain energy from compression equals strain energy from bending. Based on analysis above comparing PEbending to PEaxial, that amount of compressive loading is much higher than any realistic load (and it is probably already obvious that machines don’t operate at compressive stress exceeding buckling... otherwise the consequences would be dramatic).


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[electricpete]

... which is the point where strain energy from compression equals strain energy from bending

Strain energy was a very bad choice of words when referring to axial tension/compression.

Although this it acts analogously to strain energy (in the case of tension), it is not stored anywhere. It is associated with energy that would be added or removed by the axial force as it moves through an axial distance when the rotor bows and straightens.

The bottom line. I am satisfied and have answered my questions. Denial was definitely right on that.

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[electricpete]

Could be viewed as potential energy if the source of the load was a weight hanging on the end of a shaft... in which case we could treat it as gravitational PE.

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[Denial]

It would need to be a massless weight -)

Thanks for sorting this out, EP. I was about to attempt to derive the result again. The axial force in the beam is assumed to NOT vary through the oscillation cycle, so while (obviously) it induces some strain energy in the beam, that strain energy is constant and so does not enter into Raleigh's formula.

But, as you point out, as the beam moves from its straight position to its bowed position its end moves axially ("conservation of length"). A compressive axial force does work on the system during this movement; a tensile axial force has work done against it.

The axial movement, to a first approximation, is
(pi*d)^2/(4*l)
where d is the midpoint lateral deflection.

 

[electricpete]

re: massless weight: The mass would need to exert an approximately constant force on the rest of the system, so all that is necessary is that the axial-direction acceleration remains far below the acceleration of gravity.

The axial distance as you say is Da = “PE” / T = Pi^2*Y0^2/{4*L}] where Y0 is the peak radial movement.

Plugging in some example values:
Y0 = 0.0005m
L = 1m
then we calculate
Da = 6.2E-7m

Peak Acceleration = (2*pi*f)^2*Da
Using f = 60hz (3600rpm), we would get
Peak acceleration = 0.088 m/sec^2
Peak acceleration ~ 0.01 * g
Peak acceleration << g, so the force exerted by the mass on the remainder due to acceleration is small compared to gravity and we can approximate the force exerted by the mass on the rest of the system as approximately constant at mass*g.

I’m not saying this is the case for all systems, but certainly for many. And please understand the the whole reason for mentioning the weight was to allow the problem to fit into the mold of Raleigh and LaGrange solution methods which deal in kinetic energy and potential energy and external applied forces (for LaGrange), but do not make any allowance for a constant applied force exchanging energy with the system.


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[Denial]

While I cannot remember exactly how I derived the formula I give above, I can vividly remember one aspect of the exercise. When I reached the result, I drew extra confidence it was right because it was so simple. When I showed it to my boss (who had asked me to attempt to derive it) he told me it must be wrong because it was too simple.

 

[spongebob007]

Didn't read the math in all of the responses, but I believe the terms you are looking for are stress stiffening and spin softening. Typically it applies to thin structures and it is common in rotordynamics. I don't have a lot of experience with that type of analysis but I do remeber from my ANSYS days that you could include prestress effects in linear modal analysis.