Consider a rigidly clamped on both ends beam vs. same beam but free in space.

Question:
Will modal ( resonant) frequencies in both cases be the same?

Mode shapes are likely to be different.

I'll appreciate if someone has examples.

Dave

Hi ,

For your given conditions the natural frequencies of both the beams will be same.

Consider both ends are free, the free ends will have antinodes and the first natural frequency will be f=c/2L
f- Frequency c-velocity of sound in that medium(steel)
L= length of the tube
2L = wavelength

For both ends are fixed, the fixed end always have nodes and the first natural frequency will be f=c/2L

if you consider one end of the beam is fixed and other end is free, in that case the fixed end will have node and free end will have antinodes and the first natural frequency will be f=c/4L .for this case it will be different from the other two

Good explanation above.

To add some more discussion: IF we don't count the zero-frequency mode (translation) which exists for the free free beam but isn't very interesting, THEN the 1st, 2nd, 3rd etc resonant frequencies of the clamped/clamped beam are the same as for the free/free beam.

It sounds non-intuitive at first since we expect the stiffer or more constrained configuration (clamped/clamped) to have a higher resonant frequency. We can reconcile this by considering that the real first mode of the free/free beam is that zero-frequency mode, and it is lower than any mode of the clamped/clamped as expected. The next mode of the free/free beam is really it's 2nd mode which is lower in frequency than the 2nd mode of the clamped/clamped, but just happens to match the first mode of the clamped/clamped in frequency for reasons explained very well by Amar.

A mental experiment should convince us that the relevant mode to compare to the 1st mode of the clamped/clamped is that zero-frequency first mode of the free free, not the 2nd mode of the free free. Add a variable-stiffness clamp on each end of a beam. Start with stiffness very low. You have a rigid body body mode on a very weak spring with frequency very close to 0 resembling the free/free first mode at zero frequency. As you slowly increase the stiffness, the frequency increases from near 0 up to the clamped/clamped resonant frequency as we increase stiffness to infinity. By this mental experiment, we have seen that it is the zero frequency first mode of the free/free which turns into the first frequency mode of the clamped clamped. So the first mode of the stiffer/constrained case (clamped/clamped) has a higher frequency than the first mode of the flexible/unconstrained case (free free- zero frequency) exactly as expected. The fact that 2nd mode of free/free happens to match first mode of clamped/clamped is a mathematical coincidence which is explained as above, but does not contradict our expectation that the real first mode of the free/free has lower resonant frequency than the first mode of fixed/fixed and real 2nd mode of free/free has lower resonant frequency than 2nd mode of clamped/clamped etc.

This message has been edited. Last edited by: electricpete,

You asked for an "example". Attached is output of transfer-matrix numerical solution of a cylindrical steel beam 1 meter long and 0.1 meter diameter.

Slide 1 shows first three mode shapes and frequencies of the clamped/clamped case.

Slide 2 shows first three non-zero mode shapes and frequencies of the free/free case.

You can see that alhtough the mode shapes are vastly different, the first three frequencies (ignoring the zero-frequency mode of the free/free) are the same as discussed above.

You should be able to calculate these for yourself using Den Hartog Appendix V (let me know if you want me to show that calculation). Even without doing the calculation, you can see from the coefficients in the appendix the matching of resonant frequencies for clamped/clamped case and free/free (ignoring that first zero-frequency mode of the free/free case)

Comparison.ppt (480 KB, 49 downloads)

Thanks a lot for the replys and examples.

I knew that clamped/clamped vs. free/free modal frequencies should be the same whith the mode shapes slightly different as shown in Pete's file. But when I have conducted a modal test ( see attached data fot the first mode only ) the shapes were produced as predicted but the frequency - different.

Short description of the test setup.
Free/free mode: The beam is suspended horizontally on ropes, impact and measurement is done in horizonatal direction ( where motion is free) .
Clamped/clamped mode: The beam is clamped to a solid foundation, impact and measurement is done in vertical direction. ( I did the impact and measurement in vertical direction consiously because clamping in this direction was much more effective and gravity force is not a factor )

The modal frequencies differ significantly. I have to go back to the formulae to see how foundation stiffness affects it. I would presume that theoretical clamped/clamped frequency value calculation considers it infinitely high.

David

Fixed_beam_Movie_1.avi (772 KB, 40 downloads)

Here is Free/free beam data.

Free_beam_1_Movie_1.avi (488 KB, 29 downloads)

Yes, you are correct. The boundary conditions affect the frequency and mode shape. The clamped/clamped is an ideal condition. It should have no lateral movement of the ends and slope at the ends is 0.

You have some lateral movement at the ends and the slope at your ends is not 0. From both of these observations we conclude your 1st natural frequency will be lower than predicted by the ideal clamped/clamped model and therefore lower than your first non-zero free/free mode.

Attached I have done some more simulations of the geometry 1 m long circlar steel beam 1" diameter with more resolution for better mode shape plot.

The first slide shows the problem definition.

The 2nd slide is free/free.

The 3rd slide is clamped/clamped. Notice that the slope starts at 0 at both boundaries for all modes.

The 4th slide is pinned/pinned with very high stiffness below each end (1E10 N/m) which for all practical purposes represents an ideal pinned/pinned for this beam. Note the first resonant frequency is lower than the clamped/clamped.

The 5th thru 8 slide show progressivley lower support stiffness below the ends of an otherwise pinned/pinned geometry. The resonant frequencies decrease of course. The modeshapes change as well. The first modeshape changes gradually from a sin shape (pure pinned/pinned 1st mode) at very high support/bearing stiffness to a straight line (similar to the first rigid rotor mode) at very low suppport/bearing stiffness. Your modeshape shown in the FixedBeamMovie.avi resembles a modeshpae in between these two extremes, like the modeshape I have drawn with bearing stiffness 1E8 N/m for this particular beam.

A final note is that if you decrease that bearing stiffnesses very far as in the 8th slide, the 3rd mode shown there is very similar in frequency and shape to the first non-zero mode of the free/free (previously shown in slide 2)

This message has been edited. Last edited by: electricpete,

Comparison2.ppt (71 KB, 33 downloads)

Out of curiosity, how did you determine mode shape from an impact test?

The part certainly will vibrate at its mode shape (or a sum of its mode shapes) following an impact, but it seems like it would be a challenge to capture that data.

This is a perfect correllation between the model and test data .

Which formulae did you use?

I wonder, for a beam supported by antifriction bearings ( having some gap in vertical dicrection and therefore is free to move at least upward ) is it closer to free/free situation or ....?

Another situation is when testing the same beam in horizonatal direction, where the beam has almost unrestricted movement in this direction, is it free/free situation ?

I'll test it if I have a chance.

quote:

Originally posted by electricpete:
Out of curiosity, how did you determine mode shape from an impact test?

The part certainly will vibrate at its mode shape (or a sum of its mode shapes) following an impact, but it seems like it would be a challenge to capture that data.

Impact test provides vectorial G/Force data for each bin of the 0-Fmax range. The MEscope generates shapes for each one but the mode I have shown is at the resonance frequency.

So if I understand correctly, you impact the beam at a point and measure ratio of acceleration response at that point to force input at that point (all at the resonant frequency of interest). Then move to the next point and repeat (new point of response measurement and same new point of force input measurement). I don't think that would create a modeshape stricly speaking.

Or did you keep the point of impact the same and move the point of monitoring? That would make more sense.

Attached is an excel file which is a transfer matrix rotordynamic solution program that I programmed in vba. That is what I used to create these graphs. I should point out it is not limited to simple beams. Can add disks etc. Can include or exclude gyroscopic effects and rotating inertia effects. Some limitations as well as discussed in the instructions tab. Note this runs on macros so if your excel security is set to high it won't work. Also it works on excel 2003 and earlier, but I suspect it won't work on excel 2007.

This message has been edited. Last edited by: electricpete,

RotoSolve1_1_MASTER.xls (526 KB, 34 downloads)

quote:

Originally posted by electricpete:

Or did you keep the point of impact the same and move the point of monitoring?

I do as in the above but in reverse, which is equivalent ( although your order is easier to comprehend ). I have a fixed location sensor and roving point of impact.

Thanks. I agree they are equivalent due to "reciprocity". There is an analogous theorem in electrical circuits. Assume we meaure transfer impedance of a passive linear circuit by injecting a current at one pair of terminals and measuring voltage at another pair of terminals. We get same answer if we swap the locations of injecting current and measuring voltage.

These are 'dual' boundary conditions. You may find some information in books dealing with energy methods. This will be in some mathematical physics books and under the calculus of variations (Euler-Lagrange equation, especially places with a good section on boundary conditions).

My memory is not so good. It seems like Lanzos' book has this and Courant and Hilbert's mathematical physics has this stuff. I have a couple of good Calculus of Variation texts with this stuff, dual boundary conditions or some such non-sense.

The dual boundary conditions is sort of like the reciprocity. One of those things you have to believe but not intuitively obvious why it is so.

Based on Amar's comments, I was thinking (hoping) that the equivalence of that free/free and clamped/clamped frequency could also be explained in terms of traveling wave theory.

We can exactly represent the standing wave of a pinned/pinned beam as the sum of a forward and reverse traveling sinusoidal wave. If we imagine that the wave "reflects" off each boundary, then the polarity of the reflection is reversed (similar to a voltage traveling wave hitting a short circuit boundary). The pinned/pinned beam is shown as the first plot of attached (you can start the animation by pushing the top button on the left).

The clamped/clamped beam is more complex than the pinned/pinned and includes sinh and cosh terms as well as sin and cos.

The second plot attached is a pseudo free/free beam simulation. It is again constructed of only a sinusoidal component traveling in forward and backward direction. If we imagine that the beam "reflects" off the boundary, then this time the polarity of the reflection remains the same (similar to a voltage traveling wave hitting an open circuit boundary).The real free/free beam would also include sinh and cosh terms which would make the very ends of the beam curl in the proper direction.

I was hoping that if we used the pinned/pinned as a simplified representation of a clamped/clamped, and if we used the pseudo free/free as a simplified representation of a free/free, then we could see the wave relationship which allows constant speed of wave propagation (that is a characteristic of the beam) to result in the same frequencies for these boundary conditions.

My illustration failed. In the pinned-pinned beam, the forward or backward wave travel the length of the beam in one half cycle of the fundamental period of the sum. In the pseudo free/free beam, the forward or backward wave travel the length of the beam in one full cycle of the fundamental period of the beam. So for the same beam and same c, these two modes (if the pseudo free/free existed) would differ in frequency by a factor of 2.

So, I think my simplified analogy is too simplified to reflect the behavior of the clamped/clamped and free/free. But they make for some good pictures, anyway.

I have a suspicion there is a fairly simple explanation for the equivalence of frequencies of clamped/clamped and free/free using simple wave theory (which is what I attempted above), but I can't quite understand it if there is one.

AnimateBothR1.xls (43 KB, 40 downloads)

I seem to have a problem downloading any attachment on this page including David's avi's. Anyone else having that problem?

Here is an alternate link to the file.
http://home.comcast.net/~elect...ms/AnimateBothR1.xls

El'Pete,

Why does the Pinned/Pinned modes (above 1st mode) have displacements at the beam ends if they are pinned and restricted from motion? I am very impressed with your animation!

Walt

Thanks Walt.

The pinned/pinned diagram shows a decomposition of a standing wave ("Total") into the sum of a forward-travelling wave and backward travelling wave.

The total (purple curve) does in fact have 0 displacement at each boundary as required by the pinned/pinned boundary condition.

The forward and reverse travelling waves at each boundary have non-zero magnitude of equal magnitude and opposite sign (which of course sums to 0). This corresponds to a reflection with inversion at the boundary. When a travelling wave hits a boundary we expect either positive reflection or negative/inverted reflection or somewhere in between depending on the impedance change at the boundary. (for example a traveling voltage wave hitting a short circuit reflects with inverted polarity similar to what is shown here).

The "pseudo clamped/clamped" animation also consists of a standing wave which is decomposed into a sum of forward and backward traveling waves, but the reflection at the boundary is not inverted.

I should mention that the motivation for viewing modeshapes as a sum of traveling waves is simply to understand the resonant frequencies as was first mentioned by Amar.

If we look at the pinned/pinned case, we can see that the wave travels twice the length of the beam in one cycle. We conclude that the resonant frequency is F = c / (2*L) where c is the speed of the wave in the beam, which is a constant (characteristic of the beam material and cross section).

If we look at the pseudo free/free case, we can see that the wave travels one times the length of the beam on one cycle of vibration. So the resonant frequency of this "mode" would be c / (1*L)

From the above comparison, we would expect that the first non-zero resonant mode of the free/free beam would be a factor of 2.0 higher than the first resonant frequency of the pinned/pinned beam. In reality, the first non-zero resonant mode of the free/free beam is a factor of 2.24 higher than the first resonant frequency of the pinned/pinned beam. What went wrong? The pseudo free/free mode shown in not an exact re-creation of the real free/free mode (hence the name pseudo)... but not too far off.