First natural frequency of a cantilevered pole 2

[OzEng80]

Hi

How do you calculate the first fundamental (natural) frequency of a cantilevered mast or pole (in my case a 13m high crucifix)? I can only find frequency formulas for members where the loading and member stiffness is parallel with the gravity loading (ie horizontal beams subject vertical loads – I have vertical beam subject to horizontal loads).

Also what would be an appropriate serviceability deflection at the tip? I was intending H/125 but I’m struggling….

Thanks!

 

[jhardy1]

Assuming the axial load is not so high that the column is close to buckling under self weight, the natural frequency and mode shapes are exactly the same for a vertical cantilever as for a horizontal cantilever. That is, natural frequency is a function of geometry, mass and stiffness, but is independent of applied loads.

Hope this helps!

 

[msquared48]

You will have to lump the extra mass at the intersection.

It will not be a stsndard, simple solution due to the extra concentrated mass. It will be a different equation, or series of them.

However, do you realize that the Greek word, as translated for the cross, is "Staros", transleted "stake" or a single pole, without the crossed member?

So...the solution to the problem would be simpler without the cross member.

Mike McCann
MMC Engineering

 

[msquared48]

Sorry, but it is also dependent where the mass is in relation to the base. Agsain, this is NOT the standard solution here.

Mike McCann
MMC Engineering

 

[OzEng80]

Thanks.

I suppose a Greek game of noughts and crosses would look like binary....!

So.. the horizontal frequency is a function of the vertically orientated mass and the horizontal stiffness, it doesn't have anything to do with the mass accelerating the motion or anything? So... a members frequency is independent of it's orientation, loadings and gravity... Just feels like there shouldn't be an 'm' term in the formula....

A bit more background... I am just trying to do a first tier analysis of the crucifix to prove that it has a frequency >1Hz and therefore avoid a heap of painful calculations imposed by my (Australian) standard.

Do you think i can go lower than the H/125?

Thanks again!

 

[abusementpark]

Isn't the exact distributed mass solution a partial differential equation?? Those are brutal to solve, but I bet someone has solved it before.

 

[Ron]

Check publications on powerline structures with cross arms. Also, as Mike noted, the analysis of the cross structure is a bit more difficult as you may have competing and complementary aeolian vibration occuring at the intersection. That's a huge crucifix.

 

[271828]

This is very, very easy.

First off, Julian is right. Assuming it's not almost unstable (that's the only way "loads" make a difference), it will have the same natural frequency upside down, on its side, up in space, whatever.

If you have a copy of the AISC Design Guide 11, you can use the equations at the start of Chapter 3 to estimate the natural frequency. *Pretend* that the crucifix is a cantilever beam turned on its side and subject to its self weight. Compute the deflection at the tip and call this Delta. The natural frequency in Hz is approximately fn=0.18*sqrt(386/Delta).

Also, any modern structural analysis package will compute this natural frequency also, and be more exact. RISA, SAP, RAM Advanse...pretty much all of them.

 

[271828]

Oh yeah, the Delta in my post must be in inches.

 

[271828]

"Isn't the exact distributed mass solution a partial differential equation??"
ALL beam, string, shaft, or frame problems end up with PDEs. The lumped mass just makes it more difficult to solve, so one ends up using some kind of approximate method. The textbooks spend a lot of time on Rayleigh-Ritz, Galerkin, etc., but we structural guys generally just use the finite element method.


"Those are brutal to solve, but I bet someone has solved it before"
Sure. We did that in class. I'd never even think of it without FEA, though.

 

[miecz]

A professor of mine once said: "if you can't solve a problem, reduce the problem to something that you can solve", or some such thing...

While this isn't a standard case, it is pretty close to a cantilever with a concentrated load at the top and a uniform mass along it's length. I would solve that case and see where I stand in relation to the 1 Hz limit.

I've attached the solution to the standard case. Try yours with "L" equal to the height of the crossing member and "W" including all the material above that point. If the result is greater than 1 Hz, I'd say you're OK.

 

[bigmig]

miecz,

Great advice, and a great help in posting a simple solution to a problem that others are making too complex....a couple questions about your solution...

1) 2000 Kips ?? Is that right?
2) I can't get 4.4 Hz from your formula for angular frequency....?? When I work through your units I'm getting nowhere...which makes me think you made a mistake or I'm missing something.

For the top I get psi x in^4 x ft/sec^2 not considering applicable conversions to inches. On the bottom I get lbs x in^3+ lb/in x in^4...????

Thanks

 

[271828]

miecz,

I cranked through your equation and it seems reasonable to me.

Using my approximate method, I get 0.84 Hz compared to your 0.70 Hz. If I take away the point mass, I get 21.5 Hz with your equation and 20.4 Hz with my approximate method.

 

[FeX32]

Hmm. Well I would agree as to the above solutions, I suggest solving the beam PDE is a more informative choice. I do this quite often and the solution to this question could be done by following a simple textbook beam PDE equation with the added lumped mass as a boundary condition to the PDE.
I by no means discount anyone's work. The above posted results are very simple and easy to follow. However, they must include the assumptions that were made in order to be a complete solution.
Try solving: d^2w(x,t)/dt^2+c^2*d^4w(x,t)/dx^4=0, where c=sqrt(EI/rho*A) for a fixed-fixed boundary condition first, then if you are comfortable try setting the mass*accel you gave at a particular point equal to the deflection force.
This should give you a complete solution.




Fe

 

[miecz]

bigmig-

W=2000kips. I just picked a number out of the air to show the formulas and give everyone an example to compare to. I get;

3EIg = 38876975 kip-ft^3/sec^2
WL^3 = 2000000 kip-ft^3
wL^4 = 9000 kip-ft^3

Do we agree to there?

 

[plasgears]

Have you considered an open SS structure to avoid Karman vortices in a strong wind? A tapered column form would be interesting and structurally attractive.

 

[271828]

plasgears brings up something interesting. I wonder if the fundamental mode could be torsional. If it's an open section, I could imagine this happening. Does the first torsional mode also have to have fn > 1 Hz?

 

[FeX32]

Good point.
Although, usually in strong wind the vortex shedding frequency is significantly higher then the first mode freq. of such a structure. It should be checked.
What is mentioned about the torsional mode is only dangerous if the torsional modal frequency coincides with that of the translational frequency. This is what we call flutter. (I have never seen a crucifix flutter however)
The uncoupled fundamental mode is not likely to be torsional because torsional stiffness is likely to be higher then that of the bending mode.
Usually one could first find out what the maximum and average wind speeds are for the area then correlate this to the vortex shedding frequency. If this simple relation comes out close or relatively close to any of your calculated natural frequencies then a more rigorous analysis should be done. This is likely to not be the case though.







Fe

 

[OzEng80]

Thanks for all your inputs.

The crucifix is an open section, a 250UC90 - pretty close to a W10x49 (I think) which results in a service deflection at the tip of about h/140 (8m cantilever). The major axis frequencies are both greater than 1, the torsional frequency is very small but has low associated deflections.
The way my code handles member design with frequencies less than 1Hz is to amplify the static load. Given that the section is obviously sized for deflection (and I was willing to go down to about 1/100) I believe there is sufficient redundancy to account for ‘unanticipated’ effects in both the strength and service states.

Thanks again for all the assistance it was very informative!

 

[miecz]

OzEng80

How did you finally determine your natural frequencies?