Resonant frequency calculation 2

[opticsman]

Hi All, I 'd like to estimate the resonant frequency of thin metal ( aluminium ) plates of small dimensions e.g. 1 X 0.5 x 0.1 mm . Can anyone suggest an approximate formula and units? If i wanted to get a resonant frequency in GHz , what would the dimensions need to be ? My thanks in advance. Opticsman.

 

[GregLocock]

That's more of a beam than a plate, and more of a solid than a beam. In that case the modes are likely to be longitudinal, in which case the formula is speed of wave= squareroot(E/density)

Not building radar absorbent material are we?

Cheers

Greg Locock

 

[hacksaw]

What mechanical modes are you interested in: longitudinal, transverse, torsional? What medium is your beam located or attached to, i.e. what are boundary conditions?

To get criticals in the ghz range(?) you'll be exciting a lot of bulk and surface waves, so you'll have to be more specific about the excitation (i.e spatial distribution).

If microwave excitation is involved, then you may need to start with Maxwells equations and the exact mechanical equations with whatever form of magneto-electric/elastic coupling that you plan to focus upon. Typically in that frequency range you'll also need to include thermal excitation of your modes.

 

[opticsman]

Hi Thanks Both, Apologies for showing my ignorance of this area. I should have mentioned that i was imagining a cantilever plate (or beam or rod) vibrating in air . My main interest was to see if there is an approximate rule of thumb formula to relate dimensions to a fundamental resonance frequency. For example if a 6 " steel rule resonates at say 30 Hz ( guess) then what dimensions does it need to be to resonate at some higher frequency? Just hoping for ball park rather than very exact values. thanks again opticsman.

 

[tomirvine]

Here is the equation for the fundamental bending frequency of a cantilever beam:


fn = [1/(2 pi)][3.5156 / L^2] sqrt( EI /rho )

where

L^2 = length squared
E = elastic modulus
I = area moment of inertia of the cross section
rho = mass/length

Now assume that the material, width, and thickness are constant. The natural frequency is inversely proportional to length squared.

Tom Irvine

http://www.vibrationdata.com

 

[seaboat]

Hi,

I am not sure about the factor 1/2pi. From Den Hartog, for a cantilever beam, we have:

freq=3.52 * {E*I/[(mass/length)*length^3]}^0.5

(mass/length is referred to as mu1 )

which is derived as:

(noting that a quarter cosine wave is a staring point, but not an exact solution)

y=y0(1-cos*pi*x/(2L)) carrying on and substituing into potential and kinetic energy integrals, we find:

Potential = Pi^4*E*I*y0^2/(64*l^3)
Kinetic = mu1*omega^2*y0^2*L*(3/4-2/pi)

Equate to find:

omega= pi^2/[8*(3/4-2/pi)^0.5]*[E*I/(mu1*L^4)]

equals"

3.66/L^2 * (E*I/mu1)^0.5; the coefficient which goes to 3.55 at an exact solution.

So, what of the 1/2pi factor?

(I jsut happened to be looking for confirmation of this myself!)

Best regards,

Bill

 

[seaboat]

Oops!

The first equation at the top, I accidentally wrote length^3 when it should read length^4:

freq=3.52 * {E*I/[(mass/length)*length^4]}^0.5

of course, you can move the length out from under the radical, and get:

3.52/L^2 * [E*I/mu1]^0.5

 

[btrueblood]

1/2pi converts the calculated frequency from radians/second (which the math gurus love, because they can plug it straight into trancendental functions) to Hertz or cycles/second (i.e. what us dumb engineers count or measure when things start to vibrate).

 

[seaboat]

Thank you btrueblood----

I see now--that Den Hartog uses omega as "circular frequency" in this case---so of course, 1/(2pi) to convert to cycles/sec.

And of course I now see it in the equations, too---like you say, no conversions from radians due to the transcendentals....

Best regards

Bill

 

[DonJuanDiego]

I'm trying to do a calculation for a Cantilever beam using PVC.

I'm using the Formula that Tom Irvine pointed out however I'm having some issues with Units..

Units are as follows...

E=lb/in^2
I=in^4
rho= lb/in

If I use the formula this gives me the square root of in^3. Can someone help me out here?

Thanks,
Juan

 

[MikeyP]

Are you perhaps confusing force in pounds (in the Youngs modulus) with mass in pounds (in the linear density)?

E has dimensions M/(LT^2).
rho has dimensions M/L.
I has dimesions L^4

fn has dimensions (1/L^2) * sqrt[ (M/(LT^2) * (L^4) * (L/M) ]
= (1/L^2) * sqrt(L^4/T^2)
= 1/T


M

--
Dr Michael F Platten

 

[DonJuanDiego]

Yes,

I am confusing Force pounds with mass pounds. This is exactly my problem, thanks!

That's what I get for learning only the metric system all the way through school and using and American for development efforts.

This information is really helpful for me to determine the dimensions for my instrument!!!

Juan

 

[Tunalover]

Did you look in Roark? I think Roark may have fundamental frequency formulas.



Tunalover