cantilever

[msonys]

hi to all,

What is the general formula to calculate the resonance frequencies for a cantilever beam, I mean a formula applicable for first, second, third onwards resonance frequencies,
msonys

 

[electricpete]

Per Mechanical Vibrations by Den Hartog page 432:

wn = an * sqrt(E*I/[mu*L^4])

where:
wn = 2*pi*fn = natural frequency in radians/time

an depends on mode number n and boundary conditions. For cantilever beam, an as follows:
a1 = 3.52
a2 = 22.0
a3 = 61.7
a4 = 121
a5 = 200

E = Young's Modulus
I = area moment of inertia
mu = mass per length
L = length

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[GregLocock]

ep, does he have the higher modes? do multiples of 11 keep turning up?



Cheers

Greg Locock

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[tomirvine]

(Periodic reply, pun intended)

I have posted a beam formula paper at:

http://www.vibrationdata.com/tutorials2/beam.pdf

Tom Irvine

http://www.vibrationdata.com

 

[electricpete]

Let an=xn^2

You can solve x from:
cos(xn)cosh(xn)=-1
The above info from Mechanical Vibrations by Rao 3rd ed page 527.

An approximate way to find x is
xn~(2*n-1)*Pi/2

The above info from Shock and vib handbook 2nd ed page 7-14

Using the S&V HBK approximation we find a1-a10 as follows:
n xn an
1.0 1.6 2.5
2.0 4.7 22.2
3.0 7.9 61.7
4.0 11.0 120.9
5.0 14.1 199.8
6.0 17.3 298.4
7.0 20.4 416.8
8.0 23.6 555.0
9.0 26.7 712.8
10.0 29.8 890.4
You can see this matches pretty well except for n=1.

Again, once you find xn, you can compute an=xn^2
All of this applies to the cantilever beam.

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[electricpete]

All of the above is from the Euler/Bernoullis beam approximation. It neglects shear effects and momentum effects and some other stuff that are included in the Timoshenko model.

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[electricpete]

Taking a look at the logic of why the approximation
xn~(2*n-1)*Pi/2
works.

Let's start with the assumption that xn is large above 10. Then cosh(xn) will be very extremely large (approx exp[xn]). We are looking for the values of cos(xn) which are very small such that cosxn = -1/cosh(xn). With large number in the denominator, these are very close to the zero crossings of cos(xn) which we know are odd multiples of pi/2.

We started with the assumption that xn are very large. That is why the approximation works less for small valeus of xn and the worst agreement occurs for n=1.

If you try to do this numerically, you will find that you are multiplying a very small number with a very large number so the extra digits beyond the decimal point (not shown in my table above) are very important.

For example if you use xn=10.9, you would get
cos(10.9)*cosh(10.9)=2854 which is very far away from -1.

But if you use a more exact number xn=10.995541, you would get
cos(10.995541)*cosh(10.995541)=2-0.99 which is very close to -1.

Where the relationship to 11 came from? Just a coincidence I think.

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[electricpete]

So with a little more knowledge under my belt, I would summarize as follows:

For n<4, x1 to x4 given by:
1.875
4.694
7.854757
10.995541

For n>4, we can calculate xn almost exactly (within assumptions of Euler) by
xn=(2*n-1)*Pi/2

Radian resonant frequencies are
wn = (xn^2) * sqrt(E*I/[mu*L^4])

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