19mm Dia x 720mm Fibreglass Tube Natural Frequency 3

[paulmech01]

I need to calculate the natural frequency of a 19mm Dia x 1.6wall x 720mm long Fibreglass Tube. I am trying to calculate the critical speed of of this member as it is developing vibrations at around 80km/h (wind speed). I would to see if the critical speed is close to this 80km/h.
If it is I will need to see what can be done to design out this problem, ie increase thickness, stiffness, length etc dampening etc...

 

[electricpete]

You referred to a critical frequency related to a speed?
Does the tube roll? Or are you talking a frequecy like Karman vortex frequency?

Where is the tube supported?

Per Den Hartog's Mechanical vibrations
wn = a * sqrt(E*I/(mu*l^4)
where E is Young's Modulus in psi, mu is linear density in lbm/inch, l is length i ninches, I is moment of inertia ...
I=(1/64)*pi*(Douter^4-Dinner^4) (units inch^4

a is a unitless constant as follows:
If the tube in cantlevered, a=3.52
If the tube is simply supported both ends, a=9.87
Free/Free => a=22
There are about 10 other boundary condition coefficients listed

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[paulmech01]

I should clarify, the tube is mounted one end, and is generating a vibration when traveling at approx 80km/h, due to wind/airflow which I believe and have been advised due to vortex shedding.

Does above formula output a frequency?

What does mu denote in the above formula?

 

[electricpete]

The above formula outputs a radian frequency. To get frequency in hertz, use
f = w/(2*Pi)

mu in the linear density in lbm/inch

For your geometry (cantilever), use a=3.52 for the first bending mode, a=22 for the 2nd bending mode, and a=121 for the 3rd bending mode.

There are acoustic resonances possible as well... the harmonic series of those will depend on whether the tube is open at one or both ends.

I think the frequency of vortex shedding if the flow is perpendicular to the tube is predicted by
f = 0.22 * V/D
For V=80km/hr and D=19, that would give f = 257hz.
You might want to check this formula from another source...vortex stuff is not my thing.


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[paulmech01]

The tube is closed at both ends.

The angle of the airflow can change depending on where this is mounted.

Position a = perpendicular flow.
Position b = 45deg up from perpendicular.

 

[electricpete]

Do you know the E (Young's Modulus)? If so it will be an easy matter to compute those first three modes, assuming perpendicular flow. I don't know how to account for the 45 degree flow.



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[electricpete]

Correction - we need E and (mu or rho)

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[paulmech01]

The material is a prepreg E-Glass.

E = 30000 Mpa
Density = 1663 kg/m3

Not sure how this equates to a Linear density in lbs/in

 

[paulmech01]

I have plugged all these into an excel spreadsheet and I am now playing around with units metric/imperial.

Youngs modulus of E= 30000Mpa converts to 4351132lbs/in2.

I have corrected to rod OD to 16mm ID 14mm
Which has an I = 1328.9 metric = 0.00319 imperial

Density of 1663kg/m^3 = 0.06007969 lbs/in^3

Length L = 720mm = 28.34in

Putting all this info in gives the following values:

1st Mode 55.26 rads/sec = 8.79Hz
2nd Mode 345.43 rads/sec = 54.97Hz
3rd Mode 1899.87 rads/sec = 302.37Hz

Calculating the vortex shedding frequency from the given formula f = 0.22 * V/D with:
D = 16 mm
V = 22.22 (80km/h)
gives:
f = 305.53 Hz (Pretty close to the 3rd mode)

also if velocity = 22 m/s then freq = 302.5 Hz

Now I just need to verify these results somehow!

I'm open to sugestions on this, what do people think?

 

[electricpete]

I get frequencies 23hz, 145hz, 798hz, which do not match the vortex frequency calculated above.

Feel free to check my math:

Ishaft:=Pi/64*(Douter^4-Dinner^4);
Ishaft:=pi/64*((20.6*mm)^4-(19.0*mm)^4);
Ishaft := .245e4*mm^4

mu:=A*rho=Pi*D*t*rho
mu:=pi*19*mm*1.6*mm*(1663*kg/m^3)*(m/(1000*mm))^2;
mu := .1588238449*kg/m

f1 := 1/(2*Pi)*a1 * sqrt(E*Ishaft/(mu*L^4));
f1 := 1/(2*pi)*3.52 * sqrt(30000E6*Pascal*Ishaft/(mu*(720*mm)^4) *kg/(m*second^2*Pascal));
f1 := 23.2/second

f3:=f1*22/3.52;
f3 := 145./second

f3:=f1*121/3.52;
f3 := 798./second


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[electricpete]

Whoops. I was typing while you were posting. I didn't see the change in paramters.

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[electricpete]

You mentioned verifying the results. I'm not sure if you were talking about by calculation or by test. I assume by test.

One way to confirm the results would be using a strobe if the vibrations are big enough to see. If you can freeze the shaft at that frequency then you can confirm the frequency and the shape.

Direct measurement of the frequency with an accelerometer and data collector would be nice as long as the accel is not so heavy that it changes the system. Also you could move the accel up and down to try to map the operating deflection shape.

I tried the calc again with revised numbers and still different thatn your results...I may have made an error.

Ishaft:=Pi/64*(Douter^4-Dinner^4);
IshaftVALUE:=subs({Douter=16*mm,Dinner=14*mm,Pi=pi},Ishaft);
IshaftVALUE := .133e4*mm^4

mu:=Pi*(Douter^2-Dinner^2)/4*rho;
muVALUE:=evalf(subs({Douter=16*mm,Dinner=14*mm,rho=1663*kg/m^3,Pi=pi},mu)*(m/(1000*mm))^2);
muVALUE := .780e-1*kg/m

ConversionVALUE:=kg/(m*second^2*Pascal);
EVALUE:=30000E6*Pascal;
LVALUE:=720*mm;

f1 := 1/(2*pi)*a1 * sqrt(E1*Ishaft1/(mu1*L1^4)*conversion);
f1:=subs({E1=EVALUE,Ishaft1=IshaftVALUE,conversion=ConversionVALUE,mu1=muVALUE,L1=LVALUE,a1=3.52},f1);
f1 = 24.3 hz

f2:=f1*22/3.52;
f2 = 152. hz

f3:=f1*121/3.52;
f2 = 836. hz

Anyone else have comments on the calcs or ways to confirm?


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[electricpete]

For one thing, I think my unit conversion on pascals was wrong.

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[electricpete]

Disregard my last post - those units were right.

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[tomirvine]

Here is a beam natural frequency calculation program:

http://www.vibrationdata.com/beam.exe

Tom Irvine

http://www.vibrationdata.com

 

[paulmech01]

Thank Tom, that's handy.

Is this working on the same theory that we have been discussing.
ie.

Per Den Hartog's Mechanical vibrations
wn = a * sqrt(E*I/(mu*l^4)

also is this formula correct for calculating the vortex shedding frequency for a closed end circular hollow section dia 16mm x 720mm long, f = 0.22 * V/D.

 

[spciesla]

I'm using William T. Thompson's "Theory of Vibration with Applications" as a reference.

For the cantilever case, I get the following frequencies:
f1 = 24.4 Hz
f2 = 152.7 Hz
f3 = 427.6 Hz

The "a" factor for f3 is 61.6972 per this reference.

I calculated the vortex shedding frequency as
f = 0.21 * V / OD

For straight flow with V = 80 km/hr, fv = 291.7 Hz.
For the 45 degree angle of attack, you want the velocity normal to the cylinder, so multiply by sin(45). This results in fv = 145.8 Hz which seems pretty close to f2 (152.7 Hz).

Steve

 

[electricpete]

Hey Steve - I see your first two frequencies my first two frequencies (from my message 6 Mar 06 20:57). I was wondering if I had made a math error but now I don't think so since your's agree.

As far as the third frequency, mine doesn't agree. Mine does obeys the ratio of f3/f2 = a3/a2 as I have it listed above a3/a2 = 121/22. It must be either you made an error or I transcribed the coeeficients wrong. I'll check those coefficients when I get home tonight.

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[electricpete]

Steve - I re-read your message and see you have already identified the f factor which differs from mine and explains the difference. I'll double check...it could be I just read the wrong number.

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[paulmech01]

Thanks Steve,

I have just borrowed Thompsons, Theory of Vibration, I think I really do need a text on this topic.

The result looks good, I'll make sure it all seems Ok then I will get back to you guys with result.