Natural Frequency of Pipe

Having "played" with such things at times in my lifetime (guitar strings, for example; pun intended), I have to say that the formula you have found for calculating the natural frequency of an object may or may not be correct. I do recall that I once researched that subject, but I don't recall the formula. Why do you doubt it? Have you determined that it is not accurate by experimentation? Whack it with a hammer and measure the frequency with an accelerometer or microphone.

1.You asked in fact 3 questions, I only answered one to my knowledge. That is the natural frequency of your pipe with the offered length.

As for the other 2:

2.I tried to work out your given formula, but cannot relate that to the natural frequency. I don't see how a frequency should come out of it, when you work it out with units only. Maybe some one else can?

3.You should be able to cancel one pi in both, as a math rule.

Do you want the frequency of the pipe itself, or that of the air column enclosed by the pipe? If it is the Pipe itself, do you want the longitudinal or the transverse frequency? I'm not sure, but I suspect that the fundamental frequencies of the pipe itself will depend on the wall thickness of the pipe; I am sure that they will depend on how the ends are anchored. If anchored by knife edges, the nodes will be at those knife edges; if the pipe is embedded in concrete, the nodes will be some distance away from the support points.

thanks dkwarner

the pipe is knife edge at both the ends.

no we need the frequency of the pipe itself.

we need transverse frequency

basically this piping assembly is used in refinery application

That helps. I believe one could find the wall thickness, since you gave the weight per inch, and the material of steel. Now since it is in a refinery, I presume the pipe is to be completely filled with a viscous fluid of around 0.7 specific gravity. The weight of that fluid will affect the frequency of vibration.

If you are concerned about vibration either due to resonance with machinery or due to earthquakes, in a pipe long enough to have several supports, then distribute the support points at non-regular intervals, so each section of pipe has a different natural frequency.

I suspect the simple beam formula provided in another post will give a close enough approximation. If it is for a facility that already exists, then as someone else pointed out, use a microphone attached to a good oscilloscope. I'd use a block of wood rather than the hammer mentioned. Measurement takes all factors into account; formulzs frequently leave some out.

The pipe that I have has 4 different non-regular intervals. How Can I find the different natural frequencies for each section?

Thank you

First, welcome to CR4!

Now that last post of mine was almost 10 years ago... but I'm still here!

A lot more information would be useful, such as pipe and support dimensions and materials, whether the pipe is empty or full, if it is full, what is the viscosity of the fluid (that will have a large effect on the damping properties of the liquid), etc.

But the answer(s) remain pretty much the same: place a vibration sensor (microphone, geophone, etc.) near the center of each section and connect that sensor (through an amplifier if necessary) to an oscilloscope that has a frequency readout. Stimulate the section (eg. hit it with a soft mallet). It will probably take quite a few tries before you can get the 'scope adjusted to provide a reliable readout.

A better way would be to use a variable frequency oscillator with a loudspeaker (or just an electromagnet, if the pipe is of a magnetic material) on or very near the pipe to stimulate the pipe. The stimulation point should be some distance from the sensor. In this case you could use a sensitive voltmeter either in addition to or instead of the oscilloscope. As the frequency of stimulation approaches and passes the natural resonant frequency of the pipe, you should be able to observe a significant peak in the amplitude/voltage of the signal from the sensor.

Kaisan, you are getting close and DK is right with his arguments too. The formula you bring on makes sense. Just the wall and the diameter is missing.

About the ends and the attachment to a mass, we know nothing and the vibration will have a strong damping due to the fixation.

The air column has no energy to make the pipe resonate, my earlier post makes no sense for the question asked. I have no books here and it is too long ago. Thanks for the help. D.

it seems your formula is correct, the only this required now is λ(frequency factor). i will find and will let you know.

The factor λ= 9.87 applies to a simply supported beam (pipe). If both ends are fixed, λ= 22.4. To obtain a correct answer, one must use realistic boundary conditions. Roark is a good source for the equations.

1- We have in engineering a powerful tool named "dimensional analysis". In this case it is not necessary to stroke the tube to see if the formula is right or wrong. Let us look at it

- the frequency is in cycles/ second and thus has the dimension 1/T (T for time) to obtain this result under the root sign the dimension should be 1/T^2. Is is the case or not? Under the root we have E*J*g/(m*L^3) if we write the dimensions we obtain:

(F/L^2*L^4*L/T^2)/(F/L*T^2*L^3)=(F*L^3/T^2)/(F*L^2*T^2)= L/T^4. It does not correspond to the required result thus the formula is wrong. But if we have a second look we can find where the error is, let assume that we take off "g" then the result will be 1/T^2 exactly we expect it to be! Now let us go a bit further let assume that "m" is NOT the mass of the pipe but its WEIGHT then the "mass" = Weight/g and it is right to have "g" at the place where it is. So that either the formula was written with an error or the interpretation of it was wrong. The error can be as well in the source as by the user. who knows?

2- In case that the 2 ends CANNOT rotate (fixed ends) the wave length is equal to "L" the pipe length between the supports, the given value of 2*L is valid ONLY for rotating end and in the assumption that no pipe is present on the other side of support!

3- Pipe bending frequency has NOTHING to do with sound speed in air it depends ONLY on the supports and the pipe stiffness in bending and its specific mass per length unit.

4- The stiffness is proportional to 2 factors:

- the material factor "E" Young' s modulus and

- the geometrical factor J the inertia moment of the section which is dependent on both outer diameter and wall thickness so that we come to

5- It is not necessary to bring in the formula in an EXPLICIT form diameter and wall thickness since they are ALREADY present in an IMPLICIT form.

6- The mass per length unit is as well proportional to 2 factors

- material factor the specific mass in "mass units/volume" =and

- a geometrical one the section area

7- If we come back to the initial equation in the correct form we have E*J/(m*L^4)=

(E/ρ)*(J/A)/L^4. It is interesting to notice that the speed of sound in a material is proportional to √ E/ρ which shows again that the bending frequency is related to the sound speed in the pipe material and not in the air! The factor J/(A*L^4) is a geometry related factor which contains ALL parameters for the section as for the length. There is a further parameter which considers the supports.

The ONLY correct answer was brought by Kaizan GA to you some other answers would deserve an off topic note.

I wrote several times that we are when giving an answer taking a responsibility that it is correct, it is better if we are not sure to let others give the answer and learn than to bring an erroneous one.

There are MANY natural frequencies of a pipe. The suggestion to whack it with a hammer and capture the response with an accelerometer is your best way to evaluate it. The response waveform will be complex due to superimposed first bending, second bending, repeated roots due to lack of perfect symmetry, etc.

Use a scope with a simple FFT capability to find all of the natural frequencies. Be sure to filter the signal at about 1/4 of the sampling rate to prevent aliasing.

I believe the response will be simpler if you use a softer 'whacker' than a hammer. It will also depend on the 'Whack Location'. Around 1/3 of the length should guarantee fundamental and first harmonic vibrations.

Thanks for expanding on this.

Both suggestions are correct. The compliance of the hammer "tip" should be just enough to excite the highest frequency mode of interest. This will prevent the "double hit" phenomenon. We used to use a Big _ Hammer with screw-on heads of differing materials to tailor the excitation frequency range.

Secondly, the location of the "hit" must not be right at a node of any large mode. A choice of 1/3 would be good, because the second bending mode is near a maximum there, and the third would be pretty close to maximum also.