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Natural frequency of a wind turbine

  1. Dec 19, 2010 #1
    how do we find the natural frequency of a tower of a wind turbine if, say, the tower is circular hollow cross sectional shape (its not tapered or anything, just cylindrical all the way up). I think i know how to find it when we're ignoring its weight, but what is it when weretaking into account its weight as well?
  2. jcsd
  3. Dec 19, 2010 #2


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    There are two things that contribute to the stiffness of the structure (Well, actually more than two, but let's ignore the others for now!)

    The "elastic stiffness" comes from the properties of the material (Youngs modulus, etc). That is what you use to calulate the frequencies, ignoring any stresses in the structure.

    There is also a "stress stiffness" or "geometric stiffness", that is caused by the work you have to do when you move a body that has non-zero stress in it. A simple example is a string under tension. The string has almost zero elastic stiffness if you move the mid-point sideways, but there is also a stiffness proportional to the tension (or stress) in the string. This comes from the fact that if you move a point on the string sideways, you change the angle of the string , so a component of the tension force is now opposing the displacement of the string.

    The geometric stiffness of a string with uniform tension is simple. In your tower example it gets more complicated because the axial stress (compression not tension) is zero at the top and increases linearly to the base. The change of stiffness along the length will change the shape of the vibration mode compared with an unstressed cantilever beam, as well as changing the frequency.

    In practice you would make a computer model (e.g. a finite element analysis) to calculate the stress distribution, and then include both the geometric and elastic stiffness in the vibration calculations.

    The geometric stiffness is also important in buckling analysis. In fact if the frequency lowest frequency mode was reduced to zero by the internal stresses, the structure would buckle. (But buckling analysis is not actually done this way, because buckling doesn't depend on the mass properties of the structure that are needed to do a vibration analysis).

    For many structures this effect is small, but not always. For example the natural frequencies of large rotor blades may change by a factor of 2 or 3 times between zero RPM and the maximum operating speed.
  4. Dec 19, 2010 #3
    First of all, thank you
    well for the time being, I can ignore stresses of any kind. I know modulus of elasticity for the material. for case one (ignoring the tower's weight but taking into account mass of rotor and hub which I know of their values) I suppose I can make use of beam deflection theory and use equation wl^3/48EI (w being force) to find force and from there, using hook's F=Kx find K and ultimately find natural frequency using sqrt(K/m). (m being mass of hub and rotor).
    However it is when I need to work out natural frequency taking into account tower's own mass and weight that I am not quite sure where i'm going? can we still do the above process (ofcourse only if its right anyway) but in working out the natural frequency we use equation: natural frequency= sqrt(K/(m+mass of tower) ??
    Thanks again.
  5. Dec 20, 2010 #4


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    I think you are using "mass" and "weight" to mean the same thing. They are NOT the same thing, so that is confusing!

    If you ignore the mass of the tower, then you have a "mass on the end of a spring", where the mass is the rotor+hub and the spring is the cantilever beam.

    This is ignoring the rotational inertia of the blades. If you are thinking about a small turbine (say 5 or 10 kW) on a pole, that would be a reasonable assumption, but not for a large (say 1MW) turbine.

    Your formula is wrong because it assumes all the mass of the tower is at the top.

    The easiest way to do this is to make a computer finite element model. If you want to do it by hand, you could assume the mode shape of the vibration and find the strain and kinetic energy based on that shape (Rayleigh's method). If you assume the mode shape is the static deflected shape for a cantilever, you will finish up with a formula like
    frequency = sqrt(K/(m + b.mass of tower) where b is a constant less than 1.

    I found a reference on Google that gave b = 0.23, but I don't have a RELIABLE source for that number.
  6. Dec 28, 2010 #5
    Yes i thought my formula is wrong as it assumes all he tower's mass to be at the very top. the answer to first part is right i know (ignoring tower's mass and having rotor+hub mass at the top end). i just used 48EI/L^3 and equated it to k. once i found k, finding natural frequency was straight forward.

    can i just post the whole question first:
    The tower of a wind turbine is 54m and is constructed from a circular hollow cross section with an inner diameter of 0.3m and an outer diameter of 0.5m. The rotor and hub mass is 1.5x10^3kg and the tower is made from steel with Young’s modulus of 200Gpa.
    a) What is the natural frequency of the tower when its weight is ignored?

    b) What is the natural frequency when the towers weight is also taken into account?

    Where the area moment of inertia about x or y axis is I=(pi/4)(D^4 - d^4) where D is outter and d is inner diameter
    The equivalent mass of the tower is Meq=0.2357ρAL
    The density ρ = 7800kg/m3

    we're not considering stresses here.
    also for the second part, as we can calculate the value of moment of inertia easily, I thought I may treat it like a compound pendulum where the natural frequency is omega=sqrt(mgl/I) l being length and I moment of inertia. however it only has one mass in it where as i need to consider rotor+blades and also tower's mass.
    thank you.
  7. Dec 28, 2010 #6


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    That's the key sentence. Check out how your course defines "equivalent mass" or "effective mass".

    I would take it to mean that the actual (physical) mass of the tower is clearly M = ρAL, and for the first vibration mode it acts as like a mass Meq = 0.2357M at the top of the tower. That's the same as the factor of 0.23 I quoted before.
  8. Dec 30, 2010 #7
    I think I know what I'm doing now. Thank you......
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