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V squared damping after impact

  1. Nov 1, 2011 #1
    1. The problem statement, all variables and given/known data
    vinit = sqrt(2g*h); h = drop distance
    vfinal = 0;

    xinit = 0;
    xfinal = 100mm;

    a = g;

    Issue: non-linear damping.
    M*x'' - b*(x')^2 - k*x = 0;
    b = 128*mu*(length fluid travels)*(D^4(piston)/[(D(hydraulic)^4)(orifice opening)]


    every book I've been reading on vibrations damping says there's no solution for v^2 damping. Currently reading "Influence of Damping in Vibration Isolation" and they give an equivalent linear damping coefficient as:
    C(eq) = (D0)/(∏ω(z0)^2); D0 being energy dissipated per cycle, z0 being relative displacement.

    Then they go into equivalent damping force being: γ*F; γ= (2/sqrt(∏))*gamma((n+2)/2)/gamma((n+3)/2)...


    very long story short. Is there anyway to do a stepwise energy dissipation of a mass/spring/damper problem? Can I use something like: initial energy in - energy to compress spring - energy dissipated by damper = 0.

    the issue I think is with this is that I don't know how to figure out energy dissipated by a damper whose dependent on v^2...

    can I define a function that says: this system was deflected by 0.1mm at this time and Z amount of energy was taken away from the initial impact. W energy was taken up by the spring, and X was taken by the damper. This is how much much energy was left over at the boundary of this iteration...

    Im in analysis paralysis at the moment and I think I'm overthinking this...
     
    Last edited: Nov 1, 2011
  2. jcsd
  3. Nov 1, 2011 #2

    rude man

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    Isn't x'' = a = g = constant? So
    b(x')^2 + kx = ma?
     
  4. Nov 1, 2011 #3
    Yes. The thing is figuring out what b value I need for any drop between 25mm and 1200mm. At a height of 1000mm what b do I need to be critically damped? At a height of 400mm what b do I need to be critically damped? and so on. Vibration teacher says that the impact is making this problem very difficult to solve. Because the damping is non-linear and because the area of the orifice is also changing (see fluids orifice measurements) this problem becomes non-uniform area, non-linear damping.

    I have the non-uniform area calculations by using the hydraulic diameter: Dh = 4*(area of orifice)/(wetted perimeter). I have the damping coefficient calculations: 128*mu*L*(d(piston)^4)/(d(orifice)^4)...now I dont have the damping coefficient necessary for any given height so that I can get the orifice diameter required at that height.
     
    Last edited: Nov 1, 2011
  5. Nov 1, 2011 #4

    rude man

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    Oops, my initial assumption is incorrect, on reflection (even though you agreed!).

    I though the equation would be 1st order but it's not. And I have to cconcede that I wouldn't know how to approach it. What's the background material for this problem? I mean, numerical techniques or whatever? What's the complete statement of the problem?
     
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