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Gas Strut - Damping coeff and spring constant

  1. Jul 1, 2012 #1
    Hello everyone,
    I'm an engineering student and I have to "size" a gas strut (i.e. spring+damper) to model a trunk lid opening mechanism.
    I have a problem: on every "datasheet" (something like this http://www.strutsdirect.co.uk/components/variable-force-gas-struts.php) only forces and dimensions are written but there's nothing about damping coefficent and spring constant.
    So, I would like to ask you if someone could tell me some approximated possible damping coefficent and spring constant values or how to obtain them from a datasheet.
    Thanks.
     
  2. jcsd
  3. Jul 1, 2012 #2

    nvn

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    Maurk: Perhaps try here. E.g., for part number V06AAA0400, s = full stroke = 50 mm, P1 = initial force = 400 N, and P2 = final force = 1.2*P1 = 480 N. Therefore, spring constant is k = (P2 - P1)/s = (480 - 400)/50 = 1.60 N/mm. Therefore, when strut displacement is delta = 25 mm, the strut force would be P = P1 + k*delta = (400 N) + (1.60 N/mm)(25 mm) = 440 N.

    I currently suspect the viscous damping coefficient is relatively high, and that the strut is perhaps overdamped (zeta > 1), or only slightly underdamped (zeta > 0.7). But I do not know the damping coefficient. I hope someone else reading this forum can give you a very rough estimate of the viscous damping coefficient (c), or the viscous damping factor (zeta).
     
    Last edited: Jul 1, 2012
  4. Jul 1, 2012 #3

    AlephZero

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    This has some information on damping (some of the brochure is in German but the diagrams are labelled in English). See page 7.

    http://www.stabilus.com/fileadmin/docs/deutsch/Printmaterial/Prospekte_allgemein/Standardprogramm_2010_niedrig.pdf [Broken]

    The damping force (FR on the diagram) is effectively constant and independent of velocity, so trying to come up with equivalent viscous damping coefficient isn't very useful. A model of the dynamics is like a mass-on-a-spring, with the mass sliding on a table with a constant friction force whose direction is opposite to the velocity of the mass.

    When the mass is moving in one direction, the equation of motion is ##Mx'' + Kx = +F## and in the other direction it is ##Mx'' + Kx = -F## where ##F## is a constant force (not proportional to velocity, like a viscous damping force).

    The motion is a sequence of undamped half-cycles of simple harmonic motion, with the amplitude of each cycle reducing by a constant amount when the damping force changes direction.

    The equations of motion above don't have a unique equilibium position. The mass can stop at any position where ##|Kx| <= F##, i.e. at any position x where that the magnitude of the spring force ##Kx## is less than the "dynamic friction force" ##F##.

    For a real gas damper, the system usually "creeps" to the equilibrium position where ##F = 0## after it stops oscillating, but trying to model that final stage of the motion isn't of much practical interest.
     
    Last edited by a moderator: May 6, 2017
  5. Jul 3, 2012 #4
    I apprecitate your time and effort. Thank you everyone. I will use your suggestions.
     
  6. Jul 4, 2012 #5
    [itex]p\cdot V^{n}=p_0\cdot V^n_0=constant \\ \frac{p}{p_0}=(\frac{V_0}{V})^n \\ F(s) --> \int_{0}^{F} dF = A \cdot \int_{p0}^{p} dp \\ F=A\cdot p_0 (\frac{p}{p_0}-1) \\ F(s)=A\cdot p_0\cdot ((\frac{h}{h-s})^n -1) \\ c(s)=\frac{dF}{ds} = \frac{A\cdot dp}{ds}=\frac{A^2\cdot dp}{dV} [/itex]

    with some more integration we get
    [itex]c(s)=\frac{A\cdot p\cdot n}{h-s} [/itex]

    p=pressure
    V=volume
    h=height of cylinder
    s=h-current height of cylinder
    A=area of cylinder pushing on the gas
    n=polytropic constant

    also found this on the german wikipedia
    http://de.wikipedia.org/wiki/Gasdruckfeder
    i dont know german that well to get anything useful off this but try using google translate maybe it can help.
     
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