# Gas Strut - Damping coeff and spring constant

1. Jul 1, 2012

### Maurk

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. Jul 1, 2012

### nvn

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
3. Jul 1, 2012

### AlephZero

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

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
4. Jul 3, 2012

### Maurk

I apprecitate your time and effort. Thank you everyone. I will use your suggestions.

5. Jul 4, 2012

### spanky489

$p\cdot V^{n}=p_0\cdot V^n_0=constant \\ \frac{p}{p_0}=(\frac{V_0}{V})^n \\ F(s) --> \int_{0}^{F} dF = A \cdot \int_{p0}^{p} dp \\ F=A\cdot p_0 (\frac{p}{p_0}-1) \\ F(s)=A\cdot p_0\cdot ((\frac{h}{h-s})^n -1) \\ c(s)=\frac{dF}{ds} = \frac{A\cdot dp}{ds}=\frac{A^2\cdot dp}{dV}$

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

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.