# Rayleigh Damping

1. Apr 9, 2013

### muzialis

Hi there,

I came across the concept of Rayleigh damping. I aqm told it is unrelated to viscoelasticity and as a result is unable to reproduce the behaviour of real materials under harmonic excitation.

I can not understand why.

Considering for simplicity a 1D setting, a ball of mass $$M$$ linked to a rigid wall by a spring of elastic constant $$K$$. Rayleigh damping dictates to model losses via a matrix (in our case, a scalar) given by $$D = \alpha M + \beta K$$.
The motion of the ball under an applied harmonic force is represented by the solution of the ODE
$$M\ddot{x}+D\dot{x}+Kx = F_0 cos(\omega t)$$.
Well it seems to be that this is analogous to considering the material as a Kelvin-type (spring and newtonian dampener in parallel) viscoelastic one. Not the best representation for real materials, but not too bad in some instances. Is all this correct?

Thanks

2. Apr 9, 2013

### AlephZero

One issue is the way the damping varies as the forcing frequency changes.

For example, at a fixed amplitude $x$, the physical damping mechanism may dissipate a constant amount of energy per cycle of the motion, independent of the frequency. In that case a better model of the damping for a steady state response is $M\ddot x + K(1 + i \eta)x = F_0 \cos(\omega t)$ where $\eta$ is the damping parameter (and $i = \sqrt{-1}$).

That model has some fundamental differences from viscous or Rayleigh damping. For example the phase angle between the applied force and the displacement is different.

Another issue is a damped multi degree of freedom system where the response has several resonance peaks at different frequencies. For Rayleigh damping you only have two parameters, so you can only get the correct damping at two of the resonances.

3. Apr 10, 2013

### muzialis

AlephZero,

I understand that the first model you mention, with the complex modulus $$K(1+\eta i)$$ might be more appropriate for some physical situations, and I appreciate your remark on the possibility of multiple resonating frequencies.

But in 1D, assuming that a Kelvin viscous model is physically appropriate, am I right in saying that for such situation Rayleigh damping is fully equivalent?

Many thanks

4. Apr 10, 2013

### AlephZero

For one degree of freedom, M D and K are just scalars (not matrices). If D has the value you want, it doesn't matter how you get that value.

For a multi degree of freedom system, you have $n^2$ matrix entries in D, and if you specify D by a "simple" formula like Rayleigh damping, they are not all independent.

5. Apr 10, 2013

### muzialis

AlephZero, thank you for your reply. I understand the differences between the scalar and tensorial case, I was checking my understanding of Rayleigh damping on a case I am particularly interested in, having been misguided by a statement from a colelgaue stating that Rayleigh damping is different from viscoelasticity even in 1D.