Coulomb Damping
 Coulomb Damping Viscous Damping Single DOF In order to prevent spam, users must register before they can edit or create articles.

## 1 Introduction to Coulomb Damping

Figure 1: Single DOF model with Coulomb Damping

Steady friction forces occur in many systems when relative motion takes place between adjacent members. These forces are independent of amplitude and frequency; they always oppose the motion and their magnitude may, to a first approximation, be considered constant. Dry friction can, of course, just be one of the damping mechanisms present; however, in some systems it is the main source of damping. In these cases the damping can be modelled as in Figure 1.

The constant friction force Fd always opposethe motion, so that if the body is displaced a distance x0 to the right and released from rest we have, for motion from right to left only,

 $LaTeX: m\ddot{x}=F_{d}-kx$ or $LaTeX: m\ddot{x}+kx=F_{d}$ (2.11)

The solution to the complementary function is

 $LaTeX: x=A\mbox{sin}\left(\omega t\right)+B\mbox{cos}\left(\omega t\right)+\frac{F_{d}}{k}$ (2.12)

where

 $LaTeX: \omega=\sqrt{\frac{k}{m}}$ rad/s

Note: The particular integral may be found by using the D-operator. Thus Eqn (2.11) is

 $LaTeX: \left(D^2+\omega^2\right)x=\frac{F_{d}}{m}}$

so

 $LaTeX: x=\left(\frac{1}{\omega^2}\right) \left [ 1 + \left(\frac{D^2}{\omega^2}\right) \right ]^{-1}\frac{F_{d}}{m}$

or

 $LaTeX: x=\left [ 1 - \left(\frac{D^2}{\omega^2}\right) + \hdot \right ] \frac{F_{d}}{m\omega^2}=\frac{F_{d}}{k}$

The initial conditions were $LaTeX: x=x_{0}$ at $LaTeX: t=0$, and $LaTeX: \dot{x}=0$ at $LaTeX: t=0$. Substitution into Eqn. (2.12) gives

 $LaTeX: A=0$ and $LaTeX: B=x_{0}-\frac{F_{d}}{k}$

Hence

 $LaTeX: x=\left(x_{0}-\frac{F_{d}}{k}\right)\mbox{cos}\left(\omega t\right)+\frac{F_{d}}{k}$ (2.13)

At the end of the half cycle right to left, $LaTeX: \omega t=\pi$ and

 $LaTeX: x_{\left(t=\pi / \omega\right)}=-x_{0}+\frac{2F_{d}}{k}$

That is, there is a reduction in amplitude of $LaTeX: \frac{2F_{d}}{k}$ per half cycle.

From symmetry, for motion from left to right when the friction force acts in the opposite direction to the above, the initial displacement is therefore $LaTeX: \left(x_{0}-\frac{4F_{d}}{k}\right)$, that is the reducion in amplitude is $LaTeX: \frac{4F_{d}}{k}$ per cycle. This oscillation continues until the amplitude of the motion is so small that the maximum spring force is unable ot overcome the friction force Fd. This can happen whenever the amplitude is $LaTeX: \le \pm \frac{F_{d}}{k}$. The motion is sinusoidal for each half cycle, with successive half cycles centered on points distant $LaTeX: +\frac{F_{d}}{k}$ and $LaTeX: -\frac{F_{d}}{k}$ from the origin. The oscillation ceases with $LaTeX: \abs{x} \le \frac{F_{d}}{k}$. The zone $LaTeX: x=\pm \frac{F_{d}}{k}$ is called the dead zone.

To determine the frequency of oscillation we rewrite Eqn. (2.11) as

 $LaTeX: m\ddot{x}+k\left(x-\left(\frac{F_{d}}{k}\right)\right)=0$

Now if $LaTeX: x^{'}=x-\frac{F_{d}}{k}$, $LaTeX: \ddot{x}^{'}=\ddot{x}$ so that $LaTeX: m\ddot{x}^{'}+kx^{'}=0$, from which the frequency of oscillation is $LaTeX: \left(\frac{1}{2\pi}\right)\sqrt{\left(\frac{k}{m}\right)}$ Hz. That is, the frequency oscillation is not affected by Coulomb friction.

## 2 Vibration with both Viscous and Coulomb Damping

The free vibration of dynamic systems with viscous damping is characterized by an exponential decay of the oscillation, whereas systems with Coulomb damping possess a linear decay of oscillation. Many real systems have both forms of damping, so that their vibration decay is a combination of exponential and linear functions.

The two damping actions are sometimes amplitude dependent, so that initially the decay is exponential, say, and only towards the end of the oscillation does teh Coulomb effect show. In the anlyses of the cases the Coulomb effect can easily be separated from the total damping to leave the viscous damping alone. The exponential decay with viscous damping can be checked by plotting the amplitudes on logarithmic-linear axes when the decay should be seen to be linear.

If the Coulomb and viscous effects cannot be separated in this way, a mizture of linear and exponential decay functions have to be found by trial and error in order to conform with the experimental data.