Coulomb damping

Coulomb damping is the damping that occurs due to dry friction when two surfaces slide against one another. Coulomb damping can be there result of a mass sliding on a dry surface, axle friction in a journal bearing, belt friction etc. The case of a mass sliding on a dry surface is analyzed here, but the qualitative results apply to all form of Coulomb damping.

Imagine the simple system that consists of block of mass m and the spring with stiffness k. The spring is a connection between the block and the wall. The block of mass m slides on the dry surface and a friction force that resists the motion develops between the mass and the surface.

 

Figure 1 – a) Mass slides on a surface with a kinetic coefficient of friction µ; b) free-body diagrams at an arbitrary instant of time when x’>0; c) Free body diagrams at an arbitrary instant of time with x’<0 .="" span="">

The Coulomb’s law states that the friction force is proportional to the normal force developed between the mass and the surface. The constant of proportionality is called the kinetic coefficient of friction. Since the friction force always resists the motion, its direction depends on the sign and the velocity.

So by applying Newton’s law to the Free Body Diagram Method we derived the following equation which describes the motion of the system.

$$\begin{align} & m\ddot{x}+kx=-\mu mg\text{ \dot{x}}>~\text{0} \\ & m\ddot{x}+kx=\mu mg\text{ \dot{x}0} \\ \end{align}$$

 

Previous equations are generalized by using a single equation.

$$m\ddot{x}+kx=-\mu mg\frac{\left| {\dot{x}} \right|}{x}$$

The right hand side of previous equation is a nonlinear function of the generalized coordinate. Thus the free vibrations of a one-degree-of-freedom system with Coulomb damping are governed by a nonlinear differential equation. However, an analytical solution exists and is obtained by solving previous equation.

Without loss of generality, assume that free vibrational of the system shown in Fig 1. are initiated by displacing the mass a distance δ to the right, from equilibrium, and releasing it from the rest. The spring force draws back the mass toward the equilibrium position; thus the velocity is initially negative. Equation

$$m\ddot{x}+kx=\mu mg\text{ \dot{x}0}$$

,

applies over the first half-cycle of motion, until the velocity again becomes zero. Solution of the previous equation subjected to

$$\begin{align} & x\left( 0 \right)=\delta , \\ & \dot{x}\left( 0 \right)=0 \\ \end{align}$$

gives: 

$$x(t)=\left( \delta -\frac{\mu mg}{k} \right)\cos {{\omega }_{n}}t+\frac{\mu mg}{k}$$