Structural Dynamics for Civil Engineers - SDOF Systems

Dr. Riya Catherine George

Department of Civil Engineering
Hiroshima University, Japan
Indian Institute of Technology, Kanpur

Lecture - 05
Coulomb Damped Free Vibrations

We have seen that the Damping in the structure can be modeled using Viscous Damping.
So, in viscous damping, the damping force is proportional to the velocity of the system
and the damping coefficient can be chosen, so that the energy dissipated by the viscous
damper is equal to the energy dissipated by all energy dissipation mechanism in the
structure. So, now, we will learn the different type of damping called Coulomb Damping,
so let us see how coulomb damping works.

(Refer Slide Time: 00:53)

In coulomb damping, the energy dissipation is due to the friction between two sliding
surfaces. So, the Coulomb Damped Free Vibration can be modeled as shown here.

So, we have a mass and stiffness and the damping will be due to the friction between this
mass and this surface; and the coefficient of friction in the surface is equal to mu(µ). So,
the frictional force in the surface is equal to mu(µ) multiplied by N, when N is the
normal force acting between these sliding surfaces. The direction of this frictional force
will be equal to the opposite direction of the force.
 F=𝜇N

So, if the mass is moving in this direction, the frictional force will act in the opposite
direction and the value of this frictional force is independent of velocity unlike the
viscous damping. So, once the motion is initiated, the value of force is independent of
velocity and this type of damping models are applicable when friction damping devices
are installed in the structure.

(Refer Slide Time: 02:15)

Now, let us write the equation of motion of this coulomb damped free vibration system.
So, the free vibration system is this. So, we have x is positive in this direction. So, we
have a restoring force k x will be acting on this mass in the opposite direction and an
inertia force will also be acting in the opposite direction.

So, now, to calculate the frictional force, it will depend upon the movement of this mass.
So, if the direction of motion of this mass is towards left, we have the spring force and
inertia force in this direction, weight acting in the downward direction, normal reaction
to the weight will acting in the opposite direction, the vertical direction; when the motion
is in this direction, the frictional force will be in the opposite direction.

So, now we can sum up the forces in x direction. So, we will have the inertia force plus
the spring force is equal to the frictional force. So, m x double dot plus k x is equal to F;
where F is the frictional force which is equal to coefficient of friction multiplied by the
normal reaction and the solution of this type of differential equation will be equal to A 1
cos omega n t(cosωnt) plus B 1 sin omega n t(sinωnt), this is the free vibration response
of undamped system plus the effect due to this constant force that is F by k.

So, this is equal to a static displacement, if F is the force and k is the stiffness of the
spring this is equivalent to a static displacement. So, omega n(ωn) as we know it is
square root of k by m.

So, now let us see how the equation of motion will look like, when the mass is moving in
the opposite direction that is towards right. So, the spring and inertia force will still be
acting in the same direction but when the mass is moving in this direction, the force will
be acting in the opposite direction. So, now, we have our equation of motion as m x
double dot plus k x is equal to minus F.

So, the solution of this can be written as A 2 cos omega n t(cosωnt) plus B 2 sin omega n
t (sinωnt), that is the free vibration response minus the static response due to this force F
that is minus F by k; the constants A 1 B 1, A 2 B 2 are calculated using the initial
conditions at each half cycle.

Because this condition the mass will be moving in one direction in one half cycle; and in
the next half cycle the direction of motion will be the opposite. So, these A 1, B 1 and
these constants need to be evaluated using the initial conditions at each half cycle. So,
this needs to be evaluated at each half cycle. So, we will have a separate solution in each
half cycle.
(Refer Slide Time: 06:15)

So, now let us see the solution of this free vibration response of a coulomb damped
system. So, we will assume initial conditions, displacement initial condition is equal to x
naught(X0) and we assume that the initial velocity is equal to 0.

So, using this initial condition, we will solve the equation of motions for this system. So,
what happens when t is equal to 0? So, we know the initial condition. So, we know that
the value of x at t is equal to 0 is equal to x naught(X0) and this is the initial condition.
So, here the mass is displaced towards right and then it is released.

x(t) = x0

So, the initial condition was a displacement towards right. So, when a body is displaced
towards right and it is released, it moves to the left. So, our first half cycle will be a
motion towards left. So, let us solve the equation of motion for the first half cycle.

So, the motion of the body is towards left. So, this is the equation of motion we have to
consider; m x double dot plus k x is equal to the force F. So, the solution is, this with a
plus term F by k. So, if you substitute the value of the initial displacement in this, for t is
equal to 0; we would get that x naught is equal to A 1 plus F by k, cos omega n t(cosωnt)
is 1 at t is equal to 0 and sin omega n t(sinωnt) becomes 0.So, we will get this equation
and from this we can write A 1 is equal to x naught(X0) minus F by k. Similarly, we can
differentiate this and substitute the value of initial velocity, which is equal to 0. So, if
you do that, we would get B 1 is equal to 0.
So, we have got both the constants A 1 and B 1. Now we can write this solution for the
system in the first half cycle. So, that would be x naught(X0) minus F by k cos omega n t
plus F by k.

So, it is a cosine function with an amplitude x naught(X0) minus F by k and it is shifted

in the positive y direction by F by k and this solution is valid only in the first half cycle,
that is when the time is between 0 and the half of the natural period; that is pi by omega
n (ωn) and this solution is valid only when the velocity will become 0.

So, at the end of this first half cycle, the body will reach the extreme left position and
then the velocity will be equal to 0. So, this solution is valid until then, that is during the
first half cycle.

So, now let us see the value of this displacement when at the end of this half cycle; that
is when time is equal to pi by omega n. So, if we substitute the value of t is equal to T n
by 2 or pi by omega n, we would get the value of x as minus x naught plus 2 F by k. So,
now, this is the initial displacement for the next half cycle and again at t is equal to this
value, we would have the velocity is equal to 0.

At t= , t= Tn/2

So, for the next half cycle, we will have initial displacement equal to this and initial
velocity equal to 0 and as I have mentioned earlier, this location is the extreme left
position. So, after that, the body will move towards right in the next half cycle.

So, at this position the velocity is 0.

(Refer Slide Time: 11:17)

The solution for the first half cycle looks like this; this is a cosine function with an
amplitude equal to x naught minus F by k and this cosine function is shifted towards the
positive x direction by an amount F by k. So, the velocity was 0 at the starting of the first
half cycle, that is the initial velocity and at the end of the first half cycle also the velocity
is equal to 0. Now let us see the solution in the second half cycle.

(Refer Slide Time: 11:56)

So, in the second half cycle, the time is between pi by omega n and 2 pi by omega n this
is the natural period of the system. So, in the second half cycle, the direction of motion is
towards right. So, the frictional force will act towards left.

So, the equation of motion is this and the solution is this with the negative F by k term.
So, the initial displacement for this half cycle is equal to the displacement at t is equal to
pi by omega n, which we found out from the first half cycle. So, that was, is equal to
minus x naught plus 2 F by k. So, substituting this in this expression, we can calculate
the value of A 2.

So, we can substitute the value of initial displacement and substitute the value of t in this
and we can calculate the value of A 2 as equal to x naught minus 3 F by k and here also
we can find out using the initial velocity, which is equal to 0 again; we can find out that
B 2 is 0.

So, the total solution for this second half cycle is equal to x naught minus 3 F by k plus
cos omega n t minus F by k; and this is valid only in this half cycle, and this is valid only
when the velocity becomes 0. So, when the value of t is equal to 2 pi by omega n that is
the natural frequency, that is when the second half cycle ends. So, at that point we will
have a velocity is equal to 0.

Now let us find the value of the displacement x at t is equal to T n; that is at the end of
this second half cycle. So, if you substitute t is equal to T n in this expression, we would
get x of t is equal to x naught minus 4 F by k. So, at the end of the first half cycle, the
displacement was minus x naught plus 2 F by k.

So, now at the end of second cycle, that is equal to x naught minus 4 F by k and this is
the extreme right position and now onwards, the body will start moving towards left and
that will be the starting of the next half cycle. So, similarly we can find out the response
during the next half cycle.
 At t= , t= Tn

(Refer Slide Time: 15:33)

The solution for the coulomb damped system for the second half cycle is expressed like
this, x naught minus 3 F by k cos omega n t minus F by k. So, for the second half cycle,
the solution is this, the red curve. So, in the first half cycle the response was this, and in
the second half cycle it is this. As you can see in this curve, the response at the beginning
of first half cycle was this and in one full cycle the amplitude decays.

So, we can calculate how much this decay is; and we can also see that at the end of half
cycles the velocity that is the slope of this curve is equal to 0. So, at the end of each half
cycle, we will have velocity is equal to 0 and the displacement will have a maximum
value. We have solved the coulomb damped free vibration system up to 2 half cycles;
now we will solve it for the third half cycle.
(Refer Slide Time: 17:02)

So, the thirds half cycle, the time is between 2 pi by omega n and 3 pi by omega n, so we
have calculated the response at the second half cycle. And so, there we have found the
displacement response at time is equal to 2 pi by omega n, so that would be the initial
displacement for this half cycle.

So, that is equal to x naught minus 4 F by k and in this half cycle the direction of motion
of the mass will be towards left. So, the equation of motion will be as per this relation m
x double dot plus k x is equal to F; and we will have the solution like this, plus a positive
F by k term.

Initial displacement

So, substituting the initial displacement value in this, when t is equal to 2 pi by omega n;
we can calculate the value of A 1 and that will be equal to x naught(X0) minus 5 F by k
and here also, we can substitute the value of the velocity which is equal to 0. We can
differentiate this and substitute the value of velocity is equal to 0, that will give us the
value of B 1 and which is again equal to 0.
So, this is the solution of the system in the third half cycle, x naught minus 5 F by k cos
omega n t plus F by k and this solution this a cosine function with some amplitude and
this shifted in the positive x direction by F by k; and the solution is valid only in this half
cycle, that is the third half cycle when time is between 2 pi by omega n and 3 pi by
omega n and the solution is valid only till the velocity becomes 0 next time; that means,
the velocity is 0 when the time is equal to 3 pi by omega n, so until that time, this
solution is valid. So, now, let us find out the value of the displacement at t is equal to 3 pi
by omega n, that is at the end of the third half cycle.

So, if you substitute the value of t is equal to t pi by omega n, we would get minus x
naught plus 6 F by k. So, at the end of the previous half cycle, it was x naught minus 4 F
by k. So, the, at the end of third half cycle it is minus x naught plus 6 F by k, this is the
extreme left position. So, in the third cycle, the body was moving towards left.

At t= , t= 3Tn/2

So, at the end of the third half cycle, it is at the extreme left position; and then onwards it
moves towards the right. So, that will be the fourth half cycle. So, we can continue the
solution like this.

(Refer Slide Time: 20:56)

And now let us see the solution in the third half cycle. So, this was the solution in the
first half cycle this is the second half cycle.

And now, this is the third half cycle and the solution is equal to x naught minus 5 F by k
cos omega n t plus F by k.

(Refer Slide Time: 21:21)

So, this is the solution of the coulomb damped free vibration system for an initial
displacement is equal to 1. So, this shows many cycles. So, as you can see the motion
decays at each cycle and the decay in a cycle, we can calculate and that would be equal
to 4 F by k.

So, in each cycle the amplitude of the response reduces by 4 F by k and when the
displacement response is less than the value of F by k; that is equivalent to a static
response. So, when the displacement response is less than the value of F by k, the motion
stops.

So, when at the beginning of half cycle, if or at the end of a half cycle; if the
displacement response is less than F by k, then the motion stops completely, because the
next half cycle will not be initiated. So, to initiate the next half cycle, we need a
displacement response which is greater than F by k.
So, if it is less than F by k, the motion will not be initiated, so the body stops moving. In
another way, we can say that the body stops moving, when the spring force that is k
multiplied by x is less than the friction force. So, to continue the motion of the body, we
need a spring force which is greater than the friction force; the spring force should
overcome the friction, then only the body will move.

So, we can say that, if the spring force is less than the friction force, the body will stop
moving.

(Refer Slide Time: 23:36)

Now we will solve an example problem in coulomb damped free vibration system. So,
an SDF system consisting of a weight, spring, and friction device is shown in the Figure.
So, you have a mass and you have a spring with a stiffness k and we have a friction
device as shown here.

The friction force is given as 10 percent of the weight. So, this will be equal to 0.1
multiplied by the weight of this mass and the natural vibration period of the system is
given and it is equal to 0.25 seconds. If the system is given an initial displacement of 2
inches and released, what will be the displacement amplitude after six cycles and in how
many cycles will the system come to rest? Now let us solve this problem.

So, it is already given that the friction force is equal to 10 percentage of the weight of the
system. So, that means; F is equal to 0.1 w, w is equal to mass multiplied by gravitational
acceleration and the natural period of the system is also given and it is equal to 0.25
seconds. So, we can calculate the value of F by k.

So, F is 0.1 multiplied by the weight, and k is the stiffness of this spring; the stiffness is
not given, but we have the mass, we have the weight, and the natural frequency. So, we
can rewrite it like, 0.1 mass multiplied by g divided by k and that would be equal to we
know m by k is equal to 1 by omega n square or omega n is equal to root of k by m.

So, substituting this value, we know that this is equivalent to 0.1 multiplied by g divided
by natural frequency square and this is the circular natural frequency. We know the
natural period; and we can calculate the circular natural frequency using the natural
period and omega n is equal to 2 pi by T n, so we can substitute that.

Now we know all the value in this expression, so we can calculate the value of F by k as
0.061 inches. So, we have learnt in the theory that, the reduction and displacement
amplitude per cycle is 4 F by k. So, F by k is this much. So, 4 F by k is equal to this
multiplied by for 0.244 inches. So, now, we can calculate the first question.

So, the displacement amplitude after 6 cycles, we know the initial displacement has 2
inches. So, in each cycle the amplitude will decay by this amount. So, after 6 cycles, the
amplitude will be 2 minus 6 times this. So, 2 by 2 minus 1.464 is equal to 0.536; that is
the amplitude after 6 cycles and now it is asked, in how many cycles will this system
come to rest?

We have just learned, that the system will continue vibrating until the displacement
response is less than F by k. So, we know that F by k is 0.061 and in each cycle the
displacement response will decay by 0.244 inches. So, at the end of six cycles, the
responses 0.536; so we can just guess in two more cycles, in one cycle the decay is
0.244. So, in 2 cycles it will be 0.488.

So, let us calculate the displacement amplitude after 8 cycles. So, after 8 cycles the
displacement amplitude would be; the displacement amplitude after 6 cycles minus 2
times the decay in a cycle. So, the displacement amplitude at the end of the 8th cycle is
equal to 0.048 inches, which is less than the value of F by k, which is equal to 0.061
inches.
So, at this stage there would not be any motion happening in the next half cycle. So,
therefore, the motion stops after 8 cycle. So, this way we can evaluate the response of a
coulomb damped free vibration system.