Free Vibrations of a Single Degree of Freedom (SDOF) System with Viscous Damping Video Lecture - Civil Engineering (CE)

this video will explore the single
degree of freedom problem with damping
start off by drawing a support and the
mass will call this mass M assume that
it's attached with a linear spring
Spring has stiffness K and also a linear
damper damping constant C and this is
fixed we'll assume our x coordinate is
positive downwards like that so the next
thing we need to do is draw the free
body diagram whereby we cut the mass
away from its supports and replace those
with forces some mass now we would have
F sub s the spring force and a damper
force which we'll call X sub C and we
note that the spring force F sub s is
equal to minus K times X that is it
opposes the displacement it's a
restorative force and the damping force
similarly AFC opposes the motion minus C
times X dot all right so we start off by
writing Newton's second law
second law that says ma or em x
double-dot in this context is equal to
the spring force plus the damping force
substituting minus K X minus C X dot and
we choose to rewrite it as M X double
dot plus CX dot plus KX is equal to zero
this is our equation of motion that we
need to solve it's a second-order
differential equation and mus requires
two initial conditions the initial
conditions in this case are very simply
the displacement X at time equals zero
is equal to some constant x0 and we'll
assume the velocity X dot at 0 is equal
to some known constant V 0 so we proceed
like before where X is assumed to be of
the form C we'll use C bar to keep it
different from the C two different
constant e to the RT which implies that
X derivative is equal to our C bar e to
the RT second derivative X double dot is
equal to R squared C bar e to the RT
substituting this into equation 1 gives
us our characteristic equation the stick
equation track it like this in our
squared C bar e to the RT plus C our C
bar e to the RT plus K c e to the RT is
equal to 0 now for non-trivial solutions
C bar is nonzero so that must cancel e
to the RT is never 0 so that was cancel
and you're left with the characteristic
equation that says M R squared plus C R
plus K equals 0 we just move this up
all right the solution to the
characteristic equation the roots of
it's just a second-order polynomial
roots are given by r1 + 2 equals minus C
plus or minus the square root of C
squared minus 4 M K all divided by 2n
now you might remember from differential
calculus original equations that we need
to consider three different cases based
on what this quantity inside the square
root the square root sign looks like
three possibilities either it's greater
than zero or it's equal to zero or it's
less than zero let's look at the
critical case and that's when it's equal
to zero okay so we're assuming that C
squared minus 4 n K equals zero and
we'll call this the critical case see
that you pull it very simply this
reduces to C equals the square root two
times the square root of M cap
we'll call that C critical we'll also
take the time to note that C critical
can also be written as taking the M out
to M square root of K of M but we know
from before that Omega N squared equals
K over m so therefore C critical is
equal to 2m or negative this will come
in very handy in a second okay so the
solution proceeds as before where as the
damping reduces to zero you get an under
damped case and this becomes the square
root of a negative number and it leads
to oscillations well in the map that
leads to sine a cosine and the physical
description of that is an oscillatory
motion if C if the damping is
sufficiently high then what happens is
it really retards all motion and you
don't get oscillations in fact as C goes
to infinity one of these roots goes to
zero and you find that there's no motion
it'll motions seizes up and then finally
obviously if if it's equal to zero you
get the case of critical damping and
critical damping is defined as the
damping that brings the system back to
rest most quickly if you are trying to
reduce vibrations as quickly as possible
in the system then critical damping is
what you want okay before we go through
the solution what I'm going to do is
choose to rewrite this in what I would
call non-dimensional terms Engineers
like to use certain parameters so that
when they're solving problems and
instead of solving just one problem at a
time they're solving entire families of
problems and this becomes significant
because some of these codes run for a
couple of days on computer and
you want to run it as little as you have
to anyway so we got back to our original
equation which said that MX double dot
plus CX dot plus K X equals zero this
was the equation of motion and we define
a parameter that we're going to call
Zeta and say that is very simply the
ratio of damping to your critical
damping if Zeta is equal to 1 then
you're at critical damping if Zeta is
equal to zero then you've got no damping
it's less than 1 you've got under damped
situation it's greater than one you've
got an overdamped situation so the
damping ratio also means that C can be
re-written as the damping ratio times C
critical and we know C critical yet all
right so if I take the equation of
motion and divide both sides by M I end
up with X double dot plus C over M X dot
plus K over m x equals zero but we know
what C over m is right because we know
from here we write this out we write it
in a different color
if I follow this through C is equal to
Zeta times C critical which is 2m Omega
n - M Omega n as I'm running out of
space here that means that C of M is
equal to just rewriting this - Zeta over
again so I can rewrite my equation of
motion in this form as X double dot plus
2 Zeta Omega n plus Omega N squared
equals zero when I then assume that
excess of the form see this here C bar e
to the RT plug it in I'm going to end up
with a characteristic equation that
looks like R squared plus 2 Zeta Omega n
bar plus Omega N squared equals 0 and my
roots are 1 & 2 equals minus 2 Zeta
Omega N 1 over 2 plus or minus square
root do that something spirit of B
squared so that would be 4 Zeta squared
Omega N squared minus 4 times 1 times
Omega and skate
that's all / - all right let's make a
little more space on there simplifying
this is equal to minus Zeta Omega n plus
or minus the force all cancel again end
up with the square root of Z 2 minus 1
times Omega N squared we give this
quantity in a mistake excuse me say 2
squared minus 1 we call this Omega T or
the damped frequency of vibration as
Zeta goes to 0 the frequency of
vibration vibration goes to Omega n you
can see that chair so we can rewrite
this one so once again as minus Zeta
Omega n plus or minus Omega T which
finally means that X is equal to a 1 e
to the minus Zeta Omega n
times in fact I want to take this
constant out you're just erasing some of
how we do so this is just equal to e to
minus Zeta Omega n T multiplied by we
use the constant a 1 here cosine Omega
DT plus a2 sine Omega DT
and again this is subject to the initial
conditions that X of 0 is equal to X 0
and B 0 as X dot X dot of 0 equals BC
alright and that's where we're going to
leave it for now and leave it as an
exercise to you to go and do the
substitutions here to determine the
constants in by inspection you should be
able to see that a 1 is equal to X 0 and
then you can plug that in and so forth
for a two very easily well see you in
the next video thank you