The following content is
provided under a Creative Commons license. Your support will help
MIT OpenCourseWare continue to offer high quality
educational resources for free. To make a donation or to
view additional materials from hundreds of MIT courses,
visit MIT OpenCourseWare at ocw.mit.edu. PROFESSOR: All right. I'm a little slow
getting started today, better get going. What we're going to talk
about today is a technique-- you guys done? OK, thanks. We're going to talk about
today a technique known as modal analysis,
and it's a way of analyzing things that
vibrate, essentially thinking about them one mode and a time. Though you might not
make a lot of use of the actual calculations,
doing the math, throughout your careers, I
think if you understand it conceptually it'll help
you just have a better understanding of what
vibration is all about, just give you some insight
to it that you otherwise wouldn't have. So the basic concept is
that you can model just about any structural
vibration as the summation of the individual
contributions of each what we call natural mode. So what we mean
by that is, let's start by thinking--
actually, let me say that this applies
to both continuous systems like vibrating strings or
beams or buildings as it does to finite degree of
freedom rigid body systems. We haven't talked about
continuous systems. I'll do a lecture on
it as the last lecture of the term, just kind of an
enrichment sort of lecture. But everything I say about
finite degree of freedom systems can be extended
to continuous systems. But since we've been studying
rigid bodies and systems with finite numbers
of degrees of freedom, I'll explain-- I'll go
through this analysis in the context of rigid body
finite degree of freedom systems. So in general, we can write
the equations of motion for finite degree of freedom
systems as a mass matrix. And to keep the kind
of writing down, I'm just going to underline
matrices and a squiggle under vectors so we have them. In general, we can write
the equations of motion as a mass matrix times
an acceleration vector plus a damping matrix
times a velocity vector, stiffness matrix times
a displacement vector, all equal to some external
vector of excitations. And I'm writing these as
if these are translations, but you know, like from doing
the pendulum on the cart problem, that the
equations of motion might involve rotations
and displacements. And we let them-- they mix
together here however they fall out. But just to write
them symbolically, I'm just going to refer to
all of those coordinates with an x vector. OK, now the basic
premise of modal analysis is a thing called the
modal expansion theorem. It's basically the
assertion that you can represent any
motion set of vectors-- I'll write them kind of as a
vector here for a moment-- x, as the superposition of
each contributing mode. Now each mode has
a mode shape to it, which I'm going to call
u, and up here I'll put a superscript
for what mode it is, the first mode, times
its time-dependent behavior. And this is called, what
they call in textbooks, the natural coordinates. And we'll see what
those are in a second. So mode shape one. This is the time-dependent
and amplitude part that says how much the
contribution of mode one is to this motion and what its
time dependence is, this is. And then you'd have mode
two's contribution, q2. And this goes out to the nth
mode's contribution, qn of t. And that's the
proposition, that you can represent the total
response of the system as a superposition of
the response of each of the natural
modes of the system. And if it's an n degree
of freedom system, there will be n
natural modes, so. Now something I didn't say here. This all assumes that
the system vibrates. So this is all in the
discussion of things that exhibit vibratory motion. So this is all, it should say
here, of vibrating systems, OK? So this kind of a long
and cumbersome way of writing this out. So if you notice, each one of
these is the mode shape vector. And if I put them together in
a matrix just side by side, here's a u1 over to un and
multiply it by this vector, q1 of t down to qn. That's the same
statement but said in a much more compact way. So this statement, this
modal expansion theorem, basically says the
vector of-- these are your generalized
coordinates, which we've been using all term long. These are the
generalized coordinates that you choose to derive the
equations of motion around. The vector of
generalized coordinates can be written as uq. And these are often called
the modal coordinates or sometimes called the
natural coordinates, OK? So if we can say that
x is uq, then x dot, you take the derivative of
each one of those expressions. You'll find that's
going to be uq dot. And x double dot
equals uq double dot because these
are just constants. The mode shape vectors are
just a fixed set of numbers that represent the mode
shape to the system. Now just to-- I think maybe
this is a good time to do this. You grab one end. So this is a-- and it's hard
to see black against black. My apologies for that. So this is a guitar string
or any stringed instrument. In fact, any long, slender thing
under tension will vibrate. And it has, if I
do this carefully, that's called the first
mode of vibration. And that's when you pluck
your guitar string or violin in the middle. You mostly hear that. But at twice the frequency,
if I can get it going here, there's a second
mode of vibration. And for a taut
string, it happens to be at twice the
frequency of the first. And if my hand is
well calibrated-- it may be easier if it's
a little longer-- if I get this going right,
there's a third mode, OK? So that's what
we're calling-- oh, what I meant to say
when I was doing this is these shapes for
the vibrating string, that second mode shape happens
to be one full sine wave. And the mode shape has the
form sine n pi x over l, where l's the length
of the string. n's the mode number. So first mode. Second mode is
this, when n is 2. First mode, n is 1. nth mode or something high,
you get higher modes like that. So these are the mode shapes
for a vibrating string. That's good for now. This two degree of freedom
system with the two lump masses-- and it's
going to show up there, yeah-- this is
basically two lump masses. And we idealize the
springs as being massless, but it's a pretty
good approximation. This has two modes of vibration. And Professor Gossard made
these neat little things that can make it so-- and I'm
going to come back to this, but there's mode one. And the mode shape
is as this goes down one unit, that goes down
about two times as much. I'll give you the exact
numbers in a minute. And the other mode
shape of the system-- we're going to talk about this
today and why this happens. But if I give it the
right initial conditions, I can make it vibrate only
in the second mode shape. And so it's now deflected
with the right conditions so that it'll respond
only in second mode. This mass goes up
and down a lot. That mass goes up in that
little, opposite to it, actually. The frequency is
different from the first. But if this is moving one
unit, then this down here is moving minus 0.3 or
something like that. And that ratio is constant. And that's called
the mode shape. So if you just pick one of them
and say, let its motion be one, then all of the other
masses in the system will move in a particular
ratio to the motion of that one that you arbitrarily set to one. So this is what we
mean by mode shapes and their natural frequencies. There's the natural
frequency associated with that first mode. And we can solve these
things mathematically, and we've been doing
that a little bit in the last couple of lectures. All right. So this is the relationship
between these things, the generalized coordinates
and the modal coordinates. And we now need to see how
we're going to use these. So in general, we have
our equations of motion. And I'm going to
substitute for x, x dot, and x double dot, these and
pre multiply by u transpose. Remember the
transpose of a matrix. You just take the first
column, make it the first row. Second column, make
it the second row. So if I plug in
these up here, I get muq-- I'm going to
leave some space here because I'm going to pre
multiply in a second-- plus cuq dot plus kuq equals the
external exciting forces. Now I'm going to pre
multiply by u transpose, OK? Now a remarkable thing happens. It happens that when you do this
calculation, when you multiply this matrix times that,
one row at a time-- so this has vectors in it, 1 through n. I'm going to pick
vector r, the rth one. If I take that rth vector
and multiply it one at a time by row by
row by row, then I get a new vector
that results, which I'm going to multiply by this. And so if I'm going to pick
out one of the vectors, multiply it through
times one of the rows here-- when you transpose them,
the rows are now the vectors. So I'm going to pick. If I do the calculation-- lost
my right piece of paper here. So I'm going to just pull
out one of the calculations that you end up doing if you
do this whole triple matrix multiplication, you need
to know the following fact. So for the mode s
transpose-- that's one of the rows out of
here-- times m times one of the columns,
the rth one from here, and I do this calculation,
this is 0 for r not equal to s. What that statement says is, the
only non-zero result from this is when you multiply-- when you
take the rth column from here and you use the rth from here. All the other combinations
of this thing go to 0. And the net result
of that is that this implies that u
transpose mu always equals to a diagonal vector,
which I'll call this like that. Sometimes a mass matrix
is diagonal to start with. But even if it isn't,
you do this calculation, it will produce a
diagonal matrix. And that's because these
multiplications are always 0 unless r is the same as s. And the same is true for u
transpose Ku will give you a K matrix that is diagonal. And you know,
normally the stiffness matrix we've come
up with, they've generally been full
matrices oftentimes. But you do uKu, you will get
a diagonal stiffness matrix. And there the
little problem comes because u transpose cu, well,
sometimes, this one is diagonal only for ideal
conditions of damping. So that's something you
just have to address. So only for ideal
conditions, and that's just something you
have to deal with. So why is this? Why is there this
special, wonderful thing? The natural modes of
a system-- this one is a two degree of freedom system--
form a complete and independent set of vectors. And in this case of this two
degree of freedom system, I can pick any kinematically
allowable position, like this-- stationary, static
is one of the solutions, right, to this two degree
of freedom system-- so any possible allowable
position of these two things, static or moving,
can be described by a linear combination of
the mode shapes of the system, a weighted sum of the
mode shapes of the system. And that's all it takes. So this one has two
mode shapes, one that looks like
that, one that looks like this one's going
down, this one's going up, their particular ratios. And I can take a weighted amount
of that first mode, so much of it, and a weighted
amount of the second mode and add them together and
describe any possible position of the system. The same thing is
true of that string. It has mode shapes
that are sine waves, but they're sine 1 pi x, sine
pi 2x, over and so forth. Any possible allowable
shape of that guitar string can be made up of a
weighted sum of the mode shapes of the system. And moreover, the mode
shapes, the reason this works is because
the mode shapes are orthogonal to one another. Now, you know that if you take
2 sine waves like that string and you take first
mode sine pi x over l, and second mode say
sine 2 pi x over l and you multiply them
together and integrate from 0 to l, what do you get? You'll always get 0
if the two sines are-- if they're full wavelengths,
they go to nodes at the end, you will always get 0 if the
wavelengths are different, always, right? That's a statement
of orthogonality of sine functions. All right. The same thing is true
of these simple vectors. They are orthogonal
to one another such that if you do this
multiplication, you transpose mu, you
only get contributions when you are using mode
r transpose m mode r. You only get a contribution
of each of those. That gives you the diagonals. The same is true when
you do u transpose ku. Because of orthogonality, you
only get a diagonal matrix at the end. And under the right
conditions, u transpose cu gives you a diagonal matrix. So what's that good for? Well, here was the
set of equations that we get when we
make that substitution. This is going to give
us a diagonal mass matrix times q double dot plus,
when conditions are right, a diagonal damping
matrix times q dot, plus a diagonal
stiffness matrix times q equals u transpose F, which
as a vector times a matrix gives you back a vector,
which we call capital Q. It's a function of time. And this is called
the modal force. But if you look
carefully at these, if I pick the rth one, mode
r out of this whole thing-- if I just pick any mode out of
this, any part of this vector, and complete this
multiplication, I will find that I get an Mr,
which is the rth entry here. And now I'm going to refer
to these as the modal masses, and I'll write
them with capitals and I'll give a subscript to
tell you what the mode is. This is a number. This is the modal
mass for mode r. This gives me an
equation that looks like Mrqr double dot plus crqr
dot plus Krqr equals Qr of t. And what does that remind you
of that we've done a lot of work with? AUDIENCE: [INAUDIBLE] PROFESSOR: How many
degree of freedom system? That's the equation of motion,
the generic equation of motion, of a single degree of
freedom oscillator. And you know how to
calculate the response to initial conditions for that. You know how to calculate
the steady state response for that when you
have a harmonic input, right? What I said at the beginning of
the discussion about vibration is it's really
important to understand the single degree of
freedom oscillator because it'll give you
insight as to the behavior of complicated multiple
degree of freedom systems. And here's the proof of this. This is now n uncoupled single
degree of freedom systems. This is n independent single,
one degree of freedom systems which you can solve
one at a time. Now, lots of times a vibrating
system, a complicated one, might be this thing. If I hit this, it's vibrating. And actually, it's
pretty much vibrating at a single frequency. And once I've hit it, are there
any external forces driving it? So what kind of
response are you seeing? Response to? AUDIENCE: Initial conditions. PROFESSOR: Initial
conditions, right? Now in general, each one of
the natural modes of a system has a different natural
frequency, right? So if I hit this thing
and I look at it, really, I can just see it
wiggling back and forth basically at one
frequency So if you wanted to come up with a
simple model of this system, how many natural
modes you think you'd have to include to describe
the motion of this system? AUDIENCE: One. PROFESSOR: One. Now is that a lot
easier than having to do the full general
equation of motion for all the possible
modes that this thing has? And it turns out
a lot, right, you have to deal with the
equation of motion of a single degree of freedom
system to describe this. And that's the real point. You know they built
the Hancock building across the river 35 years ago. It was losing
windows like crazy. It was a brand new building. And when the wind would
get above 40 miles an hour, the windows started falling out. 60 stories high, 60, 61 stories
high, and the wind was blowing. Where do you suppose
the windows fall out? What part of the building? AUDIENCE: [INAUDIBLE] PROFESSOR: Huh? I mean, you'd think that
when the wind is blowing it get stronger as it goes up. It was probably blowing out
the windows at the top, right? But the windows were
breaking-- as time went by, every time a window would break,
they replaced this five foot by nine foot sheet of glass
with a piece of plywood. And so you get this
statistical sampling after a while of where
the breakage was. So you had no windows
broken at the top and a few as you got further down and
more and lots of them broken out at the bottom. It turns out that that
building was vibrating mostly in its first bending mode. It was going back
and forth like this. Also happened to have
a torsional mode. Its first torsional mode
was kind of twisting around the base like that. So in fact the
moment when the wind would get about
40 miles an hour, this building would start
rocking and rolling, mostly like this with a
little of this thrown in, OK? But you can basically model
that complicated building that has millions of possible
natural modes in it by one or possibly two single degree
of freedom oscillators. So that's the power
of modal analysis. But I think the real
power of understanding that you can do this
is that it gives you this immediate
insight as to what might be going on in something. So I look at this. I don't see a
complicated thing that I have to model with a big
finite element model. I see something that's
vibrating at one frequency. And I know it has
a little decay. It'll have damping. It'll have a natural frequency. And I get immediate
insight as to its behavior by knowing this, OK. And that's the real reason why
I wanted to show this to you. So today we'll do-- there
are sort of two directions we can go with this. One is to talk about response
to initial conditions, and the other is to
talk about the response to force excitation. So we're going to begin by
doing response to ICs, OK? And then we'll move
on probably next time and talk about response
to harmonic excitations. And we're going to use
that as the example. Before I go there, once we
have broken the system down and analyzed this way,
how do we get back to the motion of the system in
our generalized coordinates, which are the ones
we're comfortable with? Because I don't know where
to take a ruler and go measure this natural coordinate. So somehow I have to
get back to putting in the real physical
measurements that I can relate to. Well, that's easy because
where did we start with this? We started by saying this
whole thing began right here. And so at the end we just
come back and say, oh, well, x here, our generalized
coordinates, is this summation of the
mode shapes ui here, summed over i of qi of t. Now the reason I wrote
it here as a summation is to remind you
that you do this. That's the beauty
of this thing, is you only have to do it
over the modes that matter. So if you've decided to
approximate the motion of this complicated
system, by just a couple of motile contributions
because you know they're the
important ones, this is a pretty short summation. This is how you get back to
your original modal coordinates. Just take the modal amplitude,
multiply it by the mode shape. And when you do that, it
says, if this turns out to be, say, sum a sine omega t,
when you multiply by the mode shape it basically
tells you how much each generalized coordinate
gets of the motion. The mode shape
distributes thing answer out proportionally in
the correct amount. So this is how you get
back to the original. So let's think about
that system and we'll do an initial conditions
kind of problem. So I think Professor
Gossard-- I think in class you sort of figured
out what the approximate ks and ms and things
were for that system. So I actually took
it apart, weighed it, measured some
natural frequencies, and have come up with
a pretty good model, or at least pretty
good set of numbers, characterizing this two
degree of freedom system. So c1, k1, m1, k2, c2, x1, x2. So these are my generalized
coordinates, measured probably from what position? Static equilibrium, right? So I don't have to mess
with gravity in this. Measured from
static equilibrium. And to try to help keep
things understandable, I tried to write the
parameters of the system as lowercase k1s,
k2s, k3s because I want to write modal stiffness
for mode one as a capital K1, so I try to be
consistent about that. And notice where I put
the dampers in the system. That's because most of
the damping in this thing comes from the upper mass rubs
against a stationary object, which is the bar here. The lower mass rubs against
a stationary object. So I'm going to model that as
a dashpot between each mass and the fixed reference frame
because the bar doesn't move. So it's an approximate
model of the damping. And so if we do our sum of
forces on each of these masses, just do Newton's
laws on the mass, we can come up with our
two equations of motion. We get two equations of motion. And let's see. I think I'll give you some
information here first. m1. And I really don't
know the damping, but we can get that by just
counting how many cycles it takes to decay and so forth. So that's basically what I
come into this problem knowing. And I'm going to write
my equations of motion in matrix form. So it's going to end
up looking like m1. Now notice the damping in
this one, the damping force, is only proportion--
it'll be c1 x1. Doesn't involve the motion
of the other object. In this one, the
damping force only involves the second motion. So this one happens to
look like a c1, 0, c2. And the stiffness matrix, well,
that's k1 plus k2, minus k2, minus k2, and k2, x1, x2. And for no external forces,
this starts off this one has nothing on the right
hand side. it's equal to 0. So those are my
equations of motion. And you know if you
multiply these out you'd get two equations. And each one would
be this result that you get by apply
Newton's law to mass one and Newton's law to mass two. But you we've done
that enough times. I'm not going to go
through that part of it. OK. And putting it in real numbers,
that's our mass matrix. I don't know this. My stiffness matrix. So this is my K matrix here. And stiffness matrices,
they're always symmetric. Although this one
happened to be diagonal, you'll find that mass matrices
and even the damping matrices for our linear
systems are symmetric. So here's my stiffness matrix. Here's my mass matrix, OK? And also in this case
here's my damping matrix, but I'm going to leave that
because it's the one that's a little bit troublesome. So what do I need to do to this
to carry out my modal analysis? So I need to go find the results
of computing u transpose m and u and transpose Ku. And let's see what we get. So we need to know a couple
things about this system. We need to know natural
frequencies and mode shapes. So if we have this
mass matrix and we have the stiffness matrix,
then we know we can cast this. We want the undamped natural
frequencies and our mode shapes. And we know that
we can transform the equations of motion into
an algebraic problem where we solve for the natural
frequencies and mode shapes. So we have, just to remind
you really quickly of that, remember our equations
look like this undamped. And you assume that x is some
form u in fact e to the i omega t. Plug it in, you get minus
omega squared m plus K u e to the i omega t equals 0. And this now is your
algebraic problem. e to this unknown set
of amplitudes is 0. These are going to turn
out to be the mode shapes. And they're not generally 0 so
that means this has to be 0. That means we know the
determinant of this matrix. And that'll give you
in this case the two natural frequencies. This gives of you the
omega ns of the system. Omega n squareds is
what you solve for, OK? And then you go back and you
get the mode shapes out of it. But this you can do
on the computer too. You can either crank out-- for
a two degree of freedom system, this gives you a
quadratic omega squared. You solve it. You plug it back in and
get the mode shapes. I'm not going to
take the time to do that today because I want to
emphasize the modal analysis part. So I'll give you the answers. Where are we here? So you get omega 1 is 5.6546. And I seem to be keeping a
lot of significant digits, and there's a reason for that. In both mode shapes
and natural frequencies you need to carry a lot
of significant digits or modal analysis doesn't
work, or at least you don't get the clean
results you expect. If you're sloppy about the
number of significant digits and you compute u
transpose mu, then the [? off ?] diagonal terms
won't quite go to 0. And it's just because you're
not carrying enough precision. OK, now that's the two
natural frequencies. Now the u matrix, the mode
shapes for this system that goes with that. u comes out to be
1.0 and 2.2667. And that's mode. I'll do this to help you. The columns are the mode shapes. That's the first mode shape. And the second mode shape
is 1 and minus 0.2236. So those are the mode shapes
for the first and second mode that go with these two
natural frequencies. So that's for this system. The top one moves one unit. The bottom one moves 2.27
times that, same direction, positive, positive. So the upper one moves one unit. The bottom one moves
the opposite direction-- that's the minus signs--
equivalent to a phase angle of 180 degrees
minus 22% of the amount that the upper one moves. So first one moves one unit. The bottom one moves
2.2 times that. And then the second mode, which
is much harder to get going. Guess the only
way I can do it is to do it the way Professor
Gossard intended here. One unit up and down, minus
0.2236, going the opposite way. So those are our mode shapes. These are the
natural frequencies. I calculated this one and
measured it with a stopwatch. This one I can do watching
it with a stopwatch. And I came within better than
1% of getting the same number. OK. So I want my model mass matrix. I carry out this calculation. And for this system,
remember, it's going to give me back a diagonal
matrix looking like this. And in fact, the numbers are
3.5562, 0, 0, and 0.3508. And when I calculate
u transpose Ku, gives me a diagonal
stiffness matrix. And I get the numbers
113.71 and 0, 0, 109.839. And that's my diagonalized
stiffness matrix. Now something had
better be true. I'm saying that
this is now going to give me my two independent
single degree of freedom equations of motion, right? So what I'm seeking here, I
want to get two equations, one that looks like m1q1 double
dot plus c1q1 dot plus K1q1 equals 0 for no external force. That's one of the
equations I'm after. And the other one will look like
m2q2 double dot plus c2q2 dot K2q2. Now one way to check that
you've gotten the right thing is now these are two independent
single degree of freedom systems. What's the natural
frequency of this system? Yeah? Actually, I heard somebody
say square root of K1 over m1. That had better be true. But numerically
what's the number? What had it better be? It better be the omega
1 of the system, right? And so a check that
you can perform is to check to see if the
omega 1 squared equals K1/m1. You found two numbers. You've got, up
here, K1 is 113.7. m1 is 3.55. Take K1/m1, and take
its square root. So K1/m1, that's
about 30 something. Square root of 30 something
is a little less than 6. Omega 1 is 5.65. And same thing,
omega 2 had better be equal to the
square root of K2/m2. So one of the things
you can always do when you do your
modal analysis, you do your calculations, u
transpose mu, u transpose Ku. If you calculate the ratios
of each one of these things, you can go back
and check that you can see that the
natural frequencies are the ones that you started with. If they are not, then you've
messed up in your arithmetic. So now we've got our two
independent equations. And the natural
frequencies check out. But we still have a couple
of things to deal with. We have to figure
out how to calculate the initial conditions,
and we have to figure out how to deal with damping. Let's do ICs first. So those of you who
were here last time, I ended kind of
right at the end. We kind of worked our
way through figuring out the initial conditions for a
two degree of freedom system doing it the hard way. You end up with four
equations and four unknowns for the a1, a2, phi 1, phi two. Remember that? I mean, it's really painful. This is incredibly easier. We're going to do the same
thing, but extremely easily. So I would never go myself
given the choice of grinding out all those phase
angles and amplitudes in simultaneous equations. I'd do the following. Generally now I know the
initial conditions are going to be specified
not in q coordinates but in what coordinate system? In your original generalized
coordinates, right? You know, your x, this one. If I'm going to set
initial conditions here, I'm not going to say q1
is equal to something. I'm going to put this one
down one unit and this one down two units and let go. This is in x1 and
x2 coordinates. But the beautiful thing here is
that we know that x equals uq. So if I know the initial
conditions on I'll call it x0 here, if I know
the initial deflections of the system,
they're going to be u times the initial values of q. And if I know a vector of
initial velocities at time 0, they're going to be uq0 dot. So if I told you
values of x0 and you know that this equation's
true, what we need is the q0s. We need the initial
conditions in the modal coordinates in order
to finish this problem. If I told you this, how
would you solve for that? Just a little
linear algebra here. AUDIENCE: Inverse matrix of u? PROFESSOR: Yeah,
do what with it? AUDIENCE: Then you
multiply x by it. PROFESSOR: Multiply it
by u inverse, right? OK, so this implies
that q0-- well, I'll write it out a
little more fully here. So if I do u inverse
x0, that's going to be equal to u inverse uq0. u inverse times u gives you? 1, basically, right? And so if I do u
inverse x0, I get q0. That's all there is to it. Yeah? AUDIENCE: [INAUDIBLE] initial
conditions, what about c? PROFESSOR: All right,
c's a problem, OK, and I'm leaving it to the end. We're going to deal with
it as the last step. And if I have initial
velocities u inverse times x initial velocities vector, I get
the initial velocities vector in the natural coordinates. So that's how simple it is
to get the initial conditions in modal coordinates. Boom, OK? And we'll do a numerical
example in a second here. We're seeking a solution
of the form to do response to initial conditions. We seek equations
that we know are right for a single degree
of freedom system response to initial conditions. So we know that for a single
degree of freedom system, x of t-- this is for
SDOF system here-- we worked out before is equal
to some e to the minus zeta omega nt. This is just a transient decay
problem of x0 cosine omega dt plus v0 plus zeta omega n x0
all over omega d sine omega dt. We know that that's what the
response of a single degree of freedom system looks like to
initial conditions x0 and v0. And for light damping,
for small damping, you can usually even
ignore this term. So it's just even simpler. This term is small compared
to that, all right? This term, contribution from x0,
is small compared to this term. So it's basically dominated
by an x0 cosine and a v0 over omega d sine. But we know that's
the exact response for a single degree of freedom
system to initial conditions. So just by analogy
to that, we're looking for mode one
in modal coordinates. It's going to look
exactly the same way. e to the minus zeta omega
nt, q0 cosine omega 1 d-- this is omega dt-- plus
q0 dot plus zeta 1 omega 1 q0. I guess I need to
do q10 like that. This is the first
mode's equation, zeta 1. And I'll call this omega 1. But now that you get multiple
degree of freedom systems, you got to keep track of what
mode you're talking about. Mode one, damping
ratio mode one, natural frequency mode one,
initial displacement mode one, initial velocity mode
one, omega 1 d like that. And mode two is going
to be exactly analogous. q2 equals, and it's
exactly similar, except you update it
with a 2 instead of a 1. And if you plug in the
initial-- you over here have found the initial values
for q10 and q20 and q1 dot 0 and so forth. You found the initial values
that plug into that equation by just doing this. And once we have this, then
we can go back to saying, how do you get to
the final answer? Well, you just
multiply the result for q times the mode
shape and add them up. And you have the answer. But we still have to deal
with the damping problem. We're going to do that one next. But I see a bunch of hands
and some puzzled looks, so it means it's a good time
to stop and talk for a second. Yeah? AUDIENCE: [INAUDIBLE] PROFESSOR: Pardon? AUDIENCE: What if
you [INAUDIBLE] PROFESSOR: I can't quite hear. AUDIENCE: The sine theta
and the sine rate of g. PROFESSOR: Yeah, what about it? AUDIENCE: Why do we lose it? PROFESSOR: Why do we use it? AUDIENCE: Lose it. PROFESSOR: Oh,
you don't lose it. I was saying, you see
this bit, it's like that. These are two pieces
that behave like sine. And see, this one depends
on initial displacement but is multiplied by
the damping ratio. And the damping ratio for
things that are interesting is usually pretty small. So here you have a term that's
x0 cosine omega 1 damped, and here you have a contribution
that's x0 small times x0 sine omega 1 damped. So you multiply the same. They're operating on
the same frequency. Two terms at the same frequency,
you add them together, it's like a cosine omega
t minus some phase angle. If this little term is small,
that phase angle's almost 0. x0 cosine plus
something x0 sine, it gives you a cosine term
that is shifted a little bit and its magnitude is different
by this little amount. I'm just saying
oftentimes this is small. But if you don't want to
make that approximation, just carry it along. Just do it. AUDIENCE: [INAUDIBLE] PROFESSOR: Mm-hmm. AUDIENCE: [INAUDIBLE] PROFESSOR: Oh, no. I'm just saying you can
throw out this piece usually. And it makes-- I keep
in my mind-- let me see. OK, now. Vibration engineering is full
of lots of approximations because it's very
hard oftentimes to get detailed quantitative
numbers on exactly everything you need to know. So I carry around
little approximations that I know is the way
the world mostly behaves. And the way the
world mostly behaves for a single degree
of freedom system is the response to initial
conditions looks like this. And this initial value here
is always approximately x0. And this initial slope here
is always approximately v0. That's the slope. Now, it turns out that this
thing is shifted just slightly. Why? Because of this term, OK? But honest, to
tell you the truth, it really rarely matters. So as a vibration
engineer, I just remember I have an x0 cosine. I have a v0 over omega d sine. And the whole thing
decays like that. But if you like to be
mathematically precise, you carry along
that a little bit. Yeah? AUDIENCE: Don't you lose
[INAUDIBLE] sine wave? PROFESSOR: You're not
going to lose the sine. AUDIENCE: [INAUDIBLE] PROFESSOR: Oh, oh, oh. Wait a minute. I just left it out. You guys are-- well,
I'm glad you're awake. This is good. Now how's that? Ah, good. Now I know why I had
so many puzzled looks. Anybody have a
different question? Just anything now about this
whole modal analysis thing? Because then we have to deal
with this awkward part that has to do with the damping. And I've got to
finish on time, OK? All right. So damping. I've gotten this far. What I need is I need estimates
for the damping for mode one and damping for mode two, right? So the problem is that
utcu does not always equal some nice
diagonalized matrix. You sometimes get these
are not always 0, OK? The orthogonality
principle just doesn't apply to the damping terms. Just doesn't. But this actually
doesn't hurt you a lot. You just got to know that
this is going to be a problem. And when the systems
are lightly damped, the approximation, even if
your true damping in the system gives you some
non-zero elements here, the first order
behavior of the system is basically going to
be-- you can just ignore the off-diagonal elements. What practical
consequence do you think it has if you have some
actual non-zero numbers here? Go back and look
at the equations that you're trying to derive. These were the equations that
we were trying to come up with. And we wanted them to be
n individual single degree of freedom systems. But if this has non-zero
off-diagonal terms, you're going to find popping
up in this single degree of freedom equation another term
that couples it through damping to the other modes. It provides a little bit
of coupling to other modes. They can talk to one
another, all right? And what that means is if
I-- this may be a good time to do the demonstration. How do I want to say this? If the initial
displacement of the system is in the shape of one
of the natural modes-- so if this is some u, this is
exactly shaped like mode r. So this looks like
the ur vector. When I carry out
this multiplication, what do you think will happen? If this is shaped
like mode r, because of orthogonality when
I do u inverse, which is all about the mode shapes
information and the mode shapes are these orthogonal
set of independent orthogonal vectors, if this
is exactly one mode and I do u inverse times
that, I will get 0 over here on the right hand side for
every mode except the mode that that's shaped like. So if this is shaped
like a particular mode, then over here all the
modal initial conditions are 0 except that mode. That means if I set this, give
its initial conditions are equal to the shape
exactly of mode one, it only responds in mode one. And if I give it initial
conditions that are exactly shaped like that of
mode two, then it only responds in mode two. And if I give it anything else,
like I move just the top one but not the bottom
and let it go, then there's-- maybe I
better do the other one. That one had too much of
one and not the other. If I hold this one,
here's its reference. I'm going to hold it
right there and I'm going to give this one a
unit deflection and let go. Now you see a get some
of both, all right? So if when I do this
first one, say first mode, I could sit here and
measure how many cycles it takes to decay halfway and
estimate the damping ratio for that mode. If it's only moving
in this mode, I can estimate its damping
directly for that mode and get zeta 1. You agree? OK. And I did the same thing with
mode two, it's too fast for me to catch it with
a stopwatch, but I could measure its damping. And as it decays, I could
get an estimate for zeta 2, for the damping ratio
for mode two, all right? All right. But somehow I have
to get damping ratio for mode one, zeta
1, and damping ratio for mode two, zeta 2. I have to somehow
get it out of this. I have to model it somehow with
these damping coefficients that come from computing
this u transpose cu, OK. So I'm going to show
you kind of damping called Rayleigh damping, OK? Lord Rayleigh, who did
lots of things in science that you've probably
run into, proposed that if you model your
damping, the c matrix as-- this is just now the
system damping matrix that you start with-- some
alpha times the mass matrix plus beta times the stiffness
matrix-- these are now the original ones in your
generalized coordinates, just your mass and
stiffness matrices. If you say, I'm going to
approximate my damping model like this, then I want to
compute u transpose cu. I'm going to get
alpha u transpose mu plus beta u transpose Ku. But we know that this gives you
the diagonalized mass matrix, known as the modal mass matrix. This gives you the
diagonalized stiffness matrix. And so this damping
model, this is guaranteed to give you a
diagonalized damping matrix which we'll call, somehow,
some capital C2, 0, 0, C2, all right? And it's going to be alpha
times the modal mass matrix plus a beta times the
modal stiffness matrix. And those alphas and
betas you adjust. They're just
parameters you adjust to get the amount of
damping you need, OK? So for a two degree
of freedom system, C1 here is alpha
m1 plus beta K1. Modal mass, alpha
times the modal mass plus beta times the
modal stiffness. That's what you get
for the first one. And C2 is alpha m2
plus beta K2, OK? And the alphas and betas
give you two free parameters you can play with. And for a two degree
of freedom system, I can manipulate alpha and
beta to get the damping that I measure. And I forced my equations
of motion a couple. Now, Mother Nature may
say, you know, Vandiver, they don't uncouple,
and there's going to be a little
crosstalk between them. But I say, yeah,
but to first order I'm going to get a pretty
good model of the system. So let's do that in this case. Let's maybe just to keep
it-- I've got numbers here, so let my notes so I don't
get completely lost here. So I'm going to just
pick one for now. I'm going to model
my damping with just beta K, beta times my
diagonal, my stiffness matrix. And let's see what happens here. So that says my modal
damping is going to be some, for mode one, beta K1. Now what's damping ratio? Zeta 1 for a single
degree of freedom system is the damping constant for
the system over 2 omega 1 m1. But that's going to be
beta K1 over 2 omega 1 m1. But m1/K1 is omega 1 squared. So I get an omega 1
squared in the numerator. Beta omega 1 squared
over 2 omega 1. Remember, the K over the m
gave me the omega one squared, so the ms are gone. You can cancel one of these. This gives me beta
omega 1 over 2. So this now gives me a way I
can fit one of the dampings. I can get exactly what I
want, say, for mode one if I pick beta to
be the right number. OK, so in this case, I
actually did some numbers. Pardon? Can't hear you. AUDIENCE: [INAUDIBLE] PROFESSOR: No. K1/m1 is omega 1 squared. Omega 1 squared
takes care of the m1. I get rid of one
of the omega 1s. I'm left with this, OK? OK. So let's just let
beta equal 0.01. And if you let
beta equal to 0.01, then zeta 1 equals
0.01 omega 1 over 2. We know omega 1 is 5.65. This when you work
it out then gives you a number of 0.0283, about 3%. And that would say that
this system when it vibrates in mode one is going
to damp out up to 50% in about three cycles. Not bad approximations. I'm just guessing
about what it is. That's a reasonable amount
of damping for mode one. Now the problem is when I only
use just beta K as my model. Now I'm stuck with whatever
happens for mode two once I pick beta
because zeta 2 is going to be beta omega 2 over 2. And omega 2 is
quite a bit larger, so now I'm stuck with a greater
value for the second mode. In this case, it's 0.0885. So if I just pick a one
parameter model for my damping, I can make one perfect. I can match it
perfectly, but then I'm stuck with whatever
the other one is. So I did this because I
could do it simply with one. But if I'd kept the full
two-parameter model, with manipulating
both alpha and beta I could actually get both
of the two measured dampings exactly right. But if I have an n
degree-- if I have three degree of freedom system,
I only have two parameters. I can fit two of
the damping ratios, but then I'm going to be stuck
with whatever it gives me for the third. But oftentimes it's just one
mode you really care about. It's the problem mode. You're at its natural frequency. It's going like crazy. Initial conditions make it
vibrate a lot in that mode. But this is what Rayleigh
damping allows you to do. It guarantees you that you
will have a diagonalized set of equations of motion. And it gives you two
parameters that you can play with to fit the
damping model however you want. Once you have damping, now
you have the complete solution for decay from
initial conditions. And there's your two models. You can solve for q1, transient
decay given initial conditions. You can solve for q2,
transient K of the second mode. And then to get
back to the initial to the response in terms
of your modal coordinates, you just add the
two together, OK? I got some numbers here
which are just instructive. u Inverse. In order to get these
initial conditions, you've got to know u inverse. Do we know u? I gave us u. Here's our set of
mode shape vectors. And I've run out of boards. So we have the u matrix. We need u inverse, so u
inverse for this problem. And we're going to
quickly do some examples. Let's let the v0s be 0. No initial conditions
on velocities. And let's do x0, the initial
displacements, be 1 and 0. So the 1 and 0,
what we're saying is the bottom one doesn't
move, unit deflection here, let it go. What are you're going to get
for the initial conditions? x0 equals 1 and 0, well,
that implies that the qs are going to be u inverse x0. So by the way, if
this is true, this implies that all q dot initial
conditions equal 0, right? No initial velocities
in generalized coordinates, no initial
velocities in modal coordinates. But we are going to have
an initial deflection. I want to then compute u
inverse x0 and see what I get. And what I get back when I do
this one is 0.0898 and 0.9102. Remember, this is q10, q20. So for that case, it says
I'm going to get 0.08 or 0.09 equal to q1 and 0.9 of q2. And I go back over here
to my transient decay. There's no velocity. So it's basically going to
look like q10 cosine omega dt, e the minus zeta
omega t, decaying, cosine. But for mode one, its initial
amplitude's less than 0.1. And mode two, it's
got a lot of mode two. So what happens? So unit deflection here, in
fact it's mostly mode two. And just quickly
I'll do one other. x0 is 0, 1. That implies that q0
that you get from that is 0.4016 and minus 0.4016. Says you get about
equal amounts. So that's this one. I don't move this one,
but I give this one a unit deflection, let go. I get about equal
amounts of each one. And of course I've told
you the answer to this one. If I let x equal mode
one's mode shape, 1 and 2.266, that implies
that q1 equals 1 and q2, when you multiply it out, is zero 0. If I deflect it in the shape of
mode one and I do u inverse x0, I will get back 0 and 1. And if I make this
the shape of mode two, I will get back 0 for mode
one and 1 for mode two. I've out of time, but that's
your intro to modal analysis. So I think it's
conceptually powerful. Yes. AUDIENCE: How did you
get from the 0.898 value to the 0.0898 value? AUDIENCE: The inverse
should be 0.0898. PROFESSOR: Oh, is this 0.08? Yeah, OK. I may have written that down. Yeah. I'll double check that. But yeah, question? AUDIENCE: Why is it for
here that we picked c to be only a function of-- PROFESSOR: Beta? AUDIENCE: A. PROFESSOR: Because I want to
get done by the end of the-- AUDIENCE: OK. PROFESSOR: --60
minutes, 80 minutes. I could have put
both of them in, manipulated both parameters
as two equation with [? two ?] modes, two target
values of dampings. I'd find an alpha
and a beta that would make both work exactly right. Actually, just this one model
is pretty good for this case. The damping for
second mode is greater than the first mode
just happens to be. This model's not bad. AUDIENCE: All right,
so you try the three and see what gets
you the best results? PROFESSOR: Yeah.
