27. Vibration of Continuous Structures: Strings, Beams, Rods, etc.

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visit MIT OpenCourseWare at ocw.mit.edu. PROFESSOR: Today we are going
to talk about the vibration of continuous systems. Not covered on
the quiz, but it’s a really important part
of real-world vibration and the most– one of the
easiest ones to demonstrate, I’ve shown you this one
before, is the taut string. But I want to show you something
unusual about– something you may not know about strings. Wait until it calms
down here a little bit. OK, so this is your guitar
string or a piano string. It’s under tension. We’ve already seen
that it exhibits natural frequencies
in mode shape, so there’s the first mode. Looks like half a sine wave. Has a particular frequency
associated with it. Get it to stop doing
that– but if I excite it at twice the frequency– I
don’t know if I can do this. There we go. That turns out to
be exactly twice the frequency of the first. The mode shaped
one full sine wave. The mode shapes for a taut
string are sine n pi x over L. But strings can do
something else kind of neat. And that is if I
hit this thing– I’m going to wait till
it calms down here. If I give this
thing just a pulse, what do you expect to see? Are you going to see vibration? Tell me what you see. What do you see happening? Something running
back and forth. Right? What you’re seeing
is wave propagation. It’s not really vibration. Vibration we see of its modes
and standing waves and things like that. Right? So the taut string satisfies
an equation of motion that’s called the wave equation. We’re going to talk quite a
bit about that this morning. And the wave equation has
its name give something away. The wave equation describes
continuous systems of a particular kind that
support travelling waves. And so the string
will both support– I can give it a little pluck. I’ll try to just place it in
a particular shape and let go. There it is. And that little pluck just goes
back and forth back and forth at a particular speed. So is there a relationship
between the speed at which things can
travel in a string and the natural
frequencies of the string? Well, we’ll get into that today. And I’m going to start
by just showing you a little something that
comes from my research and– let’s see. Let me do this. I think this will work. Hear that? As I go slower, does
frequency go up or down? It’s kind of slow, and
I’m going to speed up. Right? Goes up as the speed goes up. So that’s the result
of the phenomenon called flow-induced vibration. And I’ll give you
a very brief intro to flow-induced vibration. You have a cylinder sitting
still, flow coming by it– water or air. The cylinder is diameter D,
velocity U, for the flow. What happens in the wake of that
cylinder, vortices are formed. And just like if you’re
paddling a canoe or something and stick a paddle
in the water, you’ll see vortices shed off the side. First you get one
that’s positive and then one that’s negative
And so one full cycle of this is from here to here. There’s a frequency
to this shedding. And the shedding
frequency, FS, in hertz, can be predicted by a simple
dimensionless parameter called the Strouhal number, St U over
D. And that’s approximately 0.2 U/D for stationary cylinders. You can predict the frequency at
which these vortices are shed. Now, associated with
the shedding of vortices is a lift force. I’ll call it some
FL cosine omega t, which is 2 pi FS, times t. So at this frequency
of vortex shedding there is a transverse force. There’s actually an
inline force also, which I’ll call FD for drag. And it goes like cosine
2 omega s times t. It’s twice the
frequency of that. And so you’ll get some
inline oscillatory excitation and what we call cross-flow
oscillatory excitation. And this is the cause
of lots of things that the people who work on it
call flow-induced vibration. Now, an amazing thing happens is
if this cylinder is elastically mounted or is flexible, and
that force starts to act on it, it will begin to vibrate. And the amazing thing,
as it begins to vibrate, it correlates the
shedding of these vortices all along the cylinder. So it’s like soldiers
marching in step going across the bridge. If everybody’s walking
randomly, then the bridge doesn’t respond too much. But if everybody
marches together, you can put a pretty
good excitation into it. Well, the motion of
the cylinder itself organizes these vortex shedding
all along the cylinder, so they’re all marching in step. And that means the force is
all correlated on the length. And you can get some pretty
substantial response. So that’s the subject called
flow-induced vibration. And with that, I’m going
to show you a few slides. Let’s dim the lights a little
bit, if you could, to see this. There’s some pictures I
just want you to see better. All right. So I do flow-induced vibration. I’ve been doing
this– working on this for all my professional career. And it’s applied, primarily,
to big, flexible cylinders in the ocean. Particularly associated with the
things that the US Navy does. Long cables and things and
also the offshore oil industry. Next slide. Can we dim the lights? Can we dim the lights? I want you to be able to see. This is a tension leg platform. It’s one of the
structures that’s used in the offshore
industry to produce oil. And one of these might be
moored in 3,000 feet of water, 1,000 meters of water. Might weigh 20,000 tons. And what’s connecting
it– what holds in place– are steel cylinders
a half a meter in diameter, 3,000 feet
long, going vertically down from each of those three
pontoon legs sticking out. And they’re under
a lot of tension. And in fact, it pulls the
thing down into the water so the buoyancy
of the whole thing puts tension on these cylinders. But now, what happens
if an ocean current comes by those cylinders? Vortex shedding, and
the cylinders vibrate. And if they vibrate, over time
they will fatigue and fail. OK. Next slide. There’s a picture of a real one. That’s a bigger one
called Marco Polo. It’s on a launch ship that’ll
take it out to the site that it is. And the ship will lower
and it will slide off. So these are big. Next slide. This is a diagram of
the Gulf of Mexico. South America is at the bottom. The Yucatan Peninsula is
sticking up there right in the middle of the bottom. This is a picture of
satellite imagery of currents in the Gulf of Mexico. And there’s a current that
flows up off of South America into the Gulf of Mexico,
goes around in a loop, and then comes out. You can see Florida sticking
down in there on the right. That current comes
out of the Gulf, goes around the tip of Florida,
and goes up the Atlantic Coast, and is known as the Gulf Stream. But it starts as
a big current that comes into the Gulf of Mexic. And, every now and then, that
current pinches off an eddy. And that’s what that red
circle is in the middle. And it’s an eddy that’s many,
many kilometers in diameter with surface currents on the
order of a meter per second or more. And those are the biggest
threat for causing flow-induced vibration
failures of long members from hanging off of
offshore structures. Next. So I’ve been doing research
in this area for a long time. This is a picture taken
in the summer of 1981. It is a piece of steel
pipe about 2 inches in diameter and 75 feet long. It’s under 750
pounds of tension, and it’s pinned at each end. It behaves almost exactly
like my rubber cord here. It has natural frequencies,
and it will vibrate if a current comes by it. So this is actually a sandbar. And at low tide, we’d do
all the work putting it up. Then, as the tide
comes in, the flow is perpendicular
to the cylinder, and vortices start shedding. And as the pipe
begins to move, they get organized all
along the length. And a typical
response mode was when the vortex shedding frequency,
therefore the lift force frequency, coincided with
the natural frequency. Then you’d expect it to give
quite a bit of response. The diagram on the left
is if you cut the cylinder and looked down its axis,
this is the trajectory that you’d see
the cylinder make. It would sit there and just
make big figure eights. So up and down vertical
is its vertical motion. Flow’s coming from,
say, left to right. Its vertical motion
is up and down. In-line motion’s like this. And exactly such a phase it
just makes big beautiful figure eights. That’s the kind of
motion you’d see. OK? So then, very much what I was
talking about a minute ago, very much behavior
dominated by vibration. Vibration in the third mode,
cross flow, was a typical one. And fifth mode,
inline, was typical. But as cylinders
go in the ocean, that one’s kind of short. Third mode vibration
is sort of low. So as years have
gone by and oil is being produced in deeper
and deeper and deeper water, the cylinders we’re
putting out there get longer and longer
and longer and longer. And the modes that are
excited by currents coming by get quite high. So this is an experiment we did. It was roughly a
1/10 scale model. Model is almost 2 inches
in diameter, 500 feet long. Scale that up by a
factor of 10, you’re up around 20 inches in
diameter and 5,000 feet long, which is exactly the
size of the drilling riser that BP had hung
off the drilling ship when the blowout occurred. It’s a piece of steel pipe,
21 inches in diameter, 3/4 of an inch wall
thickness, 5,000 feet long, under a lot of tension. And when ocean
currents come by, it behaves just like this string. And so we’re out– this
is a 1/10 scale model. So we put a big weight on
the bottom of the cylinder, put it behind a boat, and
towed it in the Gulf Stream. Next picture. So there’s the boat. It’s an oceanographic vessel. It’s actually a catamaran. Next. This is a spool that had
our test cylinder on it. There’s a reddish object
down on the bottom which is– that’s a 750-pound
piece of railroad wheel, and it’s the weight
on the bottom. And so you’d spool this
thing off, lower it down, and then do your tests. Next. Top, we measured
tension inclination. And then we also had– it’s
a pin joint at the top, so it would vibrate freely. Inside, though,
was fiber optics. Next. We had eight optical fibers. And in those optical
fibers were what we call optical strain gauges. So we had 280 optical strain
gauges instrumented up and down that pipe so we could
measure its vibration. And so you’re looking at a
cross section of the pipe. There were two optical
fibers in each quadrant, and each one of those
fibers had 35 sensors on it. Next. This is typical
experimental case. This is the surface. This is 500 feet down. This is the current profile. So the flow velocity
is about 2 feet per second near the
surface, up to 4 feet per second down on the bottom. And this is the region
where most of the excitation was coming from that would drive
the flow-induced vibration. This is measured RMS strain
caused by the bending vibration in the cylinders. And peak– the maximum
strain– is about right there. Next. Typical response spectrum. Basically, the frequency content
at three different locations. Down deep, in the
middle, near the top. This is frequency. So this would be the
peak that describes the principal cross-flow
vibration at the vortex shedding frequency. Next. This is position, bottom to top. This is time, and these
are strain records from all of those strain sensors. There’s a strain sensor about
every 2 meters along here. But what you’re seeing
is– this is evidence. The red is the
amplitude and red– let’s say red is positive strain
and blue is negative strain. And so at any location on the
pipe where it’s vibrating, it’s going to go from red to
blue, red to blue, red to blue. But it’s showing you that
they’re highly correlated all along the
length, that there’s a red streak all lined up, but
it’s not parallel to the pipe. It’s inclined. This is showing you
wave propagation. The behavior of the
pipe is completely dominated by wave propagation,
not by standing wave vibration. So totally different than
that short pipe in 1981. The wave equation. Let’s imagine we have a long
pipe or a string like that, and it can carry waves
traveling along it. The position at any location on
here– here’s a coordinate x. We describe the
motion at a point by a coordinate w of x and t. So it’s a function of
where it is and time. What describes the
motion of something which obeys the wave equation
is the following equation. Partial squared w with
respect to x squared equals 1/c squared partial
squared w with respect to t squared. That’s what’s known as the
one-dimensional wave equation. And the one-dimensional
wave equation governs an incredibly broad
category of physical phenomena. Light behaves according
to the wave equation. Sound propagating
across the room to you is governed by
the wave equation. Longitudinal vibration of rods,
torsional vibration of rods– all governed by
the wave equation. So it’s worthwhile to know
a little bit about the wave equation. And what I showed
you this morning, it has this kind
of duality to it. You can have things that vibrate
with standing waves and mode shapes, but the same
system can support waves that travel along it. So let’s figure out why that is. So I’m going to do the
derivation for you of the wave equation for a
string, just so you know where it comes
from because then that general derivation applies
to all these different things. So imagine you’ve got now–
we’re interested in eventually getting to vibration. So I’m going to make this
a finite length string. And it has this position we’ll
describe as a w of x and t. It has a tension, T, a
mass per unit length, m. So this is like
kilograms per meter is the mass per unit length of
this thing which can vibrate. So tension. Mass per unit length. L, the length of it. What other parameters
do we need? That’ll do for the moment. Now– so let’s draw
it again without. In some displaced position
and what’s exciting it may be my vortex
shedding, and so I’m going to draw that excitation here. And that we’ll describe
as F of x and t, some force per unit length. So this has units of
newtons per meter. Now, in that little–
there may also be drag forces,
the fluid damping. So I’m going to cut out a
little piece of this cylinder and do a force balance on
that piece of cylinder. So basically, F equals ma. We’re just applying Newton
to this piece of cylinder. And I’ll draw it right here. A little section
of it is curved. Here’s horizontal. There’s horizontal. We need to evaluate
all the forces on it. So the tension on this
end– so like that. And the tension on this end
is some different angle. This we’ll call theta 1. This we’ll call theta 2. And along here are my
excitation forces, F of x and t. There may be some resistance–
drag forces, damping. That’ll be a damping
constant, R of x, which is force per unit
length per unit velocity, times– the force
on this would have to be multiplied
by the velocity, so the derivative of this
displacement with respect to time. That’s the force along here,
and it can vary with position. Have we accounted
for everything? Ah, well, this is position
x, and this is at x plus dx. So this little element
is dx in length. And this is all
for small motions. And if you assume
small motions, then you can say theta 1 is approximately
equal to sine theta 1. That’s also approximately
equal to tan theta 1. And that’s equal to the
derivative of w with respect to x, just the slope. We’re going to take
advantage of that. Theta 2, same thing. It’s approximately equal to
tan theta 2 here, and sin and all those things. But that, then– the slope
has changed a little bit when you go through dx. And this is equal to the
slope on the left-hand side plus the rate of change
of the slope times dx. So the slope on the left,
this is now the slope on the right-hand side. And so now, all
that’s left to do is to write a force balance
for that little piece on the element dx. So if positive, upward. We have a T sine theta. But because sine theta
is approximately tan theta is equal to
dw dx, then there’s an upward force on the
right-hand side, which is T. And this turns into partial
squared w with respect to x squared dx. So on the right-hand
side– positive upwards– you have
T times the partial of w with respect to x, plus
partial square w with respect to x squared dx. That’s the upward force
on the right-hand side. On the left-hand side, we have a
downward force, minus T partial of w with respect to x. And you notice that this one’s
going to cancel that one. We have minus R of x partial
w with respect to t– that’s the velocity– dx long. Because that’s force
per unit length. And have we missed anything? So that’s the sum of
the external forces on this little slice. And that has to be equal
to– what did Newton say? The mass, which is the mass
per unit length, times dx, is the total mass,
times the acceleration, partial squared w with
respect to t squared. So this cancels this term. And then you notice I’m
left with everything as just something dx,
something dx, something dx. Get rid of the dx’s,
and I can write– oh, I left out something. I left out my distributed
force, F of x and t dx. It’s positive as it’s drawn. It’s over here also. So this, and I
cancel out that dx. So I put them all together
now and assemble them. I can write down the equation
that governs this motion. So T partial square
w with respect to x squared minus r of x times
velocity plus f of x and t equals m partial square w
with respect to t squared. And that just says that the
sum of the forces on the object equals its mass times
its acceleration. Now, if we’re interested in
natural frequencies and mode shapes, when we’ve been doing
one and two degree of freedom systems, and we want to get
the natural frequencies in mode shapes, we temporarily let
the damping be 0 and the force be 0, right? So we want to do
the same thing now. We’re interested in how do you
find the omega n’s and what I call the psi n’s. Because now the mode
shapes are functions. And so this is a
natural frequency and the mode shape for mode n. We know there’s lots of modes. So we let r of x and
f of x and t be 0. And when we do that,
this term goes away. This term goes away. I’m just left with T partial
squared w with respect to x squared equals this. And I’m going to
divide through by t. So I get partial squared w with
respect to x squared equals 1 over T over m partial squared
w with respect to t squared. And this T/m quantity
turns out to be the speed of wave
propagation in the medium. And that is the wave equation. So we’ve just found the
wave equation for the string just by applying Newton’s law
to a little section of string. You can do that for the vibrate. You’re going to
do the same thing, cut out a little
section of a beam, do the force balance on it,
set it equal to the mass times acceleration. And for a beam, you’ll get
a fourth order differential equation. And it’s not the wave equation. It still vibrates, but
it’s not governed by what we call the wave equation. OK, so this is the one
dimensional wave equation. This quantity T/m is
the phase velocity. It’s called phase velocity. You know, that’s a
good one to remember. For a simple string, the speed
of phenomena running down the string is the square
root of the tension divided by the mass per unit length. And if you had a long string,
I put that little pluck in it, and you can see that pluck
running back and forth on it. That’s the speed it’s going at. Basically, it’s called–
well, so if I have my string, and I put a little bump
on it, and that bump goes zipping along,
your eye will see this thing propagating at c. So to get natural
frequencies in mode shapes, we basically need to
solve this equation. And it’s quite
straightforward to do. And a technique
known as separation of variables works, which
means that all you’re doing is saying, I believe
that I’m going to be able to write the
solution as some function of x only times some function of
time only, product of two terms. And that in fact– because
we’re interested in vibration. You can tell me what
the function of time is. You’re going to tell me half the
solution just from observation. What is it? Just the time dependent part. It’s the same as anything
else that vibrates. So a single degree
of freedom system, what is the time dependent
function that we substitute in to find the natural frequency? AUDIENCE: [INAUDIBLE] PROFESSOR: Say again? AUDIENCE: e to the i omega t. PROFESSOR: e to the i
omega t would be just fine. Cosine omega t works. Sine omega t works. But e to the i omega t
is pretty easy to use. Because it’s so simple
to take the derivatives. So we can guess
that this is going to be some W of x times
Ae to the i omega t. And plug it in. Plug it into our wave
equation over here. So I’ll make sure I
write it consistently. So we plug this
into the first term. It’s two derivatives
with respect to x. So this is just– and the
time-dependent part just stays outside. And on the right-hand side, when
we plug it in here, 1 over c squared, two derivatives
with respect to time, it’s going to give me
minus omega squared, so minus omega squared
over c squared. And then it gives me back
W of x Ae to the i omega t. And now I can get rid of
the Ae to the i omega t’s. And I’m left with just an
equation involving x only. And it’s an ordinary
differential equation in w of x. So it turns into d2W dx
squared plus omega squared over c squared W equals 0. And you’ve seen this
equation before. Does this not look like,
have some similarity to, Mx double dot plus kx equals 0? They’re basically
the same equation. This one’s a function of x. That one’s a function of time. And we know the solution
to this one is some x of t is some amplitude
e to the i omega t. So therefore, we can guess
that the solution to this one is W of x is going to be–
I’ll write it as some B. Now I need a function of x. But it can be just like this–
e to the i, and I’ll say kx. I know that’s going
to be a solution. So let’s plug it in. If I plug that in, I
get minus k squared Be to the ikx plus omega squared
over c squared Be to the ikx equals 0. Well, now I get rid of these. And what I found out
is that k squared is omega squared over c squared. And this has a name– k. It’s called the wave number. And it also happens to
be 2 pi over lambda. We’ll come back to that. Lambda is the wavelength. You have sinusoidal waves
running through the medium. 2 pi over lambda is the
same as omega over c. And this is called
the wave number– really important
quantity if you’re trying to understand wave
propagation in systems. And actually, this
one, this definition applies to all wave
bearing systems, whether or not they
obey the wave equation. It’ll apply to waves
traveling down a beam as well. So the definition of wave number
is frequency divided by speed, or 2 pi over the wavelength. Well, let’s see. We can’t go much further with
just the wave equation itself. In order to get the
natural frequencies, we have to invoke
other information that we know in the problem. In particular, we
know that in order to get natural frequencies,
we had to create conditions where this could vibrate. In particular, I fix that
end, and I fix this end, and I put some tension on it. And now it’ll vibrate. But it clearly has
something to do with its ends and its length. And so this is a
boundary value problem. And we have to invoke the
boundary conditions to actually finish finding the natural
frequencies and mode shapes. Apply the boundary
conditions– so I assumed here that my W of x is going to
look something like that. In order to get a little
more information out of this, I’m going to write now W of x
in an alternative form that’s equally valid. And I’ll call it B1 cosine
kx plus a B2 sine kx. And I could relate that to
e to the ikx, B to the ikx, by real and imaginary
parts, and so forth. This is a real part. I’m saying in general it could
have a cosine part and also a sine part. But now I know my boundary
conditions are W at x equals 0. W of 0 is what? What’s the displacement
at x equals 0? AUDIENCE: [INAUDIBLE] PROFESSOR: 0. That’s the pin. That’s the end
where it’s fixed at. And we started out here
with a second order partial differential equation. And a second order equation
requires two boundary conditions. A fourth order
equation for the beam will require four
boundary conditions. We only have to find two. One of them is it has
no motion on the left. So you plug in 0 for x. Cosine of 0 is 1. Sine of 0 is 0. So we find out that
this is B1 times 1. But it has to be 0 as
the boundary condition. So that implies B1 is 0. There’s no cosines
in this answer. And W at L is 0. And so that says B2
sine kL equals 0. And that’s true. That’s only true
if kL equals n pi. So now I’ve found
out that there’s, just for vibration of
a finite length string, only particular
values of k that work. So that says that there are
special values of k which I’ll call k sub n which
are equal to n pi over L. And from that, we now
have our mode shapes. Because we can say,
ah, well, there’s special solution
for this W of x that applies only when we satisfy
the boundary conditions. And that will be some
undetermined amplitude. B2 came from the sine term. And those are our mode shapes. And now the natural
frequencies– once you know mode shapes,
natural frequencies actually become pretty trivial to find. In this case, if we know
that’s the mode shape, then how do we get the
natural frequencies? Well, we know that–
what’s the definition of k? Therefore, the
particular values of k that were allowed
solutions here are going to correspond to
particular values of omega n. And therefore, omega n squared
is just kn squared c squared. And that’s n pi
over L squared T/m. That’s omega n squared. So the natural
frequencies of a string are n pi over L root T/m. And this is in
radians per second. And I like to work
in hertz sometimes. So the natural frequencies
in hertz– omega n over 2 pi. And that becomes n
over 2L root T/m. So the first natural frequency,
f1, is 1 over 2L root T/m. Now, let’s draw. What’s the mode shape
for the first mode? Well, it’s half a sine
wave, vibrates like that. It’s full wavelength. I didn’t leave myself
quite enough room. That’s half a wavelength
of a sine wave. So the full wavelength
would be like that. This is of length L. And
so is this piece over here. So the lambda is 2L for
this particular problem. Let’s see, how do I want
to pose this question? So how long does it take
for a wave or disturbance to travel the length
of this finite string? How long does it take it
to go down there and back? How would you calculate that? Distance equals rate times time. What’s the distance? 2L. What’s the speed? c. So the length of time ought
to be 2L over c, right? So the time required– and
2L divided by T over m. But f1 is T/m divided by 2L. Hmm. So the period– so there’s
a direct connection between propagation speed,
frequencies, wavelengths. They’re very closely related. So the natural frequency of
the first mode of this string, that frequency, is exactly
1 over the length of time it takes for a disturbance
to travel down and back. So with that depth
of understanding of how the wave
equation behaves, you can guess the behavior
of lots of other things that behave like that,
like my rod here. I’ll do a little demo
with it in a second. So for example, the
longitudinal vibration, stress waves running
up and down this thing, obey the wave equation. So if I take this thing
and drop it on the floor, it’ll bounce off the floor. How long does it take
to bounce off the floor? So what do you think actually–
what physics has to happen? What’s required to make this
thing bounce off the floor? So we’re going to consider
the floor infinitely rigid. It hits the floor. It actually stays there for
some finite length of time, and then it leaves. So physically, when I was
holding up my string, if I smacked the end, what happened? A pulse took off, ran down
the end, reflected, came back. And that was one round trip. What do you suppose
happens here? I put a pulse into the end. Is it a tension or compression,
the strain that’s felt? AUDIENCE: Compression PROFESSOR: Compression. So a little compression
pulse is put into the end. That compression pulse
then, when it first hits, the compression and the speed
of propagation is finite. So that compression wave
starts traveling up here. Behind the compression wave,
this rod has come to a stop. In front of the
compression wave, the rod doesn’t know
it hit the ground yet. It’s still moving down. So that compression
wave travels up, and it is decelerating
each little slice of mass as it passes through. It brings it to a stop. And so the compression
reaches the top end. The cylinder has come to a stop. The end is free. It can’t take any strain. So an equal and
opposite tension wave has to start to make the sum
of them go to 0 at the end. The boundary condition
at the end is no strain. So it reflects as
a tension wave. Now you have a tension
wave going down. And what it does is it
accelerates every atom as it goes by, as it goes past it. So everything is stopped now. Now it starts down,
and this thing starts rebounding from– the
top rebounds from the floor before the bottom does. The top starts going up. All of it– more
and more goes up. And one hits the bottom. The tension wave hits the
floor, and it jumps off. So how long does it take? Right? And what do you guess
the natural frequency of a free-free rod is? Now, it has a funny mode shape. The mode shape is not half
a sine wave like this. The displacement of the
rod, it has free ends. The ends are moving a lot. But I’ll give you a clue. [ROD RINGING] I can hold it in the
center and not damp it. What do you think the
mode shape looks like? Half a wavelength long,
ends are free– cosine, maximum displacement,
goes to zero, maximum negative displacement. So it’s half a
wavelength long, but it’s a cosine half a wavelength. And the full wavelength is 2L. So this has mode shapes. The mode shapes– I’ve applied
different boundary conditions. These are free-free
boundary conditions. The mode shapes are
cosine n pi x over L. But they have to obey
a certain other law that we know about,
conservation of momentum. Because I’ve got
gravity to deal with, I have to hang on to this thing. But I’ve picked a place to hang
onto it that you can hear it. I’m not affecting the motion. There’s no motion
where I’m holding it. So if I were out in
space, I could do this– [ROD RINGING] –and just let it hang
there in space, right? And it would sit there and ring. What is happening to the
center of mass of this system as it vibrates? AUDIENCE: [INAUDIBLE] PROFESSOR: Stationary. So half of the mass
of this thing’s got to be moving that way. And half of the mass
has to be moving that way so that the total
center of mass doesn’t move. Well, cosine mode
shape, positive here, negative there, perfectly
symmetric, center of mass doesn’t move. So there’s all sorts
of neat little problems that you can solve just by
knowing the wave equation and figuring out
boundary conditions. How many of you stand
in the shower at home and sing, and
every now and then, you hit a note, man, you
just sound great, right? And it’s just all
this reverberation. How many of you have done that? OK, right, what’s going on? AUDIENCE: [INAUDIBLE]
Natural frequency? PROFESSOR: You’ve
hit a– somebody said natural frequency. Of what? AUDIENCE: [INAUDIBLE] PROFESSOR: Huh? AUDIENCE: [INAUDIBLE] PROFESSOR: You’ve hit the
natural frequency of the shower stall itself. If the shower stall
is a meter across, pressure waves– and
you plot pressure inside of the shower,
the lowest mode if you’re plotting pressure. Well, let’s plot actually
molecular movement. What’s the boundary
condition at the wall, the molecules at the wall? They can’t move, right? 0. So the molecular
motion at resonance in the shower stall, the
molecules, the pressures making them move back and forth, looks
like back to the string again. This is L. The first
natural frequency of sound waves bouncing
off the walls in the stall is 1 over 2L root
times c, whatever c is. And c is the speed
of sound in air, which is 340 meters per second. So 340 meters per
second divided by 2L– so if it’s 1 meter
across here, it’s 340 divided by 2, 170 hertz. So that first note you
can hit in the 1 meter across shower stall is about
170 hertz– pretty low. But you can hit second mode. It’d be twice
that, and so forth. OK, what about an organ pipe? This is an organ pipe, wood. It’s got a stoppered end. Actually, let’s do it
without the stopper. Now it’s an open organ pipe. [ORGAN NOTE] Basic wave equation–
how would you model its boundary conditions? So you can talk about maybe
particle molecular motion. This is, now again, just
sound waves, so air particles. And this is now longitudinal. Things are moving inside. So what’s the boundary condition
at this end, free or fixed? Free. And here it’s quite open, so
the boundary condition on here is free. So for the molecular motion
in a free-free organ pipe, you have to get back to that
half a wavelength cosine thing. And if you wanted to
plot pressure instead, you can write the wave
equation in terms of pressure. Pressure is– this is
pressure relief here and pressure relief there. So in fact, if is
displacement of the molecules, pressure would plot like that. You’d have what’s called
a pressure relief boundary condition. But again, it’s a
half wavelength long. What do you think the
first natural frequency of this organ pipe is? The period would be 2L over c. The frequency
would be c over 2L. So the frequency for the organ
pipe open end f1 is c over 2L. [ORGAN NOTE] Check your intuition. I’m going to close the
end– still an organ pipe. Is the frequency now going
to be higher or lower? Take a vote. How many think the
frequency is going to go up? Raise your hands, commit. All right, down. We’ve got a lot of uncertainty. All right, let’s
do the experiment. [ORGAN NOTE] [LOWER ORGAN NOTE] How come? I find that actually
kind of counterintuitive. Until I learned this, I would
have guessed the opposite way. What’s going on with
pressure in a closed pipe? Well, here at the orifice
where the sound is actually generated, it’s the pressure. If we wanted to plot
pressure at the opening, that’s a pressure relief place. So it’s 0. But at the other end where
the stopper is, it’s maximum. How many wavelengths is that? A quarter. And so the length of time
it takes for the thing to go through one
complete period is going to be 4L over c, half
the frequency of the open pipe. OK, so the wave equation
is really quite powerful, governs lots of things. I’ve got 10, 15
minutes left here. I don’t want you to go away
thinking that the whole world behaves like the wave equation. Because there are some
important other physical systems that we care about. And I’m going to
show you just one. And that’s the
vibration of the beam. So here’s the cantilever beam. The whole table is moving. And you can see it
up on the screen. OK, so its first mode
vibration, tip moves maximum. It kind of looks like
a quarter wavelength. It roughly is, but not exactly. So let’s draw a cantilever. And most of you have had 2.001. So if you put a load
P out here– bends, goes through a
displacement delta. So you know that delta equals
Pl cubed over 3EI, right? And what’s this I? AUDIENCE: [INAUDIBLE] PROFESSOR: Area
moment of inertia. Now that you’ve been
doing dynamics all term, we talk about mass
moments of inertia. There’s also area
moments of inertia. So this is the area moment
of inertia of a beam. In this case, our beam is
a little rectangular cross section. And the neutral axis is
here, a little variable y at displacement. I is the integral
of y squared dA. And dA is just a little
slice of area here, dA. And the integral of y squared
dA is your cross sectional area moment of inertia in the
direction of bending. So that is I. You can also
write it as a kappa squared A. And we ran into
this in dynamics. We called it the
radius of gyration. You had the same thing with
area moments of inertia, the radius of gyration. This is going to be really
helpful in a second. So if you solve the
force balance for a beam like I did for the string,
take a little slice, do force balance for
transverse motions– I’m not going to grind it out. And temporarily neglect
external forces and damping. I want to get to the natural
frequencies and mode shapes. So the free vibration,
no damping, equation looks like EI partial
4 w with respect to x to the fourth plus rho A
partial squared w with respect to t squared equals 0. And now this is
density, mass density. And the A, this A, is the
area of the cross section. So it’s just some bh with
thickness times the width. So rho times A is a
mass per unit length. And so mass per
unit length times dx would be the
little mass associated with the element times
the acceleration should be the forces on the element. So that’s the fourth
order partial differential equation that describes
the vibration of a beam. And you have to apply
the boundary conditions. And for the string, it was
just B1 cosine B2 sine. For the beam, it’s B1 cosine
plus B2 sine plus C2 cosh plus D2 sinh x. And then you have to apply
four boundary conditions and solve for B1, B2, and
so forth, all four of those. I won’t do it. But that’s how you do it. Separation– and separation
of variables works again. So we solve this, apply
the boundary conditions. What are the
boundary conditions? Just so you understand
what I mean by the boundary conditions, what are they
for a free-free beam, zero motion at the wall? No strain at the end, no
bending moment at the end, no sheer force at
the end– so there’s no second derivative,
no third derivative. And at the wall, the slope
is 0, the first derivative. No slope comes into the wall,
but the slope is 0 there. So those are the different
kind of boundaries. So if you have a
free-free beam, you have no bending at either end
and no strain at either end. Fixed-fixed beam–
no displacement, zero slopes at both at ends,
and all different combinations. And every different
combination gives you different natural frequencies. So you apply the boundary
conditions, and for a beam, you find out that
for all beams omega n can be written as some
beta n, a parameter, squared, I’ll call it, times the
square root of EI over rho A. And this thing varies according
to the boundary conditions. Now that’s what you get shown
in every textbook in the world. And I have a very hard
time visualizing this, getting physical
intuition by that. So something you never
see in a textbook but I often do is let’s
replace I with kappa squared A. And you get a square root of E
over rho and a square root of I over A. But I is kappa
squared A. The A’s cancel. It’s the square root
of kappa squared. So this, you know
what E over rho is? E over rho, square
root of E over rho, is the sound speed
in a solid material. So the speed of stress
waves traveling up and down this thing is the square
root of E over rho. [ROD RINGING] This is aluminum. It’s about 4,000
meters a second. So if you know just the
properties of the material, you have that. And that says then
omega n for beams is some beta n squared a
parameter times kappa CL. And this thing, this
is often written CL. It’s the longitudinal
sound speed. This is sound speed for waves
traveling through the medium. So this tells you if you
make the beam twice as thick, what do you do to its
natural frequencies? Doubles– instantly
you know that. So bending properties depend a
lot on the radius of gyration. And I’ll give you a
few natural frequencies for different
boundary conditions just so you see what
they behave like. So a pin-pin beam
looks like that. So you put a plank across the
stream, rocks on both sides, you’ve got a pin-pin
beam, basically. It’s set there in rock. So some length L
has properties EI. So the natural frequencies
for a pin-pin beam, the beta n’s, are
just n pi over L. And so your natural
frequencies– omega n looks like n pi over L
quantity squared kappa CL. And for the cantilever,
the natural frequencies look like omega n pi squared
over 4L squared, is the beta n. And I’ll write it this
way again– EI over rho A. You can always
go back and do that. Or you can call it kappa CL. This is also kappa CL. But then there are
some numbers you’ve got to use here– 1.194 squared. That’s the first mode. Second mode– 2.988 squared. And then after that– 5
squared, 7 squared, 9 squared. So this is the natural
frequency of a cantilever. Pi squared over 4L
squared times 1.194 squared kappa CL, that’s
this natural frequency. And one final case,
because I can show it to you– the free-free case. So that’s a beam bending
that vibrates like that. And I happen to know on a
beam for the first mode– this is the first mode of a beam. Where these nodes are,
where there’s no motion, I should be able to hold
it there and not damp it. And that turns out to be at
about the quarter points. So whack it like that. [ROD RINGING] And do it again. [ROD RINGING] All right, so I want you to
hold it about right there. Nope, you can’t hold it
like that, though– just got to balance it. Because you’ve got to be
right where the node is. [ROD RINGING] You can hear that
little bit lower tone. That’s that free-free
bending mode. And it’s just sitting. You can feel it vibrating
a little bit but not much. When you’re right
in the right spot, you’re right on the mode shape. You can almost see it if
you hit it hard enough. So that’s the free-free beam. And the free-free beam has
natural frequencies omega n, again, pi squared over 4L
squared kappa CL 3.0112 squared, 5 squared, 7 squared,
9 squared, so as you go up in n. So those are the
natural frequencies of a free-free beam. Oh, one last fact about
beams– so this is now a steel beam under no tension. It can support its own
weight, long though. So can a beam support
waves traveling down the beam, transverse waves
traveling down the beam? What do you think? Well, if it can support
this, it can probably support waves, right? So waves will
propagate in a beam even though this is fourth order
partial differential equation. But how fast do they go? That’s the question. So this is a beam. And I want to know about
waves traveling down it. And I’m not going
to go through– this would take another hour or so to
show you where this comes from. But here’s my beam. Here’s a disturbance
traveling along it with some speed that
I’m going to call CT. It’s transverse wave speed. It’s the speed you’d see
a crest of a wave moving at running down that beam. CT for a beam– square
root of omega kappa CL. And CL, again, is the
square root of E over rho. That’s the speed of
sound in the material. That just turns up in here. So what does this tell you
about the frequency dependence of the speed? Does the speed change
with frequency? Omega kappa CL– it’s
proportional to frequency. High frequency waves go faster
than low frequency waves in a beam. I didn’t emphasize it when
we were talking about it. But the wave equation,
what was c for the string? For the wave equation, the
speed of wave propagation was square root of T/m. Was it frequency dependent? Always traveled
at the same speed. And so there’s an
important consequence. So for anything that
obeys the wave equation, the speed of propagation is
a constant and independent to frequency. So I can make any initial
shape that I make in this thing and let it go. Its initial disturbance,
that little shape will stay that shape and run
up and down the thing forever. And that shape– you could
imagine a little pluck like this to start with. You could imagine
doing a Fourier series to approximate that. It would be made up of a
bunch of different Fourier components. And yet for something that
bears the wave equation, that little pluck will just
stay the shape of that pluck and run around forever. But not so in a beam. If you did that in a beam,
if you come up and put an impulse into a
beam, all that energy would start out together. But in very brief time, the
high frequency information would get out in front of the
low frequency information. And if you were way down this
beam, and somebody up a mile away whacks one end, and
you’re down further along, you’ll see high
frequency waves past you, and then lower frequency,
and finally really slow ones coming by, the
really long waves. So that’s called dispersion. So beam waves are dispersive. Things that obey the wave
equation are non-dispersive. The energy all travels at
the same speed independent of frequency. All right, so that’s
it for the term. I’ll see you guys
on next Wednesday.


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