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PROFESSOR: Ladies and gentlemen,
welcome to this set
00:00:24.880 --> 00:00:28.760
of lectures on the finite
element method.
00:00:28.760 --> 00:00:32.640
In these lectures I would like
to give you an introduction to
00:00:32.640 --> 00:00:37.190
the linear analysis of solids
and structures.
00:00:37.190 --> 00:00:39.950
You are probably well aware that
the finite element method
00:00:39.950 --> 00:00:42.980
is now widely used for analysis
of structural
00:00:42.980 --> 00:00:45.090
engineering problems.
00:00:45.090 --> 00:00:48.700
The method is used in civil,
aeronautical, mechanical,
00:00:48.700 --> 00:00:52.970
ocean, mining, nuclear,
biomechanical, and other
00:00:52.970 --> 00:00:55.470
engineering disciplines.
00:00:55.470 --> 00:00:59.230
Since the first applications two
decades ago of the finite
00:00:59.230 --> 00:01:03.370
element method we now see
applications in linear,
00:01:03.370 --> 00:01:07.020
nonlinear, static, and
dynamic analysis.
00:01:07.020 --> 00:01:09.580
However, in this set of
lectures, I would like to
00:01:09.580 --> 00:01:13.220
discuss with you only the
linear, static, and dynamic
00:01:13.220 --> 00:01:16.000
analysis of problems.
00:01:16.000 --> 00:01:18.700
The finite element method
is used today in
00:01:18.700 --> 00:01:21.025
various computer programs.
00:01:21.025 --> 00:01:22.960
And its use is very
significant.
00:01:26.700 --> 00:01:29.780
My objective in this set of
lectures is to introduce to
00:01:29.780 --> 00:01:34.800
you the finite element methods
or some of the finite element
00:01:34.800 --> 00:01:38.220
methods that are used for linear
analysis of solids and
00:01:38.220 --> 00:01:39.340
structures.
00:01:39.340 --> 00:01:43.120
And here we understand linear
to mean that we're talking
00:01:43.120 --> 00:01:48.300
about infinitesimally small
displacements and that we are
00:01:48.300 --> 00:01:51.740
using a linear elastic
material law.
00:01:51.740 --> 00:01:54.660
In other words, Hooke's
law applies.
00:01:54.660 --> 00:01:58.070
We will consider, in this set
of lectures, the formulation
00:01:58.070 --> 00:02:01.200
of the finite element
equilibrium equations, the
00:02:01.200 --> 00:02:06.230
calculation of finite element
matrices of the matrices that
00:02:06.230 --> 00:02:09.090
arise in the equilibrium
equations.
00:02:09.090 --> 00:02:12.170
We will be talking about the
methods for solution of the
00:02:12.170 --> 00:02:16.340
governing equations in static
and dynamic analysis.
00:02:16.340 --> 00:02:21.280
And we will talk about actual
computer implementations.
00:02:21.280 --> 00:02:26.000
I will emphasize modern and
effective techniques and their
00:02:26.000 --> 00:02:27.250
practical usage.
00:02:29.760 --> 00:02:34.820
The emphasis, in this set of
lectures, is given to physical
00:02:34.820 --> 00:02:38.460
explanations of the methods,
techniques that we are using
00:02:38.460 --> 00:02:42.200
rather than mathematical
derivations.
00:02:42.200 --> 00:02:46.720
The techniques that we will be
discussing are those employed
00:02:46.720 --> 00:02:51.320
largely in the computer programs
SAP and ADINA.
00:02:51.320 --> 00:02:56.000
SAP stands for Structural
Analysis Program and you might
00:02:56.000 --> 00:03:00.990
very well be aware that there is
a series of such programs,
00:03:00.990 --> 00:03:04.430
SAP I to SAP VI now.
00:03:04.430 --> 00:03:08.540
And ADINA stands for
Automatic Dynamic
00:03:08.540 --> 00:03:11.200
Incremental Nonlinear Analysis.
00:03:11.200 --> 00:03:16.285
However, this program is also
very effectively employed for
00:03:16.285 --> 00:03:17.700
linear analysis.
00:03:17.700 --> 00:03:21.560
The nonlinear analysis being
then a next step in the usage
00:03:21.560 --> 00:03:24.250
of the program.
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In fact, the elements in ADINA,
the numerical methods
00:03:27.290 --> 00:03:30.840
that are used in ADINA, I
consider to be the most
00:03:30.840 --> 00:03:34.020
effective, the most modern state
of the art techniques
00:03:34.020 --> 00:03:36.500
that are currently available.
00:03:36.500 --> 00:03:40.980
These few lectures really
represent a very brief and
00:03:40.980 --> 00:03:43.480
compact introduction
to the field of
00:03:43.480 --> 00:03:45.210
finite element analysis.
00:03:45.210 --> 00:03:48.330
We will go very rapidly through
some or the basic
00:03:48.330 --> 00:03:52.450
concepts, practical
applications, and so on.
00:03:52.450 --> 00:03:56.230
We shall follow quite closely,
however, certain sections in
00:03:56.230 --> 00:03:59.160
my book entitled Finite
Element Procedures in
00:03:59.160 --> 00:04:03.960
Engineering Analysis to be
published by Prentice Hall.
00:04:03.960 --> 00:04:07.970
And I will be referring in the
study guide of this set of
00:04:07.970 --> 00:04:12.190
lectures extensively to this
book to the specific sections
00:04:12.190 --> 00:04:17.370
that we're considering in the
lectures in this book.
00:04:17.370 --> 00:04:22.700
The finite element solution
process can be described as
00:04:22.700 --> 00:04:24.620
given on this viewgraph.
00:04:24.620 --> 00:04:29.880
You can see here that we talk
about a physical problem.
00:04:29.880 --> 00:04:33.220
We want to analyze an actual
physical problem.
00:04:33.220 --> 00:04:36.440
And our first step, of course,
is to establish a finite
00:04:36.440 --> 00:04:38.950
element model of that
physical problem.
00:04:41.710 --> 00:04:45.410
Then, in the next step,
we solve that model.
00:04:45.410 --> 00:04:48.550
And then we have to interpret
the results.
00:04:48.550 --> 00:04:51.600
Because the interpretation of
the results depends very much
00:04:51.600 --> 00:04:54.830
on how we established the finite
element model, what
00:04:54.830 --> 00:04:56.900
kind of model we used,
and so on.
00:04:56.900 --> 00:04:59.650
And in establishing the finite
element model, we have to be
00:04:59.650 --> 00:05:03.640
aware of what kinds of elements,
techniques, and so
00:05:03.640 --> 00:05:05.610
on are available to us.
00:05:05.610 --> 00:05:11.250
Well, therefore, I will be
talking, in the set of
00:05:11.250 --> 00:05:15.990
lectures, about these three
steps basically here for
00:05:15.990 --> 00:05:19.800
different kinds of physical
problems.
00:05:19.800 --> 00:05:23.560
Once we have interpreted the
results we might go back from
00:05:23.560 --> 00:05:28.350
down here to there to revise
or refine our model and go
00:05:28.350 --> 00:05:32.510
through this process again until
we feel that our model
00:05:32.510 --> 00:05:35.520
has been an adequate one for the
solution of the physical
00:05:35.520 --> 00:05:37.610
problem of interest.
00:05:37.610 --> 00:05:41.690
Let me give you or show you
some models that have been
00:05:41.690 --> 00:05:45.100
used in actual structural
analysis.
00:05:45.100 --> 00:05:47.890
You might have seen similar
models in textbooks, in
00:05:47.890 --> 00:05:49.580
publications already.
00:05:49.580 --> 00:05:53.190
This, for example, is a model
that was used for the analysis
00:05:53.190 --> 00:05:54.650
of a cooling tower.
00:05:54.650 --> 00:05:57.930
The basic process of the finite
element method is that
00:05:57.930 --> 00:06:01.570
we are taking the continuous
system, and we are idealizing
00:06:01.570 --> 00:06:03.660
it as an assemblage
of elements.
00:06:03.660 --> 00:06:08.110
I'm drawing here a typical
three-noded triangular shell
00:06:08.110 --> 00:06:10.710
element that was used
in the analysis of
00:06:10.710 --> 00:06:13.010
this cooling tower.
00:06:13.010 --> 00:06:16.680
We talk about very many elements
in order to obtain an
00:06:16.680 --> 00:06:18.630
accurate response prediction.
00:06:18.630 --> 00:06:21.240
And, of course, that means that
we will be dealing with a
00:06:21.240 --> 00:06:24.100
large set of equations
to be solved.
00:06:24.100 --> 00:06:26.980
And there's a significant
computer effort required.
00:06:26.980 --> 00:06:28.630
I will be addressing
all of these
00:06:28.630 --> 00:06:32.360
questions in these lectures.
00:06:32.360 --> 00:06:35.400
Here you see the finite element
model of a dam.
00:06:35.400 --> 00:06:39.380
The earth below the dam was
idealized as an assemblage all
00:06:39.380 --> 00:06:42.240
such elements here, triangular
elements now.
00:06:42.240 --> 00:06:45.280
And the dam itself was
also idealized as an
00:06:45.280 --> 00:06:47.400
assemblage of elements.
00:06:47.400 --> 00:06:51.530
We will be talking about how
such assemblages are best
00:06:51.530 --> 00:06:54.320
created, what kinds of elements
to select, what
00:06:54.320 --> 00:06:56.910
assumptions are in the selection
of these elements,
00:06:56.910 --> 00:07:00.120
and then how do we solve the
resulting finite element
00:07:00.120 --> 00:07:01.820
equilibrium equations.
00:07:01.820 --> 00:07:05.390
Here you see the finite element
analysis, or the mesh
00:07:05.390 --> 00:07:08.770
that was used in the finite
element analysis of a tire.
00:07:08.770 --> 00:07:12.170
This wall is half of the tire,
as you can see, and this was
00:07:12.170 --> 00:07:13.760
the finite element mesh used.
00:07:13.760 --> 00:07:17.140
Again, we have to judiciously
choose the kinds of finite
00:07:17.140 --> 00:07:18.790
elements to be employed.
00:07:18.790 --> 00:07:22.080
And we will be talking about
that in this set of lectures.
00:07:22.080 --> 00:07:26.750
Here you see the finite element
model employed in the
00:07:26.750 --> 00:07:29.240
analysis of a spherical
cover of a laser
00:07:29.240 --> 00:07:32.010
vacuum target chamber.
00:07:32.010 --> 00:07:33.390
This is the finite element
mesh used.
00:07:33.390 --> 00:07:35.940
Again, specific elements
were employed here.
00:07:35.940 --> 00:07:39.050
And we will be talking about the
characteristics of these
00:07:39.050 --> 00:07:41.030
elements in this set
of lectures.
00:07:41.030 --> 00:07:47.270
Here you see the model of the
shell structure subjected to a
00:07:47.270 --> 00:07:48.120
pinching load.
00:07:48.120 --> 00:07:50.790
There's a load up here and
a load down there.
00:07:50.790 --> 00:07:53.670
These are the triangular
elements that were used in the
00:07:53.670 --> 00:07:58.670
idealization of that shell and
the resulting bending moments
00:07:58.670 --> 00:08:02.110
and displacements along the line
DC are plotted here that
00:08:02.110 --> 00:08:06.570
have been predicted by the
finite element analysis.
00:08:06.570 --> 00:08:09.920
Finally here you see the finite
element idealization of
00:08:09.920 --> 00:08:14.140
a wind tunnel that was used for
the dynamic analysis of
00:08:14.140 --> 00:08:15.270
this tunnel.
00:08:15.270 --> 00:08:18.740
You can see a large number of
shell elements were employed
00:08:18.740 --> 00:08:21.860
in the idealization of the
shelf of the tunnel.
00:08:21.860 --> 00:08:25.640
Then, of course, supports were
provided here for that shell.
00:08:25.640 --> 00:08:29.680
And this was a very large
system that was solved.
00:08:29.680 --> 00:08:33.010
And the eigenvalues of this
system were calculated using
00:08:33.010 --> 00:08:35.600
the subspace iteration method
that we would be also talking
00:08:35.600 --> 00:08:37.900
about in this set of lectures.
00:08:37.900 --> 00:08:41.190
Well, with this short
introduction then, I would
00:08:41.190 --> 00:08:45.850
like to go now and discuss with
you some basic concepts
00:08:45.850 --> 00:08:47.460
of engineering analysis.
00:08:47.460 --> 00:08:50.340
There's a lot of work ahead
in this set of lectures.
00:08:50.340 --> 00:08:54.310
So let me take off my jacket
with your permission, and let
00:08:54.310 --> 00:08:58.270
us just go right on with the
actual discussion of the
00:08:58.270 --> 00:09:01.650
theory of the finite
element method.
00:09:01.650 --> 00:09:07.390
The basic concepts that I
address here, in this first
00:09:07.390 --> 00:09:13.210
lecture, is summarized basically
here once more.
00:09:13.210 --> 00:09:16.430
We are talking about the
idealization of a system.
00:09:16.430 --> 00:09:18.800
We are talking about the
formulation of the equilibrium
00:09:18.800 --> 00:09:23.110
equations, then the solution of
the equations, and then, as
00:09:23.110 --> 00:09:24.290
I mentioned earlier
already, the
00:09:24.290 --> 00:09:26.110
interpretation of the results.
00:09:26.110 --> 00:09:29.180
These are really the four
steps that have to be
00:09:29.180 --> 00:09:35.090
performed in the analysis of an
engineering system or of a
00:09:35.090 --> 00:09:38.550
physical system that
we want to analyze.
00:09:38.550 --> 00:09:41.280
Now when we talk about systems,
we are really talking
00:09:41.280 --> 00:09:45.380
about discrete and continuous
systems where, however, in
00:09:45.380 --> 00:09:51.160
reality, we recognize that all
systems are really continuous.
00:09:51.160 --> 00:09:57.110
However, if the system consists
of a set of springs,
00:09:57.110 --> 00:10:01.680
dashpots, beam elements, then
we might refer to this
00:10:01.680 --> 00:10:04.650
continuous system as a discrete
system because we can
00:10:04.650 --> 00:10:11.210
see already, it is obvious, so
to say, how to idealize a
00:10:11.210 --> 00:10:17.050
system into a set of elements,
discrete elements.
00:10:17.050 --> 00:10:19.900
In that case, the response is
described by variables at a
00:10:19.900 --> 00:10:22.090
finite number of points.
00:10:22.090 --> 00:10:25.730
And this means that we have to
set up a set of algebraic
00:10:25.730 --> 00:10:29.690
questions to solve
that system.
00:10:29.690 --> 00:10:32.840
So here I'm talking about
elementary systems ofl
00:10:32.840 --> 00:10:37.720
springs, dashpots, discrete
beam elements, and so on.
00:10:37.720 --> 00:10:41.950
In the analysis of a continuous
system the response
00:10:41.950 --> 00:10:44.360
is described really by variables
at an infinite
00:10:44.360 --> 00:10:46.220
number of points.
00:10:46.220 --> 00:10:49.370
And, in this case, we really
come up with a differential
00:10:49.370 --> 00:10:52.100
equation, obviously a set of
differential equations, that
00:10:52.100 --> 00:10:54.170
we have to solve.
00:10:54.170 --> 00:11:00.030
The analysis of a complex
continuous system requires a
00:11:00.030 --> 00:11:03.630
dissolution of the differential
equations using
00:11:03.630 --> 00:11:05.990
numerical procedures.
00:11:05.990 --> 00:11:08.880
And this solution via numerical
procedures--
00:11:08.880 --> 00:11:10.940
and, of course, in this set of
lectures we will be talking
00:11:10.940 --> 00:11:14.240
about the finite element method
numerical procedures--
00:11:14.240 --> 00:11:19.130
really reduces a continuous
system to a discrete form.
00:11:19.130 --> 00:11:23.490
The powerful mechanism that we
talk about here is the finite
00:11:23.490 --> 00:11:26.970
element method implemented
on a digital computer.
00:11:26.970 --> 00:11:29.620
The problem types that I will
be talking about are
00:11:29.620 --> 00:11:33.940
steady-state problems, or static
analysis, propagation
00:11:33.940 --> 00:11:37.810
problems, dynamic analysis,
and eigenvalue problems.
00:11:37.810 --> 00:11:41.660
And these three types of
problems, of course, arise for
00:11:41.660 --> 00:11:45.240
discrete and continuous
systems.
00:11:45.240 --> 00:11:48.520
Now let us talk first about
the analysis of discrete
00:11:48.520 --> 00:11:50.430
systems in this first lecture.
00:11:50.430 --> 00:11:53.120
Because many of the
characteristics that we are
00:11:53.120 --> 00:11:56.080
using in the analysis of
discrete systems, discrete
00:11:56.080 --> 00:12:01.580
meaning, springs, dashpots, et
cetera, we can directly see
00:12:01.580 --> 00:12:04.090
the discrete elements
of the system.
00:12:04.090 --> 00:12:07.270
The steps involved in the
analysis of such discrete
00:12:07.270 --> 00:12:13.150
systems are very similar to the
analysis of complex finite
00:12:13.150 --> 00:12:14.990
element systems.
00:12:14.990 --> 00:12:16.550
The steps involved
are the system
00:12:16.550 --> 00:12:17.860
idealization into elements.
00:12:17.860 --> 00:12:21.300
And that idealization is
somewhat obvious because we
00:12:21.300 --> 00:12:24.120
have the discrete elements
already.
00:12:24.120 --> 00:12:27.780
The evaluation of the element
equilibrium requirements, the
00:12:27.780 --> 00:12:30.920
element assemblage, and the
solution of the response.
00:12:30.920 --> 00:12:33.080
Notice when we later on talk
about the analysis of
00:12:33.080 --> 00:12:37.810
continuous systems instead of
discrete systems, then the
00:12:37.810 --> 00:12:42.490
system idealization into finite
elements here is not an
00:12:42.490 --> 00:12:45.690
obvious step and needs
much attention.
00:12:45.690 --> 00:12:50.430
But these three steps here are
the same in the finite element
00:12:50.430 --> 00:12:52.460
analysis of a continuous
system.
00:12:52.460 --> 00:12:56.960
And I would like to now discuss
all of these steps
00:12:56.960 --> 00:13:00.460
here just to show you,
basically, some of the basic
00:13:00.460 --> 00:13:03.660
concepts that we're using in the
finite element analysis.
00:13:03.660 --> 00:13:08.730
Let us look at this discrete
system here as an example.
00:13:08.730 --> 00:13:13.500
And let us display the basic
ideas in the analysis of this
00:13:13.500 --> 00:13:14.620
discrete system.
00:13:14.620 --> 00:13:18.830
Here we have a set of rigid
carts, three rigid carts,
00:13:18.830 --> 00:13:21.930
vertical carts that
are supported on
00:13:21.930 --> 00:13:23.850
rollers down here.
00:13:23.850 --> 00:13:27.340
This means that each of these
carts can just roll
00:13:27.340 --> 00:13:29.580
horizontally.
00:13:29.580 --> 00:13:37.120
The carts are connected via
springs, k2, k3, k4, k5.
00:13:37.120 --> 00:13:42.640
And the first cart here is
connected via k1 to a rigid
00:13:42.640 --> 00:13:45.440
support that does not move.
00:13:45.440 --> 00:13:48.430
The displacement all of
this cart here is u1.
00:13:48.430 --> 00:13:50.970
The load applied is R1.
00:13:50.970 --> 00:13:55.110
Notice that u1 is the
displacement of each of these
00:13:55.110 --> 00:13:59.060
springs since this
cart is rigid.
00:13:59.060 --> 00:14:02.200
The displacement of this
cart here is u2.
00:14:02.200 --> 00:14:05.450
And R2 is the load applied.
00:14:05.450 --> 00:14:07.760
The displacement of
this cart is u3.
00:14:07.760 --> 00:14:10.390
And R3 is the load applied.
00:14:10.390 --> 00:14:17.080
We now want to analyze this
system when R1, R2, looking at
00:14:17.080 --> 00:14:22.270
it, we can directly see the
elements of the system k1 to
00:14:22.270 --> 00:14:28.000
k5, and we can see directly, of
course, how these elements
00:14:28.000 --> 00:14:29.760
are interconnected.
00:14:29.760 --> 00:14:32.930
The steps that we will be
talking about in the analysis
00:14:32.930 --> 00:14:37.570
of this discrete system are
really very similar to the
00:14:37.570 --> 00:14:39.930
steps that we're using in the
finite element analysis of
00:14:39.930 --> 00:14:41.330
continuous systems.
00:14:41.330 --> 00:14:44.340
What we will be doing is that
we look at the equilibrium
00:14:44.340 --> 00:14:51.390
requirements for each spring
as a first step.
00:14:51.390 --> 00:14:54.660
Then we look at the
interconnection requirements
00:14:54.660 --> 00:15:01.440
between these springs that, in
other words, the force on
00:15:01.440 --> 00:15:05.900
these springs here at this cart,
and that spring, must be
00:15:05.900 --> 00:15:07.830
balanced by R1.
00:15:07.830 --> 00:15:11.480
And then, of course, we have a
compatibility requirement that
00:15:11.480 --> 00:15:15.940
u1 is a displacement of each
of these springs here.
00:15:15.940 --> 00:15:19.410
So we are talking about the
constitutive relations, the
00:15:19.410 --> 00:15:21.880
equilibrium requirements,
and the compatibility
00:15:21.880 --> 00:15:22.730
requirements.
00:15:22.730 --> 00:15:24.860
These are, of course, the three
requirements that we
00:15:24.860 --> 00:15:28.110
also have to satisfy in the
analysis of a continuous
00:15:28.110 --> 00:15:31.200
system using, later on, finite
element methods.
00:15:31.200 --> 00:15:35.070
Notice that these springs here
are our finite elements, if
00:15:35.070 --> 00:15:37.890
you want to think of
it that way, a very
00:15:37.890 --> 00:15:40.400
simple set of elements.
00:15:40.400 --> 00:15:43.430
In a more complex analysis,
these springs here would be
00:15:43.430 --> 00:15:46.060
plane stress elements, plane
strain elements, three
00:15:46.060 --> 00:15:47.740
dimension elements,
shell elements.
00:15:47.740 --> 00:15:50.320
And we will be talking about
how we derive the
00:15:50.320 --> 00:15:52.880
characteristics of
these elements.
00:15:52.880 --> 00:15:57.090
And we will, however,
interconnect these elements,
00:15:57.090 --> 00:15:59.860
these more complex elements,
later in exactly the same way
00:15:59.860 --> 00:16:02.850
as we connect these
simple elements.
00:16:02.850 --> 00:16:06.370
So the connections between the
elements are established in
00:16:06.370 --> 00:16:09.760
the same way, and the solution
of the equilibrium equations
00:16:09.760 --> 00:16:12.830
is also performed
in the same way.
00:16:12.830 --> 00:16:17.060
But in this simple analysis,
we are given directly the
00:16:17.060 --> 00:16:18.800
spring stiffnesses.
00:16:18.800 --> 00:16:23.470
And one other important point is
that the spring stiffnesses
00:16:23.470 --> 00:16:25.610
here are exact stiffnesses.
00:16:25.610 --> 00:16:28.530
In a finite element analysis
of a continuous system, we
00:16:28.530 --> 00:16:31.585
have a choice on what kind of
interpolations we can use for
00:16:31.585 --> 00:16:32.980
an element.
00:16:32.980 --> 00:16:35.820
We have a choice on what
assumptions we want to lay
00:16:35.820 --> 00:16:37.010
down for an element.
00:16:37.010 --> 00:16:39.940
And then using different
assumptions we are coming up
00:16:39.940 --> 00:16:43.880
with different stiffnesses of
the element domain that we
00:16:43.880 --> 00:16:45.270
will be talking about.
00:16:45.270 --> 00:16:48.750
And we will also find that the
equilibrium in that element
00:16:48.750 --> 00:16:50.980
domain is not satisfied.
00:16:50.980 --> 00:16:53.800
It will only be satisfied in
the limit as the elements
00:16:53.800 --> 00:16:55.710
become smaller, and smaller,
and smaller.
00:16:55.710 --> 00:16:59.230
Whereas in the analysis of
this discrete system, the
00:16:59.230 --> 00:17:03.880
equilibrium in each spring
is always satisfied.
00:17:03.880 --> 00:17:08.020
So this is a very simple finite
element analysis if you
00:17:08.020 --> 00:17:10.490
want to think of it that way.
00:17:10.490 --> 00:17:14.150
The elements here than are k1.
00:17:14.150 --> 00:17:17.290
And notice that the equilibrium
requirement for
00:17:17.290 --> 00:17:21.390
this element says simply that
k1u1 is equal to the force
00:17:21.390 --> 00:17:22.849
applied to this node.
00:17:22.849 --> 00:17:25.839
It's a force, the external
force, applied to this node.
00:17:25.839 --> 00:17:29.360
The equilibrium requirement of
this element, k2, is written
00:17:29.360 --> 00:17:32.730
down here in matrix form.
00:17:32.730 --> 00:17:36.060
k2 is the physical stiffness
of the spring.
00:17:36.060 --> 00:17:41.740
And F1, F2 are the forces
applied at these two ends.
00:17:41.740 --> 00:17:46.210
Notice, please, that the
superscript here refers to the
00:17:46.210 --> 00:17:50.490
element number, superscript
1 here for element 1,
00:17:50.490 --> 00:17:54.190
superscript 2 here
for element 2.
00:17:54.190 --> 00:18:02.820
And notice that we would find
that F1(2) is minus F2(2)
00:18:02.820 --> 00:18:04.490
given u1 and u2.
00:18:04.490 --> 00:18:08.590
Of course, that means the
element is in equilibrium.
00:18:08.590 --> 00:18:13.450
Notice also, if you look at this
matrix closer, that if u1
00:18:13.450 --> 00:18:19.560
is greater than u2, then we
would find that, in other
00:18:19.560 --> 00:18:23.260
words, u1 greater than u2 means
that the spring is in
00:18:23.260 --> 00:18:24.260
compression.
00:18:24.260 --> 00:18:28.680
We would find that F1(2) is
positive by simply multiplying
00:18:28.680 --> 00:18:30.060
this out here.
00:18:30.060 --> 00:18:33.720
And F2(2) is negative, which
corresponds to the physical
00:18:33.720 --> 00:18:35.820
situation that we
actually have.
00:18:35.820 --> 00:18:40.710
If u1 is greater than u2, this
force here is positive, and
00:18:40.710 --> 00:18:46.230
that force is negative because
the spring is compressed.
00:18:46.230 --> 00:18:48.230
Well, similarly, we can write
down the equilibrium
00:18:48.230 --> 00:18:51.450
requirement for the spring 3.
00:18:51.450 --> 00:18:53.020
And I've written down
the matrix here.
00:18:53.020 --> 00:18:55.250
The only difference to the
equilibrium requirements for
00:18:55.250 --> 00:18:58.720
spring 2 are that we're
using now k3 here.
00:18:58.720 --> 00:19:02.980
And, of course, the superscript
now is 3.
00:19:02.980 --> 00:19:06.420
We can then proceed to write
down the equilibrium equation
00:19:06.420 --> 00:19:11.700
for spring 4, which is the same
form as before, now k4
00:19:11.700 --> 00:19:16.010
here and the 4 superscript
denoting element 4.
00:19:16.010 --> 00:19:21.080
And, finally for k5, we have
k5 here and superscripts 5
00:19:21.080 --> 00:19:24.250
here to denote element 5.
00:19:24.250 --> 00:19:30.280
Now we should also point out
one other important point.
00:19:30.280 --> 00:19:36.980
Namely, if we look at this cart
systems here, notice that
00:19:36.980 --> 00:19:42.370
this k1 spring is
connected to u1.
00:19:42.370 --> 00:19:44.440
It's connected to u1.
00:19:44.440 --> 00:19:48.450
k4 is connected to u1 and u3.
00:19:48.450 --> 00:19:54.460
So if we look at the equilibrium
requirements here,
00:19:54.460 --> 00:19:59.960
you will notice that I
have F1 here for k1
00:19:59.960 --> 00:20:02.850
because this is u1 here.
00:20:02.850 --> 00:20:04.990
That is the global
displacement u1.
00:20:04.990 --> 00:20:09.020
And looking now at k4, a more
complicated case which is
00:20:09.020 --> 00:20:21.080
connected to u1 and u3, I have
for that spring the u1 and u3
00:20:21.080 --> 00:20:22.080
denoted here.
00:20:22.080 --> 00:20:26.210
And we have F1 and F3
here, F1 and F3.
00:20:26.210 --> 00:20:30.620
So these are the forces that go
directly into the degrees
00:20:30.620 --> 00:20:34.840
of freedom 1 and 3 respectively,
and similarly
00:20:34.840 --> 00:20:36.460
for the other springs.
00:20:36.460 --> 00:20:40.810
Now if we want to assemble the
global equilibrium equations
00:20:40.810 --> 00:20:47.460
for this structure with the
unknowns u1, u2, u3, the loads
00:20:47.460 --> 00:20:52.660
R1, R2, and R3 are known, then
we have to use now the
00:20:52.660 --> 00:20:57.730
equilibrium requirement at these
degrees of freedom u1,
00:20:57.730 --> 00:21:02.740
u2, and u3, or rather at
the cart 1, 2, and 3.
00:21:02.740 --> 00:21:05.830
And that equilibrium requirement
then means that
00:21:05.830 --> 00:21:11.010
the sum of the forces acting
onto the individual springs 1,
00:21:11.010 --> 00:21:17.720
2, 3, and 4 at degree of freedom
1 must be equal to R1.
00:21:17.720 --> 00:21:20.030
Now let us look once
at this first
00:21:20.030 --> 00:21:22.430
equation back here again.
00:21:22.430 --> 00:21:29.110
Notice u1 couples into this
spring 1, spring 2, spring 3,
00:21:29.110 --> 00:21:30.670
and spring 4.
00:21:30.670 --> 00:21:34.320
And that coupling is seen
right here in spring
00:21:34.320 --> 00:21:36.830
1, 2, 3, and 4.
00:21:36.830 --> 00:21:40.490
And summing all these forces
that are acting individually
00:21:40.490 --> 00:21:43.860
onto the springs, the sum of
these forces must be equal to
00:21:43.860 --> 00:21:45.250
the external load.
00:21:45.250 --> 00:21:47.420
That is the interconnection
00:21:47.420 --> 00:21:50.730
requirement between the springs.
00:21:50.730 --> 00:21:54.010
The equilibrium requirements
within the springs are
00:21:54.010 --> 00:21:59.300
expressed by these individual
matrices here that we looked
00:21:59.300 --> 00:21:59.820
at already.
00:21:59.820 --> 00:22:01.610
These are the equilibrium
requirements for the
00:22:01.610 --> 00:22:02.820
individual springs.
00:22:02.820 --> 00:22:06.270
Now I'm talking about the
equilibrium requirement at the
00:22:06.270 --> 00:22:07.730
carts, the interconnection
00:22:07.730 --> 00:22:10.310
requirements between the springs.
00:22:10.310 --> 00:22:14.450
Similarly, we can sum the forces
that have to be equal
00:22:14.450 --> 00:22:18.890
to R2 and sum the forces that
have to be equal to R3.
00:22:18.890 --> 00:22:22.960
And these three equations then
set up in matrix form by
00:22:22.960 --> 00:22:28.140
substituting for F1(1), F1(2),
and so on from the equilibrium
00:22:28.140 --> 00:22:32.030
requirements of the springs,
we directly obtain this set
00:22:32.030 --> 00:22:35.820
off equations, KU equals
R. Notice that K
00:22:35.820 --> 00:22:38.630
now is a 3 by 3 matrix.
00:22:38.630 --> 00:22:41.540
U is a 3 by 1 vector.
00:22:41.540 --> 00:22:43.490
R is a 3 by 1 vector.
00:22:43.490 --> 00:22:47.950
I denote matrices and vectors
by bars under the symbols.
00:22:47.950 --> 00:22:51.782
As you can see here there are
bars under these symbols.
00:22:51.782 --> 00:22:54.890
Well, if we look at these
equilibrium equations, we
00:22:54.890 --> 00:23:04.480
notice that our U vector, this
vector U here contains u1, u2,
00:23:04.480 --> 00:23:06.400
and u3 as the unknowns.
00:23:06.400 --> 00:23:11.610
Notice this T here, this
superscript T means transpose.
00:23:11.610 --> 00:23:17.670
The actual vector U actually
looks this way u1, u2, u3.
00:23:17.670 --> 00:23:20.240
It lists the displacements
vertically downwards.
00:23:20.240 --> 00:23:23.110
But it is easier to write
it this way by
00:23:23.110 --> 00:23:24.350
transposing as a vector.
00:23:24.350 --> 00:23:27.340
So UT, capital T there,
means transpose.
00:23:27.340 --> 00:23:31.790
Similarly for R we have R1, R2,
and R3 as the components.
00:23:31.790 --> 00:23:37.180
And the K matrix that we have
obtained by substituting into
00:23:37.180 --> 00:23:39.680
these equations from the
element equilibrium
00:23:39.680 --> 00:23:44.160
requirements, the K matrix
is this one here.
00:23:44.160 --> 00:23:48.400
Now let us look a little closer
at how do we construct
00:23:48.400 --> 00:23:49.520
this K matrix.
00:23:49.520 --> 00:23:55.760
Well, we note that the total K
matrix can be constructed by
00:23:55.760 --> 00:24:01.820
summing all of the individual
element matrices from 1 to 5.
00:24:01.820 --> 00:24:05.820
And these individual element
matrices are, for two
00:24:05.820 --> 00:24:07.150
extremes, written down here.
00:24:07.150 --> 00:24:09.790
K1 is a 3 by 3 matrix now.
00:24:09.790 --> 00:24:12.200
Not anymore the 1
by 1 or 2 by 2.
00:24:12.200 --> 00:24:17.500
It's a 3 by 3 matrix with just
k1 in the 1,1 position.
00:24:17.500 --> 00:24:19.690
All the other elements are 0.
00:24:19.690 --> 00:24:22.470
K2 is this matrix.
00:24:22.470 --> 00:24:26.300
So what I have done then is I
have taken the 2 by 2 matrix
00:24:26.300 --> 00:24:29.180
which appeared in the element
equilibrium requirement and
00:24:29.180 --> 00:24:32.640
has blown this matrix up filling
zeroes for the third
00:24:32.640 --> 00:24:34.240
degree of freedom.
00:24:34.240 --> 00:24:37.120
Similarly we would obtain
K3 and so on.
00:24:37.120 --> 00:24:41.530
The zeroes always appear in
those rows and columns into
00:24:41.530 --> 00:24:45.320
which the element does not
couple, in other words, into
00:24:45.320 --> 00:24:47.500
those degrees of freedom that
the element does not couple.
00:24:47.500 --> 00:24:53.130
For example, k1, this element 1
here, couples only into the
00:24:53.130 --> 00:24:54.480
degree of freedom 1.
00:24:54.480 --> 00:24:58.380
So, therefore, we have the
second and third rows be 0.
00:24:58.380 --> 00:25:03.800
Element 2 couples only into
degree of freedom 1 and 2.
00:25:03.800 --> 00:25:07.080
Therefore, the third degree
of freedom contains all
00:25:07.080 --> 00:25:09.360
zeroes, and so on.
00:25:09.360 --> 00:25:13.550
This assemblage process is
called the direct stiffness
00:25:13.550 --> 00:25:19.330
method, an extremely important
concept that is very well
00:25:19.330 --> 00:25:21.200
implemented in a computer
program.
00:25:21.200 --> 00:25:25.410
It represents the basis of the
implementation of the finite
00:25:25.410 --> 00:25:30.120
element method in almost every
code that is currently in use.
00:25:30.120 --> 00:25:34.900
The direct stiffness method has
also a very nice physical
00:25:34.900 --> 00:25:35.740
explanation.
00:25:35.740 --> 00:25:39.030
And this is what I really want
to talk to you about now for
00:25:39.030 --> 00:25:40.180
the next five minutes.
00:25:40.180 --> 00:25:44.260
The steady-state analysis, of
course, then is completed.
00:25:44.260 --> 00:25:46.440
The steady-state analysis of
this system, of course, is
00:25:46.440 --> 00:25:50.660
completed by solving this system
of equations here,
00:25:50.660 --> 00:25:51.810
equations a.
00:25:51.810 --> 00:25:57.000
Once we know U we can go back to
the elements and calculate
00:25:57.000 --> 00:25:59.820
the forces in the elements
themselves by going to the
00:25:59.820 --> 00:26:02.070
element equilibrium
requirements.
00:26:02.070 --> 00:26:08.100
Well let us look then at what
we are doing when we perform
00:26:08.100 --> 00:26:14.810
this process here by summing, in
other words, the K element,
00:26:14.810 --> 00:26:17.510
the stiffnesses of the elements
into a global
00:26:17.510 --> 00:26:18.520
stiffness matrix.
00:26:18.520 --> 00:26:21.670
And let us look at what we're
doing physically.
00:26:21.670 --> 00:26:24.040
Because that really, of
course, is the direct
00:26:24.040 --> 00:26:26.210
stiffness method that
we are using here.
00:26:26.210 --> 00:26:31.040
And it is, I think, very nice if
you can clearly see what is
00:26:31.040 --> 00:26:32.590
happening in that method.
00:26:32.590 --> 00:26:36.120
Well, the basic process
is the following.
00:26:36.120 --> 00:26:39.600
Here I have drawn the carts
without any strings.
00:26:39.600 --> 00:26:41.220
Of course, our degrees
of freedom are
00:26:41.220 --> 00:26:45.160
here, u1, u2, and u3.
00:26:45.160 --> 00:26:47.280
And the loads are R1, R2, R3.
00:26:47.280 --> 00:26:49.160
I don't need to put
them in again.
00:26:49.160 --> 00:26:55.700
This system here corresponds
to a K matrix with zeroes
00:26:55.700 --> 00:26:57.410
everywhere.
00:26:57.410 --> 00:27:02.000
Blanks in these positions
here denote zeroes.
00:27:02.000 --> 00:27:06.670
So this is a system that we're
starting off with in this
00:27:06.670 --> 00:27:11.200
direct stiffness method, a
system without any elements, a
00:27:11.200 --> 00:27:14.470
matrix without any
elements also.
00:27:14.470 --> 00:27:17.370
The process, then,
is the following.
00:27:17.370 --> 00:27:21.860
We are using this cart system.
00:27:21.860 --> 00:27:23.810
And we're adding
one spring on.
00:27:23.810 --> 00:27:25.510
That is the first edition.
00:27:25.510 --> 00:27:28.680
That is spring k1.
00:27:28.680 --> 00:27:33.390
Mathematically this means that
we're going through the
00:27:33.390 --> 00:27:34.760
following process.
00:27:34.760 --> 00:27:39.110
We are taking our K matrix with
blanks everywhere, and
00:27:39.110 --> 00:27:43.370
we're adding into it this
one element, k1.
00:27:43.370 --> 00:27:47.380
Now this is a K matrix, this
stiffness matrix governing--
00:27:47.380 --> 00:27:48.980
and this is very important--
00:27:48.980 --> 00:27:53.650
governing this system.
00:27:53.650 --> 00:27:58.220
Once again, this is a K matrix
governing this system.
00:27:58.220 --> 00:28:00.460
Of course, this is not a stable
system yet because
00:28:00.460 --> 00:28:04.320
there are no connections between
these carts here.
00:28:04.320 --> 00:28:08.220
Well, with a second edition
we're adding
00:28:08.220 --> 00:28:11.350
in the second spring.
00:28:11.350 --> 00:28:13.970
And that means we are putting
this spring there.
00:28:13.970 --> 00:28:16.110
That is k2.
00:28:16.110 --> 00:28:20.520
Well in our matrix formulation
then, what that means in our
00:28:20.520 --> 00:28:25.750
direct stiffness method is that
we are going from this
00:28:25.750 --> 00:28:30.170
system over on this K matrix
to that K matrix.
00:28:30.170 --> 00:28:32.800
We are adding this
second spring
00:28:32.800 --> 00:28:35.070
stiffness into the K matrix.
00:28:35.070 --> 00:28:39.970
So this is the stiffness
matrix that governs the
00:28:39.970 --> 00:28:43.940
equilibrium of this
physical system.
00:28:47.070 --> 00:28:50.390
Notice that this spring here,
the second spring, couples
00:28:50.390 --> 00:28:52.070
into u1 and u2.
00:28:52.070 --> 00:28:56.130
And therefore we have added
these blue elements
00:28:56.130 --> 00:28:58.730
corresponding to the second
spring into degrees of
00:28:58.730 --> 00:29:01.500
freedom 1 and 2.
00:29:01.500 --> 00:29:04.750
Next in the direct stiffness
method we're adding the next
00:29:04.750 --> 00:29:09.820
spring element, and that is
spring element number 3.
00:29:09.820 --> 00:29:12.900
Again, it couples
into u1 and u2.
00:29:12.900 --> 00:29:17.930
And the stiffness matrix that
we are now talking about is
00:29:17.930 --> 00:29:18.640
the following.
00:29:18.640 --> 00:29:23.200
We're going from this stiffness
matrix to that
00:29:23.200 --> 00:29:28.260
stiffness matrix here, adding
the green k3 in there.
00:29:28.260 --> 00:29:37.230
Now next we go from this system
to add into the system
00:29:37.230 --> 00:29:42.620
the spring 4, spring 4 now.
00:29:42.620 --> 00:29:45.400
Please notice that this is
now a stable system.
00:29:45.400 --> 00:29:49.950
It is a stable system because if
I want to put u3 over here,
00:29:49.950 --> 00:29:51.690
then I have to do work
on this spring.
00:29:51.690 --> 00:29:53.690
So this is now a
stable system.
00:29:53.690 --> 00:29:56.910
In our mathematical formulation,
or in our direct
00:29:56.910 --> 00:30:00.050
stiffness method rather, what
this then corresponds to is
00:30:00.050 --> 00:30:04.050
that we are going from this
system here, or this stiffness
00:30:04.050 --> 00:30:06.220
matrix, to that stiffness
matrix.
00:30:06.220 --> 00:30:09.080
Notice we have added
a k4 into here.
00:30:09.080 --> 00:30:13.730
And that, in fact, allows us
now to solve at this level.
00:30:13.730 --> 00:30:18.360
We could solve the equations KU
equals R. Of course, this
00:30:18.360 --> 00:30:23.880
now is a stiffness matrix
corresponding to this system.
00:30:23.880 --> 00:30:26.230
We have not quite yet
reached the system
00:30:26.230 --> 00:30:28.070
that we want to analyze.
00:30:28.070 --> 00:30:33.230
But we reach it by adding
the final spring in k5.
00:30:33.230 --> 00:30:34.870
k5 now is here.
00:30:34.870 --> 00:30:41.210
And that corresponds in our
direct stiffness method to
00:30:41.210 --> 00:30:45.210
adding this spring in there.
00:30:45.210 --> 00:30:47.970
These elements here, k5.
00:30:47.970 --> 00:30:52.120
Notice that this spring now here
couples into degrees of
00:30:52.120 --> 00:30:55.500
freedom 4 and five 5, and
that's why it appears in
00:30:55.500 --> 00:30:56.775
quadrant column 4 and 5 here.
00:31:00.480 --> 00:31:07.020
And this spring.
00:31:07.020 --> 00:31:10.130
I should've have said, this
spring here couples into
00:31:10.130 --> 00:31:17.190
column 2 and 3, column 2 and
3 meaning u2 and u3.
00:31:17.190 --> 00:31:21.270
And here we see, of course, that
the spring indeed goes
00:31:21.270 --> 00:31:24.890
into degree of freedom u2
and u3, into degree of
00:31:24.890 --> 00:31:26.490
freedom u2 and u3.
00:31:26.490 --> 00:31:29.470
So this then is to final
system that we want to
00:31:29.470 --> 00:31:33.440
analyze, and this is the
final stiffness matrix
00:31:33.440 --> 00:31:35.710
that we had to obtain.
00:31:35.710 --> 00:31:38.600
Notice, once again, this matrix
has been obtained by
00:31:38.600 --> 00:31:44.235
taking the sum over all the
element stiffness matrices.
00:31:44.235 --> 00:31:47.520
We are summing from
i equals 1 to 5.
00:31:47.520 --> 00:31:50.950
And this mathematical process,
once again, which we call the
00:31:50.950 --> 00:31:56.570
direct stiffness method
has a physical analog.
00:31:56.570 --> 00:31:59.030
You can understand it physically
in the way I've
00:31:59.030 --> 00:32:00.450
shown here.
00:32:00.450 --> 00:32:05.010
Namely you're starting off
with a blank K matrix, no
00:32:05.010 --> 00:32:09.660
elements in it at all, and you
simply add one element after
00:32:09.660 --> 00:32:12.930
the other into that K matrix
filling up the
00:32:12.930 --> 00:32:14.290
K matrix that way.
00:32:14.290 --> 00:32:17.590
And the additions are
carried out--
00:32:17.590 --> 00:32:18.670
this is important--
00:32:18.670 --> 00:32:24.560
by taking the element matrices
and adding them into the
00:32:24.560 --> 00:32:28.800
appropriate columns and
rows of the K matrix.
00:32:28.800 --> 00:32:32.280
For example, this element here
couples into degree of
00:32:32.280 --> 00:32:34.850
freedom 1 and 3.
00:32:34.850 --> 00:32:39.850
And if we go once more back to
the process that we have been
00:32:39.850 --> 00:32:46.530
carrying out here, notice our k4
here corresponds to degree
00:32:46.530 --> 00:32:50.680
of freedom 1 and degree of
freedom 3, the first row and
00:32:50.680 --> 00:32:52.750
column and third
row and column.
00:32:52.750 --> 00:32:56.030
That's where these
elements appear.
00:32:56.030 --> 00:33:00.310
So there's a neat physical
explanation for the direct
00:33:00.310 --> 00:33:03.900
stiffness method which
I wanted to
00:33:03.900 --> 00:33:05.240
discuss with you here.
00:33:05.240 --> 00:33:09.750
Now as another approach, instead
of using the direct
00:33:09.750 --> 00:33:15.380
formulation of the equations KU
equals R, the equilibrium
00:33:15.380 --> 00:33:19.410
equations of the system,
we can also use
00:33:19.410 --> 00:33:20.950
a variational approach.
00:33:20.950 --> 00:33:23.940
We will be talking about that
variational approach in the
00:33:23.940 --> 00:33:24.790
second lecture.
00:33:24.790 --> 00:33:28.030
And I would like to discuss it,
or introduce it to you,
00:33:28.030 --> 00:33:31.880
now very briefly for the
analysis of this discrete
00:33:31.880 --> 00:33:33.980
system that we just looked at.
00:33:33.980 --> 00:33:37.010
The basic process here is that
we are constructing a
00:33:37.010 --> 00:33:42.320
functional pi which is equal
to u minus w where u is the
00:33:42.320 --> 00:33:46.510
strain energy of the system and
w is the total potential
00:33:46.510 --> 00:33:48.390
of the loads.
00:33:48.390 --> 00:33:53.110
The equilibrium equations that
we just looked at, KU equals R
00:33:53.110 --> 00:33:59.685
in other words, are obtained by
invoking that del-pi shall
00:33:59.685 --> 00:34:03.740
be 0, the stationality
condition on pi.
00:34:03.740 --> 00:34:10.770
And this means that del-pi,
del-ui, shall be 0 for all ui.
00:34:10.770 --> 00:34:15.360
This then gives three equations,
And these three
00:34:15.360 --> 00:34:18.805
questions are obtained
as follows.
00:34:18.805 --> 00:34:25.770
If we use u, the strain energy
of the system is given right
00:34:25.770 --> 00:34:29.340
here, 1/2 U transpose KU.
00:34:29.340 --> 00:34:33.270
If you were to multiply this out
substituting for U and for
00:34:33.270 --> 00:34:36.360
K with the values that I've
given to you, you would find
00:34:36.360 --> 00:34:39.449
that this indeed is the strain
energy in the system.
00:34:39.449 --> 00:34:43.260
The potential of the total loads
is given by U transposed
00:34:43.260 --> 00:34:49.250
R. Notice please that there
is no 1/2 here in front.
00:34:49.250 --> 00:34:52.270
Simply U transpose R is the
potential of the loads.
00:34:52.270 --> 00:34:57.110
Now if we invoke this condition
that del-pi, del-ui
00:34:57.110 --> 00:35:03.790
shall be 0, we directly
obtain KU equals R.
00:35:03.790 --> 00:35:05.830
Now there's one important
point.
00:35:05.830 --> 00:35:11.510
To obtain u and w, u and w here,
we again, can add up the
00:35:11.510 --> 00:35:14.810
contributions from all the
elements using the direct
00:35:14.810 --> 00:35:15.550
stiffness method.
00:35:15.550 --> 00:35:21.410
In other words, this K here can
be constructed as we have
00:35:21.410 --> 00:35:26.990
shown by summing over the
elements, by summing the
00:35:26.990 --> 00:35:28.890
contributions over all
of the elements.
00:35:28.890 --> 00:35:36.130
And since this is true, we can
also write this total u as
00:35:36.130 --> 00:35:40.380
being the sum of the ui's, if
you want to, the strain
00:35:40.380 --> 00:35:44.220
energies of all of the
individual elements.
00:35:44.220 --> 00:35:48.400
So here too we could use the
direct stiffness method.
00:35:48.400 --> 00:35:51.640
Of course, in actuality, in
actual practical analysis, we
00:35:51.640 --> 00:35:56.180
never form this u, we never form
that w when we want to
00:35:56.180 --> 00:36:00.470
calculate KU equals R. This is
simply a theoretical concept
00:36:00.470 --> 00:36:02.810
that I wanted to introduce to
you, a theoretical concept
00:36:02.810 --> 00:36:06.630
that we will be using later on
in the construction of KU
00:36:06.630 --> 00:36:12.000
equals R. We never really
calculate these measures if we
00:36:12.000 --> 00:36:14.460
only want to calculate
KU equals R.
00:36:14.460 --> 00:36:17.240
It might be of interest to us to
calculate this in order to
00:36:17.240 --> 00:36:20.750
find out how much strain energy
is put into individual
00:36:20.750 --> 00:36:23.730
elements in finite
element analysis.
00:36:23.730 --> 00:36:28.480
But this is only done if you
want to evaluate error bounds
00:36:28.480 --> 00:36:30.710
on the finite element
solution and so on.
00:36:30.710 --> 00:36:34.840
If we only want the calculate KU
equals R and obtain the use
00:36:34.840 --> 00:36:40.185
in other words, to be able to
predict the displacements and
00:36:40.185 --> 00:36:43.440
the stresses in the elements,
then we would not calculate
00:36:43.440 --> 00:36:45.610
these two quantities.
00:36:45.610 --> 00:36:50.090
Now this then were the essence
of the analysis of a
00:36:50.090 --> 00:36:55.650
steady-state problem for
discrete systems.
00:36:55.650 --> 00:36:58.390
I pointed out already that if
we have an extra finite
00:36:58.390 --> 00:37:00.560
element system there, of
course, many additional
00:37:00.560 --> 00:37:02.750
concepts that we have to talk
about, a selection of
00:37:02.750 --> 00:37:05.520
elements, the kinds of
interpolations to be used,
00:37:05.520 --> 00:37:08.070
and, of course, we then have to
also talk about how do we
00:37:08.070 --> 00:37:10.330
solve these equations,
and so on.
00:37:10.330 --> 00:37:11.720
We will address these questions
00:37:11.720 --> 00:37:13.340
in the later lectures.
00:37:13.340 --> 00:37:16.660
However, another class of
problems that we will be
00:37:16.660 --> 00:37:19.660
talking about are propagation
problems.
00:37:19.660 --> 00:37:23.160
The main characteristics of
propagation problems are that
00:37:23.160 --> 00:37:25.580
the response changes
with time.
00:37:25.580 --> 00:37:28.690
Therefore, we need to include
the d'Alembert forces.
00:37:28.690 --> 00:37:31.280
Now basically what we are saying
then is that we're
00:37:31.280 --> 00:37:37.490
looking at static equilibrium as
a function of time but also
00:37:37.490 --> 00:37:39.780
taking into account the
d'Alembert forces.
00:37:39.780 --> 00:37:43.560
And that, together then, makes
it a dynamic problem.
00:37:43.560 --> 00:37:49.150
Of course, if the displacement
varies very slow, in other
00:37:49.150 --> 00:37:56.400
words, the load varies very
slow, then the inertia forces
00:37:56.400 --> 00:37:59.950
can be neglected, and we would
simply have this set of
00:37:59.950 --> 00:38:03.586
equations where R of t is a
function of time and U of t
00:38:03.586 --> 00:38:04.760
would be a function of time.
00:38:04.760 --> 00:38:09.440
However, when R of t acts
rapidly or suddenly inertia
00:38:09.440 --> 00:38:12.430
conditions are applied to the
system, then the inertia
00:38:12.430 --> 00:38:14.650
forces can be very important.
00:38:14.650 --> 00:38:15.920
We have to include
their effect.
00:38:15.920 --> 00:38:18.820
And then we have a true
propagation problem, a truly
00:38:18.820 --> 00:38:20.930
dynamic problem that
has to be solved.
00:38:20.930 --> 00:38:26.690
For our example, the M matrix
here would be this 3 by 3
00:38:26.690 --> 00:38:33.750
matrix where m1 is simply the
mass of the cart 1, m2 is the
00:38:33.750 --> 00:38:37.560
mass of the cart 2, m3 is
the mass of the cart 3.
00:38:37.560 --> 00:38:39.680
Of course these masses would
have to be given.
00:38:39.680 --> 00:38:46.780
And notice that we would
evaluate them by basically
00:38:46.780 --> 00:38:51.420
saying that this total mass
here can be evaluated by
00:38:51.420 --> 00:38:54.170
taking the mass per unit volume
times the volume.
00:38:54.170 --> 00:38:56.750
And that would be the mass that
we're talking about when
00:38:56.750 --> 00:39:02.040
we accelerate that cart
into this direction.
00:39:02.040 --> 00:39:04.970
So these masses here are
very simply evaluated.
00:39:04.970 --> 00:39:08.200
When we talk later on about
actually finite elements, we
00:39:08.200 --> 00:39:12.180
will be talking about similar
mass matrices where we simply
00:39:12.180 --> 00:39:16.600
take the total volume of an
element and lump that volume
00:39:16.600 --> 00:39:18.440
of the element to its nodes.
00:39:18.440 --> 00:39:22.050
We will also talk about
consistent mass matrices where
00:39:22.050 --> 00:39:25.260
this mass matrix is a little
bit more complicated.
00:39:25.260 --> 00:39:27.170
In other words, some of
these off-diagonal
00:39:27.170 --> 00:39:29.820
elements are not 0.
00:39:29.820 --> 00:39:33.350
Finally, we will also talk about
eigenvalue problems.
00:39:33.350 --> 00:39:36.920
In the solution of eigenvalue
problems, we will be talking
00:39:36.920 --> 00:39:41.140
about generalized eigenvalue
problems, in particular, which
00:39:41.140 --> 00:39:46.650
are Av equals lambda Bv, which
can be written down in this
00:39:46.650 --> 00:39:51.920
form where A and B are symmetric
matrices of order n,
00:39:51.920 --> 00:39:56.850
v is a vector of order n,
and lambda is a scalar.
00:39:56.850 --> 00:40:00.530
As an example, for example here
in dynamic analysis, what
00:40:00.530 --> 00:40:07.150
we will see there is K phi
equals omega squared M phi
00:40:07.150 --> 00:40:09.850
where K is the stiffness
matrix that I
00:40:09.850 --> 00:40:11.280
talked about already.
00:40:11.280 --> 00:40:14.350
This which would be for the cart
system here simply as 3
00:40:14.350 --> 00:40:17.030
by 3, this 3 by 3 stiffness
matrix that I
00:40:17.030 --> 00:40:18.160
introduced to you.
00:40:18.160 --> 00:40:20.430
And it's a mass matrix
that we just had
00:40:20.430 --> 00:40:22.870
here on this viewgraph.
00:40:22.870 --> 00:40:24.020
That is the mass matrix.
00:40:24.020 --> 00:40:28.790
And phi is the vector.
00:40:28.790 --> 00:40:31.510
If we find a solution, in other
words, if this equation
00:40:31.510 --> 00:40:38.280
is satisfied, we put an i on
there and satisfy for phi i
00:40:38.280 --> 00:40:41.200
and omega i squared.
00:40:41.200 --> 00:40:44.320
Omega i squared will
be a frequency.
00:40:44.320 --> 00:40:46.880
I will be discussing it just
now a little more.
00:40:46.880 --> 00:40:49.470
And then we're talking
about an eigenpair.
00:40:49.470 --> 00:40:52.210
But notice that is a typical
problem that we will be
00:40:52.210 --> 00:40:54.340
discussing which arises,
in other
00:40:54.340 --> 00:40:56.580
words, in dynamic analysis.
00:40:56.580 --> 00:41:00.050
Notice also that what we're
really saying here is that the
00:41:00.050 --> 00:41:01.860
right-hand side is
a load vector.
00:41:04.830 --> 00:41:09.470
And if we know v, if we
know lambda, then we
00:41:09.470 --> 00:41:11.130
know the load vector.
00:41:11.130 --> 00:41:16.540
What we would calculate then
is the same v that we have
00:41:16.540 --> 00:41:17.430
substituted here.
00:41:17.430 --> 00:41:21.420
In other words, if we consider
this to be a set of loads
00:41:21.420 --> 00:41:25.430
where v is now known, lambda
is known, then we could
00:41:25.430 --> 00:41:30.170
evaluate R. In solving Av equals
R, we would get back
00:41:30.170 --> 00:41:32.320
our v that we substituted
into here.
00:41:32.320 --> 00:41:34.670
And that is the main
characteristic of an
00:41:34.670 --> 00:41:36.630
eigenvalue problem.
00:41:36.630 --> 00:41:39.610
Well, they arise in dynamic and
buckling analysis, and let
00:41:39.610 --> 00:41:43.770
us look at one example where
we actually obtain this
00:41:43.770 --> 00:41:45.030
eigenvalue problem.
00:41:45.030 --> 00:41:47.850
And the example is simply the
system of rigid carts that we
00:41:47.850 --> 00:41:50.120
considered already earlier.
00:41:50.120 --> 00:41:53.670
We obtain the eigenvalue problem
by looking at the
00:41:53.670 --> 00:41:56.980
equilibrium equations when
no loads are applied.
00:41:56.980 --> 00:42:00.860
And we call these the free
vibration conditions, free
00:42:00.860 --> 00:42:03.990
because there are no loads
applied, free of loads.
00:42:03.990 --> 00:42:09.540
If we let U be equal to phi
times sine omega t minus tau
00:42:09.540 --> 00:42:13.090
where the time dependency now
in the response is in this
00:42:13.090 --> 00:42:17.980
function here, in the sine
function only, and if we take
00:42:17.980 --> 00:42:22.700
the second derivative of U,
meaning that we get a cosine
00:42:22.700 --> 00:42:26.820
and then a minus sign in here,
and, of course, this omega
00:42:26.820 --> 00:42:30.580
twice outside, so we have
a sign change here.
00:42:30.580 --> 00:42:35.160
We have a minus omega squared
M phi sine omega t minus tau
00:42:35.160 --> 00:42:37.270
for this part here.
00:42:37.270 --> 00:42:42.940
And for this part KU we obtain
K phi sine omega t minus tau
00:42:42.940 --> 00:42:46.130
by simply substituting
from here into there.
00:42:46.130 --> 00:42:51.470
And, of course, the sum of these
two must be equal to 0.
00:42:51.470 --> 00:42:56.260
Now this equation must
hold for any time, t.
00:42:56.260 --> 00:43:00.420
So we can simply cancel out
this part and that part.
00:43:00.420 --> 00:43:04.790
And the resulting set of
equations that we are
00:43:04.790 --> 00:43:08.590
obtaining then are given on
the last viewgraph, namely
00:43:08.590 --> 00:43:12.520
those equations being K phi
equals omega squared M phi.
00:43:12.520 --> 00:43:15.000
So that is the generalized
eigenvalue problem which we
00:43:15.000 --> 00:43:17.130
obtain in dynamic analysis.
00:43:17.130 --> 00:43:21.540
We will be later on talking
about how we solve this
00:43:21.540 --> 00:43:24.690
generalized eigenvalue problem
for the eigenvalues and
00:43:24.690 --> 00:43:25.400
eigenvectors.
00:43:25.400 --> 00:43:28.500
In the case of of the 3 by 3
system that we are considering
00:43:28.500 --> 00:43:31.780
here, in other words, the
analysis of the cart system,
00:43:31.780 --> 00:43:38.030
we only have three solutions,
omega 1 phi 1, omega2 phi2,
00:43:38.030 --> 00:43:39.770
omega3 phi3.
00:43:39.770 --> 00:43:43.850
And we call each of the
solutions an eigenpair.
00:43:43.850 --> 00:43:46.910
So there are three eigenpairs
that satisfy
00:43:46.910 --> 00:43:48.410
this particular equation.
00:43:48.410 --> 00:43:51.320
Notice that this is, in other
words, the equation that I
00:43:51.320 --> 00:43:53.310
talked about here earlier.
00:43:53.310 --> 00:43:59.820
And the eigenpairs, phi i,
omega i squared are the
00:43:59.820 --> 00:44:02.110
solutions to this equation.
00:44:02.110 --> 00:44:06.220
We are really interested in
omega i because that is the
00:44:06.220 --> 00:44:10.260
frequency in radians per second,
and the eigenvalue,
00:44:10.260 --> 00:44:12.660
however, being omega squared.
00:44:12.660 --> 00:44:15.430
In general when we have
an n by n system--
00:44:15.430 --> 00:44:19.090
and I have already written down
here the n by n, let me
00:44:19.090 --> 00:44:21.130
put it bigger once more here--
00:44:21.130 --> 00:44:23.480
and we have a general n by n
system, in other words, and
00:44:23.480 --> 00:44:28.000
not being equal to 3 just as
we have in our cart system,
00:44:28.000 --> 00:44:30.410
then we have n solutions.
00:44:30.410 --> 00:44:34.080
And, however, we will find
that in finite element
00:44:34.080 --> 00:44:36.670
analysis we do not necessarily
need to
00:44:36.670 --> 00:44:38.600
calculate all n solutions.
00:44:38.600 --> 00:44:42.080
In fact, when we consider large
eigensystems where n is
00:44:42.080 --> 00:44:46.320
equal to 1,000 or even more,
then certainly we do not want
00:44:46.320 --> 00:44:47.790
to calculate all eigenvalues.
00:44:47.790 --> 00:44:51.720
It would be exorbitantly
expensive, much too expensive
00:44:51.720 --> 00:44:53.810
to calculate all of the
eigenvalues and eigenvectors.
00:44:53.810 --> 00:44:57.250
We don't need to have them
all in analysis.
00:44:57.250 --> 00:45:00.260
And, therefore, we will talk
about eigenvalue solution
00:45:00.260 --> 00:45:03.440
methods that only calculate
the eigenvalues and
00:45:03.440 --> 00:45:07.210
eigenvectors that we are
actually interested in.
00:45:07.210 --> 00:45:10.450
We also, of course, have to,
before we actually get to that
00:45:10.450 --> 00:45:13.410
topic which is the topic of the
last lecture, we will talk
00:45:13.410 --> 00:45:15.740
about how we actually construct
these K matrices,
00:45:15.740 --> 00:45:18.840
how we calculate them, construct
them for different
00:45:18.840 --> 00:45:20.500
finite element systems.
00:45:20.500 --> 00:45:24.640
Well, this then does complete
what I wanted
00:45:24.640 --> 00:45:26.000
to say in this lecture.
00:45:26.000 --> 00:45:27.850
Thank you very much for
your attention.