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MIT OpenCourseWare at ocw.mit.edu. PROFESSOR: Ladies and gentlemen,
welcome to this set of lectures on the finite
element method. In these lectures I would like
to give you an introduction to the linear analysis of solids
and structures. You are probably well aware that
the finite element method is now widely used for analysis
of structural engineering problems. The method is used in civil,
aeronautical, mechanical, ocean, mining, nuclear,
biomechanical, and other engineering disciplines. Since the first applications two
decades ago of the finite element method we now see
applications in linear, nonlinear, static, and
dynamic analysis. However, in this set of
lectures, I would like to discuss with you only the
linear, static, and dynamic analysis of problems. The finite element method
is used today in various computer programs. And its use is very
significant. My objective in this set of
lectures is to introduce to you the finite element methods
or some of the finite element methods that are used for linear
analysis of solids and structures. And here we understand linear
to mean that we're talking about infinitesimally small
displacements and that we are using a linear elastic
material law. In other words, Hooke's
law applies. We will consider, in this set
of lectures, the formulation of the finite element
equilibrium equations, the calculation of finite element
matrices of the matrices that arise in the equilibrium
equations. We will be talking about the
methods for solution of the governing equations in static
and dynamic analysis. And we will talk about actual
computer implementations. I will emphasize modern and
effective techniques and their practical usage. The emphasis, in this set of
lectures, is given to physical explanations of the methods,
techniques that we are using rather than mathematical
derivations. The techniques that we will be
discussing are those employed largely in the computer programs
SAP and ADINA. SAP stands for Structural
Analysis Program and you might very well be aware that there is
a series of such programs, SAP I to SAP VI now. And ADINA stands for
Automatic Dynamic Incremental Nonlinear Analysis. However, this program is also
very effectively employed for linear analysis. The nonlinear analysis being
then a next step in the usage of the program. In fact, the elements in ADINA,
the numerical methods that are used in ADINA, I
consider to be the most effective, the most modern state
of the art techniques that are currently available. These few lectures really
represent a very brief and compact introduction
to the field of finite element analysis. We will go very rapidly through
some or the basic concepts, practical
applications, and so on. We shall follow quite closely,
however, certain sections in my book entitled Finite
Element Procedures in Engineering Analysis to be
published by Prentice Hall. And I will be referring in the
study guide of this set of lectures extensively to this
book to the specific sections that we're considering in the
lectures in this book. The finite element solution
process can be described as given on this viewgraph. You can see here that we talk
about a physical problem. We want to analyze an actual
physical problem. And our first step, of course,
is to establish a finite element model of that
physical problem. Then, in the next step,
we solve that model. And then we have to interpret
the results. Because the interpretation of
the results depends very much on how we established the finite
element model, what kind of model we used,
and so on. And in establishing the finite
element model, we have to be aware of what kinds of elements,
techniques, and so on are available to us. Well, therefore, I will be
talking, in the set of lectures, about these three
steps basically here for different kinds of physical
problems. Once we have interpreted the
results we might go back from down here to there to revise
or refine our model and go through this process again until
we feel that our model has been an adequate one for the
solution of the physical problem of interest. Let me give you or show you
some models that have been used in actual structural
analysis. You might have seen similar
models in textbooks, in publications already. This, for example, is a model
that was used for the analysis of a cooling tower. The basic process of the finite
element method is that we are taking the continuous
system, and we are idealizing it as an assemblage
of elements. I'm drawing here a typical
three-noded triangular shell element that was used
in the analysis of this cooling tower. We talk about very many elements
in order to obtain an accurate response prediction. And, of course, that means that
we will be dealing with a large set of equations
to be solved. And there's a significant
computer effort required. I will be addressing
all of these questions in these lectures. Here you see the finite element
model of a dam. The earth below the dam was
idealized as an assemblage all such elements here, triangular
elements now. And the dam itself was
also idealized as an assemblage of elements. We will be talking about how
such assemblages are best created, what kinds of elements
to select, what assumptions are in the selection
of these elements, and then how do we solve the
resulting finite element equilibrium equations. Here you see the finite element
analysis, or the mesh that was used in the finite
element analysis of a tire. This wall is half of the tire,
as you can see, and this was the finite element mesh used. Again, we have to judiciously
choose the kinds of finite elements to be employed. And we will be talking about
that in this set of lectures. Here you see the finite element
model employed in the analysis of a spherical
cover of a laser vacuum target chamber. This is the finite element
mesh used. Again, specific elements
were employed here. And we will be talking about the
characteristics of these elements in this set
of lectures. Here you see the model of the
shell structure subjected to a pinching load. There's a load up here and
a load down there. These are the triangular
elements that were used in the idealization of that shell and
the resulting bending moments and displacements along the line
DC are plotted here that have been predicted by the
finite element analysis. Finally here you see the finite
element idealization of a wind tunnel that was used for
the dynamic analysis of this tunnel. You can see a large number of
shell elements were employed in the idealization of the
shelf of the tunnel. Then, of course, supports were
provided here for that shell. And this was a very large
system that was solved. And the eigenvalues of this
system were calculated using the subspace iteration method
that we would be also talking about in this set of lectures. Well, with this short
introduction then, I would like to go now and discuss with
you some basic concepts of engineering analysis. There's a lot of work ahead
in this set of lectures. So let me take off my jacket
with your permission, and let us just go right on with the
actual discussion of the theory of the finite
element method. The basic concepts that I
address here, in this first lecture, is summarized basically
here once more. We are talking about the
idealization of a system. We are talking about the
formulation of the equilibrium equations, then the solution of
the equations, and then, as I mentioned earlier
already, the interpretation of the results. These are really the four
steps that have to be performed in the analysis of an
engineering system or of a physical system that
we want to analyze. Now when we talk about systems,
we are really talking about discrete and continuous
systems where, however, in reality, we recognize that all
systems are really continuous. However, if the system consists
of a set of springs, dashpots, beam elements, then
we might refer to this continuous system as a discrete
system because we can see already, it is obvious, so
to say, how to idealize a system into a set of elements,
discrete elements. In that case, the response is
described by variables at a finite number of points. And this means that we have to
set up a set of algebraic questions to solve
that system. So here I'm talking about
elementary systems ofl springs, dashpots, discrete
beam elements, and so on. In the analysis of a continuous
system the response is described really by variables
at an infinite number of points. And, in this case, we really
come up with a differential equation, obviously a set of
differential equations, that we have to solve. The analysis of a complex
continuous system requires a dissolution of the differential
equations using numerical procedures. And this solution via numerical
procedures-- and, of course, in this set of
lectures we will be talking about the finite element method
numerical procedures-- really reduces a continuous
system to a discrete form. The powerful mechanism that we
talk about here is the finite element method implemented
on a digital computer. The problem types that I will
be talking about are steady-state problems, or static
analysis, propagation problems, dynamic analysis,
and eigenvalue problems. And these three types of
problems, of course, arise for discrete and continuous
systems. Now let us talk first about
the analysis of discrete systems in this first lecture. Because many of the
characteristics that we are using in the analysis of
discrete systems, discrete meaning, springs, dashpots, et
cetera, we can directly see the discrete elements
of the system. The steps involved in the
analysis of such discrete systems are very similar to the
analysis of complex finite element systems. The steps involved
are the system idealization into elements. And that idealization is
somewhat obvious because we have the discrete elements
already. The evaluation of the element
equilibrium requirements, the element assemblage, and the
solution of the response. Notice when we later on talk
about the analysis of continuous systems instead of
discrete systems, then the system idealization into finite
elements here is not an obvious step and needs
much attention. But these three steps here are
the same in the finite element analysis of a continuous
system. And I would like to now discuss
all of these steps here just to show you,
basically, some of the basic concepts that we're using in the
finite element analysis. Let us look at this discrete
system here as an example. And let us display the basic
ideas in the analysis of this discrete system. Here we have a set of rigid
carts, three rigid carts, vertical carts that
are supported on rollers down here. This means that each of these
carts can just roll horizontally. The carts are connected via
springs, k2, k3, k4, k5. And the first cart here is
connected via k1 to a rigid support that does not move. The displacement all of
this cart here is u1. The load applied is R1. Notice that u1 is the
displacement of each of these springs since this
cart is rigid. The displacement of this
cart here is u2. And R2 is the load applied. The displacement of
this cart is u3. And R3 is the load applied. We now want to analyze this
system when R1, R2, looking at it, we can directly see the
elements of the system k1 to k5, and we can see directly, of
course, how these elements are interconnected. The steps that we will be
talking about in the analysis of this discrete system are
really very similar to the steps that we're using in the
finite element analysis of continuous systems. What we will be doing is that
we look at the equilibrium requirements for each spring
as a first step. Then we look at the
interconnection requirements between these springs that, in
other words, the force on these springs here at this cart,
and that spring, must be balanced by R1. And then, of course, we have a
compatibility requirement that u1 is a displacement of each
of these springs here. So we are talking about the
constitutive relations, the equilibrium requirements,
and the compatibility requirements. These are, of course, the three
requirements that we also have to satisfy in the
analysis of a continuous system using, later on, finite
element methods. Notice that these springs here
are our finite elements, if you want to think of
it that way, a very simple set of elements. In a more complex analysis,
these springs here would be plane stress elements, plane
strain elements, three dimension elements,
shell elements. And we will be talking about
how we derive the characteristics of
these elements. And we will, however,
interconnect these elements, these more complex elements,
later in exactly the same way as we connect these
simple elements. So the connections between the
elements are established in the same way, and the solution
of the equilibrium equations is also performed
in the same way. But in this simple analysis,
we are given directly the spring stiffnesses. And one other important point is
that the spring stiffnesses here are exact stiffnesses. In a finite element analysis
of a continuous system, we have a choice on what kind of
interpolations we can use for an element. We have a choice on what
assumptions we want to lay down for an element. And then using different
assumptions we are coming up with different stiffnesses of
the element domain that we will be talking about. And we will also find that the
equilibrium in that element domain is not satisfied. It will only be satisfied in
the limit as the elements become smaller, and smaller,
and smaller. Whereas in the analysis of
this discrete system, the equilibrium in each spring
is always satisfied. So this is a very simple finite
element analysis if you want to think of it that way. The elements here than are k1. And notice that the equilibrium
requirement for this element says simply that
k1u1 is equal to the force applied to this node. It's a force, the external
force, applied to this node. The equilibrium requirement of
this element, k2, is written down here in matrix form. k2 is the physical stiffness
of the spring. And F1, F2 are the forces
applied at these two ends. Notice, please, that the
superscript here refers to the element number, superscript
1 here for element 1, superscript 2 here
for element 2. And notice that we would find
that F1(2) is minus F2(2) given u1 and u2. Of course, that means the
element is in equilibrium. Notice also, if you look at this
matrix closer, that if u1 is greater than u2, then we
would find that, in other words, u1 greater than u2 means
that the spring is in compression. We would find that F1(2) is
positive by simply multiplying this out here. And F2(2) is negative, which
corresponds to the physical situation that we
actually have. If u1 is greater than u2, this
force here is positive, and that force is negative because
the spring is compressed. Well, similarly, we can write
down the equilibrium requirement for the spring 3. And I've written down
the matrix here. The only difference to the
equilibrium requirements for spring 2 are that we're
using now k3 here. And, of course, the superscript
now is 3. We can then proceed to write
down the equilibrium equation for spring 4, which is the same
form as before, now k4 here and the 4 superscript
denoting element 4. And, finally for k5, we have
k5 here and superscripts 5 here to denote element 5. Now we should also point out
one other important point. Namely, if we look at this cart
systems here, notice that this k1 spring is
connected to u1. It's connected to u1. k4 is connected to u1 and u3. So if we look at the equilibrium
requirements here, you will notice that I
have F1 here for k1 because this is u1 here. That is the global
displacement u1. And looking now at k4, a more
complicated case which is connected to u1 and u3, I have
for that spring the u1 and u3 denoted here. And we have F1 and F3
here, F1 and F3. So these are the forces that go
directly into the degrees of freedom 1 and 3 respectively,
and similarly for the other springs. Now if we want to assemble the
global equilibrium equations for this structure with the
unknowns u1, u2, u3, the loads R1, R2, and R3 are known, then
we have to use now the equilibrium requirement at these
degrees of freedom u1, u2, and u3, or rather at
the cart 1, 2, and 3. And that equilibrium requirement
then means that the sum of the forces acting
onto the individual springs 1, 2, 3, and 4 at degree of freedom
1 must be equal to R1. Now let us look once
at this first equation back here again. Notice u1 couples into this
spring 1, spring 2, spring 3, and spring 4. And that coupling is seen
right here in spring 1, 2, 3, and 4. And summing all these forces
that are acting individually onto the springs, the sum of
these forces must be equal to the external load. That is the interconnection requirement between the springs. The equilibrium requirements
within the springs are expressed by these individual
matrices here that we looked at already. These are the equilibrium
requirements for the individual springs. Now I'm talking about the
equilibrium requirement at the carts, the interconnection requirements between the springs. Similarly, we can sum the forces
that have to be equal to R2 and sum the forces that
have to be equal to R3. And these three equations then
set up in matrix form by substituting for F1(1), F1(2),
and so on from the equilibrium requirements of the springs,
we directly obtain this set off equations, KU equals
R. Notice that K now is a 3 by 3 matrix. U is a 3 by 1 vector. R is a 3 by 1 vector. I denote matrices and vectors
by bars under the symbols. As you can see here there are
bars under these symbols. Well, if we look at these
equilibrium equations, we notice that our U vector, this
vector U here contains u1, u2, and u3 as the unknowns. Notice this T here, this
superscript T means transpose. The actual vector U actually
looks this way u1, u2, u3. It lists the displacements
vertically downwards. But it is easier to write
it this way by transposing as a vector. So UT, capital T there,
means transpose. Similarly for R we have R1, R2,
and R3 as the components. And the K matrix that we have
obtained by substituting into these equations from the
element equilibrium requirements, the K matrix
is this one here. Now let us look a little closer
at how do we construct this K matrix. Well, we note that the total K
matrix can be constructed by summing all of the individual
element matrices from 1 to 5. And these individual element
matrices are, for two extremes, written down here. K1 is a 3 by 3 matrix now. Not anymore the 1
by 1 or 2 by 2. It's a 3 by 3 matrix with just
k1 in the 1,1 position. All the other elements are 0. K2 is this matrix. So what I have done then is I
have taken the 2 by 2 matrix which appeared in the element
equilibrium requirement and has blown this matrix up filling
zeroes for the third degree of freedom. Similarly we would obtain
K3 and so on. The zeroes always appear in
those rows and columns into which the element does not
couple, in other words, into those degrees of freedom that
the element does not couple. For example, k1, this element 1
here, couples only into the degree of freedom 1. So, therefore, we have the
second and third rows be 0. Element 2 couples only into
degree of freedom 1 and 2. Therefore, the third degree
of freedom contains all zeroes, and so on. This assemblage process is
called the direct stiffness method, an extremely important
concept that is very well implemented in a computer
program. It represents the basis of the
implementation of the finite element method in almost every
code that is currently in use. The direct stiffness method has
also a very nice physical explanation. And this is what I really want
to talk to you about now for the next five minutes. The steady-state analysis, of
course, then is completed. The steady-state analysis of
this system, of course, is completed by solving this system
of equations here, equations a. Once we know U we can go back to
the elements and calculate the forces in the elements
themselves by going to the element equilibrium
requirements. Well let us look then at what
we are doing when we perform this process here by summing, in
other words, the K element, the stiffnesses of the elements
into a global stiffness matrix. And let us look at what we're
doing physically. Because that really, of
course, is the direct stiffness method that
we are using here. And it is, I think, very nice if
you can clearly see what is happening in that method. Well, the basic process
is the following. Here I have drawn the carts
without any strings. Of course, our degrees
of freedom are here, u1, u2, and u3. And the loads are R1, R2, R3. I don't need to put
them in again. This system here corresponds
to a K matrix with zeroes everywhere. Blanks in these positions
here denote zeroes. So this is a system that we're
starting off with in this direct stiffness method, a
system without any elements, a matrix without any
elements also. The process, then,
is the following. We are using this cart system. And we're adding
one spring on. That is the first edition. That is spring k1. Mathematically this means that
we're going through the following process. We are taking our K matrix with
blanks everywhere, and we're adding into it this
one element, k1. Now this is a K matrix, this
stiffness matrix governing-- and this is very important-- governing this system. Once again, this is a K matrix
governing this system. Of course, this is not a stable
system yet because there are no connections between
these carts here. Well, with a second edition
we're adding in the second spring. And that means we are putting
this spring there. That is k2. Well in our matrix formulation
then, what that means in our direct stiffness method is that
we are going from this system over on this K matrix
to that K matrix. We are adding this
second spring stiffness into the K matrix. So this is the stiffness
matrix that governs the equilibrium of this
physical system. Notice that this spring here,
the second spring, couples into u1 and u2. And therefore we have added
these blue elements corresponding to the second
spring into degrees of freedom 1 and 2. Next in the direct stiffness
method we're adding the next spring element, and that is
spring element number 3. Again, it couples
into u1 and u2. And the stiffness matrix that
we are now talking about is the following. We're going from this stiffness
matrix to that stiffness matrix here, adding
the green k3 in there. Now next we go from this system
to add into the system the spring 4, spring 4 now. Please notice that this is
now a stable system. It is a stable system because if
I want to put u3 over here, then I have to do work
on this spring. So this is now a
stable system. In our mathematical formulation,
or in our direct stiffness method rather, what
this then corresponds to is that we are going from this
system here, or this stiffness matrix, to that stiffness
matrix. Notice we have added
a k4 into here. And that, in fact, allows us
now to solve at this level. We could solve the equations KU
equals R. Of course, this now is a stiffness matrix
corresponding to this system. We have not quite yet
reached the system that we want to analyze. But we reach it by adding
the final spring in k5. k5 now is here. And that corresponds in our
direct stiffness method to adding this spring in there. These elements here, k5. Notice that this spring now here
couples into degrees of freedom 4 and five 5, and
that's why it appears in quadrant column 4 and 5 here. And this spring. I should've have said, this
spring here couples into column 2 and 3, column 2 and
3 meaning u2 and u3. And here we see, of course, that
the spring indeed goes into degree of freedom u2
and u3, into degree of freedom u2 and u3. So this then is to final
system that we want to analyze, and this is the
final stiffness matrix that we had to obtain. Notice, once again, this matrix
has been obtained by taking the sum over all the
element stiffness matrices. We are summing from
i equals 1 to 5. And this mathematical process,
once again, which we call the direct stiffness method
has a physical analog. You can understand it physically
in the way I've shown here. Namely you're starting off
with a blank K matrix, no elements in it at all, and you
simply add one element after the other into that K matrix
filling up the K matrix that way. And the additions are
carried out-- this is important-- by taking the element matrices
and adding them into the appropriate columns and
rows of the K matrix. For example, this element here
couples into degree of freedom 1 and 3. And if we go once more back to
the process that we have been carrying out here, notice our k4
here corresponds to degree of freedom 1 and degree of
freedom 3, the first row and column and third
row and column. That's where these
elements appear. So there's a neat physical
explanation for the direct stiffness method which
I wanted to discuss with you here. Now as another approach, instead
of using the direct formulation of the equations KU
equals R, the equilibrium equations of the system,
we can also use a variational approach. We will be talking about that
variational approach in the second lecture. And I would like to discuss it,
or introduce it to you, now very briefly for the
analysis of this discrete system that we just looked at. The basic process here is that
we are constructing a functional pi which is equal
to u minus w where u is the strain energy of the system and
w is the total potential of the loads. The equilibrium equations that
we just looked at, KU equals R in other words, are obtained by
invoking that del-pi shall be 0, the stationality
condition on pi. And this means that del-pi,
del-ui, shall be 0 for all ui. This then gives three equations,
And these three questions are obtained
as follows. If we use u, the strain energy
of the system is given right here, 1/2 U transpose KU. If you were to multiply this out
substituting for U and for K with the values that I've
given to you, you would find that this indeed is the strain
energy in the system. The potential of the total loads
is given by U transposed R. Notice please that there
is no 1/2 here in front. Simply U transpose R is the
potential of the loads. Now if we invoke this condition
that del-pi, del-ui shall be 0, we directly
obtain KU equals R. Now there's one important
point. To obtain u and w, u and w here,
we again, can add up the contributions from all the
elements using the direct stiffness method. In other words, this K here can
be constructed as we have shown by summing over the
elements, by summing the contributions over all
of the elements. And since this is true, we can
also write this total u as being the sum of the ui's, if
you want to, the strain energies of all of the
individual elements. So here too we could use the
direct stiffness method. Of course, in actuality, in
actual practical analysis, we never form this u, we never form
that w when we want to calculate KU equals R. This is
simply a theoretical concept that I wanted to introduce to
you, a theoretical concept that we will be using later on
in the construction of KU equals R. We never really
calculate these measures if we only want to calculate
KU equals R. It might be of interest to us to
calculate this in order to find out how much strain energy
is put into individual elements in finite
element analysis. But this is only done if you
want to evaluate error bounds on the finite element
solution and so on. If we only want the calculate KU
equals R and obtain the use in other words, to be able to
predict the displacements and the stresses in the elements,
then we would not calculate these two quantities. Now this then were the essence
of the analysis of a steady-state problem for
discrete systems. I pointed out already that if
we have an extra finite element system there, of
course, many additional concepts that we have to talk
about, a selection of elements, the kinds of
interpolations to be used, and, of course, we then have to
also talk about how do we solve these equations,
and so on. We will address these questions in the later lectures. However, another class of
problems that we will be talking about are propagation
problems. The main characteristics of
propagation problems are that the response changes
with time. Therefore, we need to include
the d'Alembert forces. Now basically what we are saying
then is that we're looking at static equilibrium as
a function of time but also taking into account the
d'Alembert forces. And that, together then, makes
it a dynamic problem. Of course, if the displacement
varies very slow, in other words, the load varies very
slow, then the inertia forces can be neglected, and we would
simply have this set of equations where R of t is a
function of time and U of t would be a function of time. However, when R of t acts
rapidly or suddenly inertia conditions are applied to the
system, then the inertia forces can be very important. We have to include
their effect. And then we have a true
propagation problem, a truly dynamic problem that
has to be solved. For our example, the M matrix
here would be this 3 by 3 matrix where m1 is simply the
mass of the cart 1, m2 is the mass of the cart 2, m3 is
the mass of the cart 3. Of course these masses would
have to be given. And notice that we would
evaluate them by basically saying that this total mass
here can be evaluated by taking the mass per unit volume
times the volume. And that would be the mass that
we're talking about when we accelerate that cart
into this direction. So these masses here are
very simply evaluated. When we talk later on about
actually finite elements, we will be talking about similar
mass matrices where we simply take the total volume of an
element and lump that volume of the element to its nodes. We will also talk about
consistent mass matrices where this mass matrix is a little
bit more complicated. In other words, some of
these off-diagonal elements are not 0. Finally, we will also talk about
eigenvalue problems. In the solution of eigenvalue
problems, we will be talking about generalized eigenvalue
problems, in particular, which are Av equals lambda Bv, which
can be written down in this form where A and B are symmetric
matrices of order n, v is a vector of order n,
and lambda is a scalar. As an example, for example here
in dynamic analysis, what we will see there is K phi
equals omega squared M phi where K is the stiffness
matrix that I talked about already. This which would be for the cart
system here simply as 3 by 3, this 3 by 3 stiffness
matrix that I introduced to you. And it's a mass matrix
that we just had here on this viewgraph. That is the mass matrix. And phi is the vector. If we find a solution, in other
words, if this equation is satisfied, we put an i on
there and satisfy for phi i and omega i squared. Omega i squared will
be a frequency. I will be discussing it just
now a little more. And then we're talking
about an eigenpair. But notice that is a typical
problem that we will be discussing which arises,
in other words, in dynamic analysis. Notice also that what we're
really saying here is that the right-hand side is
a load vector. And if we know v, if we
know lambda, then we know the load vector. What we would calculate then
is the same v that we have substituted here. In other words, if we consider
this to be a set of loads where v is now known, lambda
is known, then we could evaluate R. In solving Av equals
R, we would get back our v that we substituted
into here. And that is the main
characteristic of an eigenvalue problem. Well, they arise in dynamic and
buckling analysis, and let us look at one example where
we actually obtain this eigenvalue problem. And the example is simply the
system of rigid carts that we considered already earlier. We obtain the eigenvalue problem
by looking at the equilibrium equations when
no loads are applied. And we call these the free
vibration conditions, free because there are no loads
applied, free of loads. If we let U be equal to phi
times sine omega t minus tau where the time dependency now
in the response is in this function here, in the sine
function only, and if we take the second derivative of U,
meaning that we get a cosine and then a minus sign in here,
and, of course, this omega twice outside, so we have
a sign change here. We have a minus omega squared
M phi sine omega t minus tau for this part here. And for this part KU we obtain
K phi sine omega t minus tau by simply substituting
from here into there. And, of course, the sum of these
two must be equal to 0. Now this equation must
hold for any time, t. So we can simply cancel out
this part and that part. And the resulting set of
equations that we are obtaining then are given on
the last viewgraph, namely those equations being K phi
equals omega squared M phi. So that is the generalized
eigenvalue problem which we obtain in dynamic analysis. We will be later on talking
about how we solve this generalized eigenvalue problem
for the eigenvalues and eigenvectors. In the case of of the 3 by 3
system that we are considering here, in other words, the
analysis of the cart system, we only have three solutions,
omega 1 phi 1, omega2 phi2, omega3 phi3. And we call each of the
solutions an eigenpair. So there are three eigenpairs
that satisfy this particular equation. Notice that this is, in other
words, the equation that I talked about here earlier. And the eigenpairs, phi i,
omega i squared are the solutions to this equation. We are really interested in
omega i because that is the frequency in radians per second,
and the eigenvalue, however, being omega squared. In general when we have
an n by n system-- and I have already written down
here the n by n, let me put it bigger once more here-- and we have a general n by n
system, in other words, and not being equal to 3 just as
we have in our cart system, then we have n solutions. And, however, we will find
that in finite element analysis we do not necessarily
need to calculate all n solutions. In fact, when we consider large
eigensystems where n is equal to 1,000 or even more,
then certainly we do not want to calculate all eigenvalues. It would be exorbitantly
expensive, much too expensive to calculate all of the
eigenvalues and eigenvectors. We don't need to have them
all in analysis. And, therefore, we will talk
about eigenvalue solution methods that only calculate
the eigenvalues and eigenvectors that we are
actually interested in. We also, of course, have to,
before we actually get to that topic which is the topic of the
last lecture, we will talk about how we actually construct
these K matrices, how we calculate them, construct
them for different finite element systems. Well, this then does complete
what I wanted to say in this lecture. Thank you very much for
your attention.
