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PROFESSOR: Ladies and gentlemen, welcome to
lecture number 11.
In this lecture, I would like to continue discussing with
you the solution of the dynamic equilibrium equations.
We discussed the solution of dynamic equilibrium equations
already in lecture number 10.
In that lecture, we consider direct integration methods.
In this lecture, I would like to consider with you, discuss
with you the mode superposition analysis.
The basic idea in the mode superposition analysis is the
transformation from the U displacements into a set of
new displacements, X.
P is an n by n matrix, which is nonsingular.
If P is nonsingular, we should recognize that the vector U of
lengths n is uniquely given when we have the vector X,
also of lengths n and vice versa.
In other words, if we have the vector X, we can calculate U.
Or if we have U, we can calculate X. P, of course, has
to be a nonsingular matrix.
We call this a transformation matrix.
And we call the X displacement, that I will be
talking about, the generalized displacement.
Now, if we use this transformation, we find that
if we substitute this transformation into this
equation for U, and we postulate that P shall not be
a function of time.
So that when we find the derivative of U,
with respect to time--
let me denote it by a dot here--
we simply obtain that U dot is equal to P times X dot.
And similarly, U double dot is equal to P times X double dot.
If we then substitute for U, U dot, and U double dot--
in other words, for the displacements, velocities, and
accelerations--
into this equation, we obtain this equation, provided--
and now, there's one additional step--
after the substitution for the displacements, velocities, and
accelerations, we also premultiply the total equation
by a P transposed.
In other words, what we do is we take a P transposed here,
and we are substituting for these vectors here from that
relation here.
The P transposed, of course, gives us, also, a P transposed
R on the right-hand side.
So R curl here is defined as shown down here.
Notice that M curl, C curl, and K curl are
given as shown here.
And they involve, of course, the M, C, and K matrices.
And this P here comes via this
substitution into these vectors.
This P transposed comes from that P transposed here.
Now, what have we been doing so far?
Basically, what we have been doing is we
have changed bases.
We have come from the displacement defined by the
finite elements--
by the finite element nodal points--
to generalized displacements.
The generalized displacements now govern the equilibrium of
the system.
Now, this transformation is effective if we find that our
M curl, C curl, and K curl have a better structure than
the matrices, M, C, and K. And, of course, if it is
simpler to find the P matrix or a P matrix.
Let's look at a very simple case.
If, for example, M curl, C curl, and K curl would all be
diagonal matrices, then the solution of this set of
equations would be very simple.
In fact, this is our objective.
It is our objective to come up with a P matrix, which when
used in this transformation, gives us an M curl, C curl,
and K curl matrix that are all diagonal.
And, of course, in addition we want to find the P matrix with
a minimum amount of effort.
Once we have diagonal matrices here on the left-hand side--
in other words, M curl, C curl, and K curl--
then, we can use our direct time integration schemes, the
Newmark method, or the essential difference method.
Usually we use the Newmark method.
Or we might even be able to solve the decoupled equations.
They're decoupled because M curl, CK curl, are diagonal.
We might even solve these equations in closed form.
The question then is how do we find P or [UNINTELLIGIBLE]
P?
We want to find it cheaply.
It shall be economical to calculate P. And we want to,
of course, have that M curl, C curl, K curl are ideally
diagonal matrices.
Well, the P matrix can be constructed--
or a P matrix can be constructed, effectively, by
looking at the free vibration equilibrium
equations of the system.
If we look at these equilibrium equations here, in
which we neglect damping and we neglect the right-hand side
forces-- we're looking at the equilibrium equations of the
dynamic system without damping and without forces applied.
And if we look for a solution to this system, then, we can
assume a solution of this form.
U, of course, being our displacement vector.
Phi being all the vector that is independent now of time.
And here is the time dependency, sin, omega, omega
is a radiant per second circular frequency.
t, of course, being the time.
And t 0 is a time shift.
If we make this assumption, substitute into here--
when we do substitute, of course, remember we have to
take the second derivative of this function here with
respect to time because we're taking the second
derivative of U here.
And that gives us a minus omega squared sin, omega t
minus t 0, substituting from here into there.
We immediately get this equation because we can cancel
out sin omega t minus t 0 on both sides.
This is the resulting equation.
So what we now want is to solve for
phi and omega squared.
This is the generalized eigenvalue problem.
The generalized eigenvalue problem because M is a matrix
other than the identity matrix in general.
And we know that they are n solutions to this problem if K
and M are n by n matrices.
There are n solutions to this problem.
We call these solutions the eigensolutions.
And we call omega 1 squared, I should say, and phi 1 the
first eigenpair.
This is the second eigenpair, this is the nth eigenpair.
There are n such eigenpairs.
Each of these eigenpairs satisfies this equation here.
Therefore, these are a solution--
we call it an eigensolution--
to this equation.
And in addition, we also can prove-- we can show--
that the phi I vectors-- these vectors here--
are M orthogonal.
M orthogonal because phi I transposed M phi j--
for example, phi 1 transposed M phi 2 is equal to 0.
And phi 1 transposed M phi j--
sorry phi 1 transposed M phi 1 is equal to 1.
And because it's equal to 1, they're not only M orthogonal,
they're indeed even M ortho normal.
What we find is that if we look at this eigenproblem, and
we notice that a certain vector satisfies this
eigenproblem, then also a multiple of that vector
satisfies the eigenproblem.
And we determine this constant, this undetermined
constant, in such a way as to have phi I transposed M phi I
equal to 1.
So these are the eigenpairs and the eigenvectors, phi 1 to
phi n, satisfy this relation.
The eigenvalues, omega 1 squared to omega n squared can
be ordered in this way where we find that we might have
multiple eigenvalues.
Omega 1 might be equal to omega 2.
But if we do, we still have only a total number of n such
eigenvalues.
I will discuss in the next lecture with you the solution
techniques that we're using to calculate these eigenpairs.
Let's for the moment assume that we have calculated them.
And we want to use them.
Well, if we now define the following
matrix, phi capital phi.
It's a capital phi because we see crossbars
top and bottom here.
If we define this matrix as phi 1 to phi n--
in other words, we're taking the first eigenvector, and
list it as a first column of this matrix here.
The second eigenvector, we list as the second column in
this matrix and so on.
So that we find, of course, that this matrix here is an n
by n matrix.
And if we define a diagonal matrix, omega squared.
This is a capital omega.
Omega squared, being equal to 0s in the
off diagonal elements.
But the diagonal elements being omega 1
squared to omega n squared.
Then, we can write the solutions to the eigenproblem.
The solutions to the eigenproblem--
and I'm still talking about this
eigenproblem problem here.
We can write the solutions to this e eigenproblem in the
following form.
Notice that this omega squared here has to be on the back of
the capital phi.
In other words, this is the solution to the problem, K phi
equals omega squared M phi, where the omega squared is a
scalar and appears here in front.
We usually write it in front.
Of course, we can put it also at the back
because it's a scalar.
We can put in front or the back.
However, when we have listed the solutions to this problem
in this way, then, we have to put the capital omega squared
at the back of M times phi.
You can verify that by simply multiplying out
the right-hand side.
We also have this relationship here, which is the M ortho
normality, the phi ortho normality with M. In other
words, this here says nothing else and phi I transposed M
phi j is equal to delta I j.
We call this the chronica delta.
And this chronica delta is equal to 1 for I
being equal to j.
And it's 0 otherwise.
So we are having this relationship here.
And if we use this relationship, and we are
premultiplying this equation by phi transposed to capital
phi transposed now, both sides.
We are finding that we are getting here
the identity matrix.
And therefore, we're having that this left-hand side is
equal to omega squared.
This matrix here.
And this is an important fact.
In other words, we find that the vectors--
the eigenvectors--
satisfy these two relationships.
Now, if we look back at what we started off with in our
multiple position analysis discussion, we wanted a matrix
that we call P, which is such that P transposed M, P ideally
is a diagonal matrix.
Well, we have no such a matrix.
If we take phi being equal to P. Then, phi transposed M phi
is in fact a diagonal matrix.
In fact, a very nice diagonal matrix, the identity matrix,
which is a matrix that is diagonal.
It has just once on the diagonal.
Similarly, we wondered here that P transposed KP is a
diagonal matrix.
Well, we have achieved that by this phi matrix.
So the phi matrix is, indeed, a particular P
matrix that we can use.
So what we are then saying is let us use a multiple position
analysis this particular P matrix.
The P matrix, which is equal to the phi matrix and phi
storing the eigenvectors of the generalized eigenproblem.
Well, substituting from here, as before, into the governing
dynamic equilibrium equations, we
directly obtain this equation.
However, we notice now that this matrix here, in general,
is not yet a diagonal matrix.
We call that our C curl.
And our C curl has not been diagonalized yet.
We will have to deal with that matrix still a little later.
How do we solve these equations?
Well, if we solve these equations, we have to also
recognize, of course, that we have to include the initial
conditions to the problem.
And the initial conditions mean that the initial
displacements are defined and the initial
velocities are defined.
These, of course, are the initial conditions that are
used in the direct integration solution of the governing
dynamic equilibrium equations.
Well, if we are now performing the solution of this set of
equations, we will have to transform these initial
conditions to initial conditions on the generalized
displacements.
And that is achieved via these relations here.
Let us see how we derive these.
Well, if we look at this equation here, and if we
notice that, of course, this equation must be
applicable 4 times 0.
We can simply put a 0 there.
And we put a 0 there.
Then, of course, we also recognize that we can
premultiply this equation with this relation here.
And if we do that on the right-hand side, we notice
that phi transposed M times that phi is
the identity matrix.
And so we have this relation directly obtained.
The left-hand side in this equation is the right-hand
side in this equation.
And the right-hand side here is simply this 0 x now.
The same holds also for the initial velocity.
We can simply put dots on top of the U here and of the X.
And therefore, we have this relation here.
So the solution of this set of equations and with these
initial conditions subject to these initial conditions,
gives us a solution to of the dynamic response.
Now, I should mention here that we still have to deal
with the damping matrix.
The damping matrix here needs particular attention.
We do not have, in general, a diagonal matrix here.
And we will want to construct specific damping matrices with
which we can deal, effectively.
And yet, of course, these matrices must make,
physically, sense.
Let us look now, however, first at the case where we
neglect damping.
When we neglect damping, we have this set of equations and
individual equations of this form.
Notice that these are individual equations because
this is a vector.
This is a diagonal matrix.
And on the right-hand side, we simply have a vector off load,
of course, time dependent.
So we can look at the first row in this matrix relation.
And the first row in that matrix relation will be this
equation here with I equal to 1.
The I-th row in this relation here is the
I-th equation here.
In other words, I subscripts on the [UNINTELLIGIBLE].
And, in general therefore, we want to
solve n such equations.
I going from 1 to n, with this being the load vector and, of
course, with the initial conditions.
These initial conditions are obtained by simply looking at
this equation here and extracting the I-th row from
each of these two relations.
So this is then the problem.
Notice the following that if we solve all n equations with
these initial conditions, then, if we use, for example,
the Newmark direct integration method on these equations, we
would obtain exactly the same solution as if we had solved
the fully coupled equations.
The fully coupled equations-- let me write them down once
more here--
plus KU equals R.
In other words, whether we apply the Newmark time
integration scheme, a direct integration scheme, to this
set of equations with the initial conditions on U and U
dot or apply the Newmark direct time integration scheme
to this set of equations, or rather this one here with
these initial conditions, we obtain exactly the same
response provided we use the same time step, delta t in
both integrations.
There has been no assumption so far.
All we have done is a transformation from the U
displacements to the X displacements.
And this transformation is effective if we can find the
phi matrix that we're using in the transformation very
cheaply, very economically.
Well, the effectiveness, really, of multiple position
goes beyond what I have just described.
The important point in multiple position analysis is
that we do not need to solve all of these equations.
That we do not need to solve and look at all of the
decoupled equations.
But let us look first once also at how we can solve the
decoupled equations now exactly.
Where we can solve them exactly for example using a
Duhamel integral formulation as shown here.
X i of t is given via this relation here.
And we would, of course, substitute our ri of tau where
we now assume that this part here is given, directly.
In a computer program, very frequently r i would only be
given as a segment of straight lines.
And if that is the case, then, of course, we might have to
still evaluate all of this here numerically if the
segment is very complicated, complex, like
an earthquake analysis.
ri might be varying very rapidly.
And then, we might still want to evaluate this, numerically.
However in general, we can evaluate here this
relationship analytically.
And we don't need to use numerical integration.
Having got xi, of course, we get our U of t as shown here
by our transforming back.
Well, the important point that I now would like to get back
to you and discuss with you for a little while is that we
do not need to use or to solve all of these equations in
multiple position analysis.
In fact, multiple position analysis is only effective
when we do not need to look at all of
these decoupled equations.
If we have to solve all decoupled equations, then, we
would have to calculate, course, first all eigenvalues
and eigenvectors, which can be an enormous expense.
And then, the multiple position analysis
would not be effective.
I would like to repeat, however, if we do solve all
decoupled equations, using a specific time integration
scheme with a specific data t time step.
Then, if we use the same time integration scheme on
decoupled equations with the same time step, we must get
exactly the same response.
Then there has been no assumption so far.
Well, however, if we then do neglect some of
the equations here.
We do not solve some of the equations.
Then, of course, our response will be different that we are
predicting via this approach when we compare with the
response predicted by solving these equations.
Well, let us look once at this graph again, which we already
looked at briefly in the last lecture.
And I would like to explain to you now why it is not
necessary to solve all of the decoupled equations in
multiple position analysis.
Why it is not, in general, not necessary?
Well, if we look at a single oscillator
equation, as given here.
I include a damp in here, of course, side
can be equal to 0.
Where P is the driving frequency, omega is the
frequency of the oscillator itself.
And if we plot the maximum dynamic response observed,
which we define if the dynamic load factor when we divide
that maximum dynamic response by the static response, we get
this set of curves.
Let us look at say the curved side equal to 0
a little bit closer.
If we do that, and if we solve this equation
with side equals 0.
This term now not being there.
For certain values of P over omega, we are getting this
curve here and that curve here.
What do we notice?
Well, we notice that when P over omega is equal to 0, we
are getting the static response.
Or P over omega very small, we're
getting the static response.
This means, of course, that omega, the system frequency,
is much larger than the applied frequency, which means
really that this oscillator is extremely stiff.
Well, when compared to the frequency of load application.
So then, we are in this range.
And we notice that say P over omega in this range here,
smaller than a certain value, and this is here 0.5 say,
smaller than 0.2.
We would see that basically, the maximum dynamic response
is really just nothing else than the static response.
So we are talking in this region here really about
static response.
And in this region here, we are talking about a truly
dynamic response.
Now what does that mean?
Well, if we look at our decoupled equations here, it
means that if i consist of frequencies, basically, which
when measured on the system of the oscillator frequency are
very small, we, basically, would have a static solution.
You would, basically, have a static solution for this
equilibrium equation.
A truly dynamic solution will only be obtained when P, in
other words, the frequency of the load divided by omega,
falls into this range here.
Well, this is what we are using in most
superposition analysis.
We only need to consider those frequencies of the system.
And these are the frequencies of the systems, the omegas,
which are truly excited by the dynamic loads.
We only need to look at those decoupled equilibrium
equations for which we have that the frequencies in the
decoupled equations are truly excited by the dynamic loads.
Now, I only talked about the P value here.
Of course, in an actual analysis, we would find that
ri of P contains a number of frequencies, a spectrum of
frequencies.
Well, what we have to do is look at the highest frequency
that is contained in ri.
And we measure that highest frequency on the frequencies
of the system.
If that highest frequency divided by the frequency of
the system is falling into this range here, we are
talking about a static response, basically.
And we do not need to consider those equilibrium equations in
are multiple position analysis.
Well, so what we are then saying is let us all solve
these equations not from i equals 1 to n but from i
equals 1 to P. And then, instead of obtaining the true
response, the "exact" response, we are obtaining a
response U P. i equals 1 to P. Phi i, x i gives us U P, where
U P is an approximation now to the U displacements.
And the U displacements, once again, are the solution to
this system of equations.
How can we find out the error?
Well, if U P is an approximate solution to the system of
equations that we want to solve, then, let us substitute
back into that system of equations.
Let me write it down once more here.
We have MU double dot plus KU equals R
that we want to solve.
Well, we can also write that, of course, MU double dot plus
KU minus R. And this shall be 0.
So if we have an approximate solution to this set of
equations, why not just substitute that approximate
solution in here and see how large the right-hand side is.
If it's exactly 0, well, then we have that our U P is,
indeed, very close to U. And if it's exactly 0 because U P
would be equal to U.
We will not find this to be 0.
We will never find it to be 0 because in a computer, we
would, of course, use finite digit arithmetic to evaluate
the left-hand side.
And we would have at least round off.
But what we want is at the right-hand side calculated,
which we now might call some value shall be close to 0.
And that means really what we want to measure
is this part here.
Now what have I written down here?
I have here epsilon P of t being this vector here.
This is a vector.
And we can call this an out of balance load vector divided by
the applied load vector.
Notice that what I'm doing here, I take the norm of this
vector because I want to get a single number.
I want to get a single number here, which is representative
of all the elements in that vector.
The same applies here.
This norm is calculated by taking each value in this
vector, squaring that value, adding the squares of the
values up, and then, taking a square root out
of that final value.
In other words, if we write it down here, quickly, what we do
is if we have a vector a, and we want to find this norm.
What we do is we take a 1 squared plus a 2 squared, et
cetera, add all the individual entries up
of that vector here--
squares of the entries up--
and then, take the square root out of it.
And that gives us a single number, which is
representative of all the elements in that vector.
Notice that if one element is large, well then, this value
must be large, too, because that one element will make
this value large here.
Well, if epsilon t is close to 0, assuming that r of t is not
0, of course.
Otherwise, you couldn't use this.
You would have to use some equivalent approach.
If epsilon P is close to 0, then, we know that our
equilibrium equation has been solved quite accurately.
In general, we might find that it's not close to 0.
And what we want to do is a static correction.
This static correction takes into account the fact that we
have neglected the higher modes.
And in the higher modes, of course, we know we have
basically static response.
that's why we did not include those equations in the
multiple position analysis.
So what do we do in that static correction?
We recognize that R could be written in this form.
This RI is exactly the value that I defined earlier.
M phi i is a vector.
You see, I can take M and multiply it by phi i, and I
get a vector.
Now they are in such a vectors.
We might not have calculated all n.
But you can say that indeed, in theory there are n.
So we can write this down.
We also then can calculate the delta R, which is that amount
of the load vector, which has not been included in the
multiple position analysis.
How do we obtain it?
Well, we know R is the total load vector.
And if we now sum instead of i equals 1 to n, only from i
equals 1 to P, we are getting this part here.
But this part is exactly what we have included in the
multiple position analysis already.
So we are left with a vector on the left-hand side, which
we have not included in the multiple position analysis.
And we did not include it because we knew at most, we
can have static response, corresponding to these loads.
Let us now calculate that static response.
That's done right down here.
Now, this static response has to be added to the response
which we predicted in the multiple position analysis.
Let us now go to the case with damping included.
So far we neglected that.
Now, with damping included, we have this part here, too.
Notice phi transpose C phi, in general, is
not a diagonal matrix.
And that's why I considered first the
case without damping.
However, if we have proportionate damping, and in
structural analysis, frequently, all that is
necessary is to include proportionate damping.
Which we postulate to be of this form.
2 omega i xi i.
Delta ij is again the chronica delta, which is equal to 1
when i is equal to j and is equal to 0 otherwise.
So if we are postulating this, where omega i,
of course, is given.
That is the free vibration frequency, which is stored in
this matrix here.
xi i is a value, which we are not given yet.
We have not discussed it yet.
But if we postulate that this matrix here, basically, or
each entry in that matrix can be written in this way.
Then, indeed, we would have decoupled equations.
It turns out that in practice, as I mentioned already, this
is all that is necessary to include damping appropriately.
Well, if we know these xi i values, then, we can also
construct a C matrix with the xi i values given.
The xi i values are frequently obtained from experiments.
In other words, the structure is taken, its vibrated, it's
excited into certain vibration modes.
And from these experiments then, we
obtain the xi i value.
So once we have the xi i value given, we can
calculate the C matrix.
And one way of calculating is to use the Caughey series.
Here we have the Caughey series, which was proposed for
calculating the C matrix, which satisfies this criteria.
In other words, damping is proportional.
Well, if the psi i values are given, we can calculate the
omega i, of course are also given.
We can calculate using this equation, the constants a 0 to
ap minus 1.
Once we have these constants, we can substitute them into
here and be coming up with a C matrix.
And this C matrix here, indeed, satisfies the
normality property shown here.
So, in other words, given the psi i values from experimental
results, given the omega i values, we can calculate a C
matrix that, in fact, gives us a diagonal matrix when we
carry out phi i transposed C phi j.
A special case of this C matrix is Rayleigh damping,
used abundantly in practice.
Where we just include the first two terms, alpha and
beta being constants now.
Let us go through a particular example.
Let's say that we know the critical damping psi 1 is 0.02
and psi 2 is 0.10.
In other words, 10% critical damping in the second mode and
2% critical damping in the first mode.
The corresponding frequencies are 1 and 3.
We want to calculate our friend, beta.
Well, what we do is use this C matrix in the relation, phi i
transposed, C times phi i.
And that should give us 2 omega psi i.
Of course, we have two i values, i equals 1
and i equals 2.
So we can apply this equation twice.
Well, when we calculate it out, we get this.
Applying it twice, we directly obtain these
two equations here.
And we solve these two equations for alpha and beta.
Having alpha and beta, we can substitute now back into this
equation here and have our C matrix.
This would be, for example, a C matrix that we could use in
a direct integration analysis.
You see, if we are performing a multiple position analysis,
we might know, from experience, from experimental
results, that psi 1 and psi 2 are of these two values.
And then, if we know that we have so much critical damping
or so much damping in the first and second vibration
mode, and we want to perform a step-by-step direct
integration analysis, we would have to construct a C matrix.
And this is the procedure to construct a C matrix.
Then, of course, we would have to ask ourselves well, knowing
these psi values for omega 1 and omega 2, what are now the
psi values that we are implicitly using for the
higher modes when we employ this specific C matrix?
And we use now, this equation here.
This is the equation, which we already derived on this view
graph here by looking at this relation.
You are now using this equation because this equation
must be applicable not just for psi 1 and psi 2 but for
all psi values.
Well, then using this equation, we can solve for the
psi values.
And, of course, psi 1 and psi 2, substituting alpha and beta
in here and omega 1 and omega 2 would, indeed, be just 2%
and 10%, 0.02 and 0.10.
However, we can now use this equation to
get the high xi values.
And there's something to be observed, which is quite
important, namely we notice that the beta value that we
calculated is associated with the high
frequencies, basically.
And the alpha value goes away.
Or this contribution goes away when we're looking at the high
frequencies.
For high frequencies therefore, the beta value is a
dominant factor.
And for low frequencies, the alpha value is
the dominant factor.
In other words, once again, looking at this equation here,
in the high frequency response, beta K, really, will
determine the damping.
And in the low frequency, the response is really the alpha
value that determines the damping.
Well, the response solution now would be performed in
exactly the same way as before.
As in the case of no damping, we
solve now again P equations.
P equations only.
And we are doing that by looking at this equation and
with, of course, ri being given as shown here and the
initial conditions being given here.
Solving this equation here via numerical integration or exact
integration, using, for something, Duhamel integral,
we obtain the xi values as a function of time.
We substitute into here to get our U P, which is an
approximation to the solution of the
dynamic equilibrium equations.
The dynamic equilibrium equations now being MU double
dot plus CU dot plus KU, being equal to r.
Notice we have a C matrix in here now.
However, and this is an important point, we have a
specific C matrix constructed such that the phi i transposed
C phi j gives us a diagonal matrix with the diagonal
elements being equal to 2 omega i, psi i.
And one of these matrices, for example, would be
the Rayleigh matrix.
The matrix obtained using Rayleigh damping, rather.
The important point is, once again, that all we have been
doing so far is to make a change of bases from the
finite element coordinates to the generalized
displacements, xi.
That was the first step.
The second step was that we are neglecting the modes P
plus 1 to n by only solving the first P equations here and
using this as an approximation to the actual solution.
The important point, of course, is now that we have to
ask ourselves when is multiple position
analysis really effective?
Some considerations I have here summarized on the last
view graph.
The multiple position analysis is effective when the response
lies in only a few modes.
What does this mean?
It means that if we look at our eigenvalue problem, K phi
equals omega squared M phi.
We only need to solve this eigenvalue problem
for a very few modes.
In other words, P being much smaller than M, I will only
have to solve for phi 1 up to phi P and omega 1 squared up
to omega P squared.
This is a very important point.
I assumed here that we are taking the lowest frequencies,
omega 1 to omega P. Of course, in general analysis, it might
just be necessary to calculate certain frequencies, say omega
10 to omega 20.
We only want to find the frequencies in a certain range
because we know that the excitation frequency only lies
in a particular range.
That would be, for example, the case in vibration
excitation analysis.
This then is a requirement for multiple position analysis,
really, to be effective.
When the response is obtained over many time intervals, then
multiple position analysis also becomes effective.
We should remember here that if we perform a direct
integration analysis, and if that is an implicit direct
integration analysis, using the
Newmark method, for example.
Then, the number of operations that we talked about in the
Newmark method was 1/2 nm squared.
That is the initial triangular factorization
that we have to perform.
m being the half bandwidth.
Plus 2 nm.
n is the order, of course, of the system.
m is the half bandwidth, once again.
Times t, the number of time steps.
This is an initial expense that we have to perform, using
the direct integration Newmark method, for example.
The same holds for the Wilson Theta method.
or the Houbolt method.
This is an initial expense.
And then, we perform this number of operations
for each time step.
Now, if we have to run 5,000 time steps, then direct
integration would be
extremely, extremely expensive.
And, in fact, you might not be able to afford it, using
implicit direct integration.
Well, therefore, you can see directed by looking at this
operation column.
What should make the analysis more effective is to bring
down the bandwidths.
And, of course, this is exactly what we're doing in
multiple position analysis.
We are going from a bandwidth of m to a bandwidth of 0.
You are having a diagonal matrix.
Remember, the bandwidths we defined to be the number of
off diagonal elements here that we are having.
We did not include the diagonal.
That is a half bandwidth.
If you look at the total band here--
that total band here was equal to 2m plus 1.
So if we are reducing the bandwidths, as we do in
multiple position analysis, then we can see immediately
that this number of operations here, which is directly
proportional to the number of times steps goes down.
And that is exactly the objective in multiple position
analysis that I talked about earlier.
That is the first important objective.
We want to reduce the bandwidths for the direct
integration analysis.
The second objective, of course, is that we do not need
to even consider all of the equations.
So looking at n.
We are reducing n, also.
We are not looking anymore at n equations.
We are looking at P equations.
And we are reducing the bandwidths as I said earlier.
m goes to 0.
And n goes to P. Of course, this is an
approximate count here.
So basically, you can see that in the solution using multiple
position analysis, the number of operations in the time
integration of the decoupled equations is
relatively small, is little.
And, in fact, this solution here can be obtained very,
very effectively.
The time integration of the decoupled equations is very
effective in an actual practical analysis.
Where the expense lies in multiple position is really in
the calculation of the required
frequencies and motions.
And, of course, this is exactly the subject of the
next lecture where I want to discuss with you the effective
solution techniques for calculating these vectors and
frequencies.
Multiple position analysis is effective, altogether only,
when we can calculate these eigenvalues and eigenvectors
in an effective way.
I should also finally, once again, emphasize that it may
be important to calculate the error, epsilon pt that I
referred to.
Or to add a static correction to the predicted response in
the multiple position analysis when P is, of course, much
smaller than n.
We have neglected the high frequency response here.
That high frequency response is a static
response, in general.
And should be taken into account via say static
correction, as I have been discussing with you.
This is all I wanted to say in this lecture.
Thank you very much for your attention.