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In the previous lecture we had talked about
the objectives of the finite element method
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where it is used in an engineering analysis
and what are the basics steps involved in
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the typical finite element analysis. In this
lecture, we are going to develop our understanding
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of the method a little further by looking
at a typical model problem.
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For that model problem we are going to develop
virtual work formulation. After the virtual
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work formulation has been developed we will
also discuss another concept called concept
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of a functional. Using the functional we are
going to develop something called the variational
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formulation and we will show that for at least
the model problem of interest both the virtual
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work formulation and variational formulations
are the same. Finally, we are going to take
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the variation formulation and with respect
to this formulation, we are going to develop
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a method called the Ritz method. We will develop
the variational formulation which we are going
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to apply to the model problem that we will
consider and we will use a method called the
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Ritz method to obtain a solution to this model
boundary value problem.
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The Ritz method is going to be used here because
it is a precursor of the finite element method.
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Let us take the model problem that we had
discussed in the previous lecture.
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This is the extension of the axial bar. If
we look, the problem has been complicated
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a little bit where in we have added a variable
cross section to the bar. The bar has a variable
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cross section; it is loaded by the distributed
force f(x) and an end load P at the point
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x is equal to L. At the point x is equal to
0 the bar is fixed; that is the displacement
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at this point is set to 0.
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If I now write the model problem in a detailed
form, the boundary value problem can be given
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as equation 1(a); where we have the differential
equation given by -- d dx of EA which is now
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a function of x du dx is equal to f of x for
all points that lie in the interval 0 to L;
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u is the displacement of the bar. With the
boundary conditions as I have already mentioned,
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u at the point x is equal to 0 is equal to
0 and the force at the point x is equal to
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L i EA du dx evaluated at the point x is equal
to L is equal to the applied force at the
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end L equal to P.
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For the model problem given earlier let us
now take the differential equation and move
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both terms to one side. That is, we will get
-- d dx (EA) which is the function of x du
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dx minus f(x) is equal to 0. This term that
we have put on the left hand side is called
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the Residual. What
we are going to do is, we are going to take
this residual and multiply it by a weight
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function w(x) that is any function w(x) which
is admissible. We will define admissibility
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later on. What do we call this residual? Let
me call it by something r(x). Residual is
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a function of x given by r(x). I take r(x)
multiply it with a function w(x) and then
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I integrate it over the interval. Obviously,
this integral is going to be equal to the
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integral of the right hand side which is equal
to 0. When I take this the integral of the
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residual multiplied by w(x) over the whole
domain that is x is equal to 0 to L, this
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is called the Weighted Residual Formulation.
The question is, what do we mean by this weight
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w(x)? This w(x) is given a name that is the
weight function. This w(x) as we will see
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later on has to satisfy certain minimum smoothness
conditions in the domain and certain other
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conditions on the boundary of the domain.
Many people use this weighted residual formulation
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that we have written here to solve the problem.
What we are going to do next is that we will
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take weighted residual and let us expand and
write it again.
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In the expanded form this becomes.... wdx
minus integral of x is equal to 0 to L fw
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dx is equal to 0. We see that in this term
second derivative of u is sitting. What we
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would like to do is pass one of the derivatives
from u to w. That is, we are going to do integration
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by parts for this term. I integrate this term
by parts
to get EA du dx into dw dx and I will take
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the required terms on the right hand side.
This will be equal to x is equal to 0 to L
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fw dx plus if I look at it, EA du dx into
w evaluated at point x is equal to L minus
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the value evaluated at point x is equal to
0. What we have done, we have integrated this
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term by parts which gave us volume integral
part that is the interval plus a part which
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is the boundary term.
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Next, what we are going to do is we will look
at the boundary term which is EA du dx w evaluated
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at the point x equals to L minus EA du dx
w evaluated at the point x equal to 0. If
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we look at this, this is the boundary part
or the boundary term at the point x is equal
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to L which is the right extreme of the member
that we have taken and this one is EA du dx
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w evaluated at the point x is equal to 0 which
is the left extreme of the member that we
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have taken. In a model problem that we have
taken, at this end EA du dx is the force given
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by the value P. While at x is equal to 0 EA
du dx is an unknown term but at the end x
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is equal to 0, u is known. What we are going
to do is at the end x is equal to L where
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the force is given we are going let w be free
that is we do not put any constraint on w
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at this end. While at the point x is equal
to 0 we are going to enforce the constraint
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that u is given equal to a value 0 by making
w is equal to 0. If we make w is equal to
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0, this term vanishes. What we are left with
is the following formulation, integral x is
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equal to 0 to L EA du dx dw dx dx is equal
to integral x is equal to 0 to L fw dx plus
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P w evaluated at point x equal to L. This
is the formulation that we get. Once we have
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obtained this next what? If we stick to this
formulation this is in a way enough to work
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with, but what we will like to do is set in
our mechanics frame work. That is we are going
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to choose w to be not any function of interest,
but a function called delta u of x.
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Delta u of x is said to be the variation of
u. Let us imagine that I am giving you two
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points x is equal to 0 and L. In between this
I have this function u of x and on top of
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this if I take a function which is close to
this, cannot be very close it can be anything
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which is close to this function. We have to
define what we mean by close. This function
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we are going to call as u(x) plus delta u(x),
such that difference between these two functions
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is the function delta u(x). Whenever we are
talking of these variations of u, it is as
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if I am taking the function in the neighborhood
of the function u and taking the difference
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of those that we can call as the variation
of the function u(x). If I take w(x) equal
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to variation of the function u(x) and put
it back in our formulation that we have obtained
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by the integration of the parts, we will get
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f delta u dx plus P delta u evaluated at the
point x equal to L.
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This is something that we are familiar with.
If you look at this part this is also equal
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to the stress in the bar sigmax due to the
displacement of the bar. This part we can
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say is the strain due to delta u and the right
hand side if we look at it is work done by
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the delta u against the distributed body force
f and the end force P. What do we have here?
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We have something called the internal virtual
work and this is virtual work done by the
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external forces. Why do I call it virtual
work? It is because our equilibrium has been
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achieved under the action of the forces f
and P and we have obtained a displacement
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for u, a corresponding strain Ex and a corresponding
stress sigmax. We imagine that from the equilibrium
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state we are going to perturb the system by
a small amount delta u. This is the variation
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of u that we are talking about. This is called
the virtual displacement delta u. The term
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that we have here on this side is the work
done by this, the strain due to the virtual
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displacement against the stress that has been
built up in the bar to counter the effect
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of the external forces. While on the right
hand side it is the virtual work done by the
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virtual displacement delta u against the external
forces that we have. If I take this equation,
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this is called Principle of Virtual Work.
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This whole formulation is also called by another
name which is the Weak Formulation. Question
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is why do we call it a weak formulation? It
is called weak formulation because if we look
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at first integral on the left hand side here,
we have taken one of the derivatives from
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u and transferred it to delta u or w that
we had taken earlier. What happens in the
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differential equation we require the second
derivative of u to be defined, because it
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was the second order differential equation
and at every point in the domain we had to
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have definition of second derivative of u.
While in the initial weighted residual formulation
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instead of w I had put delta u. All that was
required was delta u had to be defined in
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order to have this integral finite. But now
if we look at this term we only need the first
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derivative of u to be defined. That is, instead
of asking for the second derivative to be
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defined we are now requiring only the first
derivative of u to be defined which is a weakening
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of the smoothness condition on u. That is
all we need is a first derivative of u should
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be given. Similarly, we have now transferred
the derivative from u to delta u. That is,
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now we want derivative of delta u to also
be defined in the domain. That is why we call
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this weak formulation.
Next, let us now look at the properties of
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this variation delta u that we are going to
use in our formulation in future. Let us look
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at the conditions that our delta u, or the
variation of u, or the virtual displacement
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has to satisfy.
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We see at the end x is equal to 0 for our
problem we have u given; when u is given the
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variation of u is set to 0. Set a boundary
where the displacement is specified and there
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variation delta u is set to 0. That is the
first condition on delta u or on w that we
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have taken the weight function w. Second,
what we have done is we have not put any condition
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on delta u at the end x equal to f. That is,
at the boundary value the force is specified
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du dx for example, in our case at x is equal
to L, will let delta u or w to be free. They
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can take any value that you wish. Third, as
we have shown in the weak formulation or the
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principle of virtual work, all we need is
that the derivative of delta u should be defined
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in the domain. So, when the derivative is
defined the integral that we have in the variational
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or the virtual work formulation or the weak
formulation are all finite.
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We have already talked about this; the equation
that we have written earlier is called the
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principle of virtual work or the weak formulation
as we would like to call it. For engineers,
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especially people in mechanics, principle
of virtual work is something that we know
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or we are aware of so we may go ahead with
that formulation; but they both mean the same
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thing. Let us now concentrate on the equations
that we have obtained earlier.
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There we see that we have this pair EA du
dx and delta u or w occurring together and
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we said that when the force is specified,
we set delta w or delta u or w to be anything
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and when the displacement is specified we
are going to set delta u or w equal to 0.
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If a force is specified at an end that boundary
condition is called the Natural, Neumann or
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Force boundary condition. At the end where
u is specified where we are going to fix delta
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u equal to 0 that end is called an Essential,
Dirichlet or Displacement boundary condition.
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This we will generalize to two or three dimensions
in the future. These definitions should be
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kept in mind that is, we are going to generally
call them by the name Neumann or natural or
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essential or dirichlet.
Why is the force boundary condition a natural
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boundary condition? If we have go back to
our equation that we have written of weak
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formulation, we see that force appears naturally
in that formulation. If it is on the right
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hand side of the equation it is called the
natural boundary condition. The displacement
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being fixed at an end is not explicitly appearing
in our weak formulation. It has to be enforced
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by forcing delta u equal to 0. That is why
it is called the essential boundary condition
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that is it has to be forced in. This can now
be rewritten whatever we have developed earlier
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using a different concept but we will get
the same set of equation that we have obtained
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earlier.
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Next, we will go to a new concept from which
we can develop our equations all over again.
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For that we will need to do a little bit of
home work; we have to look at this variation
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of u(x) - certain more properties of it. For
example, if I take variation of du dx instead
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of u of x this will be equal to derivative
of the variation of u. Similarly if I take
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variation of u square it will be equal to
2u(x) into variation of u of x and so on.
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Variation essentially works in principle like
the derivative but it is not the derivative
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because we are talking about pertaining to
a function not taking differential of the
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derivative of a function at a point. Using
these properties we are going to develop the
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next thing.
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We are going to define something called the
functional. What is the functional? The functional
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is a function of the function. It will take,
let me call, I as the functional. It will
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take a function u(x) and it will return to
me a number which is I u(x)
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. I give one function u(x) and get a number;
I give another function u of x I will get
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another number. For example, I may define
various functional like this, integral from
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x is equal to 0 to L EA du dx whole squared
minus integral x is equal to 0 to L fu dx
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minus Pu at x is equal to L. This is a functional
where P, f, EA are given to me as material
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data or input data to the boundary value problem
that has to be supplied. If I give you the
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function u then this integral is going to
return a number, this part is going to return
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another number, this part is going to return
another number and the sum of it will be I(u)
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which is the number. Change the function u
and put some other number w; then I will get
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another number. So this is called a Map of
the function to real numbers. I can also define
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another functional I(u) by another example.
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Let's take this half of x is equal to 0 to
L. We are not going to elaborate on it now
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but this is essentially related to our mechanics
problems of interest, minus I may have F at
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the end x is equal to L, w at the end is equal
to L. This can be another functional of interest.
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Given a functional we will also define the
variation of the functional delta I(u). This
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variation will be given by definition as limit
of alpha tending to 0 of I(u) plus alpha delta
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u minus I(u) whole thing divided by alpha.
What do I have? Instead of u, I put u plus
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alpha delta u into my expression for the functional,
evaluate that number, from that subtract the
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number corresponding to I(u), divide that
by alpha, take the limit of alpha going to
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0. This is called the variation of u. For
the various boundary value problems, in many
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cases not always, we can define this functional
I(u) and given this functional I(u) we will
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see that the variation of I(u) when set to
0 also gives us the solution to the boundary
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value problem.
Let us now go back to our variational formulation
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or the weak formulation that we have defined
earlier or the principle of virtual work for
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the model problem of interest and rewrite
that in the functional I(u).
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If we have that formulation, we have the integral
x is equal to 0 to L EA du dx d delta u divided
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by dx into dx. By our definitions of the variation
keeping EA as a given constant, we cannot
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vary EA because that is something that is
supplied to us. We can write it as integral
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x is equal to 0 to L EA into half of delta
of du dx whole square dx which we can write
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as half of integral 0 to L, we put delta of
EA du dx whole square dx. The first term essentially
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can be written in a more compact way; integral
delta of half of integral of x is equal to
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0 to L EA, if I want to write it using simpler
notation, EA derivative of u with respect
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to x squared dx. Similarly, the other term
which we had integral of f delta u dx at x
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is equal to 0 to L can be written as delta
of integral of x is equal to 0 to L f u dx.
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00:29:16,019 --> 00:29:23,019
The term P delta u evaluated at the point
x is equal to L is equal to delta of Pu evaluated
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00:29:26,460 --> 00:29:33,460
at x is equal to L. By collecting all the
terms together what we end up getting is delta
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00:29:35,330 --> 00:29:42,330
of half of integral x is equal to 0 to L EA
u comma x square dx minus integral x is equal
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00:29:51,779 --> 00:29:58,779
to 0 to L f u dx minus P u at x is equal to
L. This is equal to 0 from what we have set
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00:30:07,669 --> 00:30:14,669
earlier. Then this term with in the square
brackets is now our functional I(u). We said
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00:30:21,629 --> 00:30:28,629
that the variation of I(u) is equal to 0 is
also the weak formulation or the principle
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00:30:30,289 --> 00:30:36,809
of virtual work we had given earlier. Can
you tell us what is I(u) for the given problem?
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00:30:36,809 --> 00:30:43,809
We have taken I(u), if you look at this term
in I(u), corresponds to the strain energy
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00:30:44,929 --> 00:30:51,929
u for the structure. The remaining part corresponds
to the potential of the external forces. So
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00:30:55,809 --> 00:31:02,809
what we have is this I(u) is nothing but the
total potential energy pi of the structure.
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00:31:03,789 --> 00:31:10,029
We know very well that the minimizer of the
total potential energy is the function which
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00:31:10,029 --> 00:31:17,029
solves our model problem. We can pose this
again as, let us write it, in a concise way
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00:31:19,710 --> 00:31:26,710
del I(u) is equal to 0 is also equal to del
of the pi of u. Because the same equation
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00:31:31,099 --> 00:31:37,259
can be obtained by taking a variation of the
potential I(u) which in our case corresponds
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00:31:37,259 --> 00:31:44,259
to the total potential energy associated with
the system, this is also called a variational
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00:31:50,119 --> 00:31:56,489
formulation.
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00:31:56,489 --> 00:32:03,489
Let us check that we have written delta I(u)
is given by definition as limit of alpha tending
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00:32:07,109 --> 00:32:14,109
to 0 of I(u) plus alpha delta u minus I(u)
whole thing divided by alpha. Let us see if
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00:32:19,479 --> 00:32:24,119
it is really giving us that equation back
because we simply rewrote everything in terms
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00:32:24,119 --> 00:32:29,249
of potential and we said that yes it is equal
to 0. So let us take that potential we have
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00:32:29,249 --> 00:32:36,249
defined earlier I(u) and we have put u plus
alpha delta u. What will we get? This will
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00:32:37,549 --> 00:32:44,549
be equal to half integral x is equal to 0
to L EA u prime x plus alpha delta u prime
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00:32:54,690 --> 00:33:01,690
x whole square. I will rewrite it here; it
is equal to half of L equal to 0 to L EA u
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00:33:12,970 --> 00:33:19,970
prime x plus alpha delta u prime x whole square
dx and the other part will be minus integral
215
00:33:27,749 --> 00:33:34,749
x is equal to 0 to L f into u plus alpha delta
u dx.
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00:33:35,909 --> 00:33:42,909
Finally, let us go to the next page; minus
P u at L plus alpha delta u at L. From this
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00:33:50,149 --> 00:33:57,149
if we subtract I(u) what do we get? I will
get integral x is equal to 0 to L EA into
218
00:34:04,279 --> 00:34:11,279
alpha, half I will have in front, 2 alpha,
u comma x delta u comma x plus alpha squared
219
00:34:22,450 --> 00:34:29,450
delta u comma x whole squared dx minus integral
x is equal to 0 to L f alpha delta u dx minus
220
00:34:40,070 --> 00:34:47,070
P alpha delta u at x is equal to L. We said
that we are going to divide this thing by
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00:34:48,159 --> 00:34:55,159
alpha. When we divide this thing by alpha
this term will go. We will be left with a
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00:34:55,159 --> 00:35:02,159
single term alpha and here also this one will
go and this one will go. What we have when
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00:35:02,470 --> 00:35:07,570
we take the limit then? All the terms after
division by alpha, I mean alpha setting in
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00:35:07,570 --> 00:35:12,790
front, will go to 0; they will disappear.
All the terms with no alpha in front will
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00:35:12,790 --> 00:35:14,380
be what we are left with.
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00:35:14,380 --> 00:35:21,380
This will be equivalent to writing as delta
I(u) is equal to half of equal to... 0 to
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00:35:29,560 --> 00:35:36,560
L EA u comma x delta u comma x dx. The two
and half will cancel; minus integral x is
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00:35:43,010 --> 00:35:50,010
equal to 0 to L f delta u dx minus P delta
u at x is equal to L and this has to be equal
229
00:35:54,740 --> 00:36:01,740
to 0. We see that we have obtained our principle
of virtual work by this approach too. The
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00:36:02,620 --> 00:36:09,620
variation of the functional has given us our
principle of virtual work. What we have said
231
00:36:10,190 --> 00:36:17,190
is that we obtained in our formulation the
variation of I(u) is equal to 0 or we say
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00:36:18,310 --> 00:36:25,310
that the solution to the problem given by
u of x to the model problem is an extremizer
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00:36:28,810 --> 00:36:35,810
of I(u). Whether this extremization is minimization
or the maximization or a point of inflection
234
00:36:41,970 --> 00:36:48,970
is not something that we are look at. As soon
as u(x) is an extrimizer then the extrimizer
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00:36:49,380 --> 00:36:53,650
solves the model problem of interest.
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00:36:53,650 --> 00:37:00,650
Next, let us introduce few notations. We are
going to call it B (u, v) is equal to integral
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00:37:09,010 --> 00:37:16,010
x is equal to 0 to L EA du dx dv dx dx. This
term is given by this notation. This is called
238
00:37:28,040 --> 00:37:35,040
a bilinear form. Why is this bilinear? If
I look at B u plus alpha w1 v this will be
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00:37:47,030 --> 00:37:54,030
equal to B (u, v) plus alpha B(w1, v). It
is linear in the function u. Similarly, if
240
00:38:04,780 --> 00:38:11,780
I put B u v1 plus alpha v2 this will be equal
to B u v1 plus alpha B u v2 that is it is
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00:38:23,870 --> 00:38:30,530
linear and also in v. That is why it is linear
in both the functions which give us this number
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00:38:30,530 --> 00:38:35,240
d (u, v) we call the bilinear.
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00:38:35,240 --> 00:38:42,240
Similarly, we can introduce another notation
F(v) which is equal to f v dx for our model
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00:38:52,230 --> 00:38:59,230
problem plus p
v at x equal to L. This quantity is called
a linear functional just like we have defined
245
00:39:18,580 --> 00:39:25,580
the functional earlier. What f does? It takes
the function v and gives us the number F(v).
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00:39:25,620 --> 00:39:32,620
Why it is linear? It is linear because if
I take two functions alpha1 v1 plus alpha2
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00:39:36,430 --> 00:39:43,430
v2; F of that, if we put it here in the expression
that we had taken earlier, it is equal to
248
00:39:43,950 --> 00:39:50,950
alpha1 F(v1) plus alpha2 F(v2). That is why
it is called a linear functional. Now what
249
00:39:55,730 --> 00:40:02,730
happens when using this notation is that we
have made our writing of these equations . Instead
250
00:40:10,570 --> 00:40:17,570
of v if we substitute as v with delta u then
I will get Bu delta u is equal to F of delta
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00:40:18,730 --> 00:40:24,670
u which is nothing but the principle of virtual
work or the weak formulation or the variational
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00:40:24,670 --> 00:40:31,670
formulation that we have obtained.
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00:40:33,970 --> 00:40:40,970
We have obtained this as our
equation that we have to solve in order to
get u. There should be various questions one
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00:40:49,450 --> 00:40:56,450
should ask. Tell me what do we mean by delta
u in this case and how are we going to get
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00:40:58,240 --> 00:41:05,240
to u, how to obtain u
as a function of x? Obviously, if we knew
how to get a close function to the boundary
256
00:41:11,540 --> 00:41:17,030
value problem that we have considered earlier
then we would not have had to come to this
257
00:41:17,030 --> 00:41:24,030
stage at all. What we would like to do is
we would like to obtain a series solution
258
00:41:27,410 --> 00:41:34,410
to u. What we are going to do is we are going
to represent u as a series in terms of...
259
00:41:41,000 --> 00:41:48,000
in this form. In this series what we would
like to obtain is the coefficients of the
260
00:41:55,620 --> 00:42:02,620
series; provided we choose this function phii
of x which you can call as basis functions.
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00:42:09,520 --> 00:42:14,330
If we can choose these basis functions properly
then the linear combination of the basis functions
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00:42:14,330 --> 00:42:21,330
will form our solution u(x) and we will assume
for the time being that the series is convergent.
263
00:42:22,580 --> 00:42:29,580
Implicitly we have assumed that this series
converges to u(x) at every point. The job
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00:42:29,580 --> 00:42:36,080
is to find the coefficient ai. How can we
use what we have done here to find the coefficient
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00:42:36,080 --> 00:42:42,920
ai? That is the first question.
Secondly, we do not want to find in all cases
266
00:42:42,920 --> 00:42:49,020
all the infinite coefficients. We will be
happy by getting something called the approximate
267
00:42:49,020 --> 00:42:53,050
solution.
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00:42:53,050 --> 00:43:00,050
We will take instead of the full series a
truncated series where I will have N terms
269
00:43:13,380 --> 00:43:20,380
u of N of x. This is now called truncated
series. Again, in this truncated series if
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00:43:36,610 --> 00:43:43,610
we can find all these coefficients ai, if
they are known then given this function phii,
271
00:43:43,860 --> 00:43:48,710
we have obtained our approximate solution
for the truncated solution. The question is,
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00:43:48,710 --> 00:43:55,710
how do we choose these phiis? One thing that
the phiis have to satisfy is that the phiis
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00:43:56,150 --> 00:44:03,150
have to be Linearly Independent. Linearly
independent
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00:44:14,200 --> 00:44:21,200
in our case means, let us say that I take
this finite series only and I put... equal
275
00:44:26,570 --> 00:44:33,570
to the linear combination is equal to 0 for
all x which is in this range 0 to L, we can
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00:44:36,920 --> 00:44:43,920
say including this end points. Then linear
independence means that if this is what we
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00:44:44,090 --> 00:44:51,090
required then the ais will come out to be
0, all of them, then all the ais will be equal
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00:44:54,190 --> 00:45:01,190
to 0. If this happens then we say these function
phiis are linearly independent. In forming
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00:45:07,500 --> 00:45:14,500
this series, we need these linearly independent
terms. Then we would like to now put this
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00:45:15,230 --> 00:45:19,240
in our principle or our virtual work or the
variational formulation.
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00:45:19,240 --> 00:45:26,170
Let us go by the variational formulation that
is instead of asking for delta I(u) equal
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00:45:26,170 --> 00:45:33,170
to 0, we are now going to say delta I(u) of
N is equal to 0. When we say delta I(u)(N)
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00:45:40,310 --> 00:45:47,310
is equal to 0 then what do we get. This will
be written as integral x is equal to 0 to
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00:45:49,910 --> 00:45:56,910
L EA d u of N dx d delta u of N divided by
dx dx is equal to integral x is equal to 0
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00:46:11,260 --> 00:46:18,260
to L f delta u of Ndx plus p delta u of N
evaluated at the point x is equal to L. What
286
00:46:26,700 --> 00:46:31,560
we have done is we have instead of u we put
u of N. We say that we are looking for the
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00:46:31,560 --> 00:46:37,240
function u of N which is an extremizer of
this I that we have defined for our model
288
00:46:37,240 --> 00:46:39,800
problem of interest.
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00:46:39,800 --> 00:46:46,800
The question is
290
00:46:52,990 --> 00:46:59,990
what is delta u of N? Delta u of N for us
will be equal to delta of sigma of i is equal
291
00:47:03,280 --> 00:47:10,280
to 0 to N ai phii x. We have chosen what is
our phii. That is each of these phii is an
292
00:47:16,800 --> 00:47:22,190
unknown function that we are going to use
to construct the series. The delta of u of
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00:47:22,190 --> 00:47:29,190
N will become sigma i is equal to 0 to N delta
of ai into phii of x. You see what has happened
294
00:47:36,490 --> 00:47:42,860
is that the variation of u of Nis given in
terms of the linear combination of the variation
295
00:47:42,860 --> 00:47:49,860
of the ais.
What we will do next is to find if it is given
296
00:47:51,130 --> 00:47:58,130
in terms of the variation of these ais, each
of this delta ais can be varied independently.
297
00:48:04,060 --> 00:48:10,670
What does it mean? Because this delta u is
something that is under our control; it is
298
00:48:10,670 --> 00:48:17,620
something that we are specifying, we can choose
delta a1 is equal to 1, let us say and all
299
00:48:17,620 --> 00:48:24,190
other delta ais equal to 0; that will be 1
delta u. Similarly, I can choose delta a2
300
00:48:24,190 --> 00:48:30,610
is equal to 1 and everything else equal to
0; that will be another delta u of N. With
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00:48:30,610 --> 00:48:36,300
this various choices of delta u of N, by fixing
the values of delta ai we should be able to
302
00:48:36,300 --> 00:48:43,300
construct various forms of this delta u of
N. What we will do next is, vary this independently
303
00:48:47,460 --> 00:48:54,460
and end up getting variation (N plus 1) equations
which are simultaneous equations
in terms of the (N plus 1) variable ai. This
304
00:49:16,330 --> 00:49:17,840
is another thing that we should note.
305
00:49:17,840 --> 00:49:24,840
Let us take the model problem that we have
posed. For that model problem we need delta
306
00:49:26,870 --> 00:49:33,870
u of N at 0 is equal to 0. This also we have
to specify. This is another condition that
307
00:49:40,000 --> 00:49:47,000
is going to come on these delta ais. Why will
it be on delta ais? This is because delta
308
00:49:48,120 --> 00:49:55,120
u of N equal to 0 is equal to summation delta
ai phii evaluated at 0, i is equal to 0 to
309
00:50:00,150 --> 00:50:06,750
n. We are going to look at this in a detailed
way and we are going to pose something call
310
00:50:06,750 --> 00:50:13,750
the RITZ method. We are going to choose particular
forms of this phiis specifically for the model
311
00:50:18,010 --> 00:50:25,010
problem that we are concerned and we are going
to basically formulate the problem in terms
312
00:50:25,180 --> 00:50:32,180
of the truncated series and solve it by minimizing
the total potential energy with respect to
313
00:50:32,330 --> 00:50:39,330
the coefficients delta ais.
What we will do in the next class is we will
314
00:50:40,600 --> 00:50:47,600
take N is equal to 2 or N is equal to 3 with
some specific forms of f and P and solve the
315
00:50:48,700 --> 00:50:50,930
problems and see what are these coefficients.
316
00:50:50,930 --> 00:50:55,930
In the next class, we are going to develop
the RITZ method for the model problem that
317
00:50:55,930 --> 00:51:02,930
we have taken where the N term series that
we are going to use for the approximation
318
00:51:02,980 --> 00:51:09,970
of u(x) will be given in terms of very specific
functions. We have said that we would like
319
00:51:09,970 --> 00:51:16,600
our delta u of N as well as u of N to satisfy
the zero condition at x is equal to 0. Let
320
00:51:16,600 --> 00:51:23,600
us take the phiis to be functions which are
x of i that is x of 0 is equal to 1, x of
321
00:51:25,380 --> 00:51:31,310
1, is x, x of 2, x of 3 and so on up to x
of n.
322
00:51:31,310 --> 00:51:38,310
Then u of Nx is the summation of ai x of i.
When we put u of N at 0 is equal to 0 then
323
00:51:39,390 --> 00:51:46,390
automatically a0 comes out to be equal to
the 0. Once we have a0 is equal to 0 then
324
00:51:50,440 --> 00:51:56,800
our series will be written in terms of summation
of i is equal to 1 to n ai xi and what we
325
00:51:56,800 --> 00:52:03,800
need is to develop sufficient number of equations
to obtains these ais. That is we need N equations
326
00:52:04,660 --> 00:52:10,730
in terms of the N unknown ais. Once we get
them then we will get the solutions and through
327
00:52:10,730 --> 00:52:17,460
plots of solution for certain boundary value
problems that will take as typical example
328
00:52:17,460 --> 00:52:24,460
we will show that this method does very well
for certain classes of loading and also will
329
00:52:24,490 --> 00:52:31,490
bring out the drawbacks of this method which
we are going to exploit to develop the finite
330
00:52:32,190 --> 00:52:39,190
element method as a tool to overcome these
drawbacks.
331