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PROFESSOR: Ladies and gentlemen, welcome to lecture
number two.
In this lecture, I would like to discuss some basic concepts
of finite element analysis with regard to the analysis of
continuous systems.
We discussed in the first lecture already some basic
concepts of analysis of discrete systems.
However, in actuality, in the analysis of a complex system,
we are dealing with a continuous system, and there
are some additional basic concepts that are used in
analysis of continuous systems, using
finite element methods.
And those additional concept that are used, I want to
discuss in this lecture.
Well, when we talk about the analysis of a continuous
system, we can analyze that system via a differential
formulation or a variational formulation.
If we use a differential formulation or variational
formulation, of course we obtain continuous variables,
and we have an infinite number of state variables, or rather,
if we talk about displacements, a U
displacement, for example, of a rod, as I will be discussing
just now, we will have infinite values of that
displacement along the rod.
In the differential formulation or the variational
formulation, we would have to solve for that continuous
variable along the rod.
Well, we will also notice that in the analysis of a complex
system, we cannot solve the differential equations that we
are arriving at directly, and we have to resort
to numerical methods.
Now, some concepts that have been used for a long time are
the weighted residual methods.
These have been used by Galerkin least squares
approaches to solve the differential equations that
govern the equilibrium motion of the system approximately.
In the variational formulation, the Ritz method
has been used for quite a long time.
These are classical techniques, therefore,
weighted residual method and the Ritz method.
And what I want to show to you in this lecture is how the
finite element method is really an extension off these
methods, or how this method, the finite element method, is
related to these classical techniques.
Well, when we talk about the differential formulation, we
are looking at the differential equilibrium, or
the equilibrium of a differential
element of the system.
Now, I want to show to you the basic ideas of a differential
formulation by looking at, or analyzing, this rod.
Here we have a rod that is fixed at the left-hand side,
on rollers.
x is a variable along the rod.
u is the displacement of that rod into this direction.
The rod is subjected, as shown here, to a load , R0
at its right end.
Notice that in this analysis, we assume that plane sections
remain plane.
In other words, a section that was originally there at time t
greater than 0 has moved to here.
And this movement here is the u displacement.
But notice that the vertical section here remains vertical
during the motion.
So at every section, we have only 1 degree of freedom.
There's no rotation of that section.
However, this one degree of freedom, u, varies
continuously along the rod.
Therefore, the rod itself has really an infinite number of
degrees of freedom.
For this very simple system, we could
obtain an exact solution.
However, I want to show you how we proceed in analyzing
this rod via differential formulation, a variational
formulation, and so on, simply as an example.
Therefore, the basic ideas, really, that I will be putting
forth to you, that I will be discussing with you, are
really the important things that I want to expose to you.
It's not the analysis of this very specific problem.
It's really the basic idea that I want to clarify to you
by looking at this one problem.
Well, for this problem here, the governing differential
equation of motion is shown here.
Notice u once again is the displacement of a section.
That is, this displacement here at that coordinate x.
c is given here as square root E over rho, where E is Young's
modulus of the material, rho is the mass
density of the material.
t, of course, is the time variable.
Notice the cross-sectional area a here, I have also
written down here.
This cross-sectional area cancels out on both sides, as
you will see just now.
Now this question here is obtained in the differential
formulation by looking at the equilibrium of an element.
And we might consider this to be the element that I will now
focus our attention on.
Here we have that element drawn again.
This element here of length dx is subjected at its left side,
because x comes from here as a variable and brings us up to
this station.
At the left side here, the element is subjected to sigma,
the stress sigma.
At the right side, we have the stress sigma plus
partial d sigma dx.
This here really means nothing else than a d sigma, an
increment in the sigma.
Well, the equilibrium requirement for this element
here is now that the force on this side here, that is, sigma
times a on this side, which is actually this one here, I
should have pointed to this one--
and this force here, which is the force on the right side.
If we subtract these two forces, that must be equal to
the force applied.
Or rather, the d'Alembert force.
Notice that if we look at this element, there's a force on
the right, there's a force on the left.
And this being the force on the right, let's call that,
say, R1, and let's call that R2.
So I put here R1, I put here R2.
And R1 minus R2 must be equal to the d'Alembert force, which
is due to the inertia of the material.
This is the basic Newton's law applied to this
differential element.
Now, if we referred back to how we proceeded in the
analysis of a discrete system, we really proceeded in exactly
the same way.
But our element then was a discrete element, a discrete
spring element.
We now use the same concept, but apply those concepts to a
differential element.
So this is the equilibrium equation of the element.
The constitutive relation is given here--
that the stress is equal to e, the Young's
Modulus, times the strain.
This is the strain.
And notice once again, since we are considering sections to
be remaining plane, and simply move horizontally, the only
strain that we're talking about is this one.
If we combine these two equations here, we directly
obtain that equation here.
Notice as I mentioned earlier, a cancels out.
If a is constant, that's why a does not enter into this
equation, and this substitution for sigma into
here gives us the second derivative here.
Of course, this part here cancels out that part there,
and that second derivative is this one here.
The E brought over to this side gives us a 1 over c
squared, where c, once again, is defined as shown here.
The important point, really, is that we are looking here in
the differential formulation at a differential element of
length dx at a particular station x.
That we're looking at this element and we establish the
equilibrium requirement of that element.
R1 minus R2 shall be equal to the mass of the element times
the acceleration.
We also introduced a constitutive relation.
So far, clearly, we have used two conditions for the
solution of the problem.
The first one is the equilibrium condition.
The second is the constitutive condition, or constitutive
requirement.
We have to ask ourselves, where do we satisfy the
compatibility condition?
Because there are always these three conditions that we have
to satisfy.
Well, the compatibility condition is satisfied by
solving this differential equation for this rod, and
obtaining a u that is continuous.
In other words, a u that tells that all the sections have
remained together.
We did not cut that bar apart.
In the discrete system analysis of the spring system
of lecture 1, if you were to think back to it, we had to
satisfy the compatibility condition explicitly, in
establishing the equilibrium equations, because we had to
make sure that all the springs remain attached to the carts.
Here we satisfy the compatibility condition by
solving this equation for a continuous u.
Well, the boundary conditions, of course,
also have to be stated.
And here we have a boundary condition on the
left end of the rod.
Remember, please, that the rod is fixed at its left end, so
we have this condition here, and clearly u must be 0 for
all times t at x equals 0.
At the right end, we apply a load R0.
And there we have, this being here the area times the stress
at x equals L, E times a du dx.
Notice that E du dx is, of course, the stress, and so we
have here this total force at the right end
being equal to R0.
We also have initial conditions for the solution of
this equation that I showed to you, this equation here.
We have to have two spatial conditions, two boundary
conditions, one at the left and one at the right end.
The ones that I just showed to you.
We also have to have to have two initial conditions, one on
the displacement and one on the velocity.
Well, their initial conditions, in this particular
example, might be as shown here.
At time 0, all of the displacement are 0.
And at time 0, all the velocities
along the rod are 0.
So the basic differential equation given here, once
again, plus these boundary conditions, plus these initial
conditions, define the complete problem.
I also like to point out here that this boundary condition
here, which does not involve any derivative, is called an
essential, or displacement boundary condition.
An essential boundary condition because it does not
involve any derivative when, and this is important, the
highest derivative in this differential equation is 2.
The right-hand side boundary condition is called a natural
force boundary condition.
It's really involving forces.
And it involves, as a highest derivative, a derivative of
order 1 when the differential equation here involves as the
highest derivative a derivative of order 2.
So in general--
and this is a very important point--
we can say the following.
If the highest order of the spatial derivative in the
problem governing differential equation is 2m, in our case, m
is equal to 1 for our problem.
The highest order of the spacial derivative in the
essential boundary condition is m minus 1.
In other words, in this case, of order 0.
The highest order the spatial derivative in the natural
boundary conditions that I just discussed is 2m minus 1,
which is 1 in our particular case.
Then if we talk about this problem, we talk about a C m
minus 1 variational problem.
It will become apparent to you later on why we
call it this way.
C m minus 1 means continuity of order m minus 1.
In fact, in the Ritz analysis that we will be performing
later on, that I will discuss with you later on, we find
that we need, in the solution of that kind of problem, only
continuity in the Ritz functions of order m minus 1.
Well, let us now look at the variational formulation.
I mentioned earlier that we have two different approaches.
The first approach is a differential formulation, the
second approach is the variational formulation.
The variational formulation operates in much the same way
as I introduced it to you for the
analysis of discrete systems.
We talk about pi a functional, being equal to the strain
energy minus the potential of the loads.
Now for the rod, the strain energy is given here.
Notice this is 1/2 times the stress times the strain, and
integrated over the total volume of the element, or of
the rod, I should rather say.
The total potential of the loads is given here.
I could have written it this way, with a minus out there
and a plus in there--
same thing.
This, then, is really nothing other than the loads
multiplied by the total displacement.
And of course, there's an integration involved here,
because the body forces, the body loads that I introduced
here, fB, are varying along the length of the rod.
I introduce these fB body forces because I want to use,
later on, the d'Alembert principle, put these, in other
words, equal to minus the acceleration forces, and can
directly apply what I discussed now, also to the
dynamic analysis of this rod, which we considered in the
differential formulations.
Together with stating pi as shown here, we also have to
state the left-hand boundary condition, which is an
essential boundary condition.
We have to list all the essential boundary conditions
here, or the displacement boundary conditions.
Essential and displacement mean the same
thing, in that sense.
Well, then we invoke the stationality of pi.
We are saying that del pi shall be equal to 0 for any
arbitrary variations of u that satisfy, however, the
essential boundary conditions.
This boundary condition here.
So this has to hold, this statement shall hold, for any
arbitrary variations in u.
However, del u 0 shall be 0, that satisfy, in other words,
the essential boundary condition.
Well, if we apply this variation on pi, we obtain
directly this result here.
Notice that this part here is obtained by applying the
variation on this part here.
The variation operator operates, del operates, much
in the same way as a differential operator.
So this tool cancels this at 1/2, and we are left with EA
del u del x times the variation on del u del x.
And that is given right there.
The variation on this part here gives us simply a del u
times fB, and a del uL times R. Of course, uL is equal to
is the displacement at x equal to L. And this is, therefore,
the final result.
Now if we look at this, we recognize, really, that this
is the principle of virtual displacement.
It's a principle of virtual displacement
governing the problem.
In general, we can write this principle as follows.
Notice that here, we have variations in strains, which
is that part there.
The real stresses are given there, which are those.
And here we have variations in displacement, those operating
on the body forces.
There, there.
Notice, I sum here over all body forces in general.
We have three components in fB--
the x, y, and z component.
So I list these components in a vector that I call fB.
Similarly, of course, we have three displacement components
that appear here in this vector U. Tau in general has 6
components.
Del epsilon also has 6 components.
Putting their transpose on the del epsilon vector means that
we're summing the product of the
strains times these stresses.
Similarly, we are summing here the product of the
displacement components times the force components.
We also have here a contribution
due to surface forces.
fS are the surface forces.
3 components, again, these are the variations in the surface
displacement.
So this part here, del uLR, really corresponds to this
part here, involving surface forces.
So the surface forces read again, and the variations in
the surface displacements, are those here.
Now notice that this principle here, or this equation, I
should rather say, once again has to be satisfied for any
arbitrary variations in displacements that satisfy the
essential boundary conditions.
For the problem of the rod, these del u's have to satisfy
the condition that the variation at the left-hand
boundary on u is 0, because that is the
essential boundary condition.
Also, of course, notice that this variation in the u's
corresponds to these variations in the strains.
In other words, these strains here are obtained from the
variations in the displacement.
That's important.
They are linked together.
So if we impose certain variations in the
displacement, we have to impose here the corresponding
variations in strains.
Of course, these variations in the displacements give us also
variations in surface displacement.
So these here are again linked up with that.
Later on in our formulation of the finite element method, we
will write the variations in the strains here as virtual
strains, arbitrary virtual strains.
And we are talking about the strain vector with a
bar on top of it.
Similarly, a bar here, a bar here, instead of
the variation sign.
However, the meaning is quite identical--
what we are saying here, and this is the principle of
virtual displacement that we will be talking about later
on, when we formulate the finite element equations, what
we are saying here really is that this equation has to be
satisfied for any arbitrary virtual displacements and
corresponding virtual strains.
However, the virtual displacements have to satisfy
the displacement boundary conditions, the essential
boundary conditions.
Well, from this, or rather that, which of course these
equations are completely equivalent, we
can go one step further.
And if we go one step further, by applying integration by
parts and recognizing that this part here is completely
equal to that, by that I mean taking the variation on the
derivative of x is the same as taking the sorry... by taking
variation on the derivative of u with respect to x, that is
completely identical to first taking the variation on u, and
then taking the differentiation of that
variation on u with respect to x.
Well, if we recognize that these two things are
identical, then using integration by parts on this
equation here, which means, really, on that equation here,
for the special case of the rod that we're now
considering, we directly obtain this equation here.
The integration by parts is performed by integrating this
relation here first, and notice that if we do integrate
this, we want to lower the differentiation of the virtual
part here and increase the order of differentiation on
this part here.
And that then directly gives us these two terms
and that term here.
If we then also list this part here together with what we
obtain from that part, we directly obtain this equation
here, and we proceed similarly for the coefficients of del
uL, and we obtain this part here.
So this equation here is obtained by simply using
integration by parts on, I repeat, this equation here.
We have not used any assumption.
All we did is a mathematical manipulation of this equation
in a different form.
And this is the new form that we obtained.
Del u0, of course, is the variation of u at x equals 0.
Now when we look at this relation, we can extract now
the differential equation of equilibrium and the natural or
force boundary conditions.
How do we extract them?
Well, the first point is that I mentioned already earlier,
this part here is 0, because del u0 is imposed to be 0.
So that part is 0.
We can strike it out directly.
Now when we look at this part here and that part here and
recognize that del u is now arbitrary, we can directly
extract the relation that this must be 0 and that must be 0.
How this is done?
Well, first of all, we recognize that this
integration here really goes from 0 plus to L minus, if you
want to be really exact.
Because at the boundary, we have the boundary conditions.
Of course, this 0 plus means it is infinitesimally close to
0, and L minus, we are integrating up to a distance
infinitesimally small to actually L. So putting the 0
plus and L minus here is just for conceptual understanding
really necessary.
Well, if we then impose the following variations, say that
del uL is 0, is exactly 0, then this part is out.
And then we only have to look at this part.
Now we can apply any arbitrary variation on u from 0 to L
minus, and this, then, this total integration, must be
equal to 0.
It must satisfy the 0 condition.
And that can only be true when this part here, what is in the
brackets, is 0.
Because if this is not 0, I can always select a certain
delta u which, when multiplied by that and integrated as
shown here--
remember, this part is not there--
will not give us 0.
So therefore, this part must be 0.
And that is our first condition, which is the
equilibrium condition of a differential element.
It's the equilibrium condition of a differential element.
Now, if we say, let us look at the right-hand side boundary,
and put del u0 everywhere along the length of the rod,
except at x equals exactly L. Then this part would be 0, and
this part here is non-zero, provided--
or rather, this part would be present.
And the only way that this part can be 0, as it must be,
is that this part here, the coefficient on del uL, is
actually 0.
So this way, we have extracted two points, two conditions.
This part must be 0 and that part must be 0.
We extract these conditions by looking at specific
variations on u.
First we look at the variation on u where del uL is 0, and
otherwise arbitrary from x equals 0 to L minus.
And then we can directly conclude this must be 0, and
second time around, we say, let del u equal 0 from x
equals 0 to L minus, which makes all of this integral 0,
and we can focus all our attention on this part here,
and we directly can conclude that this part now
must also be 0.
So this way, then, we extracted the differential
equation of equilibrium and the natural boundary
conditions.
Very important.
Differential equation of equilibrium and the natural
boundary condition.
And now we really recognize.
of course, that if we put fB equal to the d'Alembert force,
or minus the d'Alembert force, equal to this right-hand side,
substitute from here back into there, cancel out A, we
directly obtain this differential equation.
Notice that c is E divided by rho square root
E divided by rho.
Now, the important point really is that by having
started off with this pi functional and that condition,
this condition also, and that the variations on u at x
equals 0 shall be identically 0, we directly can extract
from this pi functional the differential equation of
equilibrium and the natural boundary conditions.
The natural boundary conditions, in fact, are
contained right in there.
That's where the natural boundary conditions appear.
The essential boundary conditions are there, and have
to be satisfied by the variations.
So in general, then, we find the following points.
The important points are that by invoking del pi equals 0
and using the essential boundary conditions only, we
generate the principle of virtual displacement, an
extremely important fact.
And this will be the starting equation that we will be using
to generate our finite element equations later on.
We also can extract the problem governing
differential equation.
Therefore, the problem governing differential
equation is contained in the principle of virtual
displacement, and this one is contained in del pi equals 0,
in the del pi equals 0 condition.
Of course also satisfying the essential boundary condition.
We also can extract the natural boundary conditions.
So these are also contained, in essence, in pi, and as I
showed to you, they're contained really in w.
That's where they appear.
Now in the derivation of the problem governing differential
equation, we used integration by parts.
And the highest spatial derivative in
pi is of order m.
We used integration by parts m times, and what we're finding
is that the highest spatial derivative in the problem
governing differential equation is then 2m.
And this then puts together the complete structure of the
equations that we're talking about.
Here I have another view graph that summarizes the
process once more.
We are starting with the total potential pi of the system
being equal to the strain energy minus the total
potential of the loads.
We're using this condition and the essential boundary
condition--
very important--
to generate the principle of virtual displacement.
At this stage, we can solve the problem, and we will do so
by using finite element methods.
We can, however, also go on from the principle of virtual
displacement using integration by parts, and then we would
derive the differential equation of equilibrium and
the natural boundary conditions.
We can solve the problem.
Well, we can solve the problem at this level really only when
we can solve the differential equation of equilibrium,
subject to the natural boundary conditions.
And that we can really only do for very simple systems.
Therefore, this process here, from here onward, can really
be followed only for relatively simple systems.
For complex systems, shell structures, complicated shell
beam structures, plane stress systems, real engineering
structural analysis systems, this is not possible, and we
stop right there, and we solve our problem this way, using
finite element methods.
Now, the one important point, however, I want to make once
more clear--
when we proceed this way, we are deriving the differential
equation of equilibrium for each differential element.
And this means that when we go this route for the simple
systems that we can go this route, we satisfy the
equilibrium condition on each differential element.
However, when we go this route, we will find that we
only satisfy the equilibrium conditions in a global sense,
in an integrated sense, using finite element methods, and
that to actually satisfy the equilibrium conditions in a
local sense, meaning for each differential element, we have
to use many elements, and only then, of course, our finite
element solution will converge to the solution that we would
have obtained solving the differential equation of
equilibrium.
Therefore, when we start from the principle of virtual
displacement, if we, as I will show you, use many elements,
we really obtain the same solution as here.
However, if we do not use many elements, if we
use a coarse mesh--
we will talk about a coarse mesh later on--
then we will see that we only satisfy the equilibrium
conditions in a global sense for the complete structure, in
an integrated sense for the complete structure.
For each finite element, we will satisfy the equilibrium
conditions, but we will not satisfy the equilibrium
conditions accurately for each differential element--
dx, dy, dz being arbitrarily small--
in the continuous body.
That will become clearer, then, when we actually go
through an example.
Now I mentioned earlier that a whole class of classical
methods are the weighted residual methods.
In the weighted residual methods, we proceed in the
following way.
We consider the steady state problem, which is given here,
with these boundary conditions.
Now this operator L2m phi equals R--
this L2m is the operator that governs the problem.
Like in our particular rod problem, L2m would be equal to
delta u delta x squared.
So the highest derivative in this spacial operator being
2m, in this case, 2, for our example.
The boundary conditions can be written this way.
And the basic step, then, in the weighted residual and the
Ritz analysis is to assume a solution all of this form,
where phi bar gives us the assumed solution ai are
parameters that are unknown, and fi are bases functions.
These are functions that have to be assumed.
Well, there is, of course, considerable concern on what
kind of functions to choose.
In the weighted residual method, these functions here,
if you directly operate on this equation here, should
satisfy all boundary conditions.
Notice that in the weighted residual method, here I'm
really talking about going on this view graph through the
differential equations of equilibrium and natural
boundary conditions, deriving them, if you want, via this
route, and then trying to solve the problem here
numerically.
We will actually, as I said earlier, in the finite element
process, go this route, but we can also go that route with
the weighted residual methods.
In fact, there is a close relationship between using
weighted residual methods at this level and the Ritz method
at that level, the way I will be describing it later on.
But in the weighted residual method, this is the
assumption.
And then if we look at this equation here, we can
construct an error R, substituting
from here into there.
And that capital R error is given by this equation.
Notice that this is, of course, our trial function
that we have.
And if the right-hand side is 0, everywhere over the domain,
then of course our error would be 0, and we would have solved
our equation that we're looking at here.
That is the equation we want to solve.
Since if all of these functions fi satisfy all of
the boundary conditions, then we are satisfying these
equations, and all we have to worry about further is to
satisfy this equation.
If R is 0, we would also satisfy the differential
equation of equilibrium, and we would have,
in fact, the solution.
However, that, of course, would be a very lucky choice
on the fi functions.
In general, R will not be identically 0
all over the domain.
In fact, how much R, or how close R will be to 0 will of
course depend on the ai.
And this is then basically our objective, namely, to
calculate ai values that are making this R, this left-hand
side capital R, as close as possible to 0.
And that can be achieved via the
Galerkin method, for example.
This is the basic process.
Here we're substituting the R. These are the trial functions,
and we're integrating the product of these over the
total domain.
The domain here, in the case of our rod, would simply be
the volume of the rod.
This is the mechanism that generates to us n equations in
the trial parameters ai.
In another approach, the least squares method, we would
operate on the square of the error, and minimize the square
of the error when integrated over the total domain with
respect to the trial parameters ai.
That again gives us n equations, and we set up these
n equations just as we're doing here, to
solve for the ai.
Knowing, then, the ai, we can back substitute into this
assumption here, and now we have our approximate
solution, phi bar.
If the trial functions have been selected to satisfy the
boundary conditions, then of course phi bar will satisfy
the boundary conditions.
However, what phi bar will not satisfy exactly is this
equation here.
However, we have minimized the error in the satisfaction of
this equation in some sense using the Galerkin method or
the least squares method.
These methods can also be extended when the fi trial
functions do not satisfy all of the boundary conditions,
namely, not the natural boundary conditions.
They can be extended to their places, but classically, they
have been used with trial functions that satisfy all
boundary conditions.
When we were to extend the Galerkin method for the case
where the trial functions do not satisfy the natural
boundary conditions, then we really talk basically about
already a Ritz analysis, and that is the next procedure
that I want to introduce to you.
Now of course, if we wanted to really start deriving the Ritz
method from the Galerkin approach--
in other words, if we wanted to derive this Ritz analysis
method from the Galerkin approach, we would have to
extend, first of all, this Galerkin approach to include
the natural boundary conditions, and then we would
have to perform integrations on this equation, and we would
obtain the Ritz analysis method.
The actual way of starting, of introducing the Ritz analysis
method, is to introduce it as a separate tool.
And it is introduced in the following way.
Let pi be the functional of the C m minus 1 variational
problem that is equivalent to the differential formulation
that we talked about earlier.
Now, the differential formulation that I talked
about here is this one.
That's a differential formulation.
And as an example once again, the operator L2m is del 2 u
del x squared.
There's a constant here.
We could also put an EA in front here, but
that only is a constant.
It doesn't change the character of the operator.
But this is basically the operator for the problem that
we considered, and of course, our boundary conditions are
also there.
So if we have a pi functional that is equivalent to the
differential formulation given in those two equations that I
just pointed out to you again, then in the Ritz method, we
substitute the trial functions, phi bar--
let us look at them again, these are the trial
functions--
into pi, and we generate n simultaneous equations for the
parameters that appear in this assumption here by invoking
the stationality of pi.
Notice that by invoking that del pi is 0, we really say
that del pi ai is 0 for all i.
And that gives us the condition which we used to set
up the individual equations that we need to set up to
solve for the trial parameters ai.
Well, let us look at some of the properties.
The trial functions used in the Ritz analysis need only
satisfy the essential boundary conditions, an extremely
important fact.
In the classical weighted residual method, as I pointed
out, the trial function should satisfy all boundary
conditions.
Therefore, they can be very difficult to choose.
In the Ritz analysis method, we only need to satisfy the
essential boundary conditions.
The application of del pi equals 0 generates the
principle of virtual displacement.
I mentioned that to you earlier already.
And therefore, in effect, we use in the Ritz analysis this
principle of virtual displacement.
By invoking del pi equals 0, we minimize basically the
violation of the internal equilibrium requirements and
the violation of the natural boundary conditions.
Well, remember that invoking del pi equal to 0 and then
using integration by parts, we actually generate, we could
generate the differential equations of equilibrium and
the natural boundary conditions the way I've shown
it to you, for the simple bar structure.
Now, what I'm saying here is that we do not want to
generate these differential equations of equilibrium and
natural boundary conditions.
However, please recognize that they are contained in the
equation del pi equals 0.
Therefore, by substituting our trial functions, we violate
the internal equilibrium requirements and the natural
boundary conditions, but we will see that we are
minimizing that violation in these conditions here.
Also, we will see that we generate a symmetric
coefficient matrix, and K and the governing equations then,
our KU equals R, that we want to solve.
And that is really the basis of the finite element method
for the analysis of continuous systems.
Let me now go in detail through an example.
Here we have is simple bar structure which has an area 1
square centimeter from A to B, and from B to C, B being this
point here where the area changes and C
being that point there.
From B to C, we have a varying area.
This variation in the area is shown here.
It's 1 plus y/40 squared is the area at any station y, y
being measured from point B, as you can see here.
The length here is 100 centimeters.
This length is 80 centimeters.
The structure is subjected to a load of 100 newtons here.
Notice that this arrow really lies on top of that dashed
line we just separated out for you to understand that there
is this arrow.
So this is the load applied at the
mid-line of this structure.
We assume once again for the structure also that there is
only the following displacement mechanism.
If a section was originally there, and it is a vertical
section to the midline, then it has moved over, and I
grossly exaggerate now, to this position, where this is
the displacement that we're talking about u.
Grossly exaggerated, of course.
So we're having a bar structure subjected to a
concentrated load fixed at the left end.
And our objective now is to solve this structure, to solve
for the unknown displacement u, being 0 here, of course as
a function of x, when this structure is
subjected to that load.
Well, in the calculation of this example, I want to
display to you as many of the concepts
that we just discussed.
Here we have pi being equal to this value here.
The strain energy is given here.
Notice this is 1/2 times the stress times the strain
integrated over the volume of the structure.
The integration goes from 0 to 180, because that is the
length of that structure.
The total potential of the external load is 100, which is
the intensity of the load times the displacement at x
equal to 180... at x equal to 180.
Well, the essential boundary condition is that u is
0 at x equals 0.
I'd like to now consider two different cases
for the Ritz analysis.
In the case one I want to use a function that spans u, which
spans continuously--
and let me draw it out here--
from u, from x equals 0, to x equals 180.
That is the endpoint.
So here we have this function, this part here and that part
there, these are the two trial parameters that we
want to solve for.
And we will select them, we will calculate them, rather,
using the Ritz analysis.
Case two, I also use trial functions, but notice now that
I'm performing the following.
We have a domain AB--
let me go back once more--
a domain AB, and a domain BC.
And I want to now use one function for AB and one
function for BC.
The AB function is simply this one here, a linear variation
up to this point.
Now notice that UB is our trial parameter--
that's the one we don't know, our trial function parameter.
x/100 is simply is the function
that I'm talking about.
And notice that this function only is applicable for this
domain where, let me put down here the length that is x
equal to 100, and this is here x equal to 180.
Now for this part here, I use this function here.
Now notice what this function does.
Well, if we look at this part here in front, it involves uB,
which is also there, and it involves uC.
uB, by the way, is the physical displacement right
here into this direction.
Of course, I'm plotting u upwards here to be able to
show it to you.
But the displacement, uB, is the displacement of this point
B to the right.
uC is the displacement of this point C to the right.
Then we recognize that this part here corresponds really
to a variation such as that.
Notice when x is equal to 100, which is that point, this
function here is 1.
When x is equal to 180, which is that point there, this part
is equal to 0, because 180 minus 100 is 80, divided by 80
is 1, and 1 minus 1 is 0.
So this dashed line corresponds to
this function here.
Let me put a dashed line underneath there.
Well, if we now look at this part here, we notice that this
part is 0, or this trial function here is 0 at this
point B. And it varies linearly like that across,
where this part here, of course, denotes uC.
That is this one here, solid black line, and here also,
solid black line.
Now the superposition of both these functions, the dashed
blue and the solid black line, give us this function here.
So the actual function that I'm talking about is a linear
variation along here, and a linear variation along here,
where I plot it vertically up here--
uB and uC here.
Now, this is a specific case that I want to draw your
attention on, because this really corresponds, as we
shall see, to a true finite element analysis.
And the reason for it is that we're talking about one domain
here and another domain there.
And both of these domains are identified as finite elements.
Well, the first step now is to use pi, invoke the
stationality condition as we did earlier, and this gives us
the principle of virtual displacement.
I mentioned it earlier already.
Our virtual strains are here.
The stresses are here.
The virtual work is on this side.
I discussed it earlier already.
We do not want to go now via this route.
We first of all want to now obtain the exact solution.
The exact solution is obtained by using integration by parts
on delta pi, being, of course, equal to 0, and extracting the
differential equation of equilibrium for each
differential element in this structure.
This means that if we are talking here about the
differential element equilibrium of each
differential element dx long anywhere along the structure,
in other words, the equilibrium of typically an
element like that.
That is a differential equation of equilibrium.
And we also, of course, have the natural boundary
conditions.
We can also derive the natural boundary conditions.
The solution to this is obtained by integration, and
this is the solution given.
Well, the stresses, then, of course are obtained by
differentiation of the u's to get strains, and multiplying
those by E, and these are the stresses in the bar.
These are the exact stresses in the bar that satisfy the
differential equations of equilibrium and the natural
boundary conditions.
This is the exact solution of this bar problem, the way I
have formulated it.
Now will perform our Ritz analysis.
In case one, we use pi equal to this.
Notice that I have substituted now our trial functions
corresponding to case one into the functional pi.
That gives us this term, that term here, and that term here.
Notice that I have broken up the integration from 0 to 100,
and 100 to 180, because the area changes from over this
length here, and only for that reason, really, I have broken
up the integrations.
Now this is pi.
And if we now invoke, we can integrate this out, and then
invoke that del pi shall be 0, we
obtain this set of equations.
We solve for a1 and a2, substitute back into our
assumption that we had earlier, and we got this u.
Notice that of course this u displacement does satisfy the
essential boundary conditions.
It does satisfy the essential boundary
conditions at x equals 0.
You can just substitute x equals 0, and you would see
that u is 0.
It does not satisfy, however, the natural boundary condition
at x equal to 180.
Sigma is given here, obtained by calculating the strains
from here and multiplying by E--
this is our approximate solution to the problem.
We are satisfying the compatibility conditions,
because the bar has remained together.
No material has been cut away from it.
Also, we are satisfying the constitutive relations, but we
do not satisfy the internal equilibrium on a differential
local elements sense.
We do not satisfy the differential equilibrium, and
we do not satisfy the natural boundary conditions.
But we satisfy them in an approximate sense.
Case two.
Here now we're talking about our two linear functions.
And here we naturally integrate from 0 to 100 for
the first linear function, and from 100 to 180 for the second
linear function.
Notice this is, again, the area, and notice that this is
here the strain squared.
It's strain squared here because our E is out there,
which would give us the stress.
And the area here, of course, is equal to 1, which we did
not write down.
The important point is that this is now our pi for these
two functions.
We again invoke del pi equal to 0.
We obtain now this set of equations.
We are solving from this set of equations uB
and uC, given here.
Having got uB and uC, of course we now have the
complete displacements along the bar, because we only need
to substitute back into our original approximations that
we looked at earlier.
Let me just get them once more here.
We had them here.
If we now substitute from uB and uC into these two
equations, we have the complete
displacement solution.
Of course, this is an approximate
displacement solution.
And similarly, our stresses are approximate.
Now on these last few graphs, I have plotted the solution.
And notice that this is here the direction x.
Here we have the point B, here we have the point C, here we
have the point A. Our exact solution, which satisfies the
constitutive relations, compatibility relations, and
the differential equations of equilibrium, and all bounded
conditions, is the solid line here.
Our solution one, case one, Ritz analysis, is the dashed
line here, and the solution two is this dashed dotted
line, down there.
Notice that we are quite close in our Ritz analysis to the
exact solution in the displacement.
However, the strains and stresses are obtained by the
differentiation of these displacement solutions, and
here I show to you the calculated stresses.
Again, point A here, point B here, point C there.
The important point is the following.
In the exact solution, we have the stress of 100 in domain
AB, and then we have this curve here, a
very high slope there.
And in our solution one, we had this variation in stress.
Notice that it goes continuously over the complete
domain, because our assumed displacement function was
continuous also over this domain, and its first
derivative was continuous over this complete domain.
So that's why our solution one is continuous there, and in
fact, we're seeing just the straight line there, because
our displacement approximation was parabolic.
Our solution two is exact here, 100, and very
approximate here for the displacement.
But notice that at the midpoint between B and C, we
get very good results.
Now the important point really is shown here on
the last view graph.
We note that in this last analysis, we use trial
functions that do not satisfy the natural boundary
condition, and I'm talking now about the piecewise linear
functions, in other words, from A to B and B to C each,
just a straight line.
We use trial functions that do not satisfy the natural
boundary conditions.
The trial functions themselves are continuous, but the
derivatives are discontinuous at point B. Notice our
stresses here are discontinuous at point B.
For a cm minus 1 variational problem, the way I've defined
it, we only need continuity in the m minus first derivatives
of the functions.
In this problem, m is 1, and therefore we only need
continuity in the functions themselves, and not in any
derivatives, because we only need continuity in the m minus
first derivative.
The domains A and B and B and C are finite elements, and in
actuality, we've performed a finite element analysis.
This is all I wanted to say in this lecture.
Thank you for your attention.