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PROFESSOR: Ladies and gentlemen, welcome to
lecture number 3.
In the previous two lectures, we discussed some basic
concepts related to finite element analysis.
In this lecture, I would like to present to you a general
formulation of the displacement-based finite
element method.
This is a very general formulation.
We use it to analyze 1D, 2D, three-dimensional problems,
plate and shell structures.
And it provides the basis of almost all finite element
analysis performed at present in practice.
We will see that the formation is really a modern an
application of the Ritz/Golerkin procedures that
we discussed in the last lecture.
We will consider in this lecture static and dynamic
conditions, but as I pointed out earlier, we will only be
concerned with linear analysis.
On this first view graph, I've prepared schematically a
sketch of a three-dimensional body.
This three-dimensional body could represent, typically, a
bridge, a shaft, a building--
whatever structure we want to analyze.
And this three-dimensional body here is subject to the
following forces-- it is subjected to concentrated
forces, forces that have components, Fxi, Fyi, Fzi, at
one point i, and there are many such points i.
The body's also subjected to body force
components, Fbx, Fby, Fbz.
These are forces per unit volume.
And we will see later on that we include in these forces the
d'Alembert forces, when we consider dynamic analysis.
The body is also subjected to distributed surface forces,
with components, Fsx, Fsy, and Fsz.
These surface forces would be, for example, distributed water
pressure in a dam, frictional forces, et cetera.
So we have, basically, concentrated surface forces,
we have distributed surface forces, and volume forces--
externally applied volume forces, body forces.
The body is, of course, also properly supported.
We have here, typically, a support that prevents
displacements in any direction.
And here, we have another such support.
Here, we have a roller support, which prevents
displacements only in this direction.
The body is defined in the coordinate system, XYZ, and
notice that I'm using here capital XYZ's.
And the displacements of the body are measured as U, V, and
W into the capital X, Y, and Z directions.
I'm using capital letters here to denote global displacements
and global coordinates.
We will later on in the finite element discretization also
introduce small lowercase x, y and z's, and u, v, and w's to
measure the displacement in the individual elements.
So the problem is, other words, that we have this body,
this general structure, subjected to certain forces,
properly constrained, and we want to calculate the
displacements of the body, the strains in the body, and the
stresses, of course, in the body.
Well, on this view graph, here, I listed the external
forces once as vectors, here's our force FB, the body force
per unit volume, with components into the x, y, and
z directions.
Here, we have the surface forces with components in the
x, y, and z directions.
And here, we have a typical concentrated force at that
point i, with components FX, FY, and FZ again.
The displacements of the body measured in the global
coordinates are U, V and W, as shown here.
Of course, notice that these U, V and W's are functions of
the capital XYZ coordinates.
The strains corresponding to these displacements, which of
course, are known, are listed here.
And the three-dimensional analysis, we have six such
known strains, from epsilon XX to gamma ZX.
Of course, the last three being the shearing strains,
and the first three being the normal strains.
The corresponding stresses are listed here.
Again, six components from tau XX to tau ZX.
Of course, if we were to actually analyze a
two-dimensional problem, such as the plane stress problem,
we would only use the appropriate quantities from
here and from there, as we'll discuss later on.
The starting point of our analysis, in which we want to
calculate the stresses, the strains, and of course, the
displacement also.
The starting point is the principle of virtual
displacements.
Now, this is the principle which we already discussed
very briefly in the last lecture.
Remember, please that we can derive it by looking at the
total potential of the system, which is given as the strain
energy, minus the potential of the external loads, W, U being
the strain energy.
If we invoke the stationarity of pi, and we use the
essential boundary conditions, which are the displacement
boundary conditions, then we can derive the governing
differential equations of equilibrium, and the force
boundary conditions, the natural boundary conditions,
as I have shown in the last lecture.
Well, we will not derive these boundary conditions and the
governing differential equations in this approach,
but rather what we do is we invoke this principle, we set
del pi equal to 0, and that gives us the principle of
virtual displacements.
And it is this principle which is a starting point all our
finite element analysis.
Let's recall once what does it mean.
Well, here we have the body forces that I applied to the
body, the surface forces that I applied to the body.
These are externally applied loads and concentrated forces
that are also applied to the body at the points, i.
These forces are in equilibrium with
the stresses, tau.
Let's assume that we know the stresses, at this point.
Then the principle states the following--
if we subject the body to any arbitrary virtual
displacements, listed in here--
and I'm saying any arbitrary virtual displacements--
excuse me, that however satisfy the essential boundary
conditions, and that means just the displacement boundary
conditions.
Then the work done by the loads, and that total work is
given here.
This is a virtual work because we are taking virtual
displacements and subject the forces to these virtual
displacements.
Then the external virtual work done is equal to the internal
virtual work done, which is obtained by multiplying the
real stresses, which are in equilibrium with these
extraordinary applied forces.
Multiplying the real stresses by the virtual strains, which
correspond to the virtual displacements.
So let me use here a different color.
These virtual strains correspond to these virtual
displacements, and of course, these virtual displacements
over the body--
these are, of course, a function of x, y and z.
These virtual displacements over the body give us also
virtual displacements on the surface of the body, which are
listed in here.
So let us put another arrow in there.
And these virtual displacement also give us concentrated
virtual displacements at those points where we have
concentrated load supply.
So once again, if we take the body and subject that body,
who is in equilibrium under Fb, Fs, and Fi, with tau--
tau being the real stresses.
If you take that body and subject it to any arbitrary
virtual displacement that satisfy the displacement
boundary conditions, then the external virtual work is equal
to the internal virtual work.
The internal virtual work being obtained by taking the
real stresses, times the virtual strains, which
correspond to the virtual displacements here, and
integrating that product over the volume of the body.
And the external pressure work is obtained by taking the real
forces, multiplying these by the virtual displacements, and
integrating these contributions over the
complete body.
Physically, what does this mean?
Here, we have again, our general body.
Let's see once, pictorially, what we're doing.
Well, let's take a certain virtual displacement, which I
depict here.
Now, here, we have a boundary condition, so this point, P
can only move over to there.
It could not move this way because we have to satisfy, in
the virtual displacements, the actual displacement boundary
conditions.
Here, the point cannot move at all, and here, this point can
also not move.
So a typical set of virtual displacement
might look like that.
Just sketched in here.
This roller out has moved over there.
So there's our new roller right there.
What I satisfy are the order displacement conditions.
Only horizontal movement was possible here.
No movement here, no movement here.
The virtual displacement vector here is that one. u
bar, for a particular point.
And that is the point that I'm looking at.
Then what the principle says, once again, is that if I take
these virtual displacements, multiply them by the real
forces, integrate that product over the total body--
that is my external virtual work, and that external
virtual work shall be equal to the internal virtual work,
which is obtained by taking the real stresses, which are
in equilibrium with these externally applied loads.
And multiplying these real stresses by the virtual
strains that correspond to these virtual displacements,
and integrating that product, as the internal virtual work
over the whole body.
This is an extremely powerful principle and an extremely
important principle, and provides a basis of our finite
element formulation.
In our finite element analysis, we are proceeding in
the following way-- we say, well, let us idealize this
complete body as an assemblage of elements, and what I've
done here is to draw one typical element.
This is an 8-node element, a brick element, a distorted
brick element, to make it a little bit more general.
It is an 8-node element because we have four nodes on
the top surface, and four nodes at the bottom surface.
There's another node here.
This element here undergoes certain
displacements, of course.
And what I will be doing is I will express the displacements
in that element as a function of the letter coordinates
system, x, y, and z.
The displacement in the element being
lower u, v, and w.
If we idealize the total body as an
assemblage of such elements--
in other words, there's another element coming in from
the top, and another element coming in from the sides, from
the four sides, and another element
coming in from the bottom.
So if we idealize the total body as an assemblage of such
brick elements that lie next to each other, et cetera.
And we express the displacement in each of these
brick elements, as a function of the nodal point
displacements, of the displacements of the corners
of the bricks, then we can, of course, express the total
displacement in the body as a function of the nodal point
displacement.
And that is the important step in the
finite element analysis.
That the displacements in each of these sub-domains or
elements are expressed in terms of nodal point
displacements.
These corner nodes, as shown here, for the brick element.
And then since the total body is made up of an assemblage of
such brick elements, we can express the total displacement
in the body as a functional of these nodal point
displacements.
And invokes the principle of virtual displacements.
Now let's go into the actual specifics.
Well, for element m, this might be element 10, m in that
case would be equal to 10.
Then we have the following relationship--
and this is the important assumption of the finite
element discretization.
We say that the displacements--
there are three displacements, U, V, and W, of course, now.
For element m, U, V and W are listed in this vector u, are
equal to a displacement interpolation matrix, Hm,
which is a function of x, y, and z, times the nodal point
displacements.
And what I'm listing in this u hat vector are all the nodal
point displacements that I've called in the finite element
discretization.
For this brick element here, we have eight nodes, and 24
nodal point displacements.
At each node, we have U, V and W. But notice, once again,
there are other brick elements on top of it, on the sides,
and below of it.
And each of these brick elements, of course, has a set
of such nodal point displacements.
We notice, however, that the element below it here--
if I take my pen here, and draw in another element, we
notice that that element has the same
node as the top element.
In other words, this node here is common to this top element
and the bottom element.
And that's where we have the coupling between elements.
We will see that more distinctly later.
So what I'm doing here is I express the displacements of
element m as a function of all the nodal point displacements,
and I'm listing here in u hat these displacements for N,
capital N nodal points.
We have this vector.
Now, in general, later on, we would simply call all of these
components, Ui's, and so our u hat here, would be
written in this way.
Notice I use the transpose, the capital T here, to denote
the transpose of a vector.
So this UN is equal to that W, capital N. That's just for
ease of notation.
This is our major assumption.
This is the major assumption in the
finite element analysis.
We will have to define for each element this displacement
interpolation matrix.
We notice that when we define it, there will be many columns
that are simply 0's because only certain displacements in
this vector listed here, in this vector, really affect the
displacement in an element.
In other words, typically, for this element here, if we look
at this node, then the displacement at this node do
not affect the displacement in this element because this node
does not belong to the element.
That is recognized by the fact that in this Hm matrix there
will be many columns that are simply 0's.
In fact, the only non-zero columns in this Hm matrix are
those that correspond to nodal points in this vector, or
nodal point displacements in this vector that
belong to element m.
Well, having laid down this assumption, and we will
define, once again, later on the Hm matrix
for specific elements--
1D, 2D, 3D elements and so on.
Having laid down this assumption, we can derive the
strains and the strains are simply given by this
relationship, with the Bm matrix is the strain
interpolation matrix, of element m.
Notice that the rows in this matrix are obtained by using
the rows in the Hm.
And differentiating these rows and combining these rows in
the appropriate way.
I show examples later on.
There is no more assumption in this step.
This Bm matrix is simply obtained from the Hm matrix.
By recognizing what strains we are talking about, and by
recognizing that we can simply use the rows here,
differentiate them, linearally combine them, if necessary, to
obtain the Bm matrix.
Of course, we also have to use our stress strain law to
obtain stresses from the strains, and the stresses in
element m are given as shown here.
These are the strains that I talked about here already.
This is our stress strain law, which can vary
from element to element.
I also introduce here an initial stress, which might
already be in the body.
This might be due to overburden pressure in an
underground structure, as an example, and so on.
So, this is our stress strain law, which we have to satisfy
for the body, of course.
Our compatibility conditions in the analysis
will also be satisfied.
The strain compatibility conditions are satisfied
because we are deriving the strains from continuous
displacements, within the element.
We'll impose on to the different elements that they
remain compatible under deformations.
By that, I mean if we have an element coming in here and
another element going out there that the elements
underloading, when the top element here, and the bottom
element, both of them are loaded.
No gap is opening up here so that displacement
compatibility between the elements is satisfied.
So if we look at the three conditions that we have to
satisfy in an analysis--
the first one being the stress strain law.
That is satisfied because we are using this equation.
The second one being compatibility.
That is satisfied because we using this relationship here
to calculate our strains from the Hm, via the Bm matrix.
And we are satisfying, of course, that the elements
remain together, so no gaps opening up.
We will be talking about it later on when we talk about
the convergence requirements also of
finite element analysis.
And finally, our equilibrium condition has to be satisfied.
That is the third condition where that equilibrium
condition is embodied in the principle of virtual work.
Here, I have it once again written down.
The equilibrium conditions are in this
principle of virtual work.
And as I stated earlier, that if this equation is satisfied
for any and all the arbitrary virtual displacements that
satisfy the displacement boundary conditions, the real
displacement boundary conditions, then tau is in
equilibrium with Fb, Fs, and Fi.
Well, what we will be doing is we will be applying this
principle of virtual displacements for our finite
element discretization, which means that in an integral
sense, we satisfy equilibrium.
However, if we look into an element, then within the
element, we will only satisfy the differential equations of
equilibrium in an approximate way.
We will not satisfy them exactly.
However, if we have a proper finite element discretization,
and by that, I mean if we satisfy all the convergence
requirements that have to be satisfied in order to obtain a
valid solution, or a reliable solution in a finite element
analysis, then we know that as our elements become smaller
and smaller and smaller, we will be finally--
and always, of course, applying the principle of
virtual displacements--
we will be finally satisfying, also, the differential
equations of equilibrium locally within each element.
So stress strain law is satisfied, compatibility is
satisfied, both of them exactly.
The equilibrium requirements are only satisfied in an
integral sense, if we have a coarse finite element mesh,
but as the finite elements become more and more, as we
refine our finite element mesh, we will be satisfying
the equilibrium requirements.
Also locally, within an element, always closer and
closer, and we will be approximating, or we will be
getting closer to the satisfaction of the
differential equation of equilibrium.
The first step now is to rewrite this principle of
virtual displacement, in this form, namely as a sum of
integrations over the elements.
There's no assumption yet.
All I've done is since our total body is idealized as a
sum of volumes, namely the volumes over the elements, I
can rewrite the total integral as a the sum
over the element integrals.
And that's what I have done here.
Notice we have now here, m, denoting element m and we are
summing over all of the elements.
There's no assumption here yet.
Now, however, I can substitute our assumption.
Namely, that Um is given in this way, and epsilon m is
given that way.
Once again, this is actually not an assumption.
This is the major assumption.
That epsilon m follows from this assumption.
This equation follows from that equation entirely.
Substituting now from here, these equations into the
principle of virtual displacements, we directly
obtain the following equations.
Here, I have on the left hand side--
let's go through this equation in detail.
On the left hand side, I have the following part.
I have an integral over all of the elements, that is the
integral here, summing over element m and integrating over
each of the element.
This part here, Bm transposed, times U hat T, is equal to the
epsilon bar, mT.
So notice that our epsilon bar m, is a virtual strain, is
given in this way.
Before, I talked only about the real strains.
In other words, the bar was not there.
I put that now into brackets here.
This bar was not there.
Well, what we're doing in the finite element analysis is to
use the same assumption for the virtual strains as we use
for real strains.
In other words, we use this equation without the bar and
with the bar, on top of the epsilon and on
top of the u hat.
So here, we have our epsilon bar transposed.
This part here is the stress.
Tau m equals Cm epsilon m--
that is our cm.
The Bm here comes in from the epsilon m.
So Bm times U hat is equal to epsilon m, once written down
again here.
So this total part here is nothing else than--
let's go back once more to the previous view graph--
nothing else than this part here, than that part there.
But with the initial stress, tau i, m being on
the right hand side.
Since we do know the initial stress, we put that one, of
course, on the right hand side.
It is a load contribution.
And here, you see it as a minus sign because we put on
the right hand side, and this part here.
This part times this u hat, notice there's
a big bracket here.
That u hat bar here operating on that Bm transposed, there's
a transposed here, too, gives us again the epsilon bar m
transposed.
So let us look now at what we have on the right hand side.
On the right hand side, I want to have
discretized this part here.
Well, what we do is we substitute here from our
displacement interpolation, here from our displacement
interpolation, and each of these integrals can directly
be expressed, in terms of the nodal point displacements.
And that's what we have done here.
The Hm times the u hat bar is the u bar m transposed.
Here, we have the u bar s.
That is the u hat bar times the Hsm.
Notice that this part, you should not have been
written that far.
In other words, there is the end.
The u bar s m transposed only goes up to there and it
embodies this Hs m transposed and the u hat bar transposed.
This part we talked about already.
So what we have done then is to rewrite--
this is the important part--
is to rewrite this principle of virtual displacements, in
which we had no assumption yet.
We rewrote this in terms of the nodal point displacements
and element interpolation matrices that we use for our
finite element discretization.
Now, we of course, have the assumption that the
displacements within each element are given by the Hm
matrix, the strains are given by the Bm matrix.
So this is the result that we have obtained.
And at this point, we now invoke the principle of
virtual displacements.
Since this principle here shall hold for any arbitrary
virtual displacements that satisfy the displacement
boundary conditions, we can now invoke
this principle n times.
And by that, I mean, once by imposing a unit displacement
at the first displacement degree of freedom, and leaving
all the others 0.
Second time around, imposing a unit displacement at the
second degree of freedom, all the others being 0.
Third time around, imposing a unit displacement at the third
degree of freedom, all the other displacements
being 0, and so on.
And that really amounts to then saying that this vector
here becomes an identity matrix.
And similarly, this one here becomes an identity matrix.
And therefore, we can take those two out, and our
resulting equation then is simply what we have left.
Taking these two identity matrices out, and that is our
finite element equilibrium equation, or I should rather
say that these are n finite element equilibrium equations,
namely corresponding to the n nodal point displacements that
we are considering.
In general, what one does most effectively is to really
derive these corresponding to all displacements.
Having removed the boundary conditions, and then later on,
one imposes the known bounded displacements.
And that's what I want to discuss a little bit later in
more detail.
So this is the result then that we obtained.
And we obtained really in shorthand, Ku equals r.
Where K is this matrix.
It's the structural stiffness matrix.
Notice that I'm summing here over the elements.
This being here, the element stiffness matrix.
Notice that this element stiffness matrix here is an n
by n matrix, has the same order as this K matrix.
However, we will recognize that a large number of columns
and rows in this matrix are simply 0.
In fact, all those columns and rows are filled with 0's that
do not correspond to a nodal point displacement degree of
freedom of element m.
I show you later on some examples.
However, by using this Bm here, and making the element
stiffness matrix of the same order as this total structural
stiffness matrix, we can directly sum over all of the
elements stiffness matrices.
And that, of course, is our direct stiffness procedure,
which already I pointed out to you earlier.
The direct stiffness procedure means that we are adding the
element stiffness matrices into the total stiffness
matrix via this summation here.
So the Km here, being what I have here in the blue
brackets, must be, of course, of the same order as K in
order to be able to do that method
theoretically, at least.
Later on, we will see that we indeed only work with the
non-zero rows and columns in the K matrix, and then use
connectivity arrays to assemble Km effectively into
the actual K matrix.
For the RB vector, we have this part.
Once again, we're using the direct stiffness procedure to
add the contributions of all the elements in order to
obtain the total RB vector.
Once again, the rows now, or rather, the elements because
this of course, is a vector here of n long now.
Those elements that do not correspond to nodal point
degrees of freedom will all be 0's.
The RS vector, similarly, is obtained as shown here.
Now, we of course, sum the element contributions as they
arise from the surface forces and the RI vector is obtained
as shown here, and the concentrated load vector is
simply a vector listing all the concentrated forces in F.
Notice that this HSM matrix here is directly obtained from
this Hm matrix.
Sometimes one has difficulties visualizing what this
matrix really is.
Well, we will see later on if this is an element here, and
our coordinate system, say, lies in that element this way,
then the Hm matrix, of course, gives us a displacement within
the element.
Whereas the HSM matrix gives us, say, the displacement on
this surface element, if it is that surface
that we want to consider.
Now, to get the displacement on the surface of the element
when we know the displacement within the total volume of the
element, well, what we simply have to do is we have to
substitute the coordinates of the surface in the Hm here to
obtain the HSM.
I will show you later on some examples.
Now, in dynamic analysis, of course, the loads are time
dependent and if we are considering a truly dynamic
analysis, then we have to include inertia forces.
And the inertia forces can directly be taken care of, or
can directly be included in analysis if we use the
d'Alembert principle.
Here, we have the body loads, which are the externally
applied forces per unit volume.
And if we split these up into those forces that are
externally applied, and those that are arising due to the
d'Alembert forces, as shown here.
Then we directly have the inertia
effect in the analysis.
We now can, of course, express our accelerations in the
element in terms of nodal point accelerations again, and
we are using here the same Hm matrix that we use already for
the displacement interpolations.
If we substitute from here and here into the RB which I had
written down here.
If we substitute into this RB here, this equation for fB,
then we directly can write down this equation here, M U
double dot plus Ku equals R, where the M matrix now is
obtained as shown here.
Notice that this R vector now only contains this RB part.
In other words, not anymore the fB here, but
rather an fB curl.
Notice also that in this analysis now, or in this view
graph, I've dropped the hat on the u.
These are the nodal point displacements, these are the
nodal point accelerations.
I've dropped the hat just for convenience.
I had already dropped it actually here also.
We earlier had the hat there.
Here, we had the hat still because I wanted to
distinguish the actual nodal point displacements from the
continuous displacements in the structure, or in the body.
So here, the hat still being there.
And here, I dropped the hat already, just for convenience
of writing, and from now on, when we have this vector, U
here, then that means that we're talking about the
concentrated nodal point displacements, or the actual
note point displacements of the finite element mesh.
I mentioned earlier that it is most convenient to include in
the formulation all of the nodal point displacements,
including those that actually might be 0.
In other words, for our three-dimensional body, to
make a quick sketch here.
If we have here our support in an actual analysis, it is most
effective to say well, let us remove the support and assign
a node there with three unknown displacements.
Once we have derived these equations of equilibrium, of
course, we now will have to impose the fact that the
displacements are 0's there.
And that is then done effectively, for example, as
shown here.
We have here the general equations, M U double dot plus
Ku equals R. And what we are doing is we are listing the
displacements and accelerations into vectors, U
double dot A, U A, and U double dot b, and Ub, where
the b components of the displacements and
accelerations are known.
And now they might be 0, as I showed here in this particular
example, or they might be actual values
that we want to impose.
If they are known, well, we can look at the first
equation, as shown here, and put all the known quantities
on the right hand side, substitute for Ub and U double
dot b, and we know then the right hand side load vector.
Thus, we can calculate Ua, and U double dot a.
Having now calculated the velocities, the accelerations,
and displacements, we can go back and get the reactions.
The reactions, of course, being Rb.
Here, I assumed that the displacements which we are
talking about in the vector here are actually the ones
that we also might want to impose.
Well, in some cases, of course, we might have defined
in our finite element formulations the U and V
displacement, as shown here.
But a displacement that we want to impose is actually
this one here, namely, that one might have to be
restrained, and this one here might have to be free.
In that case, if our finite element formulation has used
the U and V displacement, we have to make a transformation
as shown here.
A well known transformation from the U to the U bar
displacements, and this, in a more general sense, is written
down here once again.
When we have many more degrees of freedom, our T matrix would
look as shown here.
It's an identity matrix with the little transformation
matrix that I've shown here.
The cosine minus sine sine cosine matrix.
Now, put into the appropriate rows and column.
The i'th column, j'th column, i'th row, and j'th row would
carry these 2x2 matrix.
And otherwise, we just have 1's on the diagonal.
So this is the more general transformation that we are
using when we have many more degrees of freedom than just
the two that we want to modify.
Substituting from here into our equations of equilibrium M
U double dot plus Ku equals R. We directly obtained this
equation, where M bar now is shown here, K bar is shown
here, and R bar is shown here.
Let me mention here that this looks like a former matrix
multiplication.
In fact, two former matrix multiplications.
M times T, And then the product should be taken times
T transposed, pre-multiplied by T transpose.
Well, in actuality, of course, all we need to do is combine
rows and columns.
The i'th and j'th rows and columns to obtain directly our
M bar matrix, similarly for the K bar
and the R bar matrices.
Another procedure that is also used in practice--
can be very effective--
is an application of the penalty method.
In this procedure, we impose, basically, physically a spring
of very large stiffness, where K is much larger
than K bar i i.
And then we supplement our basic equations that are shown
here by this equation here.
So if this K is much larger than K bar i i, and if we
supplement this equation or add this equation into this
equation here, then we notice that the spring stiffness will
wipe out basically the other stiffnesses that come into
this degree of freedom, and our solution will simply be
that U i is equal to b, which is the one that we want.
So this penalty method can be used to impose displacement
degrees of freedom, and it really physically amounts to
adding a spring into the degree of freedom where we
want to impose a certain displacement.
And it's important, however, that if we use that method
that we are always dealing with single degree of freedoms
being imposed.
And by that, I mean the following--
if we had a system, like this one here, and our original
displacement degrees of freedom are these, then we
first have to make the transformation onto that
finite element element system to these
degrees of freedom here.
And now we add our spring in.
In other words, we want to add our spring into this system of
equations because now there's no coupling from this degree
of freedom that we want to be impose into other degrees of
freedom, through that spring.
This spring only enters on the diagonal, and now it is a
numerically stable process.
However, if we were not to perform this transformation--
in other words, if we were still to deal with these
degrees of freedom, and then add our spring in--
of course, that spring now would introduce coupling
between these two degrees of freedom, and numerical
difficulties may arise in the solution.
So basically, then in summary, if we do have degrees of
freedom to be imposed, we first go through this
transformation to obtain the M bar, U double dot bar, K bar,
U bar equals R bar system of equations where the
displacement that we're talking about are containing
those displacements that we actually want to impose.
We then can impose these displacements using the
penalty method, which is this one.
Or we can impose these displacements using the more
conventional procedure, using, in other words, simply this
procedural of imposing Ub and rewriting the equations into
two equations.
The first equation we solved for the Ua now, of course, in
this particular case, we would now have all bars on there.
And if we have done a transformation, and in the
second equation then, we obtain the reactions.
Let me now go through a simple example to show you the
application of what I have discussed.
This is a very simple example, but a very
illustrative example.
In particular, it is also the example that we talked about
already earlier in lecture 2, when we did a Ritz Analysis on
this problem.
In fact, our finite element analysis that we are now
pursuing, using the general equation that I presented to
you is really nothing else than a Ritz Analysis.
And in fact, if you look at the earlier
solutions that we obtained--
solution 2, in the Ritz analysis, corresponds to the
finite element solution that I will be
discussing with you now.
So here is a problem once again.
We have a bar of unit area, from here to there, and then
of changing area, from here to there.
The length here is 100, the length here is 80.
The bar in actuality, is supported here, but as I
mentioned earlier, we remove that support in our finite
element formulation, and introduce, in fact, a
displacement degree of freedom there.
So here, I want to put down the first node.
Now, the first step in any finite element analysis must,
of course, be the step of idealizing the total structure
as an assemblage of elements.
And there are generally choices--
how many elements to take, what type of element to
take, and so on.
In this particular case, I know that there's a
discontinuity in area here and for that reason, intuitively,
I will put one element from here to there
with a constant area.
Also, let us consider for the moment, this also as being one
element, and this then will correspond to the Ritz
Analysis that we performed earlier.
Notice that we have here a bar of unit area, a bar of
changing area.
This total bar assemblage is subjected to a load of 100, a
concentrated load of 100, as shown here.
The only strains that this bar can
develop are normal strains.
In other words, if a section originally is here, that
section we move over a certain amount and by that amount.
That is U. In the coordinate system that we are using, the
y-coordinate being in this direction, this is the
y-coordinate here.
This would be our U of Y. However, since we are dealing
with two elements to analyze this bar, what I will do is I
will introduce a little coordinate system here.
And I used little y in this particular case.
So we put a little y here.
There's also, for this element, a little y.
And the area, in this particular element, is given
as 1 plus y divided by 40 squared.
So this is the changing area in this domain.
However, also, remember please, if in our analysis, if
an original section in this area was vertical like that,
after deformations, due to the load here, it will still be
vertical, and now our displacements in this element
will also be given by a u.
And that u is a function of--
if we look at this little y, if you use that little y of
that y as this u here is a function of this y.
This y corresponds to the y in this element, that y
corresponds to the y in this element because we use
different coordinate systems for each element.
Now, this is actually an important point that we can
use for each element, a different coordinate system.
We could use Cartesian coordinate systems for each
element, different ones.
In fact, that is most generally done.
If we have specific geometries, we might use
cylindrical coordinates systems for certain elements,
Cartesian coordinate systems for other elements, and so on.
This is an extremely important point that we can have
different coordinate systems for different elements because
that eases the calculation of the
element stiffness matrices.
So here, our element 1, here our element 2.
And the displacements that we are talking about are U 1 at
this node, U 2 at this node, U 3 at that node.
These three displacements shall give us the displacement
distributions.
Of course, only an approximate displacement distribution in
the complete element mesh.
Two elements make up our element--
complete element idealization or complete element mesh.
Well, the equation that, of course I will now be operating
on is this one, KU equals R. In this particular case, we
recognize that we want to calculate our stiffness
matrix, K. Here, we have two elements, so M, in this
particular case, will be equal to 1 and 2.
We don't have a body force vector, we don't have a
surface force vector, we don't have an initial stress vector.
However, we have a concentrated load vector.
So what I will want to do then is calculate our K matrix, and
establish our concentrated load vector.
The K matrix embodies the strain displacement
interpolations, which are obtained from the element
displacement interpolations.
Now, as I already pointed out, we have
now here, two elements.
The first element shown here, second element shown here.
Notice that the U1, U2 correspond to these two
displacements, U1, U2.
The U2, U3 correspond to these displacements, U2, U3.
Notice that there's a coupling between the elements because
U2 is here a displacement of that element, and is here the
displacement of element 2.
The length of the element 1 is 100, length of
element 2 is 80.
Well, if we have two displacements to describe the
displacement in an element, then we recognize immediately
that all that we can have is a linear variation in
displacement between the two end points,
between these two nodes.
And so the element displacement interpolations
must involve these functions.
For unit displacement at this end of an element, Y over L is
the interpolation of the displacement.
For unit displacement at this end of the element, this is
the interpolation.
Notice that the actual displacement, of course, goes
into this direction, but I'm plotting it upwards to show
you the magnitude of that displacement.
If we have established these interpolation functions,
recognizing, of course, that for element 1, L is 100, for
element 2, L is equal to 80.
Then we can directly write down our H1 and H2 matrices.
They are given as shown here.
Notice that Um is equal to Hmu, where U is equal to--
I write down the transpose--
U1, U2, U3.
Now, we notice that element 1--
let's go back once more for element 1.
Only U1 and U2 influence the displacements in that element.
And that is shown here.
U3 has 0 and does not influence the displacement in
the element.
Similarly for H2, U1 does not influence the displacement in
that element.
So we have these displacements.
Notice also that I've written here, Um, of course, but that
Um here, for our specific case is simply this
displacement, Vm.
We have only one displacement component.
Taking the derivative of these relations here, we get
directly the strains.
Notice that here, we should have probably put an m there.
This is the normal strain and we obtain these matrices by
simply taking the derivatives.
Well, now we have the components that we need to
evaluate the K matrix.
And as I mentioned earlier, that is the total K matrix
obtained by summing the
contributions over the elements.
This is coming from element 1, this is coming from element 2.
Notice that the area is 1, Young's modulus of stress
strains law.
We are integrating from 0 to 100.
This is here, B1 transpose, that is B1.
This is here, B2 transpose, that is B2.
This is the area that I pointed out to you earlier.
Evaluating these two matrices, we directly
obtain these matrices.
And notice the following--
that there's no coupling from the third degree of freedom
into element 1.
Similarly, there's no coupling from the first degree of
freedom into element 2.
In fact, what we will do later on is simply calculate the
non-zero parts.
We call these the compacted element stiffness matrices.
And knowing these non-zero parts and knowing into which
degrees of freedom they have to put in the assemblage phase
to obtain the total stiffness matrix, we can directly
assemble the stiffness matrix.
In other words, if I know this part here and I know that the
first column corresponds to the first column of the global
stiffness matrix, the second column corresponds to the
second column in the global stiffness matrix, then I can
just add this contribution into this part here.
Similarly, I can simply add this contribution here into
that part there, without carrying
always these 0's along.
And that is, of course, a very important
computational aspect.
However, in theory, we are really still performing these
additions as shown here, we really still perform the
additions as we pointed them out in the
direct stiffness procedure.
In other words, we still perform this summation, as I
pointed out to you earlier.
The important point, however, is that we now have
established the K matrix, corresponding to the system.
Our R vector is simply, in this
particular case, 0, 0, 100.
Because we only have 100 applied at the
third degree of freedom.
We now have to impose that U1 is 0, we simply set U1 equal
to 0 in the equilibrium equations, as
I pointed out earlier.
We solve for U2 and U3, and having obtained U2 and U3, we
know the displacement in each of these parts, and we know,
therefore, the strains and the stresses in
each of these parts.
The solution is plotted in the example that I discussed with
you in lecture 2.
This example, really, showed some off the basic points of
finite element analysis.
Of course, we have to discuss much more how we actually
obtain the Hm matrices for more complex, more complicated
elements that I use in actual practical analysis.
However, this is all I wanted to mention in this lecture.
Thank you for your attention.