The following content is
provided under a Creative
Commons license.
Your support will help MIT
OpenCourseWare continue to
offer high quality educational
resources for free.
To make a donation, or view
additional materials from
hundreds of MIT courses, visit
MIT OpenCourseWare at
ocw.mit.edu.
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.
