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PROFESSOR: Ladies and gentlemen, welcome to lecture
number seven.
In this lecture, I would like to present to you the
formulation of structural elements.
We will be discussing beam, plate, and shell elements, and
I would like to introduce to you the isoparametric approach
for interpolations.
There are two approaches in the
formulation that we can follow.
The first one is a strength of materials approach, in which
case we look at a straight beam element, we use a beam
theory including shear effects.
If you look at a plate element, we use a plate theory
including shear effects, also.
The names associated with these theories that we're
using are the names of Reissner and Mindlin.
In the second approach, we have the continuum mechanics
approach, in which we use the general principle of virtual
displacement, but we exclude the stress components not
applicable.
For example, in a plate, we set the stress through the
thickness of the plate equal to zero.
In addition, also, we have to impose in the use of the
principle of virtual displacement the kinematic
constraints for particles on sections or originally normal
to the mid-surface.
Namely, we have to put the constraint into the structure
that the particles remain on a straight line during
deformation.
Well, as examples, I've plotted here, I've shown here
two structures, a beam and a shell.
Let's look first at a beam.
In this case, we have that the original particles normal to
the mid-surface, or the neutral axis of the beam, are
on this orange line.
I've shown here a large number of particles.
The kinematic constraint that we're talking about is that
during deformations, these particles remain
on a straight line.
They move over to the yellow line here.
In other words, point A goes to point a prime.
Another particle here goes over to this particle here.
This particles goes over to that particle, and so on, and
these particles remain on a straight line.
That is the basic kinematic assumption.
However,
We should also notice that there's a right angle between
the mid-surface, or neutral axis off the beam, and this
line of particles initially.
This right angle is not preserved during deformation.
In other words, this angle here is not a right angle
after deformations anymore.
In the case of the shell, the kinematic
constraint is quite similar.
Here, we have now the mid-surface of the shell shown
as a dashed line.
The particles on a line normal to that mid-surface are shown
here again in orange.
This is the initial line.
And during deformation, these particles remain
on a straight line.
Now they have come to be the yellow line here, and we
notice that, again, there's a right angle initially here,
but that right angle is not preserved during deformations
because, in each case, we are including shear effects.
Here, we include shear effects, and similarly here,
we also include shear effects.
Well, I've prepared some view graphs to show these facts a
little bit more distinct.
Here, we have a first view graph on which I show the
assumptions of the basic Bernoulli-Euler beam theory
that is used in the development of conventional
beam elements.
We have here the original beam element with its neutral axis
in a dashed line.
And that beam element, during deformation, becomes this.
It goes into this shape here.
We notice-- and this is important--
that this section here, which I now mark in blue, goes over
into that section, and that the displacement is w at the
mid-surface, and that the slope here at right angles to
that section is nothing else than dw dx.
In other words, this angle here is really nothing else
than that angle, dw dx.
This is the Bernoulli-Euler beam theory excluding shear
deformations.
The important point is that when we use this beam theory,
we have to match between two elements, w, in other words,
the displacement at the mid-surface has to be the same
for element one and element two.
And in addition, this slope has to be the same for both
elements, dw dx for element one on the left-hand side must
be equal to dw dx on the right-hand side.
This is the conventional beam theory that is used to develop
the Hermitian beam elements.
And what I'd like to introduce to you now is the beam theory
that we're using in the isoparametric formulation,
namely in the development of modern beam elements, pipe
elements, shell elements, and plate elements.
I should say at this point that the conventional
Hermitian beam element that you're probably familiar with
is more effective in engineering analysis than the
beam element that I'm talking about here when we just look
at a straight beam.
However, I want to look at the straight beam here as an
example to introduce to you the formulation of the
structural elements that I'm talking about.
The formulation is very well displayed, very well
demonstrated.
The basic features are very well demonstrated looking at
the beam element, although the application of this
formulation to a straight beam element is not as effective as
just the usage of a Hermitian beam element.
However, if we talk about curved beam elements, pipe
elements, then this formulation
is indeed very effective.
And, of course, for plates and shells, it
is really very effective.
Let me show to you, then, the basic points--
the basic important points--
that are being used in the formulation.
Here we have, again, our original beam element.
The neutral access is shown here, and this beam element
now moves over into this piece, into that shape, during
deformations.
We have a section here, and as I pointed out earlier, that
section has to remain straight during deformations.
In fact, it moves right to that section here.
Now notice that what we're talking about is a slope, dw
dx here, which is this angle here, plus a shear deformation
angle, gamma, and dw dx minus gamma is this angle.
And that is the angle beta.
In fact, this is the rotation of this line of particles.
In other words, we are not talking just about one state
variable, w, as we do in the Hermitian formulation, but we
talk about two state variables now, beta and w.
Beta and w both are independent, and we will see
that, later on, we will interpolate them as
independent quantities.
The important point, then, that if you look at two
elements, if you do interpolate w and beta
independently, then we need, between two elements,
continuity in w and continuity in beta.
We do not talk about continuity in dw dx.
We do not talk about that, and that is the important point,
particularly when we talk about the formulation of plate
elements and shell elements.
The fact that we'll talk about independent interpolations of
w and beta, including shear effects, of course, in an
approximate way, but including shear affects, that fact
alleviates us of many difficulties that
we encounter otherwise.
In other words, that we encounter if we use a
classical plate theory excluding shear deformations.
A good starting point for the development of the elements is
the use of the total potential, pi, of an element.
That total potential I've written down here, and we have
set pi is equal to a contribution from the bending
part plus a contribution from the shearing part, or the
shearing deformations, and, of course, there is the external
work due to distributed pressure, p, on a beam
element, and moments, externally applied moments, m,
onto the beam element.
Now, notice the quantities that I'm using here.
The bending part is given by dw dx squared, of course with
the flex or rigidity in front, and this part here is given
only in terms of the section rotation beta, which is
independent of the translation of the neutral axes, w.
Here, we have the shearing part, dw dx, minus beta As,
the shear strength.
And they have been written down here once more.
If we look at this equation, here we get dw dx minus beta
is equal to gamma.
The other quantities, of course, that I used in the
derivation of this--
pi, the stress being equal to v over as, v being the shear
force on the section, As is the shear area.
Notice that we are assuming the shear strength through the
thickness of the beam element to be constant.
They are constant because of this equation here, basically.
w, of course, varies along the length of the beam.
Beta varies along the length of the beam, but that means
gamma is constant through to the thickness of the beam.
Now, since gamma is constant through the thickness of the
beam, we have to say, of course, that also our shear
stress is constant through the thickness of the beam, and we
have to introduce a shear correction factor, k, which is
equal to As over A, where As is an equivalent shear area.
Using these quantities we obtain this pi functional,
where, once again, we simply add the bending contribution,
bending strain energy to the shear strain energy, and we
subtract the total potential of the external loads.
If we invoke the stationarity of this functional, in other
words, we invoke that delta pi is equal to zero, we obtain
the principle of virtual work, or principle of virtual
displacement, which I have discussed with you in an
earlier lecture.
The result of invoking that del pi is equal to zero, or
invoking the stationarity of pi is this equation here.
Now notice that in this equation we have, if we want
to interpret it once physically here, basically the
real stress part, here, the virtual strain part.
Similarly here, the real stress part and the virtual
strain part, and, of course, the virtual work of the
external loads.
The important point is that we only integrate along the
length of the beam, and not through the thickness anymore,
because we talk about quantities stress resultants
over the thickness.
Well, once we have arrived at this equation, we can proceed
in much the same way as we have been proceeding in the
development of continuum isoparametric elements.
Here, we look at a particular case.
Let us say we have a beam elements such as shown here.
Of course, this is the loading applied, P. The bending moment
that they're talking about here is shown here.
It is a distributed bending moment over
any part of the beam.
Similarly, P is only applied over a
certain part of the beam.
The depth of the beam is b, the width of the beam is a.
As a shear factor for a rectangular beam, the sheer
fact k that I introduced to you briefly is 5/6.
Of course, I, the moment of inertia, is ab
cubed divided by 12.
The interpolations that we would use for such a beam
element are one dimensional interpolations.
We only integrate along the length, x.
And the one dimension interpolations we discussed
already earlier.
We simply use the same that we have been using already in the
formulation of truss elements.
Two point interpolation would be this element here.
Here, we use a three point interpolation.
Notice--
and this is important--
that we're talking about w and beta, the section rotations,
as independent quantities.
So for a three point, or three nodal point beam, we would
have, in just a plane analysis, we would have six
degrees of freedom.
For a cubic element, we have eight degrees of freedom.
Of course, for a Hermitian beam element, we would have
only these degrees of freedom and those degrees of freedom,
w and dw dx here, and w, dw dx here.
So, that is the reason why the Hermitian beam
element is more effective.
However, we can use these interpolations directly to
develop curved beam elements, pipe elements, and then, of
course, this approach is-- even for beam elements--
more effective, as I mentioned earlier.
Well, let us then write down the basic interpolations that
we are using.
Here, we have w being the sum of hi wi.
hi, we discussed earlier already.
And these, of course, are nodal point transverse
displacement.
These are the nodal point rotations of
the sections, beta.
In other words, I could have used here, as a notation, beta
i, but I chose to use theta i.
If you write these equations in matrix form, we directly
obtain this relation here, where hw simply list hi, u
lists the displacements and section rotations, and similar
here, beta is given in terms of an h beta matrix.
We take the differentiations of hw and h beta, and we get
dw dx equal to Bw times u d beta dx equal beta times u.
Once we have the principle virtual work established for
the element that we're considering and have chosen
our interpolations, the approach of developing the
stiffness matrices, the load vectors, is exactly the same
as in the development of continuum elements.
Well here, I have written down the various quantities.
U transposed lists, as I said earlier, the displacement
vector nodal points and the rotation vector nodal points.
Hw simply gives the interpolation functions, q, of
course, being equal to the number of nodal
points we are using.
And here, we have H beta.
The Bw is given right here.
Notice our J inverse comes in there because we have to
transform from r to x, x being the actual physical coordinate
along the length.
Similarly, B beta being given here.
Again, the J inverse here to transform from r to x
coordinates.
Once we have written down these, we can directly
substitute into the principal of virtual work, and we come
up with the stiffness matrix, given here, and the load
vector given here.
Let me point out a few important things here.
Of course, this B beta matrix here-- just to remind you--
comes from d beta dx.
It's in fact, really, d beta dr, but because we're
integrating from -1 to plus 1, however, we have our
determinant J there to take into account the volume
transformation.
Here, of course, we have d beta dx transposed.
This comes from the virtual strains, and that comes from
the real strains.
Of course, we have the stresses.
Here, so these strains times the stress strain law, E,
being the Young's modulus, gives us the stresses.
Here, we talk about shearing deformations.
Notice that we have here dw dx, that is, the
Bw minus the beta.
So we have the derivative in here of w, but no derivative
in H beta because we are simply interpolating here the
beta values.
Again, of course, a transformation to x, and
therefore we have a determine J in there.
The load vector looks just the same way as in the development
of continuum elements.
This is the transverse loading applied P. This is, of course,
interpolating--
this [UNINTELLIGIBLE]
interpolates the virtual transverse displacements.
Here, we have the moment loads, the real moment loads,
and this interpolates here the section rotations, beta, along
the lengths of the beam.
Of course, again, the volume transformation from r to x.
This is really a straightforward application of
what we discussed earlier.
There is one important point, however, now, that I have to
point out to you.
If we consider the functional pi, as I mentioned earlier,
there's a bending part here and there's a
shearing part here.
Notice that in this development I have divided
through by EI over 2, so that I introduce a value alpha
there, and that alpha is really GAK divided by EI.
Now, if we look at that alpha value, and let's look at the
value for a rectangular section, we would see that the
a value, of course, is a times b, and the I value gives us
really an ab cubed, if this is, here, b, and that is a.
An ab cubed over 12, of course.
And we have, also, the GK that gives us these, and of course
an E in front here.
But the important point that I want to now concentrate on is
really this b over b cubed.
We can see that as the element gets thinner and thinner, as
the element gets thinner and thinner, that alpha gets
larger and larger.
If alpha gets larger and larger, this term here will be
predominant.
This term will be predominant.
Now this means, however, that if we want to finally converge
to a beam in which the shear strengths are negligible, in
other words, in which the shear strengths are extremely
small, what we would have to be able to represent in the
formulation is that this value here goes to zero.
And what happens in the formulation, really, is that
as this value gets larger and larger, any error introduced
in the formulation, due to the fact that this is not exactly
zero in the finite element interpolation, that error is
largely magnified.
Is magnified and, in fact, can introduce a very large error
if this value is not zero due to the fact that alpha becomes
larger and larger if the element
becomes thinner and thinner.
In other words, in summary once more, if we are talking
about a beam element that gets thinner and thinner for which
we know the shear strengths should becomes smaller and
smaller, our finite element interpolation must be able to
represent this fact.
Now, if we look at the shear strengths, and this is, of
course here, nothing else then basically the shear strength
squared, we now identify that dw dx minus beta, when
interpolated using our interpolation functions, must
be able to be very, very, very small.
And that is a restriction on the formulation.
So what we have to do, really, is use high enough order
interpolations so that dw dx minus beta can be smaller for
thin elements.
Of course for thick beam elements, that is not really a
constraint because we know that there are shearing
deformations, and the shear deformations can be quite
significant.
However, for thin elements, we must be able to represent the
fact that gamma is small, and therefore, we have to use high
order interpolations.
In fact, the parabolic interpolation is really the
lowest interpolation that one can recommend.
It would be better to use cubic interpolation.
In fact, we use the cubic interpolation in practice.
In that case, this gamma value can be small, and we run into
no difficulties and the element can be
very, very, very thin.
Another approach would be to use a discrete Kirchhoff
theory or reduced numerical integration.
These approaches have been developed
for low order elements.
The discrete Kirchhoff theory approach is very effective.
The reduced numerical integration can also be
effective, but has to be used with care.
In particular, as I will point out in the next lecture, we
have to be careful that we do not introduce spurious rigid
body modes into the system.
The development that I just talked about really is
applicable to an element that has a rectangular section and
the element was also lying in a plane.
We looked at a straight element.
Let us now see how we can generalize these concepts
directly to the formulation of general curved beam elements.
And for that purpose I've shown here--
I'm showing here--
a more general beam element that lies in a
three-dimensional space.
It's still rectangular, however, we could also have a
circular section instead.
In fact, when we look at a pipe element, we do talk
about-- we have, of course, a circular section.
Well, in this particular beam element, notice I'm looking at
node one here, node two there, and generally node three here
and node four here, because we want to pick up the curvature
of the element.
I have this element lying in a three dimensional
space, x, y, z.
Notice that that element has local coordinates r, s, t.
These are the isoparametric coordinates.
And psi, eta, zeta, these are the actual continuous physical
coordinates in the beam element.
We define normals at each nodal point.
The normal in the t direction here is 0 Vt 1.
The normal into the s direction is 0 Vs 1.
And similarly, we have two normals at each nodal point.
Notice that for this particular rectangular beam
element, I have the thickness, a1, here and b1, there, and a2
here, b2 there.
These thicknesses can be different.
Now what I want to use are the same basic assumptions that we
have familiarize ourselves already with when we looked at
the special case of a straight beam element in planar
deformations.
I want to use those basic assumption now in the
development of this more general beam element.
And I want to use a continuum approach.
Well, what we are doing, then, is the following.
We interpolate the coordinates, x, y, and z,
along the beam element in terms the nodal point
coordinates of the nodes that lie on the neutral axes of the
beam plus an effect that comes in due to the
thickness of the beam.
Now let us go and look in detail at the x
interpolations.
The L denotes 0 or 1, 0 being the initial configuration, 1
being the final configuration.
So let's put simply a 0 in there, think in terms of a 0
there, and let's look at the initial configuration first.
Well, here we have the initial x-coordinates of the nodal
points, and there are q of them.
These are the one dimensional interpolation functions, just
the same that we use for a truss element, for example.
Here, we have the t-axis.
This is the t-axis into the direction of the normal into
the t direction.
In other words, this is a t-axis here.
Notice that that t-axis here corresponds
to this normal here.
That s axis here corresponds to this normal here.
So here, we have t/2, and ak being the total thickness of
the beam corresponding to that t direction, hk being the one
dimensional interpolation functions again, and these are
the direction cosines of the normal in the t direction.
Here, I should really say, this is a direction cosine
corresponding to the x-axis, corresponding to the x-axis.
When we talk later about the y and z-axis, then we use the y
direction cosine and the z direction cosine.
This part comes in because the beam basically has a thickness
into the t direction.
Now, we have also to introduce the s direction part, and here
we s/2, the thickness into s direction, the one dimensional
interpolation functions, and the direction cosine in the x
direction of the normal in the s direction.
Well, if we want to find, in other words, the coordinates
of any point in the beam element, and let us look, once
more, back at the beam element.
If I want to find the coordinate of a point, p,
lying in that beam element here-- there's, say, point p--
what I have to do is I have to identify the r, s, and t
coordinates of that point, p, and then substitute these r,
s, and t-coordinates into this part here.
And I would get the corresponding x-coordinate.
I proceed similarly with the y and z-coordinates.
Notice, as I pointed out earlier, we are talking still
about the thickness ak hk here, here, and here, but
we're using the x direction cosines, the y direction
cosines, and the z direction cosines here.
And similarly for the s direction, we are using
similar quantities.
So this is the interpolation of the beam element, and using
these three formerly, we can directly obtain--
we can directly obtain--
the x, y, and z-coordinates in this system of axes of any
point in the beam element.
This is the most important fact.
The interpolations that I've listed here are the starting
point of the development of the strain displacement
matrices and displacement interpolation matrices.
Now, let us identify that if these are the original
coordinates for L being 0, then we can also apply, of
course, after the deformation, the same interpolation, and we
put L equal to 1.
If we subtract the x1 minus zero x, we should get the
displacements, u.
Notice that the displacements, u, are in the directions of
the x-axis.
Well, this is exactly how we proceed.
We use these interpolations for before deformation and
after deformation.
And we subtract these interpolations as shown here
and directly obtain the u, v, and w displacements, of
course, as a functional of r, s, and t.
Notice that if I proceed this way--
notice that if I proceed this way, I have used the basic
assumption that plane sections remain plane during
deformation.
This was the assumption that I pointed out to your earlier.
And we are using it in this general information just in
the same way as in the special application that
I showed you earlier.
Well, having then--
just to refresh your memory, the u, v, and w, in terms of
r, s, and t, of course, via these subtractions, we obtain
directly these equations here.
Notice that here we have, now, the nodal point displacements,
uk, vk, wk, and we have the change in
the direction cosines.
These are the changes in the direction cosines.
Well, these changes in the direction cosines we want to
express in terms nodal points rotations.
And that is achieved as shown on this view graph.
We can express these changes in the direction cosines
directly by taking the cross product of a vector of nodal
points rotations, and I've listed here this vector, this
is a nodal point rotation about the x, y, and z-axis at
nodal point, k.
You're taking the cross product of this vector times
the original normal.
Time the original normal.
And we get, then, the change in the normal, of course, for
the t direction and for the s direction.
If we substitute this relation here, of course, remember that
these quantities are known.
They are given.
So the unknowns now are theta k.
If you substitute from here into our relation here that I
developed earlier, we directly obtain the displacements of
any point, p, in the beam in terms of nodal point
displacements and nodal points rotations.
Because these quantities here now have been eliminated and
have been expressed in terms of nodal point rotations.
Well, now we have all the quantities that we need to
develop our strain displacement matrix for the
beam element.
Remember, all we need are the coordinate interpolations,
which I have developed already, and the displacement
interpolations.
With those two quantities, we can immediately calculate, via
the procedures that I discussed with you earlier,
the strain displacement transformation matrix.
And here it is given.
Notice that we are now talking about strain into the eta
directions.
In other words, the eta, psy, and zeta directions.
These are the directions that I pointed out to you earlier,
which are the physical coordinate
directions along the beam.
Let me show it to you once more, the picture.
The r, s, and t-coordinates are the isoparametric
coordinates.
The eta, psy, and zeta coordinates are the physical
coordinates along the beam.
In other words, for the one dimensional beam that we
looked at, this eta axis was, in fact, the x-axis.
Of course, our x, y, and z-axes are now global
Cartesian axes.
Well then, with that information, we can directly
calculate the strain displacement matrix here.
This is done effectively using numerical integration, as I
will be discussing in the next lecture.
The transformations that are necessary are also done on the
integration point level.
Notice that the uk here lists the nodal point displacements
and the nodal point rotations, and, of course, we have to
also remember one important fact, that for the beam we are
talking about stresses into the eta, psy, and zeta
directions, normal stresses, shear stresses here, that are
related via this stress strain law to the normal strains and
shear strains.
Notice that there's again the shear correction factor, k,
which we want to also include in the formulation because we
have assumed constant shearing strains through the thickness
of the beam, whereas we know that for a rectangular beam,
for example, we have parabolic shear strain distributions if
the beam is straight.
I'd like to now go on with the development of plate elements.
Here, we are talking basically about the same approach.
As I mentioned already once, the beam element that I'm
talking about here--
that I have been talking about-- is really only an
effective formulation when we talk about, and we want to
develop, a curved beam element.
For pipe elements also, in the case of pipe elements, I
should briefly mention that, of course, we have to
introduce, also, an ovalization degree of freedom.
That ovalization degree of freedom interpolates basically
the ovalization along the curved pipe.
That is an additional degree of freedom that has to be
introduced in the curved beam element formulation.
So, the beam element formulation is effective for
curved beams, pipes.
However, it does show the basic procedure that we are
following also in the development of
plate and shell elements.
And here, we have a typical plate element.
In other words, a flat shell.
The u, v, and w displacements are now
interpolated in this way.
Notice u be the displacement into the x direction, v the
displacement into the y direction, w being the
transverse displacement, and again, we're talking about
section rotations.
Beta x being the section rotation about the y-axis.
That is, the beta x section rotation.
Beta y is a section rotation about the x-axis.
So that our v, measuring z positive upwards.
Notice, our [? v4 ?] point here is negative, and that's
why we have a negative sign there.
Well, with that then given, we can immediately develop our
strains by using the strength of material equations that
tell us epsilon xx is del u del x, epsilon yy is del v del
y, et cetera.
And, of course, we're getting also our shear strengths.
Having developed the strains, we also recognize that our
stresses are given in terms of these formulae, where we have
now the stress strain law for planed stress analysis,
because we are looking at the plate as an assemblage of thin
elements, plane stress elements, lying
on top of each other.
The stress through the plate, of course, is 0, and this is,
therefore, the plane stress [? material ?]
that we have been putting in here.
And the z times this vector here gives us the strains.
The shearing stresses are given here, and our functional
pi that I used also for the beam element
already is given here.
Notice that we now have to integrate through this
thickness of the plate element.
Here's our shear correction factor again, which is
introduced just the same way as in the beam element.
Notice here we have the work, or the total potential of the
external loads, I should say.
It is convenient now to integrate through the
thickness because we can integrate prior to
interpolating the quantities, and that then yields this
value for pi, where our Cb parts and Cs part here, these
two matrices, embody the fact that we have integrated
through the thickness, so we have the
following definitions here.
Kappa simply lists basically the bending strains, or I
should say the rotations of the sections.
Of course here, we have the shearing strains.
Gamma lists the shearing strains.
Cb now is a function of h cubed, just like in classic
plate theory.
Of course, we also have an h cubed entering into the
formulation.
And this Cb matrix embodies the fact that we have
integrated through the thickness.
Here is our Cs.
Of course, in the Cs, we also have the k part.
Notice that, again, we have an h cubed here and an h there.
Therefore, to use our interpolation for plate and
shell elements, we will have to use high enough shell
interpolations to be able to represent the fact that the
shear strains go to 0 for thin plates if we want to use this
formulation for thin plates and shells.
Well, invoking now the fact that pi shall be stationary,
we directly obtain this equation here.
And this, of course, is nothing else than the
principal of virtual displacement
for the plate element.
Notice that from this point onwards, we simply need to
substitute only our interpolations.
The interpolations that we are using are now interpolations
for w, beta x, and beta y.
And, of course, we also interpolate x and y.
These interpolations, beta x and beta y, are independent
from the interpolations of w, and that is the important
points, as I mentioned in the
development of the beam element.
The fact that we are dealing, here, with three
interpolations of course means that an each nodal point, we
have three unknowns, w's, and section rotations.
Let us now look very briefly at shell elements.
The same concept that we used to develop the general beam
element after having discussed the special beam element is
also employed now in the development of what you might
call a general shell element, versus the special plate
element that I just discussed, or the special shell element
that I just discussed, because a flat shell is of course
nothing else than a plate, if we also don't
have membrane forces.
Well here, I'm showing a shell element, a
nine-noded shell element.
And notice that in this case, now, we are talking about this
normal only.
In the beam, we had two normals, vt and vs. Now we
only have one normal.
At a nodal point, we are defining the membrane
displacements, uk, vk, wk, being a transverse
displacement, and the rotations, alpha k and beta k.
These rotations are defined about the axis v1 k and v2 k.
Now notice that these two rotations, alpha k and beta k,
will give us the change in vn during deformations.
And this is really how we use alpha k and beta k.
We express a change in vn in terms of alpha k and beta k.
The procedure is the same as in the
case of the beam element.
First, we express our x, y, and z-coordinates.
And we're using here the original normal, vn, the x, y
and z direction cosines.
These are, here, the x, y and z-coordinates of the
nodal point, k.
We have our two dimensional interpolation functions, hk,
now here, because we talk about a two dimensional
surface, the mid-surface of the shell.
Of course, at each nodal point, the shell can have a
different thickness, and that is denoted by using a
different ak value at each nodal point.
Applying this interpolation here to the initial
configuration and the final configuration and subtracting
0x from 1x, and similar for y and z, we directly obtain the
displacements u, v, and w.
Notice that the displacements now are involving the nodal
point displacements and the change in the direction
cosines of the normal, denoted here.
These changes in the direction cosines of the normal can
directly be expressed in terms of the rotations,
alpha k and beta k.
Now, notice here that once we have done this, of course here
we involve now, as I mentioned earlier, the v1 and v2
directions.
And these v1 and v2 directions are arbitrarily selected.
In fact, they are--
for our shell element here-- selected as shown here.
But once we have selected v1 and v2 at each nodal point,
and they can vary from nodal point to nodal point, then we
can use this relation to attain directly the change in
the direction cosines of the normal during deformations,
when the deformations are alpha k and beta k.
So with this equation, then, and the earlier equation that
I've given to you, we can directly obtain the
displacement interpolation matrix and the strain
interpolation matrix.
One important point that I should briefly mention is
that, of course, for the shell, we have 0 stresses
through the thickness.
So, we have to use this stress strain law here.
Notice there are 0's in this row and column.
And this is, here, the plane stress part for the bending,
and that is the shear part here.
This is the stress strain law defined in a local convected
coordinate system where we are talking about the stresses
through the thickness being this direction here.
And these other stresses are aligned with
the coordinate system.
We have to transform this one here to the global coordinate
system in order to be able to use it directly in our
formulation.
And that transformation is achieved via these
transformation matrices.
Now, this element has been effectively implemented in the
ADINA computer program, and we want to use it using high
order interpolation, as I mentioned earlier, as the
basic element that therefore is very useful, which can be
used as a flat element, a curved element, all curved
element, or it can have curvature in both directions.
Also, this is the basic element that is being used.
We can also collapse nodes and derive other elements.
As I pointed out earlier, the low order elements should only
be used in very special cases.
I would not recommend these elements.
Although they can be used in principle, the element that is
really useful is this one and that one.
Both, of course, can be used as curved elements as I
pointed out.
Another feature, and this is the final view graph that I
wanted to show in this lecture, is that we can use
these elements also in transition regions.
Namely, here we have the shell element, now being flat--
that I discussed--
and we can directly couple this element into another
element, which we call a transition element, which has
the shell degrees of freedom at these nodes, but
translational degrees of freedom only here.
In other words, a continuum element degrees of freedom
right here.
Notice, three degrees of freedom at this node, only
translations, whereas here, we would have five degrees of
freedom-- three translations, two rotations.
Similarly, here we have a curved shell going into a
solid, and again here, we have a transition element.
Here, we show the five degrees of freedom at a shell node and
the three degrees of freedom at a continuum element node.
This is an effective approach to be able to couple director
shell elements into solid elements.
I have not talked, of course, about the actual derivation of
the matrices used in the formulation of
the transition element.
That is a little bit beyond what I wanted to present in
this lecture.
But the basic concepts are those that we discussed
already in the earlier lecture for continuum element and in
this lecture for structural elements.
Thank you very much for your attention.