WEBVTT
00:00:00.000 --> 00:00:02.470
The following content is
provided under a Creative
00:00:02.470 --> 00:00:03.880
Commons license.
00:00:03.880 --> 00:00:06.920
Your support will help MIT
OpenCourseWare continue to
00:00:06.920 --> 00:00:10.570
offer high quality educational
resources for free.
00:00:10.570 --> 00:00:13.470
To make a donation or view
additional materials from
00:00:13.470 --> 00:00:17.872
hundreds of MIT courses, visit
MIT OpenCourseWare at
00:00:17.872 --> 00:00:19.122
ocw.mit.edu.
00:00:21.290 --> 00:00:23.770
PROFESSOR: Ladies and gentlemen,
welcome to lecture
00:00:23.770 --> 00:00:25.060
number seven.
00:00:25.060 --> 00:00:27.260
In this lecture, I would like
to present to you the
00:00:27.260 --> 00:00:29.740
formulation of structural
elements.
00:00:29.740 --> 00:00:34.040
We will be discussing beam,
plate, and shell elements, and
00:00:34.040 --> 00:00:38.450
I would like to introduce to you
the isoparametric approach
00:00:38.450 --> 00:00:40.570
for interpolations.
00:00:40.570 --> 00:00:42.380
There are two approaches in the
00:00:42.380 --> 00:00:44.170
formulation that we can follow.
00:00:44.170 --> 00:00:47.870
The first one is a strength of
materials approach, in which
00:00:47.870 --> 00:00:51.560
case we look at a straight beam
element, we use a beam
00:00:51.560 --> 00:00:53.990
theory including
shear effects.
00:00:53.990 --> 00:00:57.120
If you look at a plate element,
we use a plate theory
00:00:57.120 --> 00:00:59.040
including shear effects, also.
00:00:59.040 --> 00:01:01.990
The names associated with these
theories that we're
00:01:01.990 --> 00:01:06.900
using are the names of
Reissner and Mindlin.
00:01:06.900 --> 00:01:09.850
In the second approach, we have
the continuum mechanics
00:01:09.850 --> 00:01:13.360
approach, in which we use the
general principle of virtual
00:01:13.360 --> 00:01:18.210
displacement, but we exclude
the stress components not
00:01:18.210 --> 00:01:19.390
applicable.
00:01:19.390 --> 00:01:22.460
For example, in a plate, we
set the stress through the
00:01:22.460 --> 00:01:25.480
thickness of the plate
equal to zero.
00:01:25.480 --> 00:01:29.560
In addition, also, we have to
impose in the use of the
00:01:29.560 --> 00:01:32.240
principle of virtual
displacement the kinematic
00:01:32.240 --> 00:01:37.730
constraints for particles on
sections or originally normal
00:01:37.730 --> 00:01:39.320
to the mid-surface.
00:01:39.320 --> 00:01:43.730
Namely, we have to put the
constraint into the structure
00:01:43.730 --> 00:01:47.160
that the particles remain on
a straight line during
00:01:47.160 --> 00:01:48.410
deformation.
00:01:48.410 --> 00:01:51.650
Well, as examples, I've plotted
here, I've shown here
00:01:51.650 --> 00:01:57.020
two structures, a beam
and a shell.
00:01:57.020 --> 00:01:59.120
Let's look first at a beam.
00:01:59.120 --> 00:02:05.020
In this case, we have that the
original particles normal to
00:02:05.020 --> 00:02:08.210
the mid-surface, or the neutral
axis of the beam, are
00:02:08.210 --> 00:02:09.630
on this orange line.
00:02:09.630 --> 00:02:12.880
I've shown here a large
number of particles.
00:02:12.880 --> 00:02:16.630
The kinematic constraint that
we're talking about is that
00:02:16.630 --> 00:02:20.380
during deformations, these
particles remain
00:02:20.380 --> 00:02:21.420
on a straight line.
00:02:21.420 --> 00:02:24.710
They move over to the
yellow line here.
00:02:24.710 --> 00:02:28.040
In other words, point A
goes to point a prime.
00:02:28.040 --> 00:02:31.400
Another particle here goes over
to this particle here.
00:02:31.400 --> 00:02:34.620
This particles goes over to that
particle, and so on, and
00:02:34.620 --> 00:02:37.210
these particles remain
on a straight line.
00:02:37.210 --> 00:02:39.970
That is the basic kinematic
assumption.
00:02:39.970 --> 00:02:40.420
However,
00:02:40.420 --> 00:02:44.660
We should also notice that
there's a right angle between
00:02:44.660 --> 00:02:49.080
the mid-surface, or neutral axis
off the beam, and this
00:02:49.080 --> 00:02:52.970
line of particles initially.
00:02:52.970 --> 00:02:56.140
This right angle is not
preserved during deformation.
00:02:56.140 --> 00:03:01.700
In other words, this angle
here is not a right angle
00:03:01.700 --> 00:03:04.310
after deformations anymore.
00:03:04.310 --> 00:03:06.400
In the case of the shell,
the kinematic
00:03:06.400 --> 00:03:08.470
constraint is quite similar.
00:03:08.470 --> 00:03:11.230
Here, we have now the
mid-surface of the shell shown
00:03:11.230 --> 00:03:12.860
as a dashed line.
00:03:12.860 --> 00:03:17.700
The particles on a line normal
to that mid-surface are shown
00:03:17.700 --> 00:03:20.670
here again in orange.
00:03:20.670 --> 00:03:22.600
This is the initial line.
00:03:22.600 --> 00:03:26.000
And during deformation,
these particles remain
00:03:26.000 --> 00:03:27.100
on a straight line.
00:03:27.100 --> 00:03:32.110
Now they have come to be the
yellow line here, and we
00:03:32.110 --> 00:03:36.280
notice that, again, there's a
right angle initially here,
00:03:36.280 --> 00:03:40.350
but that right angle is not
preserved during deformations
00:03:40.350 --> 00:03:44.530
because, in each case, we are
including shear effects.
00:03:44.530 --> 00:03:47.270
Here, we include shear effects,
and similarly here,
00:03:47.270 --> 00:03:50.010
we also include shear effects.
00:03:50.010 --> 00:03:57.340
Well, I've prepared some view
graphs to show these facts a
00:03:57.340 --> 00:03:59.010
little bit more distinct.
00:03:59.010 --> 00:04:03.590
Here, we have a first view
graph on which I show the
00:04:03.590 --> 00:04:07.880
assumptions of the basic
Bernoulli-Euler beam theory
00:04:07.880 --> 00:04:10.290
that is used in the development
of conventional
00:04:10.290 --> 00:04:11.620
beam elements.
00:04:11.620 --> 00:04:19.160
We have here the original beam
element with its neutral axis
00:04:19.160 --> 00:04:20.990
in a dashed line.
00:04:20.990 --> 00:04:26.220
And that beam element, during
deformation, becomes this.
00:04:26.220 --> 00:04:28.740
It goes into this shape here.
00:04:28.740 --> 00:04:30.840
We notice-- and this
is important--
00:04:30.840 --> 00:04:36.320
that this section here, which I
now mark in blue, goes over
00:04:36.320 --> 00:04:42.840
into that section, and that the
displacement is w at the
00:04:42.840 --> 00:04:49.600
mid-surface, and that the slope
here at right angles to
00:04:49.600 --> 00:04:54.520
that section is nothing
else than dw dx.
00:04:54.520 --> 00:04:57.690
In other words, this angle here
is really nothing else
00:04:57.690 --> 00:05:01.030
than that angle, dw dx.
00:05:01.030 --> 00:05:04.550
This is the Bernoulli-Euler beam
theory excluding shear
00:05:04.550 --> 00:05:06.130
deformations.
00:05:06.130 --> 00:05:11.040
The important point is that when
we use this beam theory,
00:05:11.040 --> 00:05:16.410
we have to match between two
elements, w, in other words,
00:05:16.410 --> 00:05:19.475
the displacement at the
mid-surface has to be the same
00:05:19.475 --> 00:05:22.780
for element one and
element two.
00:05:22.780 --> 00:05:26.620
And in addition, this slope has
to be the same for both
00:05:26.620 --> 00:05:31.190
elements, dw dx for element one
on the left-hand side must
00:05:31.190 --> 00:05:34.750
be equal to dw dx on the
right-hand side.
00:05:34.750 --> 00:05:39.750
This is the conventional beam
theory that is used to develop
00:05:39.750 --> 00:05:42.770
the Hermitian beam elements.
00:05:42.770 --> 00:05:47.620
And what I'd like to introduce
to you now is the beam theory
00:05:47.620 --> 00:05:52.980
that we're using in the
isoparametric formulation,
00:05:52.980 --> 00:05:58.290
namely in the development of
modern beam elements, pipe
00:05:58.290 --> 00:06:01.400
elements, shell elements,
and plate elements.
00:06:01.400 --> 00:06:05.130
I should say at this point
that the conventional
00:06:05.130 --> 00:06:09.090
Hermitian beam element that
you're probably familiar with
00:06:09.090 --> 00:06:12.520
is more effective in engineering
analysis than the
00:06:12.520 --> 00:06:15.670
beam element that I'm talking
about here when we just look
00:06:15.670 --> 00:06:17.660
at a straight beam.
00:06:17.660 --> 00:06:22.240
However, I want to look at the
straight beam here as an
00:06:22.240 --> 00:06:26.890
example to introduce to you
the formulation of the
00:06:26.890 --> 00:06:29.560
structural elements that
I'm talking about.
00:06:29.560 --> 00:06:32.000
The formulation is very well
displayed, very well
00:06:32.000 --> 00:06:32.980
demonstrated.
00:06:32.980 --> 00:06:36.040
The basic features are very well
demonstrated looking at
00:06:36.040 --> 00:06:39.920
the beam element, although
the application of this
00:06:39.920 --> 00:06:44.000
formulation to a straight beam
element is not as effective as
00:06:44.000 --> 00:06:47.230
just the usage of a Hermitian
beam element.
00:06:47.230 --> 00:06:50.310
However, if we talk about curved
beam elements, pipe
00:06:50.310 --> 00:06:52.140
elements, then this formulation
00:06:52.140 --> 00:06:55.230
is indeed very effective.
00:06:55.230 --> 00:06:57.540
And, of course, for plates
and shells, it
00:06:57.540 --> 00:06:59.320
is really very effective.
00:06:59.320 --> 00:07:02.080
Let me show to you, then,
the basic points--
00:07:02.080 --> 00:07:04.360
the basic important points--
00:07:04.360 --> 00:07:07.980
that are being used in
the formulation.
00:07:07.980 --> 00:07:12.250
Here we have, again, our
original beam element.
00:07:12.250 --> 00:07:16.020
The neutral access is shown
here, and this beam element
00:07:16.020 --> 00:07:22.120
now moves over into this piece,
into that shape, during
00:07:22.120 --> 00:07:23.630
deformations.
00:07:23.630 --> 00:07:29.250
We have a section here, and as
I pointed out earlier, that
00:07:29.250 --> 00:07:33.420
section has to remain straight
during deformations.
00:07:33.420 --> 00:07:40.770
In fact, it moves right
to that section here.
00:07:40.770 --> 00:07:45.970
Now notice that what we're
talking about is a slope, dw
00:07:45.970 --> 00:07:51.690
dx here, which is this angle
here, plus a shear deformation
00:07:51.690 --> 00:07:58.260
angle, gamma, and dw dx minus
gamma is this angle.
00:07:58.260 --> 00:08:01.390
And that is the angle beta.
00:08:01.390 --> 00:08:09.080
In fact, this is the rotation
of this line of particles.
00:08:09.080 --> 00:08:12.050
In other words, we are not
talking just about one state
00:08:12.050 --> 00:08:16.810
variable, w, as we do in the
Hermitian formulation, but we
00:08:16.810 --> 00:08:21.620
talk about two state variables
now, beta and w.
00:08:21.620 --> 00:08:25.950
Beta and w both are independent,
and we will see
00:08:25.950 --> 00:08:28.370
that, later on, we will
interpolate them as
00:08:28.370 --> 00:08:30.560
independent quantities.
00:08:30.560 --> 00:08:33.330
The important point, then,
that if you look at two
00:08:33.330 --> 00:08:37.350
elements, if you do interpolate
w and beta
00:08:37.350 --> 00:08:41.900
independently, then we need,
between two elements,
00:08:41.900 --> 00:08:48.620
continuity in w and continuity
in beta.
00:08:48.620 --> 00:08:53.290
We do not talk about continuity
in dw dx.
00:08:53.290 --> 00:08:57.560
We do not talk about that, and
that is the important point,
00:08:57.560 --> 00:09:00.480
particularly when we talk about
the formulation of plate
00:09:00.480 --> 00:09:02.750
elements and shell elements.
00:09:02.750 --> 00:09:06.430
The fact that we'll talk about
independent interpolations of
00:09:06.430 --> 00:09:11.160
w and beta, including shear
effects, of course, in an
00:09:11.160 --> 00:09:15.990
approximate way, but including
shear affects, that fact
00:09:15.990 --> 00:09:18.745
alleviates us of many
difficulties that
00:09:18.745 --> 00:09:21.700
we encounter otherwise.
00:09:21.700 --> 00:09:24.080
In other words, that we
encounter if we use a
00:09:24.080 --> 00:09:29.600
classical plate theory excluding
shear deformations.
00:09:29.600 --> 00:09:35.220
A good starting point for the
development of the elements is
00:09:35.220 --> 00:09:40.550
the use of the total potential,
pi, of an element.
00:09:40.550 --> 00:09:45.000
That total potential I've
written down here, and we have
00:09:45.000 --> 00:09:49.640
set pi is equal to a
contribution from the bending
00:09:49.640 --> 00:09:56.250
part plus a contribution from
the shearing part, or the
00:09:56.250 --> 00:10:01.680
shearing deformations, and, of
course, there is the external
00:10:01.680 --> 00:10:05.530
work due to distributed
pressure, p, on a beam
00:10:05.530 --> 00:10:09.330
element, and moments, externally
applied moments, m,
00:10:09.330 --> 00:10:11.100
onto the beam element.
00:10:11.100 --> 00:10:14.070
Now, notice the quantities
that I'm using here.
00:10:14.070 --> 00:10:19.890
The bending part is given by dw
dx squared, of course with
00:10:19.890 --> 00:10:26.030
the flex or rigidity in front,
and this part here is given
00:10:26.030 --> 00:10:29.720
only in terms of the section
rotation beta, which is
00:10:29.720 --> 00:10:36.370
independent of the translation
of the neutral axes, w.
00:10:36.370 --> 00:10:41.000
Here, we have the shearing part,
dw dx, minus beta As,
00:10:41.000 --> 00:10:43.140
the shear strength.
00:10:43.140 --> 00:10:46.720
And they have been written
down here once more.
00:10:46.720 --> 00:10:50.840
If we look at this equation,
here we get dw dx minus beta
00:10:50.840 --> 00:10:53.880
is equal to gamma.
00:10:53.880 --> 00:10:56.690
The other quantities, of course,
that I used in the
00:10:56.690 --> 00:10:59.030
derivation of this--
00:10:59.030 --> 00:11:05.770
pi, the stress being equal to
v over as, v being the shear
00:11:05.770 --> 00:11:10.410
force on the section, As
is the shear area.
00:11:10.410 --> 00:11:14.760
Notice that we are assuming the
shear strength through the
00:11:14.760 --> 00:11:18.130
thickness of the beam element
to be constant.
00:11:18.130 --> 00:11:23.140
They are constant because of
this equation here, basically.
00:11:23.140 --> 00:11:27.010
w, of course, varies along
the length of the beam.
00:11:27.010 --> 00:11:31.220
Beta varies along the length
of the beam, but that means
00:11:31.220 --> 00:11:34.350
gamma is constant through to
the thickness of the beam.
00:11:34.350 --> 00:11:37.035
Now, since gamma is constant
through the thickness of the
00:11:37.035 --> 00:11:42.330
beam, we have to say, of course,
that also our shear
00:11:42.330 --> 00:11:46.330
stress is constant through the
thickness of the beam, and we
00:11:46.330 --> 00:11:50.800
have to introduce a shear
correction factor, k, which is
00:11:50.800 --> 00:11:54.770
equal to As over A, where As is
an equivalent shear area.
00:11:57.360 --> 00:12:03.570
Using these quantities we obtain
this pi functional,
00:12:03.570 --> 00:12:08.150
where, once again, we simply add
the bending contribution,
00:12:08.150 --> 00:12:12.480
bending strain energy to the
shear strain energy, and we
00:12:12.480 --> 00:12:17.330
subtract the total potential
of the external loads.
00:12:17.330 --> 00:12:21.150
If we invoke the stationarity
of this functional, in other
00:12:21.150 --> 00:12:25.650
words, we invoke that delta pi
is equal to zero, we obtain
00:12:25.650 --> 00:12:29.640
the principle of virtual work,
or principle of virtual
00:12:29.640 --> 00:12:33.690
displacement, which I have
discussed with you in an
00:12:33.690 --> 00:12:35.330
earlier lecture.
00:12:35.330 --> 00:12:40.340
The result of invoking that del
pi is equal to zero, or
00:12:40.340 --> 00:12:45.250
invoking the stationarity of
pi is this equation here.
00:12:45.250 --> 00:12:49.090
Now notice that in this equation
we have, if we want
00:12:49.090 --> 00:12:52.400
to interpret it once physically
here, basically the
00:12:52.400 --> 00:12:59.970
real stress part, here, the
virtual strain part.
00:12:59.970 --> 00:13:04.690
Similarly here, the real stress
part and the virtual
00:13:04.690 --> 00:13:07.360
strain part, and, of course,
the virtual work of the
00:13:07.360 --> 00:13:10.020
external loads.
00:13:10.020 --> 00:13:12.900
The important point is that we
only integrate along the
00:13:12.900 --> 00:13:15.270
length of the beam, and not
through the thickness anymore,
00:13:15.270 --> 00:13:20.780
because we talk about quantities
stress resultants
00:13:20.780 --> 00:13:22.660
over the thickness.
00:13:22.660 --> 00:13:28.330
Well, once we have arrived at
this equation, we can proceed
00:13:28.330 --> 00:13:31.370
in much the same way as we have
been proceeding in the
00:13:31.370 --> 00:13:35.160
development of continuum
isoparametric elements.
00:13:35.160 --> 00:13:38.230
Here, we look at a
particular case.
00:13:38.230 --> 00:13:41.340
Let us say we have a beam
elements such as shown here.
00:13:41.340 --> 00:13:45.040
Of course, this is the loading
applied, P. The bending moment
00:13:45.040 --> 00:13:47.650
that they're talking about
here is shown here.
00:13:47.650 --> 00:13:49.980
It is a distributed bending
moment over
00:13:49.980 --> 00:13:51.260
any part of the beam.
00:13:51.260 --> 00:13:53.010
Similarly, P is only
applied over a
00:13:53.010 --> 00:13:54.610
certain part of the beam.
00:13:54.610 --> 00:14:00.030
The depth of the beam is b, the
width of the beam is a.
00:14:00.030 --> 00:14:04.040
As a shear factor for a
rectangular beam, the sheer
00:14:04.040 --> 00:14:08.160
fact k that I introduced
to you briefly is 5/6.
00:14:08.160 --> 00:14:10.930
Of course, I, the moment
of inertia, is ab
00:14:10.930 --> 00:14:14.230
cubed divided by 12.
00:14:14.230 --> 00:14:17.300
The interpolations that we
would use for such a beam
00:14:17.300 --> 00:14:19.860
element are one dimensional
interpolations.
00:14:19.860 --> 00:14:23.050
We only integrate along
the length, x.
00:14:23.050 --> 00:14:26.220
And the one dimension
interpolations we discussed
00:14:26.220 --> 00:14:27.860
already earlier.
00:14:27.860 --> 00:14:31.800
We simply use the same that we
have been using already in the
00:14:31.800 --> 00:14:34.090
formulation of truss elements.
00:14:34.090 --> 00:14:38.460
Two point interpolation would
be this element here.
00:14:38.460 --> 00:14:41.010
Here, we use a three point
interpolation.
00:14:41.010 --> 00:14:41.800
Notice--
00:14:41.800 --> 00:14:42.970
and this is important--
00:14:42.970 --> 00:14:47.610
that we're talking about w and
beta, the section rotations,
00:14:47.610 --> 00:14:49.550
as independent quantities.
00:14:49.550 --> 00:14:54.820
So for a three point, or three
nodal point beam, we would
00:14:54.820 --> 00:15:00.710
have, in just a plane analysis,
we would have six
00:15:00.710 --> 00:15:02.240
degrees of freedom.
00:15:02.240 --> 00:15:06.960
For a cubic element, we have
eight degrees of freedom.
00:15:06.960 --> 00:15:10.420
Of course, for a Hermitian beam
element, we would have
00:15:10.420 --> 00:15:14.150
only these degrees of freedom
and those degrees of freedom,
00:15:14.150 --> 00:15:17.710
w and dw dx here, and
w, dw dx here.
00:15:17.710 --> 00:15:20.200
So, that is the reason why
the Hermitian beam
00:15:20.200 --> 00:15:22.050
element is more effective.
00:15:22.050 --> 00:15:25.650
However, we can use these
interpolations directly to
00:15:25.650 --> 00:15:30.460
develop curved beam elements,
pipe elements, and then, of
00:15:30.460 --> 00:15:32.990
course, this approach is--
even for beam elements--
00:15:32.990 --> 00:15:35.060
more effective, as I
mentioned earlier.
00:15:35.060 --> 00:15:38.680
Well, let us then write down the
basic interpolations that
00:15:38.680 --> 00:15:39.360
we are using.
00:15:39.360 --> 00:15:43.870
Here, we have w being
the sum of hi wi.
00:15:43.870 --> 00:15:46.350
hi, we discussed earlier
already.
00:15:46.350 --> 00:15:49.980
And these, of course, are
nodal point transverse
00:15:49.980 --> 00:15:51.800
displacement.
00:15:51.800 --> 00:15:55.390
These are the nodal
point rotations of
00:15:55.390 --> 00:15:57.830
the sections, beta.
00:15:57.830 --> 00:16:00.330
In other words, I could have
used here, as a notation, beta
00:16:00.330 --> 00:16:04.390
i, but I chose to use theta i.
00:16:04.390 --> 00:16:07.860
If you write these equations
in matrix form, we directly
00:16:07.860 --> 00:16:14.900
obtain this relation here, where
hw simply list hi, u
00:16:14.900 --> 00:16:19.550
lists the displacements and
section rotations, and similar
00:16:19.550 --> 00:16:23.300
here, beta is given in terms
of an h beta matrix.
00:16:23.300 --> 00:16:27.570
We take the differentiations of
hw and h beta, and we get
00:16:27.570 --> 00:16:35.490
dw dx equal to Bw times u d beta
dx equal beta times u.
00:16:35.490 --> 00:16:40.110
Once we have the principle
virtual work established for
00:16:40.110 --> 00:16:44.170
the element that we're
considering and have chosen
00:16:44.170 --> 00:16:47.160
our interpolations, the approach
of developing the
00:16:47.160 --> 00:16:51.680
stiffness matrices, the load
vectors, is exactly the same
00:16:51.680 --> 00:16:56.740
as in the development of
continuum elements.
00:16:56.740 --> 00:17:01.790
Well here, I have written down
the various quantities.
00:17:01.790 --> 00:17:06.560
U transposed lists, as I said
earlier, the displacement
00:17:06.560 --> 00:17:09.900
vector nodal points and the
rotation vector nodal points.
00:17:09.900 --> 00:17:13.450
Hw simply gives the
interpolation functions, q, of
00:17:13.450 --> 00:17:15.170
course, being equal to
the number of nodal
00:17:15.170 --> 00:17:17.200
points we are using.
00:17:17.200 --> 00:17:19.930
And here, we have H beta.
00:17:19.930 --> 00:17:22.380
The Bw is given right here.
00:17:22.380 --> 00:17:25.640
Notice our J inverse comes in
there because we have to
00:17:25.640 --> 00:17:29.950
transform from r to x, x being
the actual physical coordinate
00:17:29.950 --> 00:17:31.160
along the length.
00:17:31.160 --> 00:17:33.390
Similarly, B beta being
given here.
00:17:33.390 --> 00:17:37.450
Again, the J inverse here
to transform from r to x
00:17:37.450 --> 00:17:39.450
coordinates.
00:17:39.450 --> 00:17:43.020
Once we have written down
these, we can directly
00:17:43.020 --> 00:17:49.020
substitute into the principal
of virtual work, and we come
00:17:49.020 --> 00:17:53.610
up with the stiffness matrix,
given here, and the load
00:17:53.610 --> 00:17:55.120
vector given here.
00:17:55.120 --> 00:17:58.370
Let me point out a few important
things here.
00:17:58.370 --> 00:18:02.870
Of course, this B beta matrix
here-- just to remind you--
00:18:02.870 --> 00:18:07.100
comes from d beta dx.
00:18:07.100 --> 00:18:12.800
It's in fact, really, d beta
dr, but because we're
00:18:12.800 --> 00:18:16.660
integrating from -1 to plus
1, however, we have our
00:18:16.660 --> 00:18:19.720
determinant J there to take
into account the volume
00:18:19.720 --> 00:18:22.190
transformation.
00:18:22.190 --> 00:18:26.870
Here, of course, we have
d beta dx transposed.
00:18:26.870 --> 00:18:29.730
This comes from the virtual
strains, and that comes from
00:18:29.730 --> 00:18:32.120
the real strains.
00:18:32.120 --> 00:18:33.460
Of course, we have
the stresses.
00:18:33.460 --> 00:18:37.870
Here, so these strains times
the stress strain law, E,
00:18:37.870 --> 00:18:41.980
being the Young's modulus,
gives us the stresses.
00:18:41.980 --> 00:18:44.700
Here, we talk about shearing
deformations.
00:18:44.700 --> 00:18:52.261
Notice that we have here
dw dx, that is, the
00:18:52.261 --> 00:18:55.510
Bw minus the beta.
00:18:55.510 --> 00:19:00.320
So we have the derivative in
here of w, but no derivative
00:19:00.320 --> 00:19:04.780
in H beta because we are simply
interpolating here the
00:19:04.780 --> 00:19:06.890
beta values.
00:19:06.890 --> 00:19:12.270
Again, of course, a
transformation to x, and
00:19:12.270 --> 00:19:15.970
therefore we have a determine
J in there.
00:19:15.970 --> 00:19:21.660
The load vector looks just the
same way as in the development
00:19:21.660 --> 00:19:23.260
of continuum elements.
00:19:23.260 --> 00:19:27.600
This is the transverse loading
applied P. This is, of course,
00:19:27.600 --> 00:19:28.730
interpolating--
00:19:28.730 --> 00:19:29.230
this [UNINTELLIGIBLE]
00:19:29.230 --> 00:19:33.270
interpolates the virtual
transverse displacements.
00:19:33.270 --> 00:19:39.230
Here, we have the moment loads,
the real moment loads,
00:19:39.230 --> 00:19:43.610
and this interpolates here the
section rotations, beta, along
00:19:43.610 --> 00:19:45.700
the lengths of the beam.
00:19:45.700 --> 00:19:51.800
Of course, again, the volume
transformation from r to x.
00:19:51.800 --> 00:19:54.960
This is really a straightforward
application of
00:19:54.960 --> 00:19:56.840
what we discussed earlier.
00:19:56.840 --> 00:19:59.990
There is one important point,
however, now, that I have to
00:19:59.990 --> 00:20:01.910
point out to you.
00:20:01.910 --> 00:20:08.390
If we consider the functional
pi, as I mentioned earlier,
00:20:08.390 --> 00:20:10.940
there's a bending part
here and there's a
00:20:10.940 --> 00:20:13.970
shearing part here.
00:20:13.970 --> 00:20:18.680
Notice that in this development
I have divided
00:20:18.680 --> 00:20:24.010
through by EI over 2, so that
I introduce a value alpha
00:20:24.010 --> 00:20:29.770
there, and that alpha is really
GAK divided by EI.
00:20:29.770 --> 00:20:37.090
Now, if we look at that alpha
value, and let's look at the
00:20:37.090 --> 00:20:44.430
value for a rectangular section,
we would see that the
00:20:44.430 --> 00:20:54.680
a value, of course, is a times
b, and the I value gives us
00:20:54.680 --> 00:21:00.896
really an ab cubed, if this is,
here, b, and that is a.
00:21:00.896 --> 00:21:04.190
An ab cubed over
12, of course.
00:21:04.190 --> 00:21:10.120
And we have, also, the GK that
gives us these, and of course
00:21:10.120 --> 00:21:11.540
an E in front here.
00:21:11.540 --> 00:21:14.950
But the important point that I
want to now concentrate on is
00:21:14.950 --> 00:21:18.580
really this b over b cubed.
00:21:18.580 --> 00:21:23.840
We can see that as the element
gets thinner and thinner, as
00:21:23.840 --> 00:21:28.380
the element gets thinner and
thinner, that alpha gets
00:21:28.380 --> 00:21:29.660
larger and larger.
00:21:32.500 --> 00:21:36.790
If alpha gets larger and larger,
this term here will be
00:21:36.790 --> 00:21:39.240
predominant.
00:21:39.240 --> 00:21:41.060
This term will be predominant.
00:21:41.060 --> 00:21:47.940
Now this means, however, that if
we want to finally converge
00:21:47.940 --> 00:21:52.540
to a beam in which the shear
strengths are negligible, in
00:21:52.540 --> 00:21:54.800
other words, in which the shear
strengths are extremely
00:21:54.800 --> 00:21:58.650
small, what we would have to
be able to represent in the
00:21:58.650 --> 00:22:03.550
formulation is that this value
here goes to zero.
00:22:03.550 --> 00:22:06.150
And what happens in the
formulation, really, is that
00:22:06.150 --> 00:22:11.430
as this value gets larger and
larger, any error introduced
00:22:11.430 --> 00:22:15.520
in the formulation, due to the
fact that this is not exactly
00:22:15.520 --> 00:22:20.080
zero in the finite element
interpolation, that error is
00:22:20.080 --> 00:22:21.340
largely magnified.
00:22:21.340 --> 00:22:26.570
Is magnified and, in fact, can
introduce a very large error
00:22:26.570 --> 00:22:33.020
if this value is not zero due to
the fact that alpha becomes
00:22:33.020 --> 00:22:35.600
larger and larger
if the element
00:22:35.600 --> 00:22:38.410
becomes thinner and thinner.
00:22:38.410 --> 00:22:42.500
In other words, in summary once
more, if we are talking
00:22:42.500 --> 00:22:47.270
about a beam element that gets
thinner and thinner for which
00:22:47.270 --> 00:22:51.340
we know the shear strengths
should becomes smaller and
00:22:51.340 --> 00:22:56.260
smaller, our finite element
interpolation must be able to
00:22:56.260 --> 00:22:58.950
represent this fact.
00:22:58.950 --> 00:23:01.600
Now, if we look at the shear
strengths, and this is, of
00:23:01.600 --> 00:23:05.100
course here, nothing else then
basically the shear strength
00:23:05.100 --> 00:23:13.650
squared, we now identify that
dw dx minus beta, when
00:23:13.650 --> 00:23:19.300
interpolated using our
interpolation functions, must
00:23:19.300 --> 00:23:23.570
be able to be very,
very, very small.
00:23:23.570 --> 00:23:27.270
And that is a restriction
on the formulation.
00:23:27.270 --> 00:23:31.390
So what we have to do, really,
is use high enough order
00:23:31.390 --> 00:23:36.850
interpolations so that dw dx
minus beta can be smaller for
00:23:36.850 --> 00:23:38.230
thin elements.
00:23:38.230 --> 00:23:41.630
Of course for thick beam
elements, that is not really a
00:23:41.630 --> 00:23:44.360
constraint because we know
that there are shearing
00:23:44.360 --> 00:23:46.690
deformations, and the shear
deformations can be quite
00:23:46.690 --> 00:23:47.800
significant.
00:23:47.800 --> 00:23:52.550
However, for thin elements, we
must be able to represent the
00:23:52.550 --> 00:23:57.220
fact that gamma is small, and
therefore, we have to use high
00:23:57.220 --> 00:23:58.580
order interpolations.
00:23:58.580 --> 00:24:02.810
In fact, the parabolic
interpolation is really the
00:24:02.810 --> 00:24:05.460
lowest interpolation that
one can recommend.
00:24:05.460 --> 00:24:08.650
It would be better to use
cubic interpolation.
00:24:08.650 --> 00:24:11.990
In fact, we use the cubic
interpolation in practice.
00:24:11.990 --> 00:24:18.520
In that case, this gamma value
can be small, and we run into
00:24:18.520 --> 00:24:22.480
no difficulties and
the element can be
00:24:22.480 --> 00:24:24.520
very, very, very thin.
00:24:24.520 --> 00:24:26.880
Another approach would be to
use a discrete Kirchhoff
00:24:26.880 --> 00:24:30.270
theory or reduced numerical
integration.
00:24:30.270 --> 00:24:32.860
These approaches have
been developed
00:24:32.860 --> 00:24:34.970
for low order elements.
00:24:34.970 --> 00:24:38.710
The discrete Kirchhoff theory
approach is very effective.
00:24:38.710 --> 00:24:41.240
The reduced numerical
integration can also be
00:24:41.240 --> 00:24:43.300
effective, but has to
be used with care.
00:24:43.300 --> 00:24:47.250
In particular, as I will point
out in the next lecture, we
00:24:47.250 --> 00:24:51.710
have to be careful that we do
not introduce spurious rigid
00:24:51.710 --> 00:24:53.050
body modes into the system.
00:24:55.990 --> 00:25:03.540
The development that I just
talked about really is
00:25:03.540 --> 00:25:10.350
applicable to an element that
has a rectangular section and
00:25:10.350 --> 00:25:13.110
the element was also
lying in a plane.
00:25:13.110 --> 00:25:14.570
We looked at a straight
element.
00:25:14.570 --> 00:25:19.450
Let us now see how we can
generalize these concepts
00:25:19.450 --> 00:25:23.640
directly to the formulation of
general curved beam elements.
00:25:23.640 --> 00:25:28.630
And for that purpose
I've shown here--
00:25:28.630 --> 00:25:30.310
I'm showing here--
00:25:30.310 --> 00:25:33.300
a more general beam element
that lies in a
00:25:33.300 --> 00:25:34.920
three-dimensional space.
00:25:34.920 --> 00:25:38.160
It's still rectangular, however,
we could also have a
00:25:38.160 --> 00:25:40.410
circular section instead.
00:25:40.410 --> 00:25:43.350
In fact, when we look at a
pipe element, we do talk
00:25:43.350 --> 00:25:45.630
about-- we have, of course,
a circular section.
00:25:48.470 --> 00:25:53.190
Well, in this particular beam
element, notice I'm looking at
00:25:53.190 --> 00:25:56.660
node one here, node two there,
and generally node three here
00:25:56.660 --> 00:25:59.810
and node four here, because we
want to pick up the curvature
00:25:59.810 --> 00:26:01.740
of the element.
00:26:01.740 --> 00:26:05.010
I have this element lying
in a three dimensional
00:26:05.010 --> 00:26:07.990
space, x, y, z.
00:26:07.990 --> 00:26:13.810
Notice that that element has
local coordinates r, s, t.
00:26:13.810 --> 00:26:15.980
These are the isoparametric
coordinates.
00:26:15.980 --> 00:26:23.260
And psi, eta, zeta, these are
the actual continuous physical
00:26:23.260 --> 00:26:26.010
coordinates in the
beam element.
00:26:26.010 --> 00:26:32.390
We define normals at
each nodal point.
00:26:32.390 --> 00:26:38.430
The normal in the t direction
here is 0 Vt 1.
00:26:38.430 --> 00:26:42.770
The normal into the s
direction is 0 Vs 1.
00:26:42.770 --> 00:26:47.720
And similarly, we have two
normals at each nodal point.
00:26:47.720 --> 00:26:50.330
Notice that for this particular
rectangular beam
00:26:50.330 --> 00:26:54.760
element, I have the thickness,
a1, here and b1, there, and a2
00:26:54.760 --> 00:26:56.020
here, b2 there.
00:26:56.020 --> 00:26:58.360
These thicknesses can
be different.
00:26:58.360 --> 00:27:05.370
Now what I want to use are the
same basic assumptions that we
00:27:05.370 --> 00:27:09.440
have familiarize ourselves
already with when we looked at
00:27:09.440 --> 00:27:13.050
the special case of a straight
beam element in planar
00:27:13.050 --> 00:27:13.930
deformations.
00:27:13.930 --> 00:27:15.990
I want to use those basic
assumption now in the
00:27:15.990 --> 00:27:19.450
development of this more
general beam element.
00:27:19.450 --> 00:27:22.870
And I want to use a continuum
approach.
00:27:22.870 --> 00:27:26.080
Well, what we are doing,
then, is the following.
00:27:26.080 --> 00:27:34.160
We interpolate the coordinates,
x, y, and z,
00:27:34.160 --> 00:27:37.750
along the beam element in
terms the nodal point
00:27:37.750 --> 00:27:45.095
coordinates of the nodes that
lie on the neutral axes of the
00:27:45.095 --> 00:27:52.950
beam plus an effect that
comes in due to the
00:27:52.950 --> 00:27:55.120
thickness of the beam.
00:27:55.120 --> 00:27:58.460
Now let us go and look
in detail at the x
00:27:58.460 --> 00:27:59.590
interpolations.
00:27:59.590 --> 00:28:05.020
The L denotes 0 or 1, 0 being
the initial configuration, 1
00:28:05.020 --> 00:28:07.750
being the final configuration.
00:28:07.750 --> 00:28:12.240
So let's put simply a 0 in
there, think in terms of a 0
00:28:12.240 --> 00:28:16.410
there, and let's look at the
initial configuration first.
00:28:16.410 --> 00:28:21.980
Well, here we have the initial
x-coordinates of the nodal
00:28:21.980 --> 00:28:25.760
points, and there
are q of them.
00:28:25.760 --> 00:28:30.160
These are the one dimensional
interpolation functions, just
00:28:30.160 --> 00:28:34.240
the same that we use for a truss
element, for example.
00:28:34.240 --> 00:28:38.030
Here, we have the t-axis.
00:28:38.030 --> 00:28:46.440
This is the t-axis into the
direction of the normal into
00:28:46.440 --> 00:28:48.340
the t direction.
00:28:48.340 --> 00:28:50.560
In other words, this
is a t-axis here.
00:28:50.560 --> 00:28:52.980
Notice that that t-axis
here corresponds
00:28:52.980 --> 00:28:54.760
to this normal here.
00:28:54.760 --> 00:28:59.430
That s axis here corresponds
to this normal here.
00:28:59.430 --> 00:29:07.350
So here, we have t/2, and ak
being the total thickness of
00:29:07.350 --> 00:29:13.125
the beam corresponding to that
t direction, hk being the one
00:29:13.125 --> 00:29:16.350
dimensional interpolation
functions again, and these are
00:29:16.350 --> 00:29:26.020
the direction cosines of the
normal in the t direction.
00:29:26.020 --> 00:29:28.250
Here, I should really say, this
is a direction cosine
00:29:28.250 --> 00:29:32.180
corresponding to the x-axis,
corresponding to the x-axis.
00:29:32.180 --> 00:29:38.180
When we talk later about the y
and z-axis, then we use the y
00:29:38.180 --> 00:29:40.960
direction cosine and the
z direction cosine.
00:29:43.820 --> 00:29:48.770
This part comes in because the
beam basically has a thickness
00:29:48.770 --> 00:29:51.670
into the t direction.
00:29:51.670 --> 00:29:57.490
Now, we have also to introduce
the s direction part, and here
00:29:57.490 --> 00:30:02.160
we s/2, the thickness into s
direction, the one dimensional
00:30:02.160 --> 00:30:06.920
interpolation functions, and the
direction cosine in the x
00:30:06.920 --> 00:30:12.950
direction of the normal
in the s direction.
00:30:12.950 --> 00:30:17.130
Well, if we want to find, in
other words, the coordinates
00:30:17.130 --> 00:30:21.010
of any point in the beam
element, and let us look, once
00:30:21.010 --> 00:30:23.390
more, back at the
beam element.
00:30:23.390 --> 00:30:26.440
If I want to find the coordinate
of a point, p,
00:30:26.440 --> 00:30:30.490
lying in that beam element
here-- there's, say, point p--
00:30:30.490 --> 00:30:35.000
what I have to do is I have to
identify the r, s, and t
00:30:35.000 --> 00:30:42.000
coordinates of that point, p,
and then substitute these r,
00:30:42.000 --> 00:30:47.320
s, and t-coordinates into
this part here.
00:30:47.320 --> 00:30:51.720
And I would get the
corresponding x-coordinate.
00:30:51.720 --> 00:30:56.430
I proceed similarly with the
y and z-coordinates.
00:30:56.430 --> 00:30:59.940
Notice, as I pointed out
earlier, we are talking still
00:30:59.940 --> 00:31:07.300
about the thickness ak hk here,
here, and here, but
00:31:07.300 --> 00:31:10.950
we're using the x direction
cosines, the y direction
00:31:10.950 --> 00:31:14.680
cosines, and the z direction
cosines here.
00:31:14.680 --> 00:31:20.110
And similarly for the s
direction, we are using
00:31:20.110 --> 00:31:22.020
similar quantities.
00:31:22.020 --> 00:31:30.780
So this is the interpolation of
the beam element, and using
00:31:30.780 --> 00:31:34.270
these three formerly, we
can directly obtain--
00:31:34.270 --> 00:31:36.300
we can directly obtain--
00:31:36.300 --> 00:31:42.010
the x, y, and z-coordinates in
this system of axes of any
00:31:42.010 --> 00:31:45.030
point in the beam element.
00:31:45.030 --> 00:31:49.150
This is the most
important fact.
00:31:49.150 --> 00:31:53.240
The interpolations that I've
listed here are the starting
00:31:53.240 --> 00:31:58.870
point of the development of
the strain displacement
00:31:58.870 --> 00:32:02.970
matrices and displacement
interpolation matrices.
00:32:02.970 --> 00:32:09.800
Now, let us identify that if
these are the original
00:32:09.800 --> 00:32:15.930
coordinates for L being 0, then
we can also apply, of
00:32:15.930 --> 00:32:19.860
course, after the deformation,
the same interpolation, and we
00:32:19.860 --> 00:32:22.610
put L equal to 1.
00:32:22.610 --> 00:32:29.690
If we subtract the x1 minus
zero x, we should get the
00:32:29.690 --> 00:32:31.400
displacements, u.
00:32:31.400 --> 00:32:35.830
Notice that the displacements,
u, are in the directions of
00:32:35.830 --> 00:32:37.020
the x-axis.
00:32:37.020 --> 00:32:39.560
Well, this is exactly
how we proceed.
00:32:39.560 --> 00:32:45.200
We use these interpolations
for before deformation and
00:32:45.200 --> 00:32:46.670
after deformation.
00:32:46.670 --> 00:32:50.500
And we subtract these
interpolations as shown here
00:32:50.500 --> 00:32:54.890
and directly obtain the u, v,
and w displacements, of
00:32:54.890 --> 00:32:59.360
course, as a functional
of r, s, and t.
00:32:59.360 --> 00:33:03.320
Notice that if I proceed
this way--
00:33:03.320 --> 00:33:06.820
notice that if I proceed this
way, I have used the basic
00:33:06.820 --> 00:33:11.750
assumption that plane sections
remain plane during
00:33:11.750 --> 00:33:12.820
deformation.
00:33:12.820 --> 00:33:17.210
This was the assumption that I
pointed out to your earlier.
00:33:17.210 --> 00:33:20.280
And we are using it in this
general information just in
00:33:20.280 --> 00:33:22.610
the same way as in the special
application that
00:33:22.610 --> 00:33:24.190
I showed you earlier.
00:33:24.190 --> 00:33:28.140
Well, having then--
00:33:28.140 --> 00:33:32.120
just to refresh your memory, the
u, v, and w, in terms of
00:33:32.120 --> 00:33:38.950
r, s, and t, of course, via
these subtractions, we obtain
00:33:38.950 --> 00:33:41.970
directly these equations here.
00:33:41.970 --> 00:33:48.090
Notice that here we have, now,
the nodal point displacements,
00:33:48.090 --> 00:33:53.790
uk, vk, wk, and we have
the change in
00:33:53.790 --> 00:33:55.370
the direction cosines.
00:33:55.370 --> 00:33:57.520
These are the changes in
the direction cosines.
00:34:00.460 --> 00:34:03.690
Well, these changes in the
direction cosines we want to
00:34:03.690 --> 00:34:07.810
express in terms nodal
points rotations.
00:34:07.810 --> 00:34:13.020
And that is achieved as shown
on this view graph.
00:34:13.020 --> 00:34:16.860
We can express these changes
in the direction cosines
00:34:16.860 --> 00:34:21.550
directly by taking the cross
product of a vector of nodal
00:34:21.550 --> 00:34:25.889
points rotations, and I've
listed here this vector, this
00:34:25.889 --> 00:34:30.340
is a nodal point rotation about
the x, y, and z-axis at
00:34:30.340 --> 00:34:32.780
nodal point, k.
00:34:32.780 --> 00:34:35.940
You're taking the cross product
of this vector times
00:34:35.940 --> 00:34:38.880
the original normal.
00:34:38.880 --> 00:34:40.590
Time the original normal.
00:34:40.590 --> 00:34:43.560
And we get, then, the change in
the normal, of course, for
00:34:43.560 --> 00:34:45.870
the t direction and for
the s direction.
00:34:48.469 --> 00:34:55.080
If we substitute this relation
here, of course, remember that
00:34:55.080 --> 00:34:56.449
these quantities are known.
00:34:56.449 --> 00:34:57.910
They are given.
00:34:57.910 --> 00:34:59.920
So the unknowns now
are theta k.
00:35:03.100 --> 00:35:08.560
If you substitute from here into
our relation here that I
00:35:08.560 --> 00:35:13.740
developed earlier, we directly
obtain the displacements of
00:35:13.740 --> 00:35:18.770
any point, p, in the beam
in terms of nodal point
00:35:18.770 --> 00:35:21.630
displacements and nodal
points rotations.
00:35:21.630 --> 00:35:25.470
Because these quantities here
now have been eliminated and
00:35:25.470 --> 00:35:28.710
have been expressed in terms
of nodal point rotations.
00:35:28.710 --> 00:35:32.530
Well, now we have all the
quantities that we need to
00:35:32.530 --> 00:35:36.060
develop our strain displacement
matrix for the
00:35:36.060 --> 00:35:37.110
beam element.
00:35:37.110 --> 00:35:41.130
Remember, all we need are the
coordinate interpolations,
00:35:41.130 --> 00:35:46.290
which I have developed already,
and the displacement
00:35:46.290 --> 00:35:47.270
interpolations.
00:35:47.270 --> 00:35:50.820
With those two quantities, we
can immediately calculate, via
00:35:50.820 --> 00:35:53.110
the procedures that I discussed
with you earlier,
00:35:53.110 --> 00:35:55.740
the strain displacement
transformation matrix.
00:35:55.740 --> 00:35:57.570
And here it is given.
00:35:57.570 --> 00:36:02.180
Notice that we are now talking
about strain into the eta
00:36:02.180 --> 00:36:03.000
directions.
00:36:03.000 --> 00:36:06.240
In other words, the eta, psy,
and zeta directions.
00:36:06.240 --> 00:36:10.940
These are the directions that I
pointed out to you earlier,
00:36:10.940 --> 00:36:12.510
which are the physical
coordinate
00:36:12.510 --> 00:36:14.650
directions along the beam.
00:36:14.650 --> 00:36:17.740
Let me show it to you once
more, the picture.
00:36:17.740 --> 00:36:21.590
The r, s, and t-coordinates
are the isoparametric
00:36:21.590 --> 00:36:22.590
coordinates.
00:36:22.590 --> 00:36:25.910
The eta, psy, and zeta
coordinates are the physical
00:36:25.910 --> 00:36:27.670
coordinates along the beam.
00:36:27.670 --> 00:36:29.820
In other words, for the one
dimensional beam that we
00:36:29.820 --> 00:36:33.980
looked at, this eta axis was,
in fact, the x-axis.
00:36:33.980 --> 00:36:37.330
Of course, our x, y, and
z-axes are now global
00:36:37.330 --> 00:36:40.110
Cartesian axes.
00:36:40.110 --> 00:36:46.950
Well then, with that
information, we can directly
00:36:46.950 --> 00:36:51.270
calculate the strain
displacement matrix here.
00:36:51.270 --> 00:36:55.760
This is done effectively using
numerical integration, as I
00:36:55.760 --> 00:36:57.790
will be discussing in
the next lecture.
00:36:57.790 --> 00:37:02.820
The transformations that are
necessary are also done on the
00:37:02.820 --> 00:37:05.400
integration point level.
00:37:05.400 --> 00:37:10.660
Notice that the uk here lists
the nodal point displacements
00:37:10.660 --> 00:37:14.410
and the nodal point rotations,
and, of course, we have to
00:37:14.410 --> 00:37:18.070
also remember one important
fact, that for the beam we are
00:37:18.070 --> 00:37:20.980
talking about stresses into
the eta, psy, and zeta
00:37:20.980 --> 00:37:25.670
directions, normal stresses,
shear stresses here, that are
00:37:25.670 --> 00:37:29.640
related via this stress strain
law to the normal strains and
00:37:29.640 --> 00:37:31.160
shear strains.
00:37:31.160 --> 00:37:35.030
Notice that there's again the
shear correction factor, k,
00:37:35.030 --> 00:37:38.690
which we want to also include in
the formulation because we
00:37:38.690 --> 00:37:43.480
have assumed constant shearing
strains through the thickness
00:37:43.480 --> 00:37:46.230
of the beam, whereas we know
that for a rectangular beam,
00:37:46.230 --> 00:37:51.810
for example, we have parabolic
shear strain distributions if
00:37:51.810 --> 00:37:53.730
the beam is straight.
00:37:56.360 --> 00:38:01.800
I'd like to now go on with the
development of plate elements.
00:38:01.800 --> 00:38:06.440
Here, we are talking basically
about the same approach.
00:38:06.440 --> 00:38:11.220
As I mentioned already once,
the beam element that I'm
00:38:11.220 --> 00:38:12.650
talking about here--
00:38:12.650 --> 00:38:15.440
that I have been talking about--
is really only an
00:38:15.440 --> 00:38:18.100
effective formulation when we
talk about, and we want to
00:38:18.100 --> 00:38:21.780
develop, a curved
beam element.
00:38:21.780 --> 00:38:25.280
For pipe elements also, in the
case of pipe elements, I
00:38:25.280 --> 00:38:28.950
should briefly mention that,
of course, we have to
00:38:28.950 --> 00:38:32.250
introduce, also, an ovalization
degree of freedom.
00:38:32.250 --> 00:38:38.580
That ovalization degree of
freedom interpolates basically
00:38:38.580 --> 00:38:44.300
the ovalization along
the curved pipe.
00:38:44.300 --> 00:38:46.320
That is an additional degree
of freedom that has to be
00:38:46.320 --> 00:38:50.710
introduced in the curved beam
element formulation.
00:38:50.710 --> 00:38:54.450
So, the beam element formulation
is effective for
00:38:54.450 --> 00:38:57.000
curved beams, pipes.
00:38:57.000 --> 00:39:01.260
However, it does show the basic
procedure that we are
00:39:01.260 --> 00:39:03.240
following also in the
development of
00:39:03.240 --> 00:39:04.880
plate and shell elements.
00:39:04.880 --> 00:39:09.390
And here, we have a typical
plate element.
00:39:09.390 --> 00:39:11.110
In other words, a flat shell.
00:39:11.110 --> 00:39:14.940
The u, v, and w displacements
are now
00:39:14.940 --> 00:39:16.480
interpolated in this way.
00:39:16.480 --> 00:39:20.170
Notice u be the displacement
into the x direction, v the
00:39:20.170 --> 00:39:22.410
displacement into the y
direction, w being the
00:39:22.410 --> 00:39:26.100
transverse displacement, and
again, we're talking about
00:39:26.100 --> 00:39:29.580
section rotations.
00:39:29.580 --> 00:39:34.580
Beta x being the section
rotation about the y-axis.
00:39:34.580 --> 00:39:37.550
That is, the beta x
section rotation.
00:39:37.550 --> 00:39:42.090
Beta y is a section rotation
about the x-axis.
00:39:42.090 --> 00:39:46.290
So that our v, measuring
z positive upwards.
00:39:46.290 --> 00:39:50.540
Notice, our [? v4 ?] point here
is negative, and that's
00:39:50.540 --> 00:39:52.790
why we have a negative
sign there.
00:39:52.790 --> 00:39:56.950
Well, with that then given, we
can immediately develop our
00:39:56.950 --> 00:40:02.830
strains by using the strength
of material equations that
00:40:02.830 --> 00:40:07.785
tell us epsilon xx is del u del
x, epsilon yy is del v del
00:40:07.785 --> 00:40:08.850
y, et cetera.
00:40:08.850 --> 00:40:11.960
And, of course, we're getting
also our shear strengths.
00:40:11.960 --> 00:40:17.040
Having developed the strains,
we also recognize that our
00:40:17.040 --> 00:40:22.080
stresses are given in terms of
these formulae, where we have
00:40:22.080 --> 00:40:27.870
now the stress strain law for
planed stress analysis,
00:40:27.870 --> 00:40:34.020
because we are looking at the
plate as an assemblage of thin
00:40:34.020 --> 00:40:36.540
elements, plane stress
elements, lying
00:40:36.540 --> 00:40:37.760
on top of each other.
00:40:37.760 --> 00:40:41.570
The stress through the plate, of
course, is 0, and this is,
00:40:41.570 --> 00:40:45.650
therefore, the plane stress
[? material ?]
00:40:45.650 --> 00:40:47.690
that we have been
putting in here.
00:40:47.690 --> 00:40:52.650
And the z times this vector
here gives us the strains.
00:40:52.650 --> 00:40:57.790
The shearing stresses are given
here, and our functional
00:40:57.790 --> 00:41:01.370
pi that I used also for
the beam element
00:41:01.370 --> 00:41:03.370
already is given here.
00:41:03.370 --> 00:41:06.150
Notice that we now have to
integrate through this
00:41:06.150 --> 00:41:08.970
thickness of the
plate element.
00:41:08.970 --> 00:41:11.650
Here's our shear correction
factor again, which is
00:41:11.650 --> 00:41:16.190
introduced just the same way
as in the beam element.
00:41:16.190 --> 00:41:20.980
Notice here we have the work, or
the total potential of the
00:41:20.980 --> 00:41:24.580
external loads, I should say.
00:41:24.580 --> 00:41:27.450
It is convenient now to
integrate through the
00:41:27.450 --> 00:41:30.220
thickness because we can
integrate prior to
00:41:30.220 --> 00:41:35.820
interpolating the quantities,
and that then yields this
00:41:35.820 --> 00:41:41.580
value for pi, where our Cb parts
and Cs part here, these
00:41:41.580 --> 00:41:44.800
two matrices, embody the fact
that we have integrated
00:41:44.800 --> 00:41:49.350
through the thickness,
so we have the
00:41:49.350 --> 00:41:51.800
following definitions here.
00:41:51.800 --> 00:41:58.580
Kappa simply lists basically
the bending strains, or I
00:41:58.580 --> 00:42:04.350
should say the rotations
of the sections.
00:42:04.350 --> 00:42:07.390
Of course here, we have
the shearing strains.
00:42:07.390 --> 00:42:12.290
Gamma lists the shearing
strains.
00:42:12.290 --> 00:42:17.440
Cb now is a function of h cubed,
just like in classic
00:42:17.440 --> 00:42:18.000
plate theory.
00:42:18.000 --> 00:42:21.190
Of course, we also have an
h cubed entering into the
00:42:21.190 --> 00:42:22.250
formulation.
00:42:22.250 --> 00:42:26.560
And this Cb matrix embodies
the fact that we have
00:42:26.560 --> 00:42:28.660
integrated through
the thickness.
00:42:28.660 --> 00:42:30.400
Here is our Cs.
00:42:30.400 --> 00:42:34.440
Of course, in the Cs, we
also have the k part.
00:42:34.440 --> 00:42:39.030
Notice that, again, we have an
h cubed here and an h there.
00:42:39.030 --> 00:42:43.340
Therefore, to use our
interpolation for plate and
00:42:43.340 --> 00:42:46.350
shell elements, we will have
to use high enough shell
00:42:46.350 --> 00:42:49.460
interpolations to be able to
represent the fact that the
00:42:49.460 --> 00:42:54.070
shear strains go to 0 for thin
plates if we want to use this
00:42:54.070 --> 00:42:56.780
formulation for thin
plates and shells.
00:42:56.780 --> 00:43:02.470
Well, invoking now the fact that
pi shall be stationary,
00:43:02.470 --> 00:43:05.850
we directly obtain this
equation here.
00:43:05.850 --> 00:43:07.770
And this, of course, is
nothing else than the
00:43:07.770 --> 00:43:09.610
principal of virtual
displacement
00:43:09.610 --> 00:43:11.880
for the plate element.
00:43:11.880 --> 00:43:16.050
Notice that from this point
onwards, we simply need to
00:43:16.050 --> 00:43:18.810
substitute only our
interpolations.
00:43:18.810 --> 00:43:22.740
The interpolations that we are
using are now interpolations
00:43:22.740 --> 00:43:27.870
for w, beta x, and beta y.
00:43:27.870 --> 00:43:30.860
And, of course, we also
interpolate x and y.
00:43:34.250 --> 00:43:37.870
These interpolations, beta x
and beta y, are independent
00:43:37.870 --> 00:43:41.080
from the interpolations of w,
and that is the important
00:43:41.080 --> 00:43:43.510
points, as I mentioned in the
00:43:43.510 --> 00:43:44.960
development of the beam element.
00:43:47.590 --> 00:43:51.480
The fact that we are dealing,
here, with three
00:43:51.480 --> 00:43:56.130
interpolations of course means
that an each nodal point, we
00:43:56.130 --> 00:44:02.330
have three unknowns, w's,
and section rotations.
00:44:02.330 --> 00:44:06.440
Let us now look very briefly
at shell elements.
00:44:06.440 --> 00:44:10.730
The same concept that we used
to develop the general beam
00:44:10.730 --> 00:44:15.690
element after having discussed
the special beam element is
00:44:15.690 --> 00:44:18.620
also employed now in the
development of what you might
00:44:18.620 --> 00:44:22.410
call a general shell element,
versus the special plate
00:44:22.410 --> 00:44:25.790
element that I just discussed,
or the special shell element
00:44:25.790 --> 00:44:31.150
that I just discussed, because
a flat shell is of course
00:44:31.150 --> 00:44:33.450
nothing else than a plate,
if we also don't
00:44:33.450 --> 00:44:37.950
have membrane forces.
00:44:37.950 --> 00:44:42.280
Well here, I'm showing
a shell element, a
00:44:42.280 --> 00:44:44.000
nine-noded shell element.
00:44:44.000 --> 00:44:48.190
And notice that in this case,
now, we are talking about this
00:44:48.190 --> 00:44:49.190
normal only.
00:44:49.190 --> 00:44:54.660
In the beam, we had two normals,
vt and vs. Now we
00:44:54.660 --> 00:44:56.900
only have one normal.
00:44:56.900 --> 00:44:59.730
At a nodal point, we are
defining the membrane
00:44:59.730 --> 00:45:04.250
displacements, uk, vk, wk,
being a transverse
00:45:04.250 --> 00:45:11.070
displacement, and the rotations,
alpha k and beta k.
00:45:11.070 --> 00:45:18.550
These rotations are defined
about the axis v1 k and v2 k.
00:45:18.550 --> 00:45:23.830
Now notice that these two
rotations, alpha k and beta k,
00:45:23.830 --> 00:45:32.560
will give us the change in
vn during deformations.
00:45:32.560 --> 00:45:35.460
And this is really how we
use alpha k and beta k.
00:45:35.460 --> 00:45:41.110
We express a change in vn in
terms of alpha k and beta k.
00:45:41.110 --> 00:45:43.290
The procedure is the
same as in the
00:45:43.290 --> 00:45:45.490
case of the beam element.
00:45:45.490 --> 00:45:50.660
First, we express our x,
y, and z-coordinates.
00:45:50.660 --> 00:45:57.530
And we're using here the
original normal, vn, the x, y
00:45:57.530 --> 00:46:00.660
and z direction cosines.
00:46:03.180 --> 00:46:08.820
These are, here, the x, y
and z-coordinates of the
00:46:08.820 --> 00:46:10.300
nodal point, k.
00:46:10.300 --> 00:46:13.770
We have our two dimensional
interpolation functions, hk,
00:46:13.770 --> 00:46:16.160
now here, because we talk
about a two dimensional
00:46:16.160 --> 00:46:19.490
surface, the mid-surface
of the shell.
00:46:19.490 --> 00:46:22.330
Of course, at each nodal point,
the shell can have a
00:46:22.330 --> 00:46:25.340
different thickness, and that
is denoted by using a
00:46:25.340 --> 00:46:30.250
different ak value at
each nodal point.
00:46:30.250 --> 00:46:35.440
Applying this interpolation
here to the initial
00:46:35.440 --> 00:46:41.520
configuration and the final
configuration and subtracting
00:46:41.520 --> 00:46:48.220
0x from 1x, and similar for y
and z, we directly obtain the
00:46:48.220 --> 00:46:51.050
displacements u, v, and w.
00:46:51.050 --> 00:46:54.740
Notice that the displacements
now are involving the nodal
00:46:54.740 --> 00:46:59.910
point displacements and the
change in the direction
00:46:59.910 --> 00:47:04.410
cosines of the normal,
denoted here.
00:47:04.410 --> 00:47:08.890
These changes in the direction
cosines of the normal can
00:47:08.890 --> 00:47:13.840
directly be expressed in
terms of the rotations,
00:47:13.840 --> 00:47:17.100
alpha k and beta k.
00:47:17.100 --> 00:47:24.440
Now, notice here that once we
have done this, of course here
00:47:24.440 --> 00:47:32.550
we involve now, as I mentioned
earlier, the v1 and v2
00:47:32.550 --> 00:47:33.600
directions.
00:47:33.600 --> 00:47:37.530
And these v1 and v2 directions
are arbitrarily selected.
00:47:37.530 --> 00:47:38.640
In fact, they are--
00:47:38.640 --> 00:47:43.610
for our shell element here--
selected as shown here.
00:47:43.610 --> 00:47:48.110
But once we have selected v1
and v2 at each nodal point,
00:47:48.110 --> 00:47:52.670
and they can vary from nodal
point to nodal point, then we
00:47:52.670 --> 00:47:56.770
can use this relation to attain
directly the change in
00:47:56.770 --> 00:48:03.930
the direction cosines of the
normal during deformations,
00:48:03.930 --> 00:48:08.330
when the deformations are
alpha k and beta k.
00:48:08.330 --> 00:48:12.530
So with this equation, then, and
the earlier equation that
00:48:12.530 --> 00:48:18.850
I've given to you, we can
directly obtain the
00:48:18.850 --> 00:48:21.580
displacement interpolation
matrix and the strain
00:48:21.580 --> 00:48:22.800
interpolation matrix.
00:48:22.800 --> 00:48:25.740
One important point that I
should briefly mention is
00:48:25.740 --> 00:48:28.980
that, of course, for the shell,
we have 0 stresses
00:48:28.980 --> 00:48:31.390
through the thickness.
00:48:31.390 --> 00:48:37.510
So, we have to use this stress
strain law here.
00:48:37.510 --> 00:48:41.600
Notice there are 0's in
this row and column.
00:48:41.600 --> 00:48:45.450
And this is, here, the plane
stress part for the bending,
00:48:45.450 --> 00:48:49.800
and that is the shear
part here.
00:48:49.800 --> 00:48:54.890
This is the stress strain law
defined in a local convected
00:48:54.890 --> 00:48:58.970
coordinate system where we are
talking about the stresses
00:48:58.970 --> 00:49:03.570
through the thickness being
this direction here.
00:49:03.570 --> 00:49:05.965
And these other stresses
are aligned with
00:49:05.965 --> 00:49:07.490
the coordinate system.
00:49:07.490 --> 00:49:10.810
We have to transform this one
here to the global coordinate
00:49:10.810 --> 00:49:14.900
system in order to be able
to use it directly in our
00:49:14.900 --> 00:49:16.330
formulation.
00:49:16.330 --> 00:49:18.820
And that transformation
is achieved via these
00:49:18.820 --> 00:49:21.160
transformation matrices.
00:49:21.160 --> 00:49:25.160
Now, this element has been
effectively implemented in the
00:49:25.160 --> 00:49:29.270
ADINA computer program, and we
want to use it using high
00:49:29.270 --> 00:49:32.860
order interpolation, as I
mentioned earlier, as the
00:49:32.860 --> 00:49:36.160
basic element that therefore is
very useful, which can be
00:49:36.160 --> 00:49:39.890
used as a flat element, a curved
element, all curved
00:49:39.890 --> 00:49:47.010
element, or it can have
curvature in both directions.
00:49:47.010 --> 00:49:50.000
Also, this is the basic element
that is being used.
00:49:50.000 --> 00:49:54.290
We can also collapse nodes and
derive other elements.
00:49:54.290 --> 00:49:59.010
As I pointed out earlier, the
low order elements should only
00:49:59.010 --> 00:50:02.950
be used in very special cases.
00:50:02.950 --> 00:50:05.020
I would not recommend
these elements.
00:50:05.020 --> 00:50:10.570
Although they can be used in
principle, the element that is
00:50:10.570 --> 00:50:13.960
really useful is this
one and that one.
00:50:13.960 --> 00:50:17.650
Both, of course, can be used
as curved elements as I
00:50:17.650 --> 00:50:19.190
pointed out.
00:50:19.190 --> 00:50:22.650
Another feature, and this is
the final view graph that I
00:50:22.650 --> 00:50:26.960
wanted to show in this lecture,
is that we can use
00:50:26.960 --> 00:50:31.830
these elements also in
transition regions.
00:50:31.830 --> 00:50:36.460
Namely, here we have the shell
element, now being flat--
00:50:36.460 --> 00:50:37.550
that I discussed--
00:50:37.550 --> 00:50:40.320
and we can directly couple
this element into another
00:50:40.320 --> 00:50:44.320
element, which we call a
transition element, which has
00:50:44.320 --> 00:50:47.450
the shell degrees of freedom
at these nodes, but
00:50:47.450 --> 00:50:50.090
translational degrees of
freedom only here.
00:50:50.090 --> 00:50:53.200
In other words, a continuum
element degrees of freedom
00:50:53.200 --> 00:50:53.890
right here.
00:50:53.890 --> 00:50:57.110
Notice, three degrees of freedom
at this node, only
00:50:57.110 --> 00:51:00.490
translations, whereas here, we
would have five degrees of
00:51:00.490 --> 00:51:03.600
freedom-- three translations,
two rotations.
00:51:03.600 --> 00:51:07.620
Similarly, here we have a curved
shell going into a
00:51:07.620 --> 00:51:11.200
solid, and again here, we have
a transition element.
00:51:11.200 --> 00:51:14.700
Here, we show the five degrees
of freedom at a shell node and
00:51:14.700 --> 00:51:18.670
the three degrees of freedom at
a continuum element node.
00:51:18.670 --> 00:51:21.870
This is an effective approach to
be able to couple director
00:51:21.870 --> 00:51:24.930
shell elements into
solid elements.
00:51:24.930 --> 00:51:29.150
I have not talked, of course,
about the actual derivation of
00:51:29.150 --> 00:51:31.870
the matrices used in
the formulation of
00:51:31.870 --> 00:51:33.510
the transition element.
00:51:33.510 --> 00:51:36.210
That is a little bit beyond what
I wanted to present in
00:51:36.210 --> 00:51:37.110
this lecture.
00:51:37.110 --> 00:51:40.280
But the basic concepts are
those that we discussed
00:51:40.280 --> 00:51:44.530
already in the earlier lecture
for continuum element and in
00:51:44.530 --> 00:51:46.510
this lecture for structural
elements.
00:51:46.510 --> 00:51:48.200
Thank you very much for
your attention.